I try to visualize this in terms of something analogous to vector spaces in linear algebra because thats the only way I can understand the fourier series.

You have a basis {$\cos(0x), \sin(0x), \cos(x), \sin(x), \cdots, \cos(nx), \sin(nx), \cdots $}

and you can represent any function as a linear combination of these (whats analogous to base vectors in linear algebra) if you can find sensible coefficients of the base vectors so you get a valid linear combination.

Now the coefficients exist if the integrals that derive $a_0$, $a_n$ and $b_n$ exist. But when I evaluate the integral to find these coefficients its doesn't matter what I choose as the bounds for the integrals as long they go from some $-L$ to $L$. You always get the same formula for $a_0, a_n$ and $b_n$ regardless of what $L$ you choose because most of the integrals become zero. Which means I should be free to choose whichever L I want.

This, in turn, should mean that I can take any function $f(x)$ and represent it as a linear combination of my base functions of sines and cosines and it will work regardless of what L I choose when I find the coefficients, or should it? I don't really know? Why does L have to correspond to the period of $f(x)$, why is even the period of $f(x)$ of any importance?

  • $\begingroup$ Could you show an example where you get the same coefficients for different $L$? You must be doing something wrong when you conclude that the integration bounds don't matter. $\endgroup$ – Henning Makholm Aug 25 '15 at 12:29
  • $\begingroup$ (And $\sin(0\cdot x)$ is the zero function and certainly doesn't belong in a basis for anything). $\endgroup$ – Henning Makholm Aug 25 '15 at 12:31
  • $\begingroup$ @HenningMakholm So the base vectors can only represent the function within the interval in which its periodic? $\endgroup$ – Arnold Doveman Aug 25 '15 at 12:34
  • $\begingroup$ @HenningMakholm I wasent refering to any particular examples, I was refering to derivation of $a_0, a_n, b_n$. When you put $f(x)$ as a linear combination of sines and cosines. And when you want to forexample derive $a_0$ you integrate both sides from $-L$ to $L$. $f(x)$ can be anything and $L$ can be anything. And yes $sin(0x)$ can simply be ignored $\endgroup$ – Arnold Doveman Aug 25 '15 at 12:42
  • $\begingroup$ x @Arnold: You seem to be claiming that you get the same coefficients no matter which $L$ you're using. I'm saying that this claim is wrong and you must have a mistake in whichever reasoning made you think this. I think if you try it for a concrete example of $f$ and two different $L$s you will see that they do not come out equal. $\endgroup$ – Henning Makholm Aug 25 '15 at 12:46

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