Fourier Series: How to interpret the function?

[NicolaiTheDane]

This is a rather simple question, but am I understanding the following correctly?

1. Homework Statement

The Attempt at a Solution

This isn't really the problem, but I have a feeling my problem the assignments, is me misunderstanding the function description. I don't see how this 2 pi periodic, but I also do not understand how else to interpret it.

 

[BvU]

So it's $4\pi$ periodic, a simple typo. What comes after your snippet ?

 

[NicolaiTheDane]

So it's $4\pi$ periodic, a simple typo. What comes after your snippet ?

I should have mentioned that it is dealing with Fourier series, so it HAS to be $2*\pi$ periodic. Is it correct to assume, that I then have to make it that way. That is fill in the span between $(\pi,3*\pi)$ in such a way as to make the fuction into a $2*\pi$ function?

 

[BvU]

What comes after your snippet ? O, I asked that already...

it HAS to be $2*\pi$ periodic

So, compress with $x'= x/2$ and proceed with $x'$

 

[NicolaiTheDane]

What comes after your snippet ? O, I asked that already...So, compress with $x'= x/2$ and proceed with $x'$

Doing that wouldn't fit with the last part of the interval. As for the other assignments:

This is easily shown using a Theorem in my book, and the points seem to be $n*\pi, n = 1,2,3..$ Now ofc this is only right if my assumption about function is correct (not the one I have written above), which is:

However this is what I aren't sure about, but its the only option that I can think of, which also satisfies the next assignment

Now I know this part has to be $2*\pi$. Assuming that it means the sum at the mentioned discontinousities, this would be mean:

$$\frac{1}{2}*a_0 + \sum_{n=0}^\infty \left(a_n*\cos(x_0*n)+b_n*\sin(x_0*n) \right) = \frac{1}{2}*\left(\lim_{n \rightarrow +x} {x_0} + \lim_{n \rightarrow -x} {x_0}\right)$$

Which means the sum for each point is just the limit going from both sides added and divided by 2. If my assumption is correct, then both at $\pi$ and at $0$ the sum becomes $2*\pi$ as it should. However like I said at the very beginning, my problem is whether I understand how to interpret the function correctly or not.

 

[LCKurtz]

Assuming we are talking about the function in post #5, you are thinking correctly but expressing it poorly. You have the given function $f(x)$ and its FS. Let's call the sum of the FS $s(x)$. What you have is that $f(x) = s(x)$ at all points where $f$ is continuous. At any points where $f$ has a step discontinuity ($n\pi$ in this case) the theorem you are using might be better stated that$$
s(n\pi) =\frac{ \lim_{x\to n\pi^+} f(x) + \lim_{x\to n\pi^-} f(x)}{2} = 2\pi $$The right side of your last equation doesn't make much sense.

 

[Ray Vickson]

I should have mentioned that it is dealing with Fourier series, so it HAS to be $2*\pi$ periodic. Is it correct to assume, that I then have to make it that way. That is fill in the span between $(\pi,3*\pi)$ in such a way as to make the fuction into a $2*\pi$ function?

No, it does NOT have to be $2\pi$ periodic; it just needs to be periodic. If it is periodic with period $T$ (say on a base interval $[a,a+T]$) then there are standard formulas for the Fourier series and its coefficients. Every textbook has this material, and a Google search of "Fourier series" will produce many web pages giving the details.

 

[LCKurtz]

Well, one thing I learned from this thread was the discont = 'true' option in Maple when plotting piecewise continuous functions to get nice plots including correct end points:
> restart: with(plots):
> f := piecewise(0 <= x and x <= 1, 1, x > 1 and x <= 3, x-1);
> plot(f, x = -1 .. 4, discont = 'true', scaling = constrained);

Complete with the dots in the right places. Thanks to @NicolaiTheDane

 

[NicolaiTheDane]

Well, one thing I learned from this thread was the discont = 'true' option in Maple when plotting piecewise continuous functions to get nice plots including correct end points:
> restart; with(plots);
> f := piecewise(0 <= x and x <= 1, 1, `and`(x > 1, x <= 3), x-1);
> plot(f, x = -1 .. 4, discont = 'true', scaling = constrained);
View attachment 233610
Complete with the dots in the right places. Thanks to @NicolaiTheDane

Happy i could give something back, even if its trivial xD

No, it does NOT have to be $2\pi$ periodic; it just needs to be periodic. If it is periodic with period $T$ (say on a base interval $[a,a+T]$) then there are standard formulas for the Fourier series and its coefficients. Every textbook has this material, and a Google search of "Fourier series" will produce many web pages giving the details.

I don't doubt your right, you shown me on other occations you know your stuff. However it hasn't been covered in the lectures, and assignments never demand that we know something, that hasn't been covered. That said do you think its wrong to assume as I did in post #5?

 

[Ray Vickson]

Happy i could give something back, even if its trivial xD
I don't doubt your right, you shown me on other occations yoIf

Happy i could give something back, even if its trivial xD
I don't doubt your right, you shown me on other occations you know your stuff. However it hasn't been covered in the lectures, and assignments never demand that we know something, that hasn't been covered. That said do you think its wrong to assume as I did in post #5?

I am responding to your original problem in post #1, where you have a period of $4 \pi.$ BVU in post #4 showed you how to do it, but in case you did not understand his answer, let me expand on it a little bit. You have some function $f(t)$ defined on $[0,4 \pi]$, whose Fourier series you want. However, you only know how to get Fourier series for functions on $[0, 2 \pi].$ You can set $t = 2 \tau$ and define $g(\tau) = f(2\tau)$, $\tau \in [0,2 \pi].$ Now just get the Fourier series for $g(\tau).$ Then put back $\tau = t/2$ in the final $\tau$-series formula.

Basically, that is how everybody obtains Fourier series on intervals like $t \in [a , a+T]$: just let
$$ t = a + \frac{b}{2 \pi} \tau$$ and then use the standard Fourier formulas for $\tau \in [0, 2 \pi].$ Once you have done that, put back the definition of $\tau$ in terms of $t$ to get the formula for the Fourier series in terms of the original variable $t.$

 

[BvU]

Doing that wouldn't fit with the last part of the interval

How so ?

And: What comes after your snippet ? O, I asked that already... twice
Please show the full problem statement, or better: type it

 

[NicolaiTheDane]

How so ?

And: What comes after your snippet ? O, I asked that already... twice
Please show the full problem statement, or better: type it

It doesn't fit, because the assignment clearly states, that the function is $2\cdot\pi$ periodic. If you simply rewrite the equation to be so, on the first $\pi$ interval by half stepping, the function does not fit on $(3\cdot\pi,4\cdot\pi)$ as the function value will be still have to be $f(3\cdot\pi) = 3\cdot\pi$ for example, as it is defined to be $f(x)=x$ on that interval. The only way to make it fit the $2\cdot\pi$ periodic scheme, AND fit with the given intervals for $x$, is to treat it as 2 halves of a $2\cdot\pi$ periodic, thus leading to what I showed in post #5

However if you don't think that's right (and I think its silly my course leader thinks its funny to make the problem statements intentionally vague), doing half steps won't fit the solutions provided for assignment (ii) and (iii), which are:

(ii) $n\cdot\pi, n\in\mathbb{N}$
(iii) $2\cdot\pi$

As per your asking for what comes after BvU. I do not want to be rude, but I have already shown you all there is. I even say "as for the other assignments" in post #5. I left out assignment (i), because its merely asking me to sketch the graph, but that is it. For convenience I'll show it all in one snip.

 

[BvU]

I do not want to be rude

No offence taken; thanks. You even provide the solution -- which is great: it can sometimes unveil what was intended.

From that solution I favour the picture in #5 over that in #1: it satisfies the answers (both of them)!

How the intended function follows from $f(x) = x \quad\forall \ x\in [0,\pi[ \; \cup \; [3\pi, 4\pi [ \ \$ is a mystery to me. Perhaps something to do with the left "[" instead of the right "]", but I doubt it.

That function description, when interpreted as in #1 falls short anyway: it doesn't tell you what the function value is for $x\in [\pi,\;3\pi]$ and you have to assume something (like 0 as you did).

Assuming 0, the exercise can still be done, but the answers come out different ( I think ):
(ii) $n\pi\cup (2n+1)\pi$
(iii) $2\pi$ and $\pi/2$

 

[NicolaiTheDane]

No offence taken; thanks. You even provide the solution -- which is great: it can sometimes unveil what was intended.

From that solution I favour the picture in #5 over that in #1: it satisfies the answers (both of them)!

How the intended function follows from $f(x) = x \quad\forall \ x\in [0,\pi[ \; \cup \; [3\pi, 4\pi [ \ \$ is a mystery to me. Perhaps something to do with the left "[" instead of the right "]", but I doubt it.

That function description, when interpreted as in #1 falls short anyway: it doesn't tell you what the function value is for $x\in [\pi,\;3\pi]$ and you have to assume something (like 0 as you did).

Assuming 0, the exercise can still be done, but the answers come out different ( I think ):
(ii) $n\pi\cup (2n+1)\pi$
(iii) $2\pi$ and $\pi/2$

Exactly what I got as well. Honestly I'll be having a chat with my course leader on monday, as this isn't the first time he is intentionally vague, or outright deceptive in his assignment descriptions. Thanks for all the inputs though, I'll make an additional reply, if it turns out he meant something completely different.