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A simple proof that 22/7 exceeds pi


The rational number 22/7 is a common approximation of the transcendental value π. It is a convergent in the simple continued fraction expansion of π. Its value is greater than π, as can be readily seen in the initial, established decimal places for these values:

22/7 \cong 3.142857\dots\,
\pi \cong 3.14159\dots\,

Although many people know this numerical value of π as a result of having read it in innumerable books, far fewer know how to compute it.

What follows is a mathematical proof that 22/7 > π. It is simple in that it is short and straightforward, and requires only an introductory-level understanding of calculus.

Contents

The idea

0<\int_0^1\frac{x^4(1-x)^4}{1+x^2}\,dx=\frac{22}{7}-\pi.

The details

That the integral is positive follows from the fact that the integrand is a quotient whose numerator and denominator are both nonnegative, being sums or products of even powers of real numbers. So the integral from 0 to 1 is positive.

It remains to show that the integral in fact evaluates to the desired quantity:

0<\int_0^1\frac{x^4(1-x)^4}{1+x^2}\,dx
=\int_0^1\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}\,dx
=\int_0^1 \left(x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}\right) \,dx
=\left.\frac{x^7}{7}-\frac{2x^6}{3}+ x^5- \frac{4x^3}{3}+4x-4\arctan{x}\,\right|_0^1
=\frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4-\pi\ (recall that arctan(1) = π/4)
=\frac{22}{7}-\pi.

Now we see the difference between 22/7 and π is greater than zero, so 22/7 > π.

Appearance in the Putnam Competition

The evaluation of this integral was the first problem in the 1968 Putnam Competition. If it seems trivially routine for a Putnam Competition problem, one may perhaps surmise that its inclusion was motivated by the conjunction of the punch line (summarized by the title of this article) with the fairly nice pattern in the integral itself.

See also

External links

Last updated: 01-04-2007 01:18:57
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