Hi, this might be trivial (or uninteresting enough to not have any elegant solution) but how do I solve for a unitary operator $U$ provided that $UOU^\dagger = aO+b\mathbb{I}$ for a given self-adjoint operator $O$ and given real numbers $a,b$?
There may be something if $a\lambda +b$ is always one of the diagonal entries for every diagonal $\lambda$. This may be possible in special circumstances. Check it out.
semigroups (where the binary operation is associative) are also really common, although it's also quite common for semigroups to come with an identity.
or a countable union of open rectangles, even. i'm not sure what there is to "visualize." any open ball contains a rectangle with rational coordinates that contains the center of the ball. i guess that's a visualization of the concept, if not the consequences of it.
note that it's fairly difficult to 'visualize' an arbitrary open or closed subset of R^n. other than being open/closed, they can look pretty goofy. but for a lot of topological results, you tend not to need to rely on what a set 'looks like,' too much.
or if you do, you're using more structure than just the topology.
@TedShifrin OK, that's an interesting possible special case but I think no such special case can exist for finite-dimensional operators. Maybe there is a simpler explanation but here's what I thought: let's say there are $k$ eigenvalues that we write as $e_1, e_2,\dots, e_k$ such that $e_{i+1}=ae_i+b$, for $i=1,\dots, k-1$ and $e_1=ae_k+b$. Either $k$ could be equal to $d=\mathrm{Tr}(\mathbb{I})$ or it could be some smaller number and there would be multiple such cyclic systems. [...]
[...] Now, using the linear recurrence formula, $e_1=ae_k+b$ can be arranged for some $k$ only if it holds for all $k$ and $e_1=b/(1-a)$, i.e., $O=(b/(1-a))\mathbb{I}$.
dvij: if the spectrum and spectral multiplicity function of O is symmetric about b/2 (b real) then O and -O + bI will be unitarily equivalent, but this might be the only interesting case
e.g. if P is a projection and ker P and ran P have the same dimension, P and I-P are unitarily equivalent
i agree that if |a| isn't 1 you run into trouble in the finite dimensional case and maybe always, but a = 1 (forcing b = 0) is the boring case of O being unitarily equivalent to itself, and a = -1 is the case i just mentioned
@leslietownes Ahh, yes, that makes sense. Turns out I carelessly canceled some factors in manipulating the result of the recurrence formula which missed these non-trivial interesting cases.
sureta it depends, both on the type of question and the people here and general activity in the chat at the time. lately there has also been fairly low activity here, which may have kicked off a feedback loop of having even chat regulars check in less often.
i'm only here because a custom alert alarm bell went off on my phone when somebody said "unitary operator"
so, my students ran into the following integral on their quantum quiz today: Let $\phi_n(x,L)=\sqrt{2/L}\sin(n \pi x/L)$ for positive integer $n$ and positive $L$. They needed to compute $\int_0^L \phi_n(x,L)\partial_ L \phi_n(x,L)\,dx$
this is not hard to write out by hand, but what surprised me is that this integral apparently vanishes identically and I don't really know why
closest i can see atm is that you can rewrite it as $\frac12 \int_0^L \partial_L [\phi_n(x,L)^2]\,dx$
but that's a total derivative w/r/t $L$, not $x$
oh. i guess i see it: Leibniz integral rule says $$\partial_L \int_0^L \phi_n(x,L)^2\,dx = \phi_n(L,L)+\int_0^L \partial_L [\phi_n(x,L)^2]\,dx$$
but $\phi_n(L,L)=0$ and $\int_0^L \phi_n(x,L)^2\,dx = 1$, both of which were required from the underlying construction
as in: given a particular mathematical framework (whatever that precisely means), are there an infinite number of ways to give axioms for it?
(my own guess would be "no" in general, but "yes" for anything actually interesting. i.e., you can come up with frameworks which are finite but that these aren't going to be all that useful)
@robjohn If you ask WolframAlpha to expand $\operatorname{Li}_{s}(1)$ in a series at $s=0$, it returns a Taylor series. At first I was very confused, but then I realized that what it was returning was the Taylor series of $\zeta(s)$ at $s=0$. Apparently it thinks the identity $\operatorname{Li}_{s}(1)= \zeta(s)$ holds for all $s$, but I'm pretty sure it only holds for $\Re(s) >1$.
If $\operatorname{Li}_s(1)=\zeta(s)$ for $\operatorname{Re}(s)\gt1$, then the analytic continuation of both for $\operatorname{Re}(s)\le1$ should be the same.
Perhaps Wolfram doesn’t consider analytic continuation in parameters.
@RandomVariable no, nor does $\zeta(s)$. They have Laurent expansions, however.
If you allow the analytic continuation
Unless I am missing something.
Oh, wait, I am thinking of $s=1$
Then, the analytic continuations have a Taylor expansion at $s=0$. If you don't look at the analytic continuation, then neither exist as naively defined for $\mathrm{Re}(s)\lt1$
That is, $\sum\limits_{n=1}^\infty\frac1{n^s}$ does not converge for $\mathrm{Re}(s)\lt1$.
@robjohn So if you allow the analytic continuation, both $\zeta(s)$ and $\operatorname{Li}_{s}(1)$ have the same Taylor series at $s=0$? Wouldn't that mean that $\operatorname{Li}_{0}(1)=- \frac{1}{2}$?
Let $U\subseteq \mathbb{R}^n$ be open, bounded, and connected. If $f\in \overline{C_{c}^{\infty}(U)}$ is constant ( in the $||.||_{W^{1,2}(U)}$ norm ) then $f\equiv 0$.
My attempt:
Choose a sequence of test functions $\phi_{n}\in C_{c}^{\infty}(U)$ such that $\phi_n\rightarrow f$ . So, $||\phi_{...
How can I motivate Lagrange's remainder for Taylor series? Once one has the guess that $R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-c)^{n+1}$ for some $c$ between $x$ and $x_0$ the proof is very simple, just applying the mean value theorem again and again. But I wouldn't be able to start the proof if I hadn't first guessed that this is the form it should take. I thought of asking this question but it seems it would be a duplicate as it has already been asked before:
https://math.stackexchange.com/questions/3268002/intuitive-understanding-of-taylors-inequality-lagranges-remainder But it didn't get any answers. There is one helpful comment but I don't find it too satisfying.
Yeah, I like the proof using Cauchy's MVT. It's very straightforward. My problem is how one would have guessed this formula in the first place. You mention generalizing the mean value theorem. I'm guessing you mean something like that if at $x_0$ we have that $f$ is $n+1$ times differentiable with its first $n$ derivatives vanishing then for every $x\ne x_0$ there is some $c$ between $x$ and $x_0$ such that $\frac{f(x)}{g(x)}=\frac{f^{(n+1)}(c)}{g^{(n+1)}(c)}$.
This isolates part of the problem, but it would still not be clear to me why I should use $g(x)=(x-x_0)^n$ or why it should be obvious to me that this result should be applied in order to find the remainder of the Taylor series.
I'm not looking for very deep motivation either. Is there a way I could play around with simple Taylor series and find this guess by myself after encountering enough examples?
You compare the error term in the case of the zeroth T.P. to the form of the first degree T.P. You expect the error in the $n-1$ case to be of the corresponding form of the next term: $(x-x_0)^n/n!$, but, as in the zeroth case, the coefficient is $f^{(n)}(c)$ instead of $f^{(n)}(x_0)$.
@Koro I'm guessing the intention is that you should imagine that there is a separator between $1$ and $j$? So for $j=10$ you'd get $x_{1,10}$ or something like that. Like when indexing a matrix using $a_{ij}$ where $i$ is row and $j$ is a column, if you substituted $i=10$,$j=11$ you'd use some separator between them.
@Snaw I don’t think that there’s a separator. given a sequence $(x_n)$, it’s being rewritten as $\{x_{kj}\}_{j=1}^\infty$ and it is being claimed that this rewritten sequence is a subsequence of $\{x_{k-1,j}\}_{j=1}^\infty$.
maybe they're wrong and it's not a subsequence. maybe it's a subsequence depending on properties specific to x_whatever that you haven't mentioned. but if the unexplained concatenation in x_{1j} was intended to mean digit concatenation i expect that somebody would have said that.
Imagine how you would do it with the usual subsequence subscripts if you take subsequences of subsequences, etc. You can't write $x_{i_{j_k}}$ with an arbitrary number of subscripts
No, the original sequence is $\{x_1, x_2, x_3,…\}$.
So the theorem being proven is “If X is a totally bounded metric space, then every sequence in X has a Cauchy subsequence.” And what I asked above is essential for the diagonal argument that the proof is going to use now.
@leslie @robjohn Have you seen this guy? I first encountered several posts where he answered (with horrible answers) and advertised an applied linear algebra book he's written. I wonder if he's an utter charlatan.
i'm not one to conspiracy theorize, but a person sock puppeting to get their rep over a threshold seems more likely than a person authoring a book and then also asking questions like that.
Ah, I recall that today I came to know that one of our moderators here on mse is writing a real analysis book but my office internet blocked the site so couldn’t check it out. I’ll try that.
It's very awkward for people outside academia to write a textbook. I couldn't have written any of my four without teaching the courses with the efforts at the books and revising for a few years.
koro: the totality of the circumstances just looks weird. very new accounts, someone who apparently claimed to author a book asking questions on the very elementary subject matter of that book, and also answering them.
i don't know.
mathematically the question and answer are fine. i didn't upvote, though. :D
ted: if there was a coyote in your front yard, but not close enough that you think you'd bother it, would you take a toddler outside to have a look at it? if you can't easily see it from a window.
DogAteMy: Consider the Minkowski sum of two line segments (thought of as the limit of a slightly thickened line segment). The sum is a parallelogram, whose boundary has double the sum as its perimeter. But each line segment is really doubled to start with.
Looking at the farthest point in a certain direction (which I suppose means the point with the largest dot product with a given unit vector) seems to be the key idea
There's a neat result called Barbier's theorem which says every curve of constant width $d$ has perimeter $\pi d$
This Minkowski fact gives a quick proof
So does the Buffon's needle problem
(filling in the details are left as exercises for the interested chatroom member)
@copper.hat: for a minute I thought that having being in the wars at MSE against the Vogons (or curators) your copper hat did not protected you well enough caused you to go into the dr i.am.great band-wagon. Anyway, good to know it was a joke.
ok. this is the difficulty with being forced into a formalism, the exact contours of which we might not know. if there isn't something that you 'already have' that you can prove this thing equal to, there might be a question as to what types of answers count as 'allowable' or not.
e.g. can you just make up some similar recursive thing and prove it's equal to that. or is it OK to introduce some non-recursive things and assume their properties.
@leslietownes The usual definition makes no reference to the algebraic structure. I think I discovered the formula for max only in an exercise in Spivak's Calculus many years ago.
Oh, OK, @robjohn. Thanks for helping.
@Jack Leslie is referring to a formula for $\max(x,y)$ in terms of algebraic operations in the real numbers.
ted: yeah. the issue here is how to prove something is "equal to max" when maybe multiplication and addition are the only primitives and there's no other thing singled out as 'the' max that we can prove it equal to.
ted: they don't have the order on naturals, apparently.
it wouldn't be uncommon for someone working in a formal system to simply say, for example, that something given by an arcane recursive formula is the definition of max. it kind of reverses our understanding of what definitions are good for.
