[umanwizard]

A curve with the properties you describe is not smooth, by definition, since it is not differentiable.

> At whatever scale you look at the curve the derivative is always wrong: you zoom in on the sine, at the peak it's got a cosine, so the d/dx is -1; but zoom in and the cosine has a sine at the crossing point, so the d/dx is 0; but zoom in and ...

This is pretty much the definition of something not being differentiable. "Differentiable" means that the approximations to the derivative (i.e., difference quotients) converge to some fixed value as the scale they're measured at approaches the infinitely small.

You might be interested in the Weierstrass function, which seems to be the sort of thing you're getting at with your idea: https://en.wikipedia.org/wiki/Weierstrass_function . Continuous everywhere, but differentiable nowhere.

Edit: the specific function you wrote down is not differentiable (at least not everywhere). For example, at x=0, its derivative, if it had one, should be cos(0) + cos (a * 0) + cos (a^2 * 0) + ... , but that series clearly diverges.

[pbhjpbhj]

cos(0) being 1, sin (0) begin 0; that series is 0 + 10e-3 + 0 + 10e-9 + 0 + 10e-15 + ... when x=0 (excuse my sloppy notation, I'm on a phone). Looks convergent to me, somewhere between 0.001000001000001 and 0.001000001000002 ?

Have you studied fractal dimensions formally, might I know the background you're speaking from?

Yes, I'm imagining, as a first example, something akin to a Weierstrass function but without the discontinuities.

https://www.desmos.com/calculator/c6tbl4zr9j

[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.