You need to prove that this improper integral is finite. Otherwise the solution is not valid.

Thanks for this video. By the way, are you Korean? I sense a Korean accent...

I have my doubts about determining uncountable amount of equivalence classes. For each prisoner we may get 1 class, so, at the end, we wont be able to cover all the classes. This problem would work for rational numbers though

...a favorite # of late is 1103011 because: a) It has 7 digits, b) it is a palindrome, c) it is a multiple of 7, and d) the sum of digits is 7.

thanks a lot for rewvising my concepts again

thank you for this..good revision

Thank u so much sir ❤🙏🏻

I wanted to listen but this guy can't even speak english...

Isnt this wrong? 15:00 you can only apply power rule to constants not functions it cancels anyway so the answer is still not wrong.

81+18+01=00

... plus a constant

Amazing thing about the factors 5, 13, 37, 109 is that they are four consecutive values of the formula 4*3^n+1, from n=0 to n=3

is this supposed to be easier lol, i dunno man

Thanks for the work mate, God bless!!!

Here is a similar problem: There is a house with 100 rooms. Each room contains an infinite collection of boxes, indexed with the natural numbers. Each box contains a random real number, which is the same over all the rooms - that is, for every natural number n, box n contains the same real number in every room. In this house, 100 set theorists play a game. Each player goes into a unique room and opens as many boxes as they like (perhaps infinitely many) as long as they leave at least one box in their room unopened. Then, each player picks an unopened box in their room and guesses what real number is inside of it. In order to win the game, at least 99 players must guess correctly. The players can discuss a strategy beforehand, but after they go into their respective rooms, no more communication is allowed. Design a winning strategy.

Crazy!!! Great Job

BUT PLEASE IF YOUNHAVE NEVER HEARSOF ROOTS OF.UNITY then surely you can solve.WITHOUT TBIS..AND YOU flrgot tomfactor oit the 3 from.tje 3n factorial.in the denominator didnt you? And can't you solve this without knowing roots of unity??

Whybisnt double facptiral.just equal to the FACTORIAL.OF.A FACTORIAL..OBVIOUSLY LOGICALLY.THAT SHOULD BE WHAT IT MEANS..Doesnt everyone else.agree??

I don't think this solution works, but feel free to correct me. Because yes, each person agrees on the same equivalence class of sequences. But each person is also at a finite position when there's a countable number of people. So as you can imagine, there's no way the first 5 people gets the right number. But because this set of sequesnces is uncountable, this will be the case for every single person. Everyone is at a finite point, and part of "only the first n people" which will for sure get it wrong, and therefore everyone will get it wrong. Because the number of people is countable, the number of people who get it right is given by the limit of n -> infinity of how many get it right for the first n people, which is 0 for all n. So the answer is 0 people get it right. Where does my logic fail?

Possibly the least convoluted example of the Axiom of Choice Had some major alarm bells ringing at 7:04

🤯🤯🤯…

Please show that there is some way that allows a prisoner to determine which equivalence class any observed sequence of numbers belongs to. (Seems like one would need some kind of axiom-of-choice-like assumption that I am not sure can be granted.)

The neerest thing i got to a solution is this: Suppose that you only allow each prisoner to display a single bit of information, by either standing up or sitting down. Here's a procidure that allows them all to know their number: First, the prisoners agree to map all the real numbers to the interval (0,1) (by way of an s curve for example). They will later reverse the map when they make the guess. Let p(n) denote the bit displayed by prisoner n. The prisoners agree to the following: p(1)=1 ( or 0, does'nt matter) p(n)=(sum from i=1 to n-1 of the (n-i)th binary digit of prisoner i's hat) all mod 2 This way every prisoner can deduce every digit of their hat, by subtracting away the effect of all the prisoners with an index smaller than the priaoner holding the rellevant digit. They can do that because they can see all the other hats.

Maybe you could reformulate the banach-tarski paradox in terms of the prisoner hat problem. Suppose each number on the prisoners hat were assigned acording to a uniform uncorralated distribution on the interval [0,1]. Since the numbers are uncorralated, any strategy based on the other hats will still have probability zero. Therefore the probability of all the prisoners guessing correctly in an arbitrary sequence of prisoners is 0×0×0×0...=0. The probability that any sequence containing all but a finite number is 0+0+0+0...=0. I think this argument fails because an arbitrary set of sequences of numbers [0,1] is not generally measurable.

Does each person know whether they are person n? Or is their order randomized? Because if the order is randomized then I don’t think this solution works; as each person might try to fill in the role of the same swapped element of the sequence

What is unclear to me is how a countable number of people can go through an uncountable number of equivalence classes in this solution you presented?

Side comment: I always found the name "countable" so annoying. It should be called "enumerable".

Euler substitution can work too Sir. ❤

Nah man. Long before you get to a hat with graham number on it, simply the information density of writing the number on the hat would collapse the prisoner into a black hole.

I think there is a way to guarantee that at the most one will be incorrect as long as the prisoners know the order they will be picked. The way is for the first prisoner to use the function f(n) = 1 / (n^2 + |h(n)|) where n is the number of the prisoner and h(n) is the number on their hat on each of the prisoners, and then multiply the result by -1 if h(n) < 0, then sum up all of the results, I'm pretty sure that the sum will converge and now every prisoner will be able to calculate their hat number like the solution to the finite number of prisoners.

As a computer scientist, I think the reason this makes me so uncomfortable is that the steps are presented as an algorithm, but each step involves a computation of infinite data that will take infinite time (uncountably infinite time to itemize all the equivalence classes beforehand, and then infinite counting to inspect all the hats and determine their class). Even in a theoretical land of pure math and infinite time, could the prisoners really distinguish between sets of numbers that differ in arbitrarily many ways, and are thereby equivalent, vs those that differ in truly infinite ways?

I understand what you are talking about. However, you did a poor job of explaining what numbers get assigned to each person. You made it sound like any random real number could be assigned in the beginning of the video. If that was the case, with countably infinite people, then the likely number of correct guesses will be 0. This is because every number comes from the uncountable infinite set of the Reals.

This seems very similar to red-blue hat puzzle from coding theory!

that is not log base x bruv

Why do you have to do all This? 1÷9 is a rational, QED.

you didn't prove it

Wonderful

The number of diagonals from a single vertex is n-3 .pls explain this concept

I really enjoy your classes! They are so clever! Thank you for offering them.

4 sin^3x formula also can use, more easy mayb

Your video was phenomenal! The explication you gave was so well thought out and put. The way you explained this was also utterly endearing to watch too, this is my first time ever I haven't felt intimidated whilst watching a math related video. This was actually very enjoyable to watch and just aw-so-informative! You made me understand concepts I've had a hard time grasping for years by now; I just want to give my sincere thanks, you helped me a lot this day:)

Beautiful

Hey will you give us the homework

another mathematical paradox for the Basel problem

I would be very pleased if you don't stop making videos like this 😄

Wow such a great explanation of the formula 👏 ❤

Thank you sir.

Nice Explanation👏👏

Helped alot thanks

Amazing explanation sir