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Converging infinite series

  1. Oct 19, 2004 #1
    The following series can be shown to converge, but exactly what does it converge to? Euler was supposed to have proven it to sum to pi^2/6, but how?

    1 + 1/4 + 1/9 + 1/16 + ... + 1/(r^2) as r -> infinity

    The following is a small maths puzzle that was asked in another forum, but which I know of no answer to:

    Assume there exists an omnipotent being who decides to play a game with a lamp. After a minute has passed, he switches it on, 1/2 a min after this, he turns it off, 1/4 of a min later he switches on, 1/8 min later off, 1/16 min later on... After an arbitrarily long time period (1 hour for example, is the lamp switched on or off? What state would it be in?
     
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  3. Oct 19, 2004 #2

    Galileo

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    One proof I know of [itex]\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}[/itex] is by using the fourier series of [itex]\frac{1}{2}\pi x - \frac{1}{4}x^2[/itex].

    Another is by evaluating the (improper) double integral:
    [tex]\int_0^1\int_0^1\frac{1}{1-xy}dx[/tex]
    It's not hard to show that the integral is equal to [itex]\sum_{n=1}^{\infty}\frac{1}{n^2}[/itex], by expanding the integrand as a geometric series.

    I don't know what clever method Euler used to prove it.

    Since it can be shown he will not flip switches for longer than [itex]\frac{\pi^2}{6}\approx 1.64[/itex] minutes, asking for it's state after t minutes is a meaningless question if [itex]t>\frac{\pi^2}{6}\approx 1.64[/itex]
     
    Last edited: Oct 19, 2004
  4. Oct 19, 2004 #3
    Thanks for the response to the first question. As for the second, can it be said that since he isn't flipping switches after 1.64 min, he must have therefore have left the lamp in only one possible state, since the lamp cannot be anything other than on or off?
     
  5. Oct 19, 2004 #4
    he simply used his famous sine product ..

    -- AI
     
  6. Oct 19, 2004 #5

    Galileo

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    Never heard of that.
     
  7. Oct 19, 2004 #6
    yeah i've never heard of that either, the only solution i know of to this problem is the fourier series one. i wonder if there's an easier or more elementary way of doing it...
     
  8. Oct 20, 2004 #7
    Could someone show Euler's method?
     
  9. Oct 20, 2004 #8
    hmm i guess euler's sine product isn't as famous i expected it to be ..

    The following identity is called as the euler's sine product ..
    [tex]sin(x) = x * \prod_{n=1}^{\infty}1-\frac{x^2}{n^2\pi^2}[/tex]

    Multiply out the first few terms and guess what is the coefficient of x^3??

    now consider the series expansion of sin(x) ... what is the coefficient of x^3 here?

    they should be equivalent .... so we get ....... <fill in the steps>

    -- AI
     
  10. Oct 20, 2004 #9

    Galileo

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    Coefficient of [itex]x^3[/itex] on the left side is [itex]-\frac{1}{6}[/itex]. On the right side: [itex]-\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}[/itex].

    So how do you prove this sine product?
     
  11. Oct 21, 2004 #10
    Consider,
    1-((a+b)/ab)x+(1/ab)x^2
    this factors as,
    (1-x/a)(1-x/b)
    Here a and b are the roots ...

    Look at the solution, its format is more important ....

    Now to euler's sine product ...
    now sin(x) can be written as a power series (an infinite polynomial**),
    its of the form x(1+........)

    it has its zeroes at 0,+/-pi,+/-2pi,.....
    So i can factor sin(x) as x(1-x/pi)(1+x/pi)(1-x/2pi)(1+x/2pi).....
    i.e x(1-x^2/pi^2)(1-x^2/2^2*pi^2) ....

    This isn't a highly rigorous proof i know but still intuitive and pretty easy to understand ...

    -- AI

    ** - an infinite polynomial isn't taken in a mild way by many mathematicians .... i apologise them to all ... :tongue2:
     
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