good evening

good evening to you

good night

Howdy @copper

Bonsoir @TedShifrin!

I'm currently reading Narahimhan's complex analysis textbook. It's pretty good to me

As I remember, it had some really unusual stuff in it. I taught part of it, but didn’t read most.

Corona theorem for example?

any complex text that starts with the Wirtinger derivatives is not elementary in my mind

It’s definitely a graduate text.

I used his differential geometric proof of big Picard.

Makes sense. My complex development stopped at Nevanlinna Pick.

@RE60K @RE60K The support of a function $e$ is the set $\{x\in dom(e)\mid e(x)\neq 0\}$. Probably will depend on the context of the paper, but if for each vertex $v$ there's an associated function $e$ I'd guess $e$ maps each vertex to the number of edges connecting $v$ to the vertex. Then the support of $e$ (function $e$ chosen when working with vertex $v$) would be the set of vertices connected to $v$

Does there exists any injective ring homomorphism from $C([0, 1], \mathbb{R}) $ to $C(K, \mathbb{R}) $ ? ($K\subset [0, 1] $ is the Cantor set)

is X\wedge n=max(X,n) or min(X,n)?

really up to whoever is defining it, but min would be standard

okey thank you

@TedShifrin There's a complex analysis textbook by Krantz that has a geometric viewpoint. But I think you already aware of it.

I read the answer of poission distribution but I am bit confused about the parameter lambda

It is mean right?

I don't get what h is and why is lambda*h a probability

and poission distribution is a pmf and lambda*h is also a pmf

I don't get it

I see some article mentioning lambda is rate

@SouravGhosh There is a surjective continuous map $K\rightarrow[0,1]$. a) Do you know this? b) Can you see what this has to do with your question?

@NotTfue In the answer I mentioned in my previous reply, it is mentioned that $v$ was the "expected number of decay events ... in a second". In the Poisson process, there is a given sample space (in the decay example, it is a second).

Given the expected number of occurrences in the sample space, $\lambda$, Poisson's Formula says that the probability of $n$ occurrences happening in the sample space is $\frac{\lambda^n}{n!}e^{-\lambda}$.

If the sample space is an amount of time, $\lambda$ is the expected rate of occurrences in that amount of time.

@anak do I have to specify $e(-1)=0$ and $e(1)=0$ for the embedding?

@Wave please use mathjax, otherwise it is hard to read

@copper.hat sorry I don't know what mathjax is and how to write it and how to read it

G’day @robjohn

@TedShifrin hello. We are back from our vacation. The temperature is much hotter here than in Mammoth.

96° here and 68° in Mammoth

Yeah, the heat wave is back ;(

Did you drive to Manmoth or fly?

We drove. We like the scenery and the drive. First time for our dog, and she liked it.

hi, i wanted to ask about a proof that is in my real analysis book to prove the archimedean property. is takes the approach to, for contradiction, assume that N is bounded above. then, say we have b := sup(N), then b-1 cannot be an upper bound, so there exists an m in N such that m > b-1, then add one to both sides and this contradicts b being an upper bound

however, i dont understand why there has to be an m in N st m > b-1

What does it mean that $b-1$ is NOT.

An upper bound. Can’t edit. Grrr.

sorry i dont quite understand your question

You said $b-1$ cannot be an upper bound. What does that mean?

i think the idea is if b is the supremum, b - 1 cant be an upper bound

where being an upper bound is something like there is no element in the set larger than the upper bound - lemme look up the exact def

Agreed. So figure out what failure to be an upper bound means.

if there exists a b in S such that x is less than or equal to b for all x in E , then we say E is bounded above and b is an upper bound of E

is the exact def

now read what I typed above

oh oh is it like if b had nothing above it then it would be an upper bound

$b-1$, right.

got it got it thanks

good.

@robjohn OK I see I misunderstood slightly I think I get it now.