0
$\begingroup$

Hi this is my first post :)This is the question

I am not sure how to do part b. You get the infinite series of $\displaystyle c_n\cdot \sin(\frac{n\cdot \pi\cdot x}{L})$ from $n=1$ to infinity

And this is equal to $\displaystyle \cos(\frac{\pi \cdot x}{L})$

So we need to find $C_n$. I am really not sure, and am unfamiliar with this 'fourier trick'

Thanks

I attached the picture

$\endgroup$
1
$\begingroup$

The fourier trick is the technique used generally to find the coefficients.

Here the idea is to utilise the fact that $sin\theta,sin2\theta,..... etc$ are all orthogonal to each other with respect to the inner product $\int_0^T fg$. So summarising when you multiply on both sides by $sin \hspace{1mm}n\theta$ and then integrate over one period all terms but $c_n$ will vanish and hence we can evaluate each term.

$c_n = \int_{0}^{T} f(x) \sin{n\theta} dx$ is the formula for evaluating each sine term

$\endgroup$
1
  • $\begingroup$ Thanks for the hints. I am not familiar with this idea. I take it that θ is (pi*x)/L So you get the infinite series of sin(nθ) equals to cos(θ)/C_n How did you get the integral form of C_n? $\endgroup$ – user136069 Mar 17 '14 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.