Question

Help!
I have to find Fourier Serie

Answer 1

Any Fourier Series?

Answer 2

i came to a0=pi/2
an=1/(pi*n^2)(-1-cos(npi)
bn= -3cos(npi)/n

Answer 3

i just have to put this all together right?

Answer 4

Yes, and you want to evaluate the terms with cos in them so they don't appear in your answer.

Namely, cos(n.pi) = (-1)^n

Answer 5

so bn=-3*(-1)^n/n?

Answer 6

based on your b_n above, yes.

Answer 7

Usually we write something like that as
\[ b_n = 3\frac{(-1)^{n+1}}{n} \]

Answer 8

so now i just use formula fo furier serie and that is it?

Answer 9

Assuming you've done everything else correctly yes.

(And you've re-written the a_n using the same property of cos(n.pi) )

Answer 10

yes

Answer 11

\[\pi/4-2/\pi \sum_{n=1}^{\infty}\cos (2n-1)x/(2n-1)^2+\int\limits_{n=1}^{\infty}(-1)^(n-1)\sin nx/n\]

Answer 12

how can get this from a_n, b_n and a_0?

Answer 13

I assume you mean sum for the last term.

If that is the answer, then you read off the Fourier coefficients
\[ a_0 = \pi/4 \] \[ a_n = \frac{-2}{\pi (n-1)^2}, \ n \geq 1 \]
\[ b_n = \frac{(-1)^{n-1}}{n} \]
So if this the answer you're aiming for, check your calculations and make sure the coefficients match this.

Answer 14

sorry, that coefficient for a_n isn't right. It is...

Answer 15

\[ a_n = \frac{1}{\pi} \frac{(-1 -(-1)^n)}{n^2} \]
It is zero when n is even and it is not zero when n is odd.

Answer 16

a1 = -2/pi
a2 = 0
a3 = -2/pi . 1/3^2
a4 = 0
a5 = -2/pi . 1/5^2
etc.

Answer 17

so because its 0, i get for a_n only -2/pi(n-1)^2

Answer 18

actually you're a_n for n > 0 is exactly right. Your a_n is
a_n = 1/(pi*n^2)(-1-cos(npi))
= -2/pi. 1/n^2 , when n is odd because cos(n.pi) = -1 when n is odd
= 0 , when n is even because cos(n.pi) = 1 when n is even

Answer 19

As for your a_0, your answer of pi/2 is exactly right. But remember that when you right down the Fourier series it's a_0/2 + ....

Answer 20

yes

Answer 21

So the only problem you have left is the 3 which appears in the b_n.

Answer 22

so b_n is not correct?

Answer 23

your b_n doesn't correspond to the answer you wrote down; they differ by a factor of 3

Answer 24

\[1/\pi \int\limits_{-\pi}^{0}x*\sin(nx)dx+2/\pi \int\limits_{0}^{\pi}x*\sin(nx)dx\]

Answer 25

this is b_n and i think is ok if i have 3

Answer 26

What is the function your are finding the Fourier series of?

Answer 27

x, -pi<x<0
2x, 0<x<pi

Answer 28

mmm....it's
2x, 0<=x<=pi

Answer 29

does this change anything?

Answer 30

changing the function at a point doesn't change anything. But check your b_n calculation.

Answer 31

can you tell me what is wrong here?

Answer 32

yes, you're right. I fell into the --- (pied up minus signs) trap. -3 it is.
Note to self: drink more coffee before calculating Fourier coefficients.

Answer 33

:)