Mathematical problem solving

Revision as of 13:48, 22 June 2006 by Paladin8 (talk | contribs) (An Historical Example)

The idea behind The Art of Problem Solving as well as many math competitions is the use of creative methods to solve problems. In a way, students are discouraged to use rote memorization as opposed to creative spontaneous thinking. Mathematical problem solving involves using all the tools at one's disposal to attack a problem in a new way.

An Historical Example

An interesting example of this kind of thinking is the calculation of the sum of the series $\frac11 + \frac14 + \frac19 + ... + \frac{1}{n^2} + ...$
The famous mathematician Leonhard Euler used the fact that:
$\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$
The zeros of $\sin{x}$ are at $0$, $\pm \pi$, $\pm{2\pi}$, etc. so Euler made the leap of claiming that the polynomial on the right hand side can be factored as
$x-\frac{x^3}{3!}+\frac{x^5}{5!}-...=x(1-\frac{x}{\pi})(1+\frac{x}{\pi})(1-\frac{x}{2\pi})(1+\frac{x}{2\pi})...$
since both sides are 0 at the same places. Dividing both sides by x and simplifying the right side, we get
$1-\frac{x^2}{3!}+\frac{x^4}{5!}-...=(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})...$
The constant terms of both sides agree, both being 1, so this crazy procedure might be valid. Setting the $x^2$ coefficients equal, we have
$-\frac16 = -\frac{1}{\pi^2}-\frac{1}{4\pi^2}-\frac{1}{9\pi^2}-...$
or, multiplying both sides by -$\pi^2$,
$\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots$
-Quoted from Art of Problem Solving Volume 2 page 258

Invalid username
Login to AoPS