Slicing Up A Pi

Calculus Level 3

We know that

$\frac{1-a^n}{1-a}=1+a+a^2+a^3+...+a^n$

whenever $|a|<1$ . It is not hard to find that in these circumstances,

$\lim_{n\to\infty}\frac{1-a^n}{1-a}=\frac{1}{1-a}=1+a+a^2+a^3+...+a^n...$

Substituting $a=-(x^2)$ into the expression gives us

$\frac{1}{1+x^2}=1-x^2+x^4-x^6+...+(-1)^n x^{2n}...$

Does the left-hand expression seem familiar? If it does, then evaluate the following:

$\int \frac{1}{1+x^2} dx = \int 1-x^2+x^4-x^6+...+(-1)^n x^{2n}... dx$

If $x=1$ , what does the right-hand expression approximately equal?

1 solution

Andrei Li
Jul 27, 2018

The key is to realize that $\frac{1}{1+x^2}=\frac{d}{dx}\arctan(x)$ . Therefore,

$\int \frac{1}{1+x^2} dx = \int 1-x^2+x^4-x^6+...+(-1)^n x^{2n}... dx$

$\arctan(x)=x-\frac{1}{3}x^3+\frac{1}{5}x^5-\frac{1}{7}x^7...(-1)^n\frac{x^{2n+1}}{2n+1}...=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$

Recall that $\arctan(1)=\frac{\pi}{4}$ . Then,

$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}...(-1)^n\frac{1}{2n+1}...=\sum_{n=0}^\infty(-1)^n\frac{1}{2n+1}$

which is approximately equal to 0.785. QED!

There is typo in the first line. $\dfrac{1}{1+x^2}$ .

Naren Bhandari - 2 years, 10 months ago

Thanks! I've fixed the problem.

Andrei Li - 2 years, 10 months ago