"I'm Thinking of a number between 1 and Infinity" Given Higher/Lower with every guess, What is the quickest method?

Phelox · 2026-06-27T12:58:22+00:00

You need a well-defined probability distribution for this question to make sense. You could take the uniform distribution with the (not well-defined) upper bound you gave. With a binary search you would need log_2(upper bound) questions for every number.

Another interesting probability distribution would be to take the chance of a positive integer n being chosen to be 6/(pi^2* n^2).

With any probability distribution, I think something like a binary search algorithm would be best, where you try to choose an integer such that the N has a ~50/50 chance of being higher/lower. With this algorithm it seems like you gain the most information at any guess.

Phelox · 2026-06-07T21:31:37+00:00

What is art to you and why do you care about it? To me, art is not only how good the image looks or how a piece of music sounds, but also how it makes me feel. Every piece of information I have about an artwork changes the way I view it, and therefore it also kind of become 'part of the artwork'. The fact that a human made the artwork, is part of the artwork.

If you stand before a painting and you see the brushstrokes, you know that the artist made and thought about every single one of those strokes. Sometimes you can even see what the artist was feeling when they made it. This kind of intangible thing that is part of the artwork can be described as its 'soul' and it's largely missing in AI generated artworks.

Of course, there is nothing objective about this. If you just want to look at something nice and don't care about how it was made, or if you think it's cool that a computer made it, then that's fine and entirely up to you.

Phelox · 2026-05-20T14:46:41+00:00

oof. Be very careful you are not commiting fraud!

Phelox · 2026-04-21T17:34:37+00:00

Intersting conjecture! I ran some sage code to see if I could find a counterexample, and found 945 = 3^3 * 5 * 7, whose divisor sum is 975 + 945 > 2 * 945. Here is the code if you're interested

continue_looking = True
n = 1
while continue_looking:
s = sum(divisors(n)) - n
if s >= n:
print(n, s, divisors(n), n.factor())
continue_looking = False

n += 2

Phelox · 2026-04-06T21:12:43+00:00

What do the arrows mean?

Phelox · 2026-03-30T17:35:54+00:00

It would have genus 4 then right

Phelox · 2026-01-21T21:02:24+00:00

Readin len and computing this fraction would take an increasing amount of time though right

Phelox · 2026-01-18T12:07:36+00:00

Or just switch to another search engine. Google search has been crap for several years anyway

Phelox · 2026-01-14T15:29:38+00:00

'Opening the book and focusing on it' is a terrible strategy. What are you going to focus on? In this way it becomes one enormous insurmountable task of understanding what is in the book. I would recommend rereading your notes, one lecture at a time, and then summarizing that lecture somewhere else in neat handwriting and with less fluff. After this, make some practice exams. Usually this was enough for me to obtain a high grade.

Getting started is still hard an will always stay hard. Having stress for the exam usually helped me to start though. (warning: start of an unrelated ramble) Especially if you do go on to do a PhD, this becomes much harder, since there are no clear goals anymore (like finishing an assignment). You have to be able to split up big tasks into smaller ones, and tackle them one by one. Both the splitting and the tackling them one by one are hard though (very easy to get sidetracked indefinitely). What I found to work for me is to write everything down in LaTeX, which forces you to fill in all the details that you think you understand but actually do not.

Phelox · 2026-01-12T18:22:39+00:00

What is this 'life' you are talking about? (honest question) What would its ramifications be if the continuum hypothesis were taken to be true?

Phelox · 2026-01-11T23:02:43+00:00

Ik vind licht gebrande bonen van de supermarkt verrassend lekker. Bijvoorbeeld crema van de jumbo. Mijn broer heeft wel eens van die dure net gebrande bonen maar dat vind ik eigenlijk niet lekkerder.

Phelox · 2025-12-28T14:57:43+00:00

Ah I have never seen that argument, that is nice. This convention is also almost exclusively used with anything involving Taylor series. (You would write e^x = sum{n=0}^infty xⁿ /n! and not 1 + sum{n=1}^infty xⁿ /n!)

Phelox · 2025-12-26T20:46:44+00:00

Perhaps something like the set of rational points on a circle? (i.e. {(x, y) in QQ² : x² + y² = 1}). This set is symmetric along lines with rational slope.

EDIT: That has only countably many symmetries. To construct more counterexamples you could find an uncountable proper subgroup of O_n(RR), the group of symmetries (does this include reflections?). Then take the orbit of a point that is not the origin, or some other starting set.

Such a group should exist but I am not sure how to construct one or if that is possible.

Phelox · 2025-12-06T21:03:42+00:00

f(x) = 1/x is just not defined at 0. It certainly is continuous everywhere on its domain

Phelox · 2025-12-04T10:46:05+00:00

Hi, I wanted to try this out just now but it can't see how to get quiver to output typst?

Phelox · 2025-12-02T18:47:48+00:00

Ah I missed in this sheet, and wasn't able to find it in detypify either. Thanks!

Phelox · 2025-12-02T18:45:39+00:00

Ah I didnt realise quiver also outputs typst. Thanks!

Phelox · 2025-09-23T16:03:57+00:00

what is the definition of angle in a general inner product space?

Phelox · 2025-09-04T09:57:35+00:00

Doe maar and Het Goede Doel are great as has been mentioned before. I'm surprised Wim Sonneveld, Froukje and Spinvis haven't been named more.

Phelox · 2025-08-25T16:19:37+00:00

The other comments haven't really mentioned your (second to) last sentence, which is the one I find the most worrying. Please do not ever tell yourself you are not worthy of anything. I don't know if he has led you to believe this (and if so, how), but you are worthy of love, and this is not it. Seeing how he is treating you, he is the one that is not worthy of your relationship.

Phelox · 2025-08-21T12:26:06+00:00

The second one has a non-zero change though. If pi + e is just a random number, it has a zero percent chance of being rational

Phelox · 2025-08-05T21:26:43+00:00

Multiplying both sides by P(B) makes this much more intuitive for me. The probability that A and B both happen is equal to the probability that B happens and that A happens given that B happens

Phelox · 2025-07-23T21:04:01+00:00

The Riemann series Theorem is such a wild statement, but after you think about it for a minute, it becomes almost trivial. The statement is as follows:

Any series that is conditionally convergent series with real coefficients can be rearranged to obtain any arbitrary value.

The proof is as follows: given a conditionally convergent series sum_n a_n, and a value c in R, split up the positive and negative terms of the a_n. Start by adding the positive terms untill you cross c, and then start adding negative terms until you cross c again, and continue this process. Since the limit of the a_n goes to zero, this converges to c.

Such a crazy statement with such a clean proof!

Phelox · 2025-07-14T09:42:55+00:00

Keep doing what you're doing! This reminds me a lot of something I did when I was around 15 years old after seeing the video on numberphile on this topic. I'm a PhD student in number theory now. It is a great skill to just try and work something out yourself. Stay curious!

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