# Homework Help: Finding 2012th term

1. Oct 26, 2013

### Dassinia

Hello,
This is not a homework, just trying to solve the previous exams and i'm stuck at one of them :
1. The problem statement, all variables and given/known data

Consider
Fn(x)=1/(2*pi) * [ sin(n*x/2) / sin(x/2) ]
Calculate:

∫ F2012(x) from -pi to pi

2. Relevant equations

3. The attempt at a solution
I've tryed replacing the sine by the exponential expression, but it doesn't really help !
Thanks

2. Oct 27, 2013

### Zondrina

$\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{sin(1006x)}{sin(\frac{x}{2})} dx$

I think a Taylor expansion would help:

$$\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{ \sum_{n=0}^{∞} (-1)^n \frac{ (1006x)^{2n+1} }{ (2n+1)! } }{ \sum_{n=0}^{∞} (-1)^n \frac{ (\frac{x^{2n+1}}{2^{2n+1}}) }{ (2n+1)! }} dx$$

I believe you can integrate that.

3. Oct 27, 2013

### Dassinia

I completly forgot the properties of Ʃ
Can we simplify by eliminatig the same terms in the numerator and denominator ?

4. Oct 27, 2013

### LCKurtz

Does $\frac {1+3+5}{2+3+6} =\frac {1+5}{2+6}$?

5. Oct 27, 2013

### Dassinia

Oh, of course not.
So, what are the steps to calculate the integral ?
I don't know how calculate that

6. Oct 27, 2013

### haruspex

From the structure of the problem, I would try obtaining a recurrence relation between the integrals of successive Fn. Integration by parts should help.

7. Oct 28, 2013

### D H

Staff Emeritus
I would try looking at the difference between successive *even* terms of Fn and using trigonometric relationships. The integration becomes trivial.

Hint: $\frac{\sin(nx) - \sin((n-1)x)} {\sin(x/2)}$ can be expressed in a very simple form.

8. Oct 28, 2013

### Dassinia

Hello,

F2N=sin²(nx)/sin²(x/2)
F2N+2=sin²((n+1)x)/sin²(x/2)

F2N-F2N+2=[sin²(nx)-sin²((n+1)x)]/sin²(x/2) = [ cos(2nx)-cos(2n+2)x)]/(cos(x)-1)

But I don't get the idea, how is it supposed to help me calculate F2012 ?

9. Oct 28, 2013

### D H

Staff Emeritus
No! Successive terms of FN means developing FN in terms of FN-1.

10. Oct 28, 2013

### Dick

Ah, I see what you are doing. But isn't it the difference between successive odd terms that comes out to be REALLY simple? The difference between even terms is somewhat simple, but you have to sum an awful lot of those simple things to get up to 2012.

11. Oct 28, 2013

### Dassinia

You said "even" terms I took two even successive terms
Difference betweend successive odd terms gives :
[cos((2n+1)x)-cos((2n+3)x)]/(cos(x)-1)

Last edited: Oct 28, 2013
12. Oct 28, 2013

### D H

Staff Emeritus
Since 2012 is even, that's the problem you have to address. Yes, you'll get a sum of 1006 terms. That sum can be expressed very simply.

13. Oct 28, 2013

### Dassinia

Apparently I didn't understand the exersise your way.
I think Fn IS a function that depends of the n we choose. But are you saying that Fn is a serie and that F2012 is the sum from n=0 to n=2012 ?

14. Oct 28, 2013

### D H

Staff Emeritus
Fn is exactly what you specified in the original post: $F_n(x) = \frac 1 {2\pi} \frac {\sin(nx/2)}{ \sin(x/2) }$. That's nice and compact, but how are you going to integrate it? One answer to solve this is to express Fn a different way, for example, as a series, each element of which is easily integrated.

15. Oct 28, 2013

### Dassinia

I made a mistake in my initial post it is Fn(x)=1/(2*pi) * [ sin(n*x/2) / sin(x/2) ]²
I just don't get it.
I am supposed to calculate the integral of the difference of the two even successive terms for the first 1006 terms ?

Ʃ [ cos(2nx)-cos(2n+2)x)]/(cos(x)-1) dx

The integral from -pi to pi
The sum starts at n=0 to n=1006

16. Oct 28, 2013

### D H

Staff Emeritus
Grrrr. Do not do that!

We have been helping you with Fn(x) as defined in the OP. How were we supposed to know you omitted that little superscript 2? This is a very different problem.

17. Oct 28, 2013

### Dick

Ok, while I'm discarding the work I did on the original problem since it's now changed, the only way to express that simply is using sigma notation. I was actually originally wondering if somebody didn't take this question from a test in 2011 to make a new test in 2012, and then decided to 'update' the question by changing '2011' to '2012', without realizing what a difference it would make.

18. Oct 28, 2013

### Dassinia

sigma notation ?
Even with the " ² " we can write it simply as I did or is it false ?
I was looking at my course notes
Can we simplify it like Fejer's Kernel ?
Kn(x)= 1/(n+1) * [ sin((n+1)*x/2) / sin(x/2) ]² = Ʃ (n+1-|r|)/(n+1) eirx sum from -n to n

Last edited: Oct 28, 2013
19. Oct 28, 2013

### Dick

No, I should stop chirping in here. I was talking to DH and that's just confusing you. I was talking about the originally (wrongly!) stated problem. For this one you should be able to actually get a simple numerical answer for $F_{2012}$ without having any series notation at all. Keep working.

20. Oct 28, 2013

### D H

Staff Emeritus
NO!

We were helping you with $\int_{-\pi}^{\pi} \sin(nx/2) / \sin(x/2)\, dx$ because that is what you asked us to help you with in the original post. You never corrected your mistake in your original post until just recently.

You do not know how annoying this is. Most of us who help with homework solve the problem before we post anything.

You need to forget everything that has been said so far because we were solving that original problem. Your new problem is *very* different. This problem is *easy*.

Last edited: Oct 28, 2013
21. Oct 29, 2013

### Dassinia

I calculate F1 F2 F3... And I found that
∫Fn(x) dx = n

To demonstrate it we can use induction proof but how do I get to Fn+1 or Fn-1 from Fn ?
Thanks

22. Oct 30, 2013

### D H

Staff Emeritus
Denote ΔFn(x) as the difference between consecutive functions in the sequence of functions: ΔFn(x)=Fn+1(x)-Fn(x). If you can develop a formula for ΔFn(x) then you can find Fn+1(x) in terms of Fn(x) and ΔFn(x).

Take this one step further: What is the difference between consecutive pairs of differences? In other words, what is Δ2Fn(x)=ΔFn+1(x)-ΔFn(x)? How does this help you find $I_n = \int_{-\pi}^{\pi} F_n(x) \, dx$ ?