Derivation for the Wallis Formula for pi

by | Jul 13, 2017 | Pure Maths | 1 comment

I’m doing a research project in the School of Physics at Monash University, looking at whether \pi is really found in quantum mechanics, which was suggested by a paper published in 2015 where they presented a quantum mechanical derivation of the Wallis product.

I’m finishing up now, and thought I’d present here a derivation of the Wallis Product. It involves Gamma Functions, the squeeze theorem and some good old fashion algebra. The start of this is actually the original derivation done by Wallis in 1655, but whereas he used recursion relations to expand the values of Wallis Integrals,

    \begin{equation*}W_n=\int_{o}^{\frac{\pi}{2}}\sin^n(x)dx,\end{equation*}

I used the gamma function representations of these integrals. So if n is odd, i.e. n=2p+1, we get

(1)   \begin{equation*} W_{2p+1}=\frac{p!\Gamma(\frac{1}{2})}{(2p+1)\Gamma(p+\frac{1}{2})} \end{equation*}

and if n is even, n=2p, the Wallis integral is

(2)   \begin{equation*} W_{2p}=\frac{\Gamma(\frac{1}{2})\Gamma(p+\frac{1}{2})}{2\Gamma(2p+1)}. \end{equation*}

Firstly, we shall use integration by parts on W_n=\int_{0}^{\pi} \sin^nxdx as follows:

    \[\int_{0}^{\frac{\pi}{2}} \sin^nx dx = \int_{0}^{\frac{\pi}{2}} \sin^{n-1}x \sin{x} dx\]

Let u=\sin^{n-1}x and dv=\sin{x} dx and doing the integration by parts we get:

    \begin{align*} \int_{0}^{\frac{\pi}{2}} \sin^nx dx &=\bigg[-\sin^{n-1}(x)\cos(x)\bigg]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\cos^2(x)(n-1)\sin^{n-2}(x)dx\\ &=(n-1)\int_{0}^{\frac{\pi}{2}}\sin^{n-2}(x)\cos^2(x)\\ \therefore W_n&=(n-1)\int_{0}^{\frac{\pi}{2}}\sin^{n-2}(x)dx-(n-1)\int_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx\\ \end{align*}

which is

    \[W_n = (n-1)W_{n-2}-(n-1)W_n.\]

(3)   \begin{equation*} \therefore \frac{W_n}{W_{n-2}}=\frac{n-1}{n}. \end{equation*}

Now note that for x \in [0,\frac{\pi}{2}], we have 0\leq \sin{x} \leq1, which means the following inequalities are true for x \in [0,\frac{\pi}{2}]

(4)   \begin{equation*} 0 \leq \sin^{2n+2}x \leq \sin^{2n+1}x \leq \sin^{2n}x. \end{equation*}

In terms of W_n notation, the inequalities in equation 4 are:

(5)   \begin{equation*} 0 \leq W_{2n+2} \leq W_{2n+1} \leq W_2n. \end{equation*}

Dividing the inequalities in equation 5 by W_{2n} we get:

(6)   \begin{equation*} 0 \leq \frac{W_{2n+2}}{W_{2n}} \leq \frac{W_{2n+1}}{W_{2n}} \leq 1. \end{equation*}

From equation 3, we find that

    \[\frac{W_{2n+2}}{W_{2n}}=\frac{2n+1}{2n+2},\]

which means that the inequalities in 6 become

(7)   \begin{equation*} 0 \leq \frac{2n+1}{2n+2} \leq \frac{W_{2n+1}}{W_{2n}} \leq 1. \end{equation*}

Now,

    \[\lim_{n\rightarrow\infty}\frac{2n+1}{2n+2}=1\]

and

    \[\lim_{n\rightarrow\infty}1=1,\]

so by the inequality in equation 7 and the  squeeze theorem, we get that

(8)   \begin{equation*} \lim_{n\rightarrow\infty}\frac{W_{2n+1}}{W_{2n}}=1. \end{equation*}

Substituting equations 1 and 2 into 8, we get:

    \begin{align*} \frac{W_{2n+1}}{W_{2n}} &= \frac{n!\Gamma(\frac{1}{2})}{(2n+1)\Gamma(n+\frac{1}{2})} \times \frac{2\Gamma(2n+1)}{\Gamma(\frac{1}{2})\Gamma(n+\frac{1}{2})}\\ &= \frac{2n!(2n)!}{(2n+1)\big[\Gamma(n+\frac{1}{2})\big]^2} \\ \end{align*}

From 8, it follows that

(9)   \begin{equation*} \lim_{n \rightarrow \infty} \frac{2n!(2n)!}{(2n+1)\big[\Gamma(n+\frac{1}{2})\big]^2} = 1, \end{equation*}

and since

    \[\Gamma(n+\frac{1}{2})=\frac{1.3.5...(2n-1)}{2^n}\sqrt{\pi},\]

we can re-write 9 as

    \[\lim_{n \rightarrow \infty} \frac{2^{n+1}n!(2n)!}{(2n+1)\big[1.3.5...(2n-1)\sqrt{\pi}\big]^2} = 1,\]

which can be re-arranged to give:

(10)   \begin{equation*} \displaystyle \prod_{n=1}^{\infty} \frac{(2n)(2n)}{(2n-1)(2n+1)}=\frac{\pi}{2}. \end{equation*}

1 Comment

  1. John Tawfik
    John Tawfik on July 24, 2017 at 3:23 am

    Wow I never really thought about it that way. Ur a bad ass 🔥🔥

    Reply

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