Is there an essential difference between Cartwright's and Niven's proofs of the irrationality of $\pi$?

Question

In an appendix to the third edition of Scientific Inference, Harold Jeffreys wrote:

The following was set as an example in the Mathematics Preliminary Examination at Cambridge in 1945 by Dame Mary Cartwright, but she has not traced its origin.

. . . and there follows a short and simple proof of the irrationality of $\pi.$

In 1947 a proof of the same proposition by Ivan Niven was published in the Bulletin of the American Mathematical Society.

See Wikipedia's article about these: Cartwright, Niven

Niven considers the function $f(x) = \dfrac{x^n(a-bx)^n}{n!}$ for $0\le x\le \pi = \dfrac a b$ and the integral $\displaystyle\int_0^\pi f(x) \sin x\, dx.$ Note that $\sin x$ is $0$ if $x$ is at either endpoint.

Cartwright works with the integral $\displaystyle \int_{-1}^1 (1-x^2)^n \cos(\alpha x)\,dx.$ Note that $(1-x^2)^n$ is transformed to $x^n(a-bx)^n$ by an affine transformation of the domain, taking $[-1,1]$ to $[0,\pi].$ And the trigonometric function $\cos(\alpha x)$ is $0$ at both endpoints if $\alpha=1.$ Later in the argument Cartwright divides a multiple of this function of $\alpha$ by $n!,$ paralleling what Niven did.

Are these proofs essentially the same?

Maybe I'll post my own answer if I figure this out before better answers are posted.

Answer 1

It's hard to say without knowing what differences should be considered essential.

Their approach certainly isn't fundamentally different. Both use the same mechanic of bounding something that's supposed to be a positive integer by something that vanishes. The vanishing quantity is exponential over factorial in both cases, too.

Both get to the key inequality by crunching out a formula using integration by parts. Since the functions being integrated are the same up to a transformation, the main difference seems to be whether you want to crunch a formula out of Niven or Cartwright's integrals. That calculation doesn't differ much because the role played by the recurrence relation for $J_n$ in Cartwright's proof is encapsulated in Niven's definition of $F$.

From the above, I would say the difference between the proofs is mostly presentation.

Answer 2

Here is an argument for one possible answer to this question, followed by “However . . .”

Niven has $\pi = a/b$ and $$ f(x) = \frac{x^n(a-bx)^n}{n!} = \frac{x^n(\pi-x)^n}{n!} b^n \quad \text{for } 0\le x\le \pi. $$ He shows that $$ 0 < \int_0^\pi f(x)\sin x \, dx \in \mathbb Z \tag 1 $$ but that $$ 0 < f(x) \sin x < \frac{\pi^n a^n}{n!}, $$ so that $(1) \to 0$ as $n\to\infty,$ thereby getting a contradiction.

Cartwright as reported by Jeffreys has $\frac 1 2 \pi = \dfrac b a$ and $$ I_n = \int_{-1}^1 (1-x^2)^n \cos\left( \tfrac \pi 2 x\right) \, dx, $$ which is the same integral as in $(1)$ modulo an increasing affine mapping that takes $[0,\pi]$ to $[-1,1],$ except that she hasn't yet divided by $n!.$

Then she shows that $$ 0 < \frac{b^{2n+1}}{n!} I_n \in\mathbb Z, \tag 2 $$ and clearly $(2)\to0$ as $n\to\infty.$

The above makes it look as if they both did the same thing, except for trivial details.

However, the method of showing that the thing reported to be a member of $\mathbb Z$ is in fact a member of $\mathbb Z$ may or may not be different in some way that is of interest. So if I update this preliminary answer later, I may address that.