Finding 2012th Term of F(x) Integral from -π to π

[Dassinia]

Hello,
This is not a homework, just trying to solve the previous exams and I'm stuck at one of them :

Homework Statement

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

∫ F₂₀₁₂(x) from -pi to pi

Homework Equations

The Attempt at a Solution

I've tryed replacing the sine by the exponential expression, but it doesn't really help !
Thanks

 

[STEMucator]

Hello,
This is not a homework, just trying to solve the previous exams and I'm stuck at one of them :

Homework Statement

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

∫ F₂₀₁₂(x) from -pi to pi

Homework Equations

The Attempt at a Solution

I've tryed replacing the sine by the exponential expression, but it doesn't really help !
Thanks

If I'm reading correctly this is your integral:

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

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.

 

[Dassinia]

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

 

[LCKurtz]

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

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

 

[Dassinia]

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

 

[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.

 

[D H]

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

I would try looking at the difference between successive *even* terms of F_n 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.

 

[Dassinia]

Hello,

F_2N=sin²(nx)/sin²(x/2)
F_2N+2=sin²((n+1)x)/sin²(x/2)

F_2N-F_2N+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 ?

 

[D H]

No! Successive terms of F_N means developing F_N in terms of F_N-1.

 

[Dick]

I would try looking at the difference between successive *even* terms of F_n 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.

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.

 

[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)

 

[D H]

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.

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.

 

[Dassinia]

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

 

[D H]

F_n 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 F_n a different way, for example, as a series, each element of which is easily integrated.

 

[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

 

[D H]

I made a mistake in my initial post it is Fn(x)=1/(2*pi) * [ sin(n*x/2) / sin(x/2) ]²

Grrrr. Do not do that!

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

 

[Dick]

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.

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.

 

[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) e^irx sum from -n to n

 

[Dick]

sigma notation ?
Even with the " ² " we can write it simply as I did or is it false ?

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.

 

[D H]

sigma notation ?
Even with the " ² " we can write it simply as I did or is it false ?

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*.

 

[Dassinia]

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

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

 

[D H]

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

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

Take this one step further: What is the difference between consecutive pairs of differences? In other words, what is Δ²F_n(x)=ΔF_n+1(x)-ΔF_n(x)? How does this help you find $I_n = \int_{-\pi}^{\pi} F_n(x) \, dx$ ?