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1:27 AM
If the proofs are valid, I suppose that would make it credible
 
@copper I think your proof was ultimately fine. The currently posted proof is isomorphic to yours.
 
2:14 AM
@TedShifrin Yeah, it was fine, I did a quick, incorrect computation when checking this morning. That's what I get when computing during a work meeting :-).
 
2:44 AM
I’m not sure why it wasn’t clear to me last night. I thought I had sign issues.
 
@copper.hat If you wish you can repost it
> Here is a tedious answer.

It is straightforward to see that for $n=1$ there at at most $2n$ vectors satisfying the conditions.

Suppose it is true for $1,..,n$.
Further, suppose we have $a_1,...,a_{2(n+1)+1}$ points in $\mathbb{R}^{n+1}$ that satisfy the conditions. This will lead to a contradiction.

Let $b_j = \operatorname{proj}_{ \operatorname{sp} \{ a_1 \}^\bot } a_j$, then the $b_j$ are all in $a_1^\bot$ (which is essentially $\mathbb{R}^{n}$). It is straightforward to verify that $\langle b_i, b_j \rangle \le 0$ for $i \neq j$, hence the inductive hypothesis shows that $b_2,...,b
 
It needs some editing.
 
3:08 AM
It cannot be undeleted because a monitor deleted it :-)
I'll ask @XanderHenderson in the morning.
 
3:27 AM
Right but you can post it a second time
 
done :-)
 
3:48 AM
thx
 
0
Q: Rudin real and complex analysis Exercise 15.7

one potato two potato Suppose $f$ is an entire function, $f(\sqrt{n}) = 0$ for $n =1,2,3,\ldots$, and there is a positive constant $\alpha$ such that $|f(z)|<\exp(|z|^\alpha)$ for all large enough $|z|$. For which $\alpha$ does it follow that $f(z) = 0$ for all $z$? [Consider $\sin(\pi z^2)$.] Since $f$ is entire, u...

 
4:04 AM
@copper.hat Did you make the major correction re projection?
 
Indeed, I rewrote that part, it was missing a $\bot$ on the projection operator parameter.
 
Ah. Cool, Nice solution!
 
still, i think there is a cleaner proof.
 
Keep me posted.
 
i will :-). i thought i could use some convexity results to bring a quick closure, but nothing obvious came to me.
 
5:04 AM
I also found that Rudin denotes $T$ as the boundary circle of the unit disk $U$.
 
5:42 AM
that's not unusual.
 
 
5 hours later…
10:29 AM
I seem to not get any answers for this question: math.stackexchange.com/questions/4538686/… so I just wanted to ask if there is anything wrong with it, is there too much context/did I include too many attempts or is this a hard question? It's my second one so I'm not really sure, I got answers for my first much faster
Is there something I should edit into it?
 
10:49 AM
Hi everyone, I need to understand what method the authors use to analyze the singularities of a matrix (i.e. singularities occur when the matrix loses its maximal rank at a particular configuration). Take a look at the following picture
They are using the minors but as far as I know minors are used for determinants which are only for square matrices. Any explanation please.
 
11:50 AM
the rank of any matrix is the maximal dimension of a non-vanishing minor
 
Let $P$ be an $N \times N$ transition matrix which is irreducible and aperiodic. Perron-Frobenius theorem states that $1$ is the unique largest eigenvalue of $P$ and for all other eigenvalues $\lambda$ of $P$ we
have $|\lambda|<1$. Let $\pi=(\pi_1, \pi_2, \ldots, \pi_N)$ be the left eigenvector of $P$ corresponding to $1$ such that $\sum_{i=1}^N \pi_i=1$. Use Perron-Frobenius theorem and linear algebra techniques (e.g. Jordan Canonical Form) to prove that as $t \to \infty$, $(P^t)_{ij} \to \pi_j \text{ for all } i,j\, .$
How do I show this?
@Thorgott Can you give me a hint?
 
12:10 PM
in complex geometry why are polydisks important? why can't we just use higher dimensional balls?
 
12:29 PM
@Roby5 no
@monoidaltransform it's more convenient, e.g. consider how you can state the Cauchy integral formula in higher dimensions
 
oh right, ofcourse
 
12:44 PM
@copper.hat Do what now?
 
1:04 PM
If $f:M\rightarrow N$ is a holomorphic map between complex manifolds then the differential of $f$ is simply the complexification of the linear map induced by $f$ (as a smooth map between smooth manifolds)?
 
@XanderHenderson I was going to ask you to undelete an answer but I just added another. Excuse the ping.
 
No worries. Pings mean that someone cares!
 
@monoidaltransform yes
 
 
2 hours later…
3:14 PM
@Thorgott thank you.
 
3:59 PM
If only wikipedia offered problems to math topics
 
4:32 PM
@monoidaltransform Be aware they are not biholomorphic. Non-obvious and non-trivial.
 
Let $S=\{s\mid s\in S\}$
 
4:50 PM
la cage aux folles
 
@copper.hat have you heard of the website thoughtmaybe.com?
 
maybe
:-), no i had not
 
my taste in movies is fairly simplistic.
le carre is about the limit of my subtlety
reality is depressing. i read the brothers (about the dulles brothers) a few years ago and am still depressed
 
:(
Forman 2017 will cheer you up, guaranteed!
 
5:00 PM
i do like cabernet sauvignon
 
Sorry, I meant the movie George Foreman (2017)
 
:-) i know, just messing
 
however, forman 2017 is a nice wine, albeit this year's market shenanigans have put it out of my reach, unfortunately
its in the $100/bottle range, sinful but tasty
 
post pandemic shenanigans
or perhaps, price gouging
 
5:12 PM
@TedShifrin surely dilation in each coordinate is biholomorphic?
 
The walking dead tv show will end after this season.
 
finally.
 
It shouldn't be ending, I think. It should run for few more years. I love that show.
:)
 
Dead Man walking is better.
 
5:27 PM
@Thorgott I mean the disk and the polydisk.
 
oh, interesting
I've not thought about that
 
It took several weeks in the several complex variables course I took centuries ago.
 
i went to a talk by steve krantz about some SCV stuff and immediately felt as though i was falling down a well
he's a good lecturer i think the subject is just like that
 
Big theorem: Pseudoconvex domain iff domain of holomorphy.
 
> In the past year Krantz has collaborated with Arni S. R. Rao of Augusta University to study the COVID-19 epidemic. They have more than twenty papers and book chapters as well as several virtual seminars on the topic
 
6:06 PM
https://www.ams.org/journals/proc/1975-049-02/S0002-9939-1975-0372586-2/S0002-9939-1975-0372586-2.pdf
I found a theorem of one of our users here
Gerald Edgar
Has a mention on wikipedia under extreme points
 
 
2 hours later…
8:02 PM
Dr Edgar wrote a really nice little introduction to fractal geometry.
I've had that book for far longer than I've been active here. :D
 
Ain’t never heard of him. Just shows to go.
It turns out I have met one of his Ph.D. students, though.
 
 
2 hours later…
10:31 PM
Complex analysis help
It seems that $\int_{|z|=1}e^{z}dz=0$ and $\int_{|z|=1}e^{1/z}dz=2\pi i$
Shouldn't they at most differ by a minus sign? It's traversing the same set but in the opposite order
Ohh no I get it I think
 
10:45 PM
Did you write the change of variables?
 
Yeah it's $\int_{|z|=1}e^{1/z}dz=\int_{|z|=1}e^{z}/{z^2}dz$ and that makes all the difference
In any case I had a mild panic of "oh no I think I was marking incorrect answers as correct"
which I think it well-founded so let me go back and double-check
So one of the questions was to prove that $\left\lvert\int_{|z|=1}e^{1/z}dz\right\rvert\le2\pi e$
and the intended solution is to use the length-maximum inequality
 
@AkivaWeinberger sign ?
 
but a lot of people showed (correctly) that $|e^{1/z}|\le e$ for $z$ in the unit circle, and concluded$$\left\lvert\int_{|z|=1}e^{1/z}dz\right\rvert\le\int_{|z|=1}\left\lvert e^{1/z}\right\rvert dz\le\int_{|z|=1}e~dz=2\pi e$$
@TedShifrin Right yes sorry
Of course, the issue that you may notice is that $\int_{|z|=1}e~dz$ in fact equals $0$.
 
That is wrong. You need $|dz|$.
 
And $\left\lvert\int_{|z|=1}e^{1/z}dz\right\rvert\le0$ is pure nonsense.
@TedShifrin Indeed. And that's what I've been writing on people's p-sets.
But a lot of people made that same mistake, which makes me worry that the professor did not explain this point clearly enough.
 
10:59 PM
OK … $dz$ is still complex-valued, no inequalities.
These are important points.
 
Yup
At least these weren't the person who tried showing that $\sup_{|z|=1}|e^{1/z}|$ was $e$ by writing, "By way of contradiction, suppose $|e^{1/z}|>e$. Then $\frac1z>1$. Then $z<1$. Then $|z|<|1|$. But this contradicts that we're on the unit circle."
So like out of 10 points how many do I take off for that? At least they correctly identified that the length of the contour is $2\pi$...
I'll discuss with the professor tomorrow.
 
11:13 PM
Re: "a lot of people made that same mistake", turns out it's 5 people
out of 30
so maybe on balance not so many
Grading is driving me crazy.
 
why not $\int_C e^{1 \over z}dz = \int_C \sum_k {1 \over k!} {1 \over z^k} dz = \int_C {1 \over z} dz = 2\pi i$.
 
11:55 PM
@copper.hat as in all the other ones die 'cause exponent not $-1$?
 
yup
 
I mean if someone explicitly evaluated the integral and said "and $|2\pi i|<2\pi e$, QED", I'd give them full points, but I don't think anyone did that
 
seems the easiest approach to me
 
I think it's 'cause they were still new to it
 
prob
 
11:57 PM
the fact that $\int_Cz^n~dz=\begin{cases}0,&n\ne-1,n\in\Bbb Z\\2\pi i,&n=-1\end{cases}$
On the other hand, $\int_{|z|=1}z^n~|dz|=\begin{cases}0,&n\ne0,n\in\Bbb Z\\2\pi ,&n=0\end{cases}$, an exercise which I think they could benefit from doing
 

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