Can chatgpt accurately calculate expected lengths in Pascal's triangle?

[Office_Shredder]

#13) Prove or disprove: Every infinite dimensional Banach space has a linearly independent sequence that is not a basis.

 

[Muu9]

TL;DR Summary: Prove you are smarter than the bot!

Chatgpt is actually pretty good at generating math problems. It's awful at solving them. I guarantee every question posted here cannot be solved by chatgpt, but maybe can be solved by a human? My plan is to spend a couple minutes getting a question I think it's cool and then posting it here - I don't know if I'll actually do it each day.

Since chatgpt is bad at knowing whether things are true or not, and I'm not going to try to solve all of these before posting, most will be of the form prove or find a counterexample
#5.) .I take a piece of string of length 1, and do the following k times: cut the string into a ratio of 2 to 1 (so a string of length 1 is cut into a string of length 2/3 and a string of length 1/3) I then throw out one of the pieces, leaving myself with a single piece that I can repeat this process on, until I have made k cuts.

What is the expected length of the final piece of the string in terms of k, if
a.) Each time I pick one of the two pieces to throw out randomly with equal probability
b.) Each time I "grab" a random spot on the string and throw that piece out - i.e. I have a 2/3 probability of throwing out the longer piece

For a, imagine a pascal's triangle of height k+1, but instead of the standard digits we have 1 at the top, then (2/3) and (1/3) on the second row, then (2/3)^2, (2/3)(1/3), and (1/3)^2 on the third row, and so on.
The k+1th row will have be (2/3)^k*(1/3)^0, then (2/3)^(k-1)*(1/3)^1, then (2/3)^(k-2)*(1/3)^2, and so on until finishing with (2/3)^0*(1/3)^k. To adjust for the fact that the middle numbers are more likely, each of these values is multiplied by the corresponding value of the normal Pascal's triangle. The mean of these adjusted values is the expected value of the final length. Anyone know how to express this in a more traditional, concise form?

For b, do the above but square each value before multiplying by the corresponding value of the normal Pascal's triangle.

Which prompt are you using to generate these?

 

[nuuskur]

A sandwich semigroup is a semigroup $S$ such that $S=SES = \{ses' \mid s,s'\in S, e\in E(S)\}$, where $E(S)$ is the subset of idempotents of $S$. An idempotent $e\in S$ is an element that satisfies $e^2=e$.

A semigroup $S$ is said to be closed if the map $S\otimes _SS\to S,\ s\otimes s'\mapsto ss'$ is bijective.

Claim. All sandwich semigroups are closed.
(False)

 

[Office_Shredder]

Which prompt are you using to generate these?

It's a bit of messing around. A common one for the more advanced questions is something like "string together random words from topology to generate a question of the form prove or disprove". The lower level ones I have asked it to generate brand new high school math competition problems. I probably only use about 20% of the responses, and they usually need some tweaking (e.g. the one about a holomorphic function which is real on the circle, was originally about a holomorphic function which was real on the real axis, which is, uh, less interesting). The string cutting one was also pretty different from what the bot actually generated, though I don't remember the changes I made.

I stopped making these because it seemed like no one was interested except for Infrared and fresh who had stopped showing up as much, if there is renewed interest I can make some more.

 

[fresh_42]

I stopped making these because it seemed like no one was interested except for Infrared and fresh who had stopped showing up as much, if there is renewed interest I can make some more.

I made so many ridiculous mistakes that I thought a break would be a good idea.

 

[Office_Shredder]

@Muu9 for the first one consider expanding $(2/3+1/3)^k$

 

[Fred Wright]

For a, imagine a pascal's triangle of height k+1, but instead of the standard digits we have 1 at the top, then (2/3) and (1/3) on the second row, then (2/3)^2, (2/3)(1/3), and (1/3)^2 on the third row, and so on.
The k+1th row will have be (2/3)^k*(1/3)^0, then (2/3)^(k-1)*(1/3)^1, then (2/3)^(k-2)*(1/3)^2, and so on until finishing with (2/3)^0*(1/3)^k. To adjust for the fact that the middle numbers are more likely, each of these values is multiplied by the corresponding value of the normal Pascal's triangle. The mean of these adjusted values is the expected value of the final length. Anyone know how to express this in a more traditional, concise form?

For b, do the above but square each value before multiplying by the corresponding value of the normal Pascal's triangle.

Which prompt are you using to generate these?

For brevity of notation let $a=\frac{2}{3}$, $b=\frac{1}{3}$, $C_{k,n}$ be the binomial coefficient of the k'th row and n'th column of Pascal's triangle and $l_k$ be the expected length for the k'th row of Pascal's triangle. For part a of the problem the expected length of the k'th cycle is
$$
E[l_k]=\frac{\sum_{n=0}^{k}C_{k,n} a^{k-n}b^{n}}{\sqrt{\sum_{n=0}^{k}C_{k,n}^2}}
$$
For part b of the problem the expected length is
$$
E[l_k]=\frac{\sum_{n=0}^{k}C_{k,n} a^{2(k-n)}b^{2n}}{\sqrt{\sum_{n=0}^{k}C_{k,n}^2}}
$$