Proofs of the mathematical result that therational number⁠22/7⁠ is greater thanπ (pi) date back to antiquity. One of these proofs, more recently developed but requiring only elementary techniques from calculus, has attracted attention in modern mathematics due to itsmathematical elegance and its connections to the theory ofDiophantine approximations. Stephen Lucas calls this proof "one of the more beautiful results related to approximatingπ".^[1]Julian Havil ends a discussion ofcontinued fraction approximations ofπ with the result, describing it as "impossible to resist mentioning" in that context.^[2]

The purpose of the proof is not primarily to convince its readers that⁠22/7⁠(or⁠3+1/7⁠) is indeed bigger than π. Systematic methods of computing the value ofπ exist. If one knows thatπ is approximately 3.14159, then it trivially follows thatπ < ⁠22/7⁠, which is approximately 3.142857. But it takes much less work to show thatπ < ⁠22/7⁠ by the method used in this proof than to show thatπ is approximately 3.14159.

⁠22/7⁠ is a widely usedDiophantine approximation ofπ. It is a convergent in the simplecontinued fraction expansion ofπ. It is greater thanπ, as can be readily seen in thedecimal expansions of these values:

The approximation has been known since antiquity.Archimedes wrote the first known proof that⁠22/7⁠ is an overestimate in the 3rd century BCE, although he may not have been the first to use that approximation. His proof proceeds by showing that⁠22/7⁠ is greater than the ratio of theperimeter of aregular polygon with 96 sides to the diameter of a circle it circumscribes.^{[note 1]}

The proof first devised by British electrical engineer Donald Percy Dalzell (1898–1988) in 1944^[4] can be expressed very succinctly:

Therefore,⁠22/7⁠ > π.

The evaluation of this integral was the first problem in the 1968Putnam Competition.^[5]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. This integral has also been used in the entrance examinations for theIndian Institutes of Technology.^[6]

That theintegral is positive follows from the fact that theintegrand isnon-negative; the denominator is positive and the numerator is a product of nonnegative numbers. One can also easily check that the integrand is strictly positive for at least one point in the range of integration, say at⁠1/2⁠. Since the integrand is continuous at that point and nonnegative elsewhere, the integral from 0 to 1 must be strictly positive.

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

(Seepolynomial long division.)

InDalzell (1944), it is pointed out that if 1 is substituted forx in the denominator, one gets a lower bound on the integral, and if 0 is substituted forx in the denominator, one gets an upper bound:^[7]

Thus we have

hence 3.1412 <π < 3.1421 in decimal expansion. The bounds deviate by less than 0.015% from π. See alsoDalzell (1971).^[8]

As discussed inLucas (2005), the well-known Diophantine approximation and far better upper estimate⁠355/113⁠ forπ follows from the relation

${\displaystyle \frac{355}{113}=3.14159292\dots ,}$

where the first six digits after the decimal point agree with those ofπ. Substituting 1 forx in the denominator, we get the lower bound

${\displaystyle {\int }_0^1\frac{x^8{\left(1-x\right)}^8\left(25+816x^2\right)}{6328}dx=\frac{911}{5261111856}=0.000000173\dots ,}$

substituting 0 forx in the denominator, we get twice this value as an upper bound, hence

In decimal expansion, this means3.141 592 57 <π <3.141 592 74, where the bold digits of the lower and upper bound are those of π.

The above ideas can be generalized to get better approximations of π; see alsoBackhouse (1995)^[9] andLucas (2005) (in both references, however, no calculations are given). For explicit calculations, consider, for every integern ≥ 1,

where the middle integral evaluates to

involving π. The last sum also appears inLeibniz' formula forπ. The correction term anderror bound is given by

where the approximation (the tilde means that the quotient of both sides tends to one for largen) of thecentral binomial coefficient follows fromStirling's formula and shows the fast convergence of the integrals to π.

Calculation of these integrals: For all integersk ≥ 0 andℓ ≥ 2 we have

Applying this formula recursively2n times yields

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

Furthermore,

where the first equality holds, because the terms for1 ≤j ≤ 3n – 1 cancel, and the second equality arises from the index shiftj →j + 1 in the first sum.

Application of these two results gives

For integersk,ℓ ≥ 0, usingintegration by partsℓ times, we obtain

Settingk =ℓ = 4n, we obtain

${\displaystyle {\int }_0^1{x}^{4n}(1-x{)}^{4n}dx=\frac{1}{(8n+1)(\genfrac{}{}{0ex}{}{8n}{4n})}.}$

Integrating equation (1) from 0 to 1 using equation (2) andarctan(1) =⁠π/4⁠, we get the claimed equation involving π.

The results forn = 1 are given above. Forn = 2 we get

${\displaystyle \frac{1}{4}{\int }_0^1\frac{x^8(1-x{)}^8}{1+x^2}dx=\pi -\frac{47171}{15015}}$

and

${\displaystyle \frac{1}{8}{\int }_0^1{x}^8(1-x{)}^{8}dx=\frac{1}{1750320},}$

hence3.141 592 31 <π <3.141 592 89, where the bold digits of the lower and upper bound are those of π. Similarly forn = 3,

${\displaystyle \frac{1}{16}{\int }_0^1\frac{x^{12}{\left(1-x\right)}^{12}}{1+x^2}dx=\frac{431302721}{137287920}-\pi }$

with correction term and error bound

${\displaystyle \frac{1}{32}{\int }_0^1{x}^{12}(1-x{)}^{12}dx=\frac{1}{2163324800},}$

hence3.141 592 653 40 <π <3.141 592 653 87. The next step forn = 4 is

${\displaystyle \frac{1}{64}{\int }_0^1\frac{x^{16}(1-x{)}^{16}}{1+x^2}dx=\pi -\frac{741269838109}{235953517800}}$

with

${\displaystyle \frac{1}{128}{\int }_0^1{x}^{16}(1-x{)}^{16}dx=\frac{1}{2538963567360},}$

which gives3.141 592 653 589 55 <π <3.141 592 653 589 96.

• Approximations ofπ
• Chronology of computation ofπ
• Lindemann–Weierstrass theorem (proof thatπ is transcendental)
• List of topics related toπ
• Proof thatπ is irrational

• The problems of the 1968 Putnam competition, with this proof listed as question A1.

• Pi
• Article proofs

• Articles with short description
• Short description is different from Wikidata