Knowing $\exp(i\pi) = -1$, can we find out the value of $\pi$? We know $\exp(i\pi) = -1$ or $e^{i\pi} = -1$.
Can we solve for this infinite polynomial in $\pi$ to obtain the value of $\pi$?
We already know the value of $e$ from two methods:
$e = \exp(1) = 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots$
or from:
$ e = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n} $
So we should in theory be able to solve for the other unknown $\pi$.
We can also try:
$\log(-1) = i\pi$
But it is weird that:
$\log(1 + (-2)) = (-2)-\frac{(-2)^2}{2}+\frac{(-2)^3}{3}-\frac{(-2)^4}{4} + \cdots$
doesn't look like $i\pi$, whatever else it could be made in to.
 A: I would say take real or imaginary part, then try to solve.  For example, define $\pi$ as the least positive zero of $\sin(x)$ and use the Taylor series for $\sin$.
Even better is $\tan(\pi/4) = 1$ and use the series for $\arctan$ evaluated at $1$:
$$
\frac{\pi}{4} = 1 - \frac{1}{3}+\frac{1}{5}-\frac{1}{7} +\dots
$$
This converges slowly, but it converges.
A: $$\exp(i\pi) = 1+i\pi-\frac{\pi^2}{2} + \cdots = -1$$
we get for the real part:

*

*$\frac{\pi^2}{2} + \cdots = 2$
giving $\pi \approx 2 $

*$\frac{\pi^2}{2} - \frac{\pi^4}{24} + \cdots = 2$
giving $\pi^2 = \left|{\frac{-\frac{1}{2}\pm i\sqrt{\frac{1}{3}-\frac{1}{4}}}{-1/12}}\right| = 
\left|{\frac{\sqrt{\frac{1}{4}+\frac{1}{12}}}{-1/12}}\right| = 6.93, \pi \approx 2.63$

*$\frac{\pi^2}{2} + \frac{\pi^4}{24} - \frac{\pi^6}{720}  \cdots = 2$
$\cdots$
Might take a while, but converges to $\pi$ oscillating and meandering like a river over the complex plane, but you can prove that the oscillations dies down as we go further down the line.
As far as the logarithmic one:
    $\log(-1) = i\pi$
has conditional sequence:
    $(−2)−\frac{(−2)^2}{2}+\frac{(−2)^3}{3}−\frac{(−2)^4}{4} + \cdots$
We know in theory a conditional converging series can be made to converge to any number.
Usually we think that it converges only to a real number but in this case it is a bit weird, $i\pi$, but useless information if we want to calculate the value of $\pi$.
Of course, Archimedes had smarter ideas – if you bound the circle between two squares (one from inside and one from outside) and take the average of the two: $\pi \approx \frac{4+2\sqrt{2}}{2} = 3.41$, just 90% there in one shot and gets better with pentagon, hexagon etc.
