Fandom

Scratchpad

Proof of Leibniz's formula for π

219,432pages on
this wiki
Add New Page
Discuss this page0 Share


The Leibniz' formula for π, due to Gottfried Leibniz, states that

\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots = \frac{\pi}{4}.

Proof

Consider the infinite geometric series


1 - x^2 + x^4 - x^6 + x^8 - \cdots = \frac{1}{1+x^2}, \qquad |x| < 1.

It is the limit of the truncated geometric series


G_n(x)=1 - x^2 + x^4 - x^6 + x^8 -+ \cdots - x^{4n-2}= \frac{1-x^{4n}}{1+x^2}, \qquad |x| < 1.

Splitting the integrand as


 \frac{1} {1+x^2}=\frac{1-x^{4n}}{1+x^2}+\frac{x^{4n}}{1+x^2}=G_n (x)+ \frac{x^{4n}}{1+x^2}

and integrating both sides from 0 to 1, we have


\int_{0}^{1}  \frac{1} {1+x^2}\, dx=  \int_{0}^{1}G_n(x)\, dx+\int_{0}^{1}\frac{x^{4n}}{1+x^2}\, dx  \ .

Integrating the first integral (over the truncated geometric series  G_n (x)\, ) termwise one obtains in the limit the required sum. The contribution from the second integral vanishes in the limit  n \rightarrow \infty as


\int_{0}^{1}\frac{x^{4n}}{1+x^2} \, dx< \int_{0}^{1} x^{4n}\, dx=\frac{1}{4n+1} \ .

The full integral


\int_{0}^{1}  \frac{1} {1+x^2}\, dx

on the left-hand side evaluates to arctan(1) − arctan(0) = π/4, which then yields


 \frac{\pi}{4} = \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots.

Q.E.D.

Remark: An alternative proof of the Leibniz formula can be given with the aid of Abel's theorem applied to the power series (convergent for  |x|<1 )


 \arctan x =\sum_{n \ge 0} (-1)^n {x^{2n+1}\over {2n+1}}

which is obtained integrating the geometric series ( absolutely convergent for |x|<1)


1 - x^2 + x^4 - x^6 + x^8 - \cdots = \frac{1}{1+x^2}

termwise.

Ad blocker interference detected!


Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.

Also on Fandom

Random wikia