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Another Viete problem

  • Thread starter ehrenfest
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  • #1
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[SOLVED] another Viete problem

Homework Statement


Prove that for every positive integer n,

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


Homework Equations


http://en.wikipedia.org/wiki/Viète's_formulas


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. :(
 

Answers and Replies

  • #2
tiny-tim
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Hi ehrenfest! :smile:

Hint: use de Moivre's theorem. :smile:
 
  • #3
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Hint: use de Moivre's theorem. :smile:
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. :(
 
  • #4
tiny-tim
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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. :(
: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:
 
  • #5
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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.
 
  • #6
George Jones
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[tex]\cos \theta + i \sin \theta = ?[/tex]
 
  • #7
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[tex]\cos \theta + i \sin \theta = ?[/tex]
[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:
  • #8
George Jones
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[tex]e^{i \theta}[/tex]

But how does that help simplify the RHS?
Substituting this shows that this series is a special type of series.

What type of series?
 
  • #9
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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!
 
  • #10
tiny-tim
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Hi ehrenfest! :smile:

You're getting confused.

You've used ∑ instead of ∏. Your:
[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]
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:
 
  • #11
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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:
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:
  • #12
tiny-tim
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… 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. …
ah, but if you use de Moivre again, then the n+1 cancels, and you get … ? :smile:
 
  • #13
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ah, but if you use de Moivre again, then the n+1 cancels, and you get … ? :smile:
What? There is a 2n+1 not an n+1 in the denominator of the cosine and the sine...
 
  • #14
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1
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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