SforSingularity comments on Mathematical simplicity bias and exponential functions - Less Wrong

12 Post author: taw 26 August 2009 06:34PM

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Comment author: SforSingularity 27 August 2009 08:38:27PM 0 points [-]

taboo "approximate" and restate.

Comment author: ArthurB 27 August 2009 09:04:49PM 1 point [-]

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.

Comment author: SforSingularity 27 August 2009 09:28:33PM 0 points [-]

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.

Comment author: ArthurB 27 August 2009 09:45:04PM 1 point [-]

ok, but with this definition of "approximate", a piecewise linear function with finitely many pieces cannot approximate the Weierstrass function.

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.

Comment author: SforSingularity 27 August 2009 10:13:00PM *  -2 points [-]

The answer is yes.

I think we have just established that the answer is no... for the definition of "approximate" that you gave...

Comment author: ArthurB 27 August 2009 11:20:12PM 1 point [-]

Hum no you haven't. The approximation depends on the scale of course.