The Wallis Product


I had seen the Wallis product, but I did not know a derivation of it until a few weeks ago, when I discovered this gem when reading the wonderful book Sources in the Development of Mathematics by Ranjan Roy. I’m storing this example to use next time I teach calculus, as it requires nothing more than integration by parts.

Like in the case of many other such things in mathematics, I don’t know what the Wallis product actually is; I can only discover it by going through a derivation. So, I won’t give away the answer until we get there. But the goal is to write down an infinite product for \pi.

In order to do this, we evaluate the integral I_n = \int_0^\pi \sin^n x\, dx. The first few are easy:

I_0 = \int_0^\pi dx=\pi

I_1 = \int_0^\pi \sin x\, dx = 2.

I can keep working out more of these, but now it’s time to tackle the general case, using integration by parts. We assume n\ge 2, and we compute \int_0^\pi u\, dv, where u=\sin^{n-1} x and v=\sin x\, dx, so that

I_n = \int_0^\pi \sin^n x\, dx = \left[-\sin^{n-1} x\cos x\right]_0^\pi +\int_0^\pi (n-1)\sin^{n-2} x\cos^2 x\, dx = \int_0^\pi (n-1)\sin^{n-2}x(1-\sin^2 x)\, dx = (n-1)(I_{n-2}-I_n).

Solving for I_n, we get I_n = \frac{n-1}{n}I_{n-2}. Now it’s easy to compute more of the I_n‘s:

I_2 = \pi\times\frac{1}{2}

I_3 = 2\times\frac{2}{3}

I_4 = \pi\times\frac{1}{2}\times\frac{3}{4}

I_5 = 2\times\frac{2}{3}\times\frac{4}{5},

and in general

I_{2n} = \pi\times\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{2n-1}{2n}

I_{2n+1} = 2\times\frac{2}{3}\times\frac{4}{5}\times\cdots\times\frac{2n}{2n+1}.

Now, note that for any n, I_{n-1}>I_n>I_{n+1}, since \sin^{n-1} x\ge \sin^n x\ge\sin^{n+1} x whenever 0\le x\le\pi. Furthermore, since \frac{I_{n+1}}{I_{n-1}}=\frac{n}{n+1}, this means that eventually I_{2n-1} and I_{2n} get very close together. So

\pi\times\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{2n-1}{2n}\approx 2\times\frac{2}{3}\times\frac{4}{5}\times\cdots\times\frac{2n}{2n+1},

or

\frac{\pi}{4}\approx \frac{2}{3}\times\frac{4}{3}\times\frac{4}{5}\times\frac{6}{5}\times\frac{6}{7}\times\cdots\times\frac{2n}{2n+1}.

(If we wanted, we could be more precise, and give two inequalities for \frac{\pi}{4}, but it doesn’t add much value, in my opinion, to do that here.)

Taking the limit, we get

\frac{\pi}{4}= \frac{2}{3}\times\frac{4}{3}\times\frac{4}{5}\times\frac{6}{5}\times\frac{6}{7}\cdots.

This is the Wallis product.

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About Simon

Hi. I'm Simon Rubinstein-Salzedo. I'm a mathematics postdoc at Dartmouth College. I'm also a musician; I play piano and cello, and I also sometimes compose music and study musicology. I also like to play chess and write calligraphy. This blog is a catalogue of some of my thoughts. I write them down so that I understand them better. But sometimes other people find them interesting as well, so I happily share them with my small corner of the world.
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