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Commonly Used Taylor Series

series when is valid/true 1

1x= 1 +x+x2+x3+x4+:::note this is the geometric series.

just think ofxasr= 1X n=0x nx2(1;1)e x= 1 +x+x22! +x33! +x44! +:::so: e= 1 + 1 +12! +13! +14! e (17x)=P1 n=0(17x)nn!=1X n=017 nxnn!= 1X n=0x nn!x2Rcosx= 1x22! +x44! x66! +x88! :::notey= cosxis an evenfunction (i.e.,cos(x) = +cos(x)) and the taylor seris ofy= cosxhas only evenpowers. 1X n=0(1)nx2n(2n)!x2Rsinx=xx33! +x55! x77! +x99! :::notey= sinxis an oddfunction (i.e.,sin(x) =sin(x)) and the taylor seris ofy= sinxhas only oddpowers. 1X n=1(1)(n1)x2n1(2n1)!or=1X n=0(1)nx2n+1(2n+ 1)!x2Rln(1 +x) =xx22 +x33 x44 +x55 :::question: isy= ln(1 +x)even, odd, or neither?= 1X n=1(1)(n1)xnn or=1X n=1(1)n+1xnn x2(1;1]tan

1x=xx33

+x55 x77 +x99 :::question: isy= arctan(x)even, odd, or neither?= 1X n=1(1)(n1)x2n12n1or=1X n=0(1)nx2n+12n+ 1x2[1;1]1

Math 142Taylor/Maclaurin Polynomials and SeriesProf. GirardiFix an intervalIin the real line (e.g.,Imight be (17;19)) and letx0be a point inI, i.e.,

x 02I :

Next consider a function, whose domain isI,

f:I!R and whose derivativesf(n):I!Rexist on the intervalIforn= 1;2;3;:::;N. Denition 1.TheNth-order Taylor polynomialfory=f(x) atx0is: p

N(x) =f(x0) +f0(x0)(xx0) +f00(x0)2!

(xx0)2++f(N)(x0)N!(xx0)N;(open form) which can also be written as (recall that 0! = 1) p

N(x) =f(0)(x0)0!

+f(1)(x0)1! (xx0)+f(2)(x0)2! (xx0)2++f(N)(x0)N!(xx0)N -a nite sum, i.e. the sum stops:

Formula (open form) is in open form. It can also be written in closed form, by using sigma notation, as

p

N(x) =NX

n=0f (n)(x0)n!(xx0)n:(closed form) Soy=pN(x) is a polynomial of degree at mostNand it has the form p

N(x) =NX

n=0c n(xx0)nwhere the constantsc n=f(n)(x0)n! are specially chosen so that derivatives match up atx0, i.e. the constantscn's are chosen so that: p

N(x0) =f(x0)

p (1)

N(x0) =f(1)(x0)

p (2)

N(x0) =f(2)(x0)

p (N)

N(x0) =f(N)(x0):

The constantcnis thenthTaylor coecientofy=f(x) aboutx0. TheNth-order Maclaurin polynomialfory=f(x) is just theNth-order Taylor polynomial fory=f(x) atx0= 0 and so it is p

N(x) =NX

n=0f (n)(0)n!xn:

Denition 2.

1TheTaylor seriesfory=f(x) atx0is the power series:

P

1(x) =f(x0) +f0(x0)(xx0) +f00(x0)2!

(xx0)2++f(n)(x0)n!(xx0)n+:::(open form) which can also be written as P

1(x) =f(0)(x0)0!

+f(1)(x0)1! (xx0)+f(2)(x0)2! (xx0)2++f(n)(x0)n!(xx0)n+::: -the sum keeps on going and going: The Taylor series can also be written in closed form, by using sigma notation, as P

1(x) =1X

n=0f (n)(x0)n!(xx0)n:(closed form) TheMaclaurin seriesfory=f(x) is just the Taylor series fory=f(x) atx0= 0.1

Here we are assuming that the derivativesy=f(n)(x) exist for eachxin the intervalIand for eachn2N f1;2;3;4;5;:::g.

2 Big Questions 3.For what values ofxdoes the power (a.k.a. Taylor) series P

1(x) =1X

n=0f (n)(x0)n!(xx0)n(1)

converge (usually the Root or Ratio test helps us out with this question). If the power/Taylor series in formula (1)

does indeed converge at a pointx, does the series converge to what we would want it to converge to, i.e., does

f(x) =P1(x) ? (2)

Question (2) is going to take some thought.

Denition 4.TheNth-order Remainder termfory=f(x) atx0is: R

N(x)def=f(x)PN(x)

wherey=PN(x) is theNth-order Taylor polynomial fory=f(x) atx0. So f(x) =PN(x) +RN(x) (3) that is f(x)PN(x) within an error ofRN(x):

We often think of all this as:

f(x)NX n=0f (n)(x0)n!(xx0)n -a nite sum, the sum stops atN :

We would LIKE TO HAVE THAT

f(x)??=1X n=0f (n)(x0)n!(xx0)n -the sum keeps on going and going:

In other notation:

f(x)PN(x) and the question isf(x)??=P1(x) wherey=P1(x) is the Taylor series ofy=f(x) atx0. Well, let's think about what needs to be forf(x)??=P1(x), i.e., forfto equal to its Taylor series. Notice 5.Taking the limN!1of both sides in equation (3), we see that f(x) =1X n=0f (n)(x0)n!(xx0)n -the sum keeps on going and going: if and only if limN!1RN(x) = 0: Recall 6.limN!1RN(x) = 0 if and only if limN!1jRN(x)j= 0 .

So 7.If

limN!1jRN(x)j= 0 (4) then f(x) =1X n=0f (n)(x0)n!(xx0)n:

So we basically want to show that (4) holds true.How to do this? Well, this is where Mr. Taylor comes to the rescue!2

2

According to Mr. Taylor, his Remainder Theorem (see next page) was motivated by coeehouse conversations about works of Newton

on planetary motion and works of Halley (ofHalley's comet) on roots of polynomials. 3

Taylor's Remainder Theorem

Version 1: for a xed pointx2Iand a xedN2N.3

There existscbetweenxandx0so that

R N(x)def=f(x)PN(x)theorem=f(N+1)(c)(N+ 1)!(xx0)(N+1):(5) So eitherxcx0orx0cx. So we do not know exactly whatcis but atleast we know thatcis betweenxandx0 and soc2I.

Remark: This is aBig Theoremby Taylor. See the book for the proof. The proof uses the Mean Value Theorem.

Note that formula (5) implies that

jRN(x)j=f(N+1)(c)(N+ 1)!jxx0j(N+1):(6)

Version 2: for the whole intervalIand a xedN2N.3

Assume we can ndMso that

the maximum off(N+1)(x)on the intervalIM ; i.e., max c2I f(N+1)(c)M : Then jRN(x)j M(N+ 1)!jxx0jN+1(7) for eachx2I.

Remark: This follows from formula (6).

Version 3: for the whole intervalIand allN2N.4

Now assume that we can nd a sequencefMNg1N=1so that max c2I f(N+1)(c)MN for eachN2Nand also so that lim N!1M

N(N+ 1)!jxx0jN+1= 0

for eachx2I. Then, by formula (7) and the Squeeze Theorem, lim

N!1jRN(x)j= 0

for eachx2I. Thus, by So 7, f(x) =1X n=0f (n)(x0)n!(xx0)n for eachx2I.3 Here we assume that the (N+ 1)-derivative ofy=f(x), i.e.y=f(N+1)(x), exists for eachx2I.

4Here we assume thaty=f(N)(x), exists for eachx2Iand eachN2N.

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