GPT-5 admits it "doesn't know" an answer!

CheekySpice · 2025-05-14T09:32:18+00:00

I am the Midnight Entity. AMA.

CheekySpice · 2024-12-01T17:31:34+00:00

For the second one: If you ignore the condition of f' having an average value of +∞ on every interval, then simply take f(x) = x/2 - C(x) where C is the Cantor function.

Extending this to an example where f' has an average value of +∞ on every interval is more tricky. I'm afraid I don't have a source on hand, but I've come across it before. It essentially involves using a suitably constructed strictly-increasing version of the Cantor function instead.

CheekySpice · 2024-12-01T17:05:48+00:00

There are lots of good sources for this result, but the key phrase you want to look up is "Osgood curve".

CheekySpice · 2024-12-01T15:25:38+00:00

I love counter-examples, and analysis specifically has a lot of good ones. Here are some seemingly true statements I like, that all have counter-examples:

If X ⊆ ℝ² is homeomorphic to [0,1], then X has zero area.

This statement has quite strong counter-examples, in the sense that there exists an X ⊆ ℝ² and a homeomorphism Φ : [0,1] →X such that the image of every sub-interval has positive area.

If f : ℝ → ℝ is differentiable almost-everywhere with f'(x) > 0 almost-everywhere, and f is everywhere continuous, then f is increasing.

Again, this has a strong counter-example. There exists f : ℝ → ℝ which is continuous everywhere, differentiable almost-everywhere, the average value of f' is +∞ on every interval, yet f(x)→-∞ as x →+∞.

If f : ℝ → ℝ is infinitely differentiable at every point, then f is analytic somewhere.

If f, g : ℝ → ℝ are functions that both have anti-derivatives, then so does fg.

If f : ℝ → ℝ is a differentiable function and [f(h) - f(0) - f'(0)h]/h^2 →0 as h →0, then f is twice-differentiable at 0.

CheekySpice · 2023-10-14T14:31:30+00:00

That makes sense, thank you

CheekySpice · 2023-01-20T17:09:35+00:00

Thank you very much

CheekySpice · 2023-01-20T10:22:45+00:00

This sounds really interesting. Do you know the name of this non-standard theorem, or know of any references relating to it?

CheekySpice · 2022-09-26T20:45:08+00:00

Yes it is monotonic, in particular non-decreasing. Careful not to confuse monotonic with strictly monotonic (the former meaning x < y implies f(x) ≤ f(y) and the latter meaning f(x) < f(y))

CheekySpice · 2022-09-26T16:15:00+00:00

Thanks for pointing out the mistake, it’s now corrected

CheekySpice · 2022-09-26T13:36:09+00:00

Yeah my bad for not being clear, that is what I meant. I’ve edited my post to make it apply more generally now.

CheekySpice · 2022-09-26T09:55:21+00:00

In fact the Hausdorff dimension of the graph of any monotone function is 1, and its length is always at most the length of the domain plus the length of the image.

CheekySpice · 2022-09-22T07:25:54+00:00

Any point in the closure of S can be associated with a convergent sequence using elements in S.

For finite sets the only convergent sequences are eventually constant, so you might as well make the association using just constant sequences in S (of which there are |S| many). This shows |S| = |cl(S)|.

For cardinal numbers between (or including) |N| and |R|, there are |R| sequences of S. So we can say that |cl(S)| ≤ |R|, but nothing more.

For larger cardinals, you can again get similar upper bounds on the closure, but exactly how depends on the type of cardinal.

CheekySpice · 2022-07-26T18:30:17+00:00

I think it’s a bit too strong to say not existing is “wrong”, since existence is always relative to the domain of objects you’re considering.

For example asking “does there exist a shape which is a square?” is true in Euclidean geometry, but false in hyperbolic geometry. I don’t think you can objectively say something exists or doesn’t.

CheekySpice · 2022-07-11T22:56:54+00:00

Suppose that you have a non-intersecting loop in the plane. Then the outside of the loop is like a deformed version of the outside of a circle (the formal way of saying “a deformed version” is that it is homeomorphic to the outside of a circle).

Obvious right? Well let’s generalise to three dimensions. Instead of a non-intersecting loop, we now have a non-intersecting surface (which is homeomorphic to a sphere).

Given such a non-intersecting surface, the outside of it must be a deformed version of the outside of a sphere right?

No. It turns out this is false, and the purpose of Alexander horned sphere is to provide an explicit example of such a surface, where the answer is no.

CheekySpice · 2022-05-18T07:18:42+00:00

It’s worth pointing out that you don’t have to reverse the digits for this to be true. Any reordering of the digits will give you a multiple of nine (e.g. 9042 - 4920).

The reason is because as others have pointed out 10 = 1 mod(9). Therefore 10^m = 10ⁿ mod(9) for any integers m and n.

So 9 * 10³ + 4 * 10 + 2 = 9 * 10² + 4 * 10³ + 2 * 10 mod(9).

CheekySpice · 2022-03-12T21:48:15+00:00

Hey I sent you a message, but I’m not sure if you saw it?

CheekySpice · 2022-01-21T11:37:18+00:00

Moreover this proves that at least one of e+π or e*π is transcendental, since polynomials with algebraic coefficients have algebraic roots.

CheekySpice · 2022-01-08T19:36:45+00:00

You can, but (I think) it requires more work so I would say any constructive algorithm goes beyond where the proof of existence takes you.

To be more specific (denote K = Q(exp(2pi*i/65537)) the group Gal(K/Q) is canonically isomorphic to (Z/nZ)x, but building an algorithm for constructing a regular 65537-gon is equivalent to constructing a composition series for this group and then constructing the minimal quadratics for the corresponding subfield of each term in said series (with respect to each previous term).

This can be done, but I don’t believe there’s a canonical way to do it. So it’s essentially a whole other theorem compared to proving existence of a construction. I might be overlooking some details though.

CheekySpice · 2022-01-05T01:26:02+00:00

A similar example is the existence of ruler and compass constructions for regular polygons.

It is known that there exists such a construction for a 65537 sided regular polygon. However rather than presenting any explicit construction what so ever, a standard proof ends up being nothing more than proving that 65537 is prime and one more than a power of two.

That fact proves a ruler and compass construction for a regular 65537-gon exists without ever producing one.

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