Math 104 - Calculus I

August 11

(but first, a quick review…)

Power seriesLast week's project was to try and sum series

using your calculator or computer. The answers correct to ten decimal places are:

Sum((-1)^n/(2*n+1),n=0..infinity) = evalf(sum((-1)^n/(2*n+1),n=0..

infinity));

Sum(1/factorial(n),n=0..infinity)=evalf(sum(1/

factorial(n),n=0..infinity));

0

12)1( 7853981635.

nn

n

0

!1 718281828.2

nn

Power series (cont.)Sum(1/n^2,n=1..infinity)=evalf(sum(1/n^2,n=1..infinity));

Sum((-1)^(n+1)/n,n=1..infinity)=evalf(sum((-1)^

(n+1)/n,n=1..infinity));

We can recognize these numbers as

1

1 6644934068.12

nn

1

)1( 6931471806.)1(

nn

n

).2(ln and , , 64

2 e

Two directions:

1. Given a number, come up with a series that has the number as its sum, so we can use it to get approximations.

2. Develop an extensive vocabulary of "known" series, so we can recognize "familiar" series more often.

Geometric series revisited

1). (provided -1

...

:series geometric the

friend, oldour begin with We

432

0

rr

a

ararararaarn

n

r as a variableChanging our point of view for a minute (or a

week, or a lifetime), let's think of r as a variable. We change its name to x to emphasize the point:

So the series defines a function (at least for certain values of x).

)1(for 1

...)(f0

32

xx

aaxaxaxaaxx

n

n

Watch out...We can identify the geometric series when we see it,

we can calculate the function it represents and go back and forth between function values and specific series.

We must be careful, though, to avoid substituting values of x that are not allowed, lest we get nonsensical statements like

!!1...168421...22221 432

Power seriesIf you look at the geometric series as a function, it

looks rather like a polynomial, but of infinite degree.

Polynomials are important in mathematics for many

reasons among which are:

1. Simplicity -- they are easy to express, to add, subtract, multiply, and occasionally divide

2. Closure -- they stay polynomials when they are added, subtracted and multiplied.

3. Calculus -- they stay polynomials when they are differentiated or integrated

Infinite polynomials

So, we'll think of power series as "infinite polynomials", and write

0

33

2210 ...

n

nn xaxaxaaxa

Three (or 4) questions arise...

1. Given a function (other than ), can it be expressed

as a power series? If so, how? 2. For what values of x is a power series representation valid?

(This is a two part question -- if we start with a function f(x) and form "its" power series, then

(a) For which values of x does the series converge?

(b) For which values of x does the series converge to f(x) ?

[There's also the question of "how fast".]

x

ax

1)(f

continued

3. Given a series, can we tell what function it came from?

4. What is all this good for?

As it turns out, the questions in order of difficulty, are 1, 2(a), 2(b) and 3. So we start with question 1:

The power series of a function of f(x)

Suppose the function f(x) has the power series...)(f 4

43

32

210 xaxaxaxaax

Q. How can we calculate the coefficients a from a knowledge of f(x)?

A. One at a time -- differentiate and plug in x=0!

i

Take note...

(0)/2f have should weso

...,201262)(f

:derivativeAnother

(0)f

have should weso ,...432)(f

write toreasonable seemsIt :(first?) Second

f(0) that have weso

...000)0(f :zeroth)(or First

2

35

2432

1

34

2321

0

03

32

210

a

xaxaxaax

a

xaxaxaax

a

aaaaa

Continuing in this way...

n!by divided

zero,at evaluated f of derivativenth

:generalIn

etc.... (0)/24,f (0)/6,f 43

na

aa

ExampleSuppose we know, for the function f, that f(0)=1 and

f ' = f.

Then f '' = f ', f ''' = f '' etc... So f '(0) = f ''(0) = f '''(0) = ... = 1.

From the properties of f we know on the one hand that So we get that...

)!previously did wesums theof oneget to1(Set x

!...

!4!3!21

0

432

n

nx

n

xxxxxe

xex )(f

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 we just calculated...

x.allfor

converges series theso and x,allfor one than less

islimit ratio theSo .01

lim have weis,

t xmatter wha No .1)!1(

!

:follows as test ratio out thecarry we,for )1(

1

n

x

n

x

xn

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

1

1

nn

.convergent

ally)(condition be toknow which weseries, harmonic

galternatin the, becomes series the1,For x1

)1(

n

n

n

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:

:fun more is 2 methodbut way, thisseriesother do could We

)!12(

)1(...

!7!5!3)sin(

!...

!4!3!21

:)sin( and for series found ve We'(0)/n! formula the Use1.

: ways twoare There

)12(753

0

432

n

xxxxxx

n

xxxxxe

xefa

nn

n

nx

xnn

The other way

0

)2(8642)(

2

!...

!4!3!21

:get toseries in the

everywhere for Substitute :Examples

:othersget on tomanipulati analytic

and/or algebraic use and seriesknown fromStart 2.

2

n

nx

x

n

xxxxxe

e

xx

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...

...!37

1

!25

1

3

11...

!3!21

by dapprximate is e :answer the toconverges

that series numerical a us give willThis it. gintegratin and

,efor knowalready weseries in the for - ngsubstituti

61

0

42

1

0

)(

2

2

dxxx

x

dx

xx

x

x

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

Application of series (cont.)

3. Differential equations : Another important application of series is to find "solutions" to differential equations (when other methods fail).

Earlier, we found the series for based on the differential equation it satisfies (y ' = y ), together with the initial condition y(0)=1.

xe

Airy functionsThe equation y '' + xy = 0 has no "elementary"

solution (the solutions of this equation are called "Airy functions", named after a famous British Royal Astronomer). But we can find series for the solution that satisfies y(0)=1, y '(0)=0, as follows:

From the initial conditions, we can assume that the series for y(x) begins:

......1)( 44

33

22 n

n xaxaxaxaxy

Airy (continued)

Then the series for series:

...)2)(1(

...201262)(

2

35

2432

n

n xann

xaxaxaaxy

y

So...

...))2)(1((

...)30()20(12)16(2

12

436

325

2432

n

nn xaann

xaaxaaxaxaaxyy

We work our way from left to right -- since y’’+ xy is supposedto be zero, every coefficient must be zero.

Therefore:

etc. , ,0 , ,0 1801

65461

32 aaaaa

So the series for y(x) begins as follows...

term.) theis termnonzero

next that theknowactually (we

...1)(

9

1806

63

x

xy xx

What is the denominator of the x term of this series?

A. 720

B. 1080

C. 1440

D. 4440

E. 12960

9

Application of series (cont.)4. Algebraic equations depending on a parameter ("regular perturbations").

Consider the equation . We could calculate the solutions of this equation using the quadratic formula, but it will be instructive to think of the two solutions (x) as functions of the parameter a .

We will find the power series of one of these functions.

Keep in mind that we are thinking of x as a function of a (and not the other way around)!

Write x = f(a).

First, if a=0, there are two roots, x = 2 and x = -2. We'll calculate the series for the larger root, so f(0)=2.

042 axx

Work it out...

.n higher tha a of powers containing terms theof all havet don' weBecause

04)2()(2

:being) timefor the ... thesupress ll(we' becomes 04equation The

f(a). x Remember, ts.coefficien fewfirst thecalculate and

... 2 f(x)

can write that weassume will weequation, aldifferenti for the did weAs

3

33

221

233

221

2

33

221

a

abababaababab

axx

ababab

Right to left

term)

theis next term theknow we(where ... 2)(f

Therefore 0.finally and , then ,

:get that and timeleft this right to fromWork

0)24()42()42(

:yields ofpower by termscollecting andout gMultiplyin

4

162

3161

221

1

12

2113

2321

2

a

a

bbb

ababbbabbbb

a

aa

Graph

To see how good the approximation is, we plot the "actual" solution (in blue) and our approximation (in red) on the same graph

And….

What is the beginning of the series for the other root?

A max/min problem

During World War II, the blood of thousands of recruits had to be tested for various diseases. The incidence of the diseases was quite low (less than one per hundred). Assume that the probability that a random person has the disease is p (for some p between 0 and 1, but much closer to 0 than to 1). Let q = 1 - p be the probability that the person does not have the disease.

Because the tests were costly, someone had the bright idea to pool several (x) samples at a time. If the pooled sample tested negative, then all x samples were negative. But if the pooled sample tested positive, then each individual sample must be tested again.

max/min

recruits. x for the required be will tests1y that probabilit theis This

.1 is positive testssamples x in the someoney that probabilit The

recruits. x for the required be test willoneonly y that probabilit theis This

. is positive testssamples x in the one-noy that probabilit The

x

q

q

x

x

max/min

A. minimize order toin

for x, solve and zero toequal set this toneed Weln

is x respect to with A(x) of derivative The

q.given A, minimize todArmy wante theObviously,

.1 )(

be llrecruit wiper performed testsofnumber average theSo,

21

111

x

x

xx

x))(x-q(

xq

(q)q

qxAxx

Rewrite

.lim that meanswhich

,10 become toseemsequation the0,pFor

p. offunction a asfor x solve want to weRemember,

0)1ln(

:p-1by q replacing little, aequation thisrewrite llWe'

0

2

112

x

x

xp

p

-p)( x

Undaunted...

we try expanding the terms in powers of p. We know that

ln(1-p) = -p + ... and

If we substitute this much into the equation, we get:

The solution of this is (which is at least consistent with the limit!).

What we have here is an example of an asymptotic series , or, a series of powers other than whole numbers.

We can calculate more coefficients by expanding the equation in higher powers of p.

...1)1(

1 xp

012 px

px 1

Good night….

Don’t Forget:

- office hours

- hand in homework

- problems of the day

See you Monday!