29. The case of aperiodic function.

Often we have to do with aperiodic function, sometimes defined only in [-π, π]. With purpose to apply our theory to such function let’s consider auxiliary function f*(x) defined as follows: (1), then we assume thatand others values are taken on in order of periodic rule. We can use now proved theorem for function f*(x). But if we talk about point lain between π and –π we have to do only with initial function f(x). The same reason result in that we may use formulas

(2) using directly function f(x). The ends of interval claim special attention. Ifthen if some test of convergence is hold then sum of series will be, different both from f(-π) and f(π). Such function may be expanded only in open interval (-π, π).

If function is defined somewhere else besides interval (-π, π) then out of this interval sum of Fourier series doesn’t coincide with function value.

Also we can take any interval instead of.

30. The case of arbitrary interval.

Suppose that function f(x) is defined in interval [-l, l] with arbitrary length 2l, l>0. If we use substitution , then we have got functionof y in interval . And we can apply considered things to this function, i.e. we can expand it in Fourier series. Coefficients are defined as follows:

Let’s return to old variable assuming. Then we got expanding of f(x) in changed trigonometric series:

(3). There are sines and cosines are taken of angles divisible not by x but . Formulas for coefficients also may be rewrote as

(4). In relation to interval ends ±l previous notes for ±π are kept. Of course, interval may be changed by any other with length 2l, for example, by interval [0,2l]. In this case formulas (4) must be rewrote as:

(4*)

31. Expanding only by sines or cosines.

If given function f(x) is odd then for equaltakes place, because. Analogically, for even function. Now let’s suppose that we have even function f(x) integrated in . Then product f(x)sin(x) will be odd function and.

So, Fourier series of even function contains only cosines:

(5)

Because in this case f(x)cosnx will be even function too then coefficients a_n may be calculated as follows:

(6).

If function f(x) will be odd then function f(x)cosnx will be odd too, so

. We may conclude that Fourier series of odd function contains only sines: (7). And(7)

Besides that, let’s note – any function f(x) defined in may be represented as sum of odd and even function:, where , . Then Fourier series of this function will contain expanding in cosines of f₁(x) and expanding in sines of f₂(x).

Next let’s suppose that function f(x) is defined only in [0,π]. Wishing to expand this function in Fourier series let’s complete it arbitrarily in [-π, 0) and then let’s apply considered notes. Arbitrary way of completeness allows us to get different trigonometric series. We may get expanding only in sines or only in cosines. If we assume that for (8) then we have got even function in interval with period 2. Coefficients of expanding may be calculated by formula (6). Analogically if we complete function using condition f(-x)=-f(x) (9) (for 0<x) then function will be odd. And their expanding will contain only sines with coefficients calculated by formula (7). Let’s note that points x=0 and x= claim special attention. If we expand by cosines then continuity will be hold in these points. But when we use expanding only by sines then continuity will be destroyed unless function value in these pint will be equal to zero.

If function is defined in interval [0,l] (l>0) then using change of variables (see before last theme) we can reduce problem to function expanding by cosines

or by sines . And coefficients are calculated accordingly by formulas

(9) or (10)