Proof that 22/7 exceeds π

Proofs of the famous mathematical result that the rational number ²²⁄₇ is greater than π date back to antiquity. What follows is a modern mathematical proof that ²²⁄₇ > π, requiring only elementary techniques from calculus. The purpose is not primarily to convince the reader that ²²⁄₇ is indeed bigger than π; systematic methods of computing the value of π exist. Unlike some elementary proofs, the calculus-based proof presented here is straightforward; its elegance results from its connections to the theory of diophantine approximations. Stephen Lucas calls this proposition "One of the more beautiful results related to approximating π". Julian Havil ends a discussion of continued fraction approximations of π with the result, describing it as "impossible to resist mentioning" in that context.

Background

²²⁄₇ is a widely used Diophantine approximation of π. It is a convergent in the simple continued fraction expansion of π. It is greater than π, as can be readily seen in the decimal expansions of these values:

${\begin{aligned}{\frac {22}{7}}&\approx 3.14285714\dots \\\pi \,&\approx 3.14159265\dots \end{aligned}}$

The approximation has been known since antiquity. Archimedes wrote the first known proof that ²²⁄₇ is an overestimate in the 3rd century BC, although he did not necessarily invent the approximation. His proof proceeds by showing that ²²⁄₇ is greater than the ratio of the perimeter of a circumscribed regular polygon with 96 sides to the diameter of the circle. A more accurate approximation of π is 355/113.

The basic idea

The basic idea behind the proof can be expressed very succinctly:

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

Therefore ²²⁄₇ > π.

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 powers of nonnegative real numbers. Since the integrand is positive, the integral from 0 to 1 is positive because the lower limit of integration is less than the upper limit of integration (0 < 1).

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

$0\,$	$<\int _{0}^{1}{\frac {x^{4}(1-x)^{4}}{1+x^{2}}}\,dx$
	$=\int _{0}^{1}{\frac {x^{4}-4x^{5}+6x^{6}-4x^{7}+x^{8}}{1+x^{2}}}\,dx$	(expanded terms in numerator)
	$=\int _{0}^{1}\left(x^{6}-4x^{5}+5x^{4}-4x^{2}+4-{\frac {4}{1+x^{2}}}\right)\,dx$	(performed polynomial long division, an important aspect of formulating algebraic geometry)
	$=\left.{\frac {x^{7}}{7}}-{\frac {2x^{6}}{3}}+x^{5}-{\frac {4x^{3}}{3}}+4x-4\arctan {x}\,\right|_{0}^{1}$	(definite integration)
	$={\frac {1}{7}}-{\frac {2}{3}}+1-{\frac {4}{3}}+4-\pi \$	(substitute one for x, then zero for x, and subtract them, use arctan(1) = π/4)
	$={\frac {22}{7}}-\pi .$	(addition)

Appearance in the Putnam Competition

The evaluation of this integral was the first problem in the 1968 Putnam Competition. It is easier than most Putnam Competition problems, but the competition often features seemingly obscure problems that turn out to refer to something very familiar.

Quick upper and lower bounds

In Dalzell (1944), it is pointed out that if 1 is substituted for x in the denominator, one gets a lower bound on the integral, and if 0 is substituted for x in the denominator, one gets an upper bound:

${1 \over 1260}<\int _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx<{1 \over 630}.$

Thus we have

${22 \over 7}-{1 \over 630}<\pi <{22 \over 7}-{1 \over 1260}.$

Also see Dalzell (1971).

Extensions

The above ideas can be generalized to get better approximations of π. More precisely, for every integer n ≥ 1,

${\frac {1}{2^{2n-1}}}\int _{0}^{1}x^{4n}(1-x)^{4n}\,dx<{\frac {1}{2^{2n-2}}}\int _{0}^{1}{\frac {x^{4n}(1-x)^{4n}}{1+x^{2}}}\,dx<{\frac {1}{2^{2n-2}}}\int _{0}^{1}x^{4n}(1-x)^{4n}\,dx,$

where the middle integral evaluates to

${\begin{aligned}&{\frac {1}{2^{2n-2}}}\int _{0}^{1}{\frac {x^{4n}(1-x)^{4n}}{1+x^{2}}}\,dx\\&\qquad =\sum _{j=0}^{2n-1}{\frac {(-1)^{j}}{2^{2n-j-2}(8n-j-1){\binom {8n-j-2}{4n+j}}}}+(-1)^{n}{\biggl (}\pi -4\sum _{j=0}^{3n-1}{\frac {(-1)^{j}}{2j+1}}{\biggr )}\end{aligned}}$

involving π. The last sum also appears in Leibniz' formula for π. The correction term and error bound is given by

${\begin{aligned}{\frac {1}{2^{2n-1}}}\int _{0}^{1}x^{4n}(1-x)^{4n}\,dx&={\frac {1}{2^{2n-1}(8n+1){\binom {8n}{4n}}}}\\&\sim {\frac {\sqrt {\pi n}}{2^{10n-2}(8n+1)}},\end{aligned}}$

where the approximation (the tilde means that the quotient of both sides tends to one for large n) follows from Stirling's formula and shows the fast convergence of the integrals to π.

Calculation of these integrals

For all integers k ≥ 0 and ℓ ≥ 2 we have

${\begin{aligned}x^{k}(1-x)^{\ell }&=(1-2x+x^{2})x^{k}(1-x)^{\ell -2}\\&=(1+x^{2})\,x^{k}(1-x)^{\ell -2}-2x^{k+1}(1-x)^{\ell -2}.\end{aligned}}$

Applying this formula recursively 2n times yields

$x^{4n}(1-x)^{4n}=(1+x^{2})\sum _{j=0}^{2n-1}(-2)^{j}x^{4n+j}(1-x)^{4n-2(j+1)}+(-2)^{2n}x^{6n}.$

Furthermore,

${\begin{aligned}x^{6n}-(-1)^{3n}&=\sum _{j=1}^{3n}(-1)^{3n-j}x^{2j}-\sum _{j=0}^{3n-1}(-1)^{3n-j}x^{2j}\\&=\sum _{j=0}^{3n-1}{\bigl (}(-1)^{3n-(j+1)}x^{2(j+1)}-(-1)^{3n-j}x^{2j}{\bigr )}\\&=-(1+x^{2})\sum _{j=0}^{3n-1}(-1)^{3n-j}x^{2j},\\\end{aligned}}$

where the first equality holds, because the terms for 1 ≤ j ≤ 3n − 1 cancel, and the second equality arises from the index shift j → j + 1 in the first sum.

Application of these two results gives

${\begin{aligned}{\frac {x^{4n}(1-x)^{4n}}{2^{2n-2}(1+x^{2})}}&=\sum _{j=0}^{2n-1}{\frac {(-1)^{j}}{2^{2n-j-2}}}x^{4n+j}(1-x)^{4n-2j-2}\\&\qquad {}-4\sum _{j=0}^{3n-1}(-1)^{3n-j}x^{2j}+(-1)^{3n}{\frac {4}{1+x^{2}}}.\qquad (*)\end{aligned}}$

For integers k, ℓ ≥ 0, using integration by parts ℓ times, we obtain

${\begin{aligned}\int _{0}^{1}x^{k}(1-x)^{\ell }\,dx&={\frac {\ell }{k+1}}\int _{0}^{1}x^{k+1}(1-x)^{\ell -1}\,dx\\&=\cdots \\&={\frac {\ell }{k+1}}{\frac {\ell -1}{k+2}}\cdots {\frac {1}{k+\ell }}\int _{0}^{1}x^{k+\ell }\,dx\\&={\frac {1}{(k+\ell +1){\binom {k+\ell }{k}}}}.\qquad (**)\end{aligned}}$

Setting k = ℓ = 4n, we obtain

$\int _{0}^{1}x^{4n}(1-x)^{4n}\,dx={\frac {1}{(8n+1){\binom {8n}{4n}}}}.$

Integrating (*) from 0 to 1 using (**) and arctan(1) = π/4, we get the claimed equation involving π.

The results for n = 1 are given above. For n = 2 we get

${\frac {1}{4}}\int _{0}^{1}{\frac {x^{8}(1-x)^{8}}{1+x^{2}}}\,dx=\pi -{\frac {47\,171}{15\,015}}$

and

${\frac {1}{8}}\int _{0}^{1}x^{8}(1-x)^{8}\,dx={\frac {1}{1750320}},$

hence 3.14159231 < π < 3.14159289, where the bold digits of the lower and upper bound are those of π. Similarly for n = 3,

${\frac {1}{16}}\int _{0}^{1}{\frac {x^{12}(1-x)^{12}}{1+x^{2}}}\,dx={\frac {431\,302\,721}{137\,287\,920}}-\pi$

with correction term and error bound

${\frac {1}{32}}\int _{0}^{1}x^{12}(1-x)^{12}\,dx={\frac {1}{2\,163\,324\,800}},$

hence 3.14159265340 < π < 3.14159265387. The next step for n = 4 is

${\frac {1}{64}}\int _{0}^{1}{\frac {x^{16}(1-x)^{16}}{1+x^{2}}}\,dx=\pi -{\frac {741\,269\,838\,109}{235\,953\,517\,800}}$

with

${\frac {1}{128}}\int _{0}^{1}x^{16}(1-x)^{16}\,dx={\frac {1}{2\,538\,963\,567\,360}},$

which gives 3.14159265358955 < π < 3.14159265358996.

See also

• Numerical approximations of π
• Computing π
• Chronology of computation of π
• Proof that π is irrational
• Proof that π is transcendental
• List of topics related to π

External links

• The problems of the 1968 Putnam competition, with this proof listed as question A1.
• The Life of Pi by Jonathan Borwein—see page 5 for this integral.