Today I Learned

June 2nd

Today I learned a proof that $e^q$ is irrational for any $q\in\QQ\setminus\{0\},$ following Hermite. Note that we can write $q=n/m$ for $n$ a positive integer and $m$ an integer, and $e^{n/m}$ rational implies $\left(e^{n/m}\right)^m=e^n$ rational, so it suffices to show (by contraposition) that $e^n$ is irrational for positive integers $n.$

This proof is truly magical. Our result is going to come from bounding: we are going to create an arbitrarily small integral that the rationality of $e$ forces to be an integer, and that will yield contradiction. Of course, we have to use something about $e,$ so we are going to use the fact that\[\frac d{dx}e^{nx}=ne^{nx}.\]Note this is a defining property of $e,$ so we can be sure we are using its full power. We would like to use this to make a (small) integral, so we have to introduce a (small) auxiliary differentiable function $F:\RR\to\RR$ here: the product rule tells us that\[\frac d{dx}F(x)e^{nx}=F'(x)e^{nx}+nF(x)e^{nx}.\]We promised an integral, and indeed, our integral is going to be\[\int_0^1\left(F'(x)+nF(x)\right)e^{nx}\,dx=F(x)e^{nx}\bigg|_0^1=F(1)e^n-F(0).\]When we suppose that $e^n$ is rational (which we will postpone as long as possible), we might hope that $F(0)$ and $F(1)$ are integers so that the above integral can be made into an integer when multiplied by the supposed denominator of $e^n.$

To proceed, we recognize we will need better control of $F'(x)+nF(x)$—after all, we have to make its integral arbitrarily small. So we define, for now, $f(x):=F'(x)+nF(x)$ and attempt to solve for $F$ in terms of $f.$ Well, we see\[F(x)=\frac1nf(x)-\frac1nF'(x).\]Of course, we don't know what $F'$ is directly, but we can solve for it by differentiating the above into $F'(x)=\frac1nf'(x)-\frac1nF''(x),$ so we see\[F(x)=\frac1nf(x)-\frac1{n^2}f'(x)+\frac1{n^2}F''(x).\]We can continue this process inductively and just write\[F(x)=\sum_{k=0}^\infty\frac{(-1)^k}{n^{k+1}}f^{(k)}(x),\]assuming everything converges properly. We can check that, assuming nice enough convergence (which we will have), this does indeed give $f(x)=F'(x)+nF(x).$

We said above that it would be nice for $F(0)$ and $F(1)$ to be integers, which makes choosing an $f(x)$ to define $F(x)$ a bit annoying: after all, there's that infinite sum to deal with. Our way out is to make $f(x)$ a polynomial, thus making the sum finite, and then multiplying $f$ with a very large power of $n$ to accommodate for the given denominators.

Now we introduce the star of our story. Choose $N$ very large, to be fixed later, making\[f_0(x):=\frac{x^N(1-x)^N}{N!}\]very small. Then we define $f(x):=n^{2N+1}f_0(x)$ to deal with the denominators so that\[F(x):=\sum_{k=0}^\infty\frac{(-1)^k}{n^{k+1}}f^{(k)}(x)=\sum_{k=0}^{2n}n^{2N-k}f_0^{(k)}(x).\]It might be totally obvious that $F(0)$ and $F(1)$ are integers, but they are: we can write\[f_0(x)=\sum_{k=N}^{2N}\binom N{2N-k}(-x)^k.\]Reading this off like a Taylor expansion, we see $k \lt N$ gives $f_0^{(k)}(0)=0,$ and for $k\ge N,$ we get $f^{(k)}(0)=(-1)^kk!\binom N{2N-k},$ and $k!$ does enough to cover the denominator of $N!.$ This shows the $f_0^{(k)}(0)$ are integers, so $F(0)$ is an integer. Further,\[f_0(x)=f_0(1-x),\]so $f_0^{(k)}(1)=(-1)^kf_0^{(k)}(0)$ is also an integer for any $k,$ meaning that $F(1)$ is an integer.

Now surely $f(x)=F'(x)+nF(x)$ still holds from our work above, for we have the nicest converge, from a finite sum. We recall our integral as\[\int_0^1f(x)e^{nx}\,dx=F(x)e^{nx}\bigg|_0^1=F(1)e^n-F(0).\]However, we can bound the integral because $f(x)=n^{2N+1}f_0(x) \lt n^{2N+1}/N!$ for $x\in(0,1).$ (Our lower-bound is that the integral is positive.) Thus,\[0 \lt \int_0^1f(x)e^{nx}\,dx \lt \frac{n^{2N+1}}{N!}\cdot e^n.\]This can be made arbitrarily small because $N!$ is faster than exponential, which is a problem because $F(1)e^n-F(0)$ has its denominator bounded by the denominator of $e^n.$

To formalize, we finally suppose for the sake of contradiction that $e^n=p/q.$ Then\[q\int_0^1f(x)e^{nx}\,dx=pF(1)-qF(0)\]is an integer, but\[0 \lt q\int_0^1f(x)e^{nx}\,dx \lt \frac{pn^{2N+1}}{N!}.\]Taking $N$ large enough, we can force $N! \gt pn^{2N+1},$ making the upper-bound less than $1.$ This gives us an integer between $0$ and $1,$ which is our contradiction. This finishes. $\blacksquare$

To close off loose ends, we show that we can in fact make $\frac{n^{2N+1}}{N!}$ arbitrarily small as $N\to\infty.$

For all $N,$ note that we can say\[\log N!=\sum_{k=1}^N\log k \gt \sum_{k=\ceil{N/2}}^N\log k \gt \frac N2\log\frac N2.\]There is a technical detail here that there are at least $N/2$ integers between $\ceil{N/2}$ and $N$ inclusive, but this is true. (If $N$ is even, there are $N/2+1$ integers; if $N$ is odd, there are $N/2+1/2$ integers.) Anyways, this implies that\[N! \gt (N/2)^{N/2}.\]If we let $N=2n^6k$ for some large positive integer $k,$ we see\[\frac{n^{2N+1}}{N!}\le\frac{n^{3N}}{(N/2)^{N/2}}=\frac{n^{6n^6k}}{n^{6n^3k}\cdot k^{n^6k}}=\frac1{k^{n^6k}}.\]This is less than $\frac1k,$ for example, so we can make the original quantity as small as we please, finishing. $\blacksquare$

As commentary, I have tried my best to motivate this proof, but it is truly magical. There is a pretty large jump to the introduction of the $f_0,$ and from there everything comes together. I really don't know where this proof comes from.

In other news, today was senior salute. It was a good day.