Split (1)

train · ~4.74M rows (showing the first 1.13M)

text stringlengths 256 16.4k
So I've been interested in computing homotopy classes of maps $T^2=S^1\times S^1$ to $\mathbb{R}\mathbb{P}^2$. So first, we can decompose $T^2$ into a cell complex with one zero cell, $S^1\vee S^1$ and a disc $D^2$ glued down to $S^1\vee S^1$ using the commutator. Since the commutator vanishes in $\pi_1(\mathbb{R}\mathbb{P}^2)$ given any map $S^1\vee S^1\rightarrow \mathbb{R}\mathbb{P}^2$, one can extend this to a map $T^2\rightarrow \mathbb{R}\mathbb{P}^2$ by the map $D^2\rightarrow T^2$, but since the gluing map is nullhomotopic, this is the same thing as a map $S^2\rightarrow \mathbb{R}\mathbb{P}^2$ and there are $\mathbb{Z}$ homotopy classes of maps like this. Therefore we can write (as a set) $[T^2,\mathbb{R}\mathbb{P}^2] = \{a,b,c,d\}\times \mathbb{Z}$ where each of the $\{a,b,c,d\}$ is a homotopy class $S^1\vee S^1\rightarrow \mathbb{R}\mathbb{P}^2$. My question is given an arbitrary map $f:T^2\rightarrow \mathbb{R}\mathbb{P}^2$ is there some sort of computation we can do in $\mathbb{R}\mathbb{P}^2$ to determine which homotopy class $f$ belongs to or at least which of the $a,b,c,d$ it belongs to? The set of homotopy classes $[T^2,\mathbb{RP}^2]$ actually consists of the following: The set $[S^1\vee S^1,\mathbb{RP}^2]$ consists of four elements, which I'll call $(1,1)$, $(-1,1)$, $(1,-1)$ and $(-1,-1)$. The notation refers to which element in $\pi_1(\mathbb{RP}^2)$ each $S^1$ maps to. The homotopy classes $[T^2,\mathbb{RP}^2]$ which restrict to $(1,1)$ are indeed in one-to-one correspondence with the integers if you are looking at based homotopy and the natural numbers if you are looking at free homotopy. The classes which restrict to $(-1,1)$, $(1,-1)$ and $(-1,-1)$ fall into only two distinct homotopy classes, corresponding to even/odd parity of the degree of that map $S^2\rightarrow \mathbb{RP}^2$ that you mentioned. You get the same count for both free and based homotopy here. There is however a hands-on way to see this: The following Pontryagin-Thom type construction is a "computation in $T^2$" rather than in $\mathbb{RP}^2$ as you asked for but I hope it may help shed some light on these maps. These ideas are also very related to Ryan's last paragraph in his answer. A version of this written up for a physics audience is in section 2.3 of my thesis. Recall that the Pontryagin-Thom construction give us bijections between the homotopy classes of maps from a manifold into another space and bordism classes of framed submanifolds. Most people who are familiar with this are familiar with the case of maps into spheres, but there are versions which work for maps into other spaces as well. In particular, since $\mathbb{RP}^2$ is the one-point compactification of a real line bundle (the Mobius strip), just as $S^n$ is the one-point compactification of $\mathbb{R}^n$ viewed as a bundle over a point, we get something relatively pretty here. Given $f:T^2\rightarrow\mathbb{RP}^2$, we consider the inverse image of a copy of $\mathbb{RP}^1$ in $\mathbb{RP}^2$. For instance, a geometrical interpretation of $f$ is as an assignment of a line in $\mathbb{R}^3$ to each point on the torus; then we consider the locus of points on the torus such that the line field has no $z$-component. Generically, "$f^{-1}(\mathbb{RP}^1)$" (abusing notation) is a set of curves on the torus (this is where the business of transversality comes into play). Furthermore, these curves carry maps to $\mathbb{RP}^1$ induced by $f$. I like to visualize this map by a rainbow coloring on the curves, i.e. points on the torus mapping to the same point on $\mathbb{RP}^1$ are colored the same. Here's a picture of $\mathbb{RP}^2$ with a copy of $\mathbb{RP}^1$ colored. We also need to keep track of how $f$ maps the normal bundle of these curves into the Mobius strip (I mean here the real line bundle version of the Mobius strip, which arises here from being the normal bundle of $\mathbb{RP}^1$ in $\mathbb{RP}^2$). That is, $f$ also gives a "Mobius-framing" to these colored curves. One way to do keep track of this pictorially on $T^2$ is to do the following. First choose some point on $\mathbb{RP}^1$, say the point which we have decided to color blue; the fiber of the Mobius strip bundle over this point is a copy of $\mathbb{R}$ and we choose one ray to be the "positive" ray. Now, on each colored curve on $T^2$, whenever you see the color blue, draw an arrow pointing towards the side of the curve which points in the positive direction. In this picture, I show the Mobius-framed colored curves in $T^2$ (the rectangle has opposite sides identified, of course) corresponding to the inverse image of $\mathbb{RP}^1$ from a certain map $f$. Note that I am marking the "positive side" of the normal bundle of the curves with + signs. These Mobius-framed colored curves should be considered up to bordism. In terms of the pictures I've been describing, this works out to saying that we can isotope the colored curves, homotope the colorings on the curves (i.e. the maps to $\mathbb{RP}^1$) as well as performing the two following local operations: For instance, the colored curves I showed above are related by a bordism to the following: A version of the Pontryagin-Thom construction thus yields that the bordism classes of Mobius-framed colored curves on $T^2$ are in one-to-one correspondence with the elements of $[T^2,\mathbb{RP}^2]$. My reference for this is the last chapter of tom Dieck's recent book Algebraic Topology. It's quite late here so I'm going to leave the details of the classification on $T^2$ for you to play around with (there's also more in 2.3.1 of my thesis). Apologies for cutting this off so abruptly, feel free to ask for clarification in the comments. For dimension reasons, the set of homotopy classes of maps between these spaces can be computed by using their fundamental crossed module. Despite the question has already been answered twice, let me explain this approach, just for those who may like it. A crossed module $C_$ is a group homomorphism $\partial\colon C_2\rightarrow C_1$ together with an action of $C_1$ on $C_2$, that we write exponentially $c_2^{c_1}$, such that $$\begin{array}{rl} \partial(c_2^{c_1})&=-c_1+\partial(c_2)+c_1,\\\ c_2^{\partial(c_2')}&=-c_2'+c_2+c_2'. \end{array}$$ Here we use additive notation despite the groups may be nonabelian. The canonical (topological) example of a crossed module is the boundary map in the long exact sequence of a pair of spaces: $$\partial\colon\pi_2(X,Y)\longrightarrow\pi_1(X).$$ The fundamental crossed module of CW-complex $X$ is obtained in this way for the pair formed by $X$ and its $1$-skeleton. $$\partial\colon\pi_2(X,X^1)\longrightarrow\pi_1(X^1).$$ Notice that $\ker\partial\cong\pi_2(X)$. The image of $\partial$ is normal in any crossed module, and in this case $\operatorname{coker} \partial\cong\pi_1(X)$. A morphism of crossed modules $f_{}\colon C_{}\rightarrow D_{}$ is a commutative square $$\begin{array}{rcccl} &C_1&\stackrel{\partial}\longrightarrow&C_2\\\ {\scriptstyle f_2}&\downarrow&&\downarrow&{\scriptstyle f_1}\\\ &D_1&\stackrel{\partial}\longrightarrow&D_2 \end{array}$$ such that $f_2(c_2^{c_1})=f_2(c_2)^{f_1(c_1)}$. Notice that any cellular map between CW-complexes $X\rightarrow Y$ induces a morphism between their fundamental crossed modules. Two such morphisms $f_{},g_{}\colon C_{}\rightarrow D_{}$ are homotopic if there exists a map $H\colon C_1\rightarrow D_2$ such that $$\begin{array}{rl}H(c_1+c_1')&=H(c_1)^{f(c_1')}+H(c_1'),\\\\partial H(c_1)&=-f_1(c_1)+g_1(c_1),\\\ H\partial(c_2)&=-f_2(c_2)+g_2(c_2).\end{array}$$ If $X$ and $Y$ are CW-complexes and $X$ is $2$-dimensional the set of homotopy classes $[X,Y]$ can be computed as the set of algebraic homotopy classes between their fundamental crossed modules. The fundamental crossed module of $T^2$ is isomorphic to $\partial\colon G'\hookrightarrow G=\langle a,b\rangle$. Here $G$ os a free group on two generators, $G'$ is its commutator subgroup, and $\partial$ is the inclusion. The action of $G$ on $G'$ is by conjugation. The fundamental crossed module of $\mathbb{R}P^2$ is even easier: $\partial=2\cdot\varepsilon\colon \mathbb{Z}[\mathbb{Z}/2]\rightarrow \mathbb{Z}$. Here $\varepsilon\colon \mathbb{Z}[\mathbb{Z}/2]\rightarrow \mathbb{Z}$ is the augmentation map of the group ring and $\mathbb{Z}$ acts on $\mathbb{Z}[\mathbb{Z}/2]$ via the natural projection $\mathbb{Z}\twoheadrightarrow \mathbb{Z}/2$. It is a beautiful exercise to compute the set of homotopy classes of maps between these two crossed modules. It's easy since the second one is made of abelian groups! If you're given a map $T^2\rightarrow\mathbb{R}P^2$ and you manage to find a homotopic cellular map (e.g. using the proof of the cellular approximation theorem), then you can say what homotopy class you started with by looking at the induced morphism on the level of crossed modules. Keep in mind that $\pi_2 \mathbb RP^2$ as a module over $\pi_1 \mathbb RP^2$ is non-trivial, and the generator of $\pi_1$ acts by negation on $\pi_2$. So technically your description $$[T^2, \mathbb RP^2] = \{a,b,c,d\} \times \mathbb Z$$ should be $$[T^2, \mathbb RP^2] = \{a,b,c,d\} \times \mathbb N$$ Or are you using based maps? Given such a map, a standard way to determine the homotopy-class is via transversality. You can also think of this as studying various induced maps on homology (or cohomology) plus some covering-space theory. Determining $\{a,b,c,d\}$ amounts to computing the homology class of the transverse intersection with $\mathbb RP^1 \subset \mathbb RP^2$. Some minimal cover of $T^2$ composed with your map admits a lift to the universal cover of $\mathbb RP^2$, the degree of this lift allows you to compute the natural number. Since you don't have preserved base-points the degree is only well-defined up to covering transformation, so it's not sign-determined.
I recently came across this in a textbook (NCERT class 12 , chapter: wave optics , pg:367 , example 10.4(d)) of mine while studying the Young's double slit experiment. It says a condition for the formation of interference pattern is$$\frac{s}{S} < \frac{\lambda}{d}$$Where $s$ is the size of ... The accepted answer is clearly wrong. The OP's textbook referes to 's' as "size of source" and then gives a relation involving it. But the accepted answer conveniently assumes 's' to be "fringe-width" and proves the relation. One of the unaccepted answers is the correct one. I have flagged the answer for mod attention. This answer wastes time, because I naturally looked at it first ( it being an accepted answer) only to realise it proved something entirely different and trivial. This question was considered a duplicate because of a previous question titled "Height of Water 'Splashing'". However, the previous question only considers the height of the splash, whereas answers to the later question may consider a lot of different effects on the body of water, such as height ... I was trying to figure out the cross section $\frac{d\sigma}{d\Omega}$ for spinless $e^{-}\gamma\rightarrow e^{-}$ scattering. First I wrote the terms associated with each component.Vertex:$$ie(P_A+P_B)^{\mu}$$External Boson: $1$Photon: $\epsilon_{\mu}$Multiplying these will give the inv... As I am now studying on the history of discovery of electricity so I am searching on each scientists on Google but I am not getting a good answers on some scientists.So I want to ask you to provide a good app for studying on the history of scientists? I am working on correlation in quantum systems.Consider for an arbitrary finite dimensional bipartite system $A$ with elements $A_{1}$ and $A_{2}$ and a bipartite system $B$ with elements $B_{1}$ and $B_{2}$ under the assumption which fulfilled continuity.My question is that would it be possib... @EmilioPisanty Sup. I finished Part I of Q is for Quantum. I'm a little confused why a black ball turns into a misty of white and minus black, and not into white and black? Is it like a little trick so the second PETE box can cancel out the contrary states? Also I really like that the book avoids words like quantum, superposition, etc. Is this correct? "The closer you get hovering (as opposed to falling) to a black hole, the further away you see the black hole from you. You would need an impossible rope of an infinite length to reach the event horizon from a hovering ship". From physics.stackexchange.com/questions/480767/… You can't make a system go to a lower state than its zero point, so you can't do work with ZPE. Similarly, to run a hydroelectric generator you not only need water, you need a height difference so you can make the water run downhill. — PM 2Ring3 hours ago So in Q is for Quantum there's a box called PETE that has 50% chance of changing the color of a black or white ball. When two PETE boxes are connected, an input white ball will always come out white and the same with a black ball. @ACuriousMind There is also a NOT box that changes the color of the ball. In the book it's described that each ball has a misty (possible outcomes I suppose). For example a white ball coming into a PETE box will have output misty of WB (it can come out as white or black). But the misty of a black ball is W-B or -WB. (the black ball comes out with a minus). I understand that with the minus the math works out, but what is that minus and why? @AbhasKumarSinha intriguing/ impressive! would like to hear more! :) am very interested in using physics simulation systems for fluid dynamics vs particle dynamics experiments, alas very few in the world are thinking along the same lines right now, even as the technology improves substantially... @vzn for physics/simulation, you may use Blender, that is very accurate. If you want to experiment lens and optics, the you may use Mistibushi Renderer, those are made for accurate scientific purposes. @RyanUnger physics.stackexchange.com/q/27700/50583 is about QFT for mathematicians, which overlaps in the sense that you can't really do string theory without first doing QFT. I think the canonical recommendation is indeed Deligne et al's Quantum Fields and Strings: A Course For Mathematicians , but I haven't read it myself @AbhasKumarSinha when you say you were there, did you work at some kind of Godot facilities/ headquarters? where? dont see something relevant on google yet on "mitsubishi renderer" do you have a link for that? @ACuriousMind thats exactly how DZA presents it. understand the idea of "not tying it to any particular physical implementation" but that kind of gets stretched thin because the point is that there are "devices from our reality" that match the description and theyre all part of the mystery/ complexity/ inscrutability of QM. actually its QM experts that dont fully grasp the idea because (on deep research) it seems possible classical components exist that fulfill the descriptions... When I say "the basics of string theory haven't changed", I basically mean the story of string theory up to (but excluding) compactifications, branes and what not. It is the latter that has rapidly evolved, not the former. @RyanUnger Yes, it's where the actual model building happens. But there's a lot of things to work out independently of that And that is what I mean by "the basics". Yes, with mirror symmetry and all that jazz, there's been a lot of things happening in string theory, but I think that's still comparatively "fresh" research where the best you'll find are some survey papers @RyanUnger trying to think of an adjective for it... nihilistic? :P ps have you seen this? think youll like it, thought of you when found it... Kurzgesagt optimistic nihilismyoutube.com/watch?v=MBRqu0YOH14 The knuckle mnemonic is a mnemonic device for remembering the number of days in the months of the Julian and Gregorian calendars.== Method ===== One handed ===One form of the mnemonic is done by counting on the knuckles of one's hand to remember the numbers of days of the months.Count knuckles as 31 days, depressions between knuckles as 30 (or 28/29) days. Start with the little finger knuckle as January, and count one finger or depression at a time towards the index finger knuckle (July), saying the months while doing so. Then return to the little finger knuckle (now August) and continue for... @vzn I dont want to go to uni nor college. I prefer to dive into the depths of life early. I'm 16 (2 more years and I graduate). I'm interested in business, physics, neuroscience, philosophy, biology, engineering and other stuff and technologies. I just have constant hunger to widen my view on the world. @Slereah It's like the brain has a limited capacity on math skills it can store. @NovaliumCompany btw think either way is acceptable, relate to the feeling of low enthusiasm to submitting to "the higher establishment," but for many, universities are indeed "diving into the depths of life" I think you should go if you want to learn, but I'd also argue that waiting a couple years could be a sensible option. I know a number of people who went to college because they were told that it was what they should do and ended up wasting a bunch of time/money It does give you more of a sense of who actually knows what they're talking about and who doesn't though. While there's a lot of information available these days, it isn't all good information and it can be a very difficult thing to judge without some background knowledge Hello people, does anyone have a suggestion for some good lecture notes on what surface codes are and how are they used for quantum error correction? I just want to have an overview as I might have the possibility of doing a master thesis on the subject. I looked around a bit and it sounds cool but "it sounds cool" doesn't sound like a good enough motivation for devoting 6 months of my life to it
You actually do recover the convolution, but as it is discussed in the comments, there is a normalization issue due to discretization. According to the documentation, fft is implemented like this: $$ A_k = \sum_{m=0}^{n-1} a_m \exp \{ - 2\pi i \frac{mk}{n} \} $$ with $A_k$ being the Fourier-coefficients, $a_m$ the $m$-th element of your signal vector and $n$ the length of the signal. Squaring this gives you $$ A_k^2 = \sum_{m=0}^{n-1} \sum_{m'=0}^{n-1} a_m a_{m'} \exp\{ - 2\pi i \frac{(m+m')k}{n} \} \} $$ Now, applying ifft to the squared Fourier-transform gives you, using the ifft-definition from the documentation: $$ \text{ifft}(A_k^2)_{m''} = \frac{1}{n} \sum_{k=0}^{n-1} \sum_{m=0}^{n-1} \sum_{m'=0}^{n-1}a_m \exp\{ - 2\pi i \frac{(m+m'-m'')k}{n} \} \}$$ With the observation, that $$ \frac{1}{n} \sum_{k=0}^{n-1} \exp\{ - 2\pi i \frac{(m+m'-m'')k}{n} \} = \delta_{m+m', m''} $$ you end up with $$ \text{ifft}(A_k^2)_{m''} = \sum_{m=0}^{n-1} \sum_{m'=0}^{n-1} a_m a_{m'} \delta_{m+m', m''} = \sum_{m=0}^{n-1} a_m a_{m'' - m}$$ This is actually how np.convolve is defined (except for some padding). If you use np.convolve on your data, you end up with the same result (except for some padding), so within the numpy-world, you did exactly what you set out to do, i.e. verify, the convolution property of the Fourier transform. As noted in the comments however, neither fft nor convolve "know" anything about your descretization, so you have to take care of that manually by multiplying the results with dt.
I recently came across this in a textbook (NCERT class 12 , chapter: wave optics , pg:367 , example 10.4(d)) of mine while studying the Young's double slit experiment. It says a condition for the formation of interference pattern is$$\frac{s}{S} < \frac{\lambda}{d}$$Where $s$ is the size of ... The accepted answer is clearly wrong. The OP's textbook referes to 's' as "size of source" and then gives a relation involving it. But the accepted answer conveniently assumes 's' to be "fringe-width" and proves the relation. One of the unaccepted answers is the correct one. I have flagged the answer for mod attention. This answer wastes time, because I naturally looked at it first ( it being an accepted answer) only to realise it proved something entirely different and trivial. This question was considered a duplicate because of a previous question titled "Height of Water 'Splashing'". However, the previous question only considers the height of the splash, whereas answers to the later question may consider a lot of different effects on the body of water, such as height ... I was trying to figure out the cross section $\frac{d\sigma}{d\Omega}$ for spinless $e^{-}\gamma\rightarrow e^{-}$ scattering. First I wrote the terms associated with each component.Vertex:$$ie(P_A+P_B)^{\mu}$$External Boson: $1$Photon: $\epsilon_{\mu}$Multiplying these will give the inv... As I am now studying on the history of discovery of electricity so I am searching on each scientists on Google but I am not getting a good answers on some scientists.So I want to ask you to provide a good app for studying on the history of scientists? I am working on correlation in quantum systems.Consider for an arbitrary finite dimensional bipartite system $A$ with elements $A_{1}$ and $A_{2}$ and a bipartite system $B$ with elements $B_{1}$ and $B_{2}$ under the assumption which fulfilled continuity.My question is that would it be possib... @EmilioPisanty Sup. I finished Part I of Q is for Quantum. I'm a little confused why a black ball turns into a misty of white and minus black, and not into white and black? Is it like a little trick so the second PETE box can cancel out the contrary states? Also I really like that the book avoids words like quantum, superposition, etc. Is this correct? "The closer you get hovering (as opposed to falling) to a black hole, the further away you see the black hole from you. You would need an impossible rope of an infinite length to reach the event horizon from a hovering ship". From physics.stackexchange.com/questions/480767/… You can't make a system go to a lower state than its zero point, so you can't do work with ZPE. Similarly, to run a hydroelectric generator you not only need water, you need a height difference so you can make the water run downhill. — PM 2Ring3 hours ago So in Q is for Quantum there's a box called PETE that has 50% chance of changing the color of a black or white ball. When two PETE boxes are connected, an input white ball will always come out white and the same with a black ball. @ACuriousMind There is also a NOT box that changes the color of the ball. In the book it's described that each ball has a misty (possible outcomes I suppose). For example a white ball coming into a PETE box will have output misty of WB (it can come out as white or black). But the misty of a black ball is W-B or -WB. (the black ball comes out with a minus). I understand that with the minus the math works out, but what is that minus and why? @AbhasKumarSinha intriguing/ impressive! would like to hear more! :) am very interested in using physics simulation systems for fluid dynamics vs particle dynamics experiments, alas very few in the world are thinking along the same lines right now, even as the technology improves substantially... @vzn for physics/simulation, you may use Blender, that is very accurate. If you want to experiment lens and optics, the you may use Mistibushi Renderer, those are made for accurate scientific purposes. @RyanUnger physics.stackexchange.com/q/27700/50583 is about QFT for mathematicians, which overlaps in the sense that you can't really do string theory without first doing QFT. I think the canonical recommendation is indeed Deligne et al's Quantum Fields and Strings: A Course For Mathematicians , but I haven't read it myself @AbhasKumarSinha when you say you were there, did you work at some kind of Godot facilities/ headquarters? where? dont see something relevant on google yet on "mitsubishi renderer" do you have a link for that? @ACuriousMind thats exactly how DZA presents it. understand the idea of "not tying it to any particular physical implementation" but that kind of gets stretched thin because the point is that there are "devices from our reality" that match the description and theyre all part of the mystery/ complexity/ inscrutability of QM. actually its QM experts that dont fully grasp the idea because (on deep research) it seems possible classical components exist that fulfill the descriptions... When I say "the basics of string theory haven't changed", I basically mean the story of string theory up to (but excluding) compactifications, branes and what not. It is the latter that has rapidly evolved, not the former. @RyanUnger Yes, it's where the actual model building happens. But there's a lot of things to work out independently of that And that is what I mean by "the basics". Yes, with mirror symmetry and all that jazz, there's been a lot of things happening in string theory, but I think that's still comparatively "fresh" research where the best you'll find are some survey papers @RyanUnger trying to think of an adjective for it... nihilistic? :P ps have you seen this? think youll like it, thought of you when found it... Kurzgesagt optimistic nihilismyoutube.com/watch?v=MBRqu0YOH14 The knuckle mnemonic is a mnemonic device for remembering the number of days in the months of the Julian and Gregorian calendars.== Method ===== One handed ===One form of the mnemonic is done by counting on the knuckles of one's hand to remember the numbers of days of the months.Count knuckles as 31 days, depressions between knuckles as 30 (or 28/29) days. Start with the little finger knuckle as January, and count one finger or depression at a time towards the index finger knuckle (July), saying the months while doing so. Then return to the little finger knuckle (now August) and continue for... @vzn I dont want to go to uni nor college. I prefer to dive into the depths of life early. I'm 16 (2 more years and I graduate). I'm interested in business, physics, neuroscience, philosophy, biology, engineering and other stuff and technologies. I just have constant hunger to widen my view on the world. @Slereah It's like the brain has a limited capacity on math skills it can store. @NovaliumCompany btw think either way is acceptable, relate to the feeling of low enthusiasm to submitting to "the higher establishment," but for many, universities are indeed "diving into the depths of life" I think you should go if you want to learn, but I'd also argue that waiting a couple years could be a sensible option. I know a number of people who went to college because they were told that it was what they should do and ended up wasting a bunch of time/money It does give you more of a sense of who actually knows what they're talking about and who doesn't though. While there's a lot of information available these days, it isn't all good information and it can be a very difficult thing to judge without some background knowledge Hello people, does anyone have a suggestion for some good lecture notes on what surface codes are and how are they used for quantum error correction? I just want to have an overview as I might have the possibility of doing a master thesis on the subject. I looked around a bit and it sounds cool but "it sounds cool" doesn't sound like a good enough motivation for devoting 6 months of my life to it
I have a stress matrix in cartesian coordinates : $\begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix}$. How can I convert it to spherical coordinates ? One way to conceptualize the stress matrix is to view it as a tensor. In general, your matrix $$T = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}$$ should be thought of in terms of how it relates on a displacement vector $v^T = (\mathrm{d}x, \mathrm{d} y, \mathrm{d} z)$. The stress tensor tells you that the energy change associated to this small displacement vector is $$\delta E = v^T T v = a {\mathrm{d}x}^2 + b {\mathrm{d}y}^2 + c {\mathrm{d}z}^2$$ Now, let's consider what happens if we change into spherical coordinates. Recall that in spherical coordinates $(r,\phi,\theta)$ $$ x = r \cos \phi \sin \theta \\ y = r \sin \phi \sin \theta \\ z = r \cos \theta $$ This gives the relations $$ \mathrm{d}x = \mathrm{d}r (\cos\phi \sin\theta) + \mathrm{d}\phi (- r\sin\phi \sin\theta) + \mathrm{d}\theta(r \cos\phi \cos\theta)\\ \mathrm{d}y = \mathrm{d}r (\sin\phi \sin\theta) + \mathrm{d}\phi (r \cos\phi \sin\theta) + \mathrm{d}\theta(r \sin\phi \cos\theta)\\ \mathrm{d}z = \mathrm{d}r (\cos\theta) + \mathrm{d}\theta(-r \sin\theta) $$ which can be written in matrix form as $$\begin{bmatrix} \mathrm{d}x \\ \mathrm{d}y \\ \mathrm{d}z \end{bmatrix} = J \begin{bmatrix} \mathrm{d}r \\ \mathrm{d}\phi \\ \mathrm{d}\theta \end{bmatrix}$$ Where J is the "Jacobian Matrix" (or change of coordinates) $$J = \begin{bmatrix} \cos\phi \sin\theta & -r \sin\phi \sin\theta & r \cos\phi \cos\theta \\ \sin\phi \sin\theta & r \cos\phi \sin\theta & r \sin \phi \cos\theta \\ \cos \theta & 0 & -r \sin \theta \end{bmatrix}$$ This means that for the displacement vector $\tilde{v}^T = (\mathrm{d}r, \mathrm{d}\phi,\mathrm{d}\theta)$ written in spherical coordinates, the vectors in two different coordinate systems can be related to each other by $$v = J \tilde{v}$$ We'll end up having that the energy change can be written as $$\delta E = (J \tilde{v})^T T (J \tilde{v}) = \tilde{v}^T (J^T T J) \tilde{v}$$ which is some pretty complicated expression in terms of the $\mathrm{d}r, \mathrm{d}\phi, \mathrm{d}\theta$. So since $(J^T T J)$ is the matrix that generates the same energy change for the vector in different coordinates, the stress matrix $\tilde{T}$ in spherical coordinates is really $$\tilde{T} = J^T T J$$ This is a pretty general lesson that will let you express the stress matrix in any coordinate system, not just spherical ones. And note that this transformation rule is different $T \rightarrow J^T T J$ is different than that for a linear transformation, which is $A \rightarrow J^{-1} A J$. This means that T is a tensor quantity, and not a linear transformation. The defining property of a tensor is that is defines a length, sending a vector $v$ to a number, $v \rightarrow v^T T v$. In our case is the energy change $\delta E$ defined above, and the goal is to keep that length the same irrespective of the coordinate system. On the other hand, a linear transformation is defined as sending vectors to vectors, $v \rightarrow A v$. So after I found the better link(http://solidmechanics.org/text/AppendixD/AppendixD.htm), I went ahead and worked the full thing out. Here are the results. Note the results assume the tensor is symmetric (ie. $S_{ij}=S_{ji}$). $$ \begin{align} S_{rr} &= \mathbf{c}_{\theta}^2S_{xx} + 2\mathbf{c}_{\theta}\mathbf{s}_{\theta}S_{xy} + \mathbf{s}_{\theta}^2S_{yy} \\ S_{r\theta} &= \mathbf{c}_{\theta}\mathbf{s}_{\theta}(S_{yy}-S_{xx}) + (\mathbf{c}_{\theta}^2- \mathbf{s}_{\theta}^2)S_{xy} \\ S_{rz} &= \mathbf{c}_{\theta}S_{xz} + \mathbf{s}_{\theta}S_{yz} \\ S_{\theta\theta} &= \mathbf{s}_{\theta}^2S_{xx} - 2\mathbf{c}_{\theta}\mathbf{s}_{\theta}S_{xy} + \mathbf{c}_{\theta}^2S_{yy} \\ S_{\theta z} &= -\mathbf{s}_{\theta}S_{xz} + \mathbf{c}_{\theta}S_{yz} \\ S_{zz} &= S_{zz} \end{align} $$ where $$ \begin{align} \cos \theta &= \mathbf{c}_{\theta} \\ \sin \theta &= \mathbf{s}_{\theta} \end{align} $$
I'm working with a simple local level model in a textbook \begin{align} y_t &= \alpha_t + \epsilon_t, \qquad \epsilon_t \sim N(0, \sigma_\epsilon^2) \\ \alpha_{t+1} &= \alpha_t + \eta_t, \qquad \eta_t \sim N(0, \sigma_\eta^2) \end{align} The conditional distribution of $\alpha_t$ given $Y_{t-1}$ (the set of all observations $y_j$ where $1 \leq j \leq t-1$) is $N(a_t, P_t)$, where we have $a_t = E(\alpha_t \mid Y_{t-1})$ and $P_t = Var(\alpha_t \mid Y_{t-1})$. The book includes this calculation: \begin{equation} E[\alpha_t(\alpha_t - a_t)] = E[Var(\alpha_t \mid Y_{t-1})] = P_t \end{equation} I don't understand where this first equality comes from. Working backwards from the law of conditional variance, I know that \begin{align} Var(\alpha_t \mid Y_{t-1}) &= E[ (\alpha_t - E[\alpha_t \mid Y_{t-1}])^2\mid Y_{t-1}] \\ &= E[ (\alpha_t - a_t)^2 \mid Y_{t-1}] \\ &= E[ (\alpha_t^2 - 2 \alpha_t a_t + a_t^2) \mid Y_{t-1}] \\ &= E[\alpha_t^2 \mid Y_{t-1}] - 2E[\alpha_t a_t \mid Y_{t-1}] + E[a_t^2 \mid Y_{t-1}] \\ \end{align} but I don't see how to get $\alpha_t (\alpha_t - a_t)$ from this, which would give me the correct value inside the expectation.
Difference between revisions of "SageMath" (→Install Sage package: rm section, cf. talk) m (add note to help identify optional sagemath dependencies available through pacman) Line 19: Line 19: == Installation == == Installation == − {{Warning\|Most if not all of the [http://doc.sagemath.org/html/en/installation/standard_packages.html standard sage packages] are available as Arch packages and exposed as (optional) dependencies of {{pkg\|sagemath}}, so there is no need to install them with {{ic\|sage -i}}.}} + {{Warning\|Most if not all of the [http://doc.sagemath.org/html/en/installation/standard_packages.html standard sage packages] are available as Arch packages and exposed as (optional) dependencies of {{pkg\|sagemath}} , so there is no need to install them with {{ic\|sage -i}}.}} * {{Pkg\|sagemath}} contains the command-line version; * {{Pkg\|sagemath}} contains the command-line version; Revision as of 11:13, 23 February 2016 SageMath (formerly Sage) is a program for numerical and symbolic mathematical computation that uses Python as its main language. It is meant to provide an alternative for commercial programs such as Maple, Matlab, and Mathematica. SageMath provides support for the following: Calculus: using Maxima and SymPy. Linear Algebra: using the GSL, SciPy and NumPy. Statistics: using R (through RPy) and SciPy. Graphs: using matplotlib. An interactive shellusing IPython. Access to Python modulessuch as PIL, SQLAlchemy, etc. Contents Installation Warning:Most if not all of the standard sage packages are available as Arch packages and exposed as (optional) dependencies of (you may see which by running pacman -Si sagemath), so there is no need to install them with sage -i. contains the command-line version; for HTML documentation and inline help from the command line. includes the browser-based notebook interface. The optional dependencies for various features that will be disabled if the needed packages are missing.package has number of Usage SageMath mainly uses Python as a scripting language with a few modifications to make it better suited for mathematical computations. SageMath command-line SageMath can be started from the command-line: $ sage For information on the SageMath command-line see this page. Note, however, that it is not very comfortable for some uses such as plotting. When you try to plot something, for example: sage: plot(sin,(x,0,10)) SageMath opens a browser window with the Sage Notebook. Sage Notebook A better suited interface for advanced usage in SageMath is the Notebook. To start the Notebook server from the command-line, execute: $ sage -n The notebook will be accessible in the browser from http://localhost:8080 and will require you to login. However, if you only run the server for personal use, and not across the internet, the login will be an annoyance. You can instead start the Notebook without requiring login, and have it automatically pop up in a browser, with the following command: $ sage -c "notebook(automatic_login=True)" Jupyter Notebook SageMath also provides a kernel for the Jupyter notebook. To use it, install and , launch the notebook with the command $ jupyter notebook and choose "SageMath" in the drop-down "New..." menu. The SageMath Jupyter notebook supports LaTeX output via the %display latex command and 3D plots if is installed. Cantor Cantor is an application included in the KDE Edu Project. It acts as a front-end for various mathematical applications such as Maxima, SageMath, Octave, Scilab, etc. See the Cantor page on the Sage wiki for more information on how to use it with SageMath. Cantor can be installed with the official repositories.package or as part of the or groups, available in the Documentation For local documentation, one can compile it into multiple formats such as HTML or PDF. To build the whole SageMath reference, execute the following command (as root): # sage --docbuild reference html This builds the HTML documentation for the whole reference tree (may take longer than an hour). An option is to build a smaller part of the documentation tree, but you would need to know what it is you want. Until then, you might consider just browsing the online reference. For a list of documents see sage --docbuild --documents and for a list of supported formats see sage --docbuild --formats. Optional additions SageTeX If you have installed TeX Live on your system, you may be interested in using SageTeX, a package that makes the inclusion of SageMath code in LaTeX files possible. TeX Live is made aware of SageTeX automatically so you can start using it straight away. As a simple example, here is how you include a Sage 2D plot in your TEX document (assuming you use pdflatex): include the sagetexpackage in the preamble of your document with the usual \usepackage{sagetex} create a sagesilentenvironment in which you insert your code: \begin{sagesilent} dob(x) = sqrt(x^2 - 1) / (x * arctan(sqrt(x^2 - 1))) dpr(x) = sqrt(x^2 - 1) / (x * log( x + sqrt(x^2 - 1))) p1 = plot(dob,(x, 1, 10), color='blue') p2 = plot(dpr,(x, 1, 10), color='red') ptot = p1 + p2 ptot.axes_labels(['$\\xi$','$\\frac{R_h}{\\max(a,b)}$']) \end{sagesilent} create the plot, e.g. inside a floatenvironment: \begin{figure} \begin{center} \sageplot[width=\linewidth]{ptot} \end{center} \end{figure} compile your document with the following procedure: $ pdflatex <doc.tex> $ sage <doc.sage> $ pdflatex <doc.tex> you can have a look at your output document. The full documentation of SageTeX is available on CTAN. Troubleshooting TeX Live does not recognize SageTex If your TeX Live installation does not find the SageTex package, you can try the following procedure (as root or use a local folder): Copy the files to the texmf directory: # cp /opt/sage/local/share/texmf/tex/* /usr/share/texmf/tex/ Refresh TeX Live: # texhash /usr/share/texmf/ texhash: Updating /usr/share/texmf/.//ls-R... texhash: Done.
Suppose $\phi$ is a real function on $\mathbb{R}$ such that $$\phi \Bigl(\int_0^1f(x)\,dx\Bigr)\leq \int_0^1 \phi(f)\,dx$$ for every real bounded measurable function $f$. Prove that $\phi$ is then convex. This one is in Rudin's Real and Complex Analysis, page 74, and, to be quite honest, I am unsure what my first step to solve this problem should even be. To help think of a general methodology, I attempted this similar question, from page 71 of the same book: Assume $\phi$ is a continuous real valued function on $(a,b)$ such that $$\phi\Bigl(\frac{x+y}{2}\Bigr)\leq\frac{\phi(x)+\phi(y)}{2}$$ for all $x, y\in (a,b).$ Prove that $\phi$ is then convex (the conclusion does not follow if continuity is omitted from the hypothesis). I take it that the latter question is just a special case of the former (assuming $\phi$ is bounded on $(a,b)$), so I attempted to solve the latter problem first. My attempt was to note that \begin{align} \phi\big(\lambda x + (1-\lambda)y\big) &\leq \phi\Bigl(\frac{2\lambda x + 2(1-\lambda)y}{2}\Bigr) \\ &\leq \frac{1}{2}\Bigl(\phi\bigl(2\lambda x\bigr)+\phi \bigl(2(1-\lambda)y\bigr)\Bigr) \\ & \leq \phi\bigl(\lambda x\bigr)+\phi\bigl((1-\lambda)y\bigr) \end{align} with the latter inequality justified simply by applying the hypothesis again twice to the previous expression. From here, however, I am stuck; I cannot justify "popping out" the $\lambda$ and the $(1-\lambda)$ in the last inequality via some bound afforded by the hypothesis and/or the continuity of $\phi$.
To answer Question 1: Yes, there do exist integrally closed abstract number rings with infinite class group. By factorization of ideals, for $R$ to be an abstract number ring it is enough that it is a Dedekind domain with finite residue field $R/\mathfrak{p}$ at each prime $\mathfrak{p}$. Theorem B of the paper mentioned by Hagen Knaf in his answer actually gives what you ask for (R. C. HEITMANN, PID’S WITH SPECIFIED RESIDUE FIELDS, Duke Math. J. Volume 41, Number 3 (1974), 565-582). Theorem B: Let G be a countable abelian torsion group. Then there is a countable Dedekind domain of characteristic 0 whose class group is G, and whose residue fields are those of the integers (i.e. one copy of $\mathbb{Z}/p\mathbb{Z}$ for each prime $p$). As such rings have finite residue fields, this gives an integrally closed abstract number ring with class group any countable torsion group you like.We can do much better than this though. After thinking about your question for a bit, I see how we can construct the following, so that all countable abelian groups occur as the class group of such rings. Let G be a countable abelian group. Then, there is a Dedekind domain $R$ with finite residue fields such that $\mathbb{Z}[X]\subseteq R\subseteq\mathbb{Q}(X)$ and ${\rm Cl}(R)\cong G$. I see some surprise mentioned in the comments below that it is enough to look at over-rings of $\mathbb{Z}[X]$ to find Dedekind domains with any countable class group. In fact, over-rings of $\mathbb{Z}[X]$ are very general in terms of prime ideal factorization, and can show the following. I'll use ${\rm Id}(R)$ for the group of fractional ideals of $R$ and $R_{\mathfrak{p}}$ for the localization at a prime $\mathfrak{p}$, with $\bar R_{\mathfrak{p}}$ representing its completion (which is a compact discrete valuation ring (DVR) in this case). Let $R$ be a characteristic zero Dedekind domain with finite residue fields. Then, there is a Dedekind domain $R^\prime$ with $\mathbb{Z}[X]\subseteq R^\prime\subseteq\mathbb{Q}(X)$ and a bijection $\pi\colon {\rm Id}(R)\to{\rm Id}(R^\prime)$ satisfying $\pi(\mathfrak{ab})=\pi(\mathfrak{a})\pi(\mathfrak{b})$. $\pi(\mathfrak{a})$ is prime if and only if $\mathfrak{a}$ is. $\pi(\mathfrak{a})$ is principal if and only if $\mathfrak{a}$ is. If $\mathfrak{p}\subseteq R$ is a nonzero prime then $\bar R_{\mathfrak{p}}\cong\bar R^\prime_{\pi(\mathfrak{p})}$. In particular, the class groups are isomorphic, ${\rm Cl}(R)\cong{\rm Cl}(R^\prime)$. The idea is that we can construct Dedekind domains in a field $k$ by first choosing a set $\{v_i\colon i\in I\}$ of discrete valuations on $k$ and, letting $k_v=\{x\in k\colon v(x)\ge0\}$ denote the valuation rings, we can take $R=\bigcap_ik_{v_i}$. Under some reasonably mild conditions, this will be a Dedekind domain with the valuations $v_i$ corresponding precisely to the $\mathfrak{p}$-adic valuations, for prime ideals $\mathfrak{p}$ of $R$. In this way, we can be quite flexible about constructing Dedekind domains with specified prime ideals (and, with a bit of work, specified principal ideals and class group). Constructing discrete valuations $v$ on $k=\mathbb{Q}(X)$ is particularly easy. Given a compact DVR $R$ of characteristic 0 and field of fractions $E$, every extension $\theta\colon k\to E$ gives us a valuation $v(f)=u(f(X))$ where $u$ is the valuation in $E$. To construct such an embedding only requires choosing $x\in E$ which is not algebraic over $\mathbb{Q}$ and, if we want the localization $k_v$ to have completion isomorphic to $R$, then we just need $\mathbb{Q}(x)$ to be dense in $E$. There's plenty of freedom to choose $x\in R$ like this. In fact, there's uncountably many $x$, as they form a co-meagre subset of $R$. So, we have many many valuations on $\mathbb{Q}(X)$ corresponding to any given compact DVR. In this way, we have a lot of flexibility in constructing Dedekind domains in over-rings of $\mathbb{Z}[X]$. I've written out proofs of these statements. As it is much too long to fit here, I'll link to my write-up: Constructing Dedekind domains with prescribed prime factorizations and class groups. Hopefully there's no major errors. I'll also mention that this is an updated and hopefully rather clearer write-up than my initial link (which were very rough notes skipping over many steps). I think also that my linked proof can be modified to show that you can simultaneously choose any prescribed unit group of the form $\{\pm1\}\times U$ where $U$ is a countable free abelian group.
In the previous lesson, we learned the mathematical definition of a gradient. We saw that the gradient of a function was a combination of our partial derivatives with respect to each variable of that function. We saw the process of gradient descent involves moving in the negative direction of the gradient. For example, if the direction of ascent of a function is a move up and to the right, the descent is down and to the left. In this lesson we will apply gradient descent to our cost function to see how we can move toward a best fit regression line by changing the variables of $m$ and $b$. Think about why gradient descent applies so well to a cost function. Initially, we said that the cost of our function, meaning the difference between what our regression line predicted and the dataset, changed as we altered the y-intercept or the slope of the function. Remember that mathematically, when we say cost function, we use the residual sum of squares where $$ RSS = \sum_{i=1}^n(actual - expected)^2 = \sum_{i=1}^n(y_i - \hat{y_i})^2 = \sum_{i=1}^n(y_i - (mx_i + b))^2$$ for all $x$ and $y$ values of our dataset. So in the graph directly below, $x_i$ and $y_i$ would be our points representing a movie's budget and revenue. Meanwhile, $mx_i + b $ is our predicted $y$ value for a given $x$ value, of a budget. And RSS takes the difference between $mx_i + b$, which is $ \hat{y_i} $ (the $y_i$ value our regression line predicts), and our actual $y_i$, represented by the length of the red lines. Then we square this difference, and sum up these squares for each piece of data in our dataset. That is the residual sum of squares. We have plotted how RSS changes as we change one variable of our regression line, $m$ or $b$. We also noted how this plot looks like a curve. Therefore, we call it our cost curve. import plotlyfrom plotly.offline import init_notebook_mode, iplotfrom graph import m_b_trace, trace_values, plotinit_notebook_mode(connected=True)b_values = list(range(70, 150, 10))rss = [10852, 9690, 9128, 9166, 9804, 11042, 12880, 15318]cost_curve_trace = trace_values(b_values, rss, mode="lines", name = 'RSS with changes to y-intercept')plot([cost_curve_trace]) In two dimensions, we decrease our RSS simply by moving forwards or backwards along the cost curve which is the equivalent of changing our variable, in this case y-intercept. So the cost curve above indicates that changing the regression line from having a y-intercept of 70 to 80 decreases our cost, the RSS. Allowing us to change both variables, $m$ and $b$ means calculating how RSS varies with both $m$ and $b$. Because the RSS is a function of how we change our values of $m$ and $b$, we can express this relationship mathematically by saying the cost function, $J$ is the following: $$J(m, b) = \sum_{i=1}^{n}(y_i - (mx_i + b))^2$$ In the function above, $J$ is a function of $m$ and $b$. $J$ just represents the residual sum of squares, which varies as the $m$ and $b$ variables of our regression line are changed. Just like the other multivariable functions we have seen thus far, we can visualize it in three dimensions, and it looks like the following. The three-dimensional graph above shows how the cost associated with our regression line changes as the slope and y-intercept values are changed. Let's explore using gradient descent to determine how to change our regression line when we can alter both $m$ and $b$ variables. When applied to a general multivariable function $f(x,y)$, gradient descent answers how much to move the $x$ variable and the $y$ variable to produce the greatest decrease in output. Now that we are applying gradient descent to our cost curve $J(m, b)$, the technique should answer how much to move the $m$ variable and the $b$ variable to produce the greatest decrease in cost, or RSS. In other words, when altering our regression line, we want to know how much of this change should be derived from a move in the slope versus how much should be derived from a change in the y-intercept. As we know, the gradient of a function is simply the partial derivatives with respect to each of the variables, so: $$ \nabla J(m, b) = \frac{\partial J}{\partial m}, \frac{\partial J}{\partial b}$$ In calculating the partial derivatives of our function $J(m, b) = \sum_{i=1}^{n}(y_i - (mx_i + b))^2$, we won't change the result if we ignore the summation until the very end. We'll do that to make our calculations easier. Ok, so let's take our partial derivatives of the following: $$\frac{\partial J}{\partial m}J(m, b) = \frac{\partial J}{\partial m}(y - (mx + b))^2$$ $$\frac{\partial J}{\partial b}J(m, b) = \frac{\partial J}{\partial b}(y - (mx + b))^2$$ Let's start with taking the partial derivative with respect to $m$. $$\frac{\partial J}{\partial m}J(m, b) = \frac{\partial J}{\partial m}(y - (mx + b))^2$$ Now this is a tricky function to take the derivative of. So we can use functional composition followed by the chain rule to make it easier. Using functional composition, we can rewrite our function $J$ as two functions: $$g(m,b) = y - (mx + b)$$ $$J(g(m,b)) = (g(m,b))^2$$ Now using the chain rule to find the partial derivative with respect to a change in the slope, gives us: $$\frac{\partial J}{\partial m}J(g) = \frac{\partial J}{\partial g}J(g(m, b))\frac{\partial g}{\partial m}g(m,b)$$ One good way to check yourself is to note that the partial derivatives yield: $$\frac{\partial J}{\partial m} = \frac{\partial J}{\partial g}\frac{\partial g}{\partial m} = \frac{\partial J}{\partial m}$$ Our next step is to solve these derivatives individually: $$\frac{\partial J}{\partial g}J(g(m, b)) = \frac{\partial J}{\partial g}g(m,b)^2 = 2g(m,b)$$ $$\frac{\partial g}{\partial m}g(m,b) = \frac{\partial g}{\partial m} (y - (mx +b)) = \frac{\partial g}{\partial m}y - \frac{\partial g}{\partial m}mx - \frac{\partial g}{\partial m}b = -x $$ Each of the terms are treated as constants, except for the middle term. Now plugging these back into our chain rule we have: $$\frac{\partial J}{\partial g}J(g(m,b)) \frac{\partial g}{\partial m}g(m,b) = (2g(m,b)) -x = 2(y - (mx + b))-x $$ So $$\frac{\partial J}{\partial m}J(m, b) = 2(y - (mx + b)) -x = -2x(y - (mx + b )) $$ Ok, now let's calculate the partial derivative with respect to a change in the y-intercept. We express this mathematically with the following: $$\frac{\partial J}{\partial b}J(m, b) = \frac{\partial J}{\partial b}(y - (mx + b) )^2$$ Then once again, we use functional composition following by the chain rule. So we view our cost function as the same two functions $g(m,b)$ and $J(g(m,b))$. $$g(m,b) = y - (mx + b)$$ $$J(g(m,b)) = (g(m,b))^2$$ So applying the chain rule, to this same function composition, we get: $$\frac{\partial J}{\partial b}J(g(m, b)) = \frac{\partial J}{\partial g}J(g(m, b))\frac{\partial g}{\partial b}g(m,b)$$ Now, our next step is to calculate these partial derivatives individually. From our earlier calculation of the partial derivative, we know that $\frac{\partial J}{\partial g}J(g(m,b)) = \frac{\partial J}{\partial g}g(m,b)^2 = 2g(m,b)$. The only thing left to calculate is $\frac{\partial g}{\partial b}g(m,b)$. $$\frac{\partial g}{\partial b}g(m,b) = \frac{\partial g}{\partial b}(y - (mx + b) ) = -1$$ Now we plug our terms into our chain rule and get: $$ \frac{\partial J}{\partial g}J(g) \frac{\partial g}{\partial b}g(m,b) = 2g(m,b) -1 = -2(y - (mx + b)) $$ Ok, so now we have our two partial derivatives for $\nabla J(m, b)$: $$ \frac{\partial J}{\partial m}J(m,b) = -2x(y - (mx + b )) $$ $$ \frac{\partial J}{\partial b}J(m,b) = -2(y - (mx + b)) $$ And as $mx + b$ = is just our regression line, we can simplify these formulas to be: $$ \frac{\partial J}{\partial m}J(m,b) = -2x(y - \hat{y}) = -2x\epsilon$$ $$ \frac{\partial J}{\partial b}J(m,b) = -2(y - \hat{y}) = -2\epsilon$$ Remember, error = actual - expected, so we can replace $y - \hat{y}$ with $\epsilon$, our error. As we mentioned above, our last step is adding back the summations. Since $-2$ is a constant, we can keep this outside of the summation. Our value for $x$ changes depending upon what x value we are at, so it must be included inside the summation for the first equation. Below, we have: $$ \frac{\partial J}{\partial m}J(m,b) = -2\sum_{i=1}^n x(y_i - \hat{y} i) = -2\sum{i=1}^n x_i\epsilon_i$$ $$ \frac{\partial J}{\partial b}J(m,b) = -2\sum_{i=1}^n(y_i - \hat{y} i) = -2\sum{i=1}^n \epsilon_i$$ So that is what what we'll do to find the "best fit regression line." We'll start with an initial regression line with values of $m$ and $b$. Then we'll go through our dataset, and we will use the above formulas with each point to tell us how to update our regression line such that it continues to minimize our cost function. In the context of gradient descent, we use these partial derivatives to take a step size. Remember that our step should be in the opposite direction of our partial derivatives as we are descending towards the minimum. So to take a step towards gradient descent we use the general formula of: current_m = old_m $ - \frac{\partial J}{\partial m}J(m,b)$ current_b = old_b $ - \frac{\partial J}{\partial b}J(m,b) $ or in the code that we just calculated: current_m = old_m $ - (-2\sum_{i=1}^n x_i\epsilon_i )$ current_b = old_b $ - ( -2*\sum_{i=1}^n \epsilon_i )$ In the next lesson, we'll work through translating this technique, with use of our $\nabla J(m, b)$, into code to descend along our cost curve and find the "best fit" regression line. In this section, we developed some intuition for why the gradient of a function is the direction of steepest ascent and the negative gradient of a function is the direction of steepest decent. Essentially, the gradient uses the partial derivatives to see what change will result from changes in the function's dimensions, and then moves in that direction weighted more towards the partial derivative with the larger magnitude. We also practiced calculating some gradients, and ultimately calculated the gradient for our cost function. This gave us two formulas which tell us how to update our regression line so that it descends along our cost function and approaches a "best fit line".
Let $L$ be the language of set theory, and for each countable ordinal $\kappa$, define $L_\kappa$ as follows: $L_0 = L$, and $L_{\alpha+1}$ is obtained by adding countably many new class constants $c_\phi$ to the signature of $L_\alpha$ to stand for each formula $\phi(x)$ of one free variable. Define $L_\alpha$ for $\alpha$ a limit ordinal to be the union of $L_\beta$ for $\beta < \alpha$. Given some set theory such as ZF, it is possible to add axioms for the new constants: let $T_0 = ZF$ and $T_{\alpha+1}$ be the theory got from $T_\alpha$ by adding $$x \in c_\phi \iff \phi(x)$$ as an axiom where $\phi(x)$ ranges over the countably many formulas in $L_\alpha$. Let $T_\alpha$ for $\alpha$ a limit ordinal be the union of $T_\beta$ for $\beta<\alpha$. This is ordinary expansion by definitions. So, for each countable ordinal $\kappa$, $T_\kappa$ is countably axiomatizable. Define $L_\Omega$ and $T_\Omega$ as the union of $L_\kappa$ and $T_\kappa$ as $\kappa$ ranges over countable ordinals. Is $T_\Omega$ countably axiomatizable? What relation does MK have to $T_\Omega$, if any? New question: is every formula of $L_\Omega$ equivalent to a formula of $L_1$ modulo $T_1$? (seems likely) please see: Is every formula of LΩ equivalent to a formula of L1 modulo T1?
The preface to Mark Srednicki's "Quantum Field Theory" says that to be prepared for the book, one must recognize and understand the following equations: $$\frac{d\sigma}{d\Omega} = \|f(\theta,\phi)\|^2, \qquad (1)$$ $$a^{\dagger}\|n\rangle = \sqrt{n+1} \space \|n+1\rangle, \qquad (2)$$ $$J_{\pm} \|j,m \rangle = \sqrt{j(j+1)-m(m\pm 1)} \mid j,m \pm 1 \rangle, \qquad (3)$$ $$A(t) = e^{+iHt/\hbar}Ae^{-iHt/\hbar}, \qquad (4)$$ $$H = p\dot{q}-L, \qquad (5)$$ $$ct'=\gamma (ct-\beta x), \qquad (6)$$ $$E=(\mathbf{p}^2c^2+m^2c^4)^{1/2}, \qquad (7)$$ $$\mathbf{E} =-\mathbf{\dot{A}}/c-\mathbf{\nabla} \varphi. \qquad (8)$$ I am certainly not ready to dive into this book, so I would like some help identifying these equations and learning more about their fundamental usefulness. I don't recognize (1), but (2) looks like a quantum mechanical creation operator? I thought those were only really useful in the context of the harmonic oscillator problem, but maybe everything is just a complicated HO problem? (3) has to do with angular momentum? (4) is a plane wave solution to the Schrodinger Eqn? (5) is the classical mechanics Hamiltonian, with cannonical coordinates? (6) is the relativistic Lorentz transformation. (7) is the general form of mass energy equivalence from SR. (8) is the electric field expressed as vector and scalar potentials? Is that really the only E&M machinery required? Any insight as to why these particular expressions are relevant / important / useful to QFT is also appreciated. Also, where are the statistical mechanics ideas hiding? In the QM?
Blast wave attenuation by lightly destructable granular materials Abstract Terrorist bombings are a dismal reality nowadays. One of the most effective ways for protection against blast overpressure is the use of lightly compacted materials such as sand [1] and aqueous foam [2] as a protective envelope or barrier. According to [1], shock wave attenuation in a mine tunnel (one-dimensional case) behind a destroyed object is given by where $$ q_e\approx q\frac{1} {{1 + 4(S/q)^{1/6} b\rho _{mat} /L^{1/3} }} $$ (1) — effective charge, qe S— exposed area of the obstacle, q— TNT equivalent (grams), L— distance between charge and obstacle, b— obstacle thickness and ρ — material density. This empirical equation is applicable only in a one-dimensional case but not for a less confined environment. Another way of protecting a structure against blast is to coat the surface with a sacrificial layer. In [3] full-scale experiments were carried out to investigate the behaviour of a covering of aluminum foam under the effect of a blast wave. mat KeywordsBlast Wave Detonation Product Aluminum Foam Sacrificial Layer Cylinder Wall Thickness These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves. Preview Unable to display preview. Download preview PDF. References 1.G.I. Pokrovsky: Explosion. (Nedra, Moscow 1980)Google Scholar 2.A.A. Borisov, B.E. Gelfand, V.M. Kudinov, B.I. Palamarchuk, V.V. Stepanov, E.I. TimofeevGoogle Scholar S.V. Khomik: Shock waves in water foams. Acta Astron. 5(11–12), 1027 (1978)Google Scholar 3. 4.H.J. Melosh: Impact Cratering. A Geologic Process. (Oxford, New York, 1989)Google Scholar 5. 6. 7. Copyright information © Tsinghua University Press and Springer-Verlag Berlin Heidelberg 2005
Real Analysis Exchange Real Anal. Exchange Volume 37, Number 1 (2011), 177-188. BV -Functions and Change of Variable p p Abstract In this note we discuss some interconnections between the space $BV_p[a,b]$ ($1\leq p\lt\infty$) of functions of bounded $p$-variation (in Wiener's sense) and the space $Lip_\alpha[a,b]$ ($0\lt\alpha\leq 1$) of Hölder continuous functions. In particular, we show that $f\in BV_p[a,b]$ if and only if $f=g\circ \tau$, with $g\in Lip_{1/p}[a,b]$ and $\tau$ being monotone, and that $f\in BV_p[a,b] \cap C[a,b]$ if and only if $f=g\circ \tau$, with $g\in Lip_{1/p}[a,b]$ and $\tau$ being a homeomorphism. Article information Source Real Anal. Exchange, Volume 37, Number 1 (2011), 177-188. Dates First available in Project Euclid: 30 April 2012 Permanent link to this document https://projecteuclid.org/euclid.rae/1335806770 Mathematical Reviews number (MathSciNet) MR3016858 Citation Merentes, N.; Sánchez, J. L. BV p -Functions and Change of Variable. Real Anal. Exchange 37 (2011), no. 1, 177--188. https://projecteuclid.org/euclid.rae/1335806770
In my story, set in the far(ish) future, humanity discovers a way to extract energy from black holes (perhaps via the Penrose process?), which sparks conflict as various factions attempt to take control over previously useless and dangerous but now suddenly valuable black holes. This energy is extremely valuable because warp drives for faster-than-light travel require massive amounts of power, which made them impractical before the proliferation of Penrose process power plants. In order for this to make sense, black holes would have to be a better source of energy than whatever humanity was using before, which I assume would be Dyson spheres/rings/swarms/etc. gathering energy radiated from stars. To see if this is plausible, I attempted to compare the energy output of a Penrose power plant vs a Dyson sphere around the sun. According to the Wikipedia article about the Penrose process, up to 29% of a black hole's mass energy can be contained in its angular momentum. So, assuming a black hole with the mass of the sun (which I know is too small to become a black hole, but I wanted to compare apples to apples), the total amount of energy that can be harvested from it is $(0.29\cdot1.989\times10^{30})c^2 = 5.18\times10^{46}$ joules. For comparison, the sun emits a total of $3.9\times10^{26}$ watts. 2 Over the 5 billion remaining years of the sun's life, assuming its energy output remains constant, it will emit a total of $(5,000,000,000\cdot60\cdot60\cdot24\cdot365)(3.9\times10^{26}) = 6.15\times10^{43}$ joules. So the black hole has over 800 times as much energy to harvest! But the real question is, how quickly can that energy be harvested? According to Wikipedia, an object's energy can be increased by up to 20.7% using the Penrose process. That means the equation for how much energy we can get from a given mass is $0.207mc^2$. So, the amount of mass we need to toss into the black hole per second to match the sun is $\frac{3.9\times10^{26}}{0.207c^2} = 2.1\times10^{10}$ kilograms per second. But only half of that mass actually falls into the black hole, the rest can be re-used. So only $1.05\times10^{10}$ kilograms are consumed each second. How much mass is that? That's about 1.75 Great Pyramids of Giza every second. At that rate, the mass of the moon would be used in $\frac{7.3\times10^{22}}{1.05\times10^{10}} = 6.95\times10^{12}$ seconds, or about 220,000 years. For the sun, it would be $\frac{1.989\times10^{30}}{1.05\times10^{10}} = 1.89\times10^{20}$ seconds, or a little under 6 trillion years. Based on the math, it seems plausible. (Assuming you can toss stuff in fast enough.) So my questions are: If a civilization is not advanced enough to harvest energy from black holes, are stars the best source of energy? If a civilization isadvanced enough to harvest energy from black holes, are black holes the best source of energy? Is the Penrose process the best way to harvest energy from a black hole? By "best" I mean produces usable energy the quickest. I know matter-antimatter annihilation would release energy extremely quickly, but you have to create the antimatter first, which uses a bunch of energy. (In my universe, humanity never figures out how to create antimatter efficiently enough to get a net gain of energy by annihilating it.)
K C Gupta Articles written in Proceedings – Mathematical Sciences Volume 96 Issue 2 November 1987 pp 125-130 A theorem which reveals an interesting relationship between originals of related functions in 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 Volume 105 Issue 2 May 1995 pp 187-192 In this paper we first solve a convolution integral equation involving product of the general class of polynomials and the 3. We also give exact references of two results recently obtained by Srivastava Current Issue Volume 129 \| Issue 5 November 2019 Click here for Editorial Note on CAP Mode
So what are spin-networks? Briefly, they are graphs with representations ("spins") of some gauge group (generally SU(2) or SL(2,C) in LQG) living on each edge. At each non-trivial vertex, one has three or more edges meeting up. What is the simplest purpose of the intertwiner? It is to ensure that angular momentum is conserved at each vertex. For the case of four-valent edge we have four spins: $(j_1,j_2,j_3,j_4)$. There is a simple visual picture of the intertwiner in this case. Picture a tetrahedron enclosing the given vertex, such that each edge pierces precisely one face of the tetrahedron. Now, the natural prescription for what happens when a surface is punctured by a spin is to associate the Casimir of that spin $ \mathbf{J}^2 $ with the puncture. The Casimir for spin $j$ has eigenvalues $ j (j+1) $. You can also see these as energy eigenvalues for the quantum rotor model. These eigenvalues are identified with the area associated with a puncture. In order for the said edges and vertices to correspond to a consistent geometry it is important that certain constraints be satisfied. For instance, for a triangle we require that the edge lengths satisfy the triangle inequality $ a + b \lt c $ and the angles should add up to $ \angle a + \angle b + \angle c = \kappa \pi$, with $\kappa = 1$ if the triangle is embedded in a flat space and $\kappa \ne 1$ denoting the deviation of the space from zero curvature (positively or negatively curved). In a similar manner, for a classical tetrahedron, now it is the sums of the areas of the faces which should satisfy "closure" constraints. For a quantum tetrahedron these constraints translate into relations between the operators $j_i$ which endow the faces with area. Now for a triangle giving its three edge lengths $(a,b,c)$ completely fixes the angles and there is no more freedom. However, specifying all four areas of a tetrahedron does not fix all the freedom. The tetrahedron can still be bent and distorted in ways that preserve the closure constraints (not so for a triangle!). These are the physical degrees of freedom that an intertwiner possesses - the various shapes that are consistent with a tetrahedron with face areas given by the spins, or more generally a polyhedron for n-valent edges. Some of the key players in this arena include, among others, Laurent Friedel, Eugenio Bianchi, E. Magliaro, C. Perini, F. Conrady, J. Engle, Rovelli, R. Pereira, K. Krasnov and Etera Livine. I hope this provides some intuition for these structures. Also, I should add, that at present I am working on a review article on LQG for and by "the bewildered". This post imported from StackExchange Physics at 2014-04-01 16:52 (UCT), posted by SE-user user346 I reserve the right to use any or all of the contents of my answers to this and other questions on physics.se in said work, with proper acknowledgements to all who contribute with questions and comments. This legalese is necessary so nobody comes after me with a bullsh*t plagiarism charge when my article does appear :P
NOTICE: Citizendium is still being set up on its newer server, treat as a beta for now; please see here for more. Citizendium - a community developing a quality comprehensive compendium of knowledge, online and free . Click here to join and contribute—free CZ thanks our previous donors. Donate here. Treasurer's Financial Report -- Thanks to our content contributors. -- Difference between revisions of "ABC conjecture" (→Statement: added Baker's version and Stewart & Yu result) m (→Statement: Baker (1998)) (3 intermediate revisions by the same user not shown) Line 11: Line 11: :<math> \|A\|, \|B\|, \|C\| < \kappa(\epsilon) r(ABC)^{1+\epsilon} \ . </math> :<math> \|A\|, \|B\|, \|C\| < \kappa(\epsilon) r(ABC)^{1+\epsilon} \ . </math> − + conjecture states that :<math> (\|A\| \cdot \|B\| \cdot \|C\|)^{1/3} < \kappa(\epsilon) r(ABC)^{1+\epsilon} \ . </math> :<math> (\|A\| \cdot \|B\| \cdot \|C\|)^{1/3} < \kappa(\epsilon) r(ABC)^{1+\epsilon} \ . </math> Line 17: Line 17: If we define If we define − :<math> \kappa(\epsilon) = \inf_{A+B+C=0,\ (A,B)=1} \frac{\max\{\|A\|,\|B\|,\|C\|\}}{N^{1+\epsilon}} \ , + :<math> \kappa(\epsilon) = \inf_{A+B+C=0,\ (A,B)=1} \frac{\max\{\|A\|,\|B\|,\|C\|\}}{N^{1+\epsilon}} \ , − then it is known that <math>\kappa \rightarrow \infty</math> as <math>\ + then it is known that <math>\kappa \rightarrow \infty</math> as <math>\\rightarrow 0</math>. − Baker introduced a more refined version of the conjecture in + Baker introduced a more refined version of the conjecture in . Assume as before that <math>A + B + C = 0</math> holds for coprime integers <math>A,B,C</math>. Let <math>N</math> be the radical of <math>ABC</math> and <math>\omega</math> the number of distinct prime factors of <math>ABC</math>. Then − :<math> \|A\|, \|B\|, \|C\| < + :<math> \|A\|, \|B\|, \|C\| < (\epsilon^{-\omega} N)^{1+\epsilon} \ . </math> This form of the conjecture would give very strong bounds in the [[method of linear forms in logarithms]]. This form of the conjecture would give very strong bounds in the [[method of linear forms in logarithms]]. Latest revision as of 18:18, 13 January 2013 In mathematics, the ABC conjecture relates the prime factors of two integers to those of their sum. It was proposed by David Masser and Joseph Oesterlé in 1985. It is connected with other problems of number theory: for example, the truth of the ABC conjecture would provide a new proof of Fermat's Last Theorem. Statement Define the radical of an integer to be the product of its distinct prime factors Suppose now that the equation holds for coprime integers . The conjecture asserts that for every there exists such that A weaker form of the conjecture states that If we define then it is known that as . Baker introduced a more refined version of the conjecture in 1998. Assume as before that holds for coprime integers . Let be the radical of and the number of distinct prime factors of . Then there is an absolute constant such that This form of the conjecture would give very strong bounds in the method of linear forms in logarithms. Results It is known that there is an effectively computable such that
I've tried to explain the solution using as little math as possible, and to give some intuition as to what makes it tick. Nonetheless there will be a little mathematical notation at the end. First steps: going beyond the obvious solution in the simplest case (N=2) The statement of the puzzle as presented here doesn't make this very clear, but the puzzle relies on the prisoners not knowing anything about which name is located in which box (until they get into the room, after which they cannot communicate anymore). If every prisoner checks 50 boxes at random, then each prisoner has a ½ chance of finding his own name. If all the prisoners choose a set of boxes at random, independently, then the probability that they all find their own name is ½ × … × ½ = 1/2 100 — infinitesimal. Making independent choices is a waste, though. If anyone gets it wrong, the situation isn't worse than if everybody gets it wrong. Rather than make independent choices, they can make correlated choices; the idea is to try to arrange that either everybody gets it right, or many get it wrong. Let's consider the simpler case when there are two prisoners. If they both choose at random, then they have ½ × ½ = ¼ chance of surviving. But there's an obvious waste: suppose prisoner #1 opens the left-hand box and finds his name: then prisoner #2 will not find his name in the left-hand box. So the prisoners can decide that #1 will look at the box on the left and #2 looks at the box on the right: that way, either they both get it right or both get it wrong, and they have a ½ chance of survival. Incidentally, note that another assumption that wasn't clearly stated here is that the prisoners get to formulate their strategy in secret. If the warden knows which prisoner chose which box, he can arrange for the prisoners to fail by putting prisoner #1's name in the right-hand box. The next step: N=4 The obvious way to generalize this to more prisoners is to assign each prisoner a fixed set of boxes that he'll open. However, I won't pursue this further, because it doesn't take advantage of an important ability: after a prisoner has opened a first box, he can base his decision on which box to open next on the content of the first box, and so on. Consider the case with 4 prisoners and 4 boxes. I'll use numbers for the prisoners' names, and assume that the boxes are numbered as well. Intuitively, it is preferable for each prisoner to pick a different box to open first, since otherwise some common choices are wasted. So prisoner #1 opens box #1 and finds a name (number). Now what? If he finds his own name (#1) (¼ chance), of course, he can stop. If he finds some different name (say 2) (¾ chance), what information does this provide? Well, since each box contains a different name, prisoner #1 now knows that box #2 does not contain 2, so prisoner #2 will not be lucky the first time either. Furthermore, the strategy should favor arranging for prisoner #2 to pick box #1 next. To simplify the analysis, I'll only look at cases where all prisoners follow the same strategy. (I don't have an intuitive argument as to why breaking the symmetry wouldn't be advantageous.) Either they all open the box whose number they found in the first box, or they all open a different box. If #1 opens box #2 and finds his name there, then either boxes #3 and #4 contain 3 and 4 respectively, or they contain 4 and 3. Either way, with the strategy of using the name in the first box, if one prisoner is lucky the second time then every prisoner is lucky! If #1 opens box #3 instead and finds his name there, then there is a ½ chance that prisoner #2 will find his name in box #2, and a ½ chance that he'll find his name in box #4. But what about prisoner #3? He finds the name of prisoner #1 in box #3, which doesn't give any clue as to where 3 might be instead. So let's concentrate on the strategy where each prisoner opens the second box whose number is what he found in the first box. What arrangement of numbers in boxes make it work? There are 4 ways to choose which box contains the number 1. The number 2 can go into any of the 3 remaining boxes. The number 3 can go in either of the 2 remaining boxes, and the number 4 must go into the one remaining box. So there are 4×3×2 = 24 different arrangements. The following arrangements lead to success because each number is either in its own box or swapped with another number: 1234 1243 1324 1432 2134 2143 3214 3412 4231 4321 That's 10 successful arrangements out of 24. The chance of success isn't very far from the theoretical maximum of ½, which is encouraging. Note that in order for the chance of success to be 10/24, we need to know that the arrangements have an equal chance of being chosen. If the warden is nasty and arranges the numbers as 2341, the prisoners are sure to lose. This is where the fact that the prisoners choose a strategy in secret comes in. In my analysis, I used numbers for prisoners — but fact the prisoners are names, not numbers, and they can pick a random assignment of names to numbers as part of their secret strategy (in fact, this assignment is the only secret part, since the warden may have looked up the solution of the puzzle). General analysis Let's explore a strategy that generalizes what we explored for 4 boxes: each prisoner opens the box with his own number, then the box whose number is contained in the first box, and so on. Consider the sequence of numbers that a certain prisoner encounters: $x_0$ (the inital box numbered with the prisoner's own number), $x_1$ (number contained in box $x_0$), $x_2$ (number contained in box $x_1$), … Since each number is contained in only one box, this sequence cannot contain any repeated element as long as it doesn't loop back to $x_0$. Eventually the sequence has to loop back to $x_0$ since it will run out of numbers. At that point, the prisoner has found his own name. The critical problem for the prisoner is whether the loop completes before or after the prisoner has opened the maximum of 50 boxes. From now on, let me use the proper mathematical vocabulary. A way to arrange distinct numbers into as many boxes is called a permutation. Opening box number $k$ and looking at the number that it contains is called applying that permutation. Repeated applications of a permutation eventually runs into a loop; such a loop is called a cycle. The prisoners succeed if all of the cycles for the permutation have a length of at most 50. Let's call a cycle long if it contains 51 or more elements. Observe that there cannot be more than one long cycle (if one cycle has at least 51 elements, then there are only 49 or fewer elements to share between the other cycles). So we can count the losing configurations by adding up the permutations of 100 elements that have a cycle of length 51, 52, …, 100. Lemma: there are $n! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot (n-1) \cdot n$ distinct permutations of $n$ elements. Proof: there are $n$ ways to pick the image of the first element, $n-1$ remaining ways to pick the image of the second elements, etc., down to a single way to pick the image of the last element. Now let's count the number of permutations that have a cycle of length $c$ (for $c \ge 51$, so that there's a single such cycle). We're actually going to count each permutation $c$ times, once for each element of the cycle. Pick the first element in the cycle: there are 100 possibilities. There are 99 possibilities for the second element, and so on, until we've picked $c$ elements. So far, that's $100 \times 99 \times \ldots \times (100-c+1)$ possibilities. There are $100-c$ remaining elements, and they can be permuted in any way, so there are $(100-c)!$ possibilities as per the lemma above. That's a total of $(100 \times 99 \times \ldots \times (100-c+1)) \times ((100-c) \times \ldots \times 2 \times 1)$ possibilities, which nicely collapses to $100!$. Recall that we counted each permutation $c$ times, since we counted it once per element in the cycle. So the number of permutations with a cycle of length $c$ is $100! / c$. The number of permutations with a long cycle is thus$$ \frac{100!}{100} + \frac{100!}{99} + \ldots + \frac{100!}{51} $$That's out of a total of $100!$ permutation. The proportion of failing permutations is thus$$ \frac{1}{100} + \frac{1}{99} + \ldots + \frac{1}{51} $$Numerically, this is about 0.6882, i.e. the chance of success is about 31.18%, a little over the requisite 30%. In general, the proportion of failing permutations for $N$ prisoners is $H_N - H_{N/2}$ where$$ H_n = 1 + \frac{1}{2} + \ldots + \frac{1}{n} $$is called the $n$th harmonic number. For large values of $n$, $H_n \approx \ln n + C$ for some number C, and the series $H_N - H_{N/2}$ converges to $\ln 2 \approx 0.6931$ from below. (I will not provide an elementary proof of that). This gives a lower limit to the chance of success for large numbers of prisoners: 30.68% is achievable.
The theorem that inkspot refers to in his answer is originally from Inoue's paper New Surfaces with No Meromorphic Functions, II which seems like a more complete reference for this question. In particular, Inoue gives an explicit example of a compact complex surface which has an analytic subvariety but no compact complex submanifolds. If $x$ is a real quadratic irrationality (i.e. a real irrational solution of a real quadratic equation), denote it's conjugate by $x'$. Let $M(x)$ be the free $\mathbb{Z}$-module generated by $1$ and $x$, then set $U(x) = \{\alpha \in \mathbb{Q}(x) \mid \alpha > 0, \alpha\cdot M(x) = M(x)\}$ and $U^+(x) = \{\alpha \in U(x) \mid \alpha\cdot\alpha' > 0\}$. Both $U(x)$ and $U^+(x)$ are infinite cyclic groups and $[U(x) : U^+(x)] = 1$ or $2$. If $\omega$ is a real quadratic irrationality such that $\omega > 1 > \omega' > 0$, then $\omega$ is a purely periodic modified continued fraction; that is, $\omega = [[\overline{n_0, n_1, \dots, n_{r-1}}]]$ where $n_i \geq 2$ for all $i$, $n_j \geq 3$ for at least one $j$, and $r$ is the smallest period. For every such $\omega$, Inoue constructs a compact complex surface $S_{\omega}$ which is now known as an Inoue-Hirzebruch surface. There are compact subvarieties $C$ and $D$ of $S_{\omega}$ with irreducible components $C_0, \dots, C_{r-1}$ and $D_0, \dots, D_{s-1}$ respectively; here $s$ is the smallest period of the modified continued fraction expansion of another element $\omega^*$ related to $\omega$ (alternatively, $s$ can be determined from the modified continued fraction expansion of $\frac{1}{\omega}$). When $r \geq 2$, $C$ is a cycle of non-singular rational curves, and when $r = 1$, $C$ is a rational curve with one ordinary double point. Proposition $5.4$ shows that $C_0, \dots, C_{r-1}, D_0, \dots, D_{s-1}$ are the only irreducible curves in $S_{\omega}$. In the case where $[U(\omega) : U^+(\omega)] = 2$, we have $r = s$. Furthermore, there is an involution $\iota$ such that $\iota(C_i) = D_i$ for $i = 0, \dots, r - 1$. The quotient of $S_{\omega}$ by $\iota$ is denoted $\hat{S}_{\omega}$ and is now known as a half Inoue surface. Note that the images of $C_0, \dots, C_{r-1}$ are the only irreducible curves in $\hat{S}_{\omega}$. If we can find a real quadratic irrationality $\omega$ such that $\omega > 1 > \omega' > 0$, $r = 1$, and $[U(\omega) : U^+(\omega)] = 2$, then $\hat{S}_{\omega}$ is a compact complex surface containing a unique curve, namely a rational curve with one ordinary double point. In particular, it provides an example of a compact complex manifold with a subvariety but no compact complex submanifolds. One such $\omega$ was given in the paper (end of section 6). Example. Take $\omega = (3 + \sqrt{5})/2$. Then $[U(\omega) : U^+(\omega)] = 2$ and $\alpha_0 =\ \text{a generator of}\ U(\omega) = (1 + \sqrt{5})/2$, $\alpha = \alpha_0^2 = (3 + \sqrt{5})/2$, $\omega = [[\overline{3}]]$, $r = 1$. In this case, $b_2(\hat{S}_{\omega}) = 1$ and $\hat{S}_{\omega}$ contains exactly one curve $\hat{C}$. Moreover, $\hat{C}$ is a rational curve with one ordinary double point and $(\hat{C})^2 = -1$. For those interested in the details, in addition to Inoue's paper, it may also be worth reading the earlier paper Hilbert modular surfaces by Hirzebruch. As mentioned in his paper, Inoue used some methods from Hirzebruch's paper (which gives some indication of why the resulting surfaces are jointly named).
I need to compute the two-sided (bilateral) Laplace transform of a numerically given function $F$,$$I(t) = \int_{-\infty}^{+\infty} {dx} \, e^{-x} \, F(x + t) ~,$$where $F(x)$ has some sharp features, e.g., at $ x = \{ 0, x_p, \cdots \} $, has finite support, and vanishes sufficiently fast for $ x < 0 $ to prevent a divergence of the integral. Due to the sharp features, I have to divide the integration region to some subregions, for instance, \begin{align} I(t) &= \int_{-\infty}^{-\Delta} {dx} \, e^{-x} \, F(x + t) + \int_{-\Delta}^{+\Delta} {dx} \, e^{-x} \, F(x + t) \\ &\quad + \int_{+\Delta}^{x_p - \Delta} {dx} \, e^{-x} \, F(x + t) + \int_{x_p - \Delta}^{x_p + \Delta} {dx} \, e^{-x} \, F(x + t) \\ &\quad + \int_{x_p + \Delta}^{+\infty} {dx} \, e^{-x} \, F(x + t) ~, \end{align} where $\Delta > 0$. I'd like to know the best and most precise way to perform such an integral numerically (quadrature methods). Is there something like Gauss-Laguerre quadrature for such an integration? I have also considered a transformation of the integration variable, $$ u = e^{-x} ~, $$ to absorb the exponential factor, $$ I(t) = \int_{0^+ = \, u(x \rightarrow +\infty)}^{+\infty = \, u(x \rightarrow -\infty)} {du} \, F(-\ln(u) + t) ~, $$ and performing the transformed integration by a Tanh-Sinh quadrature; yet I am not sure if this is the best method. As an example, one can take $$ f(x,\omega) = e^{-\alpha (x - \omega)} \, \Theta(x - \omega) $$ where $\Theta$ denotes the Heaviside step function and $\alpha > 0$. Hence, the ‘width’ of the peak at $\omega$ can be varied by $\alpha$. The two-sided Laplace transform of $f$ will be $$ \mathcal{L}[f](\omega) := \int_{-\infty}^{+\infty} dx \, e^{-x} \, f(x, \omega) = \frac{ e^{-\omega} }{1 + \alpha} ~. $$
Hello, Now I have a question about a formulation and want to seek for help. For a facility location problem, I have come up with several constraints. The decision variable $y$ is binary. But it cannot be automatically guaranteed to take on binary values, if relaxed. The problem is described as: a network has $n$ customers and $m$ facilities. Due to uncertain supply, each customer has to be assigned to these facilities level by level in increasing distance. For example, customer $i$ may be assigned to facility $j$ on level $r$, assigned to facility $k$ on level $r+1$, in this way, $y_{ijr} = y_{ik,r+1}=1$, and the distance $d_{ij} < d_{ik}$. Or we can set parameter $a_{ijk}=1$. Without capacity component, some constraints are as follows: $\sum\limits_{j}{y_{ijr}}=1 \forall i,r$ each customer has to be served in each level. $\sum\limits_{r}{y_{ijr}} \leq 1 \forall i,j$ a facility cannot be assigned more than one level $y_{ijr} + y_{ik,r+1} \leq 1 + a_{ijk} \forall i,j,k,r$ the closer facility should be assigned on the lower level. But with these constraints, in the solver output, fractional solutions of $y$ can often be obtained. For example, $y_{010} = y_{011} = y_{020} = y_{021} = 0.5$, which is not I want. If define $y$ to be binary in CPLEX or GUROBI, binary values can be always guaranteed using B&C, but it is really time consuming. Without introduce a penalty into the objective function (since it will affect my original objective function), how can we reformulate these constraints to guarantee the integrity of variables $y$? Not introduce other binary variables and big $M$, if possible. Thank you for your help. If you use a solver for Mixed Integer Programming and define the variables as binary, the result will be 0 or 1 always. Gurobi and CPLEX have this ability and also some open source solvers. If you want to solve your problem by only handling LPs in the solver, you have to use or implement your own Branch-And-Bound scheme (which will probably be much slower). Linear relaxations may always give non-integer solutions, unless the constraint matrix is totally unimodular and you use the simplex method. By using branch&bound techniques or cutting plane methods, which require you to solve a series of linear programs, you end up with an optimal integer solution. answered optimizer
The inversion barrier in $\ce{NH3}$ is approximately $5~\mathrm{kcal~mol^{-1}}$ and that of $\ce{PH3}$ is $35~\mathrm{kcal~mol^{-1}}$. This has well-known stereochemical consequences in that amines are not chiral whereas phosphines can be. How can the larger inversion barrier be explained? Ammonia is the classic system for $\ce{sp^3}$ hybridisation save methane. The lone pair (and each of the $\unicode[Times]{x3C3}$-bonds) has almost $25~\%$ s-character which corresponds nicely to $\ce{sp^3}$. However, the whole system can also swing around, changing its hybridisation to $\ce{sp^2}$ and back; a process during which the lone pair is temporarily in a p-type orbital and the s-character of the bonding orbitals increases. Elements outside of the second period show a much smaller tendency to involve the s-orbital in bonding. The bonding orbitals only have an s-character of approximately $16~\%$. This also means that the phosphorous lone pair has a much higher s-character of approximately $50~\%$, while the bonding $\unicode[Times]{x3C3}$-orbitals have a larger p-character. (This also explains the much smaller $\ce{H-P-H}$ bonding angle of approximately $90^\circ$.) An orbital of high s-character has a long way to go to turn into a p-type orbital, and the three largely p-type $\unicode[Times]{x3C3}$-orbitals have an equally long way to go to give $\ce{sp^2}$-type orbitals so the interconversion and thereby the inversion of $\ce{PH3}$ is strongly hindered. Disclaimer: This answer neglects quantum tunnelling effects, which are significant in such compounds. The gist of it is that because nitrogen is smaller and lighter than phosphorus, the rate of tunnelling and hence inversion of chirality is much faster. This issue is discussed to some extent in this question. I've attached a Walsh diagram for the $D_\mathrm{3h} \leftrightarrow C_\mathrm{3v}$ transition below, which corresponds to pyramidalisation of a planar structure. (Image taken from Orbital Interactions in Chemistry by Albright, Burdett & Whangbo, 3rd ed., p 186.) Now, in order for inversion of stereochemistry to occur at N or P, the molecule has to go from a pyramidal structure, through a planar transition state, and back to the other pyramidal structure. It is kind of like flipping an umbrella inside out! The energy barrier to inversion is, of course, the difference between the equilibrium geometry (pyramidal) and the transition state (planar). Now, $\ce{NH3}$ and $\ce{PH3}$ are both 8 valence electron species. This means that the HOMO will be the $2a_1$ orbital in $C_\mathrm{3v}$ symmetry, which becomes the $\mathrm{a_2''}$ orbital in $D_\mathrm{3h}$ symmetry. As you can see, due to a loss of bonding interactions of the central $\mathrm{p}_z$ orbital with the hydrogens, the HOMO is destabilised upon planarisation. Now, let's assume that the HOMO is the predominant orbital that determines the inversion barrier. That's quite usual in discussion of Walsh diagrams. This means that the steeper the $\mathrm{2a_1} \rightarrow \mathrm{a_2''}$ slope, the harder it is for planarisation to occur, and the larger the inversion barrier. The question therefore becomes: how much is the $\mathrm{2a_1}$ orbital stabilised upon pyramidalisation? If it is stabilised more, then planarisation is harder and the inversion barrier is larger. It turns out that the stabilisation of the $\mathrm{2a_1}$ orbital can be seen as a second-order Jahn-Teller distortion. This means that, in going from $D_\mathrm{3h}$ to $C_\mathrm{3v}$, there is a loss of molecular symmetry that causes two orbitals that originally do not have the same symmetry to become the same symmetry. In this case, the two orbitals are the HOMO and the LUMO: $$\begin{array}{c\|c\|c} \text{Orbital} & \text{Symmetry in }D_\mathrm{3h} & \text{Symmetry in }C_\mathrm{3v} \\ \hline \text{HOMO} & a_2'' & a_1 \\ \text{LUMO} & a_1' & a_1 \\ \end{array}$$ It turns out that the strength of the Jahn-Teller effect is inversely proportional to the difference in energy between the two unmixed orbitals. The idea is that orbitals close in energy interact more strongly with each other - and that should be a somewhat familiar idea, because it is a central principle of MO theory. Anyway, it means that if the HOMO-LUMO gap in $D_\mathrm{3h}$ is small, there will be a large stabilisation of the HOMO upon pyramidalisation, and hence a large inversion barrier. Finally, we have come to the last part of the explanation. The $\mathrm{a_2''}$ orbital is a nonbonding $n\mathrm{p}$ orbital on the central atom, and the $\mathrm{2p}$ orbital of $\ce{N}$ is lower in energy than the $\mathrm{3p}$ orbital of $\ce{P}$. The $\mathrm{2a_1'}$ orbital is higher in energy in $\ce{NH3}$ because the nitrogen $\mathrm{2s}$ orbital has strong overlap with the hydrogen $\mathrm{1s}$ orbitals, leading to a more strongly antibonding orbital. This means that the energy gap between $\mathrm{a_2''}$ and $\mathrm{2a_1'}$ is larger in $\ce{NH3}$. As a result: there is a smallersecond-order JTE in $\ce{NH3}$ leading to lessstabilisation of the HOMO upon pyramidalisation which corresponds to a smallerenergy barrier to planarisation a smallerinversion barrier and more rapidinversion in $\ce{NH3}$. There is a little bit more discussion in the aforementioned textbook: Orbital Interactions in Chemistry, 3rd ed. pp 187-9. Your question deals more with hybridization. First we have to note that $\ce{NH3}$ has a lot more hybridization than the other molecules with hydrogen of the group e.g. $\ce{PH3}$. This means for example that $\ce{PH3}$ and $\ce{AsH3}$ have bonding angles near 90°. Because the tetraedric angle needs a $sp^3$ hybridization. The transition state of the inversion can be stabilized by a $sp^2$ hybridization which again practically happens only for the ammoniak.
My question concerns some nomenclature on page 4 of this paper. I've always viewed the Fourier transform as a continuous analogue of Fourier coefficients: the Fourier coefficients usually defined by a sum, and the Fourier transform by an integral. However, a paper I'm reading doesn't seem to reflect that idea, and I'm not sure why. Consider a lattice $\Gamma := (2 \pi \mathbb{Z})^d$ as a subset of $\mathbb{R}^d$. Then the lattice analytically dual to $\Gamma$ is given by $\Gamma^{\ast} = \mathbb{Z}^d.$ We take the fundamental domains of the lattices $\Gamma$ and $\Gamma^{\ast}$ to be $\mathcal{O} := [0, 2\pi)^d$ and $\mathcal{O}^{\ast} := [0, 1)^d$, respectively. Let $\mathrm{det}(\Gamma)$ denote the volume of the lattice $\Gamma$. The paper then says this (on page 4): For any $u \in L^2(\mathcal{O})$ and $f \in L^2(\mathbb{R}^d)$, define the Fourier coefficients and Fourier transform respectively: $$\displaystyle\hat{u}(\mathbf{\theta}) = \frac{1}{\sqrt{\mathrm{det}(\Gamma)}} \int_{\mathcal{O}}e^{-i \langle\mathbf{\theta}, \mathbf{x} \rangle}u(\mathbf{x})d\mathbf{x}, \ \ \mathbf{\theta} \in \mathcal{O}^{\ast},$$ $$\displaystyle (\mathcal{F}f)(\mathbf{\xi}) = \frac{1}{(2\pi)^{\frac{d}{2}}} \int_{\mathbb{R}^d}e^{-i \langle \xi, \mathbf{x}\rangle} f(\mathbf{x})d\mathbf{x}, \ \ \mathbf{\xi} \in \mathbb{R}^d.$$ Could someone explain the reasoning behind these differing definitions? I can understand the definition of the Fourier transform here, but I don't understand why the Fourier coefficients of $u$ have been defined in this way. Aren't they both Fourier transforms?
For the backward Euler discretization in time: $$ \left( \frac{u^{(k)}-u^{(k-1)}}{\Delta t}, v\right) + a(u^{(k)},v) = \ell(v) $$ where $a(\cdot,\cdot)$ is the bilinear operator associated with the discretization in space. I’m considering linear elliptic problems in 2D. So I rearrange the above equation: $$ \left(\frac{1}{\Delta t}u^{(k)},v\right) + a(u^{(k)},v) = \ell(v) + \left(\frac{1}{\Delta t}u^{(k-1)},v\right) $$ since $u^{(k-1)}$ is a known quantity from the previous time step. For non-time dependent problems, the convergence in space is typically on the order of $\mathcal{O}(h^{p+1})$, with $p$ being the basis degree. Now if I have a time-dependent problem as given above, how small does my time step have to be in order to test for convergence in space? If I take a large time step, the method is stable but yields error that is large, on the order of $10^{3}$. Taking my time step extremely small yields proper convergence rates, but I don’t think this tells me anything since for small $\Delta t$, the time-dependent terms will dominate and I’m essentially solving $$ \left(\frac{1}{\Delta t}u^{(k)},v\right) = \left(\frac{1}{\Delta t}u^{(k-1)},v\right) $$ and if the time step is small, we can most likely expect $u^{(k)} \approx u^{(k-1)}$, so solving the system will yield the “correct solution”. I’m just wondering if my implementation error of Backwards euler is incorrect. I know typically we solve the linear system: $$ (\mathbf{M} + \Delta t\mathbf{K})u^{(k)} = \mathbf{M}u^{(k-1)} + \Delta t\mathbf{f} $$ with $\mathbf{M}$ and $\mathbf{K}$ being the mass and stiffness matrices and $\mathbf{f}$ being the source term. But is the implementation I gave above incorrect? I originally implemented the first way I listed, then tried the second for comparison but they yield the same result in both cases, so I'm guessing that's not the issue.
№ 9 All Issues Engibaryan N. B. Ukr. Mat. Zh. - 2014. - 66, № 8. - pp. 1092–1105 We study the problems of analytic theory and the numerical-analytic solution of the integral convolution equation of the second kind $$ \begin{array}{cc}\hfill {\varepsilon}^2f(x)+{\displaystyle \underset{0}{\overset{r}{\int }}K\left(x-t\right)f(t)dt=g(x),}\hfill & \hfill x\in \left[0,r\right)\hfill \end{array}, $$ where $$ \begin{array}{cccc}\hfill \varepsilon >0,\hfill & \hfill r\le \infty, \hfill & \hfill K\in {L}_1\left(-\infty, \infty \right),\hfill & \hfill K(x)={\displaystyle \underset{a}{\overset{b}{\int }}{e}^{-\left\|x\right\|s}d\sigma (s)\ge 0.}\hfill \end{array} $$ The factorization approach is used and developed. The key role in this approach is played by the V. Ambartsumyan nonlinear equation.
CRC Part 1: Programmer vs Compiler C compilers are quite intelligent when it comes to performing optimization tricks. When compiling code with something like GCC, you get man-centuries worth of optimization passes applied before seeing the output. Unfortunately, the compiler isn't always smart enough in the way we might want and it can't optimize code in a way we might expect, forcing the programmer to perform some "obvious" tricks by hand. Sometimes, these hand tricks result in a less desirable end result. In this post, we will explore one instance where worse-looking code (to a human) results in better-looking code to a CPU. To assist us in this exploration, we are going to be discussing my favorite error-detection code: CRC32. This wonderful little algorithm shows up in all sorts of places, from ethernet, to PNG, to iSCSI, to btrfs, and beyond. It also has a number of properties which make it ideal for a blog post. It is easy to understand the requirements. There are SSE instructions to optimize it. All functions will have a signature which looks like this (sans documentation): /** Compute the CRC-32C (Castagnoli polynomial) for data in range [first, last). * * \param first Beginning of the data range. * \param last One past the end of the data. * \returns The computed checksum.*/std::uint32_t crc32(const char first, const char* last); There are two things to take note of here. First is the use of half-open range [first, last) to behave like a proper C++ function. Second is the word "Castagnoli," which is both a fried Italian dough ball and the name of a guy who invented a better transformation for 32-bit CRC (CRC is a family of algorithms, all of which use a polynomial transformation; for example, CRC-1 with polynomial $x + 1$ is commonly known as a "parity bit"). CRC-32C has better error-detection capabilities than the "classic" CRC-32 polynomial, so you will find it used more frequently in "modern" systems like ext4 and Ceph. Implementations Implementation performance will be measured through microbenchmarking. The input ranges will be of sizes $\{2^i : 8 \leq i \leq 20\}$ (aka $\{256, 512, 1024, \ldots, 1048576\}$). All tests were run on my Intel Core i7 6700K, a 4.0 GHz processor with a Skylake core. Unless otherwise specified, the compiler used was g++ 6.3.1 with the flags -O3 and -fwhole-program. The source code is available on Gist as crc32.cpp. Boost The simplest algorithm is one that has already been written for you. To that end, let's implement our CRC in terms of boost::crc_optimal. #include <boost/crc.hpp>std::uint32_t crc32_boost(const char* first, const char* last){ boost::crc_optimal<32, 0x1edc6f41, 0xffffffff, 0xffffffff, true, true> machine; machine.process_block(first, last); return machine.checksum();} Internally, boost::crc_optimal<...>::process_block uses a simple lookup table constructed on the first use of boost::crc_optimal for the provided parameters. Removing some of the generic-ness and compiler workarounds, crc32_boost ends up looking something like this: std::uint32_t crc32_boost(const char* first, const char* last){ static std::uint32_t table[256] = compute_crc32_lookups(32, 0x1edc6f41, 0xffffffff); std::uint32_t rem = ~0U; for (const char* p = first; p < last; ++p) { int lookup_index = std::uint8(rem ^ p); rem >>= 8; rem ^= table[lookup_index]; } return ~rem;} Alright, let's look at the results in terms of nanoseconds per byte of processed data. The first takeaway is CRC32 is pretty fast -- only around 2 nanoseconds per processed byte. As usual with these sorts of things, there is a steep drop as the input size grows. There is an odd spike at 32 KiB, which likely comes from my CPU having a 32 KiB L1 cache. A: Naïve crc32b Let's start improving the performance of the algorithm! To do this, we are going to use the CPU instructions specific to CRC-32C. SSE 4.2 adds a crc32 instruction, which AT&T assembly exposes in 4 instructions: crc32b (8-bit bytes), crc32w (16-bit words), crc32l (32-bit longs) and crc32q (64-bit quads). To start, lets just use the byte-sized version. #include <cstdint>#include <smmintrin.h>__attribute__((target("sse4.2")))std::uint32_t crc32a(const char first, const char* last){ std::uint32_t code = ~0U; for ( ; first != last; ++first) code = _mm_crc32_u8(code, first); return ~code;} The use of __attribute__((target("sse4.2"))) allows the crc32a function to be compiled with support for the sse4.2 target (similar to specifying -msse4.2 on the command line). This can be useful for providing fallback mechanisms when running on systems which do not support SSE 4.2. So...it's approximately 2.5 times faster than the boost::crc_optimal implementation. It has the same shape at the Boost implementation, with the small hump at 32 KiB. B: std::for_each with crc32b Let's talk templates! #include <algorithm>#include <cstdint>#include <smmintrin.h>__attribute__((target("sse4.2")))std::uint32_t crc32b(const char first, const char* last){ std::uint32_t code = ~0U; std::for_each(first, last, [&] (const char& x) { code = _mm_crc32_u8(code, x); } ); return ~code;} If you're following along at home with GCC 6.3.1 (and many other versions), you will not be able to compile that code. It will fail with a semi-aggressive compilation error that looks a bit like this: In file included from crc32.cpp:26:0:/usr/lib64/gcc/x86_64-suse-linux/6/include/smmintrin.h: In lambda function:/usr/lib64/gcc/x86_64-suse-linux/6/include/smmintrin.h:827:1: error: inlining failed in call to always_inline ‘unsigned int _mm_crc32_u8(unsigned int, unsigned char)’: target specific option mismatch_mm_crc32_u8 (unsigned int __C, unsigned char __V)crc32.cpp:65:60: note: called from here [&] (const char& x) { code = _mm_crc32_u8(code, x); } The issue here is that while the crc32b function is compiled with __attribute__((target("sse4.2"))), the lambda function crc32b(char const, char const)::{lambda(char const)#1}::operator()(char const) const is not. Unfortunately in GCC, lambda functions do not inherit the function-specific compilation options of the containing function (this works properly in Clang). The fix is to add the attribute to the lambda function as well: [&] (const char& x) __attribute__((target("sse4.2"))) { code = _mm_crc32_u8(code, x); } They're basically the same shape...I'm willing to attribute any differences to sampling error. It looks like the use of std::for_each has no effect on performance! C: Opportunistic Strides Going byte-by-byte might be easy, but this is not the most efficient way to use the SSE instructions. The entire point of SSE is to perform single instructions to process multiple data. Here, the "multiple data" refers to multiple bytes; the "single instructions" to perform this processing are the crc32_ instructions which accept larger parameters than a byte. If we want to make our CRC implementation faster, we need to process larger chunks of input at a time. Let's build on crc32a and advance in 2-, 4-, and 8-byte increments whenever it is possible. #include <cstdint>#include <smmintrin.h>__attribute__((target("sse4.2")))std::uint32_t crc32c(const char* first, const char* last){ std::uint32_t code = ~0U; for ( ; first < last; /* inline /) { if (reinterpret_cast<std::uintptr_t>(first) % 8 == 0 && first + 8 <= last) { code = _mm_crc32_u64(code, reinterpret_cast<const std::uint64_t>(first)); first += 8; } else if (reinterpret_cast<std::uintptr_t>(first) % 4 == 0 && first + 4 <= last) { code = _mm_crc32_u32(code, reinterpret_cast<const std::uint32_t>(first)); first += 4; } else if (reinterpret_cast<std::uintptr_t>(first) % 2 == 0 && first + 2 <= last) { code = _mm_crc32_u16(code, reinterpret_cast<const std::uint16_t>(first)); first += 2; } else // if (reinterpret_cast<std::uintptr_t>(first) % 1 == 0 && first + 1 <= last) { code = _mm_crc32_u8(code, reinterpret_cast<const std::uint8_t>(first)); first += 1; } } return ~code;} The algorithm is simple: Opportunistically stride by the largest legal increment or fall through to a smaller increment. The if after the final else is commented out because it will always be true, as $x \in \mathbb{N} \rightarrow x \equiv 0 \mod 1$ and $n + 1 \leq m \rightarrow n < m$. This is around 6 times faster than the byte-by-byte implementation. The 32 KiB spike is imperceptible at this point. The speedup is in the realm of reasonable expectations, as we are executing approximately $\frac{1}{8}$ the instructions. Unfortunately, it is not 8 times faster...why? D: Explicit Strides It seems like algorithm C is wasting a lot of cycles reprocessing the same information. Adding 8 to a first which matched reinterpret_cast<std::uintptr_t>(first) % 8 == 0 will continue to match the predicate. Things get worse if we are not 8 byte aligned -- if the input is aligned to byte 1, we have to perform the same checks a number of times. A Sufficiently Smart Compiler would figure this out and use that information for optimization purposes. Is GCC sufficiently smart? Let's assume GCC isn't and be explicit about the logic that we as Sufficiently Smart Computer Programmers know. #include <cstdint>#include <smmintrin.h>__attribute__((target("sse4.2")))std::uint32_t crc32d(const char first, const char* last){ std::uint32_t code = ~0U; while (first + 1 <= last && (reinterpret_cast<std::uintptr_t>(first) % 2 != 0 \|\| first + 2 > last) ) { code = _mm_crc32_u8(code, reinterpret_cast<const std::uint8_t>(first)); first += 1; } while (first + 2 <= last && (reinterpret_cast<std::uintptr_t>(first) % 4 != 0 \|\| first + 4 > last) ) { code = _mm_crc32_u16(code, reinterpret_cast<const std::uint16_t>(first)); first += 2; } while (first + 4 <= last && (reinterpret_cast<std::uintptr_t>(first) % 8 != 0 \|\| first + 8 > last) ) { code = _mm_crc32_u32(code, reinterpret_cast<const std::uint32_t>(first)); first += 4; } while (reinterpret_cast<std::uintptr_t>(first) % 8 == 0 && first + 8 <= last ) { code = _mm_crc32_u64(code, reinterpret_cast<const std::uint64_t>(first)); first += 8; } while (first + 4 <= last) { code = _mm_crc32_u32(code, reinterpret_cast<const std::uint32_t>(first)); first += 4; } while (first + 2 <= last) { code = _mm_crc32_u16(code, reinterpret_cast<const std::uint16_t>(first)); first += 2; } while (first + 1 <= last) { code = _mm_crc32_u8(code, reinterpret_cast<const std::uint8_t>(first)); first += 1; } return ~code;} Why all the whiles? Clearly, entering a block under the condition first + 1 <= last && (reinterpret_cast<std::uintptr_t>(first) % 2 != 0 \|\| first + 2 > last), then unconditionally performing first += 1 will always make it fail the condition check immediately afterward. Okay, maybe that's only clear to the compiler. It's definitely true, though...so why not just say if? The answer will come when we genericify this function (saves for part 2). While CRC offers instructions for every power of 2 along the way, there are a number of SSE instructions which only offer 8 byte varieties. In these cases, we would actually need a while loop. Don't worry -- the compiler is smart enough to figure this out (if you're curious or suspicious, you can replace the non- _mm_crc32_u64 whiles with ifs and benchmark the result). It looks like they are about the same speed. If anything, algorithm C is faster, although only by an infinitesimal amount and only for very small buffers. To figure out why, lets look at the disassembly of crc32c. _Z6crc32cPKcS0_:cmp %rsi,%rdimov $0xffffffff,%eaxjne 401799 <_Z6crc32cPKcS0_+0x29>jmp 4017eb <_Z6crc32cPKcS0_+0x7b>nopl 0x0(%rax)lea 0x8(%rdi),%rdxcmp %rdx,%rsijb 40179f <_Z6crc32cPKcS0_+0x2f>mov %eax,%eaxcrc32q (%rdi),%raxmov %rdx,%rdicmp %rdi,%rsije 4017bb <_Z6crc32cPKcS0_+0x4b>test $0x7,%dilje 401780 <_Z6crc32cPKcS0_+0x10>test $0x3,%diljne 4017c0 <_Z6crc32cPKcS0_+0x50>lea 0x4(%rdi),%rdxcmp %rdx,%rsijb 4017c0 <_Z6crc32cPKcS0_+0x50>crc32l (%rdi),%eaxmov %rdx,%rdicmp %rdi,%rsijne 401799 <_Z6crc32cPKcS0_+0x29>not %eaxretq xchg %ax,%axtest $0x1,%diljne 4017e0 <_Z6crc32cPKcS0_+0x70>lea 0x2(%rdi),%rdxcmp %rdx,%rsijb 4017e0 <_Z6crc32cPKcS0_+0x70>crc32w (%rdi),%eaxmov %rdx,%rdijmp 401794 <_Z6crc32cPKcS0_+0x24>nopw 0x0(%rax,%rax,1)crc32b (%rdi),%eaxadd $0x1,%rdijmp 401794 <_Z6crc32cPKcS0_+0x24>xor %eax,%eaxretq xchg %ax,%ax The key part of the assembly is between lea 0x8(%rdi),%rdx ( %rdx = %rdi + 8) and the je after test $0x7,%dil ( %test = 0x7 & %dil aka %dil % 8), inside of which you will find the crc32q instruction. This is where the vast majority of processing time is spent -- a 31 byte block of code. The usages of crc32l, crc32w, and crc32b all jump to the same test $0x7,%dil. In other words, the compiler is not smart enough to unroll the one-time uses for us. Or perhaps it is smart enough to do so but is so smart that it knows there would be a performance penalty. In the worst case, these blocks will only be entered 6 times (7 byte offsets at the front and end), so being a little more expensive in these cases doesn't hurt. On the other hand, algorithm D always checks for alignment of 1, 2, and 4 before getting to the core crc32q loop. This pessimism comes with a price, even for those who provide correctly-aligned ranges; algorithm C does not penalize you unless your ranges are unaligned. The resulting assembly for crc32c is a meager 128 bytes, while crc32d weighs in at a massive 560 bytes. This difference means algorithm C has less of an penalty on your instruction cache in the real world. Furthermore, C is shorter and more readable (to me). Basically, C is better than D in every aspect. Stay tuned for part 2, we will be making a generic helper algorithm to the tune of std::for_each to make writing these SSE striding functions easy.
My class was recently given an assignment by our teacher which was to find the atomic weight of Magnesium by adding $\ce{HNO3}$ and heating it until it solidified. We did it in the laboratory as an example, with $\pu{0.21g Mg}$ and found the atomic weight to be $15.272$ and the %error was $14.272$. Afterwards, he asked us how much $\ce{HNO3}$ we should theoretically have added for $\pu{0.21g Mg}$. Most people in my class have done chemistry of this level in school but I haven't and I don't understand how to do it. If someone could help me I'd appreciate it, thanks in advance. This question appears to be off-topic. The users who voted to close gave this specific reason: " Homework questionsmust demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: How do I ask homework questions on Chemistry Stack Exchange?" – Mithoron, andselisk, airhuff, bon, Jan The reaction of $\ce{HNO3}$ and $\ce{Mg}$ is $\ce{Mg + 4HNO3 -> 2NO2 + 2H2O + Mg(NO3)2}$. That being said, you had $\pu{0.21 g}$ of $\ce{Mg}$. $\pu{Mg}$ has a mean atomic mass of approximately $24.305$ (so, you should probably check your experiment). That means $\pu{0.21 g}$ of $\ce{Mg}$ is: $\frac{0.21\pu{g}}{1}\times\frac{1\pu{mole}}{24.305\pu{g}} \approx \pu{0.0086401 moles of Mg}$. For the reaction to run to completion, you need four times the mole amount for $\ce{HNO3}$ (see the reaction), which is $\pu{0.0345604 moles of }\ce{HNO3}$. Converting the moles to grams using $\pu{63.01 \frac{g}{mol}}$ gives $\fbox{2.179 grams}$ (The formula weight on nitric acid is given only to four significant figures).
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:
so Im given a discrete sum $$x[n] = \sum\limits_{r=\infty}^{+\infty}\delta[n-rN]$$ how do I calculate its discrete Fourier series coefficients? Thank you. edit: this is what i've come to up to now: Signal Processing Stack Exchange is a question and answer site for practitioners of the art and science of signal, image and video processing. It only takes a minute to sign up.Sign up to join this community $\sum\limits_{r=-\infty}^{\infty}\delta[n-rN]$ is a periodic Kronecker delta with period $N$, but the sum to get the fourier coefficients only "uses" one period $\sum\limits_{n = 0}^{N-1}$ so you don't have to worry about the periodic extension of $\delta$ and take just $x[n] = \delta[n]$ as your function. This is going to cancel out every exponential in the sum except when $n=0$, so the result will be independent of the particular coefficient and be just $a_k = 1/N$. If you draw the sequence $x[n]$ you'll see right away how it works. You have an impulse at $n=0$, at $n=-N$, at $n=N$, etc. But if you sum over only one period, no matter which period you choose, there's always only one impulse in that period. So you're left with only one sum index contributing to the final result. I'm sure you can take it from here.
Let $\mathcal{V_1}$ and $\mathcal{V_2}$ be cocomplete symmetric monoidal categories, each endowed with a cosimplicial object $\Delta^\bullet=\Delta^\bullet_{\mathcal{V}_i}:\Delta \to \mathcal{V}_i$. Denote by $\|-\|=\|-\|_{\mathcal{V}_i}:s\mathcal{V}_i\to \mathcal{V}_i$ the functor tensor product $-\otimes_\Delta \Delta^\bullet_{\mathcal{V}_i}$, i.e. the coend $$\|X_\bullet\|=X_\bullet \otimes_\Delta \Delta^\bullet = \int^n X_n \otimes \Delta^n$$ for $X_\bullet\in s\mathcal{V}_i$ a simplicial object in $\mathcal{V}_i$. Let $F:\mathcal{V}_1\to \mathcal{V}_2$ be a lax symmetric monoidal functor which is a left adjoint and such that $F(\Delta^\bullet_{\mathcal{V}_1})\cong \Delta^\bullet_{\mathcal{V}_2}$. Then there is an induced natural transformation $$\tau:\|F-\|\Rightarrow F\|-\|$$ of functors $s\mathcal{V}_1\to \mathcal{V}_2$, since $F$ preserves coends and the chosen cosimplicial objects. Now, the category $s\mathcal{V}_i$ has a symmetric monoidal structure given by the pointwise formula $(X \otimes Y)_n = X_n \otimes Y_n$, and it is such that the induced functor $F:s\mathcal{V}_1\to s\mathcal{V}_2$ is lax symmetric monoidal. Suppose further that $\|-\|:s\mathcal{V}_i\to \mathcal{V}_i$ is lax symmetric monoidal. Therefore $\tau$ is a natural transformation between lax symmetric monoidal functors, so it makes sense to ask: Is $\tau$ a lax symmetric monoidal transformation? i.e., do the following diagrams commute? I am under the impression that it won't be the case in general... Are there reasonable extra conditions to impose such that it will hold? Here is an example of the above scenario, where I'm hoping the answer is affirmative (and it may thus serve as an inspiration to extract additional conditions to ensure a general $\tau$ above to be lax symmetric monoidal). Consider $\mathcal{V}_1$ to be simplicial sets and $\mathcal{V}_2$ to be topological spaces. Endow the first one with the Yoneda embedding as a cosimplicial simplicial set, and endow the second one with the standard cosimplicial topological space. These objects yield internal geometric realizations: of a bisimplicial set into a simplicial set (which one can prove to be the diagonal functor), and of a simplicial space into a space (the standard one). As a functor $F$, we will consider the standard "extrinsic" geometric realization of a simplicial set into a topological space, which satisfies all the hypotheses above. So the question in this case is whether the natural isomorphism $\tau_{X_{\bullet,\bullet}}: \|\mathrm{diag} X_{\bullet,\bullet}\|\cong \|[n]\mapsto \|X_{n,\bullet}\|\|$ in $X_{\bullet, \bullet}\in ss\mathrm{Set}$ is (cartesian) symmetric monoidal.
In a paper by Joos and Zeh, Z Phys B 59 (1985) 223, they say:This 'coming into being of classical properties' appears related to what Heisenberg may have meant by his famous remark [7]: 'Die "Bahn" entsteht erst dadurch, dass wir sie beobachten.'Google Translate says this means something ... @EmilioPisanty Tough call. It's technical language, so you wouldn't expect every German speaker to be able to provide a correct interpretation—it calls for someone who know how German is used in talking about quantum mechanics. Litmus are a London-based space rock band formed in 2000 by Martin (bass guitar/vocals), Simon (guitar/vocals) and Ben (drums), joined the following year by Andy Thompson (keyboards, 2001–2007) and Anton (synths). Matt Thompson joined on synth (2002–2004), while Marek replaced Ben in 2003. Oli Mayne (keyboards) joined in 2008, then left in 2010, along with Anton. As of November 2012 the line-up is Martin Litmus (bass/vocals), Simon Fiddler (guitar/vocals), Marek Bublik (drums) and James Hodkinson (keyboards/effects). They are influenced by mid-1970s Hawkwind and Black Sabbath, amongst others.They... @JohnRennie Well, they repeatedly stressed their model is "trust work time" where there are no fixed hours you have to be there, but unless the rest of my team are night owls like I am I will have to adapt ;) I think u can get a rough estimate, COVFEFE is 7 characters, probability of a 7-character length string being exactly that is $(1/26)^7\approx 1.2\times 10^{-10}$ so I guess you would have to type approx a billion characters to start getting a good chance that COVFEFE appears. @ooolb Consider the hyperbolic space $H^n$ with the standard metric. Compute $$\inf\left\{\left(\int u^{2n/(n-2)}\right)^{-(n-2)/n}\left(4\frac{n-1}{n-2}\int\|\nabla u\|^2+\int Ru^2\right): u\in C^\infty_c\setminus\{0\}, u\ge0\right\}$$ @BalarkaSen sorry if you were in our discord you would know @ooolb It's unlikely to be $-\infty$ since $H^n$ has bounded geometry so Sobolev embedding works as expected. Construct a metric that blows up near infinity (incomplete is probably necessary) so that the inf is in fact $-\infty$. @Sid Eating glamorous and expensive food on a regular basis and not as a necessity would mean you're embracing consumer fetish and capitalism, yes. That doesn't inherently prevent you from being a communism, but it does have an ironic implication. @Sid Eh. I think there's plenty of room between "I think capitalism is a detrimental regime and think we could be better" and "I hate capitalism and will never go near anything associated with it", yet the former is still conceivably communist. Then we can end up with people arguing is favor "Communism" who distance themselves from, say the USSR and red China, and people who arguing in favor of "Capitalism" who distance themselves from, say the US and the Europe Union. since I come from a rock n' roll background, the first thing is that I prefer a tonal continuity. I don't like beats as much as I like a riff or something atmospheric (that's mostly why I don't like a lot of rap) I think I liked Madvillany because it had nonstandard rhyming styles and Madlib's composition Why is the graviton spin 2, beyond hand-waiving, sense is, you do the gravitational waves thing of reducing $R_{00} = 0$ to $g^{\mu \nu} g_{\rho \sigma,\mu \nu} = 0$ for a weak gravitational field in harmonic coordinates, with solution $g_{\mu \nu} = \varepsilon_{\mu \nu} e^{ikx} + \varepsilon_{\mu \nu}^* e^{-ikx}$, then magic?
Quantization of the free electro-magnetic field has essential differences in comparison to quantization of say scalar or massive vector fields. In fact there are different approches to it. One of them is a covariant approach due to Gupta and Bleuler, see Section 3-2-1 in "Quantum Field Theory" by Itzykson and Zuber. First one modifies the usual Lagrangian $-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ to $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} +\frac{\lambda}{2}(\partial\cdot A)^2.$$ Then the conjugate momenta to the four components of $A$ are $$\pi^\rho=\frac{\partial \mathcal{L}}{\partial(\partial_0A_\rho)}=F^{\rho 0}-\lambda g^{\rho 0}(\partial\cdot A).$$ Then one assumes canonical commutation relations (see (3-102) in the book): $$[A_\rho(\vec x,t),\pi_\nu(\vec y,t)]=i g_{\mu\nu}\delta^3(\vec x-\vec y).$$ For $\mu=\nu=0$ this prescription is different from the standard one known from QM as the commutator has the wrong sign. In other words the standard prescription tells to take $\delta_{\mu\nu}$ in the right hand side rather than $g_{\mu\nu}$. This new prescription seems somewhat arbitrary and unmotivated to me. I would be happy to get any explanation. REMARK. In quantization of the Dirac field one also changes the prescription known from QM by replacing commutators with anti-commutators. However this significant change is usually motivated in text books, including the one mentioned above.
Given certain function $f(X)$ which is quadratic in $X\in\mathbb{R}^{n\times d}$, $$\frac{1}{2}tr(X^TAX) - tr(Y^TBX)$$ for positive definite weighted Laplacian matrices $A, B\in\mathbb{R}^{n\times n}$, i.e. the diagonal elements are negative summation of the negative off-diagonal elements plus a constant for strict diagonal dominance: $A=\left(\begin{array}{ccccc} c_1+\sum a_{1s} & & & & \\ & \ddots & & -a_{ij} \\ & & \ddots & & \\ & -a_{ij} & & \ddots & \\ & & & & c_n+\sum a_{ns} \end{array}\right) \in\mathbb{R}^{n\times n},$ and an arbitrary $Y\in\mathbb{R}^{n\times d}$. In order to obtain its minimim, I set the gradient to $0$, $$\nabla f(X) = AX-BY \tag 1$$ so, the solution to the following linear system $$AX=BY \tag 2$$ with the unknown $X$ would give me the minimizer of $f(X)$. Since $A$ is strictly diagonally dominant, the system is solved by Jacobi iteration known to converge to exact solution in this case. However, I'm interested if only $a$ $single$ iteration of Jacobi method on some arbitrary initialization $X_0$ would yield result $X_1$ that satisfies $f(X_1)<f(X_0)$. Note that I tried many books on Jacobi method, but I was unable to make the connection with the optimization problem. The Jacobi method is known to yield "progressive better results" to the linear system, but I'm not sure what this implies precisely, and what implications it might have on the above optimization attempt. On one place I found that the Jacobi iterands $\{X_0, X_1, \dots, X_{k-1}, X_{k}\}$ satisfy $$\lVert X-X_k\rVert_2 < \lVert X-X_{k-1} \rVert_2 \tag 3$$ where $X$ is the true solution to (2). Does the proof on the convergence of Jacobi actually imply (3)? And, if so, does that imply progressively lower values of $f(X_k)$, $k\in\{0, 1, \dots, \}$? One way I tried is $f(X_1)-f(X_0)<0$, with the replacement as in Jacobi $$X_1=D^{-1}(BY-RX_0)$$ with the splitting $A=D+R$, where $D=diag(A)$, but was unable to regroup the terms. I would appreciate if the answers would contain supporting references, so that I might explore the connection more deeply.
We know that the longest common substring of two strings can be found in $\mathcal O(N^2)$ time complexity. Can a solution be found in only linear time? Let $m$ and $n$ be the lengths of two given strings, Linear time assuming the size of the alphabet is constant. Yes, the longest common substring of two given strings can be found in $O(m+n)$ time, assuming the size of the alphabet is constant. Here is an excerpt from Wikipedia article on longest common substring problem. The longest common substrings of a set of strings can be found by building a generalized suffix tree for the strings, and then finding the deepest internal nodes which have leaf nodes from all the strings in the subtree below it. Building a generalized suffix tree for two given strings takes $O(m+n)$ time using the famous ingenious Ukkonen's algorithm. Finding the deepest internal nodes that come from both strings takes $O(m+n)$ time. Hence we can find the longest common substring in $O(m+n)$ time. For a working implementation, please take a look at Suffix Tree Application 5 – Longest Common Substring at GeeksforGeeks (Improved!) Linear time In fact, the longest common substring of two given strings can be found in $O(m+n)$ time regardless of the size of the alphabet. Here is the abstract of Computing Longest Common Substrings Via Suffix Arrays by Babenko, Maxim & Starikovskaya, Tatiana. (2008). Given a set of $N$ strings $A = \{\alpha_1,\cdots,\alpha_N\}$ of total length $n$ over alphabet $\Sigma$ one may ask to find, for each $2 \le k\le N$, the longest substring $\beta$ that appears in at least $K$ strings in $A$. It is known that this problem can be solved in $O(n)$ time with the help of suffix trees. However, the resulting algorithm is rather complicated (in particular, it involves answering certain least common ancestor queries in $O(1)$ time). Also, its running time and memory consumption may depend on $\|\Sigma\|$. This paper presents an alternative, remarkably simple approach to the above problem, which relies on the notion of suffix arrays. Once the suffix array of some auxiliary $O(n)$-length string is computed, one needs a simple $O(n)$-time postprocessing to find the requested longest substring. Since a number of efficient and simple linear-time algorithms for constructing suffix arrays has been recently developed (with constant not depending on $\|\Sigma\|$), our approach seems to be quite practical. Here is the general idea of the algorithm in the paper above. Let string $\alpha$ be concatenation of all $\alpha_i$ with separating sentinels. Construct the suffix array for $α$ as well as its longest-common-prefix array. Apply a sliding window technique to these arrays to obtain the longest common substrings. Yes. There's even a Wikipedia article about it! https://en.wikipedia.org/wiki/Longest_common_substring_problem In particular, as Wikipedia explains, there is a linear-time algorithm, using suffix trees (or suffix arrays). Searching on "longest common substring" turns up that Wikipedia article as the first hit (for me). In the future, please research the problem before asking here. (See, e.g., https://meta.stackoverflow.com/q/261592/781723.)
Thank you for using the timer!We noticed you are actually not timing your practice. Click the START button first next time you use the timer.There are many benefits to timing your practice, including: Does GMAT RC seem like an uphill battle? e-GMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT (1) $\sqrt{x^2}=x$ --> $\|x\|=x$ --> $x\geq{0}$. Not sufficient. (2) $y=\sqrt{2-x}$ --> y is equal to the square root of some number, thus $y\geq{0}$. 2-x is under the square root, thus $2-x\geq{0}$ --> $x\leq{2}$. Not sufficient. (1)+(2) $0\leq{x}\leq{2}$ and $y\geq{0}$. If $x=y=1$, then the answer is YES but if $x=2$ and $y=0$, then the asnwer is NO. Not sufficient. Answer: E. I'm a bit unclear about this. Doesn't this simply mean x=x? I really didn't know what to do with this statement... if we had -4, then$\sqrt{(-4)^2}$ would still give a -4, wouldn't it? You really need to brush up fundamentals on roots and absolute values. This is basic staff! First, of all, when the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. So, $\sqrt{(-4)^2}=\sqrt{16}=4$, not -4 and not +/-4, ONLY 4! Next, about $\sqrt{x^2}=\|x\|$. The point here is that as square root function can not give negative result then $\sqrt{some \ expression}\geq{0}$. So $\sqrt{x^2}\geq{0}$. But what does $\sqrt{x^2}$ equal to? Let's consider following examples: If $x=5$ --> $\sqrt{x^2}=\sqrt{25}=5=x=positive$; If $x=-5$ --> $\sqrt{x^2}=\sqrt{25}=5=-x=positive$. So we got that: $\sqrt{x^2}=x$, if $x\geq{0}$; $\sqrt{x^2}=-x$, if $x<0$. What function does exactly the same thing? The absolute value function: $\|x\|=x$, if $x\geq{0}$ and $\|x\|=-x$, if $x<0$. That is why $\sqrt{x^2}=\|x\|$. (1) $\sqrt{x^2}=x$ --> $\|x\|=x$ --> $x\geq{0}$. Not sufficient. (2) $y=\sqrt{2-x}$ --> y is equal to the square root of some number, thus $y\geq{0}$. 2-x is under the square root, thus $2-x\geq{0}$ --> $x\leq{2}$. Not sufficient. (1)+(2) $0\leq{x}\leq{2}$ and $y\geq{0}$. If $x=y=1$, then the answer is YES but if $x=2$ and $y=0$, then the asnwer is NO. Not sufficient. Answer: E. I'm a bit unclear about this. Doesn't this simply mean x=x? I really didn't know what to do with this statement... if we had -4, then$\sqrt{(-4)^2}$ would still give a -4, wouldn't it? You really need to brush up fundamentals on roots and absolute values. This is basic staff! First, of all, when the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. So, $\sqrt{(-4)^2}=\sqrt{16}=4$, not -4 and not +/-4, ONLY 4! Next, about $\sqrt{x^2}=\|x\|$. The point here is that as square root function can not give negative result then $\sqrt{some \ expression}\geq{0}$. So $\sqrt{x^2}\geq{0}$. But what does $\sqrt{x^2}$ equal to? Let's consider following examples: If $x=5$ --> $\sqrt{x^2}=\sqrt{25}=5=x=positive$; If $x=-5$ --> $\sqrt{x^2}=\sqrt{25}=5=-x=positive$. So we got that: $\sqrt{x^2}=x$, if $x\geq{0}$; $\sqrt{x^2}=-x$, if $x<0$. What function does exactly the same thing? The absolute value function: $\|x\|=x$, if $x\geq{0}$ and $\|x\|=-x$, if $x<0$. That is why $\sqrt{x^2}=\|x\|$. thank you for taking the time out to explain... I'm studying math after a very very long time so I'm extremely rusty and most of my questions involve very very basic concepts. I just reversed the OG13 DS question one concept that when x^2=4, x can be + or-2. then perhaps underoot 4 can be + or -ve 2 as well. I read this post, and perhaps this mixed me up (or I suppose I fail to understand rule 3). Could someone please comment? http://www.manhattangmat.com/blog/2012/ ... -the-gmat/ I'm a bit unclear about this. Doesn't this simply mean x=x? I really didn't know what to do with this statement... if we had -4, then$\sqrt{(-4)^2}$ would still give a -4, wouldn't it? You really need to brush up fundamentals on roots and absolute values. This is basic staff! First, of all, when the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. So, $\sqrt{(-4)^2}=\sqrt{16}=4$, not -4 and not +/-4, ONLY 4! Next, about $\sqrt{x^2}=\|x\|$. The point here is that as square root function can not give negative result then $\sqrt{some \ expression}\geq{0}$. So $\sqrt{x^2}\geq{0}$. But what does $\sqrt{x^2}$ equal to? Let's consider following examples: If $x=5$ --> $\sqrt{x^2}=\sqrt{25}=5=x=positive$; If $x=-5$ --> $\sqrt{x^2}=\sqrt{25}=5=-x=positive$. So we got that: $\sqrt{x^2}=x$, if $x\geq{0}$; $\sqrt{x^2}=-x$, if $x<0$. What function does exactly the same thing? The absolute value function: $\|x\|=x$, if $x\geq{0}$ and $\|x\|=-x$, if $x<0$. That is why $\sqrt{x^2}=\|x\|$. thank you for taking the time out to explain... I'm studying math after a very very long time so I'm extremely rusty and most of my questions involve very very basic concepts. I just reversed the OG13 DS question one concept that when x^2=4, x can be + or-2. then perhaps underoot 4 can be + or -ve 2 as well. I read this post, and perhaps this mixed me up (or I suppose I fail to understand rule 3). Could someone please comment? http://www.manhattangmat.com/blog/2012/ ... -the-gmat/ Rule 3 there says that $\sqrt{x^2}=3$ means that $x=3$ or $x=-3$. $\sqrt{x^2}=3$ --> square: $x^2=9$ --> $x=3$ or $x=-3$. Or: $\sqrt{x^2}=3$ --> $\|x\|=3$ --> $x=3$ or $x=-3$. I'm sorry, I still don't see it. if they had written option one =-x (in the original question), then i could safely say x is negative, right? Since they say it equals x, then x must be positive. Is that the jist of what your saying? Also, is it reasonable to assume (as mgmat says) that when we have an actual number under the square root, take only the positive root. but if we have a variable, take positive and negative both? I'm sorry, I still don't see it. if they had written option one =-x (in the original question), then i could safely say x is negative, right? Since they say it equals x, then x must be positive. Is that the jist of what your saying? Also, is it reasonable to assume (as mgmat says) that when we have an actual number under the square root, take only the positive root. but if we have a variable, take positive and negative both? If it were $\sqrt{x^2}=-x$, then it would mean that \|x\| = -x, thus $x\leq{0}$. Try to play with numbers there and it might become clearer._________________ (1) $\sqrt{x^2}=x$ --> $\|x\|=x$ --> $x\geq{0}$. Not sufficient. (2) $y=\sqrt{2-x}$ --> y is equal to the square root of some number, thus $y\geq{0}$. 2-x is under the square root, thus $2-x\geq{0}$ --> $x\leq{2}$. Not sufficient. (1)+(2) $0\leq{x}\leq{2}$ and $y\geq{0}$. If $x=y=1$, then the answer is YES but if $x=2$ and $y=0$, then the asnwer is NO. Not sufficient. Answer: E. Can't y be the root of a negative number? Let's say x = 5 the y = $\sqrt{2-5}$ = $\sqrt{-3}$ I know this concept is tested on GMAT. The GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers so even roots from negative number are undefined on the GMAT: $\sqrt[{even}]{negative}=undefined$, for example $\sqrt{-25}=undefined$._________________ How does this lead you to believe that y must be positive? Couldn't it be undefined as you said in another comment? Hi, As you say y can be +ive or undefined number.. BUT GMAT uses only real numbers and undefined numbers are not tested in GMAT.. so we don't cater for any unreal numbers and assume/accept only the REAL values .. if you take x=3.. y=$\sqrt{2-3}$=$\sqrt{-1}$.. But what happens to y, y becomes undefined number.. however if a variable is given, it has to be real so this vakue is not correct.._________________ Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email._________________
Search Now showing items 1-2 of 2 Search for new resonances in $W\gamma$ and $Z\gamma$ Final States in $pp$ Collisions at $\sqrt{s}=8\,\mathrm{TeV}$ with the ATLAS Detector (Elsevier, 2014-11-10) This letter presents a search for new resonances decaying to final states with a vector boson produced in association with a high transverse momentum photon, $V\gamma$, with $V= W(\rightarrow \ell \nu)$ or $Z(\rightarrow ... Fiducial and differential cross sections of Higgs boson production measured in the four-lepton decay channel in $\boldsymbol{pp}$ collisions at $\boldsymbol{\sqrt{s}}$ = 8 TeV with the ATLAS detector (Elsevier, 2014-11-10) Measurements of fiducial and differential cross sections of Higgs boson production in the ${H \rightarrow ZZ ^{*}\rightarrow 4\ell}$ decay channel are presented. The cross sections are determined within a fiducial phase ...
I want to calculate the fourier-legendre series coefficients $C_n$ of a bump function $f(x)$ defined as: \begin{equation} f(x)= \{\begin{array}[ll] a\exp[-\dfrac{1}{0.25-x^2}] & \text{if } \|x\|<0.5 \\ \quad \quad \quad 0 & \text{if } \|x\|\geq 0.5\\ \end{array} \end{equation} for: \begin{equation} f(x)=\sum_{n}C_nP^m_n \end{equation} where $P^m_n$ are the asociated legendre polynomials. In the code below, I have set m=0, and am attempting to calculate the first 15 coefficients in the sum. Here is my code: g[x_] := Piecewise[{{Exp[-1/(0.25 - x^2)], Abs[x] < 0.5}, {0, Abs[x] >= 0.5}}] Plot[g[x], {x, -2, 2}, PlotRange -> Full] j = 0; imax = 15; cn = ParallelTable[Integrate[g[x]LegendreP[i, j, x], {x, -1, 1}]/(Sqrt[Integrate[LegendreP[i, j, x]LegendreP[i, j, x], {x, -1, 1}]])^2, {i, j, imax}]; approx = Sum[cn[[i]]*LegendreP[i + j - 1, j, x], {i, 1, imax - j}]; Plot[approx, {x, -1, 1}, PlotRange -> Full] DiscretePlot[cn[[i]], {i, j, imax}] My problem is that it returns an error saying "Invalid integration variable or limit(s) in {-0.999959,-1,1}." I've tried changing the limits of integration to not include this strange value but to no avail. I changed the piecewise function to something much less complicated but still remaining piecewise and it integrated just fine, so it isn't the fact that it's piecewise, I don't think. I also tried changing the inequalities to include the value 0.5 in the non zero part of the piecewise function but that also didn't help. Does it just not like integrating this specific function? I assume it's something strange that mathematica has a hard time handling. Any suggestions? Any help is appreciated.
I am working on generating a (complex) solenoidal vector field $\mathbf{A}$ from a prescribed (complex) vector $\mathbf{a}$ and the gradient of a scalar, $b$, such that $$\mathbf{A} = \mathbf{a} + \nabla b.$$ The condition that I require is then $$\nabla\cdot\mathbf{A} = \nabla\cdot\mathbf{a} + \nabla^2 b = 0 \qquad \Longrightarrow \qquad \nabla^2 b = - B,$$ where $B = \nabla\cdot\mathbf{a}$. To complicate this slightly, I want to decompose $\mathbf{A},\mathbf{a}, B$ and $b$ into $$\mathbf{A} = \hat{\mathbf{A}} \exp[i(\omega t - \mathbf{k}\cdot\mathbf{x})] = \hat{\mathbf{A}} \exp[i\phi],\\ \mathbf{a} = \hat{\mathbf{a}} \exp[i\phi],\\ B = \hat{B} \exp[i\phi],\\ b = \hat{b} \exp[i\phi].$$ With this decomposition, $B = (\nabla\cdot\hat{\mathbf{a}} - i\mathbf{k}\cdot\hat{\mathbf{a}}) \exp[i\phi]$, and the Poisson equation above becomes $$(\dagger)\qquad\nabla^2 b = (\nabla^2 \hat{b} - 2i\mathbf{k}\cdot\nabla b - \lvert\mathbf{k}\rvert^2 b) \exp[i\phi] = -\hat{B} \exp[i\phi].$$ This is then solved for $\hat{b}$ iteratively using the following solver scheme (given for 1-dimension, but it is easily extended to higher dimensions): $$(\ddagger)\qquad \hat{b}_j = \frac{\Delta x^2 \hat{B}_j + \hat{b}_{j+1} + \hat{b}_{j-1} - ik\Delta x(\hat{b}_{j+1} - \hat{b}_{j-1})}{2 + k^2 \Delta x^2},$$ where $j \in \{2,\ldots,N-1\}$ indicates the discrete grid location, the position $x_j = j \Delta x$, and $\Delta x$ is the grid spacing. The boundary $j=1,N$ is fixed at zero. $\hat{B}_j$ is evaluated as $\hat{B}_j = (\hat{a}_{j+1} - \hat{a}_{j-1})/2\Delta x -ik\hat{a}_j$. The numerical solution to $\hat{b}$ produced satisfies $$\nabla^2 \hat{b} - 2i\mathbf{k}\cdot\nabla b - \lvert\mathbf{k}\rvert^2 b = -\hat{B}$$ very well, however if I construct $$\hat{A}_j = \hat{a}_j + \frac{\hat{b}_{j+1} - \hat{b}_{j-1}}{2\Delta{x}} - ik\hat{b}_j$$ and evaluate $$\nabla \cdot \mathbf{A} \simeq \Big(\frac{\hat{A}_{j+1} - \hat{A}_{j-1}}{2\Delta x} - ik \hat{A}_j\Big) \exp[i\phi],$$ then this is not only non-zero but also large. Putting the expression for $\hat{A}_j$ in terms of $\hat{a}$ and $\hat{b}$, one finds that the above equation reads $$()\qquad \nabla \cdot \mathbf{A} \simeq \Big( \frac{\hat{a}_{j+1} - \hat{a}_{j-1}}{2\Delta x} - ik \hat{a}_j + \frac{\hat{b}_{j+2} - 2\hat{b}_{j} + \hat{b}_{j-2}}{4\Delta x^2} - 2ik\frac{\hat{b}_{j+1} - \hat{b}_{j-1}}{2\Delta x} - k^2\hat{b}_j \Big) \exp[i\phi]\\ = \Big( \hat{B} + \nabla_2^2\hat{b} - 2ik \frac{\partial\hat{b}}{\partial x} - 4k^2 \hat{b} \Big) \exp[i\phi].$$ This is just the 1-dimensional version of $(\dagger)$, but with the Laplacian operator evaluated as $$\nabla_2^2 \hat{b} = \frac{\hat{b}_{j+2} - 2\hat{b}_j + \hat{b}_{j-2}}{4\Delta x^2},$$ suggesting that it is this difference which is responsible for $\nabla \cdot \mathbf{A} \neq 0$. To overcome this, I attempted to invert ($$) rather than use ($\ddagger$), this time with boundary conditions $\hat{b}_j = 0$ for $j = 1, 2, N-1, N$, but the (iterative) solution to the Poisson equation diverges. I was wondering what is responsible for this? Could my setting the two boundary cells to zero cause such a problem? The solver scheme for point $j$ involves $j+2, j+1, j-1, j-2$. Given the non-solenoidal vector $\mathbf{a}$, is there a better way to find a scalar $b$ such that $\nabla\cdot(\mathbf{a} + \nabla b) = 0$? Thank you for any help or suggestions!
The feature that makes LaTeX the right editing tool for scientific documents is the ability to render complex mathematical expressions. This article explains the basic commands to display equations. Contents Basic equations in LaTeX can be easily "programmed", for example: The well known Pythagorean theorem $x^2 + y^2 = z^2$ was proved to be invalid for other exponents. Meaning the next equation has no integer solutions: \[ x^n + y^n = z^n \] As you see, the way the equations are displayed depends on the delimiter, in this case \[ \] and . LaTeX allows two writing modes for mathematical expressions: the inline mode and the display mode. The first one is used to write formulas that are part of a text. The second one is used to write expressions that are not part of a text or paragraph, and are therefore put on separate lines. Let's see an example of the inline mode: In physics, the mass-energy equivalence is stated by the equation $E=mc^2$, discovered in 1905 by Albert Einstein. To put your equations in inline mode use one of these delimiters: , $ $ or \begin{math} \end{math}. They all work and the choice is a matter of taste. The displayed mode has two versions: numbered and unnumbered. The mass-energy equivalence is described by the famous equation \[E=mc^2\] discovered in 1905 by Albert Einstein. In natural units ($c$ = 1), the formula expresses the identity \begin{equation} E=m \end{equation} To print your equations in display mode use one of these delimiters: \[ \], \begin{displaymath} \end{displaymath} or \begin{equation} \end{equation} Important Note: equation* environment is provided by an external package, consult the amsmath article. Below is a table with some common maths symbols. For a more complete list see the List of Greek letters and math symbols: description code examples Greek letters \alpha \beta \gamma \rho \sigma \delta \epsilon $$ \alpha \ \beta \ \gamma \ \rho \ \sigma \ \delta \ \epsilon $$ Binary operators \times \otimes \oplus \cup \cap Relation operators < > \subset \supset \subseteq \supseteq Others \int \oint \sum \prod The mathematics mode in LaTeX is very flexible and powerful, there is much more that can be done with it:
The answer to the connected component version is yes, though the only proof I can come up with seems like more trouble than it would've been worth in math camp homework! Two steps: If there there are finitely many components of $S^2 \smallsetminus Z$, produce a finite-diameter bipartite tree on which the antipodal map on $S^2$ induces an involution fixing a point. Approximate any $f$ by functions satsifying step 1. Most of the work seems to be a part of Step 1 that vaguely sounded like the Jordan curve theorem: Key Fact. Let $X$ be a connected, compact, locally path-connected metric space with zero first homology. For an open subset $U$, inclusion induces a bijection between the connected components of $\partial U = \bar{U} \smallsetminus U$ and those of $X \smallsetminus U$. Proof of Key Fact: Mayer-Vietoris and some point-set debris Given $\epsilon > 0$, let $U_\epsilon$ be the set of points less than $\epsilon$ away from the closure $\bar{U}$ of $U$; and let $V_\epsilon$ be the set less than $\epsilon$ away from $X \smallsetminus U$. The Mayer-Vietoris sequence in reduced homology ends with $$ H_1(X) \to \tilde{H}_0(U_\epsilon \cap V_\epsilon) \to \tilde{H}_0(U_\epsilon) \oplus \tilde{H}_0(V_\epsilon) \to \tilde{H}_0(X) , $$ which gives a bijection between the path components of $U_\epsilon \cap V_\epsilon$ and those of $V_\epsilon$. Pass to connected components of their closures using: A connected open set $W$ in a locally path-connected space is path-connected. Proof. Each point of $W$ is has a path-connected open neighborhood in $W$; so the path components form a partition of $W$ by open sets. The closure of a connected open set $W$ is connected. Proof. In a partition of $\bar{W}$ by clopen subsets, whichever one contains $W$ contains $\bar{W}$. Then send $\epsilon$ to $0$. That is, since $$\begin{align}X \smallsetminus U &= \bigcap_{\epsilon > 0} \bar{V}_\epsilon & &\text{and} &\partial U &= \bigcap_{\epsilon > 0} \overline{U_\epsilon \cap V_\epsilon} ,\end{align}$$ the Key Fact is proven if we can biject connected components of $X \smallsetminus U$ with nested sequences of components $C_n$ of $\bar{V}_{1/n}$ (and similarly for $\partial U$): Lemma. In a compact space $X$, the intersection $C$ of nested closed connected sets $C_1 \supseteq C_2 \supseteq \cdots$ is connected. Proof. Given disjoint open subsets $U$ and $V$ of $X$ that cover $C$, their union and the sequence $X \smallsetminus C_n$ form an open cover of $X$—which has a finite subcover, so some $C_n$ lies in $U \cup V$. Since $C_n$ is connected, it lies entirely in one of $U$ or $V$; and so does $C$. Step 1: Turn connectivity data into a tree Claim. Let $Z$ be the zero locus of a continuous odd $f: S^2 \to \mathbb{R}^2$. If $S^2 \smallsetminus Z$ has finitely many connected components, then the antipodal map on $S^2$ sends some component of $Z$ to itself. Let $G$ be the graph where: Vertices are connected components of $Z$ and $S^2 \smallsetminus Z$, the sets of which we respectively denote $V_0$ and $V_{\neq 0}$. Two vertices are joined by an edge if their union is a connected set in $S^2$—equivalently, if $K \in V_0$ meets the closure of $U \in V_{\neq 0}$. $G$ is bipartite with parts $V_0$ and $V_{\neq 0}$, and the assumption of Step 1 is that $V_{\neq 0}$ is finite; so—since $S^2$ is connected—$G$ is connected with finite diameter. By the Key Fact, every vertex in $V_{\neq 0}$ is a cut vertex, which makes $G$ a tree. Every involution of a finite-diameter tree preserves either an edge or a vertex; and since the antipodal map can't exchange the parts of $G$ or fix any member of $V_{\neq 0}$, it has to fix exactly one vertex in $V_0$. This is the connected component of $Z$ we're after. Step 2: Infinitely many components in $S^2 \smallsetminus Z$ is okay Approximate $f$ by a sequence $f_n$ where $f_n$ is equal to $f$ on the connected components of $\{f \neq 0\}$ admitting an open ball of radius $\pi/n$ and zero everywhere else. The area of an open ball of radius $\pi/n$ on $S^2$ is at least $4\pi/n^2$ (the area of a sphere of circumference $2\pi/n$). So each $f_n$ has no more than $n^2$ components to its nonzero locus; and running them through Step 1 yields a sequence of nested closed connected sets $C_1 \supseteq C_2 \supseteq \cdots$ preserved by the antipodal map. Then the intersection of the sets $C_n$ is preserved by the antipodal map, in $Z$ (every component of $S^2 \smallsetminus Z$, being open, contains an open ball), and connected (by the Lemma before Step 1).
Background For definiteness (even though this is a categorical question!) let's agree that a vector space is a finite-dimensional real vector space and that an associative algebra is a finite-dimensional real unital associative algebra. Let $V$ be a vector space with a nondegenerate symmetric bilinear form $B$ and let $Q(x) = B(x,x)$ be the associated quadratic form. Let's call the pair $(V,Q)$ a quadratic vector space. Let $A$ be an associative algebra and let's say that a linear map $\phi:V \to A$ is Clifford if$$\phi(x)^2 = - Q(x) 1_A,$$where $1_A$ is the unit in $A$. One way to define the Clifford algebra associated to $(V,Q)$ is to say that it is universal for Clifford maps from $(V,Q)$. Categorically, one defines a category whose objects are pairs $(\phi,A)$ consisting of an associative algebra $A$ and a Clifford map $\phi: V \to A$ and whose arrows$$h:(\phi,A)\to (\phi',A')$$are morphisms $h: A \to A'$ of associative algebras such that the obvious triangle commutes:$$h \circ \phi = \phi'.$$Then the Clifford algebra of $(V,Q)$ is the universal initial object in this category. In other words, it is a pair $(i,Cl(V,Q))$ where $Cl(V,Q)$ is an associative algebra and $i:V \to Cl(V,Q)$ is a Clifford map, such that for every Clifford map $\phi:V \to A$, there is a unique morphism$$\Phi: Cl(V,Q) \to A$$extending $\phi$; that is, such that $\Phi \circ i = \phi$. (This is the usual definition one can find, say, in the nLab.) Question I would like to view the construction of the Clifford algebra as a functor from the category of quadratic vector spaces to the category of associative algebras. The universal property says that if $(V,Q)$ is a quadratic vector space and $A$ is an associative algebra, then there is a bijection of hom-sets $$\mathrm{hom}_{\mathbf{Assoc}}(Cl(V,Q), A) \cong \mathrm{cl-hom}(V,A)$$ where the left-hand side are the associative algebra morphisms and the right-hand side are the Clifford morphisms. My question is whether I can view $Cl$ as an adjoint functor in some way. In other words, is there some category $\mathbf{C}$ such that the right-side is $$\mathrm{hom}_{\mathbf{C}}((V,Q), F(A))$$ for some functor $F$ from associative algebras to $\mathbf{C}$. Naively I'd say $\mathbf{C}$ ought to be the category of quadratic vector spaces, but I cannot think of a suitable $F$. I apologise if this question is a little vague. I'm not a very categorical person, but I'm preparing some notes for a graduate course on spin geometry next semester and the question arose in my mind.
Suppose $\Omega\subset \mathbb{R}^n$ is bounded Lipschitz domain. When $1\leq p < n$, apparently $p< p^* = np/(n-p)$, hence $$ W^{1,p}(\Omega )\subset \subset L^{p}(\Omega ), $$ and for any $u\in W^{1,p}(\Omega)$ $$ \\|u\\|_{L^p}\leq C \\|u \\|_{W^{1,p}}\;. $$ To bound using the seminorm like the following Poincare inequality $$ \\|u\\|_{L^p}\leq C \|u \|_{W^{1,p}} := C \left(\int_{\Omega} \|\nabla u\|^p\right)^{1/p}\;,\tag{$\dagger$} $$ we need some extra conditions like $$ u = 0 \;\text{ on }\partial \Omega,\tag{1} $$ or $$ \int_{\Omega}u = 0,\tag{2} $$ or $$ \int_{U} u =0, \text{ where } U\subset \Omega, \text{ and }\operatorname{meas}_{n}(U) \neq 0. \tag{3} $$ The proof is pretty standard by using the sequential compactness to reach a contradiction. Normally the conditions like (1), (2), or (3) avoid using pointwise value because $W^{1,p}(\Omega)$ may not necessarily embedded in $C^{0,\alpha}$ spaces. Here if we assume the extra continuous condition like $u$ being continuous, I could prove the following: Proposition 1: $u\in V:= W^{1,p}(\Omega)\cap C^0(\overline{\Omega})$, $u(z) = 0$ for some $z\in \overline{\Omega}$, then ($\dagger$) is true: $$ \\|u\\|_{L^p}\leq C \|u \|_{W^{1,p}}\;.\tag{$\dagger$} $$ Proof: Assume otherwise, then there exists a sequence $\{v_n\}\subset V$ such that $$\\|v_n\\|_{L^p} = 1, \;\text{ and }\; \|v_n \|_{W^{1,p}} < \frac{1}{n}.$$Now by Rellich-Kondrachov compactness theorem, there exists a subsequence $\{v_m\}\subset \{v_n\}$ such that$$v_m\to v \; \text{ in } \\|\cdot\\|_{L^p}.$$Moreover, the $L^p$ convergence further implies there exists a subsequence $\{v_j\}\subset \{v_m\}$ converging a.e., and by $V$ itself being $C^0$, we have$$v_j\to v \; \text{ in } \\|\cdot\\|_{L^p}, \; \text{ and }v(z) = 0.$$By triangle inequality it is straightforward to verify that $$\\|v\\|_{L^p} = 1.\tag{4}$$ Now, by integrating against a smooth test function $\phi\in C^{\infty}_c(\Omega)$, $\|v_j\|_{W^{1,p}}<1/j$, and bounded convergence theorem, we have $$ \int_{\Omega} v\,\partial_{x_i} \phi = \int_{\Omega} \lim_{j\to \infty}v_j\,\partial_{x_i} \phi \lim_{j\to \infty}\int_{\Omega} v_j\,\partial_{x_i} \phi =-\lim_{j\to \infty}\int_{\Omega} \partial_{x_i} v_j\,\phi = 0. $$ As a result, the limit $v$ is a constant, also $v$ now must be 0 due to $v(z) = 0$. This contradicts with (4). Q.E.D. The result first does not look quite right, since no point values are allowed in general (as aforementioned conditions (1), (2), and (3) do). So now my question is: Is Proposition 1 correct? If not, is there any counterexample in simple domains like a ball?
Suppose I have a linear system $$ \left\lbrack \begin{array}{cc} M_1& S\\ S^{\mathrm{T}}& M_2 \end{array} \right\rbrack \left\lbrack \begin{array}{c} X\\ Y\end{array} \right\rbrack= \left\lbrack \begin{array}{c} F\\G\end{array}\right\rbrack, $$ where $M_1\in \mathrm{R}^{n\times n}$ is symmetric positive definite, $M_2\in \mathrm{R}^{m\times m}$ is symmetric positive semi-definite. $S$ is a full-rank matrix of compatible size. $X,Y,F,G$ are all vectors of appropriate size. What factors are relevant in deciding between using GMRES on the big system and using a Schurr complement technique? Factors that I would guess are relevant are how expensive $M_1$ is to invert, the condition number of $S$, and whether $M_2$ is the zero matrix. Any links to the literature are appreciated. Edit I am asking this question because I need to solve the following system. $$ \left\lbrack \begin{array}{cccc} M_1& S &0 &\Gamma^{\mathrm{T}} \\ -S^{\mathrm{T}}& 0 &\Xi^{\mathrm{T}}& 0\\ 0 & \Xi & 0 &0 \\ \Gamma&0&0&0 \end{array} \right\rbrack \left\lbrack \begin{array}{c} x_1\\x_2\\x_3\\x_4 \end{array} \right\rbrack= \left\lbrack \begin{array}{c} f_1\\f_2\\f_3\\f_4 \end{array} \right\rbrack $$ Right now, I solve this system using the Schurr complement method but it requires three iterative CG solves. I am wondering if it would be better to just "brute force it" and use GMRES.
Challenge Given an integer, \$s\$, as input where \$s\geq 1\$ output the value of \$\zeta(s)\$ (Where \$\zeta(x)\$ represents the Riemann Zeta Function). Further information \$\zeta(s)\$ is defined as: $$\zeta(s) = \sum\limits^\infty_{n=1}\frac{1}{n^s}$$ You should output your answer to 5 decimal places (no more, no less). If the answer comes out to be infinity, you should output \$\infty\$ or equivalent in your language. Riemann Zeta built-ins are allowed, but it's less fun to do it that way ;) Examples Outputs must be exactly as shown below Input -> Output1 -> ∞ or inf etc.2 -> 1.644933 -> 1.202064 -> 1.082328 -> 1.0040819 -> 1.00000 Bounty As consolation for allowing built-ins, I will offer a 100-rep bounty to the shortest answer which does not use built-in zeta functions. (The green checkmark will still go to the shortest solution overall) Winning The shortest code in bytes wins.
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 """ Definition of the variance-constrained semi-grand canonical ensemble class. """ import numpy as np from ase import Atoms from ase.data import atomic_numbers, chemical_symbols from ase.units import kB from typing import Dict, Union, List from .. import DataContainer from ..calculators.base_calculator import BaseCalculator from .thermodynamic_base_ensemble import ThermodynamicBaseEnsemble class VCSGCEnsemble(ThermodynamicBaseEnsemble): """Instances of this class allow one to simulate systems in the variance-constrained semi-grand canonical (VCSGC) ensemble (:math:`N\\phi\\kappa VT`), i.e. at constant temperature (:math:`T`), total number of sites (:math:`N=\\sum_i N_i`), and two additional dimensionless parameters :math:`\\phi` and :math:`\\kappa`, which constrain average and variance of the concentration, respectively. The below examples treat the binary case, but the generalization of to ternaries and higher-order systems is straight-forward. The probability for a particular state in the VCSGC ensemble for a :math:`2`-component system can be written .. math:: \\rho_{\\text{VCSGC}} \\propto \\exp\\Big[ - E / k_B T + \\kappa N ( c_1 + \\phi_1 / 2 )^2 \\Big], where :math:`c_1` represents the concentration of species 1, i.e. :math:`c_1=N_1/N`. (Please note that the quantities :math:`\\kappa` and :math:`\\phi` correspond, respectively, to :math:`\\bar{\\kappa}` and :math:`\\bar{\\phi}` in [SadErh12]_.). The :math:`\\phi` may refer to any of the two species. If :math:`\\phi` is specified for species A, an equivalent simulation can be carried out by specifying :math:`\\phi_B` as :math:`-2-\\phi_A`. In general, simulations of :math:`N`-component systems requires the specification of :math:`\\phi` for :math:`N-1` elements. Just like the :ref:`semi-grand canonical ensemble <canonical_ensemble>`, the VCSGC ensemble allows concentrations to change. A trial step consists of changing the identity of a randomly chosen atom and accepting the change with probability .. math:: P = \\min \\{ 1, \\, \\exp [ - \\Delta E / k_B T + \\kappa N \\Delta c_1 (\\phi_1 + \\Delta c_1 + 2 c_1 ) ] \\}. Note that for a sufficiently large value of :math:`\\kappa`, say 200, the probability density :math:`\\rho_{\\text{VCSGC}}` is sharply peaked around :math:`c_1=-\\phi_1 / 2`. In practice, this means that we can gradually change :math:`\\phi_1` from (using some margins) :math:`-2.1` to :math:`0.1` and take the system continuously from :math:`c_1 = 0` to :math:`1`. The parameter :math:`\\kappa` constrains the fluctuations (or the variance) of the concentration at each value of :math:`\\phi_1`, with higher values of :math:`\\kappa` meaning less fluctuations. Unlike the :ref:`semi-grand canonical ensemble <vcsgc_ensemble>`, one value of :math:`\\phi_1` maps to one and only one concentration also in multiphase regions. Since the derivative of the canonical free energy can be expressed in terms of parameters and observables of the VCSGC ensemble, .. math:: k_B T \\kappa ( \\phi_1 + 2 \\langle c_1 \\rangle ) = - \\frac{1}{N} \\frac{\\partial F}{\\partial c_1} (N, V, T, \\langle c_1 \\rangle ), this ensemble allows for thermodynamic integration across multiphase regions. This means that we can construct phase diagrams by directly comparing the free energies of the different phases. This often makes the VCSGC ensemble more convenient than the :ref:`semi-grand canonical ensemble <sgc_ensemble>` when simulating materials with multiphase regions, such as alloys with miscibility gaps. When using the VCSGC ensemble, please cite Sadigh, B. and Erhart, P., Phys. Rev. B 86, 134204 (2012) [SadErh12]_. Parameters ---------- structure : :class:`Atoms <ase.Atoms>` atomic configuration to be used in the Monte Carlo simulation; also defines the initial occupation vector calculator : :class:`BaseCalculator <mchammer.calculators.ClusterExpansionCalculator>` calculator to be used for calculating the potential changes that enter the evaluation of the Metropolis criterion temperature : float temperature :math:`T` in appropriate units [commonly Kelvin] phis : Dict[str, float] average constraint parameters :math:`\\phi_i`; the key denotes the species; for a N-component sublattice, there should be N - 1 different `\\phi_i` (referred to as :math:`\\bar{\\phi}` in [SadErh12]_) kappa : float parameter that constrains the variance of the concentration (referred to as :math:`\\bar{\\kappa}` in [SadErh12]_) boltzmann_constant : float Boltzmann constant :math:`k_B` in appropriate units, i.e. units that are consistent with the underlying cluster expansion and the temperature units [default: eV/K] user_tag : str human-readable tag for ensemble [default: None] data_container : str name of file the data container associated with the ensemble will be written to; if the file exists it will be read, the data container will be appended, and the file will be updated/overwritten random_seed : int seed for the random number generator used in the Monte Carlo simulation ensemble_data_write_interval : int interval at which data is written to the data container; this includes for example the current value of the calculator (i.e. usually the energy) as well as ensembles specific fields such as temperature or the number of atoms of different species data_container_write_period : float period in units of seconds at which the data container is written to file; writing periodically to file provides both a way to examine the progress of the simulation and to back up the data [default: np.inf] trajectory_write_interval : int interval at which the current occupation vector of the atomic configuration is written to the data container. sublattice_probabilities : List[float] probability for picking a sublattice when doing a random flip. The list should be as long as the number of sublattices and should sum up to 1. Example ------- The following snippet illustrate how to carry out a simple Monte Carlo simulation in the variance-constrained semi-canonical ensemble. Here, the parameters of the cluster expansion are set to emulate a simple Ising model in order to obtain an example that can be run without modification. In practice, one should of course use a proper cluster expansion:: from ase.build import bulk from icet import ClusterExpansion, ClusterSpace from mchammer.calculators import ClusterExpansionCalculator from mchammer.ensembles import VCSGCEnsemble # prepare cluster expansion # the setup emulates a second nearest-neighbor (NN) Ising model # (zerolet and singlet ECIs are zero; only first and second neighbor # pairs are included) prim = bulk('Au') cs = ClusterSpace(prim, cutoffs=[4.3], chemical_symbols=['Ag', 'Au']) ce = ClusterExpansion(cs, [0, 0, 0.1, -0.02]) # set up and run MC simulation structure = prim.repeat(3) calc = ClusterExpansionCalculator(structure, ce) phi = 0.6 mc = VCSGCEnsemble(structure=structure, calculator=calc, temperature=600, data_container='myrun_vcsgc.dc', phis={'Au': phi}, kappa=200) mc.run(100) # carry out 100 trial swaps """ def __init__(self, structure: Atoms, calculator: BaseCalculator, temperature: float, phis: Dict[str, float], kappa: float, boltzmann_constant: float = kB, user_tag: str = None, data_container: DataContainer = None, random_seed: int = None, data_container_write_period: float = np.inf, ensemble_data_write_interval: int = None, trajectory_write_interval: int = None, sublattice_probabilities: List[float] = None) -> None: self._ensemble_parameters = dict(temperature=temperature, kappa=kappa) self._boltzmann_constant = boltzmann_constant # Save ensemble parameters for sym, phi in phis.items(): if isinstance(sym, str): chemical_symbol = sym else: chemical_symbol = chemical_symbols[sym] phi_sym = 'phi_{}'.format(chemical_symbol) self._ensemble_parameters[phi_sym] = phi super().__init__( structure=structure, calculator=calculator, user_tag=user_tag, data_container=data_container, random_seed=random_seed, data_container_write_period=data_container_write_period, ensemble_data_write_interval=ensemble_data_write_interval, trajectory_write_interval=trajectory_write_interval, boltzmann_constant=boltzmann_constant ) # Save phis (need self.configuration to check sublattices so # we do it last) self._phis = get_phis(phis) # Check that each sublattice has N - 1 phis for sl in self.sublattices.active_sublattices: count_specified_elements = 0 for number in sl.atomic_numbers: if number in self._phis.keys(): count_specified_elements += 1 if count_specified_elements != len(sl.atomic_numbers) - 1: raise ValueError('phis must be set for N - 1 elements on a ' 'sublattice with N elements') 216 ↛ 219line 216 didn't jump to line 219, because the condition on line 216 was never false if sublattice_probabilities is None: self._flip_sublattice_probabilities = self._get_flip_sublattice_probabilities() else: self._flip_sublattice_probabilities = sublattice_probabilities def _do_trial_step(self): """ Carries out one Monte Carlo trial step. """ sublattice_index = self.get_random_sublattice_index( probability_distribution=self._flip_sublattice_probabilities) return self.do_vcsgc_flip( phis=self.phis, kappa=self.kappa, sublattice_index=sublattice_index) @property def temperature(self) -> float: """ temperature :math:`T` (see parameters section above) """ return self.ensemble_parameters['temperature'] @property def phis(self) -> Dict[int, float]: """ phis :math:`\\phi_i`, one for each species but their sum must be :math:`-2.0` (referred to as :math:`\\bar{\\phi}` in [SadErh12]_) """ return self._phis @property def kappa(self) -> float: """ kappa :math:`\\bar{\\kappa}` constrain parameter (see parameters section above) """ return self.ensemble_parameters['kappa'] def _get_ensemble_data(self) -> Dict: """ Returns a dict with the default data of the ensemble. This includes atom counts and free energy derivative. """ data = super()._get_ensemble_data() # free energy derivative data.update(self._get_vcsgc_free_energy_derivatives(self.phis, self.kappa)) # species counts data.update(self._get_species_counts()) return data def get_phis(phis: Dict[Union[int, str], float]) -> Dict[int, float]: """Get phis as used in the vcsgc ensemble. Parameters ---------- phis the phis that will be transformed to the format the ensemble use. """ if not isinstance(phis, dict): raise TypeError('phis has the wrong type: {}'.format(type(phis))) # Translate to atomic numbers if necessary phis_ret = {} for key, phi in phis.items(): if isinstance(key, str): atomic_number = atomic_numbers[key] phis_ret[atomic_number] = phi 283 ↛ 279line 283 didn't jump to line 279, because the condition on line 283 was never false elif isinstance(key, int): phis_ret[key] = phi return phis_ret
CryptoDB Yanbin Pan Affiliation: Academy of Mathematics and Systems Science, CAS Publications Year Venue Title 2019 CRYPTO New Results on Modular Inversion Hidden Number Problem and Inversive Congruential Generator 📺 The Modular Inversion Hidden Number Problem (MIHNP), introduced by Boneh, Halevi and Howgrave-Graham in Asiacrypt 2001, is briefly described as follows: Let $${\mathrm {MSB}}_{\delta }(z)$$ refer to the $$\delta $$ most significant bits of z. Given many samples $$\left( t_{i}, {\mathrm {MSB}}_{\delta }((\alpha + t_{i})^{-1} \bmod {p})\right) $$ for random $$t_i \in \mathbb {Z}_p$$, the goal is to recover the hidden number $$\alpha \in \mathbb {Z}_p$$. MIHNP is an important class of Hidden Number Problem.In this paper, we revisit the Coppersmith technique for solving a class of modular polynomial equations, which is respectively derived from the recovering problem of the hidden number $$\alpha $$ in MIHNP. For any positive integer constant d, let integer $$n=d^{3+o(1)}$$. Given a sufficiently large modulus p, $$n+1$$ samples of MIHNP, we present a heuristic algorithm to recover the hidden number $$\alpha $$ with a probability close to 1 when $$\delta /\log _2 p>\frac{1}{d\,+\,1}+o(\frac{1}{d})$$. The overall time complexity of attack is polynomial in $$\log _2 p$$, where the complexity of the LLL algorithm grows as $$d^{\mathcal {O}(d)}$$ and the complexity of the Gröbner basis computation grows as $$(2d)^{\mathcal {O}(n^2)}$$. When $$d> 2$$, this asymptotic bound outperforms $$\delta /\log _2 p>\frac{1}{3}$$ which is the asymptotic bound proposed by Boneh, Halevi and Howgrave-Graham in Asiacrypt 2001. It is the first time that a better bound for solving MIHNP is given, which implies that the conjecture that MIHNP is hard whenever $$\delta /\log _2 p<\frac{1}{3}$$ is broken. Moreover, we also get the best result for attacking the Inversive Congruential Generator (ICG) up to now. 2008 EPRINT Cryptanalysis of the Cai-Cusick Lattice-based Public-key Cryptosystem In 1998, Cai and Cusick proposed a lattice-based public-key cryptosystem based on the similar ideas of the Ajtai-Dwork cryptosystem, but with much less data expansion. However, they didn't give any security proof. In our paper, we present an efficient ciphertext-only attack which runs in polynomial time against the cryptosystem to recover the message, so the Cai-Cusick lattice-based public-key cryptosystem is not secure. We also present two chosen-ciphertext attacks to get a similar private key which acts as the real private key.
Probability Seminar Spring 2019 Thursdays in 901 Van Vleck Hall at 2:25 PM, unless otherwise noted. We usually end for questions at 3:15 PM. If you would like to sign up for the email list to receive seminar announcements then please send an email to join-probsem@lists.wisc.edu January 31, Oanh Nguyen, Princeton Title: Survival and extinction of epidemics on random graphs with general degrees Abstract: We establish the necessary and sufficient criterion for the contact process on Galton-Watson trees (resp. random graphs) to exhibit the phase of extinction (resp. short survival). We prove that the survival threshold $\lambda_1$ for a Galton-Watson tree is strictly positive if and only if its offspring distribution has an exponential tail, settling a conjecture by Huang and Durrett. On the random graph with degree distribution $D$, we show that if $D$ has an exponential tail, then for small enough $\lambda$ the contact process with the all-infected initial condition survives for polynomial time with high probability, while for large enough $\lambda$ it runs over exponential time with high probability. When $D$ is subexponential, the contact process typically displays long survival for any fixed $\lambda>0$. Joint work with Shankar Bhamidi, Danny Nam, and Allan Sly. Wednesday, February 6 at 4:00pm in Van Vleck 911 , Li-Cheng Tsai, Columbia University Title: When particle systems meet PDEs Abstract: Interacting particle systems are models that involve many randomly evolving agents (i.e., particles). These systems are widely used in describing real-world phenomena. In this talk we will walk through three facets of interacting particle systems, namely the law of large numbers, random fluctuations, and large deviations. Within each facet, I will explain how Partial Differential Equations (PDEs) play a role in understanding the systems.. Title: Fluctuations of the KPZ equation in d\geq 2 in a weak disorder regime Abstract: We will discuss some recent work on the Edwards-Wilkinson limit of the KPZ equation with a small coupling constant in d\geq 2. February 14, Timo Seppäläinen, UW-Madison Title: Geometry of the corner growth model Abstract: The corner growth model is a last-passage percolation model of random growth on the square lattice. It lies at the nexus of several branches of mathematics: probability, statistical physics, queueing theory, combinatorics, and integrable systems. It has been studied intensely for almost 40 years. This talk reviews properties of the geodesics, Busemann functions and competition interfaces of the corner growth model, and presents some new qualitative and quantitative results. Based on joint projects with Louis Fan (Indiana), Firas Rassoul-Agha and Chris Janjigian (Utah). February 21, Diane Holcomb, KTH Title: On the centered maximum of the Sine beta process Abstract: There has been a great deal or recent work on the asymptotics of the maximum of characteristic polynomials or random matrices. Other recent work studies the analogous result for log-correlated Gaussian fields. Here we will discuss a maximum result for the centered counting function of the Sine beta process. The Sine beta process arises as the local limit in the bulk of a beta-ensemble, and was originally described as the limit of a generalization of the Gaussian Unitary Ensemble by Valko and Virag with an equivalent process identified as a limit of the circular beta ensembles by Killip and Stoiciu. A brief introduction to the Sine process as well as some ideas from the proof of the maximum will be covered. This talk is on joint work with Elliot Paquette. Title: Quantitative homogenization in a balanced random environment Abstract: Stochastic homogenization of discrete difference operators is closely related to the convergence of random walk in a random environment (RWRE) to its limiting process. In this talk we discuss non-divergence form difference operators in an i.i.d random environment and the corresponding process—a random walk in a balanced random environment in the integer lattice Z^d. We first quantify the ergodicity of the environment viewed from the point of view of the particle. As consequences, we obtain algebraic rates of convergence for the quenched central limit theorem of the RWRE and for the homogenization of both elliptic and parabolic non-divergence form difference operators. Joint work with J. Peterson (Purdue) and H. V. Tran (UW-Madison). Wednesday, February 27 at 1:10pm Jon Peterson, Purdue Title: Functional Limit Laws for Recurrent Excited Random Walks Abstract: Excited random walks (also called cookie random walks) are model for self-interacting random motion where the transition probabilities are dependent on the local time at the current location. While self-interacting random walks are typically very difficult to study, many results for (one-dimensional) excited random walks are remarkably explicit. In particular, one can easily (by hand) calculate a parameter of the model that will determine many features of the random walk: recurrence/transience, non-zero limiting speed, limiting distributions and more. In this talk I will prove functional limit laws for one-dimensional excited random walks that are recurrent. For certain values of the parameters in the model the random walks under diffusive scaling converge to a Brownian motion perturbed at its extremum. This was known previously for the case of excited random walks with boundedly many cookies per site, but we are able to generalize this to excited random walks with periodic cookie stacks. In this more general case, it is much less clear why perturbed Brownian motion should be the correct scaling limit. This is joint work with Elena Kosygina. March 21, Spring Break, No seminar March 28, Shamgar Gurevitch UW-Madison Title: Harmonic Analysis on GLn over finite fields, and Random Walks Abstract: There are many formulas that express interesting properties of a group G in terms of sums over its characters. For evaluating or estimating these sums, one of the most salient quantities to understand is the character ratio: $$ \text{trace}(\rho(g))/\text{dim}(\rho), $$ for an irreducible representation $\rho$ of G and an element g of G. For example, Diaconis and Shahshahani stated a formula of this type for analyzing G-biinvariant random walks on G. It turns out that, for classical groups G over finite fields (which provide most examples of finite simple groups), there is a natural invariant of representations that provides strong information on the character ratio. We call this invariant rank. This talk will discuss the notion of rank for $GL_n$ over finite fields, and apply the results to random walks. This is joint work with Roger Howe (Yale and Texas AM). April 4, Philip Matchett Wood, UW-Madison Title: Outliers in the spectrum for products of independent random matrices Abstract: For fixed positive integers m, we consider the product of m independent n by n random matrices with iid entries as in the limit as n tends to infinity. Under suitable assumptions on the entries of each matrix, it is known that the limiting empirical distribution of the eigenvalues is described by the m-th power of the circular law. Moreover, this same limiting distribution continues to hold if each iid random matrix is additively perturbed by a bounded rank deterministic error. However, the bounded rank perturbations may create one or more outlier eigenvalues. We describe the asymptotic location of the outlier eigenvalues, which extends a result of Terence Tao for the case of a single iid matrix. Our methods also allow us to consider several other types of perturbations, including multiplicative perturbations. Joint work with Natalie Coston and Sean O'Rourke. April 11, Eviatar Procaccia, Texas A&M Title: Stabilization of Diffusion Limited Aggregation in a Wedge. Abstract: We prove a discrete Beurling estimate for the harmonic measure in a wedge in $\mathbf{Z}^2$, and use it to show that Diffusion Limited Aggregation (DLA) in a wedge of angle smaller than $\pi/4$ stabilizes. This allows to consider the infinite DLA and questions about the number of arms, growth and dimension. I will present some conjectures and open problems. April 18, Andrea Agazzi, Duke Title: Large Deviations Theory for Chemical Reaction Networks Abstract: The microscopic dynamics of well-stirred networks of chemical reactions are modeled as jump Markov processes. At large volume, one may expect in this framework to have a straightforward application of large deviation theory. This is not at all true, for the jump rates of this class of models are typically neither globally Lipschitz, nor bounded away from zero, with both blowup and absorption as quite possible scenarios. In joint work with Amir Dembo and Jean-Pierre Eckmann, we utilize Lyapunov stability theory to bypass this challenges and to characterize a large class of network topologies that satisfy the full Wentzell-Freidlin theory of asymptotic rates of exit from domains of attraction. Under the assumption of positive recurrence these results also allow for the estimation of transitions times between metastable states of this class of processes. April 25, Kavita Ramanan, Brown April 26, Colloquium, Kavita Ramanan, Brown Title: Tales of Random Projections Abstract: The interplay between geometry and probability in high-dimensional spaces is a subject of active research. Classical theorems in probability theory such as the central limit theorem and Cramer’s theorem can be viewed as providing information about certain scalar projections of high-dimensional product measures. In this talk we will describe the behavior of random projections of more general (possibly non-product) high-dimensional measures, which are of interest in diverse fields, ranging from asymptotic convex geometry to high-dimensional statistics. Although the study of (typical) projections of high-dimensional measures dates back to Borel, only recently has a theory begun to emerge, which in particular identifies the role of certain geometric assumptions that lead to better behaved projections. A particular question of interest is to identify what properties of the high-dimensional measure are captured by its lower-dimensional projections. While fluctuations of these projections have been studied over the past decade, we describe more recent work on the tail behavior of multidimensional projections, and associated conditional limit theorems.
I'm quite curious about this: In a basis set (I'll just use minimal-basis STO-nG basis sets for convenience), the basis functions are written as a linear combination of primitive GTOs. Are the GTOs normalised first, like this: $$\Psi^\mathrm{STO} = \sum_i N_i c_i \phi_i^\mathrm{GTO}$$ where $\phi_i^\mathrm{GTO}$ are un-normalised GTO functions, and $N_i^{2} \left<\phi_i\left\|\phi_i\right.\right> = 1$ or are they normalised like this (GTOs are not normalised): $$\Psi^\mathrm{STO} = N_\mathrm{STO} \sum_i c_i \phi_i^\mathrm{GTO}$$ where $N_\mathrm{STO}^{2} \left<\Psi^\mathrm{STO}\left\|\Psi^\mathrm{STO}\right.\right> = 1$? I thought it was the former and not the latter, but after poking around the internet I'm not so sure anymore.
I am a student studying cryptography by reading "Introduction to Modern Cryptography". I have some confusion about encryption using PRP (e.g., AES). Briefly speaking, a keyed deterministic permutation $F_{k}$ is a strong PRP if no efficient adversary (who is given oracle access to $F_k$ and $F_k^{-1}$) can distinguish whether it is interacting with $F_k$ (for uniform $k$) or $f$, where $f$ is chosen uniformly from the set of all permutations having the same domain and range. I think the definition above says nothing about the relation between two ciphertexts which are generated by a PRP. Now, I consider the following experiment $PrivK_{\mathcal{A},\Pi}^{eav}(n)$ for any encryption scheme $\Pi= (Gen,Enc,Dec)$. The adversary $\mathcal{A}$ is given input $1^n$, and outputs a pair of messages $m_0$,$m_1$, with $\|m_0\| = \|m_1\|$. A key $k$ is generated by running $Gen(1^n)$, and a uniform bit $b \in \{0,1\}^n$ is chosen. Ciphertext $c \gets Enc_k(m_b)$ is computed and given to $\mathcal{A}$. $\mathcal{A}$ outputs a bit $b^{'}$. The output of the experiment is defined to be 1 if $b^{'}=b$, and 0 otherwise. A private-key encryption scheme has indistinguishable encryptions in the presence of an eavesdropper if for all PPT adversaries $\mathcal{A}$ there is a negligible function $negl$ such that, for all $n$, $Pr[PrivK_{A,\Pi}^{eav}(n)] <= \frac{1}{2} + negl(n)$. Does the encryption scheme has indistinguishable encryptions in the presence of an eavesdropper if $Enc = F_k$ and $Dec = F_k^{-1}$, where $F_k$ is a strong PRP? In other word, when I encrypt two messages by using PRP, adversaries can be able to tell whether the two messages are identical since PRP is deterministic. However, can I ensure that it is infeasible for adversaries to learn any partial infomation about the plaintext from the cipertext?
I wish to solve a coupled system of non-linear equation of this form, $$ u_t = -(\mathcal{F})_x + f(x,u,w) \\ w_t = \mathcal{F} + g(x,u,w) $$ by stepping the equations forward in time. The first equation is the advection-diffusion equation with non-linear reaction term, and where the flux has the definition, $$\mathcal{F}=au-du_x$$ First equation $$ \int_{\Omega}u_t~dx = -\int_{\Omega}(\mathcal{F})_x~dx + \int_{\Omega}f(x,u,w)~dx \\ \tilde{u}_t = \frac{1}{h_j}\left( -\mathcal{F}_{j+\frac{1}{2}} + \mathcal{F}_{j-\frac{1}{2}} \right) + \tilde{f}(x_j,u_j,w_j) $$ where $h_j$ is the width (not volume because I am considering 1D only) of the $j$-th cell. Second equation However, how should the second equation be written, can I do the following? $$ \int_{\Omega}w_t~dx = \int_{\Omega}\mathcal{F}~dx + \int_{\Omega}g(x,u,w)~dx \\ \tilde{w}_t = \frac{1}{h_j}\mathcal{F}_jh_j + \tilde{g}(x_j,u_j,w_j) $$ Moreover, if we assume that the flux in the $j$-th cell is constant across the cell then the integral becomes, $$ \int_{\Omega}\mathcal{F}~dx = \mathcal{F}_jh_j $$ This is simply base $\times$ height of the cell, but because I want to know the cell average values I divide again by the cell width. Finally I get the following for the second equation, $$ \tilde{w}_t = \mathcal{F}_j + \tilde{g}(x_j,u_j,w_j) $$ Q1. Is this correct? In terms of implementation I know 3 of the terms in $\mathcal{F}_j$, they are: $a_j$, the cell advection velocity, $d_j$, the cell diffusion coefficient, $u_j$, the cell mass concentration, However, $u_x$ is not known. Q2. Can I simply write $u_x$ using the values of concentration interpolated to the cell faces? For example, $$ u_x = \frac{1}{h_j}\left( u_{j+\frac{1}{2}} - u_{j-\frac{1}{2}} \right) $$ This doesn't seem quite right because I was taught that in the finite-volume method, concentrations are defined in the cells and fluxes are defined on the faces. Update Regarding definition of fluxes. For the first equation I define the fluxes as, $$ \mathcal{F}_{j+\frac{1}{2}} = a_{j+\frac{1}{2}}\left( \frac{h_{j+1}}{2h_{+}} u_j + \frac{h_j}{2h_{+}} u_{j+1} \right) - d_{j+\frac{1}{2}} \frac{u_{j+1}-u_j}{h_{+}} $$ $$ \mathcal{F}_{j-\frac{1}{2}} = a_{j-\frac{1}{2}}\left( \frac{h_{j}}{2h_{-}} u_{j-1} + \frac{h_{j-1}}{2h_{-}} u_{j} \right) - d_{j-\frac{1}{2}} \frac{u_{j}-u_{j-1}}{h_{-}} $$ where $h_j$ is the cell width, and $h_{\pm}$ are the centroid distances in the forward and backward direction. Here linear interpolation is used to define the variable at the cell faces $u(x_{j-1/2})$ and $u(x_{j+1/2})$, and forward difference and backward difference is used to define $u_x(x_{j+1/2})$ and $u_x(x_{j-1/2})$. This gives a central discretisation for both the advection and diffusion terms (however, upwinding can still be introduced easily). However for the second equation the flux is not defined on the cell's left and right face, it is defined as the cell average, so I need to use a different definition for, $$ \mathcal{F}_j = a(x_j)u(x_j) - d(x_j)u_x(x_j) $$ Clearly, $a(x_j)$, $u(x_j)$, and $d(x_j)$ are already known. The only term than needs defining for the second equation is $u_x(x_j)$. So maybe I could use, $$ u_x(x_j) = \frac{1}{h_j}\left( u_{j+1/2} - u_{j-1/2} \right) $$ with linear interpolation, which as you pointed out would give a second order accurate central difference. Could I also use? $$ u_x(x_j) = \frac{1}{h_{+}}\left( u_{j+1} - u_{j} \right) $$ In my comment below, this is what I was getting at. How do define the $u_x(x_j)$ for the second equation, what are the pros. and cons. of using a central or upwinding scheme for this term.
This post is meant to teach new folks how to use MathJax and mhchem formatting on chem.SE Getting started On chemistry.SE, we use MathJax to format mathematical as well as chemical equations and similar expressions in questions, answers, and comments. MathJax allows us to typeset expressions using $\LaTeX$ notation. To use MathJax, enclose your math within single ( $...$) or double ( $$...$$) dollar signs. Single dollar signs make the math inline, for example, Let $x$ be a variable gives: Let $x$ be a variable. On the other hand, double dollar signs make the math a block element. It gets its own line, and is slightly larger. For example, The equation of motion is as follows: $$v=u+at$$ It is a SUVAT equation gives: The equation of motion is as follows: $$v=u+at$$ It is a SUVAT equation Note that, in math mode, MathJax ignores the spaces you type, e.g. $a b$ yields $a b$. MathJax formats expressions the way it is common in mathematics texts. However, the printing rules for signs and symbols used in the natural sciences and technology may require additional spaces (in particular, between the numerical value and the unit symbol). In math mode, use \ (backslash space) or ~ (tilde) if you want the equivalent of space in normal text. Where separation of numbers into groups of three digits is used, the groups shall be separated by a thin space \, (backslash comma); e.g. $299\,792\,458$ yields $299\,792\,458$. Basic chem We use the mhchem package for chemistry. It lets us easily format chemical formulas and reactions without typing too much. There really is only one command you need to know here: \ce{...}. \ce{...} takes its parameters and automatically formats it. For example, $\ce{HCl}$ dissociates in water as follows:$$\ce{H2O + HCl <=> H3O+ + Cl-}$$ Renders as $\ce{HCl}$ dissociates in water as follows: $$\ce{H2O + HCl <=> H3O+ + Cl-}$$ Note that spaces are very important for mhchem to separate super/subscripts from normal text. \ce{H3O+} will display $\ce{H3O+}$, but \ce{H2O +} will display $\ce{H2O +}$. When typesetting ions with more than a single charge, the argument has to be raised using the caret (^; also known as the circumflex accent) character, e.g. $\ce{Cu^2+}$ renders as $\ce{Cu^2+}$, while $\ce{Cu2+}$ would incorrectly render to $\ce{Cu2+}$. (Also see basic math below.) Various types of reaction arrows are supported, including ->, <=>, <==>>, etc. It also supports various types of bonds, via the \bond{..} command (to be called inside \ce{...}). You need not call \bond for normal bonds. Eg: \ce{H\bond{->}A-B=C#D\bond{~}E\bond{~-}F\bond{...}G\bond{<-}E} displays: $$\ce{H\bond{->}A-B=C#D\bond{~}E\bond{~-}F\bond{...}G\bond{<-}E}$$ Basic math Superscripts and subscripts You can denote superscripts via the ^ character, and subscripts via _. For example, x^2 renders as $x^2$, x_1 renders as $x_1$, and x_1^3 renders as $x_1^3$. If you want to include more than one character in the super/sub script, enclose it in curly braces ( {...}). For example, x^10 renders as $x^10$, but x^{10} renders as $x^{10}$ Fractions and square roots Fractions can be easily displayed using \frac{..}{..}. For example, $$\frac{a+b^c}{de+f}$$ renders as $$\frac{a+b^c}{de+f}$$ However, where it is necessary to include fractions in the body text, they shall, where possible, be reduced to a single level by using a solidus (/) or, where applicable, the negative index. For example, instead of $\frac{1}{\sqrt 2}$ write $1/\sqrt 2$ or $2^{-1/2}$; instead of $C_{\mathrm m,p} = 33.58\ \mathrm{\frac{J}{K \cdot mol}}$ write $C_{\mathrm m,p} = 33.58\ \mathrm{J/(K \cdot mol)}$ or $C_{\mathrm m,p} = 33.58\ \mathrm{J\cdot K^{-1} \cdot mol^{-1}}$. Protip: You can exclude the braces for single-character numerators/denominators (if the first character is a letter, you need to use a space after \frac, though). For example \frac12 renders as $\frac12$, and \frac ab renders as $\frac ab$. Square roots can be added in a similar manner, via \sqrt{....}. For example, \sqrt{x+y} renders as $\sqrt{x+y}$. Greek letters Greek letters can be added using a backslash ('\'), followed by the name of the letter. Captialise the first letter of the name for greek capital letters. Eg \alpha \beta \gamma \Omega renders as $\alpha \beta \gamma \Omega$. Make sure that you put spaces after these if you are typing normal alphabet characters. Eg e^{\pii} gives an error, you need to use e^{\pi i}. Note that there are special commands \varepsilon , \varsigma , \varrho , and \varpi to distinguish between the lunate Greek letters. MathJax extensions MathJax has available a variety of extensions enabling other features: strikeout lines, enclosures, text/background coloration, interactive equations, etc. Information about these extensions and how to enable them can be found at this meta post. Further reading Wikipedia TeX help page (extremely useful as a reference, useless as a tutorial) TeX/LaTeX Stack Exchange site Harvard intro to TeX LaTeX wikibook, Math section LaTeX wikibook, Advanced Math section If you would like to know more, you can continue reading about which symbols are written in italic (sloping) type and which are printed in roman (upright) type.
We show that for $r$ red cards and $b$ black cards your chance of winning is $\frac{b}{r + b}$ and that this is true for all possible strategies. We do this by induction on $T = r + b$, the total number of cards left. This means we need to show that the chance to win is $\frac{b}{T}$. For $T = 1$ you must, according to the rules of the game, call "bet". If $b = 1$, only a black card remains and your chance of winning is $1$. If $b = 0$, only a red card remains and your chance of winning is $0$. Your chance to win is $\frac{b}{T}$ in both cases. Now for the induction step, we assume that your chance to win with $b$ black cards remaining in $T - 1$ cards is $\frac{b}{T - 1}$ (for any $b \leq T - 1$). We calculate the chance to win with $b$ black cards remaining in $T$ cards (for any $b \leq T$). If you choose to call "bet" now, your chance of winning is clearly $\frac{b}{T}$. If you wait for the next card instead then your chance of winning = (chance of red on next card $\times$ chance of winning with 1 less red) $+$ (chance of black on next card $\times$ chance of winning with 1 less black). The chances to win after removing a card are given by our induction hypothesis.$$\frac{r}{T} \times \frac{b}{T - 1} + \frac{b}{T} \times \frac{b - 1}{T - 1}\\= \frac{rb + b(b - 1)}{T(T - 1)}\\= \frac{b(r + b - 1)}{T(T - 1)}\\= \frac{b(T - 1)}{T(T - 1)}\\= \frac{b}{T}$$Regardless of whether you bet now or wait, your chance of winning is $\frac{b}{T}$. This completes the induction, and so your chance of winning is always $\frac{b}{T} = \frac{b}{r + b}$. For the values given in the problem ($r = 26$, $b = 26$), this chance is $0.5$.
I'd like to learn how the Raviart-Thomas (RT) element works. To that end I'd like to analytically describe how the basis functions look on the reference square. The goal here is not to implement it myself, but rather just to get an intuitive understanding of the element. I'm largely basing this work off of the triangular elements discussed here, perhaps extending it to quadrilaterals is a mistake in itself. That said, I can define the basis functions for first RK element RK0: $$\mathbf{\phi}_i(\mathbf{x}) = \mathbf{a} + \mathbf{b}\mathbf{x} = \begin{pmatrix} a_1 + b_1 x\\a_2 + b_2 y\end{pmatrix}$$ for $i = 1,\dots,4.$ The conditions on $\mathbf{\phi}_i$ are that: $$\mathbf{\phi}_i(\mathbf{x}_j)\cdot\mathbf{n}_j = \delta_{ij}$$ where $\mathbf{n}_j$ is the unit normal shown below, and $\mathbf{x}_j$ is its coordinate. This is the reference square $[-1,1]\times[1,1]$, so this leads to a system of equations for each basis function. For $\mathbf{\phi}_1$ this is: $$\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & -1 & 0 & 1\\ -1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\end{pmatrix}\begin{pmatrix}a_1 \\ a_2\\ b_1\\ b_3\end{pmatrix} = \begin{pmatrix} 1\\0\\0\\0\end{pmatrix}$$ which can be solved to give: $$\mathbf{\phi}_1(\mathbf{x}) = \frac{1}{2}\begin{pmatrix} 1 + x\\ 0\end{pmatrix}$$ The other basis functions can be found similarly. Assuming this is correct, the next step is to find the basis functions for RK1. This is where I'm getting a little unsure of myself. According to the link above, the space we are interested in is: $$P_1(K) + \mathbf{x}P_1(K)$$ A basis for $P_1$ would be $\{ 1, x, y\}$ I think this means the RK1 basis functions should take the form: $$\mathbf{\phi}_i(\mathbf{x}) = \begin{pmatrix} a_1 + b_1 x + c_1 y + d_1 x^2 + e_1 xy\\ a_2 + b_2 x + c_2 y + d_2 xy + e_2 y^2\end{pmatrix}$$ This leaves 10 unknowns for each basis function. If we apply the same conditions as in the RK0 case, namely: $$\mathbf{\phi}_i(\mathbf{x}_j)\cdot\mathbf{n}_j = \delta_{ij}$$, where $\mathbf{n}_j$ is the unit normal as shown below: this gives us 8 equations. The other 2 I think can be found from some moments. I'm not really sure how exactly. The link above talks about integrating against a basis for $[P_1]^2$, but I'm having trouble figuring out what that means. Am I on the right track, or have I completely missed something here?
Search Now showing items 1-1 of 1 Anisotropic flow of inclusive and identified particles in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV with ALICE (Elsevier, 2017-11) Anisotropic flow measurements constrain the shear $(\eta/s)$ and bulk ($\zeta/s$) viscosity of the quark-gluon plasma created in heavy-ion collisions, as well as give insight into the initial state of such collisions and ...
I am trying to prove the integer square root theorem $\forall x: \mathbb{N}, \exists y : \mathbb{N}((y^2 \leq x) \land (x < (y+1)^2))$ for $\lfloor \sqrt{x} \rfloor$. In words: for any natural number $x$, there exists a natural number $y$ which is the integer part of the square root of $x$. I would like to use proof by induction in the Fitch system. The outline of proof is based on and Kreitz and goes as follows: Show that the predicate holds for the base case $x=0$ Assume the predicate holds for some arbitrary $x=k$ Prove the predicate holds for $x=k+1$ Case 1: If the new number $k+1$ is less than the $k^{th}$ root plus one squared then let the $k^{th}$ root be a witness. That is if $(k+1)<(\lfloor \sqrt{k}\rfloor+1)^2$ then $\lfloor\sqrt{k}\rfloor$ Case 2: If the new number $k+1$ is is greater or equal to the $k^{th}$ root plus one squared then the witness is $(\lfloor\sqrt{x}\rfloor + 1)$. That is if $(k+1) \geq (\lfloor\sqrt{k}\rfloor+1)^2$ then $(\lfloor\sqrt{k}\rfloor + 1)$. Below is my flawed attempt at the proof using the Fitch software. I am trying to follow the outline proof above. The lemmas are taken to be theorems from arithmetic. As can be seen the attempt ends universals only. I do not know how to prove the theorem with the existential quantifier. Assuming that my interpretation of the original outline is correct, what Fitch specific strategies do I need that will help me provide a correct proof? I have taken the hints in the comments. I think the proof is OK,
Homework disclaimer... The task: We are using the following algorithm to solve the quadratic equation $x^2+2px+q=0$: $x_1=\|p\|+\sqrt{p^2-q}\mathtt{;}$ $\mathtt{if}\,p>0\,\mathtt{then}\,x_1=-x_1\mathtt{;}$ $x_2=q/x_1\mathtt{;}$ Point out two places where running this algorithm can meet significant problems. How to mitigate these problems? That's my guess: The first place is this: $p^2-q$, we're risking catastrophic cancellation here; and the second place is this: $q/x_1$, we're getting a NaN if $x_1$ is $0$. The second one seems kind of easy: the only way for the absolutey larger root to be $0$ is when both roots are $0$, which entails so $p=q=0$; we can test for this case. I don't know how to mitigate the first problem, though. I can't think of a way to solve a quadratic equation without computing its discriminant, and computing its discriminant does require subtraction. Which, of course, risks catastrophic cancellation. Is there any way to avoid catastrophic cancellation when computing the discriminant of a quadratic function? EDIT: As per request from comments, moving additional info from answer to question: Hah. Regarding $x_1=\|p\|+\sqrt{p^2-q}$. Another student was arguing today that this cannot be an issue: According to him, catastrophic cancellation can only occur if $p^2\approx q$; but then $\sqrt{p^2-q}\approx 0$ and moreover $\|p\|\gg\sqrt{p^2-q}$; so the discriminant itself and any errors introduced by computing it loose significance when computing $x_1$. The Professor said he would have to think about it; I'll post an update here if/when he's done thinking. However, there is one more problem I failed to notice. Namely, if $p$ is large, then $p^2$ may overflow; but then $\sqrt{p^2-q}$ will also overflow even if its accurate value would fit in the range of the number format; so $x_1$ will be spuriously set to $\infty$. To avoid this, the Professor said, $p$ needs to be rescaled if its large enough.
Published October 2001,October 2004,February 2011. (The ideas used in this article are also used here to find the volume of a sphere and the, impressively named, mouhefanggai shape.) You need to be familiar with the formulae for the area of a square, the area of a circle, and the volume of a cuboid. We will also use ideas of enlargement and ratio. When we think of a pyramid, we usually think of one where the vertex (the top point) is above the centre of the base, a right pyramid . However in this article, we will begin with a yangma. Yangma is an ancient Chinese name for a rectangular-based pyramid whose vertex is vertically above one of the vertices of the base. We will take a yangma with a square base of side length $a$, andheight also length $a$. Three of these can be fitted together to form a cube. Since we know that the volume of the cube is $a \times a \times a = a^3$, it follows that the volume of each of these yangmas is $\frac{a^3}{3}$. Consider, for a moment, a cube with edges 1 unit long. You know that the volume is $1 \times 1 \times 1$. Now stretch that cube in a horizontal direction so that it measures a by 1 by 1. Its volume is now $a \times 1 \times 1$. If we imagine the cube sliced horizontally, there will be the same number of slices, but each will be a times as long. If we stretch the cube in a perpendicular direction, so that it measures a by b by 1, the volume will be $a \times b \times 1$, or $b$ times larger. If we stretch the cube in the third perpendicular direction, by a scale factor of $c$, the volume will be $a \times b \times c$, which we know as the formula for the volume of a cuboid. There are three independent directions in which we could enlarge a 3D shape. If we enlarge it by a scale factor $k$ in one of these directions, the volume will be $k$ times as large. Suppose we want the volume of a yangma whose height is different to the base lengths, perhaps height $h$ instead of a. Well, this is just an enlargement in the vertical direction. The scale factor? If we are turning $a$ into $h$, we have multiplied by $\frac{h}{a}$. So the volume of our new yangma is $\frac{h}{a} \times \frac{a^3}{3}$, or $\frac{a^2h}{3}$. We now have a formula for the volume of any square-based pyramid whose vertex is above one of the vertices of the base. What about if the vertex is somewhere else - the middle, for instance? What we are going to do is to imagine the pyramid cut into lots of slices horizontally. We are going to slide these slices across, so that the top of the pyramid is above the middle of the base. If we had an infinite number of slices, our pyramid would have nice straight edges. You can probably imagine that with more slices, it would look smoother than in this illustration. Has the area of any of the slices changed? So has the volume of the shape changed?We can now see that the volume of any square-based pyramid is $\frac{a^2h}{3}$. We will now look at a cone. We'll start with a right cone , whose vertex is above the centre of the base. In fact, by slicing it as in the previous section, we can show that the same formula applies for any cone.Imagine a cone whose base is a circle radius $r$, and height is $h$. This cone will fit exactly inside a square-based pyramid with base length $2r$ and height $h$. Suppose we take a slice of the pyramid with the cone inside, from some way up the pyramid. This will look like a square with a circle fitting inside. We don't know the radius of the cone at this point, so we'll call it $x$. The same is true for every slice we take: the area of the circle is $\frac{\pi}{4}$ of the area of the square. We can use the same principles to find the volume of any pyramid. If we have a pyramid with rectangular base measuring $a$ by $b$, and height $h$, then this can be obtained by stretching our square-based pyramid by scale factor $\frac{b}{a}$. The new volume will be $\frac{b}{a} \times \frac{a^2 h}{3} = \frac{a b h}{3}$. If the base of the pyramid is a triangle with base $a$ and perpendicular height $b$, it will fit exactly in the rectangular pyramid above.Any slice will look like this: Area of rectangle = $k{a} \times k{b} = k^2 a{b}$ Area of triangle = $\frac{1}{2}{k}{a} \times {k}{b} = \frac{1}{2}k^2{a}{b} = \frac{1}{2} \times \mbox{area of rectangle}$. If each triangle is half the size of the rectangle, the volume of the triangular-based pyramid will be half the volume of the rectangular-based pyramid, or $a{b}h/6$.
By Murray Bourne, 28 Jan 2006 How cool is this? Mathematical tattoos. For the mathematics behind this, see fractals. I might get one of these on my left nipple. Juz kidding... See the 4 Comments below. Posted in Mathematics category - 28 Jan 2006 [Permalink] Tags: Culture im looking 4 a fractal tatoo artist willing to do work on my canvas.skin that is free canvas to get there name out....I live in missouri I'm seeking a Tattoo Artist's, ((possible nightmare)) of a feat... A Fractal. If not is there any technology capable of rendering a Fractal Tattoo? Thank-you, from Onterrible, Canada. Hi Jo. I don't know of any technology that will produce the tattoo. But if you need one for just the fractal, this works nicely: http://www.fractalsciencekit.com/ i also would want to get a fractal tattoo. but i dunno how to know what's best for me * Name (required) * E-Mail (required - will not be published) Your blog URL (can be left blank) Notify me of followup comments via e-mail Your comment: Preview comment HTML: You can use simple tags like <b>, <a href="...">, etc. To enter math, you can can either: qq `a^2 = sqrt(b^2 + c^2)` $ \int g dx = \sqrt{\frac{a}{b}} $ NOTE: You can mix both types of math entry in your comment. Sign up for the free IntMath Newsletter. Get math study tips, information, news and updates twice a month. Join thousands of satisfied students, teachers and parents! Given name: * required Family name: email: * required See the Interactive Mathematics spam guarantee. (External blogs linking to IntMath) From Math Blogs Blog⊗ Categories IntMath top
Let's split the usual time horizon $[0,T]$ like $0=T_{0}<T_{1}<\dots<T_{n}=T$ and consider the bond price $P(t,T_{i})$ for $i=1,...,n$. We assume $$\frac{dP(t,T_{i})}{P(t,_{i})}=r_{t}dt+\xi_{i}(t)dB_{t}$$ By Ito we can recall $$P(t,T_{i})=P(0,T)\exp(\int_{0}^{t}r_{s}ds+\int^{t}_{0}\xi_{i}(s)dB_{s}-\frac{1}{2}\int^{t}_{0}\|\xi_{i}(s)\|^{2}ds)$$ Now, I am supposed to proof using Girsanov I theorem, that the process $W_{t}^{i}=B_{t}-\int^{t}_{0}\xi_{i}(s)ds$ is a standard Brownian motion under the forward measure $Q_{T_{i}}$ using $P(t,T)$ as a numeraire for $i=1,...,n$. The question states $$dW_{t}^{i}=dB_{t}-\frac{1}{N_{t}}dN_{t}dB_{t}=...=dB_{t}-\xi_{i}(t)dt$$ "Complete the ... part and look at the $\frac{1}{N_{t}}dN_{t}$, what is it and why do we use it here?" I cannot find how Girsanov is used for nond pricing and this expression with the ... part is derived from this? Thanks to the Girsanov theorem, we have the following relationship between the forward measure $\mathbb{Q}^{T_i}$ and the historical measure $\mathbb{P}$.\begin{align}\left.\frac{d\mathbb{Q}^{T_i}}{d\mathbb{P}}\right\|_{\mathcal{F}_t} &= e^{-\int_t^Tr_sds}\frac{P_t(T_i)}{P_0(T_i)} \\&= \exp\left(\int^{T_i}_{t}\xi_{i}(s)dB_{s}-\frac{1}{2}\int^{T_i}_{0}\|\xi_{i}(s)\|^{2}ds\right)\\&=\frac{N_{T_i}}{N_t}\end{align}where $N_t = \exp\left(\int^{t}_{0}\xi_{i}(s)dB_{s}-\frac{1}{2}\int^{t}_{0}\|\xi_{i}(s)\|^{2}ds\right)$. This process is an exponential martingale widely known as the Doléans-Dade exponential. By the Ito formula, the dynamic of $N_t$ is \begin{equation} dN_t = N_t\xi_i(t)dB_t \end{equation} Giranov tells us as well that it exists a Brownian motion under $\mathbb{Q}^{T_i}$ given by : \begin{align} dW_t &= dB_t - \frac{1}{N_t}\langle N_., W_.\rangle_t\\ &=dB_t -\frac{1}{N_t}N_t\xi_i(t)\langle W_., W_.\rangle_t\\ &= dB_t - \xi_i(t)dt \end{align}
Encouraged by Hilmar, I've decided to update the answer with all the steps necessary to calculate the Reverberation Time from a scratch. Presumably, it will be useful for others interested in this area. Obviously, it is the simplest approach because more advanced are definitely beyond a scope. In the beginning, you must obtain the impulse response of a room. It can be done in various ways: Firing a starter gun, popping a balloon, etc. - basically recording any impulsive-like signal with broad frequency content and omnidirectional characteristic. This is the simplest method of obtaining impulse response. Sweep Sine measurement with a usage of either linear or exponential sweep. This method is my favorite as it allows you to extract many different parameters of your system at the same time. MLS (Maximum Length Sequence) measurement with this kind of noise and using Hadamard transform to obtain the impulse response. Some comparison with sweep sine technique can be found here. After getting the impulse response $h(t)$ of your acoustic space you are able to calculate the reverberation time and other speech related parameters such as $C_{80}$, $STI$, $D_{50}$ (Clarity, Speech Transmission Index, Definition), etc. Very first step is to filter your impulse response in an appropriate frequency band. This is due to fact that reverberation time is a function of position in the room and frequency. Obviously, we assume that sound field is diffused and at each position in a room decay rate is the same. Thus we must still consider dependency on the frequency band. We are using 1/1 octave or 1/3 octave filters. Its desired parameters are defined by standard, preferably you always wish to use the filter of class 0. For most of the time, Butterworth 3rd order filter is enough (although for frequencies below 125 Hz you might experience some problems with characteristic not meeting specifications). I did the whole implementation on my own with few tweaks, but for usual applications widely used EN 61260 MATLAB implementation is good enough. Next step after filtering the $h(t)$ and obtaining $h_f(t)$, is to make the response as smooth as possible before conversion to the logarithmic scale. For that purpose, Hilbert Transform is a widely used tool. The goal is to create the analytic signal: $$h_A(t)=h(t)+j\tilde{h}(t) $$ where $\tilde{h}(t) $ is the HT of your filtered impulse response and $j$ is a complex unit. Now, knowing that your analytic signal is complex, two things are coming from its representation: $$h_A(t)=A(t)e^{j \psi (t)} $$ where $A(t)$ is the envelope of your signal and $\psi (t)$ is an instantaneous phase. Obviously, we are interested mainly in the envelope (magnitude of the analytic signal). Below you can see overlayed filtered response and its envelope: One clue for people speaking MATLAB'ish. The hilbert function returns the analytic signal, so no need of multiplying by $j$ and adding original $h(t)$. Just use abs(hilbert(h)) to get the envelope. Now we have the smoothed version of our impulse response, but it must be smoothed further. It is optional and one might consider it unnecessary. It is based on Moving Average filter of appropriate length $M$. Everyone knows the moving average filter, it is basically a low-pass FIR filter with frequency response given by: $$H(f)=\frac{\sin (\pi f M)}{M\sin (\pi f)} $$ Plots for different lengths $M$ are shown below: And on the following figure you can observe the effect of smoothing of the previously calculated energy curve ($A(t)$ in logarithmic scale becomes $E(t)$) for two different values of $M$. For sampling frequency of $48\mathtt{kHz}$ I've used $M=5001$. So now we have the Energy Curve (not a Power) which is simply the smoothed Hilbert envelope in logarithmic scale. Conversion to decibel scale is done by equation: $$E(t) = 20\log_{10} \frac{A(t)}{\max A(t)} $$ Curve $A(t)$ can be further smoothed for calculations by using the Schroeder Integration method of your envelope (also known as inversed time integration). This method gives you maximally flat decay curve and is very easy to implement. $$L(t)=10 \log_{10} \left[ \dfrac{\int_t^{\infty}h^2(\tau )d\tau}{\int_0^{\infty}h^2(\tau )d\tau} \right] $$ Again for MATLAB people: L(td:-1:1)=(cumsum(hA(td:-1:1))/sum(hA(1:td))); Mind that limit of integration ( td) is equal to $\infty$. It is true when we do not have any environmental noise. Otherwise, you will see that decaying sound 'dives into noise level'. It is depicted below. The red curve is based on the correct limit, where decay crosses noise floor. If you choose your limit of integration to be too short, then as in green curve, the estimate is not long enough, so you will get under-estimation while doing linear interpolation. On the contrary, the orange curve has too large integration limit and you will get over-estimated. You always want to find the correct limit. If you didn't obtain curve in such way, then please do so. There are many different methods for estimation of integration limit in Schroeder integral, I think that easiest to implement is the one proposed by Lundeby et al. Regarding calculation of RT itself. It must be done by performing linear interpolation of your decay curve (or should I say Schroeder curve) with linear function: $L=A\cdot t+B$ on the correct range (which is described below). When it's done, you calculate the reverberation time from the equation. You can see that point of intercept doesn't really matter (per se), you care about the gradient of your line. $$RT = \dfrac{-60}{A} $$ where $A$ is a slope coefficient from interpolated line (in dB/s). While you have that, you can also calculate the correlation coefficient of your linear fit (or r-value), that is telling you how good is the fit. Finally - limits. You cannot really calculate $T_{60}$ having $60 \mathtt{dB}$ of dynamic range as you are trying to do - you always need more (unless you really know what you are doing). So you are doing linear fit on decay curve at following dynamic ranges, and the result is being extrapolated, according to ISO 3382-2: $EDT$ (Early Decay Time): upper limit is $0 \mathtt{dB}$ and lower is $-10 \mathtt{dB}$. This parameter correlates well with perceived reverberation time. In practice, though beginning for the sake of algorithms, people are using an interval of $-1 \mathtt{dB}$ and $-10 \mathtt{dB}$ (i.e. in Norsonic analyzers). $T_{10}$: upper limit must start at $-5 \mathtt{dB}$ to remove any fluctuations and then lower limit is taken to be $-15 \mathtt{dB}$, but it always must be at least $10 \mathtt{dB}$ above the noise floor. So in fact you need at least $25 \mathtt{dB}$ of dynamic range (or INR) to be able to calculate $T_{10}$ ($5+10+10$). $T_{20}$: upper limit at $-5 \mathtt{dB}$, lower at $-25 \mathtt{dB}$. Minimum dynamic range needed is $35\mathtt{dB}$ $T_{30}$: upper limit of $-5 \mathtt{dB}$, lower at $-35 \mathtt{dB}$, with minimum $45 \mathtt{dB}$ of dynamic range. Why are people using these values? Well, because you very rarely have $75 \mathtt{dB}$ of dynamic range to be able to estimate $T_{60}$. For very smooth decay these values should be equal. Some more detailed analysis of effects can be found in this publication: Measuring Room Impulse Responses: Impact of the Decay Range on Derived Room Acoustic Parameters If you will not perform Schroeder integration, then most probably they will diverge a lot. For the very nasty room you can get something like: In theory: $EDT$ = $T_{10}$ = $T_{20}$ = $T_{30}$. Although they are not necessarily going to be. For example, if your small room is coupled with large one (let's think of small chapel coupled with cathedral by a small doorway), then you will get very sharp decay for smaller one at the beginning of the curve, and very long tail in the end: To wrap up. According to standards, you should always estimate your reverberation time $T_{N}$ starting from $-5 \mathtt{dB}$ and down to $(-5-N) \ \mathtt{dB}$. Also, ensure that you are at least $10 \mathtt{dB}$ above the noise floor. If not, then you must use another estimate. In your case, I suggest using the $T_{30}$. Below you have example of $T_{20}$ estimation for a troublesome impulse response. $E(t)$ is the energy decay curve obtained by taking the logarithm of impulse response Hilbert envelope. $L(t)$ is the Schroeder integral. Correlation coefficient between linear fit and $E(t)$ on a given range ($-5 \mathtt{dB}$ - $-25 \mathtt{dB}$) for this case is: $cc=-0.9987$. Calculated RT is $T_{20}=1.15 \mathtt{s}$ I think that should do. If anyone will follow these steps, then he should get an accuracy of $0.1 \mathtt{s} $ against widely used Dirac. Anyway, it must be remembered that even different software tend to give different results (i.e. EASERA and Dirac), so you should be totally fine. For more complicated impulse responses with artifacts at the end of impulse response one might get worse performance, but anyway, such measurements are not reliable and reflect wrongly conducted an experiment.
In Freedman and Proeyen's text on supergravity they derive the equation of motion for the gravitino using the second order formalism. However, I'm not exactly clear as to how they use partial integration and the identities given to manipulate it into the final form. They give the gravitino action as (9.1), $$S_{3/2}=-\frac{1}{2\kappa^2}\int d^Dx~e\bar{\psi}_\mu\gamma^{\mu\nu\rho}D_\nu\psi_\rho,\tag{9.1}$$ and go on to say on p. 189, "In the second order formalism, partial integration is valid, so it is sufficient to vary $\delta\bar{\psi}_\mu$ and multiply by 2 obtaining, $$\delta S_{3/2}=-\frac{1}{\kappa^2}\int d^Dx~e(D_\mu\bar{\epsilon})\gamma^{\mu\nu\rho}D_\nu\psi_\rho$$ $$=\frac{1}{\kappa^2}\int d^Dx~e\bar{\epsilon}\gamma^{\mu\nu\rho}D_\mu D_\nu\psi_\rho=\frac{1}{8\kappa^2}\int d^Dx~e\bar{\epsilon}\gamma^{\mu\nu\rho}R_{\mu\nu ab}\gamma^{ab}\psi_{\rho}.\tag{9.7}$$ We integrated by parts and used (8.37) to move to the second line$^4$ and then used the Ricci Identity (7.123) to obtain the last expression." The relevant equation they mention in the text are: $$\nabla_\mu\gamma_\nu=0,\tag{8.37}$$ and $$[D_\mu,D_\nu]\Phi=\frac{1}{2}R_{\mu\nu ab}M^{ab}\Phi,$$ $$[D_\mu,D_\nu]V^a =R_{\mu\nu cb}\eta^{ac}V^b,$$ $$[D_\mu,D_\nu]\Psi=\frac{1}{4}R_{\mu\nu ab}\gamma^{ab}\Psi,\tag{7.123}$$ where the last equation is for a spinor. This is all in section 9.1 in the textbook.
I have implemented the Richardson extrapolation of the Euler-Maruyama method to 4th order, to estimate the moments of SDE. The Euler-Maruyama works, and I would expect the Richardson extrapolation to work too, because when I suppress the noise contribution it gives the correct solution in the resulting ordinary differential equations. However, when I add the noise term, I don't get the moments (the variance, in particular) I would expect, as I do with the regular Euler method, so apparently I am doing something wrong. This is my MATLAB code for the extrapolation: function [t, y] = Richardson_weak_4th_order(a, b, t_interval, y0, h)[t1, y1] = euler_maruyama(a, b, t_interval, y0, h);[~, y2] = euler_maruyama(a, b, t_interval, y0, h/2);[~, y3] = euler_maruyama(a, b, t_interval, y0, h/4);[~, y4] = euler_maruyama(a, b, t_interval, y0, h/8);% Extrapolationy2 = y2(1:2:end,:);y3 = y3(1:4:end,:);y4 = y4(1:8:end,:);t = t1;y = 1/21(64y4 - 56y3 + 14y2 - y1); For simplicity, I am testing with an Ornstein-Uhlenbeck process $$ \mathbb{d} X_t = (m - X_t)\mathbb{d}t + \sigma \mathbb{d}W_t $$ for which $\mathbb{E}(X) \simeq m$ and $\operatorname{Var}(X) \simeq \frac{\sigma^2}{2}$. The results are OK with the Euler method, but the variance is quite bigger with the extrapolation. What am I missing? Edit: The problem is Richardson extrapolation was not implemented on the estimated moments ( as it should), but to realizations of the process.
The Annals of Probability Ann. Probab. Volume 11, Number 3 (1983), 760-764. The Reliability of $K$ Out of $N$ Systems Abstract A system with $n$ independent components which functions if and only if at least $k$ of the components function is a $k$ out of $n$ system. Parallel systems are 1 out of $n$ systems and series systems are $n$ out of $n$ systems. If $\mathbf{p} = (p_1, \cdots, p_n)$ is the vector of component reliabilities for the $n$ components, then $h_k(\mathbf{p})$ is the reliability function of the system. It is shown that $h_k(\mathbf{p})$ is Schur-convex in $\lbrack(k - 1)/(n - 1), 1\rbrack^n$ and Schur-concave in $\lbrack 0, (k - 1)/(n - 1)\rbrack^n$. More particularly if $\prod$ is an $n \times n$ doubly stochastic matrix, then $h_k(\mathbf{p}) \geq (\leq)h_k(\mathbf{p}\prod)$ whenever $\mathbf{p} \in \lbrack(k - 1)/(n - 1), 1\rbrack^n(\lbrack 0, (k - 1)/(n - 1)\rbrack^n)$. This Theorem is compared with a result on Schur-convexity and -concavity by Gleser [2] which in turn extends work of Hoeffding [4]. Article information Source Ann. Probab., Volume 11, Number 3 (1983), 760-764. Dates First available in Project Euclid: 19 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aop/1176993520 Digital Object Identifier doi:10.1214/aop/1176993520 Mathematical Reviews number (MathSciNet) MR704562 Zentralblatt MATH identifier 0519.62086 JSTOR links.jstor.org Citation Boland, Philip J.; Proschan, Frank. The Reliability of $K$ Out of $N$ Systems. Ann. Probab. 11 (1983), no. 3, 760--764. doi:10.1214/aop/1176993520. https://projecteuclid.org/euclid.aop/1176993520
I tried asking a similar question in SE.Physics, and I got some information regarding the abstract side of this, but I figured I should post here to get more complete information about the numerical benefits of variational formalisms. Assuming I am able derive a functional representation for any dynamical system (dissipative, nonlinear, fractional, PDE, ODE, discontinuous, etc), why would such a result or capability be useful? What are some practical consequences? In essence, what are functionals/variational formalisms used for in practice? Keep in mind, the functionals I'm referring to might be of non-standard form (i.e, non-conservative systems), for example: Any system with that is a potential or conservative will have a functional representation as: $$ F[\mathbf{x}]=\int^{t}_0\left(\frac{1}{2}m\dot{\mathbf{x}}(\tau)^2-V(\mathbf{x}(\tau))\right)\,\text{d}\tau $$ Where taking the first variation of this functional yields the dynamics of the system, along with a condition that effectively states that the initial configuration should be similar to the final configuration (variation at the boundaries is zero). But, you can construct other functionals for non-conservative systems, such as this convolutional functional: $$ F[\mathbf{x}]=\frac{1}{2}[\mathbf{x}^{\text{T}} * D(\mathbf{x})]-\frac{1}{2}[\mathbf{x}^{\text{T}} * \mathbf{Ax}]-\frac{1}{2}\mathbf{x}'(0)\mathbf{x}(t) $$ With $\mathbf{A}$ symmetric and $\mathbf{x}(0)$ being the initial condition, and: $$ [\mathbf{f}^{\text{T}} * \mathbf{g}]=\int^{t}_0 \mathbf{f}^{\text{T}}(t-\tau)\mathbf{g}(\tau)\,\text{d}\tau $$ If we take the first variation and assume only that the initial variation is zero, the functional is stationary with respect to:$$\frac{d\mathbf{x}(t)}{dt}= \mathbf{Ax}(t)$$ Point being that one should also consider the implications of functionals which are not inner product based. I know that these functionals are useful in the context of FEM (but how?), but I'm also interested in whether or not they may be useful in other contexts (parameter estimation, simulation, data assimilation, etc).
Search Now showing items 1-10 of 26 Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV (Elsevier, 2017-12-21) We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ... Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV (American Physical Society, 2017-09-08) The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ... Online data compression in the ALICE O$^2$ facility (IOP, 2017) The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ... Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV (American Physical Society, 2017-09-08) In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ... J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV (American Physical Society, 2017-12-15) We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($\|y\| < 0.9$) ... Enhanced production of multi-strange hadrons in high-multiplicity proton-proton collisions (Nature Publishing Group, 2017) At sufficiently high temperature and energy density, nuclear matter undergoes a transition to a phase in which quarks and gluons are not confined: the quark–gluon plasma (QGP)1. Such an exotic state of strongly interacting ... K$^{}(892)^{0}$ and $\phi(1020)$ meson production at high transverse momentum in pp and Pb-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 2.76 TeV (American Physical Society, 2017-06) The production of K$^{}(892)^{0}$ and $\phi(1020)$ mesons in proton-proton (pp) and lead-lead (Pb-Pb) collisions at $\sqrt{s_\mathrm{NN}} =$ 2.76 TeV has been analyzed using a high luminosity data sample accumulated in ... Production of $\Sigma(1385)^{\pm}$ and $\Xi(1530)^{0}$ in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV (Springer, 2017-06) The transverse momentum distributions of the strange and double-strange hyperon resonances ($\Sigma(1385)^{\pm}$, $\Xi(1530)^{0}$) produced in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV were measured in the rapidity ... Charged–particle multiplicities in proton–proton collisions at $\sqrt{s}=$ 0.9 to 8 TeV, with ALICE at the LHC (Springer, 2017-01) The ALICE Collaboration has carried out a detailed study of pseudorapidity densities and multiplicity distributions of primary charged particles produced in proton-proton collisions, at $\sqrt{s} =$ 0.9, 2.36, 2.76, 7 and ... Energy dependence of forward-rapidity J/$\psi$ and $\psi(2S)$ production in pp collisions at the LHC (Springer, 2017-06) We present ALICE results on transverse momentum ($p_{\rm T}$) and rapidity ($y$) differential production cross sections, mean transverse momentum and mean transverse momentum square of inclusive J/$\psi$ and $\psi(2S)$ at ...
Precision-Recall¶ Example of Precision-Recall metric to evaluate classifier output quality. Precision-Recall is a useful measure of success of prediction when the classes are very imbalanced. In information retrieval, precision is a measure of result relevancy, while recall is a measure of how many truly relevant results are returned. The precision-recall curve shows the tradeoff between precision and recall for different threshold. A high area under the curve represents both high recall and high precision, where high precision relates to a low false positive rate, and high recall relates to a low false negative rate. High scores for both show that the classifier is returning accurate results (high precision), as well as returning a majority of all positive results (high recall). A system with high recall but low precision returns many results, but most of its predicted labels are incorrect when compared to the training labels. A system with high precision but low recall is just the opposite, returning very few results, but most of its predicted labels are correct when compared to the training labels. An ideal system with high precision and high recall will return many results, with all results labeled correctly. Precision ($P$) is defined as the number of true positives ($T_p$) over the number of true positives plus the number of false positives ($F_p$). $P = \frac{T_p}{T_p+F_p}$ Recall ($R$) is defined as the number of true positives ($T_p$) over the number of true positives plus the number of false negatives ($F_n$). $R = \frac{T_p}{T_p + F_n}$ These quantities are also related to the ($F_1$) score, which is defined as the harmonic mean of precision and recall. $F1 = 2\frac{P \times R}{P+R}$ Note that the precision may not decrease with recall. The definition of precision ($\frac{T_p}{T_p + F_p}$) shows that lowering the threshold of a classifier may increase the denominator, by increasing the number of results returned. If the threshold was previously set too high, the new results may all be true positives, which will increase precision. If the previous threshold was about right or too low, further lowering the threshold will introduce false positives, decreasing precision. Recall is defined as $\frac{T_p}{T_p+F_n}$, where $T_p+F_n$ does not depend on the classifier threshold. This means that lowering the classifier threshold may increase recall, by increasing the number of true positive results. It is also possible that lowering the threshold may leave recall unchanged, while the precision fluctuates. The relationship between recall and precision can be observed in the stairstep area of the plot - at the edges of these steps a small change in the threshold considerably reduces precision, with only a minor gain in recall. Average precision (AP) summarizes such a plot as the weighted mean ofprecisions achieved at each threshold, with the increase in recall from theprevious threshold used as the weight: $\text{AP} = \sum_n (R_n - R_{n-1}) P_n$ where $P_n$ and $R_n$ are the precision and recall at thenth threshold. A pair $(R_k, P_k)$ is referred to as an operating point. Precision-recall curves are typically used in binary classification to study the output of a classifier. In order to extend the precision-recall curve and average precision to multi-class or multi-label classification, it is necessary to binarize the output. One curve can be drawn per label, but one can also draw a precision-recall curve by considering each element of the label indicator matrix as a binary prediction (micro-averaging). Note In binary classification settings¶ Create simple data¶ Try to differentiate the two first classes of the iris data from sklearn import svm, datasetsfrom sklearn.model_selection import train_test_splitimport numpy as npiris = datasets.load_iris()X = iris.datay = iris.target# Add noisy featuresrandom_state = np.random.RandomState(0)n_samples, n_features = X.shapeX = np.c_[X, random_state.randn(n_samples, 200 * n_features)]# Limit to the two first classes, and split into training and testX_train, X_test, y_train, y_test = train_test_split(X[y < 2], y[y < 2], test_size=.5, random_state=random_state)# Create a simple classifierclassifier = svm.LinearSVC(random_state=random_state)classifier.fit(X_train, y_train)y_score = classifier.decision_function(X_test) Compute the average precision score¶ Out: Average precision-recall score: 0.88 Plot the Precision-Recall curve¶ from sklearn.metrics import precision_recall_curveimport matplotlib.pyplot as pltfrom inspect import signatureprecision, recall, _ = precision_recall_curve(y_test, y_score)# In matplotlib < 1.5, plt.fill_between does not have a 'step' argumentstep_kwargs = ({'step': 'post'} if 'step' in signature(plt.fill_between).parameters else {})plt.step(recall, precision, color='b', alpha=0.2, where='post')plt.fill_between(recall, precision, alpha=0.2, color='b', step_kwargs)plt.xlabel('Recall')plt.ylabel('Precision')plt.ylim([0.0, 1.05])plt.xlim([0.0, 1.0])plt.title('2-class Precision-Recall curve: AP={0:0.2f}'.format( average_precision)) In multi-label settings¶ Create multi-label data, fit, and predict¶ We create a multi-label dataset, to illustrate the precision-recall in multi-label settings from sklearn.preprocessing import label_binarize# Use label_binarize to be multi-label like settingsY = label_binarize(y, classes=[0, 1, 2])n_classes = Y.shape[1]# Split into training and testX_train, X_test, Y_train, Y_test = train_test_split(X, Y, test_size=.5, random_state=random_state)# We use OneVsRestClassifier for multi-label predictionfrom sklearn.multiclass import OneVsRestClassifier# Run classifierclassifier = OneVsRestClassifier(svm.LinearSVC(random_state=random_state))classifier.fit(X_train, Y_train)y_score = classifier.decision_function(X_test) The average precision score in multi-label settings¶ from sklearn.metrics import precision_recall_curvefrom sklearn.metrics import average_precision_score# For each classprecision = dict()recall = dict()average_precision = dict()for i in range(n_classes): precision[i], recall[i], _ = precision_recall_curve(Y_test[:, i], y_score[:, i]) average_precision[i] = average_precision_score(Y_test[:, i], y_score[:, i])# A "micro-average": quantifying score on all classes jointlyprecision["micro"], recall["micro"], _ = precision_recall_curve(Y_test.ravel(), y_score.ravel())average_precision["micro"] = average_precision_score(Y_test, y_score, average="micro")print('Average precision score, micro-averaged over all classes: {0:0.2f}' .format(average_precision["micro"])) Out: Average precision score, micro-averaged over all classes: 0.43 Plot the micro-averaged Precision-Recall curve¶ plt.figure()plt.step(recall['micro'], precision['micro'], color='b', alpha=0.2, where='post')plt.fill_between(recall["micro"], precision["micro"], alpha=0.2, color='b', step_kwargs)plt.xlabel('Recall')plt.ylabel('Precision')plt.ylim([0.0, 1.05])plt.xlim([0.0, 1.0])plt.title( 'Average precision score, micro-averaged over all classes: AP={0:0.2f}' .format(average_precision["micro"])) Plot Precision-Recall curve for each class and iso-f1 curves¶ from itertools import cycle# setup plot detailscolors = cycle(['navy', 'turquoise', 'darkorange', 'cornflowerblue', 'teal'])plt.figure(figsize=(7, 8))f_scores = np.linspace(0.2, 0.8, num=4)lines = []labels = []for f_score in f_scores: x = np.linspace(0.01, 1) y = f_score * x / (2 * x - f_score) l, = plt.plot(x[y >= 0], y[y >= 0], color='gray', alpha=0.2) plt.annotate('f1={0:0.1f}'.format(f_score), xy=(0.9, y[45] + 0.02))lines.append(l)labels.append('iso-f1 curves')l, = plt.plot(recall["micro"], precision["micro"], color='gold', lw=2)lines.append(l)labels.append('micro-average Precision-recall (area = {0:0.2f})' ''.format(average_precision["micro"]))for i, color in zip(range(n_classes), colors): l, = plt.plot(recall[i], precision[i], color=color, lw=2) lines.append(l) labels.append('Precision-recall for class {0} (area = {1:0.2f})' ''.format(i, average_precision[i]))fig = plt.gcf()fig.subplots_adjust(bottom=0.25)plt.xlim([0.0, 1.0])plt.ylim([0.0, 1.05])plt.xlabel('Recall')plt.ylabel('Precision')plt.title('Extension of Precision-Recall curve to multi-class')plt.legend(lines, labels, loc=(0, -.38), prop=dict(size=14))plt.show() Total running time of the script: ( 0 minutes 0.064 seconds)
I am puzzled by a riddle to which I have been told the answer and I have loads of difficulties to believe in the result. The riddle goes as follows: "imagine you are shrunk to the size of a coin (i.e. you are, say, scaled down by two orders of magnitude) but your density remains the same. You are put into a blender of hight 20 cm. The blender will start working in 60 seconds, what to you do?" One of the best answers is apparently: "I jump out of the blender to escape (yes the blender is still open luckily)." This seems ultra non-intuitive to me and I have tried to find flaws in this answer but it seems to be fairly robust. There are two ways you can think of it: the mass scales as $\sim \: L^3$ and therefore it will be $10^6$ times smaller. If we imagine that the takeoff velocity $v_{toff}$ is the same as before being rescaled $v_{big}$. We get then the hight at which a mini us can jump by equating the takeoff kinetic energy and the potential energy i.e. $mv_{toff}^2/2=mgh $ $\Rightarrow$ $h_{mini} = v^2_{big}/(2g) = h_{big} \sim 20 \rm cm$ The second way to see it is to look more in details on how the power produced by muscles scales with the size of muscles. Basically, the power scales with the cross section of the muscle i.e. with the number of parallel "strings" pulling on the joints to contract the muscle. This implies that $P_{mini}=P_{big}/\alpha^2$ ($\alpha$ being the factor bigger than 1 by which you have been rescaled). We know that the takeoff kinetic energy will be given by $P \Delta t$. We assume now that $\Delta t \sim L/v_{big}$ so that $\Delta t_{mini} = L/(\alpha v_{big})$. In the end, this calculation tells us that $E_{mini} \sim E_{big}/\alpha^3$. However, equating again with the potential energy to get the hight we have $h_{mini} \sim E_{mini}/(m_{mini}g) = (E_{big}/\alpha^3)/(gm_{big}/\alpha^3)=E_{big}/(m_{big}g) = h_{big} \sim 20\:\rm cm$ These two reasonings seem fair enough and yet I don't trust the result they lead to. I would like to know if I am experiencing pure denial because of my prejudices or if there is some kind of flaw in the reasonings above (e.g. the fact that it is always assumed that the speed is unchanged when changing scale). I also know that some tiny animals can jump more or less as high as human beings but it seems that most of the time these species have to use some kind of "trick" to store elastic energy in their body so as to generate enough kinetic energy at the takeoff to effectively jump super high. If anyone has any thought on this, that would be very much welcome.
Let $G_q$ be a Paley graph on $q$ vertices, where $q=1 \text{ (mod 4)}$, i.e., the vertices of $G_q$ are the elements of the finite field $\mathbb{F}_q$, and there is an edge between vertices $a,b \in \mathbb{F}_q$ if and only if $a-b$ is the square of an element in $\mathbb{F}_q$. It is known that $G_q$ is a strongly regular graph with parameters \begin{align} (v,k,\eta,\mu) = \left( q, \frac{q-1}{2}, \frac{q-5}{4}, \frac{q-1}{4}\right), \end{align} and thus the second eigenvalue (in absolute value) of $G_q$ is $\frac{-1+\sqrt{q}}{2}$. Consider the random subgraph $\tilde G_q$ of $G_q$, which is the induced subgraph generated by eliminating the vertices i.i.d. with probability $1-p$. I am trying to show that the scaling of the second eigenvalue of $\tilde G_q$ for large $q$ is $O(\sqrt{pq})$. The reason I think this might be true is that because $G_q$ is strongly regular, for large $q$, $\tilde G_q$ will be "close" to a strongly regular graph with parameters $\left( pq, p\frac{q-1}{2}, p\frac{q-5}{4}, p\frac{q-1}{4}\right)$, and thus its spectrum should behave like this reduced strongly regular graph. I have been thinking of upper bounding the maximum eigenvalue of the "perturbation matrix" that would be required to convert the adjacency matrix of $\tilde G_q$ to the reduced strongly regular graph, and then use Weyl's inequality to show that the spectra are close, but I could not get it to work. Any ideas will be appreciated. Update: Adding experimental plot comparing the second eigenvalue of the random induced subgraph, second eigenvalue of Erdos-Renyi $G(pq, 0.5)$ graph, and $\sqrt{p}\frac{-1+\sqrt{q}}{2}$. For $p > 0.5$, the second eigenvalue converges to its value for the full Paley graph, which can be shown by the interlacing theorem and the fact that the eigenvalues of the full Paley graph are given by $\frac{1}{2}(q-1)$ with multiplicity 1, $\frac{-1+\sqrt{q}}{2}$, with multiplicity $\frac{1}{2}(q-1)$, $\frac{-1-\sqrt{q}}{2}$, with multiplicity $\frac{1}{2}(q-1)$. However, the interlacing theorem does not help when $p<0.5$, which is the regime I am more interested, and in this regime the second eigenvalue behaves close to that of a $G(pq, 0.5)$ graph. Update 2: Following Furedi & Komlos '80 result on the eigenvalues of i.i.d. random matrices, I have done the following: Let $A$ be the matrix such that $A_{ii}=0$, and for $i \neq j$, $A_{ij}=1$ if $\{i,j\}$ is an edge in $G_q$, $A_{ij}=-1$ otherwise (it is easy to map the eigenvalues of this matrix to the eigenvalues of the adjacency matrix of $G_q$). Define $J_i$ as the 0-1 indicator random variable representing whether vertex $i$ survives in $\tilde G_q$. Then, setting $\tilde A_{ij} = \chi(i-j)J_i J_j$, \begin{align} \mathbb{E}\left[ \sum_{i=1}^q \lambda_i^k \right] &= \mathbb{E}\left[\text{tr}(\tilde A^k) \right] = \sum_{i_1=1}^q \sum_{i_2=1}^q \cdots \sum_{i_k=1}^q \mathbb{E}\left[ \tilde A_{i_1 i_2} \tilde A_{i_2 i_3}\cdots \tilde A_{i_k i_1} \right] \notag\\ &= \sum_{i_1=1}^q \sum_{i_2=1}^q \cdots \sum_{i_k=1}^q \chi(i_1-i_2)\chi(i_2-i_3) \cdots \chi(i_k-i_1) \mathbb{E} \left[ J_{i_1}J_{i_2}\cdots J_{i_k} \right] \notag\\ &=\sum_{s=1}^k p^s\sum_{\substack{i_1,i_2,\dots,i_k \\ \left\| \{ i_1,i_2,\dots,i_k\}\right\|=s}} \chi(i_1-i_2)\chi(i_2-i_3) \cdots \chi(i_k-i_1) \;\;\;\;\;\;(), \end{align} where $s$ is the number of unique values in the set $\{ i_1,i_2,\dots,i_k\}$, and $\chi(\cdot)$ is the quadratic residue character in $\mathbb{F}_q$, i.e., \begin{align} \chi(x) = \left\{ \begin{array}{ll} 0, & \text{if $x=0$}, \\ 1, & \text{if $x \neq 0$ is a quadratic residue in $\mathbb{F}_q$}, \\ -1, & \text{otherwise} \\ \end{array} \right. \end{align} The goal is to eventually use Markov inequality in the following way \begin{align} \mathbb{P}\left( \max_i \lambda_i > a\right) = \mathbb{P}\left( \max_i \lambda_i^k > a^k\right) \leq \mathbb{P}\left( \sum_i \lambda_i^k > a^k\right) \leq \frac{\mathbb{E}\left[ \sum_i \lambda_i^k\right]}{a^k}, \end{align} where $k$ is an appropriately chosen function of $n$. What I am trying to show is that in the sum over $s$ in $()$, the dominating term should be the one with $s=k/2$, as in the paper with Furedi & Komlos. However, I don't know how to deal with the character sum term, since I don't know much about character sums or number theory. I am aware of Weil's bound, I am not sure if it's useful here. Any help/pointers would be appreciated. Update 3: I have realized the following identities (can be proven by induction), which should be helpful in analyzing the trace expression above. With these identities in hand, this should reduce to a combinatorial counting problem, but I still could not finalize the argument despite many efforts. Any help will be greatly appreciated. \begin{align} \sum_{x_1,\dots, x_{k-1} \in \mathbb{F}_q} \chi(a-x_1)\chi(x_1-x_2)\dots\chi(x_{k-1}-b) = \left\{ \begin{array}{ll} -q^{\frac{k-2}{2}}, & a \neq b\\ (q-1)q^{\frac{k-2}{2}}, & a=b \end{array} \right. \end{align} if $k$ is even, and \begin{align} \sum_{x_1,\dots, x_{k-1} \in \mathbb{F}_q} \chi(a-x_1)\chi(x_1-x_2)\dots\chi(x_{k-1}-b) = \chi(a-b) \end{align} if $k$ is odd.
There is no non-trivial one-dimensional representation of $\mathrm{U}(1)$ on a scalar field $\mathbb{R}^4\to\mathbb{R}$, but on complex fields $\mathbb{R}^4\to\mathbb{C}$, we have the one-dimensional "phase" representations by $$\phi\mapsto\mathrm{e}^{e\mathrm{i}\chi}\phi$$for $e\in\mathbb{Z},\chi\in\mathfrak{u}(1)\cong\mathbb{R}$ for $\mathrm{U}(1)$ parametrized as $\chi\mapsto \mathrm{e}^{\mathrm{i}\chi}$ (the unit circle in the complex plane). Since $\mathrm{U}(1)$ is the archetypical example of a continuous (gauge) symmetry (think of electromagnetism), complex scalar fields are an important (toy) model in QFT. Of course, every complex scalar field may equivalently be replaced by two real scalar fields being its real and imaginary part, so they are not actually needed, but using only real fields may complicate the actual calculations and notations immensely. When switching from a complex scalar $\phi$ to two real ones $\mathrm{Re}(\phi),\mathrm{Im}(\phi)$, we observe that$$ \mathrm{e}^{e\mathrm{i}\chi}\phi = (\cos(e\chi) + \mathrm{i}\sin(e\chi))(\mathrm{Re}(\phi) + \mathrm{i}\ \mathrm{Im}(\phi))$$and so, writing the real vector $\widetilde{\phi} = \left( \begin{matrix} \phi_1 := \mathrm{Re}(\phi) \\ \phi_2 := \mathrm{Im}(\phi)\end{matrix}\right)$, we see that the complex one-dimensional representation of $\mathrm{U}(1)$ turns into a two-dimensional real one with$$ \widetilde{\phi}\mapsto R_e(\chi)\widetilde{\phi}$$with the rotation matrix$$ R_e(\chi) := \left(\begin{matrix}\cos(e\chi) & -\sin(e\chi) \\ \sin(e\chi) & \cos(e\chi)\end{matrix}\right)$$which is now looking more like a representation of the real 2D rotations $\mathrm{SO}(2)$ (the usual one for $e = 1$). As a real representation, this is irreducible (you cannot diagonalize all rotation matrices at once), so you cannot reduce the degrees of freedom and still have a non-trivial representation of $\mathrm{U}(1)\cong\mathrm{SO}(2)$. Two real d.o.f. are the minimum to have some kind of non-trivial continuous symmetry going on, since $\mathrm{U}(1)$ is the simplest Lie group apart from the un-exciting $\mathbb{R},+$.
I'm sorry this must be an elementary question. I spent a good deal of time searching through webs including this site for the problem but I got none.Here's the problem:Say we have a binomial tree ... I wonder if anyone here has read the following paper by Paul Kupiec in which he approximates a loss rate distribution for a portfolio composed of (possibly) concentrated bond positions.https://www.... The JKY ABMC Model (taken from Jabbour, et al. 2001) parameterizes the binomial model (in a risk-neutral world) such that,$u = e^{r\Delta t} + e^{r\Delta t}\sqrt{e^{\sigma^2\Delta t} - 1}$$d = e^{... I have a call option with expiry in two years. In my case the option is bermudan style with first 9 months w/o ability to exercise (i.e. European) and after exercise at any time (i.e. American), but I ... I'm coursing a financial engineering course and must build a binomial tree for a security that is valued 1 in state and time 0. I dont get the formula given by the instructors, and trying to apply the ...
The logistic map is the discrete case of the logistic equation, given by: $\frac {\mathrm{d}y}{\mathrm{d}t}=ry(1-\frac{y}{Y})$ We then approximate to deduce the discrete case:$$ \begin{align} \frac{y_{n+1}-y_n}{\Delta t} &\approx ry_n\left(1-\frac{y_n}{Y}\right) \\ y_{n+1} &\approx r \Delta t y_n \left(1-\frac{y_n}{Y}\right)+y_n \\ y_{n+1}&=(1+r \Delta t)y_n-\frac {r\Delta t}{Y}{(y_n)}^2 \\ y_{n+1}&=(1+r \Delta t)y_n\Bigg( 1-\bigg(\frac{r\Delta t}{1+r\Delta t}\bigg)\frac{y_n}{Y}\Bigg) \end{align} $$ Let $\lambda=1+r\Delta t$ and $x_n=\frac {r\Delta t}{1+r \Delta t} \frac {y_n}{Y}$ . Then our equation becomes: $$x_{n+1}=\lambda x_n (1-x_n) $$ This is the logistic map. We can also think of it as a function $x_{n+1}=f(x_n)$. Question: A fixed point implies $x_{n+1}=x_n$ . Find the fixed points by solving $$ \lambda x_n (1-x_n) = x_n $$ To determine the stability of these points, we are going to find the stability, by investigating the function for values nearby the equilibrium points. Start by supposing that $x_n=X$ is a fixed point. This means that $f(X)=X$. To find a value near the equilibrium point, let $x_n=X+\epsilon{_n}$ where $\epsilon_n < < 1$. Then using the Taylor expansion: $$ \begin{align} x_{n+1}&=f(x_n) \\ X+\epsilon_{n+1} &= f(X+\epsilon_n) \\ &=f(X)+\epsilon_n f'(X)+... \end{align}$$ We neglect the higher-order terms to get: $$X+\epsilon_{n+1}=f(X)+\epsilon_n f'(X)$$ Now from above we saw that $f(X)=X$ , so we can simplify to get: $$\epsilon_{n+1} \approx f'(X) \epsilon_n$$ A fixed point, X, is then stable if: $\Bigg\|\frac{\epsilon_{n+1}}{\epsilon_n}\Bigg \| =\Bigg \|f'(X)\Bigg \| < 1$ Question: Given that $f'(x)=\lambda-2\lambda x$ , find the stability of the fixed points $x_n=0$ and $x_n=1-\frac{1}{\lambda}$ Below are some graphs of the logistic map for different values of $\lambda$ . Case 1: $\lambda< 1$ Only fixed point is 0, which is stable: Case 2: $1< \lambda < 2$ Unstable fixed point at 0 and stable fixed point at $1-\frac{1}{\lambda}$ Question: Can you find the stability for the case $2< \lambda < 3$ ? Below is a picture of some fantastic fractal behaviour which occurs for $3< \lambda< 4$. Question: Can you relate these values of $\lambda$ to what would actually be occuring in a population of organisms?
Answer 3.5 percent Work Step by Step We use the margin of error equation to find: $$=100\times \frac{1}{\sqrt{n}}\\ = 100 \times \frac{1}{\sqrt{800}} \\ =3.5\text{%}$$ You can help us out by revising, improving and updating this answer.Update this answer After you claim an answer you’ll have 24 hours to send in a draft. An editorwill review the submission and either publish your submission or provide feedback.
To evaluate detection performance, we plot the miss-rate $mr(c) = \frac{fn(c)}{tp(c) + fn(c)}$ against the number of false positives per image $fppi(c)=\frac{fp(c)}{\text{#img}}$ in log-log plots. $tp(c)$ is the number of true positives, $fp(c)$ is the number of false positives, and $fn(c)$ is the number of false negatives, all for a given confidence value $c$ such that only detections are taken into account with a confidence value greater or equal than $c$. As commonly applied in object detection evaluation the confidence threshold $c$ is used as a control variable. By decreasing $c$, more detections are taken into account for evaluation resulting in more possible true or false positives, and possible less false negatives. We define the log average miss-rate (LAMR) as shown, where the 9 fppi reference points are equally spaced in the log space: $\DeclareMathOperator*{\argmax}{argmax}LAMR = \exp\left(\frac{1}{9}\sum\limits_f \log\left(mr(\argmax\limits_{fppi\left(c\right)\leq f} fppi\left(c\right))\right)\right)$ For each fppi reference point the corresponding mr value is used. In the absence of a miss-rate value for a given f the highest existent fppi value is used as new reference point. This definition enables LAMR to be applied as a single detection performance indicator at image level. At each image the set of all detections is compared to the groundtruth annotations by utilizing a greedy matching algorithm. An object is considered as detected (true positive) if the Intersection over Union (IoU) of the detection and groundtruth bounding box exceeds a pre-defined threshold. Due to the high non-rigidness of pedestrians we follow the common choice of an IoU threshold of 0.5. Since no multiple matches are allowed for one ground-truth annotation, in the case of multiple matches the detection with the largest score is selected, whereas all other matching detections are considered false positives. After the matching is performed, all non matched ground-truth annotations and detections, count as false negatives and false positives, respectively. Neighboring classes and ignore regions are used during evaluation. Neighboring classes involve entities that are semantically similar, for example bicycle and moped riders. Some applications might require their precise distinction ( enforce) whereas others might not ( ignore). In the latter case, during matching correct/false detections are not credited/penalized. If not stated otherwise, neighboring classes are ignored in the evaluation. In addition to ignored neighboring classes all persons annotations with the tags behind glass or sitting-lying are treated as ignore regions. Further, as mentioned in Section 3.2, EuroCity Persons Dataset Publication, ignore regions are used for cases where no precise bounding box annotation is possible (either because the objects are too small or because there are too many objects in close proximity which renders the instance based labeling infeasible). Since there is no precise information about the number or the location of objects in the ignore region, all unmatched detections which share an intersection of more than $0.5$ with these regions are not considered as false positives. Note that submissions with provided publication link and/or code will get priorized in below list (COMING SOON). Method▲ User LAMR (reasonable) LAMR (small) LAMR (occluded) LAMR (all) External data used Publication URL Publication code Submitted on Faster R-CNN ECP Team 0.101 0.196 0.381 0.251 ImageNet yes no 2019-04-01 17:06:33 View HRNet Hongsong Wang 0.061 0.138 0.287 0.183 ImageNet no no 2019-08-05 17:11:04 View R-FCN (with OHEM) ECP Team 0.163 0.245 0.507 0.330 ImageNet yes no 2019-04-01 17:10:03 View SSD ECP Team 0.131 0.235 0.460 0.296 ImageNet yes no 2019-04-02 13:56:14 View YOLOv3 ECP Team 0.097 0.186 0.401 0.242 ImageNet yes no 2019-04-01 17:08:05 View YOLOv3_640 HUI_Tsinghua-Daim... 0.273 0.564 0.623 0.456 no no 2019-05-17 04:56:27 View Method▲ User LAMR (reasonable) LAMR (small) LAMR (occluded) LAMR (all) External data used Publication URL Publication code Submitted on Faster R-CNN ECP Team 0.201 0.359 0.701 0.358 ImageNet yes no 2019-05-02 10:10:01 View FasterRCNN with M... Qihua Cheng 0.150 0.253 0.653 0.295 ImageNet no no 2019-07-08 08:48:13 View HRNet Hongsong Wang 0.079 0.156 0.265 0.153 ImageNet no no 2019-08-05 17:11:04 View
Following on from the basic branching processes introduction, we now calculate the expected number of individuals at the nth generation. As in the above example, let $Z_n$ be the number of individuals in generation n, and X be a random variable describing the number of offspring an individual has, with $E[X]=\mu$ and $Var[X]=\sigma^2$ $$ \begin{align} \therefore E[Z_n] & = G'_n(1) \\ &= G'_{n-1} \Big (G(1) \Big)G'(1) \\ &= G'_{n-1}(1)G'(1) \\ &= E[Z_{n-1}] \mu \\ &= \mu^2 E[Z_{n-2}] =... \\ &= \mu^n E[Z_0] \\ &= \mu^n \end{align} $$ Clearly the eventual population size is highly dependent on the value of $\mu$ So if each individual is expected to have more than one offspring, then the population will increase. If each individual is expected to have either one or no offspring, then the population will remain constant or decrease and eventually die out. Question: Why do elephants not die out, if the above comment on mean family size holds? What are the limitations of our model in representing the reproductive lifespan of an elephant? By evaluating the mean, we see that ultimate extinction is certain only when the mean family size is $\mu \leq 1$. To find this probability exactly, we let the probability of extinction at the nth generation be $\theta_n=P(Z_n=0)$ . So the probability of ultimate extinction is $\theta = lim_{n \to \infty} \theta_n=lim_{n \to \infty} P(X_n=0)$ . $$ \begin{align} \theta_n & = G_n(0) \\ & = G_{n-1} \Big(G(0)\Big) \\ &= G\Big(G\big(...(s)...\big)\Big) \\ &= G\Big(G_{n-1}(s)\Big) \\ &= G(\theta_{n-1}) \end{align} $$ So as $n \rightarrow \infty$ , we have $\theta_n\rightarrow \theta$ and $G(\theta_{n-1}) \rightarrow G(\theta)$ . And so we can find $\theta$ by solving $$\theta=G(\theta)$$ Now there may be other roots to this equation, so we show $\theta$ is the smallest by supposing $\alpha$ is also a root. Then $\theta_1=G(0) \leq G(\alpha)=\alpha \Rightarrow \theta_2=G(\theta_1) \leq G(\alpha)=\alpha$ And so proceeding by induction, $\theta =lim_{n \to \infty} \theta_n \leq \alpha$ , which shows $\theta$ is indeed the smallest non-negative root. The dependence of $\theta$ on the value of the mean family size, is shown in the diagrams below. The first case being $\mu \leq 1$ and the second case $\mu > 1$ . Example: In the previous elephant example, we now solve for $\theta$ in the equation $$ \begin{align} \theta & = G(\theta) \\ &= (1-p^n)+p^n \theta\\ \theta(1-p^n) &= (1-p^n) \\ \therefore \theta &=1 \end{align}$$
also, if you are in the US, the next time anything important publishing-related comes up, you can let your representatives know that you care about this and that you think the existing situation is appalling @heather well, there's a spectrum so, there's things like New Journal of Physics and Physical Review X which are the open-access branch of existing academic-society publishers As far as the intensity of a single-photon goes, the relevant quantity is calculated as usual from the energy density as $I=uc$, where $c$ is the speed of light, and the energy density$$u=\frac{\hbar\omega}{V}$$is given by the photon energy $\hbar \omega$ (normally no bigger than a few eV) di... Minor terminology question. A physical state corresponds to an element of a projective Hilbert space: an equivalence class of vectors in a Hilbert space that differ by a constant multiple - in other words, a one-dimensional subspace of the Hilbert space. Wouldn't it be more natural to refer to these as "lines" in Hilbert space rather than "rays"? After all, gauging the global $U(1)$ symmetry results in the complex line bundle (not "ray bundle") of QED, and a projective space is often loosely referred to as "the set of lines [not rays] through the origin." — tparker3 mins ago > A representative of RELX Group, the official name of Elsevier since 2015, told me that it and other publishers “serve the research community by doing things that they need that they either cannot, or do not do on their own, and charge a fair price for that service” for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing) @EmilioPisanty > for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing) @0celo7 but the bosses are forced because they must continue purchasing journals to keep up the copyright, and they want their employees to publish in journals they own, and journals that are considered high-impact factor, which is a term basically created by the journals. @BalarkaSen I think one can cheat a little. I'm trying to solve $\Delta u=f$. In coordinates that's $$\frac{1}{\sqrt g}\partial_i(g^{ij}\partial_j u)=f.$$ Buuuuut if I write that as $$\partial_i(g^{ij}\partial_j u)=\sqrt g f,$$ I think it can work... @BalarkaSen Plan: 1. Use functional analytic techniques on global Sobolev spaces to get a weak solution. 2. Make sure the weak solution satisfies weak boundary conditions. 3. Cut up the function into local pieces that lie in local Sobolev spaces. 4. Make sure this cutting gives nice boundary conditions. 5. Show that the local Sobolev spaces can be taken to be Euclidean ones. 6. Apply Euclidean regularity theory. 7. Patch together solutions while maintaining the boundary conditions. Alternative Plan: 1. Read Vol 1 of Hormander. 2. Read Vol 2 of Hormander. 3. Read Vol 3 of Hormander. 4. Read the classic papers by Atiyah, Grubb, and Seeley. I am mostly joking. I don't actually believe in revolution as a plan of making the power dynamic between the various classes and economies better; I think of it as a want of a historical change. Personally I'm mostly opposed to the idea. @EmilioPisanty I have absolutely no idea where the name comes from, and "Killing" doesn't mean anything in modern German, so really, no idea. Googling its etymology is impossible, all I get are "killing in the name", "Kill Bill" and similar English results... Wilhelm Karl Joseph Killing (10 May 1847 – 11 February 1923) was a German mathematician who made important contributions to the theories of Lie algebras, Lie groups, and non-Euclidean geometry.Killing studied at the University of Münster and later wrote his dissertation under Karl Weierstrass and Ernst Kummer at Berlin in 1872. He taught in gymnasia (secondary schools) from 1868 to 1872. He became a professor at the seminary college Collegium Hosianum in Braunsberg (now Braniewo). He took holy orders in order to take his teaching position. He became rector of the college and chair of the town... @EmilioPisanty Apparently, it's an evolution of ~ "Focko-ing(en)", where Focko was the name of the guy who founded the city, and -ing(en) is a common suffix for places. Which...explains nothing, I admit.
I discovered that the mathematics behind the two papers are not exactly the same. Definition of $Q$ and $N_k$ by Brown in 1991 and 1992 $Q=\frac{f_k}{\Delta f_k}=\frac{f_k}{f_{k+1}-f_k}=\frac{1}{2^{\frac{1}{b}}-1}$ ----(1A), where $b$ is the number of bins per octave. $N_k=\frac{f_s}{\Delta f_k}=\frac{f_s}{f_k}Q$ ----(2A), where $f_s$ is the sampling rate. Definition of $Q$ and $N_k$ by Schorkhuber in 2010 $Q = \frac{f_k}{\Delta f_k} = \frac{N_k f_k}{\Delta \omega f_s}$ ----(1B) $Q = \frac{q}{\Delta \omega (2^{\frac{1}{b}}-1)}$ ----(2B) By combining equation (1B) and (2B), you can get the window length $N_k$ as below $N_k = \frac{qf_s}{f_k(2^{\frac{1}{b}}-1)}$ ----(3B) where $\Delta \omega$ is the equivalent noise bandwidth, $q$ is the scaling factor. Here's the problem that I have If you try to equate (2A) and (3B) together, you will get another expression for Q. $\frac{f_s}{f_k}Q = \frac{qf_s}{f_k(2^{\frac{1}{b}}-1)}$ After simplifying, you will get the follow expression. $Q = \frac{q}{2^{\frac{1}{b}}-1}$. Which looks very similar to equation (2B) with a missing $\Delta \omega$ factor. So, there's 3 different expressions for $Q$ now? To me, it seems a contradiction. Either of the papers must be wrong in their definition on $Q$. Another mystery is where does the term $\Delta \omega$ come from? I don't see any necessity in adding this term. It seems to me the source of the contradiction.
To evaluate detection performance, we plot the miss-rate $mr(c) = \frac{fn(c)}{tp(c) + fn(c)}$ against the number of false positives per image $fppi(c)=\frac{fp(c)}{\text{#img}}$ in log-log plots. $tp(c)$ is the number of true positives, $fp(c)$ is the number of false positives, and $fn(c)$ is the number of false negatives, all for a given confidence value $c$ such that only detections are taken into account with a confidence value greater or equal than $c$. As commonly applied in object detection evaluation the confidence threshold $c$ is used as a control variable. By decreasing $c$, more detections are taken into account for evaluation resulting in more possible true or false positives, and possible less false negatives. We define the log average miss-rate (LAMR) as shown, where the 9 fppi reference points are equally spaced in the log space: $\DeclareMathOperator*{\argmax}{argmax}LAMR = \exp\left(\frac{1}{9}\sum\limits_f \log\left(mr(\argmax\limits_{fppi\left(c\right)\leq f} fppi\left(c\right))\right)\right)$ For each fppi reference point the corresponding mr value is used. In the absence of a miss-rate value for a given f the highest existent fppi value is used as new reference point. This definition enables LAMR to be applied as a single detection performance indicator at image level. At each image the set of all detections is compared to the groundtruth annotations by utilizing a greedy matching algorithm. An object is considered as detected (true positive) if the Intersection over Union (IoU) of the detection and groundtruth bounding box exceeds a pre-defined threshold. Due to the high non-rigidness of pedestrians we follow the common choice of an IoU threshold of 0.5. Since no multiple matches are allowed for one ground-truth annotation, in the case of multiple matches the detection with the largest score is selected, whereas all other matching detections are considered false positives. After the matching is performed, all non matched ground-truth annotations and detections, count as false negatives and false positives, respectively. Neighboring classes and ignore regions are used during evaluation. Neighboring classes involve entities that are semantically similar, for example bicycle and moped riders. Some applications might require their precise distinction ( enforce) whereas others might not ( ignore). In the latter case, during matching correct/false detections are not credited/penalized. If not stated otherwise, neighboring classes are ignored in the evaluation. In addition to ignored neighboring classes all persons annotations with the tags behind glass or sitting-lying are treated as ignore regions. Further, as mentioned in Section 3.2, EuroCity Persons Dataset Publication, ignore regions are used for cases where no precise bounding box annotation is possible (either because the objects are too small or because there are too many objects in close proximity which renders the instance based labeling infeasible). Since there is no precise information about the number or the location of objects in the ignore region, all unmatched detections which share an intersection of more than $0.5$ with these regions are not considered as false positives. Note that submissions with provided publication link and/or code will get priorized in below list (COMING SOON). Method User▲ LAMR (reasonable) LAMR (small) LAMR (occluded) LAMR (all) External data used Publication URL Publication code Submitted on Faster R-CNN ECP Team 0.101 0.196 0.381 0.251 ImageNet yes no 2019-04-01 17:06:33 View YOLOv3 ECP Team 0.097 0.186 0.401 0.242 ImageNet yes no 2019-04-01 17:08:05 View R-FCN (with OHEM) ECP Team 0.163 0.245 0.507 0.330 ImageNet yes no 2019-04-01 17:10:03 View SSD ECP Team 0.131 0.235 0.460 0.296 ImageNet yes no 2019-04-02 13:56:14 View YOLOv3_640 HUI_Tsinghua-Daim... 0.273 0.564 0.623 0.456 no no 2019-05-17 04:56:27 View HRNet Hongsong Wang 0.061 0.138 0.287 0.183 ImageNet no no 2019-08-05 17:11:04 View Method User▲ LAMR (reasonable) LAMR (small) LAMR (occluded) LAMR (all) External data used Publication URL Publication code Submitted on Faster R-CNN ECP Team 0.201 0.359 0.701 0.358 ImageNet yes no 2019-05-02 10:10:01 View FasterRCNN with M... Qihua Cheng 0.150 0.253 0.653 0.295 ImageNet no no 2019-07-08 08:48:13 View HRNet Hongsong Wang 0.079 0.156 0.265 0.153 ImageNet no no 2019-08-05 17:11:04 View
I am reading "Analysis, Geometry and Modeling in Finance". In section 2.10.2 which derives the quanto adjustment, it states that (in page 46) by definition the process $S_t^{d/f}S_t^f$ is the foreign asset valued in the domestic currency and therefore should be driven under $\mathbb{P}^d $ by $\frac{dS_t^{d/f}S_t^f}{S_t^{d/f}S_t^f} = r_ddt+\sigma_S.dW^d_t+\sigma_{d/f}.dW_t^d$ My question is: 1) What is the "$.$" in $\sigma_S.dW^d_t$ and $\sigma_{d/f}.dW_t^d$? I cannot find it in "Symbol Description" of the book. If it is just multiplication, why there is no correlation in the the multi-dimension Ito's lemma? (theorem 2.2 in page 21) It say $dW^i_t.dW^j_t=\delta_i^jdt$ 2) Why can we formulate the $S_t^{d/f}S_t^f$ like this? Is it an assumption?
This is essentially an addition to the list of @4tnemele I'd like to add some earlier work to this list, namely Discrete Gauge Theory. Discrete gauge theory in 2+1 dimensions arises by breaking a gauge symmetry with gauge group $G$ to some lower discrete subgroup $H$, via a Higgs mechanism. The force carriers ('photons') become massive which makes the gauge force ultra-short ranged. However, as the gauge group is not completely broken we still have the the Aharanov-Bohm effect. If H is Abelian this AB effect is essentially a 'topological force'. It gives rise to a phase change when one particle loops around another particle. This is the idea of fractional statistics of Abelian anyons. The particle types that we can construct in such a theory (i.e. the one that are "color neutral") are completely determined by the residual, discrete gauge group $H$. To be more precise: a particle is said to be charged if it carries a representation of the group H. The number of different particle types that carry a charge is then equal to the number of irreducible representations of the group H. This is similar to ordinary Yang-Mills theory where charged particles (quarks) carry the fundamental representation of the gauge group (SU(3). In a discrete gauge theory we can label all possible charged particle types using the representation theory of the discrete gauge group H. But there are also other types of particles that can exist, namely those that carry flux. These flux carrying particles are also known as magnetic monopoles. In a discrete gauge theory the flux-carrying particles are labeled by the conjugacy classes of the group H. Why conjugacy classes? Well, we can label flux-carrying particles by elements of the group H. A gauge transformation is performed through conjugacy, where $ \|g_i\rangle \rightarrow \|hg_ih^{-1}\rangle $ for all particle states $\|g_i\rangle$ (suppressing the coordinate label). Since states related by gauge transformations are physically indistinguishable the only unique flux-carrying particles we have are labeled by conjugacy classes. Is that all then? Nope. We can also have particles which carry both charge and flux -- these are known as dyons. They are labeled by both an irrep and a conjugacy class of the group $H$. But, for reasons which I wont go into, we cannot take all possible combinations of possible charges and fluxes. (It has to do with the distinguishability of the particle types. Essentially, a dyon is labeled by $\|\alpha, \Pi(g)\rangle$ where $\alpha$ is a conjugacy class and $\Pi(N(g))$ is a representation of the associated normalizer $N(\alpha)$ of the conjugacy class $\alpha$.) The downside of this approach is the rather unequal setting of flux carrying particles (which are labeled by conjugacy classes), charged particles (labeled by representations) and dyons (flux+compatible charge). A unifying picture is provided by making use of the (quasitriangular) Hopf algebra $D(H)$ also known as a quantum double of the group $H$. In this language all particles are (irreducible) representations of the Hopf algebra $D(H)$. A Hopf Algebra is endowed with certain structures which have very physical counterparts. For instance, the existence of a tensor product allows for the existence of multiple particle configurations (each particle labeled by their own representation of the Hopf algebra). The co-multiplication then defines how the algebra acts on this tensored space. the existence of an antipode (which is a certain mapping from the algebra to itself) ensures the existence of an anti-particle. The existence of a unit labels the vacuum (=trivial particle). We can also go beyond the structure of a Hopf algebra and include the notion of an R-matrix. In fact, the quasitriangular Hopf Algebra (i.e. the quantum double) does precisely this: add the R-matrix mapping. This R-matrix describes what happens when one particle loops around another particle (braiding). For non-Abelian groups $H$ this leads to non-Abelian statistics. These quasitriangular Hopf algebras are also known as quantum groups. Nowadays the language of discrete gauge theory has been replaced by more general structures, referred to by topological field theories, anyon models or even modular tensor categories. The subject is huge, very rich, very physical and a lot of fun (if you're a bit nerdy ;)). Sources: http://arxiv.org/abs/hep-th/9511201 (discrete gauge theory) http://www.theory.caltech.edu/people/preskill/ph229/ (lecture notes: check out chapter 9. Quite accessible!) http://arxiv.org/abs/quant-ph/9707021 (a simple lattice model with anyons. There are definitely more accessible review articles of this model out there though.) http://arxiv.org/abs/0707.1889 (review article, which includes potential physical realizations)This post imported from StackExchange Physics at 2015-11-01 19:23 (UTC), posted by SE-user Olaf
Cluster expansions¶ This page provides a short introduction to cluster expansions. For more extensive descriptions, please consult for example [SanDucGra84], [Wal09] or [AngMunRah19]. A simpler introduction to the subject can be found in the first part of this tutorial. Formalism¶ In the following, we are concerned with configurations corresponding to a distribution of $M$ different species over $N$ sites that can be written as a vector $\boldsymbol{\sigma}$, whose elements can assume $M$ different values, e.g., from $S=\{-m, -m+1, \ldots m-1, m\}$ where $M=2m$ (for even $M$) or $M=2m+1$ (for odd $M$) [SanDucGra84]. One now seeks to represent a property $Q$ of the system, such as the total energy, as a function of $\boldsymbol{\sigma}$, i.e. $Q = f(\boldsymbol{\sigma})$. To this end, one can construct a complete orthonormal basis of cluster functions $\Gamma_{\alpha}(\boldsymbol{\sigma})$ [SanDucGra84] [San10], which allows one to express $Q$ in the form [Wal09] Here, the sum extends over all symmetry equivalent clusters (orbit)$\alpha$, $m_{\alpha}$ denotes the multiplicity [1] whereas thecoefficients $J_{\alpha}$ are the so-called effective clusterinteractions (ECIs). The last term in the above expression represents theaverage over cluster functions $\Gamma_{\alpha}(\boldsymbol{\sigma})$belonging to symmetry equivalent clusters (orbits). The clusterfunctions themselves are obtained as a product over point functions$\Theta$. In icet, the point functions from [Wal09] areused: In icet, the formalism is handled internally by the ClusterSpace class. [1] Note that some authors include $m_{\alpha}$ in the symmetrized product over cluster functions $\left<\Gamma_{\alpha'}(\boldsymbol{\sigma})\right>_{\alpha}$. Cluster expansion construction¶ The task of training a CE can be formally written as a linear problem where $\boldsymbol{Q}$ is the vector of target properties, the vector $\boldsymbol{J}$ represents the ECIs, and $\mathbf{\bar{\Pi}}$ is a matrix that is obtained by stacking the vectors that represent each structure (also referred to in this documentation as cluster vectors) of the training set. This problem can be approached by choosing the number of structures $n_{\boldsymbol{Q}}$ (and thus the dimensionality of $\boldsymbol{Q}$), to be (much) larger than the number of ECIs $n_{\boldsymbol{J}}$ (and thus the dimensionality of $\boldsymbol{J}$, ($n_{\boldsymbol{Q}}>n_{\boldsymbol{J}}$). The set of equations is thus overdetermined. The optimal set of ECIs for fixed training set and cluster function basis is then obtained by minimizing the $l_2$-norm of $\mathbf{\bar{\Pi}} \boldsymbol{J} - \boldsymbol{Q}$ Common algorithms [Wal09] then proceed by generating a series of CEs corresponding to different basis set choices, i.e. different values of $n_{\boldsymbol{J}}$. By comparing the performance of each CE by means of a cross validation (CV) score the best performing CE is selected. An alternative approach is to choose the number of ECIs larger than the number of structures ($n_{\boldsymbol{J}}>n_{\boldsymbol{Q}}$). The resulting problem is then underdetermined, but can be solved by means of compressive sensing [CanWak08] [NelHarZho13] [NelOzoRee13]. In practice, one obtains a sparse solution, i.e., a solution in which many ECIs are zero. icet provides extensive functionality for thepresent task, see further under Optimizers.
I did cointegration test on two identical time series, and the result shows that they are not cointegrated, but intuitively, I think they are. Can anyone share some thoughts on this? Thanks! Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It only takes a minute to sign up.Sign up to join this community Let us test that $x$ and $y$ are co-integrated, say that $x_t, y_t \sim I(1)$. In the Engle-Granger we test stationarity of the error term in $$y_t = \alpha + \beta x_t + u_t$$ which we estimate as $$\hat u_t = y_t - \hat \alpha - \hat \beta x_t$$ and find that $\hat \alpha =0$, $\hat \beta = 1$, and $\hat u_t = 0 \; \forall t$. So now when we Dickey-Fuller test residuals in something like $$\Delta \hat u_t = \gamma_0 + \gamma_1 \hat u_{t-1} + \epsilon_t$$ nothing will be significant and we won't find any co-integration. I am not precisely schooled in this theory, so I'm not sure if this means these series can't be referred to as "co-integrated" (clearly they have the same drift) or if this is just a trivial case where the test fails, Two integrated series $X_t$ and $Y_t$ are cointegrated if their linear combination (some, not any) $\alpha X_t+\beta Y_t$ is stationary. If you have $P(X_t=Y_t)=1$ for all $t$, then $P(\alpha X_t+\beta Y_t=(\alpha+\beta) X_t)=1$. So according to definition of cointegration $(\alpha+\beta) X_t$ should be stationary, which is identical to $X_t$ being stationary. And here we get the contradiction, since $X_t$ is integrated, hence not stationary. This was a basic explanation why you received your result. However a lot depends on how the actual statistic is computed. For other statistics or their software implementations you might get that two identical series are cointegrated, but that will not mean that they are. Two identical time series are the degenerate case which no-one checks against, and with degenerate cases you can always get unexpected results. Your intuition is correct. $X_t$ and $Y_t$ are cointegrated if there exists some linear combination $\alpha X_t + \beta Y_t$ that is stationary (or more generally, of lower cointegration index --- see for example, Hamilton, pag 571). If $X_t = Y_t$, the above linear combination is zero (hence stationary) whenever $\alpha = -\beta$. On the other hand, most tests exclude this particular case. The exact reasons depend on the specific test you are using. Here is an empirical strategy to test for cointegration. FIRST, check whether both $X_t$ and $Y_t$ contain an unit root. If they are both non-stationary, and hence $I(1)$, then test for co-integration: Identical? So $Y_t=X_t$ for all $t$? Then the difference is zero which is more than just a stationary time-series. They are perfectly collinear. Now, it depends on the cointegration test you use whether high collinearity will show up as cointegration. If you use Engle-Granger, you do a regression first, which will find $\alpha=0,\beta=1$ and $\epsilon_t=0$. I'm not sure what an ADF test on $0$ will actually do, and this could lead to a numeric instability. I believe that a Johansen test will also lead to a numeric instability. If this is a serious problem for some automated system, then probably the first thing to do is look for collinearity (take the timeseries covariance matrix and look at the condition number). If they are highly collinear or identical, you should catch it before running it through the cointegration engine.
In a scientific paper, I need for my Bachelor thesis, the differential cross section for photo pair production $\gamma\gamma\to e^{+}e^{-}$ is given as follow: $\mathrm{d}\sigma = \frac{r_{0}^{2}\mathrm{d}\vec{p}_{+}\mathrm{d}\vec{p}_{-}}{2\epsilon_{+}\epsilon_{-}\omega_{1}\omega_{2}(1-\cos(\theta))}(B+4A-4A^{2})\delta(\vec{k}_{1}+\vec{k}_{2}-\vec{p}_{+}-\vec{p}_{-})\delta(\omega_{1}+\omega_{1}-\omega_{2}-\epsilon_{+}-\epsilon_{-})$ where $r_{0}$ is the classic electron radius, $A=\big(\frac{1}{x_{1}}+\frac{1}{x_{2}}\big )$ and $B=\big(\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{1}}\big )$. $x_{1}$ and $x_{2}$ are relativistic invariants: $x_{1}=-2p^{(4)}_{+}k^{(4)}_{1}$ and $x_{1}=-2p^{(4)}_{+}k^{(4)}_{2}$ with the four momenta: positron: $p_{+}^{(4)}=(\vec{p}_{+},i\epsilon_{+})$, electron: $p_{-}^{(4)}=(\vec{p}_{-},i\epsilon_{-})$, photon 1: $k_{1}^{(4)}=(\vec{k}_{1},i\omega_{1})$, photon 2: $k_{2}^{(4)}=(\vec{k}_{2},i\omega_{2})$ The problem now is that these quantities are written in natural units $m=c=\hbar=1$, but I need this in SI-Units. But how can I rewrite this formula in SI units? A first step is to write for example $p_{+}^{(4)}=(\vec{p}_{+},i\epsilon_{+}/c)$ and $\hbar k_{1}^{(4)}=(\hbar\vec{k}_{1},i\omega_{1}\frac{\hbar}{c})$. But this is not enough since B is dimensionless, but A is not. How is A+B defined as an example? The paper cites the book "Quantum electrodynamics" from Akhiezer (1965), but I don't find the book on the internet or in a library. Can anybody help me?
It's well-known that in System F, you can encode binary products with the type $$ A \times B \triangleq \forall\alpha.\; (A \to B \to \alpha) \to \alpha $$ You can then define projection functions $\pi_1 : A \times B \to A$ and $\pi_2 : A \times B \to B$. This isn't so surprising, even though the natural reading of the F type is of a pair with a let-style elimination $\mathsf{let}\;(x,y) = p \;\mathsf{in}\; e$, because the two kinds of pair are interderivable in intuitionistic logic. Now, in a dependent type theory with impredicative quantification, you can follow the same pattern to encode a dependent record type $\Sigma x:A.\; B[x]$ as $$ \Sigma x:A.\;B[x] \triangleq \forall\alpha.\; (\Pi x:A.\; B[x] \to \alpha) \to \alpha $$ But in this case, there isn't a simple way of defining the projective eliminators $\pi_1 : \Sigma x:A.\;B[x] \to A$ and $\pi_2 : \Pi p:(\Sigma x:A.\;B[x]).\; B[\pi_1\,p]$. However, if the type theory is parametric, you can use parametricity to show that $\pi_2$ is definable. This appears to be known --- see, for example, this Agda development by Dan Doel in which he derives it without comment --- but I can't seem to find a reference for this fact. Does anyone know a reference for the fact that parametricity allows defining projective eliminations for dependent types? EDIT: The closest thing I've found so far is this 2001 paper by Herman Geuvers, Induction is not derivable in second order dependent type theory, in which he proves that you can't do it without parametricity.
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 """Definition of the canonical ensemble class.""" import numpy as np from ase import Atoms from ase.units import kB from typing import List from .. import DataContainer from ..calculators.base_calculator import BaseCalculator from .thermodynamic_base_ensemble import ThermodynamicBaseEnsemble class CanonicalEnsemble(ThermodynamicBaseEnsemble): """Instances of this class allow one to simulate systems in the canonical ensemble (:math:`N_iVT`), i.e. at constant temperature (:math:`T`), number of atoms of each species (:math:`N_i`), and volume (:math:`V`). The probability for a particular state in the canonical ensemble is proportional to the well-known Boltzmann factor, .. math:: \\rho_{\\text{C}} \\propto \\exp [ - E / k_B T ]. Since the concentrations or equivalently the number of atoms of each species is held fixed in the canonical ensemble, a trial step must conserve the concentrations. This is accomplished by randomly picking two unlike atoms and swapping their identities. The swap is accepted with probability .. math:: P = \\min \\{ 1, \\, \\exp [ - \\Delta E / k_B T ] \\}, where :math:`\\Delta E` is the change in potential energy caused by the swap. The canonical ensemble provides an ideal framework for studying the properties of a system at a specific concentration. Properties such as potential energy or phenomena such as chemical ordering at a specific temperature can conveniently be studied by simulating at that temperature. The canonical ensemble is also a convenient tool for "optimizing" a system, i.e., finding its lowest energy chemical ordering. In practice, this is usually achieved by simulated annealing, i.e. the system is equilibrated at a high temperature, after which the temperature is continuously lowered until the acceptance probability is almost zero. In a well-behaved system, the chemical ordering at that point corresponds to a low-energy structure, possibly the global minimum at that particular concentration. Parameters ---------- structure : :class:`Atoms <ase.Atoms>` atomic configuration to be used in the Monte Carlo simulation; also defines the initial occupation vector calculator : :class:`BaseCalculator <mchammer.calculators.ClusterExpansionCalculator>` calculator to be used for calculating the potential changes that enter the evaluation of the Metropolis criterion temperature : float temperature :math:`T` in appropriate units [commonly Kelvin] boltzmann_constant : float Boltzmann constant :math:`k_B` in appropriate units, i.e. units that are consistent with the underlying cluster expansion and the temperature units [default: eV/K] user_tag : str human-readable tag for ensemble [default: None] data_container : str name of file the data container associated with the ensemble will be written to; if the file exists it will be read, the data container will be appended, and the file will be updated/overwritten random_seed : int seed for the random number generator used in the Monte Carlo simulation ensemble_data_write_interval : int interval at which data is written to the data container; this includes for example the current value of the calculator (i.e. usually the energy) as well as ensembles specific fields such as temperature or the number of atoms of different species data_container_write_period : float period in units of seconds at which the data container is written to file; writing periodically to file provides both a way to examine the progress of the simulation and to back up the data [default: np.inf] trajectory_write_interval : int interval at which the current occupation vector of the atomic configuration is written to the data container. sublattice_probabilities : List[float] probability for picking a sublattice when doing a random swap. This should be as long as the number of sublattices and should sum up to 1. Example ------- The following snippet illustrate how to carry out a simple Monte Carlo simulation in the canonical ensemble. Here, the parameters of the cluster expansion are set to emulate a simple Ising model in order to obtain an example that can be run without modification. In practice, one should of course use a proper cluster expansion:: from ase.build import bulk from icet import ClusterExpansion, ClusterSpace from mchammer.calculators import ClusterExpansionCalculator from mchammer.ensembles import CanonicalEnsemble # prepare cluster expansion # the setup emulates a second nearest-neighbor (NN) Ising model # (zerolet and singlet ECIs are zero; only first and second neighbor # pairs are included) prim = bulk('Au') cs = ClusterSpace(prim, cutoffs=[4.3], chemical_symbols=['Ag', 'Au']) ce = ClusterExpansion(cs, [0, 0, 0.1, -0.02]) # prepare initial configuration structure = prim.repeat(3) for k in range(5): structure[k].symbol = 'Ag' # set up and run MC simulation calc = ClusterExpansionCalculator(structure, ce) mc = CanonicalEnsemble(structure=structure, calculator=calc, temperature=600, data_container='myrun_canonical.dc') mc.run(100) # carry out 100 trial swaps """ def __init__(self, structure: Atoms, calculator: BaseCalculator, temperature: float, user_tag: str = None, boltzmann_constant: float = kB, data_container: DataContainer = None, random_seed: int = None, data_container_write_period: float = np.inf, ensemble_data_write_interval: int = None, trajectory_write_interval: int = None, sublattice_probabilities: List[float] = None) -> None: self._ensemble_parameters = dict(temperature=temperature) # add species count to ensemble parameters symbols = set([symbol for sub in calculator.sublattices for symbol in sub.chemical_symbols]) for symbol in symbols: key = 'n_atoms_{}'.format(symbol) count = structure.get_chemical_symbols().count(symbol) self._ensemble_parameters[key] = count super().__init__( structure=structure, calculator=calculator, user_tag=user_tag, data_container=data_container, random_seed=random_seed, data_container_write_period=data_container_write_period, ensemble_data_write_interval=ensemble_data_write_interval, trajectory_write_interval=trajectory_write_interval, boltzmann_constant=boltzmann_constant) 163 ↛ 166line 163 didn't jump to line 166, because the condition on line 163 was never false if sublattice_probabilities is None: self._swap_sublattice_probabilities = self._get_swap_sublattice_probabilities() else: self._swap_sublattice_probabilities = sublattice_probabilities @property def temperature(self) -> float: """ Current temperature """ return self._ensemble_parameters['temperature'] def _do_trial_step(self): """ Carries out one Monte Carlo trial step. """ sublattice_index = self.get_random_sublattice_index(self._swap_sublattice_probabilities) return self.do_canonical_swap(sublattice_index=sublattice_index)
Let $Y$ be a smooth projective connected curve of genus $g>0$ over $\overline{\mathbf{Q}}$. Let $h_{\textrm{Fal}}(Y)$ be the Faltings height of $Y$. Question 1. Can one classify or describe the curves $Y$ such that $h_{\textrm{Fal}}(Y) \geq 1$? Question 2. For any $g>0$, does there exist a curve $Y$ of genus $g$ such that $h_{\textrm{Fal}}(Y) <1$? Essentially, I would like to know which curves one is excluding by looking at curves $Y$ such that $h_{\textrm{Fal}}(Y) \geq 1$. A result of Bost says that the stable Faltings height of an abelian variety $A$ over $\overline{\mathbf{Q}}$ of dimension $g$ is bounded from below by $-\frac{1}{2}\log(2\pi)g$. By the Northcott property of the Faltings height, the set of curves of genus $g$ with $h_{\textrm{Fal}}(Y) <1$ is finite. This means that I'm looking at the finite set of curves of genus $g$ with Faltings height not in the interval $$[-\frac{1}{2}\log(2\pi)g,1)\subset[-2/5g, 1).$$ Added: To answer Junkie's question, I'm aware of only one definition of the Faltings height of a curve over $\overline{\mathbf{Q}}$. There are several equivalent definitions, though. Let $X$ be a smooth projective curve of genus $g>0$ over $\overline{\mathbf{Q}}$. Let $K$ be a number field such that $X$ has a semi-stable regular model $p:\mathcal{X}\to \mathrm{Spec} O_K$ over the ring of integers $O_K$ of $K$. Then, the Faltings height $h_{\mathrm{Fal}}(X)$ of $X$ is the arithmetic degree $$h_{\mathrm{Fal}}(X):=\frac{\widehat{\mathrm{deg}} Rp_\ast \mathcal{O}_{\mathcal{X}}}{[K:\mathbf{Q}]},$$ where we endow the determinant of cohomology with the Arakelov-Faltings metric. This is well-defined, i.e., independent of the field $K$. By Serre duality, it coincides with $$h_{\mathrm{Fal}}(X)=\frac{\widehat{\mathrm{deg}} p_\ast \mathcal{\omega}_{\mathcal{X}/O_K}}{[K:\mathbf{Q}]}.$$ It also coincides with the Faltings height of the Jacobian. All of this is explained in Section 4.4 of For a curve over a number field, there is also the important relative Faltings height. This invariant depends on the number field $K$, though.
Search Now showing items 1-1 of 1 Anisotropic flow of inclusive and identified particles in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV with ALICE (Elsevier, 2017-11) Anisotropic flow measurements constrain the shear $(\eta/s)$ and bulk ($\zeta/s$) viscosity of the quark-gluon plasma created in heavy-ion collisions, as well as give insight into the initial state of such collisions and ...
Search Now showing items 1-2 of 2 D-meson nuclear modification factor and elliptic flow measurements in Pb–Pb collisions at $\sqrt {s_{NN}}$ = 5.02TeV with ALICE at the LHC (Elsevier, 2017-11) ALICE measured the nuclear modification factor ($R_{AA}$) and elliptic flow ($\nu_{2}$) of D mesons ($D^{0}$, $D^{+}$, $D^{⁎+}$ and $D^{s+}$) in semi-central Pb–Pb collisions at $\sqrt{s_{NN}} =5.02$ TeV. The increased ... ALICE measurement of the $J/\psi$ nuclear modification factor at mid-rapidity in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV (Elsevier, 2017-11) ALICE at the LHC provides unique capabilities to study charmonium production at low transverse momenta ( p T ). At central rapidity, ( \|y\|<0.8 ), ALICE can reconstruct J/ ψ via their decay into two electrons down to zero ...
Search Now showing items 21-30 of 209 Measurement of the top quark mass in the $t\bar t \to {\rm lepton+jets}$ and $t\bar t \to {\rm dilepton}$ channels using $\sqrt{s}=7$ TeV ATLAS data (Springer, 2015-07) The top quark mass was measured in the channels $t\bar{t} \to \mathrm{lepton+jets}$ and $t\bar{t} \to \mathrm{dilepton}$ (lepton=$e, \mu$) based on ATLAS data recorded in 2011. The data were taken at the LHC with a ... Constraints on the off-shell Higgs boson signal strength in the high-mass $ZZ$ and $WW$ final states with the ATLAS detector (Springer, 2015-07) Measurements of the $ZZ$ and $WW$ final states in the mass range above the $2m_Z$ and $2m_W$ thresholds provide a unique opportunity to measure the off-shell coupling strength of the Higgs boson. This paper presents a ... Search for heavy lepton resonances decaying to a $Z$ boson and a lepton in $pp$ collisions at $\sqrt{s}=8$ TeV with the ATLAS detector (Springer, 2015-09) A search for heavy leptons decaying to a $Z$ boson and an electron or a muon is presented. The search is based on $pp$ collision data taken at $\sqrt{s}=8$ TeV by the ATLAS experiment at the CERN Large Hadron Collider, ... Evidence for the Higgs-boson Yukawa coupling to tau leptons with the ATLAS detector (Springer, 2015-04-21) Results of a search for $H \to \tau \tau$ decays are presented, based on the full set of proton--proton collision data recorded by the ATLAS experiment at the LHC during 2011 and 2012. The data correspond to integrated ... Search for new phenomena in final states with an energetic jet and large missing transverse momentum in $pp$ collisions at $\sqrt{s}=8~$TeV with the ATLAS detector (Springer, 2015-07) Results of a search for new phenomena in final states with an energetic jet and large missing transverse momentum are reported. The search uses 20.3 fb${}^{-1}$ of $\sqrt{s}=8$ TeV data collected in 2012 with the ATLAS ... Search for low-scale gravity signatures in multi-jet final states with the ATLAS detector at $\sqrt{s}$ = 8 TeV (Springer, 2015-07) We search for evidence of physics beyond the Standard Model in the production of final states with multiple high transverse momentum jets, using 20.3 $fb^{-1}$ of proton-proton collision data recorded by the ATLAS detector ... Search for non-pointing and delayed photons in the diphoton and missing transverse momentum final state in 8 TeV $pp$ collisions at the LHC using the ATLAS detector (American Physical Society, 2014-12-10) A search has been performed, using the full 20.3 fb$^{-1}$ data sample of 8 TeV proton--proton collisions collected in 2012 with the ATLAS detector at the LHC, for photons originating from a displaced vertex due to the ... Search for squarks and gluinos with the ATLAS detector in final states with jets and missing transverse momentum using √s=8 TeV proton-proton collision data (Springer, 2014-09) A search for squarks and gluinos in final states containing high-pT jets, missing transverse momentum and no electrons or muons is presented. The data were recorded in 2012 by the ATLAS experiment in √s = 8TeV proton-proton ... Search for scalar diphoton resonances in the mass range 65-600 GeV with the ATLAS detector in pp collision data at √s = 8 TeV (American Physical Society, 2014-10-20) A search for scalar particles decaying via narrow resonances into two photons in the mass range 65–600 GeV is performed using 20.3 fb⎻¹ of s√=8 TeV pp collision data collected with the ATLAS detector at the Large Hadron ... Measurement of the b-hadron production cross section using decays to D*+ μ − X final states in pp collisions at √s = 7 TeV with the ATLAS detector (Elsevier, 2012-11-21) The b-hadron production cross section is measured with the ATLAS detector in pp collisions at √s = 7 TeV, using 3.3 pb−1 of integrated luminosity, collected during the 2010 LHC run. The b-hadrons are selected by partially ...
Pentation Penetration is Fifth of the Ackermann function. Pentation is denoted with symbol pen; $\mathrm{pen}_b(z)=A_{b, 5}(z)$ where $A_{b,5}$ is Fifth Ackermann function to base $b$. The explicit plot and the complex map of the natural pentation are shown in figures at right. General properties Pentation satisfies the transfer equation $\mathrm{pen}_b(z\!+\!1) = \mathrm{tet}_b\Big( \mathrm{pen}_b(z)\Big)$ where $\mathrm{tet}_b$ is tetration to base $b$. The additional condition, common for all ackermann functions is assumed, $\mathrm{pen}_b(0)=1$ Pentation is holomorphic at least in the part of the complex plane, while the real part of the argument does not exceed some constant. For $b\!=\!\mathrm e$, this constant is about $-2.5$ The range of holomorphism of pentation includes alto the real axis. For the real base $b$, larger that unity, pentation is real–holomorphic, $\mathrm{pen}_b(z^)=\mathrm{pen}_b(z)^$ Pentation has the exponential asymptotic at large negative values of the real part of the argument, $\mathrm{pen}_b(z) = L + \varepsilon + O(\varepsilon^2)$ $\varepsilon=\exp(k (z-X))$, parameters $k$ and $X$ depend on the base $b$. For base $b\!>\!1$, the increment $k$ is positive. This increment is determined by the derivative of tetration at its lowest real fixed point, and in this case, pentation is periodic function; the period $P=2 \pi \mathrm i /k$. Properties of the natural pentation The base $b$ of pentation is indicated as subscript, but it it is e, the subsctipt can be omitted, as in the case of tetration, arctetration, exponential and logarithm. Natural pentation has the following properties: the limiting value at minus infinity is fuxed point of the natural tetration, $L_{\mathrm e, 4,0} \!\approx\! -1.8503545290271812$ Incement at negative valies of real part is $k\approx 1.86573322821$ Period $P=2\pi \mathrm i /k \approx 3.36767615657879\, \mathrm i$ The period is pure imaginary, so, the map reproduces at the translations along the imaginary axis. A little bit more than two periods are covered by the range of the complex map at right. Natural pentation has the countable set of logarithmic singularities. The cut lines divide the right hand side of the complex plane to the "independent" strips of width $P$. There is no way to direct these cut lines to the left, because more singularities are located at each of them. Along the real axis, at minus infinity, the function pen grows exponentially from its asymptotic value $L_0\approx -1.8503545290271812$ The fraphic passes through points with icoordinates $(-3,-1)$, $(-1,0)$, $(0,1)$, $(1,\mathrm e)$. Visually, $\mathrm{pen}(2)$ looks as infinity, and, perhaps, $\mathrm{pen}(3)$ can be considered a "numerical approximation" of the fake Mizugadro number. As tetration has singularity and the branch point at $-2$, this singularity determines the set of singularities in vicinity of the real axis. The pair of these singularities are seen at points with coordinates approximately $(3.1,\pm 3.1)$; the zooming of the map allows to see them in details. References http://article.sciencepublishinggroup.com/pdf/10.11648.j.acm.20140306.14.pdf http://mizugadro.mydns.jp/PAPERS/2014acker.pdf D.Kouznetsov. Evaluation of holomorphic ackermanns. Applied and Computational Mathematics. Vol. 3, No. 6, 2014, pp. 307-314. http://www.ams.org/mcom/2009-78-267/S0025-5718-09-02188-7/home.html http://www.ils.uec.ac.jp/~dima/PAPERS/2009analuxpRepri.pdf D.Kouznetsov. Analytic solution of F(z+1)=exp(F(z)) in complex z-plane. Mathematics of Computation, v.78 (2009), 1647-1670. http://www.ils.uec.ac.jp/~dima/PAPERS/2010vladie.pdf D.Kouznetsov. Superexponential as special function. Vladikavkaz Mathematical Journal, 2010, v.12, issue 2, p.31-45. https://www.morebooks.de/store/ru/book/Суперфункции/isbn/978-3-659-56202-0 http://mizugadro.mydns.jp/BOOK/202.pdf D.Kouznetsov. Superfunctions (in Russian): Д.Кузнецов. Суперфункции. Lambert Academic Publishing, 2014.
Consider the integer lattice in $2d$, namely the set $\mathbb{Z}^2 = \{(x,y): x,y\in \mathbb{Z}\}$, and let $u:\mathbb{Z}^2 \to \mathbb{R} $ be a function defined on some bounded subset of $\mathbb{Z}^2$. The discrete (graph) Laplace of $u$, denote it by $\Delta^1$, at point $(x,y)$ is defined as $$ \tag{1} \Delta^1 u(x,y) = u(x + 1, y) + u(x-1,y) + u(x, y+1) + u(x, y-1) - 4u(x,y), $$ i.e. we sum over all lattice neighbors of $(x,y)$ and subtract the value of the function at the given point multiplied by $4$ (which is the $2d$ since dimension $d=2$). To compute the discrete Laplace numerically, I represent the lattice (part of it, of course) as a $2d$-array, and thus the values of $u$ are being stored in a $2d$ matrix replicating the grid. Given the nature of $\Delta^1$, when I need to compute it at some index $(i,j)$ of my array, then I need to jump over a row of the array to access the values of neighbors of $(i,j)$. This is obviously not an efficient way of accessing the array, as we force the pointer to repeatedly jump over large chunks of memory locations. Indeed, for certain problems when computing the Laplace of a given function excessively (in some iterative process, for example), even if the size of the matrix is moderately small (say $512\times512$ ) the computation time is enormous compared to the one with the same number of operations but without array access. The question: is there an efficient way (a data structure) to represent a $2d$ grid to optimize for array access operation of the form described above? Say, specifically for access pattern as in $(1)$ above. I am aware of spacial and temporal optimization of arrays, as well as that repeatedly accessing the same memory locations the compiler might promote those pointers into registry. However, the access pattern in my case is not very regular. The question is rather naive of course, but as a $2d$ grid is a ubiquitous object, I'd much appreciate if you can share your insights on efficient data structures concerning it.
The action $\int d\theta$ has a gauge symmetry $d\theta \mapsto d\theta + df$. The gauge equivalence classes are discrete (labelled by the winding number), so the equations of motion don't say anything (and the gauge field is automatically flat for dimensional reasons). The action is integer-valued, so you get a different "TQFT" for every parameter $\alpha \in U(1)$ by $S_\alpha = \alpha \int d\theta$.
Could anyone help with this question: Edit: Reducing it to one of these options would be great! 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 This question appears to be off-topic. The users who voted to close gave this specific reason: I will assume that everything is well defined and real, not paying attention to whether logarithms are taken of positive numbers and such. We have $y=a^{x^y}$ (not $y=(a^x)^y$ which would mean $y=a^{{xa}^{{xa}^{\cdots}}}$). Let me denote the derivative operator by $D$. First we need a little lemma: If $f$ and $g$ are functions of $x$, then $$ D(f^g) = De^{g\log(f)} = e^{g\log(f)}D(g\log(f)) = f^g(g'\log(f)+gf'/f). $$ (There is an easy way to memorize this. If $f$ is constant, the derivative is $f^g\log(f)g'$. If $g$ is constant, the derivative is $gf^{g-1}$. The full derivative is the sum of these two. This argument can be formalized, but it'd be a sidetrack here.) Applying this to $f(x)=x$ and $g(x)=y(x)$ gives $$ D(x^y) = x^y(y'\log(x)+y/x). $$ This will be useful soon. Taking the derivative gives \begin{align} y' &= D(a^{x^y}) \\&= a^{x^y}\log(a)D(x^y) \\&= y\log(a)D(x^y) %\\&= %y\log(a)[yx^{y-1}+\log(x)x^yy'] \\&= y\log(a)x^y[y/x+\log(x)y'] \\&= y\log(a^{x^y})[y/x+\log(x)y'] \\&= y\log(y)[y/x+\log(x)y'] . \end{align} This gives $$ [1-\log(x)y\log(y)]y' = y^2\log(y)/x, $$ from which we can solve $$ y'=\frac{y^2\log(y)}{x[1-\log(x)y\log(y)]}. $$ This does not seem to match any of the options given. Here is an alternative form, using $x^y=\log(y)/\log(a)$: $$ y'=\frac{y^2\log(y)}{x\left[1-\log\left(\frac{\log(y)}{\log(a)}\right)\log(y)\right]}. $$ Perhaps I miss a way to manipulate the formula, or perhaps the problem is mistaken; as others have pointed out, taking $y=(a^x)^y$ leads to option C. \begin{align} y&=a^{x^{a^{x\cdots}}}\\ \implies y&=a^{x^y}\\ \implies \ln y&=x^y\ln a\hspace{25pt}\cdots\text{(i)}\\ \implies \dfrac{1}{y}\dfrac{dy}{dx}&=\ln a\cdot\dfrac{d}{dx}\left(x^y\right)\\ &=\ln a\cdot\dfrac{d}{dx}\left(e^{y\ln x}\right)\\ &=\ln a\cdot e^{y\ln x}\cdot\dfrac{d}{dx}(y\ln x)\\ &=\ln a\cdot x^y\cdot\left(\dfrac{y}{x}+\ln x\cdot\dfrac{dy}{dx}\right)\\ \implies \dfrac{dy}{dx}&=y\cdot x^y\ln a\cdot\left(\dfrac{y}{x}+\ln x\cdot\dfrac{dy}{dx}\right)\\ \implies \dfrac{dy}{dx}\left(1-y\cdot x^y\ln a\cdot\ln x\right)&=\dfrac{y^2}{x}\cdot x^y\ln a\\ \implies \dfrac{dy}{dx}\left(1-y\cdot\ln y\cdot\ln x\right)&=\dfrac{y^2}{x}\cdot\ln y\hspace{25pt}\text{ as from (i) }[x^y\ln a=\ln y]\\ \implies \dfrac{dy}{dx}&=\dfrac{y^2\ln y}{x\left(1-y\cdot\ln y\cdot\ln x\right)} \end{align} So, this doesn't matches any answer. Building blocks: $$ D\left( a^{t(x)}, x\right) = a^x \ln (a) t'(x), \qquad D\left( x^x, x\right) = x^x (1 + \ln(x)) $$ First chains $$ \begin{align} D\left( a^{x}, x\right) &= a^x \ln (a) \\ D\left( a^{x^{x}}, x\right) &= x^x a^{x^x} \ln (a) (\ln (x)+1) \\ D\left( a^{x^{x^{x}}}, x\right)&= a^{x^{x^x}} x^{x^x} \ln (a) \left(x^{x-1}+x^x \ln (x) (\ln (x)+1)\right) \\ D\left( \left( \left( \left( a^x \right)^x \right)^x \right)^x, x \right) &= \left(\left(\left(a^x\right)^x\right)^x\right)^x \left(x \left(x \left(\ln \left(a^x\right)+x \ln (a)\right)+\ln \left(\left(a^x\right)^x\right)\right)+\ln \left(\left(\left(a^x\right)^x\right)^x\right)\right) \end{align} $$
At the beginning, you may start with four (component) functions $A_{\mu}(x,y,z,t)$ with $\mu=0,1,2,3$. But then you may choose one function $\lambda(x,y,z,t)$ (pronounce "lambda") and gauge transform $A_\mu$ to$$ A'_\mu (x,y,z,t) = A_\mu (x,y,z,t) +\frac{\partial}{\partial x^\mu} \lambda $$which produces a new configuration $A'_\mu$ that we must consider physically identical. So we have 4 numbers per point up to 1 number that doesn't matter, leaving us with three. However, there is one more reduction because one of the Maxwell's equations, Gauss' law$$ {\rm div}(\vec D) = \rho $$doesn't really contain time derivatives, so it restricts the initial states. Equivalently, the time derivative of Gauss' law may be derived from the other Maxwell equations, so there is another number at each point that is "redundant". This leaves us with two physical off-shell fields per point. It's perhaps clearer to count the "on-shell" degrees of freedom, the number of actual waves and their polarizations. Considering waves moving in the vacuum with wave vector $\vec k$ and frequency $\omega$. They are proportional to$$ \exp(i\vec k \cdot \vec x - i\omega t) $$where $\omega=c \|\vec k\|$. The full $A_\mu$ is$$ A_\mu = \epsilon_\mu \exp(i\vec k \cdot \vec x - i\omega t) $$and there are a priori four independent solutions. However, because of the $\lambda$-related gauge symmetry, if you change $\epsilon^\mu$ by a multiple of $k^\mu$, you get a physically identical electromagnetic wave. These $\epsilon\sim k^\mu$ waves are "pure gauge". The other degree of freedom is lost because we may impose the condition$$ \epsilon_\mu k^\mu = 0$$the Lorentz gauge, without a loss of generality. This reduces "another" non-transverse degree of freedom because the null vectors $\epsilon^\mu$ that are orthogonal to the null $k^\mu$ (and therefore allowed) are proportional to it, while these values of $\epsilon^\mu$ proportional to $k^\mu$ are exactly those that are unphysical by the gauge symmetry. To summarize, only the two transverse polarizations are physical. For example, if the wave is moving in the $z$-direction, it's only the $x$- and $y$- linearly polarized waves (or the two circular polarizations which are their combinations) that are physical. Both the longitudinal and the timelike wave are unphysical because of the "two applications" of the gauge symmetry above.
I came across this expression while doing exercises and I was wondering if it was a 'real' expression. $$\mathcal{L}=\frac{1}{2}\partial_\mu \phi \partial^\mu \phi -\frac{m^2}{2}\phi ^2 +\overline{\psi}(i\gamma^\mu\partial_\mu -M)\psi + i\lambda \phi \overline{\psi}\gamma_5\psi $$ I can recognize the first term is a kinetic term, the second as a potential term and the middle one as the Dirac Lagrangian. The last one is an interaction term, so it involves the interaction between electrons and a spin 0 particle, which one? Does this Lagrangian density represent a specific theory?
14.1. Generative Adversarial Networks¶ Throughout most of this book, we’ve talked about how to make predictions. In some form or another, we used deep neural networks learned mappings from data points to labels. This kind of learning is called discriminative learning, as in, we’d like to be able to discriminate between photos cats and photos of dogs. Classifiers and regressors are both examples of discriminative learning. And neural networks trained by backpropagation have upended everything we thought we knew about discriminative learning on large complicated datasets. Classification accuracies on high-res images has gone from useless to human-level (with some caveats) in just 5-6 years. We’ll spare you another spiel about all the other discriminative tasks where deep neural networks do astoundingly well. But there’s more to machine learning than just solving discriminative tasks. For example, given a large dataset, without any labels, we might want to learn a model that concisely captures the characteristics of this data. Given such a model, we could sample synthetic data points that resemble the distribution of the training data. For example, given a large corpus of photographs of faces, we might want to be able to generate a new photorealistic image that looks like it might plausibly have come from the same dataset. This kind of learning is called generative modeling. Until recently, we had no method that could synthesize novel photorealistic images. But the success of deep neural networks for discriminative learning opened up new possibilities. One big trend over the last three years has been the application of discriminative deep nets to overcome challenges in problems that we don’t generally think of as supervised learning problems. The recurrent neural network language models are one example of using a discriminative network (trained to predict the next character) that once trained can act as a generative model. In 2014, a breakthrough paper introduced Generative adversarial networks (GANs) [Goodfellow.Pouget-Abadie.Mirza.ea.2014], a clever new way to leverage the power of discriminative models to get good generative models. At their heart, GANs rely on the idea that a data generator is good if we cannot tell fake data apart from real data. In statistics, this is called a two-sample test - a test to answer the question whether datasets $X=\{x_1,\ldots,x_n\}$ and $X'=\{x'_1,\ldots,x'_n\}$ were drawn from the same distribution. The main difference between most statistics papers and GANs is that the latter use this idea in a constructive way. In other words, rather than just training a model to say “hey, these two datasets don’t look like they came from the same distribution”, they use the two-sample test to provide training signal to a generative model. This allows us to improve the data generator until it generates something that resembles the real data. At the very least, it needs to fool the classifier. And if our classifier is a state of the art deep neural network. The GANs architecture is illustrated in Fig. 14.1.1. As you can see, there are two pieces to GANs - first off, we need a device (say, a deep network but it really could be anything, such as a game rendering engine) that might potentially be able to generate data that looks just like the real thing. If we are dealing with images, this needs to generate images. If we’re dealing with speech, it needs to generate audio sequences, and so on. We call this the generator network. The second component is the discriminator network. It attempts to distinguish fake and real data from each other. Both networks are in competition with each other. The generator network attempts to fool the discriminator network. At that point, the discriminator network adapts to the new fake data. This information, in turn is used to improve the generator network, and so on. The discriminator is a binary classifier to distinguish if the input $x$ is real (from real data) or fake (from the generator). Typically, the discriminator outputs a scalar prediction $o\in\mathbb R$ for input $\mathbf x$, such as using a dense layer with hidden size 1, and then applies sigmoid function to obtain the predicted probability $D(\mathbf x) = 1/(1+e^{-o})$. Assume the label $y$ for true data is $1$ and $0$ for fake data. We train the discriminator to minimize the cross entropy loss, i.e. For the generator, it first draws some parameter $\mathbf z\in\mathbb R^d$ from a source of randomness, e.g. a normal distribution $\mathbf z\sim\mathcal(0,1)$. We often call $\mathbf z$ the latent variable. It then applies a function to generate $\mathbf x'=G(\mathbf z)$. The goal of the generator is to fool the discriminator to classify $\mathbf x'$ as true data. In other words, we update the parameters of the generator to maximize the cross entropy loss when $y=0$, i.e. If the discriminator does a perfect job, then $D(\mathbf x')\approx 1$ so the above loss near 0, which results the gradients are too small to make a good progress for the discriminator. So commonly we minimize the following loss which is just feed $\mathbf x'$ into the discriminator but giving label $y=1$. Many of the GANs applications are in the context of images. As a demonstration purpose, we’re going to content ourselves with fitting a much simpler distribution first. We will illustrate what happens if we use GANs to build the world’s most inefficient estimator of parameters for a Gaussian. Let’s get started. %matplotlib inlineimport d2lfrom mxnet import nd, gluon, autograd, initfrom mxnet.gluon import nn 14.1.1. Generate some “real” data¶ Since this is going to be the world’s lamest example, we simply generate data drawn from a Gaussian. X = nd.random.normal(shape=(1000, 2))A = nd.array([[1, 2], [-0.1, 0.5]])b = nd.array([1, 2])data = nd.dot(X, A) + b Let’s see what we got. This should be a Gaussian shifted in some rather arbitrary way with mean $b$ and covariance matrix $A^TA$. d2l.set_figsize((3.5, 2.5))#d2l.plt.figure(figsize=())d2l.plt.scatter(data[:100,0].asnumpy(), data[:100,1].asnumpy());print("The covariance matrix is", nd.dot(A.T,A)) The covariance matrix is[[1.01 1.95] [1.95 4.25]]<NDArray 2x2 @cpu(0)> batch_size = 8data_iter = d2l.load_array((data,), batch_size) 14.1.2. Generator¶ Our generator network will be the simplest network possible - a single layer linear model. This is since we’ll be driving that linear network with a Gaussian data generator. Hence, it literally only needs to learn the parameters to fake things perfectly. net_G = nn.Sequential()net_G.add(nn.Dense(2)) 14.1.3. Discriminator¶ For the discriminator we will be a bit more discriminating: we will use an MLP with 3 layers to make things a bit more interesting. net_D = nn.Sequential()net_D.add(nn.Dense(5, activation='tanh'), nn.Dense(3, activation='tanh'), nn.Dense(1)) 14.1.4. Training¶ First we define a function to update the discriminator. # Save to the d2l package.def update_D(X, Z, net_D, net_G, loss, trainer_D): """Update discriminator""" batch_size = X.shape[0] ones = nd.ones((batch_size,), ctx=X.context) zeros = nd.zeros((batch_size,), ctx=X.context) with autograd.record(): real_Y = net_D(X) fake_X = net_G(Z) # Don't need to compute gradient for net_G, detach it from # computing gradients. fake_Y = net_D(fake_X.detach()) loss_D = (loss(real_Y, ones) + loss(fake_Y, zeros)) / 2 loss_D.backward() trainer_D.step(batch_size) return loss_D.sum().asscalar() The generator is updated similarly. Here we reuse the cross entropy loss but change the label of the fake data from $0$ to $1$. # Save to the d2l package.def update_G(Z, net_D, net_G, loss, trainer_G): # saved in d2l """Update generator""" batch_size = Z.shape[0] ones = nd.ones((batch_size,), ctx=Z.context) with autograd.record(): # We could reuse fake_X from update_D to save computation. fake_X = net_G(Z) # Recomputing fake_Y is needed since net_D is changed. fake_Y = net_D(fake_X) loss_G = loss(fake_Y, ones) loss_G.backward() trainer_G.step(batch_size) return loss_G.sum().asscalar() Both the discriminator and the generator performs a binary logistic regression with the cross entropy loss. We use Adam to smooth the training process. In each iteration, we first update the discriminator and then the generator. We visualize both losses and generated examples. def train(net_D, net_G, data_iter, num_epochs, lr_D, lr_G, latent_dim, data): loss = gluon.loss.SigmoidBCELoss() net_D.initialize(init=init.Normal(0.02), force_reinit=True) net_G.initialize(init=init.Normal(0.02), force_reinit=True) trainer_D = gluon.Trainer(net_D.collect_params(), 'adam', {'learning_rate': lr_D}) trainer_G = gluon.Trainer(net_G.collect_params(), 'adam', {'learning_rate': lr_G}) animator = d2l.Animator(xlabel='epoch', ylabel='loss', xlim=[1, num_epochs], nrows=2, figsize=(5,5), legend=['generator', 'discriminator']) animator.fig.subplots_adjust(hspace=0.3) for epoch in range(1, num_epochs+1): # Train one epoch timer = d2l.Timer() metric = d2l.Accumulator(3) # loss_D, loss_G, num_examples for X in data_iter: batch_size = X.shape[0] Z = nd.random.normal(0, 1, shape=(batch_size, latent_dim)) metric.add(update_D(X, Z, net_D, net_G, loss, trainer_D), update_G(Z, net_D, net_G, loss, trainer_G), batch_size) # Visualize generated examples Z = nd.random.normal(0, 1, shape=(100, latent_dim)) fake_X = net_G(Z).asnumpy() animator.axes[1].cla() animator.axes[1].scatter(data[:,0], data[:,1]) animator.axes[1].scatter(fake_X[:,0], fake_X[:,1]) animator.axes[1].legend(['real', 'generated']) # Show the losses loss_D, loss_G = metric[0]/metric[2], metric[1]/metric[2] animator.add(epoch, (loss_D, loss_G)) print('loss_D %.3f, loss_G %.3f, %d examples/sec' % ( loss_D, loss_G, metric[2]/timer.stop())) Now we specify the hyper-parameters to fit the Gaussian distribution. lr_D, lr_G, latent_dim, num_epochs = 0.05, 0.005, 2, 20train(net_D, net_G, data_iter, num_epochs, lr_D, lr_G, latent_dim, data[:100].asnumpy()) loss_D 0.693, loss_G 0.694, 743 examples/sec 14.1.5. Summary¶ Generative adversarial networks (GANs) composes of two deep networks, the generator and the discriminator. The generator generates the image as much closer to the true image as possible to fool the discriminator, via maximizing the cross entropy loss, i.e., $\max \log(D(\mathbf{x'}))$. The discriminator tries to distinguish the generated images from the true images, via minimizing the cross entropy loss, i.e., $\min - y \log D(\mathbf{x}) - (1-y)\log(1-D(\mathbf{x}))$.
Electronic Journal of Probability Electron. J. Probab. Volume 22 (2017), paper no. 93, 26 pp. Renormalizability of Liouville quantum field theory at the Seiberg bound Abstract Liouville Quantum Field Theory (LQFT) can be seen as a probabilistic theory of 2d Riemannian metrics $e^{\phi (z)}\|dz\|^2$, conjecturally describing scaling limits of discrete $2d$-random surfaces. The law of the random field $\phi $ in LQFT depends on weights $\alpha \in \mathbb{R} $ that in classical Riemannian geometry parametrize power law singularities in the metric. A rigorous construction of LQFT has been carried out in [3] in the case when the weights are below the so called Seiberg bound: $\alpha <Q$ where $Q$ parametrizes the random surface model in question. These correspond to studying uniformized surfaces with conical singularities in the classical geometrical setup. An interesting limiting case in classical geometry are the cusp singularities. In the random setup this corresponds to the case when the Seiberg bound is saturated. In this paper, we construct LQFT in the case when the Seiberg bound is saturated which can be seen as the probabilistic version of Riemann surfaces with cusp singularities. The construction involves methods from Gaussian Multiplicative Chaos theory at criticality. Article information Source Electron. J. Probab., Volume 22 (2017), paper no. 93, 26 pp. Dates Received: 28 November 2016 Accepted: 25 September 2017 First available in Project Euclid: 1 November 2017 Permanent link to this document https://projecteuclid.org/euclid.ejp/1509501716 Digital Object Identifier doi:10.1214/17-EJP113 Mathematical Reviews number (MathSciNet) MR3724561 Zentralblatt MATH identifier 06827070 Subjects Primary: 81T40: Two-dimensional field theories, conformal field theories, etc. 81T20: Quantum field theory on curved space backgrounds 60D05: Geometric probability and stochastic geometry [See also 52A22, 53C65] Citation David, François; Kupiainen, Antti; Rhodes, Rémi; Vargas, Vincent. Renormalizability of Liouville quantum field theory at the Seiberg bound. Electron. J. Probab. 22 (2017), paper no. 93, 26 pp. doi:10.1214/17-EJP113. https://projecteuclid.org/euclid.ejp/1509501716
$\begin{array}{r\|lll} &\text{Name}&\text{Surname}&\text{Dessert}\\ 1&\text{Agatha}&\text{Greed}&\text{cream puffs}\\ 2&\text{Bugsy}&\text{Forager}&\text{trifle}\\ 3&\text{Delilah}&\text{Eatalot}&\text{cheesecake}\\ 4&\text{Chuck}&\text{Hunk}&\text{ice cream} \end{array}$ Four friends held a competition to see who could eat the greatest amount of dessert in weight. So they went to the Pigout Diner, ordered their preferred dessert in abundance, and the battle commenced. The results are recorded above; however, the friends ended up insensible after having crammed themselves with the richness. So, although each item is in the correct column above, the friends only managed to write down ONE item in each column in the correct position. The following facts are true about the correct order. Chuck is one place above ice cream. Trifle is not above Delilah. Greed is two places below Delilah. Trifle is one place above Forager. This is an insanely scarce amount of information, and I believe that this requires logic at a higher level than I understand, but I don't know what that logic is. Either that, or I'm missing something very important. Here's what I have so far (the logic grid): $$\begin{align} &\begin{array}{r\|cccc\|cccc\|cccc\|} &1&2&3&4&A&B&C&D&E&F&G&H\\ \hline Cp&\quad&\quad&\quad&\quad&\quad&\quad&\quad&\quad&\quad&\quad&\quad&\quad\\ Tf&&&&\times&&&&&&\times\\ Cc\\ Ic&\times&&&&&&\times\\ \hline \end{array}\\[-4pt] &\begin{array}{r\|cccc\|cccc\|} \phantom\_E&\quad&\quad&\quad&\quad&\quad&\quad&\quad&\quad\,\\ F&\times&&&&&&&\times\\ G&\times&\times&&&&&&\times\\ H\\ \hline \end{array}\\[-4pt] &\begin{array}{r\|cccc\|} \phantom\_A&\quad&\quad&\quad&\quad\,\\ B\\ C&&&&\times\\ D&&&\times&\times\\ \hline \end{array} \end{align}$$ The following list of position comparisons assumes position 1 to be 4, position 2 to be 3, etc. for simplicity's sake, so that someone in a lower position is truly "less than". $4 \geq D \geq 3\\ T \leq D\\ C = Ic + 1\\ G = D - 2\\ T = F + 1\\ 2 \geq G \geq 1\\ 3 \geq F \geq 1\\ 4 \geq T \geq 2\\ F + 1 \leq D\\ G + 2 = D\\ F \leq G + 1$
Optimization Objective Within supervised learning, the performance of many supervised learning algorithms will be pretty similar and when that is less more often be whether you use learning algorithm A or learning algorithm B but when that is small there will often be things like the amount of data you are creating these algorithms on. That’s always your skill in applying this algorithms. Seems like your choice of the features that you designed to give the learning algorithms and how you choose the regularization parameter and things like that. Support Vector Machine ( SVM) : sometimes gives a cleaner and sometimes more powerful way of learning complex nonlinear functions. Alternative view of logistic regression $h_\theta (x) = \frac {1}{1 + e^{-\theta ^{T}x}}$ Cost of example : $-(ylogh_\theta (x) + (1-y)log(1-h_\theta(x))) $ $= -ylog\frac {1}{1 + e^{-\theta ^{T}x}} + (1-y)log(1-\frac {1}{1 + e^{-\theta ^{T}x}})$ Support vector machine Logistic regression : $\underset{\theta }{min} \frac{1}{m} [\sum_{i=1}^{m}y^{(i)}(-logh_\theta(x^{(i)})) + (1-y^{(i)})(-log(1-h_\theta (x^{(i)})))] + \frac{\lambda }{2m}\sum_{j=1}^{n} \theta _j^2$ Support vector machine : $\underset{\theta }{min} C [\sum_{i=1}^{m}y^{(i)}cost_1(\theta ^Tx^{(i)}) + (1-y^{(i)})cost_0(\theta ^Tx^{(i)})] + \frac{1}{2}\sum_{i=1}^{n} \theta _j^2$ Large Margin Intuition Sometimes people talk about support vector machines, as large margin classifiers. The margin of the support vector machine and this gives the SVM a certain robustness, because it tries to separate the data with as a large a margin as possible. So the support vector machine is sometimes also called a large margin classifier. $min \frac{1}{2} \sum _{j=1}^{n} \ \theta _j^2 \ s.t \ \left\{\begin{matrix} \theta^Tx^{(i)} \geq \ 1 \ \ if \ y^{(i)} = 1\\ \theta^Tx^{(i)} \leq -1 \ \ if \ y^{(i)} = 0\\ \end{matrix}\right.$ In practice when applying support vector machines, when C is not very very large like that, it can do a better job ignoring the few outliers. Mathematics Behind Large Margin Classification $\left \\| u \right \\| = \sqrt{u_1^2 + u_2^2}$ p is the length of the projection of the vector V onto the vector U. so $u^Tv = p \cdot \left \\| u \right \\|$ and $u^Tv = u_1 \times v_1 + u_2 \times v_2$ so $p \cdot \left \\| u \right \\| = u_1 \times v_1 + u_2 \times v_2$ SVM Decision Boundary $\underset {\theta }{min} \frac{1}{2} \sum _{j=1}^{n} \ \theta _j^2 \ s.t \ \left\{\begin{matrix} \theta^Tx^{(i)} \geq \ 1 \ \ if \ y^{(i)} = 1\\ \theta^Tx^{(i)} \leq -1 \ \ if \ y^{(i)} = 0\\ \end{matrix}\right.$ if n == 2: $\begin{Vmatrix} u \end{Vmatrix} = \sqrt{u_1^2 + u_2^2}$ $\frac{1}{2} (\theta _1^2 + \theta _2^2) = \frac{1}{2} (\sqrt {\theta _1^2 + \theta _2^2})^2 = \frac{1}{2}\left \\| \theta \right \\| ^2$ So all the support vector machine is doing in the optimization objective is it’s minimizing the squared norm of the square length of the parameter vector theta. $p^{(i)}$ : a projection of the i-th training example onto the parameter vector $\theta $. $\theta ^Tx^{(i)} = \theta_1 \cdot x_1^{(i)} + \theta_2 \cdot x_2^{(i)} $ SVM Decision Boundary $\underset {\theta }{min} \frac{1}{2} \sum _{j=1}^{n} \ \theta _j^2 \ s.t \ \left\{\begin{matrix} p^{(i)} \cdot \left \\| \theta \right \\| \geq \ 1 \ \ if \ y^{(i)} = 1\\ p^{(i)} \cdot \left \\| \theta \right \\| \leq -1 \ \ if \ y^{(i)} = 0\\ \end{matrix}\right.$ If we can make the norm of theta smaller and therefore make the squared norm of theta smaller, which is why the SVM would choose this hypothesis on the right instead. And this is how the SVM gives rise to this large margin certification effect. Mainly, if you look at this green line, if you look at this green hypothesis we want the projections of my positive and negative examples onto theta to be large, and the only way for that to hold true this is if surrounding the green line. There’s this large margin, there’s this large gap that separates positive and negative examples is really the magnitude of this gap. The magnitude of this margin is exactly the values of P1, P2, P3 and so on. And so by making the margin large, by these tyros P1, P2, P3 and so on that’s the SVM can end up with a smaller value for the norm of theta which is what it is trying to do in the objective. And this is why this machine ends up with enlarge margin classifiers because itss trying to maximize the norm of these P1 which is the distance from the training examples to the decision boundary. Kernels complex polynomial features $\theta_0 + \theta_1x_1 + \theta_2x_2 + \theta_3x_1x_2 + \theta_4x_1^2 + \theta_5x_2^2 + \cdots$ $f_1 = x_1, \ f_2 = x_2, \ f_3 = x_1x_2, \ f_4 = x_1^2, \ f_5 = x_2^2$$h_\theta (x) = \theta_1f_1 + \theta_2f_2 + \cdots + \theta_nf_n$ Gaussian Kernel $f_1 = similarity(x, l^{(1)}) = e^{(-\frac {\left \\| x – l^{(1)} \right \\| ^ 2}{2\sigma ^2})}$ $\left \\| x – l^{(1)} \right \\| ^ 2 = \sum _{j=1}^{n}(x_j – l_j^{(1)})^2$ We define some extra features using landmarks and similarity functions to learn more complex nonlinear classifiers. We use new features that are computed by Kernels, not original features. How the landmarks are chosen Choose the the location of my landmarks to be exactly near the locations of my m training examples. Given $(x^{(1)}, y^{(1)}), (x^{(2)}, y^{(2)}), \cdots , (x^{(m)}, y^{(m)})$ choose $l^{(1)} = x^{(1)}, l^{(2)} = x^{(2)}, \cdots \cdots , l^{(m)} = x^{(m)}$ Given example x : $f_1 = similarity(x, l^{(1)})$$f_2 = similarity(x, l^{(2)})$$\cdots $ Cost Function : $minC\sum _{i=1}^{m}[y^{(i)}cost_1(\theta^Tf^{(i)}) + (1-y^{(i)})cost_0(\theta^Tf^{(i)})]+\frac{1}{2}\sum_{j=1}^{n=m}\theta_j^2$ Using kernels with logistic regression is going too very slow. How to choose $C \ and \ \sigma$ $C = 1 \ / \ \lambda$ $C$ big: overfitting, higher variance $C$ small: underfitting, higher bias $\sigma$ big : lower variance, higher bias $\sigma$ small : lower bias, higher variance Using An SVM Some library function : There are a few things need to do : parameter’s C choose the kernel (If we decide not to use any kernel. And the idea of no kernel is also called a linear kernel) Logistic Regression or Support Vector Machine n : number of features m : number of training examples n >> m, Logistic Regression or Linear Kernel n small, and m middle, like 1 < n < 1000, 10 < m < 10000, SVM with Gaussian Kernel n small, and m big, like 1 < n < 1000, 50000 < m, SVM is slower, so try to munually create more features and then use logistic regression or an SVM without the Kernel Well for all of these problems, for all of these different regimes, a well designed neural network is likely to work well as well. The algorithm does matter, but what often matters even more is things like, how much data do you have. And how skilled are you, how good are you at doing error analysis and debugging learning algorithms, figuring out how to design new features and figuring out what other features to give you learning algorithms and so on.
It is easy to see that graph isomorphism(GI) is in NP. It is a major open problem whether GI is in coNP. Are there any potential candidates of properties of graphs that can be used as coNP certificates of GI. Any conjectures that imply $GI \in coNP$ ? What are some implications of $GI \in coNP$ ? If $GI$ is in $coNP$, then we would have the result: $GI$ is not $NP$-complete unless $NP=coNP=PH$. (Currently known: $GI$ is not $NP$-complete unless $\Sigma_2 P = \Pi_2 P = PH$). Since $GI$ is in $coAM$, obviously derandomizing $coAM$ (doi link) would put $GI \in coNP$, but I don't know of any candidate graph properties for putting $GI \in coNP$ otherwise. I look forward to more answers though! Interestingly, in that paper they also show that Graph Non-Isomorphism has subexponential size proofs -- that is, $GI \in co NSUBEXP$ -- unless $PH = \Sigma_3 P$. This is at least headed in the direction of showing conditionally that $GI \in coNP$. How about the range (i.e. list, one entry per edge) of effective resistances? The effective resistance of an edge is the probability that the edge is in a random spanning tree. Effective resistances can be found using algorithms of Spielman and Teng, although I don't know how easy it is to actually implement (if one wanted to do experiments). Suppose we have two strongly regular graphs, which have the same eigenvalues (and we know that eigenvalues do not necessarily distinguish between non-isomorphic graphs). Then if the effective resistances (i.e. the lists, again) are the same, they can not be used to distinguish the graphs. But why would two co-spectral graphs have the same distribution of their edges in random spanning trees? Is there a known connection between the graph spectrum and the effective resistances of a graph? i.e. knowning the graph spectrum, can we compute the effective resistances? It could be interesting to point out that if GI is not in coNP then P ≠ NP. 1)If GI is not in coNp then GI ≠ NGI 2)If GI ≠ NGI then GI ≠ P 3)If GI ≠ P then P ≠ NP As a corollary of the upon propositions we have: if GI is not in coNP then P ≠ NP. The same holds if NGI is not in NP.

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.