I have been reading David Angell's lovely book, Irrationality and Transcendence in Number Theory, which has given me some fresh insights even with some of the easier proofs. But the book reminds me of something that I've long been puzzled by, which the book hasn't cleared up for me. In the book's proof that $e^r$ is irrational for nonzero rational $r$, the starting point is to consider an integral of the form $$\tag{*}\label{star}\int_0^r f(x) e^x \thinspace dx$$ for some suitably chosen polynomial $f(x)$. Lest you think this is an isolated trick used only in this one proof, let me mention that perhaps the shortest known proof of the irrationality of $\pi$ is based on the same idea (with $\sin x$ in place of $e^x$) as are even more general irrationality/transcendence proofs. My main question is:

Why would you think that multiplying by a polynomial and integrating is a fruitful idea here?

To give a better idea of what I'm asking, let me make a few remarks about how far I've gotten with trying to answer my own question. It makes sense that for an irrationality proof, we should be looking for good approximations to $e^x$. Angell attributes the idea for his proof to Hermite, and it occurs to me that Hermite is also well known for his work on orthogonal polynomials. So maybe the thought process could go something like this: If Taylor series aren't working for us (Fourier's proof of the irrationality of $e$ does use the Taylor series, but if you try to mimic it naïvely to prove the irrationality of $e^r$, you run into some difficulties when $r>1$), then maybe we can try to approximate $e^x$ using orthogonal polynomials. This idea would at least get integration into the picture. So I have a side question:

Is there any historical evidence that Hermite's proof of the transcendence of $e$ was partially inspired by his work on orthogonal polynomials?

One problem with this line of heuristic reasoning is that the best choices for the polynomial $f(x)$ seemingly have nothing to do with classical orthogonal polynomials. For the proof of the irrationality of $e^r$ or Niven's proof of the irrationality of $\pi$, the appropriate choice of $f(x)$ turns out to be something of the form $$\tag{**}\label{starstar}f(x) = {x^n (a-bx)^n\over n!}$$ where $a/b$ is the rational number that we are trying to prove can't exist. Well, if you get as far as thinking that Formula \eqref{star} might be helpful, then it's not so hard, by staring at what you get via integration by parts, to "optimize" the choice of $f(x)$. So maybe that's all there is to it: motivated by a vague analogy with the theory of orthogonal polynomials, we are led to consider Formula \eqref{star}, and then we find the optimum choice of $f(x)$ for our purposes. However, I'm still left wondering if there's some general theory that can be used to justify that polynomials of the form \eqref{starstar} are interesting. Here's one observation to suggest that something more might be hiding behind the scenes. Suppose we evaluate $$ \int_0^1 {x^n(1-x)^n\over n!} e^x \thinspace dx.$$ We find that we get (some of) the continued fraction convergents for $e$—or more precisely, we get $p-qe$ for convergents $p/q$. (By contrast, if we truncate the Taylor series for $e$, we don't generally get convergents for $e$.) So a third question might be:

Is there a way to see a priori, without "just working it out," that we should expect the continued fraction convergents to emerge from $e$ in this manner?

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    $\begingroup$ Regarding the last question, have you seen Henry Cohn's arxiv.org/abs/math/0601660 ? $\endgroup$ Apr 3 at 18:42
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    $\begingroup$ @DavidESpeyer Excellent pointer...thanks! I'll have to read it more carefully, but it looks very relevant. $\endgroup$ Apr 3 at 19:43
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    $\begingroup$ I know it isn't the heart of your question, but could you yell a bit more on how lovely the book is? $\endgroup$
    – PseudoNeo
    Apr 3 at 20:00
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    $\begingroup$ @PseudoNeo In transcendental number theory, some highly nontrivial results can be proved with surprisingly concise arguments. But it's often unclear how someone would come up with the arguments in the first place. My previous favorite book, Making Transcendence Transparent, does a fine job of motivation. But I think Angell may become my new favorite book, because I find that his motivations are typically even clearer. YMMV, of course. But I should point out that Angell commits the common error of saying that Cantor's proof of the existence of transcendental numbers is non-constructive. $\endgroup$ Apr 3 at 20:18
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    $\begingroup$ Maybe the key background was not orthogonal but rather orthonormal polynomials — the exact values of the normalizing constants are given by identities like $\int_0^\infty x^n e^{-x}dx=n!$ that Hermite would have known better than most of us, and are useful for the proofs of irrationality. (This is one thing I learned by working through details at mathoverflow.net/questions/345810/…, but even after I worked that out the inspiration never looked obvious.) $\endgroup$
    – user44143
    Apr 3 at 20:36

2 Answers 2


Here's an exposition of Niven's proof that makes the connection to orthogonal polynomials explicit. We start with an observation, easily proven by induction, that if $P\in \mathbb{Z}[x]$, then $\int_0^\pi P(x)\sin x=Q(\pi)$, where $Q\in \mathbb{Z}[x]$, and $\deg Q\leq\deg P$. If we can find a sequence of polynomials $P_n\in \mathbb{Z}[x]$, $\deg P_n=n$, such that $\int_0^\pi P_n\sin x$ tends to zero super-exponentially, then we are done, since then the assumption that $\pi=a/b$ implies that $b^n\int_0^\pi P_n(x)\sin x$ is an integer (and non-zero for infinitely many $n$, since $\sin$ is not a polynomial).

Legendre polynomials on $(0,\pi)$ are perhaps the most natural try: $P_n$ is orthogonal on $(0,\pi)$ to all polynomials of degree $\leq n-1$, and hence to the Taylor polynomial of $\sin x$ of degree $n-1$. The remainder decays super-exponentially, so that $$ \left|\int_0^\pi P_n(x)\sin(x)\,dx\right|\leq\frac{\pi^{n+1}}{n!}||P_n||_2. $$ From that point on, the only fact that you need to know about Legendre's polynomials to see that the proof is bound to work is that if you normalize them to have integer coefficients, the norm will be only exponentially large. Indeed, by Rodrigues' formula, we have for Legendre's polynomials on $(0,\pi)$ $$ P_n(x)=\frac{\pi^{-n}}{n!}\frac{d^n}{dx^n}((x-\pi)x)^n=\frac{1}{n! a^n}\frac{d^n}{dx^n}((bx-a)x)^n, $$ and those will have norm $\sqrt{\frac{\pi}{2n+1}}$ and have integer coefficients after multiplication by $a^n$, so we are done.

To obtain Niven's integral, plug in the above expression into $\int_0^\pi P_n(x)\sin (x)\, dx$ and integrate by parts $n$ times, replacing $\sin$ by $\cos$ if $n$ is odd. (Note that since all the derivatives of $(bx-a)^nx^n$ up to $(n-1)$-st vanish at the endpoints of the interval, there will be no boundary terms.) This gives an easy bound on the integral without reference to orthogonality and Taylor expansion, and also positivity. At the same time, it conceals that Legendre's polynomials were ever there :).

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    $\begingroup$ This is really interesting! Where can I find more details about this (in particular, how exactly to relate Legendre polynomials to Niven's argument)? $\endgroup$ Apr 5 at 0:00
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    $\begingroup$ @TimothyChow, I don't know a refence - I spelled out a bit more details in the post, hope that helps. $\endgroup$
    – Kostya_I
    Apr 5 at 7:48
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    $\begingroup$ Thanks for providing more details. I like this approach, but I'm still confused. In what sense are the Legendre polynomials "orthogonal to the first $n$ terms in the Taylor expansion of $\sin x$"? This sounds like a claim that, e.g., $P_3(x)$ times $x - x^3\!/3!$ integrates to zero over some natural interval, but I can't see how to make this work. Also, I can see that integrating $P_n(x)$ (as you've normalized it) $n$ times gives you Niven's polynomial, but if I integrate by parts $n$ times as you suggest, don't I get a finite series of terms involving other polynomials? $\endgroup$ Apr 5 at 15:12
  • $\begingroup$ @TimothyChow, I'm not sure what was exactly was the source of confusion, but I added a bit more details. $\endgroup$
    – Kostya_I
    Apr 12 at 11:12
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    $\begingroup$ Thanks...I have accepted your answer. I figured out what I was doing wrong...I forgot to change variables to "normalize" the Legendre polynomials to the interval $(0,\pi)$. $\endgroup$ Apr 14 at 22:33

Hermite's approximations to values of $e^x$ are based on good rational function approximations to $e^x$, which nowadays go under the name of Padé approximations (a name that came much later: Padé was 10 years old when Hermite proved $e$ is transcendental). Padé approximations are related to continued fraction expansions of functions. See Waldschmidt's notes on this topic here, here, and here.

You ask if there is a connection to orthogonal polynomials, since at first glance it doesn't seem like there is one. This is a good opportunity to point out that the way everyone learns about orthogonal polynomials today (via linear algebra, or more specifically Gram-Schmidt for some inner product on a function space) is not how orthogonal polynomials were originally discovered. They were initially studied (by Gauss, Jacobi, Chebyshev, Stieltjes, et al.) as denominators of continued fractions! Here is a theorem describing that relation.

Theorem. When $w(x)$ on $(a,b)$ has moments $\mu_n = \int_a^b x^n w(x)\,dx$ and the formal power series $\sum_{n \geq 0} \mu_nt^n$ is written as a suitable generalized continued fraction, the denominators $q_n(t)$ of its convergents are orthogonal for $w$: $\int_a^b q_m(x)q_n(x)w(x)\,dx = 0$ if $m \not= n$.

There is a whole book on the relation between orthogonal polynomials and continued fractions by S. Khrushchev: Orthogonal Polynomials and Continued Fractions from Euler’s Point of View, Cambridge Univ. Press, 2008. See also T. H. Kjeldsen, “The Early History of the Moment Problem,” Historia Mathematica 20 (1993), 19–44.

Another unexpected role for continued fractions in the development of analysis is that work by Stieltjes on convergence of analytic continued fractions is what led him to create the Stieltjes integral.

To address your first question, which is also in the subject line, I am pretty sure the first time anyone used integration against polynomials to prove transcendence or irrationality was Hermite in his proof of the transcendence of $e$. Having shown such a method was capable of leading to new results, others were inclined to try it out on similar problems, e.g., Lindemann's proof of the transcendence of $\pi$ is modeled on Hermite's proof for $e$. The modern short proofs of irrationality of $\pi$ using carefully chosen integrals could be regarded as descendants of Hermite's proof too. So the popularity of using sneaky integrals in such proofs just illustrates the common phenomenon that after a new idea solves one open problem, others will try out the idea on similar problems, and as it continues to be successful the original trick just becomes yet another method.

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    $\begingroup$ Thanks for spelling out these connections. I was vaguely aware of some connection to Padé approximants, which go a long way toward addressing my third question. But I guess it still confuses me that Hermite came up with his proof before Padé approximants were a thing. Did Hermite spell out some of these connections in his original paper? $\endgroup$ Apr 3 at 20:24
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    $\begingroup$ I will have to read Hermite's paper, which I see is reprinted in Pi: A Source Book. Time to brush up on my French! $\endgroup$ Apr 4 at 0:56
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    $\begingroup$ Would it be correct to say that the very similar proofs by Ivan Niven and Mary Cartwright are both simplifications of Hermite's proof? $\endgroup$ Apr 4 at 14:15
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    $\begingroup$ @MichaelHardy Hermite's proof of irrationality of $\pi$ on the page en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational looks similar to the proofs by Cartwright and Niven, and the end of that page's account of Niven's proof does say "Niven's proof is closer to Cartwright's (and therefore Hermite's) proof than it appears at first sight" and it shows how. $\endgroup$
    – KConrad
    Apr 4 at 14:37
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    $\begingroup$ "similar differences" remind me of the quote attributed to Dale Berra: "You can't compare me to my father – our similarities are different." $\endgroup$ Apr 6 at 1:04

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