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Chemical Forums aims at helping chemistry students. That means many of the questions posted require use of formulae - either chemical ones (like H 2 PO 4 - ) or mathematical ones (like [itex]E = E_0 + \frac {RT}{nF}\ln Q[/itex]). You may even need to use a structural formula like To make your questions easier to understand for others, please do your best to properly format your posts, it increases your chances of getting a correct and fast answer. Entering structural formulae that are displayed as images is covered in the next post in this thread. In general the best way to format chemical formulae is to use subscripts and superscripts. For example to enter formula of water you can enter it as H2O, mark 2 and click sub button placed in the second row of icons above the edit field. It will add [sub][/sub] tags around 2 so that it will be displayed as 2 , and water formula will be displayed as H 2 O. Similarly you can use sup button to format charges with [sup][/sup] tags. To make things faster some of the most often used numbers are also present in the line Other symbols: above the edit field. While technically it is possible to enter chemical formulae using LaTeX, it is not recommended. To enter some simple symbols like ° for temperature, middle dot for water of crystallization, Greek letters in the text and arrows, just click their symbols above the edit field. Mathematical formulae should be entered using LaTeX. To enter formula that will stay in its own line surround it with [tеx][/tеx] tags, to enter formula that will stay inline surround it with [itеx][/itеx] tags. (It is also possible to use double $ and double # delimiters you may know from the LaTeX, but once again, we don't recommend it). There are many LaTeX tutorials on the web, and many quite comprehensive documents describing LaTeX use (like this one: http://en.wikibooks.org/wiki/LaTeX/Mathematics ), but the basics are very simple. For example if you enter ax^2 + bx +c = 0 (surrounded by the [tеx][/tеx] tags) it will be rendered as [tex]ax^2 + bx + c = 0[/tex] ^ is for superscript, _ is for subscript of the following character (or group of characters). In LaTeX formulae you can easily enter Greek letters (enter the name preceded by the backslash - \Delta for Δ and \delta for δ, Greek letters listed above the edit field won't work here), group symbols using curly braces, create fractions and root symbols with LaTeX codes \frac (two following symbols or symbol groups will land in the nominator and denominator) and \sqrt (again, following symbol or symbols group will land under the root symbol) and so on: [tex]x_{1,2} = \frac {-b\pm\sqrt\Delta}{2a}[/tex] [tex]\frac 1 {2I} \left(\frac\hbar i\right)^2 \left[\frac 1 {\sin\theta} \frac \delta {\delta \theta} (\sin \theta \frac \delta {\delta\theta})+\frac 1 {\sin^2\theta} \frac {\delta^2}{\delta \phi ^2}\right]\Psi=E\Psi[/tex] [tex]C \iiint^{+\infty}_{-\infty}e^{-\frac \beta {2m} (p^2_x+p^2_y+p^2_z)}dp_x dp_y dp_z = 1[/tex] Please right-click the above equations and select Show Math As->TeX Commands to see how it was entered. \pm generates ± symbol, \hbar generates ħ and so on. If you want to test your LaTeX expressions before posting them at chemicalforums, you can do so at the forkosh sandbox - just remember to add tex tags before posting your LaTeX here. Please don't mix LaTeX and tags in a single formula, it is never necessary and never makes sense. You can mix them in a single post, but entering sulfuric acid as H[itex]_2[/itex]SO[itex]_4[/itex] (LaTeX subscript in normal text) instead of H 2 SO 4 (tagged subscript in normal text) makes it look awkward and is not guaranteed to work correctly. Note: [B] is the opening tag used to mark bolded parts of the text. Unfortunately, exactly the same symbol is quite often used to denote concentration of a compound 'B', especially when dealing with equilibrium and kinetic equations. In effect in your post there will be no single [B] symbol present, instead, whole post from the first occurrence of [B] to the end will be bolded. There are two possible solutions. First, simple and fast one, is to enter additional space after the opening square bracket - that is, to use [ B] instead of [B]. Second it to use [nоbbc][/nоbbc] tags - whatever is between them is not treated as UBBC code, so [nоbbc][B][/nоbbc] is rendered as [B]. If you plan on using LaTeX you can be forced to use nobbc trick (outside of tex tags) as well. And don't cry "hеlp!" in your posts. We know you need help, that's why you came here and posted. Your "hеlp!" will be replaced by "delete me*". It is an old joke built into forum many years ago.
Excerpt from textbook: According to Archimedes' law the weight of a body of mass $m$ and density $\rho$ inside air is: $$G=mg\left(1 - \frac{\rho_v}{\rho}\right)$$ Where $\rho_v$ is the density of air. A "weight" of mass $m_t$ and density $\rho_t$ in air weights $$G_t = m_tg\left(1-\frac{\rho_v}{\rho_t}\right)$$ If using a balance scale one determined $G=G_t$ one would get: $$m=m_t\frac{(1 - \rho_v/\rho_t)}{(1-\rho_v/\rho)}$$ i.e. The mass of the weight and the mass of the body are not equal. In real measurments $\rho_v/\rho_t\ll1$ and $\rho_v/\rho\ll1$ so the previous equation can be approximately written as: $$m=m_t\left(1+\frac{\rho_v}{\rho} - \frac{\rho_v}{\rho_t}\right)$$ How was the last approximation made? I tried deriving the 4th equation starting from the 3rd equation: $$m- m_t - m\frac{\rho_v}{\rho} + m_t\frac{\rho_v}{\rho_t} = 0$$Using approximation magic (perhaps the fact $\rho_v/\rho_t\ll1$ and $\rho_v/\rho\ll1$) I may turn $m$ into $m_t$ in the 3rd addend and factor $m_t$ :$$m- m_t - m_t\frac{\rho_v}{\rho} + m_t\frac{\rho_v}{\rho_t} = 0$$$$m- m_t\left(1 + \frac{\rho_v}{\rho} - \frac{\rho_v}{\rho_t}\right) = 0$$That gives me the formula I wanted. But it doesn't make any sense and probably another line of logic was used.
The key to the efficiency of an antenna (whether for transmitting or receiving - the two processes are essentially reciprocal) is resonance, and impedance matching with the source / receiver. The size also matters in terms of the relationship between power and current. A nice analysis of the impact of size of an antenna on the power/current relationship is given at this site. Summarizing: The current in a dipole antenna goes linearly from a maximum at the center to zero at the end. Because the amplitude of the generated E-field from a given point is proportional to the current at that point, the average power dissipated is (equation 3A2 from the above link): $$\left<P\right>=\frac{\pi^2}{3c}\left(\frac{I_0 \ell}{\lambda}\right)^2$$ (note - this is in cgs units... more about that later). For the same current, as you double the length of your (much shorter than $\lambda/4$) antenna, you quadruple the power. Directly related to this concept of power is the concept of radiation resistance: if you think of your antenna as a resistor into which you are dissipating power, then you know that $$\left<P\right> = \frac12 I^2 R$$ and combining that with the above equation for power, we see that we can get an expression for the radiation resistance $$R = \frac{2\pi^2}{3c}\left(\frac{\ell}{\lambda}\right)^2$$ This is still in cgs, which will drive most electrical engineers nuts. Converting to SI units (so we get resistance in Ohms) we just need a scale factor of $10^{9}/c^2$ (with $c$ in cgs units...); thus we get a simple approximation for radiation resistance in SI units (I now go from $c=2.98\times10^{10}~\rm{cm/s}$ to $c=2.97\times10^8~\rm{m/s}$): $$R = \frac{2\pi^2 c}{3\times 10^{-7}}\left(\frac{\ell}{\lambda}\right)^2$$ which agrees nicely with the expression given at this calculator for an electrically short dipole (note - their expression is for $\ell_{eff}$ which is $\ell/2$ for a short dipole; and they use slightly rounded numbers which is OK since there are some approximations going on anyway). But if we are driving with a 50 Ohm cable, and our antenna represents a much smaller "resistance", then most of the power would be reflected and we don't get a good coupling of power into the antenna (remember - because of reciprocity, everything I say about transmitting is true for receiving... but intuitively the transmission case is so much easier to grasp). So to get good efficiency, we need to make sure there is an impedance match between our antenna and the transmitter / receiver. If you know what frequency you are working at, impedance matching can be done with a simple LC circuit: the series LC represent a low impedance to the antenna, but a high impedance to the receiver. In the process, they convert the large current in the antenna into a large voltage for the receiver (source of image and detailed explanation) This is an example of resonant matching: it works well at a specific frequency. One can use signal transformers to achieve the same thing over a wider range of frequencies - but this loses you some of the advantages of resonance (all frequencies are amplified equally). It remains to be shown what the real effect is of reducing antenna size on the received signal. For this, the most extensive reference I could find was this MIT open course lecture. Starting on page 121, this shows that the effective length of a dipole determines how much of the incoming energy can be "harvested", and it again shows that the power is proportional to the square of the size. So an antenna that is twice as short will collect four times less power. But that means it will also collect four times less noise. As long as most of the noise in the system comes from "outside", the ratio (SNR) will be the same, and you don't suffer from the smaller antenna. This changes once the antenna becomes so small that other sources of noise become significant. It is reasonable to think that this will happen when the conductive (lossy) resistance of the antenna becomes comparable to the reactive (radiation resistance). But since the former scales with the length of the antenna, and the latter with the square of the length, it is obvious there will be a size at which the non-ideal effects will dominate. The better the conductors, and the better the amplifiers, the smaller the antenna can be. Summary So yes, the power transmitted drops with the square of the length, making a short antenna less efficient as a transmitter (and therefore, as a receiver). Much of the time, though, you care about signal to noise ratio - is there more signal than noise coming from your antenna? For this, we need to look at the Q of the antenna (bandwidth). The higher the Q, the more gain you have at just the frequency of interest (because of resonance); while "noise" is a wide-band phenomenon, "signal" is a narrow-band one, so a high Q amplifies the signal without amplifying (all the) noise. If we can make an antenna with a high Q, then it doesn't matter so much that it is short.
Search Now showing items 1-10 of 33 The ALICE Transition Radiation Detector: Construction, operation, and performance (Elsevier, 2018-02) The Transition Radiation Detector (TRD) was designed and built to enhance the capabilities of the ALICE detector at the Large Hadron Collider (LHC). While aimed at providing electron identification and triggering, the TRD ... Constraining the magnitude of the Chiral Magnetic Effect with Event Shape Engineering in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV (Elsevier, 2018-02) In ultrarelativistic heavy-ion collisions, the event-by-event variation of the elliptic flow $v_2$ reflects fluctuations in the shape of the initial state of the system. This allows to select events with the same centrality ... First measurement of jet mass in Pb–Pb and p–Pb collisions at the LHC (Elsevier, 2018-01) This letter presents the first measurement of jet mass in Pb-Pb and p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV and 5.02 TeV, respectively. Both the jet energy and the jet mass are expected to be sensitive to jet ... First measurement of $\Xi_{\rm c}^0$ production in pp collisions at $\mathbf{\sqrt{s}}$ = 7 TeV (Elsevier, 2018-06) The production of the charm-strange baryon $\Xi_{\rm c}^0$ is measured for the first time at the LHC via its semileptonic decay into e$^+\Xi^-\nu_{\rm e}$ in pp collisions at $\sqrt{s}=7$ TeV with the ALICE detector. The ... D-meson azimuthal anisotropy in mid-central Pb-Pb collisions at $\mathbf{\sqrt{s_{\rm NN}}=5.02}$ TeV (American Physical Society, 2018-03) The azimuthal anisotropy coefficient $v_2$ of prompt D$^0$, D$^+$, D$^{*+}$ and D$_s^+$ mesons was measured in mid-central (30-50% centrality class) Pb-Pb collisions at a centre-of-mass energy per nucleon pair $\sqrt{s_{\rm ... Search for collectivity with azimuthal J/$\psi$-hadron correlations in high multiplicity p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 and 8.16 TeV (Elsevier, 2018-05) We present a measurement of azimuthal correlations between inclusive J/$\psi$ and charged hadrons in p-Pb collisions recorded with the ALICE detector at the CERN LHC. The J/$\psi$ are reconstructed at forward (p-going, ... Systematic studies of correlations between different order flow harmonics in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV (American Physical Society, 2018-02) The correlations between event-by-event fluctuations of anisotropic flow harmonic amplitudes have been measured in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The results are ... $\pi^0$ and $\eta$ meson production in proton-proton collisions at $\sqrt{s}=8$ TeV (Springer, 2018-03) An invariant differential cross section measurement of inclusive $\pi^{0}$ and $\eta$ meson production at mid-rapidity in pp collisions at $\sqrt{s}=8$ TeV was carried out by the ALICE experiment at the LHC. The spectra ... J/$\psi$ production as a function of charged-particle pseudorapidity density in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV (Elsevier, 2018-01) We report measurements of the inclusive J/$\psi$ yield and average transverse momentum as a function of charged-particle pseudorapidity density ${\rm d}N_{\rm ch}/{\rm d}\eta$ in p-Pb collisions at $\sqrt{s_{\rm NN}}= 5.02$ ... Energy dependence and fluctuations of anisotropic flow in Pb-Pb collisions at √sNN=5.02 and 2.76 TeV (Springer Berlin Heidelberg, 2018-07-16) Measurements of anisotropic flow coefficients with two- and multi-particle cumulants for inclusive charged particles in Pb-Pb collisions at 𝑠NN‾‾‾‾√=5.02 and 2.76 TeV are reported in the pseudorapidity range \|η\| < 0.8 ...
PDE: $$ u_t (x,t) - u_{xx} (x,t) = 0, \quad u(0,t)=0, \quad u_x(2,t)=0, \quad u(x,0)=\delta(x-x_0) $$ where $0<x_0<2$ is a given constant. I worked out my solution by using separation of variables $$u(x,t)=X(x)T(t) \implies \frac{X''(x)}{X(x)}=\frac{T'(t)}{T(t)}=\lambda$$ For $\lambda>0$ and $\lambda=0$ we get trivial solutions, but for $\lambda=-k^2<0$ we get the solutions $$ X(x)=c_1 sink_nx, \quad k_n=\frac{\pi}{2}(\frac{1}{2}+n), \quad n=0,1,... $$ $$ T(t)=c_2e^{-k_n^2t} $$ Superposition gives the general solution to the PDE as $$ \sum_{n=0}^\infty c_n sin(k_nx)e^{-k_n^2t}, \quad k_n=\frac{\pi}{2}(\frac{1}{2}+n)$$ Now for my question, I don't understand how to handle the initial condition to calculate the fourier coefficients $c_n$. I have the solution for it though, and what they have done is this (I understand this step); $$ u(x,0)=\sum_{n=0}^\infty c_n sin(k_n x) = \delta(x-x_0) \quad (1)$$ This is the step I don't understand; $$ (1) \implies c_n = \frac{\int_0^2 \delta(x-x_0)sin(k_nx)dx}{\int_0^2 sin^2(k_nx)dx}=sin(k_n x_0) $$ Any explanation as to why and how they got that expression for $c_n$, and also why it equals $sin(k_n x_0)$ (it's like they're discounting the denominator and only looking at the numerator?) would be greatly appreciated.
I'm trying to calculate the SO-coupling for a single p-electron ($l=1$, $s=\frac{1}{2}$) in the uncoupled representation. This comes down to calculating these matrix elements: $$\left\langle nlm_lsm_s\left\|\hat{\vec l}.\hat{\vec s}\right\|nlm_l^{'}sm_s^{'}\right\rangle$$ It can be easily found that: $$\hat{\vec l}.\hat{\vec s} = \frac{1}{2}(\hat{ l_+}\hat{ s_-}+\hat{ l_-}\hat{ s_+})+\hat{l_z}\hat{s_z}$$ Which gives me the following matrix representation of $\hat{\vec l}.\hat{\vec s}$: \begin{bmatrix} \frac{1}{2} & 0 & 0 & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & \frac{1}{\sqrt{2}} & 0 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 0 & \frac{1}{\sqrt{2}} & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2} \end{bmatrix} I recognize this is a diagonal block matrix of the form \begin{bmatrix} A & 0 \\ 0 & A \end{bmatrix} And I've found the eigenvalues this way ($\lambda=-1$ and $\lambda=\frac{1}{2}$) Now I've tried solving for the eigenvectors but I think I'm doing something wrong because the equations I get are trivially equal to zero (The eigenket is represented by a column vector $(x,y,z)$, the calculation is for the $\lambda = \frac{1}{2}$ eigenvalue): $(A-\lambda I)\ \|\phi\rangle=0$ leads to the following equations: $$-y+z\frac{1}{\sqrt 2}=0$$ $$\frac{1}{\sqrt 2}y-\frac{1}{2}z=0$$ Which in turn gives the trivial equations: $$z=\frac{2}{\sqrt 2}y$$ $$y= 0, z=0$$ I don't see what I'm doing wrong, I get similar trivial equations when calculating the other eigenvalue.
(Math Stack Exchange suggested that the same question I posted there be migrated here; The one at Math Stack Exchange was thus deleted. The recommendation message of migration can be found here, though the page is now deleted.) Whenever students are asked to derive or prove some vector calculus identities such as $$\nabla^2\vec{A}=\nabla(\nabla\cdot \vec{A})-\nabla \times (\nabla \times \vec{A})$$ they are often asked to expand in index notation and rearrange to give the required expressions. This caused me to wonder how these identities were first derived (since not all of them are consequence of the product rule of derivatives). A brief search of the history of the topic seemed to suggest the following: Complex numbers $\to$ quaternions $\to$ Grassmann (exterior algebra) $\to$ Vector analysis. Another brief search on the topic of tensors and Albert Einstein showed that the two periods of development of vector analysis and tensor analysis have some overlap thus it might be possible they might be aware of the development of each other. However, since the vector analysis history account above does not mentioned about tensors nor index notation, how were the vector calculus identities originally derived, possibly without index notation available at that time?
In this particular case, I am talking about an extremely flexible human. In my story, this human catches another's head with his legs and manages to snap his neck. I know a human can't simply snap another's neck (They don't have enough strength), but could it be done if the attacker were able to pinch their opponent's neck (to weaken it), and then twist and snap it? It is possible in theory, but not the way it is shown in the movies. First of all, the human neck is both strong and flexible, and unless the person is taken unaware (like in a car accident), will automatically tense up in response to pressure, making it even stronger. Second, even if you could muster enough strength to break a neck, unless the rest of the body was restrained you would just end up twisting their entire body around. The only way to pull off a neck snap realistically is by getting the victim in a full-body grapple, then stretching and twisting the neck until it breaks. You'd need to be very strong to do it though; flexibility alone isn't going to cut it. It may, however, be possible to break someone's neck with an impact instead of a twist. A strong punch or kick may generate enough force, but only if the body was restrained somehow, so not really practical in a fight. It may be possible to get them in a hold and then throw yourself and them onto the floor with enough force to twist their neck to the breaking point, but this would involve more luck than anything. This may be the best option to use in a story though - make it basically a lucky fluke. It won't be an instant-kill in any event; even if you fully paralyze the person from the neck down they'll still be alive until they suffocate. More likely than not they won't be fully paralyzed and you'll instead be left with a paraplegic who is perfectly capable of talking (or screaming). Think about the possible results of car accident neck injuries. All in all, it's not a very effective killing maneuver. I decided to do a little research into neck injuries in sports (especially contact sports) to get an idea of the ways in which people can (accidentally in these cases, we hope) break each other's necks. It looks like one common way is via whiplash, rather than twisting of the neck – in this video (warning, a wrestler dies) you can see that it's either the blow to the neck or the whiplash caused by the force of the blow which caused a broken neck in this case. As IndigoPhoenix wrote, if the person's body is immobilised, then their neck is much more likely to be broken – e.g. if a person is on the ground and someone steps/falls on their neck, as experienced by this person during a game of rugby. The other means by which a fighter could break an opponent's neck with (relative) ease, which I left until last because it's so obvious, is via a "piledriver" or similar move that involves driving the opponent head-first into the ground. These moves incur a foul in professional Mixed Martial Arts, and are banned in professional wrestling, because the risk of a broken neck is simply far too high. Lifting an opponent off the ground so that their centre of balance is high enough to flip them head-first before they hit the ground is also an illegal tackle in rugby (not sure about gridiron, sorry, but it might be legal over there, the armour makes those guys crazy). In the context of fighting narratives this seems to me to be a "power" move rather than a flexibility and dexterity move, so it might not be quite your character's style, but perhaps instead of "catching the opponent's head with their legs" and twisting to break, your character could instead grab with their legs and somehow flip/drive the opponent head first into the ground? Alternatively, maybe your character could wrap their legs around the opponents neck and then fall with them to the ground, so that when they both hit the ground the force of your characters fall is directed in a way that puts their full weight/force on the opponent's neck. It takes 1000-1250 foot-pounds (1350-1700 Newton-metres) of torque to break a human neck. That kind of force is generally achieved in a 5-9 foot drop. You can also work out exactly how far you need to drop someone of mass $x$ to break their neck: $$ d = \frac{1}{g}\cdot\frac{T}{x} = \frac{T}{gx} $$ That is, the required drop distance (in metres) is equal to the torque required to break their neck (assume $1700 \text{ Nm}$) divided by the gravitational field strength (assume $9.81 \text{ N}\cdot\text{kg}^{-1}$ on Earth) multiplied by the person's mass, $x \text{ kg}$. The higher the mass, the smaller the distance required. So that's one way a human can break another's neck: drop them out of a first-floor window onto their head. The torque from the impact will snap their neck. The same can be said of any impact: if you hit hard enough, their neck will break. We can use the formula above to find a figure for force. Assuming an average person of around 85 kilograms, you need $\frac{T}{d} = gx$ Newtons of force: $$ gx = 9.81 \times 85 = 833.85 \text{ N} $$ You can use this figure to construct an impact that will have the required force to break a human neck. Using the fact that in an impact, impulse $Ft$ is equal to the change in momentum $x\Delta v$: $$ Ft = x\Delta v $$ $$ \frac{Ft}{x} = \Delta v $$ Assuming a stopping time of around 0.1 seconds (typical of many impacts), we can work out the required delta-V: $$ \Delta v = \frac{833.85 \times 0.1}{85} = 0.981 $$ So, you need an impact with a deceleration of $9.81 \text{ms}^{-2}$ to break a neck. Not surprising, given that the dropping method is, essentially, exactly that impact. You can, of course, use the force figure to work out any number of other creative ways to break your target's neck. N.B.: All these figures require that the force is applied in a non-axial plane. Applying the force directly to the top of their head won't work, as it will all transfer down the spine and spread across the body like it's meant to. You need to effectively push sideways.
I'm asked to find if the fixed-point iteration $$x_{k+1} = g(x_k)$$ converges for the fixed points of the function $$g(x) = x^2 + \frac{3}{16}$$ which I found to be $\frac{1}{4}$ and $\frac{3}{4}$. In this short video by Wen Shen, it's explained how to find these fixed-points and to see if a fixed-point iteration converges. My doubt is related to find if a fixed point iteration converges for a certain fixed point. At more or less half of the video, she comes up with the following relation for the error $$e_{k+1} = \|g'(\alpha)\| e_k$$ where $\alpha \in (x_k, r)$, by the mean value theorem, and because $g$ is continuous and differentiable. If $\|g'(\alpha)\| < 1$, then the fixed-point iteration converges. I think I agree with this last statement, but when she tries to see if the fixed point iteration converges for a certain root of a certain function, she simply finds the derivative of that function and plugs in it the root. I don't understand why this is equivalent to $$e_{k+1} = \|g'(\alpha)\| e_k$$ Can someone fire up some light on my brain? (lol, can I say this?)
Assume there exist infinitely many $x$ such that: $$3x^2+3x+1 = 3p-2$$ Where $p$ is prime. Can it be shown there exist infinitely many $y$ such that: $$3y^2+3y+1=q$$ Where $q$ is prime? I believe that it cannot be shown as our assumption tells us nothing of which primes exist such that $3p-2 = 3x^2+3x+1$ and so knowing there exist infinitely many primes of the form $3p-2$ does not help us, but maybe I am wrong (perhaps it can be shown I am wrong with a relevant proof). For your first equation, simply adding $2$ to both sides yields $$x^2+x+1=p$$ so it suffices to prove that there exists two integers $a_x$ and $b_x$ such that $a_xb_x=1-p$ and $a_x+b_x=1$. $$\therefore p=a_x+b_x-a_xb_x$$ Same for how $$y^2+y+1=\frac{q+2}{3}$$ where $q=a_y+b_y-3a_yb_y$. Clearly $ab\neq 0$ so if $a+b=1$ then one of $a$ or $b$ is negative. Now since all primes $p>2$ are odd, let $a$ and $b$ both be odd, or opposite parity. Then note that if $p>3$ then $p\equiv \pm 1\pmod 6$ and it should pretty much be trivial hereafter. Here's my take, if, $q$ replaces $p$ we get by:$$3p-2=3(p-1)+1$$ that, $$9y^2+9y+1=3x^2+3x+1\implies 3y^2+3y=x^2+x$$ which has solutions: $$(x,y)\in\{(2,1),(9,5),(35,20)\}$$ less than $(100,100)$ possible values of $p-1$ are: 0,2,6,12,20,30,42,... because $p-1$ must be the product of consecutive integers. If the $q$ replacing $p$ case works then $9y+10\equiv 3x+4 \bmod r$ for any value pairs $x,y$ both congruent to powers with exponent $r-1\over 2$ mod prime $r$. We also get $$3y+1\equiv x+1\bmod r$$ via the $$ax^2+ax+c\equiv ax+(c+a)$$ reduction with constant term $c=0$. The first reduction, gives us $$x\equiv 3y+2\bmod r\implies x-1\equiv 3y+1\bmod r$$ contradicting the second coming from the same algebra.
I was thinking about hadrons in general Yang-Mills theories and I have some doubts that I'd like to discuss with you. Suppose that we have a Yang-Mills theory that, like QCD, tend to bind quarks into color singlet states. So far nothing strange, even QED tend to bind electromagnetic charges to form neutral systems. The twist in Yang-Mills theories comes when we consider the running of the coupling constant: $$ \alpha(\mu)\simeq\frac{1}{ln(\mu/\Lambda)} $$ which present asymptotic freedom, i.e. it decrease when the energy scale $\mu$ increase. This formula also suggest that when the energy scale approach the energy $\Lambda$ the interactions between two colored objects becomes very strong, so we assume that we cannot observe colored objects on scales greater than $1/\Lambda$. Here my first question: We cannot see free quark in our world because the scale $1/\Lambda$ happen to be smaller than hadronic typical dimension so we cannot pull quarks enough outside hadrons to see them in non hadronic environment? And, if yes, could exist Yang Mills theories where this not happen, i.e. where the hadron typical scale is smaller than $1/\Lambda$ and then we can see free quarks? In QCD, being the hadrons system of strongly coupled quarks, we cannot study them by perturbative approach. However the presence of a spontaneously broken chiral symmetry allows us to study some of their properties. Here my second question: We do not need this SSB to form hadrons, right? There could be some theory were we have hadrons but not this SSB? If something is unclear tell me and I'll try to explain better.
I am reading Nakahara's Geometry, topology and physics and I am a bit confused about the non-coordinate basis (section 7.8). Given a coordinate basis at a point on the manifold $\frac{\partial}{\partial x^\mu}$ we can pick a linear combination of this in order to obtain a new basis $\hat{e}_\alpha=e_\alpha^\mu\frac{\partial}{\partial x^\mu}$ such that the new basis is orthonormal with respect to the metric defined on the manifold i.e. $g(\hat{e}_\alpha,\hat{e}_\beta)=e_\alpha^\mu e_\beta^\nu g_{\mu\nu}=\delta_{\alpha \beta}$ (or $=\eta_{\alpha \beta}$). I am a bit confused about the way we perform this linear transformation of the basis. Is it done globally? This would mean that at each point we turned the metric into the a diagonal metric, which would imply that the manifold is flat, which shouldn't be possible as the change in coordinates shouldn't affect the geometry of the system (i.e. the curvature should stay the same). So does this mean that we perform the transformation such that the metric becomes the flat metric just at one point, while at the others will also change, but without becoming flat (and thus preserving the geometry of the manifold)? Then Nakahara introduces local frame rotations, which are rotations of these new basis at each point, which further confuses me to why would you do that, once you already obtained the flat metric at a given point. So what is the point of these new transformations as long as we perform the first kind in the first place? Sorry for the long post I am just confused. This transformation is done locally, i.e. so that $g_{\alpha\beta} = \eta_{\alpha\beta}$ in a neighbourhood rather than a single point. I believe that due to topological effects, we cannot in general do it globally. However, even though the new basis is orthonormal, we have to remember that it is (in general) non-holonomic, i.e that there is no set of functions $y^\alpha$ satisfying$$e^\mu_\alpha = \frac{\partial x^\mu}{\partial y^\alpha}.$$This means that the curvature does not (in general) vanish. Indeed, it holds that$$T_{\alpha\beta\ldots\gamma} = e^\mu_\alpha e^\nu_\beta e^\sigma_\gamma T_{\mu\nu\ldots\sigma},$$for all tensors $T_{\mu\nu\ldots\sigma}$, as is obvious by the linearity of $e^\mu_\alpha$, and thus in particular for $R_{\mu\nu\sigma\tau}$. You may be confused if you have learned that the connection coefficients are given by$$\Gamma_{\mu\nu\sigma} = \frac{1}{2}\left(g_{\mu\nu,\sigma} + g_{\mu\sigma,\nu} - g_{\nu\sigma,\mu}\right),$$but this is only valid in a holonomic frame. More generally$$\Gamma_{\alpha\beta\gamma} = \frac{1}{2}\left(g_{\alpha\beta\|\gamma} + g_{\alpha\gamma\|\beta} - g_{\beta\gamma\|\alpha} + C_{\gamma\alpha\beta} + C_{\beta\alpha\gamma} - C_{\alpha\beta\gamma}\right),$$where $f_{\|\alpha} \equiv e_\alpha(f)$, $[e_\alpha,e_\beta] = C^\gamma{}_{\alpha\beta}e_{\gamma}$, and $C_{\gamma\alpha\beta} \equiv g_{\gamma\delta}C^\delta{}_{\alpha\beta}$. In particular, in an orthonormal frame:$$\Gamma_{\alpha\beta\gamma} = \frac{1}{2}\left(C_{\gamma\alpha\beta} + C_{\beta\alpha\gamma} - C_{\alpha\beta\gamma}\right).$$Unless I misunderstand you, the rotations you refer to are local Lorentz transformations, and we are interested in them because each orthonormal frame corresponds to the instantaneous rest frame of an observer (with velocity field equal to $e_0$), and thus these rotations transform between different rest frames. Note that the Transformation you mentioned $g_{\mu \nu} \mapsto g(\hat{e_\alpha},\hat{e_\beta})$ by the function $e_\alpha^{\mu}$ (also called the tetrad) is only a Change of coordinates if it holds: $e_\alpha^{\mu} = \frac{\partial u^\mu}{\partial x^\alpha}$. Here, $u^\mu(x)$ is a vector function and you can choose this function such that at some Point $x_0$ you have a flat metric. If you have done a suitable choice of $u^\mu$, metric tensors in a neighborhood of $x_0$ will not be flat in General. The Transformation $g_{\mu \nu} \mapsto g(\hat{e_\alpha},\hat{e_\beta}):=\frac{\partial u^\mu}{\partial x^\alpha}\frac{\partial u^\nu}{\partial x^\beta}g_{\mu \nu} ()$ will preserve all geometric Tensors like curvature Tensor, Torsion Tensor, etc. up to a Change of coordinate Basis. More precisely, if $X_{\mu_1 \mu_2 \dots \mu_n}$ is a geometric Tensor like the Riemann Tensor, it will Change under the Transformation () as $X_{\mu_1 \mu_2 \dots \mu_n} \mapsto \frac{\partial u^{\mu_1}}{\partial x^{\nu_1}} \frac{\partial u^{\mu_2}}{\partial x^{\nu_2}} \dots \frac{\partial u^{\mu_n}}{\partial x^{\nu_n}}X_{\mu_1 \mu_2 \dots \mu_n}=X'_{\nu_1 \dots \nu_n}$ like a Tensor should transform. Another Transformation that does not Change geometric quantities are local rotations $\Lambda^{\mu}_{\nu}(x)$. In this case you have $e^{\mu}_{\nu} = \Lambda^{\mu}_{\nu}$. Such rotation matrices have the property that its inverse matrices is simply the transpose of it (from which it follows that $det(\Lambda) = \pm 1$). A geometric Tensor will transform similary: $X_{\mu_1 \mu_2 \dots \mu_n} \mapsto \Lambda^{\mu_1}_{\nu_1} \dots \Lambda^{\mu_n}_{ \nu_n} X_{\mu_1 \mu_2 \dots \mu_n}=X''_{\nu_1 \dots \nu_n}$ Other forms of the Tensors $e_\alpha^\mu$ (not coordinate transformations or rotations) change the intrinsic geometry!
I've readed in Jech's book that the existence of the $\omega-$Erdös cardinal $\kappa(\omega)$ (that is the minimumm cardinal $\kappa$ for which $\kappa\rightarrow (\omega)^{<\omega}$) it is compatible with the axiom of constructibility $V=L$. More precisely, it can be shown that if $\kappa=\kappa(\omega)$ exists then $L\vDash\kappa\rightarrow (\omega)^{<\omega}$. In this regard, I would like to ask some things about some of the details of that proof. First of all assuming the existence of a monochromatic set $H$ (in $V$) for a colouring $c:[\kappa(\omega)]^{<\omega}\longrightarrow 2$ in $L$, we want to see that there exists a monochromatic set $H'\in L$ for $c$. For this purpouse Jech defines a tree $T$ formed by finite increasing sequences $\langle \alpha_1,\ldots,\alpha_{n-1}\rangle$ such that for every $k\leq n$ then $[\{\alpha_1,\ldots,\alpha_{n-1}\}]^k$ is monochromatic, and endows it with the reverse order $\supset$. I feel that I understand the reasons why $T\in L$ and $c$ has a monochromatic set iff $T$ is illfounded. Despite of this, I'm not sure of which is the role played by the term ''increasing sequences'' and $\supset$. More precisely, my questions are the following: Why couldn't we take as the elements of the tree any finite sequence like before paying no attention to their order and $\subset$ instead of $\supset$? If $\alpha<\omega_1^L$ then is it true that the existence of $\kappa(\alpha)$ implies $L\models\kappa(\alpha)\to(\alpha)^{<\omega}$? If this would be true, how could I prove it? Best, Cesare.
Let $X = \mathrm{Spec}\ A$, and for each prime ideal $\mathfrak{p} \subseteq A$, let $A_{\mathfrak{p}}$ be the localization of $A$ at $\mathfrak{p}$. Now, for each open set $U \subseteq X$, define $\mathcal{O}(U)$ to be the set of functions $$s: U \rightarrow \coprod_{\mathfrak{p} \in U} A_{\mathfrak{p}}$$ such that $s(\mathfrak{p}) \in A_{\mathfrak{p}}$ for each $\mathfrak{p}$, and such that $s$ is locally a quotient of elements of $A$, that is, for each $\mathfrak{p} \in U$, there is some neighborhood $V_{\mathfrak{p}}$ of $\mathfrak{p}$ with $V_{\mathfrak{p}} \subseteq U$ and elements $a,f \in A$ such that for each $\mathfrak{q} \in V$ and $f \notin \mathfrak{q}$, $s(\mathfrak{q} )= a/f$ in $A_{\mathfrak{q}}$. My question is Why is the sheaf of rings $\mathcal{O}$ defined above a sheaf of rings? While it seems clear that $\mathcal{O}$ is a presheaf, I fail to see why it satisfies the extra axioms that make it a sheaf. Hartshorne says "it is clear from the local nature of the definition that $\mathcal{O}$ is a sheaf", but while trying to prove it formally, I couldn't show that it is actually a sheaf. Also, I've just been fiddling around with the definitions for the moment trying to gain some insight about why this "provides a systematic way of keeping track of local algebraic data on $\mathrm{Spec}\ A$", but for now, it's just as mysterious as it was at the beginning. So if anyone could provide me some insight about why this is what we want to be our definition of structure sheaf, I'd be very grateful for that. Thank you in advance!
Measure here is Jordan Measure. $d(A,B)=\inf\{\vert x-y:x\in A,y\in B\}$ So I can show the first direction fairly easily because it doesn't use the condition. $\leq$ Since $S_i$ are measurable there exists poly rectangles $P_i$ such that $\mu S_i\leq\vert P_i\vert\leq \mu (S_i)+\frac{\epsilon}{2}$. Since $S_i\subseteq P_i$, $S_1\cup S_2\subseteq P_1\cup P_2$ thus $\mu(S_1\cup S_2\leq \vert P_1\cup P_2\vert=\vert P_1\vert+\vert P_2\vert\leq \mu S_1+\mu S_2+\epsilon$ so $\mu (S_1\cup S_2)\leq \mu S_1 +\mu S_2$ But to show $\mu (S_1\cup S_2) \geq \mu S_1 + \mu S_2$ I'm not sure. Obviously since the distance between any 2 points between them is always positive they don't share any points. Which I believe should mean I can construct poly rectangles $P_i$ for each $S_i$ which have lengths less then $d(S_1,S_2)$ such that $P_1\cup P_2$ should still cover $S_1 \cup S_2$
Interesting idea to use a grid. I doubt it will work for this question, since writing things in that grid format in a sense splits into arithmetic progressions modulo $p$, and usually saying things about the primes in a progression is harder. However, I cannot help but mention that using a grid like that was applied brilliantly by Maier to prove a counter-intuitive result, (using the Prime Number Theorem for Arithmetic Progessions) and the idea is now called the Maier Matrix Method. (A small digression, but it is interesting!) The Maier Matrix Method There are questions regarding primes in short intervals, and we can ask ourselves what does $$\pi(x+y)-\pi(x)$$ look like? To say anything meaningful, $y$ cannot be too small, but here lets suppose $y=\log^B(x)$ for some $B>2$. Selberg proved that under the Riemann Hypothesis, we have $$\pi(x+y)-\pi(x)\sim \frac{y}{\log x}$$ as $x\rightarrow \infty$ for almost all $x$. (A set with density $\rightarrow 1$) It was then conjectured that this asymptotic must hold for all $x$ which are sufficiently large. (This conjecture was made for several reasons, one of which is that it is true under Cramer's probabilistic model) In a surprising turn of events, Maier showed it was false, and that there exists $\delta>0$ and arbitrarily large values of $x_1$ and $x_2$ such that both $$\pi(x_1 +\log^B (x_1))-\pi (x_1)> (1+\delta)\log^{B-1}(x_1)$$and $$\pi(x_2 +\log^B (x_2))-\pi (x_2)< (1-\delta)\log^{B-1}(x_2)$$hold, despite the fact that the asymptotic holds for a set of density $1$. He proved this using a method which is now called the "Maier Matrix Method." Essentially, it is just drawing a grid which is similar to the one above, and then applying a clever combinatorial argument. The columns are arithmetic progressions, and by PNT4AP, we can easily say things about them to understand the number of primes in the grid. There is a little trick with oscillation of the Dickman Function, but then basically by the pigeon hole principle the question is solved. I definitely think you might find this expository article by Dr.Andrew Granville to be interesting. (It is quite readable, and gives an a more in depth, and very clear explanation)
Question. Is there a smooth proper scheme $X\to\operatorname{Spec}(\mathbb{Z})$ such that $X(\mathbb{Q}_v)\neq\emptyset$ for every place $v$ of $\ \mathbb{Q}$ ( including $v=\infty$), and yet $X(\mathbb{Q})=\emptyset$ ? I believe the answer is No. The evidence is flimsy : $X$ cannot be a curve or a torsor under an abelian variety (Abrashkin-Fontaine) or a twisted form of $\mathbb{P}_n$. I haven't gone through a list of smooth projective $\mathbb{Q}$-varieties which contradict the Hasse principle to check if any of them has good reduction everywhere, but the chances of such a thing are slim. Colliot-Thélène and Xu give a systematic treatement of many known examples of quasi-projective $\mathbb{Z}$-schemes $Y$ such that $Y(\mathbb{Z}_p)\neq\emptyset$ for every prime $p$ and $Y(\mathbb{R})\neq\emptyset$, but $Y(\mathbb{Z})=\emptyset$. Some of these schemes might even be smooth over $\mathbb{Z}$, but none of them is proper. Fontaine's letter to Messing (MR1274493) might be of some relevance here.
Hello, I've never ventured into char before but cfr suggested that I ask in here about a better name for the quiz package that I am getting ready to submit to ctan (tex.stackexchange.com/questions/393309/…). Is something like latex2quiz too audacious? Also, is anyone able to answer my questions about submitting to ctan, in particular about the format of the zip file and putting a configuration file in $TEXMFLOCAL/scripts/mathquiz/mathquizrc Thanks. I'll email first but it sounds like a flat file with a TDS included in the right approach. (There are about 10 files for the package proper and the rest are for the documentation -- all of the images in the manual are auto-generated from "example" source files. The zip file is also auto generated so there's no packaging overhead...) @Bubaya I think luatex has a command to force “cramped style”, which might solve the problem. Alternatively, you can lower the exponent a bit with f^{\raisebox{-1pt}{$\scriptstyle(m)$}} (modify the -1pt if need be). @Bubaya (gotta go now, no time for followups on this one …) @egreg @DavidCarlisle I already tried to avoid ascenders. Consider this MWE: \documentclass[10pt]{scrartcl}\usepackage{lmodern}\usepackage{amsfonts}\begin{document}\noindentIf all indices are even, then all $\gamma_{i,i\pm1}=1$.In this case the $\partial$-elementary symmetric polynomialsspecialise to those from at $\gamma_{i,i\pm1}=1$,which we recognise at the ordinary elementary symmetric polynomials $\varepsilon^{(n)}_m$.The induction formula from indeed gives\end{document} @PauloCereda -- okay. poke away. (by the way, do you know anything about glossaries? i'm having trouble forcing a "glossary" that is really an index, and should have been entered that way, into the required series style.) @JosephWright I'd forgotten all about it but every couple of months it sends me an email saying I'm missing out. Oddly enough facebook and linked in do the same, as did research gate before I spam filtered RG:-) @DavidCarlisle Regarding github.com/ho-tex/hyperref/issues/37, do you think that \textNFSSnoboundary would be okay as name? I don't want to use the suggested \textPUnoboundary as there is a similar definition in pdfx/l8uenc.def. And textnoboundary isn't imho good either, as it is more or less only an internal definition and not meant for users. @UlrikeFischer I think it should be OK to use @, I just looked at puenc.def and for example \DeclareTextCompositeCommand{\b}{PU}{\@empty}{\textmacronbelow}% so @ needs to be safe @UlrikeFischer that said I'm not sure it needs to be an encoding specific command, if it is only used as \let\noboundary\zzznoboundary when you know the PU encoding is going to be in force, it could just be \def\zzznoboundary{..} couldn't it? @DavidCarlisle But puarenc.def is actually only an extension of puenc.def, so it is quite possible to do \usepackage[unicode]{hyperref}\input{puarenc.def}. And while I used a lot @ in the chess encodings, since I saw you do \input{tuenc.def} in an example I'm not sure if it was a good idea ... @JosephWright it seems to be the day for merge commits in pull requests. Does github's "squash and merge" make it all into a single commit anyway so the multiple commits in the PR don't matter or should I be doing the cherry picking stuff (not that the git history is so important here) github.com/ho-tex/hyperref/pull/45 (@UlrikeFischer) @JosephWright I really think I should drop all the generation of README and ChangeLog in html and pdf versions it failed there as the xslt is version 1 and I've just upgraded to a version 3 engine, an dit's dropped 1.0 compatibility:-)
If $f : \mathbb{R}^{n} \longrightarrow \mathbb{R}$ is $L$-Lipschitz (w.r.t. the $\\|\cdot\\|_2$ norm), it is a fact that if $x \sim N(0, I)$, then $$ \mathbb{P}( f(x) - \mathbb{E}f(x) \geq t) \leq e^{-t^2/(2L^2)} \:. $$ I'm reading the proof (from Theorem 4.3 of https://galton.uchicago.edu/~lalley/Courses/386/Concentration.pdf) of this fact using Gaussian log Sobolev inequalities, and I'm a bit confused as to the important of $\mathbb{E} f(x)$ in the proof. Specifically, I'm wondering why the same proof does not imply the (obviously incorrect fact) that for any $M \in \mathbb{R}$, $$ \mathbb{P}( f(x) - M \geq t) \leq e^{-t^2/(2L^2)} \:. \:\:\: () $$ I'm going to post my logic here, and hopefully somebody can point out the flaw in the argument. Fix an $M \in \mathbb{R}$ and an $f$ which is $L$-Lipschitz and differentiable and define $g(x) := f(x) - M$. Since $f$ is $L$-Lipschitz, then so is $g$. Suppose we want to show $()$. By a standard Chernoff argument, it suffices to show that there exists finite positive constants $C, A$ such that the MGF $\mathbb{E} e^{\lambda g} \leq C e^{A \lambda^2}$ for all $\lambda > 0$ (c.f. Lemma 2.3 of Lalley's notes). I will now repeat the argument of Theorem 4.3. Define $h = e^{\lambda g/2}$. By the Gaussian log Sobolev inequality, $$ \mathrm{Ent}(h^2) \leq 2 \mathbb{E} \\|\nabla h \\|^2_2 = 2 \mathbb{E}\\| \nabla e^{\lambda g/2} \\|_2^2 = \frac{\lambda^2}{2} \mathbb{E} e^{\lambda g} \\| \nabla g \\|_2^2 \leq \frac{\lambda^2 L^2}{2} \mathbb{E} e^{\lambda g} \:. $$ Now, define the function $F(\lambda) := \mathbb{E} e^{\lambda g}$. We have that $F'(\lambda) = \mathbb{E} g e^{\lambda g}$. Hence, $$ \mathrm{Ent}(h^2) = \mathbb{E} \lambda g e^{\lambda g} - (\mathbb{E} e^{\lambda g}) \log(\mathbb{E} e^{\lambda g}) = \lambda F'(\lambda) - F(\lambda) \log(F(\lambda)) \leq \frac{\lambda^2 L^2}{2} F(\lambda) \:. $$ But this inequality is the same differential inequality which shows up in the standard proof, from which we conclude that $F(\lambda) \leq e^{L^2 \lambda^2 / 8}$. Notice how $M$ does not enter into these constants. But then we have just constructed a bound on the MGF for $g(x)$, and can apply the Chernoff argument, yielding $(*)$. Where does this argument break down?
If $M$ is a smooth closed $n$-dimensional Riemannian manifold which is Riemannian embedded in $\mathbb R^{n+1}$, then there exists a point $p \in M$ such that the sectional curvatures at $p$ are all positive. Can any one give me a hint for this problem? I was considering the maximum $p$ of function $\|x\|^2$ on $M$, then near $p$, $M$ is "wrapped" by some $S^n$ and has the same tangent space as $S^n$. But I am stuck there. I have made some progress: We consider the functions $L_q(x)=\|x-q\|^2$. Then we have a maximum $p$ of $L_q$ and we fix the unit vector $v=\frac{p-q}{\|p-q\|}$ throughout, so $v$ is the normal vector at $p$. Now if we set $q(t)=p+tv$, then when $t\leq-\|p-q\|$, $p$ is always the maximum of function $L_{q(t)}$ (this is true if we draw a ball at $q(t)$ with radius $\|p-q(t)\|$, then all $M$ is contained in this ball). Therefore when $t$ sufficiently tends to $-\infty$, the Hessian $L_{q(t)}$ is always semi-positive definite. Now if we fix a coordinate neighborhood aroud $p$, then Hessian matrix $H$ of $L_{q(t)}$ at $p$ is given by $H=2(F-tS)$ where $F,S$ are the first and second fundamental forms of $M$ at $p$. So we conclude that $S$ has to be semi-positive definite. But how can we move further to say $S$ is positive definite?
Given a (Hausdorff separated) locally convex space $X$ what can we say about a proper convex function $f:X\to\mathbb{R}$ whose domain $\emptyset\neq D(f):=\{x\in X\mid f(x)<+\infty\}$ has a non-empty (topological) interior? Recall that when $X$ is barreled the non-empty domain interior spells that the function is continuous on the interior of its domain. So what can happen when the space is not barreled other than continuity? My take so far in this is to consider the simplest type of a convex function: a sublinear function (positively homogeneous and subadditive). I took $B\subset X^$ (the topological dual) and $f=\sigma_B$ the support function of $B$; $\sigma_B(x):=\sup_{x^\in B}x^*(x)$, $x\in D(\sigma_B)$ (the barrier cone of $B$). My reformulation of the problem in this case is: What can one say about a set whose barrier cone has a non-empty interior? To simplify even more, in case $0\in {\rm core}\ {D(\sigma_B)}$ (core denotes here the algebraic interior) we get that $D(\sigma_B)=X$ because $D(\sigma_B)$ is an absorbing cone. We find that $B$ is (weak-star) bounded.
I have been asking a rather few questions of this nature lately, maybe I'm starting to realise math notation isn't as uniform as I initially thought it would be... Question: Does this notation$$\frac{\partial(y_1,\dots,y_m)}{\partial(x_1,\dots,x_n)}$$refer to the Jacobian matrix$$ J = \begin{bmatrix} \dfrac{\partial y_1}{\partial x_1} & \cdots & \dfrac{\partial y_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \dfrac{\partial y_m}{\partial x_1} & \cdots & \dfrac{\partial y_m}{\partial x_n} \end{bmatrix},$$or the Jacobian determinant $\det J$? I am aware of the ambiguity of "Jacobian" being used to refer to either the determinant or the matrix itself, is this a similar case? It's really a bit annoying because when I see things like $$ \left\| \frac{\partial(y_1,\dots,y_m)}{\partial(x_1,\dots,x_n)} \right\| $$ I don't know if it means the absolute value of the Jacobian determinant, or the determinant of the Jacobian matrix.
Search Now showing items 1-10 of 34 Search for top squark pair production in final states with one isolated lepton, jets, and missing transverse momentum in √s = 8 TeV pp collisions with the ATLAS detector (Springer, 2014-11) The results of a search for top squark (stop) pair production in final states with one isolated lepton, jets, and missing transverse momentum are reported. The analysis is performed with proton-proton collision data at s√ ... Search for supersymmetry in events with large missing transverse momentum, jets, and at least one tau lepton in 20 fb−1 of √s = 8 TeV proton-proton collision data with the ATLAS detector (Springer, 2014-09-18) A search for supersymmetry (SUSY) in events with large missing transverse momentum, jets, at least one hadronically decaying tau lepton and zero or one additional light leptons (electron/muon), has been performed using ... Measurement of the top quark pair production charge asymmetry in proton-proton collisions at √s = 7 TeV using the ATLAS detector (Springer, 2014-02) This paper presents a measurement of the top quark pair ( tt¯ ) production charge asymmetry A C using 4.7 fb−1 of proton-proton collisions at a centre-of-mass energy s√ = 7 TeV collected by the ATLAS detector at the LHC. ... Measurement of the low-mass Drell-Yan differential cross section at √s = 7 TeV using the ATLAS detector (Springer, 2014-06) The differential cross section for the process Z/γ ∗ → ℓℓ (ℓ = e, μ) as a function of dilepton invariant mass is measured in pp collisions at √s = 7 TeV at the LHC using the ATLAS detector. The measurement is performed in ... Measurements of fiducial and differential cross sections for Higgs boson production in the diphoton decay channel at √s=8 TeV with ATLAS (Springer, 2014-09-19) Measurements of fiducial and differential cross sections are presented for Higgs boson production in proton-proton collisions at a centre-of-mass energy of s√=8 TeV. The analysis is performed in the H → γγ decay channel ... Measurement of the inclusive jet cross-section in proton-proton collisions at $ \sqrt{s}=7 $ TeV using 4.5 fb−1 of data with the ATLAS detector (Springer, 2015-02-24) The inclusive jet cross-section is measured in proton-proton collisions at a centre-of-mass energy of 7 TeV using a data set corresponding to an integrated luminosity of 4.5 fb−1 collected with the ATLAS detector at the ... ATLAS search for new phenomena in dijet mass and angular distributions using pp collisions at $\sqrt{s}$=7 TeV (Springer, 2013-01) Mass and angular distributions of dijets produced in LHC proton-proton collisions at a centre-of-mass energy $\sqrt{s}$=7 TeV have been studied with the ATLAS detector using the full 2011 data set with an integrated ... Search for direct chargino production in anomaly-mediated supersymmetry breaking models based on a disappearing-track signature in pp collisions at $\sqrt{s}$=7 TeV with the ATLAS detector (Springer, 2013-01) A search for direct chargino production in anomaly-mediated supersymmetry breaking scenarios is performed in pp collisions at $\sqrt{s}$ = 7 TeV using 4.7 fb$^{-1}$ of data collected with the ATLAS experiment at the LHC. ... Search for heavy lepton resonances decaying to a $Z$ boson and a lepton in $pp$ collisions at $\sqrt{s}=8$ TeV with the ATLAS detector (Springer, 2015-09) A search for heavy leptons decaying to a $Z$ boson and an electron or a muon is presented. The search is based on $pp$ collision data taken at $\sqrt{s}=8$ TeV by the ATLAS experiment at the CERN Large Hadron Collider, ... Evidence for the Higgs-boson Yukawa coupling to tau leptons with the ATLAS detector (Springer, 2015-04-21) Results of a search for $H \to \tau \tau$ decays are presented, based on the full set of proton--proton collision data recorded by the ATLAS experiment at the LHC during 2011 and 2012. The data correspond to integrated ...
well, it is absolutely in agreement with theory. the correlation as measured by Pearson's coefficient $\rho$ is linear measure in the sense that the bounds [-1,1] are obtained only when transformations of our variables are linear, so if we have variables $X$ and $Y$ then something like $aX+bY+c$ where $a,b\in\mathbb{R^*}$, $c\in\mathbb{R}$ will have ... A constant maturity swap (CMS) rate for a given tenor is referenced as a point on the Swap curve. A swap curve itself is a term structure wherein every point on the curve is the effective par swap rate for that tenor. This is analogous to a 3m LIBOR curve represents 3m forward rates for a given tenor.A swap rate can be considered as a weighted-average of ... The main thing to keep in mind with all these different option combination strategies is that you are really trading option greeks! I think the answer to why the calender spread is so popular lies in the special combination of gamma and vega risk:Calendar spreads are the one type of trade where gamma can be negative while vega is positive (and vice versa ... In a vanilla swap, the IR on the floating leg usually depends on the reset period/swap frequency. If frequency is 6m, 6m LIBOR is used for reset, 3m LIBOR for quarterly resets etc. In a floating CMS leg, the rate used is the CMS rate, regardless of the reset frequency e.g: 10yr CMS leg will use the 10 yr CMS rate, regardless of whether the reset happens semi-... It depends on the frequency and the horizon. For instance, I got a similar looking chart when I used annual log returns as the input to the log normal distribution and went out 250 years. With daily log returns over a few years, there isn't nearly as much of a decay. However, when you go out 250 years with daily returns you still see the pattern. It depends on the exact structure. E.g., a butterfly can be bought or sold and every market participant understands which individual options are bought or sold given knowledge of the agreed spot level and distance of the wing from spot in regards to agreed strikes.Please note that a butterfly can be structured as a combination of calls but also through ... The issue is what exchanges recognize as acceptable complex option orders. This is governed by FINRA margin rules like rule 4210Brokers can only execute orders that are recognized by options exchanges. So what brokers can put on a single order ticket is limited. The most complex spreads defined by FINRA rules would be butterfly spreads, box spreads and ... The best approximation that I know isLi, Deng and Zhou, 2006. It's an analytic approximation where the price is expressed as a direct formula, so easy to implement.If you want to be VERY accurate, here's my paper,J Choi (2018) (Arxiv). It handles the options on any linear combination of assets such as basket and Asian options as well as spread option. ... ES is supposed to update every 100 msec now in TWS. I imagine the FOPs update slower than that, probably every 300 msec. IB is not a real time feed like others, they aggregate the data and send it on schedule. It's good because it doesn't lag, bad because you don't get every tick.What Matt said is true and good but if you want to stick with Excel you ... I've been analysing the same problem and i think that the way to go it's calibrating an interest rate model. Think of it as an option on a bond, there is plenty of literature about that.Also you can look at Quantlib implementation of callable bonds to get an idea of how can it be implemented. In simple terms: An ordinary swap might be a 10 year swap of Libor vs a fixed rate; this fixed rate is determined in the marketplace every day and is published by Reuters, Bloomberg etc. as the '10 year swap rate'. Once you enter into the swap this rate remains fixed for you, of course, that is why it is called a fixed rate. But every day Reuters publishes a ... A single look CMS spread option is simply an option on the difference between the two forward CMS rates and a chosen strike $K$ on a single expiry date $t$. A CMS spread cap is then a strip of options and pays on each $t_i$ from $t_1$ to $t_n$. Both types are quoted by brokers such as Tullet. Both products allow the investor a view on the shape of the yield ... After some testing, it is clear that a value of 0.005 is the most reasonable minimum volatility to use with this model.It is small enough to be a reasonable starting point (extremely unlikely that the real volatility of an option will be lower than this), while still being sufficiently high enough to avoid numerical issues when calculating implied ... Well, the probabilities implied by the market are not equal. If you believe they should be equal, then go ahead and express yourself in the market. The point is , it is not an objective fact that it must be 50/50- that's your subjective opinion. There are a few extra things to consider here where you'll get a different answer if you ask a quant or a trader.If we have a european digital that pays \$1 if the underlying is above 120 ($S_0 = 100$) at expiry, then yes i can hedge it with a call spread. This can be approximated with a call spread (with a notional of $\frac{1}{\mathrm{d}K}$, this was ... For such problems, you may consider the moment matching approach.For example, you can approximate the combination of terms where the coefficients have the same sign by a log-normal random variable, and then use the approach you mentioned. If all coefficients have the same sign,you can approximate the whole combination by a shifted log-normal random ... prepayment model could be very complicated as there are so many variables could affect the prepayment rates, such as demographic, seasonality, location etc. My experience on this was I knew many companies just directly use some prepayment data from SIFMA instead of building their own model. Typical text bool models are CPR, PSA, SMM. Nowdays, machine ... Your formula, as it stands, is incorrect, at least is if $E$ means the "expected value under real-world probabilities".I wrote a blog post explaining the basic rationale behind risk-neutral pricing where you will see that if the Fundamental Theorem of Asset Pricing theorem holds, you can write:Let $X_t=S_{1,t}-S_{2,t}$$$e^{-rt} X_t = \mathbb{E}_\mathbb{...
Fitting a Damped Sine Wave A damped sine wave is described by $$ x_{(k)} = A \cdot e^{\alpha \cdot k} \cdot cos(\omega \cdot k + p) \tag{1}$$ with frequency $\omega$ , phase p , initial amplitude A and damping constant $\alpha$ . The $x_{(k)}$ are the samples of the function at equally spaced points in time. With $x_{(k)}$ given, one often has to find the unknown parameters of the function. This can be achieved for instance with nonlinear approximation or with DFT – methods. I present a method to find the unknown parameters which is based on linear algebra only. The equation (1) is the solution to the difference equation $$x_{(k+2)} = 2 \cdot \sqrt b \cdot cos(\omega) \cdot x_{(k+1)} - b \cdot x_{(k)} \tag{2}$$ This difference equation is a linear equation system for the unknown constants b and $2 \cdot \sqrt b \cdot cos(\omega) $. The overdetermined linear system $$ \left( \begin{matrix} x_{(k+2)} \\ x_{(k+3)} \\ . \\ . \end{matrix} \right) = \left( \begin{matrix} x_{(k+1)} & x_{(k )} \\ x_{(k+2)} & x_{(k+1)} \\ . & .\\ . & . \end{matrix} \right) \cdot \left( \begin{matrix} 2 \cdot \sqrt b \cdot cos(\omega) \\ -b \end{matrix} \right) \tag{3}$$is solved using linear regression analysis. There is a pitfall in this calculation: For high oversampling rates, i.e. small resulting $\omega$, the numerical results with additional noise are poor. Performance is best near half the Nyquist rate, which results in approximately 4 samples per sine wave. The $x_{(k)}$ can be rearranged accordingly without losing samples or accuracy, such as taking every 8th sample for the $x_{(k)},x_{(k+1)}, ... $ The resultant $\omega$ has to be corrected by this integer factor. The $b$ is related to $\alpha$ by $$e^\alpha\ = \sqrt b \tag{4}$$ For the derivation of this relation see the appendix. So the $\omega$, $\alpha$ and b are known. The amplitude A and the phase $p$ are still unknown and calculated by 'mixing with a quadrature carrier' : The system of equations $$ \left( \begin{matrix} x_{(0)} \\ x_{(1)} \\ . \\ . \end{matrix} \right) = \left( \begin{matrix} e^{\alpha \cdot 0} \cdot \sin{(\omega \cdot 0)} & e^{\alpha \cdot 0} \cdot \cos{(\omega \cdot 0)} \\ e^{\alpha \cdot 1} \cdot \sin{(\omega \cdot 1)} & e^{\alpha \cdot 1} \cdot \cos{(\omega \cdot 1)} \\ . & .\\ . & . \end{matrix} \right) \cdot \left( \begin{matrix} u \\ v \end{matrix} \right) \tag{5}$$ is solved for u and v. With the u, v interpreted as the complex number $u+iv$ we obtain the amplitude and phase $$A=\|u+iv\|$$ $$p=\sphericalangle{(u-iv)}$$ In real setups the samples $x_{(k)}$ suffer from additional noise. It turns out that the results for frequency $\omega$ and phase $p$ are very stable in noise. That is not the case for amplitude A nor for the damping factor $\alpha$. These parameters are adjusted with a gradient iteration. Amplitude and damping are slightly varied until the best fit to the sampled data is achieved. I published a corresponding Matlab script in the 'Mathworks file exchange' system. You can find it by searching for 'matlab fit damped sine wave'. Download is free and you do not need a registration. I hope that this algorithm is useful and download numbers will rise. Cheers Detlef Appendix Replacing $x_{(k)}$ in (2) with its definition in (1) yields $$ \begin{align} & A e^{\alpha \cdot (k+2)} \cos {(\omega (k+2) + p)} = \\ & \\ & 2 \sqrt{b} \cos{(\omega) A e^{\alpha(k+1)}} \cdot \cos{(\omega (k+1)+p)} \\ & - b \cdot A \cdot e^{\alpha \cdot k} \cdot \cos{(\omega k + p)} \\ \end{align} $$ Division by $A \cdot e^{\alpha \cdot k}$ and application of addition theorems for cos-functions gives $$\begin{align} & e^{2 \alpha} [\cos{(\omega k + p)} \cos{(2 \omega)}-\sin{(\omega k + p)} \sin{(2 \omega)} ] = \\ \\ & 2 e^{\alpha}\sqrt{b}\cos{\omega} [\cos{(\omega k + p)} \cos{(\omega)}-\sin{(\omega k + p)} \sin{(\omega)}] \\ & - b \cos{(\omega k+p)} \\ \end{align} $$ Rearranging terms $$\begin{align} 0=& & \cos{(\omega k + p)} [e^{2 \cdot \alpha} \cos{(2 \omega)}-2e^\alpha \sqrt b \cos ^2 {\omega} +b ]- \\ & & \sin{(\omega k + p)} [e^{2 \cdot \alpha} \sin{(2 \omega)}-2e^\alpha \sqrt b \sin {\omega}\sin {\omega} ]= \\ \\ 0= & & \cos{(\omega k + p)} [e^{2 \cdot \alpha} \cos{(2 \omega)}-2e^\alpha \sqrt b \cos ^2 {\omega} +b ]- \\ & & \sin{(\omega k + p)}\sin{(2 \omega)} \cdot [e^{2 \alpha } -e^{ \alpha }\sqrt b ] \\ \end{align} $$ The last equation can only hold for any k if the expressions in the squared brackets equal 0: $$ 0= e^{2 \alpha } -e^{ \alpha }\sqrt b $$ $$ \Rightarrow e^{\alpha } = \sqrt b $$ Kaz $\omega$ is a normalized frequency, it is related to the sample frequency. So for a given signal, if the sample rate gets higher, the $\omega$ is going down. From the linear system of equations the $\cos \omega$ term is calculated. This function is flat for small $\omega$, a small change in the $\cos \omega$ will result in a big change for $\omega$. The situation ist best near $\pi / 2$, because the $\cos \omega$ function ist steepest there, and this sample rate is corresponding to Nyquist / 2 . And yes, higher oversampling rate gives better results, you use all samples, but you have to rearrange the samples. Take a look at the Matlab-script. Cheers Detlef Regards, [-Rick-] thanks for the hint. For Google 'matlab fit damped sine wave', the MathWork Exchange link is the first hit, at least for my Google results. I hope, that your efforts to understand my blog were crowned with success. I'd be more than glad to give some additional clarifications. Cheers Detlef To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments. Registering will allow you to participate to the forums on ALL the related sites and give you access to all pdf downloads.
Difference between revisions of "Algebra and Algebraic Geometry Seminar Spring 2018" (→John Lesieutre) (→John Lesieutre) Line 212: Line 212: ===John Lesieutre=== ===John Lesieutre=== − Some higher-dimensional cases of the Kawaguchi-Silverman conjecture + Some higher-dimensional cases of the Kawaguchi-Silverman conjecture Given a dominant rational self-map f : X -->X of a variety defined over a number field, the first dynamical degree $\lambda_1(f)$ and the arithmetic degree $\alpha_f(P)$ are two measures of the complexity of the dynamics of f: the first measures the rate of growth of the degrees of the iterates f^n, while the second measures the rate of growth of the heights of the iterates f^n(P) for a point P. A conjecture of Kawaguchi and Silverman predicts that if P has Zariski-dense orbit, then these two quantities coincide. I will prove this conjecture in several higher-dimensional settings, including for all automorphisms of hyper-K\"ahler varieties. This is joint work with Matthew Satriano. Given a dominant rational self-map f : X -->X of a variety defined over a number field, the first dynamical degree $\lambda_1(f)$ and the arithmetic degree $\alpha_f(P)$ are two measures of the complexity of the dynamics of f: the first measures the rate of growth of the degrees of the iterates f^n, while the second measures the rate of growth of the heights of the iterates f^n(P) for a point P. A conjecture of Kawaguchi and Silverman predicts that if P has Zariski-dense orbit, then these two quantities coincide. I will prove this conjecture in several higher-dimensional settings, including for all automorphisms of hyper-K\"ahler varieties. This is joint work with Matthew Satriano. Revision as of 21:11, 24 April 2018 The seminar meets on Fridays at 2:25 pm in room B113. Here is the schedule for the previous semester. Contents 1 Algebra and Algebraic Geometry Mailing List 2 Spring 2018 Schedule 3 Abstracts Algebra and Algebraic Geometry Mailing List Please join the AGS Mailing List to hear about upcoming seminars, lunches, and other algebraic geometry events in the department (it is possible you must be on a math department computer to use this link). Spring 2018 Schedule Abstracts Tasos Moulinos Derived Azumaya Algebras and Twisted K-theory Topological K-theory of dg-categories is a localizing invariant of dg-categories over [math] \mathbb{C} [/math] taking values in the [math] \infty [/math]-category of [math] KU [/math]-modules. In this talk I describe a relative version of this construction; namely for [math]X[/math] a quasi-compact, quasi-separated [math] \mathbb{C} [/math]-scheme I construct a functor valued in the [math] \infty [/math]-category of sheaves of spectra on [math] X(\mathbb{C}) [/math], the complex points of [math]X[/math]. For inputs of the form [math]\operatorname{Perf}(X, A)[/math] where [math]A[/math] is an Azumaya algebra over [math]X[/math], I characterize the values of this functor in terms of the twisted topological K-theory of [math] X(\mathbb{C}) [/math]. From this I deduce a certain decomposition, for [math] X [/math] a finite CW-complex equipped with a bundle [math] P [/math] of projective spaces over [math] X [/math], of [math] KU(P) [/math] in terms of the twisted topological K-theory of [math] X [/math] ; this is a topological analogue of a result of Quillen’s on the algebraic K-theory of Severi-Brauer schemes. Roman Fedorov A conjecture of Grothendieck and Serre on principal bundles in mixedcharacteristic Let G be a reductive group scheme over a regular local ring R. An old conjecture of Grothendieck and Serre predicts that such a principal bundle is trivial, if it is trivial over the fraction field of R. The conjecture has recently been proved in the "geometric" case, that is, when R contains a field. In the remaining case, the difficulty comes from the fact, that the situation is more rigid, so that a certain general position argument does not go through. I will discuss this difficulty and a way to circumvent it to obtain some partial results. Juliette Bruce Asymptotic Syzygies in the Semi-Ample Setting In recent years numerous conjectures have been made describing the asymptotic Betti numbers of a projective variety as the embedding line bundle becomes more ample. I will discuss recent work attempting to generalize these conjectures to the case when the embedding line bundle becomes more semi-ample. (Recall a line bundle is semi-ample if a sufficiently large multiple is base point free.) In particular, I will discuss how the monomial methods of Ein, Erman, and Lazarsfeld used to prove non-vanishing results on projective space can be extended to prove non-vanishing results for products of projective space. Andrei Caldararu Computing a categorical Gromov-Witten invariant In his 2005 paper "The Gromov-Witten potential associated to a TCFT" Kevin Costello described a procedure for recovering an analogue of the Gromov-Witten potential directly out of a cyclic A-inifinity algebra or category. Applying his construction to the derived category of sheaves of a complex projective variety provides a definition of higher genus B-model Gromov-Witten invariants, independent of the BCOV formalism. This has several advantages. Due to the categorical invariance of these invariants, categorical mirror symmetry automatically implies classical mirror symmetry to all genera. Also, the construction can be applied to other categories like categories of matrix factorization, giving a direct definition of FJRW invariants, for example. In my talk I shall describe the details of the computation (joint with Junwu Tu) of the invariant, at g=1, n=1, for elliptic curves. The result agrees with the predictions of mirror symmetry, matching classical calculations of Dijkgraaf. It is the first non-trivial computation of a categorical Gromov-Witten invariant. Aron Heleodoro Normally ordered tensor product of Tate objects and decomposition of higher adeles In this talk I will introduce the different tensor products that exist on Tate objects over vector spaces (or more generally coherent sheaves on a given scheme). As an application, I will explain how these can be used to describe higher adeles on an n-dimensional smooth scheme. Both Tate objects and higher adeles would be introduced in the talk. (This is based on joint work with Braunling, Groechenig and Wolfson.) Moisés Herradón Cueto Local type of difference equations The theory of algebraic differential equations on the affine line is very well-understood. In particular, there is a well-defined notion of restricting a D-module to a formal neighborhood of a point, and these restrictions are completely described by two vector spaces, called vanishing cycles and nearby cycles, and some maps between them. We give an analogous notion of "restriction to a formal disk" for difference equations that satisfies several desirable properties: first of all, a difference module can be recovered uniquely from its restriction to the complement of a point and its restriction to a formal disk around this point. Secondly, it gives rise to a local Mellin transform, which relates vanishing cycles of a difference module to nearby cycles of its Mellin transform. Since the Mellin transform of a difference module is a D-module, the Mellin transform brings us back to the familiar world of D-modules. Eva Elduque On the signed Euler characteristic property for subvarieties of Abelian varieties Franecki and Kapranov proved that the Euler characteristic of a perverse sheaf on a semi-abelian variety is non-negative. This result has several purely topological consequences regarding the sign of the (topological and intersection homology) Euler characteristic of a subvariety of an abelian variety, and it is natural to attempt to justify them by more elementary methods. In this talk, we'll explore the geometric tools used recently in the proof of the signed Euler characteristic property. Joint work with Christian Geske and Laurentiu Maxim. Harrison Chen Equivariant localization for periodic cyclic homology and derived loop spaces There is a close relationship between derived loop spaces, a geometric object, and (periodic) cyclic homology, a categorical invariant. In this talk we will discuss this relationship and how it leads to an equivariant localization result, which has an intuitive interpretation using the language of derived loop spaces. We discuss ongoing generalizations and potential applications in computing the periodic cyclic homology of categories of equivariant (coherent) sheaves on algebraic varieties. Phil Tosteson Stability in the homology of Deligne-Mumford compactifications The space [math]\bar M_{g,n}[/math] is a compactification of the moduli space algebraic curves with marked points, obtained by allowing smooth curves to degenerate to nodal ones. We will talk about how the asymptotic behavior of its homology, [math]H_i(\bar M_{g,n})[/math], for [math]n \gg 0[/math] can be studied using the representation theory of the category of finite sets and surjections. Wei Ho Noncommutative Galois closures and moduli problems In this talk, we will discuss the notion of a Galois closure for a possibly noncommutative algebra. We will explain how this problem is related to certain moduli problems involving genus one curves and torsors for Jacobians of higher genus curves. This is joint work with Matt Satriano. Daniel Corey Initial degenerations of Grassmannians Let Gr_0(d,n) denote the open subvariety of the Grassmannian Gr(d,n) consisting of d-1 dimensional subspaces of P^{n-1} meeting the toric boundary transversely. We prove that Gr_0(3,7) is schoen in the sense that all of its initial degenerations are smooth. The main technique we will use is to express the initial degenerations of Gr_0(3,7) as a inverse limit of thin Schubert cells. We use this to show that the Chow quotient of Gr(3,7) by the maximal torus H in GL(7) is the log canonical compactification of the moduli space of 7 lines in P^2 in linear general position. Alena Pirutka Irrationality problems Let X be a projective algebraic variety, the set of solutions of a system of homogeneous polynomial equations. Several classical notions describe how ``unconstrained the solutions are, i.e., how close X is to projective space: there are notions of rational, unirational and stably rational varieties. Over the field of complex numbers, these notions coincide in dimensions one and two, but diverge in higherdimensions. In the last years, many new classes of non stably rational varieties were found, using a specialization technique, introduced by C. Voisin. This method also allowed to prove that the rationality is not a deformation invariant in smooth and projective families of complex varieties: this is a joint work with B. Hassett and Y. Tschinkel. In my talk I will describe classical examples, as well as the recent progress around these rationality questions. Nero Budur Homotopy of singular algebraic varieties By work of Simpson, Kollár, Kapovich, every finitely generated group can be the fundamental group of an irreducible complex algebraic variety with only normal crossings and Whitney umbrellas as singularities. In contrast, we show that if a complex algebraic variety has no weight zero 1-cohomology classes, then the fundamental group is strongly restricted: the irreducible components of the cohomology jump loci of rank one local systems containing the constant sheaf are complex affine tori. Same for links and Milnor fibers. This is joint work with Marcel Rubió. Alexander Yom Din Drinfeld-Gaitsgory functor and contragradient duality for (g,K)-modules Drinfeld suggested the definition of a certain endo-functor, called the pseudo-identity functor (or the Drinfeld-Gaitsgory functor), on the category of D-modules on an algebraic stack. We extend this definition to an arbitrary DG category, and show that if certain finiteness conditions are satisfied, this functor is the inverse of the Serre functor. We show that the pseudo-identity functor for (g,K)-modules is isomorphic to the composition of cohomological and contragredient dualities, which is parallel to an analogous assertion for p-adic groups. In this talk I will try to discuss some of these results and around them. This is joint work with Dennis Gaitsgory. John Lesieutre Some higher-dimensional cases of the Kawaguchi-Silverman conjecture Given a dominant rational self-map f : X -->X of a variety defined over a number field, the first dynamical degree $\lambda_1(f)$ and the arithmetic degree $\alpha_f(P)$ are two measures of the complexity of the dynamics of f: the first measures the rate of growth of the degrees of the iterates f^n, while the second measures the rate of growth of the heights of the iterates f^n(P) for a point P. A conjecture of Kawaguchi and Silverman predicts that if P has Zariski-dense orbit, then these two quantities coincide. I will prove this conjecture in several higher-dimensional settings, including for all automorphisms of hyper-K\"ahler varieties. This is joint work with Matthew Satriano.
Find if the series converges or diverges, $a_n=\sum_{n=1}^\infty\frac{1}{1+\ln (n)}$. Comparing it with another series $b_n=\frac{1}{\ln(n)}$. Dividing both the series and taking limits, we get $\lim_{n\to\infty}\frac{\ln(n)}{1+\ln(n)}$. Since it is the $\infty/\infty$ form, applying H'opitals rule, we get, $\lim_{n\to\infty}\frac{1/n}{1/n}=1$. Now, $\lim_{n\to\infty}\frac{1}{ln(n)}=0, \Rightarrow b_n$ converges $\Rightarrow a_n$ converges. But the answer is, Comparing it with $\frac{1}{n}$(divergent harmonic series) we get,$\lim_{n\to\infty}\frac{n}{1+\ln(n)}=\lim_{n\to\infty}\frac{1}{1/n}=\lim_{n\to\infty}n=\infty \Rightarrow a_n $ diverges, what is wrong with my comparison? Your mistake is when you say that: $$\lim_{x \to \infty} b_n = 0$$ implies that $\sum b_n$ is convergent. That doesn't hold in general. Note further that $n > \ln(n)$ so $\frac{1}{n} < \frac{1}{\ln(n)}$. So by comparison $\sum \frac{1}{\ln(n)}$ is divergent. So as mentioned in the comment above, your comparison is fine, the conclusion is just that the series is divergent.
I'm currently working through the book Heisenberg's Quantum Mechanics (Razavy, 2010), and am reading the chapter on classical mechanics. I'm interested in part of their derivative of a generalized Lorentz force via a velocity-dependent potential. I understand the generalized force that they derive from a Lagrangian of the form $L = \frac{1}{2}m\|\vec v\|^2 - V(\vec r,\vec v,t)$ $$F_i = -\frac{\partial V}{\partial x_i} + \frac{d}{dt}\left(\frac{\partial V}{\partial v_i}\right)$$ However, in the next (critical) step of the derivation, the author cites a theorem from Helmholtz saying ...according to Helmholtz, for the existence of the Lagrangian, such a generalized force can be at most a linear function of acceleration, and it must satisfy the Helmholtz identities. The three Helmholtz identities are then listed as: $$\frac{\partial F_i}{\partial \dot{v_j}} = \frac{\partial F_j}{\partial \dot{v_i}},$$ $$\frac{\partial F_i}{\partial v_j} + \frac{\partial F_j}{\partial v_i} = \frac{d}{dt}\left(\frac{\partial F_i}{\partial \dot{v_j}} + \frac{\partial F_j}{\partial \dot{v_i}}\right),$$ $$\frac{\partial F_i}{\partial x_j} - \frac{\partial F_j}{\partial x_i} = \frac{1}{2}\frac{d}{dt}\left(\frac{\partial F_i}{\partial v_j} - \frac{\partial F_j}{\partial v_i}\right).$$ I'm trying to understand where this theorem comes from. Razavy cited a 1887 paper by Helmholtz. I was able to find a PDF online, but it is in German, so I could not verify whether or not it proved the theorem. Additionally, I could not find it in any recent literature. I searched online and in Goldstein's Classical Mechanics. The only similar concept that I can find is in the Inverse problem for Lagrangian mechanics where we have three equations known as Helmholtz conditions. Are these two concepts one in the same? If so, how should I interpret the function $\Phi$ and the matrix $g_{ij}$ that appear in the Helmholtz conditions I found online? If the cited theorem from Razavy does not relate from the inverse Lagrangian problem, could I have some help finding the right direction?
Search Now showing items 1-6 of 6 Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV (Springer, 2015-05-20) The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ... Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV (Springer, 2015-06) We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ... Multiplicity dependence of two-particle azimuthal correlations in pp collisions at the LHC (Springer, 2013-09) We present the measurements of particle pair yields per trigger particle obtained from di-hadron azimuthal correlations in pp collisions at $\sqrt{s}$=0.9, 2.76, and 7 TeV recorded with the ALICE detector. The yields are ... Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV (Springer, 2015-09) Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ... Coherent $\rho^0$ photoproduction in ultra-peripheral Pb-Pb collisions at $\mathbf{\sqrt{\textit{s}_{\rm NN}}} = 2.76$ TeV (Springer, 2015-09) We report the first measurement at the LHC of coherent photoproduction of $\rho^0$ mesons in ultra-peripheral Pb-Pb collisions. The invariant mass and transverse momentum distributions for $\rho^0$ production are studied ... Inclusive, prompt and non-prompt J/ψ production at mid-rapidity in Pb-Pb collisions at √sNN = 2.76 TeV (Springer, 2015-07-10) The transverse momentum (p T) dependence of the nuclear modification factor R AA and the centrality dependence of the average transverse momentum 〈p T〉 for inclusive J/ψ have been measured with ALICE for Pb-Pb collisions ...
Consider two random variables $X$ and $Y$ and their joint cumulative probability distribution function $F$. I'm attempting to show that $P\{a_1<X<a_2\mathrm{,}\, b_1<Y<b_2\} = F(a_2, b_2)+F(a_1,b_1)-F(a_2,b_1)-F(a_1,b_2)$ by simplifying the left hand side but I am not quite getting there. My attempt $$P\{a_1<X<a_2\mathrm{,}\, b_1<Y<b_2\} = P(\{a_1<X<a_2\}\cap\{b_1<Y<b_2\}) = $$$$P(\{X>a_1\} \cap \{X<a_2\} \cap \{Y>b_1\} \cap \{Y<b_2\}) = $$$$1 - P((\{X>a_1\} \cap \{X<a_2\} \cap \{Y>b_1\} \cap \{Y<b_2\})^c) = $$$$1 - P(\{X>a_1\}^c \cup \{X<a_2\}^c \cup \{Y>b_1\}^c \cup \{Y<b_2\}^c) = $$$$1 - P(\{X\leq a_1\} \cup \{X\geq a_2\} \cup \{Y\leq b_1\} \cup \{Y\geq b_2\}) $$ So here I suppose I must apply the Inclusion–exclusion principle and becuase $\{X\leq a_1\}$, $\{X\geq a_2\}$ and $\{Y\leq b_1\}$, $\{Y\geq b_2\}$ are respectively mutually exclusive we have that the terms will only be probabilities defined on single sets and on the intersection of two non-mutually exclusive sets. Despite of this it still seems a bit too cumbersome and I am not sure how to simplify the probabilities of intersections. $$1 - P(\{X\leq a_1\}) - P(\{X\geq a_2\}) - P(\{X\leq b_1\}) - P(\{X\geq b_2\}) $$ $$+ \,P(\{X\leq a_1\}\cap \{Y\leq b_1\}) + P(\{X\leq a_1\}\cap \{Y\geq b_2\}) $$ $$+ \,P(\{X\geq a_2\}\cap \{Y\leq b_1\}) + P(\{X\geq a_2\}\cap \{Y\geq b_2\}) $$ It just seems messy and I don't know how to continue. Am going in the right direction at least or am I completely off?
First, if $a=b$ then$$b_1=a=a_1$$Now, let $a=1$ and $b=2$. Then$$a_1=1.5 \ \text{ and }b_1=\sqrt{2} \approx 1.414$$Hence $a \ne b$ and $b_1 \leq a_1$. There's a problem with what you wrote isn't it ? It might be $a_1 \leq b_1$ iff $a=b$. Indeed, suppose$$a_1 \leq b_1$$Then using the fact that $x \mapsto x^2$ is increasing$$a+b \leq 2\sqrt{ab} \Rightarrow a^2+2ab+b^2 \leq 4ab \Rightarrow \left(a-b\right)^2 \leq 0$$With the last inequality, we clearly see that the only possibility is $a=b$. ( which justifies directly the iff ) So in other term, we ALWAYS have $b_1 \leq a_1$. The sequences you mention are well-known and closely linked to elliptical integrals ( discovery of Abel and Bernoulli ) because it is also linked with the length of the lemniscate. You can show that for $a=1$ and $b=\sqrt{2}$ both $(a_n)$ and $(b_n)$ converge to the same limit which is the length of a lemniscate of Bernoulli of parameter $1$ ( if i remember well ). You can prove it by induction, but you can wisely use the property of "adjacent" sequences. https://fr.wikipedia.org/wiki/Th%C3%A9or%C3%A8me_des_suites_adjacentes ( sorry it's french, does not know the name in english ). You can show that $(a_n)$ increases, $(b_n)$ decreases and $a_n-b_n \underset{n \rightarrow +\infty}{\rightarrow}0$. In fact, they both converges to the same limit which is often wrote as $M\left(a,b\right)$ that stands for arithmetico-geometric mean. You cannot find explicitly the value of the common limit $M\left(a,b\right)$. However for fun, you have beautiful properties$$-M\left(a,b\right)=M\left(b,a\right)$$$$-M\left(\alpha a,\alpha b\right)=\alpha M\left(a,b\right)$$$$\text{min}\left(a,b\right) \leq \sqrt{ab} \leq M\left(a,b\right) \leq \frac{a+b}{2} \leq \text{max}\left(a,b\right)$$And last but not least$$\frac{2}{\pi}M\left(a,b\right)=\int_{0}^{\pi/2}\frac{\text{d}t}{\sqrt{x^2\cos^2(t)+y^2\cos^2(t)}}$$
I am reading a book on Skyrmions, and I am at the part where the interaction of skyrmions with electrons is discussed. The chapter speaks of Spin-Transfer Torque (STT) and makes the following statement about magnetic metals: In magnetic metals, the outer-shell electrons of magnetic atoms participate in the formation of the localized moment $\mathbf{n}$, as well as acting as mobile carriers of the electric current. The two types of electrons (i.e., localized and itinerant) originating from the same atom are bound by the Hund’s coupling $$ -J_H \mathbf{n} \cdot \left( \psi^\dagger \mathbf{\sigma} \psi \right) $$ In the above, $\psi$ is the spinor field, given by $$\psi = (c_{\uparrow} \text{ } c_{\downarrow})^T$$ and $\mathbf{\sigma}$ is the vector of Pauli matrices. $J_H$ is a coupling constant. My questions are: How can there be both an electron which creates a localized moment while another acts as a mobile carrier in the same shell of the same atom of the material? How can one see that indeed the term referenced is the term that enforces Hund's rules? (I'm assuming this is what is meant by Hund's coupling)
I have been studying about the SU(2) symmetry in Heisenberg Hamiltonian with a paper 'SU(2) gauge symmetry of the large U limit of the Hubbard model' written by Ian Affleck et al(Phys. Rev. B 38, 745 – Published 1 July 1988). In the paper, they represent the spin in terms of fermionic operators with constraint that number of particle at each lattice is 1. $$S_{x}=\frac{1}{2}c^{\alpha\dagger}_{x}\sigma_{\alpha}^{\beta}c_{x\beta}, \\constraint: c^{\dagger\alpha}c_{\alpha}=1$$ And they re-express the Hamiltonian using matrix $\Psi_{\alpha\beta}\equiv\left(\begin{array}{cc}c_{1}&c_{2}\\c_{2}^{\dagger}&-c_{1}^{\dagger}\end{array}\right)$(where number denotes the spin up and down) to show the SU(2) symmetry of the Heisenberg Hamiltonian explicitly. Also the constraint can be re-expressed as follows: $$\frac{1}{2}tr\Psi^{\dagger}\sigma^{z}\Psi=\frac{1}{2}(c^{\dagger}_{1}c_{1}+c_{2}^{\dagger}c_{2}-c_{1}c_{1}^{\dagger}-c_{2}c_{2}^{\dagger})=c^{\dagger}_{1}c_{1}+c_{2}^{\dagger}c_{2}-1=0$$At this time $c^{\dagger}, c$ are operators. And they write the lagrangian of this Hamiltonian. $$L=\frac{1}{2}\sum_{x}tr\Psi^{\dagger}_{x}(id/dt+A_{0x})\Psi_{x}-H$$where $A_{0}=\frac{1}{2}\mathbf{\sigma}\cdot\mathbf{A_{0}}$. Here components of $\mathbf{A_{0}}$ are Lagrangian multipliers. This lagrangian has time dependent gauge symmetry. Here I have a problem. As far as i know $c^{\dagger}, c$ in Lagrangian are not operators anymore but grassmann variables. Therefore constraint of $(\mathbf{A_{0}})_{z}$ in the Largrangian becomesThis post imported from StackExchange Physics at 2014-08-25 11:29 (UCT), posted by SE-user Iksu Jang $$\frac{1}{2}tr\Psi^{\dagger}\sigma^{z}\Psi=\frac{1}{2}(c^{\dagger}_{1}c_{1}+c_{2}^{\dagger}c_{2}-c_{1}c_{1}^{\dagger}-c_{2}c_{2}^{\dagger})=c^{\dagger}_{1}c_{1}+c_{2}^{\dagger}c_{2}=0$$ This is different from the original constraint condition!! I am really confused about this. Can anyone help me to solve this problem?
Last Updated: May 5, 2019 When we solve a partial differential equation with a source term, we must pay attention to how to treat it numerically for better convergence. The is available in OpenFOAM so that we can handle a linearized source term in numerically stable way. SemiImplicitSource fvOption Keywords Source term linearization in OpenFOAM SemiImplicitSource Source Term Linearization If the source term under consideration is a non-linear function of a conserved variable, the linearization of it is a fundamental approach to discretizing it in the finite volume method. This topic is precisely covered in the famous book “ ” by Suhas V. Patankar: Numerical Heat Transfer and Fluid Flow In Section 4.2.5, the concept of the linearization of the source term was introduced. One of the basic rules (Rule 3) required that when the source term is linearized as \begin{equation} S = S_C + S_P \phi_P, \tag{7.6} \label{eq:sourceTerm} \end{equation} the quantity $S_P$ must not be positive. Now, we return to the topic of source-term linearization to emphasize that often source terms are the cause of divergence of iterations and that proper linearization of the source term frequently holds the key to the attainment of a converged solution. Rule 3: Negative-slope linearization of the source termIf we consider the coefficient definitions in Eqs. (3.18), it appears that, even if the neighbor coefficients are positive, the center-point coefficient $a_P$ can become negative via the $S_P$ term. Of course, the danger can be completely avoided by requiring that $S_P$ will not be positive. Thus, we formulate Rule 3 as follows: When the source term is linearized as $\bar{S} = S_C + S_P T_P$, the coefficient $S_P$ must always be less than or equal to zero. Linearized Source Specification for a Scalar Field in OpenFOAM In OpenFOAM, the linearized source terms can be specified using SemiImplicitSource fvOption. Its settings are described in the constant/fvOptions file. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 /--------------------------------- C++ -----------------------------------\ \| ========= \| \| \| \\ / F ield \| OpenFOAM: The Open Source CFD Toolbox \| \| \\ / O peration \| Version: dev \| \| \\ / A nd \| Web: www.OpenFOAM.org \| \| \\/ M anipulation \| \| \---------------------------------------------------------------------------/ FoamFile { version 2.0; format ascii; class dictionary; location "constant"; object fvOptions; } // * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * // heatSource { active true; scalarSemiImplicitSourceCoeffs { selectionMode all; // all, cellSet, cellZone, points cellSet c1; volumeMode specific; // absolute; injectionRateSuSp { T (0.1 0); } } } // ************************************************************************* // : [ selectionMode Required] Domain where the source is applied ( / all / cellSet / cellZone ) points : [ injectionRateSuSp Required] Conserved variable name and two coefficients of linearized source term 1 2 3 4 injectionRateSuSp { variable_name (Sc Sp); } : [ volumemode Required] Choice of how to specify the two coefficients – : values are given as [variable] absolute – : values are given as [variable]/${\rm m^3}$ specific Example 1. Heat Conduction with Heat Source Let’s consider a simple 1D heat conduction problem in a solid rod with a thermal energy generation per unit volume and time $\dot{q}_v {\rm [W/m^3]}$. In this case, the steady heat conduction equation in 1D is given by \begin{align} \alpha \frac{d^2 T}{d x^2} + \frac{\dot{q}_v}{\rho c} = 0. \tag{1} \label{eq:conductionEqn} \end{align} Here, we assume that the thermal diffusivity $\alpha {\rm [m^2/s]}$ is constant. We can obtain the general solution for this equation \eqref{eq:conductionEqn} by integrating it twice. \begin{equation} T(x) = -\frac{S_C}{2\alpha}x^2 + C_1 x + C_2, \tag{2} \label{eq:generalSolution} \end{equation} where $C_1$ and $C_2$ are integration constants and we put $S_C {\rm [K/s]} := \dot{q}_v/\rho c$. If we assume that the temperatures at both ends of the rod are maintained at constant temperatures $T(0) = T_1, T(L) = T_2$, we can determine the values of two integration constants as \begin{align} C_2 &= T_1, \tag{3} \\ C_1 &= \frac{T_2 – T_1}{L} + \frac{S_C L}{2\alpha}. \tag{4} \end{align} Then, the temperature distribution in the rod is expressed by the following equation: \begin{equation} T(x) = -\frac{S_C}{2\alpha}x^2 + \left( \frac{T_2 – T_1}{L} + \frac{S_C L}{2\alpha} \right)x + T_1. \tag{5} \label{eq:solution} \end{equation} If we confine the heat source region to one-quarter of the whole region as shown in the following picture, the quadratic temperature distribution is limited to the region and the linear profile is obtained in the source free region. Example 2. Passive Scalar Transport
Why can we write an arbitrary object $v_{a \dot{b} }$ our transformations in this basis act on as $$ v_{a \dot{b} } = v_{\nu} \sigma^{ \nu}_{a \dot{b} } = v^0 \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} + v^1 \begin{pmatrix} 0&1 \\ 1&0 \end{pmatrix} +v^2 \begin{pmatrix} 0&-i \\ i&0 \end{pmatrix} + v^3 \begin{pmatrix} 1&0\\0&-1 \end{pmatrix} $$ Formulated differently: How do we know that the vector space for the $(\frac{1}{2},\frac{1}{2})= (\frac{1}{2},0) \otimes (0,\frac{1}{2})$ representation of the Lorentz group is the space of hermitian $2\times2$ matrices? The vector space for the $(\frac{1}{2},0)$ representation is $\mathbb{C}^2$ and I guess the same is true for the $(0,\frac{1}{2})$ representation, but I can't put it together to end up with hermitian matrices. EDIT: I found in the book Symmetry and the Standard Model: Mathematics and Particle Physics by Matthew Robinson the following explanation. Recall that just as any real matrix can be written as the sum of a symmetric matrix and an antisymmetric matrix, any complex matrix can be written as the sum of a Hermitian matrix and an anti-Hermitian matrix. However, the two indices on our matrix $v^{a \dot b}$ transform under representations of $SU(2)$. Notice that in the generators of these copies of $SU(2)$, both sets of generators $N^-$ and $N^+$ are Hermitian (cf. (3.229)). So, we’ll limit our discussion to the case where $v^{a \dot b}$ is a Hermitian $2 \times 2$ matrix. If anyone could help me understand this line of thought my problem would be solved. Why does this allow us to " limit our discussion to the case where $v^{a \dot b}$ is a Hermitian $2 \times 2$ matrix"? I understand that our representation here acts on complex $2\times2$ matrices. But I don't understand why we can restrict to hermitian matrices.
$\newcommand{\supp}{\operatorname{supp}}$ We are given with distributions $f,g \in D'(\Bbb R)$. If $\supp f\subset (-\infty,a)$ and $\supp(g)\subset(b,\infty)$ then prove that $fg$ is well defined distribution, where $a$ and $b$ are real numbers. What I know: $\langle fg\rangle=\langle f(x)g(x),\phi(x+y)\rangle$ [Above defined convolution will be well defined if $\supp(\phi(x+y))$ has a compact intersection with the support of $f(x)g(x)$. ] $=\int_\Bbb R\int_\Bbb Rf(x)g(x)\phi(x+y)\,dx\,dy=\int_\Bbb Rf(x)[\int_\Bbb Rg(x)\phi(x+y)\,dy]\,dx$ Here, $\int_\Bbb Rg(x)\phi(x+y)\,dy$ is well defined since $\phi$ has a compact support. $\implies \supp(\phi)\cap \supp(g)\neq\phi$ Similary, $\supp(\phi)\cap \supp(f)\neq\supp\phi$ $\implies \supp(\phi)\cap \supp(fg)\neq\supp\phi$, since $\supp(fg)\subset \overline{ \supp(g)+\supp(f)}$ Pleas help me to prove it.
The total wavefunction of an electron $\psi(\vec{r},s)$ can always be written as $$\psi(\vec{r},s)=\phi(\vec{r})\zeta_{s,m_s}$$ where $\phi(\vec{r})$ is the space part and $\zeta_{s,m_s}$ is the spin part of the total wavefunction $\psi(\vec{r},s)$. In my notation, $s=1/2, m_s=\pm 1/2$. Question 1 Is the above statement true? I am asking about any wavefunction here. Not only about energy eigenfunctions. Now imagine a system of two electrons. Even without any knowledge about the Hamiltonian of the system, the overall wavefunction $\psi(\vec{r}_1,\vec{r}_2;s_1,s_2)$ is antisymmetric. I think (I have this impression) under this general conditions, it is not possible to decompose $\psi(\vec{r}_1,\vec{r}_2;s_1,s_2)$ into a product of a space part and spin part. However, if the Hamiltonian is spin-independent, only then can we do such a decomposition into space part and spin part. Question 2 Can someone properly argue that how this is so? Please mention about any wavefunction of the system and about energy eigenfunctions.
Introduction Energy and Power Basic Operations Practice Problems Transformation of signals defined piecewise Periodic signals Even and Odd Signals Commonly encountered signals Table of Contents What is a signal? The word 'signal' has been used in different contexts in the English language and it has several different meanings . In this class, we will use the term signal to mean a function of an independent variable that carries some information or describes some physical phenomenon. Often (not always) the independent variable will be time, and the signals will describe phenomena that change with time. Such a signal can be denoted by ${x}(t)$, where $t$ is the independent variable and ${x}(t)$ denotes the function of $t$. Notice that this is slightly in contrast to the notation that you may have been used to from your calculus courses. There, you may have used $y=f(x)$ to denote a function of $x$, where $x$ is the dependent variable and $y$ is the independent variable. In this course, since signals will be referred to as ${x}(t)$, ${x}$ refers to the dependent variable typically. Here are two examples of such signals. More detailed examples can be found We will also encounter signals that describe some phenomena that change with frequency. Such signals will be denoted by $X(\omega)$ where $\omega$ is the independent variable and the dependent variable $X$ changes with frequency. Here is an example. This notation, even though fairly standard in the literature, is potentially confusing since $x(t)$ is used to refer to two related but different things. Consider the sentence "A recording of John's speech will be denoted by $x(t)$ and a recording of Adele's music will be denoted by $y(t)$". Here $x(t)$ and $y(t)$ refer to the entire signals, i.e., the audio waveforms. However, if you consider the sentence "find all values of $t$ for which $x(t) < 2$", here $x(t)$ refers to the value taken by the signal at time $t$. To elaborate further, it is the function $x$ evaluated at time $t$. In the Example 2 above $x(\pi)= \pi \cos\pi = -\pi$. Continuous-time (CT) and Discrete-time (DT) signals We will encounter two classes of signals in this course. The first class of signals are those for which the independent variable changes in a continuous manner or, equivalently, the signal $\underline{x}(t)$ is defined for every real value (or a continuum of values) of $t$ in the range $(a,b)$ ($a$ can be $-\infty$ and $b$ can be $\infty$). The two examples considered above are examples of CT signals. In contrast, we will also be interested in signals which are defined only for integer values of the independent variable. These signals are called discrete-time (DT) signals and will be denoted by $\underline{x}[n]$. Such signals arise in two situations - (i) the phenomenon that is being modeled is naturally one for which the independent variable takes only integer values or (ii) A CT signal is ‘sampled’ periodically by keeping their value only for discrete time instants typically spaced uniformly. For e.g., we can choose to keep only values of the signal $\underline{x}(t)$ at time instants $nT_s, \forall n$ for a fixed sampling interval $T_s$. From the sampled values we can construct a DT signal $\underline{x}[n]$ by assigning $x[n] = x(nT_s)$. The following two examples elaborate on these two methods. How to specify or describe signals There are two ways in which we will specify or describe signals in this course. The first way is to provide an explicit mathematical description of the signals such as $x(t) = \sin(200\pi t)$ or $x(t) = e^{-t}$. Sometimes, these signals may have to be described piecewise. Often, it will be easier to describe signals by sketching the function described by the signals or "drawing a picture of the signal". One of the skills that a student should develop from this part of the course is to be able to write a mathematical description for a signal defined pictorially and vice versa. The following examples illustrates these ideas. Practical Examples Signals are everywhere in modern life. Here are a few examples. MATLAB ExercisesExercise 1 - Create your own audio file that is at least 4 seconds long (the exact time duration is not really important, but do not make the file too long). Use the wavrecord command in MATLAB and a sampling frequency of 10000 Hz. You can also try to find a wav file online. Here is one that I like samplewavfile . Use the sound command in MATLAB to play the sound. Make sure the recording is fine. Exercise 2 - Using the wavread command, read the signal into a vector called x. Also read the sampling frequency in to a variable called Fs. Make sure you understand what this sampling frequency means. Plot the received signal as a function of time. Your time axis must have units in seconds. Exercise 3 - Using the stem command in MATLAB plot the signal. What is the difference between this plot and the plot in Example 2
Please assume that this graph is a highly magnified section of the derivative of some function, say $F(x)$. Let's denote the derivative by $f(x)$.Let's denote the width of a sample by $h$ where $$h\rightarrow0$$Now, for finding the area under the curve between the bounds $a ~\& ~b $ we can a... @Ultradark You can try doing a finite difference to get rid of the sum and then compare term by term. Otherwise I am terrible at anything to do with primes that I don't know the identities of $\pi (n)$ well @Silent No, take for example the prime 3. 2 is not a residue mod 3, so there is no $x\in\mathbb{Z}$ such that $x^2-2\equiv 0$ mod $3$. However, you have two cases to consider. The first where $\binom{2}{p}=-1$ and $\binom{3}{p}=-1$ (In which case what does $\binom{6}{p}$ equal?) and the case where one or the other of $\binom{2}{p}$ and $\binom{3}{p}$ equals 1. Also, probably something useful for congruence, if you didn't already know: If $a_1\equiv b_1\text{mod}(p)$ and $a_2\equiv b_2\text{mod}(p)$, then $a_1a_2\equiv b_1b_2\text{mod}(p)$ Is there any book or article that explains the motivations of the definitions of group, ring , field, ideal etc. of abstract algebra and/or gives a geometric or visual representation to Galois theory ? Jacques Charles François Sturm ForMemRS (29 September 1803 – 15 December 1855) was a French mathematician.== Life and work ==Sturm was born in Geneva (then part of France) in 1803. The family of his father, Jean-Henri Sturm, had emigrated from Strasbourg around 1760 - about 50 years before Charles-François's birth. His mother's name was Jeanne-Louise-Henriette Gremay. In 1818, he started to follow the lectures of the academy of Geneva. In 1819, the death of his father forced Sturm to give lessons to children of the rich in order to support his own family. In 1823, he became tutor to the son... I spent my career working with tensors. You have to be careful about defining multilinearity, domain, range, etc. Typically, tensors of type $(k,\ell)$ involve a fixed vector space, not so many letters varying. UGA definitely grants a number of masters to people wanting only that (and sometimes admitted only for that). You people at fancy places think that every university is like Chicago, MIT, and Princeton. hi there, I need to linearize nonlinear system about a fixed point. I've computed the jacobain matrix but one of the elements of this matrix is undefined at the fixed point. What is a better approach to solve this issue? The element is (24x_2 + 5cos(x_1)x_2)/abs(x_2). The fixed point is x_1=0, x_2=0 Consider the following integral: $\int 1/4(1/(1+(u/2)^2)))dx$ Why does it matter if we put the constant 1/4 behind the integral versus keeping it inside? The solution is $1/2\arctan{(u/2)}$. Or am I overseeing something? it should be du instead of dx in the integral *and the solution is missing a constant C of course Is there a standard way to divide radicals by polynomials? Stuff like $\frac{\sqrt a}{1 + b^2}$? My expression happens to be in a form I can normalize to that, just the radicand happens to be a lot more complicated. In my case, I'm trying to figure out how to best simplify $\frac{x}{\sqrt{1 + x^2}}$, and so far, I've gotten to $\frac{x \sqrt{1+x^2}}{1+x^2}$, and it's pretty obvious you can move the $x$ inside the radical. My hope is that I can somehow remove the polynomial from the bottom entirely, so I can then multiply the whole thing by a square root of another algebraic fraction. Complicated, I know, but this is me trying to see if I can skip calculating Euclidean distance twice going from atan2 to something in terms of asin for a thing I'm working on. "... and it's pretty obvious you can move the $x$ inside the radical" To clarify this in advance, I didn't mean literally move it verbatim, but via $x \sqrt{y} = \text{sgn}(x) \sqrt{x^2 y}$. (Hopefully, this was obvious, but I don't want to confuse people on what I meant.) Ignore my question. I'm coming of the realization it's just not working how I would've hoped, so I'll just go with what I had before.
I think while doing the calculations, I have found the answer. @GK gave an answer but I think there is a subtle mistake in that approach because both q and $\tau$ are coupled transformations of x and t. Thus the total derivative is not equal to the partial one. So this is my solution. $q$ and $\tau$ both can be expressed a function of $(x,t)$. From the definition of a total derivative $$\frac{dq}{dt} = \Big(\frac{\partial q}{\partial x}\Big)\frac{dx}{dt} + \frac{\partial q}{\partial t}$$ \begin{equation}\Rightarrow \frac{dq}{dt} = \cosh(\psi) \frac{dx}{dt} + \sinh(\psi) \end{equation} Similarly we have $$\frac{d\tau}{dt} = \Big(\frac{\partial \tau}{\partial x}\Big)\frac{dx}{dt} + \frac{\partial \tau}{\partial t}$$ \begin{equation}\Rightarrow \frac{d\tau}{dt} = \sinh(\psi) \frac{dx}{dt} + \cosh(\psi) \end{equation} From (1) and (2) we have $$ \frac{dq}{d\tau} = \dfrac{\cosh(\psi) \frac{dx}{dt} + \sinh(\psi)}{\sinh(\psi) \frac{dx}{dt} + \cosh(\psi)}$$ $$\Rightarrow 1- \Big(\frac{dq}{d\tau}\Big)^{2}= 1- \Bigg(\dfrac{\cosh(\psi) \frac{dx}{dt} + \sinh(\psi)}{\sinh(\psi) \frac{dx}{dt} + \cosh(\psi)}\Bigg)^{2}$$ Going through the algebra and simplifying using (2) we get $$\Rightarrow 1- \Big(\frac{dq}{d\tau}\Big)^{2}= \dfrac{(\sinh^{2}(\psi) - \cosh^{2}(\psi))(\frac{dx}{dt})^{2}+(\cosh^{2}(\psi) - \sinh^{2}(\psi))}{(\frac{d\tau}{dt})^{2}} $$ Using the well know trig identity $\cosh^{2}(\psi) - \sinh^{2}(\psi) = 1$ we have, $$\Rightarrow 1- \Big(\frac{dq}{d\tau}\Big)^{2}= \dfrac{1-(\frac{dx}{dt})^{2}}{(\frac{d\tau}{dt})^{2}} $$ We know the Lagrangian $L^{\prime} = - \sqrt{1- \Big(\frac{dq}{d\tau}\Big)^{2}} $ Thus $$L^{\prime} = (\frac{dt}{d\tau})L $$ $$ \Rightarrow L^{\prime}(\dot{q}) d\tau = L(\dot{x}) dt $$ $$ \Rightarrow S^{\prime} [L^{\prime}] = S[L]$$ Even though the Lagrangian doesn't stay the same, the action does !
Cartesian Form Polar or Exponential Form Euler's Identities Conjugate Operations on two complex numbers nth power and nth roots of a complex number Functions of a complex variable Complex functions of a real variable Magnitude and Phase Plot Examples and References History Why use complex numbers? We will use the letter $j$ to refer to the imaginary number $\sqrt{-1}$. Even though $j$ is not a real number, we can perform all arithmetic operations such as addition, subtraction, multiplication, division with $j$ using the algebra of real numbers. A complex number $z$ is any number of the form $z = x+jy$, where $x$ is called the real part of $z$ and $y$ is called the imaginary part of $z$. Note: The imaginary part is not $jy$, rather it is only $y$. It is important to stick to this terminology, otherwise computations can go wrong. Often, it is useful to think of a complex number $z = x + j y$ as a vector in a two-dimensional plane as shown in the figure below, where $x$ is the $X$-coordinate and $y$ is $Y$-coordinate of the vector. Due to this relationship between a complex number and the corresponding vector, we will abuse the terminology and use the terms complex number and vector interchangeably, if the context should resolve any possible ambiguity. For example, a complex number is said to lie in the first quadrant (or, second quadrant etc) if the corresponding vector lies in the first quadrant (or, second quadrant etc). When a complex number is thought of as a vector in two dimensions, the $X$ coordinate $x$ and the $Y$ coordinate $y$ can be expressed in terms of the length of the vector $r$ and the angle made by this vector with the positive $X$-axis, namely $\theta$. Since $x = r \cos \theta$ and $y = r \sin \theta$, $z$ can be expressed as(1) where $\theta$ can be in degrees or radians (usually radians) and recall that $2 \pi \mbox{ rad } = 360\,^\circ$. $r$ is called the magnitude of $z$, denoted by $\|z\|$ and $\theta$ is called the phase of the complex number $z$, denoted by $\mbox{arg}{z}$ or $\angle z$. Using Euler's identities $z$ can be written as(2) This is known as the polar form or exponential form and it is very important to be able to convert a complex number from cartesian form to exponential form and vice versa. It is easy to see that $x,y,r$ and $\theta$ are related according to(3) Example: It is very useful to know the polar form for often used complex numbers such as $1,j,-j,-1$. They are given by $1 = e^{j0}, -1 = e^{j \pi}, j = e^{j \frac{\pi}{2}}, -j = e^{-j \frac{\pi}{2}}$. Caution: $r$ must be positive in the above expression. For example, if $z = -2 e^{j \frac{\pi}{4}}$, we must rewrite $z$ as $z = 2 e^{j \frac{5 \pi}{4}}$ and interpret $r$ as $2$ instead of $-2$. Caution: The above expression for $\theta$ in (3) does not identify $\theta$ uniquely, since $\tan(\theta) = \frac{y}{x}$ also implies that $\tan(\theta \pm \pi) = \frac{y}{x}$. It is best think of the vector $(x,y)$ and determine which quadrant this vector lies in based on the signs of $x,y$ and then make sure $\theta$ corresponds to an angle in the correct quadrant. Example: Suppose $z_1 = \frac{\sqrt{3}}{2}+j\frac{1}{2}$ and $z_2 = -\frac{\sqrt{3}}{2}-j\frac{1}{2}$. It is easy to see that $\tan^{-1}\left(\frac{y_1}{x_1} \right)=\tan^{-1}\left(\frac{y_2}{x_2} \right)$. However, $z_1$ is complex number in the first quadrant, whereas $z_2$ is a complex number is the 3rd quadrant. Therefore, $\theta_1$ should be $\pi/6$ and $\theta_2$ should be $7 \pi/6$. One important aspect of the polar form for a complex number is that adding $2 \pi$ to the angle does not change the complex number. Particularly,(4) This fact will be repeatedly used in the course. An immediate example of where this is useful is given in section of nth root of a complex number. Example: Express $e^{j 2 \pi}, e^{-j \pi}, e^{j \frac{3 \pi}{2}}, e^{j \frac{9 \pi}{2}}$ in Cartesian form. In the expansion of $e^{j\theta}$ if one replaces $\theta$ by $j \theta$ and $-j \theta$, we get the following two equations, respectively.(5) From the above equations and along with the expansions of $cos \theta$ and $sin \theta$ the following relationship can be seen to be true:(6) The conjugate of a complex number $z = x + j y$ is given by $z^* = x - j y$. When $z$ is written in polar form as $z = r e^{j \theta}$, the complex conjugate is given by $z^* = r e^{-j \theta}$. In general, to compute the conjugate of a complex number, replace $j$ by $-j$ everywhere. Let $z = x + j y = r e^{j \theta}$, $z_1 = x_1 + j y_1 = r_1 e^{j \theta_1}$ and $z_2 = x_2 + j y_2 = r_2 e^{j \theta_2}$(7) Based on the above operations, the following facts about complex number can be verified.(8) Let $z_0 = x_0 + j y_0 = r_0 e^{j \theta_0}$. For any integer $n$, the $n$th power of $z$, $z^n$ is simply obtained by using multiplication operation $n$ times. In the polar form, $z_0^n = r_0^n e^{j n \theta_0}$. Just like how the two real numbers $1$ and $-1$ have the same square, different complex numbers can have the same $n$th power. Consider the set of distinct complex numbers $z_k = e^{j \theta_0 + \frac{2 \pi k}{n}}$. All the $z_k$s are different have the same $n$th power for $k=0,1,2,\ldots,n-1$. We can see this by raising $z_k$ to the $n$th power to get(9) The $n$th root of $z$ is a bit more interesting and tricky. Any complex number $z$ which is the solution to the $n$th degree equation $z^n - z_0 = 0$ is an $n$th root of $z_0$. The fundamental theorem of algebra states that an $n$th degree equation has exactly $n$ (possibly complex) roots. Hence, every complex number $z_0$ has exactly $n$, $n$th roots. These roots can be found by using the fact $e^{j \theta} = e^{j (\theta + 2 \pi k)}$.(10) Clearly, computing $n$th roots is much easier in the polar form than in the cartesian form. Example: Find the third roots of unity $\sqrt[3]{1}$. Since $1 = 1 e^{j0}$, this corresponds to $r_0 = 1, \theta_0 = 0$. Hence, the three roots of unity are given by $r = 1, \ \ \theta = 0,\frac{2\pi}{3},\frac{4\pi}{3}$ In cartesian coordinates, they are $(1+j0)$, $\left(-\frac{1}{2}+j\frac{\sqrt{3}}{2}\right)$,$\left(-\frac{1}{2}-j\frac{\sqrt{3}}{2}\right)$. These are referred to as $1,\omega,\omega^2$ sometimes. The three roots are shown in the figure. Let $f(z)$ be a complex function of a complex variable $z$, i.e., for every $z$, $f(z)$ is a complex number. Note that a real number is also considered as a complex number and, hence, $f(z)$ could have a zero imaginary part. Examples of functions include $f(z) = \|z\|, f(z) = \mbox{arg}(z), f(z) = z^n, f(z) = \exp(z)$, etc. The exponential functions can be interpreted using Euler's identity as follows. $f(z) = \exp(z) = e^xe^{jy} = e^x\cos y + je^x\sin y$ The real part of $f(z)$ is plotted as a function of the real part of $z$, namely $x$ for the case $x<0$ in Figure. The logarithm of a complex number $\ln z$ can be also interpreted using Euler's identity as $\ln(z) = \ln\left(r e^{j (\theta + 2 k \pi) } \right) = \ln r + j (\theta + 2 k \pi)$ It can be seen from the above expression that $\ln z$is not a function of $z$. However, if we set $k=0$ in the above expression, then we get what is called the principal value of $\ln z$, denoted by Ln $z$, which is a function. Similarly, it is important to realize that for any integer $n$, the $n$th power of a complex number is a function of the complex number, i.e., for every complex number $z$, there is only one complex number $z^n$. However, for an integer $n$, the $n$th root of a complex number is not uniquely defined and hence, is not a function. Often, one may take the root corresponding to $k=0$ in the previous section as the default root and hence the principal value. Then, the principal value becomes a function. This is similar to square roots of positive real numbers being defined as the positive numbers. There are interesting examples where careless use of just the principal value as the $n$th root can lead to fallacious arguments. You may be used to dealing with functions of a variable such as $y = f(x)$, where $x$ is called the independent variable and $y$ is called the dependent variable and typically, $y$ takes real values when $x$ takes real values. In this course, we will be interested in complex functions of a real variable. Often the real variable will represent time or frequency. Such a function, normally denoted as $x(t)$ or $X(\omega)$ is a function which takes a complex value for every real value of the independent variable $t$ or $\omega$. Pay attention to the notation carefully - $t$ or $\omega$ now becomes the independent variable and $x(t)$ or $X(\omega)$ now becomes the dependent variable. We can also think of the complex function as the combination of two real functions of the independent variable, one for the real part of $x(t)$ and one for the imaginary part of $x(t)$. When dealing with real functions of a real variable, you may be used to plotting the function $x(t)$ as a function of $t$. However, when $x(t)$ is a complex function, there is a problem in plotting this function since for every value of $t$, we need to plot a complex number. In this case, we do one of two things - either we plot the real part of $x(t)$ versus $t$ and plot the imaginary part of $x(t)$ versus $t$, or we plot $\|x(t)\|$ versus $t$ and $\mbox{arg}(x(t))$ versus $t$. Either of these is fine, but we do need two plots to effectively understand how $x(t)$ changes with $t$. Example: Consider the function $x(t) = e^{j2\pi t} = \cos {2\pi t} + j \sin {2\pi t}$ for all real values of $t$. This is clearly a complex function of a real variable $t$. $\Re\{x(t)\}, \Im\{x(t)\}, \|x(t)\|, \mbox{arg}(x(t))$ are all real functions of the real variable $t$. Hence, we can plot $\Re\{x(t)\}$ versus $t$ and $\Im\{x(t)\}$ versus $t$ or we can plot $\|x(t)$ versus $t$ and $\mbox{arg}(x(t))$ versus $t$ as shown in figure. Example: Consider the function $H(\omega) = \frac{1}{1+j\omega}$, where $\omega$ is a real variable. Roughly sketch the magnitude and phase of $H(\omega)$ as a function of $\omega$.(11) A plot of $\|H(\omega)\|$ versus $\omega$ and $\angle(H(\omega))$ versus $\omega$ is shown in Figure. Example: Consider the function $X(\omega) = \frac{j \omega}{1+j\omega}$, where $\omega$ is a real variable. Roughly sketch the magnitude and phase of $X(\omega)$ as a function of $\omega$.(12) A plot of $\|X(\omega)\|$ versus $\omega$ and $\angle(X(\omega))$ versus $\omega$ is shown in figure. To plot the magnitude and phase of $H(j \omega) = e^{j a_1 \omega} + e^{j a_2 \omega}$ vs $\omega$, one of the tricks is to express $e^{j a_1 \omega} + e^{j a_2 \omega}$ as follows(13) Now, it is easy to see that $\|H(j \omega)\| = 2 \|e^{j \left(\frac{a_1+a_2}{2} \omega \right)}\| \cdot \|\cos \left[ \left( \frac{a_1 - a_2}{2} \right) \omega \right]\|$ which is simply $2 \|\cos \left[ \left( \frac{a_1 - a_2}{2} \right) \omega \right]\|$. 1. Let $z_1 = 2 e^{j\pi/4}$ and $z_2 = 8 e^{j\pi/3}$. Find a) $2z_1-z_2$ b) $\frac{1}{z_1}$ c) $\frac{z_1}{z_2^2}$ d) $\sqrt[3]{z_2}$ 2. What is $j^j$? 3. Let $z$ be any complex number. Is it true that $(e^z)^\star= e^{z^\star}$? 4. Plot the magnitude and phase of the function $X(f) = e^{j\pi f}+e^{j 3 \pi f}$, for $-1 \leq f \leq 1$. 5. Prove that $\int e^{ax} \ \cos(bx) \ dx = \frac{e^{ax}}{a^2+b^2} \left(a \cos(bx) + b \sin(bx) \right)$ References: A good online reference for complex numbers is the wiki page http://en.wikipedia.org/wiki/Complex_number. A short history of complex numbers can be found here.
The unit of illumination is the lux, lumens per square meter. What is the minimum lux required for reading? How many lux does the Sun provide at distance D? What is the minimum lux required for reading? You can plug all sorts of numbers into this depending on how good your eyes are, how big the print is, and how close you hold it to your face. I'm going to use civil twilight which is 3.4 lux. Others may like 1 foot-candle, 1 lumen at one foot, or about 10.764 lux. Whatever value you prefer, you can plug it into the formula below. How many lux does the Sun provide at distance D? To calculate the lux the Sun puts out we first have to calculate how much solar radiation (raw energy) it puts out at distance D? You take the luminosity of the Sun and distribute it over the surface of a sphere of radius D. $$ \text{solar radiation in} \, \frac{Watts}{m^2} = \frac {3.846 \times 10^{26} W}{4 \pi D^2} $$ Not all of that is visible light. We need to convert that into lux, lumens per square meter. To do that you need to integrate the power output over the curve of visible light output using the Luminosity Function which, fortunately, someone has already done coming out with a luminous efficacy of 93 lumens per W. $$ \text{illumination in lux at} \, D = \frac{93 \frac{lumens}{W} \times 3.846 \times 10^{26} W}{4 \pi D^2} $$ At the Earth where $ D = 1.5 \times 10^{11} \, m $ we get $ 1.27 \times 10^{5} \, lux $. This is higher than direct sunlight at the Earth's surface because it does not account for atmosphere. This is good because New Horizons is in a vacuum. How many lux are available to New Horizons? Pluto is currently 32.6 AU from the Sun, or $ D = 4.89 \times 10^{12} \, m $. Plugging that in we get $ 1.19 \times 10^2 $ or $ 119 \, lux $. Plenty! Or is it? That's how much light New Horizons is receiving from the Sun, but how much is bouncing off Pluto? Pluto has an albedo of about .6 so a little more than half the Sun's light is reflected back to New Horizons, or about 70 lux which is about the same as your average office hallway. Not great, but plenty for a long exposure. This paper on LORRI agrees, " At Pluto encounter, 33 AU from the Sun, the illumination level is ~1/1000 that at Earth". How far out from the Sun is visible light still sufficient to read a book? We need to solve for D. $$ I = \frac{93 \frac{lumens}{W} \times 3.846 \times 10^{26} W}{4 \pi D^2} $$ $$ I D^2 = \frac{93 \frac{lumens}{W} \times 3.846 \times 10^{26} W}{4 \pi} $$ $$ D^2 = \frac{93 \frac{lumens}{W} \times 3.846 \times 10^{26} W}{4 \pi I} $$ $$ D = \sqrt{ \frac{93 \frac{lumens}{W} \times 3.846 \times 10^{26} W}{4 \pi I}} $$ Plug in $I = 3.4 \, lux$ and get $D = 2.8 \times 10^{13}$ or $186 AU$ which puts us past the Kuiper Belt and well into the Scattered Disc. When using a different number for reading light note that lux changes with the square root of the distance. If you double the lux required you decrease the distance by 1.4. If we triple I to 10 lux (aka the foot-candle), the distance drops by 1.7 to about 108 AU. Still very far out. Could we expect colors at this distance? At Pluto you will see colors fine. At 186 AU you will see colors about as well as you can see colors at civil twilight. New Horizons has two instruments measuring visible light. LORRI is a long range panchromatic camera, meaning it acts like a regular digital camera and captures an approximation of what the human eye sees. The other is the Ralph telescope. It is a multispectral visible and infrared imager meaning it takes multiple images at several different wavelengths. These will appear grey, the grey is a measurement of the light intensity at a specific wavelength. This is usually how spacecraft "see" color because scientists aren't interested in pretty pictures with multiple wavelengths smeared together, they want data about specific wavelengths. NASA public relations people mix the images together to approximate what matches what the human eyeball would see for press releases. They can't always get it quite right. Phil Plat discusses it in detail with the Mars landers. To make these single-filter images into a color composite is not easy. If the red filter lets in less total light than the blue, you need to compensate for that when you add the images together. If the red filter is wider (lets in a wider range of reds) than the blue filter, you have to compensate for that, and so on. Short version: pictures from LORRI are closer to what you would see than from Ralph.
This question arises from the one asked here about a bound on moment generating functions (MGFs). Suppose $X$ is a bounded zero-mean random variable taking on values in $[-\sigma, \sigma]$ and let $G(t) = E[e^{tX}]$ be its MGF. From a bound used in a proof of Hoeffding's Inequality, we have that $$G(t) = E[e^{tX}] \leq e^{\sigma^2t^2/2}$$ where the right side is recognizable as the MGF of a zero-mean normal random variable with standard deviation $\sigma$. Now, the standard deviation of $X$ can be no larger than $\sigma$, with the maximum value occurring when $X$ is a discrete random variable such that $P\{X = \sigma\} = P\{X = -\sigma\} = \frac{1}{2}$. So, the bound referred to can be thought of as saying that the MGF of a zero-mean bounded random variable $X$ is bounded above by the MGF of a zero-mean normal random variable whose standard deviation equals the maximum possible standard deviation that $X$ can have. My question is: is this a well-known result of independent interest that is used in places other than in the proof of Hoeffding's Inequality, and if so, is it also known to extend to random variables with nonzero means? The result that prompts this question allows asymmetric range $[a,b]$ for $X$ with $a < 0 < b$ but does insist on $E[X] = 0$. The bound is $$G(t) \leq e^{t^2(b-a)^2/8} = e^{t^2\sigma_{max}^2/2}$$ where $\sigma_{\max} = (b-a)/2$ is the maximum standard deviation possible for a random variable with values restricted to $[a,b]$, but this maximum is not attained by zero-mean random variables unless $b = -a$.
ISSN: 1935-9179 eISSN: 1935-9179 All Issues Electronic Research Announcements January 2011 , Volume 18 Select all articles Export/Reference: Abstract: We present a novel class of functions that can describe the stable and unstable manifolds of the Hénon map. We propose an algorithm to construct these functions by using the Borel-Laplace transform. Neither linearization nor perturbation is applied in the construction, and the obtained functions are exact solutions of the Hénon map. We also show that it is possible to depict the chaotic attractor of the map by using one of these functions without explicitly using the properties of the attractor. Abstract: We describe the structure of the automorphism groups of algebras Morita equivalent to the first Weyl algebra $ A_1(k) $. In particular, we give a geometric presentation for these groups in terms of amalgamated products, using the Bass-Serre theory of groups acting on graphs. A key rôle in our approach is played by a transitive action of the automorphism group of the free algebra $ k< x, y>$ on the Calogero-Moser varieties $ \CC_n $ defined in [5]. In the end, we propose a natural extension of the Dixmier Conjecture for $ A_1(k) $ to the class of Morita equivalent algebras. Abstract: Based on the classic multiplicative ergodic theorem and the semi-uniform subadditive ergodic theorem, we show that there always exists at least one ergodic Borel probability measure such that the joint spectral radius of a finite set of square matrices of the same size can be realized almost everywhere with respect to this Borel probability measure. The existence of at least one ergodic Borel probability measure, in the context of the joint spectral radius problem, is obtained in a general setting. Abstract: An element of a free Leibniz algebra is called Jordan if it belongs to a free Leibniz-Jordan subalgebra. Elements of the Jordan commutant of a free Leibniz algebra are called weak Jordan. We prove that an element of a free Leibniz algebra over a field of characteristic 0 is weak Jordan if and only if it is left-central. We show that free Leibniz algebra is an extension of a free Lie algebra by left-center. We find the dimensions of the homogeneous components of the Jordan commutant and the base of its multilinear part. We find criterion for an element of free Leibniz algebra to be Jordan. Abstract: In this note we announce the computation of the triangular spectrum (as defined by P. Balmer) of two classes of tensor triangulated categories which are quite common in algebraic geometry. One of them is the derived category of $G$-equivariant sheaves on a smooth quasi projective scheme $X$ for a finite group $G$ which acts on $X$. The other class is the derived category of split super-schemes. Abstract: Let $T:C^1(RR)\to C(RR)$ be an operator satisfying the derivation equation $T(f\cdot g)=(Tf)\cdot g + f \cdot (Tg),$ where $f,g\in C^1(RR)$, and some weak additional assumption. Then $T$ must be of the form $(Tf)(x) = c(x) \, f'(x) + d(x) \, f(x) \, \ln \|f(x)\|$ for $f \in C^1(RR), x \in RR$, where $c, d \in C(RR)$ are suitable continuous functions, with the convention $0 \ln 0 = 0$. If the domain of $T$ is assumed to be $C(RR)$, then $c=0$ and $T$ is essentially given by the entropy function $f \ln \|f\|$. We can also determine the solutions of the generalized derivation equation $T(f\cdot g)=(Tf)\cdot (A_1g) + (A_2f) \cdot (Tg), $ where $f,g\in C^1(RR)$, for operators $T:C^1(RR)\to C(RR)$ and $A_1, A_2:C(RR)\to C(RR)$ fulfilling some weak additional properties. Abstract: Zapolsky's inequality gives a lower bound for the $L_1$ norm of the Poisson bracket of a pair of $C^1$ functions on the two-dimensional sphere by means of quasi-states. Here we show that this lower bound is sharp. Abstract: This is an announcement of the proof of the inverse conjecture for the Gowers $U^{s+1}[N]$-normfor all $s \geq 3$; this is new for $s \geq 4$, the cases $s = 1,2,3$ having been previously established. More precisely we outline a proof that if $f : [N] \rightarrow [-1,1]$ is a function with \|\|$f$\|\| $U^{s+1}[N] \geq \delta$ then there is a bounded-complexity $s$-step nilsequence $F(g(n)\Gamma)$ which correlates with $f$, where the bounds on the complexity and correlation depend only on $s$ and $\delta$. From previous results, this conjecture implies the Hardy-Littlewood prime tuples conjecture for any linear system of finite complexity. In particular, one obtains an asymptotic formula for the number of $k$-term arithmetic progressions $p_1 < p_2 < ... < p_k \leq N$ of primes, for every $k \geq 3$. Abstract: Let $X \rightarrow\ S$ be a smooth projective surjective morphism, where $X$ and $S$ are integral schemes over $\mathbb C$. Let $L_0\, L_1\, \cdots \, L_{n-1}\, L_{n}$ be line bundles over $X$. There is a natural isomorphism of the Deligne pairing 〈$L_0\, \cdots\, L_{n}$〉with the determinant line bundle $Det(\otimes_{i=0}^{n} (L_i- \mathcal O_{X}))$. Abstract: The purpose of this note is to announce complete answers to the following questions. (1) For an essential simple loop on a 2-bridge sphere in a 2-bridge link complement, when is it null-homotopic in the link complement? (2) For two distinct essential simple loops on a 2-bridge sphere in a 2-bridge link complement, when are they homotopic in the link complement? We also announce applications of these results to character varieties and McShane's identity. Abstract: We characterize order preserving transforms on the class of lower-semi-continuous convex functions that are defined on a convex subset of $\mathbb{R}^n$ (a "window") and some of its variants. To this end, we investigate convexity preserving maps on subsets of $\mathbb{R}^n$. We prove that, in general, an order isomorphism is induced by a special convexity preserving point map on the epi-graph of the function. In the case of non-negative convex functions on $K$, where $0\in K$ and $f(0) = 0$, one may naturally partition the set of order isomorphisms into two classes; we explain the main ideas behind these results. Abstract: The degree zero part of the quantum cohomology algebra of a smooth Fano toric symplectic manifold is determined by the superpotential function, $W$, of its moment polytope. In particular, this algebra is semisimple, i.e. splits as a product of fields, if and only if all the critical points of $W$ are non-degenerate. In this paper, we prove that this non-degeneracy holds for all smooth Fano toric varieties with facet-symmetric duals to moment polytopes. Readers Authors Editors Referees Librarians Email Alert Add your name and e-mail address to receive news of forthcoming issues of this journal: [Back to Top]
The number of storks and the number of human babies delivered are positively correlated (Matthews, 2000). This is a classic example of a spurious correlation which has a causal explanation: a third variable, say economic development, is likely to cause both an increase in storks and an increase in the number of human babies, hence the correlation. 1 In this blog post, I discuss a more subtle case of spurious correlation, one that is not of causal but of statistical nature: completely independent processes can be correlated substantially. AR(1) processes and random walks Moods, stockmarkets, the weather: everything changes, everything is in flux. The simplest model to describe change is an auto-regressive (AR) process of order one. Let $Y_t$ be a random variable where $t = [1, \ldots T]$ indexes discrete time. We write an AR(1) process as: where $\phi$ gives the correlation with the previous observation, and where $\epsilon_t \sim \mathcal{N}(0, \sigma^2)$. For $\phi = 1$ the process is called a random walk. We can simulate from these using the following code: The following R code simulates data from three independent random walks and an AR(1) process with $\phi = 0.5$; the Figure below visualizes them. As we can see from the plot, the AR(1) process seems pretty well-behaved. This is in contrast to the three random walks: all of them have an initial upwards trend, after which the red line keeps on growing, while the blue line makes a downward jump. In contrast to AR(1) processes, random walks are not stationary since their variance is not constant across time. For some very good lecture notes on time-series analysis, see here. Spurious correlations of random walks If we look at the correlations of these three random walks across time points, we find that they are substantial: I hope that this is at least a little bit of a shock. Upon reflection, however, it is clear that we are blundering: computing the correlation across time ignores the dependency between data points that is so typical of time-series data. To get more data about what is going on, we conduct a small simulation study. In particular, we want to get an intuition of how this spurious correlation behaves with increasing sample sizes. We therefore simulate two independent random walks for sample sizes $n \in [50, 100, 200, 500, 1000, 2000]$ and compute their Pearson correlation, the test-statistic, and whether $p < \alpha$, where we set $\alpha$ to some an arbitrary value, say $\alpha = 0.05$. We repeated this 100 times and report the average of these quantities. We observe that the average absolute correlation is very similar across $n$, but the test statistic grows with increased $n$, which naturally results in many more false rejections of the null hypothesis of no correlation between the two random walks. To my knowledge, Granger and Newbold (1974) were the first to point out this puzzling fact. 2 They regress one random walk onto the other instead of computing the Pearson correlation. (Note that the test statistic is the same). In a regression setting, we write: where we assume that $\epsilon_i \sim \mathcal{N}(0, \sigma^2)$ (see also a previous blog post). This is evidently violated when performing linear regression on two random walks, as demonstrated by the residual plot below. Similar as above, we can have an AR(1) process on the residuals: and test whether $\delta = 0$. We can do so using the Durbin-Watson test, which yields: This indicates substantial autocorrelation, violating our modeling assumption of independent residuals. In the next section, we look at the deeper mathematical reasons for why we get such spurious correlation. In the Post Scriptum, we relax the constraint that $\phi = 1$ and look at how spurious correlation behaves for AR(1) processes. Inconsistent estimation The simulation results from the random walk simulations showed that the average (absolute) correlation stays roughly constant, while the test statistic increases with $n$. This indicates a problem with our estimator for the correlation. Because it is slightly easier to study, we focus on the regression parameter $\beta_1$ instead of the Pearson correlation. Recall that our regression estimate is where $\bar{x}$ and $\bar{y}$ are the empirical means of the realizations $x_t$ and $y_t$ of the AR(1) processes $X_t$ and $Y_t$, respectively. The test statistic associated with the null hypothesis $\beta_1 = 0$ is where $\hat{\sigma}$ is the estimated standard deviation of the error. In simple linear regression, the test statistic follows a t-distribution with $n – 2$ degrees of freedom (it takes two parameters to fit a straight line). In the case of independent random walks, however, the test statistic does not have a limiting distribution; in fact, as $n \rightarrow \infty$, the distribution of $t_{\text{statistic}}$ diverges (Phillips, 1986). To get an intuition for this, we plot the bootstrapped sampling distributions for $\beta_1$ and $t_{\text{statistic}}$, both for the case of regressing one independent AR(1) process onto another, and for random walk regression. The Figure below illustrates how things go wrong when regressing one independent random walk onto the other. In contrast to the estimate for the AR(1) regression, the estimate $\hat{\beta}_1$ does not decrease in the case of a random walk regression. Instead, it stays roughly within $[-0.75, 0.75]$ across all $n$. This shines further light on the initial simulation results that the average correlation stays roughly the same. Moreover, in contrast AR(1) regression for which the distribution of the test statistic does not change, the distribution of the test statistic for the random walk regression seems to diverge. This explains why we the proportion of false positives increases with $n$. Rigorous arguments of the above statements can be found in Phillips (1986) and Hamilton (1994, pp. 577). 3 The explanations feature some nice asympotic arguments which I would love go into in detail; however, I’m currently in Santa Fe for a summer school that has a very tightly packed programme. On that note: it is very, very cool. You should definitely apply next year! In addition to the stimulating lectures, wonderful people, and exciting projects, the surroundings are stunning 4. Conclusion “Correlation does not imply causation” is a common response to apparently spurious correlation. The idea is that we observe spurious associations because we do not have the full causal picture, as in the example of storks and human babies. In this blog post, we have seen that spurious correlation can be due to solely statistical reasons. In particular, we have seen that two independent random walks can be highly correlated. This can be diagnosed by looking at the residuals, which will not be independent and identically distributed, but will show a pronounced autocorrelation. The mathematical explanation for the spurious correlation is not trivial. Using simulations, we found that the estimate of $\beta_1$ does not converge to the true value in the case of regressing one independent random walk onto another. Moreover, the test statistic diverges, meaning that with increasing sample size we are almost certain to reject the null hypothesis of no association. The spurious correlation occurs because our estimate is not consistent, which is a purely statistical explanation that does not invoke causal reasoning. I want to thank Toni Pichler and Andrea Bacilieri for helpful comments on this blog post. Post Scriptum Spurious correlation of AR(1) processes In the main text, we have looked at how the spurious correlation behaves for a random walk. Here, we study how the spurious correlation behaves as a function of $\phi \in [0, 1]$. We focus on sample sizes of $n = 200$, and adapt the simulation code from above. The Figure below shows that the issue of spurious correlation gets progressively worse as the AR(1) process approaches a random walk (i.e., $\phi = 1$). While this is true, the regression estimate remains consistent. References Granger, C. W., & Newbold, P. (1974). Spurious regressions in econometrics. Journal of Econometrics, 2(2), 111-120. Hamilton, J. D. (1994). Time Series Analysis. P. Princeton, US: Princeton University Press. Kuiper, R. M., & Ryan, O. (2018). Drawing conclusions from cross-lagged relationships: Re-considering the role of the time-interval. Structural Equation Modeling: A Multidisciplinary Journal, 25(5), 809-823. Phillips, P. C. (1986). Understanding spurious regressions in econometrics. Journal of Econometrics, 33(3), 311-340. Matthews, R. Storks deliver babies (p = 0.008) (2000). Teaching Statistics 22(2), 36–38. Footnotes Thanks to Toni Pichler for drawing my attention to the fact that independent random walks are correlated, and Andrea Bacilieri for providing me with the classic references. ↩ Moreover, one way to avoid the spurious correlation is to differencethe time-series. For other approaches, see Hamilton (1994, pp. 561). ↩ This awesome picture was made by Luther Seet. ↩
@egreg It does this "I just need to make use of the standard hyphenation function of LaTeX, except "behind the scenes", without actually typesetting anything." (if not typesetting includes typesetting in a hidden box) it doesn't address the use case that he said he wanted that for @JosephWright ah yes, unlike the hyphenation near box question, I guess that makes sense, basically can't just rely on lccode anymore. I suppose you don't want the hyphenation code in my last answer by default? @JosephWright anway if we rip out all the auto-testing (since mac/windows/linux come out the same anyway) but leave in the .cfg possibility, there is no actual loss of functionality if someone is still using a vms tex or whatever I want to change the tracking (space between the characters) for a sans serif font. I found that I can use the microtype package to change the tracking of the smallcaps font (\textsc{foo}), but I can't figure out how to make \textsc{} a sans serif font. @DavidCarlisle -- if you write it as "4 May 2016" you don't need a comma (or, in the u.s., want a comma). @egreg (even if you're not here at the moment) -- tomorrow is international archaeology day: twitter.com/ArchaeologyDay , so there must be someplace near you that you could visit to demonstrate your firsthand knowledge. @barbarabeeton I prefer May 4, 2016, for some reason (don't know why actually) @barbarabeeton but I have another question maybe better suited for you please: If a member of a conference scientific committee writes a preface for the special issue, can the signature say John Doe \\ for the scientific committee or is there a better wording? @barbarabeeton overrightarrow answer will have to wait, need time to debug \ialign :-) (it's not the \smash wat did it) on the other hand if we mention \ialign enough it may interest @egreg enough to debug it for us. @DavidCarlisle -- okay. are you sure the \smash isn't involved? i thought it might also be the reason that the arrow is too close to the "M". (\smash[t] might have been more appropriate.) i haven't yet had a chance to try it out at "normal" size; after all, \Huge is magnified from a larger base for the alphabet, but always from 10pt for symbols, and that's bound to have an effect, not necessarily positive. (and yes, that is the sort of thing that seems to fascinate @egreg.) @barbarabeeton yes I edited the arrow macros not to have relbar (ie just omit the extender entirely and just have a single arrowhead but it still overprinted when in the \ialign construct but I'd already spent too long on it at work so stopped, may try to look this weekend (but it's uktug tomorrow) if the expression is put into an \fbox, it is clear all around. even with the \smash. so something else is going on. put it into a text block, with \newline after the preceding text, and directly following before another text line. i think the intention is to treat the "M" as a large operator (like \sum or \prod, but the submitter wasn't very specific about the intent.) @egreg -- okay. i'll double check that with plain tex. but that doesn't explain why there's also an overlap of the arrow with the "M", at least in the output i got. personally, i think that that arrow is horrendously too large in that context, which is why i'd like to know what is intended. @barbarabeeton the overlap below is much smaller, see the righthand box with the arrow in egreg's image, it just extends below and catches the serifs on the M, but th eoverlap above is pretty bad really @DavidCarlisle -- i think other possible/probable contexts for the \over*arrows have to be looked at also. this example is way outside the contexts i would expect. and any change should work without adverse effect in the "normal" contexts. @DavidCarlisle -- maybe better take a look at the latin modern math arrowheads ... @DavidCarlisle I see no real way out. The CM arrows extend above the x-height, but the advertised height is 1ex (actually a bit less). If you add the strut, you end up with too big a space when using other fonts. MagSafe is a series of proprietary magnetically attached power connectors, originally introduced by Apple Inc. on January 10, 2006, in conjunction with the MacBook Pro at the Macworld Expo in San Francisco, California. The connector is held in place magnetically so that if it is tugged — for example, by someone tripping over the cord — it will pull out of the socket without damaging the connector or the computer power socket, and without pulling the computer off the surface on which it is located.The concept of MagSafe is copied from the magnetic power connectors that are part of many deep fryers... has anyone converted from LaTeX -> Word before? I have seen questions on the site but I'm wondering what the result is like... and whether the document is still completely editable etc after the conversion? I mean, if the doc is written in LaTeX, then converted to Word, is the word editable? I'm not familiar with word, so I'm not sure if there are things there that would just get goofed up or something. @baxx never use word (have a copy just because but I don't use it;-) but have helped enough people with things over the years, these days I'd probably convert to html latexml or tex4ht then import the html into word and see what come out You should be able to cut and paste mathematics from your web browser to Word (or any of the Micorsoft Office suite). Unfortunately at present you have to make a small edit but any text editor will do for that.Givenx=\frac{-b\pm\sqrt{b^2-4ac}}{2a}Make a small html file that looks like<!... @baxx all the convertors that I mention can deal with document \newcommand to a certain extent. if it is just \newcommand\z{\mathbb{Z}} that is no problem in any of them, if it's half a million lines of tex commands implementing tikz then it gets trickier. @baxx yes but they are extremes but the thing is you just never know, you may see a simple article class document that uses no hard looking packages then get half way through and find \makeatletter several hundred lines of trick tex macros copied from this site that are over-writing latex format internals.
Please assume that this graph is a highly magnified section of the derivative of some function, say $F(x)$. Let's denote the derivative by $f(x)$.Let's denote the width of a sample by $h$ where $$h\rightarrow0$$Now, for finding the area under the curve between the bounds $a ~\& ~b $ we can a... @Ultradark You can try doing a finite difference to get rid of the sum and then compare term by term. Otherwise I am terrible at anything to do with primes that I don't know the identities of $\pi (n)$ well @Silent No, take for example the prime 3. 2 is not a residue mod 3, so there is no $x\in\mathbb{Z}$ such that $x^2-2\equiv 0$ mod $3$. However, you have two cases to consider. The first where $\binom{2}{p}=-1$ and $\binom{3}{p}=-1$ (In which case what does $\binom{6}{p}$ equal?) and the case where one or the other of $\binom{2}{p}$ and $\binom{3}{p}$ equals 1. Also, probably something useful for congruence, if you didn't already know: If $a_1\equiv b_1\text{mod}(p)$ and $a_2\equiv b_2\text{mod}(p)$, then $a_1a_2\equiv b_1b_2\text{mod}(p)$ Is there any book or article that explains the motivations of the definitions of group, ring , field, ideal etc. of abstract algebra and/or gives a geometric or visual representation to Galois theory ? Jacques Charles François Sturm ForMemRS (29 September 1803 – 15 December 1855) was a French mathematician.== Life and work ==Sturm was born in Geneva (then part of France) in 1803. The family of his father, Jean-Henri Sturm, had emigrated from Strasbourg around 1760 - about 50 years before Charles-François's birth. His mother's name was Jeanne-Louise-Henriette Gremay. In 1818, he started to follow the lectures of the academy of Geneva. In 1819, the death of his father forced Sturm to give lessons to children of the rich in order to support his own family. In 1823, he became tutor to the son... I spent my career working with tensors. You have to be careful about defining multilinearity, domain, range, etc. Typically, tensors of type $(k,\ell)$ involve a fixed vector space, not so many letters varying. UGA definitely grants a number of masters to people wanting only that (and sometimes admitted only for that). You people at fancy places think that every university is like Chicago, MIT, and Princeton. hi there, I need to linearize nonlinear system about a fixed point. I've computed the jacobain matrix but one of the elements of this matrix is undefined at the fixed point. What is a better approach to solve this issue? The element is (24x_2 + 5cos(x_1)x_2)/abs(x_2). The fixed point is x_1=0, x_2=0 Consider the following integral: $\int 1/4(1/(1+(u/2)^2)))dx$ Why does it matter if we put the constant 1/4 behind the integral versus keeping it inside? The solution is $1/2\arctan{(u/2)}$. Or am I overseeing something? it should be du instead of dx in the integral *and the solution is missing a constant C of course Is there a standard way to divide radicals by polynomials? Stuff like $\frac{\sqrt a}{1 + b^2}$? My expression happens to be in a form I can normalize to that, just the radicand happens to be a lot more complicated. In my case, I'm trying to figure out how to best simplify $\frac{x}{\sqrt{1 + x^2}}$, and so far, I've gotten to $\frac{x \sqrt{1+x^2}}{1+x^2}$, and it's pretty obvious you can move the $x$ inside the radical. My hope is that I can somehow remove the polynomial from the bottom entirely, so I can then multiply the whole thing by a square root of another algebraic fraction. Complicated, I know, but this is me trying to see if I can skip calculating Euclidean distance twice going from atan2 to something in terms of asin for a thing I'm working on. "... and it's pretty obvious you can move the $x$ inside the radical" To clarify this in advance, I didn't mean literally move it verbatim, but via $x \sqrt{y} = \text{sgn}(x) \sqrt{x^2 y}$. (Hopefully, this was obvious, but I don't want to confuse people on what I meant.) Ignore my question. I'm coming of the realization it's just not working how I would've hoped, so I'll just go with what I had before.
I'll skip irrationals, but can anyone give an interpretation of the imaginary number $\sqrt{-x}$? or of the complex number $a + \sqrt{-x}$? That is, can you form a sentence that makes sense, which points to the square root of a negative number? I am explicitly not just looking for a way to complex number operations Sense is a very subjective thing. A familiar thing makes more sense than an unfamiliar one. So, perhaps, if my choice of examples does not strike you as sensible, I hope, you would have a bit of patience with my point of view. Continuing with your analogy, let us first have a one for real numbers. Real Numbers - Imagine an infinite line, with some point predefined as origin and some unit of measurement. A real number can be used to mark some position on the line. Now, consider $x \epsilon R$ and $y \epsilon R$. What is a good way to understand a pair of real numbers? Pair of real numbers - Imagine an infinite plane with some point predefined as origin and some unit of measurement. A pair of real numbers $(x, y)$ can be used to mark some position on the plane. This is how I make sense of complex numbers. Normally, when I do this, most of my students cry foul! "This is not complex numbers, this is coordinate geometry". So, let me elaborate. Another system which makes use of this representation of pair of real numbers is vectors. When I write a vector (2, 3), you would automatically assume an arrow on a plane or a point on a plane. But that is not enough for something to be a vector. A vector also needs to have some operations defined, namely '+', '$\times$' and '$\cdot$'. Without these operations, we won't be able to call $(x, y)$ to be a vector. Vector - A vector is tuple of real numbers such that the operations '+', '$\times$' and '$\cdot$' follow some specific rules. Some of these operations are more sensible than others. The '+' and '$\cdot$' operations are good, but the '$\times$' is a very difficult one to intuit. The point being, after the basic type pair of real number, if we want to understand further specializations like vectors and complex numbers, we need to understand the operations which they entail. With this in mind, let us define a new class - Not yet complex number (NYCN) - A pair of real numbers $(a, b)$ with operations '+' and '$\cdot$' defined as follows - $(a, b) + (c, d) = (a + c, b + d)$ and $(a, b) \cdot (c, d) = (ac - bd, ad + bc)$ How do we interpret the '$\cdot$' operation? It has two functions, rotation and scaling. If you take an arrow $(x, y)$ and want to rotate it anti-clockwise by say $\theta$, you could get the new point by performing $(x, y) \cdot (\cos \theta, \sin \theta)$. This rotation operation is what differentiates us from vectors. The problem with operation is in its pedagogy - it is so different from everything we have done yet (unless you count cross product in vectors). So, how do we reconcile these operations with our previous versions of '$\cdot$'. What follows is cosmetic surgery - Complex Numbers - $x + iy$ is a complex number for ($x, y \epsilon R$), if $(x, y)$ is a NYCN. $i$ is a constant which does not 'interact' with a real number much like $\hat{i}$ in vectors. Now, we want complex numbers to have our real number like addition and multiplication rules. Addition - $a + ib + c + id = (a + c) + i(b + d)$. So the new NYCN is $(x_1 + x_2, y_1 + y_2)$ which is good for us. Multiplication - $(a + ib) \cdot (c + id) = ac + i^2bd + i(ad + bc)$. This is very close to our requirement of NYCN. The $y$ term is perfect - $(ad + bc)$. But the $x$ term has $i^2$ instead of -1. So, for complex numbers to have same functionality as NYCN, we need to take $i^2 = -1$. "But this does not make sense!! Why take $i^2 = -1$?"Don't. Work with NYCNs. You would be good. The $i^2 = -1$ is a tool which makes things cleaner and equations become more beautiful. If you don't like makeup, don't put it on. The thing is, a bit of makeup does wonders to one's looks.
Is there an Intercept theorem (from Thales, but don’t mistake it with the Thales theorem in a circle) in hyperbolic geometry? Euclidean Intercept Theorem: Let S,A,B,C,D be 5 points, such that SA, SC, AC are respectively parallel to SB, SD, BD; and SB > SA, SD > SC. Then, we have SB/SA = SD/SC = BD/AD (see wiki illustration) Hyperbolic Intercept Theorem? The concept of parallel lines is tricky here… Let’s place ourselves in the Klein model, consider a triangle $ 0xy$ , two positive real numbers $ p,q$ , and construct the Einstein midpoints $ x’=m(x,0;p,q)$ and $ y’=m(y,0;p,q)$ , defined by $ $ m(a,b;p,q)=\dfrac{p\gamma_a a+q\gamma_b b}{p\gamma_a +q\gamma_b},$ $ with $ \gamma_x=1/\sqrt{1-\|\|x\|\|^2}$ the Lorentz factor. The analogue of “parallel lines” here resides in that we use the same $ (p,q)$ for $ x$ and $ y$ . Then, I can prove that $ $ \dfrac{\sinh d(x’,0)}{\sinh d(x,x’)}= \dfrac{\sinh d(y’,0)}{\sinh d(y,y’)},$ $ and I can also prove using hyperbolic sine law hat \begin{equation} \sinh d(x’,y’) \leq\min \left(\dfrac{\sinh d(x’,0)}{\sinh d(x,0)}\dfrac{\sin \alpha}{\sin \alpha’},\dfrac{\sinh d(y’,0)}{\sinh d(y,0)}\dfrac{\sin \beta}{\sin \beta’} \right)\sinh d(x,y). \end{equation} and I want to prove that either $ \dfrac{\sin \alpha}{\sin \alpha’} \leq 1$ or $ \dfrac{\sin \beta}{\sin \beta’}\leq 1$ . Here $ \alpha,\beta$ are the angles at $ x$ and $ y$ in $ Oxy$ , and $ \alpha’,\beta’$ at $ x’$ and $ y’$ in $ Ox’y’$ . Exploiting negative curvature, I could already prove that $ \alpha+\beta < \alpha’+\beta’$ , but that’s not enough… I would also be happy with a proof that Einstein midpoint operations are contracting (in $ \sinh$ of distance), when we use the same weights $ (p,q)$ .
Please, help me make equivalent transformations with this formula (A∨C→B)(A→C)(¬B→¬A∧C)(¬A→(C→B))(B→¬C→¬A). Thanks. You cannot "mix" in this way different "conventions" regarding symbols. If you want to use propositional connectives (like : $\lor, \land$) instead of boolean operators (like : $\cdot, +$) you have to rewrite your formula with $\lor$ in place of $+$ and $\land$ in place of $\cdot$ (justaxposition). If so, I think that your formula must be : $((A∨C)→B) \land (A→C) \land (¬B→(¬A \land C)) \land (¬A→(C→B)) \land (B→(¬C→¬A))$ having restored some missing parentheses. Then we eliminate $\rightarrow$, through the equivalence between $P \rightarrow Q$ and $\lnot P \lor Q$. In this way, splitting the problem, we have five conjuncts to consider: (i) $((A∨C)→B)$ is : $\lnot (A∨C) \lor B$ which, by De Morgan, is : $(\lnot A \land \lnot C) \lor B$, which in turn is equivalent to : $(\lnot A \lor B) \land (B \lor \lnot C)$, by distributivity. In boolean form is : $(\bar {A} + B)(B + \bar {C})$. (ii) $(A→C)$ is simply : $(\lnot A \lor C)$. In boolean : $(\bar {A} + C)$. (iii) $(¬B→(¬A \land C))$ is : $B \lor (\lnot A \land C)$, using double negation, which in turn is equivalent to : $(\lnot A \lor B) \land (B \lor C)$, by distributivity. In boolean form is : $(\bar {A} + B)(B + C)$. (iv) $(¬A→(C→B))$ is : $A \lor (\lnot C \lor B)$. In boolean : $(A + B + \bar {C})$. (v) $(B→(¬C→¬A))$ is : $\lnot B \lor (C \lor \lnot A)$. In boolean : $(\bar {A} + \bar {B} + C)$. Now, we can "reassemble" the conjuncts (i) to (v) without redundant terms : $(\bar {A} + B)(B + \bar {C})(\bar {A} + C)(B + C)(A + B + \bar {C})(\bar {A} + \bar {B} + C)$. Now we start with boolean simplification, "inserting" the missing terms [i.e. : $(\bar {A} + B)$ is rewritten as : $(\bar {A} + B + C\bar {C})$ i.e. $(\bar {A} + B + C)(\bar {A} + B + \bar {C})$ ] : $(\bar {A} + B + C)(\bar {A} + B + \bar {C})(A + B + \bar {C})(\bar{A} + B + \bar {C})(\bar {A} + B + C)(\bar {A} + \bar {B} + C)(A + B + C)(\bar {A} + B + C)(A + B + \bar {C})(\bar {A} + \bar {B} + C)$ and cancel the redundant terms : $(\bar {A} + B + C)(\bar {A} + B + \bar {C})(A + B + \bar {C})(\bar {A} + \bar {B} + C)(A + B + C)$. This is the standard Product-of-Sums (POS) Form.
How to Find a Fast Floating-Point atan2 Approximation ContextOver a short period of time, I came across nearly identical approximations of the two parameter arctangent function, atan2, developed by different companies, in different countries, and even in different decades. Fascinated with how the coefficients used in these approximations were derived, I set out to find them. This atan2 implementation is based around a rational approximation of arctangent on the domain -1 to 1: $$ atan(z) \approx \dfrac{z}{1.0 + 0.28z^2}$$ Arctangent [-1,1] Looking at the arctangent, it visually resembles a cubic polynomial. So in addition to deriving the rational approximation, I set out to find a polynomial version too. It turns out there is a 3rd order polynomial that is a close match to the rational approximation: $$ atan(z) \approx 0.9724x - 0.1919x^3$$ The rest of this post will cover how to reproduce the polynomial coefficients, how to implement them in the atan2 context, and wrapping up with some other thoughts. (Note: I find it odd that this 3rd order polynomial did not come up in my research. I believe this is because the rational approximation was developed for fixed-point math and was adopted for floating-point later, but I did not find evidence of this [3].) Arctangent Approximation The Remez Method is an iterative, minimax algorithm that for a given function converges to a polynomial (or rational approximation) which minimizes the max error. Conveniently, the Boost libraries includes a robust implementation in the math::tools namespace with the class remez_minimax [4]. The class's numerator() and denominator() contain the coefficients at the end of the iterative process. Although I planned to target single precision floats, it was necessary to use double precision to get useful coefficients. Below is the code necessary to initialize the remez_minimax class for atan and run iterations to find the coefficients for the polynomial approximation. At a certain point, max_change stays constant or becomes periodic and the process can stop. math::tools::remez_minimax<double> approximate_atan( [](const double& x) { return atan(x); }, // double precision required 3, 0, // 3rd degree polynomial -1, 1, // -1 to 1 range false, 0, 0, 64); // other params do { approximate_atan.iterate(); } while (approximate_atan.max_change() > 0.00001); const math::tools::polynomial<double> coefficients = approximate_atan.numerator(); const double max_error = approximate_atan.max_error(); After running this, the coefficients of {0.0, 0.97239411, 0.0, -0.19194795} were isolated, and the max error was just slightly less than 0.005 radians, which was comparable to rational version. I confirmed the results twice: On Wolfram Alpha with input: sup(abs(atan(x) - 0.9724x + 0.1919(x^3)) from -1 to 1 Testing every single precision floating point value between -1 and 1 compared against the atan in MSVC. The other metric I looked at was average error. Although the max error was comparable, the average error was 40% higher. If the max error is acceptable for any value (about 0.28 degrees), then it is likely also acceptable for all values. Measuring the average error of atan polynomial and rational approximation functions, the polynomial version is nearly 40% higher. Implementing atan2Set with a simple approximation of arctangent over the range [-1,1], the next step is integrating it into an atan2 implementation. To handle values outside the [-1,1] range, the following arctan identity is useful: $$ arctan(y/x) = \pi/2 - arctan(x/y), \text{if} \left\|y/x\right\| > 1.$$ To handle the different permutations of signs of y and x, I followed the definition of atan2 from Wikipedia [1]. There are comments to call out each of the 7 cases. The two extra cases handle the the y term being greater than x in which the above identity is needed.atan // Polynomial approximating arctangenet on the range -1,1. // Max error < 0.005 (or 0.29 degrees) float ApproxAtan(float z) { const float n1 = 0.97239411f; const float n2 = -0.19194795f; return (n1 + n2 z * z) * z; }atan2 float ApproxAtan2(float y, float x) { if (x != 0.0f) { if (fabsf(x) > fabsf(y)) { const float z = y / x; if (x > 0.0) { // atan2(y,x) = atan(y/x) if x > 0 return ApproxAtan(z); } else if (y >= 0.0) { // atan2(y,x) = atan(y/x) + PI if x < 0, y >= 0 return ApproxAtan(z) + PI; } else { // atan2(y,x) = atan(y/x) - PI if x < 0, y < 0 return ApproxAtan(z) - PI; } } else // Use property atan(y/x) = PI/2 - atan(x/y) if \|y/x\| > 1. { const float z = x / y; if (y > 0.0) { // atan2(y,x) = PI/2 - atan(x/y) if \|y/x\| > 1, y > 0 return -ApproxAtan(z) + PI_2; } else { // atan2(y,x) = -PI/2 - atan(x/y) if \|y/x\| > 1, y < 0 return -ApproxAtan(z) - PI_2; } } } else { if (y > 0.0f) // x = 0, y > 0 { return PI_2; } else if (y < 0.0f) // x = 0, y < 0 { return -PI_2; } } return 0.0f; // x,y = 0. Could return NaN instead. } Notice that only finite floating point values are handled. Permutations of +/- infinity and +/- 0 all have defined values under the IEEE standard. I decided to not include these for this this function. Looking for a Bit More Speed The goal of the approximation is to be faster than the precise implementation and with the math simplified (at least as much as I could show), the remaining optimization targeted the branching and reducing the amount of floating point comparisons. It is possible to eliminate the three branches that determined the +/-PI or +/-PI/2 offsets to atan with bit manipulation. The nice thing, and it is even suggested by the disassembly, was that some of the logic could be performed on the ALU and FPU in parallel. Overall this made an impact of about 5%, but at the cost of being harder to read. float ApproxAtan2(float y, float x) { const float n1 = 0.97239411f; const float n2 = -0.19194795f; float result = 0.0f; if (x != 0.0f) { const union { float flVal; uint32 nVal; } tYSign = { y }; const union { float flVal; uint32 nVal; } tXSign = { x }; if (fabsf(x) >= fabsf(y)) { union { float flVal; uint32 nVal; } tOffset = { PI }; // Add or subtract PI based on y's sign. tOffset.nVal \|= tYSign.nVal & 0x80000000u; // No offset if x is positive, so multiply by 0 or based on x's sign. tOffset.nVal = tXSign.nVal >> 31; result = tOffset.flVal; const float z = y / x; result += (n1 + n2 z * z) * z; } else // Use atan(y/x) = pi/2 - atan(x/y) if \|y/x\| > 1. { union { float flVal; uint32 nVal; } tOffset = { PI_2 }; // Add or subtract PI/2 based on y's sign. tOffset.nVal \|= tYSign.nVal & 0x80000000u; result = tOffset.flVal; const float z = x / y; result -= (n1 + n2 * z * z) * z; } } else if (y > 0.0f) { result = PI_2; } else if (y < 0.0f) { result = -PI_2; } return result; } Example disassembly of ALU and FPU operations being interleaved: ... tOffset.nVal \|= tYSign.nVal & 0x80000000u; and ecx, 80000000h or ecx, dword ptr[tOffset] result = tOffset.flVal; float z = y / x; divss xmm4, xmm5 tOffset.nVal = tXSign.nVal >> 31; shr eax, 1Fh imul ecx, eax result += (n1 + n2 z * z) * z; ... SIMD Optimization For work, I re-wrote ApproxAtan2 using SSE2 intrinsics to calculate 4 atan2 results simultaneously. Taking out any overhead of packing the 4 sets of x and y values, it ran about 40% faster than 4 separate calls to ApproxAtan2 run in serial. The function ended up being 38 intrinsic instructions which intuitively feels a bit more than necessary, but I limited my time to look at it. I might go back and try writing it again with the goals of being: more readable, use less instructions, and allow myself to use newer instruction sets. Results The tests I ran were a number of different inputs representing some different contexts with about 1 billion pairs of values. The values below are in seconds and have had the baseline, or time to generate inputs, removed. (I only ran one comparable test of the SIMD version vs the polynomial approximation.) The tests are inputs of what I expected in typical geometric applications. I also included a contrived test to demonstrate how the DIVSS operation on i7 can have dramatically different performance based on the input. The polynomial approximation with reduced branching seems to be 25% to 50% faster. Test std::atan2 rational approx polynomial approx polynomial SIMD 1 100.3 24.1 18.4 46.0 (11.5) 2 70.1 14.4 7.4 na 3 70.5 14.3 8.1 na 4 8.3 51.2 50.8 na Limitations of Remez Minimax The Boost class for the Remez minimax algorithm is quite nice in that you can test a number of polynomial and rational combinations to find the best set of coefficients for the required performance. I have tried this for arccos and tanh as well. However the algorithm only minimizes the error for polynomial and rational approximations when there are other types of function approximations. Modern FPU architectures can perform sqrt comparably fast to division for example. Recursive functions or continued fractions are also often decent approximations. In the case of arccos there are long known approximations using both sqrt and recursion which are superior to polynomial and rational functions found using the Remez method. There are also cases where the Remez algorithm can get stuck or become unstable, although there are work arounds. Links [1] Atan2 - Handling Zero and Planes: https://en.wikipedia.org/wiki/Atan2#Definition_and_computation [2] The Remez Method: http://www.boost.org/doc/libs/1_46_1/libs/math/doc/sf_and_dist/html/math_toolkit/backgrounders/remez.html [3] Performing Efficient Arctangent Approximation: http://www.embedded.com/design/other/4216719/Performing-efficient-arctangent-approximation [4] Boost Remez source code: http://www.boost.org/doc/libs/1_55_0/boost/math/tools/remez.hpp So I noticed that while you approximated $\arctan(z)$ over $[-1,1]$ you were only actually applying it to $[0,1]$ - which means we can do better by fitting a cubic on the proper interval. This does cost one extra multiply and two extra adds, as the cubic won't be symmetric anymore. (Coefficients are approxomate, but yield similar error.) \[\arctan(z) \approx -0.060317z^3 -0.198146z^2 + 1.044261z - 0.002178, z \in [0,1]\] for two more multiplies and another add a quartic does even better \[\arctan(z) \approx 0.141499z^4 - 0.343315z^3 - 0.016224z^2 + 1.003839z - 0.000158\] Method RMS Error Max Error Time $z \over 1+az$ 0.002348 0.007385 0.716351 $az^3+cz$ 0.003488 0.004952 0.449856 $az^3+bz^2+cz+d$ 0.0006843 0.002178 0.563637 $az^4+bz^3+cz^2+dz+e$ 0.00006394 0.0002427 0.654011 libm 1.407e-17 1.110e-16 3.020070 Error estimates are in radians over 20k evenly spaced points on the unit circle.Time is to sum up arctagents of $2^{26}$ points evenly distributed in the unit square. I couldn't get the tiny number run to replicate, except libm did get faster (LLVM 3.9.1 on an i5-3337). Hi imuli, Give me a week to pull up this project again and refresh myself. But two quick questions: 1. How did you come to the conclusion that I was only using the range [0,1]? 2. What did you use to locate the coefficients? As I mentioned I was using the boost implementation of remez_minimax. (I will also call out that the rms in my image left out a divide by 2, which you have picked up on.) Thanks, very interest to follow up on this. The extra add/multiply would be well worth the precision. 1. Apparently by misreading something, because your code obviously is using the full range! We don't need to though, because $\arctan(-z) = -\arctan(z)$. double atan2_approx(double y, double x){ double ay = fabs(y), ax = fabs(x); int invert = ay > ax; double z = invert ? ax/ay : ay/ax; // [0,1] double th = atan_approx(z); // [0,π/4] if(invert) th = M_PI_2 - th; // [0,π/2] if(x < 0) th = M_PI - th; // [0,π] th = copysign(th, y); // [-π,π] } The three branches can be eliminated with some cleverness, but I'm going for clarity here. 2. I used gnuplot's fit function. It actually gave slightly better coefficients for the symmetric cubic case, but they weren't that much better. Spent some time looking at your solution. What I like the most is how succinct and more readable it becomes when dealing with the absolute values of x and y. If I use your approach to update my SIMD implementation, I'll be sure to post up here. Thank you again, this was very insightful! Great, thank you. I'm going to process this next week when I have some free time. I noticed the title of your post and at first thought to myself, "Uh oh, another attempt at numerical approximation" -- but it is very well-done and I applaud your work. I wish there were a good Remez rational approximation feature in Python so I could experiment without having to drop back down into C++ hell. Thank you. It is interesting you mention Python. When I was looking for an implementation of Remez approximation, my goal was to find an R package. At least at the time I was doing this research, I only found the Boost implementation and Matlab (which I no longer have a license for). Can you be more specific? I looked at your pages, and these seem to be different approximations (but I could be mistaken). So I am not sure what you are asking to be credited for. The rational approximation at the very beginning of the article? - Nic Please play around with my values and you may get more precise results :) I see. Thank you. Yes, next time I come back to benchmark atan2, I can try to adapt your equation to work with in my test. At this time, none of the projects I am on are bottle-necked on atan2, so it is a lower priority. But any new results, I will post back to this thread. To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments. 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X Search Filters Format Subjects Library Location Language Publication Date Click on a bar to filter by decade Slide to change publication date range 2014 Web Resource Microelectronics Reliability, ISSN 0026-2714, 05/2018, Volume 84, pp. 1 - 6 This paper discusses the reliability performance of Wolfspeed GaN/AlGaN high electron mobility transistor (HEMT) MMIC released process technologies, fabricated... DC-ALT \| GaN \| HEMT \| SiC \| RF-ALT \| Intrinsic reliability \| Wear-out \| PHYSICS, APPLIED \| NANOSCIENCE & NANOTECHNOLOGY \| ENGINEERING, ELECTRICAL & ELECTRONIC DC-ALT \| GaN \| HEMT \| SiC \| RF-ALT \| Intrinsic reliability \| Wear-out \| PHYSICS, APPLIED \| NANOSCIENCE & NANOTECHNOLOGY \| ENGINEERING, ELECTRICAL & ELECTRONIC Journal Article PROCEEDINGS OF THE NATIONAL ACADEMY OF SCIENCES OF THE UNITED STATES OF AMERICA, ISSN 0027-8424, 02/2019, Volume 116, Issue 9, pp. 3811 - 3816 The ability to detect environmental cold serves as an important survival tool. The sodium channels Na(V)1.8 and Na(V)1.9, as well as the TRP channel Trpm8,... GCaMP \| CELLULAR BASIS \| PAIN \| CHANNEL \| NOCICEPTORS \| MULTIDISCIPLINARY SCIENCES \| Trpm8 \| nociception \| cold \| Na(V)1.8 \| SENSATION \| REVEALS GCaMP \| CELLULAR BASIS \| PAIN \| CHANNEL \| NOCICEPTORS \| MULTIDISCIPLINARY SCIENCES \| Trpm8 \| nociception \| cold \| Na(V)1.8 \| SENSATION \| REVEALS Journal Article 2002, ISBN 9780851692722 Book Medicine & Science in Sports & Exercise, ISSN 0195-9131, 02/2017, Volume 49, Issue 2, pp. 265 - 273 Increases in maximal oxygen uptake (V˙O2max) frequently occur with high-intensity interval training (HIIT), yet the specific adaptation explaining this result... FICK EQUATION \| STROKE VOLUME \| INTERVAL EXERCISE \| THORACIC IMPEDANCE \| MAXIMAL OXYGEN UPTAKE \| CYCLE ERGOMETRY \| Cardiac Output - physiology \| Young Adult \| Pulmonary Gas Exchange - physiology \| Humans \| Adult \| Female \| Male \| Hemodynamics - physiology \| Oxygen Consumption - physiology \| Body Composition - physiology \| High-Intensity Interval Training \| Index Medicus FICK EQUATION \| STROKE VOLUME \| INTERVAL EXERCISE \| THORACIC IMPEDANCE \| MAXIMAL OXYGEN UPTAKE \| CYCLE ERGOMETRY \| Cardiac Output - physiology \| Young Adult \| Pulmonary Gas Exchange - physiology \| Humans \| Adult \| Female \| Male \| Hemodynamics - physiology \| Oxygen Consumption - physiology \| Body Composition - physiology \| High-Intensity Interval Training \| Index Medicus Journal Article ASTROPHYSICAL JOURNAL SUPPLEMENT SERIES, ISSN 0067-0049, 11/2014, Volume 215, Issue 1, pp. 7 - 7 New radiative lifetimes are reported for 168 levels of V I ranging in energy from 18086 cm(-1) to 47702 cm(-1), and for 31 levels of V II ranging in energy... 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The sodium channels Na 1.8 and Na 1.9, as well as the TRP channel Trpm8, have... Journal Article Advanced Materials, ISSN 0935-9648, 06/2014, Volume 26, Issue 22, pp. 3755 - 3760 Novel solar cells, based on dense arrays of InGaAs nanowires, are grown directly on graphene. Here, graphene serves as the conductive back contact and growth... InGaAs \| nanowires \| MOCVD \| graphene \| photovoltaics \| HIGH-PERFORMANCE \| LAYER GRAPHENE \| PHYSICS, CONDENSED MATTER \| DESIGN \| PHYSICS, APPLIED \| MATERIALS SCIENCE, MULTIDISCIPLINARY \| SILICON NANOWIRES \| CHEMISTRY, PHYSICAL \| NANOSCIENCE & NANOTECHNOLOGY \| CHEMISTRY, MULTIDISCIPLINARY \| INAS NANOWIRES \| CHANNEL \| PHOTOVOLTAIC APPLICATIONS \| GROWTH \| GAAS NANOWIRES \| EFFICIENCY \| Graphene \| Epitaxy \| Graphite \| Solar cells \| Photovoltaic cells \| Nanostructure \| Epitaxial growth \| Arrays \| Conversion \| Assembly \| Nanowires InGaAs \| nanowires \| MOCVD \| graphene \| photovoltaics \| HIGH-PERFORMANCE \| LAYER GRAPHENE \| PHYSICS, CONDENSED MATTER \| DESIGN \| PHYSICS, APPLIED \| MATERIALS SCIENCE, MULTIDISCIPLINARY \| SILICON NANOWIRES \| CHEMISTRY, PHYSICAL \| NANOSCIENCE & NANOTECHNOLOGY \| CHEMISTRY, MULTIDISCIPLINARY \| INAS NANOWIRES \| CHANNEL \| PHOTOVOLTAIC APPLICATIONS \| GROWTH \| GAAS NANOWIRES \| EFFICIENCY \| Graphene \| Epitaxy \| Graphite \| Solar cells \| Photovoltaic cells \| Nanostructure \| Epitaxial growth \| Arrays \| Conversion \| Assembly \| Nanowires Journal Article 10. P–V–T equation of state of synthetic mirabilite (Na2SO4·10D2O) determined by powder neutron diffraction Journal of Applied Crystallography, ISSN 0021-8898, 04/2013, Volume 46, Issue 2, pp. 448 - 460 Neutron powder diffraction data have been collected from Na 2 SO 4 ·10D 2 O (the deuterated analogue of mirabilite), a highly hydrated sulfate salt that is... neutron diffraction \| ice IV \| mirabilite \| equations of state \| Ice IV \| Equations of state \| Neutron diffraction \| Thermodynamics \| Powders \| Solar system \| Sulfates \| Anisotropy \| Synthetic products \| Materials science \| Diffraction \| Crystallography \| Chemical compounds \| Elastic anisotropy \| Compressibility \| Icy satellites \| Deuteration \| Outer solar system neutron diffraction \| ice IV \| mirabilite \| equations of state \| Ice IV \| Equations of state \| Neutron diffraction \| Thermodynamics \| Powders \| Solar system \| Sulfates \| Anisotropy \| Synthetic products \| Materials science \| Diffraction \| Crystallography \| Chemical compounds \| Elastic anisotropy \| Compressibility \| Icy satellites \| Deuteration \| Outer solar system Journal Article 2001, Pushkinskai͡a Premʹera., ISBN 5850800267, 108 Book 12. 12.5 pm/V hybrid silicon and lithium niobate optical microring resonator with integrated electrodes Optics Express, ISSN 1094-4087, 11/2013, Volume 21, Issue 22, pp. 27003 - 27010 Journal Article 13. Measurement of prompt and nonprompt charmonium suppression in $$\text {PbPb}$$ PbPb collisions at 5.02$$\,\text {Te}\text {V}$$ TeV The European Physical Journal C, ISSN 1434-6044, 6/2018, Volume 78, Issue 6, pp. 1 - 27 The nuclear modification factors of $${\mathrm {J}/\psi }$$ J/ψ and $$\psi \text {(2S)}$$ ψ(2S) mesons are measured in $$\text {PbPb}$$ PbPb collisions at a... Nuclear Physics, Heavy Ions, Hadrons \| Measurement Science and Instrumentation \| Nuclear Energy \| Quantum Field Theories, String Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Astronomy, Astrophysics and Cosmology Nuclear Physics, Heavy Ions, Hadrons \| Measurement Science and Instrumentation \| Nuclear Energy \| Quantum Field Theories, String Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Astronomy, Astrophysics and Cosmology Journal Article 14. Performance of reconstruction and identification of τ leptons decaying to hadrons and v τ in pp collisions at s=13 TeV Journal of Instrumentation, ISSN 1748-0221, 10/2018, Volume 13, Issue 10, pp. P10005 - P10005 Here, the algorithm developed by the CMS Collaboration to reconstruct and identify $\tau$ leptons produced in proton-proton collisions at $\sqrt{s}=$ 7 and 8... Particle identification methods \| Performance of High Energy Physics Detectors \| Large detector-systems performance \| PHYSICS OF ELEMENTARY PARTICLES AND FIELDS Particle identification methods \| Performance of High Energy Physics Detectors \| Large detector-systems performance \| PHYSICS OF ELEMENTARY PARTICLES AND FIELDS Journal Article 15. Search for a heavy resonance decaying into a Z boson and a vector boson in the v(v)over-barq(q)over-bar final state JOURNAL OF HIGH ENERGY PHYSICS, ISSN 1029-8479, 07/2018, Issue 7 Journal Article 16. The mouse immunoglobulin heavy chain V-D intergenic sequence contains insulators that may regulate ordered V(D)J recombination Journal of Biological Chemistry, ISSN 0021-9258, 03/2010, Volume 285, Issue 13, pp. 9327 - 9338 During immunoglobulin heavy chain (Igh) V(D)J recombination, D to J precedes V to DJ recombination in an ordered manner, controlled by differential chromatin... B-CELL DEVELOPMENT \| ALLELIC EXCLUSION \| 3' END \| LYMPHOCYTE DEVELOPMENT \| CHROMATIN-STRUCTURE \| BIOCHEMISTRY & MOLECULAR BIOLOGY \| BONE-MARROW \| GENE REARRANGEMENT \| INTRONIC ENHANCER \| IGH LOCUS \| CTCF SITES \| Chromatin - metabolism \| Promoter Regions, Genetic \| Computational Biology - methods \| Bone Marrow Cells - cytology \| Mice, Inbred C57BL \| Immunoglobulin Heavy Chains - chemistry \| DNA, Intergenic - genetics \| Genes, Immunoglobulin Heavy Chain \| Models, Immunological \| Animals \| Oligonucleotides, Antisense - genetics \| T-Lymphocytes - metabolism \| Recombination, Genetic \| Gene Rearrangement \| Alleles \| Mice \| Models, Genetic \| Immunoglobulin Heavy Chains - genetics \| Index Medicus \| Gene Regulation \| Transcription Regulation \| Chromatin \| Insulator \| RNA \| V(D)J Recombination \| Antisense RNA \| DNase \| Immunoglobulin B-CELL DEVELOPMENT \| ALLELIC EXCLUSION \| 3' END \| LYMPHOCYTE DEVELOPMENT \| CHROMATIN-STRUCTURE \| BIOCHEMISTRY & MOLECULAR BIOLOGY \| BONE-MARROW \| GENE REARRANGEMENT \| INTRONIC ENHANCER \| IGH LOCUS \| CTCF SITES \| Chromatin - metabolism \| Promoter Regions, Genetic \| Computational Biology - methods \| Bone Marrow Cells - cytology \| Mice, Inbred C57BL \| Immunoglobulin Heavy Chains - chemistry \| DNA, Intergenic - genetics \| Genes, Immunoglobulin Heavy Chain \| Models, Immunological \| Animals \| Oligonucleotides, Antisense - genetics \| T-Lymphocytes - metabolism \| Recombination, Genetic \| Gene Rearrangement \| Alleles \| Mice \| Models, Genetic \| Immunoglobulin Heavy Chains - genetics \| Index Medicus \| Gene Regulation \| Transcription Regulation \| Chromatin \| Insulator \| RNA \| V(D)J Recombination \| Antisense RNA \| DNase \| Immunoglobulin Journal Article NATURE, ISSN 0028-0836, 04/2011, Volume 472, Issue 7342, pp. 186 - 190 Loss of function of the gene SCN9A, encoding the voltage-gated sodium channel Na(v)1.7, causes a congenital inability to experience pain in humans. Here we... SYSTEM \| ORGANIZATION \| PAIN \| MOUSE OLFACTORY-BULB \| RAT \| MULTIDISCIPLINARY SCIENCES \| RECEPTOR \| EXPRESSION \| DELETION \| SUBUNIT \| REVEALS SYSTEM \| ORGANIZATION \| PAIN \| MOUSE OLFACTORY-BULB \| RAT \| MULTIDISCIPLINARY SCIENCES \| RECEPTOR \| EXPRESSION \| DELETION \| SUBUNIT \| REVEALS Journal Article Nature Chemical Biology, ISSN 1552-4450, 2012, Volume 8, Issue 10, pp. 855 - 861 Among bacterial toxin-antitoxin systems, to date no antitoxin has been identified that functions by cleaving toxin mRNA. Here we show that YjdO (renamed GhoT)... CELLS \| REGULATOR \| BACTERIAL PERSISTENCE \| PROTEIN \| MULTIDRUG TOLERANCE \| BIOCHEMISTRY & MOLECULAR BIOLOGY \| ESCHERICHIA-COLI \| GENERAL STRESS-RESPONSE \| EXPRESSION \| NMR STRUCTURE DETERMINATION \| BIOFILM FORMATION \| RNA, Messenger - genetics \| Antitoxins - chemistry \| RNA, Messenger - metabolism \| Reverse Transcriptase Polymerase Chain Reaction \| Bacterial Toxins - chemistry \| Hydrolysis \| Bacterial Toxins - metabolism \| Biofilms \| Nuclear Magnetic Resonance, Biomolecular \| Bacterial Toxins - genetics \| Protein Conformation \| Antitoxins - metabolism \| Antitoxins - genetics \| Bacteria \| Membranes \| Toxins \| Ribonucleic acid--RNA \| Alanine \| Index Medicus \| Analytical chemistry \| Chemical Sciences CELLS \| REGULATOR \| BACTERIAL PERSISTENCE \| PROTEIN \| MULTIDRUG TOLERANCE \| BIOCHEMISTRY & MOLECULAR BIOLOGY \| ESCHERICHIA-COLI \| GENERAL STRESS-RESPONSE \| EXPRESSION \| NMR STRUCTURE DETERMINATION \| BIOFILM FORMATION \| RNA, Messenger - genetics \| Antitoxins - chemistry \| RNA, Messenger - metabolism \| Reverse Transcriptase Polymerase Chain Reaction \| Bacterial Toxins - chemistry \| Hydrolysis \| Bacterial Toxins - metabolism \| Biofilms \| Nuclear Magnetic Resonance, Biomolecular \| Bacterial Toxins - genetics \| Protein Conformation \| Antitoxins - metabolism \| Antitoxins - genetics \| Bacteria \| Membranes \| Toxins \| Ribonucleic acid--RNA \| Alanine \| Index Medicus \| Analytical chemistry \| Chemical Sciences Journal Article
Given an $n$-vector $y$ (responses) and a design matrix $X$, I wish to fit them with a simple linear regression model $$y=X\beta+e,$$ where $e\sim\mathcal{N}(0, \sigma^2I)$. Then, we have $$y\sim\mathcal{N}(X\beta, \sigma^2I).$$ Then the maximum likelihood estimations (MLE) of $\beta$ and $\sigma^2$ are just $$\hat\beta=(X^TX)^{-1}X^Ty$$ and $$\hat{\sigma^2}=\frac{(y-X\hat\beta)^T(y-X\hat\beta)}{n}.$$ I understand that $\sigma^2$'s estimator is biased in that $E\{\hat{\sigma^2}\}\neq\sigma^2$. On Page 4 of this lecture notes, the author claims $E\{\hat{\sigma^2}\}=\frac{n-r}{n}\sigma^2$ without specifying what $r$ is. I guess $r$ is the DoF loss while estimating $\beta$, i.e., the dimension of $\beta$. But how do we derive it? I tried to prove it myself but got confused at "over which distribution are we taking the expectation?"
I would like to fit a GLM to the rate underlying a Poisson process, for data with variable exposure (period of measurement) - and the question is about aggregating/grouping the data before fitting or not. So the model is $$\mu = \exp(\beta_0+\beta_1 X_1+\beta_2 X_2)$$ $$ Y_i \sim Poisson(\mu_i t_i) $$ So $\mu_i$ is the rate, $t_i$ is the time over which the observations were recorded, and $Y_i$ is the Poissonian distributed number of counts measured in $t_i$. The exposure ($t_i$) should therefore, by my understanding, appear as both an offset in the GLM and a weight (longer observations get more weight). I code this up in R as the following: #generate the data:numsamples<-50000x1<-sample(1:20,numsamples,replace=T)x2<-sample(1:10,numsamples,replace=T)t<-1/sample(1:10,numsamples,replace=T) #exposure timemu_rate<- exp(0.1+0.04x1+0.025x2) #log linear rate#generate the count data:y<-rpois(numsamples,mu_rate*t)#combine the data into a data framedf <- data.frame(y=y,x1=x1,x2=x2,t=t)#fit a glm:glm1<-glm(y~x1+x2,data=df,family=poisson(link ="log"), offset=log(df$t),weights=df$t/max(df$t))#aggregate data with identical variables - sum both y and tdf_agg<-aggregate(cbind(y,t)~x1+x2,data=df,FUN=sum)#fit a glm to the aggregated dataglm1<-glm(y~x1+x2,data=df_agg,family=poisson(link ="log"), offset=log(df_agg$t),weights=df_agg$t/max(df_agg$t)) Here I have fit to the raw data, and also aggregated data with identical values of $X_1$ and $X_2$ by summing the total count $Y$ and exposure $t$. Now, my understanding is that aggregating the data should make no difference to the fit (provided that one offsets and weights appropriately by the newly aggregated exposure $t$) - see, for example, page 10 of http://data.princeton.edu/wws509/notes/c4.pdf . However, the two glm fits give different coefficients (e.g.: 0.1051 vs 0.1065 for the intercept). Admittedly, here, the difference is less than the standard error in the coefficients - however a) I would have expected no difference at all except for machine precision errors, and b) on more complicated data sets which I can't replicate here, the discrepancy is considerably larger than the standard error. Increasing the maximum iterations and decreasing the tolerence (epsilon) seemed to have no effect. So I guess, my question boils down to a) is there something wrong with the way I have offset/weighted my data or done the aggregation? and b) is there something wrong with my expectation of obtaining the identical fit parameters with the aggregated data? Thanks, in advance.
CryptoDB Changmin Lee Affiliation: Seoul National University, Republic of Korea Publications Year Venue Title 2019 JOFC Cryptanalysis of the CLT13 Multilinear Map In this paper, we describe a polynomial time cryptanalysis of the (approximate) multilinear map proposed by Coron, Lepoint, and Tibouchi in Crypto13 (CLT13). This scheme includes a zero-testing functionality that determines whether the message of a given encoding is zero or not. This functionality is useful for designing several of its applications, but it leaks unexpected values, such as linear combinations of the secret elements. By collecting the outputs of the zero-testing algorithm, we construct a matrix containing the hidden information as eigenvalues, and then recover all the secret elements of the CLT13 scheme via diagonalization of the matrix. In addition, we provide polynomial time algorithms to directly break the security assumptions of many applications based on the CLT13 scheme. These algorithms include solving subgroup membership, decision linear, and graded external Diffie–Hellman problems. These algorithms mainly rely on the computation of the determinants of the matrices and their greatest common divisor, instead of performing their diagonalization. 2019 CRYPTO Statistical Zeroizing Attack: Cryptanalysis of Candidates of BP Obfuscation over GGH15 Multilinear Map We present a new cryptanalytic algorithm on obfuscations based on GGH15 multilinear map. Our algorithm, statistical zeroizing attack, directly distinguishes two distributions from obfuscation while it follows the zeroizing attack paradigm, that is, it uses evaluations of zeros of obfuscated programs.Our attack breaks the recent indistinguishability obfuscation candidate suggested by Chen et al. (CRYPTO’18) for the optimal parameter settings. More precisely, we show that there are two functionally equivalent branching programs whose CVW obfuscations can be efficiently distinguished by computing the sample variance of evaluations.This statistical attack gives a new perspective on the security of the indistinguishability obfuscations: we should consider the shape of the distributions of evaluation of obfuscation to ensure security.In other words, while most of the previous (weak) security proofs have been studied with respect to algebraic attack model or ideal model, our attack shows that this algebraic security is not enough to achieve indistinguishability obfuscation. In particular, we show that the obfuscation scheme suggested by Bartusek et al. (TCC’18) does not achieve the desired security in a certain parameter regime, in which their algebraic security proof still holds.The correctness of statistical zeroizing attacks holds under a mild assumption on the preimage sampling algorithm with a lattice trapdoor. We experimentally verify this assumption for implemented obfuscation by Halevi et al. (ACM CCS’17). 2018 CRYPTO Cryptanalyses of Branching Program Obfuscations over GGH13 Multilinear Map from the NTRU Problem 📺 In this paper, we propose cryptanalyses of all existing indistinguishability obfuscation (iO) candidates based on branching programs (BP) over GGH13 multilinear map for all recommended parameter settings. To achieve this, we introduce two novel techniques, program converting using NTRU-solver and matrix zeroizing, which can be applied to a wide range of obfuscation constructions and BPs compared to previous attacks. We then prove that, for the suggested parameters, the existing general-purpose BP obfuscations over GGH13 do not have the desired security. Especially, the first candidate indistinguishability obfuscation with input-unpartitionable branching programs (FOCS 2013) and the recent BP obfuscation (TCC 2016) are not secure against our attack when they use the GGH13 with recommended parameters. Previously, there has been no known polynomial time attack for these cases.Our attack shows that the lattice dimension of GGH13 must be set much larger than previous thought in order to maintain security. More precisely, the underlying lattice dimension of GGH13 should be set to $$n=\tilde{\varTheta }( \kappa ^2 \lambda )$$n=Θ~(κ2λ) to rule out attacks from the subfield algorithm for NTRU where $$\kappa $$κ is the multilinearity level and $$\lambda $$λ the security parameter. 2015 EPRINT 2015 EUROCRYPT
Unitary elements of a Banach space have been defined in this paper as follows: Let $A$ be a Banach space and $a\in A, \\|a\\|=1$. Let $S_{a}=\{f\in A':\\|f\\|=1=f(a)\}$. Then $a$ is said to be (geometrically) unitary if $A'=\text{ span }S_{a}$. Here, $A'$ is the dual space of $A$. We note that this property is true for unitary operators on a Hilbert space. Unitary operators on Hilbert spaces are invertible, and their spectra lie on the unit circle. Now, suppose $A$ is a Banach algebra. Can we say that if $a\in A$ is geometrically unitary, then it is invertible? I have been trying to find a counter-example, but have been unsuccessful so far. I considered the Banach algebra $l^{1}(\mathbb{Z})$, with convolution as the multiplication. The dual of $l^{1}(\mathbb{Z})$ is $l^{\infty}(\mathbb{Z})$. An element $f$ of $l^{1}(\mathbb{Z})$ is invertible iff $f(z)\neq 0 \, \forall z\in \mathbb{T}$, where $\mathbb{T}$ is the unit circle in $\mathbb{C}$. Let us take, for example, $f=(\cdots,0,\frac{1}{2},0,\frac{i}{2},0,\cdots)$, where the central $0$ is in the $0^{th}$ position. Then $f$ is of norm $1$ and can be shown to be not invertible. We now consider elements $\phi$ of $S_{f}\subseteq l^{\infty}(\mathbb{Z})$. $\phi$ must satisfy the following: $ \\|\phi\\|_{\infty}=1\\$ and $\phi(-1)\frac{1}{2}+\phi(1)\frac{i}{2}=1$. One possible solution is $\phi=(\cdots,1,x,-i,\cdots)$, where $x$ and the dots can be any scalar less than or equal to $1$. Is there another possible solution? In that case, we would have a unitary element that is not invertible. Alternately, is it indeed true that unitary elements described in this geometric fashion are invertible? I'd be grateful for help with this.
Is it true for $n=1$? $\sum\limits_{i=1}^1(6i-3) = 6\cdot 1 - 3 = 3 = 3\cdot 1^2$ Yes, it is true for $n=1$. Now, supposing that it is true that $\sum\limits_{i=1}^n(6i-3)=3n^2$ for some $n\geq 1$, we are curious whether or not it follows that it will necessarily also be true for $n+1$ I.e. we want to show that $\sum\limits_{i=1}^n(6i-3)=3n^2\Rightarrow \sum\limits_{i=1}^{n+1}(6i-3)=3(n+1)^2$. To do so, we start at one side and manipulate it using a series of equalities, eventually at the very end arriving at the right hand side, showing the left side equals the right side. ( do NOT begin with the equality you wish to prove and simply reach a tautology, that is circular logic and invalid) $\sum\limits_{i=1}^{n+1}(6i-3) = 6(n+1)-3 + \sum\limits_{i=1}^{n}(6i-3)$ by simply adding the final term of the summation individually. Now, we recognize the smaller summation and we know something about it based on our induction hypothesis... $\dots=6(n+1)-3+3n^2$ by induction hypothesis Continuing to simplify: $\dots = 6n+6-3+3n^2 = 3n^2+6n+3 = 3(n+1)^2$ which is exactly what we wanted to show. Thus, by the principle of mathematical induction, the result is true for all $n\geq 1$
Let $S$ be a non-empty set. Consider the statement$$\forall A\subseteq S \quad \forall B\subseteq S\quad\exists D\subseteq S \ (D\neq \emptyset \ \wedge \ D\cap (A\cup B)=\emptyset) $$The negation is$$\exists A \subseteq S \quad \exists B\subseteq S \quad \forall D \subseteq S \ (D=\emptyset \ \vee \ D\cap (A \cup B)\neq \emptyset). $$I have to determine which statement is true. I had a long think and thought that the second is true since I think the first is false. Let $S = \{1,2,3\}$. Then if I pick $A= \{1,2\}$ and $B=\{2,3\}$. If $D$ is empty then the statement is false. If $D$ is not empty, then $D$ must be a set which has common members with $A$ or $B$ and so the intersection is nonempty. Is this correct or is there a better way to think about it? Let $S$ be a non-empty set. Consider the statement$$\forall A\subseteq S \quad \forall B\subseteq S\quad\exists D\subseteq S \ (D\neq \emptyset \ \wedge \ D\cap (A\cup B)=\emptyset) $$The negation is$$\exists A \subseteq S \quad \exists B\subseteq S \quad \forall D \subseteq S \ (D=\emptyset \ \vee \ D\cap (A \cup B)\neq \emptyset). $$I have to determine which statement is true. You are correct. To see how this works for any $S$: Pick $A=B=S$. Then $A \subseteq S$, $B \subseteq S$, and $A \cup B =S$. Hence, there cannot be any non-empty $D \subseteq S$ that does not share any elements with $A \cup B$.
Funny problem. There is a number of ways it can be treated, starting from elementary (working with real & imaginary parts) & proceeding to more advanced (including some more geometric approaches). Your questions 1 & 2 read pretty much the same to me, & the only sensible answer I can provide is ' because working w/ z & $\bar{z}$ concurrently allows us to use Linear Algebra' - without getting messy, at least, which is something that the solution using real & imaginary parts cannot claim. That aside, it's clearly a trick (but then again, what isn't?). I understand your question about the determinant condition working out great in $\mathbb{C}^2$ but possibly being an overkill here, & it's a valid question. I propose to explain why the condition is in fact both sufficient & necessary using subspaces. I'll warn you that this boils down to exactly working w/ real & imaginary parts, albeit after having recast your problem in the $2\times2$ matrix form you give. From a certain POV, that's keeping the worst of both worlds: passing to that form is unintuitive & working with real/imaginary parts is cumbersome. Nevertheless, I find it far more intuitive than either solution, for reasons that might become clear once I stop running my mouth & do some math. Let's consider any two complex numbers $z = z_R + {\rm i} z_I$ & $z' = z'_R + {\rm i} z'_I$. I embed (z,z') into $\mathbb{R}^4$ by writing $(z_R,z_I,z'_R,z'_I)$ for $(z,z')$ - this is a homomorphism between $\mathbb{C}^2$ and $\mathbb{R}^4$. Then, the pair $(z,\bar{z})$ becomes $(z_R,z_I,z_R,-z_I) = z_R(1,0,1,0) + z_I(0,1,0,-1) = z_R {\rm e_1} + z_I {\rm e_i}$, with ${\rm e_{1,i}}$ the obvious vectors in $\mathbb{R}^4$. Hence, all possible pairs $(z,\bar{z})$ constitute a $2-$D subspace $\mathbb{E}$ of $\mathbb{R}^4$, spanned by ${\rm e_{1,i}}$. Write, now, $T : \mathbb{C}^2 \to \mathbb{C}^2$ for the linear operator represented by the matrix $\left(\begin{array}{cc}a&b\\\bar{b}&\bar{a}\end{array}\right)$. Then, the restriction $\left.T\,\right\vert_\mathbb{E} : \mathbb{E} \to \mathbb{E}$ is well defined, that is, $T$'s range falls indeed within $\mathbb{E}$. Indeed, we have (easy calculation) $$T {\rm e_1} = (a_R+b_R){\rm e_1} + (a_I+b_I){\rm e_i} ,\\T {\rm e_i} = (b_I-a_I){\rm e_1} + (a_R-b_R){\rm e_i} .$$ Equipping $\mathbb{E}$ with the basis ${\rm e_{1,i}}$ and expressing $T$ in that basis, then, we obtain the $2\times2$ matrix $$\left(\begin{array}{cc}a_R+b_R & b_I-a_I\\a_I+b_I & a_R-b_R\end{array}\right) ,$$ which is precisely what you'd obtain had you chosen to recast the original equation in terms of real & imaginary parts. (The reason is obvious, I believe.) Its determinant is, naturally, $\|a\|^2-\|b\|^2$. Note, finally, that the equation for $(z,\bar{z})$ you report has a right-hand side that also falls in $\mathbb{E}$ (also for obvious intuitive reasons). Hence, the equation is solvable as long as $\left.T\,\right\vert_\mathbb{E} : \mathbb{E} \to \mathbb{E}$ is a bijection, i.e., as long as the aforementioned determinant is nonzero. & that's why the determinant condition is also necessary, not just sufficient.
Search Now showing items 1-1 of 1 Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (Elsevier, 2014-09) Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, \|y\| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
I'm trying to solve a two part problem. The set up is as follows: consider a bag with $\theta$ red marbles and $7-\theta$ blue marbles, with $\theta$ being unknown. Let $x$ denote the number of red marbles found in a sample of 3. If we sample without replacement, what is a maximum likelihood estimator for $\theta$ (the number of red marbles), based on our sample?If we sample with replacement, what is the maximum likelihood estimator for $\theta$. I'd also like to check if these estimators are unbiased, and which has the smaller variance (I'd expected sampling without replacement to be superior). 1.) Sampling without replacement yields a hypergeometric distribution, with the likelihood function $\large L(\theta)=\frac{\binom{\theta}{x}\binom{7-\theta}{3-x}}{\binom{7}{3}}$. I think what I want to do is look at the ratio $\large \hat{L(\theta)}=\frac{L(\theta)}{L(\theta+1)}$, since this should be increasing up to the point $ \hat{L(\theta)}>1$ and decreasing afterwards, and we can take the MLE to be this point of inflection. A bit of algebra shows this point to be $\frac{8x-3}{3}$. When this fraction is an integer, we can see that $ \hat{L(\theta)}=1$, so $\frac{8x-3}{3}+1$ is also an MLE (it is not unique). If it is not an integer, we take the floor and see that the MLE is $[\frac{8x}{3}]$, where the brackets represent the floor function (since $[\frac{8x-3}{3}]<[\frac{8x}{3}]$, we only have one MLE in this case). 2.) The binomial case is more confusing to me, though perhaps I'm just thinking about it incorrectly. I can easily take the maximum likelihood estimate of P (which is, in this case, $\frac{\theta}{7}$), and show that this MLE is simply $\frac{x}{3}$ (this follows from taking the derivative of the log likelihood and setting it to zero). Then solving the equation $\frac{\theta}{7}=\frac{x}{3}$ yields $\theta=\frac{7x}{3}$ (this result is similar to the hypergeometric result, which is reassuring). It is not, however, a whole number and I can't tell if I should take the floor or the ceiling in this case (or perhaps both provide MLEs)? Finally, assuming that my logic is sound up to this point, I'm not sure how to check whether either of these are unbiased, or how to compare the variances (I think this should be relatively easy - I'm just drawing a blank!) Thanks so much for reading - any help is greatly appreciated!
It is safe to say that there will always be intermolecular forces at play. At the time where you will consider these you should already have a good idea about the molecules involved in your system. Based on the composition and molecular structures you can make certain assumptions. In a molecule it is straight forward to estimate (bond) polarities based on electronegativities, then infer from these how they might arrange. I will work an example based on the interactions between adenine and thymine later. [1] With the advent of the information age, tools that every chemist has at her or his disposal have become more sophisticated. We have access to many digital resources like databases, [2] or publication servers [3] to retrieve a vast amount of information. Molecular modelling, [4] or even more sophisticated quantum chemical calculations, [5] have become more important; and free tools are available for everyone to use. With that being said, here are some points that might help you identifying hydrogen bonds and other non-covalent interactions. For that purpose, let's have a look at our example molecules: Immediately we can formulate a couple of assumptions based on the schematic representation. In adenine there are five nitrogen atoms, which have a higher electronegativity than carbon. Negative partial charges will therefore be located at mostly at these. Hydrogens in organic compounds usually carry positive partial charges; albeit $\ce{C-H}$ tend to be a lot less polarised than $\ce{N-H}$ bonds. Whenever a hydrogen is involved in a bond, that bond can potentially act as a hydrogen-bond donor (see below). Similar observations can be made for thymine. Here we have two terminal oxygen atoms, which will carry a negative partial charge, since they have one of the highest electronegativities. These are often able to accept hydrogen bonds. On the other hand we also have $\ce{N-H}$ bonds, which can act as hydrogen bond donors. Charges & Electrostatic Potential Surfaces For many molecules structures are readily available. If not, some molecular editors give you the possibility to use implemented force fields to optimise built (guessed) structures. Based on those you can already do a few analyses. One tool that is quite powerful for various tasks is Avogadro, it let's you read crystal structures, perform basic calculations and much more. If you are just playing around, this is a really good choice. For example, I have imported the crystal structure of adenine into Avogadro, optimised it, and calculated the electrostatic potential. Or after extracting Cartesian coordinates, Molden let's you easily calculate the charges. [6] Hydrogen Bonds Many molecular editors try to guess hydrogen bonds based on their implemented cutoff values. That certainly is very helpful, but not everything can be automatised in this way. And especially weaker interactions won't be found. One has to go a bit deeper then. As a nice and concise example I have picked an intermolecular 2:1 complex between adenine and thymine, for which the crystal structure is available. [1] There are two principle structural parameters to decide about hydrogen bonds: (a) The distance of the hydrogen $\ce{H}$ and the hydrogen-bond acceptor $\ce{Y}$ is significantly shorter than the sum of their respective van-der-Waals radii, $d(\ce{XH\bond{...}Y})<r_\mathrm{vdW}(\ce{H})+r_\mathrm{vdW}(\ce{Y})$. [7] (b) The angle around the hydrogen is nearly linear, $\angle(\ce{XH\bond{...}Y}) \approx 180^\circ$. For weakly polarised $\ce{XH}$ bonds, isotropic dispersion forces become more important (while the directional electrostatic and covalent contributions decrease), therefore the angle becomes more flexible. We easily see that the bond angles are close to what we expect for hydrogen bonds. I have reproduced a few values from Batsanov's paper below, with the caveat that the value for hydrogen strongly varies depending on the chemical environment from $\pu{110 - 161 pm}$, so I used the classic from Bondi. [7c] Since all the distance are around $\pu{200 pm}$, they are well below the threshold we set earlier. $$\begin{array}{lr}\text{Element }\ce{Y} & r_\mathrm{vdW}(\ce{Y})/\pu{pm}\\\hline\ce{H} & \approx 120\\\ce{C} & 196 \\\ce{N} & 179 \\\ce{O} & 171 \\\hline\end{array}\hspace{2ex}\begin{array}{lr}\\\text{H-Bond }\ce{XH\bond{...}Y} & \sum r_\mathrm{vdW}(\ce{Y},\ce{H})/\pu{pm}\\\hline\ce{CH} & 316 \\\ce{NH} & 299 \\\ce{OH} & 291 \\\hline\end{array}$$ A quite interesting approach of revealing non-covalent interactions was presented by Johnson et. al., and the corresponding program is easy to use and only requires Cartesian coordinates. [8] The surfaces between the molecules represent these interactions, where green represents weak interactions, typically found for dispersion. Blue represents stronger interactions, typically found for hydrogen-bonds. Red displays repulsive forces, typically found within ring or cage systems. If you have access to quantum chemical software, then you can obtain this plot also for wave function files .wfn. Another possibility is to analyse the electron density in terms of the quantum theory of atoms in molecules (QTAIM). [9] For this you do need a wave function file. The analysis, however, is straight forward and will yield a bond path or not. If there is a bond path, we can estimate the strengths of these bond with the methodology developed by Espinosa et. al.. According to this the bondstrength is approximately half the value of the potential energy density at the bond critical point.$$E_\mathrm{H-Bond} = \frac{1}{2}V(r_{\mathrm{BCP}[\ce{XH\bond{...}Y}]})$$ I have performed such a calculation on the DF-B97D3/def2-TZVPP level of theory with Gaussian 09. The optimised geometry will be at the end. \begin{array}{lr}\text{H-Bond} & E_\mathrm{H-Bond}/\pu{kJ mol-1}\\\hline\mathrm{N(37)H \cdots O(1)} & 46.6\\\mathrm{N(5)H \cdots N(36)} & 38.5\\\mathrm{C(41)H \cdots O(2)} & 3.2\\\mathrm{N(20)H \cdots O(2)} & 50.5\\\mathrm{N(3)H \cdots N(24)} & 29.9\\\end{array} A general warning shall be applied to the above. Absolute values of these are only approximate, but fall within the range of what is expected. A very nice side effect of this methodology is, that it can be applied to intramolecular hydrogen bonds, too. Concluding remarks Dispersive interactions and hydrogen bonds become more and more important in rational reaction design. Be it for understanding of molecular structure of biomolecules, or as a guiding principle for catalyst-substrate interactions.With further development of computer technology, it should become more accessible to everyone. I hope this post demonstrates that gaining more insight can actually be quite easy (and free). Notes and References (a) Based on the structure from S. Chandrasekhar, T. R. Naik, S. K. Nayak, T. N. Row, Bioorg. Med. Chem. Lett. 2010, 20 (12), 3530-3533. DOI: 10.1016/j.bmcl.2010.04.131 PMID:20493694 CSD: 739016 (b) Adenine, CSID: 185(c) S. Mahapatra, S. K. Nayak, S. J. Prathapa, T. N. Guru Row, Cryst. Growth Des. 2008, 8 (4), 1223–1225. DOI: 10.1021/cg700743w, CSD: 652573 (d) Thymine, CSID: 1103 (e) G. Portalone, L. Bencivenni, M. Colapietro, A. Pieretti, F. Ramondo, Acta Chemica Scand. 1999, 53, 57-68. DOI: 10.3891/acta.chem.scand.53-0057, CSD: 136916 (a) The Cambridge Structural Database (CSD), https://www.ccdc.cam.ac.uk/(b) Crystallography Open Database (COD), http://www.crystallography.net/cod/(c) Computational Chemistry Comparison and Benchmark DataBase, http://cccbdb.nist.gov/ (Only for 1799 small molecules and atoms)(d) Handbook of Chemistry and Physics, http://hbcponline.com/faces/contents/ContentsSearch.xhtml(e) ... (a) SciFinder, https://www.cas.org/products/scifinder(b) Google Scholar, https://scholar.google.de/(c) Web of Science, (formerly known as Web of Knowledge) http://www.webofknowledge.com/(d) ... (a) MolCalc, http://molcalc.org/(b) Pitt Quantum Repository, https://pqr.pitt.edu/ (At the time of writing it was dead.) Github: pittquantum(c) Many open source molecular editors include the possibility to use force field calculations. For example: Avogadro, molden(d) For more on molecular modelling in the open source domain see S. Pirhadi, J. Sunseri, D. R. Koes, J. Mol. Graph. Model. 2016, 69, 127-143. An updated online version of this catalog can be found at https://opensourcemolecularmodeling.github.io. (a) For an extensive, but not necessarily complete, list of quantum chemistry software see Wikipedia.(b) For the purpose of this demonstration I will be using the proprietary software Gaussian. (c) To view crystal structures, Mercury can be obtained (for free) from the Cambridge Crystallographic Data Centre (CCDC), which also hosts CSD. https://www.ccdc.cam.ac.uk/solutions/csd-system/components/mercury/ (a) Tutorial for Avogadro (b) Tutorial for molden (a) A concise (and as far as I can tell newest) list of van-der-Waals radii of many elements can be found in S. S, Batsanov, Inorg. Mat. 2001, 37, 871-885. DOI: 10.1023/A:1011625728803 (mirrored pdf)(b) A list of van-der-Waals radii can also be found on Wikipedia(c) A. Bondi, J. Phys. Chem. 1964, 68, 441-451. doi: 10.1021/j100785a001 (a) The original publication: E. R. Johnson, S Keinan, P. Mori-Sánche§, J. Contreras-García, A. J. Cohen, W. Yang, J. Am. Chem. Soc. 2010, 132, 6498-6506. DOI: 10.1021/ja100936w (b) The presentation of the program: J. Contreras-Garcia, E. R. Johnson, S. Keinan, R. Chaudret, J-P. Piquemal, D. N. Beratan, W. Yang, J. Chem. Theory Comput. 2011, 7, 625-632. DOI: 10.1021/ct100641a (c) Download the code: http://www.lct.jussieu.fr/pagesperso/contrera/nciplot.html (d) You'll also need VMD (Visual Molecular Dynamics) from the University of Illinois (a) A very brief introduction can be found on Wikipedia. The corresponding book: Bader, Richard (1994). Atoms in Molecules: A Quantum Theory. USA: Oxford University Press. ISBN 978-0-19-855865-1. (publisher)(b) Multiwfn - A Multifunctional Wavefunction Analyzer; http://sobereva.com/multiwfn/ corresponding paper: T. Lu, F. Chen, J. Comput. Chem. 2012, 33, 580-592. DOI: 10.1002/jcc.22885 (c) Startup script (and examples) for Linux version: https://github.com/polyluxus/runMultiwfn.bash (shameless self-plug)(d) E. Espinosa, E. Molins and C. Lecomte, Chem. Phys. Lett., 1998, 285, 170–173. Appendix Optimised Structure of the adenine-thymine 2:1 complex calculated at DF-B97D3/def2-TZVPP in Gaussian 09 Rev. E.01 45 E(RB97D3/def2TZVPP/W06) = -1388.51095169 O 19.03780 11.79565 1.63996 O 14.74303 13.36808 1.71115 N 15.07940 11.09762 1.64043 H 14.04669 10.93403 1.64128 N 16.88238 12.55083 1.67499 H 17.23771 13.53695 1.70257 C 17.83067 11.52964 1.63864 C 17.29332 10.17534 1.60048 C 15.51719 12.40225 1.67775 C 15.94232 10.03610 1.60357 H 15.46489 9.06076 1.57654 C 18.24710 9.02139 1.56030 H 18.89656 9.08263 0.68030 H 18.90669 9.02997 2.43483 H 17.71085 8.06866 1.53480 N 8.28060 10.09397 1.65692 H 7.68055 9.28386 1.63653 N 8.91986 12.24943 1.71837 N 10.48101 9.01368 1.60709 N 11.91467 12.94472 1.71757 H 12.92612 13.11883 1.71578 H 11.26613 13.71385 1.74609 C 10.03193 11.42408 1.68464 N 12.26612 10.63929 1.64378 C 11.41986 11.70069 1.68284 C 9.65939 10.07380 1.64587 C 7.89700 11.42267 1.70074 H 6.85459 11.71169 1.71748 C 11.75955 9.39245 1.60917 H 12.49648 8.59107 1.57896 N 21.22700 16.46826 1.76840 N 17.31372 17.43807 1.81488 H 16.80939 18.31038 1.84233 N 19.61652 18.27309 1.82815 C 18.69253 17.30751 1.80467 N 17.71219 15.24141 1.74956 N 20.64990 14.21957 1.70686 H 19.98577 13.44513 1.68592 H 21.63800 14.02512 1.69520 C 20.27365 15.50797 1.74532 C 16.77892 16.17095 1.78070 H 15.71578 15.97320 1.77996 C 18.91965 15.92447 1.76371 C 20.85479 17.75435 1.80728 H 21.67003 18.47621 1.82413
Let's assume that $f\colon\mathbb R\to\mathbb R$ is continuous and hence Lebesgue measurable. Then, the Lebesgue integral $\int_{(0,\infty)}f(x)\,d\lambda(x)$ makes sense (but of course can be equal to $\pm\infty$). Given such $f$, how can we find this Lebesgue integral? Will it always be the case that $$\int_{(0,\infty)} f(x)\,d\lambda(x)=\int_0^\infty f(x)\,dx=\lim_{R\to\infty}\int_0^Rf(x)\,dx,\tag{$$}$$where the second and third integrals are Riemann integrals that can possibly be $\pm\infty$? I know that if $f$ is such that the improper Riemann integrals $\int_0^\infty f(x)\,dx$ and $\int_0^\infty\|f(x)\|\,dx$ converge then $f$ is Lebesgue integrable and $\int_{(0,\infty)}f(x)\,d\lambda(x)=\int_0^\infty f(x)\,dx$ us finite. However, I am asking something different here. If we don't know anything about the convergence of the improper Riemann integrals, do we still have the equality in ($$)? I know that it is true for bounded interval, and I have studied this result, but I don't see how it generalises. For a concrete example, let $f(x)=e^x$. This is continuous and hence Lebesgue measurable. Is the following reasoning valid? The function is Riemann integrable on $(0,R)$ for all $R>0$. Also, since it is Lebesgue measurable its Lebesgue integral makes sense and we have $$\int_{(0,\infty)}f(x)\,d\lambda(x)=\int_0^\infty f(x)\,dx=e^x\bigg\|_0^\infty =\infty .$$ In pretty much every source authors interchange between the Riemman integral and the Lebesgue integral automatically. But then, if $()$ holds we have $$\int_{(0,\infty)}\frac{\sin x}{x}\,d\lambda(x)=\int_0^\infty \frac{\sin x}{x}\,dx=\frac{\pi}{2},$$ when I've seen written that the Lebesgue integral of $\sin x/x$ over $(0,\infty)$ does not exist. I know that it is not Lebesgue integrable because the integral of $\|\sin x/x\|$ is unbounded, but why does it not exist? I think my confusion has something to do with the phrases integral exists, makes sense, and Lebesgue integrable. I use the first two to say that our function has an integral whose value belongs to $[-\infty,\infty]$ and the latter to say that the absolute value of our function has an integral whose value is a non-negative real. To summarise, if we accept that a Lebesgue integral may be equal to $\pm \infty$, is ($$) always true for continuous functions $f$? I am not interested in Lebesgue integrability here, I only care for the value of the Lebesgue integral over $(0,\infty)$. I want to use this fact in calculations, as in the example with $f(x)=e^x$. As you can see I am pretty confused. Any explanation will be appreciated.
Mathematics - Functional Analysis and Mathematics - Metric Geometry Abstract The following strengthening of the Elton-Odell theorem on the existence of a $(1+\epsilon)-$separated sequences in the unit sphere $S_X$ of an infinite dimensional Banach space $X$ is proved: There exists an infinite subset $S\subseteq S_X$ and a constant $d>1$, satisfying the property that for every $x,y\in S$ with $x\neq y$ there exists $f\in B_{X^*}$ such that $d\leq f(x)-f(y)$ and $f(y)\leq f(z)\leq f(x)$, for all $z\in S$. Comment: 15 pages, to appear in Bulletin of the Hellenic Mayhematical Society Given a finite dimensional Banach space X with dimX = n and an Auerbach basis of X, it is proved that: there exists a set D of n + 1 linear combinations (with coordinates 0, -1, +1) of the members of the basis, so that each pair of different elements of D have distance greater than one. Comment: 15 pages. To appear in MATHEMATIKA
OWN 1.25d Is the even part of $x(t) = \cos \left( 4 \pi t\right) \ u(t)$ periodic? If it is, what is the time period? OWN 1.25e Is the even part of $x(t) = \sin \left( 4 \pi t\right) \ u(t)$ periodic? If it is, what is the time period? OWN 1.26b Is the discrete-time signal $x[n] = \cos \left(\frac{n}{8}\right)$ periodic? If it is, what is the time period? OWN 1.26c Is the discrete-time signal $x[n] = \cos \left(\frac{\pi n^2}{8}\right)$ periodic? If it is, what is the time period? OWN 1.27a Is the system $y(t) = x(t-2)+x(2-t)$ memoryless, linear, stable, time-invariant and causal? OWN 1.27c Is the system $y(t) = \int_{-\infty}^{2t} x(\tau) \ d \tau$ memoryless, linear, stable, time-invariant and causal? OWN 1.27d Is the system $y(t) = \begin{cases} 0, \ \ t < 0 \\ x(t)+x(t-2), t \geq 0 \end{cases}$ memoryless, linear, stable, time-invariant and causal? OWN 1.27g Is the system $y(t) = \frac{d}{dt}x(t)$ memoryless, linear, stable, time-invariant and causal? Consider the system defined by $y(t) = x(t) \ u(t)$. Is this linear, time invariant, stable, invertible? OWN 2.4 Compute the convolution of(1) \begin{align} x[n] = \left\{ \begin{array}{ll} 1, & 3 \leq n \leq 8 \\ 0, & \hbox{otherwise.} \end{array} \right. \ \hbox{and} \ h[n] = \left\{ \begin{array}{ll} 1, & 4 \leq n \leq 15 \\ 0, & \hbox{otherwise.} \end{array} \right. \end{align} OWN 2.6 Compute the convolution of $x[n] = \left(\frac{1}{3} \right)^{-n} \ u[-n-1]$ and $h[n] = u[-n-1]$ Review done on Sunday, December 10, 2017 before the final exam Compute Discrete-time Fourier transform of $x[n] = \sin \left( \frac{\pi n/5}{\pi n}\right) \ \cos \left(\frac{5 \pi n}{2} \right)$
Find all continuous functions $f$ which satisfy the functional equation $$ f(x)\,f(y)-f(x+y)=\sin x\,\sin y, $$ for all $x,y\in\mathbb R$. I can prove that $f(n\pi)=\cos\left(n\pi\right)$ for all $n\in\mathbb Z$. First attempt. I have tried to prove that: $$f\left(\frac{\pi}n\right)=\cos\left(\frac{\pi}n\right),\quad\text{for all}\,\,\, n\in\mathbb Z\smallsetminus\{0\} \tag{1},$$ but I have failed. If I prove $(1)$, then the functional equation will be solved completely using the continuity of $f$. So how do we solve this functional equation?
The Cauchy problem for a tenth-order thin film equation II. Oscillatory source-type and fundamental similarity solutions 1. Universidad Carlos III de Madrid, Av. Universidad 30, 28911-Leganés, Spain 2. Department of Mathematical Sciences, University of Bath, Bath BA2 7AY, United Kingdom, United Kingdom nonlinear eigenvalues($\gamma$ is a multiindex in $\mathbb{R}^N$) of the tenth-order thin film equation (TFE-10) \begin{eqnarray} \label{i1a} u_{t} = \nabla \cdot (\|u\|^{n} \nabla \Delta^4 u) \quad in \quad \mathbb{R}^N \times \mathbb{R}_+ \,, \quad n>0, (0.1) \end{eqnarray} are studied. The present paper continues the study began in [1]. Thus, the following questions are also under scrutiny: ( I) Further study of the limit $n \to 0$, where the behaviour of finite interfaces and solutions as $y \to \infty$ are described. In particular, for $N=1$, the interfaces are shown to diverge as follows: $$ \|x_0(t)\| \sim 10 ( \frac{1}{n}\sec ( \frac{4\pi}{9} ) )^{\frac 9{10}} t^{\frac 1{10}} \to \infty as n \to 0^+. $$ ( II) For a fixed $n \in (0, \frac 98)$, oscillatory structures of solutions near interfaces. ( III) Again, for a fixed $n \in (0, \frac 98)$, global structures of some nonlinear eigenfunctions $\{f_\gamma\}_{\|\gamma\| \ge 0}$ by a combination of numerical and analytical methods. Keywords:source-type global similarity solutions of changing sign., Thin film equation, the Cauchy problem. Mathematics Subject Classification:35G20, 35K65, 35K35, 37K5. Citation:P. Álvarez-Caudevilla, J. D. Evans, V. A. Galaktionov. The Cauchy problem for a tenth-order thin film equation II. Oscillatory source-type and fundamental similarity solutions. Discrete & Continuous Dynamical Systems - A, 2015, 35 (3) : 807-827. doi: 10.3934/dcds.2015.35.807 References: [1] P. Álvarez-Caudevilla, J. D. Evans and V. A. Galaktionov, [2] P. Álvarez-Caudevilla, J. D. Evans and V. A. Galaktionov, [3] P. Álvarez-Caudevilla and V. A. Galaktionov, Local bifurcation-branching analysis of global and "blow-up" patterns for a fourth-order thin film equation,, 18 (2011), 483. doi: 10.1007/s00030-011-0105-6. Google Scholar [4] P. Álvarez-Caudevilla and V. A. Galaktionov, Well-posedness of the Cauchy problem for a fourth-order thin film equation via regularization approaches,, [5] [6] [7] M. Chaves and V. A. Galaktionov, On source-type solutions and the Cauchy problem for a doubly degenerate sixth-order thin film equation. I. Local oscillatory properties,, 72 (2010), 4030. doi: 10.1016/j.na.2010.01.034. Google Scholar [8] [9] Yu. V. Egorov, V. A. Galaktionov, V. A. Kondratiev and S. I. Pohozaev, Global solutions of higher-order semilinear parabolic equations in the supercritical range,, 9 (2004), 1009. Google Scholar [10] J. D. Evans, V. A. Galaktionov and J. R. King, Blow-up similarity solutions of the fourth-order unstable thin film equation,, 18 (2007), 195. doi: 10.1017/S0956792507006900. Google Scholar [11] J. D. Evans, V. A. Galaktionov and J. R. King, Source-type solutions of the fourth-order unstable thin film equation,, 18 (2007), 273. doi: 10.1017/S0956792507006912. Google Scholar [12] [13] [14] [15] V. A. Galaktionov, Countable branching of similarity solutions of higher-order porous medium type equations,, 13 (2008), 641. Google Scholar [16] [17] V. A. Galaktionov, E. Mitidieri and S. I. Pohozaev, Variational approach to complicated similarity solutions of higher-order nonlinear evolution equations of parabolic, hyperbolic, and nonlinear dispersion types,, 9 (2009), 147. doi: 10.1007/978-0-387-85650-6_8. Google Scholar [18] V. A. Galaktionov, E. Mitidieri and S. I. Pohozaev, Variational approach to complicated similarity solutions of higher-order nonlinear PDEs. II,, 12 (2011), 2435. doi: 10.1016/j.nonrwa.2011.03.001. Google Scholar [19] M. A. Krasnosel'skii, [20] [21] [22] X. Liu and C. Qu, Existence and blow-up of weak solutions for a sixth-order equation related to thin solid films,, 11 (2010), 4214. doi: 10.1016/j.nonrwa.2010.05.008. Google Scholar [23] [24] M. A. Vainberg and V. A. Trenogin, show all references References: [1] P. Álvarez-Caudevilla, J. D. Evans and V. A. Galaktionov, [2] P. Álvarez-Caudevilla, J. D. Evans and V. A. Galaktionov, [3] P. Álvarez-Caudevilla and V. A. Galaktionov, Local bifurcation-branching analysis of global and "blow-up" patterns for a fourth-order thin film equation,, 18 (2011), 483. doi: 10.1007/s00030-011-0105-6. Google Scholar [4] P. Álvarez-Caudevilla and V. A. Galaktionov, Well-posedness of the Cauchy problem for a fourth-order thin film equation via regularization approaches,, [5] [6] [7] M. Chaves and V. A. Galaktionov, On source-type solutions and the Cauchy problem for a doubly degenerate sixth-order thin film equation. I. Local oscillatory properties,, 72 (2010), 4030. doi: 10.1016/j.na.2010.01.034. Google Scholar [8] [9] Yu. V. Egorov, V. A. Galaktionov, V. A. Kondratiev and S. I. Pohozaev, Global solutions of higher-order semilinear parabolic equations in the supercritical range,, 9 (2004), 1009. Google Scholar [10] J. D. Evans, V. A. Galaktionov and J. R. King, Blow-up similarity solutions of the fourth-order unstable thin film equation,, 18 (2007), 195. doi: 10.1017/S0956792507006900. Google Scholar [11] J. D. Evans, V. A. Galaktionov and J. R. King, Source-type solutions of the fourth-order unstable thin film equation,, 18 (2007), 273. doi: 10.1017/S0956792507006912. Google Scholar [12] [13] [14] [15] V. A. Galaktionov, Countable branching of similarity solutions of higher-order porous medium type equations,, 13 (2008), 641. Google Scholar [16] [17] V. A. Galaktionov, E. Mitidieri and S. I. Pohozaev, Variational approach to complicated similarity solutions of higher-order nonlinear evolution equations of parabolic, hyperbolic, and nonlinear dispersion types,, 9 (2009), 147. doi: 10.1007/978-0-387-85650-6_8. Google Scholar [18] V. A. Galaktionov, E. Mitidieri and S. I. Pohozaev, Variational approach to complicated similarity solutions of higher-order nonlinear PDEs. II,, 12 (2011), 2435. doi: 10.1016/j.nonrwa.2011.03.001. Google Scholar [19] M. A. Krasnosel'skii, [20] [21] [22] X. Liu and C. Qu, Existence and blow-up of weak solutions for a sixth-order equation related to thin solid films,, 11 (2010), 4214. doi: 10.1016/j.nonrwa.2010.05.008. Google Scholar [23] [24] M. A. Vainberg and V. A. Trenogin, [1] Guirong Liu, Yuanwei Qi. Sign-changing solutions of a quasilinear heat equation with a source term. [2] [3] [4] Yuanxiao Li, Ming Mei, Kaijun Zhang. Existence of multiple nontrivial solutions for a $p$-Kirchhoff type elliptic problem involving sign-changing weight functions. [5] Daniel Ginsberg, Gideon Simpson. Analytical and numerical results on the positivity of steady state solutions of a thin film equation. [6] Guofu Lu. Nonexistence and short time asymptotic behavior of source-type solution for porous medium equation with convection in one-dimension. [7] Shao-Yuan Huang. Global bifurcation and exact multiplicity of positive solutions for the one-dimensional Minkowski-curvature problem with sign-changing nonlinearity. [8] M. Ben Ayed, Kamal Ould Bouh. Nonexistence results of sign-changing solutions to a supercritical nonlinear problem. [9] [10] [11] Yuxin Ge, Monica Musso, A. Pistoia, Daniel Pollack. A refined result on sign changing solutions for a critical elliptic problem. [12] [13] [14] Lorena Bociu, Petronela Radu. Existence of weak solutions to the Cauchy problem of a semilinear wave equation with supercritical interior source and damping. [15] Eric A. Carlen, Süleyman Ulusoy. Localization, smoothness, and convergence to equilibrium for a thin film equation. [16] [17] [18] Sergey Degtyarev. Classical solvability of the multidimensional free boundary problem for the thin film equation with quadratic mobility in the case of partial wetting. [19] Minhajul, T. Raja Sekhar, G. P. Raja Sekhar. Stability of solutions to the Riemann problem for a thin film model of a perfectly soluble anti-surfactant solution. [20] Lihua Min, Xiaoping Yang. Finite speed of propagation and algebraic time decay of solutions to a generalized thin film equation. 2018 Impact Factor: 1.143 Tools Metrics Other articles by authors [Back to Top]
Hilbert transform Partial Function definiendum $H: (\mathbb C\to\mathbb C)\to(\mathbb C\to\mathbb C)$ definiendum $H(f):=y\mapsto \frac{1}{\pi}\cdot\mathcal P\int_{-\infty}^\infty\frac{f(x)}{y-x}\,\mathrm dx$ Discussion $(H(f))=-f$ The Hilbert transform commutes with the Fourier transform up to a simple factor and is an anti-self adjoint operator relative to the duality pairing between $L^p(\mathbb R)$ and the dual space $L^q(\mathbb R)$. It is also used in the Kramers–Kronig relation/Sokhotski–Plemelj theorem to express the imaginary part of an analytic function in terms of its real part (or the other way around). This works because they get “mixed up” in the Cauchy integral formula which introduces a factor of $\frac{1}{i}$. The principal value is taken to push the complex line integral on the real line. Reference Wikipedia: Hilbert transform
I am looking at the problem of a coupled oscillator, whereby we have 3 springs connected between two walls in the following way: wall, then spring (k), then mass (m), then spring (2k), then mass (m), then spring (k), then wall. I have calculated the characteristic frequencies (I think) by using the fact that we will have $$m \dfrac{d^2x}{dt^2}=-kx+2k(y-x)=-3kx+2ky$$ and $$m\dfrac{d^2y}{dt^2}=-ky-2k(y-x)=-3ky+2kx.$$ Thus we have in a new coordinate system $$m\pmatrix{\ddot{X}\\\ddot{Y}}=k\pmatrix{-3&-2\\2&-3}\pmatrix{x\\y}.$$ Evaluating the matrix in the previous expression I obtain Eigenvalues $\mu_1=-5$ and $\mu_2=-1$. The the unit Eigenvectors will be $$\hat{e}_1=\dfrac{1}{\sqrt{2}}\pmatrix{1\\1}$$ and $$\hat{e}_2=\dfrac{1}{\sqrt{2}}\pmatrix{1\\-1}.$$ So in the new coordinate system I have $$\pmatrix{\ddot{X}\\\ddot{Y}}=\dfrac{k}{m}\pmatrix{\mu_1&0\\0&\mu_2}\pmatrix{X\\Y}.$$ This implies that $$\ddot{X}=-\dfrac{5k}{m}X$$ and $$\ddot{Y}=-\dfrac{k}{m}Y.$$ Solving these equations we obtain $$X=Asin(\omega_1t+\alpha)$$ and $$Y=Bsin(\omega_2t+\beta),$$ where $\omega_1=\sqrt{\dfrac{5k}{m}}$ and $\omega_2=\sqrt{\dfrac{k}{m}}$. Now I need to find the characteristic frequency and the mode of vibration. (Please do not give any answers as this is a homework assignment that I would like to solve myself.) Q1. Am I correct in thinking that the characteristic frequency is simply $f=\omega/2\pi$, or am I misunderstanding what the characteristic frequency is? Q2. What is the mode of vibration? I understand what this is for a longitudinal wave oscillating between two fixed points, but I think this system is a transverse one, and I am struggling to think what the modes of vibration could be. Please can someone verify whether my understanding of what the characteristic frequency is is correct, and if not could you explain what it is. Also, can someone explain what the mode of vibration is in a transverse system like this one.
Many popular indices such as the S&P 500, Nasdaq 100 and the Wilshire 5000 are value weighted - a company’s weight in the index is proportional to its market-capitalization (price shares outstanding). This post explores alternative weighting methods and their implications. The Nasdaq Given previous posts on the Nasdaq, I decided to test alternative weighting schemes using only stocks traded on the Nasdaq exchange. For all schemes, I did the following: 1) Only include stocks that were traded at the end of the previous month 2) Adjust for delisting returns to prevent survivorship bias 3) Rebalance all portfolios on the first day of the month 4) Weights are constant over any given month The weight under scheme , for company , at time , is denoted Value Weight and Equal Weight Value weights and equal weights are by far the most used weighting schemes. Value puts the most weight on the largest companies and is calculated as follows:\begin{equation}w_{i,t}^{val}=\frac{mkt ~ cap_{i,t}}{\sum\limits_{j\in J} mkt ~cap_{j,t}}\end{equation}where represents the set of all stocks trading at the end of the previous month. Some possible reasons for using value weights are that the largest companies are less sensitive to idiosyncratic risk and have the most liquid shares. Equal weights are exactly what you would expect, giving each stock equal weight! They are calculated as: \begin{equation} w_{i,t}^{eq}=\frac{1}{J} \end{equation} As we will see below, equal weights can avoid some of the “momentum” built into value weights. Weighting by Inverse Volatility Suppose you wanted to put the most weight on the least volatile stocks. Let denote the standard deviation of stock ’s returns over the previous 12 months. I calculated inverse volatility weights as: \begin{equation} w_{vol,i}=\frac{vol_i^{-1}}{\sum\limits_{j\in J} vol_j^{-1}} \end{equation} Given that Sharpe ratio (see the post on Betting Against Winners) is decreasing in volatility, this might be a good way to construct a high Sharpe ratio index. The problem is, this is more like a trading strategy than an index. Unlike value weights, it doesn’t give a good indication of how the economy as a whole is performing. Weighting by Beta Suppose you wanted to put the most weight on the stocks that were the most responsive to movements in the S&P 500 (seems very risky!). Run the following regression for each stock : \begin{equation} r_{i,t}=\alpha+\beta^i r_{m,t} + \epsilon_{i,t} \end{equation} where is stock ’s returns and is the market return, as measured by the S&P 500 index. To calculate beta weights, I ran this regression over the previous 12 months using monthly data, and computed weights as follows: \begin{equation} w_{\beta,i}=\frac{\|\beta^i\|}{\sum\limits_{j\in J} \|\beta^j\|} \end{equation} Again, this seems more like a trading strategy than an index, but I thought including a high volatility strategy would be a good point of comparison to the inverse volatility weighting scheme. Results Imagine investing 100 dollars in each of these 4 indices in 1996, the plot below shows the performance of these portfolios: Recall, we are only looking at stocks traded on the Nasdaq index. The value weighted and beta weighted portfolios seem to pick up the technology “bubble”, while the inverse volatility weighted and equal weighted do not. This makes sense - as the “bubble” stocks became more “overvalued”, their market capitalization increased, so a value weighted index would put more weight on these stocks. Similarly, these stocks were very sensitive to overall market movements (recall their extreme daily volatility), so these stocks had high betas. Below I compare the performance of these portfolios the S&P 500 over the same period: Unsurprisingly, the inverse volatility weighted portfolio has the highest Sharpe ratio (it’s approximate because it uses ). I was surprised, however, that despite its high average returns, the beta weighted portfolio has a lower Sharpe ratio than the equal weighted portfolio. As for why the equal weighted outperformed the value weighted - my guess is it may be related to the small stock effect (and this will be the topic of a future post). Update: 9/4/2018 When re-reading this post, a recent paper came to mind: Replicating Anomalies, by Hou, Xue and Zhang. (SSRN Link). The paper shows that using value weights and NYSE breakpoints makes many anomalies insignificant. It turns out weights are more important than I ever could have expected when I originally wrote this post!
Hi I am trying to prove this $$ I:=\int_0^{\pi/4}\log\left(\tan\left(x\right)\right)\, \frac{\cos\left(2x\right)}{1+\alpha^{2}\sin^{2}\left(2x\right)}\,{\rm d}x =-\,\frac{\pi}{4\alpha}\,\text{arcsinh}\left(\alpha\right),\qquad \alpha^2<1. $$ What an amazing result this is! I tried to write $$ I=\int_0^{\pi/4} \log \sin x\frac{\cos 2x}{1+\alpha^2\sin^2 2x}-\int_0^{\pi/4}\log \cos x \frac{\cos 2x}{1+\alpha^2\sin^2 2x}dx $$ and played around enough here to realize it probably isn't the best idea. Now back to the original integral I, we can possibly change variables $y=\tan x$ and re-writing the original integral to obtain $$ \int_0^{\pi/4}\log \tan x \frac{\cos 2x}{1+{\alpha^2}\big(1-\cos^2 (2x)\big)}dx=\int_0^1 \log y \frac{1-y^2}{1+y^2}\frac{1}{1+{\alpha^2}\big(1-(\frac{1-y^2}{1+y^2})^2\big)}\frac{dy}{1+y^2}. $$ Simplifying this we have $$ I=\int_0^1\log y \frac{1-y^2}{1+y^2}\frac{(1+y^2)^2}{(1+y^2)^2+4\alpha^2y^2}\frac{dy}{1+y^2}=\int_0^1\log y \frac{1-y^2}{(1+y^2)^2+4\alpha^2y^2}dy $$ Another change of variables $y=e^{-t}$ and we have $$ I=-\int_0^\infty \frac{t(1-e^{-2t})}{(1+e^{-2t})^2+4\alpha^2 e^{-2t}} e^{-t}dt $$ but this is where I am stuck...How can we calculate I? Thanks. Integrate by parts; then you get that $$I(\alpha) = \left [\frac1{2 \alpha} \arctan{(\alpha \sin{2 x})} \log{(\tan{x})} \right ]_0^{\pi/4} - \int_0^{\pi/4} dx \frac{\arctan{(\alpha \sin{2 x})}}{\alpha \sin{2 x}}$$ The first term on the RHS is zero. To evaluate the integral, expand the arctan into a Taylor series and get $$I(\alpha) = -\frac12 \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \alpha^{2 k} \int_0^{\pi/2} du \, \sin^{2 k}{u} = -\frac{\pi}{4} \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \binom{2 k}{k} \left (\frac{\alpha}{2} \right )^{2 k}$$ A little manipulation leads us to $$\alpha I'(\alpha) +I(\alpha) = -\frac{\pi}{4} \sum_{k=0}^{\infty} (-1)^k \binom{2 k}{k} \left (\frac{\alpha}{2} \right )^{2 k} = -\frac{\pi}{4} \frac1{\sqrt{1+\alpha^2}}$$ The LHS is just $[\alpha I(\alpha)]'$, so the solution is $$I(\alpha) = -\frac{\pi}{4} \frac{\operatorname{arcsinh}(\alpha)}{\alpha} $$ My first step is similar to Ron Gordon but then I took a different route. From integration by parts, the given integral can be written as: $$-\frac{1}{a}\int_0^{\pi/4} \frac{\arctan(\alpha \sin(2x))}{\sin (2x)}\,dx$$ Consider $$I(a)=\int_0^{\pi/4} \frac{\arctan(a \sin(2x))}{\sin (2x)}\,dx$$ Differentiate both the sides wrt $a$ to obtain: $$I'(a)=\int_0^{\pi/4} \frac{1}{1+a^2\sin^2(2x)}\,dx$$ Use the substitution $a\sin(2x)=t$ to obtain: $$I'(a)=\frac{1}{2}\int_0^a \frac{dt}{\sqrt{a^2-t^2}(1+t^2)}$$ Next use the substitution $t=a/y$ to get: $$I'(a)=\frac{1}{2}\int_1^{\infty} \frac{y}{\sqrt{y^2-1}(a^2+t^2)}\,dy$$ With yet another substitution which is $y^2-1=u^2$, $$I'(a)=\frac{1}{2}\int_0^{\infty} \frac{du}{u^2+a^2+1}$$ The final integral is trivial, hence: $$I'(a)=\frac{\pi}{4\sqrt{a^2+1}}$$ Integrate both sides wrt $a$ to get: $$I(a)=\frac{\pi}{4}\sinh^{-1}a+C$$ It is easy to see that $C=0$, hence with $a=\alpha$, $$-\frac{1}{a}\int_0^{\pi/4} \frac{\arctan(\alpha \sin(2x))}{\sin (2x)}\,dx=-\frac{\pi}{4\alpha}\sinh^{-1}\alpha$$ $\blacksquare$ I use on Ron's results from integrating by parts: $$\alpha I(\alpha) = - \int_0^{\pi/4} dx \frac{\arctan{(\alpha \sin{2 x})}}{\sin{2 x}},$$ As an alternative way to complete the problem, use the method differentiating under the integral sign: $$\frac{d}{d\alpha}(\alpha I(\alpha)) = -\int_{0}^{\pi/4}\frac{dx}{\alpha^2\sin^2{2x}+1}=-\frac{\pi}{4\sqrt{1+\alpha^2}}\\ \implies \alpha I(\alpha) = -\frac{\pi}{4}\int_{0}^{\alpha}\frac{d\tilde\alpha}{\sqrt{1+\tilde\alpha^2}} = -\frac{\pi}{4}\sinh^{-1}{\alpha}.$$
From the question https://www.physicsoverflow.org/32208, Mr Ryan Thorngren said in the answer that the the framing anomaly of the gravitational Chern-Simons action $$I(g)=\frac{1}{4\pi}\int_{M}\mathrm{Tr}(\omega\wedge d\omega+\frac{2}{3}\omega\wedge\omega\wedge\omega)$$ i.e. it changes under a twist of framing on $M$ by $I(g)\rightarrow I(g)+2\pi s$ with $s\in\mathbb{Z}$, is related with the group $H_{3}Spin(3)=\mathbb{Z}$. 1. What is this group $H_{3}Spin(3)$? 2. Why is it isomorphic to $\mathbb{Z}$? 3. How exactly is it related with the change of Pontryagin class under a change of framing on $M$? They also talked about $\Omega_{3}^{fr}=\mathbb{Z}_{24}$. 4. What exactly is this $\Omega_{3}^{fr}$?
I cannot claim to be an expert on AQFT, but the parts that I'm familiar with rely on local fields quite a bit. First, a clarification. In your question, I think you may be conflating two ideas: local fields ($\phi(x)$, $F^{\mu\nu}(x)$, $\bar{\psi}\psi(x)$, etc) and unobservable local fields ($A_\mu(x)$, $g_{\mu\nu}(x)$, $\psi(x)$, etc). Local fields are certainly recognizable in AQFT, even if they are not used everywhere. In the Haag-Kastler or Brunetti-Fredenhagen-Verch (aka Locally Covariant Quantum Field Theory or LQFT), you can think of algebras assigned to spacetime regions by a functor, $U\mapsto \mathcal{A}(U)$. These could be causal diamonds in Minkowski space (Haag-Kastler) or globally hyperbolic spacetimes (LCQFT). You can also have a functor assigning smooth compactly supported test functions to spacetime regions, $U\mapsto \mathcal{D}(U)$. A local field is then a natural transformation $\Phi\colon \mathcal{D} \to \mathcal{A}$ between these two functors. Unwrapping the definition of a natural transformation, you find for every spacetime region $U$ a map $\Phi_U\colon \mathcal{D}(U)\to \mathcal{A}(U)$, such that $\Phi_U(f)$ behaves morally as a smeared field, $\int \mathrm{d}x\, f(x) \Phi(x)$ in physics notation. This notion of smeared field is certainly in use in the algebraic constructions of free fields as well as in the perturbative renormalization of interacting LCQFTs (as developed in the last decade and a half by Hollands, Wald, Brunetti, Fredenhagen, Verch, etc), where locality is certainly taken very seriously. Now, my understanding of This post has been migrated from (A51.SE) unobservable local fields is unfortunately much murkier. But I believe that they are indeed absent from the algebras of observables that one would ideally work with. For instance, following the Haag-Kastler axioms, localized algebras of observables must commute when spacelike separated. That is impossible if you consider smeared fermionic fields as elements of your algebra. However, I think at least the fermionic fields can be recovered via the DHR analysis of superselection sectors. The issue with unobservable fields with local gauge symmetries is much less clear (at least to me) and may not be completely settled yet (though see some speculative comments on my part here).
I am trying to understand how a perturbation of a skew symmetric matrix by another skew symmetric matrix affects the dominant eigenvector corresponding to $\lambda=0$. Specifically, let $S$ be an $n \times n$ real-valued skew symmetric matrix , with $n$ being odd. Assume $S$ has eigenvectors $(e_1,...,e_n)$ and corresponding eigenvalues $(\lambda_1,...,\lambda_n)$, sorted such that $e_1=0$ (which is guaranteed because $n$ is odd). Thus, $e_1$ is the fixed point solution of $S$. Let $M$ be another skew symmetric matrix of size $n\times n$, with eigenvalues $(b_1,...,b_n)$ and eigenvalues $(\theta_1,...,\theta_n)$. For some small $0<\epsilon\ll1$, define: $B=S+\epsilon M$ Then $B$ is likewise skew symmetric, and can be thought of as the perturbation of $S$ by the matrix $M$. Can we approximate or calculate the leading eigenvector of $B$, corresponding to $\lambda=0$, using the eigenvalue distribution of $B$ and $M$? In other words, can we approximate how much the fixed point of $S$ changes when perturbed by $M$?
We have $$\sum_{k = 1}^n \cos \sqrt{k} = \frac{\cos 1 + \cos \sqrt{n}}{2} + \int_1^n \cos \sqrt{t}\,dt - \int_1^n \bigl(\lbrace t\rbrace - \tfrac{1}{2}\bigr)\frac{\sin \sqrt{t}}{2\sqrt{t}}\,dt.$$ The first term is bounded, and the second integral is $O(\sqrt{n})$ as one sees from the boundedness of $\sin \sqrt{t}$ and $\lbrace t\rbrace - \frac{1}{2}$. It thus remains to estimate \begin{align}\int_1^n \cos \sqrt{t}\,dt &= 2\int_1^{\sqrt{n}} u\cos u\,du \\&= 2u\sin u \biggr\rvert_1^{\sqrt{n}} - 2\int_1^{\sqrt{n}} \sin u\,du \\&= 2\sqrt{n}\sin \sqrt{n} - 2\sin 1 + 2\cos \sqrt{n} - 2,\end{align} which clearly is in $O(\sqrt{n})$. Generally, for $0 < \alpha < 1$, we obtain $$\sum_{k = 1}^n \cos k^{\alpha} = \frac{\cos 1 + \cos n^{\alpha}}{2} + \int_1^n \cos t^{\alpha}\,dt - \alpha\int_1^n p_1(t)t^{\alpha-1}\sin t^{\alpha}\,dt,\tag{$\ast$}$$ where $p_1(t) = \lbrace t\rbrace - \frac{1}{2}$. We can estimate the first integral substituting $u = t^{\alpha}$ and integrating by parts: \begin{align}\int_1^n \cos t^{\alpha}\,dt &= \frac{1}{\alpha} \int_1^{n^{\alpha}} u^{\frac{1}{\alpha}-1}\cos u\,du \\&= \frac{u^{1/\alpha-1}\sin u}{\alpha}\biggr\rvert_1^{n^{\alpha}} - \frac{1}{\alpha}\biggl(\frac{1}{\alpha}-1\biggr)\int_1^{n^{\alpha}} u^{\frac{1}{\alpha}-2}\sin u\,du.\end{align} The first term is $\frac{1}{\alpha}\bigl(n^{1-\alpha}\sin n^{\alpha} - \sin 1\bigr)$, and using $\lvert \sin u\rvert \leqslant 1$ we see that the remaining integral also belongs to $O(n^{1-\alpha})$. For the integral $$\alpha\int_1^n p_1(t)t^{\alpha-1}\sin t^{\alpha}\,dt,$$ we immediately obtain an $O(n^{\alpha})$ bound using the boundedness of $p_1$ and $\sin$. For $\alpha \leqslant \frac{1}{2}$, this is smaller than the bound on the other integral. For $\alpha > \frac{1}{2}$, we still have an $O(n^{\alpha})$ bound for the sum, which suffices to conclude that $$\sum_{n = 1}^{\infty} \frac{\cos k^{\alpha}}{k}$$ converges. But we can lower the bound using integration by parts: $$\int_1^n p_1(t) t^{\alpha-1} \sin t^{\alpha} = p_2(t)t^{\alpha-1}\sin t^{\alpha}\biggr\rvert_1^n - (\alpha-1)\int_1^n p_2(t)t^{\alpha-2}\sin t^{\alpha}\,dt - \alpha \int_1^n p_2(t)t^{2(\alpha-1)}\cos t^{\alpha}\,dt$$ where the first two terms on the right are bounded, and the last integral is elementarily bounded by $C\cdot n^{2\alpha - 1}$. Continuing integration by parts, we find that the last integral in $(\ast)$ belongs to $O(n^{1 - m(1-\alpha)})$ for every $0 < m \leqslant \frac{1}{1-\alpha}$, and then eventually we obtain $$\int_1^n p_1(t)t^{\alpha-1}\sin t^{\alpha}\,dt \in O(1),$$ so the sum belongs to $O(n^{1-\alpha})$ for every $\alpha \in (0,1)$ [this is also true for $\alpha = 0$ and $\alpha = 1$]. Noting that $\cos k^{\alpha} \geqslant \frac{1}{2}$ for $$\bigl(2m - \tfrac{1}{3}\bigr)\pi \leqslant k^{\alpha} \leqslant \bigl(2m + \tfrac{1}{3}\bigr)\pi,$$ we see that the bound above is sharp. For $n \approx \bigl((2m+\frac{1}{3})\pi\bigr)^{1/\alpha}$, we have $\Theta(n^{1-\alpha})$ successive terms that are $\geqslant \frac{1}{2}$, whence the partial sum must have had at least order $\ell^{1-\alpha}$ for some $\ell \leqslant n$.
Let $\mu_n$ be a sequence of positive radon measures on $\mathbb{R}^n$ weakly converging (as dual of continuous compactly supported functions) to a measure $\mu$. Assume that $f_n(z)$ is a sequence of positive, compacly supported functions such that: they are uniformily supported in a ball, i.e.: for every $n$ their support is contained in a ball $B_r(0)$ and they are uniformily bounded with $\lvert f_n(z)\rvert\leq C$. Moreover $f_n(z)\to 0$ for any $z\in \mathbb{R}^n$. Is it true that at least for a subsequence $n_j$, that we have: \begin{equation} \lim_{j\to\infty}\int f_{n_j}(z) d\mu_{n_j}(z)=0? \end{equation}
Question Suppose I observe a vector $\mathbf{x}=[X_1 \ldots X_n]$, where each $X_i=m_i+n_i$, with $n_i$ being an independent zero-mean Gaussian random variable with variance $\sigma^2$ (i.e. $n_i\sim\mathcal{N}(0,\sigma^2)$ i.i.d.) and $\mathbf{m}=[m_1 \ldots m_n]$ is an unknown vector with a known Euclidean length to the origin $l(\mathbf{m})=\\|\mathbf{m}\\|_2=\sqrt{\sum_{i=1}^n m_i^2}$. Suppose that the Euclidean length of the unknown vector $\mathbf{m}$ can equally likely be either $l_0$ or $l_1$, and, without loss of generality, $l_0<l_1$. Now, given the vector of noisy observations $\mathbf{x}$ I need to decide whether the length of $\mathbf{m}$ is $l_0$ or $l_1$. Note that I do not know (nor do I need to know for my answer) the values in $\mathbf{m}$, just the two possibilities for length. The intuitive procedure is to compute the Euclidean length of the observed vector $S=\sqrt{\sum_{i=1}^nX_i^2}$, and then select whichever of $l_0$ or $l_1$ is closer to $S$ (based on some threshold that is related to the probabilities of error that you are willing to tolerate). Since $S^2$ is a non-central chi-square random variable, one can (numerically) obtain (at least bounds) on the probabilities of error of this approach. However, is $S$ the "best" test statistic (in the sense of Neyman-Pearson optimality) for this problem? If it is not optimal, is a better test known? If $S$ is optimal, is there a proof of its optimality in the literature? My prior effort This seems like a problem that should be well-studied. In fact, it is related to the problem of non-coherent detection in communications theory, where $S^2$ is used by the square-law detector. John Proakis has a proof on pages 304-306 of "Digital Communications" (4th edition) that $S$ as defined is the N-P test statistic for when $n=2$ (detection of a complex-valued symbol with arbitrary phase offset). He essentially projects the problem into polar coordinate system, and takes the expectation of the test statistic over the uniform distribution for the angle. I tried a naive approach of extending Proakis' proof to $n$-spherical coordinate system (since $\mathbf{m}$ can be thought of a coordinate on an $n$-sphere with radius $l_k$) and computing the likelihood function (from N-P) by taking the expectation over the uniform distribution on all the angles. However, while I can get a closed form expression $n=3$ which yields the N-P test statistic defined by $S$, the integration gets really nasty for $n>3$. Perhaps there is something more clever that one can do, maybe using the circular symmetry of joint distribution of i.i.d. Gaussians. This question is related to my previous question.
I have the following system of equations that involve some rather complicated exponents: $$ \left\{\frac{\mathit{a} \sigma ^{41/34}}{\mathit{m}^{8/119} \mathcal{M}^{28/17}}=1,\frac{\mathit{b}\mathit{q}\mathit{m}^{2/3} \mathcal{M}^3}{\sqrt{\sigma }}=1,\frac{\mathit{f} \mathit{c} \mathit{m}^{436/357}}{\sigma ^{48/17} \mathcal{M}^{38/51}}=1\right\} $$ I'd like Mathematica to find the positive, real solutions to this equation for $\mathcal{M}$, $\sigma$, and $q$, ($a$, $b$, $c$, $m$, and $f$ remain undefined) and Solve is able to do this, but it literally takes ~30 minutes to solve the expression: FullSimplify[Abs[Cases[({sigma, M, q} /. Solve[expression, {sigma, M, q}]), _?(FreeQ[N[#, 16], Complex] && FreeQ[N[#, 16], Positive] &)][[1]]], Assumptions -> Join[assume, # > 0 & /@ coeffs]] I noticed if I make one simple change to the expression, multiplying the RHS by the denominators, i.e. $$ \left\{\mathit{a} \sigma ^{41/34}=\mathit{m}^{8/119} \mathcal{M}^{28/17},\mathit{b}\mathit{q}\mathit{m}^{2/3} \mathcal{M}^3 = \sqrt{\sigma },\mathit{f} \mathit{c} \mathit{m}^{436/357} = \sigma ^{48/17} \mathcal{M}^{38/51}\right\} $$ ...then the expression solves in about 30 seconds. Unfortunately, it is the first form that is spat out by my previous lines of code, and that form is not the same for each evaluation. I'm wondering why the first solve is so much more expensive than the second, and if there are strategies I can use when simplifying expressions to guarantee that Solve won't choke on them.
The cumulative probability distribution function $F_X(x)$ tells us how muchprobability mass there is to the left of $x$ or at $x$ for each $x$on the real line. (The choice of notation, though almost universally usedis truly dreadful for use in a classroom setting! How on earth does one read out aloud $F_X(x)$ or $P\{X\leq x\}$? F-sub-big X of little x? probability thatrandom variable $X$ is no larger than lower-case x?) Formally, the value of $F_X(x)$ is just $P\{X \leq x\}$.As Glen_b's comment says, you really should start by sketching the function$F_X(x)$ at the very least. When $X$ is a discrete random variable taking on values $x_1, x_2, \ldots$with probabilities $p_1, p_2, \ldots $ respectively, a little thought(instead of rote memorization of the definition) reveals that $F_X(x)$must be what can be described as a staircase function, increasing from$0$ to $1$ as $x$ increases, with steps of heights $p_1, p_2, \ldots$at points $x_1, x_2, \ldots$ etc. The function is discontinuousat each $x_i$, and is constant in each interval $[x_i, x_{i+1})$ (pleasebe sure to note the $[$ and $)$ in the description of the intervals).Note that $F_X(x_i)$ includes $p_i$ so that the value of $F_X(x)$ atthe point $x=x_i$ (where the function is discontinuous) is thevalue on the right. Since you are studying from a text intended forengineers, you might find this written as $F_X(x) = F_X(x^+)$.Thus,$$F_X(x) = P\{X \leq x\} = F_X(x^+) ~ \text{and} ~ P\{X < x\} = F_X(x^-).$$ In fact, for any random variable (not necessarily a discrete random variableor an integer-valued random variable as in Rusan's answer) and for any real numbers $a$ and $b$ such that $a \leq b$,$$\begin{align}P\{a < X \leq b\} &= F_X(b^+) - F_X(a^+) = F_X(b)-F_X(a),\tag{1}\\P\{a \leq X \leq b\} &= F_X(b^+) - F_X(a^-) = F_X(b) - F_X(a^-),\tag{2}\\P\{a \leq X < b\} &= F_X(b^-) - F_X(a^-) = F_X(b^-) - F_X(a^-),\tag{3}\\P\{a < X < b\} &= F_X(b^-) - F_X(a^+) = F_X(b^-)-F_X(a).\tag{4}\end{align}$$ For the special case when $b = a$, $(2)$ above becomes$$P\{X=a\} = F_X(a^+)-F_X(a^-),$$ that is, $P\{X=a\}$ is the jump(if any) in the value of $F_X(x)$ at $x=a$. If $F_X(x)$ is continuousat $x=a$, then $P\{X=a\}=0$. With this as prologue, note that your given $F_X(x)$ is a staircasefunction with jumps of $\frac 14, \frac 12, \frac 14$ at $x=-10, 30, 50$respectively; that is, $X$ takes on values $-10, 30, 50$ withprobabilities $\frac 14, \frac 12, \frac 14$ respectively, andonce you have that, the answers to the questions asked are easyto compute directly, or, if you prefer to read the $F_X(0^-)$etc off the graph that you have drawn as you apply $(1)$-$(4)$,that is fine too.
https://doi.org/10.1351/goldbook.C01032 The chemical potential of a substance Green Book, 2nd ed., p. 49 [Terms] [Book] PAC, 1994, PAC, 1996, Bin a mixture of substances B, C... is related to the Gibbs energy $G$ of the mixture by: \[\mu _{\text{B}}=(\frac{\partial G}{\partial n_{\text{B}}})_{T,p,n_{\text{C}\neq \text{B}}}\] where $T$ is the @T06321@, $p$ is the pressure and $n_{\text{B}}$, $n_{\text{C}}$, ... are the amounts of substance of B, C, ... . For a pure substance B, the chemical potential $\mu _{\text{B}}^{}$ is given by: \[\mu _{\text{B}}^{}=\frac{G^{}}{n_{\text{B}}}=G_{\text{m}}^{}\] where $G_{\text{m}}^{}$ is the molar Gibbs energy, and where the superscript attached to a symbol denotes the property of the pure substance. The superscript $^{\unicode{x29B5}}$ or $^{\unicode{x26ac}}$ attached to a symbol may be used to denote a @S05927@. See also: standard chemical potential Sources: Green Book, 2nd ed., p. 49 [Terms] [Book] PAC, 1994, 66, 533. ( Standard quantities in chemical thermodynamics. Fugacities, activities and equilibrium constants for pure and mixed phases (IUPAC Recommendations 1994)) on page 535 [Terms] [Paper] PAC, 1996, 68, 957. ( Glossary of terms in quantities and units in Clinical Chemistry (IUPAC-IFCC Recommendations 1996)) on page 966 [Terms] [Paper]
First of all, a topological space that is locally homeomorphic to a Euclidean space is locally compact, cf. this, and locally connected as those properties are local and preserved by homeomorphisms (and they hold for Euclidian spaces). Then $M$ second-countable $\Longrightarrow\ M$ paracompact with countably many connected components: This is Lemma 1.9 p.9 Foundations of Differentiable Manifolds and Lie Groups, Frank W. Warner or Thm 4.77 p.110 Introduction to Topological Manifolds, John Lee: The authors show in the first part that $M$ is countable at infinity/$\sigma$-compact, i.e. $M = \bigcup_{n\in \mathbb{N}} K_n$ with $K_n \subset \overset{\circ}{K}_{n+1}$ compact. Let now $ \mathcal{U}:=(U_i)_{i\in I}$ be any open cover of $M$: $K_0$ is compact and covered by the family open subsets $(U_i\cap \overset{\circ}{K}_1)_{i\in I}$ so one can extract a finite subsequence. (Similarly for $K_1$) Then by induction: for any $n \geq 2,\enspace \left(U_i\cap (\overset{\circ}{K}_{n+1}\backslash \overline{K}_{n-2})\right)_{i\in I}$ is an open cover of $\overline{K}_n\backslash K_{n-1}$, from which one extracts a finite cover. By such a construction, one will cover every $K_n$ and hence all of $M$; the open subset of the cover are of the form $U_i\cap (\overset{\circ}{K}_{n+1}\backslash \overline{K}_{n-2})$ and thus included in $U_i$ (refinement of a cover) and any point is in one of the $\overset{\circ}{K}_n$ which is covered by finitely many $U_i$ (locally finite cover), i.e. $M$ is paracompact. Edit: this last implication is not well justified here because a cover of $K_{n+1}$ or $K_{n-1}$ may also intersect with points of $K_n$ ... although the intuition is correct... In this post, one uses the fact that a second countable space is separable, together with local connectedness to show that $M$ has countably many connected components. $M$ paracompact with countably many connected components $\Longrightarrow\ M$ second-countable : Since $M$ has countably many connected components $(C_k)_{k\in \mathbb{N}}$, it suffices to show that each of these components has a countable basis ($\mathbb{N}\times \mathbb{N}$ is still countable). The idea is to use the fact that for any chart $(U,\varphi)$ (from definition of $M$ locally Euclidian), $ U$ is second-countable as it is homeomorphic to a subset of $\mathbb{R}^n$ which is second-countable. One shows again that $M$ is countable at infinity / $\sigma$-compact but starting with the hypothesis that $M$ paracompact. It is similar as above, cf. Theorem 12.11 p.38 of Topology and Geometry, Glen E. Bredon. Finally with $\sigma$-compactness, $M$ is covered by countably many open subsets, each of which are second-countable. I found these notes which also answers precisely this question link
I'm studying Linear Algebra for the second time, using Hoffmann & Kunze. Currently I'm trying to prove the following theorem: Theorem 7. If $A$ is an $n \times n$ matrix, then $A$ is row-equivalent to the $n \times n$ identity matrix if and only if the system of equations $A\vec{x} = \vec{0}$ has only the trivial solution $\vec{x} = \vec{0}$. The given proof in the textbook is, I find, obscure, and I don't understand it at all, so I came up with my own proof. However, it depends on $det(A) \neq 0$ and I'm not sure if this assumption can be made. I rewrote the theorem as: Let $A$ be a square matrix. Prove that $A$ ~ $I_n$ if and only if $A\vec{x} = \vec{0}$ has only the trivial solution. My proof: Let $det(A) \neq 0$. Then $rref(A) = I_n$. If $A\vec{x} = \vec{0}$, then $det(A\vec{x}) = 0 \rightarrow \vec{x} = \vec{0}$, so $\vec{0}$ is indeed the only solution. The problem with my proof is, even if it's correct, the authors only introduce determinants around page $200$ and this theorem is from chapter $1$. So really I'm looking for a proof that does not rely upon determinants.
Input Convex Neural Network (ICNN) is a neural network that has convexity in a subset of its input. In this post, I give a brief review of the model in the context of continuous action reinforcemen learning. Motivation Many inference problems are non-convex, especially when their objective functions are represented by deep neural networks. They are generally hard to solve efficiently because of potentially exponential number of stationary points with parsimonious saddle points. A local optimization method such as Gradient Descent can only guarantee the convergence to a stationary point, not necessarily the true global solution. To better motivate the introduction of ICNN, let us suppose a given task requires solving the following optimization problems for a sufficiently large number of times with varying input values of x and non-convex $f$. If $y$ is discrete, solving for $y^\star$ is $O(\|Y\|)$ where an exhaustive search is computationally feasible. If $y \in \|Y\|^d$ with a large $d$, or $y$ is continuous, we are quite out of luck with an exhaustive search. One may also consider discretizing continuous-valued $y$ but it requires good heuristics that are problem dependent. What is ICNN Below, we define ICNN convex fully in input and partially in input. Fully Input Convex Neural Networks (FICNN) FICNN is a feedforward neural network with additional constraints as below. where $\phi_i$ represents activation functions in layer i, $z_i$ represents the $i^{th}$ activated layer, $y$ constitutes the input and $f$ the function approximation we wish to learn, $\phi’$ represents its derivative with respect to the function’s inputs. Finally, $\mathbb{C}$ is the set of all convex functions and $W_i^{(y)}y_i$ is the term for passthrough layer. A sufficient condition for the network to be convex in $(\mathbf{y})$) is that $\phi$ is a convex and non-decreasing function and that all $W^{(z)}_{1:K}$ are non-negative. Though this certainly restricts the expressiveness of the network and the choice of activation functions, in practice, we claim this does not come up as an insurmountable challenge. With weights forced to being non-negative, the network loes its ability to learn an identity mapping between layers, thus affecting the non-linear expressive power of the standard neural networks. To remedy the restricted representation power, the paper proposed adding pass-through layers that directly connect the inputs $y$ to each hidden layer to learn an identity mapping. It is also worth noting that there are no constraints on the weights of the pass-through layers. Regarding the choice of activation functions, popular activation functions such as Rectified Linear Units(ReLU) or Scaled Exponential Linear Units (SeLU) and max pooling all fit into our criteria. Partially Input Convex Neural Networks (PICNN) Unlike FICNN that is convex over both $x, y$, we consider a generalization of FICNN that has convexity in a subset of the inputs, $y$. Hence, Partial Input Convex Neural Network (PICNN). where $u_i$, $z_i$ denote the hidden layers for the non-convex and convex channels respectively, $\odot$ denotes the Hadamard product and the function. PICNN in principle has larger expressiveness than FICNN. Continuous Control with Deep Reinforcement Learning Continuous control reinforcement learning problems require solving the following inference problems repeatedly and sequentially: where $s$ denotes state, $a$ action and $Q(s,a)$ action value function. Notice the negative sign applied to keep the problem in a consistent formulation as in the introduction. The inference is almost intractable as it is. There are two state-of-the-art methods generally considered for the problem: Deep Deterministic Policy Gradient (DDPG) and Normalized Advantage Functions (NAF). DDPG directly parametizes a deterministic policy with its corresponding performance function, defined as expected culumulative return given a policy: . With a deterministic policy, we can replace with . The paper showed the Deterministic Policy Gradient converges to the stochastic policy gradient. And the deterministic Policy Gradient can be estimated even more easily and since we are choosing actions using a policy, one can sidestep the $\arg\min Q$ bottleneck. NAF decomposes the Q-values into value function of the state and an advantage function such that $Q(s,a) = V(s) + A(s,a)$. NAF constraints $A(s,a)$ such that it is concave quadratic with respect to action and hence always gives a closed form solution for $\arg\min_{a} -Q(s,a)$. Notice $V(s)$ does not depend on $a$ and therefore can be ignored. ICNN shares the central idea of making $Q$ or its components convex with respect to action. When Is ICNN Useful? The way ICNN attempts to tackle the difficult inference task above is simple: it aims to formulate the problem as an approximate yet convex problem, hoping the solution to the approximate formulation is acceptably good. Though not explicitly mentioned in the original paper of ICNN, I think ICNN is applicable when one of the following two conditions hold. The first is when a given problem has some hidden convexity intrinsic to it–hidden in the sense the problem is convex with respect to some proper subset of the input which is hard to know a priori. The seond is when multi-modality is not really a problem–a strong global optimum (the mode) exists.
In science we learn about single, double, and sometimes triple bonds. From a quick search I have found up to sextuple bonds. Is there a maximum bond order? If yes/no, what causes this? Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.Sign up to join this community In the paper that you've mentioned$^{\ast1}$, there is something described, which is called effective bond order (EBO).$$\mathrm{EBO} = \frac{1}{2}\left(\sum_{i=1}^N \eta_{b,i}-\eta_{a,i}\right)$$... with $\eta_b = 2 − x$ as the occupation number of an occupied orbital and $\eta_a = x$ as the occupation number of the corresponding antibonding orbital. (In multireference quantum chemistry methods, the resulting averaged orbitals can have occupation values of $0 \leq x \leq 2$) In 2013 Ruiperez et al. calculated the effective bond order for several homo- and heteroatomic dimers.$^{\ast2}$ Based on their very high quality calculations, the [...] results show that the effective bond order (EBO) of the MoU dimer (5.5) is higher than that for the tungsten dimer (5.2), known to date as the molecule with the highest EBO. So somewhat arbitrarily, MoU seems to be one of the best candidates for the highest bond order. It seems that I have answered my own question: The maximum bond order achieved between two atoms in the periodic table is thus six [sextuple] and is represented by the Mo and W diatoms. From Reaching the Maximum Multiplicity of the Covalent Chemical Bond, by Bjrn O. Roos, Antonio C. Borin, and Laura Gagliardi
Implementing Impractical Digital Filters This blog discusses a problematic situation that can arise when we try to implement certain digital filters. Occasionally in the literature of DSP we encounter impractical digital IIR filter block diagrams, and by impractical I mean block diagrams that cannot be implemented. This blog gives examples of impractical digital IIR filters and what can be done to make them practical. Implementing an Impractical Filter: Example 1 Reference [1] presented the digital IIR bandpass filter shown in Figure 1(a) where H NO( z) is a FIR notch filter and variable K is a constant. Figure 1: Reference [1]'s proposed IIR bandpass filter: (a) high-level depiction; (b) detailed signal path depiction. The idea behind this bandpass filter is to subtract the output of the notch filter from the X( z) input to produce a narrow bandpass filter output. The Figure 1(a) filter's z-domain transfer function is: $$H_{\mathsf{BP}}(z) = \frac{Y(z)}{X(z)} = \frac{1}{1 + KH_{\mathsf{NO}}(z)}\tag{1}$$where the bandpass filter's H NO( z) term is a 2nd-order FIR notch filter defined by $$H_{\mathsf{NO}}(z) = 1-2\mathsf{cos}(\omega_c)z^{-1}+z^{-2}.\tag{2}$$ In Eq. (2) variable $\omega_c$, where $0<{\omega_c}<\pi$, represents the center-frequency of the FIR filter's notch measured in radians/sample. Based on Eq. (2) the proposed bandpass filter can be depicted as shown in Figure 1(b). The Figure 1 filter seems like a plausible design but is in fact impossible to implement (i.e., it's impractical). Here's why. The Problem The trouble with the above bandpass filter is that it contains a feedback signal path containing no delay element (a delay-less signal path) as shown by the bold arrows in Figure 2. Figure 2: Time-domain depiction of the impractical bandpass filter with the delay-less signal path highlighted in bold. Upon the arrival of an x( n) input sample, in order to compute a y( n) output sample we first need to compute the w( n) feedback sample. But to compute w( n) we need y( n)! That delay-less signal path means the Figure 2 filter cannot be implemented. The Solution: Algebra to the Rescue As it turns out, algebraic expansion of the H BP( z) equation produces a new filter z-domain transfer function that can be implemented. As shown in Appendix A, we can plug Eq. (2) into Eq. (1) to obtain a practical transfer function of: $$H_{\mathsf{BP,Practical}}(z) = \frac{\alpha}{1+2K{\alpha}\mathsf{cos}(\omega_c)z^{-1}+K{\alpha}z^{-2}}\tag{3}$$ where ${\alpha}=1/(1+K)$, and $K≠-1$. Unlike the Figure 2 filter, the Eq. (3) bandpass filter can be implemented with no problem as shown in Figure 3. Figure 3: Practical (implementable) version of the impractical Figure 1 bandpass filter. Implementing an Impractical Filter: Example 2 A second example of an impractical filter, proposed in Reference [2], is the narrowband digital notch filter shown in Figure 4. Figure 4: High-level depiction of Reference [2]'s proposed IIR notch filter. where C( z) is a notch filter and variable a is a constant. The z-domain transfer function for the proposed Figure 4 IIR filter, derived in Appendix B, is: $$H_\mathsf{N}(z) = \frac{Y(z)}{X(z)} = \frac{(1+\alpha)C(z)}{1+{\alpha}C(z)}.\tag{4}$$The C( z) function is a traditional 2nd-order narrowband IIR notch filter whose transfer function is: $$C(z) = \frac{1-2\mathsf{cos}(\omega_c)z^{-1}+z^{-2}}{1-2\rho\mathsf{cos}(\omega_c)z^{-1}+\rho^2z^{-2}}\tag{5}$$ where variable $\omega_c$, $0<{\omega_c}<\pi$, is the center-frequency of the filter's notch measured in radians/sample, and $\rho$ is the radii of the conjugate poles' z-plane locations. Based on Eq. (5) the proposed notch filter can be depicted in more detail as shown in Figure 5. Figure 5: Detailed signal path depiction of the H N ( z) notch filter. The Figure 5 bandpass filter contains a delay-less feedback signal path as shown by the bold arrows in Figure 6. Figure 6: Time-domain depiction of the impractical H N ( z) notch filter with the delay-less signal path highlighted in bold. That delay-less signal path makes the Figure 6 filter impractical because to compute a y( n) output sample we first need to compute the u( n) feedback sample. But to compute u( n) we need y( n). Similar to the Figure 2 filter, the Figure 6 filter cannot be implemented. Again, as we did in Example 1, algebraic expansion of the H N( z) equation produces a new filter z-domain transfer function that can be implemented. Substituting Eq. (5)'s C(z) expression into Eq. (4), as shown in Appendix C, we produce a practical transfer function of: $$H_{\mathsf{N,Practical}}(z) = \frac{1-2\mathsf{cos}(\omega_c)z^{-1}+z^{-2}}{1-2\frac{\rho+\alpha}{1+\alpha}\mathsf{cos}(\omega_c)z^{-1}+\frac{\rho^2+\alpha}{1+\alpha}z^{-2}}\tag{6}$$ where ${\alpha} ≠ -1$. This new H N,Practical( z) function is a traditional 2nd-order IIR filter that can be implemented using the block diagram shown in Figure 7. Figure 7: Practical (implementable) version of the impractical Figure 4 notch filter. Conclusion The bottom line of this blog is: When you encounter a digital network's block diagram that is impractical (i.e., impossible to implement because it contains a delay-less feedback signal path), algebraic analysis of the network's z-domain transfer function may well lead to a new and equivalent block diagram that is possible to implement. Postscript: December, 2017 In my original July 2016 version of this blog I presented the following Figure 8 network and claimed it to be impractical. Figure 8: Reference [3]'s spectrum analysis network. But, as was pointed out to me by Prof. Zsolt Kollár (Budapest University of Technology and Economics, Hungary) I was mistaken. If we draw the details of the resonators in Figure 8 we have Figure 9, and that network can indeed be implemented due to the delay elements in each path. Figure 9: Expanded version of the Figure 8 network. Upon arrival of an x[ n] input sample we compute all the outputs of the z multipliers, shift those products through their delay elements, and wait for the next k x[ n+1] input sample. No problem. As the cowboys in Texas would say, "The Figure 8 system works real fine." References [1] S.Engelberg, “Precise Variable-Q Filter Design,” IEEE Signal Processing Magazine, Sept. 2008, pp. 113–119. [2] S. Pei, B, Guo, and W. Lu, "Narrowband Notch Filter Using Feedback Structure", IEEE Signal Processing Mag azine, May 2016, pp. 115-118. [3] G. Peceli, G. Simon, "Generalization of the Frequency Sampling Method", IEEE Instr. and Meas. Conference, Brussels Belgium, June 1996. Appendix A: Derivation of Eq. (3) We derive Eq. (3) by starting with the above Eq. (1), repeated here as: $$H_{\mathsf{BP}}(z) = \frac{1}{1 + KH_{\mathsf{NO}}(z)}.\tag{A-1}$$ Substituting the notch filter's H NO( z) from Eq. (2) into Eq. (A-1) yields: $$H_{\mathsf{BP}}(z) = \frac{1}{1 + K[1-2\mathsf{cos}(\omega_c)z^{-1}+z^{-2}]}$$ $$= \frac{1}{(1 + K)-2K\mathsf{cos}(\omega_c)z^{-1}+Kz^{-2}}.\tag{A-2}$$ Next, dividing Eq. (A-2)'s numerator and denominator by (1+ K) we produce the desired expression for the Figure 1 proposed bandpass filter's equivalent and practical transfer function of: $$H_{\mathsf{BP,Practical}}(z) = \frac{\alpha}{1+2K{\alpha}\mathsf{cos}(\omega_c)z^{-1}+K{\alpha}z^{-2}}\tag{A-3}$$ where ${\alpha}$ = 1/(1+ K), and K ≠ -1. Appendix B: Derivation of Eq. (4) The following derivation was graciously supplied to me by Reference [2]'s co-author Bo-Yi Guo. Master's student Guo derived Eq. (4) by redrawing Figure 4 as shown in Figure B1. Figure B1: Reference [2]'s proposed IIR notch filter. From Figure B1, we start with $$W(z) = X(z)-Y(z)\tag{B-1}$$ and $$V(z) = X(z)+{\alpha}W(z).\tag{B-2}$$ Replacing W( z) in Eq. (B-2) with Eq. (B-1), we have: $$V(z) = X(z) +{\alpha}(X(z)–Y(z)).\tag{B-3}$$ The Y( z) output can be expressed as: $$Y(z) = V(z)C(z).\tag{B-4}$$ Replacing V( z) in Eq. (B-4) with Eq. (B-3) gives: $$Y(z) = [X(z) + {\alpha}(X(z)–Y(z))]C(z)$$ $$= (1+{\alpha})X(z)C(z)-{\alpha}Y(z)C(z).\tag{B-5}$$ Rearranging Eq. (B-5) yields our desired H N( z) transfer function as: $$H_\mathsf{N}(z) = \frac{Y(z)}{X(z)} = \frac{(1+\alpha)C(z)}{1+{\alpha}C(z)}.\tag{B-6}$$ Appendix C: Derivation of Eq. (6) We derive Eq. (6) by starting with the above Eq. (4) repeated here as: $$H_\mathsf{N}(z) = \frac{(1+\alpha)C(z)}{1+{\alpha}C(z)}\tag{C-1}$$ or after dividing through by C( z), $$H_\mathsf{N}(z) = \frac{1+\alpha}{\frac{1}{C(z)}+\alpha},\tag{C-2}$$ recalling from Eq. (5) that $$H_{\mathsf{N,Practical}}(z) = \frac{1-2\mathsf{cos}(\omega_c)z^{-1}+z^{-2}}{1-2\rho\mathsf{cos}(\omega_c)z^{-1}+\rho^2z^{-2}}.\tag{C-3}$$ Next we rewrite Eq. (C-2) by separating C( z) into its numerator and denominator parts as: $$H_\mathsf{N}(z) = \frac{1+\alpha}{\frac{1}{C_\mathsf{numer}(z)/C_\mathsf{denom}(z)}+\alpha}= \frac{1+\alpha}{\frac{C_\mathsf{denom}(z)}{C_\mathsf{numer}(z)}+\alpha}$$ $$= \frac{(1+\alpha)C_\mathsf{numer}(z)}{C_\mathsf{denom}(z)+{\alpha}C_\mathsf{numer}(z)}.\tag{C-4}$$ Substituting Eq. (C-3)'s numerator and denominator terms into Eq. (C-4) we have: $$H_{\mathsf{N}}(z) = \frac{(1+\alpha)[1-2\mathsf{cos}(\omega_c)z^{-1}+z^{-2}]}{1-2\rho\mathsf{cos}(\omega_c)z^{-1}+\rho^2z^{-2}+\alpha[1-2\mathsf{cos}(\omega_c)z^{-1}+z^{-2}]}$$ $$= \frac{(1+\alpha)[1-2\mathsf{cos}(\omega_c)z^{-1}+z^{-2}]}{(1+\alpha)-2(\rho+\alpha)\mathsf{cos}(\omega_c)z^{-1}+(\rho^2+\alpha)z^{-2}}\tag{C-5}$$ which is Reference [2]'s Eq. (15). Finally, dividing Eq. (C-5)'s numerator and denominator by (1+$\alpha$) we produce the desired equivalent and practical expression for the proposed notch filter's transfer function of: $$H_{\mathsf{N,Practical}}(z) = \frac{1-2\mathsf{cos}(\omega_c)z^{-1}+z^{-2}}{1-2\frac{\rho+\alpha}{1+\alpha}\mathsf{cos}(\omega_c)z^{-1}+\frac{\rho^2+\alpha}{1+\alpha}z^{-2}}\tag{C-6}$$ where ${\alpha} ≠ -1$. Previous post by Rick Lyons: An Astounding Digital Filter Design Application Next post by Rick Lyons: Should DSP Undergraduate Students Study z-Transform Regions of Convergence? To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments. Registering will allow you to participate to the forums on ALL the related sites and give you access to all pdf downloads.
Nothing would happen for a while, assuming we're talking about Sun just disappearing overnight. You can even demonstrably witness what happens with 12 hours of no sunlight in the Equator (or during polar night in the Arctic, for that matter). The primary reason for this is the vast amount of water. We can, however, give some ballpark estimate [] for the speed of process. If the Sun would just disappear Earth would begin to cool at a rate of roughly 300 W/m². Now, Earth is not an ideal blackbody and temperature is not uniform, but in terms of estimates, well, close enough. This is equivalent to having a 1 mm layer of water drop 1 Kelvin in temperature in ca. 14 seconds (1 mm of water per square meter = 1 kg of water). Tropics have an ocean mixing layer of roughly 1000 meters and average water temperature (in that layer) of something like 20 C, so if we exclude currents and atmospheric convection it would take $\frac{1000 \text{m}}{0.001 \text{m}} \cdot 14 \text{ s/K} \cdot 20 \text{K} \sim 10$ years before tropical oceans would begin to freeze over. Now, this is not what would happen, but it gives some insight on the speed of the process. In reality things are much more complicated: Solar energy received in Equator is partly transferred in ocean and atmospheric convection to polar latitudes. If we assume hurricane-ish type energy transfer, we could be looking at something like extra 10-100 W per square meter of ocean, add ocean currents for 200-300 watts extra, and we'd still end up with a time window that is closer to few years rather than few weeks. Now, on land, on the other hand. We might be talking of an equivalent of several meters of water. So it would take perhaps a month, considering energy transfer by the atmosphere, to turn land into inhabitable snowfield and another month or two to make it uncomfortable for the Nordics or Canadians. This is naturally talking in averages, so locally it could be better...or much worse. In terms of freezing, enthalpy of fusion of water is roughly 333 kJ per kilogram, much higher compared to heat capacity of 4.2 kJ/kg$\cdot$K. Therefore, assuming no geothermal energy, it would still take in order of centuries to have an ice-sheet in tropics that would measure in kilometers. [] Correct to few orders of magnitude...If lucky, then an order of magnitude.
Search Now showing items 1-1 of 1 Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (Elsevier, 2014-09) Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, \|y\| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Differences This shows you the differences between two versions of the page. Both sides previous revision Previous revision quantum_canonical_partition_function [2016/03/03 13:32] nikolaj quantum_canonical_partition_function [2016/03/03 13:33] (current) nikolaj Line 13: Line 13: Generally, material physics of finite (i.e. non-zero) temperature derives its macroscopic relations from small scale considerations. All observables are essentially determined by the relation between the possible microscopic states and their energy, which makes evaluation of the partition function possible. This is why the computation of energy levels $\varepsilon_r$ or dispersion relations $\hbar\omega({\bf k})$ are of central importance. Quantum mechanically, this requires computing eigenvalues of the Hamiltonian. Generally, material physics of finite (i.e. non-zero) temperature derives its macroscopic relations from small scale considerations. All observables are essentially determined by the relation between the possible microscopic states and their energy, which makes evaluation of the partition function possible. This is why the computation of energy levels $\varepsilon_r$ or dispersion relations $\hbar\omega({\bf k})$ are of central importance. Quantum mechanically, this requires computing eigenvalues of the Hamiltonian. - >todo: $\langle A\ rangle := \dfrac{\mathrm{tr}(\mathrm e^{-\beta H}A)}{\mathrm{tr}(\mathrm e^{-\beta H})} = \dfrac{1}{Z(\beta)} \mathrm{tr}({\mathrm e}^{-\beta H}A) $ + >todo: Ensemble average + >$\langle A\ rangle_H := \dfrac{\mathrm{tr}(\mathrm e^{-\beta H}A)}{\mathrm{tr}(\mathrm e^{-\beta H})} = \dfrac{1}{Z(\beta)} \mathrm{tr}({\mathrm e}^{-\beta H}A) $ ----- -----
Find if the series converges or diverges, $a_n=\sum_{n=1}^\infty\frac{1}{1+\ln (n)}$. Comparing it with another series $b_n=\frac{1}{\ln(n)}$. Dividing both the series and taking limits, we get $\lim_{n\to\infty}\frac{\ln(n)}{1+\ln(n)}$. Since it is the $\infty/\infty$ form, applying H'opitals rule, we get, $\lim_{n\to\infty}\frac{1/n}{1/n}=1$. Now, $\lim_{n\to\infty}\frac{1}{ln(n)}=0, \Rightarrow b_n$ converges $\Rightarrow a_n$ converges. But the answer is, Comparing it with $\frac{1}{n}$(divergent harmonic series) we get,$\lim_{n\to\infty}\frac{n}{1+\ln(n)}=\lim_{n\to\infty}\frac{1}{1/n}=\lim_{n\to\infty}n=\infty \Rightarrow a_n $ diverges, what is wrong with my comparison? Your mistake is when you say that: $$\lim_{x \to \infty} b_n = 0$$ implies that $\sum b_n$ is convergent. That doesn't hold in general. Note further that $n > \ln(n)$ so $\frac{1}{n} < \frac{1}{\ln(n)}$. So by comparison $\sum \frac{1}{\ln(n)}$ is divergent. So as mentioned in the comment above, your comparison is fine, the conclusion is just that the series is divergent.
I am asked to verify that the sequence $\left(\frac{1}{6n^2+1}\right)$ converges to $0$: $$\lim \frac{1}{6n^2+1}=0.$$ Here is my work: $$\left\|\frac{1}{6n^2+1}-0\right\|<\epsilon$$ $\frac{1}{6n^2+1}<\epsilon$, since $\frac{1}{6n^2+1}$ is positive $$\frac{1}{\epsilon}<6n^2+1$$ $$\frac{1}{\epsilon}-1<6n^2$$ $$\frac{1}{6\epsilon}-\frac{1}{6}<n^2$$ At this point, I am stuck. I'm not sure if I take the square root of both sides if I then have to deal with $\pm\sqrt{\frac{1}{6\epsilon}-\frac{1}{6}}$. That doesn't seem right. The book provides the answer: $$\sqrt{\frac{1}{6\epsilon}}<n$$ But I don't understand (1) what happened to the $\frac{1}{6}$, and (2) why there's not a +/- in front of the square root. Any help is greatly appreciated.
I have the following differential equation $$y''-xy=0\,.$$ By using $y=\sum\limits_{n=0}^{\infty} a_n x^n$, I show that $$(n+2)(n+1)a_{n+2}=a_{n-1}$$ for $n=1,2,...$, and $a_2=0$. (I differentiate the sum and by plugging it into the differential equation which is zero I get that result.) My question is how can I show that solutions converge for all $x \in \mathbb{R} $ since I cannot apply the ratio test. There are two linearly independent solutions $f_0$ and $f_1$. We assume that the first one $f_0(x)$ satisfies the conditions $f_0(0)=1$ and $f_0'(0)=0$, whilst the second one $f_1(x)$ satisfies the conditions $f_1(0)=0$ and $f_1'(0)=1$. Any solution $y=f(x)$ of this differential equation is written as the sum $$f(x)=f(0)\,f_0(x)+f'(0)\,f_1(x)\text{ for all }x\in\mathbb{R}\,.$$ If we show that $f_0$ and $f_1$ are analytic on the whole $\mathbb{R}$, then we are done. Now, from the result you obtain regarding the relations of the coefficients, you have that $$f_0(x)=\sum_{k=0}^\infty\,u_{k}\,x^{3k}\text{ and }f_1(x)=\sum_{k=0}^\infty\,v_{k}\,x^{3k+1}\,,$$ where $$u_k:=\frac{1}{\prod\limits_{r=1}^k\,\big((3k)(3k-1)\big)}=\frac{1}{(3k)_{k,3}\,(3k-1)_{k,3}}$$ and $$v_k:=\frac{1}{\prod\limits_{r=1}^k\,\big((3k+1)(3k)\big)}=\frac{1}{(3k+1)_{k,3}\,(3k)_{k,3}}\,.$$ Here, the falling factorial power $(x)_{n,h}$ for $x,h\in\mathbb{C}$ and $n\in\mathbb{Z}_{\geq 0}$ is defined by $$(x)_{n,h}:=x\,(x-h)\,(x-2h)\,\ldots\,\big(x-(n-1)h\big)\,,$$ with $(x)_{0,h}:=1$. To show that the radii of convergence of $f_0$ and $f_1$ are both infinite, we need to verify that $$\limsup_{k\to\infty}\,\sqrt[3k]{u_k}=0\text{ and }\limsup_{k\to\infty}\,\sqrt[3k+1]{v_k}=0\,.$$ However, this follows from the observation that $$(3k+1)_{k,3}\geq k!\,,\,\,(3k)_{k,3}\geq k!\,,\text{ and }(3k-1)_{k,3}\geq k!$$ for all $k=0,1,2,\ldots$, and the well known fact that $$\sqrt[k]{k!}\to\infty\text{ as }k\to\infty\,.$$ It turns out that $$f_0(x)=\frac{\sqrt[3]{3}^2\,\Gamma\left(\frac{2}{3}\right)}{2}\,\text{Ai}(x)+\frac{\sqrt[6]{3}\,\Gamma\left(\frac{2}{3}\right)}{2}\,\text{Bi}(x)$$ and $$f_1(x)=-\frac{\sqrt[3]{3}\,\Gamma\left(\frac{1}{3}\right)}{2}\,\text{Ai}(x)+\frac{\Gamma\left(\frac{1}{3}\right)}{2\,\sqrt[6]{3}}\,\text{Bi}(x)\,,$$ where $\text{Ai}$ and $\text{Bi}$ are Airy functions, and $\Gamma$ the usual gamma function. Alternatively, $$\text{Ai}(x)=\frac{1}{\sqrt[3]{3}^2\,\Gamma\left(\frac23\right)}\,f_0(x)-\frac{1}{\sqrt[3]{3}\,\Gamma\left(\frac13\right)}\,f_1(x)$$ and $$\text{Bi}(x)=\frac{1}{\sqrt[6]{3}\,\Gamma\left(\frac23\right)}\,f_0(x)+\frac{\sqrt[6]{3}}{\Gamma\left(\frac13\right)}\,f_1(x)\,.$$
In Carr and Madan (2005), the authors give sufficient conditions for a set of call prices to arise as integrals of a risk-neutral probability distribution (See Breeden and Litzenberger (1978)), and therefore be free of static arbitrage (via the Fundamental Theorem of Asset Pricing) These conditions are: Call spreads are non-negative Butterflies spreads are non-negative In the case that we have a full range of call prices: $C(K)$ is monotically decreasing $C(K)$ is convex Or if $C(K)$ is twice differentiable: $$C'(K) \leq 0 \tag1$$ $$C''(K) \geq 0\tag2$$ Carr and Madan do not mention the following constraints, thought they may be implied (?): $$C(K) \geq 0\tag3$$ $C(0)$ is equal to the discounted spot price $\tag 4$ Other authors do mention the constraints (1-4) together. For example Fengler and Hin (2012) call these the "standard representation of no-arbitrage constraints" In Reiswich (2010), the author presents the following condition: $$\frac{\partial P}{\partial K} \geq 0\tag{5a}$$ Or equivalently, via Put-Call Parity: $$\frac{\partial C}{\partial K} \geq \frac{C(K) - e^{-r\tau}S}{K}\tag{5b}$$ Reiswich claims that (5) is stricter than what is implied by (1-4) (i.e. there are sets of call prices which satisfy (1-4) but not (5)). Is this really true? If so, how do we reconcile this with Carr and Madan's claim of sufficiency? Edit: Alternately, if (5) must hold is a no-arbitrage setting, and if (1-4) are sufficient, then how do we derive (5) from (1-4)?
I have seen some examples of metrics on $\mathbb R$ which make it incomplete, like $d(x,y)=\|e^{-x}-e^{-y}\|$ and $d(x,y)=\arctan x - \arctan y$ where $1,2,3,\ldots$ is a Cauchy sequence which doesn't converge. I am curious about what other examples there are. Is it always infinity which turns out to be 'missing' or are there metrics with Cauchy sequences that do not converge but remain bounded? That depends on how much leeway one has in specifying the metric. Consider for example the fact that there exists a bijective map $\phi\colon\mathbb R \to \mathbb R \setminus \mathbb Q$. If you define the metric on $\mathbb R$ by setting $d(x,y)=\|\phi(x)-\phi(y)\|$, it will be a metric alright, and clearly not complete. If you restrict yourself to a bounded part of the reals,so some closed interval $[A,B]$ then all equivalent metrics on that part are defined on a compact space, so are always complete. This is the reason that an equivalent non-complete metric on $\mathbb{R}$ has its non-completeness at infinity, as it were. Cauchy sequences always remain bounded: If $\epsilon = 1$, there is $N$ such that for all $n,m>N$ $$d(x_n,x_m)<1$$ Therefore, $d(x_n,0)\leq d(0,x_{N+1})+d(x_{N+1},x_n)<d(0,x_{N+1})+1$, for $n>N$ and a fixed. This means that the whole sequence is inside the ball with center $0$ and radius $\max(d(0,x_{N+1})+1, d(x_1,0),d(x_1,0),...,d(x_N,0))$. In your examples, $1,2,3,...$ is bounded.
The normal ordening is a way to say: ''we throw away the zero-point energy'' (since it becomes infinity and wa say we only look at energy-differences), or to put it in the words of A. Zee: ''Create before you annihalite''. The chronological ordening comes in when you calculate the Feynman propagator (also called the Green's function), which is basically the "thing" that you are trying to calculate. If you would look at the Green's function, you know that if $L$ represents your Liouville-equation, for example in the case of the wave-equation:$$L=\frac{\partial^2}{c^2\partial t^2}-\Delta^2$$, the equation for the Green's function $G(r\|r')$ hence becomes:$$LG(r\|r')=\delta(r-r')$$, which we can use to solve the differential equation represented by $L$ by using a convolution (see the Green's function link). Now with the Feynman-propagator you do the same thing, and it follows an equivalent equation of the above (only now with some generalized delta-function since we don't werk at equal times) A book that goes deeper into Wick's theorem is the one of W. Greiner (Field Quantization), which is widely spread on the net. On pages 225-233 Wick's theorem is discussed. After Wick's theorem he starts with QED, this also gives an example of the use of Wick's theorem. Now I don't know what you hope to find in the answers since the proof is basically an immense ammount of bookkeeping, but basically you make an order-expansion of your product of operators. Wick-theorem states that:$$T(\hat{A_1}\hat{A_2}\hat{A_3}\hat{A_4}\cdots)=:\hat{A_1}\hat{A_2}\hat{A_3}\hat{A_4}\cdots:+\text{normal ordening with single contractions}+\text{normal ordening with double contractions}+\text{normal ordening with triple contractions}+...$$. This series goes on untill you run out of stuff to contract (note that you need an even number of operators), so basically op to $n/2$, with $n$ the number of operators. To use Wick's theorem, one goes order by order. Starting from the zero-th order (which is basically normal-ordening, then going first order (one contraction) and so on. Usually some orders and combinations will simply turn out to be zero (for example when you contract two creation of annihalation operators, or when you contract operators of different fields).You can also exclude terms by using the conservation of momentum and energy (some contractions violate this one). If you want to compare with Feynman-diagrams, then you need to look at the scattering matrix $S$. This one is developped into a series: $$\hat{S}=\mathbb{I}+\sum\limits_{n=1}^\infty \hat{S}^{(n)}$$,(which is basically the power-series decomposition of your unitary-time evolution operator) for each order, you add a vertex and hence more operators, and thus go to a higher order Feynman-diagram.
Consider the following three phenomena. Stein's paradox: given some data from multivariate normal distribution in $\mathbb R^n, \: n\ge 3$, sample mean is not a very good estimator of the true mean. One can obtain an estimation with lower mean squared error if one shrinks all the coordinates of the sample mean towards zero [or towards their mean, or actually towards any value, if I understand correctly]. NB: usually Stein's paradox is formulated via considering only one single data point from $\mathbb R^n$; please correct me if this is crucial and my formulation above is not correct. Ridge regression: given some dependent variable $\mathbf y$ and some independent variables $\mathbf X$, the standard regression $\beta = (\mathbf X^\top \mathbf X)^{-1} \mathbf X^\top \mathbf y$ tends to overfit the data and lead to poor out-of-sample performance. One can often reduce overfitting by shrinking $\beta$ towards zero: $\beta = (\mathbf X^\top \mathbf X + \lambda \mathbf I)^{-1} \mathbf X^\top \mathbf y$. Random effects in multilevel/mixed models: given some dependent variable $y$ (e.g. student's height) that depends on some categorical predictors (e.g. school id and student's gender), one is often advised to treat some predictors as 'random', i.e. to suppose that the mean student's height in each school comes from some underlying normal distribution. This results in shrinking the estimations of mean height per school towards the global mean. I have a feeling that all of this are various aspects of the same "shrinkage" phenomenon, but I am not sure and certainly lack a good intuition about it. So my main question is: is there indeed a deep similarity between these three things, or is it only a superficial semblance? What is the common theme here? What is the correct intuition about it? In addition, here are some pieces of this puzzle that don't really fit together for me: In ridge regression, $\beta$ is not shrunk uniformly; ridge shrinkage is actually related to singular value decomposition of $\mathbf X$, with low-variance directions being shrunk more (see e.g. The Elements of Statistical Learning3.4.1). But James-Stein estimator simply takes the sample mean and multiplies it by one scaling factor. How does that fit together? Sample mean is optimal in dimensions below 3. Does it mean that when there are only one or two predictors in the regression model, ridge regression will always be worse than ordinary least squares? Actually, come to think of it, I cannot imagine a situation in 1D (i.e. simple, non-multiple regression) where ridge shrinkage would be beneficial... On the other hand, sample mean is alwayssuboptimal in dimensions above 3. Does it mean that with more than 3 predictors ridge regression is always better than OLS, even if all the predictors are uncorrelated (orthogonal)? Usually ridge regression is motivated by multicollinearity and the need to "stabilize" the $(\mathbf X^\top \mathbf X)^{-1}$ term. Update:Yes! See the same thread as above. There are often some heated discussion about whether various factors in ANOVA should be included as fixed or random effects. Shouldn't we, by the same logic, always treat a factor as random if it has more than two levels (or if there are more than two factors? now I am confused)? Update:? Update: I got some excellent answers, but none provides enough of a big picture, so I will let the question "open". I can promise to award a bounty of at least 100 points to a new answer that will surpass the existing ones. I am mostly looking for a unifying view that could explain how the general phenomenon of shrinkage manifests itself in these various contexts and point out the principal differences between them.
[Previous] \| [Session 46] \| [Next] E. D. Reese, J. E. Carlstrom, J. J. Mohr, G. P. Holder (U.Chicago), M. K. Joy (NASA/MSFC), L. Grego (SAO), S. Patel (U.Alabama), W. L. Holzapfel (U.C.Berkeley) With analysis of Sunyaev-Zel'dovich effect and x-ray data it is possible to determine the distances to galaxy clusters. Cosmic microwave background (CMB) photons inverse Compton scatter off of the hot intracluster medium (ICM) causing a small distortion in the CMB spectrum, the Sunyaev-Zel'dovich (SZ) effect. The SZ effect is proportional to the pressure integrated along the line of sight ~\int n e T e d\ell and appears as a small (~1~mK) decrement at centimeter wavelengths. The ICM also emits at x-ray wavelengths with a signal ~\int n e 2 T e 1/2 d\ell for bremsstrahlung radiation. Taking advantage of the different dependencies on the electron density, n e, one can solve for the size of the cluster. One then has a ruler with which to measure the angular diameter distance to the cluster, independently of the distance ladder. We present our method and distances for a sample of our imaged SZ effect clusters. We use our interferometric cm-wave observations of the SZ effect at the BIMA and OVRO observatories and archival ROSAT x-ray data. We perform a maximum likelihood fit to the SZ and x-ray data simultaneously using an isothermal spherical \beta-model. These fits yield information about the shape and density of the ICM. We can then use these clusters to determine the Hubble constant. The largest source of observational uncertainty in H \circ, ~0%, is the x-ray determined electron temperature. Possible systematics are discussed, the largest being the assumption of isothermality for the ICM. The Chandra observatory will greatly reduce the uncertainty in T e and be able to detect temperature gradients as well as provide greater spatial resolution. ER acknowledges support by NASA GSRP Grant NGT-50173. If the author provided an email address or URL for general inquiries, it is a s follows: reese@piglet.uchicago.edu [Previous] \| [Session 46] \| [Next]
I know that the definition of an infinite series is the limit as $n \to \infty$ of its partial sums. $$\underbrace{\sum_{n=0}^{\infty} \ \ a_n}_\text{Infinite Series}\ \stackrel{\text{def}}{=} \ \lim_{n \to \infty} \ \underbrace{\sum_{i=0}^{n} \ a_i}_\text{Partial Sums}$$ But then how do you define a finite series? If a finite series is defined like I've done below, is that not recursive, as you're essentially defining something in terms of itself, as the partial sum of a finite series, is itself a finite series? $$\underbrace{\sum_{i=0}^{n} \ \ a_i}_\text{Finite Series}\ \stackrel{\text{def}}{=} \ \lim_{\gamma\ \to\ n} \ \underbrace{\sum_{i=0}^{\gamma} \ a_i}_\text{Partial Sums}$$ At some point we have to define these series as the sum of terms of a infinite sequence $\{ a_n \} _{n=1}^{\infty}$, for the infinite series (where we view the partial sums as the finite sums of terms of this infinite sequence), and as the sum of terms of a finite sequence $\{ a_i \} _{i=1}^{n}$ for the finite series (and I'm not sure how to view the partial sums here, perhaps as finite sums of terms of the finite sequence where $i \leq n$). I'm not sure how to do this. In short I'm having trouble with the definition of a finite series, and I'm having trouble making the connection between finite sequences and the definition of finite series, and how the two (sequences and series) relate to each other.
Say that a function $\,f: \mathbb{R}^{n\times n} \to \mathbb{R}$ is given in an element-wise form as $$f(A) = \sum_{i,j} A_{ij} F_{ij}.$$ When is this function convex on the whole set $\mathbb{R}^{n\times n}$? While trying to find an answer to this question I've stumbled upon a definition of quadratic forms based on matrices, where the condition for convexity is straightforward: $f$ is convex if and only if $F$ (the matrix defining the quadratic form element-wise) is positive semi-definite. Is there a correspondence with the case of a more general matrix function like above?
This problem is closely related to Hook Length Formula. The formula gives a way how to compute the number of fillings of a matrix $m\times n$ with distinct numbers $1, 2, \ldots, mn$ in such a way that all rows and all columns are increasing. Whenever you have such a matrix, you can translate it into a process of $m$ zeros jumping to the right to the position of $n$ ones. And because zeros sounds too bad, from now on I will call them heroes :-) At first, imagine the heros on the left of the matrix, on every row there is a hero at the "starting" position. Now you are saying the sequence $1, 2, 3, \ldots, mn$ and every time a hero has the pronounced number in front of them, he jumps behind it. So on the end, heroes are on the right of the matrix. Now, to make the original jumping process out of it, you just shear it to the right and flatten columns. So for instance$$\matrix{1 & 2 \\ 3 & 4}$$is translated into "the first hero makes two jumps and then the second hero makes two jumps", so it corresponds to $$0011\to0101\to0110\to1010\to1100$$and matrix$$\matrix{1 & 3 \\ 2 & 4}$$is translated into "First hero, second hero, first hero, second hero", so$$0011\to0101\to1001\to1010\to1100.$$ The statement of hook length formula (see the Wiki) is very pretty but unfortunately there is no simple proof of it. For our case it is$$(mn)\;\cdot\;!1!2!\cdots(m-1)!\over n!(n+1)!(n+2)!\cdots(n+m-1)!$$Notice that I am using $m$ zeros and $n$ ones instead of $n-m$ so you have to substitute it to get the exact answer.
The beginning of a measure theory book I am read introduces Cartesian products. It contains the following statements that confuse me: If $\{X_\alpha\}_{\alpha \in A}$ is an indexed family of sets, their Cartesian product $\prod_{\alpha \in A}X_\alpha$ is the set of all maps $f: A \rightarrow \bigcup_{\alpha\in A}X_\alpha$ s.t. $f(\alpha) \in X_{\alpha} \, \forall \alpha \in A$. It should be noted, and then promptly forgotten, that when $A = \{1,2\}$, the definition of $X_1 \times X_2$ (the set of all ordered pairs where the first term comes from $X_1$ and the second from $X_2$) is set theoretically different from the present definition $\prod^2_{i=1}X_i$. Indeed the latter concept depends on mappings, which are defined in terms of the former. I'm not understanding what distinction the book is trying to make about Cartesian products and the set of ordered pairs. Is it just that Cartesian products is defined in terms of functions?

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