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CronoDAS comments on Mathematical simplicity bias and exponential functions - Less Wrong
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Comments (82)
Any continuous function is approximately linear over a small enough scale. ;)
Which is the origin of many physical laws, since Nature usually doesn't care about scale at which nonlinear effects kick in, leaving huge areas of applicability for the laws based on linear approximation.
False. What you mean is "any differentiable function is approximately linear over a small enough scale".
See this
Heck, any linear function is approximately exponential over a small enough scale.
Do you mean "the exponential function is approximately linear over a small enough scale"?
Both are true.
Question is, what do you mean "approximately".
If you mean, for any error size, the supremum of distance between the linear approximation and the function is lower than this error for all scales smaller than a given scale, then a necessary and sufficient condition is "continuous". Differentiable is merely sufficient.
When the function is differentiable, you can make claims on how fast the error decreases asymptotically with scale.
And if you use the ArthurB definition of "approximately" (which is an excellent definition for many purposes), then a piecewise constant function would do just as well.
Indeed.
But I may have gotten "scale" wrong here. If we scale the error at the same time as we scale the part we're looking at, then differentiability is necessary and sufficient. If we're concerned about approximating the function, on a smallish part, then continuous is what we're looking for.
Indeed, you can't get a good linear approximation to that function...
Under the usual mathematical meanings of "continuous", "function" and so on, this is strictly false. See: http://en.wikipedia.org/wiki/Weierstrass_function
It might be true under some radically intuitionist interpretation (a family of philosophies I have a lot of sympathy with). For example, I believe Brouwer argued that all "functions" from "reals" to "reals" are "continuous", though he was using his own interpretation of the terms inside of quotes. However, such an interpretation should probably be explained rather than assumed. ;)
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...
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...
Yeah, you're right. I think I needed to say any analytic function, or something like that.
Mathematically he should have said "any C1 function". But if you are measuring with a tolerance level that allows a step function to be called exponential, then we can probably say that any continuous function is analytic too.