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Difference between revisions of "Mathematical problem solving"

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The zeros of <math>\sin{x}</math> are at <math>0</math>, <math>\pm \pi</math>, <math>\pm{2\pi}</math>, etc. so Euler made the leap of claiming that the [[polynomial]] on the right hand side can be factored as<br>
 
The zeros of <math>\sin{x}</math> are at <math>0</math>, <math>\pm \pi</math>, <math>\pm{2\pi}</math>, etc. so Euler made the leap of claiming that the [[polynomial]] on the right hand side can be factored as<br>
  
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<math>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</math><br>
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<math>x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=x\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\cdots</math><br>
  
 
since both sides are 0 at the same places. Dividing both sides by x and simplifying the right side, we get<br>
 
since both sides are 0 at the same places. Dividing both sides by x and simplifying the right side, we get<br>
  
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<math>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</math><br>
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<math>1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\cdots</math><br>
  
 
The constant terms of both sides agree, both being 1, so this crazy procedure might be valid. Setting the <math>x^2</math> coefficients equal, we have<br>
 
The constant terms of both sides agree, both being 1, so this crazy procedure might be valid. Setting the <math>x^2</math> coefficients equal, we have<br>
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<math>\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots</math><br>
 
<math>\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots</math><br>
  
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'''''-Quoted from [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=2 Art of Problem Solving Volume 2] page 258'''''
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'''''-Quoted from [http://www.artofproblemsolving.com/Store/viewitem.php?item=ps:aops2 Art of Problem Solving Volume 2], page 258'''''

Latest revision as of 05:45, 1 May 2014

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

$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=x\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\cdots$

since both sides are 0 at the same places. Dividing both sides by x and simplifying the right side, we get

$1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\cdots$

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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