If analytically, a function is not differentiable at a point, does it make sense to write a finite difference code for the function at that point?

Question

What would happen if I wrote a finite-difference code evaluated at a point where the function isn't differentiable analytically?

I'm trying to think analytically vs numerically.

Thanks,

Answer 1

You lose convergence order, or in the worst case convergence altogether. You can try this out: Take $$ f(x) = \begin{cases} 0 & \text{if $x<0$} \\ x^2 & \text{if $x\ge 0$}.\end{cases} $$ The function is differentiable, but not twice differentiable at $x=0$. It's exact derivative is $f'(0)=0$. Now compute the (second-order) symmetric finite difference approximation of this function: $$ D_h^\pm f(0) = \frac{f(0+h)-f(0-h)}{h} = h. $$ So for $h\to 0$, you have that $D_h^\pm f(0) \to f'(0)$, but only at a rate of $O(h)$, not $O(h^2)$ as otherwise expected. So the fact that $f$ is not twice differentiable means that you just lost an order of convergence.