Is this identity true?

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  • #1
quasar987
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Someone told my friend, who in turn told me that this identity was true. However, I can't prove it, and when I try to use it I can't get the right answer to a rather simple problem. So, is it true that

[tex]\frac{1}{2} + \sum_{j=1}^n cos(jx) = \frac{sin([n+\frac{1}{2}]x)}{sin(\frac{x}{2})}[/tex]

?? Thx!
 

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  • #2
AKG
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No, try n = x = 1.
 
  • #3
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I can't tell you off-hand whether or not that identity is true, but at least there is a very similar identity (which might be equal to the one you posted). First, note that

[tex]\cos{jx} = Re(\cos{jx} + i \sin{jx}) = Re(e^{ixj}) = Re(({e^{ix}})^j).[/tex]

The Re function is linear, which means that summing cos(jx) is equivalent to summing (e^(ix))^j and then calculating the real part of that. Hence the problem can be reduced to calculating the sum of a geometric series...
 
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  • #4
krab
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Pretty nifty identity, but you forgot a factor of 2. Since the left looks like a Fourier series, you can probably prove it by multiplying by cos(mx) and integrating
 
  • #5
Galileo
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It looks familiar. If I remember correctly, the name is the Dirichletkernel.

[tex]D_n:=\sum_{k=-n}^n e^{ikx}=1+2\sum_{k=1}^n \cos kx[/tex]

Use a geometric expansion to find a closed form. It only works if [itex]e^{ix}\not= 1[/itex]

[tex]D_n(x)=e^{-inx}\sum_{k=0}^{2n}e^{ikx}=\frac{\sin (n+1/2)x}{\sin x/2}[/tex]

If [itex]e^{ix}\not=1[/itex] you can expand the sum geometrically. After some algeblah you'll get the answer. Treat the case [itex]e^{ix}=1[/itex] seperately.
 
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  • #7
quasar987
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K, thanx, I proved it by induction like the OP.
 

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