10s Taylor Poly Series

Commonly Used Taylor Series

series 1 1−x

= =

when is valid/true

note this is the geometric series. just think of x as r

1 + x + x2 + x3 + x4 + . . . ∞ X

xn

x ∈ (−1, 1)

n=0

2

ex

=

1 + x +

3

so: 1 1 1 e = 1 + 1 + 2! + 3! + 4! + ... ∞ X P∞ (17x)n 17n xn (17x) e = n=0 n! = n!

4

x x x + + + ... 2! 3! 4!

n=0

cos x

∞ X

=

xn n! n=0

=

x4 x6 x8 x2 + − + − ... 1 − 2! 4! 6! 8!

=

∞ X

(−1)n

n=0

sin x

=

=

x∈R

x2n (2n)!

(−1)(n−1)

n=1

ln (1 + x)

= =

x − ∞ X

=

(−1)(n−1)

=

x − ∞ X

∞ xn or X xn = (−1)n+1 n n n=1

x3 x5 x7 x9 + − + − ... 3 5 7 9

(−1)(n−1)

n=1

∞ x2n−1 or X x2n+1 = (−1)n (2n − 1)! (2n + 1)! n=0

x2 x3 x4 x5 + − + − ... 2 3 4 5

n=1

tan−1 x

x∈R

x5 x7 x9 x3 + − + − ... x − 3! 5! 7! 9! ∞ X

note y = cos x is an even function (i.e., cos(−x) = + cos(x)) and the taylor seris of y = cos x has only even powers.

∞ x2n−1 or X x2n+1 = (−1)n 2n − 1 2n + 1 n=0

1

note y = sin x is an odd function (i.e., sin(−x) = − sin(x)) and the taylor seris of y = sin x has only odd powers.

x∈R

question: is y = ln(1 + x) even, odd, or neither?

x ∈ (−1, 1]

question: is y = arctan(x) even, odd, or neither?

x ∈ [−1, 1]

Math 142

Taylor/Maclaurin Polynomials and Series

Prof. Girardi

Fix an interval I in the real line (e.g., I might be (−17, 19)) and let x0 be a point in I, i.e., x0 ∈ I . Next consider a function, whose domain is I, f: I →R and whose derivatives f (n) : I → R exist on the interval I for n = 1, 2, 3, . . . , N . Definition 1. The N th -order Taylor polynomial for y = f (x) at x0 is: f (N ) (x0 ) f 00 (x0 ) (x − x0 )2 + · · · + (x − x0 )N , 2! N! which can also be written as (recall that 0! = 1) pN (x) = f (x0 ) + f 0 (x0 )(x − x0 ) +

(open form)

f (0) (x0 ) f (1) (x0 ) f (2) (x0 ) f (N ) (x0 ) + (x−x0 )+ (x−x0 )2 +· · ·+ (x−x0 )N ←- a finite sum, i.e. the sum stops . 0! 1! 2! N! Formula (open form) is in open form. It can also be written in closed form, by using sigma notation, as

pN (x) =

pN (x) =

N X f (n) (x0 ) (x − x0 )n . n! n=0

(closed form)

So y = pN (x) is a polynomial of degree at most N and it has the form pN (x) =

N X

cn (x − x0 )n

where the constants

n=0

cn =

f (n) (x0 ) n!

are specially chosen so that derivatives match up at x0 , i.e. the constants cn ’s are chosen so that: pN (x0 ) = f (x0 ) (1)

pN (x0 ) = f (1) (x0 ) (2)

pN (x0 ) = f (2) (x0 ) .. . (N )

pN (x0 ) = f (N ) (x0 ) . The constant cn is the nth Taylor coefficient of y = f (x) about x0 . The N th -order Maclaurin polynomial for y = f (x) is just the N th -order Taylor polynomial for y = f (x) at x0 = 0 and so it is pN (x) = Definition 2.

1

N X f (n) (0) n x . n! n=0

The Taylor series for y = f (x) at x0 is the power series:

P∞ (x) = f (x0 ) + f 0 (x0 )(x − x0 ) +

f (n) (x0 ) f 00 (x0 ) (x − x0 )2 + · · · + (x − x0 )n + . . . 2! n!

which can also be written as f (0) (x0 ) f (1) (x0 ) f (2) (x0 ) f (n) (x0 ) P∞ (x) = + (x−x0 )+ (x−x0 )2 +· · ·+ (x−x0 )n +. . . 0! 1! 2! n! The Taylor series can also be written in closed form, by using sigma notation, as ∞ X f (n) (x0 ) P∞ (x) = (x − x0 )n . n! n=0

(open form)

←- the sum keeps on going and going.

(closed form)

The Maclaurin series for y = f (x) is just the Taylor series for y = f (x) at x0 = 0. 1Here we are assuming that the derivatives y = f (n) (x) exist for each x in the interval I and for each n ∈ N ≡ {1, 2, 3, 4, 5, . . . } .

2

Big Questions 3. For what values of x does the power (a.k.a. Taylor) series ∞ X f (n) (x0 ) P∞ (x) = (x − x0 )n n! n=0

(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 point x, does the series converge to what we would want it to converge to, i.e., does f (x) = P∞ (x) ?

(2)

Question (2) is going to take some thought. Definition 4. The N th -order Remainder term for y = f (x) at x0 is: def

RN (x) = f (x) − PN (x) where y = PN (x) is the N th -order Taylor polynomial for y = f (x) at x0 . So f (x) = PN (x) + RN (x)

(3)

that is f (x) ≈ PN (x) We often think of all this as: N X f (n) (x0 ) (x − x0 )n f (x) ≈ n! n=0

within an error of RN (x) .

←- a finite sum, the sum stops at N .

We would LIKE TO HAVE THAT ∞ X f (n) (x0 ) ?? f (x) = (x − x0 )n n! n=0

←- the sum keeps on going and going .

In other notation: f (x) ≈ PN (x) and the question is where y = P∞ (x) is the Taylor series of y = f (x) at x0 .

??

f (x) = P∞ (x)

??

Well, let’s think about what needs to be for f (x) = P∞ (x), i.e., for f to equal to its Taylor series. Notice 5. Taking the limN →∞ of both sides in equation (3), we see that ∞ X f (n) (x0 ) f (x) = (x − x0 )n ←- the sum keeps on going and going . n! n=0 if and only if lim RN (x) = 0 .

N →∞

Recall 6. limN →∞ RN (x) = 0 if and only if limN →∞ |RN (x)| = 0 . So 7. If lim |RN (x)| = 0

(4)

N →∞

then f (x) =

∞ X f (n) (x0 ) (x − x0 )n . n! n=0

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

2According to Mr. Taylor, his Remainder Theorem (see next page) was motivated by coffeehouse conversations about works of Newton

on planetary motion and works of Halley (of Halley’s comet) on roots of polynomials.

3

Taylor’s Remainder Theorem Version 1: for a fixed point x ∈ I and a fixed N ∈ N. There exists c between x and x0 so that

3

f (N +1) (c) (x − x0 )(N +1) . (5) (N + 1)! So either x ≤ c ≤ x0 or x0 ≤ c ≤ x. So we do not know exactly what c is but atleast we know that c is between x and x0 and so c ∈ I. Remark: This is a Big Theorem by Taylor. See the book for the proof. The proof uses the Mean Value Theorem. RN (x)

def

=

theorem

f (x) − PN (x)

=

Note that formula (5) implies that |RN (x)|

(N +1) f (c)

=

(N + 1)!

|x − x0 |

(N +1)

Version 2: for the whole interval I and a fixed N ∈ N.3 Assume we can find M so that the maximum of f (N +1) (x) on the interval I

.

(6)

≤ M ,

i.e., max f (N +1) (c) ≤ M . c∈I

Then |RN (x)| ≤

M N +1 |x − x0 | (N + 1)!

for each x ∈ I. Remark: This follows from formula (6). Version 3: for the whole interval I and all N ∈ N. 4 Now assume that we can find a sequence {MN }∞ N =1 so that max f (N +1) (c) ≤ MN c∈I

for each N ∈ N and also so that

MN N +1 |x − x0 | = 0 N →∞ (N + 1)! for each x ∈ I. Then, by formula (7) and the Squeeze Theorem, lim

lim |RN (x)| = 0

N →∞

for each x ∈ I. Thus, by So 7, f (x) =

∞ X f (n) (x0 ) (x − x0 )n n! n=0

for each x ∈ I.

3Here we assume that the (N + 1)-derivative of y = f (x), i.e. y = f (N +1) (x), exists for each x ∈ I. 4Here we assume that y = f (N ) (x), exists for each x ∈ I and each N ∈ N.

4

(7)