# Thread: Fourier Series-Help is needed

1. ## Fourier Series-Help is needed

Let f be a function with 2pi period that is piecewise-continous in R.
Let's define a function g such as g(x)=-f(-x) .

Find the connection between the fourier coeffeicients of f(x) and the fourier coefficieients of g(x)...

Thanks a lot!

2. Originally Posted by WannaBe
Let f be a function with 2pi period that is piecewise-continous in R.
Let's define a function g such as g(x)=-f(-x) .

Find the connection between the fourier coeffeicients of f(x) and the fourier coefficieients of g(x)...

Thanks a lot!
Use the odd/even properties of $\displaystyle \sin$ and $\displaystyle \cos$ and the occaisional change of variable $\displaystyle x'=-x$:

$\displaystyle \int_{-\pi}^{\pi} g(x) \sin(x)\;dx=\int_{-\pi}^{\pi} -f(-x) \sin(x)\;dx=$ $\displaystyle \int_{\pi}^{-\pi} f(x') \sin(-x')\;dx'=\int_{-\pi}^{\pi} f(x') \sin(x')\;dx'$

and something similar-ish for $\displaystyle \cos$ will allow you to deduce the relationship between the coeffients for $\displaystyle f$ and $\displaystyle g$.

CB

3. But why have you taken sin(x) and not sin(nx)?
And what is the relationship you've received? You've only received relations between the coefficient of f(-x) and g(x)...

As you can see, I'll be delighted to get some further help in here...
THanks a lot

4. Originally Posted by WannaBe
But why have you taken sin(x) and not sin(nx)?
And what is the relationship you've received? You've only received relations between the coefficient of f(-x) and g(x)...

As you can see, I'll be delighted to get some further help in here...
THanks a lot
Because from what I posted you can do that yourself. That is how does what I posted change if you replace $\displaystyle \sin(x)$ by $\displaystyle \sin(nx)$ ?

But I suppose you want me to tell you the answer?

CB

5. I don't want you to tell me the answer ofcourse... But from what you've done we can't deduce anything about the needed relations... You've only showed a relation between the coefficients of f(-x) and g(x) and I've managed to get there on my own... The problem is to continue from there...

Hope you will be able to help me

Thanks a lot anyway

6. Originally Posted by WannaBe
I don't want you to tell me the answer ofcourse... But from what you've done we can't deduce anything about the needed relations... You've only showed a relation between the coefficients of f(-x) and g(x) and I've managed to get there on my own... The problem is to continue from there...

Hope you will be able to help me

Thanks a lot anyway
Actually I think I have indicated how to show the relation between the coefficients of $\displaystyle f(x)$ and $\displaystyle g(x)$ not $\displaystyle f(-x)$ and $\displaystyle g(x).$

$\displaystyle \int_{-\pi}^{\pi} g(x) \sin(n x)\;dx=\int_{-\pi}^{\pi} -f(-x) \sin(n x)\;dx=$$\displaystyle \int_{\pi}^{-\pi} f(x') \sin(- n x')\;dx'=\int_{-\pi}^{\pi} f(x') \sin(n x')\;dx'=$ $\displaystyle \int_{-\pi}^{\pi} f(x) \sin(n x)\;dx$

since as $\displaystyle x'$ is a dummy variable of integration and can be replaced by anything.

CB

7. Thanks...That's a shame I didn't see it on my own....
Thanks a lot!