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Another Viete problem

  1. May 21, 2008 #1
    [SOLVED] another Viete problem

    1. The problem statement, all variables and given/known data
    Prove that for every positive integer n,

    [tex]\Pi_{k=1}^n \tan \frac{k \pi}{2n+1} = \sqrt{2n+1}[/tex]


    2. Relevant equations
    http://en.wikipedia.org/wiki/Viète's_formulas


    3. The attempt at a solution
    I cannot figure out what polynomial I need to apply Viete to.

    Should I let

    [tex]f(x) = \Pi_{k=1}^n \left(x- \tan \frac{k \pi}{2n+1}\right)[/tex]

    ? That doesn't seem to help at all. I looked for relevant trig identities and couldn't find any. I tried induction (the case n=1 is trivial) but that failed also so alas I am stuck again. :(
     
  2. jcsd
  3. May 22, 2008 #2

    tiny-tim

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    Hi ehrenfest! :smile:

    Hint: use de Moivre's theorem. :smile:
     
  4. May 22, 2008 #3
    Hmm.

    [tex]\cos n \theta (1 + i \tan n \theta) = (\cos \theta + i \sin \theta)^n[/tex]

    We can let \theta = pi/(2n+1). We could then sum both sides over n. But somehow we need to get a polynomial and then apply Viete to it. I don't see what polynomial to use. Sorry tiny-tim. :(
     
  5. May 22, 2008 #4

    tiny-tim

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    :cry: (tears hair out) :cry:

    But the whole point of de Moivre's theorem is that it avoids using Viete! :rolleyes:

    Just do what you said … sum kπ/(2n+1) from 1 to n. :wink:

    ehrenfest … you keep doing this … you get it fixed in your head that you have to do something, even when you yourself have found a way that avoids it! :smile:
     
  6. May 22, 2008 #5
    Well I really didn't see how anything would simplify after I summed it:

    [tex]
    \sum_{k=1}^n \cos \frac{k \pi}{2n+1} (1 + i \tan \frac{k \pi}{2n+1} ) = \sum_{k=1}^n (\cos \frac{\pi}{2n+1}+ i \sin \frac{\pi}{2n+1})^k
    [/tex]

    But I'm sure you have some clever trick up your sleeve tiny-tim.
     
  7. May 22, 2008 #6

    George Jones

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    [tex]\cos \theta + i \sin \theta = ?[/tex]
     
  8. May 22, 2008 #7
    [tex]e^{i \theta}[/tex]

    But how does that help simplify the RHS. I know there must be something obvious I am missing? It is just the first n of the 4n+2 roots of unity but does that equal something nice?
     
    Last edited: May 22, 2008
  9. May 22, 2008 #8

    George Jones

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    Substituting this shows that this series is a special type of series.

    What type of series?
     
  10. May 22, 2008 #9
    I'm an idiot.

    The RHS is

    [tex]\frac{1-\exp \frac{i (n+1) \pi}{2n+1}}{1-\exp \frac{i \pi}{2n+1}}[/tex]

    So

    [tex]

    \sum_{k=1}^n \cos \frac{k \pi}{2n+1} (1 + i \tan \frac{k \pi}{2n+1} ) = \frac{1-\exp \frac{i (n+1) \pi}{2n+1}}{1-\exp \frac{i \pi}{2n+1}}

    [/tex]

    Sorry to keep prolonging this but I still don't see where this is going. I mean how are we ever going a product of tangents from this!
     
  11. May 22, 2008 #10

    tiny-tim

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    Hi ehrenfest! :smile:

    You're getting confused.

    You've used ∑ instead of ∏. Your:
    should be: [tex] \Pi_{k=1}^n \cos \frac{k \pi}{2n+1} (1 + i \tan \frac{k \pi}{2n+1} ) = \Pi_{k=1}^n (\cos \frac{\pi}{2n+1}+ i \sin \frac{\pi}{2n+1})^k = (\cos \frac{\pi}{2n+1}+ i \sin \frac{\pi}{2n+1})^{\sum{k=1}^{n} k}\\,=\,....
    [/tex]

    hmm … the LaTeX hasn't come out right … but I hope you see what I mean! :redface:
     
  12. May 22, 2008 #11
    Well you said sum in post #4. Sorry I still don't see where we are going with this. I agree with what you posted above and the sum of the first n integers is n(n+1)/2 but I cannot even see what will happen in the case of n=2.

    Then we have

    [tex]\exp i \frac{3 \pi}{5} = \cos \frac{\pi}{5} \cos\frac{2 \pi}{5}(1+i\tan \frac{\pi}{5})(1+i\tan\frac{2\pi}{5})[/tex]

    We can take the real part of both sides but then what?
     
    Last edited: May 22, 2008
  13. May 22, 2008 #12

    tiny-tim

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    ah, but if you use de Moivre again, then the n+1 cancels, and you get … ? :smile:
     
  14. May 22, 2008 #13
    What? There is a 2n+1 not an n+1 in the denominator of the cosine and the sine...
     
  15. May 23, 2008 #14
    Well, this was really bothering so I looked at the solution in my book that employs Viete. Using de Moivre was a good idea but I think you really do need Viete also.
     
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