Proving a Summation Equation using Cosine and Sine Functions | Help and Examples

[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 don't 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?

 

[Hurkyl]

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.

 

[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 don't know where to go from here.

 

[Hurkyl]

there is a rule that shows:

Can you write that rule in summation notation?

 

[VietDao29]

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)}

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))
In fact, it should read:
\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. :)

 

[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)}?

 

[shmoe]

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

 

[Hurkyl]

Induction works too.

 

[UrbanXrisis]

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

would the answer include the imaginary part?

i've found an example in mathworld but it isn't 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 canceled 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?

 

[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.