i sort of hate the UC. it's what i know, i grew up in it. it must have been wonderful in the 60s, 70s, 80s. it was getting raunchy by the time i was there.
I'm having trouble communicating the final solution of a question. The question is simply to show the tangent pplane of a cone $z^{2} = x^{2} + y^{2}$ at $(a,b,c) \neq \mathbf{0}$ intersects the cone in a line.
I've done all the mechanical stuff so I have my tangent plane and I set it equal to the cone at the given point, after all the simplification I will be left with the expression $(x-a)^2+(y-b)^2-(z-c)^2=0$.
I know that a line is expressed in the form $t(x,y,z)$ for some scalar $t \in \mathbb{R}$. What am I missing to arrive at that sort of conclusion here? [don't kill me Ted, I kno…
my wife currently teaches next to pomona. i think her school makes a good difference. coronavirus has been horrific for her students, she is basically a trauma counselor.
@leslietownes Maybe. I think that folk have been prognosticating a major pandemic for years, and have been suggesting ways to mitigate, but no one in power has taken those warnings seriously.
And, given the reaction to COVID and the vaccines in my neck of the woods, I would not be surprised to see this happen again. And again. And again.
@leslietownes Yeah. And most of what they do feels so abstract and remote. And prevention is expensive and easy to dismiss if you are not in the middle of a crisis.
(I say that, as your 500 billable hours comment reminds me of one: A lawyer appears before St Peter at the pearly gates. St Peter says "Oh, my! Welcome, and congratulations! According to our records you are the oldest person to ever walk through these gates! We've been looking over your billable hours, and, according to our reckoning, you must have been at least 1200 years old when you died!").
another is, surge billing. if what you're asking me to do sucks, maybe i don't want to do it without an inducement, which is just supply meeting demand.
surge at 3x and yes i will get on that memo right away.
When my father was working in the legal profession (which was for a period of about 8 years after finishing his phd), he worked for the state, and was on salary.
No billable hours.
But he got to craft legislation which f'd over California in the 80s, and testified in the impeachment of a sitting governor, so... there's that.
Yeah, I get the impression that a lot of folk end up in jail not because they are guilty (though they very often are), but because they are too dumb to keep their mouths shut.
i've always thought that if you remain silent in a situation where it can be reasonably believed the questioning party is not acting in your best interest, something something
it's legal for law enforcement to lie to you, and a criminal conviction can be based 1000% on a jury believing something someone said in disagreement with what you said, with no other evidence.
do not open your mouth. if you open your mouth they can disagree with it and goodbye to you.
the trouble is if you take something off the table as a technique you do sometimes lose the ability to have people brag about their crimes and then have it come in as evidence later. which seems like it should always matter.
i stay out of the criminal law for this reason.
let's talk about bounded linear functions on $\ell^{\infty}(\mathbb{N})$.
I'm not a fan of the name "A". It was a working title, because it was defined in terms of "a", and well .. I thought "b" was a ludicrous name (that I had writing by hand).
A few years ago there was a Russian oligarch visiting the Bay Area in his yacht named 'A' (apparently himself & wife shared the first initial A). Refueling the boat apparently cost in the vicinity of $500k.
So, A is good for expensive boats, but not for vectors.
Sorry, I have so many NDAs its hard to keep track...
Turns out a lad from my hometown Passage West was first mate on the 'A'. Small world.
Long time ago a friend was the chief engineer on one of the Carnival Cruise (I think) ships and when he came to SF I got all the high end food a starving student could want.
you know it's good if they named it for people trying to get the hell out of there.
sounds good.
i had a long running joke with one of my friends, i said, you're going to get a respiratory or venereal disease on one of those ships. these cruises are no good whatsoever.
she got the latter and i did not let her live it down.
Lol...I was literally about to comment your life sounds like a movie Copper......it still does in my opinion regardless of how humble you want to attempt to portray it....
I've never been in a situation to sign one NDA and here you have so many you can't keep track of....so modest.....
Well as per title, say I have the probability density function on domain $x \in [0,1] ; y \in [0,1]$
$$f(x,y) = \frac{12}{5} \left( x^2 + y^2 - xy \right)$$
Can I generate this density function from a given uniform (pseudo) random function on the same domain?
When using a single variant it's slig...
If $u$ is harmonic on $\Bbb D\setminus\{0\}$ where $\Bbb D$ is an open unit disc, such that $\lim_{|z|\to 1}u(z) = 0$ and $\lim_{|z|\to 0}u(z)/\log|z| =0$
Then I want to conclude $u$ is identically zero.
So far I can see $u(z)$ has singularity at $0$
Where does the contradiction arise when I assume $u$ is a nonconstant function?
It should be something like maximum and minimum principle for harmonic function
I have some question essentially understanding the induced orientation on the boundary of manifolds. Here is the definition given in Munkres' Analysis on manifolds
Definition. Let $M$ be an orientable $k$-manifold with non-empty boundary. Given an orientation of $M$, the corresponding induced or...
@Thorgott If $(U,\varphi)$ is a positive parametrization of manifold $M$ then $\varphi|_{U\cap\Bbb H^n}$ is a negative parametrization of $\partial M$?
your observation is not irrelevant though. note that $\varphi\vert_{U\cap\partial\mathbb{H}^n}\colon U\cap\partial\mathbb{H}^n\rightarrow\partial\mathbb{H}^n$ is an orientation-preserving diffeomorphism if we endow $\partial\mathbb{H}^n$ with the induced orientation as well. it simply is not an oriented chart if we interpret $\partial\mathbb{H}^n$ as $\mathbb{R}^{n-1}$ with the standard orientation.
@Thorgott Yes what I'm asking is that. $\varphi\vert_{U\cap\partial\mathbb{H}^n}\colon U\cap\partial\mathbb{H}^n\rightarrow\partial M$ with $\Bbb H^n$ an induced orientation is an orientation preserving diffeomorphism.
For $n$ odd case, $\Bbb H^n$ has opposite orientation and $\partial M$ also has opposite orientation to orientation of $M$
What I want to show is that if $(V\cap\Bbb\partial H^{n},\psi|_{V\cap\partial\Bbb H^n})$ is a positive orientation of $\partial M$, then $(U\cap\partial\mathbb{H}^n,\varphi\vert_{U\cap\partial\mathbb{H}^n})$ overlaps positively
So it's actually contained in the orientation of $\partial M$
note that $\varphi\vert_{U\cap\partial\mathbb{H}^n}\colon U\cap\partial\mathbb{H}^n\rightarrow\partial\mathbb{H}^n$ is an orientation-preserving diffeomorphism
Isn't this mean it's contained in the orientation?
it's an orientation-preserving diffeomorphism when you equip $\partial\mathbb{H}^n$ with the induced orientation
for it to be an oriented chart would mean that it is orientation-preserving when you equip $\partial\mathbb{H}^n=\mathbb{R}^{n-1}$ with the standard orientation
While solving a couple of questions I intend to use the approximation techniques to the best of my knowledge. Let us say we have three instances:
1.$\frac{1.66}{6.66}$
2.$\frac {3.33} {8.33}$
3.$\frac {86.66} {15}$
I would lets say approximate the first one as 1/4 and second one as 2/5. How do i ...
Given a local field $K$ complete with discrete valuation $\nu$, and $R$ its ring of integers (i.e., the ring consistings of the elements whose valuation is non-negative), say that for some $r\in K$, we have $4r^3\in R$ (and $3r^4\in R$ if we want to). It's probably silly, but why does it hold then that $r\in R$?
I guess I could use that we either have the $p$-adic valuation
in which case $\nu(4)\leq 2$
and then the argument works
or we have the valuation with formal Laurent series
In which case $\nu(4)=0$
so it holds then
Though I guess there'd be a more elementary argument
In the best fit line problems , for example : $(0,6) , (1,0) , (2,0)$ . We assume $y= mx+b$
then arrange this in matrix form $Ax=b$ where A = $ \begin{pmatrix} \ \ 0 \ 1 \ \\ 1 \ 1 \\ 2 \ 1\\ \end{pmatrix}$ Is it necessary that columns of A are linearly independent , if yes how do we ensure that we get linearly independent columns every time ?
@ShaVuklia you can just do this with the corresponding non-archimedean absolute value right? Then $R = \lbrace x \in K : \lvert x \rvert_\nu \leq 1 \rbrace$
I mean what you want to avoid is that $\bigcup A_n$ has full measure in an interval, thing about building a fat Cantor set of measure 1/2 in [0,1], then in each of the countably many gaps you build a fat Cantor set of measure half the length of the gap and so on
I might be completely out of it, but what about $\mathbb Q_2[X]/(X^2-2)$. That's a finite extension of $\mathbb Q_2$, thus local, but taking $r = X^{-1}$ you find that $4r^3$ is $X$
Suppose that 2% of people complete the task between 4 and 4.1 seconds. Approximately what proportion of people are likely to complete the task between 4 and 4.01 seconds? Assume a uniform distribution.
I proposed this examples because intuitively I thought that 4 could not have a valuation greater than 2 because it's $2^2$ and 2 is prime
This is false because the valuation of an extension restricted to the original field can be a multiple of the original valuation (so $2\nu$ on $R$ instead of $\nu$, for instance)
In particular if $4$ has valuation greater than 3, then taking $r$ to be the inverse of any uniformizer is going to yield a positive valuation for $4r^3$
Even though the inverse of a uniformizer is never positive
I came across this answer but I don't understand something: They write "Then, all conditions of the dominated convergence theorem are satisfied,... ", which implies that $h_n$ converges pointwise but is this the case? If yes, why?
To be more specific: $h_n$ is defined as $f_{n+N}$ but we only know that $f_n$ converges uniformly and not pointwise
@Thorgott I already thought so, but why is a new proof necessary for uniform functions? Or is the main-problem to prove that the functions are integrable?
the point of the question is the converngece of the integrals, not of the functions, I'm not sure what "new proof" is referring to as I don't know what you think of as old proof
The conditions are "Let $(\Omega,\mathfrak{A},\mu)$ be a measurable space and $\mu(\Omega)<\infty$. Let $(f_n)_{n\geq1}$ be a sequence of integrable measurable functions $f_n:\Omega \rightarrow [-\infty,\infty]$ converging uniformly on $\Omega$ to a function $f$." But aren't these conditions already enough to apply DNT?
Quick random question: Uniform continuity is a topological property, in that equivalent metrics produce the same uniformly continuous functions, correct? I was looking over a question that involved the infinity norm instead of Euclidean and was thinking since everything is just in terms of sufficiently close in the metrics involved, one could use an equivalent metric by just rescaling the $\epsilon, \delta$?
I was taught two metrics are equivalent if you can bound one in between the other with suitable positive constants, and that made it induce the same topology
a better notion of equivalence of metrics, which actually preserves a lot of metric properties such as uniform continuity, is that each metric is bounded uniformly by the other up to a multiple (i.e. the identity map is Bilipschitz)
yeah, for that notion of equivalence, which is also what you presumably meant by "rescaling the $\epsilon,\delta$", everything works out
and this in particular applies to $p$-norms on finite-dimensional Euclidean spaces
Identity being bilipschitz is a stronger form of quasi-isometry. In any case the point is that in this case the two metrics don't just induce the same topology, they induce the same uniformity
Actually, now that I think of it, I think I really learned about for norms, not metrics, in that proof that in finite dimensional real vector spaces, all norms are equivalent
Meanwhile I'm sitting in class while my students work on remedial mathematics, currently on the joys of one variable linear equations :). For this I did 7 years of grad school....
Howdy! I'm not a math guy so I can't really understand if this question has been asked before, but is it possible to order an infinite set of different real numbers?
I'm assuming it can't be well ordered, since Z has no least element, but would it be possible to linearly order it, even in an infinite amount of time?
Hi! I have a question regarding the foundation of mathematics (f.o.m). Why people look at set theory (or even ZFC version) as a basic of all mathematical objects? Is it because every mathematical theorem and objects can be turned into set-theoretic objects. I have this question because Godel proved his theorem based on set theory, he proved that all mathematics is not complete based on set theoretic arguments that shows that there are mathematical statement which we cannot prove or disprove
in fact, most people don't really look at foundations at all, because the precise foundations only matter occasionally in most other areas of mathematics
I don't quite understand your question, there's no such thing as "the axioms" and if you have a specific set of axioms, you can prove a stement to be independent of these axioms (only under the assumption of their consistency, mind you) by exhibiting different models of the axioms in which the statements have different truth values
@Thorgott So what I understand from you is that in mathematics we don't have "the axioms" instead, we have "axioms of the set theory" (or axioms of geometry or etc.) but we don't have "axioms of all mathematics objects", Is this what you are trying to emphasize?
Part of it, yes. Most people do use ZFC, but most people also don't really care. Those who really care are logicians and set theorists and a lot of their work explicitly revolves around understanding different axiom systems and how they interrelate.
@user777 Part of the power of Gödel's theorem is that it applies to any set of axioms that is both manageable and interesting (in some precise sense, namely being recursively enumerable and interpreting (some fragment of) PA)
As to why ZFC is used mostly historical reasons I guess
My book uses Taylor's theorem while dealing with a local field $K$ (complete w.r.t. discrete valuation). Is this allowed because analysis can be done on local fields?
Hm, a paper online seems to imply we can't hope for a version of Taylor's thm in such a case
yeah I guess a hunch in this case that they are isn't founded on much beyond 'we can really only talk about the size of sets associated with $f$ in this language', and thats not an especially robust foundation
