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ArthurB comments on Mathematical simplicity bias and exponential functions - Less Wrong
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No he's right. The Weierstrass function can be approximated with a piecewise linear function. It's obvious, pick N equally spaced points and join then linearly. For N big enough, you won't see the difference. It means that is is becoming infinitesimally small as N gets bigger.
that is because our eyes cannot see nowhere differentiable functions, so a "picture" of the Weierstrass function is some piecewise linear function that is used as a human-readable symbol for it.
Consider that when you look at a "picture" of the Weierstrass function and pick a point on it, you would swear to yourself that the curve happens to be "going up" at that point. Think about that for a second: the function isn't differentialble - it isn't "going" anywhere at that point!
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.
that's because you can't "see" the The Weierstrass function in the first place, because our eyes cannot see functions that are everywhere (or almost everywhere) nondifferentiable. When you look at a picture of the The Weierstrass function on google image search, you are looking at a piecewise linear approaximation of it. Hence, if you compare what you see on google image search with a piecewise linear approaximation of it, they will look the same...