Proof that 22/7 exceeds π

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Proofs of the famous mathematical result that the rational number ²²⁄₇ is greater than π date back to antiquity. What follows is a modern mathematical proof that ²²⁄₇ > π, requiring only elementary techniques from calculus. The purpose is not primarily to convince the reader that ²²⁄₇ is indeed bigger than π; systematic methods of computing the value of π exist. Unlike some elementary proofs, the calculus-based proof presented here is straightforward;^[1] its elegance results from its connections to the theory of diophantine approximations. Stephen Lucas calls this proposition "One of the more beautiful results related to approximating π".^[2] Julian Havil ends a discussion of continued fraction approximations of π with the result, describing it as "impossible to resist mentioning" in that context.^[3]

1 Background
2 The basic idea
3 The details
4 Appearance in the Putnam Competition
5 Quick upper and lower bounds
6 References
7 See also
8 External links

Background

²²⁄₇ 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:

Failed to parse (Missing texvc executable; please see math/README to configure.): \begin{align} \frac{22}{7} & \approx 3.14285714\dots \\ \pi\, & \approx 3.14159265\dots \end{align}

The approximation has been known since antiquity. Archimedes wrote the first known proof that ²²⁄₇ is an overestimate in the 3rd century BCE, although he did not necessarily invent the approximation. His proof proceeds by showing that ²²⁄₇ is greater than the ratio of the perimeter of a circumscribed regular polygon with 96 sides to the diameter of the circle.

The basic idea

The basic idea behind the proof can be expressed very succinctly:

Failed to parse (Missing texvc executable; please see math/README to configure.): 0<\int_0^1\frac{x^4(1-x)^4}{1+x^2}\,dx=\frac{22}{7}-\pi.

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:

Failed to parse (Missing texvc executable; please see math/README to configure.): 0\,	Failed to parse (Missing texvc executable; please see math/README to configure.): <\int_0^1\frac{x^4(1-x)^4}{1+x^2}\,dx
	Failed to parse (Missing texvc executable; please see math/README to configure.): =\int_0^1\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}\,dx	(expanded terms in numerator)
	Failed to parse (Missing texvc executable; please see math/README to configure.): =\int_0^1 \left(x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}\right) \,dx	(performed polynomial long division, an important aspect of formulating algebraic geometry)
	Failed to parse (Missing texvc executable; please see math/README to configure.): =\left.\frac{x^7}{7}-\frac{2x^6}{3}+ x^5- \frac{4x^3}{3}+4x-4\arctan{x}\,\right\|_0^1	(definite integration)
	Failed to parse (Missing texvc executable; please see math/README to configure.): =\frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4-\pi\	(substitute one for x, then zero for x, and subtract them—arctan(1) = π/4)
	Failed to parse (Missing texvc executable; please see math/README to configure.): =\frac{22}{7}-\pi.	(addition)

Appearance in the Putnam Competition

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.

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:^[5]

Failed to parse (Missing texvc executable; please see math/README to configure.): {1 \over 1260} < \int_0^1 {x^4 (1-x)^4 \over 1+x^2}\,dx < {1 \over 630}.

Thus we have

Failed to parse (Missing texvc executable; please see math/README to configure.): {22 \over 7} - {1 \over 630} < \pi < {22 \over 7} - {1 \over 1260}.

Perhaps no other method of calculating π to nearly three decimal places is both so quick and so elementary. Also see Dalzell (1971)^[6].

References

^ Contrast Hardy, G. H. and E. M. Wright, chapter 22, on the elementary proof of the prime number theorem.
(1938). An Introduction to the Theory of Numbers, Oxford University Press, USA; 5 edition (April 17, 1980) ISBN 0198531710.
^ Lucas, Stephen. "Integral proofs that 355/113 > π", Australian Mathematical Society Gazette, volume 32, number 4, pages 263–266.
This paper begins by calling this proposition "One of the more beautiful results related to approximating π."
^ Havil, Julian (2003). Gamma: Exploring Euler's Constant. Princeton University Press, p. 96. ISBN 0-691-09983-9.
^ Gerald L. Alexanderson, Leonard F. Klosinski, and Loren C. Larson, editors (2003). The William Lowell Putnam Mathematical Competition: Problems and Solutions: 1965–1984. Mathematical Association of America. ISBN 0883854635.
^ Dalzell, D. P. (1944). "On 22/7", Journal of the London Mathematical Society 19, pages 133–134.
^ Dalzell, D. P. (1971). "On 22/7 and 355/113", Eureka; the Archimedeans' Journal, volume 34, pages 10–13.