# Difference between revisions of "Mathematical problem solving"

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