"x = \sum_{j=0}^3 x_j 2^j" would be a nice way of compressing that. once we have sums involving sums and products of sums that both need indices, could get dicey.
@leslietownes I need to obtain the results related to extremum of convex functional. I don't know much about the conditions. I am like standing at the see shore with the question: what can you say about the extremum of convex functional? all the conditions and results are below the horizon. :'(
Is it possible to ask reference type question in main site?
sure, but people will ask what domains you are interested in, and what sets you are optimizing over.
they may also be curious about your background, so e.g. they know not to bring banach spaces into it, or they know not to bother with the reflexive case.
a motivating example or set of examples would be helpful.
here's a starting point. mathoverflow.net/questions/123714/… (on MO, not MSE) is a question about convex functionals. it takes some facts for granted (but sketches their proofs), references a book, and (in comments to the answer) an example of such a functional is given.
is all of that understandable to you? none of it? something in the middle?
copper.hat might know more, all of the functional analysis books i had probably didn't discuss this so much. (one of the commenters mentions brezis, which i don't have - and it seems to be in connection with a technical point and not the main question)
@TedShifrin Actually, In an entrance test of PhD(ina telegram group), I see a question: "What can you say about the extremum of convex functional?. This is the motivation behind this.
Can you suggest some reading on extremum of convex functional. I am searching for calculus like results. is there any result, extremum lies on the boundary?
$1.11$ Theorem** Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $B$ is not empty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$. Then $\alpha = \sup L$ exists in $S$, and $\alpha = \inf B$.
I am curious why is element of L is in S?
The theorem hasn't mentioned elements of L is in S but it states that in proof.
Or did rudin made a mistake?
And if L is not in S then L will not have lub property.
I wonder if this argument works: Since $F$ has a constant rank, say $r$, if $p\in M$ then we can find a smooth chart $(U,\varphi)$ at $p$ and $(V,\psi)$ at $F(p)$ s.t. with respect to this coordinate, $F$ has representation $\hat{F}(x_1,...,x_r,...,x_m)\to (x_1,...,x_r,0,...,0)$ where $m$ is the dimension of $M$ (and denote the dimension of $N$ by $n$). Since $\varphi,\psi$ are bijective, $\hat{F}$ should be surjective by assumption. Hence, $r = n$.
If $F:M\to N$ has rank $r$ then for $p\in M$, from the rank theorem, we can find a smooth chart $(U,\varphi)$ near $p$ and $(V,\psi)$ near $F(p)$ s.t. $F(U)\subset V$ and $\hat{F}(x_1,\ldots,x_r,\ldots,x_m) = (x_1,\ldots,x_r,0,\ldots,0)$. Since $F$ is open, $F(U)$ is open and as $F(U)\subset V$, $(F(U),\psi)$ is a smooth chart of $N$ near $F(p)$. Now, $$\hat{F}:\varphi(U)\xrightarrow{\varphi^{-1}}U\xrightarrow{F}F(U)\xrightarrow{\psi}\psi(F(U))$$ Now, $\varphi^{-1},\psi$ are bijective and $F$ is surjective, $\hat{F}$ is surjective.
@Unknownx Well, not sure guru is the right term for me, but anyway... That is a big question, you would have to narrow it down a little to get some reasonable response. A standard text is Rockafellar's "Convex Analysis".
Is there any good source to practice smooth manifold theory? (Problem set for example) Many of the exercise problems in Lee's book are about calculations.
@copper.hat hello, I have look in to the index of the book. it is talking about convex function. I need the results on the extremum of convex functional.
If $P \subset M$ is a submanifold of $M$ and $U\subset TM$, then the domain of normal exponential map of $P$ is defined to be $\Epsilon_P = NP \cap U$. What exactly is this space? How can normal vectors of $P$ intersect tangent vectors of $M$? Don't normal vectors of $P$ belong in $NM$?
Is there any clever way to calculate the radius of the image circle under the map $T:\Bbb R_{\infty}\to\Bbb C_{\infty}$ by $z\mapsto -{1\over c^2}{1\over z+d/c}$?
@onepotatotwopotato yeah, that does work. of course, you do need to pay attention to the fact that $\psi(F(U))$ can be any open subset of $\mathbb{R}^n$, but its openness is enough to guarantee that $r<n$ would violate surjectivity, so you do get $r=n$ indeed.
@Lemon Normal vectors of P in M are tangent vectors of M. NM doesn't even make sense here. You can only talk about normal vectors and hence a normal bundle for a submanifold of a given manifold. A normal vector needs to live in some ambient space.
@Thorgott But proving $F$ is open is hard I guess. Surjectivity should be necessary otherwise the statement is 'constant rank map is open.'
But if I follow the proof of 'smooth submersion is open'...
${z\over\sin(\sqrt{z})} = {z\over \sqrt{z}\sum_{n=0}^\infty {(-1)^n\over (2n+1)!}z^n} = {\sqrt{z}\over\sum_{n=0}^\infty {(-1)^n\over(2n+1)!}z^n}.$ Singularity of this function is a branch cut and zeros of $\sum_{n=0}^\infty{(-1)^n\over(2n+1)!}z^n$ but how can I find the zeros of that function?
Btw following the 'smooth subm open' proof strategy doesn't work. The proof uses the surjectivty of coordinate representation.
@NotTfue "Let $L$ be the set of all lower bounds of $B$ [in $S$]." I don't see where else the lower bounds of $B$ would be located. There is no superset of $S$ mentioned, and we are talking about the least-upper-bound property of $S$. It seems to be a sensible assumption, though it was not explicitly mentioned.
"He took a deep breath and pressed on." Of what did he take a deep breath?
@onepotatotwopotato the image of which circle? The unit circle centered at $0$?
It is LFT. I restricted to $\Bbb R$. The domain of this map is $\Bbb C_{\infty}$.
The only method I can come up with right now is calculating the image of three points -1,0,1 and finding the radius with some elementary geometry which is tedious.
Here’s a small silly question. The conclusion to one of my students’ discussion problems reduces to comparing tan(20 deg) to 0.35. It turns out the former is 0.364, so that’s slightly larger. What I’m curious about is how one would deduce that inequality if one lacked a calculator
Obviously 0.35 is quite arbitrary here, so this may not be the ‘best’ choice for a lower bound
@TedShifrin So the determinant corresponds to signed volume, but why do we care about ‘signed’? It gives us orientation, but why care? None of linear algebra deals with orientation… Or is it perhaps a shortcut that because requiring multi linearly and that linearly dependent vectors go to 0 gets a signed quantity….
But @TedShifrin not in lin alg, so how to motivate the def without suddenly introducing the notion of orientation?
@TedShifrin It just seems a bit weird to me to suddenly talk about signed volume. What’s a problem in linear algebra that requires signed volume to solve? I’m only talking of matrixes and other lin alg not multi lin or multi variable calc…
My question is not really about proving this fact. Instead, it is about the sense of the sentence. What does "the topology induced by the usual distance" mean? I never understood it clearly.
To prove the proposition, I thought of the Cauchy sequence defined by $a_n=\left(1+\frac{1}{n}\right)^{n}$...
@TedShifrin ok but the det is introduced in basic lin alg, so how do you motivate it? The entire thing of ‘signed’ suddenly coming is is unnatural @TedShifrin
At the low levels, @Shin, if you want to define the cross product of two vectors as a vector, the answer actually depends on the orientation of $\Bbb R^3$. It's for this reason that physicists call it a pseudo-vector, not a vector.
Signs show up all over the place. When you do projections, you get signed distances. And determinant changes sign if you switch the vectors — that's a sign.
DogAteMy, I wasn't sure what the point of Semiclassic's question was. Were his students really not allowed to use calculators? I doubt that.
@Shinrin-Yoku If you have a "path in the space of matrices" (a continuous family of matrices $M_t$, $t\in[0,1]$) where $M_0$ and $M_1$ have opposite orientations, there's some matrix in the middle (some $M_{t'}$) that's not surjective
A matrix has positive determinant if there's a path from the identity matrix $I$ to it through surjective matrices
@AkivaWeinberger My problem is that why should one care about ‘signed’ volume when solving problems about matrices, without already introducing the det because I want to motivate it.
Hmm @AkivaWeinberger so I guess the answer really is, is that a very natural derivation gives us signed volume, and in case we ever need unsigned we can take absolute value. But it ‘turns out’ that signed volume is actually a more useful quantity
and the formula is just B-A, whereas unsigned distance is |B-A|
@Shinrin-Yoku I suppose, yeah
By the way, thinking still about signed distance:
With signed distance, it's always true that the signed distance from A to B plus the signed distance from B to C equals the signed distance from A to C.
With unsigned distance, that's only true if B is in between A and C.
If B isn't in between, you need to subtract one of the distances rather than add it.
Similarly, the signed area of a triangle UVW equals (vol(U,V)+vol(V,W)+vol(W,U))/2. @Shinrin-Yoku
(Think of vol(U,V)/2 as the area of the triangle 0UV.)
With unsigned volume, that's only true if the origin is inside the triangle.
If the origin is outside of the triangle, we get some overlap. Signed volumes perfectly deal with that.
In fact, this formula works for finding the area of any polygon.
@Shinrin-Yoku You can see this in action about six minutes into this video:
@Shinrin-Yoku the determinant is more than a volume, it also gives an orientation. as a loose analogy, consider the number $-2$, we can think of it as giving a length $|{-2}|$ and a direction $-1$. similarly, the determinant gives a volume $|\det A|$ and an orientation $\operatorname{sgn} \det A$.
Note that that triple product equals vol(qi,qj,qk), or perhaps better notation would be det(qi,qj,qk)
Since the volume of the tetrahedron with vertices 0,qi,qj,qk is det(qi,qj,qk)/6, this formula isn't surprising when the origin is inside the polyhedron.
It's kinda magic that all the negative volumes cancel perfectly even when the origin isn't inside the polyhedron.
Here, the sum is over triangles ijk of the polyhedron (we're assuming all of the faces are triangles)
@Shinrin-Yoku So if you want a problem to motivate signed areas and volumes, you can use the problem of finding (unsigned!) areas and volumes of more complicated shapes.
Consider the following (Lp).
$$
\min \;\; \mathbf{c} \cdot \mathbf{x}\\
\text{s.t. } A\mathbf{x} \geq \mathbf{b} \\
\mathbf{x} \geq \mathbf{0} \\
$$
If $S$ is the feasable set of (Lp), Prove that the following definitions are equivalent.
(Lp) is unbounded if $\; ∃ \{x^v\}_{v}⊆ S $ $ ;cx^v →...
@Semiclassical 0.35=7/20, which makes the arithmetic fairly simple, assuming you're doing it totally by hand. Let $\tan\theta=7/20$, then use the identity for $\tan3\theta$, which is pretty easy to derive from the tan sum identity. A few lines of arithmetic gives $\tan3\theta=8057/5060$, which is less than $\tan60°=\sqrt3$.
if at least one of the coordinates of c is negative, you can pick the corresponding coordinate in one arbitrary x in S and make it as large as you like, building the sequence that is required
you can do this because all coordinates of x are positive.
@copper.hat But if all coordinates of c are positive, which means c>=0, that's where it gets hard :(
@copper.hat Do you think the fact that moving in reverse direction of the gradient decreases the amount of cx is useful here?
@copper.hat Another idea! If we suppose that S is closed (I'm not sure of that), then since S is non-empty there exists a number such as λ that the intersection of all x in S such that ||x|| <= λ is not empty. This intersection is closed and bounded so its compact (Heine Borell theorem) and f = cx attains its minimum on it. Since all coordinates of c are positive, this minimum is the minimum of cx for ALL x in S.
That means if all coordinates of c are positive, then LP is not unbounded! So this case doesn't occur and the only case that LP can be unbounded is that at least one coordinate of c is negative.
The only thing left for me to be sure of is whether S is closed or not. I suspect it is, because S can be obtained by a finite intersection of the solutions of linear inequalities which each of them are closed. for example solutions of ax + by + cz >= d is a closed Region in R^3. Is my reasoning true @copper.hat? I hope I'm not writing for nobody
overtherainbow: i would circle back to the definition of 'natural transformation'. a natural transformation can also be "applied" (not sure that's the right term) to objects, as part of its definition.
@Jakobian yes but you are deeper in the weeds than i ever was (or at least have been in 20 years). fire away!
How to solve for "r" rate? https://www.varokas.com/content/images/downloaded_images/Solve-Equations-numerically-with-Newton-s-Method/1-qZqxzla5Y4H68hTX37s_rA.png
So, there is something called the James characterization of reflexivity, you might recall it. It's a geometric description of it. Basically, it says that a space $X$ is reflexive iff there is some $\theta\in (0, 1)$ such that for any sequence $x_n\in S_X$ with $\inf \{\|x\|: x\in \text{conv}\{x_n\}\} \geq \theta$ there is $N$, $u\in \text{conv}\{x_1, ..., x_N\}$, $v\in \text{conv}\{x_{N+1}, ...\}$ such that $\|u-v\| \leq \theta$.
Now, I want to prove that if a space $X$ is not reflexive then there are sequences $x_n\in S_X$ and $x_n^*\in S_{X^*}$ such that $\langle x_m^*, x_n\rangle = \frac{1}{2}$ for $m\leq n$ and $0$ otherwise.
apparently it should follow from James theorem above, but I'm not seeing it
@leslietownes I know that C^{op} is the category which reverse the morphisms. But if I take a morphism $g:U\rightarrow V$ in C^{op} is it then $g:V\rightarrow U$ in $C$?
@TedShifrin banach space 'geometry' is bizarro world. in this world, i think the thought might be that it's 'geometric' because it only involves the original space (e.g. it does not refer to the dual)
@TedShifrin Re-established myself back in the private sector. Looking at academia in horror. Thinking about applying to Georgia Tech's M.S. in computer science. Data science is a mess.
This old post: https://chat.stackexchange.com/transcript/message/61974821#61974821
Basically, all of those are loss functions. You can think of loss functions as metrics which are usually dependent on a theoretical "true" value and a estimated value (hence statistics comes into the picture). If you want to do stats, you need to choose a loss function. Mathematically, it's mostly an arbitrary choice. In practice, it's more of a tradeoff between precision and explainability.
Stats spends some time going through various loss functions and describing what the "best" estimators are for each lo…
I've been only casually searching for jobs, and was rejected from a data scientist position (not a senior one, mind you) for not having professional experience with natural language processing, for example, after a phone interview
Ultimately, the reality I think is that computer science graduates are getting the jobs now
there are a number of places that will give some form of stat masters in conjunction with enough coursework, to a phd in basically any other grad program (i.e. to people who would not have been admitted to a stat department on their own). i don't know of math or CS masters that do this.
PhD students are required to complete four core course sequences before advancing to candidacy. At this moment, there are six core sequences offered: real anal, complex anal, PDE, topology, and algebra.
@XanderHenderson There is a CSU, from what I recall, which only requires Calc 1 for their undergrad stats degree. They also have an MS degree which is substantially watered down.
@XanderHenderson Having done stats in practice... you really need it. Most of stats involves taking $n \to \infty$ and unless you've worked through the theory, you'd have no idea
@TedShifrin Last I heard, there was a proposal to delay algebra to 9th grade for the sake of equity, and whoever's spearheading that is back to the drawing board, delaying the project
And I would imagine that the tools that are most useful to statisticians are more on the discrete side, e.g. limits of sequences. This topic is covered in calculus (typically, in the second semester, where calc I isn't going to help, anyway).
@XanderHenderson Not really. It's mostly hypothesis testing in the continuous case and dealing with asymptotic behavior, as well as what to do when you can't reasonably employ asymptotics
@Clarinetist I still don't see why calculus is really needed for that.
But I have long argued that, aside from making sure that math departments are funded, there is no reason why math departments should be teaching calculus to much of anyone.
@XanderHenderson It is not touched on until the masters-level courses (in the US) in a mathematical statistics course, which requires analysis as a prerequisite
@Clarinetist Sure, I've studied some of that stuff (I took some of the actuarial exams, back in the day, and my BA is officially in statistics). I guess my point is that what statisticians do is so distinct from what mathematicians do that a calculus class taught in a mathematics department is going to provide little preparation for someone who wants to study statistics.
Those are things that I had to refer to every day as a statistician in my last role
You'd have someone throwing a small-sample problem at you and using pre-packaged code to something, and then you'd have to tell them their results don't make sense because it relies on those two things, which are asymptotic in nature when they have a sample size of like, say, 6
In fact, I will openly admit that the calculus sequence was mostly pointless for me and wish I could've skipped to analysis, like every European country seems to do
@TedShifrin I mean, there's truth to this. The revenue-generators are statistics and calculus, from what I've seen
Speaking of which
I was catching up with my former department chair from 2 jobs ago
It sounds like the STEM dean is being pushed out because of his perceived lack of willingness to push faculty to increase their 1-year-college-level-math placement rates
The department chair described to me that his perception is that the VP doesn't think faculty have any interest in trying to improve developmental-math teaching
If you're not familiar with co-requisite models, as a hypothetical example, suppose you're trying to have a student take college algebra one semester from now (because that's a college-level math class)
If they are, say, two levels behind college algebra, in this current semester, you would have them enroll in two courses simultaneously so as to try to prepare them in an accelerated pace for college algebra
I believe I've seen one example of this so far in the elementary statistics sequence, for example, where they pair one section of it with an elementary algebra course, totaling 8 credits in one semester
The trick was, pick a point $R$ that's not on the plane we care about, and take some plane through $P, Q, R$ (one exists, but is not necessarily uniquely given because $R$ could be on $PQ$!); that plane has to be different from the one we care about, so intersects it in a unique line, etc etc etc.
Consider the following (Lp).
$$
\min \;\; \mathbf{c} \cdot \mathbf{x}\\
\text{s.t. } A\mathbf{x} \geq \mathbf{b} \\
\mathbf{x} \geq \mathbf{0} \\
$$
If $S$ is the feasable set of (Lp), Prove that the following definitions are equivalent.
(Lp) is unbounded if $\; ∃ \{x^v\}_{v}⊆ S $ $ ;cx^v →...
