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.


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.