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I'm not sure whether to put this here or in Linear Algebra, if any Mod feels it should go in Linear Algebra I won't mind.

I've just been introduced to Fourier Series decompositions in my Linear Algebra text, and I understand all the core concepts so far from the Linear Algebra side of it (a function space, where orthonormal bases allow you to isolate and solve for individual coefficients in a given Fourier Series).

At the end of the decomposition ones sees (nb: [itex]\pi[/itex] is the length of any given base)

[tex]b_k = \frac{1}{\pi}\int_0^{2\pi}f(x)sin(kx)dx[/tex]

where [itex]f(x) =\left\{\begin{matrix}1, \ \ 0\leq x < \pi \\-1, \ \ \pi \leq x \leq 2\pi \end{matrix}\right.[/itex]

I tried a number of approaches involving integration by parts using values of integration between certain ranges of the square wave [itex]f(x)[/itex] taken by inspection, but these attempts have lead nowhere.

The text simply tells me that [itex]\int_0^{2\pi}f(x)sin(kx)dx = \frac{4}{k}[/itex] for odd k. It doesn't really explain how it came to that result, though it did earlier mention something about [tex]1 = \frac{4}{\pi}\left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - ... \right)[/tex] but I don't see exactly how that fits in...I can see something similar happening in there, but I'm not sure what.

Furthermore, for [itex]f(x)sin(x)[/itex], I can see that they are both odd functions, and that as such, over the same range, and given the definition of [itex]f(x)[/itex] that in fact [itex]f(x)sin(x) = |sin(x)|[/itex]. But that doesn't give the needed [itex]\frac{4}{k}[/itex] and doesn't help with [itex]f(x)sin(kx)[/itex].(

How have they arrived at [itex]\int_0^{2\pi}f(x)sin(kx)dx = \frac{4}{k}[/itex], and

I've just been introduced to Fourier Series decompositions in my Linear Algebra text, and I understand all the core concepts so far from the Linear Algebra side of it (a function space, where orthonormal bases allow you to isolate and solve for individual coefficients in a given Fourier Series).

At the end of the decomposition ones sees (nb: [itex]\pi[/itex] is the length of any given base)

[tex]b_k = \frac{1}{\pi}\int_0^{2\pi}f(x)sin(kx)dx[/tex]

where [itex]f(x) =\left\{\begin{matrix}1, \ \ 0\leq x < \pi \\-1, \ \ \pi \leq x \leq 2\pi \end{matrix}\right.[/itex]

I tried a number of approaches involving integration by parts using values of integration between certain ranges of the square wave [itex]f(x)[/itex] taken by inspection, but these attempts have lead nowhere.

The text simply tells me that [itex]\int_0^{2\pi}f(x)sin(kx)dx = \frac{4}{k}[/itex] for odd k. It doesn't really explain how it came to that result, though it did earlier mention something about [tex]1 = \frac{4}{\pi}\left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - ... \right)[/tex] but I don't see exactly how that fits in...I can see something similar happening in there, but I'm not sure what.

Furthermore, for [itex]f(x)sin(x)[/itex], I can see that they are both odd functions, and that as such, over the same range, and given the definition of [itex]f(x)[/itex] that in fact [itex]f(x)sin(x) = |sin(x)|[/itex]. But that doesn't give the needed [itex]\frac{4}{k}[/itex] and doesn't help with [itex]f(x)sin(kx)[/itex].(

**edit: my bad it does give 4/k...was this correct then? It must be...ok, but is there a general case?**)How have they arrived at [itex]\int_0^{2\pi}f(x)sin(kx)dx = \frac{4}{k}[/itex], and

**how do I approach evaluating the numerator in these equations for solving Fourier Series coefficients in general?**
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