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Limit of a product cos

  1. Nov 16, 2007 #1

    gop

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    1. The problem statement, all variables and given/known data

    Calculate for [tex]x\in(0,\pi/2)[/tex]

    [tex]\lim_{N\rightarrow\infty}\prod_{n=0}^{N}cos(\frac{x}{2^{n}})[/tex]

    Hint: Use the Double-Angle Formulas for the sine.

    2. Relevant equations



    3. The attempt at a solution

    [tex]cos(x)\cdot cos(\frac{x}{2})\cdot cos(\frac{x}{4})\cdot...[/tex]
    [tex]\frac{\sin2x}{2\cdot\sin x}\cdot\frac{\sin x}{2\cdot\sin x/2}\cdot\frac{\sin x/2}{2\cdot\sin x/4}\cdot...[/tex]
    [tex]\frac{\sin2x}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot...\cdot\frac{1}{2\cdot\sin x/N}[/tex]
    [tex]\frac{\sin2x}{\sin(x/{2^N})\cdot2^{N}}[/tex]
    However, now I have to resolve the 0*infinity in the denom. But how do I resolve that.
     
    Last edited: Nov 16, 2007
  2. jcsd
  3. Nov 16, 2007 #2

    Dick

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    Shouldn't the N in sin(x/N) be like sin(x/(2^N))?
     
  4. Nov 16, 2007 #3

    gop

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    yes, of course. I updated the information.
     
  5. Nov 16, 2007 #4

    Dick

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    Then your denominator is a simple limit. It looks like limit c-> 0 of sin(c*x)/c, where c=1/(2^N).
     
  6. Nov 16, 2007 #5

    Dick

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    BTW, make sure you've counted the 2's in the denominator correctly. I'm finding an extra one.
     
  7. Nov 16, 2007 #6

    gop

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    that, of course, works. thx you a lot.
     
  8. Nov 17, 2007 #7

    Gib Z

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    An alternative method that I find quite nice is to use the fact that cos(x/2^n) is the real part of exp(ix/2^n). The product of the exponentials becomes an easy geometric series in the exponent =]
     
  9. Nov 17, 2007 #8
    You can also use a 'collapsing product'.

    Note,
    [tex]A_4 = \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8} \cos \frac{x}{16}[/tex]
    Then,
    [tex]\sin \frac{x}{16} A_4 = \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8} \cos \frac{x}{16} \sin \frac{x}{16}[/tex]
    So,
    [tex]2A_4\sin \frac{x}{16} = \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8}\sin \frac{x}{8}[/tex]
    Again,
    [tex]4A_4\sin \frac{x}{16} = \cos \frac{x}{2} \cos \frac{x}{4} \sin \frac{x}{4}[/tex]
    Again,
    [tex]8A_4\sin \frac{x}{16} = \cos \frac{x}{2} \sin \frac{x}{2}[/tex]
    Last time,
    [tex]16A_4\sin \frac{x}{16} = \sin x[/tex]
    That means (since [tex]x\in (0,\pi/2)[/tex])
    [tex]A_4 = \frac{\sin x}{2^4 \sin \frac{x}{2^4}}[/tex]
    And in general,
    [tex]A_n = \frac{\sin x}{2^n \sin \frac{x}{2^n}}[/tex]
     
  10. Nov 17, 2007 #9

    Dick

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    Did you try doing it that way? I take the product of the exponentials and I get exp(i*2*x). Now what? The real part of that doesn't have much to do with the product of the real parts of the terms. That trick doesn't ALWAYS make things easier.
     
  11. Nov 18, 2007 #10

    Gib Z

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    Damn It I assumed that the Real part of a product is equal to the product of the real part :( Never mind me =]
     
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