Proof that 22/7 exceeds π
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Proofs of the famous mathematical result that the rational number 227 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 227 (or 317) 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 < 227, which is approximately 3.142857. But it takes much less work to show that π < 227 by the method used in this proof than to show that π is approximately 3.14159.
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Background[edit]
227 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:
The approximation has been known since antiquity. Archimedes wrote the first known proof that 227 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 227 is greater than the ratio of the perimeter of a circumscribed regular polygon with 96 sides to the diameter of the circle.[note 1] Another rational approximation of π that is far more accurate is 355/113.
The proof[edit]
The proof can be expressed very succinctly:
Therefore, 227 > π.
The evaluation of this integral was the first problem in the 1968 Putnam Competition.[4] 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.[5]
Details of evaluation of the integral[edit]
That the integral is positive follows from the fact that the integrand is non-negative, being a quotient involving only sums and products of powers of non-negative real numbers. In addition, one can easily check that the integrand is strictly positive for at least one point in the range of integration, say at 12. Since the integrand is continuous at that point and non-negative 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{aligned}0&<\int _{0}^{1}{\frac {x^{4}\left(1-x\right)^{4}}{1+x^{2}}}\,dx\\[8pt]&=\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}}\\[8pt]&=\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}}&\\[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}&{\text{definite integration}}\\[6pt]&={\frac {1}{7}}-{\frac {2}{3}}+1-{\frac {4}{3}}+4-\pi \quad &{\text{with }}\arctan(1)={\frac {\pi }{4}}{\text{ and }}\arctan(0)=0\\[8pt]&={\frac {22}{7}}-\pi .&{\text{addition}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf81dbfde7aa3847ef1a23af6985e3277a1f143b)
(See polynomial long division.)
Quick upper and lower bounds[edit]
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:[6]
Thus we have
hence 3.1412 < π < 3.1421 in decimal expansion. The bounds deviate by less than 0.015% from π. See also Dalzell (1971).[7]
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