Mathematics

Balarka Sen

12:05 AM

Hi @Brody, @Ted

Akiva Weinberger

So on Day 1 I have one blob

and every day there's a $p$ chance that each blob splits in two and a $(1-p)$ chance that it dies/disappears

What are the odds that I never lose all of my blobs?

Zach Hauk

math.stackexchange.com/questions/2059244/… i don't know what answer to give for this other than "obviously, because isomorphisms preserve ring structure"

Mike Miller

you would presumably use the isomorphism to induce a bijection between the sets of prime ideals

Zach Hauk

well yes

its really really obvious

i tried to explain it like this but got downvotes:

Mike Miller

right, but you still didn't give him a proof

i saw your explanation

Zach Hauk

12:18 AM

-2

A: If rings R and S are isomorphic, prove that if R has 3 prime ideals, then S does too?

This fact is obvious, since isomorphisms preserve ring structure. We note that the $n$ prime ideals in $R$ must map to at least $n$ prime ideals under an isomorphism; a fact you stated yourself. Let $m$ be the amount of prime ideals in $S$. Since the inverse of an isomorphism is an isomorphism, ...

does this explanation make more sense?

Jessy Cat

I'm crisising again.

Mike Miller

I would be more explicit

Jessy Cat

0

Q: Inverting a Fourier transformed solution with complicated boundary condition

I am working on the following problem: Find the steady-state distribution of temperature in the half space $H = \{ y > 0 \}$ if the boundary temperature is $\displaystyle u(x,0) = \begin{cases} 1 & \text{if}\, -1<x<0 \\ 1-x & \text{if}\,0<x<1 \\ 0 & \text{otherwise}\end{cases}$. I.e., I nee...

Mike Miller

Show that if $\mathfrak p$ is prime, $\psi(\mathfrak p)$ is prime, and if $\mathfrak q$ is a prime ideal in the second ring, $\psi^{-1}(\mathfrak p)$ is prime

obviously these two operations are inverse to one another, because $\psi$ itself is a bijection, so $\psi: \text{Spec } R \to \text{Spec } S$ is a bijection

Jessy Cat

@Zach I don't see why you got two downvotes. It's a good answer.

There. I just changed the title of my post. Maybe now people will actually look at it.

Zach Hauk

12:22 AM

@JessyCat sorry i'm not good with diff eq and that stuff :(

Jessy Cat

Apparently, neither am I.

Or at least I haven't practiced enough yet.

I don't know how to deal with this boundary condition that has two nonzero branches.

Zach Hauk

@MikeMiller $\mathrm{Spec\;} R$ = set of prime ideals in $R$?

Mike Miller

y

Math is Life

math.stackexchange.com/questions/2057739/… Could anyone help me with this

Balarka Sen

@AkivaWeinberger For sake of simplicity suppose $p = 1/2$. Look at the tree where every node bursts into two more nodes. Look all the paths in the tree starting at the base node, of length $n$. There's $2^n$ many of them. Look at all the paths of length $\leq n$ starting at base node. There's $(2^0 + 2^1 + \cdots + 2^n) = 2^{n+1} - 1$ many of them. The probability of a blob surviving till $n$-th day is then $2^n/(2^{n+1} - 1)$. Take limit $n \to \infty$: that's $1/2$.

I guess you can make this work for arbitrary $p$ too.

Akiva Weinberger

12:36 AM

I don't follow your logic @BalarkaSen

You're saying every path corresponds to an equally likely world?

Jessy Cat

@Semiclassical if you get a chance, could you take a look at this: math.stackexchange.com/questions/2059270/… I'm sorry I'm always bugging you :(

Balarka Sen

The paths in the tree represent a blob's future. I tried to compute the number of paths of length $n$ (precisely the number of ways a blob can survive till the $n$-th day), compared to number of paths of length $\leq n$ (precisely the number of ways a blob's future can happen till day $n$).

@AkivaWeinberger Don't they?

Actually

Akiva Weinberger

But wouldn't the shorter paths be more likely?

The blob has a $p(1-p)=1/4$ chance of going down a one-edge path and dying there, no matter what $n$ is.

Also, it seems like you're trying to compute the expected value of blobs in the end (which is $\lim_{n\to\infty}(2p)^n~$ — or, $1$ for $p=1/2$, $0$ for less and $\infty$ for more), which is easier to find than what I want

Balarka Sen

Yeah, I seem to be computing the probability of some specific blob surviving though.

@AkivaWeinberger Oh well, I guess.

Akiva Weinberger

I might need to take my last sentence back

Also might not. I'm confusing myself.

Zach Hauk

12:48 AM

oh boy

Balarka Sen

Me too.

Zach Hauk