<s>[INST] arctan series multisection by roots of unity I'm trying to multisect the series for $\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \cdots$ using the method of roots of unity, as described in the paper linked in one of the comments in How to express a power series in closed form. I want to take every second term of this series. But I think that I'm doing something wrong, because I get some weird results :q
Here's what I do:
First, since I'll be taking every 2nd term, I need square roots of unity:
$\omega = e^{i\frac{2\pi}{2}} = e^{i\pi} = -1  \\
\omega^0 = (-1)^0 = 1   \\
\omega^1 = (-1)^1 = -1  \\
\omega^{-1} = (-1)^{-1} = \frac{1}{-1} = -1$
I assume the step $k=2$ and the initial offset $r = 0$, and the function $f(x) = \arctan(x)$.
Now I use the formula for the closed form of the multisected series:
$$F(x) = \frac{1}{k}\sum_{n=0}^{k-1}\omega^{-nr}f(\omega^nx)$$
After substituting my numbers and the roots of unity to this formula I get:
$$F(x) = \frac{1}{2}\sum_{n=0}^{1}\omega^{-n\cdot0}\arctan(\omega^nx)  \\
F(x) = \frac{1}{2}\left\{ \omega^{-0\cdot0}\arctan(\omega^0x) + \omega^{-1\cdot0}\arctan(\omega^1x)  \right\} \\
F(x) = \frac{1}{2}\left\{ \omega^0\arctan(\omega^0x) + \omega^0\arctan(\omega^1x) \right\} \\
F(x) = \frac{1}{2}\left\{ \arctan(x) + \arctan(-x) \right\} $$
Now since $\arctan(-x) = -\arctan(x)$, I get:
$$F(x) = \frac{1}{2}\left\{ \arctan(x) - \arctan(x) \right\} = \frac{1}{2}\!\cdot\!0 = 0$$
:-/
I thought that I perhaps should have used $r=1$ as the offset, so I tried this one too:
$$F(x) = \frac{1}{2}\sum_{n=0}^{1}\omega^{-n\cdot1}\arctan(\omega^nx)  \\
F(x) = \frac{1}{2}\left\{ \omega^{-0\cdot1}\arctan(\omega^0x) + \omega^{-1\cdot1}\arctan(\omega^1x)  \right\} \\
F(x) = \frac{1}{2}\left\{ \omega^0\arctan(\omega^0x) + \omega^{-1}\arctan(\omega^1x) \right\} \\
F(x) = \frac{1}{2}\left\{1\cdot\arctan(1\!\cdot\!x) + (-1)\!\cdot\!\arctan(-1\!\cdot\!x) \right\} \\
F(x) = \frac{1}{2}\left\{ \arctan(x) - \arctan(-x) \right\} \\
F(x) = \frac{1}{2}\left\{ \arctan(x) + \arctan(x) \right\} \\
F(x) = \frac{2\!\cdot\!\arctan(x)}{2} = \arctan(x)$$
So again, something seems to be wrong here: how come I obtained the same original function from taking only every second term of its power series?
Where did I make a mistake?
 [/INST] We have 
\begin{eqnarray*}
\tan^{-1}(x)= x- \frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}+\cdots
\end{eqnarray*}
and also
\begin{eqnarray*}
\tanh^{-1}(x)= x+ \frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\frac{x^9}{9}+\cdots
\end{eqnarray*}
Now just add these to obtain
\begin{eqnarray*}
\frac{\tan^{-1}(x)+\tanh^{-1}(x)}{2} = x+ \frac{x^5}{5}+\frac{x^9}{9}+\frac{x^{13}}{13}+\frac{x^{17}}{17}+\cdots
\end{eqnarray*}
</s>