1:09 AM
google used to display price charts with modifiable time ranges if you googled a stock ticker symbol. no longer, apparently. that's annoying.

1:20 AM
Done specifically to annoy you.

i think so. i've sent a complaint to google's inbox. i'm sure they've got their best people on it as we speak.

Munchkin should stomp her feet at their office.

her table manners are pretty good. you'd like that. she blows bubbles if you give her a straw but other than that.
she can eat a foil-wrapped burrito bit-by-bit, not unwrapping the whole thing at once, just like a real person.

She’ll call me names, I’m sure. BTW, Screech is unconed.

the fates have lifted the cone of shame.

1:27 AM
Thank goodness

is she OK? we got livvy shortly after all of that stuff. she had weird belly fur due to the shaving needed for the operation but seemed completely past it.

Yeah, it was her furious personality that made her destroy various claws after surgery. Now disinfected and healed. She’s back to biting and scratching me.

olivia's nightly ritual is to sleep with munchkin after munchkin is put to bed, from about 8 to 9:30. then she roams the house causing havoc until we both go to bed.

No havoc after that? How unfeline.

last night my daughter decided to aggressively pet olivia instead of going to sleep. olivia put a stop to that by biting her.
oh, she marches in maybe 2 or 3 am and howls at us. if she's not in bed pawing at my wife's face.

1:35 AM
She and Screech would be best friends.

the other thing she does is if you're sleeping on your side, she'll climb onto the side and sit exactly on top of you, so you can't move without destabilizing the cat.
i have no idea why humans domesticated these things. it seems like a raw deal.

1:49 AM
She’s far more reasonable than munchkin.

2:18 AM
Wrong forum. but just in case, what's the difference between \mathrm{foo} and \operatorname{foo} ?
I mean operationally :-)

well, don't leave us hanging. what's the difference?
i never used mathrm in real life. only operatorname, and only for names of operators.

roughly operatorname treats things like it would \sin, etc, whereas \mathrm just puts the leter in Roman

i thought it weird that in whatever flavor of tex i was using, \ker had first class treatment but \null did not. you had to do operatorname{null}. i blame a conspiracy of algebraist lizard people.
ooh, so i was doing good tex and not just the one thing i knew. brilliant. :)

since my advisor would not let me write papers in latex, i used it as much as i could for non paper stuff, and any idiosyncrasies paled in comparison to shitfroff, sorry, i mean troff

oh wow you actually had to use troff?
george bergman in the math department used troff with many customizations, i think until the late 2000s.
and you could very much tell.
i did some work with a guy who had learned on some very early tex flavor and he was always dropping insane garbage and hacks into what was fairly bog standard latex orthodoxy.

2:36 AM
troff was brilliant compared to paying my typist (coincidentally named leslie, he was great, just it added a whole process to the document creation that no longer exists) for my masters thesis (in my country of origin), but compared to the less unpredictable latex it was a pain.
if i don't follow someone else's setup i generate my own weirdness.

gentlemen.....good morrows......is it reasonable to claim that the basis of the row space of a matrix characterizes your k-dimensional manifold and the basis of the column space would characterize the set of the curve(s) that are created from intersecting surfaces?
it is still morrow on the west coast right? sun's still out.....

no sun here

no son here either. he is in santa cruz
manifolds are many, that is all i can contribute.

@leslietownes i have never used operatorname. The difference is spacing. Operators space.

a bit like myself.

2:45 AM
I should use operatorname for things like div and curl, but it’s too much typing unless I define a macro.
@dc3rd Huh?
BTW, \text is shorter than \mathrm

Thanks!

I was trying to elaborate on the idea of getting the k-dimensional manifold for the level surface. SInce we would need the rank of the $DF(a)$ matrix to to determine the amount of free variables I was just thinking to myself "since the rank is the number of linearly independent vectors in the row space which is a basis for that subspace, then would that serve as a basis for the manifold?"
.................but typing that out now I see that the row space is one subspace, but the manifold would be a subspace of the row space....I think...

what level surface?

You’re writing garbage.
Talk about the tangent space of the manifold at a point.
The manifold is a level set of a vector-valued function.

I'm trying to tie together the idea of the manifold with the rank............I'll discard the prior thought....and think about the tangent space

2:55 AM
The rank of the derivative is maximal. What does it tell you?

smacks copper

well rank will give you the dimension of the row space, number of rows that are lin-independent in the matrix and as a consequence I could get the number of "free variables"
and all of that has nothing with gradients in it....so I go back to read some more

With regard to the manifold?
The number of free variables is the dimension of the nullspace, which is the dimension of the tangent space of the manifold.

and having some vector we can "translate" our nullspace up to become the tangent space then.

3:01 AM
dualling with manifolds?

Tangent space is a subspace. Tangent plane, if you want, is the translate to the point.

That is correct.....sharper language from me is required

That distinction is mine, not universal.

what do you mean by maximal in this context? ....I know it is "something of largest amount"...

Rank equal to number of rows.
Think which dimension is larger and what the largest rank can be.

3:11 AM
which dimension?.....between the rowspace and the nullspace?

No, $m$ and $n$.

oh ok.....so for clarity let's say $m$ rows and $n$ columns. If $m < n$ wouldn't the rank have to be $m$ at most and if $m > n$ wouldn't it be that it has to be $n$ at the most?

3:43 AM
I have an unrelated question. In that IF statement, I want to say if there exists a v in \hat{V} such that ... and then I used the value of m that I expanded from v
I am working on my thesis, is it clear? if not how to express it more clearly

3:55 AM
@Node.JS i'm not sure what your question is, are you asking whether that if statement is valid?

I am asking if what I described in English and the Algorithm in the photo match

@Node.JS you're using $(v \equiv (X':m))$ as $v$ is congruent to $X'$ modulo $m$?

No. This is not math. This is about Contex-Free-Grammar
My question was more about the way I wrote that IF statement

just the antecedent of the implication or both the antecedent and the consequent? sorry I'm honestly not sure what your question is
if it's just about the antecedent: strictly speaking the comma should be a conjunction

How do you say in a formal Algorithm that: IF v is in V such that conditions X and Z has to be true

4:05 AM
if v is in V and X and Z
no comma hehe

Got it
Thanks

np
the shape is just $(A\land B \land C) \implies Q$
@Node.JS oh btw if whatever you're doing uses quantifiers, you need to add a universal quantifier in there for v

In the IF statement?

I don't know how the coding program you use works, but if it is in first-order logic, there needs to be a universally quantified $v$ over the entire implication if you use $v$ again in the consequent, and over the antecedent in all cases (if you want it to match to english statement)
but the above is only if you're working in first-order logic, otherwise programming languages take care of quantification

math.stackexchange.com/q/4288020/922120 I just add every definition needed to understand the problem. Any help will be appreciated.

4:19 AM
@TedShifrin Why are you not here? lgbtmath.org/People.html
Spectra is a professional association of LGBT mathematicians. It is a mailing list for LGBTQ+ mathematicians and their allies. This arose from a need for recognition and community for Gender and Sexual Minority mathematicians. == History == The association has its roots in meetups arranged at the Joint Mathematics Meetings (JMM) and a mailing list organized by Ron Buckmire. The association's name was coined by Robert Bryant and Mike Hill and references the mathematical concept of a spectrum as well as the rainbow flag. Its first official activity was a panel at the 2015 JMM with the title "Out...

I want to say if there exists a v in V such that conditions X and Y are true

4:41 AM
@Node.JS Please tell me how you composed this pseudocode, in latex? or did you use specialized programs?

there seems to be an absence of clarity as to what this pseudocode means. usually the point of pseudocode is to describe what is done without venturing into the syntax required by particular languages. i don't know much but am at a loss here.

there are no loops in first-order logic so i was confused

@leslietownes Indeed, the picture shows the algorithm, i.e. sequencing. It looks nice and clear. Was there a special program used for this performance? Here's what I'm interested in.
Gentlemen who are a specialist in vector matrix analysis. Help me simplify the equation, I'm puzzling until I come up with something worthwhile.

0

I want to prove that for any sequences $(a_n)$ and $(b_n)$, the following holds $$\limsup(a_n+b_n)\ge \liminf a_n+\limsup b_n \tag 1$$ I want to prove the above inequality only for the case when quantities on RHS are finite. I'm using the following definition. Definition: $\liminf_{n\to \infty} x... Is this correct? Thanks. In particular, I want to know whether the step just before$(3)$is correct or not. Thanks. I wanted to bring out using limsup, liminf a situation wherein limit rules can be applied. 5:13 AM 0 There are two vector expressions:$F_1=\frac{\frac{(v \times w)^T}{(v \times w)^Tu}\Omega}{\frac{(v \times W)^T}{(v \times W)^Tu}\omega}\tag{1}F_2=\frac{(v \times w)^T \Omega}{(v \times w)^T\omega}\tag{2}u,v,w,W,\Omega,\omega$- arbitrary vectors Is it possible to simplify$F_1$and$F_2$? 5:33 AM @Koro I didn't look at your proof because it is unnecessarily complicated. I added a simpler, direct proof. you did two infs, he's put a sup in there koro, i generally support this program. general laws about liminf and limsup are more fundamental and slightly more flexible than anything about limits. once you have a good library of them, you can avoid cases in many arguments. @copper: I have a sup also on RHS It appears that I need to actually read a question before answering. Dang. ba ba ba ba ba ba Ok, I will eat a little crow and see if I can come up with a 3 liner. That was a really sneaky sup hoeever. 5:38 AM but it can be done like the way you were saying. It's just that I wanted to know if step just before (3) is correct or not. I think I need a$\sup$of wine to continue. it's very handy to be able to apply one thing to a non-strict inequality and preserve the inequality, without any hypotheses about whether something exists. if i wrote an analysis book it might even start with lim sup and lim inf and only do limits later.$\sup (a_m+b_m)\ge a_m+b_m\ge \inf a_m +b_m$All I have to do now is to take sup on both sides again to finish and then take lim of course$\inf a_m=\inf\{a_n:n\ge m\}$But what about the step just before$(3)$in my post? Ok, I corrected my crow post. As a general rule, I find that rules with$\inf,\sup$can be managed directly with little need for$\epsilon$and${ 1 \over n}$thingys. I think I deserve an "In the fields" medal for that. Wow, I haven't even started drinking yet. A lot of stress these last few days. Tends to make me punchy. sorry to hear it. my job has been not great but home life has been OK so on balance things are fine. 5:52 AM copper, please also tell your view on step before$(3)$in the post. My second attempt was wrong. I am in$\sup$and$\inf$hell. I think you were trying to do something like this: 18 mins ago, by Koro$\sup (a_m+b_m)\ge a_m+b_m\ge \inf a_m +b_m$i was trying to do it in 3 lines, so wanted to avoid showing that It can be done in one line (as attempted above) also, I think. But I want to know if step before (3) is correct. I think that's correct.$\sup_{k \ge n} ( \inf_{m \ge k} a_m + b_k) = \liminf_n a_n + \limsup_n b_n$. it follows because$\lim_k \inf_{m \ge k} a_m = \sup_k \inf_{m \ge k} a_m$. 6:13 AM @Koro I find it a little hard to follow without all the relevant indices. I'll make the edit @copper. Thanks. @copper.hat this is indeed true but I am not sure about the equality before that. Perhaps it is best to ignore my mathematics tonight :-) I avoided indices as then I'm told that I use too many symbols :P but considering that I clarified in Definition the symbols I used, is the edit required? Well, generally I prefer less clutter, but in this case the particular details are important. Well, that is up to you, but you need to be clear what variables are implicit. the result i was trying to avoid was if$c_n \to c$then$\limsup_n (c_n+b_n) = c + \limsup_n b_n$. this is probably a good one to add to your bag of tricks. @copper.hat yes, this is an exercise problem just after the inequality exercise. 6:27 AM then you can do your problem in a straightforward manner noting that$\lim_n \inf_{k \ge n} a_k = \sup_n \inf_{k \ge n} a_k$. Edited. @copper Hello everyone, I apologice for the silly question but's really early in the morning ahaha: Suppose we have an orthonormal family$(e_n)$of functions in the Schwartz space of rapidly decreasing functions$S(\mathbb R)$, for instance the Hermite functions. We know that the linear span of that family is dense in the adjoint space$S'(\mathbb R)$of tempered distributions. If I define$P_m$to be the orthogonal projection on the span of$(e_1,...,e_n)$can I say that$P_m$is self adjoint in$S'(\mathbb R)$? Of course that for any$\eta,\xi\in S'(\mathbb R)$we have that$\langle \xi,P_m\eta\rangle=\langle P_m\xi,\eta\rangle\$ where the angle brackets denote the bilinear pairing between test functions and distributions