Newton’s Pi Simplified

We know from “arcsin” :

$\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}}.$

Integrate from $0$ to $x\; (0

$\int\limits_{0}^{x}\frac{d}{dx}\arcsin(x)\;dx = \int\limits_{0}^{x}\frac{1}{\sqrt{1-x^2}}\;dx$

gives

$\arcsin(x)\bigg|_{0}^{x} = \int\limits_{0}^{x}\frac{1}{\sqrt{1-x^2}}\;dx.$

i.e.,

$\arcsin(x) = \int\limits_{0}^{x}\frac{1}{\sqrt{1-x^2}}\;dx.\quad\quad\quad(\star)$

Rewrite the integrand $\frac{1}{\sqrt{1-x^2}}$ as

$(1-x^2)^{-\frac{1}{2}}= (1+(-x^2))^{-\frac{1}{2}}$

so that by the extended binomial theorem (see “A Gem from Issac Newton“),

$\frac{1}{\sqrt{1-x^2}}\overset{(A-1)}{=} \sum\limits_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}1^{-\frac{1}{2}-i}(-x^2)^i.$

Hence,

$\frac{1}{\sqrt{1-x^2}} = \sum\limits_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}(-1)^i x^{2i}.$

And,

$\int\limits_{0}^{x}\frac{1}{\sqrt{1-x^2}}\;dx=\int\limits_{0}^{x}\sum\limits_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}(-1)^i x^{2i}\;dx = \sum\limits_{i=0}^{\infty}\int\limits_{0}^{x}\binom{-\frac{1}{2}}{i}(-1)^i x^{2i}\;dx = \sum\limits_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}(-1)^i \frac{x^{2i+1}}{2i+1}.$

It follows that by $(\star)$,

$\arcsin(x) = \sum\limits_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}(-1)^i \frac{x^{2i+1}}{2i+1}.$

Let $x=\frac{1}{2},$ we have

$\frac{\pi}{6} = \sum\limits_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}(-1)^i \frac{(\frac{1}{2})^{2i+1}}{2i+1}.$

And so,

$\pi = 6 \sum\limits_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}(-1)^i \frac{1}{2^{2i+1} 2i+1}.$

Fig. 1

Given $0 < x <1,$ prove:

$|-x^2| < 1.\quad\quad\quad(A-1)$

proof

Since $0

Exercise-1 Compute $\pi$ by applying the extended binomial theorem to $\frac{\pi}{6} = \int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{1}{1+x^2}\;dx.$

Exercise-2 Can we compute $\pi$ by applying the extended binomial theorem to $\frac{\pi}{4}=\int\limits_{0}^{1}\frac{1}{1+x^2}\;dx?$ Explain.