Power series representation, I really :-]

[rocomath]

[SOLVED] Power series representation, I really need help! :-]

please let me know if i did this correctly

f(x)=\arctan{(\frac{x}{3})}

f'(x)=\frac{\frac{1}{3}}{1-(-\frac{x^2}{3^2})}

\frac{1}{3}\int[\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{3^{2n}}]dx

\frac{1}{3}\int[1-(\frac{x}{3})^{2}+(\frac{x}{3})^{4}-(\frac{x}{3})^{6}+(\frac{x}{3})^{8}+...]dx

\frac{1}{3}[C+x-\frac{(\frac{x}{3})^{3}}{3}+\frac{(\frac{x}{3})^{5}}{5}-\frac{(\frac{x}{3})^{7}}{7}+\frac{(\frac{x}{3})^{9}}{9}]

C+\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{3^{2n+1}(2n+1)}

 

[Avodyne]

Yes, it's correct! And of course, C=0.

To make Sigma bigger, use \sum instead of \Sigma:

C+\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{3^{2n+1}(2n+1)}

 

[rocomath]

Yes, it's correct! And of course, C=0.

To make Sigma bigger, use \sum instead of \Sigma:

C+\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{3^{2n+1}(2n+1)}

NO WAY! woohoo :-] Thanks, I really appreciate it. This concept is really weird to me, but I'm trying my best.