Dimostrazione che 22/7 è maggiore di π

Il numero razionale 22/7 è ampiamente usato cpme approssimazione di π. Esso è una convergenza della semplice espansione in frazione continua di π. È maggiore di π,come si può notare dalla espasione decimale:

${\frac {22}{7}}\approx 3.142857\dots \,$

$\pi \approx 3.141592\dots \,$

Nel seguito si dimostrerà che 22/7 è maggiore di pi greco per via puramente analitica.

L'idea

$0<\int _{0}^{1}{\frac {x^{4}(1-x)^{4}}{1+x^{2}}}\,dx={\frac {22}{7}}-\pi .$

Quindi 22/7 > π.

I dettagli

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 even powers of real numbers. So the integral from 0 to 1 is positive.

It remains to show that the integral in fact evaluates to the desired quantity:

$0\,$	$<\int _{0}^{1}{\frac {x^{4}(1-x)^{4}}{1+x^{2}}}\,dx$
	$=\int _{0}^{1}{\frac {x^{4}-4x^{5}+6x^{6}-4x^{7}+x^{8}}{1+x^{2}}}\,dx$
	$=\int _{0}^{1}\left(x^{6}-4x^{5}+5x^{4}-4x^{2}+4-{\frac {4}{1+x^{2}}}\right)\,dx$
	$=\left.{\frac {x^{7}}{7}}-{\frac {2x^{6}}{3}}+x^{5}-{\frac {4x^{3}}{3}}+4x-4\arctan {x}\,\right|_{0}^{1}$
	$={\frac {1}{7}}-{\frac {2}{3}}+1-{\frac {4}{3}}+4-\pi \$ (recall that arctan(1) = π/4)
	$={\frac {22}{7}}-\pi .$

Appearance in the Putnam Competition

The evaluation of this integral was the first problem in the 1968 Putnam Competition. If it seems trivially routine for a Putnam Competition problem, one may perhaps surmise that its inclusion was motivated by the conjunction of the punch line (summarized by the title of this article) with the fairly nice pattern in the integral itself.

Many years earlier, the result was given in D. P. Dalzell, On 22/7, Journal of the London Mathematical Society 19 (1944) 133-134.

See also

• Table of integrals
• Table of mathematical symbols

External links

• The problems of the 1968 Putnam competition, with this proof listed as question A1.