1

I was tinkering with Euler's Identity and I come to wonder if it was possible to derive $\pi$ from it.

I know $\pi$ can't be expressed as a fraction of two rational numbers but neither $i$ nor $e$ is rational.

$e^{\pi i} = -1$ (square both sides)

$e^{2\pi i} = 1$ (logarithm both sides)

$2\pi * i * \log e = \log 1 = 0$

This is how far I got before meeting a contradiction as this the left side equals roughly $9.1 i$. Is it even possible to derive $\pi$ from Euler's Identity and where have I messed up?

Hugo Gransträm
  • 13
  • 2
    Write e^{\pi i} (with curly braces) to get $e^{\pi i}$. – Akiva Weinberger Jun 05 '16 at 11:27
  • 4
    On the complex numbers, the logarithm isn't a function; rather, it's a multifunction (returns multiple values for one argument). This is how $e^{2\pi i}=e^0$ doesn't imply $2\pi i=0$ after taking logs; $\ln 1$ is all integer multiples of $2\pi i$. – Akiva Weinberger Jun 05 '16 at 11:30
  • Thank you :) This made it all much more clear to me ;) – Hugo Gransträm Jun 05 '16 at 11:33
  • 1
    There is also the $\operatorname{Log}$ function, which returns the principle value of the logarithm (base $e$) — the one whose imaginary part is greater than $-\pi$ and less than $\pi$. However, it's not continuous. – Akiva Weinberger Jun 05 '16 at 11:36
  • 2
    Unfortunately as the exponent and therefore log (multi) function was defined in terms of pi, such an attempt will be circular. You end up with 2i.pi log e = log 1 = 2i.pi. so pi = pi. – fleablood Jun 05 '16 at 15:14
  • Ahh. I see :) Learn something new everyday :-D – Hugo Gransträm Jun 05 '16 at 15:16

2 Answers2

0

Let us write $$\log(-1) = \pi i\tag{1}$$ We can make sense of this by defining $$ \log(-1) := \int_1^{-1}\frac{dz}{z} $$ where (since we cannot integrate through the pole at $z=0$) we integrate by convention along a curve in the upper half-plane. For example along the sides of a rectangle. Using numerical integration, we can compute $\pi$ numerically this way.

Historical: ($1$) appeared in the literature earlier than Euler's identity.

Note: If we use curves not remaining in the upper half plane, we may get values other than $\pi i$ for the integral.

GEdgar
  • 111,679
-1

Well, if we know that

$e^{i\pi} = -1$

Then we can use log base e, or just ln() to take out the e and just get the pi*i, so:

$\ln(-1) = i\pi$

That was the hard part -- now we just have to divide by i:

$\frac{\ln(-1)}{i} = \pi$

And now we have pi!

Morgan Wyoming
  • 1