Hi, I'm having quite a bit of difficulty solving the following recurrence and its quite urgent: $$T(n)=3T\left ( \lceil\frac{n}{\log_{2}n} \rceil \right )+\Theta \left ( n \right )$$ where $$T(1)=T(2)=1$$. Help would be greatly appreciated!
@AkivaWeinberger The operator algebra section is sort of slim. I did well on Banach algebras, but $C^*$ is a hurdle.
@AkivaWeinberger For instance I have this Let $A$ and $B$ be unital $C*$-algebras, and let $\Phi:A\to B$ be a surjectibe unit preserving $*$-homomorphism. (i) Let $x$ be a normal element in $A$. Suppose that $\sigma (x)$ is finite. Show that there exists a self-adjoint element $y$ in $A$ and $f\in C(\sigma (y))$ s.t. $x=f(y)$.
@AkivaWeinberger So through one of our exercises we have the existence of a self-adjoint $y\in A$, but the second assertion escapes me :/
If I have two maps $p, q: S^1 \to M$ where $M$ is a manifold, in what way does $$\oint p dq$$ make sense? I'm pretty sure we are integrating over the path of $q$, but I only know how to integrate functions from $M$ to $\mathbb R^n$ in this way, not functions from $S^1$ to $M$!
ok I think it looks like $M$ is actually $\mathbb R^{2n}$, in the paper they are doing things like $(p,q) \mapsto (-q,p)$ and saying things like " as the two-form $f$ is closed, it can be regarded as the curvature of an abelian gauge field $b$: $f=db$"
but on the otherhand they write things like "H will be finite-dimensional if and only if M has finite volume.", which if we consider $\mathbb R^{2n}$ is not really sensible
@mortjt the letters $p$ and $q$ are also used (in the paper) to denote coordinate charts that are canonical wrt to a symplectic form $f$ (ie $f=\sum_i dq^i\wedge dp_i$, but this is totally something else than a map from $S^1$ to $M$, which confuses me
ah, okay. just to give the simplest context i can:
suppose you take a harmonic oscillator $H=(q^2+p^2)/2$ and quantize it. Then the energy levels are just given by $E_n=n+1/2$, and if you compute the classical action $S(E)=\oint p\,dq$ at this energy you get $S(E_n)=n+1/2$ as well
and while that's exact for a harmonic oscillator, it also holds approximately for a generic potential with a single classical minimum. that's known as Bohr-Sommerfeld quantization, and it's the simplest example I know of semiclassical analysis (name drop!) in quantum mechanics.
(it's semiclassical because, if i work in terms of dimensionful quantities, i get an overall factor of $\hbar$ in both quantities, and the condition for the approximation to work is essentially that $\hbar$ be small)
not sure i'm understanding what the concern about definition is. for example, in the harmonic oscillator case one has $E=(p^2+q^2)/2$ along the classical trajectory and therefore $p\,dq = \sqrt{2E-q^2}\,dq$
if you've got more than two coordinates, it's more complicated but you should still be able to parametrize all coordinates wrt to time
yes but $p(t)$ is not a number/vector, but a point in the manifold. Eg I also don't know how to do $\sum_i p(t_i)$, as we dont have vector space structure of points on $M$
one objection to that: $p(t)$ is not a point in $M$. it can't be, since $M$ (as a symplectic manifold) is always two-dimensional, and $p(t)$ need not be
to give some of the physics language: one starts with a configuration space (usually $\mathbb{R}^n$) and then one considers the cotangent bundle of that configuration space
$q$ is a point in the configuration space, whereas $(q,p(q))$ is a point in the cotangent bundle
and that cotangent bundle is what a physicist would call a 'phase space.'
Here we start with an even dimensional manifold, a nondegenerate two form $f$, then locally there exists coordinates $p_r, q^r$ so that $f=\sum_r dp_r \wedge q^r$. But in the article this is not I think directly connected with the integral, as (to write the sentence): "The basic Feynman integral represents the trace (2.4) as an integral over maps from $S^1$ to M:" .. "Here $p_r(t)$ and $q^r(t)$ are periodic with a period of, say, 2π, so they define a map T : $S^1$ → M, or in other words a point in the free loop space U of M"
it definitely is connected with that, though. in the harmonic oscillator case i gave above, the time-dependence of the classical trajectory is given by $(q(t),p(t))=(\cos t,-\sin t)$
which is explicitly periodic.
in which case, btw, one has $\oint p\,dq=\int_0^{2\pi} \sin^2 t\,dt=\pi$
bah, typo in the above
$(q,p)=(\sqrt{2E}\cos t,-\sqrt{2E}\sin t)$
so that the action integrates to $2\pi E$
what's probably confusing here is what the heck those $Dp_r,Dq^r$ are
i'll say that, whatever other virtues this paper might have, it is not attempting to do a comprehensive review of the Feynman path integral. it's assuming you have enough background to follow the discussion.
i'll say that, while i don't recall how the notion of pushforwards / induced maps on tangent spaces work, i suspect that's not how he's using $Dp_r,Dq^r$ here
mostly because those expressions are usually understood as convenient notations for expressing infinite-dimensional integrations which are used to represent all the different paths in phase space.
might be the same, but i'd be suspicious of identifying them
btw, i think the Dirac condition he poses later down the page is basically the Bohr-Sommerfeld condition i stated above. it's definitely related to it, at any rate.
though the cases I'm familiar with all assume the configuration space to be Euclidean. things get harder if one assumes a more complicated manifold, and there my ignorance is real.
I have a strange question about calculating median... Here is the topic : I have a list of people, aged from 50 to 90, subscribing to a kind of insurance. Some of them keep insurance until they die. Some others simply leave. I know their age when they subscribe. I know their age when they "leave" (or die). I want to know the most relevant age median of the people taking this insurance... what's the most consistent way to calculate this?
What are the left cosets of the subgroup $\langle [4]\rangle = \left\{[1], [4]\right\}$ of $\mathbb{Z}_{15}^{\times}$? My answer it's the set $\left\{[1], [4]\right\}$ because the left cosets are $\left\{[1]h: h \in \langle [4]\rangle \right\} = \left\{[1],[4]\right\}$ and $\left\{[4]h: h \in \langle [4]\rangle \right\} = \left\{[1],[4]\right\}$.
But this is wrong because I'm saying the left cosets are the subgroup itself, which sounds wrong.
I'm gonna throw one out there Q: $A$ is a unital $C^*$-algebra. $A\neq \mathbb{C}1$ iff $A$ contains non-zero positive element $x,y$ s.t. $xy=0$ So if $A=\mathbb{C}1$, then there are no $x,y>0$ s.t. $xy=0$, as $A$ is a field and in particular an integral domain. Hence if $A$ contains non-zero positive element $x,y$ s.t. $xy=0$, then $A\neq \mathbb{C}1$. Now I'm unsure about the converse. A hint is to show $A$ contains a self-adjoint element whose sectrum contains more than one point. But for some $x\in A$, this can be written as $x=x_1 +i x_2$ with $x_1,x_2$ self-adjoint.
@DanielFischer please what is the difference between prove that the sequence is not Cauchy and that the sequence has no a Cauchy subsequence ? thank you
A batch of one hundred bulbs is inspected by t esting four randomly chosen bulbs. The batch is rejected if even one of the bulbs is defective. A batch typically has five defective bulbs. The probability that the current batch is accepted is _________
The denominator will have one real root but the others will be complex. It seems I have to split it into quadratics and a linear factor but I am unable to do that
Factorization of $x^5+1$(with the help of wolfram, as Mike said) is $\frac{-1}{4}(x+1)(-2x^2+(1+\sqrt{5})x-2)(2x^2+(\sqrt{5}-1)x+2)$ which is of no help
(I would cancel out the $-1/4$ with the terms you're factoring with; they're trying to avoid a $\frac{1+\sqrt{5}}{2}$, but I'm not sure it's worth a wandering 1/4.)
I think I spent 7 hours driving a total of 120 miles in/around LA, @robjohn. Other than dealing with the traffic, I had a good time :) Enjoyed some repertory theater at USC and a wonderful dinner at a great restaurant.
Visited a former student who works at Google (in Venice). Impressive working environment they have ...
"Note that $\mathbb{R}^{\times}$ is a subgroup of $\mathbb{C}^{\times}$ (the non-zero complex numbers under multiplication). Describe the left coset of $\mathbb{R} ^{\times}$ in $\mathbb{C}^{\times}$ containing $i$." Could anyone please translate this question because I don't understand it. What does containing $i$ mean?
Well if $z$ is a complex number then $z\cdot \overline{z}= |z|^2$ where |z| means the modulus of a complex number from there you can get a formula @MatsGranvik
i thought you meant the positive real numbers. but under multiplication that's a subgroup of the nonzero real numbers, which are themselves a subgroup of the nonzero complex numbers
if the problem says $\mathbb{R}^\times$, though, it's the second case (all nonzero)
in either case, it's a subgroup of $\mathbb{C}^\times$ and you can find the left coset of it in $\mathbb{C}^\times$ w/r/t $i$
Suppose i have a rational polynomial of degree>4 whose roots are all quadratic irrationals. Must that polynomial be solvable by radicals? (i would think the answer is yes, but i've no particular background)
that's what i figured, but i wanted to make sure i wasn't forgetting some obstruction.
with the analogy here being: suppose someone gave me a rational polynomial of sufficiently high degree (say, 10) whose roots are all quadratic irrationals. then i could certainly find all of those roots by radicals given enough time and energy, but i should hardly be surprised if that solution is both long and ugly.
and while mathematica may be able to get said roots without much evident work---well, mathematica is just not showing its work.
and equivalently i could say: just because someone tells me that my degree 10 polynomial factors like that, doesn't mean it's easy or quick to find them :)
I'm actually not sure what the unambiguous terminology would be, @Semiclassic. I would, for clarity, say that each root lies in a quadratic (degree 2) extension of $\Bbb Q$. :)
I am looking at the exericse: Let the finite group $G$ act transitively on the set $\Omega$. Then the action of $G$ and on $\Omega\times\Omega$ is defined as follows $(a,b)\cdot x=(a\cdot x, b\cdot x)$. Let $a\in \Omega$. Show that the number of orbits of $G$ on $\Omega\times\Omega$ is equal to the number of orbits of $G_a$ on $\Omega$.
I have done the following:
From the fact that the finite group $G$ acts transitively on the set $\Omega$, we have that there is just one orbit on $\Omega$.
We have that $aG=\{a\cdot g: g\in G\}$. This is an orbit of $a$.
@robjohn that's strictly speaking wider than the terminology i was using, since if $a\neq 1$ then the roots wouldn't be of the form $A+\sqrt{B}$ with $A,B$ integers
but that's just a matter of multiplying by a rational factor overall, so w/e
$\newcommand{\supp}{\operatorname{supp}}$
Let $u$ and $v$ be two distributions on $\mathbb{R}^n$, at least one of which has compact support. I have to show that $\supp(u \ast v)=\supp u + \supp v (5.1.4)$.
For $\supp(u \ast v) \subset \supp u + \supp v$ I have found the following proof:
First...
hello, R is endowed with the metric $|arctan(x)-arctan(y)|$ , i want to prove that $(x_n=n)$ is a Cauchy sequence, i see that $\lim_{p,q\rightarrow\infty} d(x_p,x_q)=0$ then $(x_n)$ is a cauchy sequence, but if i want to find $n_0$ , how to do for $|arctan(p)-arctan(q)|<\varepsilon$ ?
@robjohn have you an idea please ?
$|\arctan(p)-\arctan(q)|<|arctan(p)|+|arctan(q)|$ we suppose that p>q then we obtain that $arctan(p)<\varepsilon/2$ how to find n_0 ?
@Semiclassical Bye. Cheers for helping me out several times today. :D
user189740
22:12
So not directly a math question, but I am taking a course in abstract algebra and combinatorics. Its a bachelors level course. It's very interesting, it covers things like groups, rings, etc. However, it's a totally different type of math than I am used to, and I find it very difficult to keep track of all the properties related to groups/subgroups/etc. I feel this this is going to be near impossible to remember and get good at.
just a quick question @MikeMiller or @PVAL the reason that the boundary map takes $C_{n}(A)$ to $C_{n - 1}(A)$ is because the boundary map is a group homomorphism, so preserves subgroups right ?
if $A$ is a simplicial (or cellular) subcomplex of a complex X, then for the complex of groups associated to simplicial (or cellular) homology we have $C_*(A)$ is a subcomplex of $C_*(X)$.
It isn't true without some kind of subcomplex assumption.
Let us define the free strict monoidal category $\Sigma(\textbf{Pos})$ on the category of posets $\textbf{Pos}$. How may we prove that $\Sigma(\textbf{Pos})$ is a noetherian category if we know that $\textbf{Pos}$ is noetherian?
you know @TedShifrin in this algebraic topology I am having huge troubles understanding a proof if it involves some geometrical idea for a specific proof. A phd student who does algebraic topology told me to just understand stuff algebraically and don't spend a lot of time on something.
$\newcommand{\supp}{\operatorname{supp}}$
Let $u$ and $v$ be two distributions on $\mathbb{R}^n$, at least one of which has compact support. I have to show that $\supp(u \ast v)=\supp u + \supp v (5.1.4)$.
For $\supp(u \ast v) \subset \supp u + \supp v$ I have found the following proof:
First...
Let us define the free strict monoidal category $\Sigma(\textbf{Pos})$ on the category of posets $\textbf{Pos}$. How may we prove that $\Sigma(\textbf{Pos})$ is a noetherian category if we know that $\textbf{Pos}$ is noetherian?
