Fouier series from the sign function

[Susanne217]

Homework Statement

If I have the standard sign function defined on R is the corresponding Fourier series the standard Fourier series with plus/minus in front respectively?

Homework Equations

The Attempt at a Solution

 

[jbunniii]

Can you write down exactly what you mean by "standard sign function"? To me, it means

$$f(x) = \left\{\begin{array}{ll}1, & \textrm{if }x > 0 \\<br /> -1, & \textrm{if }x < 0 \end{array}\right.$$

and $f(0)$ is typically defined as either 0 or 1, though its value doesn't make any difference with regard to Fourier series.

There is no Fourier series defined for this function unless you restrict the domain to a finite interval. And the series will depend on which interval you choose.

 

[Susanne217]

Can you write down exactly what you mean by "standard sign function"? To me, it means

$$f(x) = \left\{\begin{array}{ll}1, & \textrm{if }x > 0 \\<br /> -1, & \textrm{if }x < 0 \end{array}\right.$$

and $f(0)$ is typically defined as either 0 or 1, though its value doesn't make any difference with regard to Fourier series.

There is no Fourier series defined for this function unless you restrict the domain to a finite interval. And the series will depend on which interval you choose.

Is says find the Fourier series

$$f(x) = sign(x)$$ where $$x \in ]-\pi,\pi[$$

But since my interval is finite. How do I then go about finding the Fourier series for that f(x)??

 

[jbunniii]

Can you write down the formula for finding the Fourier coefficients of $f(x)$, and then try plugging in this particular $f(x)$, so we can see how far you are able to get on your own?

 

[Susanne217]

Can you write down the formula for finding the Fourier coefficients of $f(x)$, and then try plugging in this particular $f(x)$, so we can see how far you are able to get on your own?

Yes if f(x) is even defined on the interval [-n,n]

then the Fourier series is $$n_0 + \sum_{k=1}^{\infty} n_k cos(k\pi x / n)$$


and

$$n_0 = \frac{1}{n} \int_{0}^{n} f(x) dx$$

$$n_k = \frac{1}{n} \int_{0}^{n} f(x) \cdot cos(k \pi x /n) dx$$


if f(x) is odd

then $$f(x) = \sum_{k=1}^{\infty} n_k \cdot sin(k \pi x /n)$$

where $$b_k = \frac{2}{n} \int_{0}^{n} f(x) \cdot sin(k \pi x /n)$$

Is this what you mean?

 

[jbunniii]

Yes, let's work with that. Your function is clearly odd, so use the formula for $b_k$. (By the way, there's a typo in your formula for $f(x)$: surely $n_k$ should be $b_k$.)

Now plug in $f(x)$ for this particular function. What is $n$ in this case?

 

[Susanne217]

Yes, let's work with that. Your function is clearly odd, so use the formula for $b_k$. (By the way, there's a typo in your formula for $f(x)$: surely $n_k$ should be $b_k$.)

Now plug in $f(x)$ for this particular function. What is $n$ in this case?

am I stupid or is n = 1 ??

and I then end up with two version of $$b_k$$

where $$b_{k_1} = \frac{2}{n=1} \int_{0}^{1} sin(kx) dx$$

and $$b_{k_2} = \frac{2}{n=-1} \int_{0}^{-1} sin(-kx) dx$$ ?

 

[jbunniii]

No, $n$ is the (one-sided) interval over which the function is defined.

Your function is defined on $[-\pi, \pi]$, so $n = \pi$.

The formula for $b_k$ tells you to integrate only over $[0,\pi]$. Your function is equal to ONE over that entire interval, so you simply end up with

$$b_k = 2 \int_{0}^{\pi} sin(kx) dx$$

You don't have to worry about the negative half of the function, because this formula for $b_k$ assumes that the function is odd.

There is a more general formula for Fourier coefficients which does not assume that the function is even OR odd:

$$a_k = \int_{-\pi}^{\pi} f(x) \cos(kx) dx$$
$$b_k = \int_{-\pi}^{\pi} f(x) \sin(kx) dx$$

Note that in the special case where $f(x)$ is odd, $a_k = 0$ (can you see why?), and the formula for $b_k$ can be simplified as follows:

$$\begin{align*}b_k &= \int_{-\pi}^{\pi} f(x) \sin(kx) dx \\<br /> &= \int_{-\pi}^0 f(x) \sin(kx) dx + \int_{0}^{\pi} f(x) \sin(kx) dx \\<br /> &= 2 \int_{0}^{\pi} f(x) \sin(kx) dx \end{align*}$$

which matches the formula you are using. Thus if the function is odd, this simplified formula allows you to integrate only over the positive half, $[0,\pi]$.

 

[Susanne217]

No, $n$ is the (one-sided) interval over which the function is defined.

Your function is defined on $[-\pi, \pi]$, so $n = \pi$.

The formula for $b_k$ tells you to integrate only over $[0,\pi]$. Your function is equal to ONE over that entire interval, so you simply end up with

$$b_k = 2 \int_{0}^{\pi} sin(kx) dx$$

You don't have to worry about the negative half of the function, because this formula for $b_k$ assumes that the function is odd.

There is a more general formula for Fourier coefficients which does not assume that the function is even OR odd:

$$a_k = \int_{-\pi}^{\pi} f(x) \cos(kx) dx$$
$$b_k = \int_{-\pi}^{\pi} f(x) \sin(kx) dx$$

Note that in the special case where $f(x)$ is odd, $a_k = 0$ (can you see why?), and the formula for $b_k$ can be simplified as follows:

$$\begin{align*}b_k &= \int_{-\pi}^{\pi} f(x) \sin(kx) dx \\<br /> &= \int_{-\pi}^0 f(x) \sin(kx) dx + \int_{0}^{\pi} f(x) \sin(kx) dx \\<br /> &= 2 \int_{0}^{\pi} f(x) \sin(kx) dx \end{align*}$$

which matches the formula you are using. Thus if the function is odd, this simplified formula allows you to integrate only over the positive half, $[0,\pi]$.

So basically my Fourier series for the sign function over the finite is with that b_k above?

 

[LCKurtz]

Hey Susanne217, I have missed you in your other thread:

https://www.physicsforums.com/showthread.php?p=2371059#post2371059

Just when I thought we were getting close, you disappeared. Did you get that problem solved?

 

[Susanne217]

Hey Susanne217, I have missed you in your other thread:

https://www.physicsforums.com/showthread.php?p=2371059#post2371059

Just when I thought we were getting close, you disappeared. Did you get that problem solved?

Hi I had some family troubles so I didn't get arround to finishing it.

I did the intermediate calculations for the b_k but how was suppose to compare that result to find the some for the second series that I didn't?

Is it some obvious theorem that I overlooked maybe? Compare two series?

 

[ramsey2879]

Is says find the Fourier series

$$f(x) = sign(x)$$ where $$x \in ]-\pi,\pi[$$

But since my interval is finite. How do I then go about finding the Fourier series for that f(x)??

Did you mean "sign" or --sine--?

 

[Susanne217]

Did you mean "sign" or --sine--?

sign function!

 

[LCKurtz]

Hi I had some family troubles so I didn't get arround to finishing it.

I did the intermediate calculations for the b_k but how was suppose to compare that result to find the some for the second series that I didn't?

Is it some obvious theorem that I overlooked maybe? Compare two series?

Reply in that other thread and we can continue there.

 

[jbunniii]

So basically my Fourier series for the sign function over the finite is with that b_k above?

Yes, that's right. Just evaluate the integral and simplify and that gives you the $b_k$ values. Then you can plug those into the formula for the Fourier series.