Inductive proof in complex arithmetic

  • #1
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Homework Statement



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]

Homework Equations



None I can think of.

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. :(
 
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Answers and Replies

  • #2
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Have you tried rewriting the LHS using complex notation and then using the good ol' summation formula for a geometric progression?
 
  • #3
SammyS
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Homework Statement



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

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

Homework Equations



None I can think of.

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 = 1}^{n + 1} {\cos{(kx)}} = \sum_{k = 1}^{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. :(
It looks to me like the summations should be from 0 to n, not 1 to n.
 
  • #4
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Sorry sorry sorry I mistyped the question! Certainly summation begins with k = 0.
 
  • #5
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Have you tried rewriting the LHS using complex notation and then using the good ol' summation formula for a geometric progression?

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?? ...
 
  • #6
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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?
 
  • #7
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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!
 
  • #8
Dick
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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!

Remembering cos(kx)=Re(exp(ikx)) might come in handy for the geometric series part as well.
 
  • #9
SammyS
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...

Now I could collect the denominators but it doesn't help (as far as I can see).
...
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.
 
  • #10
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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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