MHB Solving Exercise 2 & 3 in Multivariable Calculus

[nacho-man]

Hi, please refer to the attached image.

I am having trouble when doing
Exercise 2
Here is what I did:

$ \int_{-2}^{2}(f(x)\sin(\frac{m\pi x}{2}))dx = \sin(\frac{m\pi x}{2})a_{0} + \int_{-2}^{2} \sum\limits_{n=1}^\infty (a_{n}(\cos(\frac{n\pi x}{2})\sin(\frac{m\pi x}{2})+b_{n}\sin(\frac{n\pi x}{2})\sin(\frac{m\pi x}{2}))$

and we have to show that $b_n$ is equal to $0$, for all $n$. However
is it not true, that since the terms in the sum are mutually orthogonal, we will no longer have anything within the sum when it is integrated from -2 to 2?

In which case, we will not have a $b_{n}$? As I type this, I realize... that this is what makes $b_{n}$ = 0, but in case I am wrong, could someone please point me to the right direction?
Do I even need to evaluate the left hand side integral or the integral for $a_0$ for exercise 2?exercise 3
I am confused what to do after using the orthogonality again.
I am left with:

$ \int_{-2}^{2} f(x) \cos(\frac{m \pi x}{2})dx $ = $\int_{-2}^{2}a_0\cos(\frac{m \pi x}{2})dx$

how do i treat the f(x) ?
I am unsure how to get $a_n$ from this.

although, i know that since m is fixed, where m = n we will get a non-zero term. I feel like this has something to do with what I need.Thanks

 

[Dustinsfl]

Instead of me re-inventing the wheel with information already available here. You should take a look at the notes on Fourier series and the notes on Engineering Analysis.

The Engineering Analysis will show worked out problems with Fourier series and the Fourier series notes will show some problems and more theory.

http://mathhelpboards.com/math-notes-49/fourier-series-integral-transform-notes-2860.html

http://mathhelpboards.com/math-notes-49/engineering-analysis-notes-2882.html

 

[nacho-man]

Thanks for that, although

Could I just get verification if I am correct with my reasoning exercise 2?

as for exercise 3 I will check the notes. I have read through Differential Equations - Boyce et al which also had some very similar problems to my question, but am still a little unsure as to how to solve my question given the particular conditions.

 

[Dustinsfl]

The way you have it setup bn is not 0 since the length is from (-2, 2) the integral of \(\sin^2\) is \(L/2\) where \(L = 4\).

However, \(a_n = 0\) since sine and cosine are orthogonal.
\[
\int_{-\pi}^{\pi}\sin(nx)\cos(nx)dx = 0
\]
and
\[
\int_{-\pi}^{\pi}\sin(nx)\sin(nx)dx = \frac{2\pi}{2} = \pi
\]
As well as showing what you have done, you should also show the problem.

 

[nacho-man]

The question in regards to showing $b_n$ = 0 is exercise 2 of the attached image.

if you could refer to image attached to this post, could you explain why $b_n$ is not zero in my question?

It does indeed say that L = P/2 = 2, so L = 2. I think you were meaning to say that P=4 not L=4.

since this is the case,
that would make $b_n = 0$ because when integrating that term, it disappears?

The second thing I'm unsure about is how to integrate the term on the left hand side, with the $f(x)\sin(\frac{m\pi x}{2}$

Thanks once again :)

edit: OKay, I think i have the answer. I overlooked that f(x) was a hybrid function.

since the bounds are between -2 and 2, 1<|x|<2 so f(x) = 0
thus $b_n = 0 $ ?
Although using the same logic, I don't tihnk I could get an answer for exercise 3?

Could someone tell me where i am going wrong?

 

[topsquark]

You have to prove this (obviously) but the [math]b_n[/math] are, in fact, all 0. The signal function is even, so no odd parity terms will be in the Fourier expansion. Just my two cents about that comment.

-Dan

 

[nacho-man]

Thanks everyone, I had loads of fun =)

 

[Dustinsfl]

For 3, we have that
\[
\int_{-2}^2a_n\cos\left(\frac{n\pi x}{2}\right)\cos\left(\frac{m\pi x}{2}\right)dx
\]
This integral is 0 if \(m\neq n\). So the summation is only nonzero when \(n = m\) so the integral we are solving is then
\[
\int_{-2}^2a_n\cos^2\left(\frac{n\pi x}{2}\right)dx
\]
Using trig, we know that \(\cos^2(x) = \frac{1}{2} + \frac{\cos(2x)}{2}\).
\[
\int_{-2}^2\frac{a_n}{2}dx = 2a_n
\]
That is, \(2a_n = \int_{-2}^2f(x)\cos\left(\frac{n\pi x}{2}\right)dx\).
\[
a_n = \int_{-1}^1\cos\left(\frac{n\pi x}{2}\right)dx = \left.\frac{2}{n\pi}\sin \left(\frac{n\pi x}{2}\right)\right|_{-1}^1 = \frac{2}{n\pi} \left(
\sin \left(\frac{n\pi}{2}\right) - \sin \left(-\frac{n\pi}{2}\right) \right)
\]
Sine is odd so \(\sin(-x) = -\sin(x)\). Using this fact, we have
\[
a_n = \frac{4}{n\pi}\sin \left(\frac{n\pi}{2}\right).
\]