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

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".

• What do you mean by a Fourier $\text{sin/cos}$ series? Do you mean a series that contains only the sine or cosine terms? – aghostinthefigures Aug 19 '18 at 19:35
• @aghostinthefigures Sorry, I'm referring to what is known as "Fourier sine series" and "Fourier cosine series". – The Pointer Aug 19 '18 at 19:37
• Got it; thanks! – aghostinthefigures Aug 19 '18 at 19:39

## 3 Answers

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 $a_k = 0$ and thus can be fully approximated with a Fourier cosine series. This is described here and here for sine and cosine series respectively.

• Thanks for the clear answer. Can you please elaborate on this: "The Fourier sine and cosine series thus appear as individual "parts" of the "normal" Fourier series ..." I'm not totally clear on what you mean by this (How does Fourier sine/cosine series arise from the "normal" Fourier series?) – The Pointer Aug 20 '18 at 18:15
• Will update the answer shortly to clarify this. – aghostinthefigures Aug 20 '18 at 18:29
• Did you forget? :P – The Pointer Aug 23 '18 at 2:22
• The added content is there; I don’t see how I can say more in the answer without being repetitive. – aghostinthefigures Aug 23 '18 at 3:06
• Oh, sorry, I didn’t notice. – The Pointer Aug 23 '18 at 3:07

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.

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.