maybe less formally, if a and a' are different elements of the domain of f and f(a) = f(a') = b, then there's no one single choice of what f^{-1}(b) ought to be. there are at least two possibilities. if f^{-1} were a function there would be only one.
how about a replacement schema where words ending in ack, ick, ock, and uck are all changed to end in -kc instead of -ck? then we could have a mnemonic of the form 'k before c except after e'
Good evening everyone. I'm still having fun messing around with function approximation.
It seems there is a general method of approximating a function by taking an approximation of the function and either the exact inverse or an approximation of the inverse. Is there anything in the literature about this?
at that high level of generality it seems more of a vibe than something people would write papers about. but yeah, i could imagine stuff like that. if for some reason some number a is thought to make f(a) resemble f(x), and some g is thought to resemble f^(-1), maybe try g(f(a)) as a replacement for a in an iterative scheme to find things that will give you f(x).
anything resembling literature that i've seen on this is far more specific in its hypotheses, but i could see the vibe.
Well unfortunately I can't claim some "function of function approximations" as my results currently are subjective and require adaptation based on circumstances, but what remains objective for the time being is that it works for finding an approximation of a function and its inverse solely by computing the error to x itself.
I'm currently trying it with $2^x$
I did however try just the other day to see what would happen if I put different functions other than x into the recursive reciprocal function that I found. Curiously, for certain domains of x, it converges to the reciprocal of the function I put into it, whereas others went off to infinity.
how to write $f(x,y,z) = xy^{2} + \sin(xz) + e^{z} - 2 = 0$ as a function of $(x,y)$ near $(1,-1,0)$? I established it can be done.....but doing it...well nothing "nice" is happening.....
After some playing around, I got things "to work", but I only know it works because I have a solution....I don't understand why it failed for one scenario but not for another.
After having established $f(x,y,z) = xy^{2} + \sin(xz) + e^{z} - 2 = 0$ can be written in terms of $x,y$, looking at the $\sin$ term I "bounded" my expression: $xy^{2} + e^{z} - 3 \leq xy^{2} + \sin(xz) + e^{z} - 2 \leq xy^{2} + e^{z} - 1$
I was asked to find solutions for the partials of $z = \phi(x,y)$ w.r.t $x$ and $y$ at the point $(1,-1)$. Initially I used the upper bound and got the incorrect answer, but using the lower bound I got the correct answers. Both produced simple values. Is there some reason why only the lower bound solutions are shown?
dc3rd, were you asked to do this? a lot of implicit function theorem stuff just leverages existence of the function (which you can establish by checking derivative conditions) and the functional relationship. the example may have been chosen precisely because it's difficult to 'solve for z' in any useful sense.
Just for clarity after using one of the bounds I rewrote things as $xy^{2} + e^{z} - 1 = 0 \Rightarrow z = \ln(1 - xy^{2})$
I was asked specifically to find $\frac{\partial \phi}{\partial x}$ and same for $y$ at the given point $(1,-1)$. I wasn't asked to solve for the $\phi$ explicitly though. but I couldn't see any other way to compute the derivatives. And yes you're right there is no "easy" way to solve for $z$ in my expression....
how the question was asked was to establish that I can write $z$ as a $C^{1}$ function $z = \phi(x,y)$. then the second part asked me to just compute the partials at the given point. So I thought the only way that can be done is by finding said function of $z$, but as you saw it is not a fun endeavor....
Let $x=\limsup \frac{a_1+a_2+...+a_{n+1}}{a_n}$ . Let $\epsilon\gt 0$ be given. For large $n$ we have $\frac{a_1+a_2+...+a_{n+1}}{a_n}\lt x+\epsilon\implies \frac{a_n}{a_1+a_2+...+a_{n+1}}\gt \frac 1{ x+\epsilon}\implies \frac {a_n}{a_1}\gt \frac 1{x+\epsilon}$
black cat instagram is a pretty close-knit community. there's a tiny fraction of people who only want to post and see pictures of black cats, and we all follow each other.
then there's a broader population of casuals who come and go.
Using the other implication of limsup definition, i.e. there are infinitely many $n$ such that $\frac{a_1+a_2+...+a_{n+1}}{a_n}\gt x-\epsilon$ but here the problem is RHS could be zero so I think it doesn't help much :'(
one time i was interviewing someone for an internship and we had twice as much time because the previous interviewer had to run at the last minute. we got to talking about instagram, including black cat instagram. it turned out that they followed my cat already. i didn't let this affect my decision on the interview.
they got an offer and turned it down, which is a shame.
with shorthair cats it's pretty easy. it helps if you brush them. that reduces shedding. you can avoid brushing them, but you may need to sweep up every week or two.
and keep a brush or lint roller around to get hair off of strategic items of clothing.
i swept around my desk today for the first time in a few months, and most of what came up was black cat hair. it wasn't a lot, but it was there.
i'd probably notice it more if it were a bright white color. as it is, i never notice it.
hi, apparently the following is an easy result, but Im not seeing how to prove it: Let $u$ be harmonic in $\mathbb{C} - \overline{\mathbb{D}}$, such that we know $\partial_{r}u = 0$ on the boundary of the unit disk (and we know $u$ actually has a harmonic extension to some small neighbourhood of the boundary where this partial derivative is defined, say), then $u$ extends to the interior of the unit disk , by defining it as $u(\frac{1}{\overline{z}})$ in the interior
well okay, for sure that $u(\frac{1}{\overline{z}})$ is harmonic in the interior minus origin is just a computation and yes you can just use dz,d-zbar to make your life much easier and see it very quickly
ugh, i forgot the most important part of the question
by 'extends to the interior of the unit disk', I mean, since we already know $u$ is harmonic in a neighbourhood of the boundary, that it extends to the punctured plane.
that is, yes $u(\frac{1}{\overline{z}})$ is easily seen to be harmonic in the interior minus teh origin
but we need to show that this also gets you harmonicity at the boundary
hopefully I am making more sense now
and this is certainly where we will need to use more information about $u$ on the boundary, in particular that $\partial_{r} u = 0 $ there
where we're mixing and matching the formulas. that makes sense.
is this an exercise in ahlfors? i think i've seen it before. or not. i may be blending this with hazy memories of stuff like the schwarz reflection principle for analytic functions.
maybe it's in conway. except i don't remember conway spending a lot of time on harmonic functions.
hrm.
this is where i dig in my notes to see if i can just cut-paste an answer.
in the schwarz context i think one way of proving the analyticity in regions containing the reflection boundary is using something like morera's theorem. you sort of sidestep the whole pointwise thing, and decompose into averaged behavior. is there something similar we can do for harmonic functions?
averaged behavior might not be the right word. you replace emphasis on point values with emphasis on values along curves.
something like the mean value property maybe?
i'm thinking out loud and have not thought about this in 15+ years.
yeah there is a similar thing, in fact rather than using moreras theorem you can prove a stronger version of the schwarz reflection principle by starting with some harmonic function $v$ that is define in the upper half plane and continuous at the boundary (where it vanishes), then define it in the lower half plane by $\overline{v}(\overline{z})$ and the mean value property now obviously holds at the boundary because the integrals over the upper and lower semicircle are equal in magnitude
by the way, just to relate this even more with the schwarz reflection principle, let $z$ be a point on the boundary and take some small disk around it where $u$ is still harmonic (i mentioned earlier its harmonic in a small neighbourhood of the boundary), it has a conjugate $v$ in this disk, so that $u + iv$ is analytic, now $\partial_{r}u = \frac{1}{r} \partial_{\theta} v = 0$, so $arg v$ is constant on the boundary arcs
uh, well that doesnt relate it the schwarz actually, sorry, but it may still be useful
@PeterJohn I saw your post. And in Google, there is a paper about that problem which proves not for general topological space but for metric space. I'm not sure the problem statement is true for general space. Why don't you try first for simplicial complex?
And I'm also quite sure join operation is irrelevant
Hello, can someone explain to me how this user did his contour integration ? https://math.stackexchange.com/a/2372334/272494 If we want to integrate from on $[0,-iR]$, then this is the same as integrating on $[0,R] \cup \{Re^{i \theta} : \theta \in [0,\pi/2] \}$. But then I don't understand how he obtains his result.
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 points on its border...
If I had a joint pdf $$f_{X,Y}(x,y)$$ and I then wanted to transform it using the function $U = X/Y$ would I still use the Jacobian method - but I don't have a second transformation. Would I just define some random R.V and continue as normal?
@huzaifaabedeen You have posted that question three times in chat this week. Semiclassical posted an approach. You have not posted any thoughts or responses. Why?
I have one small question that's been bothering me since today's noon.
I have $\limsup x_n=x$ then I want to show that there exists an $N\in \mathbb N$ such that for all $n\ge N$, it follows that $x_n\le x$.
This seems correct when I tried examples of sequences $(x_n)$ and $(y_n)$ where $x_n=1$ when $n$ even and $x_n=2$ when $n$ odd; and $y_n=0$ when $n$ odd and $y_n=-n$ when $n$ is even.
@AlessandroCodenotti thank you so much for your help. I have also commented there based on nice counter-example suggested by you and tagged Mr. MartinR with whom I discussed this question (math.stackexchange.com/questions/4289711/…) today.
So there's this result by Hurewicz which says that if $f : X \to Y$ is a map (say, between compact metric spaces, although I am sure this can be relaxed) then $\dim X \leq \dim f^{-1}(y) + \dim Y$, yes? The equality fails by Pontryagin's example.
Right (and yes it can be relaxed a lot but I don't remember the exact assumptions, it's some technical general topology, for sure it is written in Dimension Theory of General Spaces by Pears)
Great. So I believe that in my setup, you have $\dim X = \inf_{\varepsilon} \sup_{\mathcal{U}} \dim N(\mathcal{U})$ where $\mathcal{U}$ runs over all covers of Lebesgue number $\varepsilon$ in the inner sup, and $N(\mathcal{U})$ is the nerve, and by $\dim N(\mathcal{U})$ I just mean the dimension of the simplicial complex. Obviously, this is bounded by the least $n$ such that $\check{H}_c^{* > n}(U)$ vanishes for all open subsets $U$ of $X$. If I understand correctly, these are actually equal?
(so it seems that the most general result is with T_4 X, weakly paracompact Y and a continuous closed surjection f:X->Y, and you get $\dim X\leq \mathrm{Ind}(Y)+\dim f$ where dim f is sup of dim of fibers)
The reason I ask is I think the analogous Hurewicz theorem in Cech cohomology can be proved with general nonsense. Given a map $f : X \to Y$ of locally compact Hausdorff spaces, there is an isomorphism $\check{H}^\star_c(X; \Bbb Z) \cong \check{H}^\star_c(Y; Rf_!\Bbb Z)$ where $Rf_!\Bbb Z$ is a sheaf of complexes, whose stalks have homology $\check{H}^\star(f^{-1}(y); \Bbb Z)$.
Think of it like a twisted Kunneth theorem. $\check{H}_c(X) \cong \check{H}_c(Y) \widehat{\otimes} \check{H}_c(f^{-1}(y))$ where $\widehat{\otimes}$ indicates some kind of twisted tensor product
This is general nonsense, you don't have to know the proof. It's not geometry
But it should tell you that the Cech cohomological version of dimension satisfies the Hurewicz inequality
More precisely it should give $\dim X \leq \inf_{y \in Y} \dim f^{-1}(y) + \dim Y$
I can always take disjoint union of X and Y with a point and send the new point to the new point to artificially introduce a zero dimensional fiber without affecting the dimension of X and Y
I spoke without thinking. In my $\check{H}_c(X) \cong \check{H}_c(Y) \widehat{\otimes} \check{H}_c(f^{-1}(y))$, the second component is in fact varying point to point wrt $y$
I can only bound by the largest dimension, that's right
I haven't checked all the details here but I feel like this should give a proof of Hurewicz theorem. My ultimate question is if a similar proof exists for the Bell-Dranishnikov's Hurewicz theorem for asdim
There's an analogue of this Cech stuff, where you use bounded cohomology, by Dranishnikov.
lol I don't know about the cohomological approach either
Also I realized I don't remember much about asdim either. I was asked the other day whether there are groups that have exponential growth, finite asdim and are amenable
OK, so here's the difference between what you're doing and what I'm doing. When you write $d\ell = (dx, dy, 0)$, you're writing the tangent vector to the path at the point $(x, y, 0)$ in the coordinates of the 3d-space:
$de_1 = (dx, 0, 0)$ is the infinitisimal direction in the $x$-axis, $de_2 = (0, dy, 0)$ is the infinitisimal direction in the $y$-axis, $de_3 = (0, 0, dz)$ is the infinitisimal direction in the $y$-axis. So the infinitisimal direction in the $z = 0$ plane in the $x = y$ direction is $de_1 + de_2 = (dx, dy, 0)$.
I much prefer using parametrized, where I use the fact that the path $\gamma(t) = (t, t, 0)$ has intrinsic 1D coordinates since the domain of the path $\gamma$ is a 1D space $\Bbb R$.
In that case, I just view $d\ell$ in the intrinsic coordinates, so it becomes $"d\ell =" \gamma'(t) dt$ where $\gamma'(t)$ is the genuine derivative of the path at the time $t$
And $\int_\gamma F \cdot d\ell = \int_a^b F(\gamma(t)) \cdot \gamma'(t) dt$
This is called "pullback" in differential geometry lingo. I pullback $d\ell$, which lives in $\Bbb R^3$, the target of $\gamma$, to the domain $[a,b] \subset \Bbb R$ of $\gamma$
If you don't care about the formalism, you can take $\int_\gamma F \cdot d\ell := \int_a^b F(\gamma(t)) \cdot \gamma'(t) dt$ as the definition of line integral of a vector field $F$ along the path $\gamma : [a, b] \to \Bbb R^3$.
It makes sense also, right?
I recommend doing it this way because 1 coordinate is superior to 3 coordinates, and you don't have to bother with what algebraic manipulation with infinitisimal quantities like $dx, dy, dz, dt$ etc all these mean rigorously
@fin You're scaling the actual unit tangent vector, which is $\gamma'(t) = (t', (2t)', 0') = (1, 2, 0)$ down. But try to compute line integral of $F(x, y, z) = x^2$ along two different parametrizations of the straightline joining $(0, 0, 0)$ and $(1, 2, 0)$ given by $\gamma_1(t) = (t, 2t, 0)$, $0 \leq t \leq 1$ and $\gamma_2(t) = (t/2, t, 0)$, $0 \leq t \leq 2$.
I have a question about naming variables. If N and N are two unrelated variables but I want to use the same N for both of them. Should I use: N' or \hat{N}
@AkivaWeinberger I don't think so. There's an embedding of $S^2 \times I$ in $\Bbb R^3$ such that at time 0 it's the standard sphere and at time 1 it's the Alexander horned sphere, I think
Take the Fox-Artin horned sphere ("thicken" the Fox-Artin wild arc so that the amount of thickening goes to zero fast enough near the endpoints). Its interior plus boundary should be homeomorphic to a ball
and then you can embed IxI into any ball where one edge is anything completely inside and the other edge is anything on the boundary
vrouvrou, what are the domain and codomain of f and what are the notions of strong and weak lower semicontinuity? at a high level, intuition suggests 'yes' because strong beats weak
@Vrouvrou if we assume what you're asking is true, then if we have a functional $f:X \to \Bbb R$ that is strongly continuous, then $-f$ is strongly continuous as well. So both $f$ and $-f$ are weakly lower semicontinuous, but that means that $f$ is both weakly lower semicontinuous and weakly upper semicontinuous, which implies that $f$ is weakly continuous. But not every strongly continuous functional is weakly continuous
@leslietownes terminology is a bit confusing here, but actually for a functional on (say) a locally convex vector space weak continuity is stronger than strong continuity
@AkivaWeinberger You can trap the annulus away from a collar nbhd of the base in a big bounded set, and then extend the base to infinity and cap it off at $\infty$ in $S^3 = \Bbb R^3 \cup \infty$
Actually, the definitions as I know them: $f$ is strongly continuous if $u_n \to u \Rightarrow f(u_n)\to f(u)$ (that's just the usual continuity) $f$ is weakly continuous if $u_n \rightharpoonup u \Rightarrow f(u_n) \to f(u)$
@Vrouvrou that just means that for a functional $f$, it doesn't make a difference if you write $f(u_n) \to f(u)$ or $f(u_n) \rightharpoonup f(u)$. But $u_n \to u$ or $u_n \rightharpoonup u$ still makes a difference
