[umanwizard]

Yes, the original function converges (not its derivative). The terms in the derivative are no longer divided by a^k. The a^k from the argument to cos/sin cancels then out (it becomes a multiplier, because of the chain rule).

So the series for the derivative is 0 + 1 + 0 + 1 + ..., which doesn’t converge.

Not sure what this has to do with fractal dimensions. This is a simple question of definitions. The word “smooth”, in math, literally implies, by definition, that the function is everywhere approximable by lines.

If you don’t think it does, can you state the formal definition of “smooth” that you’re using?

The Weierstrass function doesn’t have discontinuities, btw.