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




[apcalculus] Re: Help BC 1997 #24
Posted:
Apr 15, 2004 9:17 AM


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 17735083565 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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