# Fourier series - derivation

• lukaszh
In summary, the conversation discusses the derivation of the formulas for finding the coefficients a_n and b_n in a Fourier series. It is explained that these formulas can be derived by using an inner product space with an orthonormal basis of cos(nx) and sin(nx) functions. The conversation also touches on the orthogonality of these basis functions and how they lead to the given formulas. It is also mentioned that it is easier to work with exponential basis functions and a slightly different inner product.

#### lukaszh

Hello,
everywhere I can see this
$$a_n = \frac{1}{\pi}\int_{-\pi}^\pi f(t) \cos(nt)\, dt$$
$$b_n = \frac{1}{\pi}\int_{-\pi}^\pi f(t) \sin(nt)\, dt$$
etc... I can't find, how to derive this formulas. I'm really tired and a bit confused of this formulas, because I can't find possible way to derive them. I don't like only formula application, but I want to know, what is that formula about.
Thank you...

If you have an "inner product space", that is, an vector space with an inner product defined on it, together with an orthonormal basis, $v_1, v_2, ...$, that is such that $<v_i, v_j>= 0$ if $i\ne j$ and $<v_i, v_i>= 1$ for all i, and want to write v as a linear combination, $v= a_1v_1+ a_2v_2+ ...+ a_nv_n$, then $$\displaystyle a_i= <v, v_i[/itex]. What you have is a vector space with basis $cos(nx)$, $sin(nx)$ with inner product $<f, g>= \frac{1}{\pi}\int_{-\pi}^\pi f(t)g(t)dt$ which leads to the given formulas.$$

It works because the functions sin(nt) for different values of n are orthogonal to each other, that is,

$$\int^{\pi}_{-\pi} {\sin(nt) \sin (mt) dt} = 0$$

for $n \ne m$, and

$$\int^{\pi}_{-\pi} {\sin^2(nt) dt} = \pi$$

Likewise for cosines. Try a few examples if you like. Therefore if you have a function

$$f(t) = b_1 \sin (t) + b_2 \sin (2t) + b_3 \sin (3t) + ...$$

then, for example, letting n = 2:

$$\int^{\pi}_{-\pi} {f(t) \sin (2t) dt} = b_1 \int^{\pi}_{-\pi} {\sin (t) \sin (2t) dt} + b_2 \int^{\pi}_{-\pi} {\sin^2 (2t) dt} + b_3 \int^{\pi}_{-\pi} {\sin(3t) \sin (2t) dt} + ...$$

$$\int^{\pi}_{-\pi} {f(t) \sin (2t) dt} = b_1 \cdot 0 + b_2 \cdot \pi + b_3 \cdot 0 + ...$$

Thank you. Now I understand. Thanks Thanks

It is easier to work with the basis functions

e_n(x) = exp(i n x)

and define the inner product as

<f,g> = 1/(2 pi) Integral from minus pi to pi of f(x)g*(x) dx

## What is a Fourier series?

A Fourier series is a mathematical representation of a periodic function as an infinite sum of sinusoidal functions. It can be used to decompose any periodic function into its constituent frequencies.

## What is the derivation of Fourier series?

The derivation of Fourier series involves using the Fourier transform to express a periodic function as a sum of complex exponentials. This can then be simplified using trigonometric identities to obtain the final Fourier series formula.

## What is the significance of Fourier series?

Fourier series have many applications in engineering, physics, and mathematics. They are particularly useful in signal processing, where they can be used to analyze and manipulate different types of signals.

## What are the assumptions made in the derivation of Fourier series?

The derivation of Fourier series assumes that the periodic function is continuous and has a finite number of discontinuities. It also assumes that the function is piecewise smooth and has a period of 2π.

## What are some common applications of Fourier series?

Some common applications of Fourier series include signal and image processing, data compression, and solving differential equations. They are also used in fields such as acoustics, optics, and electronics.