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Introduction to Linear Algebra with Mathematica

Glossary

Preface

 

Cosine and sine Fourier transforms

There are known two spectral representations for the product of the impulse operator \( \displaystyle \left( {\bf j} \,\frac{\text d}{{\text d}x} \right)^2 = - \frac{{\text d}^2}{{\text d}x^2} . \) They can be derived from the main Fourier formula for either even function, f(-x) = f(x) or odd function, f(-x) = -f(x). Correspondingly, we obtain the cosine Fourier transformation and sine Fourier transformation.
For an integrable on the interval [0, ∞) function f, two transformations can be defined; one is called cosine Fourier transform:
\[ ℱ_c \left[ f \right] (s) = f^c (s) = \int_0^{\infty} f(x)\,\cos (sx) \,{\text d}x \qquad\mbox{and} \qquad f(x) = \frac{2}{\pi} \int_0^{\infty} ℱ_c \left[ f \right] (s)\,\cos (sx) \,{\text d}s ; \]
and sine Fourier transform:
\[ ℱ_s \left[ f \right] (s) = f^s (s) = \int_0^{\infty} f(x)\,\sin (sx) \,{\text d}x \qquad\mbox{and} \qquad f(x) = \frac{2}{\pi} \int_0^{\infty} ℱ_s \left[ f \right] (s)\,\sin (sx) \,{\text d}s . \]
These two transformations provide spectral representations for the second derivative operator with Dirichlet and Neumann boundary conditions, respectively. More precisely, we have
\begin{align*} ℱ_c \left[ \frac{{\text d}^2 f}{{\text d} x^2} \right] (k) &= \int_0^{\infty} f'' (x)\,\cos (kx) \,{\text d}x = - f' (0) - k^2 ℱ_c \left[ f \right] (k) , \\ ℱ_s \left[ \frac{{\text d}^2 f}{{\text d} x^2} \right] (k) &= \int_0^{\infty} f'' (x)\,\sin (kx) \,{\text d}x = k\, f (0) - k^2 ℱ_s \left[ f \right] (k) . \end{align*}
Mathematica has two dedicated commands to perform sine and cosine Fourier transforms: FourierSinTransform and FourierCosTransform; however, Mathematica defines its Fourier transforms as:
\begin{align*} {\bf{\text FourierSinTransform}}[f] &= \sqrt{\frac{2}{\pi}} \int_0^{\infty} f(x) \,\sin (kx) \,{\text d}x , \\ {\bf{\text FourierCosTransform}}[f] &= \sqrt{\frac{2}{\pi}} \int_0^{\infty} f(x) \,\cos (kx) \,{\text d}x . \end{align*}
Their inverse transforms become
\begin{align*} f(x) &= \sqrt{\frac{2}{\pi}} \int_0^{\infty}{\bf{\text FourierSinTransform}}[f] (k)\, \sin (kx)\,{\text d}k , \\ f(x) &= \sqrt{\frac{2}{\pi}} \int_0^{\infty}{\bf{\text FourierCosTransform}}[f] (k)\, \cos (kx)\,{\text d}k . \end{align*}
Upon introducing two generalized (not commutative) convolution rules
\begin{align*} \left( f \overset{0}{*} g \right) (x) &= \int_0^{\infty} f(y) \left[ g(|x-y|) + g(x+y) \right] {\text d}y , \\ \left( f \overset{1}{*} g \right) (x) &= \int_0^{\infty} f(y) \left[ g(|x-y|) - g(x+y) \right] {\text d}y , \end{align*}
we find their Fourier transforms:
\begin{align*} ℱ_c \left( f \overset{0}{*} g \right) (x) &= 2\,\left( ℱ_c \,f \right) \left( ℱ_c \,g \right) , \\ ℱ_s \left( f \overset{1}{*} g \right) (x) &= 2\,\left( ℱ_s \,f \right) \left( ℱ_s \,g \right) . \end{align*}
Example A: Consider the unit rectangular function
\[ r(x) = H(x-a) - H(x-b) = \begin{cases} 1 , & \ \mbox{ if } a < x < b , \\ 0, & \ \mbox{otherwise}, \end{cases} \qquad a < b, \]
where H is the Heaviside step function. Its sine and cosine Fourier transforms are
\begin{align*} ℱ_s \left[ H (x-a) - H(x-b) \right] &= \int_a^{b} \sin (kx)\, {\text d}x = \frac{\cos (ak) - \cos (bk)}{k} , \\ ℱ_c \left[ H (x-a) - H(x-b) \right] &= \int_a^{b} \cos (kx)\, {\text d}x = \frac{\sin (bk) - \sin (ak)}{k} . \end{align*}
Integrate[Sin[k*x], {x, a, b}]
(Cos[a k] - Cos[b k])/k
Integrate[Cos[k*x], {x, a, b}]
(-Sin[a k] + Sin[b k])/k

 

Example B: The unit height tent function:
\[ f(x) = \begin{cases} x/a , & \ 0 < x < a, \\ (2a-x)/a, & \ a < x < 2a, \\ 0, & \ x > 2a. \end{cases} \]
Its sine and cosine Fourier transforms are
\begin{align*} ℱ_s \left[ f(x) \right] &= \int_0^{2a} f(x)\,\sin (kx)\,{\text d}x = \frac{1}{a\,k^2}\left[ 2\,\sin (ak) - \sin (2ak) \right] , \\ ℱ_c \left[ f(x) \right] &= \int_0^{2a} f(x)\,\cos (kx)\,{\text d}x = \frac{4}{a\,k^2}\,\cos (ak)\,\sin^2 \frac{ak}{2} . \end{align*}
Integrate[x/a*Sin[k*x], {x, 0, a}] + Integrate[(2*a - x)/a*Sin[k*x], {x, a, 2*a}]
(2 Sin[a k] - Sin[2 a k])/(a k^2)
Integrate[x/a*Cos[k*x], {x, 0, a}] + Integrate[(2*a - x)/a*Cos[k*x], {x, a, 2*a}]
(4 Cos[a k] Sin[(a k)/2]^2)/(a k^2)

 

Example C: Consider an exponential function \( f(x) = e^{-\alpha\,x^2} . \) Its sine and cosine Fourier transforms are
\begin{align*} ℱ_s \left[ e^{-\alpha\,x^2} \right] &= \int_0^{infty} e^{-\alpha\,x^2}\,\sin (kx)\,{\text d}x = \frac{1}{\sqrt{\alpha}}\,\mbox{DawnsonF} \left[ \frac{k}{2\sqrt{\alpha}} \right] ; \\ ℱ_c \left[ e^{-\alpha\,x^2} \right] &= \int_0^{\infty} e^{-\alpha\,x^2} \,\cos (kx)\,{\text d}x = \frac{\sqrt{\pi}}{2\sqrt{\alpha}}\, e^{-k^2 /(4\alpha )} , \end{align*}
where DawnsonF[x] is a special function
\[ \mbox{DawnsonF} [x] = e^{-x^2} \int_0^x e^{t^2}\,{\text d}t . \]
Integrate[Exp[-a*x^2]*Sin[k*x], {x, 0, Infinity}, Assumptions -> a > 0 && k > 0]
DawsonF[k/(2 Sqrt[a])]/Sqrt[a]
Integrate[Exp[-a*x^2]*Cos[k*x], {x, 0, Infinity}, Assumptions -> a > 0 && k > 0]
(E^(-(k^2/(4 a))) Sqrt[\[Pi]])/(2 Sqrt[a])
Example D: We apply the Fourier transformations for sinc function:
\begin{align*} ℱ_s \left[ \mbox{sinc}_a (x) \right] &= \int_0^{infty} \frac{\sin (ax)}{x} \,\sin (kx)\,{\text d}x = \frac{1}{2}\,\ln \left\vert \frac{a+k}{a-k} \right\vert ; \\ ℱ_c \left[ \mbox{sinc}_a (x) \right] &= \int_0^{\infty} \frac{\sin (ax)}{x} \,\cos (kx)\,{\text d}x = \begin{cases} \pi /2 , & \ \mbox{ if } k < a , \\ \pi /4 , & \ \mbox{ if } k = a , \\ 0 , & \ \mbox{ if } k > a. \end{cases} \end{align*}
Integrate[Sin[a*x]/x*Sin[k*x], {x, 0, Infinity}, Assumptions -> a > 0 && k > 0]
1/2 Log[(a + k)/Abs[a - k]]
Integrate[Sin[a*x]/x*Cos[k*x], {x, 0, Infinity}, Assumptions -> a > 0 && k > 0]
1/4 \[Pi] (1 + Sign[a - k])
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Example A: For a positive number a, consider the Bessel function \( J_0 (ax) \) of the first kind. Its cosine and sine Fourier transforms are
\[ ℱ_c \left[ J_0 (ax) \right] = \int_0^{\infty} J_0 (ax)\,\cos (kx)\, {\text d} x = \begin{cases} \left( a^2 - k^2 \right)^{-1/2} , & \ \mbox{ if } 0 < k < a , \\ \infty , & \ \mbox{ if } k=a, \\ 0 , & \ \mbox{ if } k > a ; \end{cases} \]
\[ ℱ_s \left[ J_0 (ax) \right] = \int_0^{\infty} J_0 (ax)\,\sin (kx)\, {\text d} x = \begin{cases} 0, & \ \mbox{ if } 0 < k < a, \\ \left( k^2 - a^2 \right)^{-1/2} , & \ \mbox{ if } k > a. \end{cases} \]
In general, we have
\[ ℱ_c \left[ J_{2n} (ax) \right] = \int_0^{\infty} J_{2n} (ax)\,\cos (kx)\,{\text d} x = \begin{cases} (-1)^n \left( a^2 - k^2 \right)^{-1/2} T_{2n} (k/a), & \ \mbox{ if } 0 < k < a , \\ \infty , & \ \mbox{ if } k=a, \\ 0 , & \ \mbox{ if } k > a ; \end{cases} \]
\[ ℱ_s \left[ J_{2n+1} (ax) \right] = \int_0^{\infty} J_{2n+1} (ax)\, \sin (kx)\,{\text d} x = \begin{cases} (-1)^n \left( a^2 - k^2 \right)^{-1/2} T_{2n+1} (k/a), & \ \mbox{ if } 0 < k < a , \\ \infty , & \ \mbox{ if } k=a, \\ 0 , & \ \mbox{ if } k > a ; \end{cases} \]
where T2n is the Chebyshev polynomial of the first kind. Another transforms:
\[ ℱ_c \left[ x^{-n} J_{2n} (ax) \right] = \int_0^{\infty} x^{-n} J_{2n} (ax)\,\cos (kx)\,{\text d} x = \begin{cases} \frac{\sqrt{\pi}}{\Gamma (n+1/2)} \left( 2a \right)^{-n} \left( a^2 - k^2 \right)^{n-1/2} , & \ \mbox{ if } 0 < k < a, \\ 0, & \ \mbox{ if } k > a , \end{cases} \]
where Γ(x) is the gamma function.

 

Example B: Let \( \displaystyle He_n (x) = (-1)^n e^{x^2 /2} \frac{{\text d}^n}{{\text d} x^n} \left( e^{-x^2 /2} \right) , \quad n=0,1,2,\ldots , \) be the Hermite polynomial. The Fourier transforms are
\[ ℱ_c \left[ e^{-x^2 /2} He_{2n} (ax) \right] = \int_0^{\infty} e^{-x^2 /2} He_{2n} (ax)\,\cos (kx)\,{\text d} x = (-1)^n \sqrt{\frac{\pi}{2}}\, e^{-x^2 /2} k^{2n} ; \]

 

Example C: Let \( \displaystyle L_n (x) = \frac{1}{n!}\, e^{x} \frac{{\text d}^n}{{\text d} x^n} \left( x^n e^{-x} \right) , \quad n=0,1,2,\ldots , \) be the Laguerre polynomial. The Fourier transforms are
\[ ℱ_c \left[ e^{-x^2 /2} L_{n} (x^2 ) \right] = \int_0^{\infty} e^{-x^2 /2} L_{n} (x^2 )\,\cos (kx)\,{\text d} x = \frac{1}{n!}\,\sqrt{\frac{\pi}{2}}\, e^{-k^2 /2} \left( He_n (k) \right)^2 . \]
\[ ℱ_s \left[ e^{-x^2 /2} L_{n}^1 (x^2 ) \right] = \int_0^{\infty} e^{-x^2 /2} L_{n}^1 (x^2 )\,\sin (kx)\,{\text d} x = \frac{1}{n!}\,\sqrt{\frac{\pi}{2}}\, e^{-k^2 /2} \, He_n (k) \, He_{n+1} (k). \]
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