Integrating infinite sums and macluarin's expansion

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Homework Help Overview

The discussion revolves around the integration of the sine function using its Maclaurin series expansion. Participants are exploring the relationship between the infinite series representation of sin(x) and its integral.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to integrate the Maclaurin series for sin(x) and are questioning the correctness of their series manipulation. There are discussions about the proper form of the series after integration and the resulting terms.

Discussion Status

Some participants have offered corrections regarding the factorials in the series, while others are exploring how to manipulate the series to arrive at the expected result. The conversation is ongoing, with multiple interpretations being considered.

Contextual Notes

There is mention of a typo affecting the series representation, and participants are working within the constraints of the homework assignment, which may impose specific expectations on the form of the answer.

rock.freak667
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Homework Statement


Using the macluarin's expansion for sinx show that [itex]\int sinx dx=-cosx+c[/itex]

Homework Equations



[tex]sinx=\sum_{n=0} ^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}[/tex]

The Attempt at a Solution


Well I can easily write out some of the series and just show that it is equal to -cosx

but if I integrate the representation for the infinite series i get

[tex]\int sinx dx= \sum_{n=0} ^\infty \frac{(-1)^nx^{2n+2}}{(2n+1)!}[/tex]
shouldn't -cosx be:
[tex]\int sinx dx= \sum_{n=0} ^\infty \frac{(-1)^{n+1}x^{2n+2}}{(2n+1)!}[/tex]

and also I am supposed to get [itex]x^{2n}[/itex] not what I got
 
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Your denominators should be (2n+2)! in the last two lines.

With this correction, your last series is cos(x)-1; write out the first few terms to see.
 
ah yes I made a typo...but even if I change it
how do I manipulate

[tex]\sum_{n=0} ^\infty \frac{(-1)^nx^{2n+2}}{(2n+2)!}[/tex] (what I integrated and got)

Into

[tex]-\sum_{n=0} ^\infty \frac{(-1)^nx^{2n}}{(2n)!}[/tex] (What I am supposed to get)
 
Let j= n+1 and see what you get with j as the index.
 

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