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ANALYSIS
NUMERICAL
ENGINEERINGI"
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Definition 1 ,Any number r , for which f (r}. =. 0" is called a
root of equation (1) '01' zero of the function f
"1; •, ",
" .
...Equation' (1) is called a' transcendentalequation, if
f (x ) contains one (~r more)' transcendenla~function ,such as:
'it, ~,sinx' '{;coshx , ln x , etc.
:/ .'.where, n is a pcsitive integer, '. ';~
(2)Ij
Iof degree n. Equation (1) is. called a polvnomial equatiOll
if f (x) is, a :polynorbial .of degree n in' x. ;It' has' the form :
where, f (x) is' ahy Continuous [unction.
(1)f(x). == '0" "
The .topic of this chapter is finding the solution to anequation of the forrn :
J.. 111troduction
, i, ~. i
',.,~f:',:".~.': ... "
ttiapter 1·
.~,
.',
-:. . t.,,: ' .•, ..,
Iif....",
.*,~;'i}1~ .
i
~,~~
I
..,
~'.~}~
~
J '
,,
I .
I~~M~IkJ~~, I
ml
,l )
rU~jrI"~
r.."fi'IfV~!Ik<li
: L\'~~A"
~
I~I"I,.
·UIt.;;II('ir"I~iifr\i~~;!..\v''''I1",;;f';'~'$1,t·h~1~~:~~Ii!
3
. ::,'1;.:
* * * * *.
Property 5 of polynomial equations together with thefact that most transcendental ,,¢quatiqns cannot b~',-~Qlv.edexplicitly, show the necessity of' sffi,~ying numerical methods,in which ,the rOOIs.,: are, appr~x~m~~~?,)m:~lps~Iy;as desired.
Here, four of these numerical methods are introduced.
* :I; * * *
For equations of higher - degree. it was proved "thatit is impossible to 'find the roots through formulas.
There are formulas for finding the roots of third - andfourth - degree' polynomial equations. But theseformulas are so complicated, and;.thc), .are rarely used.
the roots arc found by the familiar quadratic formula:
a.~ + b x + C = 0'~.J.' ,.
5. For second - degree polynomial equation:
.. ,
2
~So, in case of real coefficients, equation (2) has an
, even number of complex roots (0 or 2 or 4'.' :,.J;, ~. .' " ,.._.. ~'; . . " , . . ,t.. »:: ::
Hence, a polyp.Qjpi~" equatioll. of 04d degree has: atleast one real root. .. ,.
4. If the coefficients ' 'Go , al , , an are real .and!. = a + i b. ~s ,·&"rool: pf' equatiom (2) , .then ,r = a - i b .is also a 'roof of equation (2).
. ,
If x - r is a factor of f (x ) . then r is a root ofCqu.~tj~Q:·.(1/ \\k·'· ",\~" '. t~<O!:' ~'
3.
2.
, ", '1 :: , ..: ' " , ' , ,Equation (2)' h~, n 'r<><?~;:.some of which may beequal, real' ~r complex' numbers.
If r is a r9<?i;.9f\f<l\l3.~~n:.(2),then x - r is a f~c~orof f(x). Wlien r, = r;.2' = ... = rm we say-thatrt is repeated m tim~;',9.t~1ff,multipltcity m. .' ,
1.
In ~hat fo~I~1~ft;-.:~',cJist so~~,.well - known properties, ofpolynomial equatI<),U,(1) : ..
• ':I~. .:.. .' •
Properties of a' polynomial ,equation
or zeros of the .fanction :
fix) ~,,~;:. 4 =: 0 .
For example, 2 and, -2 are roots of the equation:
5
bi - at1 r - C/,', I ~2
2If we t~e th~.midpoint c, = 81 + hi . as
.~,. . .~:j.~
I}o - ao=
So, the root r is ill the interval, ral, 'br ) ,whose le;'lgth is h,alf that of (ao, bo), that is,
~ . .
. ','.",
aJ ',i:, 'c ', 0
If ,·....f( c~) f'(b~) < 0 ,'then' r d is in'( Co I bot)Hmd, ,wb set ,,":" .'
If f (aD) / ( Co) < 0 ,then r is in(ao, co) and we set
Recall that f (aD) and f (bo) are of oppositesigns:
I( co) f( bs) ,< 0
or
Step 1: Compute f (co).,If f(c;o.} , = 0, then, r ,= Co: END.If f (co) ;t; 0 , . then" either
.:\.Ii
11
1l1!
4
Co- r' b;
ao - bo2Ir
as an approximation of r ; we have
a~ +' bo2
. ~,'.
. . . " "
":Step"O:,, Let" .as: = o. ,-;( "ho =ib,
Algorithm of tlte bisectiolt methoa:
in (a,b)f(x) = 0
" " ." ,,"" .. , ':Then, / has' at 'l~ast' one zero r in the interval (a, b).'. , We "suppose' that' r is, .the .oaly.root of .equation :
As before, let f be. a continuous function. Let a. < band suppose that f( ay' and f (b) have "opposite signs,
that is, .: !If ) . '"" Ira) ~~'b) ,'~'.~::-...: ..... ;, ."0, ,dL7~b
(Method of Hmwll/{ The Interval)
2•. B:isectioit: Method
R'~, ::tiI :~ ;\
I.§~~~"1:,
~"]':1utiI I
hi';i~~~'1>"
~~:IdO'qI I
~~r:t,[i:!Ii' ':~: Ir .~,:I
h'':I~~,~0'Ii
~;."'t~!."t'I
t~v,'1,;l'J.I
l1~;1":,~1
I
~)"~,~~.~
[email protected]';.'d\/
~~;/.(
7
zz» b-a < rot: 2&
where e is a given error. ... In: this case' we can determinethe number n of steps required .as follows:
From (3) J n must sati~fy -.
'.~
Suppose it is required to find an approximationen to the root such that: .. ' , .'
Thus. after '!' steps 'OJ the ,"Jisectlon algorithm,an approximtue: .root <;n wilt have beencomputed with error of at most (h. a) /:r.+1
(3)I r - Cn I s
absolute error in this approximation IS:
an + bn= , then the2
If we, take
1btl - an = (b 1 a )2 n· - n-l
Thus the toot r is now in (an J bl1) I whoselength is half that of (an-I, bn_l) :
i~ij.. ,·11~
~I.1
,! '
JI,jjiI
I!
1
1ti',:iI'I
6
If .....f;(e";') .f(l?n-I.J,~,<:- ,q,~, then r is in( Cn-I , b;"'j r arid. we set ,,' .
. an. ',= Gn-i' • . bll' =: C,,_I
If f (an-I) f (Cn-I ) < 0 ; then the root ris . in the interval (an-I, en-I) and we set
" -,"r .
or .
Compute f (cn-l)-If .:I( Cn-I) = 0, then,' r = Cn-l .- ENDIf f (c;"'J) :f:. 0·, then, e.i(her
f( qn-J) If en-I) . < 0
bo - aobi - ai =
2;
ai. + biCi --
j = 2,3, ... I n-l2'-,'.
I bo - aoIr - Cj ~ 2i + 1
Continuing in this way. we obtain in steps 2,3,_,,~n-I the successive approximations C2; C3 •••• , Cn-I
for r., where
Step n:
9
=> r In (1 15), .=->:,~ .~ 1(1)/{1:5) < 0
l.. .. ~
f(ee} :':~~f( J,5),. = O.~75 > 0Step 1:
1+2Co = .:.....;___"..= 1'5. 7i::',:·,"
Step 0:
, (':, ,', ,
f(Q,J = ..~i..!;"·'f(I.) ::= -1'.1. /(2) = 5, .' l.. :.. ,".:.'
Since . :,/(;.f!/I·.f(2) < o ,the root is in (1 1 2)
Solution: Since 'the...required root is positive.we. c?mp~tp f (0) : f (l) I f (2) I •,. untilthe SIgn ':enanges :, ,. '. ..
Example 2: .Find tl;~,small~st pOSitiv~' root of. " i! -,x-I. = 0 .COrrect' to two decimals.
':.~. .' -.
* * * * *
The required' number of steps isn :;;16
jI':
·8
log 0.6' + 5- '= 15.872 ....tog 211..>
.:. .' ,',,', :,' .' ', ',"',r':.:::: I i .:'. 1:';- ',:, "i:~.~:.~';.11~?~'\,:",'~
logO.6 '~; .JQ.g, lP~~r,= )og;:.Q.6,;;ti'~ log 1011. > log -2 log 2 .
.:If the base of the' :lggarithms is taken to be 10,then,
. a =, 1,2 b ~ 1.8 1 e = 5 * 10-0
. Co' .,.ApplYi[lg,f~.liffiula:J4)" ~~:·,;p~r.. ., ;~:n,\:;'
.'',' ',~"'" ',..J.~..... .:.:'~i, ~ : .We have. ',' .~".\'" ...,',"j,',' . " :,:\:
Solution":
:, ; ';' ,,~'..* ')j: "* .* *
(4)
. ~ log (b - a) ,:< n log 2 + 1(Jg2 e
0(The log~ithms' ~e with any·base)
.~ n log 2 > 'log (b -a ) ., lo'g 2 e
log (b - a) ~ log 2 e. 'log 1/' , .: :.;
,I
II~"ej .(J
'1'>1r)!
~j,.3.A1 '
.Ii
1!It
.:':'. ".Vl;r- ~~ld" determine' ';the' needed ~l:lm.b'erof steps byusing: (4) with· a '.=' J....,. b ~=="'2' "an'd'" 'i, = 5 :.: 10-3:
Since .a7 and b,7 have the"same first two decimals', therequired approximation 'is': .
C7 = 1.3243
'.' .".. ::.. : ",
. , .:1.' '..
i .\ : aj' "cr'~:L ., bi I( a.)' /ie;)': f(bi}:·0 1 . 'I.5.. 2' -1· +0.9 "+51 1 1.25· 1.,5 -1 -0.3' +0.92 1.25 1.37~.'.. 1.5·.· ~O} +0.2 +0.93 -, .1.25':: . . "'!11'31~~',.,:~.:1.'375"': -:0.3'
,..-0:05 +0.2 .. t. .'
'4 . '·'·'i.~312'5. 1 ·)·f:J:;j.B8~·;·' .1.375 ~0:'O5 '+0.08 +0.25 . ·1.3125 ];,32$2; , 1:3438 -0.05 +0.01 +0.08.6' .;.),3125 ! 1.3204' 1.3282 ~0.05 ·-0.02 +0,017 1.3204 1.3243 1.3282
I.
It is better to, arrange calculations in a table.as follows : (k = 2) .
P (X)-= X·~- X - \
:. .e •
2.). Since we are interested .only in the signs of1(<<') ,f(_b,) and I( c.} " round - off thesequantities Ito' the first nonzero decimal.
.-.: ,
10
: ?,,' ,.G = 5 :$ io:":"
,',. . ~,\" ..~.~......~~··t~..."1i If the fool is required to be eeneet to k
.dectmals , . round - ·off the values of ar, b,and '.CI to 1r+2 decimols : ....- ,'~~;'.'In this case
r t::I C7 - 1.3243IS
To obtain the rool'mIKCt to .two decimals, wemust continue. the aIgori~ .~ the Jrrst twodecimals of an. 'and . b~ me' 1bc same:These will be Q7 = 1.:3204 and bj = L318.2so that the rCq~.'oppr.;;m;iimon fbr ·the 1JH1t
t:. :" ...~, :",~!
. ~1t+' LSO ... ···. z». C2 = , :1:". '-::': =: 13,75
..~..:'!: tit",; I ;~. j~...,,':,:.
=> r in ( 1..:25., I..5!J )
=> i( 1.25.) f(1..5)· < 0
Step 2: f(cJ) = f(1.25) = -0.297
\. ....
1 + 1.5 = 1.25.2
Cl =
..Notes-:
,13
."
, ""'. ......
Cf =,.(), 7~Q39. , '',::.,'.~.,,
. I } ~.
The required v~ue,fo,r r is: .. '.
\,
.. , . ' .",,bi f( a.) f( c.) f(bj)t aj '~:.:ej
0' ~~:"m: ,":""'O:i5"" . 0:9 +0.8 "'+0.2 " -0.51 0.75 ,0.825 0.9, +0.,2 -O.? -0..5",
<.4,.", . >;,D·'IJ,.,,'Y" " QJ;_P?J.. . ,Q.,§2S. 1.+0.2, .. ,-t:Q"Q.03 -:-,0..2..3
~ '0.7875 O,8()625 0.825 :tlrp03 ",,-0..09 ,,;·(,QI~',4 0..7875 0..7968$ 0..80.625 +0.:0.0.3 -0..0.4, . -0..0.95' ",I D) fJ1j,'1op(: :i,o..f9i2[~" : (J. '/19.6'88 -.+0.0.03 .oes-. -0..0.46 0.7875 0.78985 ':0..19219:'" +0..003 -0..00.-9 _ : ,-0..0.27 10,((875, 0.78868 o.7~'985'~l:fllvOl "~O.6r)j":"::'0.0098." ''O.'787Y 0.'1.8809' , "d,711'868' »ooos +0..00.3 -ij.OO3
-.l, .... 4"',,,, •• ,_·t~'''I...,.......:..9: \':0 7~80.9 . ' '0;78-839 0..78868r ...
'"
,',.,.,
12
[ 1
(.. ",:
Si'n&1 bYr()!'6p fy(V:~?(:,;,~.',~O;Atnere i~Jlia rootirr tht';,f·Yntetvtil 'o!' (0.6, '0.9). ':!~i>' ,"", :~'
. -a; = 0.6 , bo = 0.9':, E = 5 * io: k =3
,.I
, ,.I.
f(0.6) = +0.8 , f(0.9) = ~0.5
,',' ,,,,\ .."',r" \\1'" 'j,l,. ""''''ri' "I bll" " ",,..,'""f" (' I,,'" ~ ::~ ", i.-i::S.t" .,,'" ,;" J~: 'n '_'.',",;,, x = _' e ':+ in - -, . ) .J2 ': ,H \'I,~l\'If'~~~~·'(:'\}}. ~ ,rl"~·',!.fJ\ ::~"I.,,, ','_',.'\. .. .~..
SinceSolutioll :
"',, \'
", '* * *, * *'! ',' ..
==> n=7
'''''6"~,/~ ll}g, Hr d'il lqg\)(), = 2 ~()g,10 '= 6.64 ... ''log, 2::.:.'", ~ :' ',r,}ogd2 " ,,:,loglO',:2
n >
1.J655 -''(-0.2548) ( 11655 1.5 )X4 --. - 0.2548 1.38.
= .1.1655 + 0.0521 -- 1.2176
15
t: :', " *= 1.5 - 0.3345, = 1.16~5
Example :.I: "'Apply', regu!fz.., falsi :m~(hqd to find the..' '.., t.li~\ 1'.00/ "'6tili~equilti'on: ...',
~ - 2 sinx = 0in: the interval ~(J.) 1.S.).; to three decimals.. . ..... .
* *" * * *
This' is' the..'formula of the, secam ; method. Usually it .converges I~r :~an ,th~bjjett;on ;~;ho'tl.'
.~: ,.
14: '
(4)_!(x,,) [ Xtt - Xn-l ,]
f( Xn) - f( Xn-l )Xn+l' .: X"
Continuing this procedure. we find
[X2 - Xl ]
~ X3 = x] - !(Xl) f( X2) _ f( X2 )
Xl - X2X) - X2
f( X2 )'f( xi ]=- f ( X2 )
This line intersects x :.iris at (x;. 0) , Hence,
x - X2
f ( Xl) - f ( X2 )= P\' '! ..' I "'}Jl(('Xl - X2
y ~ f( Xl )
Its equation is:
This .method 'is" also known as',the meth~d of falseposition or regula:falsi. " ." ";'" :,'
Suppose that f(XI) , and !(X2) are of opposite sign.Thecurve : y = f(x} .on (Xt J X2l .. .: 'is, replaced by a', straight line _. " .,;,:'::~o~gh (x/,!(x[))and1(x),/(x,)). ,: ~~',
__;_~~--:4--;;''-''--
(The, Fatse Position or Regula Fol$;Method)
3. SeeD"t Method,
...
']'7
(a,b)
The sequence i
{ XI, X] •.... , x~...~.}converges .to r if
Theorem:, .,' ", . ", ",
n = 0,1,] ..... \. '.: '" .,
xn+l = lJ (:~nJ.. J. ' "
..1 ••• ·•
~ ~ .. . ';'i ; :~.", .. ", ..\
. ,..: : ."~ ~ j I '"
~.'..Xl = s o«)Xl. = g(x,)
....' ...., If XC) is an ~flitial..(/fp,ro~n.IQ~lotl for .the e~stroot r , we construct . a sequence of successtveOJff..roximations .~ foIlqws: .":. ; .
~ = si»): .....• t;
r. .
Rewrite the given equation in the form:
• '. • .' • t'o • \ ~ •• ~ " • :\
where, / is continuoUs and differentiable, has onlyone root in the interval ( a, b). .' .
• • I. ~
:!
:'00: ",.; •• 0 r. ..... I, t' ~ OJ, '', .. ~\.:~.,
It. is ,)i1~9 c~~ .th~,fixed P'!!~lf' ip.e.(~od-.".[et the equation: . :.~I;;.;
f{x) = 9,
. '. : :'" . ';',: '.~.: b.:;i:,~:.'~ .' ','..' (The Fixed Point Met"od~h.
,' ..1.:._ Method of IterationI
'-.
16
.:' :
* *' * lie' *.' ~.
1.2362r =
From X6 and X7 we see that the r~ot r correct tothre(! decimals is:
- 1.2364 - 0.0002 = 1.2362
(1.2364 - 1.229 ~
X7 '"" 1.2364 - (-0.0008) 0.0008 + o.'o2~f'\)
= 1.2290 + 0.0074
. . (;,1.229 + ~..~~;~...: .),X6 = 1.2290 - (-0.028). _0::028
. ':.....\= 1.2176 + 0.0114 = {2290
. / 1.2176 - 1.1655 )Xs = 1.2176 - (-0.0714)l.;0.0714 + 0.2548
19
",'
'qorollaries:' (1) 'SinCe' x; -J r , as n -;} 00,
the error.. ..'..' I xf!. ~..r tcan be' made' small' as we 'pleasefor ~,u~cie.nt1x,4I large n ..
• •..• • !! •.
( '.* * ~, * *'. . ....
'! •+. .~. .~ '.
" ,\ ,
#I x" ~ r I = 0, ~'., Jim x" '= r11,-+ ""
limn -+ <0
and ( 6) gives:
......lim m"n .: 0,
n-+oom < I . 1
. ,,' ::.:~.' : .'
::'.: ':': ,,~~,'",:: { .:.......
(6)
,I XJ - r I ::; m I Xo - 'r I
I X2 - r I S; m 1 xi ~,r I:': ~ m1 ! Xo - r: I" ., " ' :.", . -': .::~:,~~",~;,~,;,:~:~:.'.~~.
...""~" : : ~""~,i.'.;'~'.~~:;~i'i:~;,..i..,'t;" :. ~ ~ " .
Since ,
Mcnee,
IXn+J - r I =:j~ rgl(~)1 I·(xn - 1')1 ~ m IXJt -r Ifor n := 0,/,2, .....
From (5) we have
18
and, consequently,
. \
for all x in (a I b)I s'. (x) I .< m < 1
we ,C1:l,n assume that there is a positive number m < 1such that:
Il (tt) I, < '1, ' '
on, (a, b),Since
'x~,+1. - r '= g(xni - str» = reo (x~ -;t),"" (5)
where ~ is some point im (xn, rj, 'th~t i~,,:hi', (a ~'b). '
and we have by using 'the,'"mean - value theorem that ;
r = g.:{ r)
Since r is the exact roof, then
,g(h) - st») ='gl(4J (b - a).
If a. junction g (cc) is continuous and differentiableon ( a , b). then there is at least one -point ~ . in( a J b) such that:
Proo[:'. ,.The .proof is based upon .. the mean':'value theorem 'I :
which states that:
r I < 4 *' 10-4
21
z:» - I XII
I x] - r I :$; ,1l (b - a j ''''~.. {(j.672~8l= 0.00038
By (6):
3.78922XJ = 3.5 .+ ~.:5.log.)]Q8,29" " ' 'ii', '('. •
Xl .=: 3.5 + os 'log i5" : .,1. 3.'77203
:Xi = 3.5· + 0.5 [0g"3.f1103 ~- 3.78829
3 + 4 =, 1.5 ° '.
2 . 'Xo =
Note that 'I s' (x) 1 is' md:rliflliiiz'J when x = 3.
'f
on (3,4):I s' (x) 1 < 1
Before applying ~"!e iteration method, wemust verfy that : "J
7 +' .j_ fo~' ~ ~ , g ( ;; j2 . ·2·
...•.:. X '=:
• .' "',"; .;,~'r' :~. .: "':. .. ' .. '" ',:Rewrite' the given equation-in the' form':
'. -»:•• " " ~ .• ,.,,' '. " ,'0j
, :
I ":
20
. t ~ , . ' ". . '.
'the' real root r is hi (3 J 4) °
::;:; S~nc~o _1(3) I( 4) < 0,.
I(1) 1 - log 4 ~~ 0,
I( 3) = - log 3 - 1 < 0,,f
~ I(2) = - log 2 - 3 < 0
Jt») c:: 2x - log oX - 7
Solutio" :Let
Estimate the error. (Round - off allcalculations to five decimal places).
4 x - log x =0 7
...._. ' ..
Example 5: By the method of iteration find the thirdapproximation _to the real root of
* * * * *
IXn ~ r/:$; m" (b-a) (7)
(2) Formula (6) can be used toestimate the error in Xn:
I ~I
I~~'~tl
I ~-11!
ifI '~
~>~I
I*i~-..,,~sr~~!'''a\"~lI .
1'"'d'~1~~
It~~!dI.~~
,~I
~'l~,.,I
~~c~:~
f"l;1~I
!~~~·t.. -;..~rI~)\
I !.~
~'1-j
{t1~''i"II
~4t.:~ A
,
23
,3(x.+l)213
".' .~
• . .' '.. • ",', 'I • ,; ..:; ···x ,=: i3,Jx: <.. i" == .... s (x r": '.;~.Is' (x) I ; .~':(x '~"// 2// /
#. • •• 3 ~ ",.~. :.~I', ':. ,I:' ; "
+ 1: • ", '. ,'"1
......3.
"r' ..~ t , "Is' (x) I := 3' ? I.
... \. '.'I,' •
'again 'we ·C:~6{:i~Ilpiythe 'meiliod 'tothis fonn. ' ..:
'''\, .
z:» x = iJ(xj.x1
2.
:. .' We.c~~t apply: the method to this formof the equation.
, . ,
Is' (x) I = 13~ - 1I» 1 on (1, 2)
1. x = ~~.- 1 6: g(x)".' ,:
1
'.•1 .'.
22
in"many ways :.',' .,(. '
x ,= g(x)
can be' rewritten in the form :
The equation:.' f(x) = 0
zz» the root r is in (I, 2) ~
Let' .f (x ) = ~ - x' _. 1
1(1) f(2) < 0
Solution:
....,':.correct to two decimal places .
Example, 6 :: By ~~,~method of iteration find the realpositive root of:r ~x-I '= 0
* * * * *
3.789 278 248 ...r =
rrence , X3. gives me root correct to at ':'east threedecimal places. .The exact value' of the root is:
25
* * * * *
.. .. h 'b2 '. " . b1f(x +h) = f(x) +'.-' . /1 (x)+'2 ill/xj + - JIll (')
. : 11 21 . 2 !
,,'t.fOr example. when k = 1 it takes the form :
. .... h .'.. ... 2
f(x + h) = f(x) + -. /1 (xl- +. !_ fll (()11 . 2!
. :.. ' ~For example, when Ii = 1 if takes -the forrn :
, ... '.'for some point ;. in (x I X +h ) .'I,:,','; .
~<x +. h).= f(x) -f. Ih, jl(X) . + ..... ..' 'h" hht... + - ptJ.(x) +. .' .dt+J)(r.)
:... k! ' ..' .···{k+.;O:! J' -. 'p,
If f is continuous" and differentiable (k +1) timesoil the' interval [x, X + h.I·,·. ·'then.
Throughout this course well make' use several. 'tiuies ofthe well known Taylor Formu/~' whose statement is:
. '
It is also called the Newton - Raphson meihod .
(The Newton - Raphson Method)
5. -->::Newton's Met/lod'" "
',',..24
* *
Note that 0.0019 is an upper bound for theerror, an~ not the error itself.
The error: -in. :.X4 ."may: be·.~~~tima~ed,.Qy (6):
I X" - z] ~ ~4 (b:- a) = (0.21l ~ 0.0019" . ! :.°f.:j '1"','
( The exact root is 1.3~4.J IT 9.§7 ... .l)..
;;'("I',.'.' rr = 1.3249
: \ ...... ) .Since XJ and X4' have the. same first two·decimals" the ~eqJJjre9~!aJlpr@im41"p;>1J'is :
XJ = g ( 1.5)' = 2.5 JIJ = 1.3514
xz (2.3572) 113 = 1.3309
XJ' = ". (2.330~) il) =' "/J259:
XI = (2.3259) 113 = 1.Jl49: ..:.I.•"
=.. 1.5.1 + 22
Take
The method' maybe: applied td'this form.. :::; .•.. ;:'.. ..·0,
z:» Iz' (X) I S 0.21 < 1 on (1, 2)
3 * 2213
1
27
(9)I I . M h~Xn - r < 12 f' ( Xn - 1 )'1
The convergence of Newton .s procedure' is lasterthan any of the prevlousmethods, It can be shown that:
"...... {'.." i',1 \
,,-
From here we get formula ' (8)., " .: ':,
.. Xq -. Xn + 1
"= '.:'::f(' ~D )tan 0
sco
,,'.;:., f,Ul(,,)
" "....f.
by its tangent line. at (~n , f (~n).) :, '
y = I (x)
Geometrically, this means that we replace the curve:tI
This is Newmn ' s formula ..
";'j
'j
(8)Xn+l ::. x; f( Xn)
ft ( ?Cn)
26
Now, a better approximation=te >.r is given by:
.. ....
f ( XIl) , :t:::J Xn - --'--':_:"_, f" ( Xn ) ,
r
I:i, -h
f(xn} + h fl(Xn) ~ 0
f(~n) ,f' ( ~n )
: ' ~' .
: I
Where ~"is some point ,in . { Xn , x; + h j. If h issufftctently small,' we can neglect the term containing 1t2,Hence
'I ~ No = f (r) = f (x; + h) -.I(Xn) +, h j (x,,) + 2 1 f (~)
By Taylor formula with k == I we 'have
. ,f (r) == 0',
r ,=, Xn' + h,"
r - Xn =: h ,
f (x) O.
an, approximation to the root r of
where,
so that:
.We set
,Let x; bethe equation;
-r;
2.9··
* * *' * ;.,:*
Of course, ihi~'':is ·a .go~d ,I/rior· for drilY . fwoiterations of Newton's 'procedure.
fi:t',. 7 * 1{)~
12f"(Xl)1
','!, •
"
on (1,2)." ,
If II(x) I :::'r 6x,I < 12
hi ==, Xl ,.: xi = .;OiD2263',
To estimate the error involved in Xl we usc (9):
= ' 1.32520
0.100704.449g4
XI 1.34783
1.34783
O~8755.750
= 1.5 -
I28
LetI (x) = r-x"-l
I( 1) f(2) < 0 => r In (1,2)
il (x) = 3 xl 1 ,
III (x) = 6 x
Take,,..
.. t 'Xo .i=1 + 2 1::.5~. ',=
Estimate the error involved.( ~qund - off to five decimals)
:>?-x-I = 0,
Apply Newton IS· 'method to find the.second approximation Xl to the positiveroot of:
'* * * * *
So/utioll :
Example 7:
and M Is em upper bound for 1/1 («):
hn = XII - Xn-l
where,
31
g ( r 1 s) = 0 = g ( Xn+ h , Yn+ h )
.~
f (r , s ) = 0 = f ( x; + h , Yn+ h )
.Using Taylor series for functions of, two variables,we have
',: .S = Yn + k.r=xn+h
Let {x; ,Yn) be an approximation to (r, s ). Let,
, Newtoll':'s' Method
* * * * .*
in 'il, then' the ..iterations :
Xn+ 1 = F ( Xn , Yn ) }G ( Xn + i , Yn)'n = 0,1.2, ...
Yn + 1 =
~'..I,
.will converge to [r i s },:
(12) ,. r F: !. ~ IF, I " J},.Gx I + 1 Gy·I·< 1 .
" :
If
30 ..·.
(11)x: = }:,'t.! '/' .. ,
F(x,y)y = O(x e ,y) ,
,:,;,~ewri~ ~~4~tiops",(10) , in the. form : .
'\'': : .The Method of Iteration
""
which converge to ( r , s}, may be obtained by twomajor 'methods "
Let {x; ,';16.) ': be 'an' initial approximation in flto the exact root (r IS) . Successive approximations
in some region n of the xy - plane. We supposethat f and g, and their partial derivatives up to thesecond order are continuous In n.
y = s (or (r, s ) )x = r
Suppose that this system has a single real root
(10)f(x,y)g(x,y)
Consider the ,system of two nonlinear equations:
6. Svstelns of Nonlinear Equations
33I
X:2~1.32+ cos 1.6357= 1.2551, Yl=0.85 + sin 1.2551 =1.8006
, x3=1.32 + cos 1.8006= 1.0922, Y3= 0.85 + sin 1.0922 =1.7376
= 0.9039. Yl= 0.85 + sin 0.9039 =1.6357XJ=/.32 + cos 2'
, in some region .n about the potnt ( r,2)",: we cap apply the iteriuion procedure:
,! r: /'+ J ty I = I sin y f <. I
I, Or I ~' lOr I = I cos x I < 1
Since '
" y, =:=. ,O.8~ + sin ~ =,..G (x, y).... , .,' " .
x = 1.32 + cos y = F(x,y)
Rewrite the equations in the form:
SO/lition : (1) By Iteratlon l'ileJilod'
.....
......
sin ~ = y - 0.85; , . ,
cos y = x 1.32
correct to two decimal places.,
..'~
, '::Exo1lJPle 8: ,"',,~!nd,the solutionnear .(1 I 2) of the, , ':;ystem:
32
,Yu + 1
(13)
Xo+ I
Hence a better, 4,pqrqx(tnq!.fpn."to " '(r , s ) is:
, Neglecting terms containing h2 1 hk, k2, .•• , we have
h f" + k fy = - f }h g" + k Sy = - g
where,I, g , ,Ix , /y g:r , gy
are evaluated at (x; 1 Yn). By Cramer's rule we find : ' ,~ (I
fy g - f gy f g" - fx gh = k =fx gy fy gx fx gy - fy g"
~~~,'~j
~!~}
I j;I~!.1t I:
lit I"'.~. 'f>~!I I
1
I ;
35"
. "
.:, "t" 't, ,
and so on .
.' .,"= ':"'1. 756.!f··.. ::". ,• I,.
" 1\' •
".'o', '
= L'J5il~ rl- (-(jO'O~6j{:I).~ (0,4~23'~l..I:. ." .. ': _L4248.t\
,'.-,-,"
= :1.1352
.' •.•.. ., ... -11 - . • '"
= l' 1';2-'11--4- (-1)·.(1>::oQJ)::~ '-(~O;OO~.{~.9825)• >7,. - - .;' -t:42~k.:: _'.
" " . "
...... " W" ., - .. • I( '.
(-0.3086)(..lJ), ',- (0,54fr3.) (-O.0961)!'" .. -l!-;,; 1:49'i:3~ ,~......• JI" :.~
'~,
Y I~.• I
; ,
,,. "
..j
'.,,' .
. :.:. '
: .
= '],12.17":" .
= '1 +' (-1) (-0.0961) - (-0.3085) (-0.9093)-1.4913
;. ;.... '.:, . .:~ .. :. \ ...~.. \
Xl
.......f - cos x f = -1 " g¥ = -1 , gv = - Sl.·'.1Y--.' x- I Y ~.' ~ .r
(2r:-'JJy /NiW1fYl{~s:'~th~tl((jd
f(x,y) = sin x - y + 0.85 , s t»,») = cosy - x + 1.32
". y -- .1. 7561;
The solution correct to two decimals is:.• " ,1 ,t,'. ].~.~ .' ~. . .
',I,' • l .:
x7=1,32 + cos 1.7580';''i.1339 , Y7~ 0.85'+ sin 1.1339 =1.7561
, .x,,=1.32 + cos 1.7376=L'15?0 , y~=0,85 + sin). 1540 =1.7644
x5=1.32+ cos 1.7644= I.ii16',·' YS;' 0.85 + sin 1.1276 ~1.7534
·x6=1.32 + cos 1.7534=1.1384, ,Y6=O.85+sinl.13847-1.7580 .!
"1: .:::~, '.,
.l;~.
r"
I.i,:~
I',.Ir .,.I
r. '.,
.!
37
by using
1Find the positive root correct to 2D of theequation, '
~;' ." , . .
I, ''Apply the mt..'thod of iteration to find: the root inproblem 2 correct 10 'lD" and estimate the erTOI'.inthe attained value.
'j:I'i :,
and estimate the error .
/Perform three steps of Newton's method to obtainthe smallest positive root of' the equation
Apply 4 stlfPs only, of the. bisection method andestimate the error,
and how many are needed if this root is required tobe correct 10 8D ?
correct to .four decimal places (4D) ,
froblems
/ .}jaw many s~s of the: bisection method are neededto find the smallest positive root of the equation
36
* * * * :;<
x = 1.1353
From the table we see that ,tne s'Qluti:an corsecttethree decimals is:
-/ ,-:1h = -1 -1 -1 -1, ,
gx = -1f-- ..-:;:...------'-Il----4---....:I---__:....:.j------ll: ..
gy =,: - sin y -0.909.:iJ i .10.9825: ·-0.9828:
"g,.,~P0s.y,~~ + 1.32 ,-0.096il ; ,0.:0'099 {O
. ~:
f= sin x -y + 0.85 '-0.308}· -'[)"O'O66: ,..:Oft){)Off r
, • I
Y 2 J, J.7.~83 : /1.'11567x
321o, "n .
-'1
We 'may 'arrange calculations in a table as follows:
, "
39
For each of' the following ,systems of nonlinearequations find correct to 2,D the root that is 'near thegiven point by the method of lteratlon and then byNewton's method :
c: .
Estimate the error involved in each method".I'I, .xl - 5 ~ +) = 0
: '
9. Perform three steps. of the biseaion Newton's.and the iteration method to find the largestpositive Toot of the equation;, . '" ~
What can this formula 'be used for?
b) Define a sequence'
~1 =:: Xn - tan x,, , .. :, .- I":'
with XII = 3.
What is lim ' Xn ?n .... co
where ,R is a' ,pos;thJ~"cons/tint'; was obtainedby applying Newton's'; method to some functionf(x}. What was f(xJ ?
'''::..Xn+l = Xn" }: cos x" sin Xn '+ R cos1 xn',
8. aJ The iteration formula
"
1Ii
. ~~
, :
.. "
,,'(
f'
.. .._1.... .
38
r is the exact root .where,
Estimate the":error
, .....'..
Perform ,~rw~,steps ,0j"Nf!!t'tQll'S method to find Xlfor the' ,(.oQ~tn .prohi~i!i6" ,. . .. ... ,~. "
7.
. ,...., , .. t,
has a root near .x = .ij and 'lind it correct to>0. ..../~ I,
2 D by' the iteration method and by the secantmethod,
Show that the equation
, , ..II! X .(i,t'"fj!~:x). = 1
6.
Find them correct; (0, ,3:P" ,by .Newton's imethod,
. ~' ...._. .has two positive roots:
Show that the eCJ.uatio¥
x~ + 2xi ~ 7') -+- j',~ 0
5.
(i) the bisection method(Ii) the secant method ,.:(iii) the method of iteration(iv) Newton's method
"Ii".~.
I'iI
..•.....'.", ..
.' .
,41
,.'
"', .,
* * ,,* * *
near (1, 1, 1) '.
.',\ .j,,
x:y z Xl + y'z" = 134'
xi~''Z2 ='. 0.09,,', ..
.1'J
"~
'., .
15. Perform one step only of Newton's method tofind the first ajiproXi'inatioi" to ihe solution of the
.'j;: "system
. '
\
.40
Ij .Solve the system
Xl +- y2 + f = 9 .xy z = 1" ~ .x + y ~ Z2 = 0
by iteration to obtain the root near (2.5,0.2,1.6 J.
x ~ y]. + 6.9 In x == 0
13. By Newton's method find the root near (3.4,2.2)of the system
10. x] + x - r' = 1 y - sin r = 0 . (0.5,0.5), ,
y Xl + yJ = ,1 , XZ.+y2 ;:;-""2 x , (0.5,1)
12. Xl + yI _ 4 = 0 e" + y =1 . (0.9, -2)I
,-~};r.;@:';.M
-."I .~~t1:
,I I
i:r;~,.'".f,a'~( :,lI
,~'I
; , If~;
,I~~?,{,'$,,
~lt;;
I !
Jr~w~j, ..,\.
~l;l ;;t;i':);AI , .,, .'~~;o
"l!~,
I X = A-I B
X '. = ,tf.-1. ~.t.:··,···
A-I.'1 ,by
.,' tv,..".' _',.. '.
':.
::, .,.
':['~'~:~.'g] ,";,' ':.~ • • • I ..
·0' ·0' '0· ··i· .I .=
IA I :/:.(J, the
4~ .
.it! we get
A-I A X = A-I . '.B ~
.,;,. ~ .
.'Premultiplying ."\.
. , .
. (b) Inverse Matrix Met/tod
. Since ..A js ...Q :~q~re..in~!J'i~withmatrix A has an inverse A .
where AI is obtained from' :A' by .replacing: the' columnnumber i in A by B.
~. < ': • i~1,2 .' ". ,nIA1' .',," . . .... ,'. . ~/.
". ' ','~• 'I ',~ •• '
.The' solution is. . ,. ' : -
~..~:.:. t .
(a) .. -Cramer's Rule
In this: case the unique solution. may be. obtal~ed by oneof the following methods: .' . :.
42
(2)fAI 7: O.
. It is well known that such -system may have a uniquesolution, infinitely many solutions, Or 110 solutions at all.We will' be. interested only in systems having a uniquesolution. To guarantee this we assume throughout that the .'determinant of the matrix A of coefficients is 'not equal tozero, that is,
A X = B.
or, briefly,
:;~J[~:l= [~;l'al)Jl Xn _bn .
[::.: ::.:ani 801
In matrix notation this system can be rewritten as:
(1)
Throughout this chapter we consider a system of Italgebraic equations in n unknowns ;
1. IlZtroiluction
SY._stemsof Linear Equations~1:1
,
45
Repeating the same process for the obtained determinant,we get .
~'. ::..
IAI = a1/'
is reduced to .i=2 '-2 Hence IA",J •••• J n , J - ""J n.a determinant of order (n-I) t: . :~
all
:~."
(./1 31J'alj ./ = alj - a" ...
where
". ',:',
.} .,;
alII
all ""
o ai"Y ... a~
; .
1 a12 alII
all all
IAI 0 ~21'-au all)
= a11 a21 3211 - a21all 311
0 312 all)an2 - aliI aWl - allJall all", .~
~.
44,
row;Subtract arJ times tile'" fiist row from I"r=2,3, .,. ,n :
.~. : . ',.:, ~ r~ ~ "1" J '.. •~ aD! ab2 ann '
= '(1'11'- ' 32 t' ~i2i
. , a12" ,1'!. all
alII
1. Divide the first row by a11 :
where we assume that an « O. If an = 0, 'weinterchange two: rQWS or columns and the resultingdeterminant equals 'Iil ;
, , ',~1 . :.:"lr :',
Bll ..a:22
all 312 311l
alnI A I =
These two me/hods may 'be applied to systems of 2. 3 •or at most" equations. They are not practical' forlarger systems, because they require a large number of j.arithmetic operations. For example, in Cramer's rule wehave to evaluate (n+l) determinant of" order n. Tocompute a detenninant of order II by factorization. wemust perform 'II!, multiplications. We now introduce allalgorithm for computing an: Ilh' order determinant ,which requires If multiplications:
{(:'~f:I:... :J;:J~'~~:.Consider the determinant ",
iI,~, ,,
~r...r,'.
, ", '
47
,,' ·51 -l "_2 -3 .:~" :, . 15o 3 --
2
",.; .
2
, , 3T 0 -1 ''2
, " r 52,·02 ....~ 2
o 2 -3 915o 0 .-3
31 0 -1 -2
1 2 -3 4-4 2 1 33 0 0 -3
IA! = 2
Solution:
2 0 -2 ,31 2 --3 4-4 2 1 33 0 0 -3
IAI -
':J-'.' /.
Example (1): Apply the Gaussian elimination methodto compute
~."
, ',~", '~
~"
,jr~ "
46
: '
* '" * *- *
The above method is, called Gaussian elimination..
k == 1•2 , :" •n-l iii =: k+1, k+2 . .... ,n.
a(~-l)a (!c) - a (/(-1) a (!c.1) * 2L_ .If - If - It ali-I) •
where, '"
I~\J ~"i ,j ""3 , 4 •....• n . Continuing this process, we get
, a~l)=: a (J) a (/) * _J
fj - 12, a~9'
where
1S3l a34' ,.. aYilIAI :::::: ~1I * a (1) '" ...
11." ...
ahV art '" a~
••
i"
,
,
,4Q
'''0 f fA j lJ. ] ... :i
. ,.'.iI '
" .: ';'
~~\:, ..-:l..'~:.'
:~!
.!: It is clear that these operations do not change the~: s~1~ti0nof' the system The matrix rA /. I BI J obtained~, : trom {A I Jj I.py elementary row transformations is saler} , to be equivalent to [A I B) ~,d" '~e'"write~: .'" .: ': . .:.' "'
.(iii) Adding a row multiplied. by a constant to .
another row.
(ii) Multiplying a row by a nonzero constant
(I) interchanging twQ rows
We'll make use of the following -elementary rowtransformations:
.Is called the augmented matrix of system (1). Notethat each row represents an equation from system (1). andthe augmented matrix determines ,c<;>mpletely, ,the givensystem.
..~: ,.The matrix
[ all
" qail '" aln
[A I,:'B] = all a2l a2n
anI anl aM bn
',2. Gauss Elimination Method
48'
* * * * *
;: -48
= 2 • 2 '" (-1) • 12
:;: 2 * 2 '" (-1)
:;: 2 * 2 * (-1)1 _ J~
23 _ 15
2
)ri~l,(f~I I! '~~'
li~~
I -i 54
= 2 • 2 o ·1 132
0 3 15--2
'fPJI $I_t·'. I
,
51
'. .All elements' in -the first column below the pivot 1 are
made zero by subtracting three times the first row fro.nthe second and two times the ftrst row from the third
I i • . ( rl ;_3 ri , rs - 2 rr ) :~..
fA I B 1 [r 2·,3
::: '3 *1 22 -2 -1
The augmented matrix is
,') ,Solution:
• r,f
3 Xi - Xl + 2 XJ = 12
x, + 2Xz + 3Xj = 11
Exa~n:ple(2): Use Gaussian elimination method to findthe solution of the system
We illustrate this by means of the following example:
(2) errors caused by the rounding off to a finitenumber of declruil places or of significant digits.
(1) the accuracy with which the coefficients and theconstant tenn.s 'are given. ,'.
50
Accuracy or Solution
The accuracy attainable in .~e values of the unkn9~depends on
* *' * * *
and ~. on .~;we' finally obtain all values of x, by backsubstitution.
an-I. notX,..l -
, I.an-l 0 ,xu
From: the: (n - l;Sf equation we find
From the nth equation we find"::,
TfaG I b]... aln•. . 1. / • J .
~.~a22 •.. alii
0 I' b~... ann
. -th eouivalent augmented, matrices haveThus systems WI.. ~" .' . • .the same solution. We; call them' equivalent systems. -1\ ;
~;
Th· G elimination: method is ·consisting of 'j:e auss . _ ,I'
reducing . [A I B] by elementary row transformations to iithe following eql!ivalent upper triangular form: r :
.i,
,
'i11ItIiIIIii!,, ., r, :
"i:",
.~
53
'. ,
0.68 -0.24/' . i~05 J' " ..:t'."7~".....r?~6I",J ··;.1X'" , ."'y-ll.?-I::'11'.2 '. O:t2 .' '.--.{'1;4. .:~. .., l\.iH12J' 'Cl
.....["~
'0
. .
.' ~ ..
[ 1 '0.68 ,-G.24. '~:~q: "::4"'>'[A I Hj == 4 1 1 rz - rr
3 -9.2 -D.5 -8.2 f3 - 3 rr:'>0
SoluttOfp. :The augmented matrix. is. .:-:
by Gauss elimination method ..Round - off all calculations to threes{gnijic;ant,digits , , " -'. ",',:'
..
3 XJ - 9.2 Xz -.0:5 Xi - 0.2
1.5,
Find the solution of the system~.:~~.'1.. ~lI;:.;t , ~·st {~ .
, XJ + 0.68 ;Xl - 0.24 xJ' = 1.05,.
;.i'
i;
, .'.~., ,}. !
} ;
Example (3):
52
* * * * *
Itmay be. .~~rifi~d}~flt..each equ~ti~l'\ of th~· system isexactly satiSfied.,., by ,tllis solution. Here .we obtained theexact solution because no rounding off was made.
n - 2 Xl - 3 X3 _- 3Xl =
1
11]:3, " .20' iT3 - 6 f2
T2 / 7f3 / (-1)
The solution- is
By' back· substitution :
, 3·· x, X2 ='J 1 3 = 1
[1 2 3' 11],... 0 1 1 3 .0012.
[
1 2 3o 1 1067
[
1 2.3o '-7 -7o -6 -7
IA I B )
1m,
55
Solution:The augmented matrix is
Solve, the' system in' example 3 -by Gausselimination method, with partial pivoting
, 'and by using the 'same number, of ..Significant digits. '
.. ':.,, ,Examele (4);
* * * * *
5. Repeating the same 'procedure for third I forth I ... ,
columns we reduce [A I B J to upper triangularform , from which the unknowns are obtained byback substitution. '
4. ' .Suorracting appropriate multiples of the second rowfrom other rows below, we reduce all elementsunder a:zz to ·zero.
" ,
3. Consider rows from the second to nth. Thesecond pivot element will' be the coefficient of X2
with lar.gest· absolute value placed in· place of a21.
2. Subtracting appropriate multiples of the first rowfrom .other rows, we reduce all 'elements under all. . _ t~:: "to zero.
54
.; We ~y :~~t~:~\;,rri~i~i,~c~u~~p.solution "by usi~,gmore ; .sil¢.ifioant ~i~i~" ;' ,9~"by; using .the :so -':called ,partial,:pivoting n : .,'" '" ~.
1. In the .first ~l?Jurnn we choose the coefficient of xiwith, 14rgeii' dbsoliitti I'vqlue,' and interchanging rowswe pui' .it m.,~~,pl1ice of 'all, . 111is coefficie~twith' /i/rtgisl hl>SOlure' value 'is called the first pivot ,:.elemetu ,
* * * * *. ~..\ " ".
. '.... \. ~
The, difference .betweea the approximate and exact "values of the unknowns was caused by rounding • off all 1
calc::uJati~~ "to ' three ':slgtJljktint digits .
x, = 0.15
The exact solution of"~:the'gWen' system is
XI ::: 1.05 - 0.68 Xl + 0.24 XJ = 0.145
~ ;',' ..', .. .'
1:-2:7 .;. '1.96 :x)·· = 1 01.-1.72
- 0.494
" ','.Back substitution' gives
i.'J
~;:,,I .~-,
, '
:j',I
,,,,
.:....
1.00
57
~\i: '"~"'.:
"
t,
.. i" '":'0 .
.: / .... :..:.'.::::; ';,,'"
"+ .". • .. ' r~") \.' . I
56
, .',' ., ... ,,'.
"
1 1 1.5} i,
-9.95 -1.25 -9.3~, ,+.,~ \ '049 0675" ,1'3 ~""~"95 fl.o.~~',,_.:-'J"';l' : l!' ,:. ..'.. ", ,\ ~; ~ , ,
, ', ,
,'.1.:
* * * * *
The attained solution IS cl~aj./y more accurate,
0.245
" 1'" "15 ], , , ' ,1;" , , " '~
-OA9': 1/6:675 '-1.25 -9.33
10.43-9.95
.:..~ .'
" ," '
[
4- 0
o
':'~..... ',~' .
" ., ...~. ..
["
i.s ] 14 1 t.. T2 - - n(A I B} 1 0.68 -0.24 L05 4- 3
3 -9.2 ~O.5 ..-8.2 f) - - rr4
.' ,:: .... + '\/!'
! . it.:
The" coefficient in' the ,first 'column>with largest,absolute 'value is ail -= 4':' " We 'intc'rchange' the first:and second' rows:
-0..24' 'j: ,1,05 ]".. ''''',,' fZ''"I 1.50'-0.5 : -8.2
0.68.1:
-9.2
,[ 1~.' ~
, ;.
~tJr'~l!~Fl
FJ
I~
.
"
,
59, ..:."
Interchange the second and 'third rows, and then' divicethe second row by -6.75:
[ 1 0.375 -0.125 0,25 ],... 0 ,0.615 >_ 4~125 3.750 -6.75 -1.75 4.5.. >(
0.25 J '" : 1'2 -', l'i4, ,','" "5: ' '",r) ,.-'2n
O~375 -0.125'1 ,,'-:' '4-6 '" ;.2. ':; ...,. ..' \":.
To make the first pivot element all the largest inabsolute value in the first column, we interchange thefirst and third rows ,and divide the first row by 8 tomake a11 unity (rJ H ''1, and rt 18 ) :
[
2 -6[A I B 1.= 1:.' 1
. '8 .. 3
.The. augmented matrix is
" '
Since 'the solution is required accurate to twodecimal places, we round off all calculations to fourdecimal places, that is, we use tw'o more decimal places.
Solutio" :
.58
by Gauss - Jordan method' with. partialpivoting.
XJ = 2.8 XI + 3 Xl -
Xl + ;Xl + 4 X3 = 4 ,
EXample .(5) :Findto two decimal places the sclutionof the system
2 Xl - 6 X2 -' 2 x~ =' 5 ,:, ,,,
* * * * *
Partial pivoting is normally used to preserve arithmetic ::,; !
accuracy.
x = C"so that directly
, "·t ~i;
Thus, the main aim of Gauss - Jordan method is to ;:: ;
reduce the matrix A ~~"/t:. 1fri~"'I'J:~~ ,f,:, :.:: ;::, ;.
'. ,.. .In this: method .the elements above and below the ,~:.
dittgonat' ar6'made', z~ro~.~';.,)J§d;f:'.,fuei diagonal elements are ; imade ones .at the. same time that the reduction is performed, ~I .
• : . . .'. • '. ,'( • . ,':t. • :'. '. :. ~. • •
: I.,,'1
. ~
.' :
': '
3~,"Gauss',.,-,'Jordan Elimination MeUlod,
1
where
(*)x ::::C - G X
1\ '. "';" ',4 I ,'I'
or, by using matrix notation ,
.:
. 1 n: Xl .-:= . - (. hi - : L .aU, .xJ ). " Iii " i = l,~, ~•. ,IZ
,ali : j = 1 ,,'r " '. ,I. :: j ¢ i
..or, briefly
(bll - ani XI - '" :- a". n-I Xn-l) = Fn'1
........................................................................
(l2n xn) -,' F2"'? ::::
(ill- alu Xu )
=XI Fl=(bl - a12 .x~1
We apply the method of iteration to find the solutionof system (1) . For this we .rewrite system (1) in the form
4. Jacohi and Gauss - Seidel Iteration Methods
60
",
.:.. f Of, • ~ ~':'.
* * *.* *! , •
x] == 1.OS'== '-0.94
two' declmal places is
XJ = 1.0514,Xl .== - 0.9393,
..
0.7336: J'-0.93931.0514
oo1
\ I
)
-0.22220.25933.9629
",: •..',.,
Xl = 0.7336·
[~
01
O· 0.' .
n ' 0I0
:1,:'
'The '~6futiozi is
. :,~. .:-, "'.
The solution correct to:. .. , ..
Xl = 0.73 Xl
.' ..
63
.," .
,,\(. , r , ) •.
NoteTo make condition (*.) satisfied or. nearly
". satisfied, we choose 'all CIS"tire .:l:flfgest' .coefficient in absolute value of x~·.' The
. , equation with such' all is the first equation.The coefficient all is taken. as ,the lar1f~ ,
.', .'
•••• '01'',..... ",. ,.'
. : ....."'r_ l..>' ~r: '.' .':r :..' '.1 •
3. '.This method is sd! - .eorrealng ,If an error' is made.:i=l,:J,.....n
62
t 1 aij I < 1j = 1j ~ i
11 ail ·1
+ 1 a'Fi I' + .•. + 1 a F~\'. ,< J ;a xi . a'x'n .
that is,
I~Ia XI
This is 'Jacobi method, A sufficient condition forconvergence of X( n} to the CX4ct solution is .
The adpantag~ of this' ·w.~od are
\;~il"; .: 1.. If the' ~fficl~9.!Q.l~ J:tas\I.D~Yzeros (sparsematrix), the method of iteration may be' more rapid
..':,i " " than 'Gauss and Gauss - 'Jordan methods.
~.\I .,.: .~ " '; 5 . :', f'. /:'.
\Mlt~n.this condition is satisfied, x.(") will converge toexact solution no matter what initial approximation is used.
.. .; .~"\'~!j/"":" ",'
.: .....
",
',:: i :
I ai, I >
)((n+l) = C - G Jr(~)
. ....:'~ , I,'! -:
or
'n=0,1,2, ....,
x(o), ...2
to the solution, we obtain, successive iterates for thesolution by
, ,
'Choosing an: initial approximation
and," , '
0 au aln
all 'au
G .... ....=ani an2 0-ann ann'
r'I 'I
6S
.: '\ .
So(
'X/I} :::; 0.75 r X:z(l) = 0" (1)• , X3 == 0.8333 ,
First approximation :
Here we added two guarding decimals to therequired number of decimals in the accuracy.
Where It =0, 1, 2, _ .
0.3333 x/n)A, 1667' (a)'-. XI, _
Rewrite the equations m the form:
X/II+1} = 0.75 + 0.25 xi") - 0.25 x/")
xilf+1) ;:: 0.4 '_ 0.4 Xl("} + 0.2 X/If)
condition ( •• ) is satisfied and we have theright to choose any tnittal approximation, say
X(o) = (0 0 0) T.
I "f > ." -1 I + f 1 J::.:.
I5 I > (2 j + I-1 J
161'>"11 + la l
64
Since
Soliltion:.Fitst we; rearrange eq~ati?ns according to theabove note:
. '.:
Use Jacobi method to find, correct to2 D, the solution of the system
Example (6):
* * * * *
. coefficient in absolute value 0/ X.z in theremaining equations t and so on.
B'M~~, I
Ij.IiI
~]r~Ir'~lIja.,
.. ~
.,
67
*,* '* '* '*
When this is done. ' the method is known as "Gapss-Seidel" method. Thus in GIlUSS - ~eidd .method we~fe the most recent approximations to the unknowns:
,.',!-' . 'and so on.
instead of x/It} and ,X2(ft)" ~:, the third equation,
X/,,+l} is fotmd by substituting
A betterXl (rr ) in the second equation.approximation
x/n+1) is found as before. A beaer approximation
A modijicaJ1on to JaCObi method is done asfollows :
* '* * '* '*
or (not rounded )
..
66
Xl = 0.26..
. .:~r ,;'~':'I
Thus correct to 2D ( rounded)
x/6} = 0.6359I -" :: !,
.. ")' '
.. :, XI (5') ,e ()6S~8:'', 'x'" (5) =.' , ._J,~" 1
, "
" ' \
xJ(~) ::::: 0.6350
xiS) = 0.6380
W.e,must continue this prooess until the valuesof all unknowns are the same correct to 2D intwo successive approximations:
Third , approximation :
x/J) == 0.75 + 0.25 * 0.2667 - 0.25 * 0.575 ... 0.6429
xz(J) ::= '0:4 - 0.4 * 0.6417 + 0.2 "0.575 = 0;2583
x/J) =: 0.~333:" 0:1667*0.6417- 0.3333 *0.2667= 0.6374
x/1) = 0.75 + 0.25 I< 0.4 - 0.25,* 0.83333 :0:; '0.6417
x/2) = 0.4 .,:o.« * 0.'75 + 0.2 * 0.83333 == 0.2667
x/2) = 0.8333..r- 0.1667 I< 0.75 - 0.3333 * 0.4= 0.5750
Second approximation:
69
~ '=••"
* * * * *
TIle solution 2 D ( rounded) is
'XI = 0.66 1 XI = 0.16 1 XJ = 0.64
(3) x/J) =- 0.6645 x/J) = 0.2611 , x/,I) = 0.63551
",'<#. I,
(4) xi"! = 0.6564 x/4) ==' 0.2645 , x/4) :I 0.6357
68
(2) X/2) , .; 0.75+0.25* 0.1 -0.25*0.6749 =0.6063
x/2) = 0.4 - 0•.1 .f 0.6063+0.2 (< 0.,6749= 0.2925
x/2)= 0.8333 - 0.1667 4< 0.6061- 03131* 0.2925"=0.6347
, ! ':'\" •
, '
Xj (I) = 0.8333 - 0.1667 *O.75 • 0.3333 * 0.1:::0.6749
X/I) = 0.4 - 0.4" 0.75 = 0.1
(1) X/I) = 0.75
where It = 0 I 1 , 2 , •.•
0.3333 x/n+1)Xj(n+l) = 0.8333 _ 0.1667 x/n+1)
X/,,+l) c:: 0.75 + 0.25 X2(n) _ 0.25 X/If)
Solution:The used equations, are
x( 0) == (0 0 0) T.
Solve the system in example (6) byGauss - Seidel method with
Example (71:
. 8. XI + 3 Xl + X] = 2.0307
3 XI + 6 X, 5 x] + X.( ::: 0.4811
XI - 5 Xl. + 7 X] +.2 X4 =: 2.1659
Xt + 2 X] + 4 X4 = 2.5657
71
6.
Use 5 significant digits ..._i~ your work,
6 Xl + X,
+ X4' =: -7" Xl - 3 X,
+
5.
Use " significant' digits in your work.
f.V~. I·
1.85 XI + 8.78 X, + 1.19 Xl = - 8.74
7.17 Xl - 2.53 Xl - 1.29 x] = 2.62
3.55 Xl - 4.16 Xl + 5.90 X3 = - 2.59
4.
., ....
,
70
2 XI + X2 - 6 X3 = 5"!:
. I
.xi -.4 'X2 ' + X3 = 4 correct to 3D
5 xr + X2 + 3 X3 = 3,3.
3.2 x + 7.4 ycorrect to 3D
=: 4:8 }.
= 5.3
8.7 x - 2.3 y
2..
Solve each of the following systems by Gausselemination :method and then by Gauss- Jordan methodboth witlt partial pivoting :
" ..
first without partial pivoting and then witlt partialpivoting. Comment the results.
3 x, - 4.031 X, - 3.112 X] = - 4.143
. .- 2 x, + 2.906 X2 - 5.387 XJ = - 4.481
- 0.002 xi + 4 X2 + "X.3 =: 7.998
1. Apply Gauss elimination method' wiih calculations ...rounded to 3D to find .the' solution of
rruoremsr!,I
..
73
• • • to",.~
..-. .. > :. ~ '),."''"I.IIi:lI;e the ranke [a,b], where .it is ceqUired,to tabulate the 'solutibU:oya
,''-'''I.~;.c.~. set of points "
".'
isDot easy to evaluate y for distinct values ofx, '
2 2 "'( ) 4 0x .. +y +ysm x+y - =
• I ,,"
: It mayuot be possible to find the analytic solution Of the ODE in terms, standard functions by any of the known analytical methods.
IIZ!;'~:Vlmwhen such an analytical solution Ispossible, its numericali~ll'~;:':~ei~aluLati'(in may be more difficult than to 'solve the original IVP
. For-example, when the analytical'solutioa is a complicatedfunction such as
Numerical methods for 'ODE's are necessary and practical for the'reasons:
By an initial value problem we mean a differential equation or a 's;~t~~ofequations plus initial conditions. Numerical methods for orP,.man'
~lt;;9Jl!lr.~~[ltIlllequations (Ol?W,s) are methods by which we can.construct the solution ofvalue pooblem'(lVP) in tabular.forin,'Thenumerical solution of an IVP is
valuesof the ~ytic (exact) solution ofthls IVP.· . . .. . .
1.INTRODUcrrON
, ," '
,Numerical Methods for Ordinary Differential Equations
72
* ,* * * *
8. Solve correct. to 2D, problems 2, 4 ; 6 J 7 byJacobi method and by' Gauss - Seidel method.
x = 1.45310 'J Y = -1.58919 J Z = - 0.27489
The exact solution rounded off to 5D is
c) Substitute each solution in the originalequations and observe that the left - and righthand sides match better with the (b), solution.
7. aJ Solve the system
2.51 x + 1.48 Y + 1/.53.z = '0.05
1.48 x + 0.93 y 1.30 z = 1.03
2.68 x + 3.(U y - 1.48 z = - 0.53
by Gauss, elimination carrying just threesignificant' digits and rounding off·Do not interchange rows.
b) Solve the system again using elimination witkpivoting .
I
~~I
75
..'. ~ , .
:.... ,. ~
:; :..
.,. .J,\,
: . -:5!:M" ':'1,
.:' . "
• I ..... " ~
then the error CaD bemade ~ a~we lik? by choosing h sufficiently ,smat,. The errorwill.tend to zero35b teads'to O. . .
fd'r~1tte(i.X+h)
· If y(n+l) is bounded, that is, there is a positive constant M such that
74
h h2' hn ()' (\n+1 ( J)y(x+h)""y(x)+-Y(x)+-y"(x)+ ...+-y n (x)+-y n+ (l;). (2)
11 2! n! (n+1)!
Itfollows from this theorem that ify(x+h) is approXiiriatelby. !
This.simple method is called the Euler or1:he'Euler~caU:chy;melhod.Throughout this part the well-known Taylor's theorem is used frequently
states that If the function y(x) has n+ 1 derivatives insome interval vVl"ruuw,5lJl.,". then there exists a number ~ e(x,x+h) such'that
(1)v' = f(x.y),
L One-step or stepwise method if the determination of y n+1 requires only theknowledge of yn .
2.multi-step method if the determination of yn+1 requires the knowledge of, thevalues ofy at more than the previous mesh point, for example, it requires theknowledge of Yo,y n_land Yn-]'
. To illustrate the concepts and the ideas in numerical methods for ODE's weshall begin by a very simple one-step method for solving numerically the IVP
.Numerical methods for ODE's are classified' as
lli..~ .Solve the difference equations .. ~r ·the recurrence relations to
approximations Yn of the exact values )'( x n )of the solution at the meshxn~n~,1,2,.."m, .
Step 2.Approximate the IvP on tJ1C mesh by a set of difference equations or r",,,,,",,,p'niQl~:.;
relations,
(4)
(3)n . J.' \,D,()T (x) := y(x) +_yt(x) + .»: +:-:-,'. Y . x· n . '. n.- ". ' n.·' '.
· 'then fhe error inthis .appio~jjn ~
where h is called the step size, the step length or mesh spacing.
x = a, =i:: + h, x =X + 2h, .... "X =x +nh, ...o 0 2 0 J1 0
,
.·1
77
~ r culations are done to five decimal places ina table as follows:.:: ~, "
(7)
t ..,;~~,ter, we'll use this solution to compare apprnximate and exact values of the solution"J meSh points.t; I From the Euler ~nnuIa (7) we haw ..
::i Y +J=Yn+hf =Yn+01{J-xn+4Yn),q ., n, n ,
:t !9ertftrbreVI1y we'll denote f(xu .Yu) by fn. ': fl.
I ·3 19 4x.Y:=Y(XF-x--+,-e ..., 16 16
:.'" ~.~';".{ :
"i ~~~ren=o,1,2,... .m.WeuseY~~~Yn+I'.instead of Y(xn) and 1? to"~ ·;i~ .,' r . •:~ ifiiicate that an approximation has been made. Formula (7) is known as the Euler.'~ fi;kula. In this way the IVP is replaced by the set of the recurrence relations (7).,~. ':..:::~ ...!..
~L§f.ep3.iII, Since Yois given, we use (7) with 'n~ to obtain the approximation'~~r1toy(x J)' Usmg Y, an.~fQnn~O)with n=I, we obtaia the approximation?? 10
1.~~("2)'. and so OD. We wilJ have a table of approximate values of the solution at the.; I '.
;~~eshpointsxj."2 ..... xm ..
:i ~~m Ie 1:. ,~ :', ' Apply the Euler metbod with step size b:;().l to find the numerical solution
solution, by a1.flllect to five decimal places oithe IVP'''1': ,~.
, Y'=I-x+4y '1, y(O)=l'.....___.,. .
'* ~ the interval (0,0.5). Find the '~ytical solation y(x) and compute the error~~,J:; Y(xn) - yn ineach step.l. :,':. I
76
For x"" xn we Write
Since y'(x) ""f(x,Y(x», we get
y(x +h) ""y(x) + hf(x,y(x».
~Approximate y(x+h) by TJ(x)(see (3», th.a1 is,
y(x + 0) ""y(x) + lIy'(x).
where h is called the step size, the step length or mesh spacing.
called a mesh. Usually these points are equally spaced, so that
x :: a x := X + h x = x + 2h, '" ,x = x + nh, ...o 'J 0'20 no.
,3""Xo <xl <Xl <"'<Xm =b
S~pl. , ," 'Replace the range [a.b], where it is required to tabulate thediscrete set of points ' .
The Euler m~tbodinvolves the following three steps:
(6)
subject to the initial condition
(5)s'> f(x,y).....
Consider the first-order and first-degree differential equation
2.THE EULER METHOD
79.
n~te that ~le erro- .at x=O.2. 0.4, .... 1.0 with steji'Siz,e, 0,1 is approxiiilately.luilf~th step size 0.2 . For example the error in computing y(l) with h...0.2 is 0.22995with h=O. t is 0.12453. ..Without knowing the exact solution, we se/from the .above tables that tJ~~'". ,~t x=<l.2, 0.4, ... , :.0 differ in the first digit, so it is netpossible to obtain the 'digit correctly wit\ 1?=O.I. Woe had to repeat computations .Wi~lsmallerh, 'or to '.InOf7 accur~te ~umcncal method, The reader should remember that the 'Elller
,~",""'''1II1nlS}~9t fccurate 'nollgh because it is a first order meoio'(i since its 'edor isabove by Ch. ."
78
Case 1: h=O.2
where, for brevity. we write fn for f(xn.Yn}·
y. 1 = Yn + hf =Yn + b(x.o +Yn) ; n=O.1.2, ....... ,tr+ .n
The Euler formula for the given IVP is
y(x)=-e".x-l.
SolutionThe given differential equation is linear and .so it is easy to show that the
solution of the given' IVP is
fit
in the interval [0,1] with step sizeh=O.2 and then With step size 11=0.1. Find the~Nf{~~-r,~~7-t~2~T-'+--~~~_-f~~~L+-~~~_Jsolution of the given IVP. and hence compute the total error in each step in bothWhat can you say about these errors? .carry out all computations to five digits, If the six digit is 5 or more. round up,
y' =x+y ; y(O):'=(}
E19lmple'2:Apply the Euler method to obtain the numerical-solution of the initial.
problem
We see from the above table that y 5:::6.32416 is the approximate ','>.11.,,....,,,,,, ¥"""'-,"'_'''y(0.5)=8.7l200, The error in this approximation is 2.38784, which is relativelyThus the Euler method with step size h=Q,I is unsatisfactory for the given pro U4Cl1lr1MIH",
~ (1'\\"V·lI'(J'\. .....~J''\J.J'I\.1\Ul~n\\YJ,\T(J' -'" .
t'.' Yn. Yn+J=Yn+
n Xn Yn fn '" J - xn + 4'yn' 'Yr:+-J'= y,?- +01fn Y(?<,{+'J) 'Bn+i~'., :. "
0 0 ' '1.0 5;0 ' ..... 1.5 1.60904
1 0.1 1.5 6.9 2.19 2'.505332 0.2 2.[9 9.56 3.146 3.830143 0.3 3.146 13.284 4.4744 5.794234 0.4 4.4744 18.4976 6.32416 8.71200
'. i1
,i,I
"
\
"
81,80
Thus YI"i~':ili~approximation of Y(XI~ ~Ccor~g to the Euler fo~ula, Note, that the~iierror m~e ISw.ve~by A.B. ~ that ~s error gets smaller as h gets smaller, !~ ;
One: YI IS .determined, we can compute Y; = f(x"y,), which appro~!-Ill~te~~i i1Y•(x, ),' the slope of the tangent line to Y=y(x} at the point (x I,y(XI ), Moving aJo~g~;theJine'passingthrough (x"y,) with slope f(x"yl) to (x,.Y,). we find that {;, ,
!; {
\', '"
Yl'= v,+ l(x",y,,}(x1 -x.,)= Yo +hl(x".yo),
y":"Yo= f(Xo.Yo)(X-Xo),
Moving&o~g,this tangent line to thepoint (x/s,). we&d
..~... 'Hence, the equation oflhe tangent line at ("o.Yo) is
~ig(l)o
1I"
i1'j,L:J :. '.~:...j i,'I·,,:~!':,j:l'J I...• '
x
y.~.,
...... ., The:,geom~t:ri~ .interpretatlon of the Euler method is shown in fig, (1), ~herej: ,
the exact :SO)UtJOIl IS represented ,by the dashed curve, The CUIVe of y=y(x) on] ,xo,xo+h'is replaced by the tangent line to thiscurveatthepoint(x ..,yJ,Since[; so that y" is tht( approximation of y(x,) obtained from the Euler formula (7) withx.and Yo are known, we can compute the slope by using the ODE as follows : I, n=l. We may oontinue in this manner to,obtain Ym• which is the approximation to
, the slope= Y'(xo)=f(xo.Y J. I' y(XIlI)=y(b)· It is easy to see that . ,!
','0,
Geometrical Interpretation of1he' Euler Me1hod
83.
. '.'~'.
. .: The Euler method is used to solve numerically on IO 1] each of t1 c. II .lVP'lI: . .rj eacnor rnc ro owrng
9. y' ~2y-I ; y(O)=:l .
10. y' = f-X+2Y ; y(O)=l.
(a) Estimate the local formula error in terms ofy(x)(b) Obtain a uniform bound for «1t+l on [0,1](0) Find the exact solution and usc it to obtain a more accurate bound I:
(d Iflor en+1
) ,h=O.l compute bounds for e1 and e4., ,',
11, Given Y.' ?(x+y) ; y(Of:=O.5,find y(0.5) by Euler method with hOC{)"05..C~mpare with the exact solution. What accuracy is 'obtained by usingh-O.Q2S.
, "jo'" I,
"., ._.
82,
7. y' ...Lx+2y ; y(O)=1:2
, 8. y' _.x1+yl ;y(O}=lFirstusc onlythree 'digits and1h~nuse four digits. ' .Col11pa're the results rounded to, three digits in both cases. Note tbM thedifferences between them are due 'to the rouocJ..offerror. The round-off errorbeCOme important ifthe calculation requires m3uy steps.
Apply the Euler method with 'step :si7~h=O.05 on the interva1[O.O.2]to .......~.........numerical solution of each oftbe following IVP's :
2. v' =x../Y ; y(2}'=4. Find y(2.5); use n=5.
3. y' """,2 -y;y(I}=O. Find y(L6); use n=6.4. v' =(y+l)1x; y(2)=3. Findy(2.8); usen~ andrS.5. y'=-.JX+Y ;'y(4r2.Findy(3);use n=5.
6. s'» xl +y2 ; y(tr2:Find y(O.5); use 0=:5 and D""lO.
Use the Euler method to determine the indicated value of y for each ofthe followingIVP's byta~ the indicated numberof subdivisions n. Ifpossible compare with the exact value:
1. Consider the IyPv' =2y-1 ; y(O)'"1.
(i) Find the exact soluJion y(x). '(ii) Apply the Euler method with step size iF'O.l inthe interval[O,05).
Compute the error En =y(xn}-Yn ineach step.Use four digits cnl.y inall your computations.1fthe fifth digit is5 ormore. round up.
(iii) Repeat (:Ii) with step size h=O,05.Compare tllC errors in (il) and (iii).(iv) Repeat (ii) with <i digits 1D all computationS. Co~are the errors in
(1i) and {iv}. . '
Froblems
,
\1 'I.
!,III
Ii1'1II:
IiI
II
. 85 .
(3)
. , . . ,.
We-set fn==i(xo.Yn)andjf n+J==i(xn+bY n+'): S?).improved Euler formula is
.. ,~terval[0,0.5}. ·COQlPu.tj: 'the ~q.oi17 .in each step ahd compare with that:~lIi~}\!!'''''''''''''''before in example (1) .in.the previous section.(2)
y'=I-l).+4y ; y(O)=l
the Improved Euler meili,i;>o .with stJ::psize h,;"o.l to. solve numerically
,0 .. '
"
h· .'y n+J=Yn+'2[f(xnSn)+f(xn+l>Y n+l)J
84.
It is not difficult to see that the corresponding value of Y is
~--------~~----------~--------------~xn . h xn+l Xxn·+-
2
..
y
Then we correct this value as follows:
We, approximate the exa~t solution y= y(x) on [Xn•.lCu+i] by the straight line
(xl"Yn) with slope ttxn'YTl)~'and then we continue along the straight linewith. f(xil+J,y* D+'I )uutU x reaches "n+I'
We· want to approximate the solution at themesh points 7? +nh ; n=l,2;· ..... the interval [Xo,bl.
'In each step'we first use the Euler method .to predict the auxiliary value
y' = f(x,y) ; y(xo) = Yo'. ".,
As before we consider the IVP
'is; .the required improved El4~t formula. It,is sometimes 'l&llcd die improved:c........ u•.,u,Y method or Heun's inethdd. '.' '. . .' •. . .
method is predictor-corrector method, because 'in each step we firstapproximate value for y( ~n+ /) by (2)and Jh.en w~ correct it by (3). It is also
ori~":step method, .because . at! th<- information. at .xJl'+J. .i~obtained from the
'In: tb,is section we make. an improvement of the Euler method whichmore 'accurate. It will become a second order method, that is, the IOca!l,,,;wwliS.cmmwill. be .of order?? , and consequently the accumulated formula error will be. 1 .h .
(1)
3~11IEIMPROVED EULER METI{OD
....
; y(O)=!'1
"'}")(+2y ; y(O)= I
37
Case 2 :h=().05
•... ..y +0.1 fn and y =Yo+O·OS(. fn+ f n+'})'Y n+1 Q n+l· .
the improved Euler method to the IVP. y'::: (x.+yr2 ; yel)=-} .' ','~.,y(2) numerically by choosing h=O.5, 0.2 and 0.1. '
--~~~-~-T~T~r-4~~-~~~r--r~~~~~~M~s~e~aro~tioo~aoo~~'~~Od~~~~~a?the differential equation by using the transformation x+y=v, andcompare with the obtained values.iinprpved Euler method find y(1) correct to at least two decimalfor the IVP .
y' ::!i(:e- xy ; y(\)=1.~~~~~~~~~~~~~~~~~~~-~~~~~~~~~~~:e~h~thefu~wfugI~~fudthe~a~scl~o~and~n~u~ffi~~~W~_l .L _J _L__ -:-_~~_:"":,,,~~/ Qetennille. the error when the improved Euler formula is used on [0;1].
l'roblemsCasel:h=O.l!
II
'.' The exact solution is y(x)=e" -x-I, so that y(0.5)=O.1487. The error when b=O.l112 and when h=O.05 is 0.0003. Hence, the reduction of h from 0.1 to O.OSthe error by a factor of 114.
the two obtained values are 0.1475 and 0.1484, we arc sure of the accuracy oftwo digits. The required approximation is ~.1484.
Exampte2.·· . '.,: . ' ...' AppJy.ihe Improved Ealer method.to the IVP
y' =x+y ; y(O)=O
to find y(O.5)correct to at least two digits from the J~ft:
Soltation . d tJ Iv i . ith hI2,.~u!-:'+:":=-+-"':":";=~+-~==-I--=~:'=""-+-~:'::'::":~""':'=::",:",,-+-_:::,:,,::,;:,,:,,::,::,_~We must apply the method with a suitable h. an len app y It agam WI .~~~~~~~~fub~~~~~~~~~~~~~~~~~~~~~~fr+~ffi~-~~ffi~~~~~-~~~~to at least'two dig;t.s. Ifiiot, we ~v? to appl~ ~~. ~elhod agam Wl!h smaUe: h, sayand so OD. We must also use at Ieast twomore digits than are ~ed.
Since the improved Euler method is a 'sec:ondorder oD.e,.It JS reasonable towith b=O.I; where we round up all our computations to four digits.
For the given IVP we have
~e~~~~~~~~D~.~a~CYci~~ro~~~~~k~~r~Y~n~'-~f~n-r~~-~7-~-~~r~'~Y~~~l-'is better than that of the Euler method. '. . .... . . .' ..' .
0,
I.
8988.
on [0,0.5]. Use six digits. Compare the error in evaluatingy(O.5) with thoseearlier by the Euler and the improved Euler methOds.
", ~:"', .....l(
IVP.', ','
Example!:Apply the second order Taylor method' \Y.ith h=O.l to solve numerically.. ' ',:' ..
. ',..-,
'; ...", ".'.
. . .; .,' ...where fn = f(xnSn). (fx)n::: fx(xnSn) and (fy>n= fy(xn.Jn).This is the three-term Taylor series formula .or the second OrderTaylor formuia.
'The three-term Taylor series method is a second order method just likeimproved EuJe~me~l\od.that is', th~'errorin Yriis~fthesam.e order h2: En =
"
error in evaluating y(O.5)' by this method and ~e imprOveq'EU1~llm~thod.jS'tAe.Note that f is a linea; function. Of course! the.acCw:ciGy.fP(:~e second qrd~~.
·"." ..... 1_- method is mucb better than that of the Euler method.(2)
, ,1,
where 17 . Approximating 11 by the sum of Ole first threeonly, 'where we replace n 'and expressing y'(Xn)8Dd Y"(xn) in terms ~~~,:",-:-+-.;;.;..;.~..:;_+-__;;:.;:..:..;..::..::....:;...._-t--~~;;"-'-___,~~:;;":";;'-l-:~";';;;';~1{x,y) and its partial derivatives. So we get
"'Yn+OJ2f -OiJ05I)
" :
(b=O.I)
As ',- th Eul th d based •. () By ~ect differentiation ,we"k,ve 'we xnow, e er me 0 was on approxunatmg y xp+JT,(xn);"y(xn)+ hy'(xn) 0 and then replacing Y'(Xn) ,hYYn. More accurate -x+sy ; fx(x,y) = -1 ; and fy(x,y) =,4.can be derived by approximating y(xll+/) by Tk(~) with k >1, that,is by blkiinl!~1IF:I1.pPl)(lIlg'fo~ula (2) to the giveu IyP, we get, 'or more terms of the Taylor series of y(xu+J) about x=~ .
.In this section we derive the fonnula co~ to t.aJcing three tamsTaylor series of y(x~,). This method is calledthe three-term Taylor series mdhodthe second order Taylor method. "
Assuming that y=y(x) has continuous derivatives up to order three (or thatcontinuous partial derivatives up to order ' two) on the intervaJ [xo .b], then wewrite·
y(xu+)=Y(xn)+ hy'(xu) + ~~y"(xn)+ ~~y'''~n)· (1)
4.1HE SECONDORDER TAnOR METHOD
,~}~,
,I }
II .~, ., '
II{I III·"~ r
\,I .
... ",.. " .
,
91
....
4. ?? =~; y(O)=l. Find y(O.S);use b=<1.l an\fh~.05. What i~th~accuracy' oltIieI+ x. .one can sure of?
1 ' .2. y' ='2 "X+2y; y(O)=t. Find y(O.4); use h=O.2.
, 3. Y',.,.,.:?+yi; y(I)=2. Find y(Q.5); use h=O.1.
.'
Use (i) the secondorder Taylor method(ii) the modified EUler method to determine the indicated value ofy fo~.. e.a~hof the following IVP's by takingtbe indicat~ step si~e : ',>
1. v' =x'1.; y(O)-=O.F~d y(C, ~\; use h-<l.l.
"
90
on the interval (O.O.5].Compare with the results obtained 'by the improved Eulerthe second order Taytormethods.
y'=l-x~y; y(Orl
Example!: ,Apply the modified EulC1' melbod with b=O.l to solve numerically the IVP
NoteWe say that two onmcricaJ methods an; equivalent if they have the same
Thus. the improwd Euler. the Ihree-reon Taylor series and the modified Euler metJllOdSare eqoivaleol, because each of :tb~ isof the second order. The results obtainedany of these melllods will s1igb~ differ 1iom tlJatobmin"<l by another o~e of them.
, , .-From this table we see that tM results are the same obtained by the improved Euler~e~od and by the' second order Taylor method. Ifmore digits are used, then slightdifferences may appear. .
Yn fn xn +0.05 Yn +O.o.5fn fn y n+1 '
1.00000 5.00000 0.05 1.25000. 5:9500' ;1.59.5001.59500 7.28000 0.15 1.95~00 . 8.6860 2.463602.46360 10.6544 0.25 2.99632 12.7353 3.73?13
15.6485 0.35 4.51956 18.n8Z'" 5';'6099523.0398 0,45' 6.76194 27.5978 8.36973
This is the m.odifed Euler' furmula.The-modified Eoler method is a second ordecmetbod.
. .Hence, "(xn +r~Yn +k)is ~ximate1y ~ to the expression in (3) betweendie. . h II _. .. ' ,braces {... ), whetll""i andk='2fn'_
This suggests thcfunnula ~I\
,.~
'YD+J"'Yn+hf(xn..f.1.Yn+'}fn)' (4)
On the other band, by using the Taylor formula for functions of two variables, we }l~ge
We apply formula (4), where f{x.y)=1~X+4Yand h=O.1. We will also writefor t ('-x. v) - 71
~~n ~:~~5~:~n+~.o~~D).:::' ~ '" .
.,.:• J ••
Solution
(3)
The second order Taylor formula em be rewrittm iqthe form
.5.nm MODIFIED EULER. METHOD .
.! .' w=h(lrX.f:4y)... :.."A;y·,n x y
" ,',.'.
"',. J.OQrQ9.0. ":0':500000'" . 'O,5,OQ,OO I,0 O~OO\:. ():S9£6d6'" ,. . '.' ",.
0.05 ···.i.~S090.: ,. 1.19000' "~'0:12$0'0' .. 0.05 1.29750 ~ 0'6t~6oo;' .
0.10. (1.61400) O~7j5QOO'\ .. 0',73560'If '-'~' , ..' . '. ,. .... 'J'b5360 ".
~.,";""",<, "" .'" ."" •• '
{Jd(j~O-!-~ (3.O'5;6ii);'1.6b~9~./' I •.
, 93
So .:
..• ', '1,.,
For the given problem we have,..
.; ;';AND so ON.
.1
1 'Yj =Yo +-sum, .6
sum. "
.. .~., ',...:.'~.. ,,' '" .,.... ......
'w /0)':2wi?) .
on
. ." .. :
. ". : ,,' r .' . ~ .~~. . ..'We' arrange calc~~~nS, in a tableas follows ':
. .: ,.1- •
Soiuti.on
.:' "y'=l-xit4y'~ y(O)=)
Example 1: "";' ...:':'C. ,.Apply. ~e Ruage-Kntta method to the lVP, '
• • ,,1\', ••','.*t,',' 'I•., .:.:,', t·.)
tPis is the classical Rwige-Kuua'formula. It is equivalent to (1) and hence It is a fourthozder~ethod,ic. En :;OOf') ,,"'~ "
-" ~9" lb 1)w2~-\XD+- ,Yn+-w,.2 2
.J, :;,' l~" .': :, ' •' ••• :" ',where
(2)
95
, 'Y'=X+Y;y(~)=O.Find y(O.4)~use-,h=O.2.
For each of the following IVP' s determine the indicated value of y by using the!1F.i~\In~:'e-Kutta 'I~ethod with the indicated step size. Ifpossible compare with the values~I1,~.l1fa[nedby solving the IVP exactly: '
,r....oblems
, The Runge-Kutta method is an accurate and practical method and itis frequentlysolve IVP's nwne<,ocaUy.
-,
Note that the error in evaluating y(O.S) by the Runge-Kutta,ni-cthod witli h=O." is1200~8.7092~-:=O.00275, while by any of the second order inethoitS it was'O.34227,by the Euler method it was 2.38784'. ' . ';. ":. " .''. \\ " ."
, !
/ '" 0.1172-0.122933xn +1.491,73Yn-
• it is easy to find the numerical solution as follows ~
the Runge-Kutta formula
94
;; .
"w./~hlt~~+:h~Yn+W3>" :', ~=(trp~Xn+o.l)+4(Yo+().11~.124xn +O.496,Yn)}=0:1372..:Q.1496xil +0.598:1-Y~"
\
IIi!{
'1\x,'Y"'Hc+4y;;'h=O.1.
wJ=lifn ::rO.l(1-~+4y)=OJ~.l Xu toAYn'
'., -J. ••• '.
1,', ;, O~IO 1.60893, "O}133S72, 0'.733572", ,'oiH 1.9tS12' (f87S2S8 1.75058
',' ,:':,,\·g~~6:' ':' ;~~~~~~"": ~~:~:"" ,,' ~:~~~~~,,""'J~,:,.: 5.37643, ..," , ,: 1 ':' ,:' ,, !-2:::'}~08~3+'"6(537643)=2.50500, '
The' eXact' 'Vai~eSto six digits ofilie solution are y(o.l)",i ,60904 and y(OThiS SH6W.!i;thchl~ aceiiraey ofthb Rurige-Kuttanictbod." ,(,'. 'We may solve thi!i:problem i.n aDotiier way as fQllows :We'firSf..exprcs~ the..w~~ in t~ of?1 and Yn.:and then express Yn+ / in ten:ns;oll~''xn~d:"'Yn~Thei 'obtainedformula is used to compute rapidly the succe~siveva!ll,esli;illtr;:-r-7::-T---Y::;:-,y.
For the given problem wehave
97
:, . ..For example. the second order equation x" = f(t.x,x') is reduced to the system' of two::',~t order equations ' ' , .,J"
.v ,
Xn-z::> xn_,
xn-l(t.x.x"x2.···,xn_ 2)
x'= x,.:. x} = Xl
" can be reduced to a system of-first order equations as follows:.. . . . ',"':,
(n) - f(t ' "" (n":'1))x - .x.x,'x ,...•x ,
A differential equation of order 0, where n >1, of Ole form
where t denotes the independent variable and x,y,;z., ... denote tbe unknown dependent,variables. this system is accompanied by the initial conditions
, A system of first order equations has, in general, the form
All nwnericaJ methods preseated-beforc for solving IVP's associated with asingle first order differential equation can be generalized in a very simple way to solvesystems offirst order di.ffere~tial equations or equations of higher orders. '
7.Introduction to Systems of First OnJer Differential Equations
.'. ,
: r ;
I :t·· 'VJ.'!
6. The Runge-Kuttamethod is applied to the IVPy'-ftx) ; y(xoryo'
Show that the Runge-Kuna formula in this case is reduced to. h' JY.n+'J co Yn+,-[f(xn) +:4f(xn+ -h)+ f(xc +h)]., ", . 6 .t • 2
This is Simpson's rule used for evaluating definite integrals.Use this r;osulttQ solve, .
;;";V(l~~h~.y(O)=Ofor y(l) with b=O.2. Compare with the true value.
."5. Given y,,,,,,,2 -y; y(l)=O. Use the Runge-Kutta method to find(i) y(1.2) with b=O.2.(ii) yeLl) with 11=0.1.(iii) Use the result in (ii) to compute y(1.2). ,(iv).Compare the accuracies in (iii) and (I). Compare also with the exact
solution.
1 ' ,2, Y'=Tx+2y; y(O)""L Find y(O.2); use h...O.L
3. y'=x-y·;Y(-lY;2. Find y(O.5); use h=O.2S. (1)
4. Find y(O.S) for theJVP y" =2x+y; y(O)"'1 by applyingthe Runge-Kutta methodwith b=O.25 and with b=O.l. Discuss the results.
,ti.'&" .
'_ \.
t" .:
, i
,98
.1:
Equations (1) can be used to express x'(tn).andy'(tn) in terms offand g. After'x(tn) and y(tn)are replaced by xnand Yn ' Wehave the Euler fonnuia
x(tn.+ 1) 10: x(tn)+ hx'(tn)y(tn+ j) ~ y(tn) +'hy~(tii-)
'f' •• ~," • •• •THE EULERMETHOD
Our' PWPQse is the determination of approximate'values X/xl,,,· an~?7exact solution x=x{t),.y=y(t) of the above problem-atthe mesh points tk = to +k=1,2,: ...Throughout this chapter, it. will be assumed that the considered system has asolution.
'fn'. .., gn xn+l Yn+lxn Yn1.00000 2.0000 -1.0000 -2.0000 0.90000 . . .i.sooo..
0.1 0.90000 1.8000 .0,8000 -1.6000 : "'o'~«ioO'.' l'1~64(fO'IJ0.2 0.82000 1.6400 -0.6200 _-1.240G. ' 0)58"0 '.' /:,"1: 'lf6(V' .
0.3 0.75800 . 1.5160 .-0.4580 -0.9.160 0..712~O ,; 1'.; '2.!f4'·'-,
0.4 0.71220 1.4,244 .0.3122. :-O':~244 . 0.68Q98 " 1;3620, ",,'_' ...: . ' )' ,I' ;",.
0.68098 1.3620 .:' f',.,• ·t .' ." "y •• ' - ~, .~. <", ':'• ': '~':'I' 'r, .: /.! I ~. '. " ,'(~.
(2)
with the initial conditions
(1) } n=O,1.... ,4 .. ',. . .:'i~\ •: I!,: .,
Xn+1=x.n +hfn = xn +O.l(xn -Yn+tn_)Yn+i =Yn -:hgn =Yn +O.l(~~n t;y.J\t\~2tn)
We arrange calculations ~ a tableas f6U~ws :
x.' = f(t,x,y) }y' ,;"'g(t,x~y) , .
• ',:', I.: ..' .
;': : ..",§olution .Euler formula give
Thus, we will consider only systems of first order equations.Moreover, we win for simplicity consider only systems of two equations
,~ :.!.~...;
U five digits. Compar~ the:~~ili~;W{t1rlh~V~~~Sof ~& J~isdol~ti.g~..so ., " .. '
. :.:.x'=y )y'=zz' == f(t,x,y,z).
Ej(a~Uj~s!:theEuler,methodwith ~~..1 to find ~(0!5}.~d:y(j:5j(¥~;:~~ 'tyP;.': ..... : ·:,.'·r~,:,.~:)':.': ~ ..
.x' = x-y+t, y'= -4'l(+y+2t;x(O~1 ,)'(0)=2. '
Also the third order equation x II, = f'(t.x.x' .x") is reduced to the system of threeorder equations .
Using Taylor formulawith n=I, we can write
II
Xn+l~xn+hf(tn,xn.Yn} .. ,' .. (3).Yo+ 1~ Yn +hg(tn,xn,rn)" .~, . t:·:;·i:·! I-... ~: ....
. " ' .: .: th th ,,;,,..,,It·' ;..:, 'atm'" g" "~j '1'o/y", .. J~b·Y.':"";5method i~O(h).'. It is clear at, e euu ·1Jl.eV'dlU =u+.: ' n + ~ ~ .: .: . . .
x'=y }s'» f'(t.x.y),
10 I.
, ,.1
..-,
.' .'.
The s~cond order numerical methods can be generalized in a similar manner asprevious section. Here we only give formula corresponding to different methodsIVP (1) & (2).. .
reader should remember that 111e local formula error in each of these methods is
The exact solution of the given IVP cannot be expressed in terms of; elementary functions. However, we can find its series solution:
_ 1J J J J 7. x-t+-t +-t +-t +...,3 3.5 3.5J
from whi<1b we see that x(D.5) to five digits is 0.543,83., .The error incomputing x(O.5)-,by the Euler method is 0.07355 ...Without knowing the exact solution we can be sure of the digits, which. do not change when using smaller step size.,
j5.KWlCU'\""" that the values of y in this t~ble are approximate values of the derivative x'mesh points.
(4)
'.
100,.tij.
nl(t ii)..
. xn Y'n''. fn
..Yn gn . xn+1 Yn+1o .0.0' O~OOOo.o. 1.00.00 t.eooo 0..00000 0.10000. 1.0.000
1 .0.1' .0.10000 1.0.00.0 1.0000 0.200.00 0.200002,,0.2'
1.0200·0.20000 1.0200.. .d.0200. 0.4.0400 0.30200 1.0604;: ;I'
3.0.3 0.30200 1.0604.' 1.0604 0:62012 .0.40804 1.1224 . ,
4. DA' ,0.40804 1.1224:.. 1.1224 0:8570'0' . 0:520285· 0.5' 0.520.28 1.2081
1.2081
"
}..... . .......
, xn +I= xn +. IUn-xn ·~:O.IYn"'Yn+l =Yn +hgn =Y[i+O.l(xn+tnyn).~
~~~J!\
~I
•Ili~f~,I
:' .
The iWtial condiuonsbecome-" '.. 'x(O)=O, y(O)=x.i (0):;=1. .. : .
Euler. formula are ., '.
x'=y }y'=x+.ty.
X." -tx' - X = 0; x(O) = 0, x'(O):;: 1
on the interval [0,0.5] with step size h=O.1.Use five digits.
SolutionThe equivalent system of first order equations is
Example 2:Use the Buler method to ~:o~V<inumerically the IVP
. .y(0.5)-y J= 1.4261.1.3620=0.06410.
and in evaluating y(O.5) is'. . .
]he total error i.e evaluating x(0.5) is
)«0.5)-xJ= O.71306-0.68098~.03208
103
..:•. i.. ,:;
.~,Example 4j :, ' , '
B~ethe three-term Taylor formula to fui~,x(O.5)fot)~hIVP
x:"~tX~·~x::=0 ; ;x(O)=o~'~. (o,~1.
, .Take 11=0.1and use, five digits inyoti ca1cut4tidbS~
~r~
~~I~\
~1I 'If~s',r'I',:I I
~II
"
'.-.
n 0 1 2 ' :,~ 3' 4, tn 0.0 ,9.1 0.2 ·0.3 Ok.xn LOOOOO 0.91000 0,83805 0.78244 0.74161Yn 2.0000 1.8200 ' 1.6761 1.5649 ,1.4832,; . "fn -1.00000 -O.8HlOO -0.63805 -0.48246 -0.34159gn -2.00000 -1.62000 -1.21610 -0.96486 -0.6832;4,:• 0.90000 0.82900 0.77425 0.73419 0.70745.
xn+l,. 1.80000 1.65800 1.54850 1.46840 1.41490.Yn+1
fn+l -0.80000 -0,62990 -0.47425 -0.33'f21 -0.20745
• -1.60000 -1.25800 -0.94850 -0.66836 ,-0.41490 'gn+ZxnH 0.91000 0.83805 0.78244 0.74161 ' 0..71416
Yn+l 1.8200 1.6761 1.5649 1.4832 " l.4283 '
The total error in computing x(O.5) is
x(0.5)-x;=O.71306-0.71416=0.001 1
,and in computing y(O.5) is
y(O.5)-y,5"" 1.4261-1.4283=-0.0022:
Here h=O.I.Since the number of calculation is too big to he arranged in a horizontal table, ,it is,reasonable to use a vertical table. ' ' ,The calculations in the nth step will be done in a column as shown.
,,'
102
The corrected values are" '
.. . '
, , "'n:+1 :=xn of:h(xn - Yn+ tu) and Yu+ J = Yn+ b(-4xn +Yn~}tn)'., " 4< • '. .'....' '" ' ",, , fn+ i =xn+ l-Yn+l +tn+i" gn +1'7 -4xo.'+1 +Yn+ 1+,2tn+1'
'" ,
Solution ,The predicted values are
. Compare the error with that in tile Euler method, Use five digits.
Example 3:Use the improved Euler' method with h==O,l"to find x(0.5) and y(O.5),
initial value problem " ' , :.. ' ' ,x=x-y+t , y'=4x+y+2t;
x(Orl ,y(O)~2. ,
1 I i. "'Yn+/'= Yn+hf(tn +-h,xn +-hfn.Yn +-hgn)·'-2 '2 ,",2.:""
(6)
I } J ,,' -:xn+1 == xn +h.f(tn +-h,xn +-hfn.Yn +-bgn),
2 2 J,' ' ,I
The Modified Euler method '
Xo := (ft)n + (fx)n xn +(fy )n'Yn' 'Yo -:::(gt)n +(gx)nxn + (gy)nYn'
, 'xn ee fn ' Yn =gn'
where
, , J'x 'I = x +hx' +_h1x"n+ ' n n 2 n >
The three-term. Taylor method
.. 1.;......."I : •x\=·~·
~" ,,,:::~
!j' - t: ~ -x }~~\1'-~F;-;Xi~~3
, \
x.....~ 'j .. i (.>-
104
five digits,
., .. ', : ., '.EXample 5 : . .'
. . Y-seRunge-Kutta method. with step s.@ 1\9).25 to compute x(O·.5)for the IvP ..
x" - txi~x =0 ;.x(O) = 0, x'(O).=.L·
it tn xn Yn O.OOStn O.OOStn2+.+0.1 O.ltn +1/)1
a 0.0 0.00000 : .LOP.OQ· .0.1000 1.01001 0.1 0.10000 1.0ibO 0.1005 1.02012 0.2 0.2020'1 1.0404 0.1010 1.0302.3 0.3 0.30810 1.0922 0.101'5 1.04054 0.4 0.42050 1.1677 0.1020 . };05085.0.5 0.54171 1.2699
·A3befor~ the local formula error is 0(h5). so that th~a~~u1at~d fonn~:;~crr~; is·O(h4). This method issufficiently accurate .. ' . .
We may arrange calculations in a table as follows:
Substituting expressions of x',x",y' and y" in terms oft,x andy in these formulas, weget
are. I(xn+1 ~ ~D+6' vl +2v2+.2vJ+v-/).
'. . . J. Yn+ 1== Yo+'6(w1+'2~2'+~wj +w4)
vv.here .
Th.e Runge-Kutta formulas for'the system .:x'''''ftt.x.y). y'''''g(t,x,y); X(0)=' x~,y(0)= Y9'
Differentiating these equations with respect to t, we findx" ::;y'=ty+x, .y" = ty' + y+x' =t(ty+x)+y+y
=tx+(e+2)y.Three-term Taylor formulas are
· We see before that the error in computing x(O.S}by the Eu1~r'method was 0.02355.
RUNGF ....KUITA METIlOD
0.54383-O.54171=O.0021i.x'==y } x(~:F<l,yeO)=:;x' (0)= 1.y'::::x+ty ..
From this table we see that:?? r.:().54171, while the exact value is 0.54383. Hence theerror made ia calculating x(0.5) is . . .' .
SolutionThe equivalent 'system of first order equations is
Xn+ I"'" xn + 0.1~n +~(O}i.(tnY~.+xn)-I.OQ5xn +(O.005tn+OJ)Yn .
Yn+1 =Yn +Ol(tn;1\ +Xn) +.!.(O.I)2[tnXo.+(tn2+2)Ynl. 2
=(O.005tn+ OJ)xn +(0.005102 +0.1 In+1.01)yu,'
, ,..l ,"
~!!I":'
. ;1'
t.(~I '
;-
4. x"+t2 x'+3x=t;x(O)=I, x' (0)=2. x(O:2),h=O.1.
5. x" -tx' -x=2;~(O,:o, x' (0)=1..;(0.4), b...o.2.
6. Find,x(O.2) correct to three digits for the IVPx" -tx' +3x=O~x(O)=1, X' (0)=02.
L x/=x+y+t, y'=4x-2y; x(O)=l, y(O)=9.:···x(02),y(O.2). h=O.1 '~'. "
\ 2. x'=2x+ty, y'=xy; x(O)=l,y(O)=l.x(O.4),y(O.4). h=O.2
3~x'=-tx-y-l. y'=x; x(O)=l.y(O}=1.x(O.2),y(O.2).~.1
,'\"
106
For each of the following ivp's flndthe indicated values of the solutionsthe Euler method (ii) one of the second order methods (use different methodsdifferent problems) (ill) Runge-Kutta method: (use 5 digits)
i>roblems .: '. .:
Although we took ri=2 only, the total error niIlt1e ID'cbri1putingx(0.5) is 0.000041 .which is much better than any of the second order methods with 1l=5.
x(O.S) .. x2'" 0.25526+0.28853=0.54379,',x' (0.5)=y(0.5) ::::0 Y2=1.0638+O.20813""1.2719.
1 0.250 0.25526 1.0638 0.26595. 0.13030 0.26595 0.130300.375 0.38824 1.1290 0.28225 0.20290 0.56450 0.405800.375 0.39639 1.1653 0.29133 0.20848 '0.58166 0.416960.500 0.54659 1.2723 0.31808 0.29569 ' 0.31808 0.29569
0.28853 0.20813.1
'>( t'I+ \:." 6xl=0+0.25526=O.25526 I yJ=1+0.063815=1.0638.
0.OOO'~0.(j000.0 1-:990Q·" O.~~QO.o --0.000000.125 ·0..125Qa....' £'.0000' o.zsooo , 0.06250, . '. ..~,.:0.125 0.12500 ... 1.0313 '0.2$7830 0.0634780.250 0.25783 1:0635 J '.26588 0.13093
v,
; x(O},{), y(O)=1./.'1<'
x'''''y=f }v'> x+ty= g,
)< --:..)(0 'r ~:;
~.,. ia ~ r:Solufion:
As before, the equivalent system is
h ""O. i: 2)
~ = o. 'tat: ..5 .
3)
BneJ JtV\Swt..rs:i) <i) ~(~)'= ~(' ....c?x)
<in \-.= ~~"'f+42. ,_ £r;. ....~(o •.s.)-~:s ...~.1314-(\il') ......'0 ... I.~'i'6'J. ~o -= ~(o. 5) - ~IO" -0.0 ,." t(iv) !f:; = ~. :"J4+tC .. £5 = -0.13131
Rrst:-p~o~ r·to
,,,, .. .2. """the Eu\er- MeU,QJ
Methods
108
-I"••, .I
:t,;~lil~
,~, .
I~~1t:.I
t!1
r~";~
~.~~~1~rlililr:l
ilO
r--- - ".
n Xn~I'\ f" .'j" -to •
0 0 i / 1·05
1 0.(')5 1·05 \ .j 1. I.H.
2 0.10 .{. I{ t.24- \.I":r
.3 o. \s LI'7 \.3'1 -\. .24 .
~, 0.20 -1.24
.Y(o)=~ .. h=o.cs.3 S"jrUf\c.""nt cLji6.:
I .t e::l=:>C+~ .)~. Cvt~:e. t'f.\(
)-;t" t')::: S or "''''j ,othe.t' n -tf.~, wi II b.e, no, . ~ ~
(!-H"'cr p-n cl ~ -= 4.(; 'i~-n
!:! (2.'8.) -=. 4.6~(z. s) ~ \:14 :::4.6' ;.
£' :::~(2.8)-B :::a4 ;q
NDtel::.\.,aJ,·, tl.H!:,-e is no ~.I"'Y'(lr 11"1 ~:~uJ:-'I(,,)cJ" !:J(~.BJb.~c.~.ust'. ,(..~e. Sl'\uJc\on '\.!S iJ.,~ \\n\t., ~:=:: .;J.',::x: _ I .'
rs., 'de:O\'nelc'r'icp..\ int(;~?....t,b~ion t>f ,Eule,; 4 fe'~""'JlIfa.
"l ~ !i" f" !:i., "I- ,...;....;._ ,
D 2 3 :2 3.11"
i 2.2- ~.4 .2 3, &
2 2·4 3.8 .2. -'t ..!3 2,·6' 4,2 .2 «.«ly 2.8 4.r;1-'
. Sc\ui.\o·n~ hoe
NUYI1e'ri cc;.-}(1) n:= 4-
''i', ,..
·109
E)( ~c:.i. .so\ution '.
r~ ::::(,Ell: ==5- ~ (!S .oi- I) -:::'tn ~' -+- .en cJ ~~\ ) x
~ ~+~ =C:x.X;::...2:J~~3 ,,~4 =.2C ~ C =2
4). ~(.:::
~(2.5) ~ ~5 -= b.4351 _. ~(-?,5)= (;.50(;4-E' ,.. ~(2S) - ~ = o. \3055 5
n ~. !:!., f" ~xnyg;; Y" -+- ( ."Q 2.0 + -4 4..4·
1 2. ~ 4,4 4.4-050 4.8405
2 2.2 4. &40..5 4.840.3 S.~2.45
.3 2.3 5.3.245 S.3~T2. 5. ~SS2
4 2·4 !,), B ~S2 5J8D14 6'.4.35'1
5 2.5 r;.4-3.s: 'I
_,26 :::-=- -+-.2.2
.:#
~ .2Vj" -=.~ + C:. .2+C ~ C-2
the. e.>(o.d~ SC'\..tiPIlf;; = fx d~x =.2.> !:j = Jt- ~ 4-
'.
". ,.,:
Il2
(ii)
(.I)
~\o ': 0, ":=l4="0 '(:;r't)'Ij(0: oS) :- y. ·.··-::;2E' J
I -, _
h 'x 'jn In ,> .' '~""'I,.,.
0 o O·fS 0.2,5 O,.,S.I.2!::.
) 0.(',5 e .SI.25 c>..z.U8..3 O. S'2,&''1
2 0,10 Cl.S'2(,} 0 .•3.3030 O.$43..,., ,
3 0.15 D.s4.3+ 1).:'1(8' 0.56-2.3
't- O• .20 o. .s; b'.z 30 0.428'(; O.,S!3=1
5 0.25 0.5837 o.4g(6" o. 6,9 &0
6 O.3C> O.60BI? 0.552.f ",.;?>-Sb'
'1 '0.3,5 o.6::?s£" 0'b264 O.6~orS 0.4-0 0.6(11 ' o.':J11 s Q.-=I,07:5
'I 0.4.5 O· '102.5 o . s e cit" '1).74:30'
10 n.So o ,'14-30
NUWlt:ri CI<-\ S 0\ ubion :!jn+I. = ~n + \, fn whet-e f~'.:c· 'jn(~'" -+-~")
( ,) 'n = 0 .0.5 "', I·
.~
ru.
::c "(-'? _.'.J,e't7~d~. o· ,
on 4(!.. r-i¥' C~nnoc b.:.: ev~JUAi-~
.fot'l"l. 'The:rc.. ~,O\. hlle_ tor S\lc.\,
The .\nA:~n:J,,...~ dosed. \ rU=~Y'p..I.
I
•
,~I
'i'l(I~,, I
( 'BerPClulli '.;s.-.:I I '~ -~ tt ..,..
_ ~-.:t.'j' "
( -!tnC'l'-'r" t'.~n)
'"T'->e eX II cd: se>1ut::ion I, ~!j , _::IC.j ::::.':1
~ ~~~.~'- X~-I = t"Pu't :1 == ~-I . t· :z' =~ x/·+ x: ;Z ::= -.
.:. .~ to) = O.SI , )~ -= ~(x -I- ~
n Xn !i, fn ~" -e- ,'"0 0 1 I Los,I 0.(15 .1·0.5 '.10.5 \.lo.sa2 0·11) \,; \053 \.2311 '01(",3 0.1.5 \.IUY 1.3&42 1.-~:<\6' I
4- 0 • .:10 I. Z3"
~I~i~ I,
114.
wif.h 5 $ubol.'vi.s.ions. 1:0sc,\ui-im ~(!I(") of tS,<t rvr= t
~(4-) =2 .
.FLI\ E'. r ~....,e.:!:h.cc-l~ (.2,) 1-0)", , r},,=j/".V';( ......~ .l
'~'. "
u., (:~\~Co»1.{:>u·~-e.
\
~(o)._" ',:r -x +- ~
Note,':' Ro~,~d'-:-'r,.rf .t1-11c.o"""fuf;~f.irms /;;0
p~t\.cc:.S. un\es.s. c:>therwis~ sto.kJ.\. f
I) Use, ;~t. E"'c:,\ ......;t~J to .-RnJ ~Ho.~)St~s,ize h::o.os If,, .' s
~(:?) ':"'(1) ~ y(o.s:) zx: 2'1; - Y. == 0.":1658;It' ""Tntt. eJ{.cu:.~ ~~h~e· f:c' ~\I'r c\e:ci.....-.\ ··f\A_c.c.s
~(o.,!;) = O. '7Jt:65
S~· £',.Ae.r"~~~d \4 et .f.';r..:sf= DrJ~the. et'r",," When \, =O.t>.25 is ~rr"xAmed:d~-\:hcJ:. ~"'~'t"\h =:;,0,,0 S • We . heo-ve
!:j(O.5) - t; ==.2 F (I)
.. ~(o.s) - ~ ~.E ,(2)
~C;Y'e. .~ c c.uro..i:e0= .f.o 110w.:s ~ ~
We' c.e, ('I C'bt:AI" 0.
uS'\i)j ." ~nd~.
An d 'l:: hA.ve t>C'llj tk.. ~·r.sf(.C''n.''''c." , \<Ie. {tre ..:!.ure fhat
!:J' .:.! .0. , 54-4-
5)r,ce T;"pl~c..c. In
!j(O ..s) ~ lj, =:: o."1544 (;' Y)'20 • 2
II
fi r!ij I. I
::! '.. 'I'. i, I
116.
:' .
( 2) _ 0 • \ .s .2 6
'.. "
~.
115
'N\fuout flnJ"""a t.h~ ~.I<&\ct bo\uiio" u.s.e, 1::.)..'11..
Vp.\u\'....S f.o',- !:1 (:0>,5) i:.q obtAI"" a "",croe ACCiJt"'""-Cc
~nd "~t.irn.,..te rhe P...-rcr-· whe-n h "",,-0,0.5
').
..A,??'j !1e E.:\,.4 \'er ,....,dkJ to, IV£> s 1<= ..x.~_; '., "-to ~lt1d ~n: Af,?rcx.irnAtc vit-I1.AC h'" ~ (.;>, .5) ....1t;J,
h = o , \ "'-nd chen with h= 0.05,
'118
0.<54-5 ~:-.5"g2-.i.O' 0,213.6" \. S-+-ib"
1.25.8.1
1.5201l .54'31
0\. C~ 1;-'&
~.\191
~.+'l(l3LSIS2
1.54,1(;"
"j
x !:In f.... ~: ...I.,~"Hd
,0 , 1 \.2 0·2 o.~8t:(; \.11'8''1.1'1-31- 0."'1'1(;\(1 \·3361 O'~4-" o',,Ssscr 1.310'.3i. :?I '0· SfJOy 1-4,.3'44- o :« o- 4,~,~'1, ~.41""t'"t-\.4r:J7 0:4212 ISo:?..I o.S" O.300f' 1.4~D.s
f.4~os 0'.3035 \. 55'~:2 1·0 0·2120 t .'5421
\.542\
, H-:>(' ~= e n+ " "''1+1
"1he \~\" cVI' J £u\e,r
'On-+-\ = ~fl -+ ~ (f"
find ~!j("1)' 'c..or-r.tc.-t eo two deci....,pj \::.Ip..ce.s~we
W"~( ofrl!:) t~'! me \l.,cd with S~~%('!-'!: h pnd ~':witl, .su..i..l::-o..bI~ h ~ ~nJ ti-t','" c...o~p"re !:'h~ (>\'t.,../nt>d
.vl « s!:,o..rt b~ ~=D.2 •
..
,117
~(O) =:1
Finer..\ J\n s wers :~) ,'A-okeh ..0.2 lJ( 4)- 2. 3'i 4- ' ).;J O. - '. 0, ~:.!:J (1 zx:
f"r-~el""\..s r' IS+' ,
,?~b\e"".;s ..p. I,,) ~ .2
, ;. "Fir.:>!: :
-.?....~5. "-'e 1.m..proved E~ler M'e.~C'd.
"""The. Se~(>nJ'-O,...c)er Tp.~lol'"'Met.hed. Ihe MoJ..if,~'dMe~od
120.' :-:.
~((').5.) "":'O~BbIS
1 2: S
o·t \. b035 '3.381<
o~a ~.3255 2. ~'l10
,.\ <?3·1 J. 7S~:t., hp;~~,~,0,. '}1;4-3 \''30'13 o. i~,6
, :, "" t:0.8615 :
~'"n +" I
\ .;603.5." .'1.?25~i.l~3.1·
O·cf143,\.\·2g5 .. 0\ ~6IS'
.X n
~(o. 5) ~ O. sG'.B"S
.'.'
Z. 5 ,;0:4 \.61
\',,6'1' '3.4021 . O. 2'il,S 4- I.333!", !
0.20,16', \. \ ~'20
C~7(4 o· Is,,1 a.qst'1o. 1,1 q li- O•. &6"8.s.
'e. I.
= ~ +0.0\ 'X.., -0.01 f (10 _~ ).... ' ,h;, I 0
~ 'jn+\ =- .~,,- 0.\ f.''l + 0.(\,,5 (2X,.;+2,j~f"), t= tam-IX +
U9
f = 2,;><;X·
(2) -(i) =s- ~ -"( ~ 3.£'
~ E ,...~-'\;'-/. - . _ -0. 0 00..2'3
!-;t(I)-'~ =. e'.)(.)-'( ~4£
)ldditiono..l: Usc. ~" o.nd ~ to e.s:ei~~ o..prl""'x,;~&the '(:,1"1"""" .....hen :h=o.1 AnJ (:0 O(,.fA';"
~ tn<Ire. a.ccL..Ira.te. value for ~( ..).
S· y" '-./', .r - 1;r.$1:.. ~ c Jeci~...:J\nc::..e.- ; '" na '<._ tl-A,.ve t:t'I~ UN
f'\A.C~s 'i r'\ 'C-omn-.on ~ we Ar~ .sure ,~
~(f) ::::! L 5 4( b'is C-Q\'"re·c::..C -t..o A.t- lea..:s.-i: -t::;wq deciy.1cJ ~\R-c.es..:
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122
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I2 0.10 '1.0'1",=1 o. If g4-6' 1~'-!t-8 '1
3 O.IS 1.14&'l o.C,'1D3 \.19,.4-
._. ..- --- --_ ..._--- - -- - - --------
8 0.40 1,.?,tloS t . .s.4'TO 1.4.221t;V" 0.4-5 .f. 422'r O.1:!IS,?, f.4G'?'t
10 0.50 1.4'3.":}-
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cr , 9 'fo,! ,.l.'.::n>8,I(.\!2.J(;' ~.,·H·40
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.5 0.5 \.4'4(1 .
h",o.\'\ -+ (ee +o.os)~'..,
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tj;) I\,e h1~c\,~til!.ds '. ~. ~)\+t. :. n
~ (, .: ': ~ :. 4- e4- :2".', I
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(rO"nc:kJ)
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0 0 -\ 1 o i.oSi o· e.s ',.0 S o·'lrfS" -.0.0"115 t • I2 ~.( . o.~.,.'Oof - o. 1~ ,"f I. ';4-"f..!.. ' "
3 \'.\..(.,\3 O.~J~·O. -1).21161 i..l'f'T8
4. '\. I'tT~ O.'l6'lS -o:~612 ....2+.S+-- --- --.- -- .. - ---_ -----_.,_;_
s ,~~S1lo: e, ~ ('2/ -0·.514-5' L42..3+
~ J. 4:,23..q o , l:f3.16 -1'.62.204 ' I. 4642-iO 0·5 ,t. ~C~2.
~ 'X '~I"I .f" (It
0 0 1 i, o.J ,.~ 0.'1'101 - 0,1'16"1
2 0.,2 \.1"180' O.'fe'S - 0.36"1& 1.~'123
3 0.3 . 1.2'1.23 O'~'* -0·5050 ~.~ttl~
4- 0.-4- I. ~'S \5 0.''8.''21 _0. S1+{;" ~.+6'+1-5 '·'i,.+C4:'1' ':
,j ~(Q) = ~, ,,~=---
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124
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bhe ir"\'"'l'frove.J. £"u\er ~...,et-hccl t;-nJ !:l(O,z) co.,..,.ec&af: \~(l.St -&.sQ ~de.c::l'yY\,",1 r\~~= if
.!:J' Q e:-)(.~· ,j !:lie) = 1
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.. .' W,ith.:.u l;' ~'nd in ~ -the eXa cf sc:"-'i:4>n ~ .£"de~r' involved . it'l each c~c:.· A.f>r....o.xirnet.te~Je.c:luc.e Do ~c.Y'e. A.ccur.:d:e va..fu e ~r ~ ((, s). .
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) . Us'e <: n··....(in. to obt:",.;,....· A' ,·rne...e·A~~urP.bevA-iLle fot'" B (~.+) . ' .
. Co~"", ....e ~e ~b~ineJ' v~ue..s w,t:h H,e e."Xp<c/;:veJue.
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3) ~T'\,,\~ t:he .s:eeoY')J -o~der T~IQr tn~bd ...An:f:4,en. ~ n-.oJ.;f;e,J .E"u)el" '"" e,-£:hecJ -1::0 f'ndwi t:\-. ~12oe 0.2 If 1/= ,)(+- ~ .;. s (1) :,,;2.
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128
\ 0.::/ 0....021+ O•.O..tr43. 0.04-430·3 0.043{;" o.o"~T . 0·131--40·3 0.•05S8 0.0 ':}t.:i 0-1-+ 2.4-0.4- 0.0"126" o.o'1~S O. 0185
o- it22b
~I 0 + CJf-. I 4- o.o..:n4
n x. !J \V·...hrx .....!I) A':;
0 0 0 0 0001 0 0.02·.... 0.04 ":,O. , 0.0 , D.022 0.0+4-
0.2 0.022 0.0/44-4 . 0.0444-
I 1-.·O,128~
::::< O. ~'S =
h :::P.Zt
~ = X -+ ~ :. ~ (0) = 0t)
Firs l: :P r<lb\e.P">.$ p. Z3 2: ~4
hintJ 5lhsWer.s : :,'j
2) .1.211-4 ... ~.s~ I1I.~4) ~.5 46'0 .... I·Cf+~2
..~, ,.
127.
Ilnd h:; 0.0.5 .' U .-v' ,"':::l.Q J~ .;;;J-". TIl
c+) "The eXAct va..lueis -O •..::?j:.I.8
(2 )""To..~ h=-od(3) L o 3<J6-
.'. .~.. . '
.,' ""
..
130.
:;( .
-.
. r •. '.,
~)"d :i) Nc.3. p.24"
~(o.z) ~,.o~.o's..2. .....t ~O.&'6'~ '= D·~.2.'::?'~'I' •••• .. • • . 4
.) ""'<eo. re':s\tl~ ;~.'~~)r 'An J c.\\i)., Do~ 4~ .s.a~b' ,49),:,·.b·~·~se '.R~n~~.~~ ~(;tl-. ~ e.fh.cJ:: ,.s~ '~\:,~.,: Qy!a~ .'. ,.,..,.~ :<;::~f'.~$ A: ':V~',~iCCtA~
"~~:. ~,,:"'~~:':~I~'~:::~'~: '~~.e.~" .:·~' c>Y ~ "
.. " ~u\ ..-fkH_., .~~\J: \:X:.':~ ~. ~:~en'_c.~ ~~ween. • ',6 ...... " " , • _, _
. ~t il..i; ~~"':L~.' , <~ ."-:.' ! :'~-x""e. eJ((lq .soltdlQo, is !i"')(~ _~x :2 -~.':Scr -\t~' ~:h'.;2-)'=O. 2~ ,';",;-';.2. .. :t. :J...,:./:~.~
129
f r. , 1) • '0.5.2 0.)\05 'P~1I0S
1.\S O.lbO:S ·0. Ii&2 0.23.2.1~ .1.15 O'.163~ ,0. JJ5'i O•.23H~.
1.2 0.'2.211· O.t.<'I'l 0.12('1",'
D·61G'~
(iiI')
'!:J(q.l):;:!.~-i'f 0.6'3\1:::: 0.10502
" >< ~ W .' 6~'::O.l()(~
0 1 0 o • I o: f
~.O5 O·O.s 0.)053 6.2\0,"
i.05. 0.OS27 O. I OSO 0·2100
\., . '0.1050 O.llOS O.IIOS
0.631\
~ (0 ..2.) ~
I") x .!5 W""hr ....",-.'i) .b~,0 i 0 ()•.2. 0.2
1.1' '0· I " O~ 2.:Z~ o~4-4--4I.( o.~II 0 • .21q~ 0.4'316"
(·2 0.2.1 s a 0.24-40 0.2#0
1,3..2"7-0I
5) (i)
132
~ () c:J .S\ r'\.:Swe,(,s;. ..
D 2•3 \8 2 '" .2 • '"1"l"i82) ~ = 2.S0!6"O... ~:= 2.50500
,
':r.
.,'
131
2) '~f.'f>\~ ~~ Runje_kl.lHq . .....,et~d to-';i(O.~). .f"r· {.-h~ \~tl£J v.ulAoe t.....()'ble~, . .
.. ~ =. -\ ':":J('+ -4~..:. ~()).:::.'i ..
~i4.:·:h :;:c..2. "' ......nd . ~eh Wi.l;\..a h 0:; 0.1 ..·.Y~ ..~,,'. o\"t~i.,ed . ·nM~e),*l·co.l vo..J~e.:.. ..'(.:' L;()Y.re~'f.cnc:1in~ ~ h::::: O~2 ~"d h =: 0.1 . -1::."
t.\,Ab ' I ~ 'r;. - 'r1 is a.. 'fW")oQn: a.c,c~r.:J:e v~.IS
.for .~((),.2,) •
.f~ou()d -off 0...\1 Co.\c.ul~io",.s t() .s~
'133.I,I,I'
i i;:! '11: •,0'
.;
a2~ ~,1i2' - c'
821J:1 + a22:1.2 + + x'./ .
D
'•• .,a' ..._'. '.,'
, .
Y, IS ""x +, an2"Z ... ,- .. +' I!lDilXII - bDl 1 r- -II
I
II.
'. ". ,tbat 15'·•,."~' " "I'~.t'" • <,.... :
. ":. ..~." '.,. .
matrix :", '~be'te .ti~l~:b'1.:', .: ',:1,;.. 't.!..I.~!"
II X IIan
(.: ':: . '.:' .'.: :' . '. ,; ,.,,' " ".~ ,~~~'Hi 'j ~ 0 1 . " ;" .,,~., . Va 'e~i>l,~in briefly the practically ~IItPO.rfelit taoss-Se1de1. , • . . • • \ !. : ., '.:0:: ':) ~ ~ iu ~;, : .,;:.,',.:' :"'interatioe. Giyen eo sJ-ste. of lioe_er equaCiOJls' A,!. - !!. ....' :~~ere' 'i. :.:..('a ') . is
;" , " i. ' " 1j. .
. , 'I'" '.:' . ......' '"",',Cafl6s-Se1d~1 'Hetbod
• ' • t ".:; ~ '" • , '. '.'00."
,,: ." •." t. : • -: :,i'.. I
.:". ne" purpose of t.bia, cbapter '1.s to'" .sb'ovbov. par,t1.l". ..... , .. ' " ," . .
,;"d~Htj;e.nt1~1' eqUat1o'~(PDE.) ,cn' b~ ':c~a""Ie'~t.~; ,;ajar/e": -,of, . ,'... '.\ "
.. ~J:&~bra1c' equations, ,IS,. .replacing tbe ;,ot't,181 4~ri·"atl.es
:"_'';'' :"~,t.h:e:1.i' fi?l t/~Is·i:ft:.:i:r'en~e .PP'rto'x'iii~'t~i;D'S'~'J !$IT-be,sr:~h.~:',~'F~igeb~~lc ~~u~'tic;D:S c~1i"'ih'ei{ 'jfeJ~·~'hed'i1b.e~il~wi'lJ'fir an, '
,v):;,.:iteJ,.~id~"p~o~'&ss:io ;'or.~e'r'to' 'o:b~ltB;;~~:;'ipprbx'ill"t'e:50lo-
.. " .~" :' t, ', ..\.:-"
, . _ i, ': ,,',.: ' ' ~Partial Dlff·enriUal Equat:i'ons ,';,';. . '. . ",:' , . .:. ,. . ",' '. -.~'.. . ... :.~. "e.
':.
" \ t .... .!" ~~, . 't
flUERl~,~' ',,'
, :. "" -', ".:'~'.;.. ,.', . .~.'.','
;'
135.
III - 1 . ·'i· '.\,.
(2) (2) - 61~'91,4 . ::&,(2) 62.207lt2 ~ 86.914 lt3· 4,
III ,,'2
(3) . (3)62.3S~
(3) .. 6;.427... 87.:354 , It,3 ::&'4 ,"x2.
solution of th·~·8i';eQ SJsteas'isNote that t.he exact
134.
- 0.25 ~2 - 0.25 x3 ...
+-0.25 'Xl
- 0.25 x4- SO
.r3 -·p.,25 x,- 25
-0.25 'xl +
- 50
EX8111ple
Apply the Gauss-Seidel iteration to t'he Eolloving
system starting vith 100, 100, SO, SO. (Perform 3
steps)
sol ut10n.. t 0.,1'. (0)'x2. " ., •••
C1 is obtll;inei!. Und'er certain c ondI tions, the
(0) (0) (0) (0) ..50.. 100 and :&3 .. x4we haye Xl Xl .. -
m .. 0
(0) + 0.25(0) + 50 87.50.. 0.25 x2 ' x3
(1)' ... o'.2.5.xio) 4- 50 84.3~S- 0.25 Xl
, (1) 0.2.5(0) + 2.5" .:53.3750.25 Xl + :14
0~2S 'x.il) (1) + 25 61.719+ 0.25 xJ
+ 25
+ 25
+ 0.25 xi·)+ 0.25 xlIII)
+ 50
tbe system in the for~ :
(.) + 0.25 :1:3(.)0.25 Xl
'..:, (,
...(11+1)
- 0.25 Xl
(.+1)- 0.25 Xl
O. 2~X~·+1)"'0.25xr+l)
or until a spe~if~etvalues of thejllccnt
It'I'tt- 1Il0'st n;cent 8p'pr~xim8t~p'DS fot: ,xi, t?" The proc,,,.,,,r.,,••,,,,,,,.
of sUcct;:<.sive,oPplic8,t~~flS of the last equa~10ns is c
nued· un~ll there is 0 li~tle o~ no Yariation
·ho.te t!la,t ve sub.1ltitute. :on 'the right "of these' , "
- a '. x-(lII+l).D,o-1 1l-1 }.-~.. . .., ;. ~i
r- II .x(III)}·,2D 11
. (lIl+l)~1l2:1'2 .-
....... .. ..
.,
137
2 ' .3 Q. ,,'~·'-oCIt.
+' _----,.1~_r2sin2,e
" ., '
~o~~es :
~ +'1k + _L ~ ;;~l~'_"~aJ? 'r Clr l' 3t2' '...'~~2
." .. :.
eel and sp~~~ical coor~~nates. l't' "takes th.se respecihe..• J f,.
" '
s5~~1~ flu~ds. The form of Laplace"s equation, :"i'
, ,', "aboye 8Pp,lie's, to rectangul,ar c:'oord1D~tes. I~ "~yi:iDdri-',' -;
....: tra1'itoCional. electric, and magnetic. potentiols and,
" 'f'e~9.~1.~",potentials in: irrot.ati'on~,l flo':'s of 1.n'c.olII;~~~
'cbarg,e in a, bod; ~ Laplif.c.e' s' ,equa~i-onelso B~·erIlB,
electJ;1.cal': ., .. ':
It SQ1'erns.'the' stead7-B~ate distribution of heat ,1.II',a'" '.:'., " . . . .
, "Laplace's eqaation 1s
ita ,~' ~~-r.+' 2 +.~ ... 0ax, ' aJ ~
The function II represents die 'tf!.peratu~e" ~t i:i.~ t-~'\ftt,(
the poiDt vhos:e" c:oordin~tes are (.~;1.%)
Tbe heat, equation 1.8
~2u; .l2a . i':
, ' + +,.LL tla )--r -:--r ~.h:2 tit' ..
~ , ax 31
of a particle vhose position at rest is (~.Y.%)·,
lIlth appr,opr1ate boundary cotldir:.i.o~s. .this ~q1lati~n
gO~'erns1'ibratio.lls ot a t.hree-dilDens10ul elastic body.
136.
.The function 0 represents the disp~acement at
in t~re. spatial ysriables
the physical phenome~8 thot they goyerri ar~ listed here.
1.': The V8'f'e e'quation
important partial differential equations
your results vith the exact solution 2 3, .4.
0000000
2.' Compare
ANSlIERS
1. 0.2251 • ~.30S6 • - 0.4939
'000000000
2. lOx - J - Z - 13 , x + lOy + 2 - 36 • -x-y + 10% ~
::r (0) .. ',y (0)
x1(0) • 0.125 (0) ( )• X2 -- 0.4 • x3° '- - 0.6
~. 8x1'- ~2 +,%' -'i 2x + 10 •_ ,,', _ 3 ' • .' 1 x2 - x3 ....
xi + X2 - 5x3 - 3
"Appl~.tbe Gauss-Seidel iteration to th~ !ollowfog
(Perform 3 steps) :
P.ROBLEHS
::;',
iI",.:.
JI,;...:~
il,Ii,
,i!-:
,..! I! :
,r.
1, '1r .,. ,f' ..! I'j ~
, I; !; ~,: ~~
139
The equation is paro'olic.·
+ 5u - .sin '1~2 .~
x \11.7 - e u,
4J2:1.2 - 0
abe elli'ptic , Parabdic,' .IT hrper'b?lic' a'~
»i - 4A~ is ..negot.iYe, -,i·ero·.·.o·r posi'tiTe.
138,
• It
a f1:u·id.flo\{. .Th'e functi on P is. the pressure. and the.>: ij is said to• I " :.~~:'i:\~'"
flu'id -,··is ass'umed 1:0 be incompressible but viscous. ::~...:~'.point (x, J) it':;":; ~.):
::::::::al e:;::::;.::::.:u:::::,:: ::~.: s .; "::::i' ~:::~:::"olY'also. be given. But the [lye partial 'differential eq~~~ '~~a) .... u _ £(;,;,y)
:.i~;:.~\: . xx· n "tions shovn already exhibit II great diversity. The Nav~.~";'.' Here A" 1 • B .. 0 • C _ 1 .' »2. ·...:;.~4A·C.. -4 < 0".
~:::::;:::::::n:: ::::'::'::~,:~::::r::::1:,:::~sc;:~t :e.c:. :0':':: .: U:::0: :~_.: d.. ,<tc tj ,:.tial diffe·ren.tial equations. y {;:. ;u 7J . It
To specify a' unique so~ution to a partial differen~: it ,82 - 4AC ..4 ) O. The equation is hyperbolic;;:' .;,:, ..2 ,ii'
tlan cqu~t10D e . additional conditions must be imposed uP~i'·;:t.C) "1 un h1UXY +
the solution functio.n. TYPica;ly. t he s e. conditions O{$ J':. .82_ 4,AC-{-2xy)Z
in the fOTIII of boundary value:> tbat n r e pres.cribed on am -eli.~'.:.f~:
r,;_;;
The equa~ionare functions of x and y
;,',J~
ilj: F, .v}u:re .A. B': ~.'.: C:are components of the velocity vector1I'ere u and y
a2u D·_3u +' .E _3u+ C --- + + Fa .. G •3J2 ih: ar
the form
A second order linear P.DE vitb tvo variables viII have
CJ8ss1r1carion of PfEsI·
roximate soljtion~
or part of the peri.llleterof the region in''''hich the 80-
~'. I·
lutioJi. i$ $ought. Tbe nature of the houndary and the boun-
dary yalae~ are ~suall~ the deterJQ1ning factors in sett.ill8
up aD. appropriate cumerical schem~ for obtaining the app-
for an e,lastic body."
5. The Ha ....ter-Stokes eq.ua 'I: 1.01)8 are
3u + (lP ...'2 aZao!!!+ 30.+ ~+at dX
,.CIT dX ax2
--2-a,.ch+ ih + a,. + ap _ 2 a2a...L!_+at a ax v 3y dy 2 --2-ax. 31
'solution the.,shearing. and nor ea I stresses can be
. It occurs io the study Qf ~18s~ic stress. a~d from
'.,Th~ bihar1ll9nic eqoation is
4 '4 4-L.!L+ 2 au + ~-oar4 3x2 a ,2 3.J~
.141
(x., - k)]
1 [( .:)uT(x''')~T Ii It,y + J< - U
~ A [uex.,. + ~) - u
(~.,.) , J
ux(x,,.)~ ~ (u.(r+h.,.) - u(x.,.>] :(,for\ltlrd-diUe:::-cn,ce)
UX(1.,):::;: ~h (x+h.y) - u(x-h',J)]' (central-difference.)
u (x.y)z ~ [u(x+h.,,)- 211(x',y) + u(x-h.7)] ••• (1)xx h
can deduce the ~oll~vi&8 :
I •tvo "lIriobles
~e nov exten~ the finite-differenee approximations to
partial derivatives., UsiDg t'he Taylor series e.xp8Dsioll in
f"(x)~ -\- [ r(x + II,) - 2f(x) + f(x - h) J, h
of f"(1):
Addin8. ve obtain the central-difference approximationI
I,.I,~~""to<
I 'f>':1l~I
:2.f(x. - h):::: f(x) - hfl(lt) + h2 f"(lt) :::.
Ilnd
h2f(x + h)Z f.(;':) + bfl(;,:) + T f"(x)
If \Ie truncate"Taylor series after three terms.
we
'~:.....140
.:.
f(x,- h)]-L.. [ f(x, + h)21tft(r)
"e obtain the cebtral~dlfferen~eap~roxi.ation
f(x' '+ b) ~ f(s) + hf"(z')
£roll
'f(x - h) == fe,_' - ~ii° (x)
SubtractiIlg
f(x) - fC;'; h)f' (x) Z -:'~~-h""_\=--"'::"<<-
8rrl.'"e at the back"ard-difference al!p;ro~l.ation
vhich' i~: .eli'lled the forvard-dlfference appro%hloti.oo to fl(x),;'
The h is replac~d b,.,:-b in the Ta,lor series.
f'(x) ~ l(x ~~) l(x)
Hence. we can sol~e for ftCx).
f(x + h) ~ ,f(x)+ hfo.(x)
the approximation
1£ ~e truncate this series af~er two,ter~s. we hayc
f"(x)21f(x + h) - f(x) + f~(x)h +
Re~Al1 tbe To~lor series expansion of a function fez):
4. Fjn1ee-Dif£e~cnce Approxi.stlons
F;' I!" ,,II'; .:r";1'
! .
,I
.: 143
2 0 0 < x < 1 • 0 < 1;,<,: ~ .,'V', u"
{ 0 013 tbe top of th~ 8({u~fe.IJ
".
the sides sDd botto. of the square100 on
.- " ';' ',':
'L,oploce equatioll
Consider Dirichlet proble~ in> th.e' ~~it :;9uare r, K {or
mesh points:, /'
the llIeUD of the values of u at. t;hE) ~our D~ig.hb~-
(i.r-h)
. 1 (6)" ihll:t ,o(x.,,.)the f11"e-pOint fO~"U7~ , . .'see from
b,
(It.,,),(x-h, y) .h
.:' .b
(x+h.y).h
oding to the Laplace equ8tio~ (3) ~~:
) ( h)- 4~(x.y) - 0 ••• (6)'tI(x+b;J)+ ,u(x.y+h)+.i,u!(x-h.y + u x,y- .
. ) . t the f, our neighbo-rela t~s u at .(x., ~to ,u',11•.s e q ue cLon ,..
, , i"; the foilo~1ng':figure '.poiot:.sSbO'WD ~
(:i:,J+j)
...(
142.
This is 11 differe~ce equBtion c~rre$po.ding to (4). We
call h the mesh size. The difference equation corres-
••• {5>, .2 I ,- b f(x.,.).
u(x+h,y)+u(x.y~~)+ u(.x-h,,)~ 1I(x.r-b)-4u(x.y}
,Subst~tute (1) aud (2) inco tbe Poisson equation(4)
,~here tor slmpllcl~yJc - h; ve 8~t
problem Is called Kcn.auD problem.
on the boundary C of i; is calle~ Dl~lchlet proble.. ~f
the ~ormal derivative 3u/3u is prescribed OD C•. then
u takes Ob a ~rescribed~esion E, such that
The problem ~f Bolying aDY ~f these CQuations in
catioDs.
vhi~b are the'most impor~ant elliptic equations in appl
••• (4,)
•.• (3)v2u - 11 + lIyy - 0xx
and the POSSOD equatloQ
V2,a .. 11 + uYJ f(x.r)xx
In this section ve cbnsid~r the Laplace eQuati
~. Elliptic Proble.s Solved by. the Fin~te-Differeuce Metbod
depends on the proble~.
Which approxi~atioD to use (forvard.central. or backYard)
~ [ u(x.y+k) - 2u(.x,,..) + u{lt.,. - Jc)Jk
145.
. .... ;...':.\' ..ter ~hr~e steps \Iearrive at (see ~ect1oD 1).:~Quations Bre now used for th~ Couss-Seided interS-
1f we start vith U1(10) nCo) _ 100 nCo) _,nCO') _ :;0,'21 '.1,2. .:l2"" ';'". ..~~..:
+ 25
021 O. 2.5u11 + 0.,25u22 • 50
u12 O. 25ull + o.2.su~2 + ,25'
0.25"21 + 0.25012
revrite (7} in the form.,' .
"call here, Liebmann f s JlIl~tb'od.
It is preferable, especially vh~n the number of equa
tions is larS,e. to use the Gause-Seidel method. v~l,h ve
Solving this system oC'linear 81~ebraic equations, we
obtain approximatl~Ds to the unknovn values of u at
the mesh points in R
B terms to the right-hond side
- 4nl1 • . u21..·+" 1112 200
ul1 + 4u21 + D22 200••• (7)
till - 4u12 of- a22 100
-144
Since ,0 "is known at the 'bound~r1 pOints', ve IIIOTe
022 + 013 + 002 + u11.- 4a12' - 0
u32 • 023 +,,' 1112: of- 021 - 4~22 .'.-0
u21 + 012 + COl + 1J10 ..:..;4all .. 0u3.1+ 022 + Ull '...aiD ....: '4021
The four equations .a're ISS, :fol10~s:
we obtain 8 single linear equDtion DC the form (6).
A~ ~acb of the four interior grId points ~11' P2l'
It is convenient to introd'Uce an abbrevioted
Ve cover t.he,region i by 'mesh points
%1 - ih . y - jh '0 ~ 1.,j ~ 3.i
the a.rid:has spacing h -~that is, 3
\3,. 3I'"
1 ' '11 .. 0 lb2
a-l00 R u-lOO 1
-v2u_O00 u"100 1 :x
,'.I
tl
il': l, !
II,
.... f
~r I, ~ .;' • •
5paciDg :1,s":not ·regu-
l,fi bott~~;'rik~t~lId
Irrc8.u1lJr Hesh SE!.llcing :
lie cODsi,~er the case vhell the lIIesh
lat. Let bl, h2, h3' and L be '~b~,4 .', .
-0-0-" -:
All calculacions were d.oDClto three d.ec.illlaJ· p.l~~e.s<.. :~
0.506(3) .' . (.3)0.162 , u21 ~ 0;159 • ul2
u(o) ., u(o)II 21
(3)ull
SolilDa this system by Gaus~ eli~i~aeion. vu s,t1 .'. . 10_0 159 .. __.;8....,-Q,508
ul1" - 7~-O"143 -. D21-61,- ~ . ~12 '63"-.·':'"
If ve ~olve it by Lieb~ann!s me~~od vith staifin~v'i~ej"'
17~,9'
. '.
2'u12 - '9
'7.. - '9
Or
2 2.., 1 9 ( 4 + !)+ s + 0 + 3""'9"'- '9 9:
" i)'i ..1 +.~)'....0' 2'
~: ( , .'-4,U'l2 - 3"- r~"9," '9 9
_ !'.. 9 ( 1 + !)",9 ,'9 9
en ve obtoin the~syst~~
146
1
-4 '2u - h .f(x.:J)
1
'_ "
There are three iuterior .csh P'01Dt8 PuFor.ula (5) is re~re$~Dtcd by t~~ ~~~~e'D:
I •
• sure belo~.
! :lA the fi;,ro.810D in tb e ulli t .aq'bare vitil I:t'Ber~ ~~ i5 the bODjd~rJ of R, vher~ R ~s the
OD 3li1:.
io Jl
~~••ple:-CODBlder the Dirichlet proble.
inYol~ed in .approXiR8tiDg Dl1 1s
88'.1 - 87.202 ,I ";0 0188.1 "t.,.,·
Qed, approxiJllatio~$ are accorate eoough. The relat:1Ye crror
1/3 • the obtat-.of the'~~Jd of Buch a lar~~ ~es~ size h
88.1 aDd a12 - a22 - 61.9. Thus. iuspite
.ote thit tbe exact ~alues to one dec!••l pla~e are
,
a(l + 0)'(10) . :.;.
.' .' - .
1
149
1 + b .'1
...... 0 +. bob
122 [ n(x + ah ! I) +
0(1 + ,d
r Exaa.ple if 8 .. b ...! vo han: tb.c· pottern'.2 •
[~ 4. !J'3-4 3'-23
the pattern is,( 1
bel + b)
nex,y+ lib) + Oe% - h , y).b(l'+ b) '1 + Ii"
(0 +'1I)u(x,12. ob J. (9)
U{X,I - b)1 + b+
vhere 0 < 8 • b < 1. This case
Suppose that
. Special esse
p18~e. vhen the boundary
is II curve intersecting the grid
t points that are not lIlesh point ..s, .•
e e. t~,C. figure •. In this ca se (S·) b e c orae s
adding (iii) and (iY). we get
2 . 2o' h1b3(h1+b3 [b1u(x+h3,:r) + h.3t1(x-h1,r)J
+ b2b4(2hl+hi) (h2u(x.,,+b4) + h4u(x.:r-h2)]
1 1 .,'-. 2(~ + ~)u(x,,).
1 3 ~ 4 .
(lY)
/1: .
'. ,
148.
'.:. "
11 • (x,,)- l' 1 [h2u.(:It., + h,)~h4·(·~·.·7:"h.2»'...n . 2 h2h4(h2+h,)
Sill11arlr,.I:rt..i.1
Hence
bl.(X+~3.:r)·+ h3u(x-b1, J.} - (hI + h3)·(x.J)
1 .+ 2hlh3(b1 + h3)a~~x.7)
HultiplJin·s.::.{t) by hI ond (1.1) 1>7h3• acldhg IlDd ignoring
the t~r•• arked by do~s. we set
(u)
(i.) .Ve obtain Fro. Ta:rlo~'s theore.
Q(~+h3.7)~ o(x.:r)+ h3" .. (x •.:r) + .~ h;Ox1f(X.:r)+~ ••
( ··1 2a x-hl.:t)~tI.(x.:r)- b1,,:x(:a:,:r>+:2 h1"xx(x,:r)- •••
.,
.1
top spaC1~8respect~Tel~
I •.'
III!.r:t •;" I! 1,.
0.789
151.
(2)1.722~.577 ull - 2.732 u21
0.571 0Il - 2.732 u2l of- (0.577)(0.25) ...(O.789){1)
...'(0.769)(1) - 0
e difference equation for P2i is
0.577 ,
3- V3V3 - 10.577 - 2.732-.'1
3-..[32
0.799'I- 1
difference equation for Pil is
- 4ull of- u21 ...1...0 + 1 - 0i- 4u11 + u21 1.25 (1)
.. ~2 ..; b1
h3 - b4 n ,I'P21 \Ie have b1 .. - • .. 2 -2-2
- (/3 - 1)11 so ,i:hl1~ U Db" /3 - 1. U$iDg (11) , 'lie obtDiD•
(1t)
':;",
J
1 u-l~"
u..! hIIt : Pn
'"h2
1 0 12 :2
u"O
150... -,
points. Th. result! 1~ sbovn h th.. figure.
[1 1
1J .Por Pu we haTe the usuall patte -4
1
Solut1on
'V.'c~lcul.te the Talue. ~t D at the needed boundar,
-]/2 0
':\J
u - x2 + J2 OD 3i
by finite-difference a.thod with h'- ~ • vhere R is
1u the filure,below
10 Io
EJfseple
SolTe the boundary yalue proble~
1 + II1
1+.' eel + .),11
and when b - II • it is
x)
l: .. J < 1nell,O) - aCl
o for x) 0 • Y > 0
u(O,y}- 0 • u(~,l-x)- 0
up a finite difference problem for
Compare "ith the solUtio',,'.....
,
J
4 u"'110
3iPn"~12
--0 2 ti-o
1Pn
0 1 p. '"
.e 2 ~)' le,ibmOIlO'sl2Ietbod.yf!
"'ned in 2•
o 1 2 3left edRe.
upper edge, Rnd 0
-, ,2 I--6--..t--t '\ ,
15-......."""-"1'--1
e right edge, x3 - 27z
on 32.7
the potential n(x,,)
that '0 _.3 on the
sides of a rect8~81c kept at po- Q
o sod 110 8S shown (V2a - 0).
the following proble~B numerically:
d the ulues of the e~~ctt~"tar:l~ po
D at PIl, P12, PI3, which lie 1m
bet~ecn conducting plntes oppea-
Usc the indicat@d 8rids ODdthe Gauss e11~ina~jnn to 80-
PROBLEMS
-0-0-0-0-0-
SolwiDg th~ SY8te~ (1),(2), ve get
°Il - 0.4963 , a21 - 0.73S1.
:....•,
152,
I~
: ...• ',It " ,
. " ..
"
" '
154,
-0-0-0-0-0-
3 _ 300x
'1 u-x - 243:.:?9 .......
", fyuo..
6
3 u-512
:"
0 3 6 8 ,.
values sh~~n in the figure belovo The eQuat~on2 2portion of the boundary is x + J D 100 •
'3
the boundary
of 'the curved
" '2.:So1.,.ethe Laplace equation "V' u - 0 in tbe 'region and for
2u - 0 for 2 + 2''-1u + x J <xx 11
lI'J for x2 + 2 1u -.,
vith h - t . Solve using,the sy.aet~Iof t,hc problem .
Solye
",3-!w'2 ;<> .;,upper ed,e is J
"boundary values shown in the figure. The equation of the
153,
u-o,!'roble~.~
x;'31.Problelll'6
1 2o --t--1-'-,---:2-.:J3L._----''''" .x
~Io
t r-u·o ..
3 .. ,., "~ .. x
2 3-2 x2
"1u"'9-3,
,
boundar, conditions. Here tbe equatio~ 01 tbe slo,in8
POrtion of the boundar, 1s 7 ~ ; - .x •
6. SolJe v1u - 0 in the indicated"'region subject to the ,heD:~;'ii.~\)
<.'
u(-l . ~}-lie! . ~) .. -uej. -~ ) 4(- ~ • - ~)
'IU~O.J~~ - u~o,-i~.u(- j. 0)- u(O:O)- u(~ • 0)
,reduced by using the s,~~etrJ :
starting ~ith zeros. (Three iterations are needed)~,
1Iint; TJle riub)b~r"of unknowns in Causs-eli~iDlltioll lIay 'be. ,
I ' ,j;~ by Gauss-elimination and bI Liebmann'sw1tb b
u(-l,,)c u(l,y)- 0 , u(x,-l)~ 1-x2
'fT 2,u - .x lu - 0 f 1 1 '1 < 1V or - (,x < '. - (,,' ,
5. Soll'enumerically:
interior polnts. Perfor~ three iterations.
(D) Gauss elimilUltioD
(b) ,Gauss-Seidel iteration' starting 'With zero Yalues at
Solve the finite diff~rence problem by :\lith h ..i"
,~~.:),,,, ' .., "
CHAPTER TWO":.'
'';t,
,
ISS'
." ..... :-
. ",, ,
-0-0-0-0-0-
9. - 55 •6 • 49. 2 • - 298. 5 • -.436. 3 '.
6. 2. ~ ;'1'; 2 '.
(b) 0, -O,2~O" 0.270 - 0.373 • o.3i3 ,"
5 () 0 0 0 16 16 16 16 _ ~ 89• . 8 • I I - 57 • - 57 " 57· ' 57' 22 B '"'228.'""'
16' " . 89 .','"here 5'1::: 0.280 aDd '228 =:: 0·,390
( b) 35 2.... _1_512 • 64 ' 64
1. 1.96' ,. 7.86 29.46.
AHSIlERS
..t, ~.'
-u""i
• 11"
"
".....1!'7
.....:'" .
,;i In general;.' the'nth f~rwa.rd.d~.H.eren~:ej~.:.. AnY1 ';' jf-' (~Y1)' ,,:' j-1Y'i+1 ., _t-lYj.{~·/ -,,
. . . . ,',~. : "
S1<arting fro~ auch' te.'blua.ted data' $' (:1).~..:~e·cen·eva.-luate' successi ~e differences by ~ iri..ff'e~e~c~ttabieas fo-..:~
" .' ...:'.. ;''' '::'
, ~,,'
\ ..
.'.or
~here Y1 '" f(xir , xi .:.';0 '+ ih i ='\~~"'~':'~~~n·•., It. m.ay .be desirable t~ know ·th~.·~aliiG· of f(x)': ·~~.~·~:~~evalue .Of':x not cited in 'th'e ta.ble, e.g. at a po:i..ri·(~:·~·ihe· ro~ge"• ' ~,' •••• • .. '.' ':,,:~, "': ;.:'; ,,: v'. '. ,.' •••• ':'.
ot tllbule.ted valuelll (1nt~r'poJ,at1.o~) .or, ~t):L'l>6i4~·x:.oute1de·.this rang~ (extra~o~atioI;). For this i~~:l.~·h~·~·e~~a~t;hulld up a calculus of" fini il'j dif:f'erenc~~:;->"",". :-.
• • • • '.. '••• , I' ' ••
Wedefine th~ 'ui-st .forY<a.rd-dH!el".e'n:~:e·.to be
AYi .. :11+1 - .~i. for'all Int.egfal i,'A:f(x) .."f(x + 'h) .;·t(x). "":: :.'.,""'" 'j. ,... ,,\. "
~'he a econd f'o~i'ra'·dif'~er.e~ce':'i~~eil~~~·:t~'lJe _..
. . ttYi.· ,,:'A(AYi) ... A (Y:I-+1',~.,Yi)·~.AY~;~·"'::~~.i~•., " /.' ',' ' :~?~.>. .
"ues of tho independent variable x ; h being the intervl: .:: .. " .' '····'f:'·
x Xo .'x, x2 x) ... .~,,' ...('{) ...
l(x),", "
Yo ':.Y1 :12 :I) ... s. .' ",";,,"n .' ::.\,:.',. . , ".
Often e.'Xpel"imen~a.l?-ata is obta.:!.nedin 'the form ot a
,·.ble of vnluea or the .function !(x) at equaly spaced va-' ;
. . .i.,:
".: .
.. :,,"..':J t ,
IN T E R POL A T·I O'N
§1•. 'n!!H~ ~ig~::~!!2~!!'~
. .; .
I~~i~,,'
'il I
......~ I
bl'\ !, I. i,.. .
~
Ii,~:~l,iil
: ~~l'
m. i
.. '
'159
.' :" ", "~
In fact. oonsider the polynomial in ::lI;:: ~f. ·.~·e.gree.li '. ./
............ .Pn(x) .. e.~~ ~ a,x~-1 +••• + 8~~,~·:t ~:.:.:.< ... ~ ..' .' ": • .' '. : ',,' " • ,~.,; .. , 1 •• ,:\ " : • • j
!lihue we have. . :... 1 • ". '. .v; .. .• : '::'.',....•:,.!... " i'. '.:::',:. 'p (x+n).· ~,).<~~h)n.~al(xt·h).~~\::~:.~ "i;}\1'::{i;;h':':':Ltn'.::.. '. \...... no .• -, .... . .' :'.:: .....~-: .. .'.-:. . . .~~.~:.iir.stforward <i1'f:r.er~Dee at ..-e.ny~o;n.\..:'~(.#.~~:';~:,'.:,.... '. '. :.... .'. n l' , ...:.,.... . . '.'.:' / : 6Pn·(x) .. a~{(~+_h)n"xnJ+ ·~l{(~\~.}..~:-~.·\~?j+··..~:~.~~1b .,\IIn~oh•..by. the binomia.l theorem I is f;l: .P.,?l~OIll,1!~;t:'e'!' degree)~"1:~ with leadine; .term nh~oxn':'1. '~.:":.'.: :;':'~:'~(.';"" ...-
, ' " ... : :<i .::/~ '.,". .S1milerly .Bee.onddi£~erenceB can ·'be.r&pr,~.8,~l.?,:tEidby a' .
. "." '._. ? tl 2po~yn?mial of degree (n-2) with leading t&~j~'~~:-})h 'aox :-. t.
',' " ' .... ,~,,'.': , ' .: \':, ",
" ":......... '
" .',
6.2
24. ',18' , " :,': .38 0.:..: : ,
1'8> .•.. I56 .:.0...... .
", • 4' ,':" .. "" ,
4 ..~.2~9..: ::: . 74 ;.: ~B.: '.:.)~~::>..:.i: :..'"..;..i5·:. . 401 . .' . . . : .; .'
110*,_=::"::'" ... =; ::u::::a:c:_ =t:::c:t:fl =:JS~=~I C:.'=*~~'~," '" ,'J,- •• ' ~.~~ • .:'~. ' .. '. :··:.: ..·::·i~~:~'~\',;··>',' " : ':._'. ' ..We'·see trom 'this example ·that '::tli~;"~h~-;,~~~A~*'.e.r.~rlo~.g~'.'. ....
, '. " " \' "'" ".' ,."';t~'·,'.,".. : ,,', •. , ' ,
a:~tl cOnEltlnt and e.n1.~h:l:gher orde1- 4~V~!ctl~.~:~:~.~~(·~e~.o.. :':.::.., '.: irhis is 0. speoial· '~~se of Ei ge~e~e.l;: ~:~~~l~\~:;~;:~.:r.o~ ".fi\ ,.: •.... ~ Iolinomia1 or degreo n , .tbe nth .ord~t.~f!~~~:.~:~':!d;4r:(ei'~n~e~;:.... "1
':.: :. ',~, J; : .., .:',::. ( -, ;' .. ' " . . req\lsl t~ n t hn eo ,. "'her~. -.iL~·~9...~he ~~~'':'. I
ffio..ient of xn. '.' :".'.: :'::':.'::.>~:::'" .... '. :
14
20
2
, \,
1.o
.3'."
·t. . ..F""'·=:===.,.'.="'''' ..ee e ==.. ""'''''''== ....=r=~;.4..•. x :t'(x) At . ,}f :A'Jt_· .: ,t; t.·
., .' . .'
: "
(3)(2)
158
. unit"). andx between 0 and 5 nt 1.nterval.·s 01:
form a difference table for these values.
Example (2): Tabluate the 'fUnction: :rex) '" .3x3 +jk2 .+ ,
for vaJ.uesx ::0(1)5 ,(the notation ~ean9 "for val~eB of. .
0.133 0.62.. 0.41'. -, 0.85' 0.15'~. 1.47J '. 0;56
. 1.'41' .'5 '2.88 '. .'.
'. . .1::"'==:'=,.--==-,;,,.;=.... =====,====-!.
0.28
0 0.00
0.020.Q2
2 0.180.16
0.44
=======
.. .. .. ..===~=~ ====~====~=====
...====:::~
~amEle P}: :F'onn a dirtere~ce.table of v~lueB
x I 0 1 2 .3 4 5 .:
fex) I V.OO 0.02 0.18 ·0.62 t «»~7 :::.:38
1"'.."' .... -=f""'''''''F"".z::,,,,:~ ..="""'...., :r ..~w.t.% £(7.) At: 1/£ Alt
110\'1B:-
.:
. j
161
'.: ,::' ' ..c - conet
" .'
'.. ' • < ' .' ( )'.. ~\ •• :- 5'.' , ' ::' ',~.:.:~,,"~'""We have .. ', ".
Vi(x) .... r(x - h.> .. (1 '" .~.:",~)q(~l i:;;· '.:.(6)TllUBV" 1 - E-' .or E-' ..:Y~"':'" :.'
. • 'Pro;~rti.. of Oporator. A • '7' On.. E I~ : ;',~
+t 'ie trivial '~o .prove that
1) E(r + g) .. E{.+ Eg
1i) E(ef) = 0 Ef
.' "-,used aleo'in the form
E ...4.1:fI~·
':.,/ ". ......~.(4)
' ..': t·b,:"·E·-1
E'.. 1 +1::..
This relation between the operators
Or
,,'
:Crom (1) tl:l.at· . . : :'
. At (x) ~:Er(x) - i.ex).·~ (l'r - .1.·)~(~i.: :.::....<', .'. " .: .
'.lIbue,
". ',,',:,, ' ..... ','
,", .
It is clear that
E2!(x)·.. Ef(:-c~ h) .';,'1'(:'" + 2~),.:.\,:/.:'·.·,~"::~.',:.arid,in general
Ert.(:x).'" r(i + rh)'
'..~'::;'" '.:.... ,., ,'... ~•• (2)
• •• (1).: .. . ,6. f(x) '" f,(x + b) .- rex)We define a new operator E b~ .
Er(x) ,. f(x + h)
§2. !!!!!£!'~~!!~~.:QE~·!!~.2r!!.,.·,,,.. '.
Considering.A as an operator., fie h~ve....
(5) "
160 '-0-0-0-
It is important, to note. that the' two notatiO::lB ,.tor'. .....expre~8ing i'~ite dif:ference~ 'Bi-e'two .a~·ife;ent '~f1yeofexpressing .the S6lDenumerioal quantities'. For example.
2 _'''2 J' JV'~1 =,AYo' VY2 -;-'AYo' 'VYJ '" is, Yo
".' ." .
p=="':;j"......,... F"'''''''','"'''''''====.".''''''''=': .x fex) 'If' '"l~.:'V~.f'
'lif' o.btail1 the follow.illS tOnil for' the difference table
eX'pl'esoed in terms c.f onclm'ard di,:t~el'.en-ce5:'
.. ' ....
.:' ,:""
VYi '" ;1i - Y.i~1 ~...'.. ">.';,~ .. . ~.'. :", :!~::' ..•
, . V"~j .' ;' Yi:~. 'Y1_1 '" .,,,..r;.. '.. '. .'; ";.' . .. . " ...:.•..::._':',',:",':.:"';,.:.A....~ ...... . " ..... ;: ."": . ~._ -..
. .. • ' ' '. .. . . . ' . ./ J:,:..1tt. .' .:. n.' il 1, . 0':"1 : .'
...., .'. ~.' V 1'1'= ~ : Yi -'V ' Y.i ~1
to beBaokward differenoes 0:1"successive' 'orders are defined'
(4 )
-0-0-0-'.
etc; Helice the nth d11'.f'erenc6n are cio~8tal'it and equal tono' n" ..
n(n-:l) (n-2) ••• J.2.1 b aox '" ~l h ~'Q ...Q.•E.n.
.'r .
!' :,
, 163
,.:.''''. ....
tunotion
, ~:,
, " " q (q '- 1) ,2,' ',',f(x) .. f(Xo) + '1 ~ f(Xo'> ',~ 21' , ,ll. :f(xo) +
+' ,'1(~, -. ;~ (Cl: - ~~: ?,,1.: ,~~x~',~,~,.'.:~,',:,This is Newton-Grego'ry 'e.dv~o:tri,g'~i~!~l'enc'E! ,:f'o~ula..
• I ','. r .' : • .""" ••••• -: •
, , /!!Xtmiple(2) I Fro~:tbci ":roilc!,w.i~ ,ia'b.l,(l:·,ev~luatE!:':Hi.~~>.1.2 1..4 ':~""'>l~!~":;",:>',~.8' ,.'.',.' _",,/
:'. ', ::~";:"~"". . .... : ..:.rex)' , 2 2:92, 4.08 '"5',~4,B,::'-',..7.12:'.." ..',,:',,\~':"'At first we m1,1stf'O;t'!ll ~iffer.enoa table to~ t~e giv(;r~
. '. ~..' .... .• ::.. . I ~'~. • • ,;i ..
Thus,
1,','.
"" ." '.
-'., q '"
x ="xo + '1h,Hence
'<~,,~~':,',"~',,',:::_':,:,_, \:
.~~'~>"~',",.:;.':·9.:.t':: .'';From §2. 'we have '" , ":,, " 'q .. '
f{x) ., r (xo + qh) .. Eqi (xc/ = '{~ +' 1:1.)' f_{xor,Using binomial tb;e,or~-t ' 'we g~t ' ....' -. ' , " ' "
, '1('1.:..1) '~';': 4'(q~~)~q-2) '~3, "':, ,}" ( )f(X)' .. 't1+Q.l\+ ~1"'6."+ ',: "31 +... ~ txo
Let x be a point near the'~~~iniDgof the table
e.'ndlet
§4. !!!!!1~!!:~t~G2!l~'~2:::!!~!~for Forward :~2~e2~~~!2~----------~-~~- . "
~o~o-o-·:.",~."'., "
'_.: ,,\ '"(a,b).
(7 )
..:
•
162 '
t,II" .'I
r: " 'n I
Geometrically. this me~a tha~~,_~,curve y = F(:x:)
paaa:LnU through the points (~o.Yo) ~,~:Xl'Yl}" 40, (~'Yll)~ie/~ i'ound. :,:'~,
Y
.~.f •••
It La known that a certain 1\mC.1;iony ~(xj
the values Yo'Yl' ••• 'Yn a~: 1;h~ :~;,.i:nt,~xo,xpof some interval (8,b), while ~ts t~~~sat the. ....t.>'.. N •points of the'interval Rr8 unkno1m:!to us. It is
" -~.~~~:to construct a funcUon 1"(x) de:fi.nd;,..:onthe whole int
.. .':.,{<'~,b) such that it take.i:l the flame viUue 80 f(x) nt
. ,.:·\·i.
Interpolation ruP.~B the ~o11o~~:
-0-0-0-. :.
The B&na propanics ere ,true ''!:or.A and v,OJ,
Elcampie' (4): Find 'a polynomial f(x~ ,Qt. 'the th~rd degree
such that f(1) '"' -.), !(.1~5) '" - ;~€?25 .1'(2) 2 t
f(2.5~::8.625.
',' .-o-o~o-' .,'. ','
. ,".. I .. ... ~,fl.J"- 1.4 ;:0 JXo r:: 1.4, h r:r 0.1,. q '... Q~l •
Substituting in Newton-Gr.e·gory :ro:nnul~:~'::·:we.set , ., . ~'o'~':n(-:c:7) ( 00'~6)
f(1.4J) .. 0.9523 + (0.)(~~0~38).:~ :; :> 2 :~... ':-: o. ,oJ +
f (0.J) (- ~..7)( -1.1). (o.~~~~.?).'"'.0'.9·568~.
we have.. :'" .
Thus,
S~nce "tho third differenoes e.:renot equal, we lIlua"t
take x "".4 (This is the nearest p:r~vio'us ~e.buler valueoto 1.·43).
(9)
.'bc:.CI .. -= =.u~ •• all ." •• :::~•• a.·a:lla:..... ;a~=~mace:
F===="p==== .. ===:z;:r="':: ....===''''r::..''':a .. '''r::F'''c:~=....•X rex) lIf(x) .~2!(x) jr{x)
1.0 0~a4270.0)75
-.0074 ~.1.1 0.86020.0301 .0010
1.2 0.9103 -.0064.00100~O237 '.
1.) 0.9340 .,..•.0054.:.00090.0183 ",.,
1.4 0.9523. .. ~Q.045..0.0090.0138
~"OO361.5 0.966,' ,~, . . :~46d~.0.0102
.-.002'11.6 ·0.9763 .. . .'0.9836
9.0075 • " ':~~~" <;-, • •..,,',:., .
1.7
.. '
The difference table is(8f .
164
' ..
. ErB.mp:i~' !in '~f$'lutty'lr1(1:.4'3}"i~m tlie:~Ollo~')~~le;::
. . ~ . "::1 ~'O':.:~::::)';;.}..':~",{": '.:" , :,'2 ' .., :'. ".~~~..;: .' " ~4" :~::".':.~~:""~.''. •{(x) 0.8421' p.•8~p2 ,'0.9103. 0.9340 0.9523 > .'.... ~.'., . ' . ' . ". " . -, -, ' ". . " ...:'
X ·.1:~.5.,: : :t..~..:< ... ,."7 .: ,,,',,' .
:r(x). 0.9661. 0.~·76J .0..9838,
Note: 'In this example we te.ke Xo CI 1. If we oot·;:i-.o... 1_." 'we .will get tbe o'r{meresul,t f·(1.) .. 3.47.
._.~·-~'~;;·~·;~~c.~s.az-e nqi' Ilqual we'm~qt tiu:.e the 'naueet t. . '. . . . '. " .
" 1 ·2:: ; ..
1.? '2.92' .9:92 0'.24, ',.19
1•.4- 4'.08 Cf.24l:~40
1.6 5.48 0.24'1.64
1.8 ~.·12. ,II:' "-=~?;:;-~~==-;:=:==;::'=:;===F'1'=':'=:= ...
. :
F .... ==-=:::.=="'F=::;<.:===F=:;:;===::=X' i"(X)'. .Ai.(~) l/!(x)
,tq, ,
I
I
li
~
I,,~~Il
~~
I'~~,~~i~ .I
,I
c 100.25.-
166
DO. that ',x = xn + q~
. Using (3) and (5). ot §2. we get
1'(~): ~ ,f(~ + ~~) :" Eqf{X,..)' .. (E-1 )-q r(~)
= (1 -v) :r(~) ,q (q + 1) 2
'''If+·qV+·. 2! ~+ ••• }t(~)
(q <0)
,, x':" x',q _. . n'. ·h
and let.".
-0-0-0-
§~. Newton-GrogeryFormula~2!_~~~~~~_~!~!E21!!!~~
Let x be poi,nt noar the end 01' the given'. table
8' '.
" "
16 .'8,
.,3 ' -1'
5 15
.7 399 71
515
. .: "11.' .
111' ',X. .3
f.(x) ...1
Example. (5): From ·t'he following table avalute 1(10.5)':
-0-.0-0-
. ., ',' '. .TRia ,1s NeVlton~Gt'egory b6()lcward int~rpolaiiou .formula..
+ •••
ThUBj we have" .,'• t· "' ... ,. "
" "(11).
. ~':,
-w~ have
xo" ;: f. ,h-: ./).19_. q c X - 1 2 2. 0.5:; x-:By l'tewt.on: ..:Gregory formula, we Bet
f(x} ~ - .3+ (2x'~2)(1.375) + ~(2X'- 2)(2i - 3)(2.25)
+ i(2x - 2)(2x - 3)(2x - 4)(0.75)
:= x.3 - 2x - 2.
. (10)
1.5 --,1.6251.375
2.:?503.625
.3;0000.750
2' 26.625
2.5 .,8.625
F=~~=F=====~==F==:===-======;-r~==:=~~'X f~} A!~) ~t~) 2f~)
:1 -3
'I .:'.
~.
, ,'I I:pJ[.,.' ,, .. ,
II '11L·/),.;"ii'l
" .... r
:,1,
jj :j
'; .I.~
I!
r-169'Also we have
•• ~(1)
ThuB,
. " .', '
15 6705. , ' " " " "~; " 'l:;::::=,;;;:;-;:;;':=c ==::=-=J:;:;;:;=~=~;;==;::=';;'U:=:;j==='::::!:I;;;; =~=irc,;::-=,;:::;_'
, ":" ", ' ... ,. I ' " '.
Since' the;'~~ird d1i':fer~nc~s e.r.~"equal/, .\v.~:dedu'Ce':·::that !(f) 1,B a polynomi~l ,or ,'~~e'~~i~~:;,d~gr::':,~···'r:' ,"-,, '~
Newton' a fonnuJ..a. for divided' d:[i'iE;ren~es:" " ; ,
10
6
,45~, " . :::.f:'··::';., ,':0',:.::: .:....~
116-4:2 '.71
. ,4-3 '"
116 142"'11 '" 26 ':,( I;
1'1-116 r; :- :~ '40"'26 2:: 149b - 4 :.~:14,2·.;49 10-.3"414
,.1.2ig:4i4 " 389 62-40 '2"f5::4-":;: ,;- ...
'970 ~-3~9 :::62
6105-1~70'" 94715-1
f{x) 45 116 414 ~970 6705FQ~C:;=""=~==I: -c===:.:;:::;=;::;::;:::====-r'=c:.:=t:'.:=.==t::';~i::a:;r=;::~'~~~~~;'
X f(x) Il'!(x) '~{f:'(x) , t(3t~{,". . '. .' ',." . ,:. ',. '"
, 1510643
B. table. Vieillustrate this 'by meana of -the :foll~w1ng examples:
EXample (6): Form the div1ded dif.fel'cnce table) for ,the
function .f(x)
"
(13) ,',*.
t,'(12)
: '
, '.:
Newton-Gregory "i~rplulO: for 'fo~a:t:d-~ci"b~ckviard 1ri,te- ':
l-polation deduced i~'~'§4t'§5 may be us,~~',ci~~~_'when the p ,, • ., l '.' _ '.
Xo ' :X, , ••. I xn' are; ':lil.u1cii6tan~." '~o~t~becase .wlienpoints are not cqUidi~t~t' ~e derine ..t~~::~~l:~id~ddi1:ferenc
as ,:f'ol,J,.ows: ' i: ' , t.~i:;-~:i:r'.'.: .. ,: .' .',' ,';,;.~~,..l,t,':'" : '
,Tbe :first div1a,ed:..dJ~f.e.ren~~,~:f'''~Pii~~(f.~r,xo'x, 1s
.. "
§6/ 'Divided Differences ".'--~-------~---:"'~- .
", ,
. ; .... :.,
I'i
. "j
,i·I
.sj
...~'" ..
interpolation tormu!a:..... , '.
'3.. 2x - 3x.
f(7) = 45 + (7-'~(71)+(7-~)(~-4)(~~)~{7-J)(7~4)
(7-6) (2) "" 665. ., ". "~s
f{x) == '4S+(x-3)(7f)+(x-:3)(x-4)(2i;>..+(~~3}(x-4j (~-6)(.~). . . " .
...' ........ ' ":.~.
.' . '.': '..' (15) ..''C.
-,~' : '.' ..
. 1'('2)
f(x) a:
..::.
==.:.:::...~(~82.1 lJ!liIl{.; Lagra.nge's fo.rmule. ov.el.uate .t(2) if'. x 0 1 ..~ >'t"(".' . . " '\\." . .;.rex) 8 .. 1.1" '::l~'_,.206 ..
(2-1)(2·-j)(2:"·6) '-~.t-2:"o)(2-3)(2·~6). .-- ........_._.....,...--'-' - 8 + ..... '1~O--1.)(O~3l(O~61 : ~.:;: ~.H~O~(1':'3)n..6f··." /. ./. '(2:"0)t2~1}.(2~6.)· .:-:;') '. (2--0)(2-1H2..;3): . :". ::." .. ..... ._._. . . ,. ,5- . ". ~'. . .... ' 2'Q~·· .
+::: ·O-O).(.j..·1.):(3..&fl "';.:~.~~.(6;..ciH6~f)(6-:-)· .~<.:::: "=18" :j'.j,t::" '.:', ..':'..:.,:-;' .' " ;.: :"' ..:..:;; .
,... .': :'~. ',~,,\:.\. '. ., :,,", ,.,. ....;.: ·_A·~'·M~,\ ~..... I ,,' !." :•. 0 ,0 " •. ., .
. ' : :.... . "",., ." . .,' ',:':" ': ', , .st. .inverSe' Ihhiri!oiatto}!· -,·.~ ',., '.: . ".
ID'~~raeinterpOl.at~::--~:-~::~~r:::::~·~k~~~~':.'~:.':.." '. '. .. . ... ,..~". .. . .:: .' \ ~J"'·..i·.e- :;(J~. r!1··.~h·)·'~:.;~·:u;.t\:~~').;-, . .
the vaiue of X.corresponding to tl. .. 'grvezt'V"~~'iHcO'i·.'tlri·'1"tID¢tion..; .' ..r,:";:,, :"i' \\::.~~~,.)'!\: !_; -, :{. '.
170
renee, we getSubstituting in Newton's£O~U18 for div~ded diffe-
EXemple :(7) I From the .date. .gi';en in: exam.p.i~'.(6) c;v~luate 'f(7) imd find .f'(x)~ ..
The dl~ided c11ii~re.nc~:t~~i~:1s• ' : • '. .: ,.' ~ ••••• '\", •• '. -» ,,:.
.F-~"~'~;r;)!~~;(;r:=~}2;7~;:~:;';{;)~'.,:..,. ..". , .'. """ .,., .. ' ',.' , ,
'J _ll . :':;.::":. ..: ·1.1 . .' ....
4 116 ·.:.~··~~·t,:::.:." .''6 .... ~1'4-9-:::... :"~ .. : '" _g_";
414 ., : . . lfO;:. ~.,S9. '. ".'.':'.. .. 2'.'10 1970 ' . . 62' ...9~7·· '. :."
15 6705. ' .1:...====="''''==.....==;,,====1==:;:==== ..1==:;==:...;.
.. Ol-.
. . ,
.1'(x,xO':Xl) = f(X~'~;';2) +. ~i-X2):f,(~:~~~~Xl~X2)' ;' ••• (3). . .... t .. '" :
Multiplyine (2) by (-,;;:-XO)~:..(?)..by..~~~Xci)(X-X1) •~••.. . . '. .... ":, .: :,' ~~>:;"~~'~"'''~::..: ,",.and' adding, we get ..... ':,' .: :. '::".".:..; .
f(~):. ~ f(XO)+(~-Xd)f(:X:O'~~:~;')~(;~'~)(*~~'J)~(X~'X1 ~x2)+
.. '. ..' +(X-XOf('X~±1')'(i~i~)~'<:X~.~~~~~;~i~;;>" .. :-.'.. ' .. '.. ' "'. . ..':: .~.:' ' .
. ." , .", " . , .1; ,,"! :'. .
l"(x) ~ .r(:ro)·+(X-Xo~ :A'.1'~:x~~+(~,,:,XO?·~x~i:l)d2.r(~~)' +";: ,+(:X-XP~J.x":X1)~~-;:i:2?,A").t(:Xoj·",~,.'.~.~ .. '
~h1s is the required f'onnu~a. . ..~.
... '., '.. 'we getSimilarly,
••• (2)
: , ',X -.::i:1 ., : ':'. ' .....,.:' ".':. f(;::xO). = f{xo'x1) .+ '(~-~'1)f'(X;~O"X1)
, : ... Ii .
, ~,
172'173wher.e q
. '.:'.. ' ,",' :'::: ,:.:,,':.
" . ':''':,''
':., .: \>.. .1' ',..
. , \".
The .uier.~Od of' s'uc6'e'~~·~v~.approxilIiati'dn 'or' ':Jt~rf1tion:
.. d~J~~der'Ne~~~'::Gr~go~' f'o~ula:":' " .. .': . ~(q-1), t} .' q(cj-1)(q-2) 2,'3 7. Yo + qA~o+" . 21 Yo+ 31 t:. 'Yo+
-0-0-0-
" .~,. "'.'',~:,,;'~~''.'., .
, :,,: "
3.8
th~ valu~'otx when'y ~.~O·g~v~n
x,' 1 .3 4. '... 6 .... .: .f(x) . -3 15 48 .192
· (30~15)(36-48)(JO~~~~).oX .. . '.1
. (-:'3-15) (-3-48)(-3";192.)• t ' • ',' "" •
(J0+.3) (:30-48) (':30-192').. ':t . ." '. • J· .. (15t)? ~15-4~).q?:~.'~2).: ..' " .. (jO:+-J)(,30-' 5 )(,30:"'92)· '..+'., . . ...... ".4'. :'. (48+3)(48-15)(48-192) .· . ()0+.3) (30-15)'( j6-48) '.. + . ....:'.... .. 6:. :-::(~?~-f:3).(192~.~.?-)F~9~-48)
,,',; "
Examp~~;':(9): Applying Lagrange's .f'oXD,lul~.:1nvere41y find
The use 'ofLagrange' 8 }'ormula:
I~·.Lag:rv.nge's tormu1.a if we con'slder X as a function
of y, ·.'then it may be written in th'i\ro:rm:. '.. ··(·1::'Y1) (3''-3'2)';· ~(Y-Yn) . ·,y'_":Yc:i)·(1,.':'Y2) ..••• (Y-Yn)x = xo+
~~o:-:r1) (:ro':'y2) - (.Y0-y n) (Y)::"10) (;V'i-y 2) ••• (Y1-Yn)"~"': (Y-:Yb)'(~ -Yl).~' (Y-Y~i,) . ;i'+.'.:,; ~. + )(.... : (Yn":'Yo)(Yll-Yl) ••• (Y~-~~~1) n:::-r:' " ",'
For this 'we ahall use one of the following two meth
1) Use of Lagrange's !ormul~
.'.i1) . Method ot successive appi-oximatiolls.
(1.7)a,
. ,.' .:-;":..
175
, '. .... ~ \
int e:r.PQla~.t:?n.~:fO:!"ml1:ll1 ~or. d~."l.".ded.
may.be written B~bolicnlly ashD
f(x + b) = e f(x)
(i)
, Since,,2 I2
eY '" 1 + 11 + 2T + 3! +•••
.. ,'.,'".
... ,.,','"
erent1e.tlng (1) w.r';t. x, 'weget. '.< ." .
2f(Xo'x1tx2)+ 16x-2(xo+X1+x2) }:r·(X~.x1 ,x~~''X3)::. :r."(2) := (2)(7) +. (2)(1):' =. 16. .
r:, .f'(5) .. 11 +.(8)(7) + (21){.1)·;'' 98.
';~'" :'~:':'",§8.:: ~~~!'!~2!::~g!!r~!!~~~£'l~~(1_. . ",,/. ~~". . ~.' .: . : " .
. .. . tet t~~':l~'q~j,o~~7>":f(~);')~ :gi:ve:~ by meana 01' a..:""'1'. "";':",' ,·".,I~'''~:.':,',,;' II'" i:~"'':":' '" .' ':.:, ..•..;:.. .. t~~fe".a,nd 11; ·;!.s.:~eq,i;t1·re.a.+o ·:t:1nd. :r .(x) trt; l\' certain.bi::'~l'.'I~j~{·~;·,~~~~···'~~,'~p~;o~:~~t'~",:.;(~),'~y a . po .• • .:' • ..... .' ~.•:' 'f~l~ ,~ 'I.' " ,.",. '., " .: 'j' .', .":_ " ,
'. by;.·UBi~g·anY·.:..o:(. th'!! i~te:tp;~·~ti~ntonnuis!3. ·.Then" .'. '. ~.. ',: :. . .).. ." ,:
:d,f:e'ferent'iate:the obta.in'eit pol~omial the l'eq,'lilred• ".' ,: •• "., +' , ,
of 'time's . We .111ustre:te ,,'thi~ by' Xne~s 0/ the iol;L• , '. ,'.':", I -, •
.. ..: ,:,",0-;'0-0- .
I·'.,: .
··1"",,,,,,,,~~..a, we ge~
= f(xo) + (x-xo) !(xo.xl) + (x-xo~{x-~l) f(xo'x1,.:X2''. <,,:,.;'
+ (x-xo).(x-x1) (x-x2) f(xo'Xpx?,x,3)
. tfe~ent1a> ~ng 'both eidc.G ~ th re:apect "to i; we ge~
"/ (Xo,x1)+ { (x-xo)+(X-X.,) J i'(xd,x1·x2) + , :
..t (X-;1)(X-X2)+(X-Xo)(X-X2)+(X-Xo)·(X-X1)}· .:;.' _"
f(Xo,Xl'~2'X.3) "." • ~ ••.• ' .••••••. , ...... (U .
1tuting in (1) from dif.ference, tabl~. wi.tll..x:,=t 5..,:: ..',:.~~.', "' .
. : •.. :.:::::r xo.+. qh. '.
(19)'~, ,.. ,.
:~·r:!,~~..IIr ,Ii.1I'
[:
1 ~
I;Ii:1It,~
Ji
·ii· i!".
:.1I':
.'i[II
117
Fo-ymtbe' di~ference table :ri>r the values' of ~he f'unq.tiOD
'1 '" ;x3 _ 5x2 + x - 1 at X = 1 (2),' 1', ...
. ,:{ ., .-0-0-0-0-0-
l,..!LQ_!!_1!_!;:",~L§, ", , '
, ,·t· . '
(iii) we have, " ', h2D21'(Xr '" A4~(X) -'d:r(x)' "
(O.2):f~(3::'? ~',?,:,3.10+ O•.O~~'," 0.335", .• • .~_,'. ',l,:">:'"~ £ (J~j), ':" 8.375 ' ,. ". ,',. '., .•
To find r I (3~g)'~::Lth e~r~rO(h)j/we use fO;m~~''\i,vh, '. " :,~:;'.'. V2·. " .v3 , .. '. ,':""
hD :: V.,. -"2'" '+' ""'J .',.....'.' ....: ',', "~:,,./','.i{ (O.Z)!' (3;,9) :: 1.418 + 0.~~5. ,,~, O.~2~ ~_'·1-.5~8
I . • .:.. •
~ £ (3.9) n 7.74 ..
".,' ..' . . . ~" ." ':_. " ,"
two terms in' (ii) (. ', '" .
'hDf(x) ';..6~(xJ -l A2f(X) ,',' .:
o.~f' (3.).= O.82~-1(6:310).= o.~661 . ....:.. ., ,
==t• .t' (3.3) = 3.J4
,', .
, " , : ' , ' :;3.',;, O(h2) and,1"'(.3.9) !'with ~'rro:tc' O(h" )~,
, Tbe d1fferollc,e tnb~~ for t',x) "l-~" . . .. " , . ~. , "
" F:C:~=_="==="'~~!"="'~","';"~~==?<~::i~~j==.i';'=:;,,,i:·X f<;;): 'A~(X): A2f'(X) i;?i:'(x) .
,',
.. t-, :-'40. I
(21,l"
",., ,
...(
,176 .
in which 'lD terms a~e taken into: ~9count is of
ExemEle {122 ; Given the tabie
x ).1 3.3 3.5 3.7 .3.9, f(X) 0.311 0.840 1.66) 2.796 4.214
obtain rJ.O•3) : 'with O(h2) IIerror "" l' 0.) with
It can be proved tbat the err~'r in the expansions
'derivative' in te~a:'of .to·rward·and :b~~i.ward differences
..'.. . .
. ' ..62 1:)D!(x) .. 1i (il- -r + T .;...)f(x)
Taking tbe powers of eq'uation (ii), we obtain. 2 2 2 '3 l' 4' 5 "5
b D .. A - A + T2.A .- b.6. + ••• ':3,) .' 3 J i 1 A5, . ~
.b D '" A -"2" + 4" .' ..~ . - .' • " etc. )'.Simila.rly, we can prove that' .' V. "" 1 _ e-hD
. ThU8,
.Taking the logarithms of both aided, we get .'1:J.? 1::..) .64
bD = ,(n(l +Cl.) .. A_ --r + T - ""'4 + •••
1 hDe =l+A.or
hDA-e
ThuB,
hDf(x) : (8 - 1)f(x).
On the other hand,
A:r(x) .. f(x + h)
. .'
r•rLl-f.\~.
i,r.{~l~". ','
.j
. ~".'
f(x) 48 100 294 '900. 1210 "2'028
11 ",1.3'.1075, 4
) ~1 meane, of 'NewtonIa divided ditteren~e form~l~'f1nd'." .
the vaiu~ ~f ~(2)t 1'(8) and 1'(15)·.t:r~m:.:the ·1'o.ll,Csw_:tns'
..~. 2.2
~1nh x '. 4.457'
Approximate the root
P~.in{thll .table, .." '.
": \',., '.r', J. .. j'... .: . .:. " ." ~., ,',,: . ", _...0) In prol1l~m 5 i'il;td:x: wh~ y·io;o·,·?9~·, '" .::,:.".",". :~:;"~::"':i:
. .' :',':"'" .
'Below 4040-606'1-8081-~OO
,'(>:1-120
" ,:,.~.~'r:4~~',:,,:,.:',.',-",",:.',!o':, :' ,;.
51 ". .'.( ,.:.
.3~ . :: (';: ',,',\ . 31 : -,
:Ji'Z~':of - :,>ttldlmt e 1'\:'. ': .31
MAR X S30-4040:"5{l50-6060-7070-80
FrOIDthe .:t.'ollowilig'table, firid tbe.n~Der OJ'I:,~tttdetj.'f;e
who obtained 1ee9 tan 45 inerlte •.; ~ :.'.;.~f" /".,,)"
'~'. i .
'(23)
• ,,·1
1'78 ..) ..
, .
:!'.(~)".1 '-) ·25' 129 '38.'... ..Compute f{O.S) end· f(2) :.hy ~o1ng"
i) Newton's toim~la :(1i) ~'Lagrangele formuia •.
... '4, .'J 'I. .' •
0':':" ".":.1,'.·
,',I..7.)" Given thEi table
. ,"." " . ,'.
:1[ " -2." '.' :). " '2 .'.:.....1....::r (x): :25'·" '. '.;.:8·.• ' : ":1 5. . '-'2',). ; :
• • •• ' ••••• 0 •• , "\ • •
.. Find ):'(:x:) 89' Q 'polYnomial' and find .f(0).• '.,.': : ", : • ~.'. .' " 0, "~ • I • :". .' • -',
.,\ "6) Given
. f(x)'3 11 ·'27
Pind .f(S.5).
.:4 6 8 .,',X ,2. :: :'10
83.
•• o •• "~.Glven, ", "
.3.'. . 4 .:''':40" !35:,',
4) '·F.~d .f(;j gH'en' ,'.:
':x . . O' : ...1 .. ' z':f(x) 4·" "{5'
", ..
e . r ..... .'.Find a10 2~ 15 using the tabular data
• l" ... " •
,,'Di~ 260 '= 0.4)8)7 ;. '-sin 27" ~ ·0~45.399t . sin 28° =. '. .
3)
~) Given the table' ....
x :1 '. 2 J '. 4 5:log :x: '0.000 0.)01 0•.477 . 0.602 0.699
l;valuat-e log 1.7, log 2.5 an.d log 4.6
J80
~~------------~x__-+ ~x
','.'. s.:
The lcaat-sq~aree Une 'ijas t.he. ~.qva~1c:>n. ~..
The general prOb'lem. or :!1Dt;4l~~,~q~af'1"6n~of ~p~ro
x!mating curvea 113 .call~d:curie :ti:~~~~~.~. '.~' praotic~,, ':', " .. ' . '.' .Jll~tY!'6 ot, equatio~ 1s <;,ften Sqgl§~~t.~~::Cr~m.'~\:1,41"~~~:t.te,~:
diagram. !l'huBfor tig. ,: (A) :'If/,o.~1.d~:us~'.a'f3il'6~~t l~~
:.;'1"'= a +. bx .. w~~e ~or '.f:l;~. " (11) ~e~,':e,~i~~':~~': ~:,P~~~~l:i,~r'. a fiuach-e.t1c c~n:e y ~ 'a \. b:i ... eX2~.::::~\, . : .::..: ., ': :.~?:"":": .:'. . "~:":..~~~·:~t·..:'~'.::. ',:;'.".,..:,"..;:'':..'
GenerallY., 'lI1ore than ODSeun~ ~;~ I/o:~1;:~t1p:{¢,:Li. '.: " ". .' : ,-.--. :, ..appear .~o .f1 t', a Bet o~ data.. Toe;~oi,4::..(1..Qdi~,d~a~ j#.d&!ne~tin constructing. appro:z:iIllLlting.curi.~·s:~p h8.v~ ~:~ .~~~~, 6"
.. ", . " '~y:\:" . " ... '').e'st-fitting curve. ' .' : : -.,:.." ' (x~' ~l?' ,
. . . '., ,(:it' .~Y) ,Of e,11 curvee appl'oX1D!ating . '.:. :,1 ,:'
.., ': , ~. .ot data 'PQi.D~.. ~he .. :.'."
.' '- '! •
,()ul'Ve.·'b.-Il:v~ngtho P~P~!t:r': tpat ..', :'::';:4;:. '. '_.:' ",: :'. '2 ;'. 2 2,' .' I' ' ..' .' :\:',." •. ,,42 ..' ., d, -t d2,+.!o+ ~ =,4 minixn~, ,,";.:; , .' ': " .',. '.'
. -,. ' .'..' , ..." . (x 'Y')'.'...ia the:~·.~ot-!ltting o~rv~. " " . ~.' ,2. :.,ApUNt! '1i'llvi~g'this prop~rtl' is •. : .. - ~.x .
. :. . " .'Oaid..'to· tit the data in 'tHe lea8t~B.qilEfres sense ~d je.' . 'It ' , ...
'Qil~d ~ least-squares c~~~. . ..' :... - '"
) . . " ; ','"a curve 1a c.e.lled .tl.ll. Eipprox~t1lig, curve •.'. .~ .. ,
F;o~ t'be ~ofitter ai68~ '1:-t ':i~'\5#~n:'~O~Bibl~':~'~.:visualize 11 smooth ~urve 'appro~i~a.i:iiig;'~h~:':~'a.~~..:,'::.':· ' '
.' .'7J' .. ,... ',.' " . .(,25)'
:)
i. j
"?'-. ..._.. .
Y.1.
scatter· diagram.
If 'we collect the data ahovd~g. ~orl'e8pondi~g values
01' tbe 've..ria'til~s X' (\lld1 and plat tbelli' on rectangular
cool'ciiri'ate oyetem., we shall obterin :thE'"so-called a.,:' ~
'Very often in praotice a relat1onB~iP 1~ found to
exist 'between t~ (or more) variables,' and one wisbes'
to exprena this relatiollshlp in mathematioal from byin ' . , . . .
dcte~ng an equation connecting the.y~ria~le6.
.. 18.
- )0,
-0-0-0-
f'(42.~of the" equation. f(x) '" 0
f(34)~ 13 f(JB)~ ,
15) Apply Lagrange'e formula inversely to find a root'! '.
9,1'(2)
(24) ,
14) . Use Lagr~ge'a formula to :find i.ex') i1'
x ,& 0 2 5
l'(x) 2 J 12 147,
Find a polynomial f(x).ouch tbat'
,:f(-4) 1245. ~(-1) '" 33. f(O) =:: S.
. f (5) ,'" 1335••· Also find !(1) ~
13)
~.,~j,[: ,
~( tI' 1,I :' ,
~~L~
'I. ,,; ~:~~!.
~1;: i~'"\ .~.I ;' I'}....r~ i~ hi;). \
~~Iiff~II I
, .: .'
: ,:.:. »>
..:... " ••• ·1 ...·.: ':: • .'
, ,.' ,
. ,:., ','
. i:
" .x 0 0.5
. '1, 1.95 1.55
, "
,'441
:' 24) Fit a least-squares curve ';- ",.e-bx, to the ~~a'. ,'" .
(27)
, : .
a lea~t-oquareourve '1
x I 0
J) }!'it
2) Fit a'least-equarca line to 1;h~':~e.in.',"'·5 , §" e,' ,,' ',9.'-,~:'.,:.~<"3 4 6 '5 .': 8'
4'
"•~ • ,I. •
1.8 4.9 '!;'~7''. .' _" ,..x 1;2
1) Fit a least-square parabola to tpe data
PROB'LEK.S-----.----------000-'
• ,'0
••• '7 '" 0.545 +' 0.6~6 x
.s,.which give, on solution,
'"\, '6-' ."a ee '-r:r-' :,' Ii = .'7
til' • . 1T
.~... .(26) .
182.',
. ,,64 ""566 + 52 b.
L 56 40 524 364". , ' ,~er'efore ~e l:i.'~vei'". a + bx, where
40··..: 8e +' S'6'b ,
, ,. ".' ,' ..... ',
,':
" ,"
t "'1. "1',2, ':;9 ;, " :i:',:"",4,', .1.~:,,~64 36,. 24' '5 . ~4 '. 46,.7 '81 63s 121 88·
9 1-96 1.36"."';. '. ;," "."
1)-"
, , 4', :6e9
,11-.
14
" '§21~!!2~:,We ~~~m the table i,,'::' x Y, '.x
.': ... ' .
The l'eaa't(-square's parabola hae the equa.tLon
, Y '" a' + lix,'+ 'c:' x? "where ':11,) and c are d~t.ennined frOln ';' -,
" ",',:,' .. ' .' " :~: .., '., 1:;:. '11 ee . na,+ b L- Xl + o.L._:"Xl;'":-'.,,., ,'::;I~ '. . . -' '. ' .'. . .,' , .; .. .-':~, .
, " ':'~~:,XIYi:~?' a'~ ,xi -: b ~,:'x~:':~'~:,~'xi~ ..',-:Ei ';x2y: ;.. a 1::.:x2 +' b t '~J '+ (, t.. 'i4 '~'I,~'" ii, ,i i" ,1 '';. 1.1:' " ' . :'
-0-0-0-
, '~Ple, ..:';h ~)':" "'}l'it "s +eaat.:..equ13.res,'),ttl,~',to. the, fat,a" "., . x 1.., J 4, 6 .e . ,:' 9 11 14
, ' ,':, Y '1' 2 4· 4 .. 5". :7, " ' e 9
" ...
equBt'1qns ," whe:re ·.6:,.1,·.are dete:rniin.ed from the. , "., - .., 1h,Yi ~~ + b.L xi ...
. ...' .2~,XiYl';' e.,L ~1'+ b ~'xl'
: .
.. " .
185 ,,:
.. ':" '" .5 ·0.6~~. -=c.:;:==='1:Z"==a=c:_=====~:::==-;==~==b:~CI~==::£-=====C::==;: ... . :'_", ".,
4 0 •.602
. ,,
."..
" ",
11 136 _ . ' ' ., .. '., ." :: .b;:;===-.~a=u_Iau..IOII:'D==:;=-~~;.DD1I1:IiaQ-=:;,.'
.·32 .,::::... :
20 :':.,.~...,~.:,;:·,:,:'..::s"'\ .... oiob ..""28:: '.::: 0 .r .:228,' .:,:~';'~:~8!.:..:: ."
1+~76 '.'404 :, . -r '
1 .:"4,
.J -16
5 4 . '7 104'
9. ))2
-12
t: ,"
t:>===p"=-' ",,=-,,,,;=P"''''-;';;='''='f''''"'''''~. % :i' '(;, j}',:: :b? .6,1t.
'~...' ~. . .
·An;:,we:;'s·' of. INTERPOLATION '.....
1(2) 11, tli1~.m~~ '~b8t'Xo '" i:;, h '" ?
.. .'.~.'
:"} .. ~,
·1''.
. 187,
, .-0-0-0-.,..
I ~•.! ~
• i" .. .a.:.:r... :~t.l. ".' ,:' .,': .!;,
= 0.."1926 ·,:"'·O..088G - 0.-4646 ~:0.405 .' .' '.'. . "t.· too I: . . '.~..': .'(.;:'~~.. 5'} .' ~;._;:~~~-\4.6.L~ .. ,_..4.6 .-"5 '.:'.. : _'. -. . " .', . ..' - ,
1 "'." 0~'4' . . . , _".'
. , ,". :-:.,' ' .
q - 2.5 - ) .. ".:: "0' 5- 1 ...- •
.';. 1:.(2.5) '" 0__.i!1 +(.-0.•5)(0.178).. (:..o.~~(O.5'> (~. "25) .: ,
~~, . "': .. ,. " ,," :~~:
.. I
~ c:3 ", . .:.';::::i':"2';S>""~~ .
.' '.
•
~186'
" .,;''''',
.q .,,.·r •
" " .."4.6 - 4 ~ 61 ~ •.. .
x' ..o
,. ,xo .. 2
. . .' ','.' ..· = '0"2107 + 0.0126 - b~b02.3 .· .' ·:1 .'
, .',.. • • ,..J'. .:. •
· ~ :?.·2·.tn -: 0.0023 = o'.•~g21·b·~0.221
~ L·X ilt 2.5 L.s...-.& "3':.'f':.,!.1.0g x. 1..-...:-' .JL..-.~.. . .: ., . .~-:,.; . . .2.5 ..; 2~ O'5q .. 1 ... ..
. . t,•. 'I
l ( .),' 1. (1 6) ::' (>.5 (-6.5)Of; ?·~5. ~'0:'}~·'·l.-t;·9::~5.::9.·7. ;~':"l:,' .2.. .
', '. '. +- '.(0.5)(-0.5)(- 1.:~!):. (0 ..023). '.. ::." .: ,,, "': :r,~:\.. : ~.I ::.;.:::.::.'.. .'....;.:. o.;3()1·.+' 0.08S9 +·,d.006()·+:·0~·00144 ..' ; ;.'. .. :.::.::..:~.. .', ~
... ';.,' O.J~i4·;;··":'"O.j9~·::~'·~·.: '.: .: .." :. . . .. :.5. lo ••
4 ("':~:~t:'x·i4·· 6L~ '~'·;'~~":1..0g.'~. ,._ .JL...e....•"i. ': .... ,' '; • fl.
", ~.!'.' .
.. .
where q .. 1.7.1':' 1 ';':~ :
.••• log (1,'7)"'';'0 + ;0::+'(0.•301) +. Q~?~~0.3).: .+ ...24-.,.(:. 0.3)(-1';).) ~- 0.051)
-. .' ':' :' :3"'''2 .,!::'.:
X.',1\
(~ 0.12) +
.''. .
'~'. 1·_
( J )
. .. .
:Jo = 1 . ~ 1..: ~ x ~ 1~7.1.~ .;. '" log X 4-"'1; JL~'1
rex) .. ·.f(xo) ... q A f(xo)'+ g(~.~1.) ,i· f(Xo) +
+ 9.(9-1 ~ f9 -2) 1:,,'3: .f(Xo)
• Q. (9..':', i)log(1.7)·", f(1) + q(o •.301) + --2 - (-0.125) +
~. g(g~1lfg~2){~0.051)
{ 2 ) .
,
J
.'.
'I
•··1·i
,iI
'.1·,1
I.j
.!
....:', ,': .
.. '1",',
.J
" ".~"
.....-.
, -. r : :" "i·i: .
,',
: ,",
'iB.9.: .
.3:3 "'''' .; ...•..~.:.'.:.•.,'., .;: ,!,; • ~ """
-, ~'. . :.'( .":'.':.!'. 'j ..
2):
-4 1:1
•.{~. il,
a 50u. .
10 8~
. ,'. "
,.' ,5)
.. ."':' ...where'
, ',~' :'. ", \ ".'.' . ,....,."; ','
, '.
rex)
( 5 )::' I':
188 .
·4 85-==_<t:=:_",=""",,,,,,,,I:,,,,,,,,,,-==.::!..:===,, .
45
-O:--O,'":P~ "• ,. '. f '.
. ... .
£', '. {j : ".:!
F"'''··:F''''''·''_C 1=:="i"P"~=F·'l'·... : i:('::y~'1':11,·. ';d' t~~t. /:).4:.
'. v 0.:. ,. .1 I I":- ,,,,,'
.3,. .' 1 4 '8,
11 6". :
Z. '15 14- Q!
25 s:3 ,40 20
, . • t'.. ""26 15 - .~~:.......15..' ..q. ,- -, •.
• ,: I'sin(26 15) = 0.43837+ 15 (0.Ol~~2)
x .. 'a 26 a.';_' -,:- __ ..,>' l0' ~. '
'/J', '" 0 '';/",,'.26'.15 .... LI~""-,",:,-',,,.'~, Y".~l8,~·:% ,,~.J~
. ... .. "" '. .. ,' ...._ 4:.,/:.1 !,~~.. ~i;QJ:=~F=2:D=:::.ct.==_=,~.;:Ia=t::::::::I==_=::a===a====
, '. .:, . '1
28 0.46947
',26" . '0:4)837" . ('.";t/ ~..--,
.... 9~q~~62.,.27 0.45399 --0.00014
0.01548
(: 4 )
4)
)
.,'- ',: ".. ~\ ..- ..
..•.. '. :~:: :...
1'(0) .. 1.. .. () 2 10X -+- l ,,' , , '. _. -._-,; A-'::• '. f:,:, ::= x" -', .. - .:.... .: .- '.. ....... ". "',;r,',
',' .". "'. i, -,!,' .;
'~... a .
( 7 )
'~
~190 '
.. .f(x) .. f(xo) + (x - xo) I1f(xo) + (x - xo) (x -'
- X1)j/f(xo)
f(x) 25 + (x'. 2)(- ~1) ~ t~ + 2)(x - 1).1
"" 25 + (x :..~) [- 11 + x' - ~
25 + (x + 2)(x - 12)
. ..~===C=7===-===-====-=~==C
-111 -9 1
-7 02- -15 1
-4" 4 -23
F====--..===F"'....• ="'''==P''''''''::''x y A.' .6.,2 ~3-2 25
-0-0-0-,
= 22.578125t::::.22~58
.. .
•,co ". 4 .;.l:.f{S.5) •.tI ~'J5.5 - 4 ,1.5q ~ 2 ..: ~- = 0;75 .
t(5.5) '" 11 + (0.:75)(-16) ~ . (O.75M- 0.25) , 7
+ (O.75){- 0.25)(- 1.25) .3. .31
11 + 1~ - 0.6525 + 0.234375
2}.234375 - 0.65250
.. .
( 6 )
6)
!' •
l
" !
IIi
, 'ill",,
,
19·3-
.' '.,:" '":\:., ....',;,::;.., .<, ,,{'::
q(q._'1) .. '2" .' '.rex) .. :r(xo) ... qAt(xo) + 21' . .t:;.t(xo) .......
fc+.4) 'ow~' ~ (o.')~4i)T ~qrft.>'k::.~!,.f~...· (~:).~.: .. '..-, >:":.~.... tJ ..:;_ ':'. I • ' •
+ (?..:'~~- 3·:~(-.l.6ti~~1~:~o..~}(~~1~~~(~~(~?.. .3.1+ 16:S·-I. ~s_ i.;f;:~{~1.~:,'~·Jf~.*t~:..:'.:> ..
~ 43 $-\.uck~···· . _.:". .... ":' ::.....- -, :: ., .... --.~: ' '.: r .. : ,.,'.,
_ ',4' I- ' •• ' "".:.
.'•1 _. ,',
4ft - 40 ,1,.Hi .' ':'.o.~q. ..
:ro '" 44'·· .. ....
q ..
1.,... "'." .. .,"" D~~b= "'~ "'= ..=··"'a"'="':/.
so.. . ....15~
laO
"to
"60...
'~ "
-2S, ..-.-. ",\''", '.
40 11
.',-.. (9).~:,:-"':1·
. '. reo, i»
( 8 )
.' 'i'95. 19'4
. ,.' ,.
. ;'
'1 -. 'Y'Ob,y P.:.
q ::
e.nd
"1'2D ,Sq0l.'
.,..20/00 S4o....iQ.r·O
'5·0
(11) .(10)
I
i "
'l
i. j~ .
11 I1 .) .
it :II~~
197" :t .
-o(
12)
end.' :
. ::.. ~': ',':y :'1'0' .... 5:'- 4'-457 . _ 0.54,2 •NO '0:.3816:"
;, ••... Ci, ;. '&Y.~.i" ': ..', 1.Oqg .... '.";' 1.009 ::-:- .;);.. .,. . \ . . \. ;",.,' ~'".,'
(.13) .t· '::".,.' .,';
r
, '.::.ll' 19"". 6
...=~':=-=!' ::==.::;-;::... ~:'~.-~'~.....~.." ',', .. .
y - YJ.'. ' ..1 ,(j ('<1 _. 1)AYo. 2! r ,
• , ~f
q ... .2.6 6~695==:=~.=~~f::--=';:
'.~..2 4 :A~""""" . ""'.. ..,.:1.J ,I. ~ ,I ' ••
. ; j 009' .:.''2 4- .~ "11/ -:~.., .,:. ,.. ..!' .. "lh-4- 6., ..... ' ..'~":.,',0.220;
1.229
',; .',,', "',' )\4' ". \
.'."~<+~"0.~;~~':.>- ... ",
. ' 11)
.. ,': " ~". " .<1, ~91,·"",,1t..J~A2·:y:"·.. ' .'
1 '0'.' i'. .. ';2,I'M ~'\ :/',:,7"".',"· ",.125, .,. u ...·O ' ....
62. 125 j (1.125 ) ". . .'2! '.
. '"•. ": :"" '::",:. v •
(12)
\.
.,199
,.,I r'
'J2 ..... ••• • •. '!.
.. ' .'" .... ~" 30 34 .:: 3~". 42...-J? -1.3 3· .1,~
. ,.~.
, ,
. . ,,'.'.:'.:;'::;t ,:':'::,:.~:';:.'.,!', "
. ,""
, " .-0"0-0-
. .:. ': .<=I -41- x(x-1) (x-5:~,,,,~-W- X(X,:,1)(X-:-2)
• • • \ I.." ..~. • ':. '.: " " :; ,
,15)
.', .
.,i:' .14) ....• ·'!(x)
..
'.•j'
II·I
/':I
. 4(j~ ;'" ~ .~~ : '!.. .13 2028 -r. : .~~ .,' . :." '",: .~.
, • .'" : '" : :... • J + • • ~ ,
F==_=z::..6",. "'..... , !'I..=_ ... "..~<'.",..=:.=====. . j' ();3 ,. . ;." , :.f'2)·o 52 + (2 -·4)~~?) + ~~;.;·4)~C2-.5~) ;5'+
.: + (2 - 4)(2' ':'1 ~)(~ .:.1.;;" , :' ;' :". :.>.'. . ."'''';''. ·.,;,L.,:,.· .... ..... .';
= 52 ~ 104·<..}0 -: 3.0 ,;,. 142 '- 1'4' .. 8 :, .. f,~~,:/~.'~. t: ' ":'.
:1'(8) = 452and .. ' . ":.." ..'_;..'" :.~.
.f. 15) = 52+(15-~)?2~(15~4"(1~-5)'3+·(.15-4·) (15-5) (1 .. '. . ..... " ~~'. .~ , .. ',~. . ' .= 5~ + 572 + 16~~'~8~O .... 3154
1.o
"I'
I' .01 0
525 100 15
977 294 . 21
20210 900 '27
.310 :11 1210
. ' ..:'.33
1
~='==F"" "='F~"'''-]Q=.;;'''r:...."'=P....'!'=F"'c:~·=. . . . .',2 ,.3 ,4 ;5x' l' ':~ ·fl IJ. 11 Ii4 48
" 't-,'1.:3)
I.' \ •.
I, I." ;'"
I,' ..
~~-;'l~X-2~~X-~r(2) + "(x';'O)(x-2)(X:"'S) (J)'.. ,0-1 0-2 0.- . '. '.~ .
+' ~~O).(x~') (x':'5):;:'(12)+" (~-O.)··(x'-"i (x-2} (147)(2..0).(2-1)(2-5)' (.5..,0)(5-1)(5-2)
.. -,,~ (X-1)(~-2)(~~~).·(lX·(X-:2r<-~.:.?)·:.. . ....
. t(x) = '1245 + .(x + 4) <-; 4'.4) .(~ + 4.·)(·S.i'.25.) :..'
+ (x + 4)(X+1)(~-O)(' •76J)+'(x+4)(x+'1)'(x-O) (X-2).
= 1.069
....
(15) .
.. .
• v ".
',:', .
.' .
zui .
• ,0'
2.1 "',a .. 9.8 b.'i:' 96.04 C . ". : .. ·.... (8). '. '. '. .,'.:': .::',:." . .:' ", .. '.', ,.' .' '.
Multiply each' equatt~n trom: (1)' t'~.(8) :'b'y:"~;h'e',c~~r.i,l-.': :..,', : ••• .' .• .: : .~,;:. .,.. • r , '.. . • • ::: " ':. ,','. . ",:-... '.,' ..
oieni of a and adding we g~t "" , .• " • ..., •• ...• •~ I'. •. . '2' .' ':I' .j , ' ••• : ',_ ,'.; ,:.
46.4· Ba + 42.2 x +,'291.2·x :-. '-.: '::.' ···..•.••(9) .....' . . ..'~ ':~.::,' . : . .
lrlultiply eaoh equati~n. 'trom (1)' 't~ '(8) 'by. th~ .ooet:ri-:::. :~:• '. ., "I !. • . : '~"': : '",' /~>':' . .: . '.:. .' . . .:
oient of.b and adding we got the second equat~on ~ a,b.~r . ".
and 0, alao m)llt1ply equations from '(1) t~'(8) by. '~he '.
coefficient ot c and adding wq get the'thIrd eqUation'a,b and c , t.hen Bolve these equq:h~ne~'~d ..aubBt1~ui.e in
,f . ~ l4.5 = a + 8.6 b'+ 7).96 C
•••(6)
.. ' r·.(7)a ...7.T b + 50.41 0
. '.,. "', .'.3i6.8
"., .".a ... 5.7 ~ + 32J49 c7.2. :... . ",
·:.••..,;~4)·:
.: ~(5).
. "
" .....';'.:
• :1:" ..
-: ',:'
. ._. . '"
.'. ':..'. (1). . 4.•5 '" a+ 1.2 b + 1-.44 e
5.9 .. a ... 1.8 b ... ).24 0,
7~O = a + ).1 b + 9.61 S)
, ..'7.8 :: a ... 4.9 b + ~4.01 e
y = 8 + bx .. ~x2 . ",.•Another solution.---~------------
"." "-0-0-0",,:
" ":and bolva then;. we find the values ot a.b ~d' c •
.~,, ;.
(17)
" '"
3'8.7 3)2.82
:-26 •.~6 .·25.~.:
:·4.~.28" .:. ,'.;' .
~o5B~~6q~.i. !~":"l: i··;··..:.25.~r,.:·J. §$,1 ::;:+$~~:bJ;'6:~:
;4.9:, t~·a.5.·7' ·7.2
7'~:1';' 6~.8·8~6::4:5 '73.9.8· ·2.7
3•.1 '7.0" l'
., .x . : r
" 1.2 4.51.8 5.9
y 4.5 5.9 7.0;,.7:8: 7.2: 6.8 4.5.' 2.7~": ('~::...~~:;..~.. ,:i ~ ... .:., ". .
!l!h'eleast squares. par~bola 'lia~' tJ:le'e9uiftion
b.' ''2j.~, .,..,.
.y ....a + x + ex· ..:..,~ .. wbere a, b and 0 are cont.:. .
CURVE, :PliTi'ING'". ' .....
(16)
and
. ;
....I
r.~~1,(,'2
""r.~~.t..:;
~
,.~.
, \)~ ..
r~
~I, I~! .,, I~~~i~'·1, I
r~'r,til'i~ .
'1; J
.' ~~
. .I. !
" " :.. \ ..," ....
':' .
'.'.,
'r
". .: •• ,.JJ.') . ::. " ',.
·~•••. (1).: :',' ~'.::';.:;:::."..•• :.(.2} ;':," .: s , _ : '\
=~=F~~~=D=F===F~~===~==.);, .; .'~: :::".~'f.:.I,A2·v: r...::'"J; ~.~:~- .. .-;x lor, '1 x x log· y'o '1.09'9 0 .o:). 1.87~" ':J 1.~m~.2 "~;4g6;'~'4 4~9.36.4 )~)51 ...1:~.. 15.648
=~=a-'':: !Z==t:l1::=~ =:::=:' :====~S::Z;,,~==[Z]=~;~;!~Q;l_~~;;!;~~_,,' ,.:
.~,'..
1.87? 'r . 2~4§8. ·3.9.12t , ", .... "I ',5:.,,: ..... . .
'log '1 1.099
since. --~ __ +-__+-__-+__~
. Y J 6.5 1.1.8 ~~':'.'<;;;:'::'}\:~:".'.. ",th~n we have ·anothfrr.'.'ta'l:fle·.f'.:'·:~:~:':;'PJ~:""
., • • • • \0 .'1'-" '.:r.~:~~.~, ..... :.:iC 0 1'· "::,2 . "'4
"': .. '
.1 .. , .
..(2o 4 . '."
. ~. In y 0: l1'f .:(J..+ .~b~.. ~ ....~~>:;.J.,..... , ..' ! .. ".
.'.~' .". ..
'dent of b and .addine 'we ~~t" '».:':",'.' .. \ . ,.." ' ..'i.\,:, ..: ' ...... ~. .
. 226 so 42a + 3,36b\·'...··":':'.:" .' .'. ..' " :: },' ...... .
solve (7) an'd (a) we get S8.m.~·:,~.~~uti~n.' '
-0":0-.0',':....... ..~", ". .:. " . ..~.. -:.. '. .,to .;, .' h "
28 ,. 6a + 42b .:':',' ': ' ..
. Multiply each equat Lcn from .d:(t~)6)by the co~·~~i-.
. . :.,... '." "
cicnt 01' a and nd~ing we get
Multiply each equation il'om' (1fj;.~.'('6) .~Y the c,o~i:ri-
•• ~(5r·
" •.••.(6) :. ' ..' ," "
" . .' .:'
8 .'" a + 1\b .: ''',.. '.".' ' ..
." .5 .. a + 9b
~;:.':'.. ·t. t ,
203202'
.f"
. 6 = a +' 8b
226 .... 4.2.·~ +'3.36. i>:• ~ .: 101' ~. •
which give the solut~o'~·.:aj.=:_; 1/3 .mid b ..,61rt : .to: • .' '1..•
• .• .y '", -:-1/3 +. ~ ~. :::;. ~ ':.. ." " .~:'" ' .., ... ", ':, .' . ), ',", ..: .. :.
.. - 0.3) +·0:.:71' !,ii' =» . ~ :'.'. ". . .; ", ",::.....:~.. :.... ::.. ;. ": ". Another eoJ:ution .• ;:f .:,}." .I':. " [.., ,) " .
. ":'. : :;'.:; f "~II .:, J 11.:/.:) \ -,;:,;';";"': ".,'1:lUpstitute inJi .."'·a i+ '6: ", wei get',
,,: ."},,:, J·"l~,(;;':l\··:.E:·\.~f;·:;\.~r:(,'::~t ..... , .....' . '12 '", ~.!+ib: '; '.': ; ::'(l,'t:;.'i:'· ·CJ·i.1.r:~1.:·~i ~'.:-'.:)':'I ~"\:
iJ ,:,'a '+ 5b :. .• ', .. I .."~,,: .~,~J,~.!""=,,,,::,'_',:.~.'.. : :. ' .
.4 .. a + 6b '" .. 1.1, ':' 'i.:~;.. L:l" 'thtl.:·::eq\l"'l j .\ "1' ;"
,", .Tbere1'ora we have y =: Q + bx " wiie-:G
.... ::0 .. "' .... = "'="'= ,;.=="'~..... ""L 42 :28 .J'~'~225-===="'''''''''' ....,.,:::=== ...... ''''''re
'..U .S·. ,'''21· ss9 5· err· 45
6.8I• .,.
64
2 9· ~
.3.2·5 .. 15
4 36 24
48
i'
..·"';;=;~~=;';~·F,;·,;~;",=.....
x Y .. ·:ir2.:xy
2)
(18)
•
. ,
~:.
II .'
': '204-,
.' • 0
,• 0 I':,,.', ',_.',-o:...~-o..:. .:The same idea as in
. ~:: 9 • .351 CI+4.i~:B:+·1b' .' .#_;\.,'f,
Multiply eae», eq'~;i.9n"by:,th~:~'Oar:1'1Cie~t,or"~.' : ,: .' . ., ~.. ...we get, ' " ,:'..-:',
22.4%-~j~:;t~i~~..(5) aIld (6) -az-e '.thl;~~~~eq~nt1~ the aho:ve "",.~'t'n,oa'.
:: , ;: M:.t~"~.~ ~.\ ~ ',. '; ... ",l -O-Q-O- " 0,
: '. 0 1.~ ... ' ','."+: ~5)
,U~lt1ply eacb equation by th~ ooeffioient of
.' ~d add:1.ngwe g~t '
).912 • In a ~ 4b
'J' 1.872 .. In a'+. "b:
2.468 = in a + 2b'
.'then 1.009 .. in '0 ~ O;b
"ln, Y ;" In,' a + bx. ...
" .0'7;'y .. ).063 .
-;:. , , .:."', ...., solution--------then
,~"
data witb'....., ,....
. . Find the B~r;;~t l~n,e' th~t; flatiafies tbe following9.351 ~ 4 log a,'+ 7 ~
22.456 Q 7 t~ ~'+21 b
'Bolve we get l~ a '"1~ '194 ".~ a '= 3~063
;', "
..:....., (20)' 1
'<1'", t
~'" 'hI "
.~~';:liB 'j,,b fr, "I l'r' "I "'~ililtj I,
~
j,<
Y;l£~I1:~
"I 1',
~(i.I!'(t '"'~ : ,:" I', ;' r'
'~1I,{ !.,~I '
"
!:,, '
~(',~.:~'.~:"II 'j
~ ...
" 2.07' .
. 'I.red curve.
Theae are three equations' in ~~b eild ~'. Solv e then BU'!?eti:-. • • ! .. i ..
tute in y ::: c:x2 + bX + il. :to' get th~ ~quat1on or the r~q:u1-
'~ r';55,;,&+ 225 b ... 979 c '" 1586.''.
\ ": o' ",
15 a +·55 b + 225 c = .368.. ;- .~
Sa + 15 b'+ 55 C = 91
3) Coefficient of c then adding ...• ,1,1,'
'>1< ,': • .:2) Coefficient of b
" ,' .... "
Multiply each equati?n by
1) Ooeffioient of 0 ~::', .. : '
lI. + b +. o· '" 1
a + 2b' + '40 .. 'a :a + .3b .+ '9c '" Fa+ 4b + 160 '" 25
a + 5b + 25c .,. 40
substituting we get
~.!.: find the curve y = ex2.+ bx + a that satle£~esthe following date
: I ~, : 1,~I 2t 14~
--0-0;"0-
.• •
'i', I,
(2.3)
r
.....
" '. ",':.
.'he[~r";~q~·:9D~0"J~fniU\-.~tgr~H-','
9 . 27
',1... 1~, . '0:.'.5 . . 0.5" ,., ( " .,. ':
8 16 1.•732..' .3.464'•. ''':. •rv '" ,.
131' )« -.9;..''·2' 0·•.8~6· J
" .. ..'.3 . .' .1
o'o ·0·
• • • .x' 'r
. 0 ·0
.5 0.866 1;.,."
2 jby' the !ollo'!f1ng tab;t.~
x l'
~.!.:. 'Dctenn,ine by :~.~~:jnet'h~~ ;'~~' J,~a~~. I!q~re8 the
fi'ttlng se'cond-order .polynoinia~.to the date. g:l,'V'~:p.
. "".
.. . ...,": ", . ',.:'
(22) .
.1
..;X' - Xo
·1.(x "-,")
q =
x,·?:··:. 'xo·.·::/1h' ~:.1·
the· following
.3.2
4
. 209~·.
4 3.2b~~b=c=~~===~==~=-=====
2 1.8
1.7-,·1
Q'~5: ~. Q
)'. 2.3 0.4·o.Q
;;':'
the
; 1,.7 I,~e 12~J 1
data
Find the function that satiefies
-:0-0-0-
Tb,en use 'tll'c 'least ~q:~~~~':,:~~~'~ddto: ve;i''J;Y.. ,' ,,"'",'1:.,. ,',: ...result·,? . . " . ,.. "
, 'fi ....."'......=:,;,.;;~:i=~;~·;;"CF=..=..:x: : y ~ ..i,..:,' . 62 .6.3 '
and solve the reaulti~g:t~w equations to get a & b.
.' "
(b) Coeff. of b and)dd.·
(e) Coeff ~ of tog a' atid 'add, .'
log 0.7 log a - b
log 0.5 = ~og a - 1,44'b
Multiply these equat'iolls 01-. ' ', .
log 1.95 tog a -'0
I.Of!, , 1~55 fog a - o , 25 b
(25),
':.:-
:', .Y '", Log a -, b:x: .'. _... :1og
., . ,', .'. "'.
",:.
'.''.. '"
........·(b.~·OOe1'1'.· . ,..~;, ,. 7. d I.?f :-t>i·. then add
:, .' ,.' ,: '~," 1':" ~ ,) . . .. I '.' .. : J ',' -
. :.. . Log 0.5..,,:.l~~·I!-....t-. b..:'_:-.:'.' ".: ..:."~''>:::'::~'.':,~.'.)q~:.l'.a:~:~::i~~:··a:..i\~:-· : .
..1o~'50:~::~·~.i:·~.·;·~.b· :.'.. . '., .., ..( ..i;
.~ii~P1Y"t~~~~'e~q~~~~'o~fJ::~~. . • ~:.:..:.-.': . ' """. l' tI
,"('~)'. 90e1'~~' 'ot ·,o~',s.. ;', ::·.!rhen .sdd
:. "'" .',' .. ....UU~!'1ht~~'ctl?,' g~t .':
"::::
x' ; '6 ," '1" :.' 2" '4
y . 3 ,OS 1.1.8 50
.. log -:t ... J:~8 ~. + ·b.x:. '.
if ..!:i. :. .find the curve
(24)
CHAPTER TH·REE
·t ,.
..Multlbl:r eflc)i' (l~U~~i~ ·by· .1)' ~ocr·;·.oi-:~:·~~:icrJ~.:.~:~~·.'(~).:O~~f£.· of' b' and
. ."'",' ".;. ,{:,:~:~:",":.?".J::"'~;~",::;:: '.' 'I ;." " ',' ,
',) Coef:r~ o:t::.e.: ~~ 1i~d~,~'·.~h?~... ....;...... ,,' ,"~ ',': .: t.\ ~.::.t'f~t.1.F~:~i!l,l;-I~~".:.,..': ~~:. ',',''.; ,::;.:.' ..
. ' .' ,:'-.. ·.:j54.:a:;i+.:.~(l.Q~~li'+.' 30 ci.; ....:.~q.~~.. ,.: " J' 'J'.: ',,''' ..'.: ' !." ".' " l',i.'" . .. ,,:. " ',' .: '. -,
. ..' ·<.':"1··~O~:~,;~·~..3(l.>b.::~ :1Q.. 0.:. ~. 25· .
S~1vin& we get<~f'r:~F'·~f:~,•'.':. . '.. 'a·=.O~·2!':~·;·::.b'~.:'~0.5 ,:~
. ,; r)· < •~•••2 .; \. "': " '" ,.;·:yd:.O.~ix-.,~~tO.5i+2
. '.: ': I' .... :. . .. ../).I • .. .'i.1o!= =.. ~ , ..~••~ i~~~~#e.~!. .~§§ : .. :§ '.; ....
." ,:.:210"
. '.
: '.'-•. r(x) = '1'.7; +. (:x':'1·:>,~6:t.·;':,~.·,(x~!.)~X':'~'.'(0.4).. ' .. ".:.' ' .. :... :.. "2:" ." .-·1.~7 +. 0';1 ·X:::-.<?~1t'.~..:~.(~.._'3x ·t..2) .:
. ".". . ... '~:'. ' ..::.' .':( ..... : '.
. ;.. t •.1·.+ 0.1 ·X"';".0·~·i':·.+··.0~:2:i.-.o~6x·+ 0.4...... ':, ,',: .. ,: ",' \'" .' '.
;;.' 6.2:.-t2 .,. o , 5x:;+·;2'. .... ,..",.'• , '. • t' '~:"" '.: '.' ..
. .n;,'. the le·a·.e~"eq~ei-e mrith~'d};te~'.:. .. .. .' "" -: ". ,,: . . .-~:' .;,;.:" . '. ". ,'....
..... . '1 '" '.ia. ,+:·:bx.+,: c' .
•
'. '... (26")' . ' .. ' . '... "" .'
o. \
213 -------------------------i __~I~ ~_
! I
is specified first.
#; The number in'a'mahiX are called its elements,
CO' • fiO,*' :'DmceA= L5 12 16.l· ."9, ij. has two ro;vs,and three columns, we
" (C' 't~at A has size 2 x 3 (read 2 by 3) , where the number of rowssay
*' The horizontal are rows, and the vertical arecolumns,. ,,,';
such as: A, BJ C, and so on: .~.
Which is caUecHl ~~1rix~,1,;, , , ,''(:,:'~,:,..':,~ , In'~yin'boiic~lly'"'tepfisertt.i!1gitiat~c~s:'we 'shall use ca~itallette~
f 2
1 -: ]:, "."; 5
:'9 -8
Can be described by the t~ct:anitlar'alTay
equations
2x + y - 32 == 0 ,
8x + 5y + 4z ::::0 ,
7x - 9Y' - 8z =0 ;
Matrices: finding ways to describe many situations is mathematics and
economics leads to the study of rectangular arrays of
numbers, Consider, for example, the system of linear.... ;' .:-\ ~. . . .. ,
( MATRIx ALGEBRA )
:. ~. !I
.: ~:, (j,
I:..'f;
~('
f~";1; ..
,:'.;.I~;!'
r,
,,:.'.~,r.' .~ • ,1,
, •. ',''.i' .:..'
jl'i~ ;;:,~~~~~i.. ''','~'', "
I_ ~I;:t i,t4 :;,;., I
I tI !
M
ft
I____________ ----------.114_' ~~
; :~~
A~ [
2 3 ~ n3 .. 4 ,5:
,4 5 6
215
~~mp\~; If A =I aij J has size 3 x 4 and ~;i.,~:i+j, find A?', 'I'• • ! "'. •
S_o,ltttio,J1: Here i = 1, 2, 3 a~ci'j ~1,t 2,:3,4, and A has ( 3 ) (4 )= l2
'elements. Since aiJ :::i+j •the element in row iand column] is. ~btained by' addihltth~"'~~.s~rand}ltence'al'i'~':r+" t=1,
. !..- j . ....
al2= 1+2 = 3; al3= 1+ 3=4 and son'on, Thus-.i(.
. , .',[ fQ'J!.. :.~ :
, 6o
.. a", r.'}:h"~ltibi. L:' .... I
Ii,II:!:t-i'!
r
'J .. !J OJ f{:'; ",t " , :.. : ,alJ ,au all1
all- ~~ a211" ~ ,A= I l IJ
"I '. ;~ '- " ,iI
am) aOl2 8_0
"'[
,I
Wherei= 1.2,3.;, ,..• m and J,= I, 2~.3, '" , n..• , .. : '·:,·,';.~:L,j;(~;.~.
Df!fmiQ!l.l1: An ~ X ~ matrix A _is ,3 recta~~l~ ru:r~Y,«,n;t,,d, ~ f .. ",' (9r,!c9,mpJ~)()",~~~~~:?,:~~, in mh9m..A)n~ro~
vertical column ,,~ : '-'" 'il":':" ,
For the element aI2'reac(-(:~',().I,l,~:-;-:,two).the,~~,~Jlb.:,J ,the row and, the 2n<ts~b., i, the column in which the
~. ,
appears. Generalizing, we say that the symbol aijelement, in, the- illl r~w -and -j!b, Column,' w_e now
definition of a matrix. .
Example: Construct a three - elements.column matrix' such that a21= 6 andaij = 0 otherwise.
. ',,' '~'''''''_'.f''.O\''''''':'_.***
.1III
_________________________217---.--------------~-----
,[ all .•A= '0. ,0 o
'8) If aij = 0 for i :> j" the square ,matrix A is called upper ttiaQgular. ," . ,.. :': " .' .. " . '.~: ' . " -
, -matrix.~-
, The matrix A is called Null matrix m:zero. mat~ and is denoted QYO. I
____ -==-=_ _::::::;;;:;;:;~_::::' :Z""ill:t,='''~'''l..'j,.;.;:,.;.;.'_-=__ .,'--:IIt
[0 0 ]A=, '0' ,';, 0. , . '\, .
'. ': if: .~ij=O for all values of i and j
~ ]o1
o
I' .., ,
=0; i~j
A= [~ 0 ] • A~ [ ~
The matrix A is .called unit matrix; or Ideut'itY OlatJ;ix and is denoted
b~:' .
i=ijff
[ -2 0]A= .o -2
o3
o
. The matrix A iSI:aited a~)
[
.3A= . 0
o
, I.e
A= [; ~ ] 'f"0, ,~, "
A= '0 7 0
0 0 5'. ,.', , ,::'
. . ..~ ".,aij :==,c;
.'i:=j
, ,
=0 ; ht;j
[~
0 ] [ ;c 0 nA=c , A=" .o C
'0 0
116
5) If
i& A= [ a~,o ] C" 0 0
a22 and A= ~ a22 00 8..13~--- '- -
The matrix A is called a.diagonal matrix.
~ [ 1 2 3
Jand[1 4 ]-[ 0 5 3 5
'7 4 2 .." .
The elements 1, 0; 2 and 1, 5 are main diagonals in tworeceptively.
,l;,
4) If aij -:i'n, i =j
= 0, i~j
3) If m = n, we say that A is a~ order n. and .;
elem~nts all >an> _,.. _. > ~ form the mam diagonal orA. "
---------------------- 219-'-' ------~----~-------
called scalar multiplication. and k A is a scalar multiple of A
'_' .,,§caf~rmui~i~iication12eflnition: If ,A is an m x n matrix and kis a real number (also called a.
& " " • '. .~:.,!~." : "-scalar ), then , by k A , we denote Utem x n matrix obtained~ .' .by multiplying each element .in..A:by ;k:. , This operation is
;' .
'.' . •. j: ~J.Where A, B, C, and 0 have the same size,'
• .•.c :3) A+'O=O+A=A
(~omm~ve)
( associative ) . '.
( ideittlty property )
I) A+B=B+A , ";'.... .2) A+(B+C)~(A+B)+C=A+B+C
, and
, [5and .B= ,.1
* ** * *A=[; 0 -~J-1
A+B= [~ -3 ~ll1
[-2 3 -~JA-B= 1. -3
, 'Prol!ertiesof matri~addition
12elillitio1J,: If A :;:::[ !lid and B,;:::::[bij 1are both. m x n.matrices-, then thesum A+B = [ aij-: bij] , and the subtraction
A - B = [aij-bij] = A + ( - B ) .for all i= 1 , 2 •3 •... , m , arid
j = 1,2.3,; .:. ;Ii:' .
',:.,' ':,.\'
'~'.
=:>4x.='6 . px=3f.2. '. ~ ,=>-y_ =.4 ~ y = ±2.
X-2=4-3 x
y-I ;'3
. .... . '. 1,..: : .- :' .. ' . . .~.. ~.. ~. .... ". By equating cOtrespondillg'eleii{ei1~.,~m\Jst have'
."!"~ ........' c,,· ..... ·~I ....... .;.." •
, . .' t..:: .... . ". ~-' .Fjndy~~~,~f.~cm~.:r1 ...
.,1.." ). ,,:;~.,- .,.' .' ':
1 ] = F4.-3Xi-I . . L 0. .
v , \: •• :" ,~~. • I t' . • I
A matrix equation can define a systerit of equations. For example,
have the' same size and Bij == by for each i and j (that
" 9cm~RO~~d~g,~Jet?~~~~~U&J;::.;_;.,
/}ef1i!llion: Matrices A = [~] ~ B = ['bg ] are equal ifaqd'. . .
Equality of Matrices .'
: •• ~. r-
au:0:
' ... ~;i;~,.,~,~,... , .'matrix.
...... :", .
~N~~'-~-------------------221-----------------------
=> 8;;,= -an => 2ail =0,\(' , .. , ::"',, ';i''l;',1
-c Hence ..in the skew-symmetric matrix the elements of malH 'tlla1ffinal~.e,equar zero .:
'~.
~ut' iF[A" == ;\ the ~a~x A is'·~ii~,~~~~~~'andaij = -aji for all iand j. Specially' i=j
b.f
h"01' •.. ,,', .. ,'
i'~' .J.'.;. c.: '
, [ aA;" :
______________________ 220------------------------
..Solution: Matrix A is 3x2 so AT is 2 x3. then
AT=/>!=[~ 2 ~J5
Example: if
Ddinl(jOll: 'The transpose of an m x n matrix A, denoted by AT or /.' , is he .n x m matrix whose jill row is the jib column of A ,
'Transpose of matrix
If A is a matri~, the matrix formed from A: by..:.:in~t:::or~c.:.:.l'ia=a.:.:.:Ja.;.::~:.::..::_3;I',';;t~;I~~s is calledthe transposeof'.":
'f..: ~a~ A = [ aij 1is calledt:~m_~if~ ~ ~at:is. ~, issymmetric if" it is a squarema1i1Xtor·whichaij.:;.~i.a.nai.iheelements .of 'A are symmetric wfth respect to. .the-rnaindiagonal of A.
·i~·,· ::;~:·.~::;t.'..f.: I~!,'., . "''-''''';'- '1;,: t.,;'
1) (A')'=A2) (A±B)I=A'±B~:: '.3) (k A)' =~A' ':
.. 4)",(AB)' .=B' A': ~:,., ; ....
------_..:....---------_ ...._--_ .. _._-
1) K(A+B)=kA+kB
2) (k, + k2 ) A = k, A + k2 A
3) k, (k2 A ) = (k, kl ) A4) OA=O'5) kO=OWhere A, Band 0 are thesame size, and k, k, and h arc any scalars.
.Properties of scalar multiplication
o3
o-I
6 ] ..
-12•• = [:~-3 [ 2
For elCample: .
____~---------------- 223------~--------------------'---_..:._:.:.:.:_:_:_.:___222--~_~ _
as sum of symmetric and skew - symmetric matrices.
:. . -. .·.'5;;
58
. - 10
Example: Put** * "Hennctian and skew-hermetian matrices
, • .' ." • ~;'.": t,,' .:' ', ••••' "'~'~., ~
,IfA = (a'j) matrix of elements real or complex, then' ma\:fix:,A ,=( 3ij).'1( ,.' .,.' .
',-is called conjugate matiix for.A - ': ,t' , .' " •• ,,' •
* *
..l~y, 8] [ 0 "1 . -I J8 Y, ..-+: ,_ 9/2
21' '.,0 "/2~ \':'
. 8 Y:t 15 . ", l' _ 21/2 ' ,._:,o. '. ':.
,:' : ' .. ....
Then required representation is:
9
o
',' Since
R!= ~ ( A+A' ) , =0 Y, ( A' + (A' J ')It'= y, ( A'. ~ ,~ ) = !.oS ( A +A' ) ;:;R
Then R 'symmetric, And since
S I .~ Yl (!A - A')' = !.oS ( A' - CA')I).....1/ (I J' '-; 12 A -A)=-'~(A-A/)
S' =-S" '
!~e~ S is sk~w - symmetric matrix, and.. "R +"'S:;;!~tA_-;f.~i)+ ~{(A - A')::: A.
S::=~(A-A/)=Y, l' -:2' .-21..
If A is rea..Ls.guaie matrix, them A can b .' ...."'. . . e 'A I, Of
wntten uniquely as sum of symmet . t . .___ lnc ma nx '··;··:·'·:'R,;~l1ds.k.~A;'~1 t . S -_._ .._:.'_. •. • _ ..c ma nx . . ....'.--- . ,~ .
',f "/ : • . .6.•••,
Proot Since A.= R + S, then', consider R = Yl(A + A') d S _ YlC'" . (i~~:.~. an - 1 A - A ) .r;:
- f' .o .:
8
Since '0- ,.
N=l2 -4.
9 ]5 '8 . -~O ,and
7 11 15
. [ 41 I:] UR=~(.A+A/):=:~ 1 16 =
16 ,. 30
h
A= l-':-g
225,-------.,----~--~~--------
~-i-26 .J'. '.. ,
.~
" ~., and
• and
2i
2i-4
10 i
. (4·f)h .J-I3
.' .,
2i+2
4
-2
.,".,~
-1
.i+ 1
4
2 - 2i
4+i
Where B := ~ ( A +A!) and C => ~ ( A - "A! )IWe leave tbe proof to. the student I
.' ,
can' be written uniquely as sum of bel'met ian
matrix Band skew-bennetiao matrix C.- '. . _---
If A is re!t1or compl~ square matrix, then A
. ,.zeros or purely complex.
and the clements of the main diagonal must be real.
2) If the elements of skew-Hermetion real then it is skew- ':~,
symmetric, aud the elements of the main diagonal must be <,
1) If tbe elements of hermetiaa ml\tJ:lx real, then it is symmetric, .',... ,
!J,el.lnitLq,It: A square matrix A::::( aU) is called herrnetiau if (p.}) ;:; Ai.e aji=au for all i, j = 1 > 2 , .. : , n.
But A is called skew - bermetlan if (AI) = - Ai.e 1:Ijl ;: - aij for all' i, j '-'1 , 2 , ... , n
4Where k scalar rea! or complex, A and B are the same size.2i - i.2+i
A'=[
A/=[ -~;-\" 4.
as sum of hermeuan and skew-hermetian matrices.
2i - 1.
Si+2
-1 -I- i
l).(A)=A .
2) A±B=A±B
.3) (kA)=kA4) (A) I= (A')
Propertiesof conjugates
r' 2 - i
A"'l· 3
2 i ]- i-I
.'
('- ---, __ -.,- 2,29 -----------
",
. ,',' {: ~, . ,',
b );AB and B.4 • c.' ) 2D ~ 3F ;r , : 1.
d ) CB+D; e )AB +DF ,
.;,a) A ('BD) and (AB) D, 'b) A (C+ E}~rid'XP+'AE~;:'""Ci1"EF-h2Aa) (C+E)I andC 1+E I" b )(AB)'and B'A~ 'ciB'9+ A, d) A+A'
put the foll~win~ m~trice~ ~ sum :o~Sy~~¢tric' and skew-,-.~ .~. . ',.;~f' f " .'.
symmetric matrices' ,:
";,,:...
(associative) , >'(distributive)
.. ,' ...... ""----~~--~--------·2~----------------~
- If AB=AC ;6B=CExam1!Ie:'
......
[ -2 3 J[ ~ 3 ] [ 10 ~J=2 -3 0 -10 6
T -2 -: ].[ -5 3 ] [' I~' -6 ]=2 0 0 -io 6
: ] ~,o
Example:
. "
- If AB = 0 ( where 0 is the zero matrix) then A :;I: 0 and B "* 0
Then 'AB*BA
U T,beproduct of matrices aloe not commutative.
But if AB' = }3A, then the two matrices A and B
commutative for each other.
4 J[ ~[
,2BA= 1
-I ][ ~2[
2AB;" 3
Solution:
,J [ -5 -2 A( BC )='( AB,) C'=ABCee
4 -5 7 A ( B+CY= A B' +AC
] [(A+B)C=AC+BC
-1 -1 3=
I 14 3 . ;'.
---------------------- 231----------------------
•
, ',( * * * * *
compute AB , and Anwhere n positive integer ..
,II
~
j ik~)"aiil:i "IDi 'J
~
j
,,
I
~
' ,
,
t
QrJ~:
and find A2- B2 , ( A - B )( A +B ) if
Discuss whyI) ( A +B i*A2+ 2AB +B2
II) ,( A - B )( A +B )*A2 - B2
~I I'IW1it
~
M\·u
~]I"
2 3
2 4 6
~
2 ,3
I,
----~~-------------- 230------------------~~·~~
"g') Show that At= [I~"~21] , '. h i "..' .:~~ Vi- -1 co~m\ltatlve Wit it') trans~ose.
!lIe commutative for each other,
7) Show that
[ 1
2' 3 JB~[-: -1-6 ]
A= '3 2 ' 0' 2 '9
-1 -1 -I . -1 -J -4
[3 -2i 10 ] [Fi 2 -2i 8 la) . b-' "-- '
t + 2i " 13 2+2i 1+ i,..,, 6i -1 + i 0
b 2 2 lati,ne,:
'6) Provethat the' A= t 2 the
A2-4A-SI =0 2, , .
5) put the following matrices as sum of hermetian and
matrices
5, 1 [10" I, b) "-7 9 3
·6' 7, '
. 10)'Let'A 'and B be symmetric matrices.
a) Show that A+B is symmetric.b)Show that AB is symmetric ifand only if AB:= BA.
-----------------------nl------------------~~
. P~operti'es'or deterritin~n~' .'.
I) The determinants of a.matrix and its' transpose are equal
i.e I A I ;'1ATl'
2) If matrix B results from matrix A. by interchanging two rows (col~)of A, Then I B I= - I A I. .'f.· •
.....Expansion of IA I about theIirst columu.
I.AI =13 •
3
,1
'I
- The determinant IMijI is called the minor of aij .:-The cofactor Aij of aij is defined as Aij = (-1) i+j IMijI.
'.'
=6
. " ~.:.
3 2 3 2-2 +32 1 2 1
,:.~~::~,
Definition: ~t A [a;j] be an n x n matrix let Mij be the ( n - 1 ) x ( n - I}::.
submatrix of A obtained by deleting the i.Lh roe andj.Lh
of A.
= 1(2 - 3' )'- 2 ( 4 - 9 )+3 ( 2 - 3 )=:) + '1.0 - 3=6expansion of IA I about the first roe also
+ a13 a21 an J al) an all #
s.s 2 3' .. J 2 3 2. 1.xl -2' +32 3 2· 3 2· 3 f3 2
IAI all all a21 all . a21. an= all -.aI2· +al3an all a31 . all a31 all
= all an 333- all a12 all - al2 a21 alJ + al2 aJI 'a3,
is a 3 x 3 matrix. thenthe
A= [. :~: ::
all . all
Example: If'e.
t ..
6 I .7 = 4 (7) - 6 ( 8) =28 - 48 = -20I~al~J is a 2 x 2 matrix then the IA I= all a22 - al2a22 .' .
Example: rrA =[allall
ami am2 _..•_ _ lin",
D= IA 1=
( OETERMlNANfS )
l2efJnj(ion: Let A = (lij ] be an n x n matrix we define the determinant of·\····
A :Mitten IA I
. :j.In this section we define the notion of determinant and study some of:
its properties.
mirli,
rr.r11:11~~~
L,U~:l¥~:ti .i~ltW~. :
~~:r~"'.1'1
iI
~
r.~,,
I~:,, l!VI1
Il~~--------------------235----------~----------
Xl 1)1
Xl ihI = I,' I
1 'I •Xn
bm '
"...,.,all Xl + al2 X2 + .., + alo Xn ,= blall 'XI +- all X2 + ..' + a2n Xn == b2~ L ~ ~'::"__ -- - -~.: - --~'-.:.-'- -,~.ami XI + am2 Xl + ... + !lIN>; ~\~\ ~I[',: :
The system of given equations is equivalent toa matrix equation
, .Let there be m non - homogeneous linear equations -invariables
this section we shall solve the system of linear equations,
LINEAR EQ
~J[ 1' 2J [2 -lJ [4Also AB == 3 4 1, 2 := 10
And IABl =20-30 == -10= IAIIBI:
'~ , 2~4--------~----------
All:;: (-1) 2+3\ ~ ,"': il'l~3. 'Now a31 All + a32 A22;~;~~3:A~~::::::(4)( 19)+ 5 (-'14)+ (-2) (3 )~O
, 3 \ =-14, ,:-.2
''3' '\ ., 19,A~ = (_1)2+1 11~2 ", ,4
2
35
" [- rA= ;'2, ,
;4
For example: Let
6) If a matrix A = [aij] is ,up~er (lowe~) trian&Ula;. then IA I=alla22 .. , ann'
7) [fA = [ aij ) is an n x n matrix, then
ail AId +~ Aiel+ , + ainAlat= 0; for i:;o.'; k
a'lj AIle. + alj Alle. + ' + ani Ank:;= 0; for j:;l!!<.
,1 2 3 5 ,0 9s.s2 -1 3 = 2 -1 3 =4,
1 0 1 ,0 1
obtained by adding twice the second row to its first row.
5) If B = ( bij ] is obtained from A = [ ~ij ] byadding to each element of .:;.the ~ row (column) of A. a constantctirnes the corresponding element ,<1',,,:;,,I!!!!!!!!l~'
of its slhrow (column) r :;0.'; s , then 181 = ,IA I..
The determinant of a prodo:ctl of two matrices is the product of their ,
, ,determinan~ that is, IAB I == IA II B I
; ] andB~ D ' -~JThen IA I = -2 & I ~ 1 = 5
3) If two rows (columns) of A are equal, then IA 1=0.4) ,If a row (column) of A consists of zeros, then "AI ::;;0.
~------------------- 237---------------------
-1 1 = 6 (zero) -1 (-3) + 1 (3) = 6
r ~l ....... ''.
I6 ..2 =.t {~3)~.:.6(-3)'+ 1 (-3) = 12
: . .. . .1 -1 '1
6
-1 2 .= l:(-3) -1 (-3)+ 6 (3) = 18
1
6
-----------------------136 ~ __~ ~
For a given set of n nQn-hOinogeneouslinear equations in n(unknows) each x.; i:;: 1;,2, .... , n is a quotient., thedenominat~rof
is the determinant IA I·of the .coefficient matrix A and the numerator e
which is a determinant obtained from IA I by substituting the
D.2 = D.y.=~( 2
1and
1t.3 == ~.=
2
• I
, .The equation '(1) can be solved' by two different methods:, .
The Ilrstmethod: CRAMER'S RULE
IFirst Case IThe number of equations = the number of'unknows
1& m = n.
-I
2 -1There are two different cases to. solve ~e ocn-homogeneoos.InIl)
= t (O)q (-3)+ 1(3)=6* 0----+(1)AX=B
We have the matrix equation
the matrix A is non-singular matrix i.e IA I ~0 .
e: solve by CRAMER'S rule the system of equations-?==-"'.....
x +Y + z= 6. x- Y + z> 2 •. 2x +Y- z = 1
(" Here A is called coefficient matrix of order
matrix of unknows of otder ( n xl) and B is the matrix of aterms of order ( m x -l »
•• " ' •• » ,
'. ,
for the ill! column of IA I'
'tr. I.Denoting by.'
all al2 --- AIR XI
a21 a22 A2n ' .' Xl--.........._-A= x~ andB',s>
<1m I ~!"2_.__ ..- am" Xo
. i-,,
".. :;Z.J "~~~j I~'l'tv,,!
~I!,h).';",I I~ .~;1.l~..~j~;":, .~,)I
~r
1 .
~I
~II
~'! It~
~
~
,,.:.'
i
~
,,
~
,i\Jo
iI
~,fi
Now we have A"I = I~ I adj AThe first .method: I.D.verseof a matrix in terms of its alj '.( adjoint of a matrix.) '. '.
. '.,' ,, '"',
. "
!2.W(1it!.ea:1) A square matrix A == [ a;j ] is known as singular matrix if IA I= Now~tg"~:t~::~: ~,;l,al\'YA"i:n,'g:: :i~ ..e~ements. as th~ ~o,f~C~~~:,,:~f~~::
and has no inverse (reciprocal). ' ';:\j?~:IKC(m~)Dorld
2) In case IA 1,* 0, the matrix A is kno~ as nonsingular matrix, ['6
-4-4-1 -2~2']'r,Signs"'·l.l·.;::463:,'';"...,' ':~'.~".: ? 'J trcenspo~has inverse. adjA: .,-:4. _, . 2" r .
3). ~f~~re exists a ~quare matrix B of the same order as that of A .' 3 2"
thatl A B = BA =- II Then B is s-aidto be~e inverse-of Adenoted by A01.ThuS by definition, !!;A-I =.KIA =. J I (A is.said ., Ioi[. 6 4 3 ]be invertible ). . 4 3 4
3 .-.~ z .'To calculate the inverse matrix of A we use two different mel±!ocls;;)j~.~
The second method: The equation A X= BwhenA of order (n x nof order ( n x 1 ),' 'can be solved by using'
inverse matrix ofmatrix A.
.. . '[-2 -2 7J',::U-:P.:.fl~,~~llli:-Find the inverse of the matrix A = . 4 3 -12 ,
-1 0 2 '
,.A,.... '-···IAI=-2(6)+2(8-12')-:r7(3)_=.12 - 8 +.21 = 1 * 0'
expanding about the first row.
.@~~.=. '1 A I "'" ~, then CRAMER'S method fails. Iand this is a Unique solution.
A A-I = A-IA = I and (I), we haveLl) .6. 'I. 18z = --=-=--=3/j, .6. 6
orLl2 6. 12
Y = --x-=r= -6-=2 ---~(i)
then A ( adj A ):::; IA I I = ( adj A) A
A ( I!I adj A ) = I=(I~ I adj A ) A
Lli .6.'/( . 6x=.---;;- :::;T = -6- = t. Then
If A be a square matrix of any order' say n a~d I the unit matrix of the1:·;1!I'I~I" -"t;.,'"It:',.l.i'
~]:]'-1 .
r, '-l [~
0 9 5 -2 °J [1 ·0 0 -40 161"1 -.9fJ . '
0 13 -5 '1 -3 -2 1 O~ 0~ O· CD 5.
. r2+ 3rJ ' .·0 1. 5 '-20 -2 -1' ')0
, .
Thusii'
[ 4016
9 ]A~'=. ,13 -5 ',. ;:3··' , '.. .'
. ',5 -2 -1 . v ,
2'41
.element, and eliminate the elements: Above and under it
:' - 2r> [~
2 3 ,0 0] [1 0 9 5 -2fl - 2r2
CD -3 -2, 1 O~, 0 1.. -3 -2 1f3+.2r2
fJ -, rl 0 -2 5 -1 0 I .' 0 0 -1 -5 2
. "'i"",
The computation can be carried out as follows, we choose any. . . . " ......". \',:', :.::. ";.:':,~'~;:'"
element ( it is better to be equal 1) from any row, we will call it 'piyot
o
33
8
2
5
o
1
oo
oo ]o '1[
CD'(AI1) :
25
o
matrix
------------~-------i~o----------~~----~~:
: .
':.,to'
,', :Elementary row operations as:
i) Interchange two rows.
ii) Multiply any row by a sc.ilar C, c;'·O.:
iii) Ad~ ( subtract) two rows.
, ;: ·lsl If:p·':':,Jl'~·I,(,'.t ,',,':!. ..', •
. Tile secondmetllod: 'E~eme~taryrow ir:.dnSConnntioos (operati~Ds) ," .... r; .j', ~:," d· ," .
, , We. write the matrix ( A IL), by use ele~ent&Y'row ".. :,' ,t:"i .. :" ;':"', .. ' • .: ( ...... (/ "'I'. " .':
operations we o~tai.n:,mnthe form (i I a-Y .
;i,e;.(A In/.~.o~ (I" B)the.~,~.~A·i't
...' . .;. ..~,'
\ ,- " :. :
.:[, 2, -5]. ''-4' 1:If '," ,
-1_ -=L.A ....,,14. "
WA:['~, ; 5,
Example: Find the inverse of the rt:\.«tr-j).{ A;"" :'[~ ~ ]
$.f)/u1ion: IA I = 6 - 20 ::= - 14 :;e (i
1~~--------------~~24~------------------~-
19]';;':1 (-27 -1) -2 (18 - 3)+'10 (2+9)1 '.
9 ::= - 28 - 30 + 110 = ,52
2
-3
9
10
2-3
3 ].=.10 (6 - 1) -2 (-2 - 9)~"3({~'2~),;1 , ,. ...... . . ' ....-2 "'" SO + 22 + 8.4 :f:,' 156 .
,
~ ]=1.(.2.9,) -10(-4·3)+3 (is. 3)
-2 = -11 + 70 + 45 =:' 104
'1';==::::' Solve the system of equations
x+ 2y + 3z = 10" 2x - 3y +Z = 1, 3x+y.-1z,~9;;:..;,JJ~~:' by using CRAMER'S method.
A=, [231 ~ ~ .]=1(6.1).2(-4.3)+3(2,~+9)1 -2 = 5+ L4 + 33 = 52 # 0
......
, ... "~I\ i .:',,,;~.• +
~IX==A-' B=> X=A-I B
Which is a unique solution of A X = B .
In order tosolve the system Of equations'
A X = B
by using the inverse of matrix, if we pre-multiply either side of= B byA"I, we have A-I A X= A-Of B
----------- 241 _
.#
~J~ ]0']
3/11
-5~1.lJ3/11
5/3 1/370'
~5/1 fJ.= -_1 . [ 73/ll i II -2
A-' = [7/11. -2/11
Thus
o 7111·
1 -2/11
5/3 1/3
CD -2/11
5/3 113
1113 -2/3~.2rJ [ ..~
.rl * 3/11.[~ 0
r~.513r}[ ~
OJ Jii/3 [ ..11 ~ 2o
Now we have A-I = IAll adiA = _1_ [7J ., t -2
dIJ.u.tlter im.u(i~n:
;AI[)~r~ .~
-2J [7transpose :3 I" ./ ~
2 ]'SjgnS [ 73 ~-5
~Lutiol1: IA I :::21 - 10==11 ~ 0
Example: Find the inverse of the matrix
I.
I!
~iI!
."I.;.
245---------------------
Pre-multiplying either side by A-I, we have .A'Iy=j!:1 A X= I X
X=A-I'y
Y -= A X
Which is y' s interms of xs: the required is the inverse linear
i-e x' s in terms of X' s
if.!i;I~:m!2:tl:·The given sy!rem can be written as
that we have linear transformations as
y, = -2x, - 2X2 + 7xJ
Y2 =4x, + 3Xl - 12xJ.:Yl ;:: -XI + 2X3
IIJ' 10J [156J [~']' ,5 1 ....._1_, " 104 '- . 2
l -52 - #..7 9' 52 .' 1
the solutions of the equations are I
" [5 7
1:], , ., 1 -11A.1- - 7- 52, '
II 5 -7
244
Now we.have A-';::: 11"1 a9j A .
. [5 -:-7, l'lJ"~'[?adjA: -7 -11 -5 ' 7, 11 -5" -7, ' 11
, DOW we determine KJ :J'"
"
Pre-multiply either side by A-I, we have
another solution: By using the inverse of matrix.
. 63 52z= - =-=\A. 52
. Thus '
~ ... 1').~
------------------~--,247----------------------
- The rank of a non-singular square matrix of order n is n :
- The rank of singular s'qu~re.matrix of order n is less than n .
- The rank of a matrix A is denoted by rank (A) .
,i~Thus the rank of a null-matrix is zero.
The rank of the matrix A is the greatest order of the non-zero
minor of that matrix,, '
Or
Conclusively, the rank r of a matrix is the largest integer for
which the statement II not all minors of order r are zero II isvalid ' ',\, ,
!2dJ!1lJlg,n Uj :We say that the rank of the matrix.A = [ alj] is r if:
a) All the minors of order ( r + '1) vanish.
b) There is at least one minor of order r, which does not vanish.
Rank of a matrix:
lsecondcaserl-------------------------------
If the number of equations in IA:X = B Idon't equal 'the number of
unknown's, i-e m :;t: n and if rn = n, IA I = 0 the system can be solved by
using the rank of a matrix ,
Which is the equation we want.
, ' or equivalent
y = 2x + 1 , yields
- 2 x; ~'3Y' ~ 2 (x' - y' ) + 1
y' = _i_ x' + _1_ thus (x' y') satisfi~s y = .i.x +_13 5" '35
substituting in
So x = x' _ y'y =_2 x' + 3 y' '
D:1= [: : J[; ]~ [~] ~[~ :r[;:]~ [ :] ~ [-~ ~ J[;: ]
$.ol!1ti(!J1: Since
Example: By using the inverse linear transformations for
fi' . orth, .
y' == 2x + 1 > ind the image e line y == 2 X+ 1 ',:. where (x,
be a point of the line and (x' , y' ) be its image.
Which is the inverse linear transformations we want
Thus .x, = 6Yl + 4Y2 + 3YlX2 = 4y, + 3Y2 + 4YJ
x) = 3YI + 2Y2 + 2Yl
'4
3
2
from example in page (27) , then
---~',:",,'''''::';''';...:''~"":"':';":"':''':'''--.;__-248;-"-·..-'-'-"--- ~~:."
. ~., .','~:',~:.' .
,Thus, :,1 J,')," "
= 4 _' 6 =: -2, * 0,:,,":">,1: ('.,~ , ....
, SolUtion:' a) Since: '
P',.,'1
p:L,,,
. Find the rank of the matrix
[ ~
3 4 3
9 12 9
-3 -4 -3
Since
A_[T 3 43 ] [ I
3 4 n9 12 9 r2 - 3rl 0 0 0,-:--! ..-l -3 -4 -3 ?+ r" . 'ri~ 0 0
Thus, the rank of A is Qill!,
Find the rank of th6 matrix .,if 1 .", 1 2 3
r, 3 0 3
L -2 -3 -3
1 2 3
I~,
,~w'.1l,v,~,
:t
'. = . ~• J. '. \.,' , • ',',.. .: ~,; .... ,
• ,.-,.,".~".. u"... / a) Equivalent matrix:A matrix B is called the equivalent to a matrix A if B can be
obtained from A by a' sequcQ.,ce of elementary row
trallsfonnations (operations) and"denoted B~ A.
b) Equivalent matrices have tb~"samerank.ic) The rank of the transpose of a matr'ix is the same as that 0
the original matrix. i.e r ( A ) == r ( AT)
34'" 5
, 1
.. ,,:[:l" 3'J [t 2: a)-, "",', ',,:, " ·,b) ..,2' '4"'> .""2",4
operations .I:·~·I!I. ~.I,
"':;', !Example: Find 1h~rank of th~ fo'HowingmatriCes "
" ,.;r.~ ..:. :'_ , ,
determine this number by' a series of elementary row
. .- . _. " .....
249-----------------------
~.II~
..
~.
~
.,
,
lfilI~;',iI'~
,
~~'
~I
QiJI( -1, ,
II I
~'1-t i
!1 I~1!ill'j
Til:jl;I~~. t!liJ1 rI I
~~~~Zli~,II
I'iJJ ~I
: It
t!
-------:------- 2SO--------~
, }:
".:'1
=~~. Show that the equationsX + 2y,'; Z = 3'" 3 x:'-:Y::.~,!'2.~:";:r;'1x -' 2y" F3z';: '2' ; x,,'-':y ':-t':i:bt::;:;, 1.'1-
are consistent and solve Ihenr..,. ~",.'
, '
,,,,,,,,,,,,,0/"" The system of non - homogeneous ~qua~iofls'll( X. =:~:Iis;said to be
in-consistent, if rank (A) < ~k CA*) and has no S91ution.; , ,,i(. r , " l'
Let the system of non - homogeneous linear equations be equivalent
~e matrix equation,! A X=, S I)~~ere A is of order (m x n ) , X be of
( n x L) and B be' of order ( m x r ) and we define a matrix
* ;,.(A IBJ be of order ( m x. ( n + 1) )'whi~h iscalled augmented'matrix
The system of non - homogeneous equati~ns[A X = n" is said to be , ,
consistent, if rank (A):=::rank (A * ) :-arid have a solution,a) If rahk' (A) = rank (A*)= nurJbe~ oiuriknoV\~ variabl~s, then the
, .:.. . ~.", I
solution is unique solution. . ;" ., . I " !:\ :. , ., '. .
b)ff rank (A) = rank (A*) < nuiber of unknown variables, ,9,1en,the: ',.... . .
solution is an infinite number of solutions, , '
':l'. I,
t 0 0 3M-
f, + 3rJ .. 0 1 . 0 3/4~ 0 0 -1 -3/4r2 - r)
0 0" 0 0
Thus r (A) = 3
': ": , ' ., ' ,
Solution: We have
·CD I, 2 3
3 0 3,A-
-2 ' -3 -3
2 3
2 3f1-f,
r ../ 0 2 -2 0fl - fl
0 -3 -s :-6r. - r,
0 0 0 0
1 1 2 3
/'JIlj 0 CD -1 0
0 -3 -5 -6
0 0" 0 0
0 3 3
£1 - r,2 0 -1 0r ./f3 +3r2 0 0 -8 -6
0 0 0 0
1 0 3, 3
0 1· -1 0f3/8 @~ 0 0 -J!4
0 0 0 0
251 ----------
.. :.
;dJi ~ .------------ 252------------ ...' ,------------ 253------------
-11 ['-3 2 -I]r2 - f, .
3 r ../ 0 12 4 4 'r3 + r,
6 2 21 0. ,..,. ,
, '.
:...1 ] f,-2f,. [1 -9 0 n~__;;;__/ 0 3 1fl + ri .
0 00, .
2
64
r, +5f,'T412. -5 -I 0 r./ 0 -1 0 -4
____ ./ T2 + ST,
r .; Q --1 0 TJ + 4f4 0 0 -1 -4 T~Us'r tA) = 2~ r'(i\) ~ number ofun~o~ variables
\.V 0 6 -1 0 0 -I ;i, The systemhas an infinite solution, and the system equivalentto the
~hUs, I ran; (A,*):;: 3 = rank (A) I the equations are' consistent, but ;::.' '.system '
rank (A):;: rank (A*)=3=nu~berofunknownvariables, the solution is " => x, - 9X2 =-3 , let X2 :=: a
o
, ~~aLple: investigate the system of equations x
~[~fJ 12 .o 3
[
"CDA'- :
. xl·3x2+2x3=-l,
Solution:
3
Thus r ( A ) = 1 -:J; r ( A * ) = 3
The system inconsistent, and has no soluti?n.
2
Q)-3
-3
93
4
oo
5
oo l]4
8
12
5
10
15
~ample: Investigate the syste~ of equations XI + 5X2 + 4X3 = 1 ,
2x, + IOx2 + 8x")= 3, 3xI + 15x2 + 12xJ:::; 5.
unique solution, and equal x = -1 , y = 4, andz =:,:4
o-5 5o ooo
7 0 3 5 :'5 0 0 5fl + 2r2 f2 + 3()·
-5 -1 0,/ 3 -1 2 . ./r r
T3 - 2rz @ 0(2+ 2T)
-4 0 -1 0-4 0(4 - (2 (, - ()
2 0 0 -2-2 0 -I -2
2 -1 3
3 CD 2 1A*-
2 -2 3 2
1 : -1 -1 .
Denoting the coefficient matrix by A and augmented matrix by A*we have
equation
2 -1
UJ3.
3 -1 2 = 1 .
2 -2 3 2
1 -1 1 -1
Solutio,,: The given system of equations is equivalent to the matrix
II.,
: .. '255 .: ----------
. ":'!!
'.. ,
. '_".x + 2y -,3z = 4
c=Ll'"=>l~'
-tv+ 14z=-10
2'54 -__,;~__,;~--'-'-_._----
== 3z - 4z - _1Q_ + 4. 7=v z+ k .,..
, " "7:: t :
J"
Soitl.tjOJJ,:=> x =3z • 2y + 4
19 row:from,
I,:'i) unique solution and find it
: \
ii) -an infinite solution aitd ; find them
.iii) ,no'solutions,
from 2nd row:EL'lmRle: Find" the values' of K such that the system of equations,·
x + 2y' -3z=4 ..3x-y+.5z=2>4x+Y,+(~-14)z=k+,~,:'. ~. t~:~.,.has
, Thus from the 3~ row:
(~- 16) z =k - 4
t-1 JLx=--+,k+4 7
, '," , ii) An infinite solutions if r (A) = r (A*) < numberof unknown variables
i-e tel - 16 =0; and 'i - 4=0 => Ik;'41
, 10::::':>7y:::1'4z t.. '10: , ,=> Y = 2z + -7-
12 +i 17°, '1y= k+~ _
i) a unique solution: iff (A) = r (A *) = number -ofunknown variables
; j_ e k?- - 16:t=°, k - 4 ;c 0 ',,'
=> (k-4)(k+4)-:l:0,k:f::4
-3" '4 ]14 '-10
K-16 ,k-4. '"[ :~] : [:.a-3 J = [9aa].+ [' ~:'J
x, -3a+ I, ..:j a " 1
[
XI 'J ' [ 9j "["-3 'j' ·:: = a _~ + •~.. ..•
, Where a is parameter.
------------------~~·256----------------------'----~----------------- 257~----------------------
are consistent. then find all solutionswhen
At,'111~l
=kl.3x + 2y + 2z - St
, 2x + 5y + 5z - 1St4x - y -z + 8t
=8.
=0Example: Find the relation' between k. and kl such
equationst~I
~1~~~
0 ,'= '. :~:.. , '~', i- e: , 2k2 + 8
____ .;.._;__--.:.....;.---------------- ,,' ';' 11"
that the system of', ~ 4k2 + 16 - k, - 5k1" ;' =>''. ' 22, ' ~, v.
" '
, ; ~ ~:,I kl + k2 = 161~\ln.. ' .., ", ;;rvnich is the relation we want. ;'"
:;,' ... ;.'0 ,.';'T,'. 'I', ,
kl = 9, kl::< 7. ::~:}',..: :,
. :r.
~I
iI1
iI
II
Jji1
iii) 00 solution: if r (A)< r (A*)
i-e ~ - 16 = 0 and k - 4
~(k=-4r
Thus r (A) = 2, in order' for the system' to -be consistent, thenr(A*)=2 ",:
, >: ' I
, P, ,.
t'" ir:';(~
(2kl + .8)111 - (kl ,+ 5k~)/22 ](k, + 5k2) 122
kl - (8/22) (k, + 5!<1)
'[ ] [ 8 ] [ ] [ ]
x = -a + 17 -1 sn. ~ :: + "" =a ~ + I~n
Where a is parameter,
and the solutions are,
=>x = -a + _8_7
.....", . ,"
[ : 0 0rl III
r rl 12: 0 0
-( -1
[ a0 0 0
rl .:r],0rr3 - 8 ~ '~
0 L
-1 -1 0
and from the 1$t equation, we have8
x = 3z - 2y+ 4 = -z +7-
, 10y=2a+ -7-
o,0-1
oo-1
It: 2k2:+ a ' J'22 kl + 5~t
8 k2
Then from the 21l!1equation,we have
10y=2z+ -7-
z=alet.x+ 2y - 3z =4
-7y+14z=-~O,
Thus from 111and 2w!row, we have:
I
',I
:'2" 2 -5 8 ]'5 5 -18 kl
8) -1 8 k2
'~'E~-------------------------\:~:,~( .». r' uttan:
z 0 1- Z+l!-: ..l=
6 I- t- 6+ q -'Bt-, = z# + q+ 'B=
0 I 0 q = Ii.
0 0 I e = X
~4l 'q="\'~ = x
~umnd'6 = z+'1i. + Xp'L Z = 1,+ X
------II :slJo!Jenba .Jeali!1 snoaua60WOH·"
( I X U ) lQPJO],0 X ' ( u X w )
JaplO10 v :;'JotlM'suonanba J~u!l sn(Y.)u~fJowoqs~ "0 = X v I W~lSAS
Qlp uallJ '}q.rrew OJ~ s~g X'!Jlt!w alp~E JO lu~lS,.(S~ip lI! J1: ", .
s::llqe!JUAUMOlDJun JO J~q.w~u ?' (V) J11 (Z0= ax = =·tx= IX (U(>!lAIOS onbmn) suonnlos
JB!A!lJ asrxo ~1;)1ft umt} salqe!JUA, UMOlqUn.]o J::lq~,nu = ·(V) .I 11 (I. " ..
, ." .stiO!llm\>;))OUl~~SAS;:)ql Qlu3!lSdAUI :aljlmCI::!l
( QPI- ~J~[i .i p l-v/I\i
,.' . 0, L- I- I1- £ £ Z
, ,.-_.;_---;.--------- SSt --------....,_.---
"L-
£
o
0= AI-- x [
, 0::;: Z - AE + Xl
smp , ( suonrqos ;:)l!UlJU! ue ) suourqos l'BIlI.!Jl - uou !'~4 LU;)ll:;ASu~'lU. 'S~Jq'B!JCAUA\0ID(un JO J~qwnu >Z= (V) 1 smu.
c,A. __ =X<:;:;:" t:
uouenbo 1.1UlOldt:__ =z<:;:;:'171-
[£/r;-J [ -e (£/r;~)J '= [Z ]1 u= 'B = A
£IL- e (£11..,-). == x '
"" ,
'0 ::= zp + A.p ;.+ x£ •Q.:= z£ -I- AZ + X! '
6S:~
I6- p- O l'L+"[ 01
01 ~JS'- © 0 ~' 11' P -v11£ - IJ
£ Z I ' . ( Z" '.
0=zZ1 + A01 + Xl.
i.;
----- ·261 ----------------------------------
i) x - 3y == 1 , 2x +4y == 7ii).3x-i'4y·+2z=1. 6x'+5y-2z=-4,' 9x-y+4z=~:3
."9) show that tile following equations are consistent and solve them
a) 3x - y + z= 4,. 2x +Y- 2~=-1 , x - 2y+ 3z= 5 .'b) 7x+3y+4z=1, 9x+4y+3z==-1. 5x+y+2z:::·1.
8) By using,>• I 'I~,~••
a) Cramer's method
b) The inverse matrix
Solve the following system of equations. :
a) x' s interms ofy' sb) y' s intcrms of x' s .
'Then find:
7) If XI + Xl + Xl .=YI +2Yl + 3Yl ,2Xl + 3x) = 2;Y2 +3Y3 .
5Xl + 5xi +x) = yi + 2Yl +4YJ
., . . a) Yl = 3xI - 5X2 , )/2 = 4xI + Xl
b) Xl -= YI+ 2Y2+ 3Y3;
X2 -= 3YI + 4Y2+ 4y),
Xl =- 7Yl + lOY2+ 12y)
. ~6)Find the inverse linear transformations for the following linear
transformations
.\,..-0'....... ,......r-
e)I :d] 'dlr 2 3 ]2
2 4 5 6
3 6 7. 8 9
a) [ 1:} b) [~
2 3le) [~ :]-2 2 3
2 4
4; .IfKI == [; : ] find A ?
5) Find the rank-of the following matrices:
a) [ 12 :1 U
3 4 .7
J4. ' b). 2 4 5 8
1" 2 3
26Q--
3) Find an inverse ifpossible :
I25
2) By two different methods find an inverse matrix of the matrix
________ --l( EXERCISES )~---------
1) Bytow different methods findan inverse matrix of the matrix.
[~ ;]
Thus r (A) = 3 ~ number of unknown variables.
There exist only trivial solution (unique solution) i-e x = y =z =: 0 .
o-2
o
o-2
o
,
I~
l~r···1[.I ,Y
,
:f.11. :~
Im1tm~)
.. ~{;•~~
i"
Then find a) x' s interns ofy' s b) y' s interns of x' s__ -:- ---,,--_. :tGj ...L......... .. ,,' ,
:2XJ + 3X2 + X3 = 2Yl + 3Y2, +Y3,
~I - .2X2 +4xj = 5YI .- 2Yl +4y)
2xI +X2 + 6X3 = Y3
~) If,.'
.. '
". ,.
4) If XI = 19 Yl + 2Yl - 9YJ )
X2 = -4 Yl - Y2+ iYj
Xj =-2Yl +Y3
lhew find theinverse linear transformation
,~,
and hencededuceA-I.
!i] ,'1 satisfies .the ;~ation Al ~4A - 51:: 0,212~
1'.3) Prove that A = 2
, . 2
-1 ]2 . sh.ow that7) if A == [~3 ., ] ~B= [_;a) ( A B) -I =B -I"A -I
b)(AB)'=B'A'.
c) lAB 1,= IAIIBI
, .
,l \' t
, ..
[
1.
• b) . 1aj' [ 3 l ]
2 . - 4
. }. ", ~(.p. .
~r.,
". ;;
• ,': I.
'.,' :
" :~..'
~ ••• t' 1
A X~ [: '1,= i'; -1 .3 ]' • [7 -t 3]1--1 . d. .. 3 . " "
.. , .and hence find the"soiution of'the.system
13) By usirig the elementary row operationsfind the inverso~ of "... "':6':" ~, J. . '.
[ 1 -1 OJ'-1 2 1,2 2 3
-x+2y +kz ==0_" - .. .
,ley -5z=·O.
x+ ky - 2z=O,3x - 4y= kx.
a) 3x-2y=O,b) 2x':' y - 3z="0;
c) x - 2y =ky,
1) no 'solution, 2 ) a unique solution, and 3) infinite solutionsa) x + y = 3 , x + ( tC - 8 ) y = k ' .b)' x + y''+':Z''; 2 > • 2x +3y +2z=5. 2x 't 3y+( I?- I') z=:k+ 1
.' : : (y' :. 12)'.Find' all 'value" of K' for which the following system bas non - trivial ',: :
, . , ~ohitidbs, and find these solutions for all values of K.
11) find all' values' of K 'for which the following linear systems has . ,
.: ::' f:'" .~. ~I • ." r.· ...4'", ,', -~---,--____:___;_;~..:..:.:!..:.:.:":':"":"';_:___':---' • "I ••••• • ~.' :'
....r~· ~. !:t:. .~,..;:" , 'l~,f')···i ':. l~.'bytwo different methodsfind the inverse of:
.! ...
2x-y +4t= 8.x +t:::;;O.
e) x+y+3:i'+2t=7,,f) x+y+z+t==O,
3x -Y> z="3_
-x+2y+z=3 -
5x+Sy+z=0
9x. - y - 3z =0 .2y + 6;.f.?,8 ': ,',,'x+2y+z=0 -
10) investigate the following system of equ~~ns :
a) 2x +- 2y - 3z= -2 , x - 3y+ 2z::=2, ,
b) 2x+2y+3z=7> x-y=-l,
c) x+y+z=O, 2y+3z=O,
d) 2x'4-4y' ':7Z'=='0• 7x - y - 2z ==0 >
AX=,,-X
eigen value of A if there exists a nonzero vector X such that" ··:······h{'·' .
. :. ':_:~ '1
~lWinitiolt: Let- A be an (n x n) matrix. The real number ~:js ~lled an".
:: ~':.
Thenii!
Every square matrix satisfies' its own'characteristic equation,i-e If. I A-XI I =~,+ill·X+;l2Xl+ .. ·..... +anxll==0·:
. '. :., ....:2 " -''; 1· iC!<) ~.+al,A + a2 A + _....._+ an A ·=0;_
., .' '. . .' .
. 5 4) Cayley - hamilton theorem(.J
Characteristic equation
The equation I A - X I I == 0 is said to be the ch~racter.istic equation
ofA. . :.
Characteristic polynomial
The determinant IA - X I 11s known as characteristic polynomial of A• " •• #
)
2)
(\
~ 3). i
. Befo~e discussing-the problem of eigen values and .e.i~~p'vectors, we.will write some important definitions.i
: llGljnltlon :1) Characreristic matrix
If A is square matrix, .then the matrix A ~ X I is said to be the
characteristic matrix of A .
It is very useful in all applications ...involving vibra~ons. In, • I •
raerodynamics, . elasticity, nuclear physics, mechanics, chemical: . ", . " . . " ',. . . '. t • ~ . "/' ,;'engineering, biology, differential equations, and so - on -.. ',' , . : ." . . .,':' ~~~.."...
»,
:<: f:.:..·· -----------------...--------------------------
.: r (.ru~·"kLU~.~D:~.G~,'(r:~O~)
*
2x+z-w=-0
x+y-2z=O
~;':::........---------------- 265 -------------
.~
_________________________________264----~------~---------r.'
* ***
8) Investigate the following system of equations :
a) x+y+z+w=O, x+3y+2z+4w=O,
b) . 2x - Y- 3 z =0, -x + 2y + Z=0 ,
a) x+y+ 2z=5, 3x-2y+ z=3, 5x - 5y + (I?- l)z = k
b) 3x- y + kz=4, 2x +y- 2z·=-l, x~2y+.3z=5
c) x + y - z==2, x + 2y+z=3, x+y+(IC--5)z=k
7) find all values ofK forwhich the following linear systems has :".' ,
1) No solution, 2) a unique solution, and 3) infinite solutions -
6) Investigate the following system' of equations :a) 3x - y - z.=4, 5x-2)''': 2z=3, 15x-6y+ 5z=O
b) 16x+2y+i:=:3. 7x+4y;2x=-6
,..!.,----'--------- 2.67 ----------'----
Or
~2 .4 .-"-
This means that ( 1 .,.i..H4 - .A. ) +2 :=:: 0l' i'. .
A; -5A+6.'=O=(A-2)(i..-3)
, f,1.
,..=0
1
,determinant of its coefficient matrix is zero, thus if and only if
I;~
----lo-... (4).'
S,fJlu(Jon: ,We W,ish t~ find all real numbers A ~~
X = [ Xl X2f satisfying (1) • that is
[i-A . ""'1 " J"[',(I':] .. [. 0 J:"
:':'2 '·4'-')." ':':X2'.:. =. ~ "
The linear' system in (4) J~as.3 nontrivial solution if an(l only: if.the ..,~..
alln~v~rs, .
Example: Find th~eigen.va1ues·and eigen vectors of~~ ~ A =[_~.~]
"'..(':
Equation (3) has three values on, ,!::Q AI, A2 , A,3, To determine the eigen - vectors •.sub~tute in(2),about
AI we know XI & so - on.
---+(3)
If all - A .
Which follows that every characteristic root A of a.matrix A is a root
; . of its characteristic equation IA, - A I I=0 .
. ~..:.' :
"
------;,...~ (2)
Which.takes the form,... "--[A~AI]X.=O
f·".1' • ','
. i.i'.
!,~. ~.; .:~i.. '
a21 ..: an - A, a23 'X2 :.
", ... ' ·.a31, I'" all a33 e- A . ,XJ., ,
ooo
Xf'
=
Or
all XI + all Xl + al3 X3 """Ax);' ','
-.:a~i'~I' +' a~·xl. + a2i XJ ='),;X'l.
all XI + a32X2 + al) X, =AXl '
Or. hHhe form». ;,.
(311-A)XI + 3,lX2 + au~ =0,
a21XI + (an-X)xl + 32JX3=0,
al\.xl,,'+ ~.31.~1.+ :(a.33.-i..)~7:'~... ... ..
[:: ]= [~t ~ X3 f.' and "-any
Xl .
Where the unknown X is the.vector
real number', ,
Equation (Ijcan be written as:
(1)
the equation
AX='}.._X
associated with the elgen value A.•~ " .' ~ :, ~, :' ;' I .. _..T'or proper values or characteristic values, and latent values 1 .
. . I
For.' simpli~ity, we willexplain this 'sUbject on the square rnatri<:Cs·of:'· :the ,3ed 'degree, and all the ~foimati~DS are iigtitfor ail squarematrices of":
any degre:e?Suppbse'the sqtiate matrixA = ['39] oftbe3ed degree, andlet
: ~--:......,..._;_.;.;,.;;.._:_--_.::,_.;..:..,..--------------._;.~....~.~:-;__----:-'--7-:----....,--------------•"
• : .•.. '''''' ., f " ': ',; .'. • • ., . . i 'Every nonzero vector X ~tiSty~g (1) is~lOO an agen vector of A ,~ Note that X = 0 alway~ ~*fies equation (2) for all values of A .
l~.Since X ¢O. the m.a~.[ A :'.J.. I ] is singular , so that IA - A I r =0 ..
-,----------- 269 _:__-- ..-..'-""".-_-..,---:----
: ..#. '~..Thus .
Thus
From.tWo equations only, we find X,i- e . o x; + 2x; :..xj =0,
Xi. - Xl + X3 = 0..
-: '1' 2 .-1,. I 0XI : Xl : x~·=:= -1.. 1.; - 1
-1
-4
- ..The eigen values of A are then
. or (A. - 1 ) ( A - 2 ) (A - 3) =°..This means that )..3 - 6 1,.2+ .ll '}..- 6 ::::0
i-e from I-A. 2 - 1
1 -A 1·. =0.
4 -4 5-1,. .' ..... :,~
We find the values of '}..from the"characte~~tic equation :.... . . .; . '. '. '.~: ". '
.,;.:-1 ]1.X1j" :[0] ..I . 'X2 ::::::.. () ' :....,-.::...:,.~7..·--+-(4)
5 - /.; Xl . ": 0. . r-'> .[
1-1.. ..... 2.
1 '-A
4 -4
:. .:.'~50[u(if!l!: first we write
A= [
2 -I ]..I 0 14 -4 5
268
Example: Find the eigen values and eigen vectors .ofthe matrix. _" , .
x, I. from the 1$'equation ~2xI + Xl = 0 ::::) 2xi = X2. => .::::_. X2 2
Thus anon - trivial solution .is X, =U lis an eigen vector of A
associated with 1..2 = 3 . We define the matrix of eigen vectors as:
T = [ X, xl] ~ T=.[ ~ ~]
. i
[:~..Thus a non - trivial solution is X, =[: 1Similarly, for 1..2= 3 , we obtain
We find x, and X2 from one equation
1
2
-1~2[
. .Are the eigen values of A. To find an eigen vector of A associated
with A! .::::2 ; we substitute in (4) :
1.., =2Hence
_________ ~ 271-------~------------
\ '
~,( _ ';l1- .A) \ ~ - A. - 6 \ _2 2 - 6 3 \ 2, 1"- i,, _2 ,- A. - 1 - A., - 1 -2
, 2 ' " " ,~ ( _2 - ')..,) (A - ')..,- 12 ) - 2 ( -2 A - 6 ) - 3 ( - 4 + 1 - A ) =0
~ (-2 - A,)(A-'4 )('}..,+3 )+4 A+ 12+ 9+3'A,=O
j··e from -2 - A 2 . - 3 .v.:,2 1 - A -6 =0
"":1 -2 -A
We Dnd,the values of A from the chara<:teristic equation,
2__ -'---+ (4)
[
-2 -).. 2·
-I
A= [
-2 2-3 ]
2 -6
- I -2 '0
$..olu(i~: First we 'write
Example: Find the eigcn values and eigen vectors ofthe'trtatri:X,
-:----------- 2.7(1"
[
-1 ]'x, ~ I ,~d the maun<of,eil!l'n vectors .i'
t(,
Thus
From the fi:rstand the second equations, we obtain:
", '," ': ':;'12 '~'~ll;, 1'-2' -'l}' "~2 \ 21xl,x2,xJ~'_3 "\1":'-,1;~ 1': 1, '-3 =-1 1':4
[ ;~
, '
U i : ,2-;,
-3,4 ' '
-4
, at.A3 = 3 : Substitute i~ (4) to.find X, '• I••
Thus
Thus
If we solve the first equation'with the,seco~d equation ~e obtain:
XI : X2 :, Xl = 0 : 0 : 0
, We must solve for e~ami?~ethe first equation with the 3ed equation,in order to obtain about nonzero solution
at Al = 2: Substitute in (4) to find X~, t-~ -2 - 1 ]~ [ ~
2 -1 ] [::] U]T== 1 1
-2
4 4
=-4 3 'j
(.
} '_'
..~
"'j. '1". .• '1
'. ' .. J." a" ,,!
."i .
I" I
r-------~------~~--273--.-.--------------------Thus B is similar to A .
and B.= [~ ~] == [~. ~J~J,letT= [;
-1] [1 1][1~ . -2· 4 l.
1J [2then r'=2 -1. [ 1For example: If A = -2
I .
non - singular matrix T such that
B == T' A:T
1) Oiagonalization C?f sguare. matrices
l!eQnitjan: Similar matrices
A matrix B is said to be similar" to a matrix A ifth~re is a
Someapplications of eigen values and aigen vectc?J'SI
#. ~ .. ~]~. [g. -~1 -1 .1 0 ~J
and the matrix of eigen vectors is
From the first and the second equations we obtain
XI': Xl : X3 = - 24 : - 48 : 24 = - 1 : - 2 :
t13= 5: substitute in'(4):
------------~---------.272--------~--~--------
X2 = a and X3 = b, thenPutting
from oue equation only• ','., : j'
Since
at At = -3: Substitute in (4)
1:.£ r I
2 -3 ] [::] [~], . 2 4 -6 =
, - 1 -2 3
A3=5Al = - 3 , A2= -3 ,
. ,.'. The eigen values of A are then
: ~(-2-A)(A-4)(A+3)+7(A.+3)=O
::=> ( A.+3 ) [ ( A.- 4 ) ( - ).,- 2) + 7 J = 0
::=> ( A.+ 3 ) [ - A2+ 2 A + 15 ] =0
::::> (A. + 3)( A.l- 2), - 15);: 0
::::> ( A. +3 ) ( A. + 3 )(~ - 5) =0
-----------------------275----------~--~--------
~1= 0 =:> ( 3 .-A.) (j -A.) ( 5 - A.')o ' ,+2 [- 10+ 2A.]= 0 '
5-A. '
3 -A -2
-2 3-A.
o 6
Step 1. Find the eigen values of A from the characteristic equation ,
IA - A.I I = 0 i-e from the equation ''.(
-2 0
Jj " ;0
O· 5[
·3
A=. -:
Example: fi~d the matrix.T that diagonalizes
, ,
corresponding to Xi. i= 1, 2, .. ., n.. ',' ..
as its successive diagonal elements ,where An isith~ eigen value
- The procedure for diagOl)<iI,iiitiga square matrix A of order. , : '. .(nxn).
Step 1. Find the eigen values from the charac~ristic"dliliit1'Ott:':
,IA-A.I I =·0" i-e "A.I"~f'?''''~';'':';;.;:,~~'i:Stcg 2. Find n linearly independent eigen 'Vect?l'S.:qfA.•~!.':.'~h ... ,X,,'.,Step 3. Form the'matrix T having XI , X2, ••• , Xc as its column vectors.Step 4. Find TI.
Step S. The matrix 0 =TlAT wiH then bediagonal' with A.I', ))...2', •• ,•• ~ ~. ,
. ;-: ·.'.""L.!., ; I:., "T is said to dlagonalize A "
Definition.: A square matrix A is called diagooalizable if there is an
in~ertibtc matri~ Itkuch tfj~~flA T is diag6ital; the matrix
OJ, =TDAl
:-,/l
(I)
274
[
all XI+ all Xl
AT=' .
aux, + 812 X2
Using (I) we sec that
]'[A. ' d 'J
:: .. D= ~' x[X,
T==(VcVz)=
Xl
We form the following matrices:
he the corresponding cigenvectros, i.e.
. V,= c]Let A se ( aij ) be square matrix of order 2. Let A,', 1..2 he ifi-~!;
eigenvalues and
----;..---......;__:_--- ..- 27.7 ---~--------
[::] : [~. +0 b ] ,= a: [-ll]' + b ,[ ~] ,"
!oi' Xl , b 0 1 ,, ' .x,til.· ,x'=lnThus
We will find XI : Xl : X3 from one equation ~mly.
~ from -2xl - 2X2 +O. XJ= 0
( 3-A-2 0 ]['] [~lwe have-2 3-1.. o :K-l -
0 0 5 - A x,
[ -2-2
~. ][:] [n' ,
-2 -2 =0 0,
Step 2. at Al =5 : Substitute in the eq~,tions
A.3 = 1,
=>(3-}..)(3~A.)(5-A.)-4(5-A.)=O
~ (5 - A) [(3 -}..)(3 -}..) - 4 ] = 0
~ (5 - A,) ['A,l_ 6 A.+5) =0
:;:>(5~J...)(}"-1 )(}"-5)=O
..
" ',
,276", .,",II~lI:,rf
.... J '.."
. "': ( ",..,
.."
l',-
h.~ '" ..••I,' ; , I '~•", .~;
,I'" ·t,
~I
.-:
--------_.:.-_-'-,., "',:,-~'"-:-:....-......;_----,----::....,,.-'-Ii
IlHlli~!" '
Then the transformation TIl'A T reduces A tothe diagonal form
D :T\ AT'=D' ..'
Similarly. if,A =( aij ).is a square matrix whose eigenvalues arc "'I, , '
Thus.we,p~qve~:,tk,ab;,., ,AT.7. TD
Premultiplying both sides by TI we get' " ,
, .: , ,":(,,' ,.,::: ;;ryi 'A'T"~{'D, r
;'I~, ,
,~.:mrm,
, J!,T.i. itf1.l,'i ; 'Y,
=> A. ( 1 - A.f = 0 , so the eigen values of A are '~I = 0 , 1.,2= 1 and A)= 1 .
We now determine the eigen vectors associated with A.l,= 1..3=.1 ' from'It,. . .. . . • ~) .
'219,'<"" -'-'-----------,
::i " 0',1The characteristic equation ofA is 0' 1- ',., ,2 =0
0 0 I-A.
Example: Let
[• ','1"
0 0",
1-
J~" :
A= 0" 1
'20 0 1
Ifall tbe values of A. are real and not all distiDCt, then the square"
jnatrix mayor may not be dlligonalizable:
-------'-ll VERY'IM':'ORTANT REMARK ,1----_, .....
"
= 1/2 [-01 '~ ~ J: [~: ~ : J'J 0'050".
-'I, U: nn .~~J.', [, AI ',: 0 0 ]
_, .0' A2 0
, '. 0 0 A)
. ,.'
278 ------__:_-~-
[ -I0 l ]~[I 0 i]r.,=adj T: -01 0 - 1 1 0'
-2 o 0 2'. ".
transpose[ - 1 1 OJ [ -I n,/r o 0 2 => rl= l12 ,0 0
1 ,." 1 o ' 1" '
StepS.
[I 1 .. ~1:-2 :r 0 ]r'AT='h 0 0 3 0, 1 1 0 1 0
=8:8:0='1:1:0
Thus x,=[ i) andT-[i ~ J] - [r -~ ~]Step 4. IT I=-t (- 1) -I- 1 ( I ) == 2 ~ 0
at A3 = 1
=> [ 2 -2 0 ] [::] m-2 2 0 ::
0 0 4
We will find XI : X2 : Xl by soling two equations, from the 20d and " I
, 12 o I .. 1-2 01.1-2 ~I3ed equations, we have XI : x~ : X):= 0 4 . 0 4 ',0
----------------------~. 281------------------------
• "r ,
,:
Thus
.\',.
r -2 ~]A= -2 ]:
0 o·. '
. Solution:
. Step I. .AI=5, Al=5 f: A)= I.:
. [- '] -: x,=U]Step 2. : Xl= ~. &
[-~' 0 1StepJ. '1'= 0
t 0
[ - 1 qS~e~4. ~'='h 0 0
1 1 0:..,;
Step 5. Since A6::; T 06 yl ' ":" .,
s , ;,
------------------~---- 280----------~------------
. .' .~.: .:¬ xample:Find A6 if
(.: Q.. : QkA..Il
o'"
o --- 0
0-0
t. Ii.
'k is positive inreger, and'.' I;
~ . "~1 .l"· ,.'i{DK: o A':>. i.
Where
. And .1 (k times)A"=;(TlDyli'=1TDT' ..:TDY' ... r o r"Ak=T D"TI
For any square matrix A. we haveD=Y' AT ill: A=TDY'
Thus Al~,(T DTI) (1l D T·' ).7' T Ii Tl
2) Powers of a square matr.ix.' . .c " ,.; ~
0 0 0:
diagonnlized . But the matrix A= 0 1 0 cah be diagonnlized~, " f!f, 0:",' .
Must be determined from.only one equation..' I
The rank of matrix of coefficient must be equal one, in this case
x, : '" : x, = 0 : 2 :0 I!!J.I! X" = [ ; lwhiCh are notlinearly
" independent •U X, bust b~ not equal to Xl ; A can not be,
.-1 '
~.:~I,
'ft'jf:~III
:,.
J~
[
-1O·
o
ooo
~
,1_,,':~
I
I ,I~l~I ~.II
I
283 -~--------
A-[ ~2
3]' r 2, 3 ]" ,
8) 1 2 ,9) A~:~':O -1 2 '
- 1 , -2 ? .y '. '0:: '0 2 '...
10)A=[ ~2
~ ] .. , " -:.:.~I
.'. ,
'/,',
'f .r 2 2,; , ' ,.
[ 3 -~ ,:l 111) Find A7 if A= 0 2 , ,0. " .
0 '0 '()
In the following matrices dctcnninejf A~ diago~a1~'1~. ttso, fihd'a matrix T such that TI A T = D, " , r '. ,.... '0'
~·]'2) [,~, ...~~J:;f[} -;..:l4)[; : ::}5{ ~ -: {l~l::.~:]7) [ ~ ~ :]
-2, -2 1. ,
::;. I. t
, (~.t,JFind eigen valuesand eig~~v~ctorso(:J~~~'ci~" ",i :
..
. ~\" '.
[_I 4 -2] [ 5 0 0]8) A= -3 4~ , 9) ~
5 ~ =A-3 1
·IO)A-U0
° ] [ 6
:3_12] _ A0 o ,~1) -10 -5
0 1 -4 -2 - t .
In problems froi:n5 to 7 find AS; .
282
~nthe following matricesdetermine if A is diagonalizablc, If so , find
a matrix T such thatTI A T = D
12 ] [1t 7 ,2) 6[
- 141)
- 20o ] j) [I 0 0]- I 0 1
o I
4) [~ ~ ~ ].5) [~ _°2 n6l~~~-d7) [-: ••
Find eigen values and eigen vectors of each matrix
( EXERCISES J
->,
285------------------~" .~':
... .). .k~l
Where Utcr be the component of UKwith largest absolute ~ue. ';.. ". , .'"...l:...
'1ihu.~VK.isan approximation to jlD dgcn~~~ for J>.. ass~i~~itb· ~ ~
Step4. For aD approximation to \ ; , .' . . " .- . :II·r"'fIi·
k>l
Step2. Compute AUo =UI. 1
Step3. ?efine Ve= 1--IUk .UKT
The power method consists of the following stcps.-
Step1. Choose an arbitrary vectors Uo as, ~ inil..iaLapl)l'oki.a¢:i9Il to ~.'. . l '.. ....,. \. ~ "'" :
eigenvector of A as~~ciated wiili ~ . For exan~p'J~,q ..~'(~.]~(L~:It, ...9.. ,. P. 1 , .. ' .
:' .... . -', _,'
i .eigenvalue of largest magnitude.
'j THE P.OWERME!HOO ,.
In this section we present some .methods for . .findi~g ej~e.~valuesandeigenvectors of matrices numerically ..
"
Ex (2) :- from the 1~ examples,I,
X-¥= (2, 3. 2,-1) 1(4.2, 1,3)
= 1 .4 + 3,.2 +.2." I + (-I) ·3=8+6+2-,;l
= l3.
iiI,
'. !iiit:
Xn) ,
...._----------- 284,·'-----------'----
Def:- If X(x), X2, ..... ". xn), and Y=(Y1o Y2, ......... ,yJ are vectors in RD , then
their inner product is defined ~Y:
X-y =xlYI+ )(2Yl + ... :..... + x"yD'
vector X-¥, Thus from equ. (2),
The distance between the points (2~3,2, -1) and (4, 2, 1,3) is the length of the'.'
Ex(1):- Let Xc>(2, 3,2. -1) & Y=(4. 2, 1,3)
'The~ ~~ =-j22 t.32~ 2~+(-1)2 =Jf8IIYII=·,.J4'2 + 22 + 12+32 =. ./30 .
. ·'II.Y_,·n: ..J(""'_'y)2+(X _y.)2+ +(x _y')2 (2)r ~ A y'" ~ ""I" . 1" 2 2 ....... n n········
Thus this distance is given by:
_defined as the length of the rector X-Y, where X = (XI. X2,
Y=(Yt. Yl, ...... .y,J
The distance betweenthe points (XL, Xl, Xn) and (Yh Yl, Yn) is then
We also define the,distance from the point (xi, Xl, ., •.•• XII) to the origin-by (1).. . . .
1X11=~X~4-X~+ : +X; (1)
, ,
Oef :- The length ormagnitude or normof a vector X -= (x., Xl, , .. '" Xu) in R" is
We sbaU now define the notion of length of a vector in R" by
geD~raliziDg \be corresponding idea for RI,•
, IINN~RPR90U~T!.oNIN R" I
__,_---~~---------------_--:-----..,'.:,.:1):1!t
:1. im·r'i-t;J'II ).
,I)',,
'U~
.'
'-__;__.,-- 'tS:,'-------:---------
:'(', :'
'. ,
:: A ,~(AVs'V) ~ 1.560576 +6.024,
S (V ,V ) 1.2621445 S
I
:1
V =_1_(3.0~6)=(O.512),5 6.048 6.048 1 J
~..U = AV =(4 11°·524)=(3.096)
S 4 2. 5 1 6.048
t,
., (AVA,V4) 7.670304 6 1/\, ~ "::::l ~ .0 84 (V ,V ) t .274576
4 4
, r
U = AV =(4 lX°.548)=(3.192)an·~' 11 '~_,_1_,(3.192)=(O.524J4 J' 2 5 1 6.096 4 6.096 6.096 1
" ,': .. .;.~ii!
U = AV =(4 lXO.600) 2_,(3.4) and 'V == ~(,~,A)= (0.548)3 2 2 5 1 6,2 3 6.2 '(?2 I'
____ ~~~ ~ i86~' -- __ ------------~
,... . !"t' .... . .~.~.:.
.~. I,"i! .•A.2,1'::i6.059
'j
, 1 ~ :
Since AV1 =Ul
U ~(4~/.lyp.';1.fJ~(J.856) and V -: _,_1_(3.856,) =,(°.900 )"2 2., 'SA 1, '6.428 ~'6.428 6.428 ' "t ,
; I' :
, '
., (AVI,Vt) 9..1811844)A ~ ::::l _I -(~,V;) ~,,609796~ ::::l6.081
, ' r:) (~ '~)C)=(~)_UtV ,1(5) , (0.7]4)3) Let = - = ..
t 7 1' 1
2) Compute AUo"" U I
,decimal <places and round" after each multjpli~lion and division for nineiterations (K=8) for the matrix
A =(; ~)I) L"U, =Gl= (I If
stepS. Increase K by 1
let UK+)... A VK andreturn to step3, /'
•••••••• '" I ·''I.' j. _----'------------~:-::---.......;...-----...\. ~, ' : ",', '"
j \, ' .. .' 1I '!u. I
Ex (3}:~ , !:, ICompute an' approximation to the eigenvalue of largest magnitude and I
, associated+eigeovector bythe power method. Carry out the computation to three " ,!
'-- 289 _
Iteration VK Approximation of ~ ,
0 (11)'.;~ 1 (0.714, 1)T 6.081
2 (0.600 1)' 6.0593 (0.548. 1) 6.0334 (0.524 1 I '6.0185 0.512 Il' 6.0106 0.506 1)' 6.0057 ·(0.503 1) I 6.0028 (0.501' i)' 6.001
The exact answers are /\ = 6,X1 == (0.5)I ..'. .i.
We conclude that the dominant eigenvalue is approximately 6.00 1 and that an, I
, X/\ (0.501 )eigenvector is approximately = I,.
1505004+6.002 7.507004it ~ . ~ ..8 . 1.251.001 1.251001..
V =_'l_(3.012)=(0.50IJ'8 6.006 6.006 1
(4 1)(0.'503' ) = (3.012' )
U 8 == A V 1 = 2, 5· 1 6.006
. ., '..
~-_:_-:...._:_;_-_;__--- ~~'''':'''----------
~ ~6.00f.·
. A. ~ 1.515036 +.6.006 ~ 7.521036
1 1.253009 i.253009
V =_1_(3.024)==(0.503)7 6.012 3.012 1.
1)(0.506) = (3.024)5 I 6.012
(AV6,VIS) 1.530144 +6.012/L ~ ~6 (V V) 1.256036
6' 6
. l (3.048) (0.5Q6)V6 ==6.024 6.024 == I
U =AV ~(4 lJ(0.512)==(3.048)6 . 5 2 5 I 6.024
-, '.
v _ 1 (6.004 ) _ ( 1 ')4 - 6.004l4.004 - 0.667
.'A ~ 6.004+2.674672 :::::6.001
3 1.446224I, , '
. ',".
,.,
v - 1 (6.022 ) - ( 1 J' ' ,3 - 6m 4.022 - 0.q6~ , ,
• • .. " :: ..~.:' ~',;",,: ":. -t ' '.
'"
1.454276/L ~ 6.022 + 2.710828 ~ 6.005 "2
. ,'"a'"'.. ,......
• ' J\ 1, ,I
!
""__~---, 290 ;._ __
y = ,i ,',.(6.142 )~(. 1 )':2 6J42.l4.142 0.674'.
'. (. . ( i ("U =AV ~ 4 ,3 1 '~ 6. t422 . ,1"23)0.71-41',4.142).
R> 6.027
, .' ,.,., (A V ,V )and A, ::::: 1 I
1 (V ,V )I 1
•let VI = l(7)::: ( 1 )
7 5 0.714
Compute AU. =U1
, '
In the following Example: Compute an approximation to the eigenvalue
.ofIargest magnitude and associated eigenvector by the power method. Carry out
the computation to three decimal places and round after each multiplication and
division for fifth iterations (K= 4).
Ex (4):-
A = (; ~)
let U0 == C)
'- ~193~~----------------------~
~2 = 8,;04 (~:~~:)=(0.;38 ) .. .'"J: -8210.:38 H0.138lL 10.704+0 785512
~ ((O.~38 }(~,;38))' 1.114244
..1,2 ;::: 10.311
and
'..,
·1
.~__~~~ 2~2 _/
. d'V - 1 (16) _ (.1 )An 1 -16 1 ,- 0.063
In the following Examples: Compute an approximation to the eigenvalue
of .largest magnitude and associated eigenvector by th~ power method, Carry out
::,,:tho, computation to-three deci~al places and round after each multiplication and
, division.for six iterations (K=5);
'. -.E~(5):-
: A • (! _82J
=Let U =(1)•. ' 1
it4 :::: 6.000
Al ::::6.000 and Xl =::; ( 1 ) = V0.667 4
A ::::6,001 ...2.668667 ::::J 6.0004
1.444889
r[ Il 0.667
I
'r,}~I~i 'I
~i\ -.
'I~' '.'r" ..!'e
and
V2 = ![,~,5) ~ [. 0.'125) . ,.
. A '" ~-34
~t0.~25):C\25']]2 ((-0112l}'(~~.~~;1).
- 4.25 - 0.359315 :>: -~J~8' .A2.::::: 1.015625
, (-4 21-0.5)_( 4 )ComputeU2 = AV, = 3 I '1 - ~O.5
((-4 21-0.5) (-0.5)13 IA_ 1 ' ·1 i)~-2.-0.5:::::-2.5:>:_2,OOO
A, '" ((-~.5W~l. 1.25 U5 .
and
_1(-2)=(-0.5)And V - - 1I 4 4
~'
~'I
~"
. .As ~ 9.-971
, 1\ (l }-v~ ~9.971 and Xl::::: 0.244 -, 5,
AndVs = 1 (1;~\~)= (0.;44), ' 10.t12 .
-'210 ;•• }( 0 ;44)] '" 9.952+061291'(('O.;44}(O.;4~J) . l.05953~
( ) (' t )1. 9.736 _ , '
And V4 =-- 2 '566 - 0.264'9.736 ' .
, "
ComputeU 4 = A V 3 =G _8210,;17 )= (~:~!:)
A :::::9.736 +0,556822 :::::9.830J 1.047089
, ,I
I
....___~- __ :..:....._~___:__ 29'7 ..-/
, " '~..... ._..' " '.
AI ::::1 -4.975 and X t ~ ( - t ) = V, 0.527 sI
its =:s -4.975 .
V I (- 4,86'~'Jr, ( - 1 )"And '= _' -- '- ..= '5, 4:868' ·..~!L566 ... , '.0."527· ':. . ..',. '., .
1/I.
I. ,
_______________ 296' ~
Co~pute Us = A V.4 = (-:-34 2Y I ) (-4.868)I Jl-O.4jW . 2.566
A '" [(-3'. ~l-0.~3')( -0~34l) '" -4668 ~Ll13644~-5 034A -. (( 1 }( 1 ))' 1:..88356 .
-0.434 - 0.434
d V - 1 (5.352 J-( 1 )An -4 - 5.352 -2.~,74 - -0.434
, - 5.352 -l.571024/1;3 ~ ~ -4.752
1.456976
V I (- 4.25 J (- I )And 3 = 4.25 2.875 = 0.676
: ( 4 21 1 \ __(-4.251Compute U- = A V2 = - J I, -' 3 1 -0,125 2.875)
I
Ir·~illi!~
I ',
Ir~
. '..":.:-.' .;..._.;.:,__-------------------_. i :
.~
-------------------- i99·~1 ~ /
Example (2): We obtained
X'7'C1 cos 3t +C2 sin 3t +icos~':4
To get y we multiply (i) by -3 operate on (it) by D and add to etinll~ate x: .2 '',:r (D + 9)y:=:·18 sin t
From this equation we obtain
J'=C3COs3t+C4sin3t ",,2..sint," 4'
General solution for x and y contain 4 arbitraiy constants. "
RUle: Ifthe number of'the arbitrllry,~~:mstantscontained Inx and y is greater
then the degree of the determinet, we substitute by x.y in any equation andreduce the number of the 'constants to the required one.To illustrate this rule, consider the equatio~, of.
'.
Also equals two
.' 9y = -C, sm3!+C2 cosJt - -sin/4
We see that x and y contain two arbitrary constants Ct. C2•
Note that the degree ofthe determinent
/D3 -31D :::D2_9
y= -C1 si.n3t+C2 COS3t-±sint-.2Sint
Hence, the general solution for ~ is
x = C, cos 3t + C2 sin 3t + !cost4
substituting in (i) we get
I 'y=-Dx-2sint3
~P,,I
,.'- , 298~· ~
, First we write the equation in the fromDx - 3y.;= 6 sin t (i)
3x +Oy "'"0 ... '., ,.. (ii)Operate with D on (i) and multiply (ii) by 3; and add to eliminate y:
IYx + 9x "= 6D sin t(IY+9)x=6 cos tx.:,c ""Cl cos ,3t+C2 sin 3t
I 3xl'i. ={j ---cost=-cost
D2+9 4
Example (1) : solve the eqtl~tions,!, '"d,x . d:J--3y=6smt,-+3x=O.dt .' dt
IL11(D) LI2(D)1L21(D) L22(D)
At first we eliminate y and its derivatives. then we solve the obtainedequation for x. Substituting in any equatlon we get y,
The number of arbitrary constants in x and y ~ust be equal to the degreeof the determinent
We have to find x.y in terms of the independent variable t and arbitrary
constants,
LII (D)x + LI2 O?)Y ." fl (t)L21 (D)x t. I.:n{Djy= f2 (t)
Where D == !!_and L;j (D) are polynomial from in D,dt ' ,
Consider the two simultaueeus D.E.
SIMULTANEOUS DIFFERENTI,AL EQUATION ';
)\.:I
!
'------------ .....:;·301 ------------...,
Setting here t = .0 in both sides. we get
........ .(4)(
..t tX=T¥~T e I
.. "f 0
Thus,
, . .~."(
. ..t I¥= e 1
o
'\, , . A.2(u =C1 e • V r;;C2 e '
In matrix notation
Whose geueral solutions are
: -DY-[~I ,t°Jn~U~:lThis last matrix equation represents the following equations
du dv-=J,u, -=A.ZV, .dJ. dl'
or
premultiplying this equation by the constant matrix yl ~d taking into".
considerations (3) , we get
dY =r'ATYdt '
settingx =TY , where "(= ell v)\ is the vector of ucw unknown functions
u and v in r,equation (2) becomes
:!...TY=ATYdi . ~. I
~ .'
.,I.
'- ~~~ ~~-----,jOO------------------------~
...... (2). '.'dx=/V{dJ
'where x.=1;X)~ A =(Qu' t;11Z). ly a21 a22
Let A be diagonsilized by the matrix T. i.e.T1 AT=D ' ..... :.(3)
Consider tile system of two simultaneous equation
dx-""3Ilx+a'1Y }dt . . :.(l}dy-'" 321X+ all Ydy
With the initial condition x(O) == Xo , y(O) =Yo
System (1) may be written in the following form:
SOLUTION OF SIMULTANE;OUS DIFFERENTIAL EQUATIONS
Thus, the required solution J;ory is, 9'
'J =C:t cos 3t- CI sin 3t- -Sin t. '4
We substitute in equation (ii):3 .'
3(C1cos 3t + Cl sin 3t + -cos t) + (-3C) SIn3t +, 4
+ 3C4 cos 3t - 2.cos t) '" O., 4
3(C1 + C,,)cos 3t + 3(C1'7e)) sin 3t~.
~ ~.~~'I
"'q, 1
~, !,,~,..~I I
I,I •
'~~fIf
~~
~
l "
~,!
t~,~.I,I'~l~l1\Ih"~'lI.~I
-1mI!
lim1:..~ • I
"r "' .. """,. '-'-_:--- .....;!303' ---:----------_./
The solution of the given system is
x=,2et -3e-t +Z:.(
y = -3et +s«:' -'4z =_el +Je-t
~.,
[;H-:22 II" 0 ~F-1 ~J~lI-3 -2 0 et 1
-1 -1 0 0 e t - 2 -I
[ 2e'3 - t +2] .' .- e , "
= _3et +6e-t -4_et +3e- t
T~(-:2 _12]-3-1 -I
[-I -1 ~Jrl= 2-2 -1
Hence, by (5) we have
the eigenvalues are 0,1;-1, and
dz_= -4x .,.2y - zdt
with the initial conditions x(o) '" 1 , yeo) = -1. 7.(0):a 2.
f:..= [-~9~5 _12]-4 (-2 -I
0'
f!f~
i; ,;',
~ ~ . 30i~----------------------~
The required solution is
x=2e'J.t _e-5t ,
y=eSI +2e-5t
Example (4) : Solve the system:
d:X_,_""6x + 3y + zdt
-51 )-e+2e-Sl
By formula (5) we have
Fonnula (5) remains trus ifthc number of unknown functions isgreater than 2.
Example (3): Find the solution of the system dx ..,.·3x+4y, 4Y", 4x - 3ydt , dt
sa.tisfying the initial conditions x(o) ",I , y(o) = 3.
The eigenvalues of the matrix
A-(! }3)arc 5,-5 and the corresponding eigenvectors nrc VI - (2 1)~•v2=(-I 2)~
. Thus,
T",,(21 -1)andTI.,..!..(2 I)2' 5 -1 2
..: ....(5)
Substitution in (4) yields
