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### Pi and the Basel's problem

In 1644 the Italian mathematician

The Euler's demonstration, published in its final form in 1741, is particularly interesting: Euler supposed that it's possible to apply the rules of the finite polynomials even those endless.

We start with the development in Taylor series for the sine function in 0: \[\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\] Dividing by $x$ both terms, we obtain: \[\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots\] whose roots are $\pi$, $-\pi$, $2\pi$, $-2\pi$, $3\pi$, $-3\pi$, $\ldots$ By changing the variable as $z = x^2$, the polynomial above becomes: \[\frac{\sin(\sqrt{z})}{\sqrt{z}} = 1 - \frac{z}{3!} + \frac{z^2}{5!} - \frac{z^3}{7!} + \cdots\] whose roots are $\pi^2$, $4\pi^2$, $9\pi^2$, $\ldots$

Now, given a polynomial $a_n x^n + \cdots + a_3 x^3 + a_2 x^2 + bx + 1$, for the

Last observation: in 1982 on the magazine

**Pietro Mengoli**proposed the so-called*Basel's problem*, which asked for the exact solution to the square of the sum of the reciprocals of all the natural numbers: \[\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \cdots\] The solution to the problem came in 1735 thanks to**Leonard Euler**, at the time at the beginning of his brilliant career as a problem solver. The Swiss mathematician proved that the exact sum of the series is $\pi^2 / 6$.The Euler's demonstration, published in its final form in 1741, is particularly interesting: Euler supposed that it's possible to apply the rules of the finite polynomials even those endless.

We start with the development in Taylor series for the sine function in 0: \[\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\] Dividing by $x$ both terms, we obtain: \[\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots\] whose roots are $\pi$, $-\pi$, $2\pi$, $-2\pi$, $3\pi$, $-3\pi$, $\ldots$ By changing the variable as $z = x^2$, the polynomial above becomes: \[\frac{\sin(\sqrt{z})}{\sqrt{z}} = 1 - \frac{z}{3!} + \frac{z^2}{5!} - \frac{z^3}{7!} + \cdots\] whose roots are $\pi^2$, $4\pi^2$, $9\pi^2$, $\ldots$

Now, given a polynomial $a_n x^n + \cdots + a_3 x^3 + a_2 x^2 + bx + 1$, for the

*formulas of Viète*, we have that the sum of the reciprocals of its roots has as result $-b$. Applying this result for finished polynomials to infinite polynomial in $z$ above, we get: \[\frac{1}{3!} = \frac{1}{6} = \frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \frac{1}{16\pi^2} + \cdots\] and so: \[\frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2}\] It's simple to observe the connection between Mengoli's series and Riemann's zeta \[\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}\] with $s=2$.Last observation: in 1982 on the magazine

*Eureka*, it appeared a rigorous proof of Euler's result signed by**John Scholes**, although it seems that such a demonstration circulated already to late sixties between Cambridge corridors.
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