Proof that 22/7 exceeds π

Proof

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

${\displaystyle 0<{\int }_0^1\frac{x^4{\left(1-x\right)}^4}{1+x^2}dx=\frac{22}{7}-\pi .}$

Therefore, ⁠22/7⁠ > π.

The evaluation of this integral was the first problem in the 1968 Putnam Competition.

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 ⁠1/2⁠. 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:

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

(See polynomial long division.)

Quick upper and lower bounds

In 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:

${\displaystyle \frac{1}{1260}={\int }_0^1\frac{x^4{\left(1-x\right)}^4}{2}dx<{\int }_0^1\frac{x^4{\left(1-x\right)}^4}{1+x^2}dx<{\int }_0^1\frac{x^4{\left(1-x\right)}^4}{1}dx=\frac{1}{630}.}$

Thus we have

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

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

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

${\displaystyle 0<{\int }_0^1\frac{x^8{\left(1-x\right)}^8\left(25+816x^2\right)}{3164\left(1+x^2\right)}dx=\frac{355}{113}-\pi .}$

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

where the first six digits after the decimal point agree with those of π. Substituting 1 for x 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 for x in the denominator, we get twice this value as an upper bound, hence

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

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. For explicit calculations, consider, for every integer n ≥ 1,

${\displaystyle \frac{1}{2^{2n-1}}{\int }_0^1{x}^{4n}(1-x{)}^{4n}dx<\frac{1}{2^{2n-2}}{\int }_0^1\frac{x^{4n}(1-x{)}^{4n}}{1+x^2}dx<\frac{1}{2^{2n-2}}{\int }_0^1{x}^{4n}(1-x{)}^{4n}dx,}$

where the middle integral evaluates to

${\displaystyle \begin{array}{cc}\frac{1}{2^{2n-2}}& {\int }_0^1\frac{x^{4n}(1-x{)}^{4n}}{1+x^2}dx\\ =& \sum _{j=0}^{2n-1}\frac{(-1{)}^j}{2^{2n-j-2}(8n-j-1)(\genfrac{}{}{0ex}{}{8n-j-2}{4n+j})}+(-1{)}^n\left(\pi -4\sum _{j=0}^{3n-1}\frac{(-1{)}^j}{2j+1}\right)\end{array}}$

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

${\displaystyle \begin{array}{cc}\frac{1}{2^{2n-1}}{\int }_0^1{x}^{4n}(1-x{)}^{4n}dx& =\frac{1}{2^{2n-1}(8n+1)(\genfrac{}{}{0ex}{}{8n}{4n})}\\ & \sim \frac{\sqrt{\pi n}}{2^{10n-2}(8n+1)},\end{array}}$

where the approximation (the tilde means that the quotient of both sides tends to one for large 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

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

Applying this formula recursively 2n 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,

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

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

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

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

${\displaystyle \begin{array}{cc}{\int }_0^1{x}^k(1-x{)}^{\ell }dx& =\frac{\ell }{k+1}{\int }_0^1{x}^{k+1}(1-x{)}^{\ell -1}dx\\ & ⋮\\ & =\frac{\ell }{k+1}\frac{\ell -1}{k+2}\cdots \frac{1}{k+\ell }{\int }_0^1{x}^{k+\ell }dx\\ & =\frac{1}{(k+\ell +1)(\genfrac{}{}{0ex}{}{k+\ell }{k})}.(2)\end{array}}$

Setting k = 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) 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

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

and

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

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,

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

with correction term and error bound

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

hence 3.141 592 653 40 < π < 3.141 592 653 87. The next step for n = 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 gives 3.141 592 653 589 55 < π < 3.141 592 653 589 96.