Power series for arctan(x/3)???

**pianopiano** · April 12th 2009, 08:42 PM

How does one go about finding the power series for this? I know that the derivative of arctan(x) = 1/(1+x^2), so if I think I can find the power series for arctan(x), which would just be integrating each term of the power series for 1/(1+x^2), but that (x/3) with the "/3" is throwing me off. How do I do it? Thanks.

**TheEmptySet** · April 12th 2009, 08:52 PM

Originally Posted by pianopiano

How does one go about finding the power series for this? I know that the derivative of arctan(x) = 1/(1+x^2), so if I think I can find the power series for arctan(x), which would just be integrating each term of the power series for 1/(1+x^2), but that (x/3) with the "/3" is throwing me off. How do I do it? Thanks.

$f(x)=\tan^{-1}\left( \frac{x}{3}\right)$

Now we can take a derivative to get

$\frac{df}{dx}=\frac{1}{3}\frac{1}{1+\left( \frac{x}{3}\right)^2}=\frac{1}{3}\sum_{n=0}^{\inft y}\left( -\frac{x}{3}\right)^{2n}=\sum_{n=0}^{\infty}\frac{x ^{2n}}{3^{2n+1}}$

Now we can integrate to find f

$f(x)=\int f'(x)dx=\int \sum_{n=0}^{\infty}\frac{x^{2n}}{3^{2n+1}}dx=\sum_ {n=0}^{\infty}\frac{x^{2n+1}}{3^{2n+1}(2n+1) }$

**pianopiano** · April 12th 2009, 09:11 PM

Thanks, but what happened to (x/3)^2? When you put it into the series it became (-x/3)^n, but why isn't it [(-x^2)/9]^n since (-x/3) was squared?

**TheEmptySet** · April 13th 2009, 07:51 AM

Originally Posted by pianopiano

Thanks, but what happened to (x/3)^2? When you put it into the series it became (-x/3)^n, but why isn't it [(-x^2)/9]^n since (-x/3) was squared?

Yes you are correct, I made a mistake

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