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12:05 AM
hi, i had a question about determinants. i currently find them very unmotivated. of course there is the notion of "signed volume" and the determinant measuring how this volume changes under a linear map. but i find this a bit unsatisfying, firstly because even in the real case if one is not doing e.g. integration i'm not sure why i should be thinking about volumes, and secondly because this definition works for any vector space on any field (so no measure is required)
the other definitions i know of having to do with exterior powers are just as problematic since i don't really understand why one would look at those either
hopefully i'm not coming off as impossible to satisfy!
 
12:36 AM
det(T) is just the dual of the top wedge power of T. duh.
have you thought of it as a multilinear map, divorced from geometric stuff?
math.stackexchange.com/questions/668/… the answer from benabou is roughly what i have in mind.
everyone has different notions of what constitutes motivation. if you look on that page the majority seem to like the geometrical approach.
people who just like doing stuff with algebra and weird fields and things might like just focusing on alternating multilinear maps. you don't need to introduce exterior algebras to deal with them. but many find this approach too dry
 
yeah! i have been understanding it using the map on the top exterior power. and i agree the geometrical approach is nice, sorry for the subjectivity of the question. maybe this is just a matter of getting used to alternating multilinear maps, but since i am not so experienced with a lot of math i have little experience with even the relatively common ones (e.g. commutators)
so i still find them unmotivated, but it might be one of those "motivated by its consequences" things
 
if it makes you feel any better, with many classes of operators on infinite dimensional spaces it is possible to prove the nonexistence of anything like a determinant. pick whatever properties you want to extend, they don't go through.
i think one route to intuition is, well, historically it came out of geometry and solving systems of equations, and once you have it, you end up mostly working with through its properties because the formula sucks. and then why not step back and just consider those properties.
like so many other things in math, it may have emerged out of the festering swamp called g--m-try.
 
haha
i think one other thing which makes me curious is the fact that we can speak of determinants and their properties without reference to any choice of basis, purely as properties of linear maps, but this might be more a consequence of the preference for abstraction than anything
 
12:54 AM
the geometry monster walks the land. he doesn't need a basis, he just wants blood.
 
1:05 AM
the top exterior power is pretty much just the "vector space of oriented n-dimensional parallelipipeds in V" (where n=dim(V)). this is a one-dimensional vector space and a linear map f:V->V induces a map on this exterior power, which, by one-dimensionality, is necessarily given by multiplication with a scalar. this scalar is the determinant of f, i.e. the determinant is the factor by which the linear map scales parallelipipeds of the top dimension.
this is still geometric, but perhaps it makes it clearer that this does not really require a notion of "volume" or "measure"
 
yeah, this is the definition i'm familiar with. it just seems a bit arbitrary, but this might just be my inexperience talking
 
1:21 AM
@P-addict you could look at an existing answer, selected entirely randomly math.stackexchange.com/q/2013886/27978
 
definitely i like this answer/motivation, as the multilinear alternating map definition is an interesting algebraic characterization of the determinant (and is beautifully motivated by signed volumes), but maybe my question is more, "why care about multilinear alternating maps especially if we're not doing volumes?", or "why care about exterior powers?"
of course, i know one can, to the annoyance of people trying to help, always play this "why care about XYZ" game, and i'm not trying to do that! i was just wondering if we could go "one step further" than alternating multilinear maps.
of course personally i find linear algebra interesting and beautiful, which is more than enough reason to study it :)
 
1:36 AM
we care, because it tells us something about the behavior of the map itself
namely, a linear map is invertible if and only if its determinant is
 
yeah, this is an interesting point. maybe there is some "natural" condition we could impose on a homomorphism $\text{End}(V)\rightarrow k$, $k$ the underlying field, which spits out only the determinant? this seems related to the invertibility property
oh, actually, would any such homomorphism preserve invertibility?
never mind that's not true
or maybe it is but the proof i had in mind doesn't work i think
 
the determinant is not linear, so it'd be a homomorphism $\mathrm{End}(V)\rightarrow k$ of multiplicative monoids
I'm not sure if there are good conditions uniquely characterizing it as such
$k$ has a lot of multiplicative endomorphisms, which do not seem to be easily ruled out using some normalization
 
2:13 AM
Hi, if $\frac{1}{2}t_2\leq t_1\leq\frac{1}{2}$, how can we find the bounds for $\frac{2t_1-t_2}{1+t_2}$, i.e. what would $x,y$ be in $x\leq\frac{2t_1-t_2}{1+t_2}\leq y$? Thank you!
 
not enough information to give bounds.
 
Sorry, also $t_1,t_2\in[0,1]$.
 
sorry, you had once chance.
jk
 
oops, I forgot to specify sorry!
 
$x=0, y=2$
you always have $2t_1-t_2 \ge 0$ (and the denominator is always $\ge 1$), you could pick $t_1=t_2=0$ to get this.
 
2:21 AM
Could you please describe how you figured that out?
Or the general method of solving this type of problem?
 
i don't think there is a general method..
 
think about how to make the denominator big and the numerator small, and vice versa
 
@copper.hat Hmm can't $2$ only be obtained if $t_1=1,t_2=0$ but then $t_1\leq\frac{1}{2}$ from the first restriction so this can't be possible?
 
i agree with copper. unless 'this type of problem' means something very specific
 
Good morning
 
2:29 AM
can you believe this guy? he thinks it's morning.
 
well, for rational functions and linear constraints there's probably some notion of 'method' in the realm of cylindrical algebraic decomposition
but "make one side as big as possible and one side as possible" remains the basic idea
 
Good morning leslie, to you too. I will go to sleep now. See you tomorrow
 
that's my line.
 
@Semiclassical Yes I tried, 3n + 13, and it spins on 13, 26 and 52
 
i wonder what the largest cycle someone has found on an $n\mapsto 3n+b$ type map
 
2:46 AM
is this the Veritasium wave
 
no, that's the name of my prog rock band
 
@pritchard sry, should be $y=1$.
 
3:24 AM
An $m\times n$ row reduced echelon matrix (rref) $A$ can also be described as either all entries of $A$ are zero or there exists a natural number $r$ ($1\le r\le m$) and $r$ natural numbers $c_1,c_2,\cdots, c_r$ such that 1) $c_1\lt c_2\lt ...\lt c_r$, 2) $a_{ij}=0$ for $i>r$ and $a_{ij}=0$ for $j\lt c_i$, 3)$a_{ic_i}=\delta_{ij}, 1\le i\le r, 1\le j\le r$.
$\delta$ represents Kronecker's delta. I don't understand (3).
The problem with $(3)$ is: $a_{1c_1}=\delta_{1j}$ then $j$ can vary here hence we have on RHS a set of values but LHS is only one value.
 
the a_{i c_i} is the pivot entry in row i.
the sequence c_j spells out which columns the pivots are in.
 
hi leslie!
 
you have zeros below pivots and anything in a row that comes before the next pivot.
 
I know that leslie, the problem is with $(3)$ :-(
 
putting row echelon form in symbols is insanity. i get why people do it, but whyyyyyyy.
 
3:29 AM
I think $(3)$ is wrong.
 
no it's fine. it says you've got zeros above and below the pivot.
if there's a pivot in row i, column j, then if you run down the entries of the jth column you ought to get 0 except at the pivot.
 
But $a_{ic_i}$ for a fixed $i$ is a single value, then how can it equal $\delta_{ij}$ in which j can apparently vary.
 
let j vary. if it's not i you get zero.
just like if the matrix is in RREF you hop and skip down a column with a pivot in it, you get zero unless you land on the pivot.
 
hmm, hard to digest yet. so (3) is replacement for "A is row-reduced".
 
i agree that it's very hard to read, but unless i'm missing something, it's not wrong.
some people say 'row reduced' to mean only row echelon form and not reduced row echelon form. watch out for that.
it's capturing the RREF aspect.
 
3:33 AM
leslie, this definition is from Hoffman &Kunze's section 1.4
@leslietownes here by row reduced, I mean -all columns containing pivots have their non-pivotal entries zero.
 
ok. then yes. (3) is capturing that.
but beware that in the wild, sometimes "row reduce" only means "put in row echelon form" i.e. zeros to the left of pivots but not necessarily above and below them.
i looked in my lecture notes to see if i had a good slide on this, because i want to compete with ted in promoting my own offerings.
 
@leslietownes :)
@leslietownes i understand that. some people also use "pivots" and some use "leading non zero entry" to mean the same thing :)
 
i am among these people.
 
I also say pivots since i watched professor Gilbert Strang's lectures :)
15 mins ago, by Koro
An $m\times n$ row reduced echelon matrix (rref) $A$ can also be described as either all entries of $A$ are zero or there exists a natural number $r$ ($1\le r\le m$) and $r$ natural numbers $c_1,c_2,\cdots, c_r$ such that 1) $c_1\lt c_2\lt ...\lt c_r$, 2) $a_{ij}=0$ for $i>r$ and $a_{ij}=0$ for $j\lt c_i$, 3)$a_{ic_i}=\delta_{ij}, 1\le i\le r, 1\le j\le r$.
But still, I am having a hard time digesting how can a singleton value equal a set
 
so my notes spend several slides on row echelon form, with blanks where we would have done examples. then there's one slide saying what RREF is, two blanks for two examples, and then the slide says "The book calls this 'Gauss-Jordan elimination.' You can use it, or not. I will not separately test this topic and if you can put a matrix in row echelon form and solve systems of linear equations without it that is fine with me."
i shouldn't have put that in print.
 
3:40 AM
@leslietownes I understand this is the idea but it's not visible to me from the equations alone.
 
this is one of those things that should not be put into symbols.
 
@leslietownes haha, solving by elimination and not using RREF was fine with you :)
My college professor would have deducted grades if he asked about finding rank of a matrix in exam :(
 
it should be a_{j c_i} = delta_{ij}, should it.
 
no
 
somethings are best described pictorially
 
3:45 AM
my attitude was, the arithmetic mistakes you do with backsubstitution are really no different from what you do with matrix operations, so i'm not going to ask you to go to RREF unless you want.
koro one thing to keep in mind is that people tend not to formally reason with matrices in RREF, so it is not really that helpful to symbolically express what RREF is. it's a calculational tool.
there's the fun fact that the product of RREF matrices is RREF, but who RREF'in cares, as far as i'm concerned.
 
@leslietownes hmm but still it is painful for me to not be in a position to either confirm the definition as right or wrong. :-(
 
EM4
@leslietownes hiiii!
 
how about a_{ij} = delta_{i c_j}? that feels better.
 
@leslietownes Leslie, the copy I had had this $j$ made very small so I confused that with $i$ so the problem is solved.
 
oh.
shame on paper books, they don't let you zoom in.
 
3:58 AM
So we have $a_{ic_j}$
 
i wonder how many times they use that formalism later in the textbook.
 
@leslietownes it was a softcopy so it let me zoomed in. Sorry about causing confusion :(
 
it is really never used in $\mathbb{R}$ numerical work.
 
i was more befuddled than confused.
 
After zooming in a lot, I saw it was $j$ and not $i$
 
4:00 AM
depending on the age of the book, typesetting of subscripts was really bad about things like that. i vs. j.
this is one of the things that motivated knuth to develop tex
 
how did they type before tex/latex?
 
my non numerical exposure is limited, so my perspective slightly warped, but i do not understand the huge amount of space given to rref.
 
i think it's because it saves space (else you'll have to write so many +, - signs)
 
knuth wrote a good (free!) article about it. projecteuclid.org/download/pdf_1/euclid.bams/1183544082
he takes one journal and gives examples of how its typesetting evolved over the years, pointing out the good and the bad. very fun reading if you're into it.
also, tex hadn't fully matured yet, so there's a fun work-in-progress flavor to it
copper i don't know if you saw it, but when i taught linear algebra it got one slide that ended in "you can use it, or not."
 
Also, I have seen in some books written before 1950, they write "shew that" instead of "show that"
 
4:04 AM
the selectrix had a key for doing it
other mortals had to use a type-it
 
"shew" was hopefully gone by 1950. unless people were putting on an affectation.
the US patent office is horrible at typesetting math. they will fail to subscript and superscript, they will miss fraction bars, everything. usually because the attorneys submitting the work regard all of it as noise anyway and nobody reads it.
 
haha
 
they are pretty good with typesetting sequences of nucleic acids. ACGCTAGTCACCGCTTACGATCGATCCTGGA
 
AAAAGGGGGGG
 
put some uracil in there. AAAAUUUUUGGGGGG
 
4:07 AM
copper, T and C are also important I think
not sure though
 
i'm surprised levis doesn't have a pair with some ACGTU derived name
 
there are companies that will print out any sequence you want. you could actually get a bottle full of AAAAGGGGGGGG if you wanted it.
that would be a weird birthday gift.
 
its all greek to me
actually no, i understand a smattering of greek
 
good morning leslie
 
why just the lawyer?
returning in a week for a few weeks to brush up on my accent
 
4:48 AM
0
Q: Definition of the module of compatible $\mathcal{U}$ families of $\Gamma$.

love_sodamIn Spanier AT (p.325), there is a notion $\Gamma(\mathcal{U})$ called the module of compatible $\mathcal{U}$ families of $\Gamma$ where $\mathcal{U} = \{U\}$ is a collection of open sets and $\Gamma$ is a presheaf of modules on $X$. Definition. A compatible $\mathcal{U}$ family of $\Gamma$ is an...

 
guten morgen
 
5:04 AM
Genosse
 
 
1 hour later…
6:20 AM
the totally tropical taste
 
did we lose the cold war? are people still saying genosse?
one time i was in stockholm and i was with some extreme left-wing friends, and a taxicab driver refused to drive us to the bar we wanted to go to. i didn't know at the time, but it was some kind of hotbed of swedish communsim.
probably in my FBI file now.
we were going there for the pricing of the refreshments, not the politics. i didn't plan on reading any marx while i was there.
whatever.
 
i think it is 'amusing' how many in the us view sweden as some sort of social nirvana
 
6:37 AM
i liked it when i was there although i don't think i would want to live there full-time.
my friend lived in the equivalent of a housing project. it was cool but he had like 750 sq ft to his name. i need a tiny bit more.
 
my main memory was people were not interactive and alcohol was prohibitive. returned quickly to copenhagen on the ferry where alcohol was less prohibitive.
 
copenhagen for me was like a time travel back to the 1970s. which i had not experienced.
they had smoking zones in the airport. like a little rope around where people could smoke.
that reminded me of restaurants in california circa 1985.
 
it had many similarities to ireland at the time except people were not interactive
 
except it wasn't 1985.
danes are distant up to a point and then beyond it there's no filter. there is no middle ground.
i have a gradient between 'i don't know you, go away' and 'here's the worst thing in my mind right now.'
 
yep, that would be my impression
 
6:50 AM
i inherently sympathize with any culture that focuses on the worst thing in one's mind
 
i will confess my innermost secrets to a random person but reveal nothing to those closest.
 
is that cultural?
i have this problem too.
 
i suspect something to do with potatoes
 
i met someone at a party once for about 2 hours, and dumped like half of my life's story on her. we're friends on linkedin and still check in from time to time.
she knows stuff about me that my wife doesn't know.
and i know an enormous amount of stuff about her life that her husband doesn't know. this is continents apart so there is nothing pervy about any of it.
it's just easier to talk to people if you know there won't be some kind of immediate interruption in your social circle.
 
I think it has something to do with the presence of social media and potatoes.
 
7:02 AM
that is a good observation.
 
Hello. I have a question. So in a game, if you win in the $i'th$ round you are paid $2^i$ dollars and to play the $i'th$ round you have to pay $2^{i-1}$. The probability of you winning in any round is 0.5. Whats the expected loss player incurs before profiting from the game. So if I win in round $i$ my win trumps all my losses, so I made a profit.
 
someone should do something about the potatoes.
 
So let X denote the money lost in round $i$ (such that I have never won before) and p(X) be its probability. So the expected money lost upto round $i$ would be $\frac{2^{i}-1}{2^{i}}$ and I took its limit to infinty. Am i correct?
 
7:53 AM
Hello!!
 
@Shobhit wouldn't you always win a net of $1$ at the end of the game?
 
From Gauss elimination method with pivot do we get $R=G_3^{-1}P_1G_2^{-1}P_0G_1^{-1}A$ or $G_3P_1G_2P_0G_1A=R$ ? Which is the correct one?
 
@Shobhit Beautiful problem... i dont have the mathematical skill required to solve it so I am going to brute force it.
So if you win the first round you get 2^1=2$ and loss 2^(1-1)=2^0=1$? So if you win 1st round, your win trumps all your losses and so you made a profit right?
 
@AdilMohammed see if you find my result. You always win a net of $1$ at the end.
if I understand the rules.
 
His question was "What's the expected loss player incurs before profiting from the game" and we both figured out profit occurs from first-round itself right? so the answer to the "expected loss player incurs before profiting from the game" is 1$ right?
Oh probability is 0.5 I havent looked at that part yet while brute forcing.
 
8:02 AM
@AdilMohammed oh, okay, I had just read the rules and not the full question.
@AdilMohammed That is a lot of money, it seems to me.
 
@AdilMohammed To play the i^th round you pay 2^{i-1} and if yyou win you get 2^i and if not you get 0 back.
@robjohn is the question not clear?
@robjohn if the game ends when you win, then yes net would be 1.
 
@Shobhit yes, but the expected loss before winning is very large.
 
would it not be what I wrote above? I dont know how to mark it
would it be 0.5(1) + 0.5^2( 1 + 2) + 0.5^3 (1+ 2 + 2^2)...?
 
Winning on the first round is probability $\frac12$ with a cost of $1$. winning on the second round has a probability of $\frac14$ with a cost of $3$. winning on the third round has a probability of $\frac18$ with a cost of $7$. That is the probability of winning on round $n$ is $2^{-n}$ with a cost of $2^n-1$. Adding up the losses times their probabilities gives a divergent series.
 
would it be 0.5(1) + 0.5^2( 1 + 2) + 0.5^3 (1+ 2 + 2^2)...?
ok.
Thank you.
 
8:17 AM
i'm doing my best to consume as many potatoes as i can.
 
@copper.hat spudboy
 
gimme potatoes or give me death
 
don't tread on my potato?
 
:-)
 
9:15 AM
Does someone of you have an idea about my question?
0
Q: QR decomposition with permutation matrix

Mary StarAt the $QR$ decomposition with permutation matrix is the matrix $R$ equalt to $R=G_3^{-1}P_1G_2^{-1}P_0G_1^{-1}A$ or $G_3P_1G_2P_0G_1A=R$ ? Which is the correct one? In general it holds that $QR=PA$, right?

 
 
2 hours later…
10:52 AM
for $r,s \in \mathbb{R}$, we have three possibilities:$r<s$, $s<r$, or $s=r$
what's this property called?
looks like linear ordering
but this isn't quite what I'm trying to get at...
isn't there some first order logic property here that allows, for arbitrary $r,s \in \mathbb{R}$, the possibility that $r=s$?
 
why s = $I$.$d$ and not $I$cos(theta)
In the book they said the projection of l onto the ray direction, d, is computed: s = $l$ . $d$
yesterday we agreed that the dot product is the projection I multiple by the magnitude of $d$
that's from real time rendering 4ed
 
11:18 AM
In mathematics, the law of trichotomy states that every real number is either positive, negative, or zero.More generally, a binary relation R on a set X is trichotomous if for all x and y in X, exactly one of xRy, yRx and x = y holds. Writing R as <, this is stated in formal logic as: ∀ x ∈ X ∀ y ∈ X ( [ x < y ∧ ¬ ( y < x ) ∧ ¬ ( x...
@fido9dido is d a unit vector?
 
ah, there we go. thank you very much
 
direction vector normally a unit vector yes
wait a sec
i see, my bad thanks again
 
another question, is $\forall x,y \in S: x=y \lor x \neq y$ an axiom, or is it derived?
i'm trying to derive it from zfc
 
11:35 AM
In logic, the law of excluded middle (or the principle of excluded middle) states that for every proposition, either this proposition or its negation is true. It is one of the so called three laws of thought, along with the law of noncontradiction, and the law of identity. However, no system of logic is built on just these laws, and none of these laws provide inference rules, such as modus ponens or De Morgan's laws. The law is also known as the law (or principle) of the excluded third, in Latin principium tertii exclusi. Another Latin designation for this law is tertium non datur: "no third...
 
ah, thank you so much, these links have been super useful
 
 
2 hours later…
1:32 PM
Lets say we have x points with us on a line. The cost of visiting any point is 1. After we visit any point once, it stays visited forever. Each time, we randomly visit any point on the line which has not yet been visited. We keep doing this until our points become such that out of any n consecutive points, more than one has been visited. What is the expected cost in order to achieve this state ?
Could anyone help me with this ? Can't even figure how to start :/
 
 
2 hours later…
3:04 PM
$\zeta(0)-\zeta(-s)+\frac{1}{2}\zeta(-2s)-\frac{1}{6}\zeta(-3s)+\cdot\cdot\cdot=0$ How do you solve this for $0$?
I think the only method that might shed light is a computer algebra system
 
 
1 hour later…
4:16 PM
what is it with people and the zeta function
 
4:29 PM
@geocalc33 just for clarity, would you explicitly write out the terms? it looks as if you might have intended $\sum\limits_{k=0}^\infty\frac{(-1)^k}{k!}\zeta(-ks)$
 
 
1 hour later…
5:47 PM
@geocalc33 It looks as if that is positive for $s\gt0$ and negative for $s\lt0$ and blows up near $0$
Of course, I need to look further into evaluation in $(-2,0)$
 
6:01 PM
There will be poles at $z=-\frac1n$ for $n\in\mathbb{N}$.
 
6:39 PM
Hi. I am studying PCE (Polynomial Chaos Expansion). If anybody has knowledge about this and wants to have a discussion, feel free to ping me.
This discussion can be in any form which makes the other people comfortable. (Starting from a voice call where the opponent doesnt need to talk for privacy issues but I can explain the thing and you can see holes in my understanding or ask me questions about it, to some text messages in the format mentioned above.)
 
7:33 PM
Is it possible to use the Laplacian function when solving PDEs? This is my current attempt: i.imgur.com/W9jTxbF.png
 
In your answer, @robjohn , you stress recognizing the independent variable when confronted with the little-o notation. In this particular case, you recognize both $h$ and $u$ as independent variables, correct?
 
*in Mathematica I mean
 
@schn Can you look at the definition of $o(g(x))$ and try applying it to a given situation instead of trying to abstract a calculus of little-o? It applies differently in situations where the estimate needs to be uniform in another variable from where it doesn't. It is highly context dependent, and because of the definition, it is not easy to specify uniform little-o-ness.
 
@robjohn hmm 🤔
Thanks for your comments
 
The sum is equal to $\sum_{j=1}^\infty e^{-j^s}\qquad\text{(converges for $s\gt0$)}$
and $-\frac12+\sum_{j=1}^\infty\left(e^{-j^s}-1\right)\qquad\text{(converges for $s\lt-2$)}$
in the interval $(-2,0)$ there are poles and zeroes between those poles.
 
7:49 PM
@robjohn You’re saying that this sum is equivalent to the sum of zetas?
 
yes
 
Hmm
 
@geocalc33 if the definition is as I stated above
3 hours ago, by robjohn
@geocalc33 just for clarity, would you explicitly write out the terms? it looks as if you might have intended $\sum\limits_{k=0}^\infty\frac{(-1)^k}{k!}\zeta(-ks)$
 
Yes that’s what I intended
@robjohn how many zeros do you think there are? And where do you think they are located?
 
@geocalc33 I think there should be a zero between each pair of consecutive poles.
2 hours ago, by robjohn
There will be poles at $z=-\frac1n$ for $n\in\mathbb{N}$.
 
7:57 PM
Makes sense
So countably infinite number of zeros
And these are purely real zeros?
 
@robjohn Offhand, this depends on order of pole and residues?
 
@TedShifrin the poles should all be order $1$
but the signs may be wrong.
 
I haven't analyzed the details. OK, so if the residues are of the same sign, I'll believe you.
 
Hi @TedShifrin
 
as I said, the sign are alternating so there may be no zeros between them
 
8:03 PM
Hi, a @Balarka.
 
@geocalc33: there may be no zeroes between the poles. They are order one, but the signs of the residues alternate, so it could look like cosecant where it never crosses the axis.
 
Ted seems to be playing the rôle of killjoy this afternoon.
 
@robjohn okay
What would change if gamma(1+1/s) was added to the sum of zetas? (Sorry on mobile)
I’m attempting to sketch the graph
 
8:20 PM
Scrap what I said about $x\gt0$. I made a bad assumption while summing that.
I am looking at it again.
 
@TedShifrin I have a simple topology problem if you'd like to hear
 
Simple? Coming from you? Ha.
 
Yeah, it's just fiddling with some open sets.
Let's say $S, L$ are two nonproperly embedded noncompact submanifolds (without boundary!) of a manifold $M$, and $S$ is contained in the frontier (or topological boundary) of $L$, $S \subset \partial L$.
I take a tubular neighborhood with shrinking width $\nu(S)$, $\nu(L)$ for both $S$ and $L$, respectively, let's say with tubular projections $\pi_S$ and $\pi_L$ respectively.
 
So $L$ in particular is not properly embedded ... as you need frontier.
 
Right.
I would like to choose a smaller tubular neighborhood with shrinking width, $\nu'(S) \subset \nu(S)$, such that $\nu'(S) \cap \nu(L) \subset \nu(S) \cap \pi_L^{-1}\nu(S)$.
($\pi_L^{-1} \nu(S)$, of course, means $\pi_L^{-1}(\nu(S) \cap L)$)
 
8:28 PM
My brain is not functioning well enough for all these letters.
 
Well, mine neither, which is why I am writing to clear the fog :)
It seems plausible enough but just need to make sure I have an argument.
 
So why do you need the $\nu'$ at all?
 
Here is the point. $\nu(S) \cap \nu(L)$ is wherever the pair of tubes intersect. But $\nu(S) \cap \pi_L^{-1} \nu(S)$ is a subset of $\nu(S) \cap \nu(L)$ consisting of points $p$ such that $\pi_L(p) \in \nu(S)$ as well.
Apriori this can be a strict subset. Not all points in the intersection of the pair of tubes need to map in $\nu(S)$.
 
I need a concrete example to focus on. Is the issue that the tube around $L$ is not extending nicely to $\partial L$?
 
Exactly. Let me show you a picture, I have a perfect one.
 
8:38 PM
@geocalc33 It appears that $f(0)=-\frac1{2e}$ and $f'(0)=-\frac1{2e}\log(2\pi)$, which doesn't comport with what I said earlier. So one or the other approach is not correct. I will get back to this later. Sorry.
 
$S = \{0\}$ is the fat dot, $L = \Bbb R \setminus 0$ is the complement of the fat dot in the $x$-axis.
$\nu(S)$ is a ball around the origin, $\nu(L)$ is the tube around the $x$-axis which shrinks as it approaches the origin.
Projection $\pi_L$ is not necessarily orthoprojection to the x-axis.
 
But why did $\nu(L)$ have to shrink necessarily?
 
It's a tubular neighborhood for the nonproperly embedded submanifold $\Bbb R \setminus 0 \hookrightarrow \Bbb R^2$
If it doesn't shrink you will not have a bundle structure.
It'll end up engulfing the origin, which will be a bad nbhd for the punctured real line.
 
it was cold outside.
 
Why can't it shrink to a finite but positive radius? I don't see what you're claiming.
 
8:42 PM
$\nu(L)$ has to be topologically equivalent to $L \times (0, 1)$, right?
So it must have "no fiber" over $0 \notin L$.
 
Why can't I restrict a tubular neighborhood of $\Bbb R$ to $\Bbb R-\{0\}$?
Yes, you forget the fiber over $0$.
 
Oh, sure, that's possible, I guess. Not the kind of tubular neighborhoods that I use, but yeah.
 
My brain definitely isn't working, but you were not making sense to me.
So I guess shrinking $\nu(S)$ to $\nu'$ is somehow dual to enlarging $\nu(L) if necessary.
 
Right.
 
So you're not talking about shrinking radii of normal disks. In the case of a point (as in your example), you have to take a weird open set, not a tiny disk.
 
8:48 PM
In my example, with my $\nu(L)$, or your $\nu(L)$?
 
Your $\nu(L)$. My $\nu(L)$ works just fine regardless.
 
It depends on how $\pi_L$ behaves. The picture illustrates an example of $\pi_L$ by drawing its level curves. In this case note that $\pi_L(\nu(S) \cap \nu(L)) \subset \nu(S)$.
But imagine that the level curves were arranged a little differently, then this might fail.
Reposting the image for visibility:
 
Actually, I don't see how to do your example at all.
I don't see how to get an open set containing the thick point.
 
$\nu(S)$ is just a ball around the thick point $S = \{0\}$. The projection map $\pi_S$ from that ball to $S = \{0\}$ is constant map to $0$.
 
But what part of that ball can you take to fit inside your shrunken tubular neighborhood?
It has to be missing a whole wedge coming in to $0$.
 
8:53 PM
Oh, most definitely you cannot choose $\nu(S)$ such that $\nu(S) \subset \nu(L)$. Absolutely agree.
But that was not my question.
@TedShifrin 100% agree.
 
OK, so, as I suggested, I'm lost in the notation.
I think I have some sort of COVID brain fog.
 
Nah, I'm explaining a little poorly because I was getting confused earlier too.
It's no big deal
 
9:08 PM
It's so nice that people can discuss your topics. I guess not many people are familiar with PCE right?
 
i'm not
 
9:23 PM
why don't you ask the question on main? better chances of an answer
 
9:51 PM
the discussion just goes into complete chaos when pce comes up
 
10:29 PM
The number of primes is infinite. This is clear. Since each prime number is a natural number and the number of natural numbers is $\aleph_0$, the number of primes is $\aleph_0$. However, I have heard people state that "infinity is not a quantity"; I don't get it. Their main objection is that infinity would be "quantifying the unquantifiable" - an absurdity. I tried asking about this on MSE but the question got shot down. What are your thoughts on this, folks?
 
@Koro Here is a lengthy, but not complicated, derivation of a general solution.
 
Here is the question I deleted.
 
11:00 PM
@Koro To solve $a_{n+1}=\frac 1{4-3a_n}$ using that solution multiply by $\sqrt3$ and get $\sqrt3\,a_{n+1}=\frac 1{\frac4{\sqrt3}-\sqrt3\,a_n}$ then set $b_n=\sqrt3\,a_n$.
Then solve $b_{n+1}=\frac1{\frac4{\sqrt3}-b_n}$
 
11:23 PM
Where(what room) would be my best shot at finding the answer to a probability question ?
 
0
Q: Why are these integrals equivalent to the Riemann Hypothesis?

mickRiemann Hypothesis is equivalent to the integral equation $$\int_{-\infty}^{\infty} \frac{\log \mid \zeta (1/2+it)\mid }{1+4t^2} \ dt =0$$ Many other integral equations exist that are equivalent. How to show that they are equivalent ? They usually include absolute value of a function. Why is that...

 
11:41 PM
@mick Note that $\log|z|=\mathrm{Re}(\log(z))$
 

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