0
$\begingroup$

I came across this equation: \begin{equation} e^{-a|t|}=e^{-at}u(t)+e^{at}u(-t) \end{equation} Where $u(t)$ is a step function, which is defined as \begin{equation} u(t)= \begin{cases} 1 &t\geq0\\ 0 &t<0 \end{cases} \quad \text{or} \quad u(t)= \begin{cases} 1 &t>0\\ 0 &t<0 \end{cases} \end{equation} My question is, how is the upper equation defined at $t=0$? Depending on which step function definition you use I would guess the LHS is $1$ while the RHS is $2$, or it is undefined? What am I missing here?

As a sidenote, this question arose when I came across \begin{equation} \left(\frac{1}{2}\right)^{|n|}=\left(\frac{1}{2}\right)^{n}u[n]+\left(\frac{1}{2}\right)^{-n}u[-n-1] \end{equation} which is similar, but discrete. This equation does make sense to me, as in you shift one of the step functions so that you don't end up having the coefficient twice at $n=0$.

  • 2
    It doesn't actually matter how your function behaves in a single point (in your case 0). Such a value is indeed a set with Lebesgue measure equal to 0, it wouldn't change it's Fourier transform for example, or it's norm. From such point of view such a function belongs to an equivalence class, and any function of such a class would behave in the same way. – 2017-01-24
  • 0
    @user8469759 that helps, I was thinking too graphically about this. Thanks! – 2017-01-24

0 Answers 0