Fourier Series for f(x) = x(2∏-x)

[azserendipity]

Homework Statement

Please take into account I'm not very good at maths so I would just like to make sure that what I am doing so far is correct

Obtain the Fourier transform for:

f(x) = x(2∏-x) 0<x<2∏ f(x)=f(x+2∏)

Homework Equations

f(x) = 1/2 a0 + Ʃ {an cosnx + bn sinnx} = ao/2 + a1cosx +a2cos2x + b1sinx+b2sin2x +...

The Attempt at a Solution

an = 1/x ∫x(2∏-x)cos(nωt) dt

I've attached the rest of my working out below:

 

[vela]

Your formula for a_n is incorrect. You should have $1/\pi$ in front, not 1/x. What is ω equal to in this case? What about the b_n coefficients?

I'm not sure exactly what you're doing with the integration. Why did you break it up into two pieces?
The second piece is definitely wrong. There's no cosine in it, for one thing.

 

[azserendipity]

Ah that makes more sense!

The question itself is:

Obtain the Fourier series for:

f(x)=x(2∏-x) 0<x<2∏ f(x)=f(x+2∏)

I split the integration because there was an f(x) and an f(x) so thought you were supposed to. There was also no mention of what the values for ω is and I hadn't gotten to the bn bit yet.

In the examples I have from my lecturer he has them with cosines in so I figured that this one would be the same. Please note I'm not very good at Fourier series!

 

[vela]

Where did you get the formulas for a_n and b_n? The definition of $\omega$ should have been given there. Show us the formula for a_n exactly as given (before you plug in f). Let's get all these little details straight before you dive into the problem.

 

[azserendipity]

The formulas I got the formulas for an and bn from my lecture notes. The exact formulas are written as:

Periodic functions as Fourier series

f(x)= $\frac{1}{2}$$a{0}$+$Ʃ^{∞}_{n=1}${an cosnx + bn sinnx=$\frac{a0}{2}$ + a1 cosx + a2 cos2x+...+b1 sinx + b2 sin2x +...

a0=$\frac{2}{T}$$\int$$^{T/2}_{-T/2}$ f(x)dx ie. 2xmean value of f(x) over period

an= $\frac{2}{T}$$\int$$^{T/2}_{-T/2}$ f(x) * cosnx dx ie. 2x mean value of f(x)*cosnx over period

bn= $\frac{2}{T}$$\int$$^{T/2}_{-T/2}$ f(x) * sinnx dx ie. 2x mean value of f(x)*cosnx over period


I have also attached a printscreen of the lecturers example that I have been working off of

 

[vela]

The formulas I got the formulas for an and bn from my lecture notes. The exact formulas are written as:

Periodic functions as Fourier series

f(x)= $\frac{1}{2}$$a{0}$+$Ʃ^{∞}_{n=1}${an cosnx + bn sinnx=$\frac{a0}{2}$ + a1 cosx + a2 cos2x+...+b1 sinx + b2 sin2x +...

The above formula is applicable to functions with a period of $2\pi$.

a0=$\frac{2}{T}$$\int$$^{T/2}_{-T/2}$ f(x)dx ie. 2xmean value of f(x) over period

an= $\frac{2}{T}$$\int$$^{T/2}_{-T/2}$ f(x) * cosnx dx ie. 2x mean value of f(x)*cosnx over period

bn= $\frac{2}{T}$$\int$$^{T/2}_{-T/2}$ f(x) * sinnx dx ie. 2x mean value of f(x)*cosnx over period

These formulas aren't quite right. It's mixing up the case where the period is $2\pi$ and the more general case of a period of T.

When the period is $2\pi$, you have
\begin{align*}
f(x) &= \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx) \\
a_n &= \frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos nx\,dx \\
b_n &= \frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin nx\,dx
\end{align*}

When the period is T, you have
\begin{align*}
f(x) &= \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos n\omega_0x + b_n \sin n\omega_0x) \\
a_n &= \frac{2}{T}\int_{-T/2}^{T/2} f(x)\cos n\omega_0x\,dx \\
b_n &= \frac{2}{T}\int_{-T/2}^{T/2} f(x)\sin n\omega_0x\,dx
\end{align*}where $\omega_0 = 2\pi/T$. If you set $T=2\pi$ in these formulas, you'll recover the ones above.

For your particular f(x), the period is $2\pi$, so you can use the first set of formulas.

 

[azserendipity]

Thank you so much for giving me those equations!

I have attempted them using those equations and I've attached when I've done so far.

The red circle is something that I am unsure about and everything below it is also what I want to check if it is okay?

 

[vela]

Sorry, I should have been a little more careful. The equations above assume one period of the function is defined on the interval [-T/2, T/2], but one period of f(x) in this problem is defined on [0, 2pi]. You want to use these equations instead:
\begin{align*}
f(x) &= \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx) \\
a_n &= \frac{1}{\pi}\int_0^{2\pi} f(x)\cos nx\,dx \\
b_n &= \frac{1}{\pi}\int_0^{2\pi} f(x)\sin nx\,dx
\end{align*}What the limits of the integrals are don't really matter other than they have to cover exactly one period.

 

[azserendipity]

Hi,

the answer I got for Ao was 4pi^3 / 3. I've attached a copy of the worknigs just to do a final check that I'm doing it right!

Thank you so much for helping me as well :)

Also is Fourier series similar to the complex Fourier series or would I have to use a whole new set of equations?

 

[vela]

Close. It should be $4\pi^2/3$.