Mathematics - 2024-07-10

Semiclassical

04:48

hmm. i'm looking through my Saves, and there's a Q&A with a marked answer which provides a conjectured result but doesn't actually derive it. (it's not a bad answer, it's just a pretty hard problem). since the old Q&A exists it seems like asking for a proof would comprise a duplicate

i could bounty it i guess

leslie townes

semi: whatever you do, don't share this obviously interesting question with the channel

Semiclassical

lol

2

Q: Solving a functional equation for finding equivalent resistance in an infinite ladder

I was trying out a physics question to find the equivalent resistance between two points A and B. I got a functional equation on solving, but I couldn't proceed further. Finding f(1) gives the answer to the question. \begin{gather*} \frac{( x+1) f( x+2)}{( x+1) +f( x+2)} \ +\ x=f( x) \ \\ \end{g...

i had some conjectural comments of my own but solving it was beyond me too

leslie townes

looks cool

Semiclassical

yeah, it's a neat question

particularly since the conjectured result seems correct

but, yeah, i'm not sure if a new question is appropriate or just a bounty saying "hey, someone should prove this"

leslie townes

if you had new stuff to add i think it would totally work as a new question, even if it's asking for what would amount to the same answer (if only because it feels more awkward to "answer" an existing post with a reframing of the question and further comments)

Semiclassical

05:02

hmm

leslie townes

offhand i wonder if reducing it to a single recursion actually makes it more difficult, e.g. a lot of eigenvalue problems end up looking like that if you roll up an infinite number of equations into something that you solve for the entries of a single vector in terms of its first entry

whereas an eigenvalue problem is an eigenvalue problem

Semiclassical

hmm, yeah

the difficulty of it vs a more standard resistor ladder is that the resistances keep increasing

but there should still be a way to write this as a big-ass 2-by-2 matrix product

Semiclassical

05:22

@leslietownes had to write it out again, but the matrix version of this would be: Let $M_k=\begin{pmatrix} 1 & 0 \\ -1/(2k+2) & 1 \end{pmatrix}\begin{pmatrix} 1 & -2k-1 \\ 0 & 1 \end{pmatrix}$ and let $v_{k+1}=M_k v_{k}$. For most $v_0$, $\|v_k\|\to \infty$ as $k\to\infty$. For what $v_0$ does $\|v_k\|\to 0$ instead?

the conjecture amounts to $v_0=(4,\ln 5)$

not sure that version is an improvement tho. the advantage of that formulation typically is that, if $M$ were constant, then diagonalizing $M$ makes this trivial

leslie townes

yeah even there my thought is this is v = Mv for some decidedly infinite dimensional M that you can maybe analyze in some other way than matrix algebra

Semiclassical

hmm

the other idea i had now is to impose $\|v_{k}\|^2 < \|v_0\|$ given $v_0=(1,x)$ and see what range of $x$ that requires

nickbros123

07:35

"for every point x in $\Phi$, x is a unicorn" and "for every point x in $\Phi$, x is not a unicorn" are both true

psie

08:15

Consider the problem $Y\mid X=x\in\text{Bin}(n,x)$ with $X\in U(0,1)$. I need to compute $\operatorname{E}Y$, $\operatorname{Var}Y$ and finally $\operatorname{Cov}(X,Y)$, but without finding the distribution of $Y$. The first two follow from the law of iterated expectation and variance respectively. However, the covariance I struggle with; we need to compute $$\operatorname{E}(XY)-\operatorname{E}X\operatorname{E}Y.$$ How do I compute the first term without knowing anything about $Y$?

psie

08:31

I think I've worked it out. It's another application of the law of total expectation.

CowperKettle

08:51

US presidents explain L'Hôpital's rule in under 90 seconds

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Peter