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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 1: We consider a step function (which is actually a difference of two Heaviside functions) \[ \chi_{[0, R]} (x) = H(x) - H(x - R) = \begin{cases} 1 , & \quad \mbox{if } 0 < x < R , \\ 0 , & \quad \mbox{if } x > R. \end{cases} \] We did not define the values of the step function at points x = 0 and x = R of discontinuity for two reasons. First of all, any integral transformation is not sensetive for values of the function at the discrete number of points. Second, we want to check these values with the inverse Fourier-sine transform.

The Fourier sine transform of step function is \[ f^s = \int_0^R \sin kx\,{\text d} x = \frac{1 - \cos (kR)}{k} . \]

Integrate[Sin[k*x], {x, 0, R}]
(1 - Cos[k R])/k
Its inverse transform is \[ ℱ^{-1}_s \left[ f^s \right] = \frac{2}{\pi} \int_0^{\infty} \frac{1 - \cos (kR)}{k} \,\sin (kx)\,{\text d}k , \]
Integrate[(1 - Cos[k R])/k*Sin[k*x], {k, 0, Infinity}]*2/Pi
1/2 ((R - x)/Sqrt[(R - x)^2] + (2 x)/Sqrt[x^2] - (R + x)/ Sqrt[(R + x)^2])
which is a step function \[ \chi_{[0, R]} (x) = \begin{cases} 0 , & \quad \mbox{if } x = 0 , \\ 1, & \quad \mbox{when } 0 < x < R , \\ ½ , & \quad \mbox{if } x = R , \\ , 0 , & \quad \mbox{when } R < c < \infty . \end{cases} \] We check with Mathematica the values of the inverse sine Fourier transform at point x = R, which is ½, as expected:
Integrate[(1 - Cos[k R])/k*Sin[k*R], {k, 0, Infinity}]*2/Pi
R/(2 Sqrt[R^2])
Application of Fourier-sine transform becomes 1-to-1 (bijective) when its domain includes functions that vanish at the origin (x = 0) and at other points satisfy the Dirichlet conditions.

The Dirichlet conditions are a set of sufficient conditions that guarantee a periodic function has a convergent Fourier series. These conditions are: a function must be absolutely integrable over a period, have bounded variation (a finite number of maxima and minima), and have a finite number of finite discontinuities within that period. If a function meets these criteria, its Fourier series will converge to the function's value at points of continuity and to the average of its left and right limits at points of discontinuity.    ■

End of Example 1
   

 

 

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