I dont understand it fully. My graph is in the interval $-2π<t<2π$ And Im asked to find $s(3π)$ which I know is the same as $s(π)$ because the period of a Fourier-sine function is 2π. My graph is the same as the graph in the solutions manual
But what exactly am I looking at here? Since there is discontinuity at s(3π). Dirichlets conditions tell us $s(3π)$ = $\frac {f(3π^+)+(f(3π^-)}{2}$
Where exactly is $f(3π^+) and (f(3π^-)$? Is it the points I have drawn and marked f(3π^+) and f(3π^-) ? The point were the graph is closest to t=3π from both the left hand and right hand side?
The answer says $s(3π)=\frac {f(3π^+)+(f(3π^-)}{2}$ = $\frac {e^0+(-e^0)}{2} = \frac {1+(-1)}{2} = 0$
But I see it as $s(3π)=s(π)=\frac {e^π + (-e^π)}{2}=0$ So I think I am looking at my points wrong.