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Infinite product

  1. Mar 30, 2004 #1
    work out the value of the following: :redface:

    1/2*sqrt[1/2 + (1/2)*(1/2)]sqrt[1/2 + (1/2)*sqrt[1/2 + (1/2)*(1/2)]]...

    Any help is appreciated. :confused: :smile:

    Thnak you. :biggrin:

    what mite b of interest :rolleyes: is the fact that it looks similar to the bottom of Viete's famous formula for pi (employing for the first time an infinite product):

    [ i derived the following from cos(theta/2^1)*cos(theta/2^2)*cos(theta/2^3)...cos(theta/2^n) ]

    pi=2/[sqrt[1/2]*sqrt[1/2 + (1/2)*sqrt[1/2]]sqrt[1/2 + (1/2)*sqrt[1/2 + (1/2)*sqrt[1/2]]]...]
  2. jcsd
  3. Mar 30, 2004 #2
    [tex]\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}}\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2}}}\cdots[/tex]

    How exactly does that continue? I can think of a few ways.

  4. Mar 31, 2004 #3
    how it contains

    i know how it continues...but i need to calculate the infinite product.
  5. Mar 31, 2004 #4

    matt grime

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    But we don't know how it continues so we can't help you.
  6. Mar 31, 2004 #5
    oh right! sorry...i misunderstood the intention of the commentator. Im assuming it continues from what i wrote with it being multiplied by:

    (*) sqrt[1/2 + (1/2)*sqrt[1/2 + (1/2)*sqrt[1/2 + (1/2)*(1/2)]]]

    to get the next part, you take what i've written here(*), multiply it by 1/2, add 1/2 to what you now have, then square root what you have now, then multiply it to the product.

    You follow the same procedure for the next part. Take what you last added, multiply by half, add half to this, then square root what you've then got.
  7. Apr 1, 2004 #6


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    So do you mean:

    [tex]\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\cdot\frac {1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2}}}\cdots }[/tex]

    As apposed to:

    [tex]\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\cdot\frac {1}{2}}\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2}\cdots}}[/tex] ?

    In which case is it not:

    [tex]x = \frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\cdot\frac {1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2}}}\cdots }[/tex]

    [tex]2x = \sqrt{\frac{1}{2}+\frac{1}{2}\cdot\frac {1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2}\cdots }}}[/tex]

    [tex] 4x^2 = \frac{1}{2}+\frac{1}{2}\cdot\frac {1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2}\cdots}} [/tex]

    [tex] 4x^2 - \frac{1}{2} = \frac{1}{2}\cdot\frac {1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2}\cdots}}[/tex]

    [tex] 8x^2 - 1 = x [/tex]

    Baring in mind that one of the answers for x will have been created when squaring it. (Or something to that effect I've never done anything like this before)
  8. Apr 1, 2004 #7
    how i wish it was, but its not.sorry. its the second one.
  9. Apr 1, 2004 #8


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    It looks like the terms are given by

    x_n = \left\{
    \frac{1}{2} & n = 0 \\
    \sqrt{ \frac{1}{2} + \frac{1}{2} x_{n-1} } & n > 0

    So that for [itex]n > 0[/itex] we have [itex]x_{n-1} = 2 x_n^2 - 1[/itex].

    (Hey, these look like the cosine half and double angle formulas! :smile:)
  10. Apr 1, 2004 #9
    Does that help us realise what the whole thing multiplies itself into? Maybe this'll help:
    I already derived a very similar-looking infinite product based on Viete's method. I derived from cos(theta/2^1)*cos(theta/2^2)*cos(theta/2^3)...cos(theta/2^n)

    the following:

    pi=2/[sqrt[1/2]*sqrt[1/2 + (1/2)*sqrt[1/2]]sqrt[1/2 + (1/2)*sqrt[1/2 + (1/2)*sqrt[1/2]]]...]

    I used cosine's half and double-angle formulaes to do this.

    I think Im not understanding what you're saying...sorry.
  11. Apr 1, 2004 #10


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    I'm hypothesizing that the infinite product you are trying to compute is given by

    \prod_{i=0}^{\infty} x_i

    where the [itex]x_i[/itex] are given by the relation I presented.

    So I'm trying to make sure I know what your product is before I try and tackle it. :smile:
    Last edited: Apr 1, 2004
  12. Apr 1, 2004 #11
    Dont take offence o'l boy! Dont take me wrong, I am grateful for the help, especially from a PF mentor such as yourself.
  13. Apr 5, 2004 #12
    I've been told many times to 'plug-in' (2*pi)/3 into the cos-product. But i've failed to see the significance of this....maybe someone else will not.
  14. Apr 5, 2004 #13
    Quddusaliquddus....have you tried: x = sqrt(1/2 + 1/2*x) ?
  15. Jun 18, 2004 #14

    I cheated by calculating the value using spreadsheet software, finding the reciprocal, and looking up the reciprocal in Plouffe's Inverter at
    which gives that it is 1 over 2/3*Gamma(1/3)*Gamma(2/3) (which is also sum(1/(45/2*n^2-45/2*n+5),n=1..inf) )
    Here are approximate values:
    the product: 0.4134966715663
    its reciprocal: 2.41839915231229
    Gamma(1/3): 2.67893853470775
    Gamma(2/3): 1.35411793942640
    From this page
    we get Gamma(1/3)Gamma(2/3) = 2*PI/sqrt(3) so the value is
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