OK I am going over d'Alembert solutions. And I came across the following example.
$$ f(x) = \begin{cases} \frac{3}{10}x &0 \le x \le \frac{1}{3} \\ \frac{3(x-1)}{20} & \frac{1}{3} \le x \le 1 \\ \end{cases}$$
we're supposed to determine the subsequent motion of the string, which is attached at x=0 and x=1 and length L= 1. with $c=\frac{1}{\pi}$
the thing i am struggling with is getting the odd extension right. I know it is pretty simple here -- you have a line that has a 3/10 slope up until x=1/3 and then it turns down with a -6/10 slope after that until you get to 5/3 and then it starts going up again, with a 3/10 slope, et cetera.
but when i try to derive the odd extension by going through the process of finding the Fourier sine series, i dont get a line with 3/10 slope repeating as the text does, I get something with a sine function in it.
Here's what I did:
we have $f(x) = \frac{3}{10}x$ for the interval $ 0 \le x \le \frac{1}{3}$ so I started with making a Fourier sine series with that. $b_n = \int^L_0 f(x) \sin(\frac{(n\pi x)}{L})dx $ and that gives us $b_n = \int^L_0 \frac{3}{10}x \sin(3n \pi x)dx $ since L = 1/3 in this case.
GOIng through the integration by parts (I know I could look up $\int x\sin x$ but I want to make sure here) I got: $u= x$ and $du = dx$ and $dv = \sin (3\pi n x) dx$ and $v = \frac{\cos (3 \pi nx )}{3 \pi n}$
which gets me $x \frac{\cos (3 \pi nx )}{3 \pi n} + \int \frac{\cos (3 \pi nx )}{3 \pi n}dx = x \frac{\cos (3 \pi nx )}{3 \pi n} + \frac{\sin (3 \pi nx )}{(3 \pi n)^2}| ^{\frac{1}{3}}_0$
ANd doing the algebra I get : $$\frac{-1}{3}\frac{\cos (\pi n )}{3 \pi n}+\frac{\sin ( \pi n )}{(3 \pi n)^2} = \frac{-1}{3}\frac{(-1)^n}{3 \pi n}= \frac{(-1)^{n+1}}{9 \pi n} = b_n$$
so my odd extension series is $$\frac{2}{1/3}\sum^\infty_{n=1} \frac{(-1)^{n+1}}{9 \pi n} \sin (3n \pi x)= \frac{2}{3}\sum^\infty_{n=1} \frac{(-1)^{n+1}}{\pi n} \sin (3n \pi x)$$
which looks nothing like what I thought I was supposed to get. Unless I did in fact do it right. We are looking at the graph at time$t=\pi / 3$, and the D'Alembert formula is supposed to be $\frac{1}{2}[f^*(x-ct)+f^*(x+ct)]$ which would be $\frac{1}{2}[f^*(x-1/3)+f^*(x+1/3)]$
I am just not sure how they are getting that answer, through the last step. Any help is appreciated. I just can't figure out how they get the graph shape they show in the diagram when my fourier extension -- assuming i did it right at all -- doesn't seem to yield anything like that.