# Newton’s Pi

Fig. 1

Shown in Fig. 1 is a semicircle centered at C $(\frac{1}{2}, 0)$ with radius = $\frac{1}{2}$. Its equation is $(x-\frac{1}{2})^2+y^2=(\frac{1}{2})^2, 0 \le x \le 1, y \ge 0.$

Simplifying and solving for $y$ gives $y = x^{\frac{1}{2}}\cdot(1-x)^{\frac{1}{2}}.\quad\quad\quad(1)$

We see that

Area (sector OAC) = Area (sector OAB) + Area (triangle ABC) $.\quad\quad(*)$

And, $a = BC = \frac{1}{2}-\frac{1}{4} = \frac{1}{4}, \quad\quad h = AB = \sqrt{AC^2-BC^2} = \sqrt{\frac{1}{2} - \frac{1}{4}}=\frac{\sqrt{3}}{2}.$

It means

Area (triangle ABC) $= \frac{1}{2}ah =\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{32}.\quad\quad\quad(2)$

Moreover, $\cos(\theta) = \frac{BC}{AC} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} \implies \theta = \frac{\pi}{3}.$

Since $\theta$ is one-third of the $\pi$ angle forming the semicircle, the sector is likewise a third of the semicircle. Namely,

Area (sector OAC) $= \frac{1}{3}$ Area (semicircle) = $\frac{1}{3} \cdot \frac{1}{2} \pi (\frac{1}{2})^2 = \frac{\pi}{24}.\quad\quad\quad(3)$

Area (sector OAB) is the area under the curve $y=x^{\frac{1}{2}}\cdot(1-x)^{\frac{1}{2}}$ from its starting point $0$ to the point $x=\frac{1}{4}.$ i.e.,

Area (sector OAB) $= \int\limits_{0}^{\frac{1}{4}}x^{\frac{1}{2}}\cdot(1-x)^{\frac{1}{2}}\;dx$ $= \int\limits_{0}^{\frac{1}{4}}x^{\frac{1}{2}}\cdot((-x)+1)^{\frac{1}{2}}\;dx$

By the extended binomial theorem: $(x+y)^r=\sum\limits_{i=0}^{\infty }x^iy^{r-i}, x, y, r \in R, |x|< |y|$ (see “A Gem from Isaac Newton“) $= \int\limits_{0}^{\frac{1}{4}}x^{\frac{1}{2}}\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}(-x)^i\;dx$ $=\int\limits_{0}^{\frac{1}{4}}x^{\frac{1}{2}}\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}(-1)^ix^i\;dx$ $= \int\limits_{0}^{\frac{1}{4}}\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}(-1)^ix^{\frac{2i+1}{2}}\;dx$ $= \sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}(-1)^i\int\limits_{0}^{\frac{1}{4}}x^{\frac{2i+1}{2}}\;dx$ $=\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}(-1)^i\frac{2}{2i+3}x^{\frac{2i+3}{2}}\bigg|_{0}^{\frac{1}{4}}$ $x=\frac{1}{4} \implies x^{\frac{2i+3}{2}}$ simplifies beautifully: $\left(\sqrt{\frac{1}{4}}\right)^{2i+3}=\left(\frac{1}{2}\right)^{2i+3}.$ $=\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}(-1)^i\frac{2}{2i+3}(\frac{1}{2})^{2i+3}$ $=\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}\frac{1}{4(2i+3)}(-\frac{1}{4})^i.\quad\quad\quad(4)$

Expressing (*) by (3), (2) and (4), we have $\frac{\pi}{24}=\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}\frac{(-1)^i}{4^{i+1}(2i+3)}+\frac{\sqrt{3}}{32}.$

Therefore, $\pi=24\left(\sum\limits_{i=0}^{\infty}\binom{r}{i}\frac{(-1)^i}{4^{i+1}(2i+3)}+\frac{\sqrt{3}}{32}\right).\quad\quad\quad(5)$

Observe first that $\sqrt{3} = \sqrt{4-1} = \sqrt{4(1-\frac{1}{4})}=2\sqrt{1-\frac{1}{4}}=2\sqrt{\frac{-1}{4}+1}=2(\frac{-1}{4}+1)^{\frac{1}{2}}$

and so we replace $(\frac{-1}{4}+1)^{\frac{1}{2}}$ by its binomial expansion. As a result, $\sqrt{3} = 2\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}(-\frac{1}{4})^i.\quad\quad\quad(6)$

Substituting (6) into (5) then yields $\pi = \sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}\left(-\frac{1}{4}\right)^i\left(\frac{1}{4(2i+3)} + \frac{2}{32}\right).$

Fig. 2 shows that with just ten terms (0 to 9) of the binomial expression, we have found $\pi$ correct to seven decmal places.

Fig. 2