# Summation equation help

1. Feb 14, 2006

### UrbanXrisis

I am to show that...

$$\sum_{n=-N}^{+N} cos(\alpha -nx)=cos\alpha \frac{sin(N+0.5)x}{sin(x/2)}$$

$$\sum_{n=-N}^{+N} cos(\alpha)cos(nx)+\sum_{n=-N}^+Nsin(\alpha)\frac{sin(N+0.5)x}{sin(x/2)}$$

$$\sum_{n=-N}^{+N}sin(\alpha)\frac{sin(N+0.5)x}{sin(x/2)} =0$$

$$cos(\alpha) 2 \sum_{n=0}^{+N} cos(nx)$$

I know of a rule that shows...

$$\frac{1}{2}+cos(x)+cos(2n)+...cos(nx)=\frac{sin(N+0.5)x}{2sin(x/2)}$$

but I dont see how to apply it to get my answer, since my summation is similar to equation (9) on this site: http://mathworld.wolfram.com/Cosine.html

any ideas?

2. Feb 14, 2006

### Hurkyl

Staff Emeritus
I think you've messed up in stating your problem -- it doesn't really make sense.

Anyways, you know trig identities, right? You could try applying some of them.

Or, you could always grind through an inductive proof.

3. Feb 15, 2006

### UrbanXrisis

oh wow, i totally messed up there...

$$\sum_{n=-N}^{+N} cos(\alpha -nx)=cos\alpha \frac{sin(N+0.5)x}{sin(x/2)}$$

$$cos(\alpha -nx) =cos(x)cos(nx)+sin(\alpha)sin(nx)$$

$$\sum_{n=-N}^{+N} cos(\alpha)cos(nx)+\sum_{n=-N}^{+N} sin(\alpha)sin(nx)$$

$$sin(\alpha) \sum_{n=-N}^{+N} sin(nx)=0$$ since it is an odd function

so I am left with...
$$cos(\alpha) 2 \sum_{n=0}^{+N} cos(nx)=cos\alpha \frac{sin(N+0.5)x}{sin(x/2)}$$
$$2 \sum_{n=0}^{+N} cos(nx)=\frac{sin(N+0.5)x}{sin(x/2)}$$

there is a rule that shows:
$$\frac{1}{2}+cos(x)+cos(2n)+...cos(nx)=\frac{sin(N+ 0.5)x}{2sin(x/2)}$$

I am stuck on this part and I dont know where to go from here.

4. Feb 15, 2006

### Hurkyl

Staff Emeritus
Can you write that rule in summation notation?

5. Feb 15, 2006

### VietDao29

This is wrong, counter-example: N = 0, the LHS is 2, whereas the RHS is 1, and it's true that: $$2 \neq 1$$, right?
You are wrong when assuming that:
$$\cos \alpha \sum_{n = -N} ^ {+N} (\cos (nx)) = 2 \cos \alpha \sum_{n = 0} ^ {+N} (\cos (nx))$$
$$\cos \alpha \sum_{n = -N} ^ {+N} (\cos (nx)) = 2 \cos \alpha \sum_{n = 1} ^ {+N} (\cos (nx)) + \cos \alpha \cos (0n) = 2 \cos \alpha \sum_{n = 1} ^ {+N} (\cos (nx)) + \cos \alpha$$
$$= \cos \alpha \left( 1 + 2 \left( \sum_{n = 1} ^ {+N} \cos (nx) \right) \right)$$.
Now you can go from here, right? Hint: follow Hurkyl's suggestion. :)

6. Feb 15, 2006

### UrbanXrisis

$$\frac{sin(N+ 0.5)x}{2sin(x/2)}=-\frac{1}{2}+\sum_{n = 1} ^ {+N} \cos (nx)$$

so now everything fits into place!

Just out of curiosity, how would someone derive: $$\frac{1}{2}+cos(x)+cos(2n)+...cos(nx)=\frac{sin(N+ 0.5)x}{2sin(x/2)}$$?

7. Feb 15, 2006

### shmoe

Use $$\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$$ to turn it into a geometric sum.

8. Feb 15, 2006

### Hurkyl

Staff Emeritus
Induction works too.

9. Feb 15, 2006

### UrbanXrisis

would the answer include the imaginary part?

i've found an example in mathworld but it isnt the sum from 1 to infinity, but from 0 to infinity: http://mathworld.wolfram.com/Cosine.html

how is it possible that their cosine on the numerator is cancelled out to form $$\frac{sin(N+0.5)x}{sin(x/2)}$$ when the summation is changed from (0 to inifinity is what they have) to 1 to infinity?

10. Feb 15, 2006

### shmoe

There won't be an imaginary part.

The sum on mathworld, (9)-(13) I guess you mean, is a little different from yours (note it doesn't go to infinity). You both have a term for n=0, but yours is half theirs and your argments for the sin in the numerator are different.

Their method of derivation is essentially what I suggested (though they should have said something about the x=integer multiple of 2*pi case). You might want to work with yours in the form $$\sum_{n = -N} ^ {+N} \cos (nx)$$ though.