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I know some people say that $e^{i \pi} +1= 0$ is cool because it links fundamental yet disparate constants. But to me, the existence of a "nice" equation like this is not surprising. For hundreds of years mathematicians have been trying to make everything as "nice" as possible; we define trigonometry to line up with the informal notion of the unit circle (as opposed to the $2$ units circle, or the $.352546...$ units circle), we always try to make sure our functions are differentiable or at least continous when we extend them, we define complexes so that the reals are nicely embedded in them, etc. So to me it just doesn't seem that surprising that after hundreds of years of fine-tuning our definitions to be nice, the definitions come full circle and spit out some nice equations.

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closed as primarily opinion-based by Jack, J.-E. Pin, zoli, C. Falcon, user223391 Mar 10 '17 at 7:11

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ You don't have to be fascinated by that formula, no-one forces you to be ;o). $\endgroup$ – StackTD Mar 8 '17 at 17:09
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    $\begingroup$ From a real analysis point of view, what surprises me the most is how $\sum (-1)^n\frac {x^{2n+1}}{(2n+1)!}$ and $\sum (-1)^n\frac {x^{2n}}{(2n)!}$ are connected to the unit circle. $\endgroup$ – Guacho Perez Mar 8 '17 at 17:11
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    $\begingroup$ We don't need a unit circle to define trigonometry. And it is not at all intuitive what $e^{ix}$ has to do with trigonometry. The thing is that $e$ and $\pi$ seem to come from two disparate definitions and needs, so it is nice and surprising that they fit together. However, nobody requires you to be impressed. $\endgroup$ – Thomas Andrews Mar 8 '17 at 17:13
  • $\begingroup$ @ThomasAndrews Yeah I know we formally define the trig functions in another way, but when we formally define things we almost always define them so they they line up with the informal "nice" definition. I understand that nobody is requiring me to be impressed. I just know that most people smarter than me are fascinated with this formula, so I want to know what I am missing. $\endgroup$ – Ovi Mar 8 '17 at 17:16
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    $\begingroup$ We didn't carefully define things so they all lined up, we carefully defined things so that, individually, they made sense and were useful. Then the relationship was revealed. The relationship shows that something deep is going on. To me, even knowing several different proofs of this result, I don't feel like I "understand" it. $\endgroup$ – Thomas Andrews Mar 8 '17 at 17:22
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Maybe I should post an answer, explaining how really $e^{iπ} + 1 = 0$ is forced by our choice of what we want the exponential function to be. This may either make one more fascinated by the way it is so forced, or less fascinated because now the mystery is more clearly unraveled. $\overset{..}\smile$ $ \def\nn{\mathbb{N}} \def\lfrac#1#2{{\large\frac{#1}{#2}}} \def\abs#1{\left|#1\right|} $

The deep part is that $\exp' = \exp$ on $\mathbb{C}$. This is the key property that we want $\exp$ to have, having which we would be able to solve all homogenous linear differential equations, such as arising from many real-world phenomena such as radioactive decay, simple harmonic motion and population dynamics. It turns out that we can indeed construct such a function $\exp$ on $\mathbb{C}$ via the power series, which is naturally motivated by adding one term at a time to 'eliminate' the error in the desired identity for the previous partial sum (For instance when we have reached $f(x) = (1+x+\lfrac12x^2)$ there error would be $f(x)-f'(x) = \lfrac12x^2$ and so we 'need' to add $\lfrac1{3!}x^3$ to 'cancel' it off.) Of course, one must still prove that the infinite series actually converges and has the desired property, and in fact there is an elementary way to do it without having to prove termwise differentiability! (See below section.)

From this one property and the Taylor series (which can be motivated by the same reasoning as above but using integration by parts to keep track of the error), we get $\exp(z+w) = \exp(z) \exp(w)$ for complex $z,w$, which then as described in this post forces $\exp(it)$ to trace the unit circle around the origin in the complex plane when $t$ is a real parameter. (Underlying this is that multiplying by complex numbers is equivalent to applying spiral transforms around the origin.) Finally we obtain $\lfrac{d(\exp(it))}{dt} = i\exp(it)$, so if $t$ is time then $\exp(it)$ traces the circle at a constant rate as well because $|i\exp(it)| = 1$, which implies that the the period of $\exp$ is $i$ times the total distance traced (the perimeter of the circle).

If anyone feels there is any particular point in this answer where there is still something deep going on, feel free to comment and I'll see whether I can answer!


On request, here is a sketch of my elementary proof of the convergence and derivative of the power series $\exp(z) = \sum_{k=0}^\infty \lfrac1{k!} z^k$ for any complex $z$. (It got a bit too long to put in comments.)

Let $r \in \nn$ such that $2|z| \le r$. As $m \in \nn \to \infty$, eventually $m > r$ and so $\abs{\lfrac1{m!}z^m} \le \lfrac1{r!r^{m-r}} (\lfrac12r)^m$ $= \lfrac{r^r}{r!} 2^{-m} \to 0$ and hence $\abs{ \sum_{k=m+1}^{m+n} \lfrac1{k!}z^k } \le \sum_{k=m+1}^{m+n} \lfrac1{k!}|z|^k$ $\le \lfrac{z^m}{m!} \sum_{k=1}^n \lfrac1{m^k}|z|^k$ $\le \lfrac{z^m}{m!} \sum_{k=1}^n 2^{-k}$ $\to 0$. Thus by Cauchy convergence $\exp(z) = \sum_{k=0}^\infty \lfrac1{k!} z^k$ converges.

Now we prove the key property that $\exp'(z) = \exp(z)$. Let $R_m(z) = \exp(z) - \sum_{k=0}^m \lfrac1{k!} z^k$. Take any $h \in \mathbb{R} \to 0$. Let $m \in \nn$ such that $\abs{\lfrac1{m!}z^m} < |h|$ and $|R_m(z)| < |h|^2$ and $|R_m(z+h)| < |h|^2 $. Then:

  $\abs{ \exp(z+h) - \exp(z) - \exp(z)h }$

  $< \abs{ \sum_{k=0}^m \lfrac1{k!} ((z+h)^k-z^k) - \exp(z)h } + 2|h|^2$

  $= \abs{ \sum_{k=0}^m \lfrac1{k!} \big( k z^{k-1} h + \sum_{j=2}^k \binom{k}{j} z^{k-j} h^j \big) - \exp(z)h } + 2|h|^2$

  $\le \abs{ \sum_{k=1}^m \lfrac1{(k-1)!} z^{k-1} h - \exp(z)h } + \sum_{k=0}^m \lfrac1{k!} \sum_{j=2}^k \binom{k}{j} |z|^{k-j} |h|^j + 2|h|^2$

  $\le \abs{ \Big( \sum_{k=0}^m \lfrac1{k!} z^k - \exp(z) - \lfrac1{m!}z^m \Big) h } + \sum_{k=0}^m \lfrac1{k!} \sum_{j=2}^k \binom{k}{j} (|z|+1)^k |h|^2 + 2|h|^2$

  $\le ( |h|^2 + |h|^3 ) + \sum_{k=0}^m \lfrac1{k!} 2^k (|z|+1)^k |h|^2 + 2|h|^2$

  $\le 3|h|^2 + |h|^3 + \exp(2|z|+1) |h|^2$

Therefore $\exp'(z) = \exp(z)$.

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You cannot fine tune reality.

This result is not merely a convention, like choosing to work with a unit circle or a circle with radius $.374$.

$\pi$ is a fundamental constant that determines the ratio of the circumference of a circle to its diameter. This is not "fine tuned"; this is a fundamental constant

The imaginary unit is fundamental in that expands the set of real numbers to make it so that every polynomial with coefficients in $\mathbb C$ has a root in $\mathbb C$. Note that this is not true in $\mathbb R$

The exponential function is fundamental (among other things) because is the unique function (apart from multiplicative constants) that solves the differential equation $y'=y$

All of these results have not been fine-tuned. These are fundamental constants that were all discovered in very different settings to solve very different problems.

And somehow, the pieces fall together perfectly in a beautiful formula that relates togerther all these quantities. Isn't that something?

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    $\begingroup$ I would not call maths "reality", but this is another topic. $\endgroup$ – Vincent Mar 8 '17 at 17:13
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We didn't arbitrarily define anything. Everything was defined with a purpose, and they were defined with separate purposes.

The number $\pi$ is defined as a ratio. It is a number that has been known for ages, since before Euclid.

Complex numbers, while having a messier history, did not get anything like a strong treatment until the 16th century.

The first reference known for $e$ is in the 17th century.

These numbers come from different areas of mathematics. Complex numbers from a desire to find solutions to equations, $\pi$ from a desire to understand circles, and $e$ from a desire to more easily compute logarithms, which were used to do multiplication by addition. (Back in the days before computers, people published books of lookup tables for the values of logarithms and trig functions.)

It's not even remotely obvious what $e^{z}$ means when $z$ is complex. But it turns out that you can give it essentially only one meaning, if we want $e^z$ to be remotely like $e^x$ for $x$ real.

So, somehow, there is something "deep" going on here.

There is nothing about a good definition that makes similar things just magically appear. A good definition might leave an intractable problem. The usual purpose of a definition is to make things clearer, and to reveal a pattern. When you get a result like this, something feels "revealed."

The power series for $e^{z}$ comes up again and again. The moment generating function in probability, for example, or in matrices. There is something very deep about it. It comes up in combinatorics. Something about it helps us understand things far afield of where we originally encountered it.

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