# Determine the Fourier series

## 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

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Chestermiller
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?

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?

dont know what you mean by argument?

rude man
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.

rude man
Homework Helper
Gold Member
dont know what you mean by argument?
f(x) the argument is x.

rude man
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.

Chestermiller
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.

rude man
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:
Chestermiller
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

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] ????

Chestermiller
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

∞ 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

vanhees71
Gold Member
2019 Award
Do you know, how to calculate the Fourier coefficients? If so, just calculate them ;-)).

dont know how to calculate b1

Chestermiller
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.

Chestermiller
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

It looks complicated, wouldnt know where to start!

Chestermiller
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