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ArthurB comments on Mathematical simplicity bias and exponential functions - Less Wrong
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That is because they are approximated by piecewise linear functions.
It means on any point you can't make a linear approximation whose precision increases like the inverse of the scale, it doesn't mean you can't approximate.
taboo "approximate" and restate.
I defined approximate in an other comment.
Approximate around x : for every epsilon > 0, there is a neighborhood of x over which the absolute difference between the approximation and the approximation function is always lower than epsilon.
Adding a slop to a small segment doesn't help or hurt the ability to make a local approximation, so continuous is both sufficient and necessary.
ok, but with this definition of "approximate", a piecewise linear function with finitely many pieces cannot approximate the Weierstrass function.
Furthermore, two nonidentical functions f and g cannot approximate each other. Just choose, for a given x, epsilon less than f(x) and g(x); then no matter how small your neighbourhood is, |f(x) - g(x)| > epsilon.
The original question is whether a continuous function can be approximated by a linear function at a small enough scale. The answer is yes.
If you want the error to decrease linearly with scale, then continuous is not sufficient of course.
I think we have just established that the answer is no... for the definition of "approximate" that you gave...
Hum no you haven't. The approximation depends on the scale of course.