Proof that 22/7 exceeds π
Proofs of the mathematical result that the rational number is greater than (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 ". Julian Havil ends a discussion of continued fraction approximations of 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 is indeed bigger than ; systematic methods of computing the value of exist. If one knows that is approximately 3.14159, then it trivially follows that  < , which is approximately 3.142857. But it takes much less work to show that  <  by the method used in this proof than to show that is approximately 3.14159. Background. is a widely used Diophantine approximation of . It is a convergent in the simple continued fraction expansion of . It is greater than , 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 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 is greater than the ratio of the perimeter of a regular polygon with 96 sides to the diameter of a circle it circumscribes. The proof. The proof can be expressed very succinctly: Therefore,  > . 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 . 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: Quick upper and lower bounds. In , it is pointed out that if 1 is substituted for in the denominator, one gets a lower bound on the integral, and if 0 is substituted for in the denominator, one gets an upper bound: Thus we have hence 3.1412 < < 3.1421 in decimal expansion. The bounds deviate by less than 0.015% from . See also . Proof that 355/113 exceeds. As discussed in , the well-known Diophantine approximation and far better upper estimate for follows from the relation where  the first six digits after the decimal point agree with those of . Substituting 1 for in the denominator, we get the lower bound substituting 0 for in the denominator, we get twice this value as an upper bound, hence In decimal expansion, this means , where the bold digits of the lower and upper bound are those of . Extensions. The above ideas can be generalized to get better approximations of ; see also and (in both references, however, no calculations are given). For explicit calculations, consider, for every integer , where the middle integral evaluates to involving . The last sum also appears in  Leibniz' formula for. The correction term and error bound is given by where the approximation (the tilde means that the quotient of both sides tends to one for large ) of the central binomial coefficient follows from Stirling's formula and shows the fast convergence of the integrals to . Calculation of these integrals: For all integers and we have Applying this formula recursively times yields Furthermore, where the first equality holds, because the terms for cancel, and the second equality arises from the index shift in the first sum. Application of these two results gives For integers , using integration by parts  times, we obtain Setting , we obtain Integrating equation (1) from 0 to 1 using equation (2) and , we get the claimed equation involving . The results for are given above. For we get and hence , where the bold digits of the lower and upper bound are those of . Similarly for , with correction term and error bound hence . The next step for is with which gives .