Re: Calculus problem


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Re: Calculus problem



Well...  the integral of sin(3x^2) using a Maclaurin series is:

(infinity)
__   (-1)^n * 3^(2n+1) * x^(4n+2)
\    ---------------------------- + c
/_         (2n+1)! * (4n+3)
n=0

Assuming I didn't make an error somewhere.
---
Andy Johnson
Give me ambiguity or give me something else.

On Sun, 12 Apr 1998 23:39:38 -0400 Nicholas P Konidaris Ii
<npk+@ANDREW.CMU.EDU> writes:
>>>    Someone please integrate this by hand: sin(3x^2)
>>I get -cos(x^3) in both cases.  Do you have a problem with that?
>
>Yeah, I do: It's wrong.
>
>The integral of sin(3x^2) is hard to do.  If you need an exact
>solution,
>i do not think that it is possible.  The best you can get is a numeric
>solution anyways.
>
>The best I can tell you is that the integral converges to 1/2 as
>x->infinity.
>
>nick
>

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