Well, try just doing the integral
$\displaystyle \displaystyle{c_{0}=\frac{1}{2\pi}\int_{-\pi}^{\pi}x\,dx}$
the usual Calculus II way. Do you see something in computing it that might tell you what that formula should be?
Zero. But how about this function:
$\displaystyle \displaystyle f(x)=e^{-x}},\,\,0 < x < {2\pi}$
I plotted the function $\displaystyle \displaystyle f(x)=e^{-x}}$ on my calculator and see that the function is not symmetric. The function is in the fourth- and first quadrant. Can I conclude that since it is not symmetric it is an odd function? And if I compute $\displaystyle c_{0}$ over the interwal $\displaystyle (0,2\pi)$ I will get the following:
$\displaystyle \displaystyle{c_{0}=\frac{1}{2\pi}\int_{0}^{2\pi}e ^{-x}}dx}=-\frac{1}{2\pi}(e^{-2\pi}}-e^{0}})=\frac{1}{2\pi}(1-e^{-2\pi})$
Correct.Zero.
Your exponential function has neither even nor odd symmetry. Most functions have neither. You cannot conclude that because a function is not even, that therefore it is odd, or vice versa.
I'm not exactly sure what you mean by this. I was under the impression that since your interval of interest has all positive x-values, you're either in the first or fourth quadrant, and since your exponential function is positive everywhere, therefore you are in the first quadrant.The function is in the fourth- and first quadrant.
Your computation of the $\displaystyle c_{0}$ component is correct for that function.
I meant that when I plot the function on my calculator the function is drawn over the first and fourth quadrant. But you explained that clearly that the x-values are either in the first or fourth quadrant.
I did compute $\displaystyle c_{0}$ on my latest exam but in the answers that is put up on the insitutions homepage they only compute $\displaystyle c_{k}$ and they leave out $\displaystyle c_{0}$ when computing the exponetial Fourier series for $\displaystyle \displaystyle f(x)=e^{-x}},\,\,0 < x < {2\pi}$. The value of $\displaystyle c_{0}$ is $\displaystyle {\neq 0} $. Why would $\displaystyle c_{0}$ be left out, do you think?Your computation of the $\displaystyle c_{0}$ component is correct for that function.
You might leave it out if you're only interested in the spectral information; that would be an application-specific situation, I think. For example, in filter design for RLC circuits, a DC component might be disregarded if you're only interested in filtering out a particular frequency. However, it is true that you can't re-construct the original function exactly without the constant term.