TAYLOR SERIES CHALLENGE!

234 Construct a polynomial P()x=+ a01 ax + ax 2 + ax 3 + ax 4 with the following behavior at x = 0: P(0)= 1 P′(0)= 2 P′′(0)= 3 P′′′(0)= 4 P(4)(0)= 5

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

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

P()n (0) n!

This will come in VERY HANDY in a little bit. POLYNOMIAL APPROXIMATION ln(1+ x )

th 234 Construct a 4 degree polynomial Px()=+ a01 ax + ax 2 + ax 3 + ax 4 that matches the behavior of ln(1 + x ) at x = 0 through its first 4 derivatives. POLYNOMIAL APPROXIMATION ln(1+ x )

What did we just do?

• We constructed the 4th order Taylor Polynomial for fx()= ln(1+= x )@ x 0 • If we kept going, we would improve the approximation near x = 0 • The series is called a Taylor Series. SERIES FOR SIN(X) AND COS(X)

Construct the 7th order Taylor polynomial and the Taylor series for sin (x) at x = 0.

()n Recall: The coefficients should be P (0) n!

How do we come up with the explicit form to represent the series? xxxx3579 ∞ x −+−++=...∑ ??? 3! 5! 7! 9! n=0

It’s purely coincidental that the 7th order Taylor polynomial also happens to be of 7th degree! SERIES FOR SIN(X) AND COS(X)

How do we come up with the explicit form to represent the series? xxxx3579 ∞ x −+−++=...∑ ??? 3! 5! 7! 9! n=0

Let’s start by noticing the alternating signs… 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? nth term Degree 01 13 ∞ x21n+ (1)− n 25 ∑ 37 n=0 (2n + 1)!

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?

nth term Degree 0 1 2 3 MACLAURIN & TAYLOR SERIES DEFINED 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 fff(0) (0)′′′ (0) (0) f()nk (0)∞ f () (0) (0)x −+01 (0)xx −+ (0)... −++ 2 (0)... x −+=nk∑ x 0! 1! 2!nk !k =0 !

This series is also called the Maclaurin series generated by f. APPROXIMATING A FUNCTION NEAR ZERO

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

There are really two ways we can handle this…

Using the definition Intuitively P()nnx n!

How good is this approximation? TAYLOR SERIES GENERATED 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

fa′′() f()nk () a∞ f () () a f ()()()()...()...afaxa+−+−++−+=−′ xa2 xank∑ () xa 2!nk !k =0 !

The partial sum shown here, is n ()k called the Taylor polynomial of fa() k Pxn ()=−∑ ( x a ) order n for f at x = a. k =0 k! FOR EXAMPLE…

Find Taylor series generated by fx()= ex @ x= 2

∞ 2 Note: We can verify this using either ⎛⎞e k differentiation or integration! ∑⎜⎟(2)x − k =0 ⎝⎠k! OR, FOR EXAMPLE…

Find the third order Taylor polynomial for f ()xxxx=−+− 232 3 4 5

(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 (1 + cos 2 x ) by 2 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. SERIES YOU MUST MEMORIZE!

1 ∞ =+1xx +2 + ... + xnn + ... =∑ x ( x < 1) 1− x n=0 1 ∞ =−1xx +2 − ... +− ( x )nnn + ... =∑ ( − 1) x ( x < 1) 1+ x n=0 xx2 nn∞ x exx =+1 + + ... + + ... =∑ ( allrealx ) 2!nn !n=0 ! xx35 x 21nn++∞ x 21 sinxx=− + −+− ...(1)nn + ... =∑ (1) − ( allrealx ) 3! 5! (2nn++ 1)!n=0 (2 1)! xx24 x 2nn∞ x 2 cos xallrealx=−1 + − ... +− ( 1)nn + ... =∑ ( − 1) ( ) 2! 4! (2nn )!n=0 (2 )! Series you might want to know (not as vital though!)

xx23 xnn∞ x ln(1+=−+xx ) −+− ... ( 1)nn−−11 += ...∑ ( − 1) ( −≤≤ 1 x 1) 23 nnn=1

xx35 x 21nn++∞ x 21 tan−1 xx=− + −+− ... ( 1)nn + ... =∑ ( − 1) ( x ≤ 1) 35 21nn++n=0 21 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