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Why is it easier to prove $e$ is transcendental than $\pi$?

I noticed that the proofs of $\pi$'s transcendence are much longer and have more details to check than those of $e.$ My guess is that it's simply because the infinite series for $e$ is much simpler and easier to work with than any series for $\pi$, but I wanted to see if there was more to it, so I asked my professor a few weeks ago as he was going over some bonus proofs of transcendence (including $\pi$ and $e$) thrown in at the end of a Galois theory textbook. He said he couldn't really see a philosophical justification for why one should be easier than the other.

Has this question been addressed in the literature, and if so, what do the transcendental number theory experts have to say about it?

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    $\begingroup$ @Carl-FredrikNybergBrodda Not sure what you mean, I gave my guess. Should I add more details and reasoning? $\endgroup$
    – Display name
    Commented May 31, 2021 at 9:59
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    $\begingroup$ You are asking: "why is it easier?". Do you actually mean to ask "is it easier?", or are you under the impression that it is easier, and you are wondering why? If so, what gave you this impression? $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented May 31, 2021 at 10:00
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    $\begingroup$ @Displayname One reason might be that we have the extremely simple formula for $e$, arising from the Taylor series of $e^x$. Compare with say the Taylor series for $arctan(x)$, which can be used to produce $pi/4$, but the speed of convergence is much slower. To prove trancendence, one wants an extremely rapid convergence. $\endgroup$
    – Per Alexandersson
    Commented May 31, 2021 at 10:07
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    $\begingroup$ The function $f(x)=e^x$ has very nice analytical properties and behaves very nicely with respect to both differentiation and integration. Loosely, this is what enables one to prove transcendence results about values of this function. Now the relation of this function to $e$ is obvious, it is just $f(1)$. To bootstrap this to work for $\pi$ we need to work harder, and employ a more involved relation $f(i\pi)=-1$. $\endgroup$
    – Wojowu
    Commented May 31, 2021 at 10:38
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    $\begingroup$ The fundamental reason is that $\pi$ is a period (in the sense of Kontsevich-Zagier), while $e$ is apparently not a period:-) $\endgroup$
    – Alexandre Eremenko
    Commented May 31, 2021 at 15:08

2 Answers 2

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This is not a complete answer, but I would say that the "standard" way to prove the transcendence of $\pi$ is as a corollary of the more general fact that $e^\alpha$ is transcendental for all nonzero algebraic $\alpha$. For general $\alpha$, one has to come up with a general method for dealing with those pesky algebraic numbers in the exponent. But for $\alpha=1$, clever ad hoc arguments are possible. For example, in the book Making Transcendence Transparent by Burger and Tubbs (which I highly recommend as a source for further details), they show how to write down explicitly a polynomial $\mathcal{P}_n(z)$ such that $\mathcal{P}_n(1), \mathcal{P}_n(2), \ldots, \mathcal{P}_n(d)$ provide exceptionally good rational approximations to $e, e^2, \ldots, e^d$ respectively. This proof does exploit special properties of the series $\sum_n z^n\!/n!$ so this perhaps vindicates your intuition that $e$ is easier because we have a nicer series for $e$ than for $\pi$. On the other hand, this argument is a bit circular, because isn't $${\pi \over 4} = 1 - {1\over 3} + {1\over 5} - {1\over 7} + \cdots$$ a "nice" formula for $\pi$? Well, maybe, but it's not "nice" in a way that lets us prove transcendence! Hmmm…

So I think that the answer is that we don't know of a way to prove the transcendence of $\pi$ that is significantly simpler than a proof of a more general result, whereas we do know some ad hoc tricks that work for $e$. In principle this could change in the future if, for example, someone finds an amazingly simple ad hoc proof for the transcendence of $\pi$, or (less likely) a dramatic simplification of the proof of the transcendence of $e^\alpha$ for all nonzero algebraic $\alpha$.

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    $\begingroup$ Of related interest is Direct proof that $\pi$ is not constructible: Is there a proof that $\pi$ is not constructible that doesn't essentially prove the stronger result that $\pi$ is not algebraic? As I once said in sci.math, "$\ldots$ proving that pi is transcendental is a huge over-kill for simply showing that pi is not constructible. It's like showing a certain real number is not an integer multiple of 8 by showing the real number under consideration is irrational." $\endgroup$
    – Dave L Renfro
    Commented Jun 1, 2021 at 18:40
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In Hardy & Wright, An Introduction to the Theory of Numbers, 6th edition, Theorem 204 is "$e$ is transcendental," and Theorem 205 is "$\pi$ is transcendental." For the latter, they write, "The proof is very similar to that of Theorem 204, but there are one or two slight additional complications." I think that a full answer to the question lies in going through both proofs, to see what those additional complications are, and why they are needed in the study of $\pi$ and not in the study of $e$.

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