Math 104 - Calculus I

Part 8

Power series, Taylor series

SeriesFirst .. a review of what we have done so far:

1. We examined series of constants and learned that we can say everything there is to say about geometric and telescoping series.

2. We developed tests for convergence of series of constants.

3. We considered power series, derived formulas and other tricks for finding them, and know them for a few functions.

4. We used the ratio tests to determine intervals on which power series converge, and use the other tests to check convergence at the endpoints of the intervals.

Geometric series

a/(1-r) = a + ar + ar2 + ar3 + ...

provided |r|< 1.

We often use partial fractions to detect telescoping series, for which we can calculate explicitly the partial sums sn

Tests for convergence for series of constants

Fundamental divergence test (nth term must go to zero for convergence to be possible)

Integral test

Comparison and limit comparison tests

Ratio test

Root test

Alternating series test

Power series

f(x) = a0 + a1 x + a2 x2 + a3 x3 + ...

where an = f(n)(0)/n!

We know the series for ex, sin(x), cos(x),

1/(1-x), and a few related functions.

Convergence of power seriesBefore we get too excited about finding series, let's make sure that,

at the very least, the series converge.

Next week, we'll deal with the question of whether they converge to the function we expect. But for now, we'll assume that if they converge, they converge to the function they "came from".

(Strictly speaking, this is not always true -- but it is true for a large class of functions, which includes nearly all the ones encountered in basic science and mathematics. This fact was not fully appreciated until the early part of the twentieth century.)

Fortunately, most of the question of whether power series converge is answered fairly directly by the ratio test.

Recall that...for a series of constants , we have that the

series converges (absolutely) if the the limit of the absolute value of is less than one, diverges if

the limit is greater than one, and the test is indeterminate if the limit equals one.

To use the ratio test on power series, just leave the x there and calculate the limit for each value of x. This will give an inequality that x must satisfy in order for the series to converge.

0nnb

n

n

bb 1

For the series for the exponential function...

x. allfor

converges series thex, allfor one thanless

islimit ratio theSince 01

lim have weis,

xt matter wha No 11

:follows as test ratio out thecarry we,For

11

.n

x

.n

x

x)!n(

!nx

b

b

e

n

n

)n(

n

n

x

Your turn...

Calculate the series for the function sin(x) and determine for which x the series converges.

Here’s a more interesting example

series). geometric theofnt (reminisce 1x if diverges

and 1x if converges series that theis conclusion The

simply x. isinfinity ton tends as thisoflimit The

.1

)1(

gives time test thisratio The - - )1(

1

n

nx

xn

nx

n

xn

n

n

n

What remains...

divergent. be toknow we

whichseries, harmonic the, becomes series

the1,For x -1. xand 1 xpoints check the tois

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.convergent

ally)(condition be toknow which weseries, harmonic

galternatin the, becomes series the1,For x1

)1(

n

n

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Final Conclusion

otherwise. diverges and

1x1- if converges series The1

nnxn

OK, your turn...

converge? series thedoes x of valuesFor which 1

)2(2

nn

x n

A. -1 < x < 1

B. -2 < x < 2

C. 1/2 < x < 1/2

D. -2 < x < 2

E. -1/2 < x < 1/2

One more...

converge? series thedoes x of valuesFor which 1

n

nnx

A. -1 < x < 1

B. -1 < x < 1

C. 1 < x < 1

D. -1 < x < 1

E. 0 < x < 1

From these examples,

...it should be apparent that power series converge for values of x in an interval that is centered at zero, i.e., an interval of the form [-a, a] , (-a, a], [-a, a) or (-a, a) (where a might be either zero or infinity). The interval is called the interval of convergence and the number a is called the radius of convergence .

Let’s go backTo finding series of functions:

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)!12(

)1(...

!7!5!3)sin(

!...

!4!3!21

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0

432

n

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xxxxxe

xefa

nn

n

nx

xnn

The other way

0

)2(8642)(

2

!...

!4!3!21

:get toseries in the

everywhere for Substitute :Examples

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and/or algebraic use and seriesknown fromStart 2.

2

n

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Try this...

Take the derivative of the series for sin(x) to get

0

)2(642

)!2(

)1(...

!6!4!21cos(x)

n

nn

n

xxxx

Integrate both sides of the geometric series from 0 to x to get:

...432

x)-ln(1-

...1

432

0 0

3

0

2

00-11

xxx

x

dttdtttdtdtdtx xxxx

t

Negate both sides and replace x by (-x) everywhere to get:

ln(2)). tosums seriesfourth that theshows (This

...432

)1ln(432

xxx

xx

Start from the geometric series again...

.)4

toconverges

earlier saw weseriesfirst that theshows (This

...119753

)arctan(

:get x to to0 from sidesboth integrate Now

...11

1

119753

1086422

xxxxxxx

xxxxxx

And substitute x for x everywhere it appears to get2

A challenge to think about...

How to get the other one from previously

1

1 ?)( 2

nn

1. Limits: Series give a good idea of the behavior of functions in the neighborhood of 0:

We know for other reasons that

We could do this by series:

Application of Series

1)sin(

lim0

x

xx

1...001!7!5!3

1 lim)!12(

)1(lim

)sin( lim

642

00

2

00

xxx

n

x

x

xx

n

nn

xx

This can be used on complicated limits...

Calculate the limit:

A. 0

B. 1/6

C. 1

D. 1/12

E. does not exist

)(03

e1

)sin( lim

xx

xx

Application of series (continued)

2. Approximate evaluation of integrals: Many integrals that cannot be evaluated in closed form (i.e., for which no elementary anti-derivative exists) can be approximated using series (and we can even estimate how far off the approximations are).

Example: Calculate to the nearest 0.001. dxx 1

0

)( 2

e

We begin by...

...!!

...dx!

x

!

xx

dx

xx

)x(

x

37

1

25

1

3

11

321

by edapproximat is e :answer the toconverges

that series numericala us give willThis it. gintegratin and

,efor know already weseries thein for - ngsubstituti

61

0

42

1

0

2

2

According to Maple...

The last series is an alternating series with decreasing terms. We need to find the first one that is less than 0.0005 to ensure that the error will be less than 0.001. According to Maple:

evalf(1/(7*factorial(3))), evalf(1/(9*factorial(4))),evalf( 1/(11*factorial(5)));evalf(1/(7*factorial(3))), evalf(1/(9*factorial(4))),evalf( 1/(11*factorial(5)));

evalf(1/(13*factorial(6)));evalf(1/(13*factorial(6))); .02380952381, .004629629630, .0007575757576

.0001068376068

Keep going...

So it's enough to go out to the 5! term. We do this as follows:

Sum((-1)^n/((2*n+1)*factorial(n)),n=0..5) = sum((-1)^n/((2*n+1) Sum((-1)^n/((2*n+1)*factorial(n)),n=0..5) = sum((-1)^n/((2*n+1) *factorial(n)),n=0..5);*factorial(n)),n=0..5);

evalf(%);evalf(%);

.7467291967=.7467291967

5

04158031049

!)12()1(

nnn

n

and finally...

So we get that to the nearest thousandth.

Again, according to Maple, the actual answer (to 10 places) is

evalf(int(exp(-x^2),x=0..1));evalf(int(exp(-x^2),x=0..1));

.74669241330

747.e1

0

)( 2

dxx

Try this...

Sum the first four nonzero terms to approximate

A. 0.7635

B. 0.5637

C. 0.3567

D. 0.6357

E. 0.6735

1

0)cos( dxx