10sTaylorPolySeries.pdf

Commonly Used Taylor Series

series 1 1−x

when is valid/true

=

1 + x + x2 + x3 + x4 + . . .

=



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

∞

xn

x ∈  ( −1, 1)

n=0

e

x

2

=

1 + x + ∞

=

x

=

1 −

=



4

x

+

2!

(−1)n

=

x −

=



x3

3!

x

+

(−1)

(n−1)

n=1

=

x −

=



x2

2

+

∞

(−1)

(n−1)

n=1

x

=

x −

x3

3

+

∞

=

 n=1

1 1 1 2!  + 3!  + 4!  +  . . . ∞

+ ...

4!

e(17x) =

4!

6



(17x) n=0 n!

∞

n

=

 17

n

n

x n!

n=0

−

x

x

+

6!

8!

(−1)(n−1)

x5

− ...

5!

x ∈

−

x7

7!

x9

+

x2n−1 or = (2n − 1)! x3

3

−

xn or = n x5

5

−

x4

4

+

∞



(−1)n

n=0

x5

5

(−1)n+1

n=1

x7

7

x2n−1 or = 2n − 1

+

x9

9

xn n

− ...

∞



(−1)n

n=0

x2n+1 (2n + 1)!

− ...

∞



 x 2n+1 2n + 1

1

R

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

− ...

9!

R

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

8

x2n (2n)!

∞

tan

+

3!

x ∈

n=0

−1

x

n!

∞

ln(1 + x)

x

n

2

sin x

+

2!

so: e  = 1 + 1 +

4

x n=0

cos x

3

x ∈

R

quest question ion:: is y   = ln(1 + x)   even, odd, or neither? x ∈  ( −1, 1]

quest question ion:: is y   = arctan( 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 (n)

and whose derivatives f 

: 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: 

 pN (x) =  f (x0 ) +  f  (x0 )(x − x0 ) +

f  (x0 )

2!

f (N ) (x0 ) (x − x0 ) + · · · + (x − x0 )N  , N ! 2

(open form)

which can also be written as (recall that 0! = 1) f (0) (x0 )

f (1) (x0 )

f (2) (x0 )

f (N ) (x0 ) (x−x0 ) +· · ·+ (x−x0 )N  N ! 2

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

 pN (x) =

N 

 pN (x) =

 f (

n)

(x0 ) (x − x0 )n . n!

n=0

(closed form)

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

 pN (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  c n  is the  n th Taylor coefficient of  y  =  f (x) about  x 0 . 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 N 

 pN (x) =

 f (

n)

n=0

Definition 2.

1

(0) n x . n!

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

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

f  (x0 )

2!

(x − x0 )2 + · · · +

f (n) (x0 ) (x − x0 )n + . . . n!

(open form)

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 +. . . ←  the sum keeps on going and going . n! 0! 1! 2! The Taylor series can also be written in closed form, by using sigma notation, as ∞

P ∞ (x) =

 f (

n)

n=0

(x0 ) (x − x0 )n . n!

(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 ∞

P ∞ (x) =

 f (

n)

n=0

(x0 ) (x − x0 )n 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 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  x 0 is: def 

RN (x) = f (x) − P N (x)

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

(3)

that is f (x) ≈ P N (x)

within an error of  RN (x) .

We often think of all this as: N 

f (x) ≈

 f (

n)

n=0

(x0 ) (x − x0 )n n!

←  a finite sum, the sum stops at N .

We would LIKE TO HAVE THAT ∞

??

f (x) =

 f (

n)

n=0

(x0 ) (x − x0 )n n!

←  the sum keeps on going and going .

In other notation:

??

f (x) ≈ P N (x)

f (x) =  P ∞ (x) and the question is where y  =  P ∞(x) is the Taylor series of  y  =  f (x) at x0 . ?? 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 lim N →∞  of both sides in equation (3), we see that ∞

f (x) =

 f ( n=0

n)

(x0 ) (x − x0 )n n!

←  the sum keeps on going and going .

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) =

 f ( n=0

n)

(x0 ) (x − x0 )n . 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

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 3

Version 1: for a fixed point x  ∈  I  and a fixed N  ∈ N. There exists c  between x  and x0  so that RN (x)

def 

=

f (N +1) (c) (x − x0 )(N +1) . (N  + 1)!

  theorem

f (x) − P N (x)

=

(5)

So either  x  ≤  c  ≤  x 0  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. Note that formula (5) implies that N +1)

|RN (x)|

 ( f 



(c) |x − x0 |(N +1) . (N   + 1)!

=

(6)

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)

i.e.,

 (x)

 max f (

N +1)

c∈I 

Then

|RN (x)| ≤

on the interval  I  ≤ M ,

 (c)

≤ M .

M  +1 |x − x0 |N  (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  { M N }∞  so that N =1





max f (N +1) (c) ≤ M N  c∈I 

for each N  ∈ N  and also so that lim

N →∞

M N  +1 = 0 |x − x0 |N  (N  + 1)!

for each x  ∈  I . Then, by formula (7) and the Squeeze Theorem, lim |RN (x)| = 0

N →∞

for each x  ∈  I . Thus, by So 7, ∞

f (x) =

 f ( n=0

n)

(x0 ) (x − x0 )n n!

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)