I was working on the heat equation: $$\mu_t= \alpha \mu_{xx}$$ $$\mu(0,t)=0$$ $$\mu_x(L,t)=0$$ $$\mu(x,0)=f(x)$$
I arrived at the following solution: $$\mu=\sum_{m=1}^{\infty }b_me^{\frac{-\alpha ^2(2m-1)^2\pi^2}{4L^2}t}\sin(\frac{(2m-1)\pi x}{2L})$$
So $\mu(x,0)=f(x)=\sum_{m=1}^{\infty }b_m\sin(\frac{(2m-1)\pi x}{2L})$ I am now stuck since I have no clue how to extend f(x) to a sine fourier series with new L = 2*L. So how do I calculate $b_m$ ? Thanks!
Since you are using a Fourier Sine series, you will need to use an odd $2L = 80$ periodic extension of $f(x)$. Define $$ f_o(x) = \begin{cases} f(x) & 0\leq x \leq 80 \\ -f(-x) & -80 \leq x <0 \end{cases}. $$ Notice that in general this defines an odd function about $x=0$ and because your function is already odd about $x=40$, this extension gives an odd extension about $x=\pm40$. In general the extension would have to be $$ f_0(x) = \begin{cases} f(x) & 0 \leq x <L \\ -f(-x) & -L \leq x < 0\\ -f(-x-2L) & L\leq x \leq 2L\\ f(x+2L) & -2L \leq x < -L. \end{cases} $$ Now you should multiply both sides of your equation for $\mu(x,0) = f(x)$ by $\sin(m\pi x/(2L))$ and integrate from $-2L$ to $2L$. You will have to split the integral involving $f(x)$ up into two cases, but it should turn out nonzero. The orthogonality of the sines on the other side of the equation will define your $b_m$, only for odd $m$ of course!