Power series representation

  • Thread starter nameVoid
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  • #1
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[tex]
\int \frac{x-arctanx}{x^3}dx
[/tex]
[tex]
\frac{d}{dx}( x-arctanx ) = 1-\frac{1}{1+x^2}=\frac{x^2}{x^2+1}
[/tex]
[tex]
= x^2 \sum_{n=0}^{\infty}(-1)^nx^{2n} = \sum_{n=0}^{\infty}(-1)^nx^{2n+2}
[/tex]
[tex]
\int \sum_{n=0}^{\infty}(-1)^nx^{2n+2} dx = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+3}}{2n+3}+C
[/tex]
[tex]
C=0?
[/tex]
[tex]
\int \frac{\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+3}}{2n+3}}{x^3} dx
[/tex]
[tex]
\int \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{2n+3}}dx
[/tex]
[tex]
\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+3)(2n+1)}}
[/tex]
[tex]
\sum_{n=1}^{\infty}(-1)^n\frac{x^{2n}}{(2n+2)(2n)}}+C
[/tex]
here i took d/dx converted to geomtric then integrated divided by x^3 then integrated again not sure how to deal with the ontants of integration in this case took c=0 on the first integral
 

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  • #2
vela
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[tex]
\int \frac{x-arctanx}{x^3}dx
[/tex]
[tex]
\frac{d}{dx}( x-arctanx ) = 1-\frac{1}{1+x^2}=\frac{x^2}{x^2+1}
[/tex]
[tex]
= x^2 \sum_{n=0}^{\infty}(-1)^nx^{2n} = \sum_{n=0}^{\infty}(-1)^nx^{2n+2}
[/tex]
[tex]
\int \sum_{n=0}^{\infty}(-1)^nx^{2n+2} dx = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+3}}{2n+3}+C
[/tex]
[tex]
C=0?
[/tex]
If you set x=0, you can show that C is, in fact, 0 as you assumed.
[tex]
\int \frac{\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+3}}{2n+3}}{x^3} dx
[/tex]
[tex]
\int \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{2n+3}}dx
[/tex]
[tex]
\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+3)(2n+1)}}
[/tex]
[tex]
\sum_{n=1}^{\infty}(-1)^n\frac{x^{2n}}{(2n+2)(2n)}}+C
[/tex]
here i took d/dx converted to geomtric then integrated divided by x^3 then integrated again not sure how to deal with the ontants of integration in this case took c=0 on the first integral
You're fine up until the last step. Note that in the next-to-the-last line, you have odd powers of x, and in the final line you have only even powers of x. The two expressions aren't equal to each other.
 
  • #3
241
0
right..
 

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