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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 :-)
just ignore, google answered me.
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
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?
smelly gradients?
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
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.
Q: Proof verification for proving the inequality $\limsup(a_n+b_n)\ge \liminf a_n+\limsup b_n$

KoroI 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
Q: Simplify complex vector expression

dtnThere 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
1 hour later…
8:00 AM
@AkivaWeinberger Glad to share
@TedShifrin Kind of. Some motivation for my work comes from physics, but I'm mostly interested in effective algebraic geometry
@Koro Looks good. I think it ended up being similar in that you used the above limit trick ($c_n \to c$).
8:33 AM
Suppose $X$ is a topological space with involution map $\nu:X\to X$ (i.e., $\nu\circ\nu =id_X$) and $X$ is a union of two closed subspaces $A$ and $B$ such that $\nu(A) = A$ and $\nu(B) = B$. Then $A\cup B$ can be embedded into $A\ast B$? where $A\ast B$ is a join operation.
Ignore the above question
Let $X$ is a topological space with involution map $\nu:X\to X$ (i.e., $\nu\circ\nu =id_X$) such that $\nu(x)\neq x$ for any $x\in X$. Suppose $X$ is a union of two closed subspaces $A$ and $B$ such that $\nu(A) = A$ and $\nu(B) = B$. Then $A\cup B$ can be embedded into $A\ast B$? where $A\ast B$ is a join operation.
8:55 AM
@PeterJohn I don't believe that. Consider six disjoint unit intervals $a,b,c,d,e,f$. Define an involution $\nu$ by identity map $a\mapsto b$, $c\mapsto d$ and $e\mapsto f$. Define $A = a\cup b\cup c\cup d$ and $B = c\cup d\cup e\cup f$.
@love_sodam what do you mean by identity map?
@PeterJohn Identity map on each factors $a$ to $b$, $c$ to $d$ and $e$ to $f$.
Hello everyone
I use the (very efficient !) formula searching tool approach0.xyz At first, I had no match ; I have then changed the query into $a_{n+1}=\sqrt{a_n+cn}$ which is in fact more natural :). — Jean Marie Oct 23 at 9:39
Ohh sorry wrong copy paste
Q: Is there a close form for this sequence $a^2_{n+1}=a_n+cn$?

user635988 Let $c$ be a positive real number and $(a_n)_{n\geq1}$ be the sequence defined by $a_1=1$ and the recurrence relation $$a^2_{n+1}=a_n+cn$$ $1$. Find a real number $p$ such that for any integer $n\geq1$. one has $a_n\leq p\sqrt n$ $2$. Find an equivalent of $a_n$, when $n$ tends to infinity, of t...

Could someone take a look at my question here?
9:20 AM
@leslietownes hey leslie, would you know if there's actually a statement corresponding to this, for arbitrary vector spaces that might be R^n or not?
3 hours later…
11:53 AM
@copper.hat thanks a lot @copper for confirming. :)
1 hour later…
1:08 PM
Q: Circle containing three points, maybe all collinear

jcklieWe all know that a circle is exactly defined by three distinct non-collinear points. But I need a way to solve the following problem (all in 2D): Given three points, calculate a circle with all three points on its border if it exists, else calculate a circle with minimum radius which has two poin...

1:41 PM
@huzaifaabedeen schematically, I’d proceed like so
First, check if the points are collinear, and if so figure out which one is in the middle. Then you can draw a circle which centered at the midpoint of the outer points and passes through the outer points, and this will certainly enclose the remaining point
That reduces the problem to the case of non-collinear points
For that, elementary geometry will work: draw the perpendicular bisectors of the three points and find their intersection to get the center. The radius is then the distance to any of the three points
2 hours later…
3:18 PM
@shintuku for general vector spaces (over R or C) you need an inner product to make sense of orthogonal projection but in finite dimensions the facts/proofs are roughly the same and amount to matrix algebra. in infinite dimensions you still need an inner product and get the same stuff if you work with subspaces that happen to be closed in the topology induced by the inner product.
with some fuss you can define linear maps that aren't continuous in the infinite dimensional setting, or work with not-closed subspaces, but the mental pictures we draw when we think about orthogonal projection stop working.
for example 'if this subspace is not the whole space there is room to choose something nonzero that is orthogonal to the subspace' underlies a lot of constructions/ideas and this is exactly what can fail if the subspace is not closed.
@AkivaWeinberger Autocorrect really likes the word "duck".
hah, when i read that last night, i was going to say.
my phone doesn't do that but i get it a lot from friends. somehow my phone knows what i mean.
i'm surprised my phone doesn't do that, because i do quite a lot of texting about ducks
i raise ducks and am constantly telling anxious prospective customers, sorry, i have no ducks left to give.
3:40 PM
thanks for the clarification!
@leslietownes So you don't really give a duck.
no. i sell ducks to people at the right price.
no, ... I was, never mind.
honestly there is some truth to my confusion about autocorrect not picking up on it. i do talk about actual ducks a lot on text because i visit one or two actual duck ponds every week and there is a lot of text chatter about the actual ducks.
i even have photos of actual ducks on my phone. this is not a bit.
Ok, now I'm really down.
3:48 PM
Hi all.
I want to ask one question:
$\liminf \frac 1{a_n}=\frac 1{\limsup a_n}$ holds for any positive sequence $(a_n)$. My question is why positivity of $a_n$ required here.
@leslietownes they are adorable :)
@Koro have you tried with a sequence that is not all positive?
It doesn't work for $a_n=(-1)^n$
try (-1)^n even
you read my mind
3:52 PM
$\limsup a_n$ will be positive and $\liminf\frac1{a_n}$ will be negative
you can toss in infinitely many negative things to a positive sequence without changing its lim sup but potentially changing its lim inf
But I don't know why this is happening. I understand your point professor Rob. It is a very valid point but if I think about a non empty set of non zero numbers $A$, then I claim:$\inf \frac 1 A=\frac 1{\sup A}$, where $1/A$ is the set of all numbers $1/a, a\in A$. Let $x:=\inf \frac 1A$ be non zero finite then there exists a sequence $(a_n)\in A$ such that $\frac 1{a_n}\to x$ hence $a_n\to \frac 1x$ that is $y:=\sup A\ge \frac 1x$. I claim that $\sup A=\frac 1x$ because if $\sup A\gt 1/x$ then
did you try A = {-1, 1}, where 1/A = A
i'll award 50 newly minted lesliecoins to the first person to id those ducks
there exists a sequence $(b_n)\in A$ such that $b_n\to y$ hence $\frac 1{b_n}\to \frac 1y$ so $\inf \frac 1A\le 1y\lt x$, which is a contradiction. So this establishes that $\inf \frac 1A=\frac 1{\sup A}$. My question is: here I have nowhere used positivity of numbers in A so why does the result require positivity of sequence?
@leslietownes :)
how do you deduce that 1y < x, exactly
think of A = {-1, 1} where y = 1 and x = -1
4:05 PM
Oh yes, if $-1/1<1$ then by what I have done above I should have got $1/(-1)\gt 1$, which is absurd!
Thanks a lot @Leslie :)
@Leslie: The ducks photo reminded me of the ducks at my college. It was a lovely sight to see them cross road. :)
they are american wigeons, male and female. 50 lesliecoins have been awarded to leslie, and subsequently recorded on an indelible, democratized, infallible, wealth-conferring ledger. the male shows you why some people call this wigeon "baldpate."
you sure know a lot about birds
it is good to think about the hypotheses of everything involving liminf and limsup. a lot of analysis statements about lim can sometimes be refined with liminf and limsup and these things always exist. analysis became a lot easier for me when i turned it into a game of 'find an inequality, then apply liminf or limsup.' sometimes you introduce an epsilon or something to get the inequality.
I think I can agree on this. I've been working on such inequalities :)
I have a second question also: I want to show that if $(a_n)$ is a sequence of positive numbers then $\liminf \frac {a_{n+1}}{a_n}\le \liminf a_n^\frac 1n$ using this definition: $\liminf a_n=\lim_{n\to \infty} \inf_{m\ge n} a_m$.
rudin signals this once. i think it's in the root test. you don't need a limit to exist, only a limsup. a lot of lightbulbs turned on in my head when i absorbed that.
and you seem to be asking just about that. find that portion of rudin.
4:13 PM
Hi, if $u$ is harmonic in a punctured disk at $0$, and in the open half disk punctured at $0$ I know that $u(z) - \log|z| \rightarrow 0$ as $z \rightarrow 0$ (in this open half disk), does the same necessarily hold in the other half disk apriori?
Leslie: Yes, I know but he uses $\epsilon$ argument and I want to show using the definition I stated above.
koro at some point you will have to stop imposing your One Chosen definition of liminf. it's equivalent to a million other things and you will eventually have to accept that. or condemn yourself to a life of seeing some equivalence proved in the middle of an argument just so that you have it your preferred way.
doesn't rudin prove the equivalence of a few definitions, or am i inventing that? my memory is not so good.
the post office lost my paper copy of rudin and have a poor quality scan as an 'ebook'
oh nvm, i answered my question
In fact, here is the complete inequality (and I remember I discussed that with you a long time ago when I was talking about proving $n^\frac 1n\to 1$ using the inequality): $\liminf \frac {a_{n+1}}{a_n}\le \liminf a_n^\frac 1n\le \limsup a_n^\frac 1n\le \limsup \frac {a_{n+1}}{a_n}$
no, I don't think the equality was proved in Rudin's. Definition of liminf of a sequence $(x_n)$ in Rudin's: infimum of set of all limit points of the sequence $(x_n)$.
the answer is yes, because letting $\epsilon(z) = u(z) - log|z|$ this is harmonic in the punctured centered at $0$, and $u(z) = log|z| + \epsilon(z)$ holds in this punctured half disk , so by the identity principle for harmonic functions it has to hold in the whole punctured disk..
4:21 PM
Using the definition, it was proven that: for every $\epsilon\gt 0$, there exists N such that for all n$\ge N$, we have $\liminf a_n-\epsilon \lt a_n$ and that there exist infinitely many $n$ such that $a_n\lt \liminf a_n+\epsilon$
so he does connect it to sequential behavior a little bit. OK. it shouldn't be too difficult from that to show that your definition and rudin's are equivalent.
i realize i was mis-remembering from some other analysis book which used your definition. i can't remember its name, but it was yellow.
are you sure it was yellow and not green?
copper, today's topic is waterfowl.
weirdly, yes? i don't think i owned a green book. maybe a partial green/white book.
this is exactly how my memory works. i can never remember the author or title. i can remember the color of the book and what side of the page it was on, left or right. and if it's near the bottom or top i can remember that too.
I agree that the definitions can be shown to be equivalent. :)
i have a sharp but useless memory
"how do i prove this?" "get the book that's red mostly on the cover, and it's about 2/3 in, on the left side of the page, in the middle"
when i started grad school there was a student who had a memory that worked the useful way: author, title, theorem or page number. but he sounded like a psychopath. he'd pipe up "that's in ___, page 128, theorem 3.3." OK. we'll keep you busy here while the authorities dig up your basement and find all of the missing runaways.
4:26 PM
woah!! page number too?
yes, it was bizarre.
he quit the program to do his own business. i googled him just now, he seems to be doing OK. probably earns more than most math professors, but not much more. or at least he had OK people design his website.
hello, can someone explain me : Sobolev functions, as Lebesgue functions, are defined only up to measure zero and thus we identify functions that are equal almost everywhere.
what it means "defined only up to measure zero"
in order to make many spaces of interest to ODE/PDE amenable to limit operations it is helpful to regard some functions as the same.
if f(x) = 0 for all values of x except -6, -4, 0, 676, 34543, and 4253465346, people will say, that should be the zero function.
this is a fine move but it means that if f is only a member of some function space, the point value "f(x)" might not have meaning.
for example if two "functions" are regarded as the same as long as they disagree at only finitely many points, there's no way to say what the value of the "function" is at 0. it could be anything. you can change the value at one point without changing whether it is the same as other functions.
measure zero is a subtle concept but it gets at this. in the sobolev world, two functions f and g, and by this i mean normal functions, might be regarded as 'the same,' as long as the set of x for which f(x) isn't g(x) is "small enough."
that's what is specified by "measure zero," or "a.e.," or whatever.
it's helpful for limits, and doesn't really affect integrals, but once you do this, you can no longer say what the value of a function is at a point (because you can change that without making the function 'different'). you can talk about values of integrals of a function on intervals around that point (because you can't change that without making the function 'different').
this is somewhat vague and might not be helpful. if there is a more specific query that might help more.
4:47 PM
i understand but what is : are defined only up to measure zero"
i understand that f can be not defined only on set with measur zero
someone here?
the idea is that f and g are regarded as the same if {x: f(x) isn't g(x)} has measure zero
once you do that, functions f and g that are 'the same' don't have to take the same value at one point, or 340596734 points, or even some collections of infinitely many points
so there's no longer any meaning to expressions like "f(x)" in isolation (which appears to invoke the value of f at only one point) or other expressions (e.g. "f(1) + f(2) + f(3) + ... + f(500)")
because the sets on which f has these values are measure zero
i uderstand the general idea is that we can say that two functions are equal if they are are equal a.e
but what they mean by "only up"
well if functions are "the same" if equal a.e., and f and g are "the same" (and i decide to write that as f = g), i can't conclude that f(1) = g(1)
i can conclude $\int_0^1 f(x) \, dx = \int_0^1 g(x) \, dx$ because the value of this integral doesn't depend on what f and g might do on sets of measure zero
but f(1) and g(1) do depend on what f and g might do on sets of measure zero
if the slogan bothers you, it might be helpful to focus on some specific instance where the issue is unclear
i personally don't find the slogan to be very informative
5:04 PM
i can't believe i missed a fowl discussion
i have some heron pix too
these are available for all purchasers of lesliecoin
throw your worthless money away and join the high-speed rocket ship to financial security
i think i have a pic of a coopers hawk with a little bird in it beak, just outside our house
i love coopers hawks. if you put a bird feeder up in your yard, the only bird you are feeding is the coopers hawk.
thank you @leslietownes
our city actually has a sign that they put up in the spring for the cooper's hawk. DANGER - NESTING HAWKS, MAY BE AGGRESSIVE.
i was pleased to see this a block from the house (i knew the nest already). good use of tax dollars.
side note, i don't think i've ever seen a hawk be aggressive to a human near a nest. crows do this all the time.
6:04 PM
i have never seen a hawk being aggressive towards a human. i have seen crows being aggressive, i think we discussed this before.
one of my friends has been attacked by crows several times. i think it must be because of something she did.
6:16 PM
in a weird crossover, i saw a hawk menaced by crows this morning.
A: lim sup inequality $\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n $

KoroLet's assume that $(a_n)$ and $(b_n)$ are non-negative. Suppose that both the quantities on RHS are finite. $$\limsup a_n\limsup b_n=(\lim \sup_{m\ge n}a_m) (\lim\sup_{m\ge n}b_m)=\lim(\sup_{m\ge n}a_m\sup_{m\ge n}b_m)\tag 1$$ where the second equality follows by limit rules (product of limits of...

I tried to present my understanding in the answer. I hope my understanding in the answer is correct.
Hi, there is a question that we are supposed to work with through talking to others.. COnsider the intervals $(a,b)$ and $(c,d)$ where $a<b<c<d$ if we pick $a'\in (a,b)$ and $c'\in (c,d)$ then $(a,b)\vee (c,d)$ ($\vee$ denotes wedge sum) is not a manifold, right?
You start with $1. I offer to flip a coin. If it lands heads, I multiply your money by 10. If it lands tails, I multiply your money by 0. How many times should you take my offer?
@monoidaltransform That gives an X-shape, right?
so it suffices to look at the cross in $\mathbb{R}^2$, with the subsapce topology, if $U$ is a neighborhood of the point in the middle of the cross, then $U-middle point$ has atleast $4$ path components, right? @AkivaWeinberger
6:31 PM
As a sanity check, $A\subseteq B$ then number of path components of $A$ is $\leq $ number of path components of $B$, right?
@monoidaltransform No. $[1,2]\cup[3,4]$ is a subset of $\Bbb R$ but it has two components and $\Bbb R$ has one
@AkivaWeinberger Or, more pathologically, $\mathbb{Z}$ has a countable number of path components, while $\mathbb{R}$ is path-connected.
Knowing that $A \subsetneq B$ tells us nothing about how many path components $A$ may have relative to $B$.
If your space $U$ were a manifold, there would be a neighborhood of the middle point $x$ that's homeomorphic to $\Bbb R^n$
and every neighborhood of $x$ has an open subset that's an X-shape, so that X-shape would have to be homeomorphic to an open subset of $\Bbb R^n$
but there's no open set in $\Bbb R^n$ that's connected but falls apart into four components when you remove a point
(if $n=1$ then removing a point from any connected subset of $\Bbb R$ splits it into two; if $n>1$ then removing a point from any connected subset of $\Bbb R^n$ leaves it connected)
6:57 PM
This is kinda cool
The improper integral integral of a differentiable function is not differentiable
how the ceiling function is onto if we don't include all the range of numbers in the codomain of the function?
@Avra It depends on what the codomain is
If we think of it as a function $\Bbb R\to\Bbb Z$ then it's onto; if we think of it as a function $\Bbb R\to\Bbb R$ then it's not
Most functions are onto on to their image
i missed that @AkivaWeinberger :(
then in my case it's since Z -> z
7:07 PM
@Astyx most?
$\lceil \frac n2 \rceil, \mathbb{Z} \to \mathbb{Z}$
Ah yeah so that is surjective because to hit $k\in\Bbb Z$ you can start with $n=2k$
This is the "turn a six-sided die into a three-sided die" function
we can prove it by casees @AkivaWeinberger!
If you're ever playing a board game and want a way to choose between three options equally, you can roll a die and divide by two rounding up
since we hae both odd and even
7:09 PM
(doing it mod 3 also works)
@Avra No need. You just need to find one thing that maps onto $k$; you don't need to find everything that maps onto $k$
@AkivaWeinberger Thanks <3
7:40 PM
Let me put a smale on your face
Evert that frown upside-down!
8:10 PM
:( => ):
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