TAYLOR SERIES

CHALLENGE!Construct a polynomial

Sounds hard right? But luckily, the predictability of the differentiation of polynomials helps out here!

2 3 40 1 2 3 4( )P x a a x a x a x a x= + + + +

with the following behavior at x = 0:

(4)

(0) 1(0) 2(0) 3(0) 4(0) 5

PPPPP

=′ =′′ =′′′ =

=

SEE ANY PATTERNS?Hopefully you notice that there is a pattern to the coefficients of the xn term…..

( ) (0)!

nPn

This will come in VERY HANDY in a little bit.

POLYNOMIAL APPROXIMATIONln(1 )x+

Construct a 4th degree polynomial that matches the behavior of at x = 0 through its f irst 4 derivatives.ln(1 )x+

2 3 40 1 2 3 4( )P x a a x a x a x a x= + + + +

POLYNOMIAL APPROXIMATIONln(1 )x+

What did we just do?

• We constructed the 4th order Taylor Polynomial for

• If we kept going, we would improve the approximation near

• The series is called a Taylor Series.

( ) ln(1 ) @ 0f x x x= + =

0x =

SERIES FOR SIN(X) AND COS(X)Construct the 7th order Taylor polynomial and the Taylor series for sin (x) at x = 0.

Recall: The coeff icients should be ( ) (0)

!

nPn

It’s purely coincidental that the 7th order Taylor polynomial also happens to be of 7th degree!

How do we come up with the explicit form to represent the series?

3 5 7 9

0... ???

3! 5! 7! 9! n

x x x xx∞

=

− + − + + =∑

SERIES FOR SIN(X) AND COS(X)

Let’s start by noticing the alternating signs…

How do we come up with the explicit form to represent the series?

3 5 7 9

0

... ???3! 5! 7! 9! n

x x x xx∞

=

− + − + + =∑

Next, let’s try to find a pattern in those exponents in terms of n.Can you find an expression in terms of n that explains that pattern?

nthterm Degree

0 1

1 3

2 5

3 7

2 1

0( 1)

(2 1)!

nn

n

xn

+∞

=

−+∑

Note: 3 terms is sufficient to establish a pattern.

GROUP IT UP!!!Construct the 6th order Taylor polynomial and the Taylor series for cos(x) at x = 0.

Compare your method with that of the other groups. Is there a shortcut?

nthterm Degree

0

1

2

3

MACLAURIN & TAYLOR SERIESDEFINED

If we generalize the steps we followed to construct the coefficients of the power series discussed thus far, we arrive at this definition:

Let f be a function with derivatives of all orders throughout some open interval containing 0. Then the Taylor series generated by f at x = 0 is

(0) ( ) ( )0 1 2

0

(0) (0) (0) (0) (0)( 0) ( 0) ( 0) ... ( 0) ...0! 1! 2! ! !

n kn k

k

f f f f fx x x x xn k

∞

=

′ ′′− + − + − + + − + =∑

This series is also called the Maclaurin series generated by f.This series is also called the Maclaurin series generated by f.

APPROXIMATING A FUNCTIONNEAR ZERO

There are really two ways we can handle this…

( )

!

n nP xn

Find the 4th order Taylor polynomial that approximates near x = 0.cos 2y x=

Using the definition Intuitively

How good is this approximation?

TAYLOR SERIESGENERATED AT X = A

We can match a power series with f in the same way at ANY value x = a, provided we can take the derivatives. Actually, this is where we apply all those “horizontal transformations”!

Let f be a function with derivatives of all orders throughout some open interval containing a. Then the Taylor series generated by f at x = a is

( ) ( )2

0

( ) ( ) ( )( ) ( )( ) ( ) ... ( ) ... ( )2! ! !

n kn k

k

f a f a f af a f a x a x a x a x an k

∞

=

′′′+ − + − + + − + = −∑

The partial sum shown here, is called the Taylor polynomial of order n for f at x = a.

( )

0

( )( ) ( )!

knk

nk

f aP x x ak=

= −∑

FOR EXAMPLE…Find Taylor series generated by ( ) @ 2xf x e x= =

2

0

( 2)!

k

k

e xk

∞

=

⎛ ⎞−⎜ ⎟

⎝ ⎠∑

Note: We can verify this using either differentiation or integration!

OR, FOR EXAMPLE…Find the third order Taylor polynomial for 3 2( ) 2 3 4 5f x x x x= − + −

(a) At x = 0 (b) At x = 1

COMBINING TAYLOR SERIES

On the intersection of their interval of convergence, Taylor series may be added, subtracted, and multiplied by constants and powers of x, and the results are once again Taylor series. The Taylor series for f(x) + g(x) is the sum of the Taylor series for f(x) and the Taylor series for g(x)because the nth derivative of f + g is f(n) + g(n), and so on.

We can obtain the Maclaurin series for by

substituting 2x in the Maclaurin series for cos x, adding 1, and dividing the result by 2. The Maclaurin series for sinx + cos x is the term-by-term sum of the series for sin x and cos x. We obtain the Maclaurin series for xsinx by multiplying all the terms of the Maclaurin series for sin x by x.

(1 cos 2 )2

x+

SERIES YOU MUST MEMORIZE!

2

0

2

0

2

0

3 5 2 1 2 1

0

1 1 ... ... ( 1)1

1 1 ... ( ) ... ( 1) ( 1)1

1 ... ... ( )2! ! !

sin ... ( 1) ... ( 1) ( )3! 5! (2 1)! (2 1)!

cos

n n

n

n n n

n

n nx

n

n nn n

n

x x x x xx

x x x x xx

x x xe x all real xn n

x x x xx x all real xn n

∞

=

∞

=

∞

=

+ +∞

=

= + + + + + = <−

= − + − + − + = − <+

= + + + + + =

= − + − + − + = −+ +

∑

∑

∑

∑2 4 2 2

0

1 ... ( 1) ... ( 1) ( )2! 4! (2 )! (2 )!

n nn n

n

x x x xx all real xn n

∞

=

= − + − + − + = −∑

Series you might

want to know(not as vital though!)

2 31 1

1

3 5 2 1 2 11

0

ln(1 ) ... ( 1) ... ( 1) ( 1 1)2 3

tan ... ( 1) ... ( 1) ( 1)3 5 2 1 2 1

n nn n

n

n nn n

n

x x x xx x xn n

x x x xx x xn n

∞− −

=

+ +∞−

=

+ = − + − + − + = − − ≤ ≤

= − + − + − + = − ≤+ +

∑

∑

2004 FORM B BC2 (Parts a, b, and c only)

2004 FORM B BC2 ANSWERS

2005 FORM B BC3 (Parts a and b only)

2005 FORM B BC3 ANSWERS

2006 FORM B BC6 (Parts a, b, and c only)

2006 FORM B BC6ANSWERS