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I read a bit about this equation: $e^{i\pi}=-1$ For someone knowing high school maths this perplexes me. How are these three irrational numbers so seemingly smoothly related to one another? Can this be explained in a somewhat intuitive manner? From my perspective it is hard to comprehend why these almost arbitrary looking irrational numbers have such a relationship to one another. I know the meanings and origins behind these constants.

Hans Lundmark
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Andreas Hartmann
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    Just to clarify, $i$ is not an irrational number – Eleven-Eleven Apr 06 '15 at 19:53
  • In this case, $e$ doesn't stand for the irrational number $e$ but for the trigonometric function : $\forall x, e^{ix} = \cos x + i\sin x$, where $i$ is the imaginary (not irrational) number defined as : $i^{2} = -1$. Thus in your case, $e^{i\pi} = \cos\pi + i\sin\pi = -1 + 0 = -1$. – krirkrirk Apr 06 '15 at 19:57
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    @Eleven-Eleven So you claim $i\in\mathbb Q$? – Hagen von Eitzen Apr 06 '15 at 20:00
  • While $e^{i\pi} = -1$ might look a little curious... Given what you may know about $2\pi$, does $e^{2\pi i} = 1$ make any new intuitive connections for you? – jameselmore Apr 06 '15 at 20:00
  • @krirkrirk In this case, $x\mapsto e^x$ is the natural exponential function, and the irrational number $e$ is its base. – JiK Apr 06 '15 at 20:05
  • @Eleven-Eleven $i$ is an irrational number simply because it is not a rational number. –  Mar 16 '17 at 02:31
  • @Dove: the word irrational number does not signify anything which is not rational, but rather it means a real number which is not rational. So $i$ may be perhaps called a non-rational number but it simply isn't irrational. – Paramanand Singh Jul 27 '17 at 03:08

3 Answers3

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Just think about it this way: $\pi$ is related to the circle, whose equation is $x^2+y^2=r^2$. Euler's constant e is related to the hyperbola, whose equation is $x^2-y^2=r^2$. In order to turn $y^2$ into $-y^2$ we need a substitution of the form $y\mapsto iy$.

Lucian
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    You might want to clarify how $e$ is related to the hyperbola, because it is not a priori obvious. –  Apr 06 '15 at 19:57
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    @Rahul: If we take the hyperbola $x^2-y^2=1$ and rotate it by $45^\circ$, then magnify each coordinate by $\sqrt2$, we get $y=\dfrac1x$ . At the same time, by definition, e is the number with the property that $\displaystyle\int_1^{\large e}\dfrac{dx}x~=~1$. – Lucian Apr 06 '15 at 20:04
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The most intuitive explanation I know involves a combination of three facts:

  1. If a particle has position $p(t)$ proportional to its velocity $p'(t)$, say $p'(t) = kp(t)$, then $p(t) = A e^{kt}$ for some constant $A$. We can take this as a definition of the exponential.

  2. In the complex plane, multiplication by $i$ is the same as a counter-clockwise rotation by $90$ degrees.

  3. A particle whose velocity is perpendicular to its position travels at constant speed, tracing a circle centered at the origin.

Combining these three, we find that the particle with position $p(t) = e^{it}$ travels uniformly in a circle. Since $p(0) = 1$ and $p'(0) = i$, we must have $p(t) = \cos{t} + i\sin{t}$, so that $p(\pi) = -1$.

Andrew Dudzik
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The natural definition of $e$ is not a definition of that single number, but rather of a special function $x\mapsto \exp(x)$, which has the property that $\exp(0)=1$ and that it is its own derivative: $\exp'(x)=\exp(x)$. It turns out, that $\exp$ is uniquely determined by this.

It follows that for any constant $k$, $\exp'(kx)=k\exp(kx)$ and $\exp''(kx)=k^2\exp(kx)$ and so on. Now both sine and cosine are functions with $f''(x)=-f(x)$, which matches the above if one picks $k$ so that $k^2=-1$, i.e., let $k=i$. A closer look show sthat $f(x)=\exp(ix)$ has $f(0)=1$ and $f'(0)=i$, so it looks exactly like $\cos x+i\sin x$ (and by another uniqueness argument, "looks like" means "equals" here). Therefore $\exp(i\pi)=\cos \pi+i\sin\pi=-1$ is a simple value again - in the end this is by the very definition of $2\pi$ as the period of sine and cosine.

Writing $\exp(x)$ instead of $e^x$, this may look less like magic. One can motivate by the fact that the properties and uniqueness of $\exp$ imply the functional equation $\exp(x+y)=\exp(x)\exp(y)$ (which looks like the rules for powers) that one customary uses the suggestive notation $e^x$ for $\exp(x)$ where $e$ is defined as $\exp(1)$.

Hagen von Eitzen
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