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Find the Maclaurin series for the function $\tan^{-1}(2x^2)$

Express your answer in sigma notation, simplified as much as possible. What is the open interval of convergence of the series.

I have the correct answer, but I would like to use another method to solve this.

By taking the function and its derivative, find the sum and then taking the anti derivative. This does not yield the same answer for me.

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Actually, when seeing these kind of functions, don't ever imagine to do series expansion by Taylor upon the whole function. Divide-and-conquer is a good tactics. See @AndréNicolas. –  FrenzY DT. Dec 1 '12 at 3:25
    
@Julian Kuelshammer - thank you, noted. This is not homework, I have the correct answer but I would like to use another method to solve it –  Mike Beta Dec 1 '12 at 3:29

1 Answer 1

up vote 2 down vote accepted

You probably know the series for $\tan^{-1} t$. Plug in $2x^2$ for $t$.

If you do not know the series for $\arctan t$, you undoubtedly know the series for $\dfrac{1}{1-u}$. Set $u=-x^2$, and integrate term by term.

For the interval of convergence of the series for $\tan^{-1} (2x^2)$, you probably know when the series for $\tan^{-1}t$ converges. That knowledge can be readily translated to knowledge about $\tan^{-1}(2x^2)$.

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I found the answer this way, however I am trying to figure this out by taking the derivative of the function and solving the series and then taking the anti-derivative to solve this –  Mike Beta Dec 1 '12 at 3:25
    
Getting an explicit expression for the $n$-th derivative of $\tan^{-1}(2x^2)$ is quite unpleasant. If you really want to do it that way, you can get a recurrence that expresses the $n+1$-th derivative in terms of the $n$-th derivative. That will get you a recurrence for the values at $0$. If you have some experience with recurrences, you can prove by induction that the $n$-th derivative at $0$ is what it is. It helps to know the answer, which you do by the substitution process described above. –  André Nicolas Dec 1 '12 at 3:35

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