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Here are two constructions, showing that the expectation of the min can get down to around $1+1/\sqrt{n}$, and on the other hand can get up to a value approaching $2$ as $n\to\infty$. Maybe someone can do better! Let the random variables be $X_1, X_2,\dots, X_n$. (1) Aiming to make the expected minimum small; the idea is to let each value occur once, with a small perturbation to achieve the required probability that any two variables are the same. Fix $p>0$ ($p$ will get small as $n$ gets large). Let $\tilde{X}_1, \dots, \tilde{X}_n$ be a uniform permutation of $\{1,\dots, n\}$.Independently, let $Y$ be drawn uniformly from $\{1,\dots,n\}$. Finally (and independently of $(\tilde{X}_i)$ and $Y$), choose $B_1, B_2, \dots, B_n$to be independent and each taking value $1$ with probability $p$, and $0$ with probability $1-p$. Now define $X_i$ by setting\begin{equation}X_i=\begin{cases}\tilde{X}_i& \text{ if } B_i=0\\Y&\text{ if } B_i=1\end{cases}.\end{equation} If we choose $p$ such that $p^2 + 2p(1-p)/n = 1/n$, then it's easy to check thatfor each pair $i,j$, the variables $X_i$ and $X_j$ are equal with probability $1/n$. (Namely, $X_i=X_j$ whenever $B_i=B_j=1$, or $B_i=1, B_j=0$ with $Y=\tilde{X}_j$,or $B_i=0, B_j=1$ with $Y=\tilde{X}_i$.)From this it's easy to get (using symmetry) that all the pairwise distributions are as required. The relevant $p$ is around $1/\sqrt{n}$. Then the probability that the minimum is 1 is approximately $1-p$, and more generally the distribution of the minimum is approximately geometric with parameter $1-p$, and so with mean $1/(1-p)$. Overall we obtain that the expected minimum is $1+1/\sqrt{n}$ plus smaller order terms. (2) Aiming to make the expected minimum large; the idea is to let each value occur either 0 or 2 times, again with a small perturbation to get the pairwise distributions precisely right. For convenience let $n$ be even. Let $Y_1, Y_2, \dots, Y_{n/2}$ be $n/2$ distinct values chosen uniformly at random from $\{1,2,\dots,n\}$. Now let $\tilde{X}_1, \dots, \tilde{X}_n$ be a uniform permutation of $Y_1, Y_1, Y_2, Y_2, Y_3, Y_3, \dots, Y_{n/2}, Y_{n/2}$. So each value occurs either 0 or 2 times in the sequence $(\tilde{X}_i)$. The marginal distributions are correct, but the pairwise distributions are not quite correct; we have $\tilde{X}_i=\tilde{X}_j$ with probability $1/(n-1)$, while we want this probability to be $1/n$. To correct this, let $Z_1, \dots, Z_n$ be a uniform permutation of $\{1,2,\dots, n\}$. Now with probability $1/n$, let $X_i=Z_i$ for all $i$, and with probability $1-1/n$, let $X_i=\tilde{X}_i$ for all $i$. Again it's not hard to check that the pairwise distributions are correct. Now the probability that 1 appears in the sequence is approximately $1/2$; more generally, the distribution of the min of the sequence is approximately geometric with probability $1/2$ (since the numbers of occurrences of $1,2,3,\dots$ are approximately independent). The expectation of the min is then $2+o(1)$ as $n\to\infty$. Looking to improve this: note that to get the pairwise distributions right, the number of occurrences of any value $k$ should have mean approximately $1$ and variance approximately $1$ as $n\to\infty$. However, maybe we could improve on construction (2) by one in which the number of occurrences of $1,2,3,\dots$ are not asymptotically independent; again, to get the pairwise distributions right, they have to be asymptotically uncorrelated but this does not imply independence, so maybe by playing around one could get it so that, say, the number of occurrences of 1 and the number of occurrences of 2 tend to be 0 simultaneously (which will help to push up the expected minimum).
The following question has puzzled me for some time: Let $(\Omega,\Sigma)$ be a nonempty, measurable space. Does there necessarily exist a probability measure $\mu:\Sigma\to[0,1]$? If there exists a nonempty measurable set $A$ such that no nonempty subset of $A$ is measurable (an atom), we can simply let $\mu(B)=1$ if $A\subseteq B$ and $\mu(B)=0$ otherwise. So the problem is only interesting if the $\sigma$-algebra has not atoms. This rules out every countably generated $\sigma$-algebra. An example of a $\sigma$-algebra that has no atoms but supports a probability measure is $\{0,1\}^\kappa$ for $\kappa$ uncountable, which we can endow with the coin-flipping probability measure.
I decided to start answering this question by building a galaxy (well, a model of a galaxy, but it sounds cooler the first way). A lot of research has already been done in this area, specifically, in density wave theory, which explains the winding arms of spiral galaxies. Before we begin, here’s your 60-second introduction to the structure of spiral galaxies. A spiral galaxy can be thought of as a conglomerate of three separate structures: The galactic disk, the flattened section of the galaxy lying on the galactic plane. It is in turn composed of the thin disk and the thick disk, containing relatively younger and older stars, respectively. The disk also contains spiral arms. The galactic bulge, a dense region in the center of the galaxy which extends further out of the galactic plane than the galactic disk does. The galactic halo, a roughly spherical set of stars, gas, globular clusters, and dark matter than surrounds the galaxy. The stellar component of this can be found in the galactic spheroid. The halo has, on average, a lower mean density of gas and stars, though it is rich in dark matter. Now we can construct a model of a spiral galaxy, starting with a gravitational potential denoted by $\Phi(R,\theta,z,t)$, where $R$ is the radius along the plane, $\theta$ is the azimuthal angle, $z$ is the vertical distance above the plane, and $t$ is time. We’re working in cylindrical coordinates, but we’re also taking time into account. Over hundreds of millions of years, spiral galaxies rotate, and stars move in and out of dense and not-so-dense regions. I’m not going to display any results like this because my Mathematica skills are currently limited, but it’s not too hard to do. We could choose a rather simple model for our galaxy. Power law radial density models are the simplest, where the density in the plane is$$\rho(R)=\rho_0\left(\frac{R}{R_0}\right)^{-\alpha}\tag{1}$$where $R_0$ is a reference radius and $\alpha$ is an real number. $\alpha=2$ fits many observed rotation curves, to a decent degree of accuracy. Adding in the $z$-component is then simple; this is just multiplied by one of two possible factors:$$\exp\left(\frac{-z^2}{z_0^2}\right)\quad\text{or}\quad\text{sech}^2\left(\frac{z}{z_0}\right)\tag{2a, 2b}$$with scale height height $z_0$. This seems simple enough, and the corresponding potential can be calculated without too much trouble. However, current data has led to better models using a sort of exponentially decreasing radial fit for the galactic potential. I’m strongly basing my choice here on information I gathered in this answer, using data from Antoja et al. (2011). Their equation for the potential is of the form$$\Phi(R,\theta,t)=\sum_mA_m(R)\cos(m\theta-m\theta_0-\phi_m(R)-\Omega_p t)\tag{3}$$which is the sum of terms of indices $m$. $A_m(R)$ is the radial amplitude, $\theta_0$ is some reference angle, $\phi(R)$ is a function that determines how the arms wind, and $\Omega_p$ is the pattern speed. I’ll neglect $\Omega_p$ for now, and look only at the density at $t=0$. Antoja et al. decided to keep only the $m=2$ term. Normally, the $m=0$ and $m=2$ terms dominate (with occasionally a smaller $m=4$ term providing richer structure), but this model is simpler. They used the simple radial profile$$A_2(R)=-A_{sp}Re^{R/R_{\Sigma}}\tag{4a}$$with scale length $R_{\Sigma}$. $\phi(R)$ is, in general, a little more complicated. Their choice (denoted $g(R)$) is fairly standard:$$g(R)=\left(\frac{2}{N\tan i}\right)\ln\left(1+\left(\frac{R}{R_{sp}}\right)^N\right)\tag{4b}$$where $i$ is the inclination of the arms and $R_{sp}$ is another scale length. We assume that $N$ is large. In reality, $N\to\infty$, but taking $N=100$ is good enough. All that remains is to insert their parameters. For many, there are ranges, so I’ve picked ones that are roughly average, for the Milky Way:$$\begin{array}{\|c\|c\|}\hline \text{Parameter}&\text{Best-fit value}\\\hline A_{sp} & 1000\text{ }[\text{km s}^{-1}]^2\text{ kpc}^{-1}\\\hline R_{\Sigma} & 2.5\text{ kpc}\\\hline i & 14^{\circ}\\\hline R_{sp} & 3.1\text{ kpc}\\\hline \theta_0 & 74^{\circ}\\\hline \Omega_p& 15\text{-}30\text{ km s}^{-1}\\\hline\end{array}$$Now we go to Mathematica. The density, $\rho$, can be found by Poisson’s equation:$$\nabla^2\Phi=4\pi G\rho\tag{5}$$where $G$ is the gravitational constant. It is much easier to go from potential to density than density to potential, and all we have to do for the former is use Mathematica’s Laplacian operator. Here’s the code I used, with all constants scaled to SI units: G = 6.67410^(-11) Asp = 10001000000/(310^(19)) rsig = 2.5310^19 inc = 60 (degrees) Points = 100 rsp = 3.1 310^19 theta0 = 74 (degrees) (Omega =22.53.240810^(-17)) A[r_] := AsprExp[-r/rsig] g[r_] := (2/PointsTan[inc Degree])Log[1 + (r/rsp)^Points] potential[r_, theta_, z_] := -A[r]Cos[2(theta - theta0) - g[r]]10^5 density[r_, theta_, z_] := Evaluate[(1/(4PiG))* Laplacian[potential[r, theta, z], {r, theta, z}, "Cylindrical"]] flatDensity[r_, theta_] := density[r, theta, 0] RevolutionPlot3D[ Evaluate[flatDensity[r, theta]], {r, 3310^19, 10310^19}, {theta, 0, 2*Pi}, Mesh -> None, ColorFunction -> "DarkRainbow"] There are a few things to note here. First, be careful to put the value for $i$ in degrees using the Degree option; trigonometric functions in Mathematica assume the value is in radians otherwise. Second, I’ve had to make two modifications to make the output visible. I changed the inclination to $60^{\circ}$ to make the winding clearer, and I multiplied to density (actually, the potential, as well) by a factor of $10^5$. Without that, RevolutionPlot3D and the other operations really choke. When looking at the output, then, be mindful of that factor of five orders of magnitude. Side view of the density graph. Top view of the density graph. The spiral structure should be quite evident here. However, there are two perturbing details. The first is that there is explosive growth near the center. I’ve deliberately truncated the inner radius to $3\text{ kpc}$, which is where the spiral structure really starts. A different density profile is needed there. At radii similar to the Sun’s orbital radius, our density profile is sufficient. Eventually, at large enough $R$, $\rho$ actually becomes less than zero, but we should treat that as an unphysical result and assume that the profile is truncated once $\rho=0$. This happens around $\sim8\text{ kpc}$, indicating that we need to add a value for $m=0$. The exact fitting for that can be handwaved a little, but in the spiral arms, it appears that the results match local mean densities to within a few orders of magnitude ($\sim10^{-18}\text{-}10^{-20}\text{ kg/m}^3$, which isn’t too bad). Let’s say, then, that we add this $m=0$ term to avoid negative densities. If we want $\rho>0$ out to about $12\text{ kpc}$, then we need it to be around $\sim2.45\times10^{-18}\text{ kg/m}^3$. Again, at smaller radii, this will produce larger-than-usual densities, but it is necessary to avoid unphysical results. How much of this is stars, though, and how much is gas, dust, and other objects? I’d be comfortable with approximating the stellar density as roughly our figure from above. Dark matter follows a roughly spherical halo distribution, often described by a Navarro-Frenk-White (NFW) profile. The disk density distribution, then, describes stars and other luminous matter, as well as gas and dust. From what I’ve read (see e.g. this Physics Stack Exchange question and answers), roughly 75-90% of baryonic matter in the disk is in the form of stars and related objects, which I’m really comfortable with rounding up to 100%. Stars have different masses, distributed, in general, according to an initial mass function (IMF). I’ve talked about this in more detail before, and I suspect that nobody’s too eager for me to rehash the necessary sections. Essentially, though, you calculate the total number of stars over a given mass range and then calculate the total mass of all of those. You then scale that to match the total stellar mass of the galaxy, which is done by integrating the density function over the relevant area. Doing so would require multiplying our current expression by some sort of exponentially decaying function of $z$, which I suggested earlier. Again, the specifics vary; take your pick. Once we’ve done this, we have a value of $n(R,\theta,z)$, the number density of stars at a certain point. To figure out what a stellar population looks like in a given area, simply calculate the mean inter-particle distance, $\langle r(R,\theta,z) \rangle$:$$\langle r(R,\theta,z) \rangle\propto(n(R,\theta,z))^{-1/3}\tag{6}$$At radii similar to the Sun’s orbital radius, we should see separations on the order of a few light-years. You can pretty easily, then, create a small group of stars with the same mean separation (at large radii, number density is approximately constant), and simply add some random perturbations. Distribute the masses according to an IMF, and voilà!
@JosephWright Well, we still need table notes etc. But just being able to selectably switch off parts of the parsing one does not need... For example, if a user specifies format 2.4, does the parser even need to look for e syntax, or ()'s? @daleif What I am doing to speed things up is to store the data in a dedicated format rather than a property list. The latter makes sense for units (open ended) but not so much for numbers (rigid format). @JosephWright I want to know about either the bibliography environment or \DeclareFieldFormat. From the documentation I see no reason not to treat these commands as usual, though they seem to behave in a slightly different way than I anticipated it. I have an example here which globally sets a box, which is typeset outside of the bibliography environment afterwards. This doesn't seem to typeset anything. :-( So I'm confused about the inner workings of biblatex (even though the source seems.... well, the source seems to reinforce my thought that biblatex simply doesn't do anything fancy). Judging from the source the package just has a lot of options, and that's about the only reason for the large amount of lines in biblatex1.sty... Consider the following MWE to be previewed in the build in PDF previewer in Firefox\documentclass[handout]{beamer}\usepackage{pgfpages}\pgfpagesuselayout{8 on 1}[a4paper,border shrink=4mm]\begin{document}\begin{frame}\[\bigcup_n \sum_n\]\[\underbrace{aaaaaa}_{bbb}\]\end{frame}\end{d... @Paulo Finally there's a good synth/keyboard that knows what organ stops are! youtube.com/watch?v=jv9JLTMsOCE Now I only need to see if I stay here or move elsewhere. If I move, I'll buy this there almost for sure. @JosephWright most likely that I'm for a full str module ... but I need a little more reading and backlog clearing first ... and have my last day at HP tomorrow so need to clean out a lot of stuff today .. and that does have a deadline now @yo' that's not the issue. with the laptop I lose access to the company network and anythign I need from there during the next two months, such as email address of payroll etc etc needs to be 100% collected first @yo' I'm sorry I explain too bad in english :) I mean, if the rule was use \tl_use:N to retrieve the content's of a token list (so it's not optional, which is actually seen in many places). And then we wouldn't have to \noexpand them in such contexts. @JosephWright \foo:V \l_some_tl or \exp_args:NV \foo \l_some_tl isn't that confusing. @Manuel As I say, you'd still have a difference between say \exp_after:wN \foo \dim_use:N \l_my_dim and \exp_after:wN \foo \tl_use:N \l_my_tl: only the first case would work @Manuel I've wondered if one would use registers at all if you were starting today: with \numexpr, etc., you could do everything with macros and avoid any need for \<thing>_new:N (i.e. soft typing). There are then performance questions, termination issues and primitive cases to worry about, but I suspect in principle it's doable. @Manuel Like I say, one can speculate for a long time on these things. @FrankMittelbach and @DavidCarlisle can I am sure tell you lots of other good/interesting ideas that have been explored/mentioned/imagined over time. @Manuel The big issue for me is delivery: we have to make some decisions and go forward even if we therefore cut off interesting other things @Manuel Perhaps I should knock up a set of data structures using just macros, for a bit of fun [and a set that are all protected :-)] @JosephWright I'm just exploring things myself “for fun”. I don't mean as serious suggestions, and as you say you already thought of everything. It's just that I'm getting at those points myself so I ask for opinions :) @Manuel I guess I'd favour (slightly) the current set up even if starting today as it's normally \exp_not:V that applies in an expansion context when using tl data. That would be true whether they are protected or not. Certainly there is no big technical reason either way in my mind: it's primarily historical (expl3 pre-dates LaTeX2e and so e-TeX!) @JosephWright tex being a macro language means macros expand without being prefixed by \tl_use. \protected would affect expansion contexts but not use "in the wild" I don't see any way of having a macro that by default doesn't expand. @JosephWright it has series of footnotes for different types of footnotey thing, quick eye over the code I think by default it has 10 of them but duplicates for minipages as latex footnotes do the mpfoot... ones don't need to be real inserts but it probably simplifies the code if they are. So that's 20 inserts and more if the user declares a new footnote series @JosephWright I was thinking while writing the mail so not tried it yet that given that the new \newinsert takes from the float list I could define \reserveinserts to add that number of "classic" insert registers to the float list where later \newinsert will find them, would need a few checks but should only be a line or two of code. @PauloCereda But what about the for loop from the command line? I guess that's more what I was asking about. Say that I wanted to call arara from inside of a for loop on the command line and pass the index of the for loop to arara as the jobname. Is there a way of doing that?
Although the question is easy to pose, I think some background will help to motivate it, so I'll start with it. Consider variables $X=(X_1, \ldots, X_n)$ over a field $K$ and the elementary symmetric functions $T=(T_1, \ldots, T_n)$ in $X$. In other words $X$ are the roots of the polynomial $Y^n + T_1 Y^{n-1} + \cdots + T_n$. A polynomial $f$ in $X$ is symmetric is $f(s X) = f(X)$ for any permutation $s$. Here $s X := (X_{s(1)}, \ldots, X_{s(n)})$. Then a basic fact is that if $f(X)$ is symmetric, then $f(X) = g(T)$, for some polynomial $g$. It is reasonable to define an alternating polynomial to be $f$ that satisfy $f(s X) = sign(s) f(X)$, where $sign(s) = \pm 1$ is the signature. The "elementary" alternating polynomial is the Vandermonde polynomial $V(X) = \prod_{i<j} (X_j-X_i)$, and any other alternating polynomial can be expressed as a polynomial in $T$ and $V$. Note that $V$ is a square root of the discriminant $\Delta$ of $Y^n + T_1 Y^{n-1} + \cdots + T_n$ and the discriminant has an explicit formula in terms of $T$ using the Sylvester matrix. That definition for alternating polynomials gives nothing interesting in characteristic $2$ (because then $1=-1$). The only definition that makes sense to me in characteristic $2$ is: $f$ is alternating if $f(s X) = f(X) + add.sign(s)$. Here $add.sign(s) = 0,1$ is the additive signature, i.e., equals $1$ if $s$ is odd and $0$ if $s$ is even. I already figured out what is the "elementary" alternating polynomial $u/V$ and what is the Artin-Schreir equation it satisfies: $u(X) = \sum_{s \ {\rm is\ even}} X^{n-1}_{s(1)} \cdots X^0_{s(n)}$ and it satisfies the Artin-Schreier equation $X^2 + X = \frac{u(X) u(s_0 X)}{\Delta}$, where $s_0$ is any odd permutation (e.g., transposition), and $\Delta$ is again the discriminant. (Note that $u(X) + u(s_0 X) = V$.) My question is: Does there exist a nice formula for $\frac{u(X) u(s_0 X)}{\Delta}$ in terms of $T$?
In the last few days I thought a lot about (fully) time-constructible functions and I will present what I found out by answering Q1 and Q3. Q2 seems too hard. Q3: Kobayashi in his article (the reference is in the question) proved that a function $f:\mathbb{N}\rightarrow\mathbb{N}$, for which there exists an $\epsilon>0$ s.t. $f(n)\geq (1+\epsilon)n$, is fully time constructible iff it is computable in $O(f(n))$ time. (note that it is irrelevant whether the input or output is in unary/binary since we can transform between these two representations in linear time). This makes the following functions fully time-constructible: $2^n$, $2^{2^n}$, $n!$, $n\lfloor \log n \rfloor$, all polynomials $p$ over $\mathbb{N}$ s.t. $p(n)\geq (1+\epsilon)n$ ... Kobayashi also proved fully time-constructibility for some functions that grow slower than $(1+\epsilon)n$, like $n+\lfloor\lfloor\log n\rfloor^q\rfloor$ for $q\in\mathbb{Q}^+$ ... To continue with examples of fully time-constructible functions, one can prove that if $f_1$ and $f_2$ are fully time-constructible, then $f_1+f_2$, $f_1f_2$, $f_1^{f_2}$ and $f_1\circ f_2$ are also fully time-constructible (the later follows directly from Theorem 3.1 in Kobayashi). This altogether convince me that many nice functions are indeed fully time-constructible. It is surprising that Kobayashi did not see a way to prove fully time-constructibility of the (nice) function $\lfloor n\log n\rfloor$ (and neither do I). Let us also comment the definition from Wikipedia article: A function $f$ is time-constructible, if there exists a Turing machine $M$ which, given a string $1^n$, outputs $f(n)$ in $O(f(n))$ time. We see that this definition is equivallent to our definition of fully time-constructibility for functions $f(n)\geq (1+\epsilon)n$. Q1: This question has a really interesting answer. I claim that if all time-constructible functions are fully time-constructible, then $EXP-TIME=NEXP-TIME$. To prove that, let us take an arbitrary problem $L\in NEXP-TIME$, $L\subseteq\{0,1\}^*$. Then there exists a $k\in\mathbb{N}$, s.t. $L$ can be solved by a NDTM $M$ in $2^{n^k-1}$ steps. We can assume that at each step $M$ goes in at most two different states for simplicity. Now define the function$$f(n)=\left\{\begin{array}{ll} 8n+2 & \mbox{if }\left(\mbox{first } \lfloor\sqrt[k]{\lfloor\log n\rfloor+1}\rfloor\mbox{ bits of } bin(n)\right)\in L\\ 8n+1 & \mbox{else} \end{array}\right.$$ I claim that $f$ is time-construcible. Consider the following deterministic Turing machine $T$: on input $w$ of length $n$ it computes $\left(\textrm{first }\lfloor\sqrt[k]{\lfloor\log n\rfloor+1}\rfloor\textrm{ bits of }bin(n)\right)$ in $O(n)$ time then it simulates $M$ on these bits, where the bits of $w$ determine which (formerly nondeterminisic) choices to take. accept iff $\left(M\textrm{ accepts using choices given by }w\right)$. Note that the length of $w$ ($=n$) is enough that it determines all nondeterministic choices, since $M$ on input $\left(\textrm{first }\lfloor\sqrt[k]{\lfloor\log n\rfloor+1}\rfloor\textrm{ bits of }bin(n)\right)$ makes at most $n$ steps. We can make $T$ run in at most $8n+1$ steps. Now the following Turing machine proves that $f$ is time-constructible: on input $w$ of length $n$ run $T$ and count steps in parallel so that exacly $8n$ steps are done. if $T$ rejected or would reject in the next step, go to a halting state in the next step. Else, make one more step and then go to a halting state. Now suppose that $f$ is fully time-constructible. We will prove that this leads to $EXP-TIME=NEXP-TIME$. The following algorithm solves $L$: on input $x$, let $n$ be the number with binary representation $x00\ldots 0$ ($\|x\|^{k-1}$ zeros). It follows that $x=\left(\textrm{first }\lfloor\sqrt[k]{\lfloor\log n\rfloor+1}\rfloor\textrm{ bits of }bin(n)\right)$. compute $f(n)$ in time $f(n)$ and check whether it is divisible by 2. This algorithm runs in exponential time and solves $L$. Since $L\in NEXP-TIME$ was arbitrary, $EXP-TIME=NEXP-TIME$.
I think the work of Dr. Paul Garabedian (and Dr. Schiffer)[1], and Dr. Mel'nikov (who built on Dr. Garabedian's result) are important theorems that were almost forgotten. I'll share the main theorem from Dr. Mel'nikov's work[2] as it incorporates the main result from Dr Garabedian's: Given complex numbers $z_1,\ldots,z_n$ and a parameter $r > 0$ such that $\|z_i-z_j\| > 2r$, $i \neq j$, we denote by $A = A(z_1,\ldots,z_n,r)$ the $(n \times n)$-matrix with entries $$ \alpha_{i,j} = r \sum_{k \neq i, k \neq j} \frac{1}{(z_i-z_k)\overline{(z_j-z_k)} - r^2}, 1 \leq i \leq n, 1 \leq j \leq n$$ Definition 1: We set $$\lambda_1 = \lambda_1(z_1,\ldots,z_n,r) = ((r^{-1}I + A)^{-1}(\mathbf{1}),\mathbf{1}),$$ where $I$ is the standard identity $(n\times n$)-matrix, the vector $\mathbf{1} = (1,\ldots,1)\in\mathbb{C}^n$, and $(\cdot,\cdot)$ is the standard Hermitian scalar product in $\mathbb{C}^n$. Theorem 1: For each bounded open set $G \in \mathbb{C}$:$$\gamma (G) = \sup\{\lambda_1(z_1,\ldots,z_n,r)\},$$where the supremum is taken over all finite collections of points $\{z_1,\ldots,z_n\} \subset G$ and values of $r > 0$ such that $\|z_i - z_j\| > 2r, i \neq j$, and the distance $d(z_i,\delta G) > r$ for all i (in other words, the union of disjoint discs $\Delta(z_i,r) = \{\|z-z_i\| < r\}$ is in G). I used/demonstrated/incorporated this result "by accident", and only discovered these works about 18 months ago when I sought to explain my result rigorously. I have not seen any work aside from my own that uses this result, maybe because calculation/representation of a measure is not only conceptually challenging (both in terms of implementation and visualisation), but finding empirical data that is collected in both a uniform and robust manner (such that it can be subject to these types of analyses) is also difficult. However in terms of mathematical analysis, I have a hard time thinking of other work that is of this calibre. There is a lot of meat on the bone in terms of empirical/theoretical proofs. I found the work of Dr Garabedian especially impressive because he came to be known as a Computational Fluid Dynamicist, but it seems clear from his astounding result in 1950 that he was originally a mathematician. I believe that such a successful transition is demonstrative of the empirical/theoretical boundaries encroached by this work. Very few are blessed to be able to contribute to this ("narrow") area, but I feel those who are interested in the numeric aspect of Calculus, and have good proof skills, may stand to benefit. This result is Kelvin-esque, if you ask me. [1] Garabedian, P.R.; Schiffer, M., On existence theorems of potential theory and conformal mapping, Ann. Math. (2) 52, 164-187 (1950). ZBL0040.32903., [2] Mel’nikov, M.S., Analytic capacity: Discrete approach and curvature of measure, Sb. Math. 186, No.6, 827-846 (1995); translation from Mat. Sb. 186, No.6, 57-76 (1995). ZBL0840.30008.
Let $X$ be an affine holomorphic symplectic variety of dimension $2n$, with the associated Poisson bracket { , }. Let's say it's an integrable system when there are $n$ algebraically independent holomorphic functions $I_i$ ($i=1,\ldots,n$) on $X$ such that they Poisson-commute: $\{I_i,I_j\}=0$. Pick a simple Lie algebra $\mathfrak{g}$, and pick a coadjoint orbit $O\subset \mathfrak{g}^$. As is well-known, this is naturally a holomorphic symplectic variety: $x,y\in \mathfrak{g}$ gives a function on $O$, and its Poisson bracket is then given by the Lie bracket: $\{x,y\}_{PB}=[x,y]$. I wonder which coadjoint orbit is integrable, in the sense given above. For example, any regular orbit $O$ of $\mathfrak{g}=\mathfrak{gl}(n)$ is integrable: let $X:\mathfrak{g}^\to\mathfrak{g}$ be the identification via the invariant inner product, choose a generic element $a\in \mathfrak{g}$, and consider a function on $O$ given by $P_k(t)= \mathrm{tr} (X+ta)^k$ where $t$ is a complex number. (I'm sorry for a slightly confusing notation, but $a$ here is a constant function on $O$ taking value in $\mathfrak{g}$.) After some manipulation, $P_k(t)$ and $P_{k'}(t')$ are seen to Poisson-commute. Therefore, coefficients $P_{k,i}$ of $t^i$ of $P_k(t)$ all Poisson-commute. Note that $P_{k,k}$ is a constant function on $O$, because $O$ is a coadjoint orbit. Then there are in total $1+2+\cdots+(n-1) = (\dim \mathfrak{gl}(n)-\mathrm{rank} \mathfrak{gl}(n))/2$ independent commuting operators. Let me conjecture that all coadjoint orbits are integrable. Another large class of affine holomorphic symplectic varieties are Nakajima's quiver varieties, which include all the coadjoint orbits of $\mathfrak{gl}(n)$. A similar question can be asked: which quiver variety is integrable in this sense? Update Thanks to the answers so far, I could track down references, see e.g. the end of Sec. 4 of this paper, showing that any coadjoint orbit of real compact Lie algebras is integrable. I guess the proof should carry over to the semisimple orbits (and presumably nilpotent Richardson orbits) of complex semisimple Lie algebras, in the holomorphic sense. So the question now is: how about nilpotent orbits in $\mathfrak{g}_{\mathbb{C}}$? Update 2 Indeed, A. Joseph says in this article that he essentially proved that Richardson orbits are integrable in this article (in the setting of enveloping algebras, not the associated graded.) It would be interesting e.g. non-special orbits are not integrable... but I have no idea.
The path integral formalism doesn't really replace the operator formalism. You still need to know that the state space is a Hilbert space, that probabilities are computed by norm-squaring inner products in this space, that observables are operators on this space, that time evolution is implemented by a unitary operator, and so forth. Basically, you have to start with the Dirac-von Neumann axioms (or something very similar): States are rays in a Hilbert space $\mathcal{H}$. Observables are self-adjoint operators on $\mathcal{H}$. The expectation value of an observable $A$ in a state represented by a unit vector $\psi$ is $E[A] = \langle \psi, A \psi\rangle$. You need these postulates to make any connection to experiment. The path integral doesn't provide alternatives. (You're implicitly using these basic postulates when you talk about superpositions of states and when you label the endpoints of trajectories with observable quantities.) But these postulates are rather minimal. They don't even mention time evolution. So people usually add another axiom: Time evolution is represented on the state space by unitary operators $U(t)$, for $t \in \mathbb{R}$. The path integral formalism and the canonical formalism aren't really competing sets of physical postulates. They're just different methods for setting up examples which satisfy the physical postulates above. Let's look at an example, a particle moving in a potential $V$ in 3-space. In the canonical formalism, you declare the state space is $L^2(\mathbb{R}^3)$, that the observables are functions of the standard multiplication and differentiation operators $Q$ and $P$, and that the time evolution is given by $U(t) = e^{-itH/\hbar}$ where $H$ is the Hamiltonian $P^2/2m + V(Q)$. In the path integral formalism, you start with the same input data (the configuration space $\mathbb{R}^3$, the mass $m$, and the potential $V$), and you use them to define integrals on the space of paths. Then you use these path integrals to construct a Hilbert space, observables acting on the Hilbert space, and time evolution operators $U(t)$. The path integral directly constructs the matrix elements of $U(t)$, and you then derive Schrodinger's equation. Modulo some normalization problems, this is what your equations 1 and 2 say: Your $G(x_2,t_2; x_1, t_1)$ is the matrix element $\langle x_2 \| U(t_2 - t_1) \| x_1 \rangle$, and equation 2 just says that $\psi(t_2) = U(t_2 - t_1) \psi(t_1)$. Note also: The derivation of Schrodinger's equation is not deep. All Schrodinger's equation says is that time translations act unitarily on the Hilbert space. $i\hbar \partial_t \psi(t) = H \psi(t)$ is the equivalent infinitesimal form of $\psi(t) = e^{-itH/\hbar} \psi(0)$. It's interesting if you've chosen $H$ by some classical analogy and now need to find $U(t)$ / solve the eigenstate problem. In the path integral, however, you have $U(t)$ already, and you find $H$ via $H = i\hbar \partial_t U(t)\|_{t=0}$. Likewise, you prove that the Hilbert space you get from the path integral is actually $L^2(\mathbb{R}^3)$, that the observables are the usual position and momentum operators, and that the Hamiltonian is really the one you'd expect. So, how do you actually get a Hilbert space from a path integral? Basically, you divide a time segment $[0,t]$ into $[0,s] \cup [s,t]$, and you look at how the path integral decomposes into integrals over the smaller intervals and over the field values on the boundary. The integral over the boundaries defines the Hilbert space inner product. Then observables are defined by using the path integral to compute their matrix elements. Appendix 1 of Polchinksi Vol 1 has a pretty nice explanation of this. -- Footnote -- If you're being careful about the mathematical details, you construct Euclidean path integrals, and you get the Wiener measure $d\mu(\phi) = e^{-S(\phi)} d\phi$ associated to $V$ and $m$. Then you use the Osterwalder-Schrader Reconstruction theorem to define the Hilbert space and the operators. This is covered in some detail in Chapters 3 & 6 of Glimm & Jaffe's Quantum Physics book. The basic idea is this: Wiener measures allow you to integrate certain functions on the space of paths, namely those which can be systematically approximated by functions of the form $F_{t_i}(\phi) = \phi(t_1) \phi(t_2) .. \phi(t_n)$. Wiener measures are reflection positive: Let $\mathcal{F}^+$ be the collection of functions of the form above for which the approximations only involve $t_i > 0$. Then the bilinear form defined by $B(F_{t_i},G_{t_j}) = \int \overline{F_{-t_i}(\phi)} G_{t_j}(\phi) d\mu(\phi)$ is non-negative. The Hilbert space is the quotient $\mathcal{H} =\mathcal{F}^+ / \operatorname{kernel}(B)$. Any function $F$ in $\mathcal{F}^+$ acts on $\mathcal{F}^+$ via multiplication. These multiplications descend to define observables on $\mathcal{H}$. Translations by $t > 0$ act on $\mathcal{F}^+$. This descends to the action of a semi-group $\mathbb{R}_{\geq 0}$, and this semigroup action analytically continues to a unitary action of $\mathbb{R}$. There are axioms, due to Osterwalder & Schrader, which guarantee that a Euclidean path integral measure defines a Hilbert space and observables satisfying a stronger form of the axioms above (the Wightman axioms). I don't believe the OS axioms are physical postulates, however. They don't refer to observable quantities, and they make use of imaginary time in a way which doesn't generalize to all quantum theories.
We have the matrix Laplacian matrix $G=A^TA$ which has a set of eigenvalues $\lambda_0\leq\lambda_1\leq\ldots\leq \lambda_n$ for $G\in\mathbb{R}^{n\times n}$ where we always know $\lambda_0 = 0$. Thus the Laplacian matrix is always symmetric positive semi-definite. Because the matrix $G$ is not symmetric positive definite we have to be careful when we discuss the Cholesky decomposition. The Cholesky decomposition exists for a positive semi-definite matrix but it is no longer unique. For example, the positive semi-definite matrix $$ A=\left[\!\!\begin{array}{cc} 0 & 0 \\ 0 & 1\end{array}\!\!\right],$$has infinitely many Cholesky decompositions$$ A=\left[\!\begin{array}{cc} 0 & 0 \\ 0 & 1\end{array}\!\right]= \left[\!\begin{array}{cc} 0 & 0 \\ \sin\theta & \cos\theta\end{array}\!\right] \left[\!\begin{array}{cc} 0 & \sin\theta \\ 0 & \cos\theta\end{array}\!\right]=LL^T.$$ However, because we have a matrix $G$ that is known to be a Laplacian matrix we can actually avoid the more sophisticated linear algebra tools like Cholesky decompositions or finding the square root of the positive semi-definite matrix $G$ such that we recover $A$. For example, if we have the Laplace matrix $G\in\mathbb{R}^{4\times 4}$,$$G=\left[\!\begin{array}{cccc} 3 & -1 & -1 & -1\\-1 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ -1 & 0 & 0 & 1 \\\end{array}\!\right]$$we can use graph theory to recover the desired matrix $A$. We do so by formulating the oriented incidence matrix. If we define the number of edges in the graph to be $m$ and the number of vertices to be $n$ then the oriented incidence matrix $A$ will be an $m\times n$ matrix given by $$A_{ev} = \left\{\begin{array}{lc} 1 & \textrm{if }e=(v,w)\textrm{ and }v<w \\ -1 & \textrm{if }e=(v,w)\textrm{ and }v>w \\ 0 & \textrm{otherwise},\end{array}\right.$$where $e=(v,w)$ denotes the edge which connects the vertices $v$ and $w$. If we take a graph for $G$ with four vertices and three edges,then we have the oriented incidence matrix $$A = \left[\!\begin{array}{cccc} 1 & -1 & 0 & 0\\ 1 & 0 & -1 & 0 \\ 1 & 0 & 0 & -1 \\\end{array}\!\right],$$and we can find that $G=A^TA$. For the matrix problem you describe you would construct a graph for $G$ with the same number of edges as vertices, then you should have the ability to reconstruct the matrix $A$ when you are only given the Laplacian matrix $G$. Update: If we define the diagonal matrix of vertex degrees of a graph as $N$ and the adjacency matrix of the graph as $M$, then the Laplacian matrix $G$ of the graph is defined by $G=N-M$. For example, in the following graph we find the Laplacian matrix is$$G=\left[\!\begin{array}{cccc} 3 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\\end{array}\!\right] - \left[\!\begin{array}{cccc} 0 & 1 & 1 & 1\\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\\end{array}\!\right].$$Now we relate the $G$ to the oriented incidence matrix $A$ using the edges and nodes given in the pictured graph. Again we find the entries of $A$ from $$A_{ev} = \left\{\begin{array}{lc} 1 & \textrm{if }e=(v,w)\textrm{ and }v<w \\ -1 & \textrm{if }e=(v,w)\textrm{ and }v>w \\ 0 & \textrm{otherwise},\end{array}\right..$$ For example, edge $e_1$ connects the nodes $v_1$ and $v_2$. So to determine $A_{e_1,v_1}$ we note that the index of $v_1$ is less than the index of $v_2$ (or we have the case $v<w$ in the definition of $A_{ev}$). Thus, $A_{e_1,v_1} = 1$. Similarly by the way of comparing indices we can find $A_{e_1,v_2} = -1$. We give $A$ below in a more explicit way referencing the edges and vertices pictured.$$A = \begin{array}{c\|cccc} & v_1 & v_2 & v_3 & v_4 \\ \hline e_1 & 1 & -1 & 0 & 0\\ e_2 & 1 & 0 & -1 & 0 \\ e_3 & 1 & 0 & 0 & -1 \\\end{array}.$$ Next, we generalize the concept of the Laplacian matrix to a weighted undirected graph. Let $Gr$ be an undirected finite graph defined by $V$ and $E$ its vertex and edge set respectively. To consider a weighted graph we define a weight function $$w: V\times V\rightarrow \mathbb{R}^+,$$which assigns a non-negative real weight to each edge of the graph. We will denote the weight attached to edge connecting vertices $u$ and $v$ by $w(u,v)$. In the case of a weighted graph we define the degree of each vertex $u\in V$ as the sum of all the weighted edges connected to $u$, i.e.,$$d_u = \sum_{v\in V}w(u,v).$$From the given graph $Gr$ we can define the weighted adjacency matrix $Ad(Gr)$ as an $n\times n$ with rows and columns indexed by $V$ whose entries are given by $w(u,v)$. Let $D(Gr)$ be the diagonal matrix indexed by $V$ with the vertex degrees on the diagonal then we can find the weighted Laplacian matrix $G$ just as before$$G = D(Gr) - Ad(Gr).$$ In the problem from the original post we know $$G=\left[\!\begin{array}{ccc} \tfrac{3}{4} & -\tfrac{1}{3} & -\tfrac{5}{12} \\-\tfrac{1}{3} & \tfrac{2}{3} & -\tfrac{1}{3} \\ -\tfrac{5}{12} & -\tfrac{1}{3} & \tfrac{3}{4} \\\end{array}\!\right].$$ From the comments we know we seek a factorization for $G$ where $G=A^TA$ and specify $A$ is of the form $A=I-1_nw^T$ where $w^T1_n=1$. For full generality assume the matrix $A$ has no zero entries. Thus if we formulate the weighted oriented incidence matrix to find $A$ we want the weighted adjacency matrix $Ad(Gr)$ to have no zero entries as well, i.e., the weighted graph will have loops. Actually calculating the weighted oriented incidence matrix seems difficult (although it may simply be a result of my inexperience with weighted graphs). However, we can find a factorization of the form we seek in an ad hoc way if we assume we know something about the loops in our graph. We split the weighted Laplacian matrix $G$ into the degree and adjacency matrices as follows$$G=\left[\!\begin{array}{ccc} \tfrac{5}{4} & 0 & 0 \\0 & 1 & 0 \\ 0 & 0 & \tfrac{11}{12} \\\end{array}\!\right]-\left[\!\begin{array}{ccc} \tfrac{1}{2} & \tfrac{1}{3} & \tfrac{5}{12} \\\tfrac{1}{3} & \tfrac{1}{3} & \tfrac{1}{3} \\ \tfrac{5}{12} & \tfrac{1}{3} & \tfrac{1}{6} \\\end{array}\!\right] = D(Gr)-Ad(Gr).$$ Thus we know the loops on $v_1$, $v_2$ and $v_3$ have weights $1/2$, $1/3$, and $1/6$ respectively. If we put the weights on the loops into a vector $w$ = $[\frac{1}{2}$ $\frac{1}{3}$ $\frac{1}{6}]^T$ then we can recover the matrix $A$ we want in the desired form$$A = I-1_nw^T = \left[\!\begin{array}{ccc} \tfrac{1}{2} & -\tfrac{1}{3} & -\tfrac{1}{6} \\-\tfrac{1}{2} & \tfrac{2}{3} & -\tfrac{1}{6} \\ -\tfrac{1}{2} & -\tfrac{1}{3} & \tfrac{5}{6} \\\end{array}\!\right].$$ It appears if we have knowledge of the loops in our weighted graph we can find the matrix $A$ in the desired form. Again, this was done in an ad hoc manner (as I am not a graph theorist) so it may be a hack that worked just for this simple problem.
And I think people said that reading first chapter of Do Carmo mostly fixed the problems in that regard. The only person I asked about the second pset said that his main difficulty was in solving the ODEs Yeah here there's the double whammy in grad school that every grad student has to take the full year of algebra/analysis/topology, while a number of them already don't care much for some subset, and then they only have to pass rather the class I know 2 years ago apparently it mostly avoided commutative algebra, half because the professor himself doesn't seem to like it that much and half because he was like yeah the algebraists all place out so I'm assuming everyone here is an analyst and doesn't care about commutative algebra Then the year after another guy taught and made it mostly commutative algebra + a bit of varieties + Cech cohomology at the end from nowhere and everyone was like uhhh. Then apparently this year was more of an experiment, in part from requests to make things more geometric It's got 3 "underground" floors (quotation marks because the place is on a very tall hill so the first 3 floors are a good bit above the the street), and then 9 floors above ground. The grad lounge is in the top floor and overlooks the city and lake, it's real nice The basement floors have the library and all the classrooms (each of them has a lot more area than the higher ones), floor 1 is basically just the entrance, I'm not sure what's on the second floor, 3-8 is all offices, and 9 has the ground lounge mainly And then there's one weird area called the math bunker that's trickier to access, you have to leave the building from the first floor, head outside (still walking on the roof of the basement floors), go to this other structure, and then get in. Some number of grad student cubicles are there (other grad students get offices in the main building) It's hard to get a feel for which places are good at undergrad math. Highly ranked places are known for having good researchers but there's no "How well does this place teach?" ranking which is kinda more relevant if you're an undergrad I think interest might have started the trend, though it is true that grad admissions now is starting to make it closer to an expectation (friends of mine say that for experimental physics, classes and all definitely don't cut it anymore) In math I don't have a clear picture. It seems there are a lot of Mickey Mouse projects that people seem to not help people much, but more and more people seem to do more serious things and that seems to become a bonus One of my professors said it to describe a bunch of REUs, basically boils down to problems that some of these give their students which nobody really cares about but which undergrads could work on and get a paper out of @TedShifrin i think universities have been ostensibly a game of credentialism for a long time, they just used to be gated off to a lot more people than they are now (see: ppl from backgrounds like mine) and now that budgets shrink to nothing (while administrative costs balloon) the problem gets harder and harder for students In order to show that $x=0$ is asymptotically stable, one needs to show that $$\forall \varepsilon > 0, \; \exists\, T > 0 \; \mathrm{s.t.} \; t > T \implies \|\| x ( t ) - 0 \|\| < \varepsilon.$$The intuitive sketch of the proof is that one has to fit a sublevel set of continuous functions $... "If $U$ is a domain in $\Bbb C$ and $K$ is a compact subset of $U$, then for all holomorphic functions on $U$, we have $\sup_{z \in K}\|f(z)\| \leq C_K \\|f\\|_{L^2(U)}$ with $C_K$ depending only on $K$ and $U$" this took me way longer than it should have Well, $A$ has these two dictinct eigenvalues meaning that $A$ can be diagonalised to a diagonal matrix with these two values as its diagonal. What will that mean when multiplied to a given vector (x,y) and how will the magnitude of that vector changed? Alternately, compute the operator norm of $A$ and see if it is larger or smaller than 2, 1/2 Generally, speaking, given. $\alpha=a+b\sqrt{\delta}$, $\beta=c+d\sqrt{\delta}$ we have that multiplication (which I am writing as $\otimes$) is $\alpha\otimes\beta=(a\cdot c+b\cdot d\cdot\delta)+(b\cdot c+a\cdot d)\sqrt{\delta}$ Yep, the reason I am exploring alternative routes of showing associativity is because writing out three elements worth of variables is taking up more than a single line in Latex, and that is really bugging my desire to keep things straight. hmm... I wonder if you can argue about the rationals forming a ring (hence using commutativity, associativity and distributivitity). You cannot do that for the field you are calculating, but you might be able to take shortcuts by using the multiplication rule and then properties of the ring $\Bbb{Q}$ for example writing $x = ac+bd\delta$ and $y = bc+ad$ we then have $(\alpha \otimes \beta) \otimes \gamma = (xe +yf\delta) + (ye + xf)\sqrt{\delta}$ and then you can argue with the ring property of $\Bbb{Q}$ thus allowing you to deduce $\alpha \otimes (\beta \otimes \gamma)$ I feel like there's a vague consensus that an arithmetic statement is "provable" if and only if ZFC proves it. But I wonder what makes ZFC so great, that it's the standard working theory by which we judge everything. I'm not sure if I'm making any sense. Let me know if I should either clarify what I mean or shut up. :D Associativity proofs in general have no shortcuts for arbitrary algebraic systems, that is why non associative algebras are more complicated and need things like Lie algebra machineries and morphisms to make sense of One aspect, which I will illustrate, of the "push-button" efficacy of Isabelle/HOL is its automation of the classic "diagonalization" argument by Cantor (recall that this states that there is no surjection from the naturals to its power set, or more generally any set to its power set).theorem ... The axiom of triviality is also used extensively in computer verification languages... take Cantor's Diagnolization theorem. It is obvious. (but seriously, the best tactic is over powered...) Extensions is such a powerful idea. I wonder if there exists algebraic structure such that any extensions of it will produce a contradiction. O wait, there a maximal algebraic structures such that given some ordering, it is the largest possible, e.g. surreals are the largest field possible It says on Wikipedia that any ordered field can be embedded in the Surreal number system. Is this true? How is it done, or if it is unknown (or unknowable) what is the proof that an embedding exists for any ordered field? Here's a question for you: We know that no set of axioms will ever decide all statements, from Gödel's Incompleteness Theorems. However, do there exist statements that cannot be decided by any set of axioms except ones which contain one or more axioms dealing directly with that particular statement? "Infinity exists" comes to mind as a potential candidate statement. Well, take ZFC as an example, CH is independent of ZFC, meaning you cannot prove nor disprove CH using anything from ZFC. However, there are many equivalent axioms to CH or derives CH, thus if your set of axioms contain those, then you can decide the truth value of CH in that system @Rithaniel That is really the crux on those rambles about infinity I made in this chat some weeks ago. I wonder to show that is false by finding a finite sentence and procedure that can produce infinity but so far failed Put it in another way, an equivalent formulation of that (possibly open) problem is: > Does there exists a computable proof verifier P such that the axiom of infinity becomes a theorem without assuming the existence of any infinite object? If you were to show that you can attain infinity from finite things, you'd have a bombshell on your hands. It's widely accepted that you can't. If fact, I believe there are some proofs floating around that you can't attain infinity from the finite. My philosophy of infinity however is not good enough as implicitly pointed out when many users who engaged with my rambles always managed to find counterexamples that escape every definition of an infinite object I proposed, which is why you don't see my rambles about infinity in recent days, until I finish reading that philosophy of infinity book The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. It derives its name from the problem faced by someone who is constrained by a fixed-size knapsack and must fill it with the most valuable items.The problem often arises in resource allocation where there are financial constraints and is studied in fields such as combinatorics, computer science... O great, given a transcendental $s$, computing $\min_P(\|P(s)\|)$ is a knapsack problem hmm... By the fundamental theorem of algebra, every complex polynomial $P$ can be expressed as: $$P(x) = \prod_{k=0}^n (x - \lambda_k)$$ If the coefficients of $P$ are natural numbers , then all $\lambda_k$ are algebraic Thus given $s$ transcendental, to minimise $\|P(s)\|$ will be given as follows: The first thing I think of with that particular one is to replace the $(1+z^2)$ with $z^2$. Though, this is just at a cursory glance, so it would be worth checking to make sure that such a replacement doesn't have any ugly corner cases. In number theory, a Liouville number is a real number x with the property that, for every positive integer n, there exist integers p and q with q > 1 and such that0<\|x−pq\|<1qn.{\displaystyle 0<\left\|x-{\frac {p}... Do these still exist if the axiom of infinity is blown up? Hmmm... Under a finitist framework where only potential infinity in the form of natural induction exists, define the partial sum: $$\sum_{k=1}^M \frac{1}{b^{k!}}$$ The resulting partial sums for each M form a monotonically increasing sequence, which converges by ratio test therefore by induction, there exists some number $L$ that is the limit of the above partial sums. The proof of transcendentally can then be proceeded as usual, thus transcendental numbers can be constructed in a finitist framework There's this theorem in Spivak's book of Calculus:Theorem 7Suppose that $f$ is continuous at $a$, and that $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x=a$. Suppose, moreover, that $\lim_{x \to a} f'(x)$ exists. Then $f'(a)$ also exists, and$$f'... and neither Rolle nor mean value theorem need the axiom of choice Thus under finitism, we can construct at least one transcendental number. If we throw away all transcendental functions, it means we can construct a number that cannot be reached from any algebraic procedure Therefore, the conjecture is that actual infinity has a close relationship to transcendental numbers. Anything else I need to finish that book to comment typo: neither Rolle nor mean value theorem need the axiom of choice nor an infinite set > are there palindromes such that the explosion of palindromes is a palindrome nonstop palindrome explosion palindrome prime square palindrome explosion palirome prime explosion explosion palindrome explosion cyclone cyclone cyclone hurricane palindrome explosion palindrome palindrome explosion explosion cyclone clyclonye clycone mathphile palirdlrome explosion rexplosion palirdrome expliarome explosion exploesion
Find the parametric equation for the curve. $$x^{2}+y^{2}=10$$ I haven't learned parametric equations fully yet, so I wanted to check with you guys and see if you can confirm if I'm doing this correctly and possibly go more in depth on the problem if you can? Because it's centered at (0,0) the origin, and it has a radius of sqrt(10) then the answer is this, right? $$(x(t),y(t)) = (\sqrt{10}\cos t\,,\, \sqrt{10}\sin t)$$
Let a (free) particle move in $[0,a]$ with cyclic boundary condition $\psi(0)=\psi(a)$. The solution of the Schrödinger-equation can be put in the form of a plane wave. In this state the standard deviation of momentum is $0$, but $\sigma_x$ must be finite. So we find that $\sigma_x\sigma_p=0$. Is something wrong with the uncertainty principle? This is what happens if one cares not for the subtlety that quantum mechanical operators are typically only defined on subspaces of the full Hilbert space. Let's set $a=1$ for convenience. The operator $p =-\mathrm{i}\hbar\partial_x$ acting on wavefunctions with periodic boundary conditions defined on $D(p) = \{\psi\in L^2([0,1])\mid \psi(0)=\psi(1)\land \psi'\in L^2([0,1])\}$ is self-adjoint, that is, on the domain of definition of $p$, we have $p=p^\dagger$, and $p^\dagger$ admits the same domain of definition. The self-adjointness of $p$ follows from the periodic boundary conditions killing the surface terms that appear in the $L^2$ inner product $$\langle \phi,p\psi\rangle - \langle p^\dagger \phi,\psi\rangle = \int\overline{\phi(x)}\mathrm{i}\hbar\partial_x\psi(x) - \overline{\mathrm{i}\hbar\partial_x\phi(x)}\psi(x) = 0$$ for every $\psi\in D(p)$ and every $\phi\in D(p^\dagger) = D(p)$, but not for $\phi$ with $\phi(0)\neq\phi(1)$. Now, for the question of the commutator: the multplication operator $x$ is defined on the entire Hilbert space, since for $\psi\in L^2([0,1])$ $x\psi$ is also square-integrable. For the product of two operators $A,B$, we have the rule$$ D(AB) = \{\psi\in D(B)\mid B\psi\in D(A)\}$$and $$ D(A+B) = D(A)\cap D(B)$$so we obtain\begin{align}D(px) & = \{\psi\in L^2([0,1])\mid x\psi\in D(p)\} \\D(xp) & = D(p)\end{align}and $x\psi\in D(p)$ means $0\cdot \psi(0) = 1\cdot\psi(1)$, that is, $\psi(1) = 0$. Hence we have$$ D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1]) \land \psi(1) = 0\}$$and finally$$ D([x,p]) = D(xp)\cap D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1])\land \psi(0)=\psi(1) = 0\}$$meaning the plane waves $\psi_{p_0}$ do not belong to the domain of definition of the commutator $[x,p]$ and you cannot apply the naive uncertainty principle to them. However, for self-adjoint operators $A,B$, you may rewrite the uncertainty principle as$$ \sigma_\psi(A)\sigma_\psi(B)\geq \frac{1}{2} \lvert \langle \psi,\mathrm{i}[A,B]\rangle\psi\rvert = \frac{1}{2}\lvert\mathrm{i}\left(\langle A\psi,B\psi\rangle - \langle B\psi,A\psi\rangle\right)\rvert$$where the r.h.s. and l.h.s. are now both defined on $D(A)\cap D(B)$. Applying this version to the plane waves yields no contradiction. Notice that $\psi(x)$ is defined on a circle of circumference $a$. Multiplying $x$ on this circle is really multiplying a periodic extension of $x$, i.e., the sawtooth function $x - a\lfloor x/a\rfloor$, where $\lfloor y\rfloor$ means the largest integer not greater than $y$. So, the commutator of the position and momentum operators involves the derivative of not only $x$ but also the discontinuous part $-a\lfloor x/a\rfloor$. Therefore, \begin{equation} \sigma_{x} \sigma_p \geq \frac{1}{2}\Big\|\langle \psi\|\,[\hat{x},\hat{p}]\,\|\psi\rangle\Big\| = \frac{\hbar}{2}\Bigg\|\Big\langle\psi\,\Big\|\frac{d}{dx}\big(x - a\lfloor x/a\rfloor\big)\Big\|\,\psi\Big\rangle\Bigg\| = \frac{\hbar}{2}\Big\|1-a\|\psi(0)\|^{2}\Big\|. \end{equation} For a plane wave $\psi(x) = e^{ikx}/\sqrt{a}$, the above reduces to $\sigma_{x} \sigma_p\ge0$, as desired. There are two ways to interpret the boundary conditions you are imposing. The first case is that of a system which is infinite in extent, but has a periodic regularity. This is like an electron in an idealised 1D crystal, where the periodic boundary condition is imposed by the presence of nuclei regularly spaced. In this case, the plane wave solution has $\sigma_p$ = 0 but $\sigma_x$ is infinite. The second case, is that of a particle in a ring. In this case, you can imagine the particle as being constrained within the ring by a infinitely deep potential well. The system is not actually 1D, it is 2D. Now you have to consider both $\sigma_x \sigma_{p_x}$ and $\sigma_y \sigma_{p_y}$, and even though $\sigma_x = \sigma_y \sim a$, the uncertainty in momentum will be imposed by the thickness of the ring. The plane wave solution will in fact represent angular momentum eigenstates.
This paper studies fractional integral operator for vector fields in weighted$L^1$. Using the estimates on fractional integral operator and Stein-Weissinequalities, we can give a new proof for a class of Caffarelli-Kohn-Nirenberginequalities and establish new $\divg$-$\curl$ inequalities for vector fields. A new, extended nonlinear framework of the ordinary real analysisincorporating a novel concept of {\em duality structure} and its applicationsinto various nonlinear dynamical problems is presented. The duality structureis an asymptotic property that should affect the late time asymptotic behaviourof a nonlinear dynamical system in a nontrivial way leading naturally tosignatures generic to a complex system. We argue that the present formalismwould offer a natural framework to understand the abundance of complex systemsin natural, biological, financial and related problems. We show that the powerlaw attenuation of a dispersive, lossy wave equation, conventionally deducedfrom fractional calculus techniques, could actually arise from the presentasymptotic duality structure. Differentiability on a Cantor type fractal set isalso formulated. We develop in this work a general version of paracontrolled calculus thatallows to treat analytically within this paradigm some singular partialdifferential equations with the same efficiency as regularity structures. Thiswork deals with the analytic side of the story and offers a toolkit for thestudy of such equations, under the form of a number of continuity results forsome operators. We illustrate the efficiency of this elementary approach on theexamples of the 3-dimensional generalised parabolic Anderson model equation andthe generalised KPZ equation driven by a (1+1)-dimensional space/time whitenoise. The sine process is a rigid point process on the real line, which means thatfor almost all configurations $X$, the number of points in an interval $I =[-R,R]$ is determined by the points of $X$ outside of $I$. In addition, thepoints in $I$ are an orthogonal polynomial ensemble on $I$ with a weightfunction that is determined by the points in $X \setminus I$. We prove auniversality result that in particular implies that the correlation kernel ofthe orthogonal polynomial ensemble tends to the sine kernel as the length$\|I\|=2R$ tends to infinity, thereby answering a question posed by A.I. Bufetov. Let $\mathcal{P}({\bf N})$ be the power set of ${\bf N}$. We say that afunction $\mu^\ast: \mathcal{P}({\bf N}) \to \bf R$ is an upper density if, forall $X,Y\subseteq{\bf N}$ and $h, k\in{\bf N}^+$, the following hold: (F1)$\mu^\ast({\bf N}) = 1$; (F2) $\mu^\ast(X) \le \mu^\ast(Y)$ if $X \subseteq Y$;(F3) $\mu^\ast(X \cup Y) \le \mu^\ast(X) + \mu^\ast(Y)$; (F4) $\mu^\ast(k\cdotX) = \frac{1}{k} \mu^\ast(X)$, where $k \cdot X:=\{kx: x \in X\}$; (F5)$\mu^\ast(X + h) = \mu^\ast(X)$. We show that the upper asymptotic, upper logarithmic, upper Banach, upperBuck, upper Polya, and upper analytic densities, together with all upper$\alpha$-densities (with $\alpha$ a real parameter $\ge -1$), are upperdensities in the sense of our definition. Moreover, we establish the mutualindependence of axioms (F1)-(F5), and we investigate various properties ofupper densities (and related functions) under the assumption that (F2) isreplaced by the weaker condition that $\mu^\ast(X)\le 1$ for every$X\subseteq{\bf N}$. Overall, this allows us to extend and generalize results so far independentlyderived for some of the classical upper densities mentioned above, thusintroducing a certain amount of unification into the theory. We prove a global well-posedness result for the Landau-Lifshitz equation withGilbert damping provided that the BMO semi-norm of the initial data is small.As a consequence, we deduce the existence of self-similar solutions in anydimension. In the one-dimensional case, we characterize the self-similarsolutions associated with an initial data given by some ($\mathbb{S}^2$-valued)step function and establish their stability. We also show the existence ofmultiple solutions if the damping is strong enough. Our arguments rely on thestudy of a dissipative quasilinear Schr\"odinger obtained via the stereographicprojection and techniques introduced by Koch and Tataru. The article is concerned with polynomials $g(x)$ whose graphs are "partiallypacked" between two horizontal tangent lines. We assume that most of the localmaximum points of $g(x)$ are on the first horizontal line, and most of thelocal minimum points on the second horizontal line, except several"exceptional" maximum or minimum points, that locate above or under two lines,respectively. In addition, the degree of $g(x)$ is exactly the number of allextremum points $+1$. Then we call $g(x)$ a multipartite Chebyshev polynomialassociated with the two lines. Under a certain condition, we show that $g(x)$is expressed as a composition of the Chebyshev polynomial and a polynomialdefined by the $x$-component data of the exceptional extremum points of $g(x)$and the intersection points of $g(x)$ and the two lines. Especially, we studyin detail bipartite Chebyshev polynomials, which has only one exceptionalpoint, and treat a connection between such polynomials and elliptic integrals. In this work, a generalization of pre-Gr\"{u}ss inequality is established.Several bounds for the difference between two \v{C}eby\v{s}ev functional areproved. This is a draft of my textbook on mathematical analysis and the areas ofmathematics on which it is based. The idea is to fill the gaps in the existingtextbooks. Any remarks from readers are welcome. This article studies sufficient conditions on families of approximatingkernels which provide $N$--term approximation errors from an associatednonlinear approximation space which match the best known orders of $N$--termwavelet expansion. These conditions provide a framework which encompasses somenotable approximation kernels including splines, so-called cardinal functions,and many radial basis functions such as the Gaussians and generalmultiquadrics. Examples of such kernels are given to justify the criteria, andsome computational experiments are done to demonstrate the theoretical results.Additionally, the techniques involved allow for some new results on $N$--terminterpolation of Sobolev functions via radial basis functions. We give a complete classification of analytic equivalence of germs ofparametric families of systems of linear differential equations unfolding ageneric resonant singularity of Poincar\'e rank 1 in dimension 2 whose leadingmatrix is a Jordan bloc. The moduli space of analytic equivalence classes isdescribed in terms of a tuple of formal invariants and a single analyticinvariant obtained from the trace of monodromy. Moreover, analytic normal formsare given for all such singularities. We also explain the phenomena ofconfluence of singularities and of change of order of Borel summability offormal solutions in dependence on the complex parameter. In this paper, we derive systems of ordinary differential equations (ODEs)satisfied by modular forms of level three, which are level three versions ofRamanujan's system of ODEs satisfied by the classical Eisenstein series. We obtain a two weight local Tb theorem for any elliptic and gradientelliptic fractional singular integral operator T on the real line, and any pairof locally finite positive Borel measures on the line. This includes theHilbert transform and in a sense improves on the T1 theorem by the authors andM. Lacey. Sharp weighted boundedness of Calder\'on-Zygmund operators on non-homogeneousspaces was obtained in \cite{vk}. In this article we address this problem formultilinear Calder\'on-Zygmund operators. We have adapted the method ofpointwise domination by suitable multuilinear sparse operators and as aconsequence we have sharp weighted bounds for multilinear Calder\'on-Zygmundoperators. The Whittaker-Shannon-Kotel'nikov (WSK) sampling theorem provides areconstruction formula for the bandlimited signals. In this paper, a novel kindof the WSK sampling theorem is established by using the theory of quaternionreproducing kernel Hilbert spaces. This generalization is employed to obtainthe novel sampling formulas for the bandlimited quaternion-valued signals. Aspecial case of our result is to show that the 2D generalized prolatespheroidal wave signals obtained by Slepian can be used to achieve a samplingseries of cube-bandlimited signals. The solutions of energy concentrationproblems in quaternion Fourier transform are also investigated. Let $C$ be a compact convex subset of $\mathbb{R}^n$, $f:C\to\mathbb{R}$ be aconvex function, and $m\in\{1, 2, ..., \infty\}$. Assume that, along with $f$,we are given a family of polynomials satisfying Whitney's extension conditionfor $C^m$, and thus that there exists $F\in C^{m}(\mathbb{R}^n)$ such that$F=f$ on $C$. It is natural to ask for further (necessary and sufficient)conditions on this family of polynomials which ensure that $F$ can be taken tobe convex as well. We give a satisfactory solution to this problem in the case$m=\infty$, and also less satisfactory solutions in the case of finite $m\geq2$ (nonetheless obtaining an almost optimal result for $C$ a finiteintersection of ovaloids). For a solution to a similar problem in the case$m=1$ (even for $C$ not necessarily convex), see arXiv:1507.03931,arXiv:1706.09808, arXiv:1706.02235. The quaternion Fourier transform (QFT), a generalization of the classical 2DFourier transform, plays an increasingly active role in particular signal andcolour image processing. There tends to be an inordinate degree of interestplaced on the properties of QFT. The classical convolution theorem andmultiplication formula are only suitable for 2D Fourier transform ofcomplex-valued signal, and do not hold for QFT of quaternion-valued signal. Thepurpose of this paper is to overcome these problems and establish thePlancherel and inversion theorems of QFT in the square integrable signals spaceL2. First, we investigate the behaviours of QFT in the integrable signals spaceL1. Next, we deduce the energy preservation property which extends functionsfrom L1 to L2 space. Moreover, some other important properties such as modifiedmultiplication formula are also analyzed for QFT. Athanasiadis conjectured that, for every positive integer $r$, the local$h$-polynomial of the $r$th edgewise subdivision of any simplex has only realzeros. In this paper, based on the theory of interlacing polynomials, we provethat a family of polynomials related to the desired local $h$-polynomial isinterlacing and hence confirm Athanasiadis' conjecture. This paper is devoted to a generalization of a Hadamard type inequality forthe permanent of a complex square matrix. Our proof is based on a non-trivialextension of a technique used in Carlen, Lieb and Loss (Methods andApplications of Analysis 13 (1) (2006) 1-17). We give an application tocoefficients of products of linear forms and show some auxiliary inequalities,which might be of independent interest. For data sampled from an arbitrary density on a manifold embedded inEuclidean space, the Continuous k-Nearest Neighbors (CkNN) graph constructionis introduced. It is shown that CkNN is geometrically consistent in the sensethat under certain conditions, the unnormalized graph Laplacian converges tothe Laplace-Beltrami operator, spectrally as well as pointwise. It is provedfor compact (and conjectured for noncompact) manifolds that CkNN is the uniqueunweighted construction that yields a geometry consistent with the connectedcomponents of the underlying manifold in the limit of large data. Thus CkNNproduces a single graph that captures all topological features simultaneously,in contrast to persistent homology, which represents each homology generator ata separate scale. As applications we derive a new fast clustering algorithm anda method to identify patterns in natural images topologically. Finally, weconjecture that CkNN is topologically consistent, meaning that the homology ofthe Vietoris-Rips complex (implied by the graph Laplacian) converges to thehomology of the underlying manifold (implied by the Laplace-de Rham operators)in the limit of large data. We prove in this note one weight norm inequalities for some positiveBergman-type operators. In this paper, we develop via real variable methods various characterisationsof the Hardy spaces in the multi-parameter flag setting. Thesecharacterisations include those via, the non-tangential and radial maximalfunction, the Littlewood--Paley square function and area integral, Riesztransforms and the atomic decomposition in the multi-parameter flag setting.The novel ingredients in this paper include (1) establishing appropriatediscrete Calder\'on reproducing formulae in the flag setting and a version ofthe Plancherel--P\'olya inequalities for flag quadratic forms; (2) introducingthe maximal function and area function via flag Poisson kernels and flagversion of harmonic functions; (3) developing an atomic decomposition via thefinite speed propagation and area function in terms of flag heat semigroups. Asa consequence of these real variable methods, we obtain the fullcharacterisations of the multi-parameter Hardy space with the flag structure. Given an ideal $\mathcal{I}$ on $\omega$, we prove that a sequence in atopological space $X$ is $\mathcal{I}$-convergent if and only if there exists a``big'' $\mathcal{I}$-convergent subsequence. Then, we study several propertiesand show two characterizations of the set of $\mathcal{I}$-cluster points asclassical cluster points of a filters on $X$ and as the smallest closed setcontaining ``almost all'' the sequence. As a consequence, we obtain that theunderlying topology $\tau$ coincides with the topology generated by the pair$(\tau,\mathcal{I})$. The hook length formula for $d$-complete posets states that the $P$-partitiongenerating function for them is given by a product in terms of hook lengths. Wegive a new proof of the hook length formula using $q$-integrals. The proof isdone by a case-by-case analysis consisting of two steps. First, we express the$P$-partition generating function for each case as a $q$-integral and then weevaluate the $q$-integrals. Several $q$-integrals are evaluated using partialfraction expansion identities and others are verified by computer. We give a survey of elliptic hypergeometric functions associated with rootsystems, comprised of three main parts. The first two form in essence anannotated table of the main evaluation and transformation formulas for elliptichypergeometric integeral and series on root systems. The third and final partgives an introduction to Rains' elliptic Macdonald-Koornwinder theory (in partalso developed by Coskun and Gustafson).
Short version : What is the motivation behind the local condition in the definition of a test category ? Why do we want the slice categories $A/a$ to be weak test ? Long version : I start with some background. In Pursuing stacks, Grothendieck searches for a characterization of modelizers, that is of categories $\mathcal M$ together with a class of arrows $W$ such that the localization $W^{-1}\mathcal M$ is equivalent to the category $\mathsf{Hot}$ of homotopy types. He concentrates on categories $\mathcal M$ where the weak equivalences (the elements of $W$) are arising naturally. Among them, there are the toposes $\mathcal X$, where an arrow $\varphi \colon x \to x'$ is a weak equivalence if and only if the induced geometric morphism $\mathcal X/x \to \mathcal X/x'$ is an equivalence of Artin-Mazur$^{[1]}$. Notably, if $\mathcal A$ is a small category, the weak equivalence of the topos $\widehat{\mathcal A}$ are those maps $F \to F'$ between presheaves such that the induced morphism$$ \widehat{(\mathcal A/F)} \simeq \widehat{\mathcal A}/F \to \widehat{\mathcal A}/F' \simeq \widehat{(\mathcal A/F')} $$is an equivalence of Artin-Mazur. It can be showed that those are precisely the maps $F \to F'$ such that the induced functor $\mathcal A/F \to \mathcal A/F'$ is in the class $\mathcal W_\infty$ of functors between small categories whose topological realizations of the nerve are weak homotopy equivalences. For a small category $\mathcal A$, denote $i_{\mathcal A} \colon \widehat{\mathcal A} \to \mathsf{Cat}$ the functor $F \mapsto \mathcal A/F$. The weak equivalence of $\widehat{\mathcal A}$ just described are the elements of $W_{\widehat{\mathcal A}} = i_{\mathcal A}^{-1}(\mathcal W_\infty)$. Moreover $i_{\mathcal A}$ admits a right adjoint $i_{\mathcal A}^\ast$ (namely $\mathcal C \mapsto \hom_{\mathsf{Cat}}(\mathcal A/-,\mathcal C)$). Grothendieck then defines a weak test category as a small category $\mathcal A$ such that the adjunction $i_{\mathcal A} \dashv i_{\mathcal A}^\ast$ respects weak equivalences and induces an equivalence$$ {W_{\widehat{\mathcal A}}}^{-1}\widehat{\mathcal A} \simeq \mathcal {W_\infty}^{-1}\mathsf{Cat}. $$Notably, $\widehat{\mathcal A}$ together with $W_{\widehat{\mathcal A}}$ is a modelizer. Even better, Grothendieck finds a good charaterization of weak test categories (namely, those are the categories $\mathcal A$ such that for any small category $\mathcal C$ with a final object, the functor $i_{\mathcal A}i_{\mathcal A}^\ast(\mathcal C) \to \mathcal C$ is in $\mathcal W_\infty$). Now, in every pieces of literature about the subject that I have read so far, the author write something like : "But the notion of weak test category has the inconvenient not to be local". This is the starting point for Grothendieck to refine his notion : a test category is a weak test category $\mathcal A$ such that for every object $a$, the slice category $\mathcal A/a$ is weak test. But why do we even want the slice categories to be test ? Weak test is already a sufficient condition on $\mathcal A$ for the topos $\widehat{\mathcal A}$ to be a modelizer : why the need to refine the notion ? [1] I know nothing about cohomology of toposes, but I surely am ready to admit that it is natural to think of a Artin-Mazur equivalence as a weak equivalence.
I'm interested in maximizing a function $f(\mathbf \theta)$, where $\theta \in \mathbb R^p$. The problem is that I don't know the analytic form of the function, or of its derivatives. The only thing that I can do is to evaluate the function point-wise, by plugging in a value $\theta_$ and get a NOISY estimate $\hat{f}(\theta_)$ at that point. If I want I can decrease the variability of these estimates, but I have to pay increasing computational costs. Here is what I have tried so far: Stochastic steepest descent with finite differences: it can work but it requires a lot of tuning (ex. gain sequence, scaling factor) and it is often very unstable. Simulated annealing: it works and it is reliable, but it requires lots of function evaluations so I found it quite slow. So I'm asking for suggestions/idea about possible alternative optimization method that can work under these conditions. I'm keeping the problem as general as possible in order to encourage suggestions from research areas different from mine. I must add that I would be very interested in a method that could give me an estimate of the Hessian at convergence. This is because I can use it to estimate the uncertainty of the parameters $\theta$. Otherwise I'll have to use finite differences around the maximum to get an estimate.
There is mathematical justification for setting Dirichlet boundary degrees of freedom to a value. However, you should adjust your variational form accordingly. If you are looking at a general problem, say: Find $u\in\mathcal{U}$ such that $a(u,w)=l(w) \ \ \forall w\in\mathcal{V}$ where $\mathcal{U}=\{u:\int \nabla u^2 < \infty, u=g\text{ on }\Gamma_D\}$ $\mathcal{V}=\{u:\int \nabla u^2 < \infty, u=0\text{ on }\Gamma_D\}$ Instead we can write $u = v + g$ where $v\in\mathcal{V}$ and $g$ is the Dirichlet condition. Then the variational form becomes $a(v+g,w)=l(w)$ or by using the linearity of $a(.,.)$ $a(v,w)=l(w)-a(g,w)$ In a finite element code, you can form your element stiffness matrix as if there were no boundary conditions. Then you take the column of the local matrix which corresponds to the Dirichlet boundary condition, scale it by the coefficient you want to enforce, and subtract it from the right-hand-side. This is the discrete form of what I wrote above, $-a(g,w)$. Then you zero out that column and the corresponding Dirichlet row, placing a 1 in the diagonal and the coefficient you wish to enforce. This decouples the equation from the system and yet sets the value you wish to enforce. I recommend The Finite Element Method: Linear Static and Dynamic Finite Element Analysis, by Tom Hughes. He has an expanded discussion of this issue starting on page 8.
Search Now showing items 1-10 of 27 Production of light nuclei and anti-nuclei in $pp$ and Pb-Pb collisions at energies available at the CERN Large Hadron Collider (American Physical Society, 2016-02) The production of (anti-)deuteron and (anti-)$^{3}$He nuclei in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been studied using the ALICE detector at the LHC. The spectra exhibit a significant hardening with ... Forward-central two-particle correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV (Elsevier, 2016-02) Two-particle angular correlations between trigger particles in the forward pseudorapidity range ($2.5 < \|\eta\| < 4.0$) and associated particles in the central range ($\|\eta\| < 1.0$) are measured with the ALICE detector in ... Measurement of D-meson production versus multiplicity in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV (Springer, 2016-08) The measurement of prompt D-meson production as a function of multiplicity in p–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector at the LHC is reported. D$^0$, D$^+$ and D$^{*+}$ mesons are reconstructed ... Measurement of electrons from heavy-flavour hadron decays in p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV (Elsevier, 2016-03) The production of electrons from heavy-flavour hadron decays was measured as a function of transverse momentum ($p_{\rm T}$) in minimum-bias p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with ALICE at the LHC for $0.5 ... Direct photon production in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV (Elsevier, 2016-03) Direct photon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV was studied in the transverse momentum range $0.9 < p_{\rm T} < 14$ GeV/$c$. Photons were detected via conversions in the ALICE ... Multi-strange baryon production in p-Pb collisions at $\sqrt{s_\mathbf{NN}}=5.02$ TeV (Elsevier, 2016-07) The multi-strange baryon yields in Pb--Pb collisions have been shown to exhibit an enhancement relative to pp reactions. In this work, $\Xi$ and $\Omega$ production rates have been measured with the ALICE experiment as a ... $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ production in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV (Elsevier, 2016-03) The production of the hypertriton nuclei $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ has been measured for the first time in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE ... Multiplicity dependence of charged pion, kaon, and (anti)proton production at large transverse momentum in p-Pb collisions at $\sqrt{s_{\rm NN}}$= 5.02 TeV (Elsevier, 2016-09) The production of charged pions, kaons and (anti)protons has been measured at mid-rapidity ($-0.5 < y < 0$) in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV using the ALICE detector at the LHC. Exploiting particle ... Jet-like correlations with neutral pion triggers in pp and central Pb–Pb collisions at 2.76 TeV (Elsevier, 2016-12) We present measurements of two-particle correlations with neutral pion trigger particles of transverse momenta $8 < p_{\mathrm{T}}^{\rm trig} < 16 \mathrm{GeV}/c$ and associated charged particles of $0.5 < p_{\mathrm{T}}^{\rm ... Centrality dependence of charged jet production in p-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV (Springer, 2016-05) Measurements of charged jet production as a function of centrality are presented for p-Pb collisions recorded at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector. Centrality classes are determined via the energy ...
Construct two functions $ f,g: R^+ → R^+ $ satisfying: $f, g$ are continuous; $f, g$ are monotonically increasing; $f \ne O(g)$ and $g \ne O(f)$. Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. It only takes a minute to sign up.Sign up to join this community There are many examples for such functions. Perhaps the easiest way to understand how to get such an example, is by manually constructing it. Let's start with function over the natural numbers, as they can be continuously completed to the reals. A good way to ensure that $f\neq O(g)$ and $g\neq O(f)$ is to alternate between their orders of magnitude. For example, we could define $f(n)=\begin{cases} n & n \text{ is odd}\\ n^2 & n \text{ is even}\\ \end{cases}$ Then, we could have $g$ behave the opposite on the odds and evens. However, this doesn't work for you, because these functions are not monotonically increasing. However, the choice of $n,n^2$ was somewhat arbitrary, and we could simply increase the magnitudes so as to have monotonicity. This way, we may come up with: $f(n)=\begin{cases} n^{2n} & n \text{ is odd}\\ n^{2n-1} & n \text{ is even}\\ \end{cases}$, and $g(n)=\begin{cases} n^{2n-1} & n \text{ is odd}\\ n^{2n} & n \text{ is even}\\ \end{cases}$ Clearly these are monotone functions. Also, $f(n)\neq O(g(n))$, since on the odd integers, $f$ behaves like $n^{2n}$ while $g$ behaves like $n^{2n-1}=n^{2n}/n=o(n^{2n})$, and vice-versa on the evens. Now all you need is to complete them to the reals (e.g. by adding linear parts between the integers, but this is really beside the point). Also, now that you have this idea, you could use the trigonometric functions in order to construct ``closed formulas'' for such functions, since $\sin$ and $\cos$ are oscillating, and peak on alternating points. Good illustration for me is: http://www.wolframalpha.com/input/?i=sin%28x%29%2B2x%2C+cos%28x%29%2B2x $$ f(n) =2x+sin(x) $$ $$ g(n) =2x+cos(x) $$ $$ f\neq O(g) $$ $$ g\neq O(f) $$
This question seems to have been specifically designed to make you realise that the homology groups of a space can sometimes pick out homotopical information about a space which is not contained in the fundamental group alone. In this case, we have $X$ which is homotopy equivalent to $S^2\vee S^1\vee S^1\vee S^1$ (as you rightly pointed out, although were one small step away from reaching) and then we have a space which I'll call $Y$ which is homotopy equivalent to $S^1\vee S^1\vee S^1$. Now then, $X$ and $Y$ are both connected and path connected (the first place we'd look to see if two spaces were homotopy equivalent). Also, $X$ and $Y$ have the same fundamental group by a trivial application of Van-Kampen's theorem giving $$\pi_1(X)\cong\pi_1(Y)\cong F_3$$ where $F_3$ is the free group on three generators. This also tells us that $H_0$ and $H_1$ will be isomorphic as $H_0$ counts path components and $H_1\cong\pi_1^{ab}$ the abelianisation of the fundamental group. Where is the next place to look then? There are a few approaches which I'd suggest initially. The first would be to note that $\pi_2(X)$ is non-trivial (generated by the inclusion of $S^2$ in to the wedge product) and $\pi_2(Y)$ is trivial as $Y$ is a graph. However, you may not have met the higher homotopy groups yet so disregard this approach if you haven't. Next, I would suggest just calculating homology groups. This is rather easy if you've had enough practice calculating simplicial homology and you should find that $H_2(X)$ is non-trivial (again generated by the inclusion map) whereas $H_2(Y)$ is trivial (using a standard dimension argument). This is probably the approach your text/teacher expects. The last approach I would suggest, which you may be comfortable with, is to find the universal cover of $X$ and $Y$. We know that $Y$ is a graph and so its universal cover $\tilde{Y}$ is contractible, however $X$ has a universal cover $\tilde{X}$ which is not contractible (essentially by the same argument that $S^2$ is not contractible, the identity map $S^2\rightarrow S^2$ is not homotopic to the constant map). This last approach is probably more contrived than the other two approaches though as the easiest way to show that the identity map $S^2\rightarrow S^2$ is not homotopic to the constant map is to use cellular homology.
Could any one give an example of a bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$? Thank you. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community Could any one give an example of a bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$? Thank you. This question appears to be off-topic. The users who voted to close gave this specific reason: First, note that it is enough to find a bijection $f:\Bbb R^2\to \Bbb R$, since then $g(x,y,z) = f(f(x,y),z)$ is automatically a bijection from $\Bbb R^3$ to $\Bbb R$. Next, note that since there is a bijection from $[0,1]\to\Bbb R$ (see appendix), it is enough to find a bijection from the unit square $[0,1]^2$ to the unit interval $[0,1]$. By constructions in the appendix, it does not really matter whether we consider $[0,1]$, $(0,1]$, or $(0,1)$, since there are easy bijections between all of these. There are a number of ways to proceed in finding a bijection from the unit square to the unit interval. One approach is to fix up the "interleaving" technique I mentioned in the comments, writing $\langle 0.a_1a_2a_3\ldots, 0.b_1b_2b_3\ldots\rangle$ to $0.a_1b_2a_2b_2a_3b_3\ldots$. This doesn't quite work, as I noted in the comments, because there is a question of whether to represent $\frac12$ as $0.5000\ldots$ or as $0.4999\ldots$. We can't use both, since then $\left\langle\frac12,0\right\rangle$ goes to both $\frac12 = 0.5000\ldots$ and to $\frac9{22} = 0.40909\ldots$ and we don't even have a function, much less a bijection. But if we arbitrarily choose to the second representation, then there is no element of $[0,1]^2$ that is mapped to $\frac12$, and if we choose the first there is no element that is mapped to $\frac9{22}$, so either way we fail to have a bijection. This problem can be fixed. (In answering this question, I tried many web searches to try to remember the fix, and I was amazed at how many sources I found that ignored the problem, either entirely, or by handwaving. I never did find it; I had to remember it. Sadly, I cannot remember where I saw it first.) First, we will deal with $(0,1]$ rather than with $[0,1]$; bijections between these two sets are well-known, or see the appendix. For real numbers with two decimal expansions, such as $\frac12$, we will agree to choose the one that ends with nines rather than with zeroes. So for example we represent $\frac12$ as $0.4999\ldots$. Now instead of interleaving single digits, we will break each input number into chunks, where each chunk consists of some number of zeroes (possibly none) followed by a single non-zero digit. For example, $\frac1{200} = 0.00499\ldots$ is broken up as $004\ 9\ 9\ 9\ldots$, and $0.01003430901111\ldots$ is broken up as $01\ 003\ 4\ 3\ 09\ 01\ 1\ 1\ldots$. This is well-defined since we are ignoring representations that contain infinite sequences of zeroes. Now instead of interleaving digits, we interleave chunks. To interleave $0.004999\ldots$ and $0.01003430901111\ldots$, we get $0.004\ 01\ 9\ 003\ 9\ 4\ 9\ldots$. This is obviously reversible. It can never produce a result that ends with an infinite sequence of zeroes, and similarly the reverse mapping can never produce a number with an infinite sequence of trailing zeroes, so we win. A problem example similar to the one from a few paragraphs ago is resolved as follows: $\frac12 = 0.4999\ldots$ is the unique image of $\langle 0.4999\ldots, 0.999\ldots\rangle$ and $\frac9{22} = 0.40909\ldots$ is the unique image of $\langle 0.40909\ldots, 0.0909\ldots\rangle$. This is enough to answer the question posted, but I will give some alternative approaches. According to the paper "Was Cantor Surprised?" by Fernando Q. Gouveâ, Cantor originally tried interleaving the digits himself, but Dedekind pointed out the problem of nonunique decimal representations. Cantor then switched to an argument like the one Robert Israel gave in his answer, based on continued fraction representations of irrational numbers. He first constructed a bijection from $(0,1)$ to its irrational subset (see this question for the mapping Cantor used and other mappings that work), and then from pairs of irrational numbers to a single irrational number by interleaving the terms of the infinite continued fractions. Since Cantor dealt with numbers in $(0,1)$, he could guarantee that every irrational number had an infinite continued fraction representation of the form $$x = x_0 + \dfrac{1}{x_1 + \dfrac{1}{x_2 + \ldots}}$$ where $x_0$ was zero, avoiding the special-case handling for $x_0$ in Robert Israel's solution. The Cantor-Schröder-Bernstein theorem takes an injection $f:A\to B$ and an injection $g:B\to A$, and constructs a bijection between $A$ and $B$. So if we can find an injection $f:[0,1)^2\to[0,1)$ and an injection $g:[0,1)\to[0,1)^2$, we can invoke the CSB theorem and we will be done. $g$ is quite trivial; $x\mapsto \langle x, 0\rangle$ is one of many obvious injections. For $f$ we can use the interleaving-digits trick again, and we don't have to be so careful because we need only an injection, not a bijection. We can choose the representation of the input numbers arbitrarily; say we will take the $0.5000\ldots$ representation rather than the $0.4999\ldots$ representation. Then we interleave the digits of the two input numbers. There is no way for the result to end with an infinite sequence of nines, so we are guaranteed an injection. Then we apply CSB to $f$ and $g$ and we are done. There is a bijection from $(-\infty, \infty)$ to $(0, \infty)$. The map $x\mapsto e^x$ is an example. There is a bijection from $(0, \infty)$ to $(0, 1)$. The map $x\mapsto \frac2\pi\tan^{-1} x$ is an example, as is $x\mapsto{x\over x+1}$. There is a bijection from $[0,1]$ to $(0,1]$. Have $0\mapsto \frac12, \frac12\mapsto\frac23,\frac23\mapsto\frac34,$ and so on. That takes care of $\left\{0, \frac12, \frac23, \frac34,\ldots\right\}$. For any other $x$, just map $x\mapsto x$. Similarly, there is a bijection from $(0,1]$ to $(0,1)$. First, note that the exponential function is a bijective map of $\mathbb R$ to $(0,\infty)$. Now let $G$ be the irrationals in $(0,\infty)$. I'd like a bijective map of $(0,\infty)$ to $G$. It can be done as follows: if $x = r \pi^n$ for some nonnegative integer $n$ and rational $r$, let $f(x) = \pi x$, otherwise $f(x) = x$. Finally, it suffices to find a bijective map of $G^3$ to $G$. This can be obtained using continued fractions. Each $x \in G$ can be expressed in a unique way as an infinite continued fraction $x = x_0 + \dfrac{1}{x_1 + \dfrac{1}{x_2 + \ldots}}$ where $x_0$ is a nonnegative integer and $x_1, x_2, \ldots$ are positive integers. Denote this as $[x_0; x_1, x_2, \ldots]$. We then map $(x,y,z) \in G^3$ to $[x_0; y_0+1, z_0+1, x_1, y_1, z_1, \ldots]$.
R C Soni Articles written in Proceedings – Mathematical Sciences Volume 100 Issue 1 April 1990 pp 21-24 In this paper we evaluate the inverse Laplace transform of$$\begin{gathered} s^{ - \eta } (s^{l_1 } + \lambda _1 )^{ - \sigma } (s^{l_2 } + \lambda _2 )^{ - \rho } \hfill \\ \times S_n^m [xs^{ - W} (S^{l_1 } + \lambda _1 )^{ - \upsilon } (S^{l_2 } + \lambda _2 )^{ - w} ]S_{n'}^{m'} [ys^{ - w'} (S^{l_1 } + \lambda _1 )^{ - \upsilon '} (S^{l_2 } + \lambda _2 )^{ - w_r } ] \hfill \\ \times H[z_1 s^{ - W_1 } (S^{l_1 } + \lambda _1 )^{ - \upsilon _1 } (S^{l_2 } + \lambda _2 )^{ - w_1 } ,...,z_r s^{ - w_r } (S^{l_1 } + \lambda _1 )^{ - \upsilon _r } (S^{l_2 } + \lambda _2 )^{ - w'} ] \hfill \\ \end{gathered} $$ Due to the general nature of the multivariable H-function involved herein, the inverse Laplace transform of the product of a large number of special functions involving one or more variables, occurring frequently in the problems of theoretical physics and engineering sciences can be obtained as simple special cases of our main findings. For the sake of illustration, we obtain here the inverse Laplace transform of a product of the Hermite polynomials, the Jacobi polynomials and Volume 104 Issue 2 May 1994 pp 339-349 In the present paper we derive three interesting expressions for the composition of two most general fractional integral oprators whose kernels involve the product of a general class of polynomials and a multivariable Current Issue Volume 129 \| Issue 5 November 2019 Click here for Editorial Note on CAP Mode
Concerning partial fraction decomposition I can proof the following theorem: For two polynomials $f(x),g(x) \ne 0$ with coefficients in $\mathbb C$ one can find the following unique representation: $$\frac{f(x)}{g(x)}=\sum_{i=1}^{m}\sum_{j=1}^{n_i}\frac{a_{ij}}{(x-\alpha_i)^{j}}+p(x)$$ where $\alpha_1,\dots,\alpha_m$ are zeros of $g(x)$, $n_1,\dots,n_m$ their multiplicities, $p(x)$ is another polynomial and $a_{ij} \in \mathbb C, \forall i,j$. Now I read in lecture notes that if $f(x)$ and $g(x)$ have coefficients in $\mathbb R$, then one can represent $\frac{f(x)}{g(x)}$ as a linear combination (with real coefficients) of terms of the following types: $x^k, k \in \mathbb N_{\ge0}$; $\frac{1}{(x-\alpha)^k}, k \in \mathbb N_{\ge0}$ where $\alpha$ is a real zero of $g(x)$; $\frac{1}{\big((x-\mathcal Re(\alpha))^2+\mathcal Im(\alpha)\big)^k},k \in \mathbb N_{\ge0}$ where $\alpha$ is a complex zero of $g(x)$; $\frac{x-Re(\alpha)}{\big((x-\mathcal Re(\alpha))^2+\mathcal Im(\alpha)\big)^k},k \in \mathbb N_{\ge0}$ where $\alpha$ is a complex zero of $g(x)$; I do not fully see why it should be the case, but some aspects are clear to me. Specifically, elements of the type $x^k$ come from the polynomial $p(x)$, elements of the type $\frac{1}{(x-\alpha)^k}$ one can also clearly see in the representation given in the theorem above. To get elements of the types 3 and 4 one apparently needs to add two elements which are also present in the representation given by the theorem: $$\frac{a}{(x-\overline \alpha)^k}+\frac{b}{(x-\alpha)^k}=\frac{b(x-\overline \alpha)^k+a(x- \alpha)^k}{\big((x-\mathcal Re(\alpha))^2+\mathcal Im(\alpha)^2\big)^k}$$ So I am not sure how can I turn $b(x-\overline \alpha)^k+a(x- \alpha)^k$ into $1$ or $x-\mathcal Re(\alpha)$.
I'm told to use Gauss's Theorem to compute the flux of a field $\vec F = <x,y^2,y+z>$ along the boundary of the cylindrical solid $x^2+y^2 \le 4$ below $z=8$ and above $z=x$. I know by Gauss's Theorem that: Net Flux = $\iint_{\partial D} \vec F \cdot \vec ndS = \iiint_D \nabla \cdot \vec FdV$ This computation is pretty straight forward. $\nabla \cdot \vec F = 2+2y$. But the region of integration is particularly difficult to map out. I thought to use cylindrical coordinates and setting the bounds to $0 \le \theta \le 2 \pi$, $0 \le z \le 8$, and $0 \le r \le 4$, but this seems like it would just give me the area of the cylinder of height 8--and wouldn't include the part where z=x slices through the cylinder. What would be the right way to go in terms of the bounds of integration?
This question already has an answer here: We have the result, $\displaystyle{\frac{1}{R}=lim_{n\to\infty}\|\frac{a_{n+1}}{a_n}\|}$, where $R$ is the radius of the convergence, where $\displaystyle{a_n}$ is the coefficient of the series $\displaystyle{\sum_{n=0}^{\infty}}a_nz^n$, Here we redefine the series as$\displaystyle{\sum_{n=0}^{\infty}}a_nz^n$, where $a_n=(0,1,0,1,0,2,0,6,0,24,0,.....)$, so we can not use this method? or can we?, We have other result as $\displaystyle{\frac{1}{R}=\lim \inf} \|a_n\|^{-1/n}=\lim\inf{\frac{1}{\|a_n\|^{1/n}}}$ My question is 1)Is it valid to take $a_n=0$ for infinitely many $n\in \mathbb{N}$ 2) $\displaystyle{\frac{1}{R}=\lim\inf{\frac{1}{\|n!\|^{1/n}}}}????$ Can someone help how to move further
When working out the maths and solving for the Level conditional on experience XP, we obtain: $$Level = \frac{1 + \sqrt{1 + 8 \times XP \div 50}}{2}$$ For example, what is the player's level for \$XP = 300\$? $$ \frac{1 + \sqrt{1 + 8 \times 300 \div 50}}{2} = 4 $$ As requested. Or, what is the level for XP = 100000? $$ \frac{1 + \sqrt{1 + 8 \times 100000 \div 50}}{2} = 63 $$ More generally expressed, for an arbitrary starting threshold at Level 1: $$ Level = \frac{1 + \sqrt{1 + 8 \times threshold \div 50}}{2} $$ You can also do the reverse and calculate the XP needed for any given level by solving the above formula for XP. $$ XP = \frac{(Level^2-Level) \times threshold}{2} $$ Note that the above formula works with fractions but you need to round down to the next integer value. For example in C++/C# you could use (int)Level. To obtained the above closed form formula, I used difference equations, Gauss summation and a quadratic formula. If you are interested in the solution of this formula step by step... We do an recursive algorithm by starting our considerations that ultimately Experience(next_level) = Experience(current_level) + current_level50. For example, to obtain \$XP_{Level3}\$ we have: $$ XP_{Level3} = XP_{Level2} + 2 \times 50 $$ Where, 250 comes from the OP's request that experience needed to reach the next level is current level50. Now, we substitute \$Xp_{Level2}\$ with the same logic into the formula. That is: Substitute \$XP_{Level2} = XP_{Level1} + 2 \times 50\$ into the above formula: $$ Xp_{Level3} = Xp_{Level1} + 1 \times 50 + 2 \times 50 $$ and \$ Xp_{Level1} \$ is just 50, which is our starting point. Hence $$ Xp_{Level3} = 50 + 2 \times 50 = 150 $$ We can recognize a pattern for recursively calculating higher levels and a finite chain of summations. $$ Xp_{LevelN} = 50 + 2 \times 50 + 3 \times 50 + ... + (N-1) \times 50 = \sum_{i=0}^{n-1} i \times 50$$ Where N is the level to be achieved. To get the XP for the level N, we need to solve for N. $$ Xp_{LevelN} \div 50 = \sum_{i=0}^{n-1} i $$ Now the right hand side is simply a summation from 1 to N-1, which can be expressed by the famous Gaussian summation \$ N \times (N+1)\div2-N \$. Hence $$ Xp_{LevelN} \div 50 = N(N+1) \div 2-N $$ or just $$ 2(Xp_{LevelN} - 50) \div 50 = N(N+1)-2N $$ Finally, putting everything on one side: $$ 0 = N^2-N-2 \times Xp_{LevelN} \div 50 $$ This is now a quadratic formula yielding a negative and positive solution, of which only the positive is relevant as there are no negative levels. We now obtain: $$ N = \frac{1 + \sqrt{\frac{1+4 \times 2 \times Xp_{LevelN}}{50}}}{2} $$ The current level conditional on XP and linear threshold is therefore: $$ Level = \frac{1 + \sqrt{1+8\times XP \div threshold}}{2} $$ Note Knowing these steps can be useful to solve for even more complex progressions. In the RPG realm you will see besides a linear progression as here, the actually more common fractional power or square relationship, e.g. \$Level = \frac{\sqrt{XP}}{5.0}\$. However, for game implementation itself, I believe this solution to be less optimal as ideally you should know all your level progressions beforehand instead of calculating them at runtime. For my own engine, I use therefore pre-processed experience tables, which are more flexible and often faster. However, to write those tables first, or, to just merely ask yourself what XP is needed to, let's say, obtain Level 100, this formula provides the quickest way aimed at answering the OP's specific question. Edit: This formula is fully working as it should and it outputs correctly the current level conditional on XP with a linear threshold progression as requested by the OP. (The previous formula outputted "level+1" by assuming the player started from Level 0, which was my erring--I had solved it on my lunch break by writing on a small tissue! :)
I need to calculate $$\int_{-\infty}^{+\infty} \frac{1}{\left(e^x+ e^{-x}\right)^n} e^{-\rho x^2 + a x} dx$$ where $n \in \mathbb{N}$, $\rho > 0$ and $a \in \mathbb{R}$, but I don't know how to follow. I've tried to include the expression in symbolic software trying to get a result with respect other functions, but nothing. I start thinking about approximating the integral using numerical integration, but before, I would like to be sure that the integral can not be expressed with respect other functions. Does anybody knows if I should go directly for numerical integration? I am really lost, thank you in advance. Updates: Some particular cases can be computed using Wolfram Alpha. It seems that $$\int_{-\infty}^{+\infty} \frac{1}{\left(e^x+ e^{-x}\right)} e^{-\rho x^2 + x} dx = \frac{\sqrt{\pi/\rho}}{2}$$ and $$\int_{-\infty}^{+\infty} \frac{1}{\left(e^x+ e^{-x}\right)} e^{-\rho x^2 + 3 x} dx = \int_{-\infty}^{+\infty} \frac{1}{\left(e^x+ e^{-x}\right)} e^{-\rho x^2 - 3 x} dx = (2 e^{1/\rho}-1)\frac{\sqrt{\pi/\rho}}{2}.$$ More generally (thank you JanG) we have $$\int_{-\infty}^{+\infty} \frac{1}{\left(e^x+ e^{-x}\right)} e^{-\rho x^2 + (2m+1) x} dx = \frac{\sqrt{\pi/\rho}}{2} (-1)^m \left( 1 + 2 \sum_{\ell=1}^m (-1)^\ell e^{\ell^2/\rho}\right)$$ for $m$ non-negative integer.
I don't think counting even 1-D holes is appropriate for intuitive understanding of homology. Take a torus for example: its 1st homology $H_1$ is generated by a circle along the torus (the "hole" in this case it the dohnut hole), and a circle across the torus (now the "hole" is the void inside the dohnut surface). The connection between the "holes" and homology generators is not intuitive. More importantly, homology is innate to the manifold, not its embedding, so there may be no holes to speak of. I'd recommend to look intuitively at $k$th homology $H_k$ classes that are not zero as embedding of an oriented $k-$manifold $V^k$ into your manifold $M^n$ that cannot be contracted into a point along $M^n$, although I'm sure others would promptly correct me. [Prompt correction: $V^k$ represents zero in homology if it is the boundary of a $(k{+}1)$-manifold in $M$.] Two embeddings of $V_1^k$ and $V_2^k$ represent the same class if there is an oriented manifold with boundary $W^{k+1}$ embedded in $M^n$ such that its boundary consists of $V_1$ and $V_2$ with one of them having opposite orientation. As for cohomology classes $H^k$, they are dual to $H_k$ in linear algebra sense: elements of $H^k$ are linear functionals $w:H^k\to R$. IMO the best way to look at $H^k$ is from DeRham viewpoint, where elements of $H^k$ are represented by differential $k-$forms. The duality between $H^k$ and $H_k$ is straightforward: for $w\in H^k$ and $V\in H_k$ $w(V)=\int_V w $. EDIT: Based on HJRW's comments I've got to change the above description of homology groups to differentiate from homotopy ones. The core idea in homology is the notion of two embeddings of $V^k_i$ in $M^n$ being homologous, that is, equivalent under the relation describer in the 2nd paragraph of this answer. In particular, if $V^k_1$ is homologous to $V^k_2$, and $V^k_2$ is contractible, then homology class $[V^k_1]$ represented by $V^k_1$ is also $0$, even if $V_1^k$is not contractible along $M^n$. $H_k$ is naturally a group: union of embeddings defines the sum, and reversal of orientation negates the homology class. Moreover, $H_k$ is a commutative group, unlike $\pi_1$, precisely because the equivalence relation is more flexible than homotopy. There are other nice features of $H_k$ not present in $\pi_k$, for example $H_k(M^n)=0$ for $k>n$, etc. It's often useful to take embeddings with coefficients in a ring other than $Z$; to simlify the following let's assume in the following homology with real coefficients. A bit more on duality now. Suppose that $M^n$ is a compact manifold, and $V^k$ and $W^{n-k}$ are "nice enough" embeddings in general position to each other. Count their intersections with the sign that corresponds to whether or not the "combined" orientation of the tangents at the intersection point coincides with $M^n$ orientation. This leads to the map $(V^k,W^{n-k})\to R$. It turns out that this maps depends only on the embeddings' homology classes $[V^k]$ and $[W^{n-k}]$. This makes $H^{n-k}$ dual, in linear algebra sense, to $H^k$ (disclaimer #2: we are using real coefficients). This is known as Poincare duality. Recall now the definition of cohomology as dual to homology and you get Poincare isomorphism between $H^k$ and $H_{n-k}$. Given that isomorphism a natural question arises why do we need to define cohomology at all, wouldn't homology suffice? Well there are a few answers to that. My favourite is that cohomology has a very useful product operation that makes it into a ring. From DeRham viewpoint the product of cohomology classes corresponds to the wedge product of the corresponding differential forms. And that ring structure opens a treasure box of opportunities...
Background The best rational approximations $p/q$ to an irrational $\alpha$ are defined by the property$$\left\|\alpha - \frac{p}{q}\right\| < \left\|\alpha - \frac{p'}{q'}\right\|$$for all $q' \leq q$. The approximants $p/q$ are found by simply truncating the continued fraction expansion. The "most" irrational number is the Golden ratio $\phi$, which is defined by the property that for any given $N$, it has the most good approximations which satisfy $q < N$. Furthermore, for (i) algebraic and (ii) almost all irrational numbers, they satisfy the bound $$ \left\|\alpha - \frac{p}{q}\right\| > \frac{1}{q^{2+\epsilon}} $$ for any $\epsilon > 0$ and $q$ sufficiently large. Context I am interest in known generalisations of these results to the approximation of multiple irrationals. I have found a generalisation of part of the final result, which is provided by the Subspace theorem. The subspace theorem has the following corollary: for $D$ rationally independent algebraic numbers $(\alpha_1, \alpha_2, \ldots \alpha_D)$, $$ \left\|\alpha_d - \frac{p_d}{q}\right\| > \frac{1}{q^{1+1/D+\epsilon}} $$ for any $\epsilon > 0$, and $q$ sufficiently large. Questions My questions are: Is there a commonly used corresponding definition of the best rational approximations $(p_1/q,p_2/q \ldots p_D/q)$ to the irrational tuple $(\alpha_1, \alpha_2, \ldots \alpha_D)$? (generalising the first equation above) If there is a good definition, is there a better method than exhaustive search for finding the rational approximations $p_d/q$ to the irrational tuple $\alpha_d$? (generalising the truncated continued fraction expansion) For a given $D$ is there a known "most irrational" tuple $(\alpha_1, \alpha_2, \ldots \alpha_D)$ in the sense that there are the maximal number of good approximations satisfying $q<N$ for any $N$? (generalising the Golden ratio)
Short answer: no difference between Primal and Dual - it's only about the way of arriving to the solution. Kernel ridge regression is essentially the same as usual ridge regression, but uses the kernel trick to go non-linear. Linear Regression First of all, a usual Least Squares Linear Regression tries to fit a straight line to the set of data points in such a way that the sum of squared errors is minimal. We parametrize the best fit line with $\mathbb w$ and for each data point $(\mathbf x_i, y_i)$ we want $\mathbf w^T \mathbf x_i \approx y_i$. Let $e_i = y_i - \mathbf w^T \mathbf x_i$ be the error - the distance between predicted and true values. So our goal is to minimize the sum of squared errors $\sum e_i^2 = \\| \mathbf e \\|^2 = \\| X \mathbf w - \mathbf y \\|^2$ where $X = \begin{bmatrix}— \mathbf x_1 \,— \\ — \mathbf x_2 \,— \\ \vdots \\ — \mathbf x_n \,— \end{bmatrix}$ - a data matrix with each $\mathbf x_i$ being a row, and $\mathbf y = (y_1 , \ ... \ , y_n)$ a vector with all $y_i$'s. Thus, the objective is $\min\limits_{\mathbf w} \\| X \mathbf w - \mathbf y \\|^2$, and the solution is $\mathbf w = (X^T X)^{-1} X^T \mathbf y$ (known as "Normal Equation"). For a new unseen data point $\mathbf x$ we predict its target value $\hat y$ as $\hat y = \mathbf w^T \mathbf x$. Ridge Regression When there are many correlated variables in linear regression models, the coefficients $\mathbf w$ can become poorly determined and have lots of variance. One of the solutions to this problem is to restrict weights $\mathbf w$ so they don't exceed some budget $C$. This is equivalent to using $L_2$-regularization, also known as "weight decay": it will decrease the variance at the cost of sometimes missing the correct results (i.e. by introducing some bias). The objective now becomes $\min\limits_{\mathbf w} \\| X \mathbf w - y \\|^2 + \lambda \, \\| \mathbf w \\|^2$, with $\lambda$ being the regularization parameter. By going through the math, we obtain the following solution: $\mathbf w = (X^T X + \lambda \, I )^{-1} X^T \mathbf y$. It's very similar to the usual linear regression, but here we add $\lambda$ to each diagonal element of $X^T X$. Note that we can re-write $\mathbf w$ as $\mathbf w = X^T \, (X X^T + \lambda \, I)^{-1} \mathbf y$ (see here for details). For a new unseen data point $\mathbf x$ we predict its target value $\hat y$ as $\hat y = \mathbf x^T \mathbf w = \mathbf x^T X^T \, (X X^T + \lambda \, I)^{-1} \mathbf y$. Let $\boldsymbol \alpha = (X X^T + \lambda \, I)^{-1} \mathbf y$. Then $\hat y = \mathbf x^T X^T \boldsymbol \alpha = \sum\limits_{i=1}^{n} \alpha_i \cdot \mathbf x^T \mathbf x_i$. Ridge Regression Dual Form We can have a different look at our objective - and define the following quadratic program problem: $\min\limits_{\mathbf e, \mathbf w} \sum\limits_{i = 1}^n e_i^2$ s.t. $e_i = y_i - \mathbf w^T \mathbf x_i$ for $i = 1 \, .. \, n$ and $\\| \mathbf w \\|^2 \leqslant C$. It's the same objective, but expressed somewhat differently, and here the constraint on the size of $\mathbf w$ is explicit. To solve it, we define the Lagrangian $\mathcal L_p(\mathbf w, \mathbf e ; C)$ - this is the primal form that contains primal variables $\mathbf w$ and $\mathbf e$. Then we optimize it w.r.t. $\mathbf e$ and $\mathbf w$. To get the dual formulation, we put found $\mathbf e$ and $\mathbf w$ back to $\mathcal L_p(\mathbf w, \mathbf e ; C)$. So, $\mathcal L_p(\mathbf w, \mathbf e ; C) = \\| \mathbf e \\|^2 + \boldsymbol \beta^T (\mathbf y - X \mathbf w - \mathbf e) - \lambda \, (\\| \mathbf w \\|^2 - C)$. By taking derivatives w.r.t. $\mathbf w$ and $\mathbf e$, we obtain $\mathbf e = \cfrac{1}{2} \boldsymbol \beta$ and $\mathbf w = \cfrac{1}{2 \lambda} X^T \boldsymbol \beta$. By letting $\boldsymbol \alpha = \cfrac{1}{2 \lambda} \boldsymbol \beta$, and putting $\mathbf e$ and $\mathbf w$ back to $\mathcal L_p(\mathbf w, \mathbf e ; C)$, we get dual Lagrangian $\mathcal L_d(\boldsymbol \alpha, \lambda; C) = -\lambda^2 \\| \boldsymbol \alpha \\|^2 + 2 \lambda \, \boldsymbol \alpha^T y - \lambda \\| X^T \boldsymbol \alpha \\| - \lambda C$. If we take a derivative w.r.t. $\boldsymbol \alpha$, we get $\boldsymbol \alpha = (XX^T - \lambda I)^{-1} \mathbf y$ - the same answer as for usual Kernel Ridge regression. There's no need to take a derivative w.r.t $\lambda$ - it depends on $C$, which is a regularization parameter - and it makes $\lambda$ regularization parameter as well. Next, put $\boldsymbol \alpha$ to the primal form solution for $\mathbf w$, and get $\mathbf w = \cfrac{1}{2 \lambda} X^T \boldsymbol \beta = X^T \boldsymbol \alpha$. Thus, the dual form gives the same solution as usual Ridge Regression, and it's just a different way to come to the same solution. Kernel Ridge Regression Kernels are used to calculate inner product of two vectors in some feature space without even visiting it. We can view a kernel $k$ as $k(\mathbf x_1, \mathbf x_2) = \phi(\mathbf x_1)^T \phi(\mathbf x_2)$, although we don't know what $\phi(\cdot)$ is - we only know it exists. There are many kernels, e.g. RBF, Polynonial, etc. We can use kernels to make our Ridge Regression non-linear. Suppose we have a kernel $k(\mathbf x_1, \mathbf x_2) = \phi(\mathbf x_1)^T \phi(\mathbf x_2)$. Let $\Phi(X)$ be a matrix where each row is $\phi(\mathbf x_i)$, i.e. $\Phi(X) = \begin{bmatrix}— \phi(\mathbf x_1) \,— \\ — \phi(\mathbf x_2) \,— \\ \vdots \\ — \phi(\mathbf x_n) \,— \end{bmatrix}$ Now we can just take the solution for Ridge Regression and replace every $X$ with $\Phi(X)$: $\mathbf w = \Phi(X)^T \, (\Phi(X) \Phi(X)^T + \lambda \, I)^{-1} \mathbf y$. For a new unseen data point $\mathbf x$ we predict its target value $\hat y$ as $\hat y= \mathbf \phi(\mathbf x)^T \Phi(X)^T \, (\Phi(X) \Phi(X)^T + \lambda \, I)^{-1} \mathbf y$. First, we can replace $\Phi(X) \Phi(X)^T$ by a matrix $K$, calculated as $(K)_{ij} = k(\mathbf x_i, \mathbf x_j)$. Then, $\phi(\mathbf x)^T \Phi(X)^T$ is $\sum\limits_{i = 1}^n \phi(\mathbf x)^T \phi(\mathbf x_i) = \sum\limits_{i = 1}^n k(\mathbf x, \mathbf x_j)$. So here we managed to express every dot product of the problem in terms of kernels. Finally, by letting $\boldsymbol \alpha = (K + \lambda \, I)^{-1} \mathbf y$ (as previously), we obtain $\hat y= \sum\limits_{i = 1}^n \alpha_i k(\mathbf x, \mathbf x_j)$ References
The second Newton's law of motion (fundamental principle of dynamics) say that the sum of the forces $ \vec{F} $ on an object is equal to the mass $ m $ of this object multiplied by the acceleration $ \vec{a} $ of this object: $$ \sum{\vec{F}}=m.\vec{a} $$ This post illustrates the fundamental principle of dynamic on a very basic and understandable example. In the following, we will assume a mobile body moving on an inclined plane: The aim is to calculate the body's acceleration. We will assume the following hypotheses: This first approach is the simplest, it assumes that the body can only move in the direction of the inclined plane. According to this, all the forces can be projected along the plane direction. Applying the fundamental principle of dynamics gives: $$ m.\vec{a} = \sum{\vec{F}} = \vec{R} + \vec{F_g} = \vec{F_m} $$ The forces acting on the body are the gravity force $ \vec{F_g} $ and the ground reaction force $ \vec{R} $. As $ \vec{R} $ is perpendicular to the motion, its projection along the motion direction is null. The projection of $ \vec{F} $ along the motion direction is given by: $$ \| \vec{F_{m}} \| = \|\vec{F_g}\| . sin(\Theta) $$ We can now deduce that : $$ m.\|\vec{a} \| = \|\vec{F_g}\| . sin(\Theta) = m.g.sin(\Theta) $$ Note that the previous equatio n can be divided by $ m $. This shows that the motion is independent from the mass. Most people think that the heavier the body is, the fastest it will fall. But in fact, the falling speed is only dependent of frictions and gravity. The acceleration is thus given by: $$ \| \vec{a} \| = g.sin(\Theta) $$ If the plane is horizontal, $ \Theta=0 $, the body's acceleration is null, the body doesn't move. If the plane is vertical, $ \Theta=\pi/2 $, the body's acceleration is equal to g, the body is in free fall. In the previous section, we made an assumption on the acceleration direction. In the following, we will consider that the acceleration direction is unknown. The forces acting on the body are the gravity force $ \vec{F_g} $ and the ground reaction force $ \vec{R} $. We will project each force vector along the $ \vec{x} $ and $ \vec{y} $ axes. Projection of the gravity force: $$ \vec{F_g}= \left( \begin{array}{r c l} \vec{F_x} \ \vec{F_y} \end{array} \right) = \left( \begin{array}{r c l} 0 \ -m.g \end{array} \right) $$ Projection of the ground reaction force: $$ \| \vec{R} \| = m.g.cos(\Theta) $$ $$ \vec{R}= \left( \begin{array}{r c l} \vec{R_x} \ \vec{R_y} \end{array} \right) = \left( \begin{array}{r c l} -m.g.cos(\Theta).sin(\Theta) \ m.g.cos(\Theta).cos(\Theta) \end{array} \right) $$ As previously, the fundamental principle of dynamics gives: $$ m.\vec{a} = \sum{\vec{F}} = \vec{R} + \vec{F_g} = \vec{F_m} $$ $$ m.\vec{a}= \left( \begin{array}{r c l} \vec{F_x}+\vec{R_x} \ \vec{F_y}+\vec{R_y} \end{array} \right) = \left( \begin{array}{r c l} -m.g.cos(\Theta).sin(\Theta) \ -m.g+m.g.cos(\Theta).cos(\Theta) \end{array} \right) $$ $$ \vec{a}= g.\left( \begin{array}{r c l} -cos(\Theta).sin(\Theta) \ cos(\Theta)^2-1 \end{array} \right)= g.\left( \begin{array}{r c l} -sin(\Theta).cos(\Theta) \ -sin(\Theta).sin(\Theta) \end{array} \right) $$ This result confirms that the acceleration is given by: $$ \| \vec{a} \| = g.sin(\Theta) $$
As Chandra Chekuri pointed out in a comment, you could just compute the transitive closure via fast matrix multiplication, solving the problem in O($n^\omega$) time (use your favorite method, O($n^{2.376}$) via Coppersmith and Winograd, or more practically using Strassen's O($n^{2.81}$)), and this would be good for dense graphs. Now, I claim that if you can beat this running time for your problem for dense graphs, you would obtain an algorithm for triangle detection which is more efficient than computing the product of two Boolean matrices. The existence of such an algorithm is a major open problem. I'll reduce the triangle problem to the n-pairs-DAG-reachability problem.Suppose we are given a graph G on n nodes and we want to determine whether G contains a triangle. Now, from G create a DAG G' as follows. Create four copies of the vertex set, $V_1$, $V_2$, $V_3$, $V_4$. For copies $u_i\in V_i$, $v_{i+1}\in V_{i+1}$ for $i=1,2,3$, add an edge $(u_i,v_{i+1})$ iff $(u,v)$ was in G. Now if we ask whether there is a path between any of the pairs $(u_1, u_4)$ for all $u\in $G, then this would exactly be asking whether there is a triangle in $G$. The current graph has $4n$ nodes and we are asking about $n$ pairs. However, we can add $2n$ isolated dummy nodes and have $3n$ queries instead (by adding a query for $2n$ distinct pairs $(y,d)$ where $y\in V_2\cup V_3$ and $d$ a dummy), thus obtaining a $6n$-node instance of exactly your problem.
The longitudinal diffusion coefficients $D_L$ in the range of 0.1 to 1.5 kV/cm can be expressed as defined by the Einstein relation \[ D_L=\frac{\mu\epsilon_L}{e}=\left(\frac{a_0+a_1E+a_2E^{3/2}+a_3E^{5/2}}{1+(a_1/a_0)E+a_4E^2+a_5E^3}\right)\left(\frac{b_0+b_1E+b_2E^2}{1+(b_1/b_0)E+b_3E^2}\right)\left(\frac{T}{T_0}\right)^{-3/2}\left(\frac{T}{T_1}\right), \] with the parameters given previously. The transverse diffusion coefficients $D_T$ is related to $D_L$ by: \[ \frac{D_L}{D_T} = 1+ \frac{E}{\mu}\frac{\partial \mu}{\partial E} \] The corresponding diffusion length can be calculated from the drift time $t$ by: \[ \sigma = \sqrt{2Dt} \]
I stumbled across this question and I cannot figure out how to use the value of $\cos(\sin 60^\circ)$ which would be $\sin 0.5$ and $\cos 0.5$ seems to be a value that you can only calculate using a calculator or estimate at the very best. $\cos(\sin(\pi/3))=\cos(\sqrt{3}/2)$. (You can confirm the first step by observing the 30-60-90 special triangle.) I believe this is far as you can go without the use of a calculator. Using a calculator I get the final answer to be $0.6478593448524569104400717351567034556620295591904946...$ (according to Wolfram Alpha). Notice that $\cos(60)=\cos(\pi/3)$ so to remove confusion we need to write some extra notation. These are related by $\cos_{deg}(\theta)=\cos_{rad}(\frac{\pi\theta }{180})$ In degrees, I used my calculator and obtain $\sin_{deg}(60)=0.8660254038$ and hence $$\cos_{deg}(\sin_{deg}(60))=\cos_{deg}(0.8660254038) = 0.9998857706.$$ Notice the convention of degrees or radians matters here. I assume that the cosine in question is the cosine in degrees function. This is not to be confused with the usual radian-measure based cosine. For radian-based trig functions:$$ \cos_{rad}(\sin_{rad}(\pi/3))=\cos_{rad}(0.88660254038) = 0.6478593448 $$
I'm doing some revision using a study guide before the semester starts up again, and have gotten stuck on part of a question. The question I’m having trouble with is as follows: A rock of mass $m = 1.27kg$ is tied to a string and spun in a circle as it slides on a frictionless horizontal surface. The radius of the circle the rock follows is $r = 1.04m$. At a given moment, the string lies along the direction of the arrow in the diagram when the rock is in the position shown. The magnitude of the tension in the string is $T = 18.1N$. What are the speed and rate of change of the speed of the rock at that moment? Important equations: $\text{Tangential force} = T\sin(\theta)$ $\text{Radial force} = T\cos(\theta)$ $\text{Tangential acceleration} = dv/dt$ $\text{Radial acceleration} = v^2/r$ Using these equations you can find the velocity and acceleration, but the question doesn’t supply the angle needed to use these. Working backwards from the known solutions given in the example, $v = 3.82\text{m/s}$ and $a = 2.47\text{m/s}^2$, gives the angle to be about 10 degrees, but there has to be a way to find this using only the given values of $T$, $m$ and $r$. Can anyone suggest ways to get the angle? Every way I think of either leads to incorrect answers or requires already knowing the answer to the question.
I am trying to understand the concept of group velocity of a free particle wave packet: $$\Psi(x,t) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} \phi(k)e^{ikx}e^{-\frac{i \hbar k^2 t}{2m}}dk.$$ Where $$\omega(k) = \frac{\hbar k^2}{2m}.$$ Assuming that $\phi(k)$ is narrowly peaked about some particular value $k_0$ we can Taylor expand $\omega(k)$ about $k_0$ to get $$\omega(k) = \frac{\hbar k_0}{2m} + \frac{\hbar k_0}{m}(k-k_0)+ \frac{\hbar}{2m}(k-k_0)^2.$$ Then after defining variable $\kappa := k-k_0$ we get $$\Psi(x,t) \approx \frac{1}{\sqrt{2 \pi}}e^{\frac{-i \hbar k_0^2 t}{2m}}e^{\frac{i \hbar k_0^2 t}{m}} \int^{\infty}_{-\infty}\phi(\kappa + k_0)e^{i(\kappa + k_0)(x-\frac{\hbar k_0 t}{m})}dk.$$ This integral is the superposition of waves of the form $$e^{i(\kappa + k_0)(x-\frac{\hbar k_0 t}{m})}.$$ However notice that each of these waves have the same speed $\frac{\hbar k_0}{m}$. Then in terms of the energy $$v_{\text{group}} = \frac{d \omega(k_0)}{dk} = \frac{\hbar k_0}{m} = \sqrt{\frac{2 E_0}{m}}.$$ Hence for stationary state $\Psi_{k_0}$ we have that $$v_{\text{group}} = 2 v_{\text{phase}}.$$ Apparently we can then evaluate the group velocity for each $k$, hence $v_{\text{group}} = \frac{d \omega(k)}{dk}$. Question:I understand the above calculation. But qualitatively I don't understand how the group velocity can be a function of $k$ as opposed to just a single value (independent of $k$) which describes how fast the envelope propagates? I know that each stationary state has it's own phase velocity but I thought that superimposing these waves of different speeds produces an envelope which travels at a single velocity $v_{\text{group}}$?
The example I'm trying to understand is: $ \hat{S}_{x} \begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{pmatrix} = 1/2 \begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{pmatrix} $ My interpretation of this is that the vector shows you the probabilities of a particle being spin up or spin down if you square them. And I've been told that $ \hat{S}_{x} $ gives you the spin as an eigenvalue, but how? Since its 50:50 of getting -1/2 and 1/2. $ \hat{S}_{x} $ has only given you one of them. Is it that $ \hat{S}_{x} $ only measures the magnitude of spin in the x direction?
I want to find the equation of motion that comes from the following Lagrangian density $$\mathscr{L}=\mathbf{E}\cdot\left(\nabla^{2}\mathbf{E}\right)$$ where $E_{i}=\partial_{i}\phi\;(i=x,y,z)$ . In this case the $\phi$ and its 3rd derivatives are the independent variables. The Euler-Lagrange equation contains two terms $$\partial_{i}\frac{\mathscr{\partial L}}{\partial(\partial_{i}\phi)}$$ and $$\partial_{ijk}\frac{\mathscr{\partial L}}{\partial(\partial_{ijk}\phi)}.$$ For the first term I'm getting $$\partial_{i}\frac{\mathscr{\partial L}}{\partial(\partial_{i}\phi)}=\nabla\cdot\left(\nabla^{2}\mathbf{E}\right),$$ a nice scalar function. =============== Now, Im having problem with the 3rd derivatives term. QUESTION: Are permutation of the index to be considered independent variables or the same?, e.g, it is $\partial_{xxy}\phi$ the same independent variable as $\partial_{xyx}\phi$? If different permutations are to be understood as the same, I get the following $$\partial_{ijk}\frac{\mathscr{\partial L}}{\partial(\partial_{ijk}\phi)} = \;\partial_{xxx}E_{x}+3\partial_{xyy}E_{x}+3\partial_{xzz}E_{x} +3\partial_{xxy}E_{y}+\partial_{yyy}E_{y}+3\partial_{zzy}E_{y} +3\partial_{xxz}E_{z}+3\partial_{zyy}E_{z}+\partial_{zzz}E_{z},$$ which doesn't seems to be an scalar. QUESTION: Is my result correct? If not, why not (obviously)? If yes, then the last expression should be expressed in an scalar combination of $\nabla$ that I'm not seeing it, correct? EDIT: If,on the other hand, I should only have to consider one permutation of $\partial_{xxy}\phi$, then the result seems to be $$\partial_{ijk}\frac{\mathscr{\partial L}}{\partial(\partial_{ijk}\phi)} = \;\partial_{xxx}E_{x}+\partial_{xyy}E_{x}+\partial_{xzz}E_{x} +\partial_{xxy}E_{y}+\partial_{yyy}E_{y}+\partial_{zzy}E_{y} +\partial_{xxz}E_{z}+\partial_{zyy}E_{z}+\partial_{zzz}E_{z}=\nabla\cdot\left(\nabla^{2}\mathbf{E}\right),$$
It is not an answer, but just some hints. Consider the simplest free QFT with a massless bosonic scalar, the terms in the Lagrangian are local : $\phi(x) \square \phi(x)$. Considering an interacting theory ($\phi^3, \phi^4$). You are interested in calculate scattering amplitudes with incoming particles and outcoming particles. You will use the propagator in $\frac{1}{k^2}$ which form is direcly linked to the above Lagrangian term, and you may note that this propagator has a pole when $k$ is on-shell. If you consider only a tree-level diagram, the transition amplitude is simply the product of propagators, each propagator could be written $\frac{1}{l^2}$, where $l$ is the sum of some external momenta (at each vertex, you have momentum conservation). So a pole of the scattering amplitude corresponds to the pole of the propagators, and this corresponds by putting on-shell some particular sum of the external momenta. Now, consider a loop-diagram, with dimensional regularization, like $I(q) \sim g^2 (\mu^2)^\epsilon\int d^{4-\epsilon}p \frac{1}{p^2}\frac{1}{(p-q)^2}$,where $\epsilon$ is $>0$, $q$ is an external momentum. By using the Feynmann formula $\frac{1}{ab} = \int_0^1 \frac{dz}{[az+b(1-z)]^2}$, you will get : $I(q) \sim g^2 (\mu^2)^\epsilon\int_0^1 dz \int (d^{4-\epsilon}p) \dfrac{1}{[p^2 - 2p.q(1-z)+q^2(1-z)]^2}$, and finally : $I(q) \sim g^2 (\mu^2)^\epsilon ~\Gamma(\frac{\epsilon}{2})\int_0^1 dz \dfrac{1}{[q^2 z(1-z) ]^{\large \frac{\epsilon}{2}}}$ Here, $\epsilon$ is $>0$, so we see that if $q^2=0$, the integral is not defined, so $q^2=0$ should represent a pole for the scattering amplitude. The relation with the locality could be seen as looking at the Fourier transform (taking $\epsilon=0$) of the scattering amplitude which could be written $I(x) \sim [D(x)]^2$, where $D(x)$ is the propagator in space-time coordinates. Now, we should hope that any scattering amplitude, with loops, should have poles, which corresponds to some particular sum of the external momenta being on-shell.
In this appendix we provide the proof of Proposition 26, restated here in a self-contained way. We shall follow the arguments in Sect. 4.3 (Doubling trick) in [11] almost verbatim. In that paper, the target was a k-pronged graph with a single vertex; in the present paper, the target is a graph with finitely many vertices, and the arguments are very slightly modified. For A a conformal annulus, let $\hat{A}$ be the annulus obtained by doubling the annulus A across its boundary component $\partial _-A$. That is, if we denote, as usual, the boundary components $\partial A = \partial _+ A \sqcup \partial _- A$, then we set $\hat{A}$ to be the identification space of two copies of A, where the two copies of $\partial _- A$ are identified. We write this symbolically as $\hat{A}= A \sqcup _{\partial _- A} \bar{A}$, where $\bar{A}$ refers to A equipped with its opposite orientation. Proposition 42 Let A be a conformal annulus and let $\chi $ be a metric graph with finitely many vertices and edges of finite length. Fix a continuous map $\phi :\partial _+A\rightarrow \chi $ on one boundary component that takes on each value only finitely often, and consider the solution $h:A\rightarrow \chi $ to the partially free boundary problem that requires h to agree with $\phi $ on that boundary component $\partial _+A$. Then this map h extends by symmetry to a solution $\hat{h}$ of the symmetric Dirichlet-problem on the doubled annulus $\hat{A} = A^+ \sqcup _{\partial _{-}A} A^-$ where one requires a candidate $\phi $ to be the map on both boundary components of $\hat{A}$. In particular, we have $h = \hat{h}\vert _{A^+}$. Warmup to the proof of Proposition 42 We begin by assuming that the solution h to the partially free boundary problem described above has image $h(\partial _-A)$ of the boundary component $\partial _-A$ disjoint from the vertices of $\chi $. By this assumption, near the boundary $\partial _-A$, we have that h is a harmonic map to a smooth (i.e. non-singular) target locally isometric to a segment. First, we show that for the solution of the partially free-boundary problem, the normal derivative at the (free) boundary component $\partial _-A$ vanishes. We include the elementary computation below for the sake of completeness. Consider a family of maps $u_t: A \rightarrow \mathbb {R}$ defined for $t \in (-\epsilon , \epsilon )$ . A map $u_0=h$ in this family is critical for energy if $$\begin{aligned} 0&=\frac{d}{dt}\Bigr \|_{t=0} E(u_t) \\&= \frac{d}{dt}\Bigr \|_{t=0} \frac{1}{2} \iint _A \|\nabla u_t\|^2 dvol_A \\&= \iint _A \nabla \dot{u} \cdot \nabla u_0 dvol_A \\&= -\iint _A \dot{u} \Delta u_0 dvol_A + \int _{\partial A} \dot{u} \frac{\partial }{\partial \nu } u_0 dvol_{\partial A} \end{aligned}$$ where the last equality is obtained by an integration by parts. (Here $ \frac{\partial }{\partial \nu }$ indicates the outward normal derivative to the boundary.) Thus, since $\dot{u} = \frac{d}{dt}\Bigr \|_{t=0} u_t$ is arbitrary (and vanishes on the boundary component where the value of u is fixed), we see that the necessary conditions for a solution $u_0$ to the partially free boundary value problem are that $$\begin{aligned} \Delta u_0&=0\ (\text {in the interior of }A)\nonumber \\ \frac{\partial }{\partial \nu } u_0&=0 \text { (on the free boundary).} \end{aligned}$$ (23) We then show that the partially free boundary solution h is “half” of a Dirichlet problem on a doubled annulus. We follow an approach developed by A. Huang in his Rice University thesis (see Lemma 3.5 of [14 ]). Let $\hat{h}: \hat{A} \rightarrow \chi $ denote the map defined on $\hat{A}$ that restricts to h on the inclusion $A \subset \hat{A}$ and, in the natural reflected coordinates, on the inclusion $\bar{A} \subset \hat{A}$. By the continuity of h on A and its closure, it is immediate that $\hat{h}$ is continuous on $\hat{A}$. The vanishing of the normal derivative at the boundary (23) implies that the gradient $\nabla h\|_{\partial _- A}$ is parallel to $\partial _- A$. As that gradient is continuous on A up to the boundary (see e.g. [7], Theorem 6.3.6), we see that $\hat{h}$ has a continuously defined gradient on the interior of the doubled annulus $\hat{A}$. Next, note that because $\hat{h}$ is $C^1$ on $\hat{A}$, we have that $\hat{h}_i$ is weakly harmonic on $\hat{A}$. In particular, we can invoke classical regularity theory to conclude that $\hat{h}$ is then smooth and harmonic on $\hat{A}$. Thus, since $\chi $ is an NPC space, the map $\hat{h}$ is the unique solution to the Dirichlet harmonic mapping problem of taking $\hat{A}$ to $\chi $ with boundary values $h\|_{\partial _+ A}$. This concludes the proof of the model case. Next, we adapt this argument to the general case when the image of the boundary $h(\partial _-A)$ might possibly contain a vertex of the graph $\chi $. To accomplish the extension to the singular target case, we first analyze the behavior of the level set $h^{-1}(O)$ of a vertex O within the annulus A, particularly with respect to its interaction with the free boundary $\partial _- A$.$\square $ Lemma 43 Under the hypotheses above, any connected component of the level set $h^{-1}(O)$ of a vertex O within the annulus A meets the free boundary $\partial _- A$ in at most a single point. Proof We begin by noting that the proof of the Courant-Lebesgue lemma (see Lemma 9), based on an energy estimate for h on an annulus (see, for example, Lemma 3.2 in [28]) extends to hold for half-annuli, centered at boundary points of $\partial _- A$. Applying that argument yields a uniform estimate on the modulus of continuity of the map h on the closure of A only in terms of the total energy of h. Thus there is a well-defined continuous extension of the map h to $\partial _- A$. We now study this extension, which we continue to denote by h. First note that there cannot exist an arc $\Gamma \subset A \cap h^{-1}(O)$ in the level set for O in A for which $\Gamma $ meets $\partial _-A$ in both endpoints of $\partial \Gamma $. If not, then since A is an annulus, some component of $A {\setminus } \Gamma $ is bounded by arcs from $\partial _-A$ and $\Gamma $. But as $\partial _-A$ is a free boundary, we could then redefine h to map only to the vertex O on that component, lowering the energy. This then contradicts the assumption that h is an energy minimizer. Focusing further on the possibilities for the level set $h^{-1}(O)$, we note that by the assumption on the boundary values of h on $\partial _+A$ being achieved only a finitely many times, the level set $h^{-1}(O)$ can meet $\partial _+A$ in only a finite number of points (in fact the number of them is also fixed and equal to a number K in subsequent applications, since the boundary map would be a restriction of the collapsing map for a pole of finite order). Therefore, with these restrictions on the topology of $h^{-1}(O)$ in A in hand, we see that by the argument in the previous two paragraphs, each component of $h^{-1}(O)$ then must either be completely within, or have a segment contained in $\partial _-A$, or - the only conclusion we wish to permit - connects a single point of $\partial _-A$ with a preimage of the vertex on $\partial _+A$. Consider the first case where a component of $h^{-1}(O)$ is completely contained within $\partial _-A$. A neighborhood N of a point in such a component then has image h( N) entirely within a single prong, so the harmonic map on that neighborhood agrees with a classical (non-constant) harmonic function to an interval. Thus in a neighborhood of the boundary segment, say on a coordinate neighborhood $\{\mathfrak {I}(z) = y \in [0, \delta ) \}$, the requirements from Eq. (23) and that $h(0) = O$ and non-constant imply that the harmonic function h to (i) is expressible locally as $\mathfrak {I}(az^k) + O(\|z\|^{k+1})$ for some $k \ge 1$ and some constant $a \in \mathbb {C}^*$, (ii) be real analytic, and (iii) satisfy $\frac{\partial h}{\partial y} = 0$ (where $z = x+iy$). It is elementary to see that these conditions preclude this segment $h^{-1}(O)$ from being more than a singleton: that $h^{-1}(O)$ contains a segment defined by $\{y=0, x\in (-\epsilon , \epsilon )\}$ implies that the constant a in condition (i) is real. But then $0=\frac{\partial h}{\partial y} = \mathfrak {R}(akz^{k-1}) +O(\|z\|^{k})$ also on that segment $\{y=0, x\in (-\epsilon , \epsilon )\}$: thus $a = 0$, and so the map h must be constant, contrary to hypothesis. The same argument rules out the case when the level set $h^{-1}(O)$ meets the free boundary $\partial _-A$ in a segment, and that segment is connected by an arc of $h^{-1}(O)$ to $\partial _+A$. Namely, for this situation, we apply the argument of the previous paragraph to a subsegment of $h^{-1}(O)$ on $\partial _-A$ with a neighborhood whose image meets only an open prong, concluding as above that such a segment on $\partial _-A$ is not possible. Thus the intersection of such a component of the level set $h^{-1}(O)$ with the free boundary $\partial _-A$ is only a singleton, as needed. $\square $ Conclusion of the proof of Proposition 42: it remains to consider the case when the image of the boundary $\partial _-A$ by h contains a vertex O. It is straightforward to adapt, as follows, the argument we gave in the warmup for the smooth case to the singular setting. Consider a neighborhood of a point on $h^{-1}(O) \cap \partial _-A$. Doubling the map on that half-disk across the boundary $\partial _-A$ yields a harmonic map from the punctured disk to the graph (defined everywhere except at the isolated point $h^{-1}(O) \cap \partial _-A$). That harmonic map is smooth on the punctured disk and of finite energy, and hence has a Hopf differential of bounded $L^1$-norm. The puncture is then a removable singularity for that holomorphic differential, and hence for the harmonic map. The extended map $\hat{h}$ is then harmonic on the doubled annulus, and is the (unique) solution to the corresponding Dirichlet problem, as required. (Note that the normal derivative of the map may have a vanishing gradient at the boundary prior to doubling; this results in a zero of the Hopf differential on the central circle of the doubled annulus.)$\square $
I am a little confused about where does the mass and stiff matrix come from. In Discontinuous Galerkin we divide the domain in elements, $\Omega = \cup^K_{k=1} D^k$. Then assume the solution $u$ can be locally approximated by a polynomial as $$u_h=\sum_{i=1}^{Np}\hat u_i^k \Phi_i$$ We then introduce the approximated solution into the conservation law to get the residual $$\mathcal{R} = \frac{\partial u_h}{\partial t} + \frac{\partial au_h}{\partial x} = 0$$ We define now a space of test functions $V_h = \cup^K_{k=1}V_h^k$ where a locally defined member of $V_h^k$ is defined as $$\phi_h^k=\sum_{j=1}^{N_p} \hat \phi_j^k \Phi_j$$ Finally we require that the residual to be orthogonal to all the test functions in $V_h$ yielding $$\int_{D^k}\mathcal{R}(x,t)\Phi_j dx= 0$$ Why do we say that the residual must be orthogonal to alltest functions? Shouldn't we need it to be just orthogonal to the functions locally? In the inner product between the residual and the test functions, why are we using $\Phi_j$ instead of $\phi_h^k$? Is it because we are asking to be orthogonal to all the test functions, i.e. the basis of the test functions? So we don't care about the coefficients $\hat \phi_j^k$ ? How does the mass/stiffness matrix appear? This is just notation probably, but I don't see how from the integral a matrix is built. EDIT Perhaps I didn't explain myself correctly. So, given the PDE we multiply by a test function and integrate by parts over spatial domain, gives $$\int_{D^k} \frac{\partial u_h}{\partial t} \phi_j+ \frac{\partial au_h}{\partial x} \phi_j dx$$ $$\int_{D^k} \frac{\partial u_h}{\partial t} \phi_j+ \frac{\partial \phi_j}{\partial x} au_h = -[(au_h)\phi_j]^{k+1}_{k}$$ Now suppose $u_h = \sum_{i=1}^{N_p}\hat u_h \phi_i$ I get something like $$\int_{D^k} \frac{\partial}{\partial t}(\sum_{i=1}^{N_p}\hat u_h \phi_i) \phi_j+ \frac{\partial \phi_j}{\partial x} (\sum_{i=1}^{N_p}\hat u_h \phi_i) = -[(au_h)\phi_j]^{k+1}_{k}$$ where in this last step, I'm confused about two things should I replace $u_h$ in the RHS? From there how can a matrix arise? I'm sure I am missing something in my development but I cannot see what it is.
Originally posted on stats.stackexchange, I'll pair the post down to something a bit more general. Suppose I have vectors $\{\mathbf{\delta}, \mathbf{x}_1, \ldots, \mathbf{x}_J\}$, where $\delta \in \mathbb{R}^{J}$ and $\mathbf{x}_i \in \mathbb{R}^{I}$. Furthermore, let $\mathbf{A} \in \mathbb{R}^{I\times J}$ be such that $$ \mathbf{A} = \begin{pmatrix} \delta_1 \mathbf{x}_1 \\ \delta_2 \mathbf{x}_2 \\ \vdots \\ \delta_J \mathbf{x}_J \end{pmatrix} $$ Is there any sequence of BLAS-optimized operations that can make $\mathbf{A}$ from my set of vectors? To be concrete, I understand that if I have the matrix $$ \mathbf{B} = \begin{pmatrix} \mathbf{x}_1 \\ \mathbf{x}_2 \\ \vdots \\ \mathbf{x}_J \end{pmatrix} $$ then $$ \text{diag}(\delta) \, \mathbf{B} := \begin{pmatrix} \delta_1 & 0 & \cdots & 0 \\ 0 & \delta_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \delta_j \end{pmatrix} \mathbf{B} = \mathbf{A} $$ But it's not clear to me if making $\text{diag}(\delta)$ is an efficient operation, or even if the multiplication between $\text{diag}(\delta)$ and $\mathbf{B}$ is efficient. Edit: I noticed that a similar, fortran, specific question was asked and answered. However that question addresses efficient ways to multiply diagonal matrices (in fortran, no less). Whereas, my question references the use of diagonal matrices as one way in which the solution can be approached. My thoughts are that this question is a bit more general than the aforementioned one.
I wanted to better understand dfa. I wanted to build upon a previous question:Creating a DFA that only accepts number of a's that are multiples of 3But I wanted to go a bit further. Is there any way we can have a DFA that accepts number of a's that are multiples of 3 but does NOT have the sub... Let $X$ be a measurable space and $Y$ a topological space. I am trying to show that if $f_n : X \to Y$ is measurable for each $n$, and the pointwise limit of $\{f_n\}$ exists, then $f(x) = \lim_{n \to \infty} f_n(x)$ is a measurable function. Let $V$ be some open set in $Y$. I was able to show th... I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd.I can come up with arguments for that , but I also have arguments in the opposite direction.For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd... Consider a non-UFD that only has 2 units ( $-1,1$ ) and the min difference between 2 elements is $1$. Also there are only a finite amount of elements for any given fixed norm. ( Maybe that follows from the other 2 conditions ? )I wonder about counting the irreducible elements bounded by a lower... How would you make a regex for this? L = {w $\in$ {0, 1}* : w is 0-alternating}, where 0-alternating is either all the symbols in odd positions within w are 0's, or all the symbols in even positions within w are 0's, or both. I want to construct a nfa from this, but I'm struggling with the regex part
Let $\mathcal{C}, \mathcal{D}, \mathcal{E}$ be (symmetric?) monoidal categories, and $H : \mathcal{C} \times \mathcal{D} \to \mathcal{E}$ be a functor that is monoidal in both arguments, ie. $H(C,-)$ and $H(-,D)$ are (strong) monoidal for all objects $C$ and $D$. And take two monoids $M : \Delta \to \mathcal{C}$ and $N : \Delta \to \mathcal{D}$ (where $\Delta$ is the free monoidal category on a monoid, 1). $H(M 1, N 1)$ is a monoid in two different ways, as $H(M1, -)$ and $H(-,N1)$ both preserve monoids. I'd like to use an Eckmann-Hilton style argument to prove that the two monoids coincide and are commutative. Do I need additional assumptions to prove this is true? The concrete example I had in mind is: $\mathcal{C} = h\mathrm{Top}_^{op}$, $\mathcal{D} = h\mathrm{Top}_$, $\mathcal{E} = \mathrm{Set}$ and $H = \mathrm{Hom}$, $M$ is the "cogroup" on $S^1$ used to define the fundamental group and $N$ is some topological group and I'd like to conclude that $\pi_1(N,id_N) = \mathrm{Hom}_{h\mathrm{Top}_*}((S^1,1),(N,id_N))$ is abelian. Also, is there a category $\mathcal{K}$ representing the functors like $H$, ie. $$\mathrm{Hom}_\mathrm{MonCat}(\mathcal{K},\mathcal{E}) \cong \{ H : \mathcal{C} \times \mathcal{D} \to \mathcal{E} \text{ st. } H(C,-) \text{ and } H(-,D) \text{ are monoidal for all } C, D \}$$ sort of like a tensor product? Is it of any use?
I am trying to understand intuition of ADMM (alternating direction methods of multipliers). It combines dual ascent and method of multipliers. Downside of method of multiplier is the loss of decomposability. Why does method of multipliers lose decomposability? The question is missing some important context about the form of the objective function and constraints. I believe that the OP is referring to minimizing a function $ f(x)=f_{1}(x_{1})+\cdots + f_{N}(x_{N}) $ subject to constraints $Ax=b$. Here the vector $x$ is decomposed into $N$ blocks (that can be vectors of various sizes) $x=\left[ \begin{array}{c} x_{1} \\ x_{2} \\ \vdots \\ x_{N} \end{array} \right]$. We can write $Ax=b$ as $Ax=A_{1}x_{1}+ \cdots A_{N}x_{N}=b$. The Lagrangian can be written as $L(x,\lambda)=f_{1}(x_{1})+ \cdots + f_{N}(x)+ \lambda^{T}(A_{1}x_{1}+ \cdots +A_{N}x_{N}-b)$ In the dual decomposition method, we can minimize the Lagrangian with respect to $x$ by minimizing with respect to the components $x_{1}$, $x_{2}$, $\ldots$, $x_{N}$ in sequence. This works because $f(x)$ is decomposable and the $\lambda^{T}(Ax-b)$ term is also decomposable. In the method of multipliers, the Lagrangian is augmented with a term $\ldots + (\rho/2) \\| Ax -b \\|_{2}^{2}$. This extra term is not decomposable in terms of functions of $x_{1}$, $x_{2}$, $\ldots$, $x_{N}$ because it is quadratic and involves cross-terms of $x_{i}$ and $x_{j}$. Thus the augmented Lagrangian cannot be minimized by components in the same way as the unaugmented Lagrangian.
Search Now showing items 1-10 of 32 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 ...
Special quasirandom structures¶ Random alloys are often of special interest. This is true in particular forsystems that form random solid solutions below the melting point. It is,however, not always easy to model such structures, because the system sizesthat lend themselves to, for example, DFT calculations, are often too small toaccomodate a structure that may be regarded as random; the periodicity imposedby boundary conditions introduces correlations that make the modeled structuredeviate from the random alloy. This problem can sometimes be alleviated withthe use of so-called special quasirandom structures (SQS) [ZunWeiFer90]. SQScells are the best possible approximations to random alloys in the sense thattheir cluster vectors closely resemble the cluster vectors of truly randomalloys. This tutorial demonstrates how SQS cells can be generated in icet using a simulated annealing approach. There is no unique way to measure the similarity between the cluster vector ofthe SQS cell and the random alloy. The implementation in icet usesthe measure proposed in [WalTiwJon13]. Specifically, the objective function$Q$ is calculated as Here, $\Gamma_{\alpha}$ are components in the cluster vector and$\Gamma^\text{target}_{\alpha}$ the corresponding target values. Thefactor $\omega$ is the radius (in Ångström) of the largest pair clustersuch that all clusters with the same or smaller radii have$\Gamma_{\alpha} - \Gamma^\text{target}_{\alpha} = 0$. The parameter$L$, by default 1.0, can be specified by the user. The functionality for generating SQS cells is just a special case of a moregeneral algorithm for generating a structure with a cluster vector similar to any target cluster vector. The below example demonstrates both applications. Import modules¶ The generate_sqs and/or generate_target_structure functions need tobe imported together with some additional functions from ASE and icet. It is advisable to turnon logging, since the SQS cell generation may otherwise run quietly for a fewminutes. from ase import Atomfrom ase.build import bulkfrom icet import ClusterSpacefrom icet.tools.structure_generation import (generate_sqs, generate_sqs_by_enumeration, generate_target_structure)from icet.io.logging import set_log_configset_log_config(level='INFO') Generate binary SQS cells¶ In the following example, a binary FCC SQS cell with 8 atoms will begenerated. To this end, an icet.ClusterSpace and targetconcentrations need to be defined. The cutoffs in the cluster space areimportant, since they determine how many elements are to be included whencluster vectors are compared. It is usually sufficient to use cutoffs suchthat the length of the cluster vector is on the order of 10. Targetconcentrations are specified via a dictionary, which should contain all theinvolved elements and their fractions of the total number of atoms.Internally, the function carries out simulated annealing with Monte Carlotrial swaps and can be expected to run for a minute or so. primitive_structure = bulk('Au')cs = ClusterSpace(primitive_structure, [8.0, 4.0], ['Au', 'Pd'])target_concentrations = {'Au': 0.5, 'Pd': 0.5}sqs = generate_sqs(cluster_space=cs, max_size=8, target_concentrations=target_concentrations)print('Cluster vector of generated structure:', cs.get_cluster_vector(sqs)) In this simple case, in which the target structure size is very small, it is more efficient to generate the best SQS cell by exhaustive enumeration of all binary FCC structures having up to 8 atoms in the supercell: sqs = generate_sqs_by_enumeration(cluster_space=cs, max_size=8, target_concentrations=target_concentrations)print('Cluster vector of generated structure:', cs.get_cluster_vector(sqs)) Generation of SQS cells by enumeration is preferable over the Monte Carlo approach if the size of the system permits, because with enumeration there is no risk that the optimal SQS cell is missed. Generate SQS cells for a system with sublattices¶ It is possible to generate SQS cells also for systems with sublattices. In thebelow example, an SQS cell is generated for a system with two sublattices; oneFCC sublattice on which Au, Cu, and Pd are allowed, and anotherFCC sublattice on which H and vacancies (V) are allowed. Note thattarget concentrations are specified with respect to all atoms, which meansthat the concentrations must always sum up to 1. The example generates an SQScell for a supercell that is 16 times larger than the primitive cell, in total32 atoms. The keyword include_smaller_cells=False guarantees that thegenerated structure has 32 atoms (otherwise the structure search would havebeen carried out among structures having 32 atoms or less). In this example, the number of trial steps is manually set to 50,000. Thisnumber may be insufficient, but will most likely provide a reasonable SQScell, albeit perhaps not the best one. The default number of trial steps is3,000 times the number of inequivalent supercell shapes. The latter quantityincreases quickly with the size of the supercell. primitive_structure = bulk('Au', a=4.0)primitive_structure.append(Atom('H', position=(2.0, 2.0, 2.0)))cs = ClusterSpace(primitive_structure, [7.0], [['Au', 'Cu', 'Pd'], ['H', 'V']])target_concentrations = {'Au': 6 / 16, 'Cu': 1 / 16, 'Pd': 1 / 16, 'H': 2 / 16, 'V': 6 / 16}sqs = generate_sqs(cluster_space=cs, max_size=16, include_smaller_cells=False, target_concentrations=target_concentrations, n_steps=50000)print('Cluster vector of generated structure:', cs.get_cluster_vector(sqs)) Generate a structure matching an arbitrary cluster vector¶ The SQS cell generation approach can be utilized to generate the structurethat most closely resembles any cluster vector. To do so, one can employ thesame procedure but the target cluster vector must be specified manually. Notethat there are no restrictions on what target vectors can be specified (excepttheir length, which must match the cluster space length), but the space ofcluster vectors that can be realized by structures is restricted in multipleways. The similarity between the target cluster vector and the cluster vectorof the generated structure may thus appear poor. primitive_structure = bulk('Au')cs = ClusterSpace(primitive_structure, [5.0], ['Au', 'Pd'])target_cluster_vector = [1.0, 0.0] + [0.5] * (len(cs) - 2)target_concentrations = {'Au': 0.5, 'Pd': 0.5}sqs = generate_target_structure(cluster_space=cs, max_size=8, target_cluster_vector=target_cluster_vector, target_concentrations=target_concentrations)print('Cluster vector of generated structure:', cs.get_cluster_vector(sqs)) Source code¶ examples/sqs_generation.py """This example demonstrates how to generate special quasirandom structure."""# Import modulesfrom ase import Atomfrom ase.build import bulkfrom icet import ClusterSpacefrom icet.tools.structure_generation import (generate_sqs, generate_sqs_by_enumeration, generate_target_structure)from icet.io.logging import set_log_configset_log_config(level='INFO')# Generate SQS for binary fcc, 50 % concentrationprimitive_structure = bulk('Au')cs = ClusterSpace(primitive_structure, [8.0, 4.0], ['Au', 'Pd'])target_concentrations = {'Au': 0.5, 'Pd': 0.5}sqs = generate_sqs(cluster_space=cs, max_size=8, target_concentrations=target_concentrations)print('Cluster vector of generated structure:', cs.get_cluster_vector(sqs))# Use enumeration to generate SQS for binary fcc, 50 % concentrationsqs = generate_sqs_by_enumeration(cluster_space=cs, max_size=8, target_concentrations=target_concentrations)print('Cluster vector of generated structure:', cs.get_cluster_vector(sqs))# Generate SQS for a system with two sublattices,# fcc lattices with Au, Cu, Pd on one lattice and H, V on anotherprimitive_structure = bulk('Au', a=4.0)primitive_structure.append(Atom('H', position=(2.0, 2.0, 2.0)))cs = ClusterSpace(primitive_structure, [7.0], [['Au', 'Cu', 'Pd'], ['H', 'V']])target_concentrations = {'Au': 6 / 16, 'Cu': 1 / 16, 'Pd': 1 / 16, 'H': 2 / 16, 'V': 6 / 16}sqs = generate_sqs(cluster_space=cs, max_size=16, include_smaller_cells=False, target_concentrations=target_concentrations, n_steps=50000)print('Cluster vector of generated structure:', cs.get_cluster_vector(sqs))# Generate structure with a specified cluster vectorprimitive_structure = bulk('Au')cs = ClusterSpace(primitive_structure, [5.0], ['Au', 'Pd'])target_cluster_vector = [1.0, 0.0] + [0.5] * (len(cs) - 2)target_concentrations = {'Au': 0.5, 'Pd': 0.5}sqs = generate_target_structure(cluster_space=cs, max_size=8, target_cluster_vector=target_cluster_vector, target_concentrations=target_concentrations)print('Cluster vector of generated structure:', cs.get_cluster_vector(sqs))
The complex method of interpolation, going back to Calder\'on and Coifman etal., on the one hand, and the Alexander-Wermer-Slodkowski theorem on polynomialhulls with convex fibers, on the other hand, are generalized to a method ofinterpolation of real (finite-dimensional) Banach spaces and of convexfunctions. The underlying duality in this method is given by the Legendretransform. Our results can also be interpreted as new properties of solutionsof the homogeneous complex Monge-Amp\`ere equation. It is proven an analogue of the Theorem of Moser using The ApproximationTheorem of Artin. The purpose of this note is to prove some result which may be seen as newanalogue of the Theorem of Moser. Moduli spaces of stable parabolic bundles of parabolic degree $0$ over theRiemann sphere are stratified according to the Harder--Narasimhan filtration ofunderlying vector bundles. Over a Zariski open subset $\mathscr{N}_{0}$ of theopen stratum depending explicitly on a choice of parabolic weights, areal-valued function $\mathscr{S}$ is defined as the regularized critical valueof the non-compact Wess--Zumino--Novikov--Witten action functional. Thedefinition of $\mathscr{S}$ depends on a suitable notion of parabolic bundle`uniformization map' following from the Mehta--Seshadri andBirkhoff--Grothendieck theorems. It is shown that $-\mathscr{S}$ is a primitivefor a (1,0)-form $\vartheta$ on $\mathscr{N}_{0}$ associated with theuniformization data of each intrinsic irreducible unitary logarithmicconnection. Moreover, it is proved that $-\mathscr{S}$ is a K\"ahler potentialfor $(\Omega-\Omega_{\mathrm{T}})\|_{\mathscr{N}_{0}}$, where $\Omega$ is theNarasimhan--Atiyah--Bott K\"ahler form in $\mathscr{N}$ and$\Omega_{\mathrm{T}}$ is a certain linear combination of tautological$(1,1)$-forms associated with the marked points. These results provide anexplicit relation between the cohomology class $[\Omega]$ and tautologicalclasses, which holds globally over certain open chambers of parabolic weightswhere $\mathscr{N}_{0} = \mathscr{N}$. We construct new complete Einstein metrics on smoothly bounded strictlypseudoconvex domains in Stein manifolds. This is done by deforming theK\"ahler-Einstein metric of Cheng and Yau, the approach that generalizes theworks of Roth and Biquard on the deformations of the complex hyperbolic metricon the unit ball. Recasting the problem into the question of vanishing of an$L^2$ cohomology and taking advantage of the asymptotic complex hyperbolicityof the Cheng-Yau metric, we establish the possibility of such a deformationwhen the dimension of the domain is larger than or equal to three. It is studied the classification problem for formal holomorphic embeddingsbetween Shilov Boundaries of Bounded Symmetric Domains of First Type. For a reduced pure dimensional complex space $X$, we show that if Barlet'srecently introduced sheaf $\alpha_X^1$ of holomorphic $1$-forms or the sheaf ofgerms of weakly holomorphic $1$-forms is locally free, then $X$ is smooth.Moreover, we discuss the connection to Barlet's well-known sheaf $\omega_X^1$. Loewner driving functions encode simple curves in 2-dimensional simplyconnected domains by real-valued functions. We prove that the Loewner drivingfunction of a $C^{1,\beta}$ curve (differentiable parametrization with$\beta$-H\"older continuous derivative) is in the class $C^{1,\beta-1/2}$ if$1/2<\beta\leq 1$, and in the class $C^{0,\beta + 1/2}$ if $0 \leq \beta \leq1/2$. This is the converse of a result of Carto Wong and is optimal. We alsointroduce the Loewner energy of a rooted planar loop and use our regularityresult to show the independence of this energy from the basepoint. We study the behavior of real-normalized (RN) meromorphic differentials onRiemann surfaces under degeneration. We determine all possible limits of RNdifferentials in degenerating sequences of smooth curves, and describe thelimit in terms of solutions of the corresponding Kirchhoff problem. We furthershow that the limit of zeroes of RN differentials is the set of zeroes of atwisted meromorphic RN differential, which we explicitly construct. Our main new tool is an explicit solution of the jump problem on Riemannsurfaces in plumbing coordinates, by using the Cauchy kernel on thenormalization of the nodal curve. Since this kernel does not depend on plumbingcoordinates, we are able to approximate the RN differential on a smooth plumbedcurve by a collection of meromorphic differentials on the irreduciblecomponents of a stable curve, with an explicit bound on the precision of suchapproximation. This allows us to also study these approximating differentialsat suitable scales, so that the limit under degeneration is not identicallyzero. These methods can be applied more generally to study degenerations ofdifferentials on Riemann surfaces satisfying various conditions. Brjuno and R\"ussmann proved that every irrationally indifferent fixed pointof an analytic function with a Brjuno rotation number is linearizable, andYoccoz proved that this is sharp for quadratic polynomials. Douady conjecturedthat this is sharp for all rational functions of degree at least 2, i.e., thatnon-M\"obius rational functions cannot have Siegel disks with non-Brjunorotation numbers. We prove that Douady's conjecture holds for the class ofpolynomials for which the number of infinite tails of critical orbits in theJulia set equals the number of irrationally indifferent cycles. As a corollary,Douady's conjecture holds for the polynomials $P(z) = z^d + c$ for all $d > 1$and all complex $c$. We gave an alternative short proof on the finite generation of holomorphicfunctions with polynomial growth on Riemann surfaces with nonnegativecurvature. The first proof was due to Li and Tam. We study some features of the energy of a deterministic chordal Loewnerchain, which is defined as the Dirichlet energy of its driving function. Inparticular, using an interpretation of this energy as a large deviation ratefunction for SLE$_\kappa$ as $\kappa$ tends to 0 and the known reversibility ofthe SLE$_\kappa$ curves for small $\kappa$, we show that the energy of adeterministic curve from one boundary point A of a simply connected domain D toanother boundary point B, is equal to the energy of its time-reversal ie. ofthe same curve but viewed as going from B to A in D. Given a formal flat meromorphic connection over an excellent scheme over afield of characteristic zero, in a previous paper we established existence ofgood formal structures and a good Deligne-Malgrange lattice after suitablyblowing up. In this paper, we reinterpret and refine these results byintroducing some related structures. We consider the turning locus, which isthe set of points at which one cannot achieve a good formal structure withoutblowing up. We show that when the polar divisor has normal crossings, theturning locus is of pure codimension 1 within the polar divisor, and hence ofpure codimension 2 within the full space; this had been previously establishedby Andre in the case of a smooth polar divisor. We also construct anirregularity sheaf and its associated b-divisor, which measure irregularityalong divisors on blowups of the original space; this generalizes anotherresult of Andre on the semicontinuity of irregularity in a curve fibration. Oneconcrete consequence of these refinements is a process for resolution ofturning points which is functorial with respect to regular morphisms ofexcellent schemes; this allows us to transfer the result from schemes to formalschemes, complex analytic varieties, and nonarchimedean analytic varieties. We show the existence of a complex K3 surface $X$ which is not a Kummersurface and has a one-parameter family of Levi-flat hypersurfaces in which allthe leaves are dense. We construct such $X$ by patching two open complexsurfaces obtained as the complements of tubular neighborhoods of ellipticcurves embedded in blow-ups of the projective planes at general nine points. The aim of this paper is to investigate the equivalence conditions foruniform perfectness of quasi-metric spaces. We also obtain the invariantproperty of uniform perfectness under quasim\"obius maps in quasi-metricspaces. In the end, two applications are given. A new approach in Loewner Theory proposed by Bracci, Contreras,D\'iaz-Madrigal and Gumenyuk provides a unified treatment of the radial and thechordal versions of the Loewner equations. In this framework, a generalizedLoewner chain satisfies the differential equation $$ \partial_{t}f_{t}(z) = (z- \tau(t))(1-\overline{\tau(t)}z)\partial_{z}f_{t}(z)p(z,t), $$ where $\tau :[0,\infty) \to \overline{\mathbb{D}}$ is measurable and $p$ is called aHerglotz function. In this paper, we will show that if there exists a $k \in[0,1)$ such that $p$ satisfies $$ \|p(z,t) - 1\| \leq k \|p(z,t) + 1\| $$ for all$z \in \mathbb{D}$ and almost all $t \in [0,\infty)$, then $f_{t}$ has a$k$-quasiconformal extension to the whole Riemann sphere for all $t \in[0,\infty)$. The radial case ($\tau =0$) and the chordal case ($\tau=1$) havebeen proven by Becker [J. Reine Angew. Math. \textbf{255} (1972), 23-43] andGumenyuk and the author (Math. Z. \textbf{285} (2017), no.3, 1063--1089). Inour theorem, no superfluous assumption is imposed on $\tau \in\overline{\mathbb{D}}$. As a key foundation of our proof is an approximationmethod using the continuous dependence of evolution families. We prove that the upper envelope of a family of subharmonic functions definedon an open subset of $\mathbb{R}^{N}$, $(N\geq2)$, that is finite every where,is locally bounded above outside a closed nowhere dense set with no boundedcomponents. Then we conclude as a consequence that a separately subharmonicfunction is subharmonic outside a closed nowhere dense set with no boundedcomponents. It generalizes a result due to Cegrell and Sadullaev. In a classical work of the 1950's, Lee and Yang proved that for fixednonnegative temperature, the zeros of the partition functions of aferromagnetic Ising model always lie on the unit circle in the complex magneticfield. Zeros of the partition function in the complex temperature were thenconsidered by Fisher, when the magnetic field is set to zero. Limitingdistributions of Lee-Yang and of Fisher zeros are physically important as theycontrol phase transitions in the model. One can also consider the zeros of thepartition function simultaneously in both complex magnetic field and complextemperature. They form an algebraic curve called the Lee-Yang-Fisher (LYF)zeros. In this paper we continue studying their limiting distribution for theDiamond Hierarchical Lattice (DHL). In this case, it can be described in termsof the dynamics of an explicit rational function R in two variables (theMigdal-Kadanoff renormalization transformation). We study properties of theFatou and Julia sets of this transformation and then we prove that theLee-Yang-Fisher zeros are equidistributed with respect to a dynamical(1,1)-current in the projective space. The free energy of the lattice getsinterpreted as the pluripotential of this current. We also prove a more generalequidistribution theorem which applies to rational mappings havingindeterminate points, including the Migdal-Kadanoff renormalizationtransformation of various other hierarchical lattices. We prove in this note one weight norm inequalities for some positiveBergman-type operators. We study isomonodromic deformation of Fuchsian linear q-difference systems.Furthermore, we are looking of the behaviour of the Birkhoff connexion matrixwhen q goes to $1$. We use our results to study the convergence of the Birkhoffconnexion matrix that appears in the definition of the q-analogue of the sixthPainlev\'e equation. In this paper, we develop via real variable methods various characterisationsof the Hardy spaces in the multi-parameter flag setting. Thesecharacterisations include those via, the non-tangential and radial maximalfunction, the Littlewood--Paley square function and area integral, Riesztransforms and the atomic decomposition in the multi-parameter flag setting.The novel ingredients in this paper include (1) establishing appropriatediscrete Calder\'on reproducing formulae in the flag setting and a version ofthe Plancherel--P\'olya inequalities for flag quadratic forms; (2) introducingthe maximal function and area function via flag Poisson kernels and flagversion of harmonic functions; (3) developing an atomic decomposition via thefinite speed propagation and area function in terms of flag heat semigroups. Asa consequence of these real variable methods, we obtain the fullcharacterisations of the multi-parameter Hardy space with the flag structure. Given an imaginary quadratic extension $K$ of $\mathbb Q$, we classify themaximal nonelementary subgroups of the Picard modular group$\operatorname{PU}(1,2;\mathcal O_K)$ preserving a totally real totallygeodesic plane in the complex hyperbolic plane $\mathbb H^2_\mathbb C$. Weprove that these maximal $\mathbb R$-Fuchsian subgroups are arithmetic, anddescribe the quaternion algebras from which they arise. For instance, if theradius $\Delta$ of the corresponding $\mathbb R$-circle lies in $\mathbbN-\{0\}$, then the stabilizer arises from the quaternion algebra$\Big(\!\begin{array}{c} \Delta\,,\, \|D_K\|\\\hline\mathbb Q\end{array}\!\Big)$. We thus prove the existence of infinitely many orbits of$K$-arithmetic $\mathbb R$-circles in the hypersphere of $\mathbb P_2(\mathbbC)$. In the context of the complex-analytic structure within the unit diskcentered at the origin of the complex plane, that was presented in a previouspaper, we show that a certain class of non-integrable real functions can berepresented within that same structure. In previous papers it was shown thatessentially all integrable real functions, as well as all singular Schwartzdistributions, can be represented within that same complex-analytic structure.The large class of non-integrable real functions which we analyze here cantherefore be represented side by side with those other real objects, thusallowing all these objects to be treated in a unified way. In the context of the complex-analytic structure within the unit diskcentered at the origin of the complex plane, that was presented in a previouspaper, we show that the complete Fourier theory of integrable real functions iscontained within that structure, that is, within the structure of the space ofinner analytic functions on the open unit disk. We then extend the Fouriertheory beyond the realm of integrable real functions, to include for examplesingular Schwartz distributions, and possibly other objects. In the context of the complex-analytic structure within the unit diskcentered at the origin of the complex plane, that was presented in a previouspaper, we show that singular Schwartz distributions can be represented withinthat same structure, so long as one defines the limits involved in anappropriate way. In that previous paper it was shown that essentially allintegrable real functions can be represented within the complex-analyticstructure. The infinite collection of singular objects which we analyze herecan thus be represented side by side with those real functions, thus allowingall these objects to be treated in a unified way.
Hot-keys on this page r m x p toggle line displays j k next/prev highlighted chunk 0 (zero) top of page 1 (one) first highlighted chunk from collections import OrderedDict from typing import List import numpy as np from ase import Atoms from icet import ClusterSpace from icet.core.sublattices import Sublattices from mchammer.calculators.base_calculator import BaseCalculator class TargetVectorCalculator(BaseCalculator): """ A ``TargetVectorCalculator`` enables evaluation of the similarity between a structure and a target cluster vector. Such a comparison can be carried out in many ways, and this implementation follows the measure proposed by van de Walle et al. in Calphad 42, 13 (2013) [WalTiwJon13]_. Specifically, the objective function :math:`Q` is calculated as .. math:: Q = - \\omega L + \\sum_{\\alpha} \\left\| \\Gamma_{\\alpha} - \\Gamma^{\\text{target}}_{\\alpha} \\right\|. Here, :math:`\\Gamma_{\\alpha}` are components in the cluster vector and :math:`\\Gamma^\\text{target}_{\\alpha}` the corresponding target values. The factor :math:`\\omega` is the radius of the largest pair cluster such that all clusters with the same or smaller radii have :math:`\\Gamma_{\\alpha} - \\Gamma^\\text{target}_{\\alpha} = 0`. Parameters ---------- structure structure for which to set up calculator cluster_space cluster space from which to build calculator target_vector vector to which any vector will be compared weights weighting of each component in cluster vector comparison, by default 1.0 for all components optimality_weight factor :math:`L`, a high value of which effectively favors a complete series of optimal cluster correlations for the smallest pairs (see above) optimality_tol tolerance for determining whether a perfect match has been achieved (used in conjunction with :math:`L`) name human-readable identifier for this calculator """ def __init__(self, structure: Atoms, cluster_space: ClusterSpace, target_vector: List[float], weights: List[float] = None, optimality_weight: float = 1.0, optimality_tol: float = 1e-5, name: str = 'Target vector calculator') -> None: super().__init__(structure=structure, name=name) 63 ↛ 64line 63 didn't jump to line 64, because the condition on line 63 was never true if len(target_vector) != len(cluster_space): raise ValueError('Cluster space and target vector ' 'must have the same length') self.cluster_space = cluster_space self.target_vector = target_vector if weights is None: weights = np.array([1.0] * len(cluster_space)) else: if len(weights) != len(cluster_space): raise ValueError('Cluster space and weights ' 'must have the same length') self.weights = np.array(weights) if optimality_weight is not None: self.optimality_weight = optimality_weight self.optimality_tol = optimality_tol self.orbit_data = self.cluster_space.orbit_data else: self.optimality_weight = None self.optimality_tol = None self.orbit_data = None self._cluster_space = cluster_space def calculate_total(self, occupations: List[int]) -> float: """ Calculates and returns the similarity value :math:`Q` of the current configuration. Parameters ---------- occupations the entire occupation vector (i.e. list of atomic species) """ self.structure.set_atomic_numbers(occupations) cv = self.cluster_space.get_cluster_vector(self.structure) return compare_cluster_vectors(cv, self.target_vector, self.orbit_data, weights=self.weights, optimality_weight=self.optimality_weight, tol=self.optimality_tol) def calculate_local_contribution(self, local_indices, occupations: List[int]) -> float: """ Not yet implemented, forwards calculation to calculate_total. """ return self.calculate_total(occupations) @property def sublattices(self) -> Sublattices: """Sublattices of the calculators structure.""" sl = self.cluster_space.get_sublattices(self.structure) return sl def compare_cluster_vectors(cv_1: np.ndarray, cv_2: np.ndarray, orbit_data: OrderedDict, weights: List[float] = None, optimality_weight: float = 1.0, tol: float = 1e-5) -> float: """ Calculate a quantity that measures similarity between two cluster vecors. Parameters ---------- cv_1 cluster vector 1 cv_2 cluster vector 2 orbit_data orbit data as obtained by ``ClusterSpace.orbit_data`` weights Weight assigned to each cluster vector element optimality_weight quantity :math:`L` in [WalTiwJon13]_ (see :class:`mchammer.calculators.TargetVectorCalculator`) tol numerical tolerance for determining whether two elements are exactly equal """ if weights is None: weights = np.ones(len(cv_1)) diff = abs(cv_1 - cv_2) score = np.dot(diff, weights) if optimality_weight: longest_optimal_radius = 0 for orbit_index, d in enumerate(diff): orbit = orbit_data[orbit_index] if orbit['order'] != 2: continue if d < tol: longest_optimal_radius = orbit['radius'] else: break score -= optimality_weight * longest_optimal_radius return score
This article will be permanently flagged as inappropriate and made unaccessible to everyone. Are you certain this article is inappropriate? Excessive Violence Sexual Content Political / Social Email Address: Article Id: WHEBN0000039136 Reproduction Date: The accelerating universe is the observation that the universe appears to be expanding at an increasing rate. In formal terms, this means that the cosmic scale factor a(t) has a positive second derivative,[1] so that the velocity at which a distant galaxy is receding from us should be continuously increasing with time.[2] The Supernova Cosmology Project and High-Z Supernova Search Team first discovered the accelerating expansion of the Universe through observations of distant ("High-Z") supernovae in 1998.[3][4] Other evidence for the expansion of the universe has since been found through observation of baryon acoustic oscillations, and the clustering of galaxies. The most widely accepted explanation for the accelerating expansion is the existence of dark energy, a substance with negative pressure which is spread homogeneously throughout the universe. Since Hubble's discovery of the expansion of the universe in 1929,[5] the Big Bang model has become the accepted explanation for the origin of our universe. The Friedmann equation defines how the energy in the universe drives its expansion. where the four currently hypothesized contributors to the energy density of the universe are curvature, matter, radiation and dark energy.[7] Each of the components decreases with the expansion of the universe (increasing scale factor), except perhaps the dark energy term. It is the values of these cosmological parameters which physicists use to determine the acceleration of the universe. The acceleration equation describes the evolution of the scale factor with time where the pressure P is defined by the cosmological model chosen. (see explanatory models below) Physicists at one time were so assured of the deceleration of the universe's expansion that they introduced a so-called deceleration parameter q_0.[8] Current observations point towards this deceleration parameter being negative. To learn about the rate of expansion of the universe we look at the magnitude-redshift relationship of astronomical objects using standard candles, or their distance-redshift relationship using standard rulers. We can also look at the growth of large-scale structure, and find that the observed values of the cosmological parameters are best described by models which include an accelerating expansion. The first evidence for acceleration came from the observation of Type Ia supernovae, which are exploding white dwarfs who have exceeded their stability limit. Because they all have similar masses, their intrinsic luminosity is standardizable. Repeated imaging of selected areas of sky is used to discover the supernovae, then followup observations give their peak brightness, which is converted into a quantity known as luminosity distance (see distance measures in cosmology for details).[9] Spectral lines of their light can be used to determine their redshift. For supernovae at redshift less than around 0.1, or light travel time less than 10 percent of the age of the universe, this gives a nearly linear distance-redshift relation due to Hubble's law. At larger distances, since the expansion rate of the universe has changed over time, the distance-redshift relation deviates from linearity, and this deviation depends on how the expansion rate has changed over time. The full calculation requires integration of the Friedmann equation, but a simple derivation can be given as follows: the redshift z directly gives the cosmic scale factor at the time the supernova exploded. So a supernova with a measured redshift z = 0.5 implies the Universe was 1/(1+0.5) = 2/3 of its present size when the supernova exploded. In an accelerating universe, the universe was expanding more slowly in the past than it is today, which means it took a longer time to expand from 2/3 to 1.0 times its present size compared to a non-accelerating universe. This results in a larger light-travel time, larger distance and fainter supernovae, which corresponds to the actual observations. Riess found that "the distances of the high-redshift SNe Ia were, on average, 10% to 15% farther than expected in a low mass density \Omega_M = 0.2 Universe without a cosmological constant".[10] This means that the measured high-redshift distances were too large, compared to nearby ones, for a decelerating universe.[11] In the early universe before recombination and decoupling took place, photons and matter existed in a primordial plasma. Points of higher density in the photon-baryon plasma would contract, being compressed by gravity until the pressure became too large and they expanded again.[8] This contraction and expansion created vibrations in the plasma analogous to sound waves. Since dark matter only interacts gravitationally it stayed at the centre of the sound wave, the origin of the original overdensity. When decoupling occurred, approximately 380,000 years after the Big Bang,[12] photons separated from matter and were able to stream freely through the Universe, creating the cosmic microwave background as we know it. This left shells of baryonic matter at a fixed radius from the overdensities of dark matter, a distance known as the sound horizon. As time passed and the universe expanded, it was at these anisotropies of matter density where galaxies started to form. So by looking at the distances at which galaxies at different redshifts tend to cluster, it is possible to determine a standard angular diameter distance and use that to compare to the distances predicted by different cosmological models. Peaks have been found in the correlation function (the probability that two galaxies will be a certain distance apart) at 100h^{-1} Mpc,[7] indicating that this is the size of the sound horizon today, and by comparing this to the sound horizon at the time of decoupling (using the CMB), we can confirm that the expansion of the unvierse is accelerating.[13] Measuring the mass functions of galaxy clusters, which describe the number density of the clusters above a threshold mass, also provides evidence for dark energy.[14] By comparing these mass functions at high and low redshifts to those predicted by different cosmological models, values for w and \Omega_m are obtained which confirm a low matter density and a non zero amount of dark energy.[11] Given a cosmological model with certain values of the cosmological density parameters, it is possible to integrate the Friedmann equations and derive the age of the universe. By comparing this to actual measured values of the cosmological parameters, we can confirm the validity of a model which is accelerating now, and had a slower expansion in the past.[11] The most important property of dark energy is that it has negative pressure which is distributed relatively homogeneously in space. where c is the speed of light, \rho is the energy density. Different theories of dark energy suggest different values of w, with w < -1/3 for cosmic acceleration (this leads to a positive value of \ddot{a} in the acceleration equation above). The simplest explanation for dark energy is that it is a cosmological constant or vacuum energy; in this case w = -1. This leads to the Lambda-CDM model, which has generally been known as the Standard Model of Cosmology from 2003 through the present, since it is the simplest model in good agreement with a variety of recent observations. Riess found that their results from supernovae observations favoured expanding models with positive cosmological constant (\Omega_\lambda > 0) and a current acceleration of the expansion (q_0 < 0).[10] Current observations allow the possibility of a cosmological model containing a dark energy component with equation of state: This phantom energy density would become infinite in finite time, causing such a huge gravitational repulsion that the universe would lose all structure and end in a Big Rip.[15] For example, for w = -3/2 and H_0 = 70 km·s-1·Mpc-1, the time remaining before the Universe ends in this "Big Rip" is 22 billion years.[16] Other explanations for the accelerating universe include quintessence, a proposed form of dark energy with a non-constant state equation, whose density decreases with time. Dark fluid is an alternative explanation for accelerating expansion which attempts to unite dark matter and dark energy into a single framework.[17] Alternatively, some authors have argued that the universe expansion acceleration could be due to a repulsive gravitational interaction of antimatter.[18][19][20] Another type of model, the backreaction conjecture,[21][22] was proposed by cosmologist Syksy Räsänen: the rate of expansion is not homogenous, but we are coincidentally in a region where expansion is faster than the background. In this model, inhomogeneities in the early universe cause the formation of walls and bubbles, where the inside of a bubble has less matter than on average. According to general relativity, space is less curved than on the walls, and thus appears to have more volume and a higher expansion rate. If we live inside such a bubble, then it would appear that the universe is expanding at an accelerating rate.[23] The benefit is that it does not require any new physics such as dark energy. Räsänen does not consider the model likely, but without any falsification, it must remain a possibility. As the Universe expands, the density of radiation and ordinary and dark matter declines more quickly than the density of dark energy (see equation of state) and, eventually, dark energy dominates. Specifically, when the scale of the universe doubles, the density of matter is reduced by a factor of 8, but the density of dark energy is nearly unchanged (it is exactly constant if the dark energy is a cosmological constant).[8] In models where dark energy is a cosmological constant, the universe will expand exponentially with time from now on, coming closer and closer to a de Sitter spacetime. This will eventually lead to all evidence for the Big Bang disappearing, as the cosmic microwave background is redshifted to lower intensities and longer wavelengths. Eventually its frequency will be small enough that it will be absorbed by the interstellar medium, and so be screened from any observer within the galaxy. This will occur when the universe is less than 50 times its current age, leading to the end of cosmology as we know it as the distant universe turns dark.[24] Alternatives for the ultimate fate of the universe include the Big Rip mentioned above, a Big Bounce, Big Freeze, or Big Crunch. Big Bang, Xenon, Large Hadron Collider, Milky Way, Proton Calculus, University of Cambridge, Trinity College, Cambridge, Lincolnshire, Physics Texas A&M University, Stanford University, Astronomy, Cosmology, Lick Observatory Albert Einstein, Dark matter, String theory, Physical cosmology, Epistemology
The package is a powerful tool, based on pgfplots tikz, dedicated to create scientific graphs. Contents Pgfplots is a visualization tool to make simpler the inclusion of plots in your documents. The basic idea is that you provide the input data/formula and pgfplots does the rest. \begin{tikzpicture} \begin{axis} \addplot[color=red]{exp(x)}; \end{axis} \end{tikzpicture} %Here ends the furst plot \hskip 5pt %Here begins the 3d plot \begin{tikzpicture} \begin{axis} \addplot3[ surf, ] {exp(-x^2-y^2)x}; \end{axis} \end{tikzpicture} Since pgfplot is based on tikz the plot must be inside a tikzpicture environment. Then the environment declaration \begin{axis}, \end{axis} will set the right scaling for the plot, check the Reference guide for other axis environments. To add an actual plot, the command \addplot[color=red]{log(x)}; is used. Inside the squared brackets some options can be passed, in this case we set the colour of the plot to red; the squared brackets are mandatory, if no options are passed leave a blank space between them. Inside the curly brackets you put the function to plot. Is important to remember that this command must end with a semicolon ;. To put a second plot next to the first one declare a new tikzpicture environment. Do not insert a new line, but a small blank gap, in this case hskip 10pt will insert a 10pt-wide blank space. The rest of the syntax is the same, except for the \addplot3 [surf,]{exp(-x^2-y^2)x};. This will add a 3dplot, and the option surf inside squared brackets declares that it's a surface plot. The function to plot must be placed inside curly brackets. Again, don't forget to put a semicolon ; at the end of the command. Note: It's recommended as a good practice to indent the code - see the second plot in the example above - and to add a comma , at the end of each option passed to \addplot. This way the code is more readable and is easier to add further options if needed. To include pgfplots in your document is very easy, add the next line to your preamble and that's it: \usepackage{pgfplots} Some additional tweaking for this package can be made in the preamble. To change the size of each plot and also guarantee backwards compatibility (recommended) add the next line: \pgfplotsset{width=10cm,compat=1.9} This changes the size of each pgfplot figure to 10 centimeters, which is huge; you may use different units (pt, mm, in). The compat parameter is for the code to work on the package version 1.9 or later. Since LaTeX was not initially conceived with plotting capabilities in mind, when there are several pgfplot figures in your document or they are very complex, it takes a considerable amount of time to render them. To improve the compiling time you can configure the package to export the figures to separate PDF files and then import them into the document, add the code shown below to the preamble: \usepgfplotslibrary{external} \tikzexternalize See this help article for further details on how to set up tikz-externalization in your Overleaf project. Pgfplots 2D plotting functionalities are vast, you can personalize your plots to look exactly what you want. Nevertheless, the default options usually give very good result, so all you have to do is feed the data and LaTeX will do the rest: To plot mathematical expressions is really easy: \begin{tikzpicture} \begin{axis}[ axis lines = left, xlabel = $x$, ylabel = {$f(x)$}, ] %Below the red parabola is defined \addplot [ domain=-10:10, samples=100, color=red, ] {x^2 - 2x - 1}; \addlegendentry{$x^2 - 2x - 1$} %Here the blue parabloa is defined \addplot [ domain=-10:10, samples=100, color=blue, ] {x^2 + 2x + 1}; \addlegendentry{$x^2 + 2x + 1$} \end{axis} \end{tikzpicture} Let's analyse the new commands line by line: axis lines = left. xlabel = $x$ and ylabel = {$f(x)$}. \addplot. domain=-10:10. samples=100. \addlegendentry{$x^2 - 2x - 1$}. To add another graph to the plot just write a new \addplot entry. Scientific research often yields data that has to be analysed. The next example shows how to plot data with pgfplots: \begin{tikzpicture} \begin{axis}[ title={Temperature dependence of CuSO$_4\cdot$5H$_2$O solubility}, xlabel={Temperature [\textcelsius]}, ylabel={Solubility [g per 100 g water]}, xmin=0, xmax=100, ymin=0, ymax=120, xtick={0,20,40,60,80,100}, ytick={0,20,40,60,80,100,120}, legend pos=north west, ymajorgrids=true, grid style=dashed, ] \addplot[ color=blue, mark=square, ] coordinates { (0,23.1)(10,27.5)(20,32)(30,37.8)(40,44.6)(60,61.8)(80,83.8)(100,114) }; \legend{CuSO$_4\cdot$5H$_2$O} \end{axis} \end{tikzpicture} There are some new commands and parameters here: title={Temperature dependence of CuSO$_4\cdot$5H$_2$O solubility}. xmin=0, xmax=100, ymin=0, ymax=120. xtick={0,20,40,60,80,100}, ytick={0,20,40,60,80,100,120}. legend pos=north west. ymajorgrids=true. xmajorgrids to enable grid lines on the x axis. grid style=dashed. mark=square. coordinates {(0,23.1)(10,27.5)(20,32)...} If the data is in a file, which is the case most of the time; instead of the commands \addplot and coordinates you should use \addplot table {file_with_the_data.dat}, the rest of the options are valid in this environment. Scatter plots are used to represent information by using some kind of marks, these are common, for example, when computing statistical regression. Lets start with some data, the sample below is to show the structure of the data file we are going to plot (see the end of this section for a link to the LaTeX source and the data file): GPA ma ve co un 3.45 643 589 3.76 3.52 2.78 558 512 2.87 2.91 2.52 583 503 2.54 2.4 3.67 685 602 3.83 3.47 3.24 592 538 3.29 3.47 2.1 562 486 2.64 2.37 The next example is a scatter plot of the first two columns in this table: \begin{tikzpicture} \begin{axis}[ enlargelimits=false, ] \addplot+[ only marks, scatter, mark=halfcircle, mark size=2.9pt] table[meta=ma] {scattered_example.dat}; \end{axis} \end{tikzpicture} The parameters passed to the axis and addplot environments can also be used in a data plot, except for scatter. Below the description of the code: enlarge limits=false only marks scatter meta parameter explained below. mark=halfcircle mark size=2.9pt table[meta=ma]{scattered_example.dat}; Bar graphs (also known as bar charts and bar plots) are used to display gathered data, mainly statistical data about a population of some sort. Bar plots in pgfplots are highly customisable, but here we are going to show an example that 'just works': \begin{tikzpicture} \begin{axis}[ x tick label style={ /pgf/number format/1000 sep=}, ylabel=Year, enlargelimits=0.05, legend style={at={(0.5,-0.1)}, anchor=north,legend columns=-1}, ybar interval=0.7, ] \addplot coordinates {(2012,408184) (2011,408348) (2010,414870) (2009,412156)}; \addplot coordinates {(2012,388950) (2011,393007) (2010,398449) (2009,395972)}; \legend{Men,Women} \end{axis} \end{tikzpicture} The figure starts with the already explained declaration of the tikzpicture and axis environments, but the axis declaration has a number of new parameters: x tick label style={/pgf/number format/1000 sep=} \addplot commands within this ybar parameter described below is mandatory for this to work). enlargelimits=0.05. legend style={at={(0.5,-0.2)}, anchor=north,legend columns=-1} ybar interval=0.7, The coordinates in this kind of plot determine the base point of the bar and its height. The labels on the y-axis will show up to 4 digits. If in the numbers you are working with are greater than 9999 pgfplot will use the same notation as in the example. pgfplots has the 3d Plotting capabilities that you may expect in a plotting software. There's a simple example about this at the introduction, let's work on something slightly more complex: \begin{tikzpicture} \begin{axis}[ title=Exmple using the mesh parameter, hide axis, colormap/cool, ] \addplot3[ mesh, samples=50, domain=-8:8, ] {sin(deg(sqrt(x^2+y^2)))/sqrt(x^2+y^2)}; \addlegendentry{$\frac{sin(r)}{r}$} \end{axis} \end{tikzpicture} Most of the commands here have already been explained, but there are 3 new things: hide axis colormap/cool mesh Note: When working with trigonometric functions pgfplots uses degrees as default units, if the angle is in radians (as in this example) you have to use de deg function to convert to degrees. In pgfplots is possible to plot contour plots, but the data has have to be pre calculated by an external program. Let's see: \begin{tikzpicture} \begin{axis} [ title={Contour plot, view from top}, view={0}{90} ] \addplot3[ contour gnuplot={levels={0.8, 0.4, 0.2, -0.2}} ] {sin(deg(sqrt(x^2+y^2)))/sqrt(x^2+y^2)}; \end{axis} \end{tikzpicture} This is a plot of some contour lines for the same equation used in the previous section. The value of the title parameter is inside curly brackets because it contains a comma, so we use the grouping brackets to avoid any confusion with the other parameters passed to the \begin{axis} declaration. There are two new commands: view={0}{90} contour gnuplot={levels={0.8, 0.4, 0.2, -0.2}} levels is a list of values of elevation levels where the contour lines are to be computed. To plot a set of data into a 3d surface all we need is the coordinates of each point. These coordinates could be an unordered set or, in this case, a matrix: \begin{tikzpicture} \begin{axis} \addplot3[ surf, ] coordinates { (0,0,0) (0,1,0) (0,2,0) (1,0,0) (1,1,0.6) (1,2,0.7) (2,0,0) (2,1,0.7) (2,2,1.8) }; \end{axis} \end{tikzpicture} The points passed to the coordinates parameter are treated as contained in a 3 x 3 matrix, being a white row space the separator of each matrix row. All the options for 3d plots in this article apply to data surfaces. The syntax for parametric plots is slightly different. Let's see: \begin{tikzpicture} \begin{axis} [ view={60}{30}, ] \addplot3[ domain=0:5pi, samples = 60, samples y=0, ] ({sin(deg(x))}, {cos(deg(x))}, {x}); \end{axis} \end{tikzpicture} There are only two new things in this example: first, the samples y=0 to prevent pgfplots from joining the extreme points of the spiral and; second, the way the function to plot is passed to the addplot3 environment. Each parameter function is grouped inside curly brackets and the three parameters are delimited with parenthesis. Command/Option/Environment Description Possible Values axis Normal plots with linear scaling semilogxaxis logaritmic scaling of x and normal scaling for y semilogyaxis logaritmic scaling for y and normal scaling for x loglogaxis logaritmic scaling for the x and y axes axis lines changes the way the axes are drawn. default is ' box box, left, middle, center, right, none legend pos position of the legend box south west, south east, north west, north east, outer north east mark type of marks used in data plotting. When a single-character is used, the character appearance is very similar to the actual mark. , x , +, \|, o, asterisk, star, 10-pointed star, oplus, oplus, otimes, otimes, square, square, triangle, triangle, diamond, halfdiamond, halfsquare, right, left, Mercedes star, Mercedes star flipped, halfcircle, halfcircle, pentagon, pentagon, cubes. (cubes only work on 3d plots). colormap colour scheme to be used in a plot, can be personalized but there are some predefined colormaps hot, hot2, jet, blackwhite, bluered, cool, greenyellow, redyellow, violet. For more information see:
The idea of the argument is this: an integrable function that does not tend to zero may have thinner and thinner pikes, and the integral on each pike needs to tend to zero. If the function is uniformly continuous, these pikes can't get thinner and thinner, and the function can't be integrable, as the integral on pikes does not tend to zero. I'll consider $\Bbb R^+$ here, but the same argument works with $\Bbb R^-$, hence $\Bbb R$. Suppose $f$ is uniformly continuous on $\Bbb R^+$ and $f(x)$ does not tend to zero as $x\to+\infty$. Let's prove it's not integrable. Since $f$ does not converge to zero, there is an $\varepsilon>0$ such that you can define an increasing sequence $x_n\to+\infty$ such that $\|f(x_n)\|>\varepsilon$. Since $f$ is uniformly continuous, for this $\varepsilon$, there is a $\delta>0$ such that $\|x-y\|<\delta\implies\|f(x)-f(y)\|<\varepsilon/2$.That means that on every interval $]x_n-\delta,x_n+\delta[$, $\|f(x)\|>\varepsilon/2$. Hence the area under the curve on this interval is larger that $\delta\varepsilon>0$, and that does not depend on $n$. Therefore $\int_{x_n-\delta}^{x_n+\delta}f(u)du$ does not tend to zero as $n\to+\infty$, and the function $h(x)=\int_0^{x}f(u)du$ can't converge as $x\to+\infty$, hence the function $f$ is not integrable.
I'd recommend "The Art of Molecular Dynamics Simulation" by D. C. Rapaport. The code samples are written in C. I'm not a huge fan of the programming style of the book, but at least it's not FORTRAN.Having said that, my advise would be to take any book where neighbour lists are explained (for instance the Frenkel & Smit book, which I guess it's what you are using now) and just implement the algorithm, which is usually written in pseudo-code. If you want a TL;DR, there are several types of neighbour lists. All of these rely on your interaction potential being 0 outside a certain range $r_c$. Verlet lists ($\mathcal{O}(N^2)$ - $\mathcal{O}(N^{3/2})$ but with a small-ish prefactor): do an $N^2$ sweep where for each particle you fill an array of neighbours that are closer than $r_c + r_v$, where $r_v$ is a parameter and save the current position of each particle (let's call it $\vec{r}_i(t_0)$. Evolve your simulation by using the list of neighbours to compute the forces. After each integration step check whether, for any particle $i$, $\|\vec{r}_i(t) - \vec{r}_i(t_0)\| \geq r_v / 2$. If it is then update all the lists (see step 1). Cell lists ($\mathcal{O}(N)$): Divide your simulation box in cells of linear size $l > r_c$. Use linked lists to assign particles to the cells. For each particle $i$, the list of neighbours is given by all the particles that are in $i$'s cells and in each of the (8 in 2D and 26 in 3D) neighbouring cells. Calculate the forces between $i$ and its neighbours. After each integration step check whether particles have crossed cell boundaries and update the data structure accordingly Verlet lists built with cell lists (usually the best choice, $\mathcal{O}(N)$ with a smaller prefactor than just using cells): Step 1 of the "Verlet lists" section is carried out by using cell lists. The linked lists that store the cell data structures are kept updated throughout the simulation The reason why using Verlet lists instead of cell lists is (in MD simulations) better is that the average number of neighbours is smaller for the former than the latter. The difference is due to the fact that, for each particle, cell lists give you a list of neighbours that are in a volume $(3l)^3$, which, if $r_v$ is chosen wisely, is much larger than the volume of the Verlet sphere ($4/3 \pi (r_c + r_v)^3$). Therefore, on average, you compute much fewer distances with Verlet lists.
FlashChat Actuarial Discussion Preliminary Exams CAS/SOA Exams Cyberchat Around the World Suggestions Thread Tools Search this Thread Display Modes # 1 An Introduction to MimeTex For those who wish to format math questions on the web using super and subscripting, try using the Tex tags. Tom has been kind enough to import MimeTeX to this site. One thing I cannot yet do with it is make those clever angle brackets for terms certain. You may see some good samples of how this works at the MimeTex Site , which is a public domain site and I don't believe I am hotlinking. Here's a basic intro. superscripts x^2 ex: Greek Letters \mu ex: Subscripts are done by _. You can put both into a sum or integral. \sum is for summation and \int is for integration. Put special functions like sine, cos and log with backslashes before their name to typeset them correctly. \infty is infinity ex: ex: if More examples can be found on the website, where you can click on an expression and see the tags that generate it. Grouping can be done with curly braces. ex: I hope this brief introduction, combined with the link, will make posting math questions on the web a little bit clearer and easier. __________________ Sticks and Stones may break my bones, but Markov Chains excite me. Last edited by ACCtuary; 02-12-2006 at 11:26 PM.. # 3 You can also read about LaTeX here: http://en.wikipedia.org/wiki/LaTeX if you follow all (or most) links at the bottom, you should have a much better idea as to how to use it. A lot of math faculty uses LaTeX or something similar to write all their papers / class notes. Some universities even have classes on how to use it. # 4 # 5 # 6 which is odd. for the life of me I can't imagine how I found it. I usually just use the new posts search and go through them. odd. 2007 is going to be a great year for me. I can feel it. # 7 Was this thread not stickied before? __________________ Exams How to explain actuarial exams to someone else... Good Einstein quote - "One had to cram all this stuff into one's mind for the examinations, whether one liked it or not. This coercion had such a deterring effect on me that, after I had passed the final examination, I found the consideration of any scientific problems distasteful to me for an entire year." # 8 Some more examples... Certain annuity-due: \ddot{a}_{\overline{n}\|} = Complete life expectancy: \overset{\circ}e_x = Temporary life expectancy: \overset{\circ}e_{x:\overline{n}\|} = Continuous term insurance: \bar{A}_{\overset{1}x:\overline{n}\|} = __________________ We help people pass actuarial exams. Follow us on Twitter, Facebook, and LinkedIn. # 9 When I first saw the title for this thread, I thought it said "An Introduction to Mime Tax". Thought it was going to be hilarious. That is all. __________________ GAME ON!!!!!!! Let your ness show. Join the D&D fun. Started but applications still accepted Officially assigned the role of Dictator, 9/30/09. Bow to my whims. Thread Tools Search this Thread Display Modes
Consider a Yang-Mills $U(1)$ gauge theory in $d=2+1$ flat dimensions. It is known that in $d=2+1$ the field strength $F=dA$ can be written in terms of a dual scalar field $a$, in the following way: $F=\star da$. (The star is the hodge dual of forms) My question is why $a$ is compact, i.e. its target space is a circle (which is, $a$ is invariant under $a\sim a+2\pi$ In all the reviews of $3d$ gauge dynamics this fact is simply stated, with no proof at all. To me, it is not obvius. Thank you very much.
№ 9 All Issues On the Growth of the Maximum of the Modulus of an Entire Function on a Sequence Abstract Let M f( r) and μ f( r) be, respectively, the maximum of the modulus and the maximum term of an entire function f and let Φ be a continuously differentiable function convex on (−∞, +∞) and such that x = o(Φ( x)) as x → +∞. We establish that, in order that the equality $\lim \inf \limits_{r \to + \infty} \frac{\ln M_f (r)}{\Phi (\ln r)} = \lim \inf \limits_{r \to + \infty} \frac{\ln \mu_f (r)}{\Phi (\ln r)}$ be true for any entire function f, it is necessary and sufficient that ln Φ′( x) = o(Φ( x)) as x → +∞. English version (Springer): Ukrainian Mathematical Journal 54 (2002), no. 8, pp 1386-1392. Citation Example: Filevych P. V. On the Growth of the Maximum of the Modulus of an Entire Function on a Sequence // Ukr. Mat. Zh. - 2002. - 54, № 8. - pp. 1149-1153. Full text
I have no idea how to solve the following INTEGER problem or prove its hardness. Thanks for any help/comment/open discussion! Assume there are $N$ startups. For each startup $i$, you can invest $x_i\in \{0,1,...,C_i\}$ dollars where $C_i$ is the maximal investment that it accepts, and you will get a reward as $f_i(x_i)$. The reward function $f_i(x_i)$ has the following properties for all $i$: (1) $f_i(.)$ is non-decreasing and $f_i(0)=0$ (2) $f_i(.)$ is not necessarily "convex" or "concave" (3) $0 \leq x\leq x'\leq C_i$, $f_i(x)x' \geq f_i(x')x$ (4) the function $g_i(x) = f_i(x+1) - f_i(x)$ is not necessarily non-decreasing or non-increasing. Your total budget is $C$. By selecting an INTEGER vector $(x_1,...,x_N)$, you want to maximize the total rewards $\sum_{i=1}^Nf_i(x_i)$ subject to (1) $x_i\in \{0,1,...,C_i\}$ for any $i$ and (2)$\sum_{i=1}^N x_i \leq C$.
I don't see why you would like to use subgradients for that. Correct me if I'm wrong, but when the problem is fully differentiable (kmeans actually is), the subgradient is the gradient vector $\nabla f$ itself. Then it might just equate to using an EM-like algorithm, iteratively finding the means and solving for the assignments. If you like the optimization perspective, think of what we like to do as a joint minimization over cluster centers $$J_2(c_1,\cdots,c_m,\boldsymbol{\mu}_1, ... ,\boldsymbol{\mu}_k)=\frac{1}{m}\sum\limits_i \\|\mathbf{x}_i-\boldsymbol{\mu}_{c_i}\\|^2$$where $\mathbf{c}=(c_1,\cdots,c_m)^\top$ denotes the cluster indices to which examples $\{\mathbf{x}_i\}$ are assigned. We then alternatively solve for $\mathbf{c}$ and $\boldsymbol{\mu}$. Fix $\boldsymbol{\mu}$, solve for $\mathbf{c}$:$$\forall 1\leq k\leq K : C(k)\leftarrow \{i : k={\operatorname{arg\,min}}_k\\|\mathbf{x}_i−\boldsymbol{\mu}_k\\|^2\}$$Note that recent techniques from optimization (See On Differentiating Parameterized $ \arg \min $ and $ \arg \max$ Problems with Application to Bi Level Optimization) can make this step differentiable. Otherwise, one can benefit from subgradients (see below). Fix $\mathbf{c}$ and solve for $\boldsymbol{\mu}$ by running a gradient descent using:$$\nabla_{\boldsymbol{\mu}} J_2 = \sum_{i=1} 2(\boldsymbol{\mu}-\mathbf{x})^\top = 0$$ You might just use gradient descent for this. If instead, the problem is robustly formulated via an $L_1$ norm: $$J_1(c_1,\cdots,c_m,\boldsymbol{\mu}_1, ... ,\boldsymbol{\mu}_k)=\frac{1}{m}\sum\limits_i \|\mathbf{x}_i-\boldsymbol{\mu}_{c_i} \|$$ then the estimator can be found by a median and this variant is coined K-medoids (when the center is picked on a data point) or K-medians in the case of Manhattan-distance median. It is also beneficial for discrete problems where computing a continuous mean is difficult. These problems are not differentiable everywhere and subgradients might be useful (in fact either proximal gradients or constrained minimization tools are preferred).
This is my first question. I have the following data that I'd like to approximate as a parametric function: \begin{align} y = a + (bx_1 + cx_2 + dx_3 + ex_1x_2 + fx_1x_3 + gx_2x_3 + hx_1x_2x_3 + i)(j\sin(kx_1x_2x_3) + l\cos(mx_1x_2x_3)) \end{align} testdata <- read.csv('..path_to_file/gistfile1.txt', sep = "")plot(1:35,testdata$X0, col = 'blue', pch = 19, ylim = c(0,8),type = "l",lwd = 2, xlab = "x", ylab = "y_i",cex.lab = 2)lines(1:35,testdata$X1, col = 'red', pch = 19, ylim = c(0,1),lwd = 2)lines(1:35,testdata$X1.1, col = 'green', pch = 19, ylim = c(0,1),lwd = 2)lines(1:35,testdata$X8, col = 'magenta', pch = 19, ylim = c(0,1),lwd = 2)leg.txt <- (c("y","x1","x2","x3"))legend(18,8,leg.txt, col = c('blue','red','green','magenta'), lty = 1,lwd= 2,border = 'white') Any suggestions how to modify the function shown below so as to get a closer fit?
Flexible equation of state for a hard sphere and Lennard–Jones fluid near critical temperature Permanent link: https://www.ias.ac.in/article/fulltext/pram/083/06/0955-0962 Author uses the condition in terms of contact point radial distribution function $G(\sigma, \lambda(\eta_c, \alpha))$ containing the self-consistent function $\lambda(\eta_c, \alpha)$ and condition of continuity at $\sigma/2$ = contact point, to determine equation of state, (EoS). Different EoSs in terms of built-in parameter, 𝑚, can be obtained with a suitable choice of $\lambda(\eta_c, \alpha)$ and the present EoSs have less r.m.s. deviation than Barker–Henderson BH2 for LJ fluids, and results are much closer to molecular dynamics (MD) simulations than expectations and reproduce the existing simulation data and present EoS for LJ potential, with the help of a set of minimum single-scaled parameter, $a_0(\eta_c, \alpha)$ for a given reduced temperature, $T^\ast = (1/\beta \epsilon)$= 1.4, 2, 3, 4, 5, 6. It has been found that parameter 𝛼 = 1.059128388 can be used to fix up the critical temperature parameter $T_c$ = 1.3120(7) to that of a computer simulation result. Current Issue Volume 93 \| Issue 5 November 2019 Click here for Editorial Note on CAP Mode
My original question (posted in https://math.stackexchange.com/questions/1584430/can-all-power-sets-be-limit-cardinals) was: Is it possible to create a model of ZFC, so that the cardinality of each power set is a limit cardinal (as opposed to GCH where they are always successor cardinals)? Obviously, from Easton's theorem, this is possible for regular cardinals.I also showed that in this model strong limit cardinals must also be fixed points of the $\aleph$ function (using Shelah's upper bound from PCF), and therefore such a model would mean $2^{\aleph_\delta} \geq \aleph_{\delta^+}$. But whether we can improve Shelah's upper limit from $2^{\aleph_\delta} < \aleph_{{\|\delta\|}^{+4}}$ to $2^{\aleph_\delta} < \aleph_{\delta^+}$ is an open question. So for strong limit cardinals, we don't know whether such a model is possible (even for one cardinal, let alone all of them). This leaves the singular weak limit $\delta$ case, which isn't interesting unless we assume the continuum function doesn't become constant for all $\gamma$ such that $\kappa < \gamma < \delta$. Assuming this, can we create a model of where the cardinality of the power set of singular weak limit are limit cardinals? For a concrete example, take the following rule (which satisfies Easton's theorem, and the known limitations for singular cardinals in http://www.math.tau.ac.il/~gitik/icm5.pdf): $$2^{\aleph_\alpha} = \aleph_{\aleph_{\alpha + 1}}$$
I am aware of the debate on whether Schrödinger equation was derived or motivated. However, I have not seen this one that I describe below. Wonder if it could be relevant. If not historically but for educational purposes when introducing the equation. Suppose that we have the time dependent Schrödinger equation for a free particle, $V=0$. $$-\frac {\hbar i}{2m} \nabla^2 \Psi_\beta = \frac {\partial \Psi_{\beta}}{\partial t} $$ As the particle moves its heat is diffused throughout space. Now consider that we consider Heat equation or in general Diffusion equation: $$\alpha\nabla^2 u= \frac {\partial u}{\partial t} $$ Where $u$ is temperature. Also we have particle diffusion equation due to Fick's second law. $$D \frac {\partial^2 \phi}{\partial x^2}= \frac {\partial \phi}{\partial t} $$ Where $\phi$ is concentration. Furthermore, probability density function obeys Diffusion equation. So as the free particle moves, the heat, the temperature, or the density is diffused. Now we can motivate Schrödinger equation in an intuitive way. Mathematically it is describing the same diffusion. Am I right? Have you seen more like this motivation elsewhere?
Because there are natural computational problems involving many mathematical objects, there are a bunch of implications of complexity class separations like $\mathrm{P} \neq \mathrm{NP}$. I think the first paper to investigate this idea is probably Mike Freedman's Complexity classes as mathematical axioms, which assumes a complexity class separation (namely $\mathrm{NP} \neq \mathrm{P}^{\#\mathrm{P}}$, which is stronger than $\mathrm{P} \neq \mathrm{NP}$) to prove that knots with certain properties exist. The main idea of all these arguments is to prove an implication like "If all objects of type $T$ satisfy property $P$, then there is an efficient algorithm for a problem which we assumed has no efficient algorithm." You can then deduce the existence of objects of type $T$ which satisfy property $\neg P$. (Here the meaning of "efficient" depends on the class separation you assume.) The exact thing Freedman proves is a little esoteric, so let me give two other examples that have a somewhat similar flavor. The systole of a metric manifold is the length of the shortest non-contractible loop on it (I'll also use the word for the shortest loop itself). Probably the first manifold whose systole you'd try to understand is a flat torus $\mathbb{T}^d = \mathbb{R}^d / \Lambda$ for some lattice $\Lambda = \langle v_1, \dots, v_d \rangle$, since these are pretty much the simplest metric manifolds. One natural thing might be to try to say something about the word length of the systole $\gamma$ when considered as an element of $\pi_1(\mathbb{T}^d)$ equipped with the generating set $\{ v_1^, \dots, v_d^ \}$ where $v_i^$ is the loop in $\mathbb{T}^d$ naturally associated to $v_i$. For example, maybe the systole always has a relatively short word expressing it. Say, maybe we can always write $\gamma =_{\pi_1(\mathbb{T}^d)} \sum_i n_i v_i^$ with $\sum_i \|n_i\| < \sqrt{d}$ or something. It turns out that a modest strengthening of $\mathrm{P} \neq \mathrm{NP}$ actually rules this out. Specifically, assuming that $\mathrm{NP}$-hard problems do not have time $2^{o(n)}$ bounded-error probabilstic algorithms, we can prove the following: For any $\ell(d) = o(d / \log d)$, there exist infinitely many $d$ and $\Lambda=(v_1, \dots, v_d)$ such that every systolic loop $\gamma$ on the torus $\mathbb{R}^d / \Lambda$ has word length at least $\ell(n)$ in the generating set $\{ v_1^, \dots, v_d^ \}$. The idea of the argument is just that such a bound would imply a sub-exponential time algorithm for the $\mathrm{NP}$-hard problem of computing the systole: namely just enumerate all possible words in the generators of the given word length and pick the one which is represented by the shortest loop (this is easy to compute). You can use a similar argument also to show that faithful representations of $S_n$ have dimension at least $n^{\varepsilon}$ for some $\varepsilon > 0$ (assuming some version of the strong exponential time hypothesis.) The basic idea is given here -- the argument is very simple. These arguments I think give some idea of why proving class separations should be hard. They immediately imply that mathematical objects of all kinds must have certain kinds of complexity, or else they could be used to give algorithms contradicting the class separations. So, the class separation simultaneously demonstrates the existence of such complexity in all such mathematical objects at once. Bonus: Tim Roughgarden and Inbal Talgam-Cohen have some writing along these lines as well showing that class separations imply markets in which certain kinds of equilibria do not exist.
As to the title of your question, Logistic function: where does it come from?, I can provide an intuition for the logistic function which is the common interpretation from a machine learning perspective. It seems that the underlying question has already been answered above, but I thought this interpretation could help with your intuition about logistic functions in general. Imagine you would like to model the probability of a random variable, $X$. Denote the corresponding probability density for this as $p(x)\in[0,1]$. Now given some data, we would like to model $p(x)$ in some way. The classic way to approach modelling in machine learning is via linear regression. Something of the form of $F = \beta G$, where $F$ is a vector of outputs, $\beta$ is a vector of coefficients, and $G$ a matrix of inputs (classically it would be $Y = \beta X$, but since I already used $X$ as a R.V I would like to avoid confusion). An example of a linear regression can look like: $f:\mathbb{R}^3\rightarrow \mathbb{R}: \beta_0 g + \beta_1 g^2 +\beta_2 g^3 \mapsto ( -\infty,\infty)$ i.e. a cubic function. However we cannot model a probability via regression so far! The image of the cubic function for example was $( -\infty,\infty)$, and the image for $p(x)$ was $[0,1]$. Consider the first transformation: $\frac{p(x)}{1-p(x)}\in(0,\infty)$. Now consider taking the logarithm of this: $\log\left(\frac{p(x)}{1-p(x)}\right)\in(-\infty,\infty)$. Wow, now we can directly us linear regression to model $\log\left(\frac{p(x)}{1-p(x)}\right)$, so that, \begin{align}\log\left(\frac{p(x)}{1-p(x)}\right) &= \beta G \\\left(\frac{p(x)}{1-p(x)}\right) &= e^{\beta G} \\p(x) &= \frac{1}{1+\exp(-\beta G)}\end{align} Now choose the a linear regression function for $\beta G$ to match your problem, and what you find is that you are in fact modelling a probability for linear regression!
LaTeX supports many worldwide languages by means of some special packages. In this article is explained how to import and use those packages to create documents in Spanish. Contents Spanish language has some special characters, such as the ñ and some accentuated words. For this reason the preamble of your document must be modified accordingly to support these characters and some other features. \documentclass{article} \usepackage[utf8]{inputenc} \usepackage[spanish]{babel} \begin{document} \tableofcontents \vspace{2cm} %Add a 2cm space \begin{abstract} Este es un breve resumen del contenido del documento escrito en español. \end{abstract} \section{Sección introductoria} Esta es la primera sección, podemos agregar algunos elementos adicionales y todo será escrito correctamente. Más aún, si una palabra es demasiado larga y tiene que ser truncada, babel tratará de truncarla correctamente dependiendo del idioma. \section{Sección con teoremas} Esta sección es para ver qué pasa con los comandos que definen texto \end{document} There are two packages in this document related to the encoding and the special characters. These packages will be explained in the next sections. Modern computer systems allow you to input letters of national alphabets directly from the keyboard. In order to handle a variety of input encodings used for different groups of languages and/or on different computer platforms LaTeX employs the inputenc package to set up input encoding. In this case the package properly displays characters in the Spanish alphabet. To use this package add the next line to the preamble of your document: \usepackage[utf8]{inputenc} The recommended input encoding is utf-8. You can use other encodings depending on your operating system. To proper LaTeX document generation you must also choose a font encoding which has to support specific characters for Spanish language, this is accomplished by the package: fontenc \usepackage[T1]{fontenc} Even though the default encoding works well in Spanish, using this specific encoding will avoid glitches with some specific characters. The default LaTeX encoding is OT1. To extended the default LaTeX capabilities, for proper hyphenation and translating the names of the document elements, import the babel package for the Spanish language. \usepackage[spanish]{babel} As you may see in the example at the introduction, instead of "abstract" and "Contents" the Spanish words "Resumen" and "Índice" are used. An extra parameter can be passed when importing the babel package with spanish support: \usepackage[spanish, mexico]{babel} This will set a localization for the language. By now only mexico and mexico-com are available, the latter will use a comma instead of a dot as the decimal marker in mathematical mode. Mathematical commands can also be imported specifically for the Spanish language. \section{Sección con teoremas} Esta sección es para ver que pasa con los comandos que definen texto \[ \lim x = \sen{\theta} + \max \{3.52, 4.22\} \] El paquete también agrega un comportamiento especial a <<estas marcas para hacer citas textuales>> tal como lo indican las reglas de la RAE. You can see that \sen, \max and \lim are properly displayed. For a complete list of mathematical symbols in Spanish see the reference guide. For this commands to be available you must add the next line to the preamble of your document: \def\spanishoperators{} Notice also that << and >> have a special format in Spanish, this can conflict with some packages. If you don't need these or you want to use the direct keyboard input « » set the parameter es-noquotes, comma separated inside the brackets of the babel statement. Sometimes for formatting reasons some words have to be broken up in syllables separated by a - ( hyphen) to continue the word in a new line. For example, matemáticas could become mate-máticas. The package babel, whose usage was described in the previous section, usually does a good job breaking up the words correctly, but if this is not the case you can use a couple of commands in your preamble. \usepackage{hyphenat} \hyphenation{mate-máti-cas recu-perar} The first command will import the package hyphenat and the second line is a list of space-separated words with defined hyphenation rules. On the other side, if you want a word not to be broken automatically, use the {\nobreak word} command within your document. Spanish LaTeX commands in mathematical mode LaTeX command Output \sen sen \tg tg \arcsen arc sen \arccos arc cos \arctg arc tg \lim lím \limsup lím sup \liminf lím inf \max máx \inf ínf \min mín For more information see
I think the code below meets your requirements. The \phantom{\phantom{=\int b(x)} command inserts an invisible block of the same width as its contents (i.e., of the line above), and the \smash{...} command prevents the invisible block from taking up too much vertical space. (Try the code without the \smash command to see the effect on the spacing between lines 2 and 3 of the align* environment.) Finally, the extra {} snippet inside the \phantom command is there to tell TeX to treat the (otherwise invisible) equal sign as a "normal" math-relational operator. Separately, I've eliminated a stray right parenthesis from the first line and added a couple of "thin space" commands, \,, to provide a bit of separation between the integrands and the "dx" terms; doing so tends to improve legibility (and it's a practice followed by many mathematicians, including Don Knuth -- the creator of TeX). :-) \documentclass{article} \usepackage{amsmath} \begin{document} \begin{align} f &= \int a(x) \,dx \\ &= \int b(x) \\ &\smash{\phantom{{}=\int b(x)}}\cdot c(x)\,dx \end{align} \end{document}
And I think people said that reading first chapter of Do Carmo mostly fixed the problems in that regard. The only person I asked about the second pset said that his main difficulty was in solving the ODEs Yeah here there's the double whammy in grad school that every grad student has to take the full year of algebra/analysis/topology, while a number of them already don't care much for some subset, and then they only have to pass rather the class I know 2 years ago apparently it mostly avoided commutative algebra, half because the professor himself doesn't seem to like it that much and half because he was like yeah the algebraists all place out so I'm assuming everyone here is an analyst and doesn't care about commutative algebra Then the year after another guy taught and made it mostly commutative algebra + a bit of varieties + Cech cohomology at the end from nowhere and everyone was like uhhh. Then apparently this year was more of an experiment, in part from requests to make things more geometric It's got 3 "underground" floors (quotation marks because the place is on a very tall hill so the first 3 floors are a good bit above the the street), and then 9 floors above ground. The grad lounge is in the top floor and overlooks the city and lake, it's real nice The basement floors have the library and all the classrooms (each of them has a lot more area than the higher ones), floor 1 is basically just the entrance, I'm not sure what's on the second floor, 3-8 is all offices, and 9 has the ground lounge mainly And then there's one weird area called the math bunker that's trickier to access, you have to leave the building from the first floor, head outside (still walking on the roof of the basement floors), go to this other structure, and then get in. Some number of grad student cubicles are there (other grad students get offices in the main building) It's hard to get a feel for which places are good at undergrad math. Highly ranked places are known for having good researchers but there's no "How well does this place teach?" ranking which is kinda more relevant if you're an undergrad I think interest might have started the trend, though it is true that grad admissions now is starting to make it closer to an expectation (friends of mine say that for experimental physics, classes and all definitely don't cut it anymore) In math I don't have a clear picture. It seems there are a lot of Mickey Mouse projects that people seem to not help people much, but more and more people seem to do more serious things and that seems to become a bonus One of my professors said it to describe a bunch of REUs, basically boils down to problems that some of these give their students which nobody really cares about but which undergrads could work on and get a paper out of @TedShifrin i think universities have been ostensibly a game of credentialism for a long time, they just used to be gated off to a lot more people than they are now (see: ppl from backgrounds like mine) and now that budgets shrink to nothing (while administrative costs balloon) the problem gets harder and harder for students In order to show that $x=0$ is asymptotically stable, one needs to show that $$\forall \varepsilon > 0, \; \exists\, T > 0 \; \mathrm{s.t.} \; t > T \implies \|\| x ( t ) - 0 \|\| < \varepsilon.$$The intuitive sketch of the proof is that one has to fit a sublevel set of continuous functions $... "If $U$ is a domain in $\Bbb C$ and $K$ is a compact subset of $U$, then for all holomorphic functions on $U$, we have $\sup_{z \in K}\|f(z)\| \leq C_K \\|f\\|_{L^2(U)}$ with $C_K$ depending only on $K$ and $U$" this took me way longer than it should have Well, $A$ has these two dictinct eigenvalues meaning that $A$ can be diagonalised to a diagonal matrix with these two values as its diagonal. What will that mean when multiplied to a given vector (x,y) and how will the magnitude of that vector changed? Alternately, compute the operator norm of $A$ and see if it is larger or smaller than 2, 1/2 Generally, speaking, given. $\alpha=a+b\sqrt{\delta}$, $\beta=c+d\sqrt{\delta}$ we have that multiplication (which I am writing as $\otimes$) is $\alpha\otimes\beta=(a\cdot c+b\cdot d\cdot\delta)+(b\cdot c+a\cdot d)\sqrt{\delta}$ Yep, the reason I am exploring alternative routes of showing associativity is because writing out three elements worth of variables is taking up more than a single line in Latex, and that is really bugging my desire to keep things straight. hmm... I wonder if you can argue about the rationals forming a ring (hence using commutativity, associativity and distributivitity). You cannot do that for the field you are calculating, but you might be able to take shortcuts by using the multiplication rule and then properties of the ring $\Bbb{Q}$ for example writing $x = ac+bd\delta$ and $y = bc+ad$ we then have $(\alpha \otimes \beta) \otimes \gamma = (xe +yf\delta) + (ye + xf)\sqrt{\delta}$ and then you can argue with the ring property of $\Bbb{Q}$ thus allowing you to deduce $\alpha \otimes (\beta \otimes \gamma)$ I feel like there's a vague consensus that an arithmetic statement is "provable" if and only if ZFC proves it. But I wonder what makes ZFC so great, that it's the standard working theory by which we judge everything. I'm not sure if I'm making any sense. Let me know if I should either clarify what I mean or shut up. :D Associativity proofs in general have no shortcuts for arbitrary algebraic systems, that is why non associative algebras are more complicated and need things like Lie algebra machineries and morphisms to make sense of One aspect, which I will illustrate, of the "push-button" efficacy of Isabelle/HOL is its automation of the classic "diagonalization" argument by Cantor (recall that this states that there is no surjection from the naturals to its power set, or more generally any set to its power set).theorem ... The axiom of triviality is also used extensively in computer verification languages... take Cantor's Diagnolization theorem. It is obvious. (but seriously, the best tactic is over powered...) Extensions is such a powerful idea. I wonder if there exists algebraic structure such that any extensions of it will produce a contradiction. O wait, there a maximal algebraic structures such that given some ordering, it is the largest possible, e.g. surreals are the largest field possible It says on Wikipedia that any ordered field can be embedded in the Surreal number system. Is this true? How is it done, or if it is unknown (or unknowable) what is the proof that an embedding exists for any ordered field? Here's a question for you: We know that no set of axioms will ever decide all statements, from Gödel's Incompleteness Theorems. However, do there exist statements that cannot be decided by any set of axioms except ones which contain one or more axioms dealing directly with that particular statement? "Infinity exists" comes to mind as a potential candidate statement. Well, take ZFC as an example, CH is independent of ZFC, meaning you cannot prove nor disprove CH using anything from ZFC. However, there are many equivalent axioms to CH or derives CH, thus if your set of axioms contain those, then you can decide the truth value of CH in that system @Rithaniel That is really the crux on those rambles about infinity I made in this chat some weeks ago. I wonder to show that is false by finding a finite sentence and procedure that can produce infinity but so far failed Put it in another way, an equivalent formulation of that (possibly open) problem is: > Does there exists a computable proof verifier P such that the axiom of infinity becomes a theorem without assuming the existence of any infinite object? If you were to show that you can attain infinity from finite things, you'd have a bombshell on your hands. It's widely accepted that you can't. If fact, I believe there are some proofs floating around that you can't attain infinity from the finite. My philosophy of infinity however is not good enough as implicitly pointed out when many users who engaged with my rambles always managed to find counterexamples that escape every definition of an infinite object I proposed, which is why you don't see my rambles about infinity in recent days, until I finish reading that philosophy of infinity book The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. It derives its name from the problem faced by someone who is constrained by a fixed-size knapsack and must fill it with the most valuable items.The problem often arises in resource allocation where there are financial constraints and is studied in fields such as combinatorics, computer science... O great, given a transcendental $s$, computing $\min_P(\|P(s)\|)$ is a knapsack problem hmm... By the fundamental theorem of algebra, every complex polynomial $P$ can be expressed as: $$P(x) = \prod_{k=0}^n (x - \lambda_k)$$ If the coefficients of $P$ are natural numbers , then all $\lambda_k$ are algebraic Thus given $s$ transcendental, to minimise $\|P(s)\|$ will be given as follows: The first thing I think of with that particular one is to replace the $(1+z^2)$ with $z^2$. Though, this is just at a cursory glance, so it would be worth checking to make sure that such a replacement doesn't have any ugly corner cases. In number theory, a Liouville number is a real number x with the property that, for every positive integer n, there exist integers p and q with q > 1 and such that0<\|x−pq\|<1qn.{\displaystyle 0<\left\|x-{\frac {p}... Do these still exist if the axiom of infinity is blown up? Hmmm... Under a finitist framework where only potential infinity in the form of natural induction exists, define the partial sum: $$\sum_{k=1}^M \frac{1}{b^{k!}}$$ The resulting partial sums for each M form a monotonically increasing sequence, which converges by ratio test therefore by induction, there exists some number $L$ that is the limit of the above partial sums. The proof of transcendentally can then be proceeded as usual, thus transcendental numbers can be constructed in a finitist framework There's this theorem in Spivak's book of Calculus:Theorem 7Suppose that $f$ is continuous at $a$, and that $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x=a$. Suppose, moreover, that $\lim_{x \to a} f'(x)$ exists. Then $f'(a)$ also exists, and$$f'... and neither Rolle nor mean value theorem need the axiom of choice Thus under finitism, we can construct at least one transcendental number. If we throw away all transcendental functions, it means we can construct a number that cannot be reached from any algebraic procedure Therefore, the conjecture is that actual infinity has a close relationship to transcendental numbers. Anything else I need to finish that book to comment typo: neither Rolle nor mean value theorem need the axiom of choice nor an infinite set > are there palindromes such that the explosion of palindromes is a palindrome nonstop palindrome explosion palindrome prime square palindrome explosion palirome prime explosion explosion palindrome explosion cyclone cyclone cyclone hurricane palindrome explosion palindrome palindrome explosion explosion cyclone clyclonye clycone mathphile palirdlrome explosion rexplosion palirdrome expliarome explosion exploesion
Westergaard is a British scientist. For vertical stress computation, he had proposed a formula in 1938. The formula is presented below. Westergaard's Equation for Point Loads If Q is the point load and σ z is the vertical stress due to the point load, \[\sigma_{z}=\frac{Q}{2{\pi}z^{2}}\times \frac{\sqrt{(1-2\mu)/(2-2\mu)}} {[(1-2\mu)/(2-\mu)+(r/z)^{2})]^{3/2}}= \frac{Q}{z^{2}}I_{w}\] Here, μ= Poisson's Ratio. If Poisson's Ratio is considered zero for all practical purposes, \[\sigma_{z}=\frac{Q}{{\pi}z^{2}}\times \frac{1} {[1+2(r/z)^{2})]^{3/2}}= \frac{Q}{z^{2}}I_{w}\] Where, \[I_{w}= \frac{(1/\pi)} {[1+2(r/z)^{2})]^{3/2}}\] \[I_{w}\] is known as Westergaard stress coefficient.
Discrete Gaussian Samplers over the Integers¶ This class realizes oracles which returns integers proportionally to$\exp(-(x-c)^2/(2σ^2))$. All oracles are implemented using rejection sampling.See DiscreteGaussianDistributionIntegerSampler.__init__() for which algorithms areavailable. AUTHORS: Martin Albrecht (2014-06-28): initial version EXAMPLES: We construct a sampler for the distribution $D_{3,c}$ with width $σ=3$ and center $c=0$: sage: from sage.stats.distributions.discrete_gaussian_integer import DiscreteGaussianDistributionIntegerSamplersage: sigma = 3.0sage: D = DiscreteGaussianDistributionIntegerSampler(sigma=sigma) We ask for 100000 samples: sage: from six.moves import rangesage: n=100000; l = [D() for _ in range(n)] These are sampled with a probability proportional to $\exp(-x^2/18)$. More precisely we have to normalise by dividing by the overall probability over all integers. We use the fact that hitting anything more than 6 standard deviations away is very unlikely and compute: sage: bound = (6sigma).floor()sage: norm_factor = sum([exp(-x^2/(2sigma^2)) for x in range(-bound,bound+1)])sage: norm_factor7.519... With this normalisation factor, we can now test if our samples follow the expected distribution: sage: x=0; l.count(x), ZZ(round(nexp(-x^2/(2sigma^2))/norm_factor))(13355, 13298)sage: x=4; l.count(x), ZZ(round(nexp(-x^2/(2sigma^2))/norm_factor))(5479, 5467)sage: x=-10; l.count(x), ZZ(round(nexp(-x^2/(2sigma^2))/norm_factor))(53, 51) We construct an instance with a larger width: sage: from sage.stats.distributions.discrete_gaussian_integer import DiscreteGaussianDistributionIntegerSamplersage: sigma = 127sage: D = DiscreteGaussianDistributionIntegerSampler(sigma=sigma, algorithm='uniform+online') ask for 100000 samples: sage: from six.moves import rangesage: n=100000; l = [D() for _ in range(n)] # long time and check if the proportions fit: sage: x=0; y=1; float(l.count(x))/l.count(y), exp(-x^2/(2sigma^2))/exp(-y^2/(2sigma^2)).n() # long time(1.0, 1.00...)sage: x=0; y=-100; float(l.count(x))/l.count(y), exp(-x^2/(2sigma^2))/exp(-y^2/(2sigma^2)).n() # long time(1.32..., 1.36...) We construct a sampler with $c\%1 != 0$: sage: from sage.stats.distributions.discrete_gaussian_integer import DiscreteGaussianDistributionIntegerSamplersage: sigma = 3sage: D = DiscreteGaussianDistributionIntegerSampler(sigma=sigma, c=1/2)sage: from six.moves import rangesage: n=100000; l = [D() for _ in range(n)] # long timesage: mean(l).n() # long time0.486650000000000 REFERENCES: class sage.stats.distributions.discrete_gaussian_integer. DiscreteGaussianDistributionIntegerSampler¶ A Discrete Gaussian Sampler using rejection sampling. __init__( sigma, c=0, tau=6, algorithm=None, precision='mp')¶ Construct a new sampler for a discrete Gaussian distribution. INPUT: sigma- samples $x$ are accepted with probability proportional to $\exp(-(x-c)²/(2σ²))$ c- the mean of the distribution. The value of cdoes not have to be an integer. However, some algorithms only support integer-valued c(default: 0) tau- samples outside the range $(⌊c⌉-⌈στ⌉,...,⌊c⌉+⌈στ⌉)$ are considered to have probability zero. This bound applies to algorithms which sample from the uniform distribution (default: 6) algorithm- see list below (default: "uniform+table"for $σt$ bounded by DiscreteGaussianDistributionIntegerSampler.table_cutoffand "uniform+online"for bigger $στ$) precision- either "mp"for multi-precision where the actual precision used is taken from sigma or "dp"for double precision. In the latter case results are not reproducible. (default: "mp") ALGORITHMS: "uniform+table"- classical rejection sampling, sampling from the uniform distribution and accepted with probability proportional to $\exp(-(x-c)²/(2σ²))$ where $\exp(-(x-c)²/(2σ²))$ is precomputed and stored in a table. Any real-valued $c$ is supported. "uniform+logtable"- samples are drawn from a uniform distribution and accepted with probability proportional to $\exp(-(x-c)²/(2σ²))$ where $\exp(-(x-c)²/(2σ²))$ is computed using logarithmically many calls to Bernoulli distributions. See [DDLL2013] for details. Only integer-valued $c$ are supported. "uniform+online"- samples are drawn from a uniform distribution and accepted with probability proportional to $\exp(-(x-c)²/(2σ²))$ where $\exp(-(x-c)²/(2σ²))$ is computed in each invocation. Typically this is very slow. See [DDLL2013] for details. Any real-valued $c$ is accepted. "sigma2+logtable"- samples are drawn from an easily samplable distribution with $σ = k·σ_2$ with $σ_2 = \sqrt{1/(2\log 2)}$ and accepted with probability proportional to $\exp(-(x-c)²/(2σ²))$ where $\exp(-(x-c)²/(2σ²))$ is computed using logarithmically many calls to Bernoulli distributions (but no calls to $\exp$). See [DDLL2013] for details. Note that this sampler adjusts $σ$ to match $k·σ_2$ for some integer $k$. Only integer-valued $c$ are supported. EXAMPLES: sage: from sage.stats.distributions.discrete_gaussian_integer import DiscreteGaussianDistributionIntegerSampler sage: DiscreteGaussianDistributionIntegerSampler(3.0, algorithm="uniform+online") Discrete Gaussian sampler over the Integers with sigma = 3.000000 and c = 0 sage: DiscreteGaussianDistributionIntegerSampler(3.0, algorithm="uniform+table") Discrete Gaussian sampler over the Integers with sigma = 3.000000 and c = 0 sage: DiscreteGaussianDistributionIntegerSampler(3.0, algorithm="uniform+logtable") Discrete Gaussian sampler over the Integers with sigma = 3.000000 and c = 0 Note that "sigma2+logtable"adjusts $σ$: sage: DiscreteGaussianDistributionIntegerSampler(3.0, algorithm="sigma2+logtable") Discrete Gaussian sampler over the Integers with sigma = 3.397287 and c = 0 __call__()¶ Return a new sample. EXAMPLES: sage: from sage.stats.distributions.discrete_gaussian_integer import DiscreteGaussianDistributionIntegerSampler sage: DiscreteGaussianDistributionIntegerSampler(3.0, algorithm="uniform+online")() -3 sage: DiscreteGaussianDistributionIntegerSampler(3.0, algorithm="uniform+table")() 3 algorithm¶ c¶ sigma¶ tau¶
Polya-Hurwitz program. This may become more interesting in light of the recent progress in the Polya-Jensen program by Griffin, Ono, Rolen, Zagier. We will first provide definitions of some functions involved. The Riemann Xi-function $\Xi(z)$ is related to the Riemann zeta-function $\zeta(s)$ via ([A], [B]): $\Xi(z)=\xi(\tfrac{1}{2}+iz)$,$\xi(s)=\tfrac{1}{2}s(s-1)\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s)$. Riemann $\Xi(z)$ can be expressed as a Fourier transform of a positive, fast decaying, and even kernel $\Phi(t)$ ([A], [B]): \begin{equation}\Xi(z)=2\int_0^{\infty}\Phi(t)\cos(zt)\mathrm{d}t,\tag{1}\end{equation} where \begin{equation}\Phi(t)=\sum_{k\geqslant 1}\phi_k(t)=\Phi(-t),\tag{2}\\\end{equation} \begin{equation}\phi_k(t)=\left(4\pi^2 k^4 e^{9t/2}-6\pi k^2e^{5t/2}\right)\exp\left(-\pi k^2 e^{2t}\right)\tag{3}.\end{equation} The Polya aspect of this approach is the following: Truncate the Kernel $\Phi(t)$ of (2) and/or the integration range in (1) such that the resulting Fourier transform leads to a family of entire functions which only have real roots. One such candidate is given in [C]:\begin{equation}\Phi_{\color{red}n}(t)=(1/2)\sum_{1\leqslant k\leqslant {\color{red}n}}\left(\phi_k(t)+\phi_k(-t)\right)=\Phi_{\color{red}n}(-t)\tag{4}\end{equation} \begin{equation}\Xi_{\color{red}n}(z)=2\int_0^{(1/2)\log {\color{red}n}}\Phi_{\color{red}n}(t)\cos(zt)\mathrm{d}t=\Xi_{\color{red}n}(-z),\tag{5}\end{equation} We refer to [D] and [E] for a near complete review on the zeros of entire functions as Fourier transforms. The Hurwitz aspect of this approach is the following: Corollary of Hurwitz's theorem in complex analysis (applied to our case) [F]: If $\Xi(z)$ and $\{\Xi_n(z)\}$ are analytic functions on a domain $S_{1/2}(z)=\{z: 0<Im(z)<1/2\}$, $\{\Xi_n(z)\}$ converges to $\Xi(z)$ uniformly on compact subsets of $S_{1/2}(z)$, and all but finitely many $\Xi_n(z)$ have no zeros in $S_{1/2}(z)$, then either $\Xi(z)$ is identically zero or $\Xi(z)$ has no zeros in $S_{1/2}(z)$. The functional equation for $\zeta(s)$ becomes $\Xi(-z)=\Xi(z)$. The candidate of $\Xi_n(z)$ in (5) automatically satisfies this functional equation. Another benefit of Polya-Hurwitz approach is that the entrance barrier is relatively low (comparing to other approaches that usually require the advance knowledge of analytical number theory). To get started, one only needs to know Fourier transform, basic complex analysis, some knowledge of entire functions, polynomials etc. So anyone who has math training with the college undergraduate math major may start to work on the Polya-Hurwitz approach and learn other necessary new math as he/she goes. The most difficult part of Polya-Hurwitz approach seems to be the following:(for example,) proving that all the zeros of $\Xi_n(z)$ in (5) are real in $S_{1/2}(z)$. One may need to have several iterations: guess one form of the Kernel like $\Phi_n(1,t)$ and complete the integration to get explicit expression for $\Xi_n(1,z)$. If all the zeros of $\Xi_n(1,z)$ are found not to be all real in $S_{1/2}(z)$, then move on to $\Phi_n(2,t)$ and $\Xi_n(2,z)$... References: [A] Titchmarsh,"The Theory of the Riemann Zeta-Function", (1986). [B] Edwards, "Riemann's Zeta Function", (1974). [C] Shi, "On the zeros of Riemann Xi-function", (2017) arXiv:1706.08868. [D] Dimitrov and Rusev, “ZEROS OF ENTIRE FOURIER TRANSFORMS” (2001), 108 page review paper. [E] Hallum, “ZEROS OF ENTIRE FUNCTIONS REPRESENTED BY FOURIER TRANSFORMS” (2014), Master thesis. [F] Conway, "Functions of One Complex Variable",(1978)
Background I've met this problem when I was trying to convert a elliptic PDE problem into the corresponding variational problem in order to apply finite element method. The PDE is an elliptic PDE with non-zero Dirichlet boundary condition: Denote $$ Lu=-\nabla\cdot(a\nabla u)+bu $$ Then the equation is $$ \left\{\!\! \begin{aligned} &Lu=f,x\in\Omega\\ &u\|_{\partial \Omega}=g \end{aligned} \right. $$ When $g\equiv0$, I know the corresponding variational problem is find $u\in H_0^1(\Omega)$, such that $$ a(u,v)=(f,v), \forall v\in H_0^1(\Omega) $$ where $$ \begin{aligned} a(u,v)&:=\int_\Omega a\nabla u\cdot\nabla v\,dx+\int_\Omega buv\,dx\qquad \\ (f,v)&:=\int_\Omega fv\,dx,\qquad \forall u,v\in H_0^1(\Omega) \end{aligned} $$ (This is actually the weak form of the original PDE.) Here comes my problem: For general g,if I can find a function $w\in H^1$ such that $w\|_{\partial\Omega}=g$, by letting $\tilde u=u-w$, we have$$\left\{\!\!\begin{aligned}&L\tilde u=\tilde f,x\in\Omega\\&\tilde u\|_{\partial \Omega}=0\end{aligned}\right.$$whose solution is already known. So how to find such a $w$?
I'm trying to evaluate the integral of the Chebyshev polynomials of the first kind on the interval $-1 \leq x \leq 1 $ . My idea is to use the closed form $$T_n(x) = \frac{z_1^n + z_2^n}{2}, $$ where $z_1 = (x + \sqrt{x^2 - 1})$ and $z_2 = (x - \sqrt{x^2 - 1})$, giving the following integral: $$ \int _{-1}^{1}\!1/2\, \left( x+\sqrt {-1+{x}^{2}} \right) ^{n}+1/2\, \left( x-\sqrt {-1+{x}^{2}} \right) ^{n}{}{dx} $$ I'm stuck at integrating $z_1$ and $z_2$. I tried integrating by parts n times, but i'm looking for a general formula. My calculus is pretty rusty so i'm not sure if this is the way to go. Any tips? Thanks alot. It is actually not that hard. You can derive a lot of relations on the wiki page yourself by substituting $\cos\theta$ for $x$ and use the defining relation of Chebyshev polynomials: $$T_n(\cos\theta) = \cos( n\theta)$$ For example, one have: $$\begin{align}\int T_n(x) dx = & \int T_n(\cos\theta) d\cos \theta\\ = & -\int \cos(n\theta)\sin\theta d\theta\\ = & -\frac12 \int \left(\sin((n+1)\theta) - \sin((n-1)\theta)\right)d\theta\\ = & \frac12 \left(\frac{\cos((n+1)\theta)}{n+1} - \frac{\cos((n-1)\theta)}{n-1}\right) + \text{const.}\\ = & \frac12 \left(\frac{T_{n+1}(x)}{n+1} - \frac{T_{n-1}(x)}{n-1}\right) + \text{const.} \end{align} $$ An alternative easy way to do this would be to convert everything to the complex exponential. Given $$T_n(\cos\theta) = \cos(n\theta) = \frac{1}{2}\left( e^{in\theta} + e^{-in\theta} \right) $$ and since $\frac{d\cos\theta}{d\theta} = \frac{i}{2}\left( e^{i\theta} - e^{-i\theta} \right)$ we have $$\begin{align}\int T_n(x)dx & = \int T_n(\cos\theta)d\cos\theta \\ & = \int \cos(n\theta)d\cos\theta \\ &= \frac{i}{4}\int \left( e^{in\theta} + e^{-in\theta} \right)\left( e^{i\theta} - e^{-i\theta} \right)d\theta \end{align}$$ which is straightforward to integrate. At the end of the day you will get $$\int T_n(x)dx = \frac{1}{2}\left(\frac{T_{n+1}}{(n+1)}- \frac{T_{n-1}}{(n-1)} \right) $$ after converting the complex exponential back to the trigonometric form and using the definition of the $n$-th Chebyshev polynomial as given above. Wikipedia has a nice article on the Chebychev polynomials: http://en.wikipedia.org/wiki/Chebyshev_polynomials. In particular, there is this: $\int T_n(x) dx = \frac{n T_{n+1}(x)}{n^2-1}-\frac{x T_n(x)}{n-1}$.
Here I am on thin ice but let me try: I have a feeling (please comment!) that a main difference between statistics and econometrics is that in statistics we tend to consider the regressors as fixed, hence the terminology design matrix which obviously comes from design of experiments, where the supposition is that we are first choosing and then fixing the explanatory variables. But for most data sets, most situations, this is a bad fit. We are really observing the explanatory variables, and in that sense they stand at the same footing as the response variables, they are both determined by some random process outside our control. By considering the $x$'s as "fixed", we decide not to consider a lot of problems which that might cause. By considering the regressors as stochastic, on the other hand, as econometricians tend to do, we open the possibility of modeling which try to consider such problems. A short list of problems we then might consider, and incorporate into the modeling, is: measurement errors in the regressors correlations between regressors and error terms lagged response as regressor ... Probably, that should be done much more frequently that it is done today? EDIT I will try to flesh out an argument for conditioning on regressors somewhat more formally. Let $(Y,X)$ be a random vector, and interest is in regression $Y$ on $X$, where regression is taken to mean the conditional expectation of $Y$ on $X$. Under multinormal assumptions that will be a linear function, but our arguments do not depend on that. We start with factoring the joint density in the usual way$$ f(y,x) = f(y\mid x) f(x)$$but those functions are not known so we use a parameterized model$$ f(y,x; \theta, \psi)=f_\theta(y \mid x) f_\psi(x)$$where $\theta$ parameterizes the conditional distribution and $\psi$ the marginal distribution of $X$. In the normal linear model we can have $\theta=(\beta, \sigma^2)$ but that is not assumed. The full parameter space of $(\theta,\psi)$ is $\Theta \times \Psi$, a Cartesian product, and the two parameters have no part in common. This can be interpreted as a factorization of the statistical experiment, (or of the data generation process, DGP), first $X$ is generated according to $f_\psi(x)$, and as a second step, $Y$ is generated according to the conditional density $f_\theta(y \mid X=x)$. Note that the first step does not use any knowledge about $\theta$, that enters only in the second step. The statistic $X$ is ancillary for $\theta$, see https://en.wikipedia.org/wiki/Ancillary_statistic. But, depending on the results of the first step, the second step could be more or less informative about $\theta$. If the distribution given by $f_\psi(x)$ have very low variance, say, the observed $x$'s will be concentrated in a small region, so it will be more difficult to estimate $\theta$. So, the first part of this two-step experiment determines the precision with which $\theta$ can be estimated. Therefore it is natural to condition on $X=x$ in inference about the regression parameters. That is the conditionality argument, and the outline above makes clear its assumptions. In designed experiments its assumption will mostly hold, often with observational data not. Some examples of problems will be: regression with lagged responses as predictors. Conditioning on the predictors in this case will also condition on the response! (I will add more examples). One book which discusses this problems in a lot of detail is Information and exponential families: In statistical theory by O. E Barndorff-Nielsen. See especially chapter 4. The author says the separation logic in this situation is however seldom explicated but gives the following references: R A Fisher (1956) Statistical Methods and Scientific Inference $\S 4.3$ and Sverdrup (1966) The present state of the decision theory and the Neyman-Pearson theory.
This is all just a result of sloppy language on the part of people describing quantum mechanics.The state$$ \left\lvert \Psi \right\rangle = \frac{1}{\sqrt{2}} \left( \left\lvert \uparrow \right\rangle + \left\lvert \downarrow \right\rangle\right) \tag{1}$$is a superposition of the two orthogonal states $\left\lvert \uparrow \right\rangle$ and $\left\lvert \downarrow \right\rangle$.The state is unlike either basis vector alone.A velocity vector$$\left\lvert v \right\rangle = a\left\lvert x \right\rangle + b\left\lvert y \right\rangle \tag{2}$$for some values $a$ and $b$ is also a superposition of two orthogonal velocity vectors.It is unlike either basis vector alone. Talking about $\left\lvert \Psi \right\rangle$ as "simultaneously in both states" is just plain sloppy.It's a superposition.It's not like either basis vector alone.It is, as you say, something completely distinct. However, you can often find a description of superposited states as a system being in both basis states at the same time (e.g. the spin being simultaneously up and down), but I have never found a similar description regarding vector quantities (for example velocity being along z axis and x axis at the same time if the vector was somewhere between these axes) in classical mechanics, despite both situations being a result of addition of the basis vectors. The reason for this disagreement in language comes from the fact that, in the end, quantum state vectors tell you probabilities of experimental outcomes.It really bugs people to think of the state of a physical system being fundamentally probabilistic.When it comes to measurement, the state $\left\lvert \Psi \right\rangle$ means that the system has a 1/2 probability to be measured spin up and and 1/2 probability to be measured spin down.People don't naturally think about the world around them in terms superposition states whose coefficients correspond to probability amplitudes.They'd rather think about the classical states independently and try to form some kind of notion of the system existing in combinations of classical states.Therefore, they naturally (but erroneously) say that the system is in both classical states at the same time, when really, as you said, the system is in a state that's completely different from either classical basis state.
I have a little question and need some help with the notation. So, the question goes as follows: A bond with a maturity of ten years that pays annual coupons of 8% has a price of \$90. A bond with a maturity of ten years and annual coupons of 4% has a price of \$80. What is the ten year zero rate? I don't actually know what the ten-year zero rate is. I set up a system of equations with the information that is given: Is it just all about finding $y$? \begin{align} \$90 =& \sum_{i=1}^{10} e^{-yi} (0.08 Z) &+& e^{-y\cdot 10}Z\\ \$80 =& \sum_{i=1}^{10} e^{-yi} (0.04 Z) &+& e^{-y \cdot 10}Z\\ \Leftrightarrow \$10 =& \sum_{i=1}^{10} e^{-yi} (0.04 Z)\\ \Rightarrow \$70 =& e^{-y\cdot 10}Z \end{align} And is there any way to solve for the zero rate by hand? (This is the reason why I'm wondering; to solve this system, a calculator is necessary, but all the other homework problems were solvable by hand!)
2018-2019 ICPC, NEERC, Northern Eurasia Finals (Unrated, Online Mirror, ICPC Rules, Teams Preferred) Finished You are given a positive integer $$$n$$$. Find a sequence of fractions $$$\frac{a_i}{b_i}$$$, $$$i = 1 \ldots k$$$ (where $$$a_i$$$ and $$$b_i$$$ are positive integers) for some $$$k$$$ such that: $$$$$$ \begin{cases} \text{$b_i$ divides $n$, $1 < b_i < n$ for $i = 1 \ldots k$} \\ \text{$1 \le a_i < b_i$ for $i = 1 \ldots k$} \\ \text{$\sum\limits_{i=1}^k \frac{a_i}{b_i} = 1 - \frac{1}{n}$} \end{cases} $$$$$$ The input consists of a single integer $$$n$$$ ($$$2 \le n \le 10^9$$$). In the first line print "YES" if there exists such a sequence of fractions or "NO" otherwise. If there exists such a sequence, next lines should contain a description of the sequence in the following format. The second line should contain integer $$$k$$$ ($$$1 \le k \le 100\,000$$$) — the number of elements in the sequence. It is guaranteed that if such a sequence exists, then there exists a sequence of length at most $$$100\,000$$$. Next $$$k$$$ lines should contain fractions of the sequence with two integers $$$a_i$$$ and $$$b_i$$$ on each line. 2 NO 6 YES 2 1 2 1 3 In the second example there is a sequence $$$\frac{1}{2}, \frac{1}{3}$$$ such that $$$\frac{1}{2} + \frac{1}{3} = 1 - \frac{1}{6}$$$. Name
If a polygon can be cut into $m$ as well as into $n$ triangular pieces of equal area, can it also be cut into $m+n$ triangles of equal area? (I'm editing after realizing that my conjecture that a convex equidissectable polygon must have an equidissection with all triangles meeting in a common vertex is very wrong.) Some background: An equidissection of a polygon is a dissection into triangles of equal area. The equidissection spectrum of a polygon is the set of all $n$ such that there exists an equidissection into $n$ pieces. A famous theorem of Paul Monsky states that the spectrum of the square is the set of all even positive numbers (the nontrivial part being that an equidissection into an odd number of pieces is impossible). Most quadrilaterals do not admit any equidissection, and there are quadrilaterals for which the existence of an equidissection is an open question. The wikipedia article on Equidissection describes more of the background. If a polygon can be equidissected into $n$ pieces, then it can obviously be equidissected into any multiple of $n$ by subdividing the pieces. So the spectrum is closed under multiplication by a positive integer. On the wikipedia page mentioned above, an example is given of a quadrilateral with a "non-principal" spectrum. It is equivalent to a "kite" with corners at $(-1,0), (0,-1), (2,0)$ and $(0,1)$. The two diagonals will divide the area in ratios 1 to 1 and 1 to 2 respectively, and therefore it allows equidissections into 2 and 3 pieces. My question is motivated by the observation (which can hardly be new, but is not mentioned in the wikipedia article or in this manuscript) that this kite also admits an equidissection into 5 triangles (or into any number $\geq 2$). To see this, draw straight lines from the point $(4/5, 3/5)$ to the vertices $(-1,0)$ and $(0,-1)$. This will divide the kite into triangles of relative areas $1/5$, $2/5$ and $2/5$. More generally, suppose that $P$ and $Q$ are points (of the interior or the boundary) such that the straight lines from $P$ to the corners of the quadrilateral yield triangles whose areas are multiples of $1/m$ of the total area, and similarly the lines from $Q$ give areas that are multiples of $1/n$ of the total area. Then from the point $\frac {m}{m+n}\cdot P + \frac{n}{m+n}\cdot Q$ we will get areas that are of the form $$\frac{a}{m}\cdot \frac{m}{m+n} + \frac{b}{n}\cdot \frac{n}{m+n} = \frac{a+b}{m+n}$$ and therefore multiples of $1/(m+n)$ of the total area. Starting from $P=(0,1)$ and $Q=(2,0)$, we can therefore obtain equidissections of the kite into 5, 7 etc. parts. Similarly, the (non-convex) quadrilateral with vertices in $(-1,0)$, $(0,-1)$, $-2/3,0)$ and $(0,1)$ can be equidissected into 2 (obvious) or 3 (by the vertical line from $(-2/3,-1/3)$ to $(-2/3,1/3)$) pieces. And by continuing the boundary line between $(0,1)$ and $(-2/3,0)$ to the point where it crosses the line from $(0,-1)$ to $(-1,0)$, we divide it into triangles of relative areas $3/5$ and $2/5$, showing that an equidissection into 5 pieces is also possible. Let us say that a dissection is star-like if all triangles have a common vertex. The argument above shows that If a polygon has a star-like equidissection into $m$ pieces, and another into $n$ pieces, then it also has one into $m+n$ pieces. Of course a highly non-convex polygon may have equidissections none of which are star-like, and it is easy to see that this might be the case also for convex polygons: Take a regular hexagon of unit area and label the vertices A, B, C, D, E, F. If we replace the vertex at A by a vertex at a point A' on the line through A parallel to the line BF, then the new hexagon still admits an equidissection into six pieces by cutting along the triangle BFD (and dividing the middle piece into three). But there are only countably many choices of A' for which there is some point P such that the six triangles with vertices in P and two adjacent vertices of the hexagon all have rational area. So perhaps the answer to my question is the obvious "No, why should it?", and the reason I haven't found a counterexample is just that dissections into a small number of pieces tend to be star-like. There is a short comment on the second page of this article that shows that the spectrum of a trapezoid is closed under addition, but I don't see how that would generalize.
The Metropolis computer network consists of $$$n$$$ servers, each has an encryption key in the range from $$$0$$$ to $$$2^k - 1$$$ assigned to it. Let $$$c_i$$$ be the encryption key assigned to the $$$i$$$-th server. Additionally, $$$m$$$ pairs of servers are directly connected via a data communication channel. Because of the encryption algorithms specifics, a data communication channel can only be considered safe if the two servers it connects have distinct encryption keys. The initial assignment of encryption keys is guaranteed to keep all data communication channels safe. You have been informed that a new virus is actively spreading across the internet, and it is capable to change the encryption key of any server it infects. More specifically, the virus body contains some unknown number $$$x$$$ in the same aforementioned range, and when server $$$i$$$ is infected, its encryption key changes from $$$c_i$$$ to $$$c_i \oplus x$$$, where $$$\oplus$$$ denotes the bitwise XOR operation. Sadly, you know neither the number $$$x$$$ nor which servers of Metropolis are going to be infected by the dangerous virus, so you have decided to count the number of such situations in which all data communication channels remain safe. Formally speaking, you need to find the number of pairs $$$(A, x)$$$, where $$$A$$$ is some (possibly empty) subset of the set of servers and $$$x$$$ is some number in the range from $$$0$$$ to $$$2^k - 1$$$, such that when all servers from the chosen subset $$$A$$$ and none of the others are infected by a virus containing the number $$$x$$$, all data communication channels remain safe. Since this number can be quite big, you are asked to find its remainder modulo $$$10^9 + 7$$$. The first line of input contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \leq n \leq 500\,000$$$, $$$0 \leq m \leq \min(\frac{n(n - 1)}{2}, 500\,000)$$$, $$$0 \leq k \leq 60$$$) — the number of servers, the number of pairs of servers directly connected by a data communication channel, and the parameter $$$k$$$, which defines the range of possible values for encryption keys. The next line contains $$$n$$$ integers $$$c_i$$$ ($$$0 \leq c_i \leq 2^k - 1$$$), the $$$i$$$-th of which is the encryption key used by the $$$i$$$-th server. The next $$$m$$$ lines contain two integers $$$u_i$$$ and $$$v_i$$$ each ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \ne v_i$$$) denoting that those servers are connected by a data communication channel. It is guaranteed that each pair of servers appears in this list at most once. The only output line should contain a single integer — the number of safe infections of some subset of servers by a virus with some parameter, modulo $$$10^9 + 7$$$. 4 4 2 0 1 0 1 1 2 2 3 3 4 4 1 50 4 5 3 7 1 7 2 1 2 2 3 3 4 4 1 2 4 96 Consider the first example. Possible values for the number $$$x$$$ contained by the virus are $$$0$$$, $$$1$$$, $$$2$$$ and $$$3$$$. For values $$$0$$$, $$$2$$$ and $$$3$$$ the virus can infect any subset of the set of servers, which gives us $$$16$$$ pairs for each values. A virus containing the number $$$1$$$ can infect either all of the servers, or none. This gives us $$$16 + 2 + 16 + 16 = 50$$$ pairs in total. Name
In the paper, we introduce a quantum random walk polynomial (QRWP) that can be defined as a polynomial $$\{P_{n}(x)\}$$ { P n ( x ) } , which is orthogonal with respect to a quantum random walk measure (QRWM) on $$[-1, 1]$$ [ - 1 , 1 ] , such that the parameters $$\alpha _{n},\omega _{n}$$ α n , ω n are in the recurrence relations $$\begin{aligned} P_{n+1}(x)= (x - \alpha _{n})P_{n}(x) - \omega _{n}P_{n-1}(x) \end{aligned}$$ P n + 1 ( x ) = ( x - α n ) P n ( x ) - ω n P n - 1 ( x ) and satisfy $$\alpha _{n}\in \mathfrak {R},\omega _{n}> 0$$ α n ∈ R , ω n > 0 . We firstly obtain some results of QRWP and QRWM, in which case the correspondence between measures and orthogonal polynomial sequences is one-to-one. It shows that any measure with respect to which a quantum random walk polynomial sequence is orthogonal is a quantum random walk measure. We next collect some properties of QRWM; moreover, we extend Karlin and McGregor’s representation formula for the transition probabilities of a quantum random walk (QRW) in the interacting Fock space, which is a parallel result with the CGMV method. Using these findings, we finally obtain some applications for QRWM, which are of interest in the study of quantum random walk, highlighting the role played by QRWP and QRWM. Quantum Information Processing – Springer Journals Published: Jan 5, 2014 It’s your single place to instantly discover and read the research that matters to you. Enjoy affordable access to over 18 million articles from more than 15,000 peer-reviewed journals. All for just $49/month Query the DeepDyve database, plus search all of PubMed and Google Scholar seamlessly Save any article or search result from DeepDyve, PubMed, and Google Scholar... all in one place. Get unlimited, online access to over 18 million full-text articles from more than 15,000 scientific journals. Read from thousands of the leading scholarly journals from SpringerNature, Elsevier, Wiley-Blackwell, Oxford University Press and more. All the latest content is available, no embargo periods. “Hi guys, I cannot tell you how much I love this resource. Incredible. I really believe you've hit the nail on the head with this site in regards to solving the research-purchase issue.”Daniel C. “Whoa! It’s like Spotify but for academic articles.”@Phil_Robichaud “I must say, @deepdyve is a fabulous solution to the independent researcher's problem of #access to #information.”@deepthiw “My last article couldn't be possible without the platform @deepdyve that makes journal papers cheaper.”@JoseServera
Assume $V=L$ and let $\kappa$ be a Mahlo cardinal. Let $L[G]$ be the generic extension obatined by Mitchell forcing to make $2^{\aleph_0}=\aleph_2=\kappa.$ It is known that in the extension there are no special $\aleph_2$-Aronszajn trees but there are $\aleph_2$-Aronszajn trees. Question 1.Is there any $\aleph_2$-Souslin tree in $L[G]?$ I assume the believe is that there are, but I don't know how to prove it. Surprisingly, if, instead of $\aleph_2$, we consider a cardinal $\lambda^+ > \beth_\omega,$ with $\lambda$ regular and apply the Mitchell forcing to get $2^\lambda=\lambda^{++}=\kappa,$ then the results of Assaf Rinot show that there are $\lambda^{++}$-souslin trees in the extension. My second question is motivated by the work of Laver-Shelah. Assume $V=L$ and $\kappa$ is weakly compact. In order to produce a model of $CH+$there are no $\aleph_2$-Souslin trees, Laver and Shelah, first force with Levy collapse $Col(\aleph_1, < \kappa),$ and over it do an iteration to kill all possible $\aleph_2$-Souslin trees. Question 2.Are there any $\aleph_2$-Souslin trees just after doing the Levy collapse $Col(\aleph_1, < \kappa)$?
The Connected Component Process Model Creates an instance of the Connected Component point process model which can then be fitted to point pattern data. Usage Concom(r) Arguments r Threshold distance Details This function defines the interpoint interaction structure of a point process called the connected component process. It can be used to fit this model to point pattern data. The function ppm(), which fits point process models to point pattern data, requires an argument of class "interact" describing the interpoint interaction structure of the model to be fitted. The appropriate description of the connected component interaction is yielded by the function Concom(). See the examples below. In standard form, the connected component process (Baddeley and Moller, 1989) with disc radius $r$, intensity parameter $\kappa$ and interaction parameter $\gamma$ is a point process with probability density $$ f(x_1,\ldots,x_n) = \alpha \kappa^{n(x)} \gamma^{-C(x)} $$ for a point pattern $x$, where $x_1,\ldots,x_n$ represent the points of the pattern, $n(x)$ is the number of points in the pattern, and $C(x)$ is defined below. Here $\alpha$ is a normalising constant. To define the term C(x), suppose that we construct a planar graph by drawing an edge between each pair of points $x_i,x_j$ which are less than $r$ units apart. Two points belong to the same connected component of this graph if they are joined by a path in the graph. Then $C(x)$ is the number of connected components of the graph. The interaction parameter $\gamma$ can be any positive number. If $\gamma = 1$ then the model reduces to a Poisson process with intensity $\kappa$. If $\gamma < 1$ then the process is regular, while if $\gamma > 1$ the process is clustered. Thus, a connected-component interaction process can be used to model either clustered or regular point patterns. In spatstat, the model is parametrised in a different form, which is easier to interpret. In canonical form, the probability density is rewritten as $$ f(x_1,\ldots,x_n) = \alpha \beta^{n(x)} \gamma^{-U(x)} $$ where $\beta$ is the new intensity parameter and $U(x) = C(x) - n(x)$ is the interaction potential. In this formulation, each isolated point of the pattern contributes a factor $\beta$ to the probability density (so the first order trend is $\beta$). The quantity $U(x)$ is a true interaction potential, in the sense that $U(x) = 0$ if the point pattern $x$ does not contain any points that lie close together. When a new point $u$ is added to an existing point pattern $x$, the rescaled potential $-U(x)$ increases by zero or a positive integer. The increase is zero if $u$ is not close to any point of $x$. The increase is a positive integer $k$ if there are $k$ different connected components of $x$ that lie close to $u$. Addition of the point $u$ contributes a factor $\beta \eta^\delta$ to the probability density, where $\delta$ is the increase in potential. If desired, the original parameter $\kappa$ can be recovered from the canonical parameter by $\kappa = \beta\gamma$. The nonstationary connected component process is similar except that the contribution of each individual point $x_i$ is a function $\beta(x_i)$ of location, rather than a constant beta. Note the only argument of Concom() is the threshold distance r. When r is fixed, the model becomes an exponential family. The canonical parameters $\log(\beta)$ and $\log(\gamma)$ are estimated by ppm(), not fixed in Concom(). Value An object of class "interact" describing the interpoint interaction structure of the connected component process with disc radius $r$. Edge correction The interaction distance of this process is infinite. There are no well-established procedures for edge correction for fitting such models, and accordingly the model-fitting function ppm will give an error message saying that the user must specify an edge correction. A reasonable solution is to use the border correction at the same distance r, as shown in the Examples. References Baddeley, A.J. and Moller, J. (1989) Nearest-neighbour Markov point processes and random sets. International Statistical Review 57, 89--121. See Also Aliases Concom Examples # NOT RUN { # prints a sensible description of itself Concom(r=0.1) # Fit the stationary connected component process to redwood data ppm(redwood, ~1, Concom(r=0.07), rbord=0.07) # Fit the stationary connected component process to `cells' data ppm(cells, ~1, Concom(r=0.06), rbord=0.06) # eta=0 indicates hard core process. # Fit a nonstationary connected component model # with log-cubic polynomial trend # }# NOT RUN { ppm(swedishpines, ~polynom(x/10,y/10,3), Concom(r=7), rbord=7) # } Documentation reproduced from package spatstat, version 1.59-0, License: GPL (>= 2)
I've done a little bit of research and it seems Millikan was able to measure the ratio between the charge of the electron and its mass. But how can one measure one of the two constants to get the value of the other? The mass-to-charge ratio $m/e$ of the electron was first measured by J.J. Thomson, the discoverer of the electron, using cathode rays in 1897: It should not be surprising that one may measure this ratio even without isolating "individual electrons" because the electric force acting on a charge may be written as $$ F = ma = Ee, \quad a = E\cdot \frac em $$ So what was left was just to measure the mass or charge separately. Millikan and Fletcher did the relevant oil drop experiment in 1909. The electric force $F=Ee$ acting on a single drop with charge $e$, a single extra (or deficit) electron, may be calculated when it is set equal to the drag force from hydrodynamics, $6\pi e\eta v_1$. The viscosity $\eta$ is the most difficult thing to know but otherwise all quantities are known so $e$ may be calculated. If one knows the charge and the ratio, one may calculate the mass as $m = e/ (e/m)$.
SuNem $\mathrm{Nem}_{q}(z)=z+z^3+qz^4$ It is assumed, that $q\!>\!0$, although the formula can be used for some other values of the parameters too. SuNem is specific solution of the transfer equation $\mathrm{Nem}_{q}\big( \mathrm{SuNem}_{q}(z)\big)=\mathrm{SuNem}_{q}(z\!+\!1)$. It is assumed that $\mathrm{SuNem}_{q}(0)=1$. Also, the specific asymptotic behaviour at infinity is assumed, $\mathrm{SuNem}_{q}(z) = {\displaystyle \frac{1}{\sqrt{-2z}}}\left(1+O\left(\frac{1}{\sqrt{-2z}} \right)\right)$ for any fixed phase $\mathrm{Arg}(z)$ different from zero. For any $\varepsilon>0$, the formula is valid for any large $\|z\|$ such that $\|\mathrm{Arg}(z)\|>\varepsilon$. Along the real axis, SuNem shows fast growth from zero at $-\infty$ to plus infinity at $+\infty$. Contents Asymptotic expansion Function SuNem is constructed by its asymptotical expansion. For the superfunction $F$ of the Nemtsov transfer function $T=\mathrm{Nem}_{q}$, it can be obtained from the transfer equation $T(F(z))=F(z+1)$ Keping some positive integer mumber $M$ of terms, the asymptotic solution can be written as follows: $\displaystyle \tilde F(z) = \frac{1}{\sqrt{-2z}} \left(1+ \frac{P_m(\ln(-z))} {(-2z)^{m/2}} \right)$ where $\displaystyle P_m(z)=\sum_{n=0}^{\mathrm{IntegerPart}(m/2)} A[m,n]\, z^m$ Substitution of $\tilde F$ into the transfer equation gives the coefficients $A$. These coefficients can be calculated with the mathematica code below: Mathematica generator of the algorithm The first 18 terms of the asymptotic representation of super function $F$ can be computed with Mathematica software, using the code below: T[z_]=z+z^3+q z^4 P[m_, L_] := Sum[a[m, n] L^n, {n, 0, IntegerPart[m/2]}] a[1, 0] = -q; a[2, 0] = 0; m = 2; F[m_,z_] = (-2 z)^(-1/2) (1 + Sum[P[n, Log[-z]]/(-2 z)^(n/2), {n, 1, 2}]); s[m]=Numerator[Normal[Series[(T[F[m,-1/x^2]] - F[m,-1/x^2+1]) 2^((m+1)/2)/x^(m+3), {x,0,0}]]] sub[m] = Extract[Solve[s[m]==0, a[m,1]], 1]; SUB[m] = sub[m] For[m = 3, m < 18, F[m, z_] = ReplaceAll[(-2 z)^(-1/2) (1+Sum[P[n, Log[-z]]/(-2 z)^(n/2), {n,1,m}]), SUB[m-1]]; s[m] = Numerator[Normal[Series[(T[F[m,-1/x^2]]-F[m,-1/x^2+1]) 2^((m+1)/2)/x^(m+3),{x,0,0}]]]; t[m] = Collect[Numerator[ReplaceAll[s[m], Log[x] -> L]], L]; u[m] = Table[ Coefficient[t[m] L, L^n] == 0, {n, 1, 1 + IntegerPart[m/2]}]; tab[m] = Table[a[m, n], {n, 0, IntegerPart[m/2]}]; Print[sub[m] = Simplify[Extract[Solve[u[m], tab[m]], 1]]]; SUB[m] = Join[SUB[m - 1], sub[m]]; m++]; For[m=1, m<18, For[n=0,n<(m+1)/2, A[m, n] = TeXForm[ReplaceAll[a[m, n], sub[m]]]; Print["APQ[", m, "][", n, "]=", A[m, n], ";"] n++]; m++]; Evaluation of superfunction First, the superfunction of the Nemtsov function is constructed, that does not satisfy the requirement on its value at zero. The idea is to use the asymptotival expansion $\tilde F$ of the superfunction in the area, where it provides the good approximation, displacing the argument of superfunction into this area with using of the transfer equation. Superfunction $\mathrm{SuNe}_q$ of the Nemtsov function $\mathrm{Nem}_q$ appears as limit $\displaystyle \mathrm{SuNe}_q(z)=\lim_{n \rightarrow \infty} \mathrm{Nem}_q^{\,n} (\tilde F(z\!-\!n))$ Explicit plot of function $\mathrm{SuNe}_q$ is shown in figure at right for $q=-1, -0.5, 0, 0.5, 1, 2, 3~$. Then, function SuNem appears with the appropriate displacement of the argument: $\mathrm{SuNem}_q(z)=\mathrm{SuNe}_q(x_0(q)+z)$ where displacement $x_0=x_0(q)$ is real solution of equation $F_q(x_0)\!=\!1$. This solution is shown in figure at left. In order to show the general trend of function $x_0$, the graphic is extended a little bit into the range of negative $q$. Inverse function $\mathrm{AuNem_{q}} \big( \mathrm{Nem_{q}}(z)\big) = \mathrm{Nem_{q}}(z\!+\!1)$ Iterates of the Nemtsov function $\mathrm{Nem}^n(z)=\mathrm{SuNem}(n+\mathrm{AuNem}(z))$
Search Now showing items 1-10 of 24 Production of Σ(1385)± and Ξ(1530)0 in proton–proton collisions at √s = 7 TeV (Springer, 2015-01-10) The production of the strange and double-strange baryon resonances ((1385)±, Ξ(1530)0) has been measured at mid-rapidity (\|y\|< 0.5) in proton–proton collisions at √s = 7 TeV with the ALICE detector at the LHC. Transverse ... 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 ... Inclusive photon production at forward rapidities in proton-proton collisions at $\sqrt{s}$ = 0.9, 2.76 and 7 TeV (Springer Berlin Heidelberg, 2015-04-09) The multiplicity and pseudorapidity distributions of inclusive photons have been measured at forward rapidities ($2.3 < \eta < 3.9$) in proton-proton collisions at three center-of-mass energies, $\sqrt{s}=0.9$, 2.76 and 7 ... 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 ... Measurement of pion, kaon and proton production in proton–proton collisions at √s = 7 TeV (Springer, 2015-05-27) The measurement of primary π±, K±, p and p¯¯¯ production at mid-rapidity (\|y\|< 0.5) in proton–proton collisions at s√ = 7 TeV performed with a large ion collider experiment at the large hadron collider (LHC) is reported. ... Two-pion femtoscopy in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV (American Physical Society, 2015-03) We report the results of the femtoscopic analysis of pairs of identical pions measured in p-Pb collisions at $\sqrt{s_{\mathrm{NN}}}=5.02$ TeV. Femtoscopic radii are determined as a function of event multiplicity and pair ... 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 ... Charged jet cross sections and properties in proton-proton collisions at $\sqrt{s}=7$ TeV (American Physical Society, 2015-06) The differential charged jet cross sections, jet fragmentation distributions, and jet shapes are measured in minimum bias proton-proton collisions at centre-of-mass energy $\sqrt{s}=7$ TeV using the ALICE detector at the ... Centrality dependence of high-$p_{\rm T}$ D meson suppression in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV (Springer, 2015-11) The nuclear modification factor, $R_{\rm AA}$, of the prompt charmed mesons ${\rm D^0}$, ${\rm D^+}$ and ${\rm D^{+}}$, and their antiparticles, was measured with the ALICE detector in Pb-Pb collisions at a centre-of-mass ... K(892)$^0$ and $\Phi$(1020) production in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (American Physical Society, 2015-02) The yields of the K*(892)$^0$ and $\Phi$(1020) resonances are measured in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV through their hadronic decays using the ALICE detector. The measurements are performed in multiple ...
The Full Width Half Maximum (as defined by John Rennie's answer ) can have the following physical meanings: For a Lorentzian lineshape spectrum, it is proportional to the square magnitude of the strength of the coupling between electromagnetic field and the atomic transition begetting the spectrum in the first place; For a Lorentzian lineshape spectrum, it is the reciprocal of the $1/e$ lifetime $\tau$ of the atomic transition, which, in the Lorentzian lineshape case, has the unique memoryless lifetime probability density function, namely $p(t) = \exp(-t/\tau)/\tau$; For spectrums dominated by Doppler broadening, say in thermalized gas molecules the FWHM is proportional to both the square root of temperature, but it is also proportional to the centre frequency. For more understanding of the points (1) and (2), see the section "The Shape of the Spectrum Without a Cavity" in my answer to "Why do lasers require mirror at the ends?" and the model of the excited atom coupled to the ground state electromagnetic field there, as well as the alternative and equivalent Wigner-Weisskopf theory - this is referenced in my answer too although by far the best exposition I know of (Weisskopf's / Wigner's own) is in German, there is no direct English language translation and the main English exposition I know of is: Section 6.3, M. O. Scully and M. S. Zubairy, "Quantum Optics" which is not the grandest piece of technical writing I know. Basically it works like this: the atom is coupled roughly equally to all modes of the electromagnetic field (at least, the coupling is roughly flat near the transition frequency), so it tries to couple equally to all. However, only those modes that are tuned to the transition can be coupled into successfully; for the others, the beating between the EM field mode and the atomic transition thwarts the radiation into that mode by destructive quantum interference. The stronger the coupling constant near the transition frequency, however, the more this detuning and destructive interference effect can be overcome and thus the broader the spectral line, which, in this theoretical model, is indeed exactly Lorentzian: $$H(\omega) = \frac{1}{(\omega - \omega_0)^2 \tau^2 + 1}$$ where $2/\tau$ is the angular frequency FWHM, $$\tau = \frac{1}{2\,\pi\,\|\kappa\|^2}$$ is the transition lifetime, $\kappa$ is the (complex) coupling co-efficient between each electromagnetic field mode and the transition and $\omega_0$ is the transition frequency. The probability amplitude to find the atom still excited a time $t$ after observing that it is excited is: $$\psi(t) = \left\{\begin{array}{ll}0;&t<0\\\frac{1}{\sqrt{\tau}} \exp\left(-i\,\omega_0\,t - \frac{t}{2\,\tau}\right); & t\geq0\end{array}\right.$$ A common reason for a nonthermalised line to deviate from Lorentzian lineshape is that it is owing to a sequence of transitions rather than a lone transition. In that case, the spontaneously radiating atom is not memoryless. See my answer to "Are there old aged particles?". For thermalized gas transitions, the spectrum will be broadened in a way related to the Maxwell-Boltzmann velocity distribution. See the Wikipedia page for "Doppler Broadening".
Definition: Define the $k$-HamiltonianCycles problem as the decision problem that asks if a given graph has at least $k$ distinct Hamiltonian cycles. Question: Is there some constant $k$ so that the $k$-HamiltonianCycles problem is $NP$-complete on the class of planar $4$ connected graphs? In particular, what about $k = 3$? If not, what about $k = 4$? Motivation: It is well known that $1$-HamiltonianCycles is $NP$-complete on the class of planar $3$-connected graphs. On the other hand, all planar $4$-connected graphs have a Hamiltonian cycle, by Tutte. Additionally, it is known that a $4$-connected planar graph $G$ is "Hamiltonian Connected." That is, given any two vertices, $x$ and $y$, there is a Hamiltonian $xy$-path in $G$. This implies that a $4$-connected planar graph always has at least $2$ Hamiltonian cycles. Given one cycle, say $C$, let $e$ be an edge not in $C$. If $\gamma$ is the Hamiltonian path between the endpoints of $e$, we can extend it to a Hamiltonian cycle containing $e$. If the $4$ connected planar graph $G = (V,E)$ has at least one vertex of degree $\geq 5$, then by $4 \|V\| > \sum_{v \in V} deg(v) = 2\|E\|$, $\|E\| > 2\|V\|$. This implies that there is an edge not in the union of any two Hamiltonian cycles of $G$, so that one is guaranteed a third Hamiltonian cycle by the same argument as in the previous paragraph. Since a planar graph has at most $3n - 6$ edges, one cannot repeat this to guarantee the existence of a fourth Hamiltonian cycle. I don't know if there is an argument guaranteeing that all $4$-connected planar graphs have at least $3$ Hamiltonian cycles, or if there is a different kind of argument guaranteeing that all $4$-connected planar graphs have at least $4$ Hamiltonian cycles. I am wondering if these problems, or similar ones, are $NP$-complete. This question is similar, but not the same: https://mathoverflow.net/questions/233476/what-is-the-complexity-of-finding-a-third-hamilton-cycle-in-cubic-graph
Inspired and intrigued by this question, I decided just for fun to throw in another integer into the factors and look what happens. So for $k\in\mathbb Z$, let us define $$K_r(n,k):=\prod_{\ell_1=1}^n\cdots\prod_{\ell_r=1}^n\left( 4\cos^2\left(\frac{\pi\ell_1}{2n+1}\right)+\cdots+4\cos^2\left(\frac{\pi\ell_r}{2n+1}\right)\color{red}{+k}\right).$$ (Notation: it would look better to write $K_n(r,k)$ instead, as $r$ and $k$ are variable while $n$ will be rather kept constant, but I have chosen to keep it compatible with the previous question. Thus the index $r$ corresponds to the "complexity" of the product.) Lucia's comment applies to this generalization as well, showing that the $K_r(n,k)$ are still integers. Now, most amazingly, all their prime factors seem to be about as small as for the $k=0$ case! They are even smaller for 'slightly' negative $k$, reaching as extremal cases e.g. $$K_3(4,-6)= K_3(4,-3)= K_4(4,-7)=K_4(4,-4)= K_5(4,-8)= K_5(4,-5)= 1,$$ a pattern which suggests $$K_r(4,-r-3)=K_r(4,-r)=1. $$ But note that even though this identity is "only" about $9$th roots of unity, it seems far from trivial. Any idea how to prove it? Likewise we have $K_4(2,-6)= K_5(2,-5)= K_5(2,-10)= K_6(2,-9)= 1$, but those four identities involving $5$th roots of unity seem to be sporadic. From time to time, there are some pure powers among the $K_r(n,k)$'s, like $$K_3(3,-4)=13^4,\quad K_5(2,-10)= K_5(2,-5)= 5^{17},\quad K_4(2,-8)=-2^{10}.$$ But no appearing pattern. Note that occasionally the product vanishes, e.g. $K_3(4,-6)=0.$ And what about recurring factors, as seen e.g. by comparing $K_5(3,-13)=2^{60}7^{91}13^{35}29^{1}$ with $K_5(3,-6)=-7^{91}13^{6}$? Like the $k=0$ case, the values $K_r(n,k)$ seem to be "essentially" $r$-th powers, i.e. if $m^r$ is the biggest $r$-th power contained in $K_r(n,k)$, then not very much remains for the "$r$-remainder" $\dfrac{K_r(n,k)}{m^r}$. Interesting because the product $K_r(n,k)$ has $r$ factors. What does that mean?... Numerical experiments suggest that if $r$ is a prime, the $r$-remainder of $K_r(n,k)$ always divides the polynomial value $$p_n(r,k):=\sum_{i=0}^n\binom{2n-i}ik^{n-i}r^i$$ (see oeis.org/A054142 for the coefficients and some combinatorial properties of what seems to be called Morgan-Voyce polynomials), with the quotient being some small power, which is often $\pm1$, e.g. systematically for $n=2$. (NB. Here it was really necessary to choose the somewhat incompatible notation $p_n(r,k)$ instead of $p_r(n,k)$ because $n$ does not vary.) Any idea why the Morgan-Voyce polynomials come in here? This behavior is very similar for the $2$-remainder of $K_4(n,k)$ (not for the $4$-remainder), so it possibly extends from primes $r$ to prime powers $r$. But for $r=6$ the patterns become messy. Note that all this doesn't say anything about the other prime divisors involved (meaning those of $m^r$, i.e. of $m$), but it might be a start in understanding what is happening here. And it may be a hint towards the conjecture that also the other prime divisors grow only polynomially with $r$. : If we keep $r,n$ constant, then $K_r(n,k)$ is by definition a polynomial in $k$ of degree $n^r$. Now: EDIT Numerical evidence suggests that this polynomial is always 'highly reducible', meaning that all its irreducible factors have degrees at most $n$. A proof of this would readily answer the original question why all the prime factors of $K_r(n,0)$ are relatively small. For example, $$K_3(3,k-5)=k^6 (k^3-7k-7)^3(k^3-7k+7)^3(k^3-21k+7)$$Note that the shifting of $k$ by $5$ makes the factors look nicest. I have no idea why the primes $5$ and $7$ are featured for $K_3(3,k)$. Likewise $$K_3(4,k-6)=(k-3)k^6f(k)^6 [ -f(-k+3) ]^3 (k^3 - 9k - 9 )^3 (k^3 - 9k + 9 )^3 \\ \cdot (k^3 - 3k^2-9k+3)^ 3(k^3 - 27k - 27)$$ where $f(k)=k^3 - 3k^2 + 3 $. Might there be a connection with the abelian group $\mathbb Z_n ^r$, with its structure giving insights into this polynomial factorization?
Three postdoctoral research positions are advertised at University College London: Research associate to work with Felix Schulze in geometric analysis. The post is partially funded by the project “Regularity and stability of curvature flows and their applicatons to geometric variational problems” of the German Research Foundation. Research Associate to work with Jason Lotay on the EPSRC-funded project “Special holonomy: geometric flows and boundary value problems”. Research Associate in Geometry of Moduli Spaces to work with Michael Singer on the EPSRC-funded project “Monopole moduli spaces and manifolds with corners”. More information at http://www.homepages.ucl.ac.uk/~ucahjde/geometry/positions.htm Dates: February 5-10, 2014 Place: Tokyo Institute of Technology Website: http://www.kotaroy.com/je-geom/index-en.html Curso de Doctorado: Doctoral Intensive Week in PDEs and Applications Fechas: 21-25 de Octubre de 2013 Primera sesión: 21 de Octubre a las 9:00 horas Lugar: Seminario Alberto Dou (Aula 209), Facultad de Ciencias Matemáticas, UCM Organiza: Instituto de Matemática Interdisciplinar en colaboración con el Proyecto Europeo FIRST http://www.mat.ucm.es/imi Curso de Doctorado: The partial differential equations of image processing (and their surprising applications to today’s life) Profesor: Jean-Michel Morel (Ecole Normale Sperieur de Cachan, Francia) Fechas: 21-25 de Octubre de 2013 Primera sesión: 21 de Octubre a las 9:00 horas Lugar: Seminario Alberto Dou (Aula 209), Facultad de Ciencias Matemáticas, UCM Organiza: Instituto de Matemática Interdisciplinar en colaboración con el Proyecto Europeo FIRST http://www.mat.ucm.es/imi Curso de Doctorado: Numerical Simulation and Optimization of Industrial Problems Profesor: Alfredo Bermúdez de Castro (Universidad de Santiago de Compostela) Fechas: 21-25 de Octubre de 2013 Primera sesión: 21 de Octubre a las 9:00 horas Lugar: Seminario Alberto Dou (Aula 209), Facultad de Ciencias Matemáticas, UCM Organiza: Instituto de Matemática Interdisciplinar en colaboración con el Proyecto Europeo FIRST http://www.mat.ucm.es/imi Curso de Doctorado: Some Methods of Nonlinear Functional Analysis in Nonlinear Partial Differential Equations Profesor: Jesús Hernández (Universidad Autónoma de Madrid) Fechas: 21-25 de Octubre de 2013 Primera sesión: 21 de Octubre a las 9:00 horas Lugar: Seminario Alberto Dou (Aula 209), Facultad de Ciencias Matemáticas, UCM Organiza: Instituto de Matemática Interdisciplinar en colaboración con el Proyecto Europeo FIRST http://www.mat.ucm.es/imi Curso de Doctorado: Introduction to semigroup theory applied to evolutionary equations Profesores: José M. Arrieta and Anibal Rodríguez Bernal (Universidad Complutense de Madrid) Fechas: 21-25 de Octubre de 2013 Primera sesión: 21 de Octubre a las 9:00 horas Lugar: Seminario Alberto Dou (Aula 209), Facultad de Ciencias Matemáticas, UCM Organiza: Instituto de Matemática Interdisciplinar en colaboración con el Proyecto Europeo FIRST http://www.mat.ucm.es/imi Taller de Modelado basado en ecuaciones con COMSOL Multiphysics Imparte: Prof. Jesús H. Lucio García (Universidad de Burgos) Fecha: 10 de Octubre de 2013, de 10 a 14h Lugar: Seminario 306, Facultad de Ciencias Matemáticas, UCM Organiza: Instituto de Matemática Interdisciplinar en colaboración con el Grupo Momat y Addlink http://www.mat.ucm.es/imi We have received information about an open position in Analysis/Geometric Analysis at UCSC. See details here. Dates: October 10-11, 2013 Place: IEMath-GR The topic of the thesis should fit the objectives of the research project: Geometry of Symmetric Spaces (MTM2012-34834), funded by the Ministerio de Economía y Competitividad, Spain (see a description below). Principal investigator: Joan Porti, Universitat Autònoma de Barcelona Contact e-mail: porti@mat.uab.cat How to apply and further information on http://www.idi.mineco.gob.es/portal/site/MICINN/menuitem.dbc68b34d11ccbd5d52ffeb801432ea0/?vgnextoid=68eb71d255dcf310VgnVCM1000001d04140aRCRD Deadline for applications: September 10, 2013, 15:00 (Barcelona time). Tentative starting date: January 2014. Applicants must hold a Bachelor degree in Mathematics and preferably also a Master’s degree in Mathematics. Salary: work contract of 16,422 € each year (gross) for four years. Full time dedication. Project Summary We will study different aspects of the geometry of symmetric spaces: geometric structures, actions of thin groups and integral geometry. More precisely the questions are: We rise the study of the space of representations of groups in the isometries of a symmetric spaces, either locally (deformations) and globally (moduli spaces). We want to study geometric structures associated to those representations, with special interest in projective structures. We aim to understand the action of thin groups in the ideal boundary of a symmetric space of higher rank. The goal is to describe the dynamics in terms of the Tits building structure, in particular we seek for discontinuity domains at infinity, if possible cocompact. We want to develop integral geometry in symmetric spaces, starting with rank one, taking the approach of valuations. We aim to deepen the results for complex projective and hyperbolic spaces and to extend them to quaternionic spaces (projective, hyperbolic and affine). The research group M2S2M (Mathematical Modeling and Simulation of Environmental Systems) offers two 4-year term pre-doctoral contract to prepare a Ph D Thesis. The group M2S2M performs a research activity since 1982 in Computational Fluid Dynamics, dealing with numerical modeling of environmental flows (shallow water and oceanic flows), turbulence modeling, numerical analysis, scientific computing and domain decomposition techniques. It has developed over 30 research projects with private and national and international public funding, published over 200 research papers and advised 10 Ph D Thesis. The group M2S2M is seeking young, highly motivated researchers with good formation in applied mathematics (mathematicians, physicists, engineers, architects), who meet the following requirements: Very good ability for team work and learning. A master degree in mathematics (engineering, architecture or physics might also be considered). High level of proficiency in English or Spanish. Special interest in the fields of numerical analysis and scientific computing. The two pre-doctoral contracts are funded by the Spanish Government, within two research projects: “Development of reduced models of aero-thermal flows in buildings”. The objective of the project is to build numerical models for the computation of the energetic efficiency of buildings, based upon the accurate parameterization of the thermal exchange between the air and the building. It is developed by a research team composed by applied mathematicians and thermal engineers. It is a highly challenging subject, as usual procedures for the computation of the heat exchange are un-useful due to the huge amount of degrees of freedom required.The long-term application of this research is the elaboration of computed-aided programs for the eco-efficient design of buildings, much as today structural computer-aided program allow to compute the structural charges in the process of design of building by architects. “Development, analysis, and efficient implementation of high order numerical methods for simplified fluid models with data uncertainty”. The ultimate aim of this project is the development of efficient geophysical flow simulators that are expected to become useful tools to model river and channel flows, marine currents, sedimentation/erosion processes, turbidity currents, etc. These tools play a fundamental role in operational forecasting and risk management in natural hazards, as floods, avalanches, tsunami waves, etc. To achieve this goal, simplified depth-integrated flow models will be considered whose mathematical form are nonlinear hyperbolic PDE systems that may include source terms and/or nonconservative products. Interested candidates are requested to send their CV to Prof. Tomás Chacón, chacon@us.esand to Prof. Enrique Fernández-Nieto, edofer@us.es, specifying their interest for contract 1 or 2. The deadline for the call is next September 10, 2013. Detailed information (in Spanish) including the formal submission procedure is available at:http://www.idi.mineco.gob.es/portal/site/MICINN/menuitem.dbc68b34d11ccbd5d52ffeb801432ea0/?vgnextoid=68eb71d255dcf310VgnVCM1000001d04140aRCRD Dates: September 4-7, 2013 Place: School of Mathematical Sciences (University of Valencia, Spain) Website: http://www.uv.es/~gruv/GAS-UV13/GAS-UV13.html Dates: September 2-5, 2013 Place: Evora, Portugal Website: http://www.ifwgp2013.uevora.pt/ We have recently moved our former website to a more functional, WordPress based structure that allows to share more information with users (activities calendar, news, MathJax compatibility). We hope you find it helpful. Suggestions and comments are welcome at our email address, iemath(at)ugr.es Doc Course: Some recent results in Real Analytic Geometry Prof. Francesca Acquistapace and Prof. Fabrizio Broglia Dates: Sept 10-13, 2013. First session: Sept10, at 10:30 AM. Place: Seminar 238, Facultad de Ciencias Matemáticas, UCM. Organized by: Instituto de Matemática Interdisciplinar, in collaboration with the Departament of Algebra and the research project Geometría Real (MTM2011-22435) To be held at IEMath-GR next October 3-4, 2013. Local organizers: Antonio Alarcón at alarcon[at]ugr.es José Miguel Manzano at jmmanzano[at]ugr.es More info at http://www.ugr.es/~jmmanzano/reag2013/ From now on, all pages in IEMath-GR website admit LaTeX commands in order to produce math formulas as the one below: \[ \log(x) = \int_1^x \frac{1}{u}\mathrm{d}u \]
Has something gone haywire? Let us know about it! A for awesome Posts: 1901 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 Contact: I have 64-bit Golly 2.8 on Mac, and it crashes when I try to set the rule to Code: Select all B4c4e4c4c4c3e3c5c5e6c6e2e2c2i2n2n6n6i6i3i5a3y1e1c5k7c7e4t4q4t6k4w4i88888000005i5y3a1c1e3k7e7c6e6c2c2e2n2i2i4r6i6n6n3c3e5e5c4e4c4e4e4e/S4c4e4c4c4c3e3c5c5e6c6e2e2c2i2n2n6n6i6i3i5a3y1e1c5k7c7e4t4q4t6k4w4i88888000005i5y3a1c1e3k7e7c6e6c2c2e2n2i2i4r6i6n6n3c3e5e5c4e4c4e4e4e . This came about via a script, which could be optimized, but it also happens when I try to do it through the interface, and I'm not sure what the issue is. I could probably try something to fix the rulestring, but I still want to know what's going on. x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$http://conwaylife.com/wiki/A_for_all Aidan F. Pierce Mr. Missed Her Posts: 90 Joined: December 7th, 2016, 12:27 pm Location: Somewhere within [time in years since this was entered] light-years of you. Not unique to Mac. Crashes when I try it on Windows. There is life on Mars. We put it there with not-completely-sterilized rovers. And, for that matter, the Moon, Jupiter, Titan, and 67P/Churyumov–Gerasimenko. Andrew Moderator Posts: 763 Joined: June 2nd, 2009, 2:08 am Location: Melbourne, Australia Contact: Golly 2.9b1 doesn't crash -- just reports the rule as unknown (which it is -- you definitely need to fix the script that is creating it). Download links for 2.9b1 are here: viewtopic.php?f=7&t=2554 rowett Moderator Posts: 1730 Joined: January 31st, 2013, 2:34 am Location: UK Contact: The rule is valid it's just too long. Golly has a maximum rule length of 200 characters and the rule as specified is 267 characters long. It would certainly be worth taking Andrew's advice and optimizing the script that created it to remove all of the duplication. For reference the canonical form of the rule is only 65 characters long: B012-ak3-jnqr4ceiqrtw5-jnqr6-a78/S012-ak3-jnqr4ceiqrtw5-jnqr6-a78 Andrew Moderator Posts: 763 Joined: June 2nd, 2009, 2:08 am Location: Melbourne, Australia Contact: rowett wrote:The rule is valid it's just too long. Ah, I see. Chris, feel free to increase that limit to something like 500 if you think 200 is too small. There's no particular reason why we chose 200. rowett Moderator Posts: 1730 Joined: January 31st, 2013, 2:34 am Location: UK Contact:
The unitary discrete Fourier transform (DFT) of a sequence of numbers $x_n$ to $X_k,$ with integer $0 \le n < N$ and $0 \le k < N,$ can be defined as: $$X_k = \frac{1}{\sqrt{N}} \sum_{n=0}^{N-1} x_n e^{-2\pi ikn/N}\tag{1}$$ and the inverse discrete Fourier transform (IDFT) as: $$x_n = \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} X_k e^{2\pi ikn/N}\tag{2}$$ If $x_n$ is modulated (multiplied) by a unit-magnitude zero-phase complex sinusoid $e^{-2\pi ibn/N}$ before DFT, and the IDFT output is demodulated (divided) by the same, then we get another transform pair from the family of generalized discrete Fourier transforms, parameterized by the constant $b:$ $$X_k(b) = \frac{1}{\sqrt{N}} \sum_{n=0}^{N-1} x_ne^{-2\pi ikn/N}e^{-2\pi ibn/N} = \frac{1}{\sqrt{N}} \sum_{n=0}^{N-1} x_ne^{-2\pi i(k+b)n/N}\tag{3}$$ $$x_n = e^{2\pi ibn/N}\frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} X_k(b)\cdot e^{2\pi ikn/N} = \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} X_k(b)\cdot e^{2\pi i(k+b)n/N}\tag{4}$$ Whereas DFT samples frequencies $2\pi k/N,$ the frequency-shifted transform samples frequencies $2\pi (k+b)/N.$ This can be visualized on the Z-plane: Figure 1. Z-plane representation, showing the unit circle, of the frequencies sampled by A) DFT and B) the frequency-shifted DFT, with $N = 12$ and $b = 1/3.$ The shift $2\pi b/N$ appears as the angle between the real axis and the vector from origin to one of the sampled frequencies. Question: What are the time-domain symmetry and periodicity properties of such frequency-shifted Fourier transforms when extending $n$ beyond $0\le n<N$ in the inverse transform, and how does this depend on the parameter $b?$ For DFT (or with $b = 0$) the time-domain extension has period $N$ with no time-domain symmetry imposed by the transform.
LaTeX supports many worldwide languages by means of some special packages. In this article is explained how to import and use those packages to create documents in German. Contents German language has some special characters. For this reason the preamble of your file must be modified accordingly to support these characters and some other features. \documentclass{article} %encoding %-------------------------------------- \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} %-------------------------------------- %German-specific commands %-------------------------------------- \usepackage[ngerman]{babel} %-------------------------------------- %Hyphenation rules %-------------------------------------- \usepackage{hyphenat} \hyphenation{Mathe-matik wieder-gewinnen} %-------------------------------------- \begin{document} \tableofcontents \vspace{2cm} %Add a 2cm space \begin{abstract} Dies ist eine kurze Zusammenfassung der Inhalte des in deutscher Sprache verfassten Dokuments. \end{abstract} \section{Einleitendes Kapitel} Dies ist der erste Abschnitt. Hier können wir einige zusätzliche Elemente hinzufügen und alles wird korrekt geschrieben und umgebrochen werden. Falls ein Wort für eine Zeile zu lang ist, wird \texttt{babel} versuchen je nach Sprache richtig zu trennen. \section{Eingabe mit mathematischer Notation} In diesem Abschnitt ist zu sehen, was mit Macros, die definiert worden, geschieht. \[ \lim x = \theta + 152383.52 \] \end{document} There are three packages in this document related to the encoding and the special characters. These packages will be explained in the next sections. If your are looking for instructions on how to use more than one language in a single document, for instance English and German, see the International language support article. Modern computer systems allow you to input letters of national alphabets directly from the keyboard. In order to handle a variety of input encodings used for different groups of languages and/or on different computer platforms LaTeX employs the inputenc package to set up input encoding. In this case the package properly displays characters in the German alphabet. To use this package add the next line to the preamble of your document: \usepackage[utf8]{inputenc} The recommended input encoding is utf-8. You can use other encodings depending on your operating system. To proper LaTeX document generation you must also choose a font encoding which has to support specific characters for German language, this is accomplished by the package: fontenc \usepackage[T1]{fontenc} Even though the default encoding works well in German, using this specific encoding will avoid glitches that occur if you copy the text from the generated PDF with some specific characters. The default LaTeX encoding is OT1. To extended the default LaTeX capabilities, for proper hyphenation and translating the names of the document elements, import the babel package for the German language. \usepackage[ngerman]{babel} As you may see in the example at the introduction, instead of "abstract" and "Contents" the German words "Zusammenfassung" and "Inhaltsverzeichnis" are used. The new ortographic rules approved in 1998 are supported by babel using ngerman instead of the german parameter, which supports the old ortography. Sometimes for formatting reasons some words have to be broken up in syllables separated by a - ( hyphen) to continue the word in a new line. For example, Mathematik could become Mathe-matik. The package babel, whose usage was described in the previous section, usually does a good job breaking up the words correctly, but if this is not the case you can use a couple of commands in your preamble. \usepackage{hyphenat} \hyphenation{Mathe-matik wieder-gewinnen} The first command will import the package hyphenat and the second line is a list of space-separated words with defined hyphenation rules. On the other side, if you want a word not to be broken automatically, use the {\nobreak word} command within your document. Commands enabled for the German language Command Description "a to produce the character ä, can be used with upper-case and lower-case vowels. "s and "z to produce the German character ß. Works on upper-case and lower-case. "ck for ck to be hyphenated as k-k. "ff for ff to be hyphenated as ff-f, this is also implemented for l, m, n, p, r and t. "\| disable ligature at this point. "- An explicit hyphen sign that allows hyphenation in the rest of the word. "` Left double quotes, „ "' Right double quotes, “ "< French left double quotes « "> French right double quotes » For more information see
Suppose $\gcd(n, \phi(n)) > 1$. If $n$ is not squarefree, there exists a prime $p$ such that $p^2$ divides $n$. Then $H = \mathbb{Z}_p \times \mathbb{Z}_p$ is not cyclic and neither is $H \times \mathbb{Z}_{\frac{n}{p^2}}$. If $n$ is squarefree, there exist prime divisors $p$ and $q$ of $n$ such that $q$ divides $p-1$. Then there exists a non-abelian group $H$ of order $pq$ and $H \times \mathbb{Z}_{\frac{n}{pq}}$ is not cyclic. The other direction is not so easy to answer, but I'll give you a few good references. Jungnickel and Gallian give elementary proofs (or at least rough outlines for a proof) in these two papers: Jungnickel, Dieter. On the Uniqueness of the Cyclic Group of Order $n$. Amer. Math. Monthly, Vol. 99, No. 6 (1992) JSTOR Gallian, J. A. Moulton, David. When is $\mathbb{Z}_n$ the only group of order $n$?, Elemente der Mathematik, Vol. 48 (1993) Link to article Pakianathan and Shankar have a paper that goes beyond cyclic numbers and gives a number theoretic characterization for abelian, nilpotent and solvable numbers too: Pakianathan, Jonathan. Shankar, Krishnan. Nilpotent Numbers. Amer. Math. Monthly, Vol. 107, No. 7 (2000) JSTOR
The space of ends of a finitely generated group is always homeomorphic to 0, 1, 2 points, or a Cantor set, and in which of these 4 cases it falls is governed by Stallings' characterization (wikipedia link) in terms of amalgam/HNN splittings over finite subgroups. Up to homeomorphism, this provides a complete picture. However, the space carries a canonical Hölder structure, described below. My question is: Does there exist an infinitely-ended finitely generated group whose space of ends is not Hölder-equivalent to the standard triadic Cantor set? Given a connected locally finite graph and ends $\eta,\omega$, a base-point $x$ and $B_x(n)$, define $v_x(\eta,\omega)$ as the largest $n$ such that there a connected component of $X\smallsetminus B_x(n)$ accumulating on both $\eta$ and $\omega$ (this equals $\infty$ iff $\eta=\omega$). This yields an ultrametric $d_x=2^{-v_x}$ on the space of ends. (If you're familiar to the Gromov boundary of Gromov-hyperbolic spaces, you may somewhat recognize this as an easier analogue.) The bilipschitz type of the boundary does not depend on $x$; more precisely the identity map $(X,d_x)\to (X,d_y)$ is bilipschitz for all $x,y\in X$. Furthermore, it is easy to see that quasi-isometries between such graphs induce Hölder homeomorphisms between space of ends. In particular, for a finitely generated group, this Hölder structure on the space of ends is well-defined. For the more obvious examples I can imagine (say, virtually free groups), this yields the standard Hölder structure on the Cantor space. But I do not know in general, even assuming that the group is accessible. For graphs whose space of ends is homeomorphic to a Cantor space, it is an instructive exercise to produce nonstandard Hölder structures on the space of ends (take a triadic regular tree and assign a huge length to some edges).
Can someone come up with a proof for this little theorem? Suppose that $F_a(s)$ is a Dirichlet series and $a(n)$ is its associated arithmetic function, that is: $$F_a(s)=\sum_{n=1}^{\infty}\frac{a(n)}{n^s}$$ Then the $a(n)$ are given by: \begin{equation} \label{eq:a(n)} \nonumber a(n)=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}F_a(2j)}{(2i+1-2j)!} \end{equation} The advantage of this formula is that if you know $F_a(s)$ at the even integers, you know the coefficients of its series expansion. Let me give an example, before questions rain down asking for clarification. We know that: $$\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}$$ where $\mu(n)$ is the Mobius function. Therefore: $$\mu(n)=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}\zeta(2j)^{-1}}{(2i+1-2j)!}$$ Now, one good application of this formula is the generalization of the Mobius function. Whatever the $k$, real or complex, positive or negative, we have by definition that: $$\zeta(s)^{-k}=\sum_{n=1}^{\infty}\frac{\mu_k(n)}{n^s}$$ and you may ask, what is the $\mu_k(n)$ for each $n$ that satisfies the above equation? Well, they are given by: $$\mu_k(n)=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}\zeta(2j)^{-k}}{(2i+1-2j)!}$$ An example: $\sqrt{\zeta(s)}=1+\frac{1}{2\cdot 2^s}+\frac{1}{2\cdot 3^s}+\frac{3}{8\cdot 4^s}+\frac{1}{2\cdot 5^s}+\frac{1}{4\cdot 6^s}+\frac{1}{2\cdot 7^s} +\frac{5}{16\cdot 8^s} +\frac{3}{8\cdot 9^s}+\frac{1}{4\cdot 10^s}+\frac{1}{2\cdot 11^s}+\cdots\\$ Another example, the Von Mangoldt function divided by the log is given by: $$\frac{\Lambda(n)}{\log n}=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}\log\zeta(2j)}{(2i+1-2j)!}\\$$
I know several papers that treat this, but it seems that most of these papers do things very differently with quite different conclusions, so I am confused. Basically, when one tries to do classical field theory (as in, the branch of physics) in a mathematically precise manner, one considers a field $\psi$ to be a section of some fiber bundle $\pi_E:E\rightarrow M$ over the spacetime $M$, and writes up a lagrangian $\hat L\in \Omega^n_H(J^1(E))$, which is a horizontal $n$-form on $J^1(E)$ (sometimes $J^2(E)$) such that the action is $$ S[\psi]=\int_M (j^1\psi)^\ast\hat L. $$ If one wishes to involve gravity, one seeks to consider field theories that are diffeomorphism invariant. If $\pi_E:E\rightarrow M$ is a fiber bundle whose sections are some kind of matter fields, then to have a well-defined concept of diffeomorphism-invariance, for any diffeomorphism $\phi:M\rightarrow M$ one must be able to lift this diffeomorphism into a fiber bundle automorphism $\phi_E\in\text{Aut}(E)$ in a consistent manner. For the tensor bundles (or indeed any natural bundle), there is a functorial lift given by the tangent map (at least in the case of tensor bundles). For example, if $X\in\Gamma(TM)$ is a smooth vector field, and $\phi:M\rightarrow M$ is a diffeo, then $$ \phi_E=T\phi:TM\rightarrow TM $$ is this vector bundle automorphism. On the other hand, in physics, very important fields are spinor fields, sections of the spinor bundle $\pi_S:S\rightarrow M$. This, however is not a natural bundle (to my knowledge), and I do not know if there is any "canonical" way to lift diffeomorphisms into $\text{Aut}(S)$. Since general relativity heavily involves diffeomorphism-freedom, it is extremely important to be able to do that. In particular, I have no idea how to define the stress-energy tensor of a spinor field without representing diffeomorphisms on $S$ somehow. This situation is further confusing me, since so far I have been an ardent defender of the viewpoint that in the usual local tensor calculus-based formalism, there is no essential difference between "active diffeos" (point transformations) and "passive diffeos" (coordinate transformations). However the behaviour of a "traditional" spinor field under a coordinate transformation is simple and clear, a spinor field transforms as a scalar under coordinate transformations, and "as a spinor" under changes of orthonormal frames. However in the modern, invariant viewpoint, one cannot afford this approach. For example, if at $x\in M$, one is given a spinor $\psi\in S_x$ and a vector $v\in T_xM$, and one considers the tensor product $\psi\otimes v\in S_x\otimes T_xM$, then a diffeo will move $v$ to $T\phi(v)\in T_{\phi(x)}M$, but it will not move $\psi$ at all, so this tensor product under a diffeo would become $\psi\otimes T\phi(v)\in S_x\otimes T_{\phi(x)}M$, a product of fibers taken over different base points - clearly undesirable. Is there any agreed-upon method of dealing with the action of spinors under diffeos? If so, is there a simple way of writing it down/stating it?
Range searching Published Book Section © 2018 by Taylor & Francis Group, LLC. A central problem in computational geometry, range searching arises in many applications, and a variety of geometric problems can be formulated as range-searching problems. A typical range-searching problem has the following form. Let S be a set of n points in R d $ { \mathbb R } $, and let R $ { \mathsf R } $ be a family of subsets of R d $ { \mathbb R } $ ; elements of R $ { \mathsf R } $ are called ranges. Typical examples of ranges include rectangles, halfspaces, simplices, and balls. Preprocess S into a data structure so that for a query range γ ε R $ \gamma \in { \mathsf R } $, the points in S ∩ γ $ S \cap \gamma $ can be reported or counted efficiently. A single query can be answered in linear time using linear space, by simply checking for each point of S whether it lies in the query range. Most of the applications, however, call for querying the same set S numerous times, in which case it is desirable to answer a query faster by preprocessing S into a data structure. Full Text Duke Authors Cited Authors Published Date Book Title Handbook of Discrete and Computational Geometry, Third Edition Start / End Page International Standard Book Number 13 (ISBN-13) Digital Object Identifier (DOI) Citation Source
In this post I’ll be talking about one of my all-time favorite computer games, Factorio! Specifically: how well it lends itself to automation, and how a lot of benefits can come from using external tools to find the most efficient solutions to many of the challenges the game presents. Factorio is a game all about automation. As a programmer who loves scripting and making things as efficient as possible, I love this aspect of it. Yet like any piece of software I might write, Factorio is complex enough that for most aspects of it, the “best” solution is not obvious, nor is it obvious that there even exists a “best” solution. Like many other games, Factorio is built around progression. You start out with very little: simple hand tools, basic materials available to you in the environment, and only a few simple crafting recipes. Yet, even the smallest amount of manual labor unlocks the potential to automate an otherwise tedious task. In other games, progression usually takes the form of increasing the power or capabilities of the player, but in Factorio, what the player unlocks is the ability to do more and more automation. The ultimate goal of Factorio is to build a factory that runs entirely on its own. (Technically, you “win” the game by building and launching a rocket ship, but more advanced players got bored with that 100 gameplay-hours ago.) To be more specific, here are the basic steps of your first Factorio game: Realize your goal. For new players, the game tells you this: build and launch a rocket ship. Realize that in order to do this, you’re going to need to gather a lot of resources and research many different technologies. But, you’ve got to start somewhere. How about some iron plates? Find some iron ore on the ground and start mining it with your pickaxe. Realize it’s boring and think: I want to automate this! Build some automated mining tools, like a mining drill. Realize it requires coal to run, so now you have to mine some coal as well. Build some storage for all these raw materials you now have coming in from your mining drills. Realize you need some way of centralizing your storage since the ore patches are a bit of walk from each other. Build some transport belts to ferry items halfway across the map into your storage area. Now we’re talking! Your hands are tired from crafting all these belts. Make some assembling machines to do the crafting for you! Wait a second… these assembling machines require electricity! Time to build a water pump, boilers, steam engines, and power poles to make a power grid… Time to do some research! This requires many different materials with complicated crafting recipes that require different amounts of ingredients. Oh no! All these machines are producing pollution and attracting hostile aliens! Time to build some defenses. And so on… The point here is that on the way to your ultimate goal, Factorio presents you with various problems, and usually the solution to these problems introduces two more problems. The game is a wonderful balancing act of managing those problems as they arise, but still maintaining sight of your goal. The main challenge that Factorio gives the player, and the one I really want to focus this post on, is similar to a concept in chemistry known as the limiting reagent. In a chemical reaction, reactants (reagents) are consumed to form products. But only exact amounts of each reactant can be consumed to cause each reaction. For example, take a look at this chemical reaction showing the combustion of benzene with oxygen to form carbon dioxide and water: \[2\mathrm{C}_6\mathrm{H}_6 + 15\mathrm{O}_2 \rightarrow 12\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O} \] The reaction can only occur if exactly 2 moles of benzene are present and exactly 15 moles of oxygen are present. If you have 2 moles of benzene and 2 moles of oxygen, nothing will happen. If you have 15 moles of benzene and 15 moles of oxygen, the reaction will occur, but you’ll have 13 moles of benzene left over. Recipes in Factorio are the same way. They require exact amounts of ingredients to make, and the key part of the problem is that the ratio of those ingredients is not always 1, meaning you can have differing amounts of required ingredients. Here’s what a recipe in Factorio might look like: This recipe requires 3 electronic circuits, 5 iron gear wheels, and 9 iron plates. The implication of this is that if you have equal amounts of supply of each of those ingredients, you’re going to run out of iron plates very fast. And it gets even more complicated: each of the recipes for these ingredients also has requirements. In fact, the recipes for electronic circuits and iron gear wheels both require iron plates. So how many iron plates does it really take to make an assembling machine? Going further, how many of each basic material does it to take to make an assembling machine? And in what ratio do I need to have these materials in supply in order to not run out of any of them? Before I go on, let me just say that Factorio presents many other kinds of interesting challenges besides these, like belt balancing, train scheduling/routeing and factory modularity. Maybe I’ll leave these topics for future posts. The in-game interface, as well as the official Factorio wiki have somewhat limited capability to answer these questions. They tell you the ingredients and ratio of those ingredients required for any recipe, but it’s hard for the player to determine anything further from that without doing a lot of multiplying. So I decided to save myself and other players the headache and write a tool that will do all this for me. As is the Factorio way, I automated the process. The first challenge was to make sure I had a complete description of all of the in-game recipes. I wasn’t about to enter them all by hand, so I decided to write a scraper that would parse use the game’s (relatively) opaque data files included in the installation. While the game engine itself is written in C++ and distributed in a binary, all of the game data is distributed as Lua scripts in order to allow easy modding access. I’m not too big on Lua, so I decided to use my language of choice, Python, and find a simple Lua interpreter that could parse the game data for me. Note: Lupa requires a bit more setup than simply installing the Python package. You’ll also need to install a Lua runtime. The docs have all the instructions neccessary. It took a little snooping, but after perusing the data files in the game install folder, I found the Lua scripts simply extend a large table called data with all of the item “prototypes” and recipes. Getting the data out from there was simply a matter of executing those scripts, grabbing the data object, and looking through the (quite large) returned Python dictionary for the objects I needed. Once I had the raw data, I built a tree of Recipe objects in Python representing each recipe. For this part I had to make the simplifying assumption that recipes produce only one item, and each item is produced by only one recipe, neither of which is always the case. A Recipe simply keeps track of what items it requires and in what quantities, how many of the product it produces, and how much energy the recipe requires. Energy in Factorio recipes is simply a measure of how long the recipe takes to complete (different methods of crafting can produce recipes at different energy rates). Energy actually turns out to be another important factor in determining supply and demand. Let’s say my goal is to produce electronic circuits at one item per second. The recipe for electronic circuits takes one second and requires 3 copper cables. Copper cables take 0.5 seconds to make. So in order to make one electronic circuit per second, I need: \[\frac{3~\mathrm{copper~cable}/\mathrm{electronic~circuit} \times 0.5~\mathrm{sec}/\mathrm{copper~cable}}{1~\mathrm{sec}/\mathrm{electronic~circuit}} = 1.5~\mathrm{copper~cable~factories}\] This is a simple example with a lot of one’s in the equation, so let me just show you the full equation: \[F = \frac{I_E \times I_C}{P_E \times P_C}\] where $F$ is the number of required factories, $I_E$ is the amount of energy the ingredient’s recipe takes, $I_C$ is the count of the ingredient the recipe takes, $P_E$ is the amount of energy the product’s recipe takes, and $P_C$ is the count of the products the recipe yields. This equation, when applied to each ingredient of a given recipe and recursively down the recipe tree, allows us to answer the questions we asked before. From this we can determine exactly both the number of items as well as the number of factories producing those items required to produce an item at a given rate. Next, I needed a nice concise way of showing this data to the user. I decided to use my favorite GUI library, PyQt5, and build a custom widget to display these recipe trees. The recipe tree widget builds a recursive layout that displays the product at the top and lays out the recipe trees of each ingredient in a horizontal layout below it. I then customized the widget painter to draw lines over the widget from the product icon to each ingredient icon, to further emphasize the tree relationships. Here’s what I did for displaying the actual data: Next to each item’s icon, I display the number of that item requried to produce the item above it. So if the item above it says ×3 and the recipe calls for 2 of that item, it will show a ×6. Under each item is the number of that reciperequired to produce that many items. So if the recipe produces 2 of the item and the item has a ×5 next to it, this number will show ×2.5. The icon next to this number indicates the simplest tool required to craft the recipe. Next to the recipe count is the amount of energyrequired to execute that many recipes. So if the recipe count is ×4 and each recipe takes 10 energy, this will show 40🗲. Here’s what the recipe tree looks like for electronic circuits: This one diagram tells us a lot. Here are a few facts I’ll pick out: Making 1 electronic circuit requires 3 copper cables and 1 iron plate. Making 1 electronic circuit eventually requires 1.5 copper ore and 1 iron ore. I need 1.5 factories producing copper cable and 1 factory producing iron plate for every factory producing electronic circuits. In terms of time required, the iron plates are probably going to be the limiting factor, since the iron plate requirement takes 3.5 energy and the copper cable requirement takes only 1.5 energy. This is harder to be sure about since actual time required depends on the assembling machine tier being used. Let’s see what it looks like for something a bit more complicated, like a construction robot: And here’s what it looks like for something ridiculous like power armor MK2: These diagrams are extremely useful to Factorio players because they can tell you: Exactly how many factories you need producing ingredients to support a specific recipe What ratio of ingredients you need to keep in stock to support a recipe How long certain recipes will take How many raw materials certain recipes consume The last two points require only summing values in the diagram. I added this as a sidebar in my application that shows the totals: Currently my code isn’t really in a shareable state quite yet, but once I clean it up a bit I’ll throw it on GitHub. This is a bit of an aside, but I think this talk touches on a lot of great points about how a good Factorio player acts a lot like a software developer:
Fermat left only one proof. The area of a Pythagorean triangle is never a square number. Fermat wrote , “If the area of a right-angled triangle were a square, there would exist two biquadrates (fourth powers) the difference of which would be a square number.” That is, $ a^4\, -\, b^4 = c^2$ He used the method of infinite descent which is a proof of non-existence or a form of proof by contradiction. It was not a new proof – it is recorded in Euclid’s Elements. It relies on the Well-Ordering Principle that a set of numbers contains a least element. One assumes that a solution exists in the natural numbers which would imply that a second, smaller solution also exists and so on. Since there cannot exist an infinity of of ever smaller solutions in the natural numbers, the original premise is incorrect and the assumption is contradicted. The descent step seems to construct a set of positive integers which does not have a least member. Therefore, it is the empty set. We consider the related Diophantine equation $x^4 + y^4 = z^2$ and whether it has any solutions. Let the solution be : $x = x_1,\,\, y = y_1,\,\, z = z_1$. If gcd $(x_1, y_1) = d \gt 1$, then we can write $x_1 = dx’,\, y_1 = dy’$. Substitute into the equation : $(dx’)^4 + (dy’)^4 = z_1^2$ Factorise : $d^4(x’^4 + y’^4) = z_1^2$ It follows that $d^2$ divides $z_1$ and that $z_1 = d^2z’$.$(z_1^2/d^4 = z’^2,\,\, z_1^2 = d^4z’^2,\, \,z_1 = d^2z’)$ Then : $x’^4 + y’^4 = z’^2$, where gcd$(x’, y’) = 1$ and $z’ \lt\, z$. We can assume that gcd $(x_1, y_1) = 1$. We have established that they have no common factors. We now write the equation as $(x_1^2)^2 + (y_1^2)^2 = z_1^2$ This is a primitive Pythagorean triple.$ \begin{align} x_1^2 &= m^2\, – \,n^2 \tag 1\\ y_1^2 &= 2mn \tag2\\ z_1 &= m^2 + n^2 \tag3 \end{align} $ where $m$ and $n$ are relatively prime positive integers of opposite parity and $m\gt n$. In this case, $m$ must be odd and $n$ must be even otherwise $ x_1^2 = m^2\,-\,n^2\equiv 0\, – 1 \equiv 3 \pmod 4$ which is impossible because a square number is congruent to either $0$ or $1$ in modulo $4$. Let : $n = 2r$ Modify the equations :$ \begin{align} x_1^2 &= m^2 \,- \,4r^2\tag4\\ y_1^2 &= 4mr \tag5\\ z_1 &= m^2 + 4r^2\tag6 \end{align} $ From ($4$) : $x_1^2 + 4r^2 = m^2$ so $(x_1, 2r, m)$ is a primitive Pythagorean triple. From $(5)$ : $(y_1/2)^2 = mr$ where $m$ and $r$ are relatively prime; it follows that $m$ and $r$ are both squares. Let : $m = s^2$ and $r = t^2$ We descend.$ \begin{align} x_1 &= u^2\,-\,v^2\tag7\\ 2r &= 2uv\tag8\\ m&= u^2 + v^2\tag9 \end{align} $ where $u$ and $v$ are relatively prime positive integers with opposite parity and $u\gt v$ But : $n = 2r = 2t^2$ So : $uv = t^2$ Hence, $u$ and $v$ are both squares. Let : $u = x_2^2,\,v = y_2^2$ Substitute into $(9)$ above : $s^2 = x_2^4 + y_2^4 = z_2^2 \text\,{,say} \tag{10}$ Thus, this is another solution to the original equation. It is ‘smaller’ because $0 \lt z_2 = s \le m \lt m^2 + n^2 = z_1$ We have completed the descent step. Thus, the assumption that there exists a positive solution is contradicted. The Diophantine equation $x^4 + y^4 = z^2$ has no positive solutions. Since any fourth power is necessarily a square, the method of infinite descent can be used to show that the Diophantine equation $x^4 +y^4 = z^4$ has no positive solutions. © OldTrout 2018
I have the solution to problem 2.1 from Bergersen's and Plischke's textbook which I don't quite understand. I'll post the question itself and its solution. a) Consider a harmonic oscillator with Hamiltonian $H = 1/2(p^2+q^2)$ show that any phase space trajectory $x(t)$ with energy $E$, on the average, spend equal time in all regions of the constant energy surface $\Gamma(E)$. b) Consider two linearly coupled harmonic oscillator: $$H=1/2(p_1^2+p_2^2+q_1^2+q_2^2+q_1q_2)$$Express the phase space trajectory $x=(p_1,p_2,q_1,q_2)$ in terms of the initial values of the amplitude and phase of the normal coordinates. Show that there are regions of the constant energy surface which aren't visitedfor any particular trajectory $x(t)$. Now, for the solution: a) Let $$p=r\cos \phi ; q=r\sin \phi$$ The constant energy surface $\Gamma(E)$ in the $q-p$ plane is the circle $r=const$. The equations of motion can be written $\dot{r}=0 ; \dot{\phi}=const$, i.e., the phase space point moves with constant velocity and covers the whole "surface". It thus spends equal time in all regions of $\Gamma(E)$. b) The solutions to the equations of motion can be written $$q_1=r_1\sin \phi_1 + r_2 \sin \phi_2$$ $$ q_2 = r_1\sin\phi_1 - r_2\sin\phi_2$$ where $r_1,r_2$ are constants of the motion and $$\phi_1 = \sqrt{3/2}t + \phi_1(0)$$ $$\phi_2 = \sqrt{1/2}t + \phi_2(0)$$ The projection of the constant energy surface on the $r_1,r_2$ surface is the ellipse $$3/4 r_1^2+1/4 r_2^2=const$$ However, with given initial conditions only a single point on this ellipse is visited. I don't understand the solution to part b), I assume that $p_i = r_i \cos \phi_i$, and then to derive $3/4 r_1^2+1/4 r_2^2=const$ I need to use the Hamiltonian given in b) and then equate what I get to some constant, but I get if I am not mistaken the following: $$1/2(r_1^2+2r_1^2\sin^2 \phi_1 + r_2^2) = constant$$ Which is off by far, what do I need in order to derive this ellipse equation, I don't see how did they derive it out of thin air. Any answers? Thanks.
I'm trying to find the best approximation to the function $e^x$ in the finite dimensional polynomial space $P_4$ with respect to the standard basis vectors $B=\{1,x,x^2,x^3,x^4\}$ with inner product $$(f,g)=\int_0^1 \frac{f(x)g(x)}{\left(x(1-x)\right)^{\frac{1}{2}}}dx$$ Let $w=\sum_{j=0}^4a_j\phi_j $ where $\phi_j=x^j$. Then, the best approximation of $w$ to $e^x$ satisfies the orthogonality property: $$(e^x-w,\phi_i)=0$$ for $i=0,...,4$. Hence, $$\sum_{j=0}^4 a_j (\phi_j,\phi_i) = (e^x,\phi_i)$$ for $i=0,...,4.$. This generates linear system of equations which I'm trying to solve. Unfortunately, the integrals are a bit difficult to evaluate by hand. So, I tried using a three point gaussian quadrature rule to approximate the integrals and solve the system. Unfortunately, I keep getting a nearly singular system. When I observed this, I tried using a higher order quadrature rule, but the matrix remains ill-conditioned. I looked at my code over and over again and I can't seem to find any errors in the implementation. I'm tempted to believe that these inner products simply can't be evaluated well with a gaussian quadrature rule. Is this common for Gram Matrices? Should I use a different quadrature rule? Or is it a bug in my code? I have provided my matlab code assembing this matrix system below for reference. Any help would be greatly appreciated. n=5;%Quadrature points on [-1,1]z1=-sqrt(3/5);z2=0;z3=sqrt(3/5);%Quadrature points on [0,1]x1=z1/2+1/2;x2=z2/2+1/2;x3=z3/2+1/2;A=zeros(n,n);b=zeros(n,1);%assemble gramm matrixfor i=1:n for j=1:n F1=x1^(i-1+j-1)/sqrt(x1(1-x1)); F2=x2^(i-1+j-1)/sqrt(x2(1-x2)); F3=x3^(i-1+j-1)/sqrt(x3(1-x3)); A(i,j)=(1/2)((5/9)F1+(8/9)F2+(5/9)F3); end G1=x1^(i-1)exp(x1)/sqrt(x1(1-x1)); G2=x2^(i-1)exp(x2)/sqrt(x2(1-x2)); G3=x3^(i-1)exp(x3)/sqrt(x3(1-x3)); b(i)=(1/2)((5/9)G1+(8/9)G2+(5/9)*G3);end

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