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?

The problem is that the arctangent series has already had this procedure applied once:
$$ \arctan{x} = \frac{1}{2} (-i\log{(1+ix)}-(-i\log{(1-ix)})). $$
If one applies the sifting procedure again, the answer comes out the same (each term in the sum is sifted and returns the same terms as the original, which add to give the same function:
$$ f(x) \mapsto \frac{1}{2}(f(x)-f(-x)) \mapsto \frac{1}{2}\left(\frac{1}{2}(f(x)-f(-x))\right)-\frac{1}{2}\left(\frac{1}{2}(f(-x)-f(-x))\right) = \frac{1}{2}(f(x)-f(-x)) $$
(this procedure produces a function that has a certain symmetry; applying the operation again does not produce more symmetry).
So the arctangent series is already missing terms: you actually want every fourth term of $x+0x^2-x^3/3+0x^4/4 + \dotsb$. Hence the correct sum is
$$ \frac{1}{4}(\arctan{x}+i\arctan{ix}-\arctan{(-x)})-i\arctan{(-ix)}) = \frac{1}{2}(\arctan{x}+\arg\tanh{x}) $$
by using some complex trigonometric identities.
