The fact that it's all about $P$ and $I$ !

Calculus Level 3

$\large P = \prod_{k=1}^\infty \left( \dfrac{2k}{2k-1} \cdot \dfrac{2k}{2k+1} \right)$

$\large I = \displaystyle \int_0^1 \sqrt{1-x^2} \,dx$

Let $P$ and $I$ be as defined above. What is the relation between $P$ and $I$ ?

P=2I

Either of

P

I

does not converge to a value

P= \dfrac I2

P=I

2 solutions

Rishabh Jain
Jan 25, 2017

$\large P=\prod_{k=1}^\infty \left( \dfrac{(2k)^2}{(2k-1)(2k+1)}\right)=\dfrac{\pi}2$

See Wallis formula(3) here

While for evaluating $I$ put $x=\sin \theta$ so that:

$I = \displaystyle \int_0^{\frac{\pi}2} \cos \theta \cdot \cos \theta \mathrm{d}\theta$

$= \displaystyle \int_0^{\frac{\pi}2} \dfrac{1+\cos 2\theta}{2} \mathrm{d}\theta$

$=\left[\dfrac{\theta+\frac{\sin 2\theta}{2}}{2}\right]_0^{\frac{\pi}2}=\dfrac{\pi}{4}=\dfrac{P}{2}$

$\Large \boxed{\implies P=2I}$

Kushal Bose
Jan 25, 2017

The first one is Wallis-Product

$P=\dfrac{\pi}{2}$

$I=\dfrac{\pi}{4}$

So, $P=2I$