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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?

Michael Hardy
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BarbaraKwarc
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2 Answers2

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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.

Chappers
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    OK, Somos and DonaldSplutterwit already explained the mistake to me in another answer: I used the wrong numeration for the terms (I was counting the terms of the sum instead of the exponents of the powers). – BarbaraKwarc Jul 19 '17 at 23:57
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    However, your answer adds something I'd like to ask: the connection between the logarithms and the arctangent. Although I can see how this works for the series (the series for $\ln(1+x)$ is just alternating the sign and dividing each subsequent power by its exponent, and the series for $\arctan(x)$ is alternating the signs with every odd power divided by its exponent), I'm curious how could one come up with that idea without using the roots of unity trick? Is there's some law of mathematics or geometry behind it, such as the one behind $e^{ix}=\cos(x)+i\sin(x)$? – BarbaraKwarc Jul 20 '17 at 00:05
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    From Euler's formula one has $$\tan{z}=-i\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}} = -i \frac{e^{2iz}-1}{e^{2iz}+1}. $$ Use this to solve $x=\tan{z}$ for $e^{2iz}$, take logs and you end up with $$\arctan{x} = \frac{1}{2i}\log{\frac{1+ix}{1-ix}}, $$ which is the same as the result given in the answer provided the right branches of the logarithms are chosen. The same idea works for the other inverse trigonometric and hyperbolic functions, although most of them require solving a quadratic. – Chappers Jul 20 '17 at 00:19
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    One can also use partial fractions on the $\int_0^x dy/(1+y^2)$ definition if preferred. – Chappers Jul 20 '17 at 00:20
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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*}

Donald Splutterwit
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  • Interesting trick! I'll add it to my toolbox. Nevertheless, now I'm rather focused on figuring out why the formula for series multisection doesn't work in this case for me. Did I do something wrong? Or is it that the formula cannot be applied to this series from some reason? (If this is the case, I'd like to know the reason.) – BarbaraKwarc Jul 19 '17 at 23:08
  • You need to use the factor $i$ (not $-1$) in your question in order to get your terms to cancel out. – Donald Splutterwit Jul 19 '17 at 23:13
  • Why $i$? The paper says that if I want to take every $k$th term, I have to use the powers of the $k$th primitive root of unity, that is, $\omega = e^{i\frac{2\pi}{k}}$. In my case, $k=2$, so I use $\omega = e^{i\frac{2\pi}{2}} = e^{i\pi} = -1$ and its powers, which alternate between $1$ and $-1$. There's no $i$ among them. $i$ would be a 4th root of unity. – BarbaraKwarc Jul 19 '17 at 23:18
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    You have to include the zero terms. That is, $0 + x + 0x^2 - x^3/3 + 0x^4 + x^5/5 + \dots$. When you include them you want to use a 4th root of unity. – Somos Jul 19 '17 at 23:21
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    The powers are going up in steps of $2$ ... so you need to take $k=4$ ...$\omega = e^{i\frac{2\pi}{4}} = e^{\frac{i\pi}{2}} = i$. – Donald Splutterwit Jul 19 '17 at 23:22
  • Oooooh, so that's how it is supposed to work! I thought that the numeration goes according to the terms of the sum, not the powers. Damn, they should have made that paper more clear then. Please make it an answer and I'll accept it. – BarbaraKwarc Jul 19 '17 at 23:22