4
$\begingroup$

From Wikipedia: "Euler's identity is a special case of Euler's formula from complex analysis, which states that for any real number x,

$$ e^{ix} = cos(x) + isin(x) $$

where the inputs of the trigonometric functions sine and cosine are given in radians."

If $ e^{i\pi}=-1 $, then π is calculated in radians.

There appears to be an infinite number of solutions that yield -1 because x is calculated in radians, such that any value of $x$ that is a positive or negative odd multiple of π will yield -1

Examples:

  • $$ e^{iπ} = -1 $$
  • $$ e^{i(-π)} = -1 $$
  • $$ e^{i3π} = -1 $$
  • $$ e^{i(-3π)} = -1 $$
  • $$ e^{i5π} = -1 $$
  • $$ e^{i(-5π)} = -1 $$

Question: Being that $ cos(x) + isin(x) $ is identical to $ e^{ix} $

And cos(π) + isin(π) in degrees does not equal -1

Then is there a way to calculate an answer to $ e^{i\pi} $ without using degrees or radians? As you would with a simple problem like $$ 3^2=9 $$

$\endgroup$
  • $\begingroup$ If you define $E$ to be $e^{\pi/180}\approx1.01761$, then $E^{ix}=\cos(x^\circ)+i\sin(x^\circ)$ in degrees. $\endgroup$ – Akiva Weinberger Oct 20 '15 at 0:46
  • 1
    $\begingroup$ By the way, where you write $e^{i-\pi}$, you mean $e^{i(-\pi)}=e^{-i\pi}$. Remember, the expression $i-\pi$ means i minus pi, not i times negative pi. $\endgroup$ – Akiva Weinberger Oct 20 '15 at 0:48
3
$\begingroup$

$$e^{x}=1+x+\frac{x^2}{2}+\frac{x^3}{6}\cdots$$ Based on the Taylor expansion rules $$e^{ix}=1+ix+\frac{(ix)^2}{2}+\frac{(ix)^3}{6}...$$ You plug in $ix$ for $x$ and simplify all you notice $$=1+ix-\frac{x^2}{2}-\frac{ix^3}{6}+\frac{x^4}{24}\cdots$$ which you can simplify into two [fairly recognizable] sums! $$=(1-\frac{x^2}{2}+\frac{x^4}{24}\cdots)+i(x-\frac{x^3}{6}+\frac{x^5}{120}...)$$ By recogizable I mean sin and cos taylor expansions. $$e^{ix}=\cos(x)+i\sin (x)$$ Plug in $\pi$ and you get $$e^{i(\pi)}=-1+i*0=-1$$

All you really have to know are Taylor Sums!

$\endgroup$
  • 1
    $\begingroup$ The only answer I'm really looking for is my original question in bold. Is it possible to solve this without resorting to interpreting $e^{i\pi}$ in radians or degrees. which means without applying cos(x) or sin(x) (which I imagine are both limited by either radians or degrees) i.e your answer resorts to pluging in pi into cos and sin. I'm looking for a way that doesn't do that. $\endgroup$ – Tony Oct 20 '15 at 2:01
  • 1
    $\begingroup$ No, because we would never know the definition of e^ix without using trig functions. That was the point of my answer. $\endgroup$ – user253055 Oct 20 '15 at 2:03
  • 1
    $\begingroup$ Okay cool thanks for the clarification. Is it impossible or hasn't been done? $\endgroup$ – Tony Oct 20 '15 at 2:06
  • 1
    $\begingroup$ No way without trig functions has been found, but people generally agree that there is no other way to solve $e^{ix}$ $\endgroup$ – user253055 Oct 20 '15 at 2:36
2
$\begingroup$

You can expand the Taylor polynomial for $e^{i\pi}$: $$ e^{i\pi} = \sum_{k=0}^{\infty} \frac{(i\pi)^k}{k!} = 1 + i\pi - \frac{\pi^2}{2} \cdots$$

$\endgroup$
  • $\begingroup$ I was familiar with this method, but don't know how to solve for what it converges to. $\endgroup$ – Tony Oct 20 '15 at 1:30
  • 1
    $\begingroup$ @lurker currently building a better answer :) If you wait about 5 min I will show you. $\endgroup$ – user253055 Oct 20 '15 at 1:50
2
$\begingroup$

Yes! Just write $-1$ as you did. No radians at all in this representation of $e^{i\pi}$. The only thing is that there is more than one way to write this number, by using radians. In fact, any complex number, except the zero, can be written in infinitely many ways with the form $\cos(x)+i\sin(x)$ and in only one with the form $a+ib$.

$\endgroup$
  • $\begingroup$ So $$ e^{i\pi} = -1 $$ even without the application of radians? That's cool. Is there a proof for that? $\endgroup$ – Tony Oct 20 '15 at 1:28
  • 1
    $\begingroup$ The proof of e to the ix requires taylor series, which assume that the ix part is interpreted in radians. That being said, you don't need to say 'i pi radians' because the theorem already assumes the radians part. If you said '180 i' the formula would interpret it as 180 radians, not the same thing. $\endgroup$ – Faraz Masroor Oct 20 '15 at 1:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.