I'm reading Bellard's proof for his eponymous formula computing pi digits, and I can't get past the first line.
Given that:
- $\displaystyle-\ln(1-x) = \sum_{n=1}^\infty \frac{x^n}{n}$ for $|x| < 1$,
- $\arctan(y) = \Im[-\ln(1-iy)]$ for any real $y$
(I didn't know the last one but Wolfram Alpha agrees, so okay), the author gets
$$\arctan\left(\frac{1}{a-1}\right) = \Im\left[-\ln\left(1- \frac{1+i}{a}\right) \right] = \sum_{n=0}^\infty \frac{(-1)^n2^{2n}}{a^{4n+3}}\left( \frac{a^2}{4n+1} + \frac{2a}{4n+2} + \frac{2}{4n+3}\right)$$
In this equation, I don't even get the first equality, and assuming it to be correct, I don't see how to derive the second equality… I've tried expanding using Newton's theorem and expanding the binomial coefficients, but it did not work.
Any ideas?
EDIT
A friend nearly found the solution for the second equality (up to some mysterious sign error), and the first equality is actually invalid when $a \in (0, 1)$, according to Wolfram Alpha: on this interval, we rather have $\arctan\left(\frac{1}{a-1}\right) = \Im\left[-\ln\left(1- \frac{1+i}{a}\right) \right] \color{red}{+ \pi}$ .
So I still don't know why $\arctan\left(\frac{1}{a-1}\right) = \Im\left[-\ln\left(1- \frac{1+i}{a}\right) \right]$ for $a \notin (0,1)$, but given that Bellard uses $a = 2$, let's assume it's fine. However I'm interested if someone could find the solution to this problem.
Now, from $\Im\left[-\ln\left(1- \frac{1+i}{a}\right) \right]$ to $\displaystyle\sum_{n=0}^\infty \frac{(-1)^n2^{2n}}{a^{4n+3}}\left( \frac{a^2}{4n+1} + \frac{2a}{4n+2} + \frac{2}{4n+3}\right)$:
$$\begin{align*}
-\ln\left(1 – \frac{1+i}{a}\right) &= \sum_{k=1}^\infty \frac{1}{k}\left(\frac{1+i}{a}\right)^k \\
&= \sum_{n=0}^\infty \left[ \frac{1}{4n+1} \left( \frac{1+i}{a}\right)^{4n+1} + \dotso + \frac{1}{4n+4} \left( \frac{1+i}{a}\right)^{4n+4}\right]\\
&= \sum_{n=0}^\infty\left( \frac{1+i}{a}\right)^{4n+1}\left[ \frac{1}{4n+1}+ \dotso + \frac{1}{4n+4}\left(\frac{1+i}{a}\right)^3 \right]
\end{align*}$$
We remark that $1 + i = \sqrt{2} e^{i\pi / 4}$ so :
$$\begin{align*}
-\ln\left(1 – \frac{1+i}{a}\right) &= \sum_{n=0}^\infty \frac{\left( \sqrt{2} e^{i \pi/4}\right)^{4n+1}}{a^{4n+1}}\left[ \frac{1}{4n+1}+ \dotso + \frac{1}{4n+4}\frac{\left( \sqrt{2} e^{i \pi/4}\right)^3}{a^3} \right]\\
&=\sum_{n=0}^\infty \frac{2^{2n}(-1)^n}{a^{4n+1}}\left[ \frac{\left( \sqrt{2} e^{i \pi/4}\right)}{4n+1}+ \dotso + \frac{1}{4n+4}\frac{\left( \sqrt{2} e^{i \pi/4}\right)^4}{a^3} \right]\\
\end{align*}$$
Taking the imaginary part, knowing $\sqrt{2} e^{i \pi/4} = 1+i$ and $(1+i)^2 = 2i$, $(1+i)^3 = 2 – 2i$ and $(1+i)^4 = -2$:
$$\begin{align*}
\Im \left[-\ln\left(1 – \frac{1+i}{a}\right)\right] &= \sum_{n=0}^\infty \frac{2^{2n}(-1)^n}{a^{4n+1}}\left[ \frac{1}{4n+1}+ \frac{2}{4n+2}\frac{1}{a} – \frac{2}{4n+3} \frac{1}{a^2}\right]\\
&= \sum_{n=0}^\infty \frac{2^{2n}(-1)^n}{a^{4n+3}}\left[ \frac{a^2}{4n+1}+ \frac{2a}{4n+2} \color{red}- \frac{2}{4n+3}\right]\\
\end{align*}$$
So everything is good, except for the last minus sign…
Best Answer
1. The $\operatorname{arctan}$ formulas.
Consider what it means for a complex number $u$ to be a logarithm of $z$: essentially, that $e^u = z$.
Since $e^u = e^{\Re u + i \Im u} = e^{\Re u} e^{i\Im u} = r e^{i \theta}$ you see that pretty much by definition, $\Im \ln(z) = \Im u = \theta$ is the angular component of $z$ in the complex plane.
Now, take the point $z = 1-iy$, is it an elementary trigonometric exercise to check that the angle forme between the real axis and $z$ is $\theta$ such that $\tan\theta = -y$. By the above remark this means exactly that $\Im \ln(1-iy) = \arctan(-y)$.
Since $\arctan$ is an odd function, this means that $-\Im \ln(1-iy) = \arctan(y)$.
The very same argument works for $z = 1 - (1+i)/a$. Indeed, this time we have
$$\begin{align*} \tan \theta & = \frac{\Im z}{\Re z} \\ & = \frac{-1/a}{1 - 1/a} \\ & = \frac{-1}{a - 1} \end{align*} $$ And therefore, $-\Im\ln(1 - (1+i)/a) = \arctan(1/(a-1))$.
2. The summation formula
You have it right, except that $(1+i)^3 = 2i(1+i) = -2 + 2i$. And not $2 - 2i$ as you wrote, which was the source of this erroneous sign.
Well done! For additional fun formulas and the rationale behind them (in French), you can have a look at this pdf from the olden days.