# Is this identity true?

1. Oct 4, 2005

### 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!

2. Oct 5, 2005

### AKG

No, try n = x = 1.

3. Oct 5, 2005

### 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...

Last edited: Oct 5, 2005
4. Oct 5, 2005

### 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

5. Oct 5, 2005

### 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.

Last edited: Oct 5, 2005
6. Oct 5, 2005

7. Oct 5, 2005

### quasar987

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