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Topic: [ap-calculus] Re: Help BC 1997 #24
Replies: 1   Last Post: Apr 15, 2004 9:17 AM

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Richard J Maher

Posts: 80
Registered: 12/6/04
[ap-calculus] Re: Help BC 1997 #24
Posted: Apr 15, 2004 9:17 AM
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Hello Art,

Given the Taylor Series for sin(x), the series for sin(x^2) is
x^2 -x^6/3! + x^10/5! + .... Since sin(x) equals its Taylor
Series on it interval of convergence, which is all of R,
so too f'(x) = sin(x^2) = x^2 -x^6/3! + x^10/5! + ... on R.
(This essentially is composition of functions.) When you integrate
you get f(x) = x^3/3 - x^7/(7*(3!)) + x^11/(11*(5!)) + ...
Hence the -1/42 appears as the coefficient for x^7.

Hope this helps.

Dick Maher

Richard J. Maher
Mathematics and Statistics
Loyola University Chicago
6525 N. Sheridan Rd.
Chicago, Illinois 60626
1-773-508-3565
rjm@math.luc.edu

On Wed, 14 Apr 2004, Art Stahl wrote:

The Taylor series for sin x about x = 0 is x - x^3/3! + x^5/5! -
... If f is a function such that f '(x)=sin(x^2) then the
coefficient of x^7 in the Taylor series for f (x) about x = 0 is

a) 1/7! b) 1/7 c) 0 d) -1/42 (correct answer) e)
-1/7!
I got the answer but my solution is verrrrrry long. I took the 7th
derivative (6th past f ') and evaluated it at at x = 0 and divided by
7!.. It worked!!

Thanks,

Art Stahl


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