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Introduction to Linear Algebra with Mathematica

Glossary

Preface

Definition of Fourier Series

 

Finite Partial Sums of Fourier Series

For any real x, the finite sum
\[ F_N (x) = \frac{a_0}{2} + \sum_{k=1}^{N} \left[ a_k \cos \left( \frac{k \pi x}{\ell} \right) + b_k \sin \left( \frac{k \pi x}{\ell} \right) \right] \]
or
\[ F_N (x) = \sum_{k=-N}^{N} \alpha_n e^{{\bf j}n \pi x/\ell} \]
is called a trigonometric polynomial of order N.

In calculus, you learnt that an infinite series \( \sum_{k\ge 0} a_k \) converges (or is convergent) to S if the sequence of partial sums \( S_n = \sum_{k= 0}^n a_k \) tends to S as n → ∞. In case of Fourier series, elements of series depend on real parameter x ∈ [−ℓ, ℓ], and we come to definition of its partial sums.

Theorem 1: For any real-valued function f : [−ℓ, ℓ] → ℝ and any positive integer N ∈ ℕ = { 0, 1, 2, … }, its N-th partial Fourier exponential sum
\begin{equation} \label{EqFourier.5} S_N (f;x) = \sum_{n=-N}^N \hat{f}(n)\,e^{{\bf j} n\pi x/\ell} \end{equation}
and N-th partial Fourier trigonometric sum
\begin{equation} \label{EqFourier.6} S_N (f;x) = \frac{1}{2}\,a_0 + \sum_{k=1}^N \left[ a_k \cos \left( \frac{k\pi x}{\ell} \right) + b_k \sin \left( \frac{k\pi x}{\ell} \right) \right] \end{equation}
are the same; here coefficients 𝑎k, bk and \( \displaystyle \hat{f}(n) \) are determined by the Euler--Fouier formulas \eqref{EqFourier.2} and \eqref{EqFourier.4}, respectively. Moreover, the N-th partial Fourier sum is expressed in an integral form, known as a convolution, which is denoted by star:
\[ S_N (f;x) = \frac{1}{\ell} \int_{-\ell}^{\ell} f(y)\,D_N \left( \frac{\pi (x-y)}{\ell} \right) {\text d}y = \frac{1}{\ell} \left( f \star D_N \right) (x), \]
where DN(z) is called the Dirichlet kernel, and
\[ D_N (z) = \frac{\sin \left( N + \frac{1}{2} \right) z}{2\,\sin\frac{z}{2}} . \]
Note that according to the L'Hôpital's rule, DN(0) = N + ½. It is a custom to utilize ★ for convolution of two functions:
\[ \left( f \star g \right) (x) = g \star f (x) = \int f(y) g(x-y)\,{\text d} y . \]
We are going to prove the statement for a particular case when ℓ = π in order to simplify calculations. We compute
\begin{align*} S_N (f; x) &= \sum_{n=-N}^N \hat{f}(n) \,e^{{\bf j}nx} = \sum_{n=-N}^N \frac{1}{2\pi} \int_{-\pi}^{\pi} f(y)\,e^{-{\bf j}n y} {\text d}y\, e^{{\bf j}nx} \\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(y) \sum_{n=-N}^N e^{{\bf j}n\left( x-y \right)} \\ &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(y)\,D_N (x-y) \,{\text d}y = \frac{1}{\pi} \, f \star D_N (x) . \end{align*}
Here
\[ D_N (t) = \frac{1}{2}\,\delta_N (t) = \frac{1}{2} \sum_{n=-N}^N e^{{\bf j}n t} \qquad ({\bf j}^2 =1) \]
is the Dirichlet kernel. Also, f ★ g means the convolution of f and g, which is defined as
\[ f \star g (x) = \int_{-\ell}^{\ell} f(y)\,g(x-y)\,{\text d}y = \int_{-\ell}^{\ell} f(x-y)\,g(y)\,{\text d}y = g \star f . \]
Next, we have
\begin{align*} \delta_N (x) &= \sum_{n=-N}^N e^{{\bf j} n x} = e^{-{\bf j} N x} \sum_{n=0}^{2N} e^{{\bf j} n x} \\ &= e^{-{\bf j} N x} \frac{1 - e^{-{\bf j} \left( 2N+1 \right) x}}{1 - e^{{\bf j} x}} = \frac{e^{-{\bf j} Nx} - e^{-{\bf j} \left( N+1 \right) x}}{1 - e^{{\bf j} x}} \\ &= \frac{e^{-{\bf j} \left( N + \frac{1}{2} \right) x} - e^{{\bf j} \left( N + \frac{1}{2} \right) x}}{e^{-{\bf j} x/2} - e^{-{\bf j} x/2}} \\ &= \frac{\sin \left( N + \frac{1}{2} \right) x}{\sin\frac{x}{2}} \end{align*}
because
\[ \sin\theta - \frac{1}{2{\bf j}} \left( e^{{\bf j}\theta} - e^{-{\bf j}\theta} \right) . \]

 

The formula for trigonometric partial sum is proved in Lemma 1 of convergence section.
Lemma 1: The following integrals diverge:
\[ \lim_{N\to\infty}\int_{-\pi}^{\pi} D_N (x)^2 {\text d} x = \infty , \qquad \mbox{and} \qquad \int_{-\pi}^{\pi} \left\vert D_N (x) \right\vert {\text d} x \ge 4\,\mbox{Si}(\pi ) + \frac{8}{\pi}\,\ln n , \]
where Si(x) is the sine integral.
For any small δ, we break the integral into three parts:
\[ I_N = \int_{-\pi}^{\pi} D_N (x)^2 {\text d} x = I_{\delta} (N) + I_2 + I_3 , \]
where
\[ I_{\delta} (N) = 8 \int_{0}^{\delta} \frac{\sin^2 \left( N + \frac{1}{2} \right) x}{x^2} \, {\text d} x , \]
and
\[ I_2 = \int_{|x| \le \delta} \sin^2 \left( N + \frac{1}{2} \right) x \left( \frac{1}{\sin^2 \frac{x}{2}} - \frac{4}{x^2} \right) {\text d} x , \qquad I_3 = \int_{|x| > \delta} D_N (x)^2 {\text d} x . \]
Since the kernel of integral I2 is bounded in a neighborhood of the origin,
Series[1/(Sin[x/2])^2 - 4/x^2, {x, 0, 5}]
SeriesData[x, 0, { Rational[1, 3], 0, Rational[1, 60], 0, Rational[1, 1512]}, 0, 6, 1]
both integrals, I2 and I3, tend to zero as N → ∞ according to Riemann--Lebesgue lemma. Using substitution s = (N + ½)x in integral Iδ, we get
\[ I_{\delta} (N) = 4 \left( 2N + 1 \right) \int_{0}^{\delta \left( N + 1/2\right)} \frac{\sin^2 s}{s^2}\, {\text d}s . \]
Since the latter integral converges
Integrate[(Sin[s])^2/s^2, {s, 0, Infinity}]
\[Pi]/2
\[ \int_0^{\infty} \frac{\sin^2 s}{s^2}\,{\text d}s = \frac{\pi}{2} , \]
we conclude that
\[ I_{\delta} (N) = O\left( 2N + 1 \right) \to \infty \qquad\mbox{as} \quad s N \to infty . \]
End of Proof for the first part
We have
\begin{align*} \| D_N (x) \|_1 &= \int_{-\pi}^{\pi} \left\vert D_N (x) \right\vert {\text d}x = 2 \int_{0}^{\pi} \left\vert D_N (x) \right\vert {\text d}x \ge \int_0^{\pi} \frac{\left\vert \sin [(2N+1) x/2] \right\vert}{x}\,{\text d} x \end{align*}
because \( \displaystyle \frac{2}{x} \le \frac{1}{\left\vert \sin (x/2) \right\vert} . \) In the latter integral, we make substitution \( \displaystyle t = \left( 2N+1 \right) x/2 \) to obtain
\begin{align*} \| D_N (x) \|_1 &\ge \int_0^{(2N+1) \pi /2} \frac{|\sin t|}{t}\,{\text d}t \\ &= \sum_{k=0}^{N-1} \int_{k\pi}^{(k+1)\pi} \frac{|\sin t|}{t}\,{\text d}t + \int_{N\pi}^{N\pi + \pi /2}\frac{|\sin t|}{t}\,{\text d}t \\ &\ge \sum_{k=0}^{N-1} \int_{k\pi}^{(k+1)\pi} \frac{|\sin t|}{(k+1)\pi}\,{\text d}t + \int_{N\pi}^{N\pi + \pi /2}\frac{|\sin t|}{\pi \left( N + 1/2 \right)}\,{\text d}t . \end{align*}
In each integral of the latter sum, we change the dummy variable of integration so each integration starts from zero:
\begin{align*} \| D_N (x) \|_1 &\ge \left\vert \sum_{k=0}^{N-1} \int_0^{\pi} \frac{\sin (s)} {\left( k+1 \right) \pi} \,{\text d} s \right\vert + \frac{1}{\left( N + 1/2 \right) \pi} \int_0^{\pi /2} \frac{\sin (s)}{s} \,{\text d} s \\ & \ge \frac{2}{\pi} \sum_{k=0}^{N-1} \frac{1}{k+1} + \frac{2}{\left( 2N + 1 \right) \pi} \,\mbox{Si} \left( \frac{\pi}{2} \right) \\ &= \frac{2}{\pi}\, H_N + \frac{2}{\left( 2N + 1 \right) \pi} \,\mbox{Si} \left( \frac{\pi}{2} \right) , \end{align*}
where HN is the harmonic number that approaches infinity as logarithm when N → ∞. Indeed, Hn = lnn + γ +o(n) as n → ∞, where γ = 0.5772156649 is the Euler–Mascheroni constant.
The Dirichlet kernel is related to Fourier expansion of the Dirac delta function:
\[ \delta (x-a) = \frac{1}{2\pi} \, \int_{-\infty}^{\infty} e^{{\bf j} (x-a)t} \, {\text d} t = \lim_{n\to \infty} \, \frac{1}{2\pi} \, \sum_{k=-n}^n \, e^{{\bf j} k(x-a)} \quad \left( = \frac{\sin \left[ \left( n+ \frac{1}{2} \right) (x-a) \right]}{2\pi\, \sin \frac{x-a}{2}} \right) , \]
where j is the unit vector in the positive vertical direction on the complex plane ℂ, so j² = −1.

Example 3: The following sum

\[ D_N^{\ast} = \sum_{k=1}^N \sin (kx) = \frac{\cos \frac{x}{2} - \cos \left( N + \frac{1}{2} \right) x}{2\,\sin\frac{x}{2}} \tag{3.1} \]
is known as the Dirichlet conjugate kernel. The derivation of formula (3.1) is based on extracting the imaginary part of
\[ \sum_{k=1}^N z^k = \Im \frac{z - z^{N+1}}{1-z} , \qquad z = e^{{\bf j} x} . \]
Multiplying by the complex conjugate (1 - z)*, we get
\[ D_N^{\ast} = \Im \frac{z - z^{N+1}}{1-z} \cdot \frac{(1-z)^{\ast}}{( 1- z)^{\ast}} = \Im \frac{z - z^{N+1} -|z|^2 - z^{N+1} z^{\ast}}{(1-\cos x)^2 + \sin^2 x} . \tag{3.2} \]
Extraction of the imaginary part yields
\[ D_N^{\ast} = \frac{\sin x - \sin (N+1) x - \sin Nx}{2 - 2\,\cos x} \]
Then using trigonometric identities
\begin{align*} 2\,\sin\alpha\,\cos\beta &= \sin \left( \alpha + \beta \right) + \sin \left( \alpha - \beta \right) , \qquad \alpha = 1/2, \quad \beta = N + 1/2 , \\ \sin x &= 2 \sin\tfrac{x}{2}\,\cos \tfrac{x}{2} , \\ 1- \cos x &= 2\,\sin^2 \tfrac{x}{2} , \end{align*}
we derive the required formula for the Dirichlet conjugate kernel.    ■
End of Example 3

 

 

 

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