Determine the Fourier series

agata78

Homework Statement

Determine the Fourier series for the periodic function of period 2∏ defined by:

-2 when (-∏ ) ∠ x ∠ (-∏/2)
f(x)= 2 when ( -∏/2) ∠ x ∠ (∏ /2)
-2 when (∏/2) ∠ x ∠ (∏)

how to start i?. I have already drawn it but what next.
thank you

Mentor

Homework Statement

Determine the Fourier series for the periodic function of period 2∏ defined by:

-2 when (-∏ ) ∠ x ∠ (-∏/2)
f(x)= 2 when ( -∏/2) ∠ x ∠ (∏ /2)
-2 when (∏/2) ∠ x ∠ (∏)

how to start i?. I have already drawn it but what next.
thank you

The Attempt at a Solution

The function runs from -π to +π. As yourself: is the function even in x or odd in x? If it is odd in x, then you only need to include the sin terms in the Fourier Series. if it is odd in x, then you only need to include the constant term and the cosine terms in the Fourier Series. What is the argument of sin or cosine going to be if the function runs from -π to +π?

agata78
yes, the function runs from -∏ to ∏. its from (-∏ to ∏/2) ( -∏/2 to ∏/2) and ( ∏/2 to ∏). There is no period in it, it doesnt repeat. Than it has to be even in. even in is cosinus than yes?

agata78
dont know what you mean by argument?

Homework Helper
Gold Member
yes, the function runs from -∏ to ∏. its from (-∏ to ∏/2) ( -∏/2 to ∏/2) and ( ∏/2 to ∏). There is no period in it, it doesnt repeat. Than it has to be even in. even in is cosinus than yes?

Oh yes it does repeat. That's what 'periodic' means.

Homework Helper
Gold Member
dont know what you mean by argument?

f(x) the argument is x.

Homework Helper
Gold Member
yes, the function runs from -∏ to ∏. its from (-∏ to ∏/2) ( -∏/2 to ∏/2) and ( ∏/2 to ∏). There is no period in it, it doesnt repeat. Than it has to be even in. even in is cosinus than yes?

Yes, it's an even function in the argument x.

Mentor
dont know what you mean by argument?
I was thinking that the terms in the Fourier Series should be Ancos(nx) where x runs from -π to +π, and n is the sequence of integers 1,2,3,... So, the argument is nx.

Homework Helper
Gold Member
I was thinking that the terms in the Fourier Series should be Ancos(nx) where x runs from -π to +π, and n is the sequence of integers 1,2,3,... So, the argument is nx.

You're right, the argument of each of the harmonics is nx, not x.

I wanted to say that his function was an even function of x, x being the argument of f(x).

Sorry Chet!

Last edited:
Mentor
You're right, the argument of each of the harmonics is nx, not x.

I wanted to say that his function was an even function of x, x being the argument of f(x).

Sorry Chet!
No apology necessary RM. You're the man!

Chet

agata78
Is this function is odd or even, cause im confused. When it repeats itself its called odd when its not its even.
This one repeats it self then its odd yes?

And using co- efficient formulae with 2L= 2π, than L= π.
after calculating
[a][/0] =0
[a][/ 1] = 0

dont know how to calculate [/0] ????

Mentor
Is this function is odd or even, cause im confused. When it repeats itself its called odd when its not its even.
This one repeats it self then its odd yes?

And using co- efficient formulae with 2L= 2π, than L= π.
after calculating
[a][/0] =0
[a][/ 1] = 0

dont know how to calculate [/0] ????

No. A function is odd if f(-x)=-f(x), and it is even if f(-x)=+f(x). Your function is even. So, it will be described by a cosine series. Do you have a formula for calculating the coefficients of a Fourier cosine series? If so, please show us, and show us in detail how you applied it to the case of arbitrary n.

Chet

agata78
∞ n∏x n∏x
Sf(x) = a0 +Ʃ [ an cos L + bn sin L ]
n=1

1 ∏
a0= 2∏ ∫ f(x) dx =O
-∏
1 ∏
a1= ∏ ∫ f(x) cos nx dx= 0,
-∏

what happenes if even times even = is it odd? I know when odd times odd = even.

Im sorry i dont have a clue how to use latex, i tried so many times and still doesnt work. Its hard

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Do you know, how to calculate the Fourier coefficients? If so, just calculate them ;-)).

agata78
dont know how to calculate b1

Mentor
∞ n∏x n∏x
Sf(x) = a0 +Ʃ [ an cos L + bn sin L ]
n=1

1 ∏
a0= 2∏ ∫ f(x) dx =O
-∏
1 ∏
a1= ∏ ∫ f(x) cos nx dx= 0,
-∏

what happenes if even times even = is it odd? I know when odd times odd = even.

Im sorry i dont have a clue how to use latex, i tried so many times and still doesnt work. Its hard

Your result for a1 is incorrect. The integral in the right hand side is not zero. Also, the value of n in the cosine term should be n=1. Try doing the integral for n =1 again, and please be more careful.

agata78

Mentor

$$\int_{-\pi}^{+\pi}{f(x)\cos{x}dx}=\int_{-\pi}^{-\pi/2}{(-2)\cos{x}dx}+\int_{-\pi/2}^{+\pi/2}{(2)\cos{x}dx}+\int_{\pi/2}^{\pi}{(-2)\cos{x}dx}$$

Are you able to evaluate the integrals on the right hand side of this equation?

Chet

agata78
It looks complicated, wouldnt know where to start!

Mentor
It looks complicated, wouldnt know where to start!

Before you can start working with Fourier Series, you will first have to learn how to integrate simple functions.

Chet