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There is this popular video going around that shows $\pi$ is irrational using visualization of the function, $z(\theta) = e^{i \theta} + e^{i \theta \pi}$. I understand the reason intuitively, both the functions have different periodicities so the sum is not going to be periodic unless one of the periodicities is an integral multiple of the other.

Here is my attempt outline at proving it mathematically. First find the roots of the function, my idea is if the function is going to repeat itself than it should be some integer multiple of the roots. If I can prove there is no integer multiple of the roots that has same values for the function $z(\theta)$ (other than 0) then it can't be periodic. Does this make sense?

Travis Willse
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Naren Nallapareddy
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    It does not have to be an integer multiple. Rational would do. – badjohn Oct 29 '23 at 04:42
  • Ah that does make sense but you can't get rational number from multiplication of any rational number with pi right? – Naren Nallapareddy Oct 29 '23 at 04:45
  • Yes, it is true that if the ratio is irrational then the sum will not be periodic. However, trying to prove that one period is an integral multiple of the other is too much and doomed to fail since it isn't true. Another problem is that functions can be periodic without having any roots. – badjohn Oct 29 '23 at 05:07
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    A "visualization" of a function can't prove that the function isn't periodic. If you replace $\pi$ with a rational with a large denominator, you will get a periodic function, but it might not look periodic if you don't see enough of it. – Gerry Myerson Oct 29 '23 at 06:45
  • Surely one should prove $\pi$ is irrational, then prove this function is aperiodic, not the other way round. – J.G. Oct 29 '23 at 07:48
  • @GerryMyerson Yes of course, I am not trying to prove pi is irrational rather the function is not periodic because pi is irrational. – Naren Nallapareddy Oct 29 '23 at 16:54
  • @J.G. That wasn’t my intention, I am accepting pi is irrational. Sort of like validating the hypothesis that the function is not periodic because pi is irrational although original authors intention was to create a visualization about the implication of pi being irrational. – Naren Nallapareddy Oct 29 '23 at 16:57
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    "shows $\pi$ is irrational using visualization of the function" gave me a different impression. I sincerely doubt there's any way to prove the function is aperiodic other than by using $\pi$'s irrationality. – J.G. Oct 29 '23 at 17:04
  • Absolutely, my bad. I wasn't clear. – Naren Nallapareddy Oct 29 '23 at 18:31

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Hint Suppose the that $z(\theta)$ has period $T > 0$. Then, $$2 = z(0) = z(T) = e^{iT} + e^{\pi i T} .$$ Thus, $e^{i T} = e^{i \pi T} = 1$. (Why?)

In particular there are integers $p, q$ such that $T = 2 \pi q$ and $\pi T = 2 \pi p$. Dividing gives $\pi = \frac{p}{q} ,$ which contradicts the irrationality of $\pi$.

Travis Willse
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  • Awesome, I think I ended up at the same overall expression. Obviously on a complex plane $e^{i\theta}$ intersects real number line at the most at two points. I just had additional constraint that T should be a rational multiple of root because the function does have roots. – Naren Nallapareddy Oct 29 '23 at 17:01