0

I have the function $f(t)=e^{-|t|}$

And I want to distribution derivate it to $f''(t)$.

I am aware of that the $f'(t)$ function will be:

enter image description here

But how do I derivate to $f''(t)$ ?

Elias Costa
  • 14,658
user1223333312
  • 27

2 Answers2

2

In $t=0$ you have a jump of size $-2$ so you need to add $-2\delta$ to the derivative computed pointwise.

Marcin Malogrosz
  • 1,960
1

Observe that there is a jump discontinuity in $f'$ at $0$, and that this is the total extent of the singular support. Outside of zero $f, f',f'' ...$ will all simply be successive smooth derivatives. So we just need to find it at zero. Jumps tend to correspond to Dirac-delta-like masses in the derivative, more rigorously, if we let $H(x)$ be the Heaviside function (H(x) = 0 for $x \leq 0$, H(x) = 1 for $x > 0$) then

$$f' - 2H = g$$ $$f' = g + 2H$$

Where $g$ is continuously differentiable. So $f'' = g' + 2H'$. If you know what the distributional derivative of the Heaviside is (and this is one of the first things they usually mention) you're done.

jxnh
  • 5,228