7
$\begingroup$

When I first found out that $e^{i\pi} = -1$, I was blown away. Does anyone here know one of (many I'm sure) proofs of this phenomenal equation? I can perform all of the algebra to get the $-1$. But, where does this come from? What is the derivation?

$\endgroup$
  • 9
    $\begingroup$ Prove Euler's identity $e^{i\theta} = \cos \theta + i \sin \theta$ using Taylor series. Then plug in $\theta = \pi$. $\endgroup$ – Henry T. Horton Mar 26 '13 at 0:20
  • 4
    $\begingroup$ Nitpick: Fans of Euler's identity prefer to write $e^{i\pi}+1=0$. $\endgroup$ – Yoni Rozenshein Mar 26 '13 at 0:25
  • 2
    $\begingroup$ @HenryT.Horton but some (e.g. Rudin in Real and Complex Analysis) take that equation to be the definition of $\cos$ and $\sin$ $\endgroup$ – kahen Mar 26 '13 at 0:26
  • $\begingroup$ You neeed to understand complex exponentials to understand this. $\endgroup$ – Lost1 Mar 26 '13 at 0:26
  • $\begingroup$ How about considering the derivative of $\frac{\cos x + i \sin x}{e^{ix}}$ $\endgroup$ – user27182 Mar 26 '13 at 0:41
17
$\begingroup$

Here's a slick derivation. Let $f(x)= e^{-ix}(\cos x + i\sin x)$. Taking the derivative we have that $f'(x) = -ie^{-ix}(\cos x + i \sin x) + e^{-ix}(-\sin x + i\cos x) = 0$ so $f(x)$ is constant. But $f(0) = e^0(1 + 0) = 1$ so $f \equiv 1$ so $e^{ix} = \cos x + i \sin x$. Plugging in $x = \pi$ yields the result.

Remark: This proof is very beautiful, and uses the characteristic property of everything involved ($e^x$ is the unique nontrivial function that is its own derivative, $i$ squares to $-1$, etc), but as a first-timer the Taylor series proof told me much more.

$\endgroup$
  • 1
    $\begingroup$ It makes the assumption that $\frac{{\rm d}e^{ix}}{{\rm d}x} = ie^{ix}$, a reasonable definition, but not something that can be proven. $\endgroup$ – DanielV Mar 5 '15 at 3:20
6
$\begingroup$

The identity is a special case of Euler's formula from complex analysis, which states that

$e^{ix}$=$\cos x+i\cdot\sin x$

for any real number $x$. (Note that the variables of the trigonometric functions sine and cosine are taken to be in radians, and not in degrees.) In particular, with $x = \pi$, or one half turn around the circle:

$e^{i\pi}$=$\cos\pi+i\cdot\sin\pi$

Since

$\cos\pi=-1$

and

$\sin\pi=0$

it follows that

$e^{i\pi}$ =$-1+0i$

which gives the identity

$e^{i\pi}$$+1=0$

The physical explanation of Euler's identity is that it can be viewed as the group-theoretical definition of the number $\pi$. The following discussion is at the physical level but can be made mathematically strict. The group is the group of rotations of a plane around 0. In fact, one can write:

$e^{i\pi}$=$(e^{i\delta})^{\pi/\delta}$

with $\delta$ being some small angle. The last equation can be seen as the action of consecutive small shifts along the circle caused by the application of infinitesimal rotations starting at 1 and going for the total length of the arc connecting points 1 and -1 in the complex plane. In fact, each small shift can be written as multiplication by

$1+i\delta$

and the total number of shifts is π/δ. In order to get from 1 to -1 the total transformation would be

$(1+i\delta)^{\pi/\delta}$

Now, taking the limit when $\delta \to0$, denoting $i\delta = 1/n$ and using the definition of:

$e$=$\lim\limits_{n \to \infty}(1+\frac{1}{n})^n$

we arrive at Euler's identity. The $\pi$ itself is defined as the total angle which connects $1$ to $-1$ along the arch.

Summarizing, we can say that because the circle can be defined through the action of the group of shifts which preserve the distance between a point and another point, the relation between π and e arises. This simple argument is the key to understanding other seemingly miraculous relations involving $π$ and $e$.

Source: Wikipedia

$\endgroup$
2
$\begingroup$

$$e^{ix}=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}=\sum_{k=0}^{\infty}\frac{(ix)^{2k}}{(2k)!}+\sum_{k=0}^{\infty}\frac{(ix)^{2k+1}}{(2k+1)!}=$$ $$=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k}}{(2k)!}+i\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!}=$$ $$=\cos x+i\sin x$$ since $$i^{2k}=(-1)^k,i^{2k+1}=i(-1)^k$$ for $x=\pi$ we get $$e^{i\pi}=\cos (\pi)+i\sin(\pi)=-1$$

$\endgroup$
  • $\begingroup$ This has many mistakes! Double factorial should be a factorial, and you're missing $(-1)^k$ in every expansion. $\endgroup$ – Pedro Tamaroff Mar 26 '13 at 0:34
  • 1
    $\begingroup$ No... $i^{2k}=(-1)^k$, and $i^{2k+1}=(-1)^ki$. Also, the double factorial is still wrong. $\endgroup$ – Pedro Tamaroff Mar 26 '13 at 0:37
  • 1
    $\begingroup$ Peter, why negative vote, if we see history of edits I correct the answer before your edit. The misprints are results of my hastiness to give answer soon. I kept your warning. $\endgroup$ – Adi Dani Mar 26 '13 at 1:07
  • $\begingroup$ I did not downvote, yet I do recommend you take care when answering. $\endgroup$ – Pedro Tamaroff Mar 26 '13 at 1:11
  • $\begingroup$ Then apologize. I hope in future cooperation $\endgroup$ – Adi Dani Mar 26 '13 at 1:18
2
$\begingroup$

1) The definition of powers $a^b$, where $b$ is an arbitrary rational number and $a>0$ is derived by the algebraic properties of exponents of positive integer exponents.

2) The definition of powers $a^b$, where $b$ is an arbitrary real number and $a>0$ is defined as the limit of powers with rational exponents: Choose a sequence of rational numbers $q_n\to b$, and let $a^b=\lim a^{q_n}$. (There are several things to prove here to show this makes sense, but it's all ok).

3) For the particular value $b=e$, one obtains the function $\exp (x)=e^x$. This function is infinitely differentiable with $\exp '(x)=\exp (x)$. Thus it's Taylor expansion is particularly nice (and well-known).

4) The radius of convergence of the Taylor expansion of $\exp$ is shown to be $\infty $ and to converge everywhere to $\exp(x)$. Thus, any complex number can be plugged into it, and so $\exp(z)$ is defined for all complex numbers.

5) The trigonometric functions $\sin$ and $\cos$ can be defined as follows: The point $(\cos x,\sin x)$ is that point on the unit circle in the $X-Y$ plane that forms an angle of $x$ radians with the positive $X$-axis (measuring from the $X$-axis counter clockwise).

6) From geometric considerations it can be shown that these trigonomerric functions are infinitely differentiable, with their familiar derivatives, and with their familiar Taylor expansions.

7) The Taylor expansions have infinite radius of convergence, and converges everywhere to to their respective sructures. Thus $\sin z$ and $\cos z$ are defined for all complex numbers.

8) By the general theory of power series, the three power series mentioned above can be manipulated much like finite power series (order of summation changes without affecting the result etc.).

9) When plugging in $\pi i$ into the power series expansion of $\exp$, thus computing $e^{\pi i}$, one can show (this is a good exercise) that: $$e^{\pi i}=\cos \pi + i\cdot \sin \pi.$$

10) Thus $e^{\pi i}=-1$.

Remark: More modern treatments of the trigonometric and exponential functions, in order to avoid unnecessary subtle geometric considerations and limiting mess, define these functions via their Taylor expansions.

$\endgroup$
  • 3
    $\begingroup$ I have to take exception to your claim about the Taylor expansion of $\mathbb R \ni x \mapsto e^x \in \mathbb R$. While it is true that it has radius of convergence $\infty$, it does not follow that the function is everywhere equal to its Taylor series. Recall the standard example $\varphi(x) = \begin{cases}e^{-1/x^2} & x\neq0 \cr 0 & x=0\end{cases}$. It's smooth and its Taylor series around $0$ has infinite radius of convergence, but $\varphi^{(n)}(0)=0$ for all $n$, so the Taylor series around $0$ for $\varphi$ is just the zero function. $\endgroup$ – kahen Mar 26 '13 at 1:07
  • $\begingroup$ Thanks @kahen, I clarified that in the two places where I refer to Taylor expansions. $\endgroup$ – Ittay Weiss Mar 26 '13 at 1:49
1
$\begingroup$

It's important to note that this identity is dependent upon our idea of complex exponentiation, which is to say how we decide to extend the exponential function to the complex plane. By convention, this is done by declaring the exponential function to be analytic with the same Taylor series as it has in the reals. This is why derivations of this fact all use some argument which relies on Taylor expansion or at least differentiation acting the same way for complex numbers as they do for real numbers (and it seems like these two should be equivalent (root cause?)).

$\endgroup$
0
$\begingroup$

There's a nice proof over at Math Overflow: https://mathoverflow.net/questions/51283/rigourous-proof-of-eulers-identity

$\endgroup$

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.