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A few weeks back I asked a question which lead to Euler's formula being brought up. I don't have the mathematical background to fully appreciate it's purported mathematical beauty.

Just yesterday I wondered about the graph of the function $(-1)^x$ and after some exploration and research was again lead back to Euler's formula.

I'm wondering if someone can provide an explanation of Euler's formula sufficient to help me appreciate it in the context of periodic functions.

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theideasmith
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  • It depends on what you mean by a layperson. I don't think there is a entirely satisfactory explanation that does not involve an understanding of the exponential function and/or differential equations. The approach you take comes down to how one defines $\cos,\sin$. – copper.hat Mar 27 '16 at 17:43
  • One viewpoint (not an explanation) is to consider the complex numbers with $0$ removed. Any point $z$ can be written as $z = |z| {z \over |z|}$, and $|z|$ is just a real number, and ${z \over |z|}$ is a point on the unit circle. Since $|z|>0$ we can write it as $|z|=e^x$ for some real $x$ and since ${z \over |z|}$ is on the unit circle, we can write ${z \over |z|} = \cos \theta + i \sin \theta$ for some $\theta$ (there are many $\theta$s that will solve this, corresponding to wrapping around the circle). This gives $z= e^x(\cos \theta + i \sin \theta)$. Any $z\neq 0$ can be written this way... – copper.hat Mar 27 '16 at 18:04
  • (Cont.): This would suggest the definition $e^{i \theta} = \cos \theta + i \sin \theta$, so we can now write $z=e^{x+i \theta}$. This is not, of course, an explanation, but might provide some insight to the relationship. – copper.hat Mar 27 '16 at 18:05
  • Some of the proofs here might be helpful: http://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-eit-cos-t-i-sin-t – Hans Lundmark Mar 27 '16 at 18:15
  • @HansLundmark: Nice – copper.hat Mar 27 '16 at 18:23

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So what functions solve the DE $y" = -y$? That is, what functions can you differentiate twice, take the negative, and get the function back?

They are $\sin(x), \cos(x), e^{ix}$. However, because the original question was a second order DE, there is only two solutions, not three. That must mean $e^{ix}$ can be written as $A\sin(x) + B\cos(x)$. Solving for those constants gets you Euler's formula.

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