Proof that 22/7 exceeds π

This is not a perfect 22/7 circle, because 22/7 is not a perfect representation of pi.

Template:Pi box Proofs of the mathematical result that the rational number Template:Sfrac is greater than [[pi|Template:Pi]] (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 Template:Pi". Julian Havil ends a discussion of continued fraction approximations of Template:Pi with the result, describing it as "impossible to resist mentioning" in that context.

The purpose of the proof is not primarily to convince its readers that Template:Sfrac Template:Nowrap is indeed bigger than Template:Pi. Systematic methods of computing the value of Template:Pi exist. If one knows that Template:Pi is approximately 3.14159, then it trivially follows that Template:Pi < Template:Sfrac, which is approximately 3.142857. But it takes much less work to show that Template:Pi < Template:Sfrac by the method used in this proof than to show that Template:Pi is approximately 3.14159.

Background

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

$\begin{array}{cc}\frac{22}{7}& =3.\overline{142857},\\ \pi & =3.14159265\dots \end{array}$

The approximation has been known since antiquity. Archimedes wrote the first known proof that Template:Sfrac 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 Template:Sfrac is greater than the ratio of the perimeter of a regular polygon with 96 sides to the diameter of a circle it circumscribes.Template:Refn

Proof

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

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

Therefore, Template:Sfrac > Template:Pi.

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. This integral has also been used in the entrance examinations for the Indian Institutes of Technology.

Details of evaluation of the integral

That the integral is positive follows from the fact that the integrand is non-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 at Template:Sfrac. 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:

$\begin{array}{cc}0& <\underset{0}{\overset{1}{\int }}\frac{x^4{(1-x)}^4}{1+x^2}dx\\ & =\underset{0}{\overset{1}{\int }}\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}dx& \text{expansion of terms in the numerator}\\ & =\underset{0}{\overset{1}{\int }}(x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2})dx& \text{ using polynomial long division}& \\ & ={(\frac{x^7}{7}-\frac{2x^6}{3}+x^5-\frac{4x^3}{3}+4x-4\arctan x)|}_0^1& \text{definite integration}\\ & =\frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4-\pi & \text{with }\arctan (1)=\frac{\pi }{4}\text{ and }\arctan (0)=0\\ & =\frac{22}{7}-\pi .& \text{addition}\end{array}$

(See polynomial long division.)

Quick upper and lower bounds

In Template:Harvtxt, it is pointed out that if 1 is substituted for Template:Math in the denominator, one gets a lower bound on the integral, and if 0 is substituted for Template:Math in the denominator, one gets an upper bound:

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

Thus we have

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

hence 3.1412 < Template:Pi < 3.1421 in decimal expansion. The bounds deviate by less than 0.015% from Template:Pi. See also Template:Harvtxt.

Proof that 355/113 exceeds Template:Pi

As discussed in Template:Harvtxt, the well-known Diophantine approximation and far better upper estimate [[355/113|Template:Sfrac]] for Template:Pi follows from the relation

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

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

where the first six digits after the decimal point agree with those of Template:Pi. Substituting 1 for Template:Math in the denominator, we get the lower bound

$\underset{0}{\overset{1}{\int }}\frac{x^8{(1-x)}^8(25+816x^2)}{6328}dx=\frac{911}{5261111856}=0.000000173\dots ,$

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

$\frac{355}{113}-\frac{911}{2630555928}<\pi <\frac{355}{113}-\frac{911}{5261111856}.$

In decimal expansion, this means Template:Nowrap, where the bold digits of the lower and upper bound are those of Template:Pi.

Extensions

The above ideas can be generalized to get better approximations of Template:Pi; see also Template:Harvtxt and Template:Harvtxt (in both references, however, no calculations are given). For explicit calculations, consider, for every integer Template:Math,

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

where the middle integral evaluates to

$\begin{array}{cc}\frac{1}{2^{2n-2}}& \underset{0}{\overset{1}{\int }}\frac{x^{4n}(1-x{)}^{4n}}{1+x^2}dx\\ =& \sum _{j=0}^{2n-1}\frac{(-1{)}^j}{2^{2n-j-2}(8n-j-1)(\genfrac{}{}{0ex}{}{8n-j-2}{4n+j})}+(-1{)}^n(\pi -4\sum _{j=0}^{3n-1}\frac{(-1{)}^j}{2j+1})\end{array}$

involving Template:Pi. The last sum also appears in [[Leibniz formula for pi|Leibniz' formula for Template:Pi]]. The correction term and error bound is given by

$\begin{array}{cc}\frac{1}{2^{2n-1}}\underset{0}{\overset{1}{\int }}{x}^{4n}(1-x{)}^{4n}dx& =\frac{1}{2^{2n-1}(8n+1)(\genfrac{}{}{0ex}{}{8n}{4n})}\\ & \sim \frac{\sqrt{\pi n}}{2^{10n-2}(8n+1)},\end{array}$

where the approximation (the tilde means that the quotient of both sides tends to one for large Template:Math) of the central binomial coefficient follows from Stirling's formula and shows the fast convergence of the integrals to Template:Pi.

Calculation of these integrals: For all integers Template:Math and Template:Math we have

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

Applying this formula recursively Template:Math 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,

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

where the first equality holds, because the terms for Template:Math cancel, and the second equality arises from the index shift Template:Math in the first sum.

Application of these two results gives

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

For integers Template:Math, using integration by parts Template:Math times, we obtain

$\begin{array}{cc}\underset{0}{\overset{1}{\int }}{x}^k(1-x{)}^{\ell }dx& =\frac{\ell }{k+1}\underset{0}{\overset{1}{\int }}{x}^{k+1}(1-x{)}^{\ell -1}dx\\ & ⋮\\ & =\frac{\ell }{k+1}\frac{\ell -1}{k+2}\cdots \frac{1}{k+\ell }\underset{0}{\overset{1}{\int }}{x}^{k+\ell }dx\\ & =\frac{1}{(k+\ell +1)(\genfrac{}{}{0ex}{}{k+\ell }{k})}.(2)\end{array}$

Setting Template:Math, we obtain

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

Integrating equation (1) from 0 to 1 using equation (2) and Template:Math, we get the claimed equation involving Template:Pi.

The results for Template:Math are given above. For Template:Math we get

$\frac{1}{4}\underset{0}{\overset{1}{\int }}\frac{x^8(1-x{)}^8}{1+x^2}dx=\pi -\frac{47171}{15015}$

and

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

hence Template:Nowrap, where the bold digits of the lower and upper bound are those of Template:Pi. Similarly for Template:Math,

$\frac{1}{16}\underset{0}{\overset{1}{\int }}\frac{x^{12}{(1-x)}^{12}}{1+x^2}dx=\frac{431302721}{137287920}-\pi$

with correction term and error bound

$\frac{1}{32}\underset{0}{\overset{1}{\int }}{x}^{12}(1-x{)}^{12}dx=\frac{1}{2163324800},$

hence Template:Nowrap. The next step for Template:Math is

$\frac{1}{64}\underset{0}{\overset{1}{\int }}\frac{x^{16}(1-x{)}^{16}}{1+x^2}dx=\pi -\frac{741269838109}{235953517800}$

with

$\frac{1}{128}\underset{0}{\overset{1}{\int }}{x}^{16}(1-x{)}^{16}dx=\frac{1}{2538963567360},$

which gives Template:Nowrap.

See also

• [[Approximations of π|Approximations of Template:Pi]]
• [[Chronology of computation of π|Chronology of computation of Template:Pi]]
• Lindemann–Weierstrass theorem (proof that Template:Pi is transcendental)
• [[List of topics related to π|List of topics related to Template:Pi]]
• [[Proof that π is irrational|Proof that Template:Pi is irrational]]

External links

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