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anonymous

  • 5 years ago

How does \[\sum_{x=1}^{\infty}(\frac{1}{n})^{2} = \frac{pi^2}{6}\]

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  1. anonymous
    • 5 years ago
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    See 'The Basel Problem'

  2. anonymous
    • 5 years ago
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    I saw it.

  3. anonymous
    • 5 years ago
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    Oh, I forgot, you're the little kid who hates/is jealous of me.

  4. anonymous
    • 5 years ago
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    The best (maybe only?) proof I've seen involves the Taylor series of sin(x). it goes like this. \[\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...\] I don't know how to rigorously justify this, but since sin(x) has zeros at \[0, \pm \pi, \pm 2\pi, \pm3\pi,...\], it is equivalent to a polynomial with zeros at the same points, i.e. \[\sin(x)=x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})...\] since this has zeros at precisely the same points. Also notice that if you divide both sides by x, and take the limit as x approaches 0, you get 1 on both sides. (I guess having the same zeros and being the same at some point makes them the same, given that we know sin(x) can be represented by a polynomial, from the taylor series) Anyway, we look at all the x^3 terms for sin(x) and add them up, getting the sum of the reciprocals of the squares, divided by pi^2. Because of the x^3 term in the Taylor series, this must be equal to 1/3!=1/6. If we multiply by pi^2, we get that the sum of the reciprocals of the squares is pi^2/6.

  5. anonymous
    • 5 years ago
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    Don't worry, you're in good company; Euler couldn't rigorously justify it either (well, at least not in his first proof)

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