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Inductive proof in complex arithmetic

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that for any [itex]n \in \mathbb{N}[/itex] and [itex]x \in \mathbb{R}[/itex], we have

    [tex]\sum_{k = 0}^{n} {\cos{(kx)}} = \frac{1}{2}+ \frac{\cos{(nx)} - \cos{[(n+1)x]}}{2 - 2\cos {x}}[/tex]

    2. Relevant equations

    None I can think of.

    3. The attempt at a solution

    Try induction. The result holds if n = 0. Suppose the result holds for some natural number n. Then we get

    [tex]\sum_{k = 0}^{n + 1} {\cos{(kx)}} = \sum_{k = 0}^{n} {\cos{(kx)} + \cos{[(n + 1)x]}} = \frac{1}{2}+ \frac{\cos{(nx)} - \cos{[(n+1)x]}}{2 - 2\cos {x}} + \cos{[(n + 1)x]}[/tex]

    Now I could collect the denominators but it doesn't help (as far as I can see). I'm stuck at this point. Please help me out!!

    For some perspective ... This is a third year complex analysis course and this is my first assignment. My first year algebra course covered induction and complex numbers, so I can't tell if the purpose of this question is review or if there's a deeper trick to it. The rest of the problems on this assignment are on the topology of the complex plane so this problem is kind of a sore thumb. It seems like it should be "easier" but I'm having a hard time. :(
     
    Last edited: Jan 17, 2012
  2. jcsd
  3. Jan 17, 2012 #2
    Have you tried rewriting the LHS using complex notation and then using the good ol' summation formula for a geometric progression?
     
  4. Jan 17, 2012 #3

    SammyS

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    It looks to me like the summations should be from 0 to n, not 1 to n.
     
  5. Jan 17, 2012 #4
    Sorry sorry sorry I mistyped the question! Certainly summation begins with k = 0.
     
  6. Jan 17, 2012 #5
    Which complex notation are you talking about? We actually haven't learned much about complex numbers in particular ... It's an analysis class, so the prof defined [itex]\mathbb{C}[/itex] as [itex]\mathbb{R}^2[/itex] with "funny multiplication" and then we immediately started talking about Cauchy sequences, topology, holomorphic functions, Cauchy-Riemann equations, that kind of thing. So I actually don't really know what to expect of this particular problem. It might be a trick question?? ...
     
  7. Jan 17, 2012 #6
    If you're a third year undergrad, you should know the formula [itex]\cos x = \frac{1}{2} \left( e^{ix} - e^{-ix} \right)[/itex], am I mistaken?
     
  8. Jan 17, 2012 #7
    Oh yeah! I saw it before in first year but it's been a while since then, lol. That's not really too in-depth but the only stuff I've done with complex numbers was studying inner product spaces, and all we didn't need stuff complex exponential notation to do that.

    But yeah I'll try that formula and get back to you, thanks!
     
  9. Jan 17, 2012 #8

    Dick

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    Remembering cos(kx)=Re(exp(ikx)) might come in handy for the geometric series part as well.
     
  10. Jan 17, 2012 #9

    SammyS

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    It certainly does help.

    [itex]\displaystyle \frac{\cos{(nx)} - \cos{[(n+1)x]}}{2 - 2\cos (x)} + \cos{[(n + 1)x]}[/itex]
    [itex]\displaystyle =\frac{\cos{(nx)} - \cos{[(n+1)x]}}{2(1 - \cos(x))} +\frac{2(1 - \cos(x))\cos{[(n + 1)x]}}{2(1 - \cos(x))}[/itex]

    [itex]\displaystyle =\frac{\cos{(nx)} - \cos{[(n+1)x]}+2\cos{[(n+1)x]}-2\cos(x)\cos[(n+1)x] }{2(1 - \cos(x))}[/itex]

    [itex]\displaystyle =\frac{\cos{(nx)} +\cos{[(n+1)x]}-2\cos(x)\cos[(n+1)x] }{2(1 - \cos(x))}[/itex]​
    It gives you the cos[(n+1)x] term in the numerator. All that you need to show is the the rest of the numerator gives -cos[(n+2)x]. What mr. vodka suggested will easily accomplish that.
     
  11. Jan 17, 2012 #10
    Actually the formula is [itex]\cos x = \frac{e^{ix} + e^{-ix}}{2}[/itex] but the general idea of using this formula was very helpful! It's a lot easier to show this directly by plugging in that formula and using the geometric sum formula, than to use induction.

    Cool problem though! Thanks guys.
     
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