# Is there an easy way to expand $\arctan(1+x)$ at point $x=0$

I want to expand $\arctan(1+x)$ into Taylor series at the point $x=0$, but I do not want to calculate its higher order derivatives. Is there an easy way to do this, just as we done for $\arctan(x)$ at $x=0$? I know the result should be like this: $\frac{\pi}{4} +{\frac {1}{2}}x-{\frac {1}{4}}{x}^{2}+{\frac {1}{12}}{x}^{3 }-{\frac {1}{40}}{x}^{5}+{\frac {1}{48}}{x}^{6}-{\frac {1}{112}}{x}^{7 }+{\frac {1}{288}}{x}^{9}+O \left( {x}^{10} \right)$. But actually calculating the higher order derivatives is a boring work. Please give me some hints. Thanks a lot.

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You have to differentiate it only once: $$\frac 1{(x+1)^2+1}$$ then consider its Taylor series, using $\frac1{1-z}$, then integrate it back.

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The verb corresponding to "derivative" is "differentiate". – joriki Sep 21 '12 at 11:21
Thanks, corrected – Berci Sep 21 '12 at 11:40

@BerciPecsi's suggestion is not usable for me, for I can not find a better way to estimate the order of $O((1+x)^n)$ for some $n\in\mathbb{N}$.

Actually I find a way (not so easy) to expand $\arctan(1+x)$ into Taylor's expansion with Peano's remainder at the point $x=0$. Here are the details:

First, write $\arctan(1+x)$ as follows:

\begin{gather*} \arctan(1+x)=\frac{\pi}{4}+\arctan\left(\frac{x}{2+x}\right). \end{gather*}

Then, noting that $\lim_{x\to0} \frac{x}{2+x}=0$, and

$$\frac{x}{2+x}=\frac{1}{1+x/2}\cdot(x/2)$$

leads to the expected results, and what you need to do is just do expand $1/(1+x/2)$ and $\arctan(t)$, as $t\to 0$.

I think this method is still wordy, but it is not so boring than that of using the primitive formula with calculating the high order derivatives of $\arctan(1+x)$ at $x=0$.

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