Syllabus

Numerical Analysis: Finite differences - Forward backward and central difference. Newton’s forward and backward differences interpolation formulae. Sterling’s formulae, Lagrange’s interpolation formula. Solution of non-linear equations in one variable by Newton Raphson and Simultaneous algebraic equation by Gauss and Regula Falsi method. Solution of simultaneous equations by Gauss elimination and Gauss Seidel methods. Fitting of curves (straight line and parabola of second degree) by method of least squares.

UNIT -1

Numerical Analysis

18/04/2017

Submitted by: Ms. Laxmi Peepliwal (Electrical Branch)

Submitted to : Mr. Amit Sharma (Department of Mathematics)

� Interpolation -

Interpolation lies at the heart of numerical analysis. Generally we come across situations where we wish to find some unknown values with the help of given set of observation.

Example- We are given that the number of school going children in 1970,1980, 1990, 2000, 2010…….. Then the process of finding the desired figure is known as Interpolation.

� Definition of Interpolation (Mathematically) -

The technique or method of estimating unknown values from a given set of observation is known a Interpolation .

Mathematically, let x=a , a+h ,a+2h , ………., a+nh

Y=f(x)=f(a) , f(a+h) , f(a+2h) , ………., f(a+nh) Then the method of finding f(m), for x=m is known as interpolation & where m lies outside this range , it is known as extrapolation. Example- X 0 5 10 15 20 Y 7 11 19 23 27 That the method of finding f(3) , f(7) , f(17) with the help of given data is known as interpolation & that of finding f(22) or f(-0.5) is known as extrapolation.

� Assumption For Interpolation –

1- These should be no sudden jump of falls during the interval in the value of the function that is the the value of the function f(x) should be in an increasing or decreasing order.

2- The length of the interval must be uniform .

3- When we apply the calculus of finite differences the function must be capablr of being expressed in o polynomial form.

� Methods Of Interpoltion-

1) Method of graph 2) Method of curve fitting 3) Method of finite differences

Finite Difference Caluculas -

The study of finite difference calculus has become very important

due to its wide applications in everyday life .

It deals with the change in the value of the function y (dependent

variable) due to change in x (independent variable).

Let y=f(x)

X a a+h a+2h -----------

Y=f(x) f(a) f(a+h) f(a+2h) ----------

First difference of f(x) at x=a is denoted by ∆f(a)

∆f(a) = f(a+h) – f(a)

First difference of f(x) at x= a +h is

∆f(a+h) = f(a+2h) – f(a+h) & so on

∆f(a+h) - ∆f(a) = ∆2 f(a) is known as second difference of f(x) at x=a

i.e.

∆f(a) = f(a+h) – f(a)

∆2f(a)=∆f(a+h)- ∆f(a)

∆3f(a)=∆2f(a+h)-∆2f(a)

…………………………………….

∆nf(a)=∆nf(a+h)-∆n-1f(a)

The operator ∆ is known as forward difference operator for a given function f(x) .

Forward Difference – Table –

x Y=f(x) ∆y ∆2y ∆3y ∆4y ∆5y

X0 Y0

∆y0

X1 Y1 ∆2y0

∆y1 ∆3y0

X2 Y2 ∆2y1 ∆4y0

∆y2 ∆3y1 ∆5y0

X3 Y3 ∆2y2 ∆4y1

∆y3 ∆3y2

X4 Y4 ∆2y3

∆y4

X5 Y5

Question.. - Form the Forward Difference table for the following data.

X -1 0 1 2 3 4 5

Y -13 -7 -1 11 35 77 143

Solution -

x Y=f(x) ∆y ∆2y ∆3y ∆4y ∆5y ∆6y -1 -13 6 0 -7 0 6 6 1 -1 6 0 12 6 0 2 11 12 0 0 24 6 0 3 35 18 0 42 6 4 77 24 66 5 143

Fundamental Theorem of Finite Differences –

If f(x) = a0 xn +a1x

n-1+a2xn-2+…………..+ an , a0 ≠0 is an nth degree

polynomial then the nth differences of f(x) is ∆nf(x) =a0hn

Where ‘h’ is interval of differencing and ∆n+1 f(x) =0

Question - Evaluate ∆6[(ax-1)(bx2-1)(cx3-1)] ; h=1

Solution- ∆6[(ax-1)(bx2-1)(cx3-1)]

= ∆6[abcx6 – bcx5 – acx4 + (ab+c)x3 + bx2 + ax – 1]

= ∆6(abcx6) - ∆6(bcx4)-∆6(acx4)+∆6[(c-ab)x3]+∆6(bx2)+∆6(ax)- ∆6

= abc.1.6!

= 720 abc

( Ans)

Question - Evaluate ∆tan-1x

Solution- tan-1(x+h)-tan-1x

{ F(x)=f(x+h)-f(x) }

{tan-1a – tan-1b =tan-1(���)

(���) }

= tan-1 �� ��

�(�� ).�

=tan-1

�� ���∗� …………………………..( Ans.)

� Factorial Notation of A Function (Factorial Function or Generalized Power) – The product of n consecutive factors , the first of which s equal to x and each , subsequent factor is h less than the preceding that is Xn=x(x-h)(x-2h)(x-3h)-----------------[x-(n-1)h] Where h is some fixed constant is known as factorial function or generalized power.

Differences of Factorial Function- ∆x(n)=(x+h)(n)-x(n) = [(x+h)(x)(x-h)(x-2h)------(x-(n-1)h)] = -[x(x-h)(x-2h)-----(x-(n-1)h)] =x(x-h)(x-2h)------(x-(n-2)h)[x+h-x+(n+1)h] =x(x-h)(x-2h)-----(x-(n-2)h)(nh) ∆x(n)=nh(x)(n-1) ∆2x(n)=∆(∆x(n)=∆(nhx(n-1)=nh∆x(n-1)=n(n-1)h2x(n-2) ∆nx(n)=[n(n-1)(n-2)(n-3)------------3.2.1]hn.1 ∆nx(n)=n!hn

Question- Express x3-2x2+x-1 into a factorial polynomial. Hence find ∆4f(x)=0

∆nx

(n)=n!h

n

Solution- Let F(x)= x3-2x2+x-1 Ax(3)+Bx(2)+Cx(1)D ………(A) X3-2x2+x-1=A(x)(x-1)(x-2)+B(x)(x-1)+Cx+D …(i) Put x=0; then -1=D Put x=1; then C+D=-1 C=>0 Put x=2; then 2B+2C+D =>1=>B=1 Comparing the cofficient of x3 on both side of equation 1. , we get. A=1 Hence from equation A. f(x)=x(3)+x(2)-1 ∆f(x)=∆[x(3)+x(2)-1] = ∆x(3)∆x(2)+∆(1) = 3x(2)+2x(1) ∆2f(x)=∆[3x(2)+2x(1)] =6x(1)+2x(0)=6x(1)+2 ∆3f(x)=6(x)(0)=6

∆4f(x)=0 (Ans.)

� Properties Of Forward Difference

Operator – 1. ∆C=0-0=0 ; where C is constant.

2. ∆(af(x)+bg(x)=a∆f(x)+b∆g(x) 3. ∆n(∆mf(x)=∆m+n[f(x)]= ∆n+m[f(x)] 4. ∆[f(x) g(x)]=f(x+h)∆g(x)+g(x)∆f(x)

5. ∆ �(�)�(�) =

�(�)∆�(�)��(�)∆�(�)�(�� )�(�)

6. ∆ex=ex+h-ex=ex(en-1)

7. ∆logx =log(x+h)-logx = log ��

� =log[1+ �]

8. ∆sin(ax+b)=2sin� � sin[ax+b+ah+

��]

Backward Difference Operator- The first backward difference of a function y=f(x) whose tabulated values are (x0,y0) (x1,y1)…….(xn,yn) are y1-y0,y2-y1,……..yn,yn-1 and denoted by ∇y1,∇y2……….∇yn

In general ∇yk=yk-yk-1

Backward Difference Table-

x y=f(x) ∇y ∇2y ∇3y ∇4y ∇5y ∇6y X0 Y0 ∇y0 X1 Y1 ∇2yo ∇y1 ∇3y0 X2 Y2 ∇2y1 ∇4y0 ∇y2 ∇3y1 ∇5y0

x3 Y3 ∇2y2 ∇4y1 ∇6y0

∇y3 ∇3y2 ∇5y1 X4 Y4 ∇2y3 ∇4y2 ∇y4 ∇4y3 X5 Y5 ∇2y4 ∇y5 X6 Y6

� Other Difference Operator- 1. Shift Operator(E)-

E(f(x))=f(x+h) E2(f(x))=E(E(f(x))=E[f(x+h)]=f(x+2h) En(f(x))=f(x+nh) E-1(f(x))=f(x-h) E-n(f(x))=f(x-nh)

2. Average Operator(�)-

�(f(x))=�[f [� +

�]f(x- �)

3. �[f(x)]=f(x+ �)-f(x-

�)

4. ∆f(x)=f(x+h)-f(x) 5. ∇f(x)=f(x)-f(x-h)

Relation between Difference Operators-

1. ∆=E-1 2. ∇=1-E-1 3. E=ehD 4. �=E1\2-E-1\2

5. �=�(E1\2+E-1\2)

6. ∆=� �2+�√1+�2\4

7. �2=1+�2\4

Sterling’s Interpolation Formula- Y=y0+u

Newton Gregory Forward Interpolation Formula:

Y=y0+u∆y0+�(��)

�! ∆2y0+�(��)(���)

�! ∆3y0+……………

….+�(��)………[��( �)]

! ∆ny0

Where u =���!

Question - The values of a function y=f(x) at some points are given in the following table.

X 0 1 2 3 4 5 6

Y=f(x) 176 183 197 205 215 223 231

Find the value of function y=f(x) at x=0.2

Solution - x0=0 , x=0.2

u=���!

= !.��!

= 0.2

x Y=f(x) ∆y ∆2y ∆3y ∆4y ∆5y ∆6y 0 176 7 1 183 7 14 -13 2 197 -6 -21 8 -8 -33 3 205 -2 -12 -51 10 -4 18 4 215 -2 6 8 -2 5 223 0 8 6 231

Y=176+0.2*7+!.�(!.��)

�! *7+!.�(!.��)(!.���)

�! -

13+!.�(!.��)(!.���)(!.���)

"! *-21 +!.�(!.��)(!.���)(!.���)(!.��")

#! *-33

+ !.�(!.��)(!.���)(!.���)(!.��")(!.��#)

$! *-51

Y=176+1.4-0.56-0.624+0.7056-0.168+1.041

Y=176.75 ………(Ans)

Newton Gregory Backward Interpolation formula-

Y=yn+u∇yn+�(��)

�! ∇2yn+�(��)(���)

�! ∇3yn+……………..

Where u= ���

Question- Find the approximate function given by the table and hence find f(2.5)

X 0 1 2 3

Y 1 3 7 13

Solution- xn = 3 , x=2.5

u=���!

= ���

=x-3

x Y=f(x) ∇y ∇2y ∇3y 0 1 2 1 3 2 4 0 2 7 2 6 3 13

Y=13+(x-3)6+(���)(����)

�! *2+(���)(����)(�����)

�! *0

Y=13+6x-18+(���)(���)

�! *2

Y=13+6x-18+x2-2x-3x+6

Y=x2+x+1

Now put x=2.5

Y=(2.5)2+2.5+1

Y=9.75 …………………(Ans.)

Lagrange’s Interpolation formula for unequal spaced points-

Y=(���)(����)……….(��� )

(�!��)(�!���)………….(�!�� ) y0 + (���!)(����)……………(��� )

(���!)(����)………………(��� )

y1+ …………………… + (���!)(���)…..(��� �)

(� ��!)(� ��)(� ���)…..(� �� �� �)

yn

Lagrange’s inverse interpolation formula-

X=(%�%)(%�%�)……….(%�% )

(%!�%)(%!�%�)………(%!�% ) x0 + (%�%!)(%�%�)……….(%�% )

(%�%!)(%�%�)………(%�% ) x1 +

……………+ (%�%!)(%�%)…………(%�% �)

(% �%!)(% �%)……………(% �% �) xn

Question- Given the following pairs of values of x&y

X 5 6 7 11

Y 12 13 14 16

Interpolation the value of y at x=10

Solution-

Y=(!�$)(!�&)(!�)

(#�$)(#�&)(#�) *12 + (!�#)(!�&)(!�)

($�#)($�&)($�) *13

+(!�#)(!�$)(!�)

(&�#)(&�$)(&�) *14 + (!�#)(!�$)(!�&)(�#)(�$)(�&) *16

Y="∗∗(�)∗�

�∗�"∗�$ + #∗∗�∗��∗��∗�# +

#∗"∗�∗""∗�∗�� +

#∗"∗∗$$∗#∗�

Y=�"'��" -

$## +

�'!�" +

��!$!

Y=2-4.33+11.66+5.33

Y=14.66 …………(Ans).

� Fitting of Curves (Straight line &

Parabola of Second Degree ) By method of least Squares-

Fitting of Straight Lines-

Working Rule- 1. Let the line to be fitted be y=a+bx 2. Form the normal equations

∑ )=na+b∑ � ∑ �)=a∑ �+b∑ �2

3. Solve the two normal equations to get the values of a&b.

Question- Fit a Straight line to the following data-

X 1 2 3 4 6 8

Y 2.4 3 3.6 4 5 6

Solution- Let the lne to be fitted is y=a+bx ……………1.

By the principle of least squares, we get the normal equations as.

∑ )=na+b∑ �

∑ �)=a∑ �2

X y x2 xy

1 2.4 1 2.4

2 3 4 6.0

3 3.6 9 10.8

4 4 16 16.0

6 5 36 30.0

8 6 64 48.0

∑ 24 ∑ 24 ∑ 130 ∑ 113.2

24=6a+24b ………………..(i).

113.2=24a+130b ……………………(ii).

From equation (i) and (ii) . Now the equation 1 can be multiply by 4, we get

Now the values of a,b,c are-

a=1.976

b=0.505

Best fitted straight line-

Y=1.976+0.505x ……………………(Ans)

Fitting of Parabola- WORKING RULE-

1) Let to parabola to be fitted be y=a+bx+cx2 ……………(i) 2) From the normal equation

∑ )=na +b∑ �2 +c∑ �3 ∑ �) = a∑ �+b∑ �2+c∑ �3 ∑ �2y=a∑ �2+b∑ �3+c∑ �4

3).. Solve the normal equation to get a,b & c and substitude n equation(i).

Question- Fit a parabola of second degree to the following data

X 0 1 2 3 4

Y 1 1.8 1.3 2.5 6.3

Solution - Let the parabola to be fitted be-

Y=a+ bx+ cx2

By the method of least squares the three normal equations are –

∑ )=na+b∑ �+c∑ �2

∑ �)=a∑ �+b∑ �2+c∑ �3

∑ �2y=a∑�2+b∑�3+c∑�4

x y X2 X3 X4 xy X2y 0 1 0 0 0 0 0 1 1.8 1 1 1 1.8 1.8 2 1.3 4 8 16 2.6 5.2 3 2.5 9 27 81 7.5 22.5 4 6.3 16 64 256 25.2 100.8

/ �= 10

/ )= 12.9

∑ �2=30 ∑ �3=100 ∑ �4=354 / �)= 37.1

∑ �2y=130.3

12.9=5a+10b+30c …………………(i).

37.1 =10a+30b+100c ………………..(ii).

130.3=30a+100b+354c …………………(iii)

From equation (i)&(ii) , we get.

Now multiply equation (i) by 2 and equation (ii) by 3 , we get.

Then

a=1.42

b=-1.07

c=0.55

Best fitted straight line –

Y=1.42-1.07x+0.55x2 ……………………( Ans.)

� Gauss Seidel Methods- Let us consider a system of equation in three unknowns x,y,z. a1x+b1y+c1z=d1 a2x+b2y+c2z=d2

a3x+b3y+c3z=d3 Rewrite the above equation as –

X=

� [d1-b1y-c1z] ………………..(i)

Y=

�� [d2-a2x-c2z] ………………(ii)

Z=

2� [d3-a3x-b3y] ……………..(iii).

Now,this method can be used to above equation if – a1>b1+c1

b2>a2+c2 c3>a3+b3 Now taking y(0) , z(0) as initial values for y,z ,we get- X(1) from equation 1-

X(1)=

� [d1-b1y(0)-c1z

(0)]

Now using x(1),z(0)as initial values for x,z,we get-

Y(1) from equation 2

Y(1)=

�� [d2-a2x(1)-c2

(0)]

Now using x(1),y(1) as initial values for x,y , we get- Z(1) from equation 3

Z(1) = 2�[d3-a3x

(1)-b3y(1)]

Now repeat this rocess taking x(1) , y(1) , z(1) as initial values . This process is repeated till we get the required solution. Question- Solve the following system of equation by Gauss Seidel method. 27x+6y-z=85 6x+15y+2z=72 X+y+54=110 Solution- Now check the condition 27>6+1 15>6+2 54>1+1 These condition is satisfied the equations Given system of equation can be written as

X=1

27[85-6y+z]

Y=1

15[72-6x-2z]

Z=1

54[110-x-y]

Let the initial values be y=0,z=0 First iteration-

X(1)=1

27[85-6(0)+1(0) =3.148

Y(1)= 1

15[72-6(3.148)-2(0)] = 3.5408

Z(1)=1

54[110-3.148-3.5408] = 1.9132

Second iteration-

X(2)=1

27[85-6(3.5408)+1(1.9132)] = 2.4321

Y(2)=1

15[72-6(2.4322)-2(1.9132)] = 3.5720

Z(2)=1

54[110-2.4322-3.572] = 1.9259

Third iteration-

X(3)= 1

27[85-6(3.572)+(1.9259)] = 2.4257

Y(3)=1

15[72-6(2.4257)-2(1.9259)] = 3.5729

Z(3)=1

54[110-2.457-3.5729] = 1.9259

Fourth iteration-

X(4)=1

27[85-6(3.5729)+(1.9253)] = 2.4254

Y(4)=1

15[72-6(2.4254)-2(1.9253)] = 3.5730

Z(4)=1

54[110-2.4254-3.5731] = 1.9259

Fifth iteration-

X(5)=1

27[85-6(3.5729)+(1.9253)] = 2.4254

Y(5)=1

15[72-6(2.4254)-2(1.9253)] =3.5730

Z(5)=1

54[110-2.4254-3.5731] = 1.9259

………………Ans.

� Gauss Elimination Method- Question- Solve the system of equations by Gauss elimination method.

X1+x2+x3=6

3x1+3x2+4x3=20

2x1+x2+3x3=13

Solution- Given system can be written as-

AX=B

Where

A= 1 1 13 3 42 1 3

, X= �1�2�3

, B 6

2013

Augmented matrix

C= 1 1 1:3 3 4:2 1 3:

62013

R2→R3-3R1 ,

R3→R3-2R1

C~1 1 1:0 0 1:0 −1 1:

621

R2↔R3

C~1 1 1:0 −1 1:0 0 1:

612

Z=2

-Y+Z=1

Y=1

X+Y+Z=6

X+1+2=6

X=3

Now ,

X=3 , y=1 , z= 2 ………………………. (Ans.)

� Newton Raphson Method- When the derivative of f(x) is a simple expression & easily found , the real roots of f(x)=0 can be computed by a process called the Newton Raphson Method.

Working Rule- 1) Denote the function by f(x) . Select two points so that root

lies between these points. 2) Choose the initial approximation ,say x0, sufficiently close to

the root. 3) Differentiate f(x) with respect to x to get f’(x).

4) Apply Newton Raphson formula-

Xn+1=xn-�(� )�,(� )

,n=0,1,2,3………….. To get the root correct to the desired decimal places.

Question- Find by Newton Rahson method a root of root of the equation, X3-3x-5=0 Solution- Let f(x)=x3-3x-5 F(2)=-3 , f(3)=13 Root lies between 2&3. F(2) is much closer to zero ( or f(2)<f(3) ) So we take initial approximate root x0=2 By Newton Raphson formula

Xn+1=xn-�(� )�9(� ) ……………(i)

F’(x)=3x2-3 For n=0 , the first approximate is

X1=x0-�(�:)�9(�!) = 2-

(�3)3(2)2�3

= 2.3333

For n=1, the second approximation is,

X2=x1- �(�)�9(��)

=2.3333-{(2.3333∗2.3333∗2.3333)�3(2.3333)�3}

3(2.3333∗2.3333)�3 =2.2805

For n=2 , the third approximation is ,

X3=x2-�(��)�9(��)

= 2.2805-{2.2805∗2.2805∗2.2805)�3(2.2805)�3}

3(2.2805∗2.2805)�3 = 2.2790

For n=3 , the fourth approximation .

X4=x3-�(��)�9(��)

=2.2790-{2.2790∗2.2790∗2.2790)�3(2.2790)�5}

3(2.2790∗2.2790)�3 =2.2790

…………………….(Ans.)