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.
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.
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
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.
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.....
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...
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
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?
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
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
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...
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.
@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.
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...
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 $F_2$?
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.
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.
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
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.
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 MarieOct 23 at 9:39
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...
We 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...
@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
@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.
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.
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
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?
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."
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.
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?
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'
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..
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.
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.
"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.
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.
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.
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
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
Let'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?
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
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)
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 $F_2$?
