First nonzero terms of Fourier sine series

[frenchkiki]

Homework Statement

Consider the function φ(x) ≡ x on (0, l). Find the sum of the first three (nonzero) terms of its Fourier sine series.

Reference: Strauss PDE exercise 5.1.3

Homework Equations

The Attempt at a Solution

I have found the coefficient without difficulty, it is A_{n}= -2l/n\pi cos (n\pi) + 2/n^{2}\pi^{2} sin (n\pi).

When I plug in n=1,2,3; I obtain 2l/\pi, -2l/\pi and 2l/3\pi, respectively.

However the answer is given by: 2l/\pi sin (\pix/l) - 2l/\pi sin (2\pix/l) + 2l/3\pi sin (3\pix/l)

What I found comes from A_{n} for n=1,2,3 but I do not understand which those sine terms come from.

Thanks in advance

 

[Dick]

Homework Statement

Consider the function φ(x) ≡ x on (0, l). Find the sum of the first three (nonzero) terms of its Fourier sine series.

Reference: Strauss PDE exercise 5.1.3

Homework Equations

The Attempt at a Solution

I have found the coefficient without difficulty, it is A_{n}= -2l/n\pi cos (n\pi) + 2/n^{2}\pi^{2} sin (n\pi).

When I plug in n=1,2,3; I obtain 2l/\pi, -2l/\pi and 2l/3\pi, respectively.

However the answer is given by: 2l/\pi sin (\pix/l) - 2l/\pi sin (2\pix/l) + 2l/3\pi sin (3\pix/l)

What I found comes from A_{n} for n=1,2,3 but I do not understand which those sine terms come from.

Thanks in advance

So far you've calculated the coefficients of the sine series. Even saying 'coefficients' means they should multiply something. The sum of the sine series is a function of x. The thing the coefficients multiply is exactly those sine functions your are integrating against. Look at equation 6 here http://mathworld.wolfram.com/FourierSeries.html

 

[frenchkiki]

So far you've calculated the coefficients of the sine series. Even saying 'coefficients' means they should multiply something. The sum of the sine series is a function of x. The thing the coefficients multiply is exactly those sine functions your are integrating against. Look at equation 6 here http://mathworld.wolfram.com/FourierSeries.html

Thank you for your reply.
So that would be sin (n\pix/l) for each corresponding n? Do I need multiply the coefficients by sine because I am asked for a sine series?

 

[Dick]

Thank you for your reply.
So that would be sin (n\pix/l) for each corresponding n? Do I need multiply the coefficients by sine because I am asked for a sine series?

Yes, it's asking you for the sine series. Not just the coefficients.

 

[frenchkiki]

Yes, it's asking you for the sine series. Not just the coefficients.

Alright, so if I get this right, to find the first 3 nonzero terms, I plug in my results for n=1,2,3 in equation 6 and disregard the A_{n} cos (nx) terms since I am asked for the sine series, and A_{0} vanishes so I'm left with B_{n} sin (nx) (A_{n} in my answer). Thanks again