Proving the Truth of a Mysterious Identity

[quasar987]

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

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

?? Thx!

 

[AKG]

No, try n = x = 1.

 

[Muzza]

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

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

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...

 

[krab]

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

 

[Galileo]

It looks familiar. If I remember correctly, the name is the Dirichletkernel.

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

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

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

If e^{ix}\not=1 you can expand the sum geometrically. After some algeblah you'll get the answer. Treat the case e^{ix}=1 seperately.

 

[lurflurf]

Almost
There was a thread about the correct form a while back
https://www.physicsforums.com/showthread.php?t=86735

 

[quasar987]

K, thanx, I proved it by induction like the OP.