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Mathematical problem solving

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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.

A Historical Example

An interesting example of this kind of thinking is the calculation of the sum of the series \frac11 + \frac14 + \frac19 + \cdots + \frac{1}{n^2} + \cdots
The famous mathematician Leonhard Euler used the fact that:

\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots

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!}-\cdots=x(1-\frac{x}{\pi})(1+\frac{x}{\pi})(1-\frac{x}{2\pi})(1+\frac{x}{2\pi})\cdots

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!}-\cdots=(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})\cdots

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}-\cdots

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

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Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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