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I can see the value of $e^{i\pi}$ is $-1$, this value is round without decimals. The $e$ value can be defined as the value (v) that the derivative function of $v^{x}$ is still $v^{x}$. And the $\pi$ value is defined as the value of circumference over diameter of whichever circle. And $i$ is $\sqrt{-1}$ everybody knows.

These 3 values $e$ and $i$ and $\pi$ have no direct relations, but $e^{i\pi}$ is a round value, what is the intuition (the simple explanation) behind this?

Martin Sleziak
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Dan D.
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  • I don't think of $i$ as $\sqrt{-1}$, since the latter is not uniquely defined. – Randall Jun 25 '20 at 02:10
  • I think this has already been asked: https://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-ei-varphi-cos-varphi-i-sin-varphi. Some proofs there can be considered intuitive. – Poder Rac Jul 01 '20 at 12:00

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You see, there are other seemingly unlikely identities... $${\left(\sqrt{2}^{\sqrt2}\right)}^{\sqrt{2}} = 2,$$ $$\log 2 + \log 5 = 1,$$ etc.

The fact is, $\pi$ is defined such that the identity holds. Usually $\pi$ is defined as the smallest positive number such that $2\pi$ is a period of $f(x)=e^{ix}$. So $e^{i\pi}$ must be a square root of $1$, but since $2\pi$ is the smallest positive period, you can't take to positive square root. Therefore $e^{i\pi}=-1$.

To see why this is a good definition, consider doing this the other way: defining $\pi$ with circumference requires a definition of curve length, which requires the definition of improper integrals, the smallest positive period approach only requires the definition of $e^x$ as a power series sum, which only requires the definition of limits. You can also define $e^x$ as a unique solution to a differential equation, which is also straightforward. This is the definition used in Rudin's real analysis book.

Trebor
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    @JCAA See Rudin's excellent real analysis book for my definition of $\pi$. Also, the convention in my country is $\log$ for base 10 and $\lg$ for base 2. – Trebor Jun 25 '20 at 04:30
  • Defining $\pi$ with circumference doesn't require integral calculus. The only thing needed is a proof that the curve length exists, then you define $\pi$ using that length. – Poder Rac Jul 03 '20 at 18:11
  • @PoderRac But to prove that something exists, you need to define it first. And the curve length is defined using (one way or the other) integral calculus. – Trebor Jul 04 '20 at 05:34
  • You can use polygonal paths. Then the arc length $L(f)$ of a curve defined by $f:[a,b]\to\mathbb{R}^n$ is the limit of the sum of line segment lengths for a regular partition $[a,b]$ as the number of segments approaches infinity: $L(f)=\lim_{N\to\infty}\sum_{k=1}^N |f(t_k)−f(t_{k−1})|$ where $t_k=a+\frac{k(b−a)}{N}$ for $k=0,1,\ldots N$. Now, the definition is complete and does not involve integrals nor derivatives. After introducing derivatives and integrals, the definition can be shown to be equivalent to $L(f)=\int_a^b |f'(t)|, \mathrm dt$ given that $f$ is continuously differentiable. – Poder Rac Jul 04 '20 at 11:16
  • @PoderRac That's correct. – Trebor Jul 04 '20 at 11:57
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Really what's fundamental is not the number $e$ itself but the exponential function $x \mapsto \exp(x)$. The question is what happens when we feed this function purely imaginary numbers.

Whatever it is, it had better be on the unit circle, assuming the additive rule for $\exp$ extends to complex numbers, since $\exp(it)*\overline{\exp(it)} = \exp(0) = 1.$

So $t \mapsto \exp(it)$ is a map from a line to a circle.

If we believe that complex differentiation is to work like real differentiation, then the speed is $|d/dt \exp(it)| = |i \exp(it)| = 1$.

So the exponential function is wrapping the purely imaginary line around the unit circle at a constant speed of $1$. Since we start at $\exp(0) = 1$, by the time we're halfway around the circle we get to $\exp(\pi i) = -1$.

hunter
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  • (to clarify, this isn't intended as a "proof" of $\exp(i \pi) = -1$ since you have to figure out what the extension of the exponential function to the complex plane means in the first place; it's just "why you should believe it.") – hunter Jun 25 '20 at 04:57
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This is just geometry. $\sin(\pi)=0$ and $\cos(\pi)=-1$ which are trivial statements if you draw the picture. But by your reasoning $\sin$ is a complicated function and $\pi$ is a complicated number while $\sin(\pi)$ is the round number $0$.

markvs
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    I don’t think it’s as uninteresting as you imply. The base of the natural logarithm seems defined in a way that has no connection to geometry or trigonometry, yet the power series shows that $e^{i\theta} = \cos(\theta) +i \sin(\theta)$, as you’ve alluded to. I found the connection very surprising when I first learned about it, and still find it fascinating. – Joe Jun 25 '20 at 02:30
  • I didn’t say that you said it was uninteresting (but apparently that’s what you think I said). I said that you implied it’s uninteresting. That’s what I think a statement like “this is *just* geometry” implies. I also think that’s what making your sarcastic analogy to oddness disappearing implies. I think many people *do* find Euler’s formula fascinating. Feynman called it “the most remarkable formula in mathematics.” – Joe Jun 25 '20 at 03:19
  • "I am against fetishizing formulas in general" – that sentence is just sooo weird. It gave me goosebumps. – Poder Rac Jul 01 '20 at 12:02
  • What do you mean by my 'hole'? I don't quite understand your comment. – Poder Rac Jul 01 '20 at 12:16