Calculus I – Math 104 The end is near!. 1. Limits: Series give a good idea of the behavior of functions in the neighborhood of 0: We know for other reasons.

Calculus I – Math 104

The end is near!

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

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

x

fxf

b-a

afbf

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

k

fa

(x)Rx a ...xaxaxaax

(x)R

x

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 xn

somewherefxR

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

xx

x

xxx

5th derivative

For 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.

!5)(

5

4

xxR

Your turn...

places? decimal 10 toget to together add to

need wedo series theof many terms How

-- aroundquestion turn theNow

? 1.6458333 !3

5.

!2

5.5.1

ion approximat theis accurate How32

5.

e

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 becomes

error theThus, infinity. togoesn as zeroapproach will

quantity this x,of any valuefor and --)!1(

)(

n

nn

n

n

n

xx

n

xxR

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

axa

axa

axaax

Maclaurin

Series 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

kpk

p

k

p

k

p

xk

px

If p is not a positive integer...

... 11

1 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+1

n+1na + 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 aaa

xxxx ...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

0102

xxx

aa

xaaaxxx

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

11

2

xxxx

xx

Then use partial fractions to write:

(almost)! done be will we and

for series get thecan weif So1

1

))((1

))((1

))((1

x

x

xxxx

Work it out...

...

...)1(

...

...)1(

3

3

2

3

3

2

1

321111

1

321111

xx

xxxx

xx

xxxx

x

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 the

formula aobtain will we and of esknown valu in theput weIf

thatus tellThis

...))()()(()(f 34423322

51

xxxx

5

)1( )1()1( nnn

na