This sample post demonstrates inline and display equations rendered with MathJax. For example, Euler’s identity is (e^{i\pi} + 1 = 0), and the Gaussian integral is
\[\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}.\]Fourier transform
Using angular frequency (\omega), define the Fourier transform and its inverse by
\[\mathcal{F}\{f\}(\omega) = \widehat{f}(\omega) = \int_{-\infty}^{\infty} f(t)e^{-i\omega t}\,dt, \qquad f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \widehat{f}(\omega)e^{i\omega t}\,d\omega.\]As an example, the transform of a Gaussian is another Gaussian:
\[f(t) = e^{-t^2/(2\sigma^2)} \quad\Longrightarrow\quad \widehat{f}(\omega) = \sigma\sqrt{2\pi}\,e^{-\sigma^2\omega^2/2}, \qquad \sigma > 0.\]Differentiation in time becomes multiplication by (i\omega) in the frequency domain:
\[\mathcal{F}\{f^{(n)}\}(\omega) =(i\omega)^n\widehat{f}(\omega).\]The convolution theorem turns convolution into pointwise multiplication. With ((f * g)(t) = \int_{-\infty}^{\infty} f(\tau)g(t-\tau)\,d\tau), it reads
\[\mathcal{F}\{f*g\}(\omega) = \widehat{f}(\omega)\widehat{g}(\omega).\]Laplace transform
For a causal signal, the one-sided Laplace transform is
\[\mathcal{L}\{f\}(s) = F(s) = \int_0^\infty f(t)e^{-st}\,dt.\]For (f(t)=e^{-at}), where (a>0), direct integration gives
\[\mathcal{L}\{e^{-at}\}(s) = \int_0^\infty e^{-(s+a)t}\,dt = \frac{1}{s+a}, \qquad \operatorname{Re}(s)>-a.\]The transform of a derivative also records the initial value:
\[\mathcal{L}\{f'\}(s) = sF(s)-f(0^+).\]This is why Laplace transforms are useful for initial-value problems. For example, transforming (y’(t)+ay(t)=0), with (y(0)=y_0), gives
\[sY(s)-y_0+aY(s)=0 \quad\Longrightarrow\quad Y(s)=\frac{y_0}{s+a} \quad\Longrightarrow\quad y(t)=y_0e^{-at}.\]A transform pair in action
The Fourier transform can turn a differential equation into an algebraic one. For the heat equation
\[\frac{\partial u}{\partial t} = \kappa\frac{\partial^2 u}{\partial x^2}, \qquad \kappa > 0,\]transform in (x), using frequency (\xi). Since (\mathcal{F}_x{\partial_x^2 u}=-\xi^2\widehat{u}), the equation becomes
\[\frac{\partial \widehat{u}}{\partial t} = -\kappa\xi^2\widehat{u}.\]Solving this ordinary differential equation in time yields
\[\widehat{u}(\xi,t) = \widehat{u}(\xi,0)e^{-\kappa\xi^2t}.\]The exponential factor damps high frequencies most rapidly, which is the frequency-domain view of heat smoothing out sharp variations.