# Deriving of Fourier Theory, Fourier Series

In this chapter it will bel shown, how it is possible to derive the periodic functions as a function series. First, let′s take into account the analysis of continuous functions. Next, consider the following defined and continuously differentiable function set in [a;b] interval. Furthermore, let′s consider the defined and infinitely continuously differentiable f(t) function in [a;b] interval. The question is the following. Is it possible to express f(t) as infinite function series: where It is well known, that a defined and infinitely continuously differentiable f(t) function in [a;b] interval can be expressed as power series as follow: where in current case , and is the Taylor coefficient Consequently Next let′s consider, when is a periodic function i.e. in [a;b] interval, where T ͼ R is the period of f(t) and k=1,2, .....n,.. natural numbers.
Let′s attempt to derive the periodic function in [a;b] interval as the superposition of trigonometric functions as follow: where natural numbers and let′s suppose as the fundamental period of trigonometric function.

Then apply the trigonometric addition sum theorem for the trigonometric function series: Let's use the following denote: , Thus , which means the periodic function may be approximated by a finite trigonometric series.

Consequently, the following problems have to be faced:

• Is it possible to find any series for f(t), i.e. if f(t) is a periodic function is it possible to express it as a superposition of trigonometric functions?
• Can f(t) function be expressed as a finite or infinite trigonometric series?
• If f(t) can be expressed as finite or infinite series, then how is possible to determine the , and coefficients?

Since f(t) periodic function is attempted to be expressed as the superposition of trigonometric functions, see the following denotes with the corresponding series elements:
Let functions
and trigonometric function be elements of the series.

The mathematical question: is it possible to describe f(t) periodic function by infinite trigonometric series as follow: Since the series is based on trigonometric function elements, let′s suppose the period of f(t) is 2π, i.e. f(t)=f(t+2kπ) (period of the elements of the trigonometric series are 2π too). Furthermore, let′s suppose the trigonometric series representing the f(t) periodic function is consistently convergent in [-∞;∞].
Then multiply both side of the following expression by cos(nt) and sin(nt) as shown below:
f The following two expressions are obtained after the multiplications:  Since the trigonometric series is consistently convergent and |cos(nt)|≤ 1 as well as |sin(nt)|≤ 1, thus the multiplication by cos(nt) and sin(nt) will keep the consistently convergence. Due to the consistently convergence the elements of trigonometric series can be integrated by members in [a; a+2π] interval:  ,
where , as well as due to the result of one period definite integral of the trigonometric function.
Consequently, the two integral can be described as follow:  The next step is to analyze the integrals behind the ∑ expressions. Trigonometric addition sum theorem is applied in the analysis: Within the definite integral operation you can find sin(mt) and sin(qt) type of expressions, where m=n-k and q=n+k and they have 2π period too. Consequently, these definite integrals are zero (see above).

Following a similar procedure result in: The next step is to determine the following definite integrals: and Two cases are considered. First, when n≠k:  Next case is, when n=k: and Thus the results of the definite integrals:  Consequently, if f(t) is periodic f(t)=f(t+2nπ) and consistently convergent , then it can be expressed as a trigonometric series as follow: , where the coefficients are: , and The constant can be derived as follow: Consequently: and 