What is the advantage of using Fourier Series representation rather than the function itself?

Question

Suppose we have a function $f$ defined over $[a,b]$ to the real numbers, i.e. $f: [a, b] \to \mathbb R.$.

We can approximate this function as Fourier Series.

Suppose $a_n, b_n$ is the Fourier series coefficients,

Then

$$a_n = \frac{2}{b-a} \int_a^bf(x) \cos(nx)$$

$$b_n = \frac{2}{b-a}\int_a^bf(x)\sin(nx) $$

$$f(x) \approx \frac{a_0}{2}+ \sum^\infty_{n=1} a_n\cos(nx) + b_n\sin(nx)$$

My question is, what is the advantage of using the approximate Fourier Series,

rather than the function $f(x)$ itself? When would we use the approximation over the funtion itself?

Answer 1

I think the point is the functions available might be highly irregular, and it is difficult to treat such functions. If we can decompose the function into a series which "converges" globally, then we can substitute the study of the function with its Fourier series.

For many analysis purposes like PDE, this helps to translate properties of the original function into properties of the series in a way we can reasonably control. If the original function is unknown (for example only known the boundary values and some decay conditions), then this information translation process can be extremely helpful because it makes a hard problem much easier. You can also decompose the original function into linear combination of other orthogonal sets, but the results may not be as nice.

Answer 2

Often a problem can be made easier by reducing it to individual frequencies. I've provided an example below that occurs when working with linear differential equations.

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Consider the oscillator equation with a time dependent periodic forcing function,

$$ \frac{d^2 x}{dt^2} + k x = F(t)$$

This can be a very difficult problem for an arbitrary forcing function $F$, but if we decompose $F$ into its fourier series then we really just need to solve three simple problems.

$$ \frac{d^2 x}{dt^2} + k x = a_n\cos(n\omega t)$$

$$ \frac{d^2 x}{dt^2} + k x = b_n \sin(n\omega t)$$

$$ \frac{d^2 x}{dt^2} + k x = a_0$$

All three of the above equations are easily solvable. Lets call the various solutions to these equations $x_\lambda(t)$. We can construct the solution to the original problem by just taking linear combinations of these solutions,

$$ x(t) = \sum_\lambda x_\lambda(t) $$