« first day (5097 days earlier)      last day (220 days later) » 

02:58
@XanderHenderson When I reread what I wrote, I saw that I considered the PDE to be $F(u)=0$ while giving the defn of a linear PDE by saying that the PDE $F(u)=0$ is linear if $F$ is linear.
But I now feel that this might not be complete.
To be more precise, I must say that in general for a PDE $F(u)=c$, we say that it's a linear one if $F$ is linear.
With this minor modification I think the defn of linearity of a PDE becomes more accurate. I hope you'll agree, right?
@XanderHenderson Ok, so then my understanding for this is: A PDE is quasi linear if the PDE is non linear but the coefficients infront of the highest-ordered partial derivative terms are all functions of the independent variables or lower order derivative terms.
Is the above interpretations correct?
I think with your help I have gained a fairly precise amount of understandings of all these definitions. As always, thank you so much for your time! Your references and comments really enriched my understanding.
 
1 hour later…
04:24
Hi Chat
2
Q: Absolute value of cubic polynomial roots lower than 1

MariaAssume we have a cubic polynomial $ x^3 +bx^2+xc+d=0 $, with $b,c,d$ real numbers. Let $x_1, x_2, x_3 $ be the roots, either real or complex. What is the relation of the coefficients $b,c$ and $d$ in order to have the roots inside the unit sphere, that means $ |x_i| < 1$ for $i=1,2,3$ ?

I was wondering if there is a condition which ensures all the three roots have absolute value >1
baymax, if p(z) is a polynomial of degree 3 without 0 as a root then q(z) := z^3 p(1/z) is also a polynomial of degee 3 without 0 as a root, and a nonzero number w is a root of q if and only if 1/w is a root of p. so such a p will have all roots with norm more than 1 iff q has all roots with norm less than 1. the coefficients of q relate to those of p in a simple way, so conditions from answers to the question you posted can be used to get conditions of the kind you are looking for
Yes thats nice! I also just got a similar thought just after posting this!
Thank you!
To determine the conditions for $(\lambda^3 + a \lambda^2 + b \lambda + c = 0)$ to have all roots with absolute value greater than one
$$ \lambda = \frac{1}{z} \implies P\left(\frac{1}{z}\right) = \frac{1}{z^3}(c z^3 + b z^2 + a z + 1) = 0$$
gives $c z^3 + b z^2 + a z + 1 = 0$
For the polynomial $(c z^3 + b z^2 + a z + 1)$ to have all roots inside the unit circle, the following conditions must be satisfied
using Routh--Hurwitz criterion
$
\begin{aligned}
&c > 0 \\
&b > 0 \\
&a > 0 \\
&b a - c > 0 \\
&c (b a - c) - b^2 > 0
\end{aligned}
$
Is the above correct?
05:04
0
A: Vector bundle construction theorem

Snailman$\mathbf {question 1}$ $E$ $=$ $\bigcup_{\alpha \in I} {\alpha} × U_{\alpha} × \Bbb R^{n}$/${equavalance- relation=:\rho}$. (The cup is disjoint union.) For Hausdorffness, take two elements first. Then, by the Hausdorffness of $M$, if $\pi(x) ≠ \pi(x')$, they are contained in each neighborhoods ...

0
A: Reference for theorems in Hirsch

SnailmanProof of 4.1 By definition of regularity of $f$ and $f|_{\partial M}$, $T_x f((\partial M)_x \cap (f^{-1}(y))_x)$ = $T_x f((\partial M)_x)$ $\cap$ $T_x f(f^{-1}(y))_f(x)$ = $(f|_{\partial M}(\partial M))_x$ $\cap$ ${y}_{f(x)}$ = ${{y}_{f(x)}}$ = $T_x f((f^{-1}(y))_x)$. So, $T_x f((f^{-1}(y)_x)$ $...

Why anyone doesn't give me vote? Is my proof not specific?
Or in above 2nd post, Is it 'not good' or wrong?
In second post, is there anything to be edited or add more specific information about point 'x' in the proof?
05:31
one of those was posted only an hour ago? i don't think the site moves so quickly that i would expect any votes in that time. i dunno about the other one. not everybody knows differential topology. people who don't understand the question (which has only 2 upvotes since 2015) are probably not going to read, let alone upvote, any answers.
i used to know a tiny amount of differential topology, and don't see how your answer to the question re hirsch is responsive to the question. part of this might be due to the fact that the original post is not really one question, but more like two questions (when site guidelines would suggest one question per post). or maybe it's a request for a single reference and not for two answers to the two questions. either way, i don't understand it.
i'm not saying that i doubt you, i just don't have the understanding that i would need to have in my head to upvote the answer. this is pretty high level stuff for MSE and i would not expect tons of upvotes generally.
when i answer a really old question, it is not uncommon for me to get at max 1 or 2 upvotes, and they usually don't arrive right away. just to put the site activity in perspective. a lot of the really busy stuff and the high activity is on lower level questions.
 
2 hours later…
07:22
for finite vectors spanning a vector space, we can argue that there exists a linearly independent set inside this that spans the same, by starting with one element $\alpha$, and asking algorithmically, does this span the set? if not, there is an element outside the span($\alpha$), so we append this outside element and ask the question again, and so on... Now, is thre a way to extend this argument to arbitrary sets of vectors?, how do I generalise this algorithm for infinite steps
Is the evaluation of $LM[u]$ correct?
basicall y i wanna show that, if an arbitrary set S spans V, there exists a linearly independent set inside this set that also spans V
I don't think the evaluation is correct.
There should've been an extra term $-x\frac{\partial u}{\partial y}$ in the expression for $LM[u]$
But I don't get why isn't that included there?
\ok, speaking practically we require |S|$\geq$|L| where $|S|$ is size of spannign set, and $|L|$ is size of any linearly independent set
@leslietownes Can you help me with this a bit?
07:46
@nickbros123 this requires axiom of choice
08:24
I am getting hung up on a sentence in my textbook:
> Since the indicator function of an open rectangle (i.e. $(a_1,b_1)\times (a_2,b_2)\times\cdots\times (a_n,b_n)$) is the increasing limit of a sequence of continuous functions with compact support, the probability measure $\mu$ is also determined by the values of $\int \varphi\,\mu(\mathrm{d}x)$ when $\varphi$ varies in the space $C_c(\mathbb R^n)$ of all continuous functions with compact support from $\mathbb R^n$ into $\mathbb R$.
1. Why is an indicator function on an open rectangle an increasing limit of a sequence of continuous functions with compact support?
2. I don't see how $\mu$ is determined by the values $\int \varphi\,\mu(\mathrm{d}x)$. Is this clear to someone?
09:10
This was a quite useful answer to 1. and 2., i.e. that all the nonnegative bounded continuous functions, when integrated against the measure $\mu$, determine $\mu$. Although I believe the continuous functions with compact support are a subset of the bounded nonnegative continuous functions, so the statement in the quoted passage seems to be weaker.
10:01
In the link I posted above, does anyone know if $$f_n(x)=\min\{n\cdot d(x,U^c),1\},$$where $U$ is open, have compact support?
10:21
@Thorgott Consider a fibration $M = \Sigma_2\times S^1\xrightarrow{f} S^1$ where $\Sigma_2$ is a genus 2 closed oriented surface. If there is an embedded closed oriented surface $\Sigma$ in $M$ which is transversal to a fiber $S = \Sigma_2$ and not homotopic (or isotopic) to $S$. Then $f(\Sigma) = S^1$?
I wonder in this case if it's possible that $\Sigma$ is embedded in $\Sigma_2\times (-1,1)$ once I view $S^1 = [-1,1]/\sim$ and $S = \Sigma_2\times\{0\}$.
10:35
@psie It depends on $U$. If the closure of $U$ is compact, then $f_n$ have compact support. In my case, I was considering $U$ to be an open rectangle, so indeed, they have compact support.
 
1 hour later…
11:51
@SineoftheTime , by any chance when you have time could you go and see what I wrote in "the h bar"?
I tried to use the conservation of energy method, to confirm that the result was the same, but I got 2 different results
 
1 hour later…
12:57
Fixed it anyway 👍
 
2 hours later…
15:07
@psie bounded open rectangle
15:44
@SoumikMukherjee could u hint to how i would use AC to prove this?
@nickbros123 eiter transfinite recursion or take a maximal set by Zorn's lemma
If $V$ is your vector space ten well-order $V$ and proceed wit your aloritm by transfinite recursion (wen a coice is required, pick least element wrt your well-orderin)
What happened to your typing?
On mobile?
no, I'm writin on a latop and two of my letters don't work on it
and I didn't brin te keyboard
I just don't feel like copy-pastin/usin auto-correct
16:22
@Pizza ok :)
I see, g and h aren't working
anyone want to play?lichess.org/WFElFcYz
@SineoftheTime hi
@SineoftheTime i didnt get anyone on the physics chat...
16:45
@RyderRude they know they will lose
@SineoftheTime maybe :P but they never play much
maybe we can try some other chat
im on philosophy chat sometimes
but its too empty
modern day pilosopy
no new ideas are recognized it seems
@Jakobian there r still people contributing new things.. e.g. Dennett gave some thought experiments to prove qualia r illusion
modern day philosopy sucks
@RyderRude do people actually consider tem just like tey would consider say, Nitzce
it is an interestin topic - ow istorically reconized do tins are before tey become actually reconized
16:56
@Jakobian im not sure how seriously they r taken. i personally found them flawed
@SineoftheTime maybe but i havent studied it personally
the arguments r in the qualia section en.m.wikipedia.org/wiki/Eliminative_materialism
in the Quining Qualia section
@RyderRude sounds like aruin on definitions
i think that qualia r fundamental
humans abstract out information from qualia. but an illusionist would say that information is all there is
idk, tis qualia tin doesn't sound realistic
and yet u have it :P
> Examples of qualia include the perceived sensation of pain of a headache, the taste of wine, and the redness of an evening sky.
I find te two first examples different from te tird
17:05
all three r different here. the first is pain qualia, the second is taste, the third is vision
so tey mean perception of it as red
its just tat I find te prasin to be vaue
maybe. philosophy is not about well defined things
No, as in it as two different meanins
17:08
oh
wereas taste or pain relate to wat one feels, redness feels like a fundamental property of a tin
but redness is also what one feels..
you can perceive it
just like ow you taste sometin but it also as some inerent taste based on ow its made out of
yeah.. r u saying red is also possessed by the object being perceived?
as in, tins tat make it to be perceived to be red by us, yes
for taste its similar I uess
17:10
yes.. they do possess that. but same goes for taste, like u noted
pain is different in this regard i guess
so te explanation on wikipedia could be more precise
pain is feeling ur own body. whats possessed by the body must be some sort of damage
@Jakobian maybe..
touch is another qualia
@RyderRude to be perceived as red you need to activate certain tins in te eye tat differ dependin on te colour and it sould be more complex tan pain
but it sould be similar to ow pain can be perceived across various limbs and oter tins
tere are different types of pain as well as taste and colours to be perceived
yeah..i would say pain is also pretty complex. it also has a spectrum like color does
taste doesnt seem to have 1D spectrum
"pain" just is an umbrella term I uess
17:15
yeah
taste is larely based on smell
te actual taste is muc more simpler tan we tink
if you try eatin wit your nose closed you'll see tat taste is very simple
hmm.. idk :P
well, not 100% perfect since you still can't disable your nose
unless you et covid or sometin
17:17
also, u can taste something by smelling it
must be similar brain areas activated
yea, we taste tins by smellin tem, pretty muc
wat you smell is ow it will taste
also, we can imagine qualia to activate brain areas
like imagining a memory
it can give the feeling of seeing or tasting something
@RyderRude no its just tat it all enters trou te nose and its really smell tat you are mainly perceivin wen tastin sometin
from te front and from te back
@Jakobian oh
remember ow nose is connected to te mout from te inside
17:19
yeah
I'm not sure tis is 100% true but tis is my understandin of ow taste works
but the tongue is also primarily involved in taste
it has taste buds
maybe similar to the cone and rod cells for vision
yea, but tonue only ives you te simple version
tere's like wat, 6 different tastes?
i would say tongue is primary
but undreds of different smells
17:23
@Jakobian it's just broad categories
its te smell tat ives tins teir actual taste
idk...
@RyderRude I don't tink so
we also perceive space and time
I tink its literally just tat te tonue can only distinuis between like 6 tastes
and its te nose tat does te eavy liftin
17:26
taste can diminish when nose is closed, yeah
 
1 hour later…
18:42
@Thorgott interesting characterization of normality using covers
$X$ is normal iff every point-finite open covering can be shrinked
(I forgot a lot from Dugundji, I knew that there are some properties of it relating to shrinking, this is pretty cool though)
Did you knew of this theorem?
$X$ is normal iff every point-finite/locally-finite/finite open covering can be shrinked
(from the proof of theorem 6.1)
19:24
@onepotatotwopotato it sounds plausible, but I don't have an argument
@Jakobian I knew that locally finite cover of a normal space is shrinkable
and every point-finite cover shrinks to a locally finite one, so that gives (1) => (2)
never thought or knew anything about the converse direction
@Thorgott wdym by this? In general it doesn't as there exists normal metacompact spaces which aren't paracompact.
the converse argument is easy
I'm sure you'd know about it if you ever thought about it
still, as trivial as the converse might be, its interesting to point this out
 
1 hour later…
21:42
@Jakobian I misremembered
a point-finite partition of unity induces a locally finite cover, that's what I had in mind
doesn't solve the problem at hand
@Thorgott any partition of unity induces a locally finite one iirc
I know because I was thinking about it recently
what does "partition of unity" mean to you?
a family of continuous functions $f_i$ with $0\leq f_i$ and $\sum_i f_i = 1$
this is a theorem of Mather from 1965
the proof I know is based on Engelking's proof (I modified it here and there)
don't mind that I use natural numbers, this part of proof is for arbitrary indexing sets
by "induces" I of course mean that there is a locally finite partition of unity $h_i$ with $\text{coz}(h_i)\subseteq \text{coz}(f_i)$
there are some arguments I didn't give explicitly here, namely the one after I mention Engelking
22:29
@Jakobian ok, I reserve "partition of unity" for locally finite ones
this statement is also in tom Dieck's AT btw
@Jakobian I’m glad to see you’ve recovered the use of ‘g’ and ‘h’
22:44
@robjohn no I was just copypasting letters
@Thorgott Mather's theorem says you can do either without much consequences
yeah, I didn't know it was due to Mather

« first day (5097 days earlier)      last day (220 days later) »