Question

6. The Heaviside step function is piecewise-defined as $H(x) = egin{cases} 0, & x < 0, \ 1, & x ge 0. end{cases}$ (a) Graph $H(x)$. (b) Graph $H(x - 1)$. (c) Use the integral definition of the Fourier transform and the trigonometric/exponential relation given by $sin(x) = frac{1}{2i} [e^{ix} - e^{-ix}].$ to confirm that $mathcal{F}[H(x+1) - H(x-1)] = frac{2sin(xi)}{xi}.$ (d) Graph $H(b - x)$ where $b =$ your guess as to my age in years. (e) Suppose $u(t) = sin(at)$ where $a =$ your age in years, and $u(x,t) = H(t - x)u(t - x)$. Use the definition of the Heaviside function to write down the piecewise-defined step solution $u(x, t)$.

          6. The Heaviside step function is piecewise-defined as
$H(x) = egin{cases} 0, & x < 0, \ 1, & x ge 0. end{cases}$
(a) Graph $H(x)$.
(b) Graph $H(x - 1)$.
(c) Use the integral definition of the Fourier transform and the trigonometric/exponential
relation given by
$sin(x) = frac{1}{2i} [e^{ix} - e^{-ix}].$
to confirm that
$mathcal{F}[H(x+1) - H(x-1)] = frac{2sin(xi)}{xi}.$
(d) Graph $H(b - x)$ where $b =$ your guess as to my age in years.
(e) Suppose $u(t) = sin(at)$ where $a =$ your age in years, and $u(x,t) = H(t - x)u(t - x)$.
Use the definition of the Heaviside function to write down the piecewise-defined step
solution $u(x, t)$.

$6. The Heaviside step function is piecewise-defined as H(x) = egincases 0, x < 0, 1, x ge 0. endcases (a) Graph H(x). (b) Graph H(x - 1). (c) Use the integral definition of the Fourier transform and the trigonometric/exponential relation given by sin(x) = frac12i [e^ix - e^-ix]. to confirm that mathcalF[H(x+1) - H(x-1)] = frac2sin(xi)xi. (d) Graph H(b - x) where b = your guess as to my age in years. (e) Suppose u(t) = sin(at) where a = your age in years, and u(x,t) = H(t - x)u(t - x). Use the definition of the Heaviside function to write down the piecewise-defined step solution u(x, t).$

Added by John M.

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