Integration maclariun series

However, you can integrate the terms within the sigma notation, which in this case is the Maclaurin series for sin(x).In summary, the person was having trouble integrating the Maclaurin series for sin(x) and was unsure if it was integrable. They also mentioned trying various integration methods such as u-substitution, but were unable to find a solution. The other person explained that since the series is equal to sin(x), it is indeed integrable and can be integrated term by term. They also mentioned that the person's expertise in other areas suggests that they should have no trouble integrating the series.
  • #1
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i was stumbled on a problem was i was looking over my book...i can do things like, differntial geometry, Fourier analsysis, advanced calculus, calculus etc...but i can;t figure out this problem...i don't think i can figure it out...it's from the maclariun series...as most of you know, sinx=x-x^3/3!+x^5/5!-x^7/7!+..=sigma ^infinity, n=0 (-1)^nx^2n+1/ (2n+1)!...at sin(.4)...i don;t think this is integratable...i tried everything from integration to u-substituion but it won't work...
 
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  • #2
Would it be too much trouble to tell us what the problem is?

You say "it's from the maclariun series...as most of you know, sinx=x-x^3/3!+x^5/5!-x^7/7!+..=sigma ^infinity, n=0 (-1)^nx^2n+1/ (2n+1)!...at sin(.4)...i don;t think this is integratable" so I take it you want to integrate? Integrate what? Integrate sin(x) or its Maclaurin series? What do you mean "at sin(.4)"?

Certainly sin(x) is integrable. Since the Maclaurin series for sin(x) is equal to sin(x), the Maclaurin series is. What exactly is it that you want to do?
 
  • #3
integrate the sigma part
 
  • #4
"integrate the sigma part"

That was an answer? "The sigma part" of what?

Assuming that you mean "integrate the Maclaurin series for sin(x)", the point I made before was that since the series is equal to sin(x) so it certainly is integrable. Its integral is the same as the integral of sin(x): cos(x).

You can also, of course, do it term by term: the Mclaurin series for sin(x) is, as you say x- (1/3!)x3+...+ (1/(2n+1)!)x2n+1+ ... Integrating term by term gives
(1/2)x2- (1/4!)x4+ ...+ (1/(2n+2)!)x2n+2+ ... which is, of course, the Maclaurin series for cos(x).

If you can do things like "differntial geometry, Fourier analsysis, advanced calculus", I don't see why you would think sin(x) was not integrable.
 
  • #5
no was talking about the sigma notation part but o well ok
 
  • #6
you can integrate the sigma notation no more than you can integrate the + symbol.
 

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