How Does a Fourier $\sin$/$\cos$ Series Arise From a "Normal" Fourier Series? How Does This Relate to the Generalised Fourier Series?

Question

I am told that Fourier showed that we can represent an arbitrary continuous function, $f(x)$, as a convergent series in the elementary trigonometric functions

$$f(x) = \sum_{k = 0}^\infty a_k \cos(kx) + b_k \sin(kx)$$

Also, suppose that $\{\phi_n(x)\}^\infty_{n = 0}$ is a set of orthogonal functions with respect to a weight function $w(x)$ on the interval $(a, b)$. And let $f(x)$ be an arbitrary function defined on $(a, b)$. Then the generalised Fourier series is

$$f(x) = \sum_{k = 0}^\infty c_k \phi_k (x)$$

I have the following questions relating to this:

1. How does a Fourier $\sin$/$\cos$ series arise from a "normal" Fourier series $f(x) = \sum_{k = 0}^\infty a_k \cos(kx) + b_k \sin(kx)$?

2. How does this relate to the generalised Fourier series $f(x) = \sum_{k = 0}^\infty c_k \phi_k (x)$?

I would greatly appreciate clarification on this.

EDIT: When I say Fourier $\sin$/$\cos$ Series, I'm referring to what is known as "Fourier sine series" and "Fourier cosine series".

Answer 1

The "normal" Fourier series is simply a specific case of the generalized Fourier series for which $$\phi_k = \{\sin(kx),\cos(kx)\},\ w(x) = 1$$ where $k$ is appropriately defined based on the domain of $f$ precisely to make each function in the family orthogonal to each other.

Consequently, the coefficients $\{a_k, b_k\}$ of the "normal" Fourier series are calculated in the same way as the generalized ones $c_k$, through an inner product with the "target" function $f$.

The Fourier sine and cosine series thus appear as individual "parts" of the "normal" Fourier series since

$$f(x) = \sum_{k = 0}^\infty a_k \cos(kx) + b_k \sin(kx) = \sum_{k = 0}^\infty a_k \cos(kx) + \sum_{k = 0}^\infty b_k \sin(kx)$$

and only one of them may be all that is required to approximate $f$ when either $a_k$ or $b_k$ are collectively $0$. This is equivalent to saying that $f(x)$ is orthogonal to either all $\sin(kx)$ or all $\cos(kx)$ on the function domain.

Such a situation can usually be qualitatively deduced before inner products are calculated; for example, a symmetric function $f(x) = f(-x)$ defined on a symmetric domain $(-a, a)$ will have $b_k = 0$ and thus can be fully approximated with a Fourier cosine series. This is described here and here for sine and cosine series respectively.

Answer 2

I don't know whether you are asking about compact fouier series or exponential fouier series,I will tell about compact fouier series $c_n=\sqrt {a_n^2+b_n^2}$ & $\theta_n=-\tan^{-1} {\frac {a_n}{b_n}}$ so compact fouier series becomes $x(t)=c_0+\sum_{i=1}^{\infty}c_n\ cos ({ n*\omega_o*t+\theta_n})$.

Hint:- $\cos(a+b)=\cos a*\cos b-\sin a*\sin b$

Note:- Fourier series is used for representing only periodic signals only ,for arbitrary signal we use fouier transform.

Answer 3

You can expand an integrable function $f$ on $[-\pi,\pi]$ in a Fourier series as you have written in your first equation. You can also start with a function on $[0,\pi]$ and extend it to an even function on $[-\pi,\pi]$, and the resulting Fourier series will have only $\cos$ terms. Or you can extend $f$ to be an odd function on $[-\pi,\pi]$, and the resulting Fourier series will have only $\sin$ terms. There are pointwise issues that arise at $0$ when you expand in this way, but this is not so difficult to analyze.