9.2 On Dirichlet’s Conditions and Fourier’s Theorem.
A theorem, of the type described in §9.1, concerning the expansibility of a function of a real variable into a trigonometrical series is usually described as Fourier’s theorem. On account of the length and difficulty of a formal proof of the theorem (even when the function to be expanded is subjected to unnecessarily stringent conditions), we defer the proof until §9.42 and §9.43. It is, however, convenient to state here certain sufficient conditions under which a function can be expanded into a trigonometrical series.
Let \(f(t)\) be defined arbitrarily when \(-\pi \leq t \leq \pi\) and defined for all other real values of \(t\) by means of the equation \[ f(t + 2\pi) = f(t), \] so that \(f(t)\) is a periodic function with period \(2\pi\).[1]
Let \(f(t)\) be such that \(\int_{-\pi}^{\pi} f(t) \, d t\) exists; and if this is an improper integral, let it be absolutely convergent.
Let \(a_{n}, b_{n}\) be defined by the equations[2] \[ \pi a_{n} \!=\! \int_{-\pi}^{\pi}\! f(t) \cos nt \, d t, \enspace \pi b_{n} \!=\! \int_{-\pi}^{\pi}\! f(t) \sin nt \, d t \enspace (n=0,1,2,\ldots). \]
Then, if \(x\) be an interior point of any interval \((a, b)\) in which \(f(t)\) has limited total fluctuation, the series \[ \frac{1}{2} a_{\:\!0} + \sum_{n=1}^{\infty} (a_{n} \cos nx + b_{n} \sin nx) \] is convergent, and its sum is \(\frac{1}{2} \left\{ f(x+0) + f(x-0) \right\}\).[3] If \(f(t)\) is continuous at \(t=x\), this sum reduces to \(f(x)\).
This theorem will be assumed in §§9.21–9.32; these sections deal with theorems concerning Fourier series which are of some importance in practical applications. It should be stated here that every function which Applied Mathematicians need to expand into Fourier series satisfies the conditions just imposed on \(f(t)\), so that the analysis given later in this chapter establishes the validity of all the expansions into Fourier series which are required in physical investigations.
The reader will observe that in the theorem just stated, \(f(t)\) is subject to less stringent conditions than those contemplated by Dirichlet, and this decrease of stringency is of considerable practical importance. Thus, so simple a series as \(\sum_{n=1}^{\infty} (-1)^{n-1} (\cos nx) / n\) is the expansion of the function \(\log \left| \, 2 \cos \frac{1}{2} x \, \right| \);[4] and this function does not satisfy Dirichlet’s condition of boundedness at \(\pm \pi\).
It is convenient to describe the series \(\frac{1}{2} a_{\:\!0} + \sum_{n=1}^{\infty} (a_{n} \cos nx + b_{n} \sin nx)\) as the Fourier series associated with \(f(t)\). This description must, however, be taken as implying nothing concerning the convergence of the series in question.
9.21 The representation of a function by Fourier series for ranges other than \((-\pi,\pi)\).
Consider a function \(f(x)\) with an (absolutely) convergent integral, and with limited total fluctuation in the range \(a \leq x \leq b\).
Write \(x = \frac{1}{2} (a + b) - \frac{1}{2} (a-b) \pi^{-1} x', \quad f(x) = F(x')\).
Then it is known (§9.2) that \[ \frac{1}{2} [F(x'+0) + F(x'-0)] = \frac{1}{2} a_{\:\!0} + \sum_{n=1}^{\infty} (a_{n} \cos nx' + b_{n} \sin nx'), \] and so \[ \begin{align*} & \frac{1}{2} \left\{ f(x+0) + f(x-0)\right\} \\ & \hfill = \frac{1}{2} a_{\:\!0} + \sum_{n=1}^{\infty} \left\{ a_{n} \cos \frac{n \pi (2x-a-b)}{b-a} + b_{n} \sin \frac{n \pi (2x-a-b)}{b-a} \right\}, \end{align*} \] where by an obvious transformation \[ \begin{align*} \frac{1}{2} (b-a) a_{n} =& \int_{a}^{b}\! f(x) \cos \frac{n \pi (2x-a-b)}{b-a} \, d x, \\ \frac{1}{2} (b-a) b_{n} =& \int_{a}^{b}\! f(x) \sin \frac{n \pi (2x-a-b)}{b-a} \, d x . \end{align*} \]
9.22 The Cosine Series and the Sine Series.
Let \(f(x)\) be defined in the range \((0,l)\) and let it have an (absolutely) convergent integral and also let it have limited total fluctuation in that range. Define \(f(x)\) in the range \((-l,0)\) by the equation \[ f(-x) = f(x). \]
Then \[ \frac{1}{2} \left\{ f(x+0) + f(x-0) \right\} = \frac{1}{2} a_{\:\!0} + \sum_{n=1}^{\infty} \left\{ a_{n} \cos \frac{n \pi x}{l} + b_{n} \sin \frac{n \pi x}{l} \right\}, \] where, by §9.21, \[ \begin{align*} l a_{n} =& \int_{-l}^{l}\! f(t) \cos \frac{n \pi t}{l} \, d t = 2 \! \int_{0}^{l}\! f(t) \cos \frac{n \pi t}{l} \, d t, \\ l b_{n} =& \int_{-l}^{l}\! f(t) \sin \frac{n \pi t}{l} \, d t = 0, \end{align*} \] so that when \(-l \leq x \leq l\), \[ \frac{1}{2} \left\{ f(x+0) + f(x-0) \right\} = \frac{1}{2} a_{\:\!0} + \sum_{n=1}^{\infty} a_{n} \cos \frac{n \pi x}{l}; \] this is called the cosine series.
If, however, we define \(f(x)\) in the range \((0,-l)\) by the equation \[ f(-x) = -f(x), \] we get, when \(-l \leq x \leq l\), \[ \frac{1}{2} \left\{ f(x+0) + f(x-0) \right\} = \sum_{n=1}^{\infty} b_{n} \sin \frac{n \pi x}{l}, \] where \[ l b_{n} = 2 \! \int_{0}^{l}\! f(t) \sin \frac{n \pi t}{l} \, d t; \] this is called the sine series.
Thus the series \[ \frac{1}{2} a_{\:\!0} + \sum_{n=1}^{\infty} a_{n} \cos \frac{n \pi x}{l}, \quad \sum_{n=1}^{\infty} b_{n} \sin \frac{n \pi x}{l}, \] where \[ \frac{1}{2} l a_{n} = \int_{0}^{l}\! f(t) \cos \frac{n \pi t}{l} \, d t, \quad \frac{1}{2} l b_{n} = \int_{0}^{l}\! f(t) \sin \frac{n \pi t}{l} \, d t, \] have the same sum when \(0 \leq x \leq l\); but their sums are numerically equal and opposite in sign when \(0 \geq x \geq -l\).[5]
Example 1. Expand \(\frac{1}{2} (\pi - x) \sin x\) in a cosine series in the range \(0 \leq x \leq \pi\).
[We have, by the formula just obtained, \[ \frac{1}{2} (\pi - x) \sin x = \frac{1}{2} a_{\:\!0} + \sum_{n=1}^{\infty} a_{n} \cos nx, \] where \[ \frac{1}{2} \pi a_{n} = \int_{0}^{\pi}\! \frac{1}{2} (\pi - x) \sin x \cos nx \, d x. \] ][ But, integrating by parts, if \(n \neq 1\), \[ \begin{align*} &\int_{0}^{\pi}\! 2 (\pi - x) \sin x \cos nx \, d x \\ \\ &\quad = \int_{0}^{\pi}\! (\pi - x) \left\{ \sin (n+1) x - \sin (n-1) x \right\} \, d x \\ \\ &\quad = \left[\:\! (x - \pi) \left\{ \frac{ \cos (n+1) x }{n+1} - \frac{ \cos (n-1) x }{n-1} \right\} \:\!\right]^{\pi}_{0} \\ &\qquad \qquad \qquad \quad - \int_{0}^{\pi}\! \left\{ \frac{ \cos (n+1) x }{n+1} - \frac{ \cos (n-1) x }{n-1} \right\} \, d x \\ \\ &\quad = \pi \left( \frac{1}{n+1} - \frac{1}{n-1} \right) = \frac{-2\pi}{(n+1)(n-1)} \\ \end{align*} \] Whereas if \(n=1\), we get \[\int_{0}^{\pi} 2 (\pi - x) \sin x \cos x \, d x = \frac{1}{2} \pi.\] Therefore the required series is \[ \frac{1}{2} + \frac{1}{4} \cos x - \frac{1}{1 \cdot 3} \cos 2x - \frac{1}{2 \cdot 4} \cos 3x - \frac{1}{3 \cdot 5} \cos 4x - \cdots. \] It will be observed that it is only for values of \(x\) between \(0\) and \(\pi\) that the sum of this series is proved to be \(\frac{1}{2} (\pi - x) \sin x\); thus for instance when \(x\) has a value between \(0\) and \(-\pi\), the sum of the series is not \(\frac{1}{2} (\pi - x) \sin x\), but \(-\frac{1}{2} (\pi + x) \sin x\); when \(x\) has a value between \(\pi\) and \(2 \pi\), the sum of the series happens to be again \(\frac{1}{2} (\pi - x) \sin x\), but this is a mere coincidence arising from the special function considered, and does not follow from the general theorem.]
Example 2. Expand \(\frac{1}{8} \pi x (\pi - x)\) in a sine series, valid when \(0 \leq x \leq \pi\).
[The series is \(\sin x + \dfrac{\sin 3x}{3^{3}} + \dfrac{\sin 5x}{5^{3}} + \cdots.\)]
Example 3. Shew that, when \(0 \leq x \leq \pi\), \[ \frac{1}{96} \pi (\pi - 2x) (\pi^{2} + 2 \pi x - 2 x^{2}) = \cos x + \frac{\cos 3x}{3^{4}} + \frac{\cos 5x}{5^{4}} + \cdots. \]
[Denoting the left-hand side by \(f(x)\), we have, on integrating by parts and observing that \(f'(0) = f'(\pi) = 0\),
\[ \begin{align*} \int_{0}^{\pi}\! f(x) \cos nx \, d x =&\: \frac{1}{n} \left[f(x) \sin nx \vphantom{\int} \right]_{0}^{\pi} - \frac{1}{n} \! \int_{0}^{\pi}\! f'(x) \sin nx \, d x \\ \\ =&\: \frac{1}{n^{2}} \left[f'(x) \cos nx \vphantom{\int} \right]_{0}^{\pi} - \frac{1}{n^{2}} \! \int_{0}^{\pi}\! f''(x) \cos nx \, d x \\ \\ =& -\frac{1}{n^{3}} \left[f''(x) \sin nx \vphantom{\int} \right]_{0}^{\pi} + \frac{1}{n^{3}} \! \int_{0}^{\pi}\! f'''(x) \sin nx \, d x \\ \\ =& -\frac{1}{n^{4}} \left[f'''(x) \cos nx \vphantom{\int} \right]_{0}^{\pi} = \frac{\pi}{4 n^{4}} (1 - \cos n \pi).] \end{align*} \]
Example 4. Shew that for values of \(x\) between \(0\) and \(\pi\), \(e^{s x}\) can be expanded in the cosine series \[ \begin{align*} &\frac{2 s}{\pi} \left(e^{s \pi} - 1\right) \left( \frac{1}{2 s^{2}} + \frac{\cos 2x}{s^{2} + 4} + \frac{\cos 4x}{s^{2} + 16} + \cdots \right) \\ & \qquad \qquad \qquad \qquad - \frac{2 s}{\pi} \left(e^{s \pi} - 1\right) \left( \frac{\cos x}{s^{2} + 1} \ + \frac{\cos 3x}{s^{2} + 9} + \cdots \right), \end{align*} \] and draw graphs of the function \(e^{s x}\) and of the sum of the series.
Example 5. Shew that for values of \(x\) between \(0\) and \(\pi\), the function \(\frac{1}{8} \pi (\pi - 2x)\) can be expanded in the cosine series \[ \cos x \ + \frac{\cos 3x}{3^{2}} + \frac{\cos 5x}{5^{2}}+ \cdots, \] and draw graphs of the function \(\frac{1}{8} \pi (\pi - 2x)\) and of the sum of the series.
9.3 The nature of the coefficients in a Fourier series.[6]
Suppose that (as in the numerical examples which have been discussed) the interval \((-\pi, \pi)\) can be divided into a finite number of ranges \((-\pi, k_{1})\), \((k_{1}, k_{2})\), \(\ldots, (k_{n}, \pi)\) such that throughout each range \(f(x)\) and all its differential coefficients are continuous with limited total fluctuation and that they have limits on the right and on the left (§3.2) at the end points of these ranges.
Then \[ \pi a_{m} = \int_{-\pi}^{k_{1}}\! f(t) \cos mt \, d t + \int_{k_{1}}^{k_{2}}\! f(t) \cos mt \, d t + \cdots + \int_{k_{n}}^{\pi}\! f(t) \cos mt \, d t. \]
Integrating by parts we get \[ \begin{align*} & \pi a_{m} = \left[ m^{-1} f(t) \sin mt \vphantom{\int} \right]_{-\pi}^{k_{1}} + \left[ m^{-1} f(t) \sin mt \vphantom{\int} \right]_{k_{1}}^{k_{2}} + \cdots + \left[ m^{-1} f(t) \sin mt \vphantom{\int} \right]_{k_{n}}^{\pi} \\ & \quad - m^{-1}\! \int_{-\pi}^{k_{1}}\! f'(t) \sin mt \, d t - m^{-1}\! \int_{k_{1}}^{k_{2}}\! f'(t) \sin mt \, d t \cdots - m^{-1}\! \int_{k_{n}}^{\pi}\! f'(t) \sin mt \, d t, \end{align*} \] so that \[ a_{m} = \frac{A_{m}}{m} - \frac{b_{m}^{\:\!\prime}}{m}, \] where \[ \pi A_{m} = \sum_{r=1}^{n} \sin m k_{r} \left\{ f(k_{r} - 0) - f(k_{r} + 0), \right\} \] and \(b_{m}'\) is a Fourier constant of \(f'(x)\).
Similarly \[ b_{m} = \frac{B_{m}}{m} + \frac{a_{m}'}{m}, \] where \[ \pi B_{m} = - \sum_{r=1}^{n} \cos m k_{r} \left\{ f(k_{r} - 0) - f(k_{r} + 0) \right\} - \cos m \pi \left\{ f(\pi - 0) - f(-\pi + 0) \right\}\!, \] and \(a_{m}'\) is a Fourier constant of \(f'(x)\).
Similarly, we get \[ a_{m}' = \frac{A_{m}'}{m} - \frac{b_{m}''}{m}, \quad b_{m}' = \frac{B_{m}'}{m} + \frac{a_{m}''}{m}, \] where \(a_{m}''\), \(b_{m}''\) are the Fourier constants of \(f''(x)\) and \[ \begin{align*} \pi A_{m}' =& \sum_{r=1}^{n} \sin m k_{r} \left\{ f'(k_{r}-0) - f'(k_{r}+0) \right\}, \\ \pi B_{m}' =& - \sum_{r=1}^{n} \cos m k_{r} \left\{ f'(k_{r}-0) - f'(k_{r}+0) \right\} \\ &\qquad - \cos m \pi \left\{ f'(\pi - 0) - f'(-\pi + 0) \right\}. \end{align*} \]
Therefore \[ a_{m} = \frac{A_{m}}{m} - \frac{B_{m}'}{m^{2}} - \frac{a_{m}''}{m^{2}}, \quad b_{m} = \frac{B_{m}}{m} + \frac{A_{m}'}{m^{2}} - \frac{b_{m}''}{m^{2}}, \]
Now as \(m \rightarrow \infty\), we see that \[ A_{m}' = O(1), \quad B_{m}' = O(1), \] and, since the integrands involved in \(a_{m}''\) and \(b_{m}''\) are bounded, it is evident that \[ a_{m}'' = O(1), \quad b_{m}'' = O(1). \]
Hence if \(A_{m}=0\), \(B_{m}=0\), the Fourier series for \(f(x)\) converges absolutely and uniformly, by §3.34.
The necessary and sufficient conditions that \(A_{m} = B_{m} = 0\) for all values of \(m\) are that \[ f(k_{r} - 0) = f(k_{r} + 0), \quad f(\pi - 0) = f(-\pi + 0) \] that is to say that \(f(x)\) should be continuous for all values of \(x\).[7]
9.31 Differentiation of Fourier series.
The result of differentiating \[ \frac{1}{2} a_{\:\!0} + \sum_{m=1}^{\infty} (a_{m} \cos mx + b_{m} \sin mx) \] term by term is \[ \sum_{m=1}^{\infty} \left\{ m b_{m} \cos mx - m a_{m} \sin mx \right\}. \]
With the notation of §9.3, this is the same as \[ \frac{1}{2} a_{\:\!0}' + \sum_{m=1}^{\infty} ( a_{m}' \cos mx + b_{m}' \sin mx), \] provided that \(A_{m} = B_{m} = 0\) and \(\int_{-\pi}^{\:\!\pi}\:\! f'(x) \, d x = 0\); these conditions are satisfied if \(f(x)\) is continuous for all values of \(x\).
Consequently sufficient conditions for the legitimacy of differentiating a Fourier series term by term are that \(f(x)\) should be continuous for all values of \(x\) and \(f'(x)\) should have only a finite number of points of discontinuity in the range \((-\pi, \pi)\), both functions having limited total fluctuation throughout the range.
9.32 Determination of points of discontinuity.
The expressions for \(a_{m}\) and \(b_{m}\) which have been found in §9.3 can frequently be applied in practical examples to determine the points at which the sum of a given Fourier series may be discontinuous. Thus, let it be required to determine the places at which the sum of the series \[ \sin x + \frac{1}{3} \sin 3x + \frac{1}{5} \sin 5x + \cdots \] is discontinuous.
Assuming that the series is a Fourier series and not any trigonometrical series and observing that \(a_{m} = 0\), \( b_{m} = (2m)^{-1}(1 - \cos m \pi)\), we get on considering the formula found in §9.3,[8] \[ A_{m} = 0, \quad B_{m} = \frac{1}{2} - \frac{1}{2} \cos m \pi, \quad a_{m}' = b_{m}' = 0. \]
Hence if \(k_{1},\) \( k_{2},\ldots\) are the places at which the analytic character of the sum is broken, we have \[ \begin{align*} 0 = \pi A_{m} = & \: \left[ \vphantom{\int} \sin m k_{1} \left\{ f(k_{1} - 0) - f(k_{1} + 0) \right\} \right. \\ & \qquad \left. \vphantom{\int} + \sin m k_{2} \left\{ f(k_{2} - 0) - f(k_{2} + 0) \right\} + \cdots \right]. \end{align*} \] Since this is true for all values of \(m\), the numbers \(k_{1},\) \( k_{2},\ldots\) must be multiples of \(\pi\); but there is only one multiple of \(\pi\) in the range \(-\pi < x < \pi\), namely zero. So \(k_{1} = 0\), and \(k_{2},\) \( k_{3},\ldots\) do not exist. Substituting \(k_{1} = 0\) in the equation \(B_{m} = \frac{1}{2} - \frac{1}{2} \cos m \pi\), we have \[ \pi \left(\frac{1}{2} - \frac{1}{2} \cos m \pi \right) = - \left[ \vphantom{\int} \cos m \pi \left\{ f(\pi - 0) - f(-\pi + 0) \right\} + f(-0) - f(+0) \right]. \]
Since this is true for all values of \(m\), we have \[ R\frac{1}{2} \pi = f(+0) - f(-0), \quad \frac{1}{2} \pi = f(\pi - 0) - f(-\pi + 0). \]
This shews that, if the series is a Fourier series, \(f(x)\) has discontinuities at the points \(n \pi\) (\(n\) any integer), and since \(a_{m}' = b_{m}' = 0\), we should expect \(f(x)\) to be constant in the open range \((-\pi, 0)\) and to be another constant in the open range \((0, \pi)\).[9]