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Calculus I – Math 104 The end is near!. Series approximations for functions, integrals etc. We've been associating series with functions and using them to evaluate limits, integrals and such. - PowerPoint PPT Presentation
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Calculus I – Math 104
The end is near!
Series approximations for functions, integrals etc..
We've been associating series with functions and using them to evaluate limits, integrals and such.
We have not thought too much about how good the approximations are. For serious applications, it is important to do that.
Questions you can ask--
1. If I use only the first three terms of the series, how big is the error?
2. How many terms do I need to get the error smaller than 0.0001?
To get error estimates:
Use a generalization of the Mean Value Theorem for derivatives
Derivative MVT approach:
. )(somewhere' f f(0) )f(get to thisRearrange
)0()( x)and 0between (somewhere' f
get --x b 0,aSet
)()( b) and abetween (somewhere' f
: theoremvalue-mean theRecall
xx
xfxf
b-aafbf
If you know...
If you know that the absolute value of the derivative is always less than M, then you know that
| f(x) - f(0) | < M |x| The derivative form of the error estimate for
series is a generalization of this.
Lagrange's form of the remainder:
.!
)0( where
)(f
: be remainder"" let the and f(x),for series theof toup terms theusing
obtainedion approximat theyou write Suppose
)(
33
2210
kfa
(x)Rx a ...xaxaxaax
(x)Rx
k
k
nn
n
n
n
Lagrange...
Lagrange's form of the remainder looks a lot like what would be the next term of the series, except the n+1 st derivative is evaluated at an unknown point between 0 and x, rather than at 0:
So if we know bounds on the n+1st derivative of f, we can bound the error in the approximation.
1)1(
)!1()()(
n
n
n xnsomewherefxR
Example: The series for sin(x) was:
anyhow. zero is series theof term thebecause
)()sin(
have we terms,(nonzero) first two theuse weIf...)sin(
4
4!3
!7!5!3
3
753
x
xRxx
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x
xxx
5th derivativeFor f(x) = sin(x), the fifth derivative is f '''''(x) = cos(x).
And we know that |cos(t)| < 1 for all t between 0 and x. We can conclude from this that:
So for instance, we can conclude that the approximation sin(1) = 1 - 1/6 = 5/6 is accurate to within 1/5! = 1/120 -- i.e., to two decimal places.
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5
4
xxR
Your turn...
places? decimal 10 toget to together add to
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? 1.6458333 !3
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5.5.1
ion approximat theis accurate How32
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ee
Another application...Another application of Lagrange's form of the
remainder is to prove that the series of a function actually converges to the function. For example, for the series for sin(x), we have (since all the derivatives of sin(x) are always less than or equal to 1 in absolute value):
0
12
1
)!12()1()sin( :in writing justified now are
weSo limit. in the zero and - smally arbitraril becomeserror theThus, infinity. togoesn as zeroapproach will
quantity this x,of any valuefor and --)!1(
)(
n
nn
n
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nxx
nxxR
Shifting the origin -- Taylor vs Maclaurin
So far, we've been writing all of our series as infinite polynomials and using values of the function f(x) and its derivatives evaluated at x=0. It is possible to change one's point of view and use values of the function and derivatives at other points.
As an example, we’ll return to the geometric series
1). and 1-between for x only validwasexpansion f(x) the(since 0 and 2-between for x validbe wouldexpansion This
...)1()1()1()1(1
1)1(1
1)1(f)(g
writecould then we1),f(xg(x)function new a define weIf
...11
1)(f
432
432
xxxx
xxxx
xxxxx
x
Taylor series
By taking derivatives of the function g(x) = -1/x and evaluating them at x=-1, we will discover that the expansion of g(x) we have found is the Taylor series for g(x) expanded around -1:
g(x) = g(-1) + g '(-1) (x+1) + g ''(-1) + ....
!2)1( 2x
Note:
0.a when series)Maclaurin thecalled now is(that friend oldour tosspecialize that thisNote
... 3!
)()(''' f 2!
)()('' f !1
)(' f )f( )f(
:a xaround f(x) ofexpansion Taylor thehave wegeneral,In 32
axaaxaaxaax
MaclaurinSeries expansions around points other than
zero are useful when trying to approximate function values for x far from zero, but close to a different point where much is known about the function.
But note that by defining a new function g(x) = f(x+a), you can use Maclaurin expansions for g instead of general Taylor expansions for f.
Binomial series
integer. positive a is p if )1( ofexpansion the
gives and worksThis . )!(!
!
tcoefficien binomial theis where
)1(
: theorembinomial theoftion generaliza a isIt ns.applicatiomany in arises that seriesimportant An
0
p
p
k
kp
x
kpkp
kp
kp
xkp
x
If p is not a positive integer...
... 111 series!
harmonic galternatin thegives this-1,p if instance,For ... 1 )1(
k).,binomial(pfor definition new a need weand )polynomial a of instead series a givesit (i.e., stopt doesn'it except worksexpansion same then the
32
3!3
)2)(1(2!2
)1(
xxx-x
xx p xx pppppp
Fibonacci numbers
Everyone is probably familiar with the famous sequence of Fibonacci numbers. The idea is that you start with 1 (pair of) rabbit(s) the zeroth month. The first month you still have 1 pair. But then in the second month you have 1+1 = 2 pairs, the third you have 1 + 2 = 3 pairs, the fourth, 2 + 3 = 5 pairs, etc... The pattern is that if you have a pairs in the nth month, and a pairs in the n+1st month, then you will have pairs in the n+2nd month.
The first several terms of the sequence are thus:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, etc...
Is there a general formula for a ? n
n n+1n+1n
a + a
Generating functionsThis is a common problem in many parts of
mathematics and science. And a powerful method for solving such problems involves series -- which in this case are called generating functions for their sequences.
For the Fibonacci numbers, we will simply define a function f(x) via the series:
game. theinto relation recurrence get the tohave weNow
...5321)(f
12
43233
2210
nnn aaaxxxx ...xaxaxaax
Recurrence relation
To do this, we'll use the fact that multiplication by x "shifts" the series for f(x) as follows:
Now, subtract the second two from the first -- almost
everything will cancel because of the recurrence relation!
...)(f
...)(f
...)(f
42
31
20
2
43
32
210
44
33
2210
xaxaxaxx
xaxaxaxaxx
xaxaxaxaax
The result is...
do? thisdoes goodWhat
!1
1)(f
that deduced have weSo .1 that recallBut )()(f)1(
2
10
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Further...
later).in valuesput the ll(we' and where),)((1
r denominato Factor the rescue! the tofractions Partial series. theof tscoefficien theare
they since numbers, Fibonacci for the formula a have
will then wefor series out the figurecan weIf
215
2152
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2
xxxx
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Then use partial fractions to write:
(almost)! done be will we andfor series get thecan weif So
1
1
))((1
))((1
))((1
x
x
xxxx
Work it out...
...
...)1(
...
...)1(
3
3
2
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3
2
1
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1
321111
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xFirst
And
Now, recall that...
1
215
215
2. and
5 1.
are and about factsimportant Two . and
Our series for f(x) becomes:
number Fibonaccinth for theformula aobtain will we and of esknown valu in theput weIf
thatus tellThis
...))()()(()(f 344233225
1
xxxx
5)1( )1()1(
nnn
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