Page:EB1911 - Volume 10.djvu/775

uniform, but non-uniformity of convergence of the series does not necessarily imply discontinuity in the sum.

Form of Fourier’s Series.—If it be assumed that a function ƒ(x) arbitrarily given for values of x such that 0 ≦ x ≦ l is capable of being represented in general by an infinite series of the form

${\text{A}}_{1}\sin {\frac {\pi x}{l}}+{\text{A}}_{2}\sin {\frac {2\pi x}{l}}+\dots +{\text{A}}_{n}\sin {\frac {n\pi x}{l}}+\dots ,$

and if it be further assumed that the series is in general uniformly convergent throughout the interval 0 to l, the form of the coefficients A can be determined. Multiply each term of the series by $\textstyle \sin {\frac {n\pi x}{l}}$, and integrate the product between the limits 0 and l, then in virtue of the property $\textstyle \int _{0}^{l}\sin {\frac {n\pi x}{l}}\sin {\frac {n^{\prime }\pi x}{l}}dx=0$, or $\textstyle {\frac {1}{2}}l$, according as n′ is not, or is, equal to n, we have $\textstyle {\frac {1}{2}}l{\text{A}}_{n}=\int _{0}^{l}f(x)\sin {\frac {n\pi x}{l}}dx$, and thus the series is of the form

$\textstyle {\frac {2}{l}}\sum _{1}^{\infty }\sin {\frac {n\pi x}{l}}\int _{0}^{l}f(x)\sin {\frac {n\pi x}{l}}dx$

(1)

This method of determining the coefficients in the series would not be valid without the assumption that the series is in general uniformly convergent, for in accordance with a known theorem the sum of the integrals of the separate terms of the series is otherwise not necessarily equal to the integral of the sum. This assumption being made, it is further assumed that ƒ(x) is such that $\textstyle \int _{0}^{l}f(x)\sin {\frac {n\pi x}{l}}dx$ has a definite meaning for every value of n.

Before we proceed to examine the justification for the assumptions made, it is desirable to examine the result obtained, and to deduce other series from it. In order to obtain a series of the form

${\text{B}}_{0}+{\text{B}}_{1}\cos {\frac {\pi x}{l}}+{\text{B}}_{2}\cos {\frac {2\pi x}{l}}+\dots +{\text{B}}_{n}\cos {\frac {n\pi x}{l}}+\dots$

for the representation of ƒ(x) in the interval 0 to l, let us apply the series (1) to represent the function $\textstyle f(x)\sin {\frac {\pi x}{l}}$; we thus find

${\frac {2}{l}}\sum _{1}^{\infty }\sin {\frac {n\pi x}{l}}\int \limits _{0}^{l}f(x)\sin {\frac {\pi x}{l}}\sin {\frac {n\pi x}{l}}dx,$

or

${\frac {1}{l}}\sum _{1}^{\infty }\sin {\frac {n\pi x}{l}}\int \limits _{0}^{l}f(x)\left\{\cos {\frac {(n-1)\pi x}{l}}-\cos {\frac {(n+1)\pi x}{l}}\right\}dx.$

On rearrangement of the terms this becomes

${\frac {1}{l}}\sin {\frac {\pi x}{l}}\int \limits _{0}^{l}f(x)dx+{\frac {2}{l}}\sum \sin {\frac {\pi x}{l}}\cos {\frac {n\pi x}{l}}\int \limits _{0}^{l}f(x)\cos {\frac {n\pi x}{l}}dx$

hence ƒ(x) is represented for the interval 0 to l by the series of cosines

${\frac {1}{l}}\int \limits _{0}^{l}f(x)dx+{\frac {2}{l}}\sum _{1}^{\infty }\cos {\frac {n\pi x}{l}}\int \limits _{0}^{l}f(x)\cos {\frac {n\pi x}{l}}dx$

(2)

We have thus seen, that with the assumptions made, the arbitrary function ƒ(x) may be represented, for the given interval, either by a series of sines, as in (1), or by a series of cosines, as in (2). Some important differences between the two series must, however, be noticed. In the first place, the series of sines has a vanishing sum when x = 0 or x = l; it therefore does not represent the function at the point x = 0, unless ƒ(0) = 0, or at the point x = l, unless ƒ(l) = 0, whereas the series (2) of cosines may represent the function at both these points. Again, let us consider what is represented by (1) and (2) for values of x which do not lie between 0 and l. As ƒ(x) is given only for values of x between 0 and l, the series at points beyond these limits have no necessary connexion with ƒ(x) unless we suppose that ƒ(x) is also given for such general values of x in such a way that the series continue to represent that function. If in (1) we change x into −x, leaving the coefficients unaltered, the series changes sign, and if x be changed into x + 2l, the series is unaltered; we infer that the series (1) represents an odd function of x and is periodic of period 2l; thus (1) will represent ƒ(x) in general for values of x between ±∞, only if ƒ(x) is odd and has a period 2l. If in (2) we change x into −x, the series is unaltered, and it is also unaltered by changing x into x + 2l; from this we see that the series (2) represents ƒ(x) for values of x between ±∞, only if ƒ(x) is an even function, and is periodic of period 2l. In general a function ƒ(x) arbitrarily given for all values of x between ±∞ is neither periodic nor odd, nor even, and is therefore not represented by either (1) or (2) except for the interval 0 to l.

From (1) and (2) we can deduce a series containing both sines and cosines, which will represent a function ƒ(x) arbitrarily given in the interval −l to l, for that interval. We can express by (1) the function $\textstyle {\frac {1}{2}}\left\{f(x)-f(-x)\right\}$ which is an odd function, and thus this function is represented for the interval −l to +l by

${\frac {2}{l}}\sum \sin {\frac {n\pi x}{l}}\int \limits _{0}^{l}{\frac {1}{2}}\left\{f(x)-f(-x)\right\}\sin {\frac {n\pi x}{l}}dx;$

we can also express $\textstyle {\frac {1}{2}}\left\{f(x)+f(-x)\right\}$, which is an even function, by means of (2), thus for the interval −l to +l this function is represented by

${\frac {1}{l}}\int \limits _{0}^{l}{\frac {1}{2}}\left\{f(x)+f(-x)\right\}dx+{\frac {2}{l}}\sum _{1}^{\infty }\cos {\frac {n\pi x}{l}}\int \limits _{0}^{l}{\frac {1}{2}}\left\{f(x)+f(-x)\right\}\cos {\frac {n\pi x}{l}}dx$

It must be observed that ƒ(−x) is absolutely independent of ƒ(x), the former being not necessarily deducible from the latter by putting −x for x in a formula; both ƒ(x) and ƒ(−x) are functions given arbitrarily and independently for the interval 0 to l. On adding the expressions together we obtain a series of sines and cosines which represents ƒ(x) for the interval −l to l. The integrals

$\int \limits _{0}^{l}f(-x)\cos {\frac {n\pi x}{l}}dx,\int \limits _{0}^{l}f(-x)\sin {\frac {n\pi x}{l}}dx$

are equivalent to

$-\int \limits _{0}^{-l}f(x)\cos {\frac {n\pi x}{l}}dx,\quad +\int \limits _{0}^{-l}f(x)\sin {\frac {n\pi x}{l}}dx$

thus the series is

${\frac {1}{2l}}\int \limits _{-l}^{l}f(x)dx+{\frac {1}{l}}\sum _{1}^{\infty }\cos {\frac {n\pi x}{l}}\int \limits _{-l}^{l}f(x)\cos {\frac {n\pi x}{l}}dx+{\frac {1}{l}}\sum _{1}^{\infty }\sin {\frac {n\pi x}{l}}\int \limits _{-l}^{l}f(x)\sin {\frac {n\pi x}{l}}dx,$

which may be written

${\frac {1}{2l}}\int \limits _{-l}^{l}f(x')dx'+{\frac {1}{l}}\sum _{1}^{\infty }\int \limits _{-l}^{l}f(x')\cos {\frac {n\pi (x-x')}{l}}dx'$

(3)

The series (3), which represents a function ƒ(x) arbitrarily given for the interval −l to l, is what is known as Fourier’s Series; the expressions (1) and (2) being regarded as the particular forms which (3) takes in the two cases, in which ƒ(−x) = −ƒ(x), or ƒ(−x) = ƒ(x) respectively. The expression (3) does not represent ƒ(x) at points beyond the interval −l to l, unless ƒ(x) has a period 2l. For a value of x within the interval, at which ƒ(x) is discontinuous, the sum of the series may cease to represent ƒ(x), but, as will be seen hereafter, has the value $\textstyle {\frac {1}{2}}\left\{f(x+0)+f(x-0)\right\}$, the mean of the limits at the points on the right and the left. The series represents the function at x = 0, unless the function is there discontinuous, in which case the series is $\textstyle {\frac {1}{2}}\left\{f(+0)+f(-0)\right\}$; the series does not necessarily represent the function at the points l and −l, unless ƒ(l) = ƒ(−l). Its sum at either of these points is $\textstyle {\frac {1}{2}}\left\{f(l)+f(-l)\right\}$.

Examples of Fourier’s Series.—(a) Let ƒ(x) be given from 0 to l, by ƒ(x) = c, when 0 ≦ x < 1/2l, and by f(x)= −c from 1/2l to l; it is required to find a sine series, and also a cosine series, which shall represent the function in the interval.

We have

$\int \limits _{0}^{l}f(x)\sin {\frac {n\pi x}{l}}dx=c\int \limits _{0}^{{\frac {1}{2}}l}\sin {\frac {n\pi x}{l}}dx-c\int \limits _{{\frac {1}{2}}l}^{l}\sin {\frac {n\pi x}{l}}dx$

$={\frac {cl}{n\pi }}(\cos n\pi -2\cos {\frac {1}{2}}n\pi +1).$

This vanishes if n is odd, and if n = 4m, but if n = 4m + 2 it is equal to $\textstyle {\frac {4cl}{n\pi }}$; the series is therefore

${\frac {4c}{\pi }}\left({\frac {1}{2}}\sin {\frac {2\pi x}{l}}+{\frac {1}{3}}\sin {\frac {6\pi x}{l}}+{\frac {1}{5}}\sin {\frac {10\pi x}{l}}+\dots \right)$

For unrestricted values of x, this series represents the ordinates of the series of straight lines in fig. 1, except that it vanishes at the points 0, 1/2l, l, 3/2l ...

Fig. 1.

We find similarly that the same function is represented by the series

${\frac {4c}{\pi }}\left(\cos {\frac {\pi x}{l}}-{\frac {1}{3}}\cos {\frac {3\pi x}{l}}+{\frac {1}{5}}\cos {\frac {5\pi x}{l}}-+\dots \right)$

during the interval 0 to l; for general values of x the series represents the ordinate of the broken line in fig. 2, except that it vanishes at the points 1/2l, 3/2l ...

Fig. 2.

(b) Let ƒ(x) = x from 0 to 1/2l, and f(x) = l − x, from 1/2l to l; then

$\int \limits _{0}^{l}f(x)\sin {\frac {n\pi x}{l}}dx=\int \limits _{0}^{{\frac {1}{2}}l}x\sin {\frac {n\pi x}{l}}dx+\int \limits _{{\frac {1}{2}}l}^{l}(l-x)\sin {\frac {n\pi x}{l}}dx$

$=-{\frac {l^{2}}{2n\pi }}\cos {\frac {n\pi }{2}}+{\frac {l^{2}}{n^{2}\pi ^{2}}}\sin {\frac {n\pi }{2}}+{\frac {l^{2}}{n\pi }}\left(\cos {\frac {n\pi }{2}}-\cos n\pi \right)+{\frac {l^{2}}{n\pi }}\cos n\pi -{\frac {l^{2}}{2n\pi }}\cos {\frac {n\pi }{2}}+{\frac {l^{2}}{n^{2}\pi ^{2}}}\sin {\frac {n\pi }{2}}$

$={\frac {2l^{2}}{n^{2}\pi ^{2}}}\sin {\frac {n\pi }{2}}$