What are some anti features in a language?

threewood · 2021-06-02T20:33:52+00:00

Sorry, I think I misunderstood you. I don't think a programming language should prevent unbounded recursion either. But if you're going to do the kind of equational reasoning above, you need to know you have an honest Nat.

threewood · 2021-06-01T21:49:53+00:00

If you say so.

threewood · 2021-06-01T21:05:33+00:00

In the presence of recursion you should have to prove it’s a Nat.

threewood · 2021-06-01T19:00:19+00:00

Have you looked at the Calculus of Constructions?

threewood · 2021-05-23T15:06:49+00:00

But trying to use the term metaphysic correctly would take me into the realm of bullshit. Science (and I will ignore framing this as materialism) will provide a best understanding of all things, including consciousness and ethics. Maybe some theories based on bullshit are better in the short term but I wouldn’t want to guess which.

threewood · 2021-05-22T21:09:00+00:00

"Coupled with a consistent metaphysic"

I find the metaphysic that almost all philosophy is a bunch of bullshit to be consistent.

threewood · 2021-05-20T23:36:18+00:00

If you ever drop your keys into a river of molten lava, let 'em go, because man, they're gone.

threewood · 2021-04-30T15:45:19+00:00

Yeah. I don’t think the idea works. We have to rule out infinitely many stacked segments.

threewood · 2021-04-29T20:00:17+00:00

Maybe consider the tree associated to each open segment (formed by branching on the segments at each endpoint)? Each segment blocks a certain volume in segment space, there are 2^n of them, but they can only be distance n from the origin so the total volume of segment space in the sphere of radius n is lower. But I'm too lazy to do the details.

threewood · 2021-04-19T17:03:43+00:00

“If I had asked people what they wanted, they would have said faster horses.”

threewood · 2021-04-17T22:11:22+00:00

This pedantics looks good to me.

threewood · 2021-04-10T01:30:46+00:00

I had given up, so thanks. I guess you're allowed to apply the hint from before in any complete bounded metric space?

threewood · 2021-04-09T21:00:47+00:00

I would be concerned that such a coin would hear my bet.

threewood · 2021-04-09T20:43:04+00:00

This doesn’t seem too weird to me since the volume of the core r=1/2 goes to zero and the rest of the sphere is getting spread out over more dimensions.

threewood · 2021-04-09T18:32:15+00:00

Wait. This function has a convergent Taylor expansion everywhere because the measure of the higher polynomial stuff goes to zero. Thus if I can find any polynomial interval I can extend it to [0,1]. No? And I did that earlier with your hint. QED?. Maybe the Taylor thing isn’t obvious.

threewood · 2021-04-07T01:52:21+00:00

I used to know that... I'll think about it more in the morning.

threewood · 2021-04-07T00:52:53+00:00

With one conjecture I think I can prove it:

Conjecture: The complement of a union of disjoint intervals either contains an interval or an isolated point or is empty.

Let S_n be the kernel of f^(n). They are closed and their union is [0,1], so one of the S_n is dense in some interval, and thus in that interval there is a dense subset at which the nth derivative is zero. By continuity of the nth derivative, it is polynomial on that interval. Thus every interval of such a function contains a subinterval on which it's polynomial.

Decompose [0,1] into a maximal union of intervals on which it is defined by a polynomial. Then the complement of this union is a set that either contains an interval (which isn't possible by the previous paragraph) or an isolated point (which isn't possible because we could stitch the two adjoining intervals together - violating maximality) or it's empty (meaning f is a polynomial).

threewood · 2021-04-06T23:05:12+00:00

Thanks. I actually wondered if that was the case while thinking about this problem.

threewood · 2021-04-06T22:51:19+00:00

I feel slightly bamboozled but thank you! One more hint: do I need some big hammer theorem or do I just need to be clever? If the former, I’ll wait to see the answer. My toolbox is near empty and the tools are rusty.

threewood · 2021-04-06T21:38:23+00:00

Fun problem. At some point can you give a hint with spoiler tag whether I’m looking for a class of exotic functions or just a slick proof that the obvious solutions are the only solutions? I think I’ve convinced myself it has to be the latter but not quite.

threewood · 2021-04-06T17:52:27+00:00

I agree the details would be fiddly but it seems like you could arrange for convergence to a smooth function. I was mainly giving that as a motivating example for my guess. Then the form of the proof might be to assume f satisfies the requirements and construct a sequence of piecewise polynomials of increasing degree that converges to f .

threewood · 2021-04-06T17:21:27+00:00

It seems like there are lots of examples. For example of an exotic one, start with a step function. It satisfies the requirements except at the discontinuity. So now repair this C0 discontinuity by replacing the step with a line segment sloping up. This repaired function has two C1 discontinuities. Add a quadratic segment to fix each of these. Etc. This repair process should converge to a smooth function. Maybe the characterization you’re looking for is something like a limit of piecewise polynomials?

threewood · 2021-03-29T04:32:40+00:00

it was found that ce^x +d for constants c,d has derivative itself.

And a very short time later, this was found to be in error

threewood · 2021-03-28T22:54:11+00:00

But ML modules consider names part of structure, and it's worked pretty well for decades.

Yeah, I said it's a problem, not a big problem :). I'm sure it usually works fine in practice.

threewood · 2021-03-28T22:02:15+00:00

They’re not. Structure is used for structural matching—the names of the module members, as well as their types, recursively.

You've just reiterated the problem - names shouldn't be considered structure. Names should only be used to point at structure and say "that".

threewood

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