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, &amp; \textrm{if }x &gt; 0 \\<br /> -1, &amp; \textrm{if }x &lt; 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, &amp; \textrm{if }x &gt; 0 \\<br /> -1, &amp; \textrm{if }x &lt; 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 &amp;= \int_{-\pi}^{\pi} f(x) \sin(kx) dx \\<br /> &amp;= \int_{-\pi}^0 f(x) \sin(kx) dx + \int_{0}^{\pi} f(x) \sin(kx) dx \\<br /> &amp;= 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 &amp;= \int_{-\pi}^{\pi} f(x) \sin(kx) dx \\<br /> &amp;= \int_{-\pi}^0 f(x) \sin(kx) dx + \int_{0}^{\pi} f(x) \sin(kx) dx \\<br /> &amp;= 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.