Fourier transform and derivation

[arcTomato]

Homework Statement: I don't know how can I derivation Eq.(2.2)
Homework Equations: Fourier coefficients

Homework Statement: I don't know how can I derivation Eq.(2.2)
Homework Equations: Fourier coefficients

Dear all.
I don't know how can I derivation Eq.(2.2).
Where Σk is come from??

 

[Mark44]

Moved from a HW section.
The derivation, which is described as capable of being "straightforwardly computed," doesn't seem at all obvious to me.

 

[Orodruin]

How is ##t_k## defined? I feel that more information is needed than the image you posted.

Moved from a HW section.
The derivation, which is described as capable of being "straightforwardly computed," doesn't seem at all obvious to me.

On that side note, I think words such as ”straightforwardly” or ”simply” simply do not belong in textbooks.

 

[arcTomato]

Thank you @Orodruin and I appreciate for your kindness.
This paper doesn’t refer to what ##t_k## is.
That’s is the point I can’t understand.

 

[Orodruin]

Can you give a reference to the paper?

 

[arcTomato]

Thanks for reply,@Orodruin!
https://pure.uva.nl/ws/files/2212461/47104_Fourier_techniques.pdfThis is the paper.Could you access this?
What we are talking about is in Chapter 2(Introduction).

 

[MathematicalPhysicist]

##x_k=x(t_k) = \frac{1}{N}\sum_j A_j \cos \omega_j t_k + B_j \sin \omega_j t_k## , plug the solution:
##A_j = \sum_k x_k \cos \omega_j t_k , \ \ \ B_j = \sum_k x_k \sin \omega_j t_k##, and you get:
##x_k=\frac{1}{N} \sum_j \sum_m x_m \cos \omega_j t_m \cos_j t_k + x_m \sin \omega_j t_m \sin \omega_j t_k ##, if ##m\ne k##, then the term on the RHS will vanish, since on the the LHS there are no terms of this form, so the sum over ##m## becomes ##x_k(\cos^2 \omega_j t_k + \sin^2 \omega_j t_k)=x_k##, the sum over ##j## gives you the ##N## factor that cancels with ##N## in the denominator.
That's all there is to it.

 

[arcTomato]

Thanks for reply @MathematicalPhysicist (I like your name).
I got it! This is the easy-to-understand explanation.
I appreciate for you all.