Proof that 22/7 exceeds π

Proofs of the famous mathematical result that the rational number 22/7 is greater than π (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 its mathematical elegance and its connections to the theory of diophantine approximations. Stephen Lucas calls this proof “one of the more beautiful results related to approximating π”.^[1] Julian Havil ends a discussion of continued fraction approximations of π 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 that 22/7 (or 3¹⁄₇) is indeed bigger than π; systematic methods of computing the value of π exist. If one knows that π is approximately 3.14159, then it trivially follows that π < 22/7, which is approximately 3.142857. But it takes much less work to show that π < 22/7 by the method used in this proof than to show that π is approximately 3.14159.

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:

${\begin{aligned}{\frac {22}{7}}&=3.\overline {142\,857},\\\pi \,&=3.141\,592\,65\ldots \end{aligned}}$

The approximation has been known since antiquity. Archimedes wrote the first known proof that 22/7 is an overestimate in the 3rd century BCE, although he may not have been the first to use that 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. Another rational approximation of π that is far more accurate is 355/113.

The proof

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 evaluation of this integral was the first problem in the 1968 Putnam Competition.^[3] 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 the Indian Institutes of Technology.^[4]

Details of evaluation of the integral

That the integral is positive follows from the fact that the integrand is a quotient whose numerator and denominator are both non-negative, being sums or products of powers of non-negative 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.

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

${\begin{aligned}0&<\int _{0}^{1}{\frac {x^{4}(1-x)^{4}}{1+x^{2}}}\,dx\\[8pt]&=\int _{0}^{1}{\frac {x^{4}-4x^{5}+6x^{6}-4x^{7}+x^{8}}{1+x^{2}}}\,dx\quad {\text{(expansion of terms in the numerator)}}\\[8pt]&=\int _{0}^{1}\left(x^{6}-4x^{5}+5x^{4}-4x^{2}+4-{\frac {4}{1+x^{2}}}\right)\,dx\\&{}\qquad {\text{(polynomial long division)}}\\[8pt]&=\left.\left({\frac {x^{7}}{7}}-{\frac {2x^{6}}{3}}+x^{5}-{\frac {4x^{3}}{3}}+4x-4\arctan {x}\right)\,\right|_{0}^{1}\quad {\text{(definite integration)}}\\[6pt]&={\frac {1}{7}}-{\frac {2}{3}}+1-{\frac {4}{3}}+4-\pi \quad ({\text{since }}\arctan(1)=\pi /4{\text{ and }}\arctan(0)=0)\\[8pt]&={\frac {22}{7}}-\pi .\quad {\text{(addition)}}\end{aligned}}$

(See polynomial long division.)

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:^[5]

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

Thus we have

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

hence 3.1412 < π < 3.1421 in decimal expansion. The bounds deviate by less than 0.015% from π. See also Dalzell (1971).^[6]

Proof that 355/113 exceeds π

As discussed in Lucas (2005), the well-known Diophantine approximation and far better upper estimate 355/113 for π follows from the relation

$0<\int _{0}^{1}{\frac {x^{8}(1-x)^{8}(25+816x^{2})}{3164(1+x^{2})}}\,dx={\frac {355}{113}}-\pi .$

Note that

${\frac {355}{113}}=3.141\,592\,92\ldots ,$

where the first six digits after the period agree with those of π. Substituting 1 for x in the denominator, we get the lower bound

$\int _{0}^{1}{\frac {x^{8}(1-x)^{8}(25+816x^{2})}{6328}}\,dx={\frac {911}{5\,261\,111\,856}}=0.000\,000\,173\ldots ,$

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

${\frac {355}{113}}-{\frac {911}{2\,630\,555\,928}}<\pi <{\frac {355}{113}}-{\frac {911}{5\,261\,111\,856}}\,.$

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

Extensions

The above ideas can be generalized to get better approximations of π; see also Backhouse (1995)^[7] and Lucas (2005) (in both references, however, no calculations are given). For explicit calculations, consider, 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) of the central binomial coefficient 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

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

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

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

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

${\frac 14}\int _{0}^{1}{\frac {x^{8}(1-x)^{8}}{1+x^{2}}}\,dx=\pi -{\frac {47\,171}{15\,015}}$

and

${\frac 18}\int _{0}^{1}x^{8}(1-x)^{8}\,dx={\frac 1{1\,750\,320}},$

hence 3.141 592 31 < π < 3.141 592 89, 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 3.141 592 653 40 < π < 3.141 592 653 87. 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 3.141 592 653 589 55 < π < 3.141 592 653 589 96.

See also

• Approximations of π
• Chronology of computation of π
• Proof that π is irrational
• Lindemann–Weierstrass theorem (proof that π is transcendental)
• List of topics related to π

References

1. Lucas, Stephen (2005), "Integral proofs that 355/113 > π" (PDF), Australian Mathematical Society Gazette, 32 (4): 263–266, MR 2176249, Zbl 1181.11077
2. Havil, Julian (2003), Gamma. Exploring Euler's Constant, Princeton, NJ: Princeton University Press, p. 96, ISBN 0-691-09983-9, MR 1968276, Zbl 1023.11001
3. Alexanderson, Gerald L.; Klosinski, Leonard F.; Larson, Loren C. (editors) (1985), The William Lowell Putnam Mathematical Competition: Problems and Solutions: 1965–1984, Washington, D.C.: The Mathematical Association of America, ISBN 0-88385-463-5, Zbl 0584.00003
4. 2010 IIT Joint Entrance Exam, question 38 on page 15 of the mathematics section.
5. Dalzell, D. P. (1944), "On 22/7", Journal of the London Mathematical Society, 19 (75 Part 3): 133–134, doi:10.1112/jlms/19.75_part_3.133, MR 0013425, Zbl 0060.15306.
6. Dalzell, D. P. (1971), "On 22/7 and 355/113", Eureka; the Archimedeans' Journal, 34: 10–13, ISSN 0071-2248.
7. Backhouse, Nigel (July 1995), "Note 79.36, Pancake functions and approximations to π", The Mathematical Gazette, 79 (485): 371–374, JSTOR 3618318

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.