# Infinite trigonometric series, find the constant C_n

Hi this is my first post :) 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

## 1 Answer

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

• 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? – user136069 Mar 17 '14 at 19:23