Therefore 22⁄7 > π.
The details
That the integral is positive follows from the fact that the integrand is a quotient whose numerator and denominator are both nonnegative, being sums or products of powers of nonnegative 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 (0 < 1).
It remains to show that the integral in fact evaluates to the desired quantity::
Appearance in the Putnam Competition
The evaluation of this integral was the first problem in the 1968 Putnam Competition. [cite book | title = The William Lowell Putnam Mathematical Competition: Problems and Solutions: 1965–1984 | author = Gerald L. Alexanderson, Leonard F. Klosinski, and Loren C. Larson, editors | publisher = Mathematical Association of America | isbn = 0883854635 | year = 2003 | url = http://books.google.com/books?id=HNLRgSGZrWMC&pg=PA9&dq=December-7-1968+Putnam+Mathematical-Competition&ei=DZCfR4iRJJu4sgPRu-CwCg&sig=u4-SIYyVtV3rbwN-p56c42BGUKw ] 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.
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: [Dalzell, D. P. (1944). "On 22/7", "Journal of the London Mathematical Society" 19, pages 133–134.]
:
Thus we have:
Also see Dalzell (1971) [Dalzell, D. P. (1971). "On 22/7 and 355/113", "Eureka; the Archimedeans' Journal", volume 34, pages 10–13.] .
Extensions
The above ideas can be generalized to get better approximations of π. More precisely, for every integer "n" ≥ 1,
:
where the middle integral evaluates to
:
involving π. The last sum also appears in Leibniz' formula for π. 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 "n") 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
:
Applying this formula recursively 2"n" times yields
:
Furthermore,
:
where the first equality holds, because the terms for 1 ≤ "j" ≤ 3"n" − 1 cancel, and the second equality arises from the index shift "j" → "j" + 1 in the first sum.
Application of these two results gives
:
For integers "k", "ℓ" ≥ 0, using integration by parts "ℓ" times, we obtain
:
Setting "k" = "ℓ" = 4"n", we obtain
:
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
:
and
:
hence 3.14159231 < π < 3.14159289, where the bold digits of the lower and upper bound are those of π. Similarly for "n" = 3,
:
with correction term and error bound
:
hence 3.14159265340 < π < 3.14159265387. The next step for "n" = 4 is
:
with
:
which gives 3.14159265358955 < π < 3.14159265358996.
References
ee also
*Numerical approximations of π
*Computing π
*Chronology of computation of π
*Proof that π is irrational
*Proof that π is transcendental
*List of topics related to π
External links
* [http://www.kalva.demon.co.uk/putnam/putn68.html The problems of the 1968 Putnam competition] , with this proof listed as question A1.
* [http://www.cecm.sfu.ca/%7Ejborwein/pi-slides.pdf The Life of Pi] by Jonathan Borwein—see page 5 for this integral.