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#8) Prove or give a counterexample:
Every ideal in a commutative ring is the intersection of the prime ideals containing it.
Every ideal in a commutative ring is the intersection of the prime ideals containing it.
I think representations of topological groups are by definition continuous. If you don't require continuity, this looks very false. For example, consider the group ##GL(n,\mathbb{C})## equipped with the indiscrete topology, which trivially makes it compact. Then the identity map (which is not continuous) ##GL(n,\mathbb{C})\to GL(n,\mathbb{C})## can be considered as a representation, which is certainly not unitary. The righthand side of course has the standard topology.Office_Shredder said:#7,). Prove or give a counterexample: Every irreducible complex representation of a compact group is unitary
A lot of words there. Is irreducible necessary? Does the representation need to be continuous? Maybe someone will find out!
Infrared said:I think representations of topological groups are by definition continuous. If you don't require continuity, this looks very false. For example, consider the group ##GL(n,\mathbb{C})## equipped with the indiscrete topology, which trivially makes it compact. Then the identity map (which is not continuous) ##GL(n,\mathbb{C})\to GL(n,\mathbb{C})## can be considered as a representation, which is certainly not unitary. The righthand side of course has the standard topology.
This doesn't look true to me: The matrix ##A=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}## has only a one dimensional eigenspace ##\text{Span}\left(\begin{pmatrix} 1 \\ 0\end{pmatrix}\right)## but ##\text{max}_{||x||=1} ||Ax||## occurs when ##x=\pm \begin{pmatrix} 0 \\ 1 \end{pmatrix}##. This specific matrix can't come from a compact representation because it generates a non-compact subgroup though.Office_Shredder said:there is a vector that maximizes the norm, you can show it's an eigenvalue
Infrared said:This doesn't look true to me: The matrix ##A=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}## has only a one dimensional eigenspace ##\text{Span}\left(\begin{pmatrix} 1 \\ 0\end{pmatrix}\right)## but ##\text{max}_{||x||=1} ||Ax||## occurs when ##x=\pm \begin{pmatrix} 0 \\ 1 \end{pmatrix}##. This specific matrix can't come from a compact representation because it generates a non-compact subgroup though.
Let ##f(x)=x^2\sin(1/x^2)## for ##x\neq 0## and ##f(0)=0,## let's say defined on ##[-1,1].## We see that ##f'(x)=2x\sin(1/x^2)-\frac{2}{x}cos(1/x)## for ##x\neq 0## and ##f'(0)=\lim_{h\to 0} \frac{h^2\sin(1/h^2)}{h}=0.## As ##f## is continuous on a compact domain, it is uniformly continuous. Since ##f'## is not bounded, but any differentiable Lipschitz function has derivative bounded by its Lipschitz constant, this ##f## cannot be Lipschitz.Office_Shredder said:#10) Prove or disprove: Every uniformly continuous function that is differentiable on a closed interval is necessarily Lipschitz continuous.
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.Office_Shredder said: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
Muu9 said:Which prompt are you using to generate these?
I made so many ridiculous mistakes that I thought a break would be a good idea.Office_Shredder said: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.
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 isMuu9 said: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?