Varification needed for small trigonometrical Fourier series, PRESSING

[toneboy1]

Hi,
I have x(t) = 1/2 + cos(t) + cos(2t)

so I can see that a0 = 1/2
and that it is an even function so there is no b_n
Also that T = 2pi so
a_n = 2/2pi ∫₀^2pi x(t).cos(nω₀t) dt

but when I integrate this I get a_n = 0 yet I've been told that the answer is

x(t) = 1/2 + Ʃ_{n = 1}² cos(nω₀t)

which would mean that a_n = 1, but I'm not sure how.

Anyone?

Thanks heaps!EDIT: I just found another related problem I'm struggling with if anyone's interested:
https://www.physicsforums.com/showthread.php?t=654607

 

[rude man]

This has been adequately addressed already. Your x(t) is already a Fourier series. If you computed the coefficients to be different you made a math error.

 

[toneboy1]

Yeah, after this was inactive I posted it in the homework help section: https://www.physicsforums.com/showthread.php?p=4173472&posted=1#post4173472

 

[berkeman]

Yeah, after this was inactive I posted it in the homework help section: https://www.physicsforums.com/showthread.php?p=4173472&posted=1#post4173472

Please do not multiple post like that. If you wanted this thread moved, click the Report button on your post and ask the Mentors to move it.

 

[alphysicist]

Hello,

Thank you for sharing your question and concern about verification for small trigonometrical Fourier series. It seems like you have correctly identified that x(t) is an even function, which means that all the bn terms will be equal to zero. However, the issue you are facing is with calculating the an terms.

In order to verify the small trigonometrical Fourier series, you will need to use the formula for calculating the an terms:

an = 2/T ∫0T x(t).cos(nω0t) dt

In your case, T = 2π and ω0 = 1, so the formula becomes:

an = 2/2π ∫0 2π [1/2 + cos(t) + cos(2t)].cos(nt) dt

Now, when you integrate this, you will get:

an = 1/π [sin(nt) + 1/2 sin(t) + 1/4 sin(2t)] evaluated from 0 to 2π

Since the sine function is periodic with a period of 2π, sin(nt) evaluated at 2π will be equal to sin(0) = 0. Similarly, sin(2t) evaluated at 2π will also be equal to sin(0) = 0. This leaves us with:

an = 1/π [1/2 sin(t)] evaluated from 0 to 2π

Using the fact that sin(t) is also an odd function, we know that sin(t) evaluated at 0 will be equal to 0 and sin(t) evaluated at 2π will also be equal to 0. Therefore, the an terms will be equal to 0, which is the same result you got.

However, when you use the formula for the Fourier series, you will get:

x(t) = 1/2 + Ʃn = 1^2 cos(nω0t) = 1/2 + cos(t) + cos(2t)

So, there must be a mistake in the formula you were using to calculate the an terms. I suggest double-checking the formula and making sure you are using the correct values for T and ω0.

As for the related problem you mentioned, I recommend posting it on a physics forum or discussing it with a colleague or mentor who has experience with Fourier series. They may be able to