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

I have been searching for a convincing interpretation of this. I understand how it comes about but what is it that it is telling us?

Best that I can figure out is that it just emphasizes that the various definitions mathematicians have provided for non-intuitive operations (complex exponentiation, concept of radians etc.) have been particularly inspired. Is that all that is behind the slickness of the Famous Five equation?

Any pointers?


closed as off-topic by Mikhail Katz, user1688, Steven Landsburg, HJRW, Yemon Choi Jan 21 '16 at 21:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Mikhail Katz, Community, Steven Landsburg, HJRW
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ $\exp:\mathfrak{lie}(S^1) \rightarrow S^1 \hookrightarrow \mathbb{C}$ is a pretty natural map. Anyway, I think that this is off-topic. Voting to close. $\endgroup$ – Steve Huntsman Jun 5 '10 at 3:01
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    $\begingroup$ I regard this identity as the definition of pi. $\endgroup$ – Qiaochu Yuan Jun 5 '10 at 3:40
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    $\begingroup$ Voting to close? That's unbelievable... $\endgroup$ – José Hdz. Stgo. Jun 5 '10 at 5:13
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    $\begingroup$ By the way, where does the term "famous five" come from? Mathematicians don't call it that. Does it come from some popularization? $\endgroup$ – Gerald Edgar Jun 18 '10 at 19:01
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    $\begingroup$ Obviously, this formula cannot serve as a definition of $\pi$, because it holds for infinitely many distinct values in place of $\pi$. $\endgroup$ – Emil Jeřábek Jun 23 '11 at 12:02

$$e^{i\pi}=\lim\limits_{N\to\infty}\left(1 + \frac{iπ}{N}\right)^N$$

alt text

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    $\begingroup$ This may be the best answer I've seen on all of Math Overflow. $\endgroup$ – Kevin O'Bryant Jun 5 '10 at 3:15
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    $\begingroup$ Thanks, Kevin, that's very kind of you. The credit really goes to Wikipedia :) $\endgroup$ – Andrey Rekalo Jun 5 '10 at 4:00
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    $\begingroup$ Specifically, here: en.wikipedia.org/wiki/Euler%27s_identity $\endgroup$ – Dan Ramras Jun 5 '10 at 4:18
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    $\begingroup$ Yes. And the 3D vizualization of Euler's formula is also instructive en.wikipedia.org/wiki/File:Euler%27s_Formula_c.png $\endgroup$ – Andrey Rekalo Jun 5 '10 at 4:26
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    $\begingroup$ Really nice. Reminds me of a cute argument that appears in the superb book on Complex Analysis by Tristan Needham. As Professor Needham says, an individual putting forward an answer of the type "well that's just a definition" is giving a low blow to one of Euler's greatest contributions to Math. $\endgroup$ – José Hdz. Stgo. Jun 5 '10 at 5:11

I'd like to add something to the visual answer above (or below, or wherever it ends up). It was not until I was into my 40s that I realized that there was an intuitive way of understanding that $e^{i\pi}=-1$, as opposed to the power-series derivation that seems a bit too formal somehow. (What I'm about to say comes into the category of thing that I happened to think of for myself but it's so natural that obviously many other people will have had the same thought and I absolutely do not wish to make some ludicrous priority claim. I haven't read Needham's book, but I dare say he has the same argument.)

Why should it be that the limit of $(1+i\pi/N)^N$ is equal to -1? To answer this, let's think about what multiplication by $1+i\pi/N$ does. Well, $1+i\pi/N$ has modulus very close indeed to 1 (by which I mean that it has modulus $1+O(N^{-2})$), and argument very close indeed to $\pi/N$. Therefore, multiplication by $1+i\pi/N$ is approximately rotation by $\pi/N$. So if you do it N times, then the result is approximately rotation by $\pi$, which is multiplication by -1. The approximations are good enough that one can make this argument rigorous fairly easily.

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    $\begingroup$ I really do like these two answers, but I cannot help thinking that they should be accompanied by an equally intuitive explanation of why $\lim_{n\to\infty}(1+x/n)^n=e^x$. Which, in turn, boils down to a question of understanding the exponential function in the first place. I am not saying this is hard to achieve, but it is something I rarely see done well. $\endgroup$ – Harald Hanche-Olsen Jun 5 '10 at 13:35
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    $\begingroup$ I agree. My answer to that is that it's fairly easy to prove that the function f(x) = lim (1 + x/n)^n has the property f(x+y) = f(x)f(y). $\endgroup$ – gowers Jun 5 '10 at 15:17
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    $\begingroup$ One may also interpret the limit as convergence of solutions to Euler's numerical scheme $$\frac{\phi_n-\phi_{n-1}}{\pi/n}=i\phi_{n-1},\quad \phi_0=1,$$ to a solution of the ODE $$\dot{\phi}=i\phi,\ t\in(0,\pi].$$ $\endgroup$ – Andrey Rekalo Jun 5 '10 at 16:00
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    $\begingroup$ The explanation given by Conway and Guy in The Book of Numbers is also in the spirit of these two answers. $\endgroup$ – Todd Trimble Jun 5 '10 at 16:38

The answers so far give interpretations of the exponential as a limit of discrete approximations. An alternative interpretation is that any continuous map that takes addition to multiplication on the complex line and takes reals to reals has a purely imaginary kernel isomorphic to the integers. The constant $e$ arises from a normalization for which unit speed paths on the imaginary axis are taken to unit speed paths on the unit circle, and $\pi$ then shows up as a path length. One way to emphasize the additive-multiplicative relationship is to expand the formula as: $e^{\pi i-0i} = -1/1$.

Here is a more formal treatment: Both $(\mathbb{C}^\times, \times)$ and $(\mathbb{C}, +)$ are one dimensional analytic groups, and the latter is simply connected, so there is a universal covering homomorphism $\exp: \mathbb{C} \to \mathbb{C}^\times$ from the additive group to the the multiplicative group. The homomorphism is unique once we choose a normalization, e.g., by demanding that it is analytic and satisfies the differential equation $\partial_z \exp = \exp$. The differential equation can be related to the homomorphism, after choosing coordinates, by considering the respective formal group laws, or just reasoning heuristically with infinitesimals.

Claim 1: The function $\exp$ takes purely imaginary numbers to elements of unit norm.

Proof: The function $\exp$ takes additive inverses to reciprocals (because it is a homomorphism), and complex conjugates to complex conjugates (because it is defined over the reals). By composing, we find that reflection in the imaginary axis is taken to unit circle inversion, and fixed points are taken to fixed points.

Remarks: Note that the only part of the normalization we used here was the fact that $\partial_z \exp$ is a real multiple of $\exp$. The "defined over the reals" bit may be unsatisfying to some, but the conjugation behavior can be verified directly by expanding as a power series that converges everywhere, and checking that the coefficients are real. One can also prove the claim by more direct methods, such as applying the above differential equation to grind out the identity $\frac{\partial}{\partial y} | \exp(iy) |^2 = 0$.

Claim 2: $\exp(\pi i) = -1$.

Proof: By combining the previous claim concerning unit norms with the differential equation $\partial_z \exp = \exp$, we conclude that $\exp$ takes any unit speed path on the imaginary axis to a unit speed path on the unit circle. We have $\exp(0) = 1$ by the homomorphism assumption, and the length of a minimal path from $1$ to $-1$ on the unit circle is $\pi$.

Remarks: Depending on how $z \mapsto e^z$ is defined, one may still have to check that it agrees with $\exp$, but this isn't a big deal. I tried to avoid choosing square roots of minus one as much as possible, but the statement of the identity makes it a bit difficult to maintain such discipline.


Just wanted to include this excellent illustration of Euler's formula
(it really deserves to be shown here in its own right and not just as a link in one of the comments):

alt text

Source: http://en.wikipedia.org/wiki/Euler_formula

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    $\begingroup$ Why the downvote??? $\endgroup$ – vonjd Jan 21 '16 at 13:23

Consider the exponential map for the Lie group $U(1)$.

The formula $e^{\pi i}$ means: starting from the identity element $1$, keep traveling "in the direction of $i$" (Which is "going up" at first, because the imaginary axis is of course going up. But this direction "changes" on the $\mathbb{R}^2$-plane as you move! So eventually you will be moving among the circumstance of the unit circle), and walk a distance of $\pi$.

Where do you end up with? You end up traveling half a circle (distance = $\pi$) and reach $-1$. Therefore $e^{\pi i} = -1$.

BTW, I think it is not a very good idea to simply regard the formula as the definition of $\pi$ and consider it trivial. Because then one have to explain why this $\pi$ is the same $\pi$ in the usual definition using the circumstance / area of a circle.

So, the real question is not about $\pi$, but about why something seemingly "only related to a circle" will have anything to do with an algebraic expression involving $e$ and $i$.


Though I would contribute with this animated version of vonjd's illustration:

An animation.

This animation is from here.

  • $\begingroup$ Please excuse me as I am still figuring out how to post photos here. $\endgroup$ – Aj Brown Jan 20 '16 at 17:48
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    $\begingroup$ This is a gif version of @vonjd's answer. $\endgroup$ – Chris Ramsey Jan 20 '16 at 18:20
  • $\begingroup$ I added the animation. Edit your post to see what I wrote to do so. $\endgroup$ – Joonas Ilmavirta Jan 21 '16 at 7:10

A way to introduce the exponential is by the ODE:

$$f' = \alpha f, \ \ f(0) = 1,$$

whose unique solution is $t \mapsto exp(\alpha t)$. This is common when $\alpha$ is natural, but what if $\alpha$ is complex? And, in particular, what if $\alpha = i$? Of course, $f$ must be complex-valued instead of real-valued.

Then $f' = if$, that is, $f'$ is the vector $f$ rotated by $\pi/2$ in the trigonometric direction. If we forget a moment about the complex structure, we get:

$$y' = x, \ x'=-y, \ \ f(0) = (x,y)(0) = (1,0).$$

A solution of this system is a trajectory whose speed is always tangential to its position vector. But then, the function $(x,y) \mapsto x^2+y^2$ is a first integral, so it is constant along trajectories. In particular, the trajectory starting from $(1,0)$ stays on the circle of radius $1$ centered on $0$. But then, since $|f|$ is constant, so is $|f'|$: the trajectory moves at unit speed along this circle of radius $1$.

Since the trajectory starts from $1$, after a time $\pi$, it will have traveled along a half-perimeter, so it will be in $(-1,0)$. Whence $exp(i \pi) = -1$.

@Andrey Rekalo's interpretation of this equation is Euler's method applied to this ODE.


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