What is the closed form for the summation of cos(nθ) from -N to N?

[SpaceDomain]

I am having a difficult time working with some of these infinite series. I studied them in calc 2, but that was a few years ago.

Could someone help me figure out how to find what the following sum converges to:

$$\sum_{n=-N}^N{cos(n \theta)}$$

Shouldn't there be some property or convergence test I could do to find out what it actually converges to?

 

[SpaceDomain]

This expands to:

$$cos(-N\theta)+ ... +cos(-4\theta)+ cos(-3\theta)+ cos(-2\theta) + cos(-\theta) + cos(0) <br /> + cos(\theta) + cos(2\theta) + cos(3\theta)+ cos(4\theta)+ ... + cos(N \theta)$$


And since $$cos(-\alpha) = cos(\alpha)$$ the above expansion boils down to:

$$cos(N\theta)+ ... +cos(4\theta)+ cos(3\theta)+ cos(2\theta) + cos(\theta) + cos(0) <br /> + cos(\theta) + cos(2\theta) + cos(3\theta)+ cos(4\theta)+ ... + cos(N \theta)$$

Which can be simplified as:

$$cos(0) + 2cos(\theta) + 2cos(2\theta) + 2cos(3\theta)+ 2cos(4\theta)+ ... + 2cos(N \theta)$$

 

[SpaceDomain]

So I am actually looking at the case where $$\theta = \frac{\pi}{2}$$

Therefore:

$$2cos(0) = 2$$

$$2cos( \frac{\pi}{2} ) = 0$$

$$2cos(\pi) = -2$$

$$2cos( \frac{3\pi}{2} ) = 0$$

$$2cos( 2\pi ) = 2$$

$$.$$

$$.$$

$$.$$

$$2cos(N \frac{\pi}{2})$$

So the sequence would be:
$$\{2, 0, -2, 0, 2, ... , 2cos(N \frac{\pi}{2}) \}$$

Is my logic correct?
And if so, how do I sum the sequence to find what it converges to?

 

[mathman]

So I am actually looking at the case where $$\theta = \frac{\pi}{2}$$

Therefore:

$$2cos(0) = 2$$

$$2cos( \frac{\pi}{2} ) = 0$$

$$2cos(\pi) = -2$$

$$2cos( \frac{3\pi}{2} ) = 0$$

$$2cos( 2\pi ) = 2$$

$$.$$

$$.$$

$$.$$

$$2cos(N \frac{\pi}{2})$$

So the sequence would be:
$$\{2, 0, -2, 0, 2, ... , 2cos(N \frac{\pi}{2}) \}$$

Is my logic correct?
And if so, how do I sum the sequence to find what it converges to?

Your logic is correct. The conclusion is that it doesn't converge.

 

[SpaceDomain]

Okay. Well this is part of a homework problem, so if an admin could move this to the homework section in "calculus and beyond" that would be great.

Here is the whole problem:

$$\lim_{N \rightarrow \infty}<br /> \frac{1}{2N+1}<br /> \sum_{n=-N}^N{|cos(n \frac{\pi}{4})|^2}$$Here is how far I have evaluated it:

$$\lim_{N \rightarrow \infty}<br /> \frac{1}{2N+1}<br /> \sum_{n=-N}^N{cos^2(n \frac{\pi}{4})}$$

$$\lim_{N \rightarrow \infty}<br /> \frac{1}{2N+1}<br /> \sum_{n=-N}^N{\frac{1+cos(2n \frac{\pi}{4})}{2}}$$

$$\lim_{N \rightarrow \infty}<br /> \frac{1}{2N+1}<br /> (\sum_{n=-N}^N{\frac{1}{2}} + <br /> \sum_{n=-N}^N{\frac{1}{2}cos(n \frac{\pi}{2})}})$$

$$\lim_{N \rightarrow \infty}<br /> \frac{1}{2N+1}<br /> \sum_{n=-N}^N{\frac{1}{2}} +<br /> \lim_{N \rightarrow \infty}<br /> \frac{1}{2N+1}<br /> \sum_{n=-N}^N{\frac{1}{2}cos(n \frac{\pi}{2})}}$$

$$\frac{1}{2}<br /> \lim_{N \rightarrow \infty}<br /> \frac{1}{2N+1}(2N+1)+<br /> \lim_{N \rightarrow \infty}<br /> \frac{1}{2N+1}<br /> (\frac{1}{2})<br /> \sum_{n=-N}^N{cos(n \frac{\pi}{2})}}$$

$$\frac{1}{2}<br /> \lim_{N \rightarrow \infty}(1)+<br /> \frac{1}{2}<br /> \lim_{N \rightarrow \infty}<br /> \frac{1}{2N+1}<br /> \sum_{n=-N}^N{cos(n \frac{\pi}{2})}}$$

$$\frac{1}{2}+<br /> \frac{1}{2}<br /> \lim_{N \rightarrow \infty}<br /> \frac{1}{2N+1}<br /> \sum_{n=-N}^N{cos(n \frac{\pi}{2})}}$$

 

[SpaceDomain]

The final answer is
$$\frac{1}{2}$$

So as long as I have done everything else correct so far this part should go to zero:

$$\lim_{N \rightarrow \infty}<br /> \frac{1}{2N+1}<br /> \sum_{n=-N}^N{cos(n \frac{\pi}{2})}}$$

Is that correct?
And if so, how does that go to zero?

 

[SpaceDomain]

Maybe that is as far as I can simplify:

$$\sum_{n=-N}^N{cos(n \frac{\pi}{2})}}$$

So I should evaluate the limit on the outside. That would make:
$$<br /> \lim_{N \rightarrow \infty}<br /> \frac{1}{2N+1}=0$$Therefore making the entire term:

$$\lim_{N \rightarrow \infty}<br /> \frac{1}{2N+1}<br /> \sum_{n=-N}^N{cos(n \frac{\pi}{2})}} = 0$$

Is that correct logic?

 

[mathman]

I suggest you go back to start.
cos²(nπ/4) =1 for n=0, 1/2 for n=1, 0 for n=2, 1/2 for n=3, and cycles from then on. For each set of 4 terms, the sum is 1, so the limit of the original sum is 1/4.

 

[SpaceDomain]

Could you please elaborate. I don't understand.

 

[mathman]

Could you please elaborate. I don't understand.

I'm not sure how to elaborate, except to say that I made an error in arithmetic. The sum of four terms (1 + 1/2 +0 + 1/2) is 2 (not 1), so the limit = 1/2. This agrees with the result you got.

 

[ross_tang]

Actually, you can find a close form for your summation expression. Please refer to http://www.voofie.com/concept/Mathematics/" for details:

http://www.voofie.com/content/156/how-to-sum-sinn-q-and-cosn-q/"

In short,

$$\sum _{n=-N}^N \cos (n \theta )=\cos (N \theta )+\cot \left(\frac{\theta }{2}\right) \sin (N \theta )$$

With this close form, you can solve your limit problem easily.