Zero. But how about this function:
I plotted the function 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 over the interwal I will get the following:
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 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 on my latest exam but in the answers that is put up on the insitutions homepage they only compute and they leave out when computing the exponetial Fourier series for . The value of is . Why would be left out, do you think?Your computation of the 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.