Difference between revisions of "Mathematical problem solving"
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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> | 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> | ||
The famous mathematician [[Leonhard Euler]] used the fact that:<br> | 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> | <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> | 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> | <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> | 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> | <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> | 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> | <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> | 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}+\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 [[Art of Problem Solving Volume 2]] page 258''''' | '''''-Quoted from [[Art of Problem Solving Volume 2]] page 258''''' | ||
Revision as of 15:21, 22 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 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 
The famous mathematician Leonhard Euler used the fact that:
The zeros of 
are at 
, 
, 
, etc. so Euler made the leap of claiming that the polynomial on the right hand side can be factored as
since both sides are 0 at the same places. Dividing both sides by x and simplifying the right side, we get
The constant terms of both sides agree, both being 1, so this crazy procedure might be valid. Setting the 
coefficients equal, we have
or, multiplying both sides by -
,
-Quoted from Art of Problem Solving Volume 2 page 258
