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Sophie M's user avatar
Sophie M's user avatar
Sophie M
  • Member for 7 years, 9 months
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Most elementary proof showing that exponential growth wins against polynomial growth
Thanks, corrected! And yes, that's right, so if you're comfortable arguing from $n^k \leq C_k 2^n$ down to $n^k < 2^n$, then you don't need the compression, but it's a fun trick and it felt in the spirit of the question.
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$L^p$ norm of Fourier transform of function composed with a diffeomorphism
What have you tried? One can start writing down some things using basic facts about the Fourier transform, but it would help to know where the basics are proving insufficient for your desired application.
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Non-diagonalizability of the adjacency matrix of a directed graph
Do you mean to assume that the graph is weakly connected? If not, then consider, for a counterexample, a graph $G$ with four states $A,B,C,D$, with three edges: a $2$-cycle $A \to B \to A$ and a lone edge $C \to D$. If the graph is weakly connected then I think it's probably true. (I use the convention that $A$ for the $G$ I described has $3$ entries equal to $1$ and $13$ entries equal to $0$.)
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Probabilistic problem on random spanning trees
@fedja Beautiful! That's quite similar to what I was trying in my answer below, but the analysis of your graph is a lot simpler!
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Probabilistic problem on random spanning trees
No, probably not! That seems like a good way to prove that what I'm outlining here can't work.
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Probabilistic problem on random spanning trees
Also -- do you expect this to be true or false, i.e. do you expect a proof or a counterexample? Where did it come from?
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Probabilistic problem on random spanning trees
The function $f$ is fixed when taking the expectation?
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Cohomology for extension problems in symbolic/topological dynamics?
correcting silly mistake in response to comment
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Expected number of compositions needed to get constant function
@SamHopkins Yes, $k=1$ is special because a single map generates a commutative monoid; $k \geq 2$ should be different. An average-case version of my question, closer to yours: Fix $k$ and $\ell$. Let $\{ f_1, \dots, f_k \}$ be a set of $k$ distinct maps $[n] \to [n]$ chosen uniformly at random. Let $i_1, \dots, i_{\ell}$ be chosen from $\{ 1, \dots, k \}$ uniformly and independently (both from each other and the $f_i$'s). What is the probability that $f_{i_{\ell}} \circ \cdots \circ f_{i_1}$ is a constant map?
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Expected number of compositions needed to get constant function
A refinement of the question, since the accepted answer links to a complete solution: suppose you draw your functions instead from the uniform distribution on some subset $\{ f_1, \dots, f_k \}$, where there is some sequence of these $f_i$'s that composes to a constant map. For a given $k$, which sets $\{ f_1, \dots, f_k \}$ will maximize the expected number of compositions required for a constant map? How small does $k/n^n$ have to be before this worst-case expectation is substantially larger than $2n$?
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