Difference between revisions of "Mathematical problem solving"

(Mathematical Problem Solving)
 
(A Historical Example)
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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 ones disposal to attack a problem in a new way.
 
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 ones disposal to attack a problem in a new way.
  
== A Historical Example ==
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== An Historical Example ==
An interesting example of this kind of thinking is the calculation of the sum of the series <math>\frac11 + \frac14 + \frac19 + ... + \frac{1}{n^2} + ...</math>
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An interesting example of this kind of thinking is the calculation of the sum of the series <math>\frac11 + \frac14 + \frac19 + ... + \frac{1}{n^2} + ...</math><br>
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The famous mathematician [[Leonhard Euler]] used the fact that:<br>
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<math>\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...</math><br>
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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>
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<math>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})...</math><br>
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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!}-...=(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})...</math><br>
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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>-\frac16 = -\frac{1}{\pi^2}-\frac{1}{4\pi^2}-\frac{1}{9\pi^2}-...</math><br>
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or, multiplying both sides by -<math>\pi^2</math>,<br>
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<math>\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}</math><br>
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'''''-Quoted from [[Art of Problem Solving Volume 2]] page 258'''''

Revision as of 20:56, 17 June 2006

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 ones 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 -$\pi^2$,
$\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}$
-Quoted from Art of Problem Solving Volume 2 page 258