# Thread: Exponential Fourier Series - Trouble solving

1. 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?

2. Originally Posted by Ackbeet
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?
Aha, Yes, the value of the integral is in fact zero. So, I can only double the integral if the given function is an even function when computing $\displaystyle {c_{0}$

3. All true. But you know even more than that. What should you get when you integrate an odd function over a symmetric interval?

4. Originally Posted by Ackbeet
All true. But you know even more than that. What should you get when you integrate an odd function over a symmetric interval?

$\displaystyle \displaystyle f(x)=e^{-x}},\,\,0 < x < {2\pi}$

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:

$\displaystyle \displaystyle{c_{0}=\frac{1}{2\pi}\int_{0}^{2\pi}e ^{-x}}dx}=-\frac{1}{2\pi}(e^{-2\pi}}-e^{0}})=\frac{1}{2\pi}(1-e^{-2\pi})$

5. Zero.
Correct.

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.

The function is in the fourth- and first quadrant.
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.

Your computation of the $\displaystyle c_{0}$ component is correct for that function.

6. Originally Posted by Ackbeet

I'm not exactly sure what you mean by this.
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.

Your computation of the $\displaystyle c_{0}$ component is correct for that function.
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?

7. 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.

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