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Dear Uncle Colin, I'm told that the graphs of the functions $f(x) = x^3 + (a+b)x^2 + 3x - 4$ and $g(x) = (x-3)^3 + 1$ touch, and I have to determine $a$ in terms of $b$. Where would I even start? - Touching A New Graph Except Numerically TroublingRead More →
Some while ago, I showed a slightly dicey proof of the Basel Problem identity, $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac {\pi^2}{6}$, and invited readers to share other proofs with me. My old friend Jean Reinaud stepped up to the mark with an exercise from his undergraduate textbook: The French isn't that difficult,Read More →
Dear Uncle Colin I've been asked to find $\sum_3^\infty \frac{1}{n^2-4}$. Obviously, I can split that into partial fractions, but then I get two series that diverge! What do I do? - Which Absolute Losers Like Infinite Series? Hi, WALLIS, and thanks for your message! Hey! I'm an absolute loser whoRead More →
In this month's episode of Wrong, But Useful, Colin and Dave are joined by @niveknosdunk, who is Professor Kevin Knudson in real life. Kevin, along with previous Special Guest Co-Host @evelynjlamb, has recently launched a podcast, My Favorite Theorem The number of the podcast is 12; Kevin introduces us toRead More →
This post is inspired by a Futility Closet article. Do visit them and subscribe to their excellent podcast! Suppose you're dealt a bridge hand1, and someone asks whether you have any aces; you check, and yes! you find an ace. What's the probability you have more than one ace? ThisRead More →
Dear Uncle Colin I have been asked to describe how $y = \frac{3x^2-1}{3x+2}$ behaves as $x$ goes to infinity. My first answers, "$y$ goes to infinity" and "$y$ approaches $x$", were both wrong. Any ideas? - Both Options Reasonable, Erroneous Limits Hi, BOREL, and thanks for your message! My firstRead More →
One of the more surprising results a mathematician comes across in a university course is that the infinite sum $S = 1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{n^2} + ...$ comes out as $\frac{\pi^2}{6}$. If $\pi^2$s are going to crop up in sums like that, they should beRead More →
Dear Uncle Colin, What is $\frac{1}{\infty}$? - Calculating A Number, Though Outside Reals Hi, CANTOR, and thanks for your message! The short answer: it's undefined. The longer answer: Infinity is not a number. It's not something you're allowed to divide by. The calculation doesn't make sense, and writing it downRead More →
"So," said the Mathematical Ninja, "we meet again." "In fairness," said the student, "this is our regularly-scheduled appointment." The Mathematical Ninja was unable to deny this. Instead, it was time for a demand: "Tell me the square root of 22." "Gosh," said the student. "Between four-and-a-half and five, definitely. 4.7Read More →
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Suppose the application is a Lamport signature scheme.
Is the following a secure hash $\{0,1\}^n \rightarrow \{0,1\}^n$? $$ H(x) = x \oplus P(x) $$ where $P$ is a public permutation that permutes an input of length $n$.
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Suppose the application is a Lamport signature scheme.
Is the following a secure hash $\{0,1\}^n \rightarrow \{0,1\}^n$? $$ H(x) = x \oplus P(x) $$ where $P$ is a public permutation that permutes an input of length $n$.
Lamport's signature requires a hash function $H$ that is a one-way function. So your question is whether $H(x) = x \oplus P(x)$ is one-way when $P$ is a public permutation.
I will assume that $P$ is a public
random permutation (a.k.a., an ideal permutation). In that case, yes the construction is one-way. It is very closely related to the Matyas-Meyer-Oseas (MMO) construction of a collision-resistant hash function from an ideal block cipher.
To hash a sequence of blocks $m_1 m_2 \cdots$, the MMO construction works by iterating the function $s \gets E(s, m_i) \oplus m_i$, where $s$ is the continuously updating internal state and $E$ is an ideal cipher. If you could find inverses in $H(x) = x \oplus P(x)$, for a public permutation $P$, then you could easily find collisions (even second preimages) in an MMO hash. Since MMO hash is provably secure in the ideal cipher model, this impiles that $H$ is one-way:
Consider a 2-block message $m_1 m_2$ and its hash which is computed as:
To find another message that collides with this one, pick $m_1' \ne m_1$ arbitrarily and set:
If you had an $m'_2$ that satisifed
then $m'_1 m'_2$ would be the desired collision. Note that everything in this expression is known except $m'_2$, and so $E(s'_1, \cdot)$ is a public random permutation. So the task reduces to that of finding preimages of a function of the form $x \mapsto P(x) \oplus x$, where $P(x) = E(s'_2,x)$.
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Is it possible to solve Killing equations in
Mathematica for a general vector?
I am looking for a way to create Killing equations and then find what the vectors are, but I have a problem with this.
Introduction
First of all what they are. Without going in to the all gory details of general relativity, in short, Killing vectors are vectors that satisfy Killing equations:
$\nabla_\mu X_\nu+\nabla_\nu X_\mu=0$
Killing vector, according to the dimensions we are working in (3D, 4D etc.), and what coordinates, is a list with number of elements equating the number of dimension. So If I'm working in 2D sperical coordinate system, and I'm only interested in radial and polar coordinates, I'll have a Killing vector of the form
X = { Xθ[θ,ϕ], Xϕ[θ,ϕ]}
If I'm working in 4D spherical coordinate system with coordinates $\{t,r,\theta,\phi\}$, I'll have a Killing vector with components
X = { Xt[t,r,θ,ϕ], Xr[t,r,θ,ϕ], Xθ[t,r,θ,ϕ], Xϕ[t,r,θ,ϕ]}
The above equation is given in terms of covariant derivative, and for covariant vector (with indices down) is
$\nabla_\mu X_\nu=\frac{\partial X_\nu}{\partial x^\mu}-\Gamma^\lambda_{\mu \nu}X_\lambda$
Now $x^\mu$ is just coordinate for $\mu=t,r,\theta,\phi$, so $x^t=t, x^r=r$ etc. And $\Gamma^\lambda_{\mu \nu}$ are Christofell symbols that I can easily find. Oh, and sometimes the partial derivative is noted as $\partial_\mu$.
Example
I'm working on an easy example, a 2D sphere. It's metric is given by
$\begin{pmatrix} 1 & 0\\ 0 & \sin^2\theta \end{pmatrix}$
My code is this
xIN = {θ, ϕ};n = 2;met = {{1, 0}, {0, Sin[θ]^2}};inversemetric := Inverse[met] // FullSimplifycoord = xIN;(*Christoffel symbols*)affine := affine = Simplify[ Table[(1/2) Sum[ inversemetric[[μ, ρ]] (D[met[[ρ, ν]], coord[[λ]]] + D[met[[ρ, λ]], coord[[ν]]] - D[met[[ν, λ]], coord[[μ]]]), {ρ, 1, n}], {ν, 1, n}, {λ, 1, n}, {μ, 1, n}]]listaffine := Table[{Style[ Subsuperscript[Γ, Row[{coord[[ν]], coord[[λ]]}], coord[[μ]]], 18], Style[affine[[λ, ν, μ]], 14]}, {λ, 1, n}, {ν, 1, n}, {μ, 1, n}] // FullSimplify;data = {#[[1]], "=", #[[2]], #[[3]], "=", #[[4]]} & /@ Partition[DeleteCases[Flatten[listaffine], Null], 4];data = Insert[data[[#]], #, 1] & /@ Range[Length[data]];TableForm[data](*Derivations*)der[f_, σ_] := D[f, xIN[[σ]]]derxU[xU_, μ_, ν_] := Module[{λ}, der[xU[[μ]], ν] + Sum[affine[[ν, μ, λ]] xU[[λ]], {λ, 1, 2}]] // FullSimplifyderxd[xd_, μ_, ν_] := Module[{λ}, der[xd[[μ]], ν] - Sum[affine[[ν, λ, μ]] xd[[λ]], {λ, 1, 2}]] // FullSimplifyderxUup[xU_, μ_, ν_] := Module[{λ, ρ}, Sum[inversemetric[[ν, ρ]] (der[xU[[μ]], ρ] + Sum[affine[[ρ, μ, λ]] xU[[λ]], {\λ, 1, 4}]), {ρ, 1, 4}]] // FullSimplifyderxdup[xd_, μ_, ν_] := Module[{λ, ρ}, Sum[inversemetric[[ν, ρ]] (der[xd[[μ]], ρ] - Sum[affine[[ρ, λ, μ]] xd[[λ]], {\λ, 1, 4}]), {ρ, 1, 4}]] // FullSimplify
Now, I have specified the general form of my Killing vector:
ξ = { ξθ[θ, ϕ], ξϕ[θ, ϕ]};
And I've set up Killing equations:
Killeq = Table[ derxd[ξ, ν, μ] + derxd[ξ, μ, ν] == 0, {μ, 1, 2}, {ν, 1, 2}] // Flatten
And I get my equations, in Table form
$$ \begin{array}{c} 2 \xi \theta ^{(1,0)}(\theta ,\phi )=0 \\ \xi \theta ^{(0,1)}(\theta ,\phi )+\xi \phi ^{(1,0)}(\theta ,\phi )-2 \cot (\theta ) \xi \phi (\theta ,\phi )=0 \\ \xi \theta ^{(0,1)}(\theta ,\phi )+\xi \phi ^{(1,0)}(\theta ,\phi )-2 \cot (\theta ) \xi \phi (\theta ,\phi )=0 \\ 2 \sin (\theta ) \cos (\theta ) \xi \theta (\theta ,\phi )+2 \xi \phi ^{(0,1)}(\theta ,\phi )=0 \\ \end{array} $$
And that's what I should get, so the code is working (yaaaay! :D)
Now, even though I could just specify
Killeq[[1]] = 0 and so on, is there an automatic way for
Mathematica to see if there are same, and just give me the list of the ones left (Some kind of
If statement)? The problem could be identifying which equation is which later on, but I could just look at the original form of Killeq and see it from there. This would be useful if I need to make
TeXForm later on.
And the second part that is bothering me is: How do I solve this?
I tried with
DSolve[{ Killeq[[1]], Killeq[[2]], Killeq[[4]]}, { ξθ[θ, ϕ], ξϕ[θ, ϕ]}, {θ, ϕ}]
But I got the error:
`DSolve::overdet: There are fewer dependent variables than equations, so the system is overdetermined.`
Is there a way of finding these kind of things with
Mathematica? :\ Edit
I tried by separately solving each equation
Flatten[ Table[ DSolve[ Killeq[[i]], ξθ[θ, ϕ], {θ, ϕ}], {i, 1, 4}]]
And I get this:
$$ \begin{array}{c} \xi \theta (\theta ,\phi )\to c_1(\phi ) \\ \xi \theta (\theta ,\phi )\to \int_1^{\phi } \left(2 \cot (\theta ) \xi \phi (\theta ,K[1])-\xi \phi ^{(1,0)}(\theta ,K[1])\right) \, dK[1]+c_1(\theta ) \\ \xi \theta (\theta ,\phi )\to \int_1^{\phi } \left(2 \cot (\theta ) \xi \phi (\theta ,K[1])-\xi \phi ^{(1,0)}(\theta ,K[1])\right) \, dK[1]+c_1(\theta ) \\ \xi \theta (\theta ,\phi )\to -\csc (\theta ) \sec (\theta ) \xi \phi ^{(0,1)}(\theta ,\phi ) \\ \end{array} $$
Now, given that my 2nd and 3rd equations are repeating, is there any way of solving this with
Mathematica?
I am interested, because it would greatly help me find Killing equations in higher dimensions.
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Neil Sloan's On-Line Encyclopedia of Integer Sequences, which everyone who works seriously (or recreationally) with integer sequences regards as the ultimate authority, lists the triangular numbers as sequence A000217, and unambiguously includes 0 as a triangular number:
It was brought to my attention that all the references on this blog to triangular numbers (and to all other polygonal numbers) have omitted the 0 and started at 1. Please bear this omission in mind when browsing these pages - sometimes you might want to add a zero to the beginning of the sequences (but sometimes maybe not). 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431,...
I am consoled, only somewhat, by the fact that wikipedia and mathworld (two other august authorities on sequence-related matters) also omit the zero from their triangular number lists.
This reminds me of the age-old question:
Is zero a natural number?
Putting the definition of the natural numbers on the chalkboard is always dangerous - you risk having some student insist that your definition is
wrong, wrong, wrongbecause they were taught that natural numbers included (or excluded) zero, and yours doesn't. It seems that most sources, including Sloan's OLEIS, say that 0 is notnatural (see A000027).
I suspect that (most) mathematicians do not care (much) about this - they just redefine the term "natural number" to be what they need it to be at the moment they happen to be using it. If they need a zero, they add a zero, and move on.
Wikipedia suggests that when you encounter a situation where it might matter, one should use $\mathbb{N}_0$ when zero is to be included, and $\mathbb{N}$ when it isn't (or when it doesn't matter).
The comparison between the triangulars and the naturals is not spurious. Wikipedia defines the triangulars as sums of the naturals (I, somewhat strangely, tend to think of naturals as one-dimensional triangular numbers). If this is your chosen definition then you are likely not to include zero, and you might use this formula:
\[t_n = \sum^{n}_{i=1}i.\]
However, if we want to rehabilitate this particular formula for the triangulars + 0, we just need to adjust the index:
\[t_n = \sum^{n}_{i=0}i\]
What about other triangular number formulas? Can they all include zero too? Well, the simplest, $t_n = \frac{n(n+1)}{2}$ works just fine when you let $n=0$.
Sometimes, when we are thinking of how they relate to the binomial coefficients, we might want to use this formula:
\[t_n = \left( \begin{array}{c} n +1 \\2\end{array} \right) \]
This might give you pause, because when $n = 0$ we seem to be "out of bounds." Luckily we have:
\[\left(\begin{array}{c} n \\r \end{array}\right) = 0 \mbox{ for } r > n\]
Which is exactly what we need.
As far as the posts on this blog are concerned, the only way of expressing the triangulars that needs obvious modification in order to work for the triangulars + 0 is the generating function:
\[g(x) = \frac{1}{\left( 1-x \right)^{3}} = 1 +3x + 6x^2 + 10x^3 + ...\]
which gives the triangulars as in the coefficients on the right hand side. To have a generating function for the triangulars + 0 you need to modify this to be:
\[g(x) = \frac{x}{\left( 1-x \right)^{3}} = 0 +x +3x^2 + 6x^3 + 10x^4 + ...\]
Multiplying by $x$ is the generating-function equivalent to shifting indexes, which is what we had to do for our first formula.
Thanks to Alexander Povolotsky for bringing these issues to light.
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Prove that the number of subgroups in $D_n = \tau (n) + \sigma (n)$ where $\tau (n)$ represents number of divisors of $n$ and $\sigma (n)$ represnts the sum of divisors of $n$.
Attempt: $D_n = \{e,r,r^2, \cdot \cdot \cdot, r^{n-1},s,rs,r^2s, \cdot \cdot \cdot, r^{n-1}s \}$
Then, $\{e,r,r^2, \cdot \cdot \cdot, r^{n-1}\}$ is a cyclic subgroup of $G$. There are $\tau(n)$ such cyclic subgroups.
Total number of subgroups of order $2=n+1$ if $n$ is even and $=n$ if $n$ is odd.
Hence, the sum is $\tau(n)+n$ or $\tau(n)+n+1$
Now, let $H$ denote the set of reflections, then if $a=r^is,b=r^js ~\in~H$, then $r^is~ (r^js)^{-1} = (r^is)^{-1}r^js=sr^{-i}r^js=sr^{j-i}s=s^2r^{i-j}=r^{i-j} \notin H$
Hence, the set of reflections cannot form a subgroup in themselves in any case.
Now, we must consider sets involving both rotations and reflections and find conditions for them to form subgroups.
By Lagrange's theorem, order of such a subgroup $H$ must divide $n$ as we have already found out the total number of subgroups of order $2$ So :
(a) A reflection is it's self inverse
(b) A reflection times a reflection is always a rotation
(c) Inverse of a rotation $r^i = r^{n-i}$
(d) A rotation times a rotation is always a rotation.
How do I choose such rotation and reflective elements such that they form a group. Suppose, $|H|=4$, In that case, $H$ should look something like this:
$H = \{f_1,f_2,r^i,r^{n-i}\}$ where $f_1,f_2$ are reflections and $f_1f_2=r^i$
How can I extend this to $n$ variables? A hint to move ahead would be really appreciated. Thanks
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Over the years, I've noticed some distinct trends among the physics and chemistry majors I've mentored and worked with.Those with GPAs in the 3.8-4.0…
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In this Insights article I'll present an example that shows how Intel® AVX-512 instructions can be used to read a whole row of data in a single operation,…
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This is the first of several posts that will develop some mathematical machinery for studying Fermi-Walker transport. In this first post, we focus on Minkowski…
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Here I present a simple (but to the best of my knowledge, new) derivation of the formula for the sum of the infinite geometric series. The derivation is…
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Part III: Representations10. Sums and Products.Frobenius began in ##1896## to generalize Weber's group characters and soon investigated…
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IntroductionAbacuses are commonly seen in ancient civilizations around the world as the convenient tool to do simple calculations. In this article, we…
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Learning Objectives* Gain confidence and experimental care in making accurate measurements.* Understand the relationship between force and spring stretch.*…
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Part II: Structures5. Decompositions.Lie algebra theory is to a large extend the classification of the semisimple Lie algebras…
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Part I: Basics1. Introduction.This article is meant to provide a quick reference guide to Lie algebras: the terminology, important theorems,…
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IntroductionIn the first part of this series, I talked about some fundamental notions in the world of algorithms. Beyond the definition of an algorithm,…
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IntroductionBernoulli's equation is one of the most useful equations in the field of fluid mechanics, owing to its simplicity and broad applicability.…
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We recently had a question in the relativity forums that mentioned the behavior of magnetic field lines, and reminded me of my own confusion at school…
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I am studying the deformation theory of associative algebras (and Poisson algebras) and came across a question for which I cannot find an answer:
Let $(A,\mu)$ be a commutative associative algebra over a field $\mathbb{F}$. The first order deformations of $\mu$ are classified up to equivalence by the second Hochschild cohomology group $HH_{\mu}^{2}(A)$. I'm wondering if this can be generalized in the following sense:
Given a $k^{th}$ order deformation $\mu_{(k)} = \mu + \mu_{1} h + ... + \mu_{k} h^k$ of $\mu$, in some cases it is possible to extend it to a $(k+1)^{th}$ order deformation $\mu_{(k+1)} = \mu_{(k)} + \mu_{k+1}h^{k+1}$, where $\mu_{k+1}$ is a bilinear map $A \times A \to A$. In this case it is natural to ask what are all the possible extensions up to equivalence. It turns out that the set of all possible deformations consists of the affine space $ \mu_{k+1} + \ker(\delta_{\mu}^2), $ where $\delta_{\mu}^2 : HC^{2}(A) \to HC^{3}(A)$ is the Hochschild coboundary operator. Furthermore, if two extensions correspond to cohomologous elements of $HC^2(A)$, it turns out that the extensions are equivalent (in the sense that the resulting deformed algebras are isomorphic). My question is whether the converse holds, which is to say do equivalent extensions correspond to cohomologous elements? (And hence $HH_{\mu}^2(A)$ would classify the $(k+1)^{th}$ order deformations extending a particular $k^{th}$ order deformation of $\mu$.)
Here is what I have tried so far (in the case of $k = 2$): consider two equivalent $2^{nd}$ order deformations of $\mu$:
$\mu_{(2)} = \mu + \mu_{1} h + \eta h^2, \ \ \mu_{(2)}' = \mu + \mu_{1} h + \eta' h^2,$ with isomorphism $\Phi = 1 + \phi h + \psi h^2$. Then from the requirement that $\Phi(\mu_{(2)}(F,G)) = \mu_{(2)}'(\Phi(F), \Phi(G))$ for $F, G \in A$ we get that
$\delta_{\mu}^{1}(\phi) = 0,$ and
$\eta - \eta' = \delta_{\mu}^{1}(\psi) + \delta_{\mu_{1}}^{1}(\phi) + \mu \circ \phi \otimes \phi.$
Note that I use the notation $\delta_{\mu_{1}}^{1}(\phi)(F,G) = \mu_{1}(F, \phi(G)) + \mu_{1}(\phi(F),G) - \phi(\mu_{1}(F,G)).$
I need to end up with a formula like $\eta - \eta' = \delta_{\mu}^1 ( \gamma)$ for some $\gamma \in Hom_{\mathbb{F}}(A,A)$. It doesn't seem like its going to work, but I haven't been able to think of a counter example, and I have been assured that the result should be true.
P.S. I'd like to know if the analogous statement holds for deformations of a Poisson algebra.
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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 ($|y| < 0.9$) ...
Highlights of experimental results from ALICE
(Elsevier, 2017-11)
Highlights of recent results from the ALICE collaboration are presented. The collision systems investigated are Pb–Pb, p–Pb, and pp, and results from studies of bulk particle production, azimuthal correlations, open and ...
Event activity-dependence of jet production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV measured with semi-inclusive hadron+jet correlations by ALICE
(Elsevier, 2017-11)
We report measurement of the semi-inclusive distribution of charged-particle jets recoiling from a high transverse momentum ($p_{\rm T}$) hadron trigger, for p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in p-Pb events ...
System-size dependence of the charged-particle pseudorapidity density at $\sqrt {s_{NN}}$ = 5.02 TeV with ALICE
(Elsevier, 2017-11)
We present the charged-particle pseudorapidity density in pp, p–Pb, and Pb–Pb collisions at sNN=5.02 TeV over a broad pseudorapidity range. The distributions are determined using the same experimental apparatus and ...
Photoproduction of heavy vector mesons in ultra-peripheral Pb–Pb collisions
(Elsevier, 2017-11)
Ultra-peripheral Pb-Pb collisions, in which the two nuclei pass close to each other, but at an impact parameter greater than the sum of their radii, provide information about the initial state of nuclei. In particular, ...
Measurement of $J/\psi$ production as a function of event multiplicity in pp collisions at $\sqrt{s} = 13\,\mathrm{TeV}$ with ALICE
(Elsevier, 2017-11)
The availability at the LHC of the largest collision energy in pp collisions allows a significant advance in the measurement of $J/\psi$ production as function of event multiplicity. The interesting relative increase ...
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Let f:C→Cf : \mathbb{C} \to \mathbb{C}f:C→C be holomorphic on C\mathbb{C}C. Also, suppose the following holds: ∣f(z)∣→+∞ as ∣z∣→+∞.\big|f(z)\big| \to +\infty \text{ as } |z| \to +\infty.∣∣f(z)∣∣→+∞ as ∣z∣→+∞.
Then which of the following necessarily holds?
Details and Assumptions:
Problem Loading...
Note Loading...
Set Loading...
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\[\]
\defl{Hypergeometric distribution has the following properties:}
When units are selected from a finite population without replacement and the population consists of successes and failures. The major difference between the Hypergeometric distribution and the Binomial distribution is that the probability of selecting a success is {\bf not constant and is not independent} from each draw.
\defm{Hypergeometric Distribution}
\[ P(X=x)=f(x) = \frac{{A\choose x}{{N-A}\choose {n-x}}}{{N\choose n}} \]
$X$ has a Hypergeometric distribution. $x$ is the number of successes in the sample $n$ is the sample size $A$ is the number of successes in the population $N$ is the population size $N-A$ is the number of failures in the population $n-x$ is the number of failures in the sample $\mu = E(X) = \frac{nA}{N}$ $\sigma^2 = V(X) = \left(\frac{N-n}{N-1}\right) \frac{nA(N-A)}{N^{2}}$ $\frac{N-n}{N-1}$ is called the finite population correction factor.
\defs{Examples}
The number of spades selected when 5 cards are drawn from a standard 52 deck of cards. There are 20 Sony CD players in stock at the Sony store at the Mall Bangkapi; 5 are defective. A customer buys 6 of the 20 CD players. The number of defective CD players bought of the 6 CD players. The computer lab has 20 computers and 15 of the 20 computers have illegal software on them. The number of computers selected with illegal software from a sample of size 5.
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LaTeX is a great tool for printable professional-looking documents, but can be also used to generate PDF files with excellent navigation tools. This article describes how to create hyperlinks in your document, and how to set up LaTeX documents to be viewed with a PDF-reader.
Contents
Let's start with a minimal working example, by simply importing the
hyperref package all cross-referenced elements become hyperlinked.
\documentclass{book} \usepackage[utf8]{inputenc} \usepackage[english]{babel} \usepackage{hyperref} \begin{document} \frontmatter \tableofcontents ... \end{document}
The lines in the table of contents become links to the corresponding pages in the document by simply adding in the preamble of the document the line
\usepackage{hyperref} One must be careful when importing hyperref. Usually, it has to be the last package to be imported, but there might be some exceptions to this rule.
The default formatting for links can be changed so the information in your documents is more clearly presented. Below you can see an example:
\documentclass{book} \usepackage[utf8]{inputenc} \usepackage[english]{babel} \usepackage{hyperref} \hypersetup{ colorlinks=true, linkcolor=blue, filecolor=magenta, urlcolor=cyan, } \urlstyle{same} \begin{document} \tableofcontents \chapter{First Chapter} This will be an empty chapter and I will put some text here \begin{equation} \label{eq:1} \sum_{i=0}^{\infty} a_i x^i \end{equation} The equation \ref{eq:1} shows a sum that is divergent. This formula will later be used in the page \pageref{second}. For further references see \href{http://www.sharelatex.com}{Something Linky} or go to the next url: \url{http://www.sharelatex.com} or open the next file \href{run:./file.txt}{File.txt} It's also possible to link directly any word or \hyperlink{thesentence}{any sentence} in your document. \end{document}
This is a complete example, it will be fully explained in the rest of the article. Below is a description of the commands related to the colour and styling of the links.
\hypersetup{ ... }
\colorlinks=true
\linkcolor=blue
\filecolor=magenta
\urlcolor=cyan
\urlstyle{same}
Links to a web address or email can added to a LaTeX file using the
\url command to display the actual link or
\href to use a hidden link and show a word/sentence instead.
For further references see \href{http://www.sharelatex.com}{Something Linky} or go to the next url: \url{http://www.sharelatex.com}
There are two commands in the example that generate a link in the final document:
\href{http://www.sharelatex.com}{Something Linky}
\url{http://www.sharelatex.com}
The commands
\href and
\url presented in the previous section can be used to open local files
For further references see \href{http://www.sharelatex.com}{Something Linky} or go to the next url: \url{http://www.sharelatex.com} or open the next file \href{run:./file.txt}{File.txt}
The command
\href{run:./file.txt}{File.txt} prints the text "File.txt" that links to a local file called "file.txt" located in the current working directory. Notice the text "run:" before the path to the file.
The file path follows the conventions of UNIX systems, using . to refer the current directory and .. for the previous directory.
The command
\url{} can also be used, with the same syntax described for the path, but it's reported to have some problems.
It was mentioned before that all cross-referenced elements become links once
hyperref is imported, thus we can use
\label anywhere in the document and refer later those labels to create links. This is not the only manner to insert hyperlinks manually.
It's also possible to link directly any word or \hyperlink{thesentence}{any sentence} in you document. If you read this text, you will get no information. Really? Is there no information? For instance \hypertarget{thesentence}{this sentence}.
There are two commands to create user-defined links.
\hypertarget{thesentence}{this sentence}
\hyperlink{thesentence}{any sentence}
Links in a document are created having in mind a document that will be read in PDF format. The PDF file can be further personalized to add additional information and change the way the PDF viewer displays it. Below an example:
\hypersetup{ colorlinks=true, linkcolor=blue, filecolor=magenta, urlcolor=cyan, pdftitle={Sharelatex Example}, bookmarks=true, pdfpagemode=FullScreen, }
Using the command
\hypersetup, described in the section styles and colours, accepts extra parameters to set up the final PDF file.
pdftitle={Sharelatex Example}
bookmarks=true
pdfpagemode=FullScreen
See the reference guide for a full list of options that can be passed to
\hypersetup.
Linking style options
Option Default value Description
hyperindex
true Makes the page numbers of index entries into hyperlinks
linktocpage
false Makes the page numbers instead of the text to be link in the Table of contents.
breaklinks
false Allows links to be broken into multiple lines.
colorlinks
false Colours the text for links and anchors, these colours will appear in the printed version
linkcolor
red Colour for normal internal links
anchorcolor
black Colour for anchor (target) text
citecolor
green Colour for bibliographical citations
filecolor
cyan Colour for links that open local files
urlcolor
magenta Colour for linked URLs
frenchlinks
false Use small caps instead of colours for links PDF-specific options
Option Default value Description
bookmarks
true Acrobat bookmarks are written, similar to the table of contents.
bookmarksopen
false Bookmarks are shown with all sub-trees expanded.
citebordercolor
0 1 0 Colour of the box around citations in RGB format.
filebordercolor
0 .5 .5 Colour of the box around links to files in RGB format.
linkbordercolor
1 0 0 Colour of the box around normal links in RGB format.
menubordercolor
1 0 0 Colour of the box around menu links in RGB format.
urlbordercolor
0 1 1 Colour of the box around links to URLs in RGB format.
pdfpagemode
empty Determines how the file is opened. Possibilities are UseThumbs (Thumbnails), UseOutlines (Bookmarks) and FullScreen.
pdftitle
Sets the document title.
pdfauthor
Sets the document Author.
pdfstartpage
1 Determines on which page the PDF file is opened.
For more information see
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LaTeX supports many worldwide languages by means of some special packages. In this article is explained how to import and use those packages to create documents in
Spanish.
Contents
Spanish language has some special characters, such as the
ñ and some accentuated words. For this reason the preamble of your document must be modified accordingly to support these characters and some other features.
\documentclass{article} \usepackage[utf8]{inputenc} \usepackage[spanish]{babel} \begin{document} \tableofcontents \vspace{2cm} %Add a 2cm space \begin{abstract} Este es un breve resumen del contenido del documento escrito en español. \end{abstract} \section{Sección introductoria} Esta es la primera sección, podemos agregar algunos elementos adicionales y todo será escrito correctamente. Más aún, si una palabra es demasiado larga y tiene que ser truncada, babel tratará de truncarla correctamente dependiendo del idioma. \section{Sección con teoremas} Esta sección es para ver qué pasa con los comandos que definen texto \end{document}
There are two packages in this document related to the encoding and the special characters. These packages will be explained in the next sections.
Modern computer systems allow you to input letters of national alphabets directly from the keyboard. In order to handle a variety of input encodings used for different groups of languages and/or on different computer platforms LaTeX employs the
inputenc package to set up input encoding. In this case the package properly displays characters in the Spanish alphabet. To use this package add the next line to the preamble of your document: \usepackage[utf8]{inputenc} The recommended input encoding is utf-8. You can use other encodings depending on your operating system.
To proper LaTeX document generation you must also choose a font encoding which has to support specific characters for Spanish language, this is accomplished by the
package: fontenc \usepackage[T1]{fontenc} Even though the default encoding works well in Spanish, using this specific encoding will avoid glitches with some specific characters. The default LaTeX encoding is
OT1.
To extended the default LaTeX capabilities, for proper hyphenation and translating the names of the document elements, import the
babel package for the Spanish language. \usepackage[spanish]{babel} As you may see in the example at the introduction, instead of "abstract" and "Contents" the Spanish words "Resumen" and "Índice" are used. An extra parameter can be passed when importing the babel package with spanish support: \usepackage[spanish, mexico]{babel} This will set a localization for the language. By now only mexico and mexico-com are available, the latter will use a comma instead of a dot as the decimal marker in mathematical mode.
Mathematical commands can also be imported specifically for the Spanish language.
\section{Sección con teoremas} Esta sección es para ver que pasa con los comandos que definen texto \[ \lim x = \sen{\theta} + \max \{3.52, 4.22\} \] El paquete también agrega un comportamiento especial a <<estas marcas para hacer citas textuales>> tal como lo indican las reglas de la RAE. You can see that
\sen,
\max and
\lim are properly displayed. For a complete list of mathematical symbols in Spanish see the reference guide. For this commands to be available you must add the next line to the preamble of your document:
\def\spanishoperators{} Notice also that
<< and
>> have a special format in Spanish, this can conflict with some packages. If you don't need these or you want to use the direct keyboard input « » set the parameter
es-noquotes, comma separated inside the brackets of the
babel statement.
Sometimes for formatting reasons some words have to be broken up in syllables separated by a
- (
hyphen) to continue the word in a new line. For example, matemáticas could become mate-máticas. The package babel, whose usage was described in the previous section, usually does a good job breaking up the words correctly, but if this is not the case you can use a couple of commands in your preamble.
\usepackage{hyphenat} \hyphenation{mate-máti-cas recu-perar}
The first command will import the package
hyphenat and the second line is a list of space-separated words with defined hyphenation rules. On the other side, if you want a word not to be broken automatically, use the
{\nobreak word} command within your document.
Spanish LaTeX commands in mathematical mode
LaTeX command Output
\sen
sen
\tg
tg
\arcsen
arc sen
\arccos
arc cos
\arctg
arc tg
\lim
lím
\limsup
lím sup
\liminf
lím inf
\max
máx
\inf
ínf
\min
mín
For more information see
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For those interested, here is a slightly more rigorous proof that the construction exists using measure theory.
We will define a function $H$ which is continuous from the left at ever $x \in [0,1]$ but is only continuous from the right at the irrationals. To build such a function we need the folliowing scaffolding.
First let $\psi: \mathbb{N} \to \mathbb{Q} \cap [0,1]$ be a bijection enumerating the rationals in the interval $[0,1]$. Then define $B(x) = \{n : \psi(n)< x\}$ or equivalently $B(x) = \psi^{-1}([0,x))$. Finally let $\mu: P(\mathbb{N}) \to \overline{\mathbb{R}}$ be the counting measure. We additionally define a measure $\nu: P(\mathbb{N}) \to \mathbb{R}$ such that $$\nu(A) = \int_A 2^{-n} d \mu(n).$$The measure $\nu$ has the additional property that $\nu \ll \mu$ and $\nu(\mathbb{N}) = \sum_{n \in \mathbb{N}} 2^{-n} \mu({n}) = 1 < \infty.$
We claim that the function $H(x) = \nu(B(x))$ has the properties of $f$ in the statement of the problem. We will first show that forevery $x \in [0,1]$ the function $H(x)$ is left continuous. Take a sequence of $x_k \to x$ from the left, we can then rearrange the sequence to be strict monotonic. It follows that if $k > m$ then $$B(x_m) = \{n: \psi(n) < x_m < x_k < x\} \subset \{n: \psi(n) < x_k < x\} = B(x_k).$$ By the finiteness of $\nu$ we have that by upward measure continuity$$\lim_{k \to \infty} H(x_k) = \lim_{k \to \infty} \nu(B(x_k)) = \nu\left(\bigcup_{k=1}^\infty B(x_k) \right) = \nu(\left \{n: \psi(n) < x\}\right ) = H(x).$$Note that if $ n \in \bigcup B(x_k)$ there is an $K$ so that $\psi(n) < x_k < x $ so any $n$ with $\psi(n) < x$ is in $\bigcup B(x_k).$
Next we claim that $H$ is only right continuous only when $x$ is irrational. Take a sequence of $x_k \to x$ from the right and rearrange the sequence to be strict montonic. It follows that if $k > m$ then $$B(x_m) = \{n: \psi(n) < x_m\} \supset \{n: \psi(n) < x_k < x_m\} = B(x_k).$$ By finiteness of $\nu$ and downard measure continuity$$\lim_{k \to \infty} H(x_k) = \lim_{k \to \infty} \nu(B(x_k)) = \nu\left(\bigcap_{k=1}^\infty B(x_k) \right) = \nu(\left \{n: \psi(n) < x_k\ \forall k\}\right ).$$
If $x$ is irrational then $m \in \{n: \psi(n) < x_k\ \forall k\}$ implies that $\psi(m) < x$ and if $\psi(m) < x$ then $\psi(m) < x_k$ for all $k$ so $\{n: \psi(n) < x_k\ \forall k\} = B(x)$ and $H(x_k) \to H(x)$ from the right. If $x$ is rational then $x= \psi(q)$ for some$q \in \mathbb{N}.$ Thus $x < x_k \forall k$ implies that $\{n: \psi(n) < x_k\ \forall k\} = B(x) + \{q\} = D$. It follows that $\nu(D) = \nu(B(x)) + 2^{-q} > H(x)$. So $H(x_k) \to H(x) + 2^{-q} \neq H(x)$ from the right, and so $H$ is not right continuous at the rationals.
We have thus shown that for any $x \in [0,1] \setminus \mathbb{Q}$, any sequence $x_k \to x$ has the property $\lim H(x_k) = x$ from the left and the right, and if $x \in [0,1] \cap \mathbb{Q}$ then if $x_k \to x$, $\lim H(x_k)$ does not exist. Therefore $H$ is continuous at every irrational and discontinuous at every rational.
You can then repeat this construction along the whole real line by adding the nearest integer each time.
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For one of my applications it is necessary to know the geopotential at each model level of the ERA5 data. Since ERA5 and ERAI only store the geopotential at the surface, it may not be immediately obvious how to arrive at the remaining values. This post deals with how to achieve this. While shown for ERA5, the procedure is similar for ERA-Interim.
Model levels in ERA5
Essentially the vertical coordinate system in ERA5 is chosen as such that closer to the surface the model levels follow the topography while farther up they follow pressure levels with a transitional region in between. Some of the details of this are summarized nicely here (for ERA-Interim). In ERA5 the levels are numbered from top to bottom, with 1 being the topmost layer and 137 the one closest to the surface. This is illustrated by the following plot:
The levels shown above are actually referred to as
full model levels. Each of these full model levels is defined and bounded by a half-level above and a half-level below. Note that the full model levels are associated with a volume while the half-levels correspond to an area – the interface between two full model levels. Calculating pressure at half model levels
While not necessary right now we’ll need information about the pressure at the half-levels (interfaces between full model levels) later on. The equation that defines the pressure at each half-level is
\[ p_{k+1/2} = a_{k+1/2} + b_{k+1/2}\,p_s \]
where $p_{k+1/2}$ denotes the pressure at half-level $k+1/2$, the surface pressure and $a_{k+1/2}$ and $b_{k+1/2}$ constants that define the vertical coordinate.
Where do we get the values of the constants $a_{k+1/2}$ and $b_{k+1/2}$ from? Luckily the ECMWF lists them all in a nice table, complete with some exemplary values of quantities at half levels (pressure) and full-levels (pressure, geopotential altitude, geometric altitude, temperature and density).
Where is the half-level $k+1/2$ located? An easy way to think of this is to remember that levels are numbered from top to bottom. E.g. the smaller the value the higher up the associated model level will be. Adding k+1/2 then is the half-level below full model level k while k-1/2 is the half-level above.
The pressure at a full model-level k is given by the average of the pressure between the two bounding half-levels:
\[p_k = \frac{1}{2}(p_{k-1/2} + p_{k+1/2})\]
The vertical discretisation and the remaining equations we need are documented in Part 3 of the IFS documentation (IFS 2015).
Geopotential at full model levels
The starting point is equation (2.22):
\[ \phi_k = \phi_{k+1/2} + \alpha_k R_{\text{dry}}(T_v)_k \]
Here $\phi_{k+1/2}$ is the geopotential at half-level $k+1/2$, $R_{\text{dry}}$ the gas constant for dry air and $(T_v)_k$ the virtual temperature at full-level $k$. The coefficient $\alpha_k$ is defined as $\alpha_1 = \ln{2}$ and for $k > 1$ as
\[ \alpha_k = 1-\frac{p_{k-1/2}}{\Delta p_k}\ln{\left(\frac{p_{k+1/2}}{p_{k-1/2}}\right)} \]
To calculate $T_v$ directly from fields available in ERA5 the following equation is applicable:
\[ T_v = T \left\{ 1 + \left[ (R_{\text{vap}}/R_{\text{dry}} ) – 1 \right] q \right\} \]
Here $T$ is the temperature, $q$ the specific humidity and $R_\text{vap}$ the gas constant for water vapour.
This leaves us with $\phi_{k+1/2}$ as the only remaining quantity. It is given by equation 2.21 in IFS (2015):
\[ \phi_{k+1/2} = \phi_S + \sum_{j=k+1}^N R_\text{dry} (T_v)_j \ln{\left( \frac{p_{j+1/2}}{p_{j-1/2}}\right)} \]
with $\phi_S$ as the geopotential at the surface and $N$ as the number of full model-levels, which is $137$ in case of ERA5.
ERA5 data required
To actually carry out the calculation, the following ERA5 fields are required (parameter number in brackets)
geopotential at the surface (129.128) pressure at the surface (134.128) temperature (130.128) specific humidity (133.128)
Here’s an example netcdf file that contains exactly this data for a region encompassing the South Island of New Zealand over two days (only model levels 30-137), plus a .csv file that contains the model-level defining constants $a_k$ and $b_k$.
Python code example
The following code calculates the geopotential of a given ERA5 model level based on the data above. To run it you’ll need to download this library of mine from github [geopotcalc on github] and place it in a folder named ‘lib’ within the directory you’ll be running the following code from. Furthermore the python packages xarray, numpy and pandas should also be installed.
import lib.gpcalc as gpcimport xarray as xaimport pandas as pdera5_ds = xa.open_dataset('./era5_ml_example.nc')mldf = pd.read_csv('./akbk.csv',index_col='n')gpc.set_data(era5_ds, mldf, 137)gpc.get_phi(130) # calculates the geopotential at model level 130.
The output should look like this:
<xarray.DataArray (time: 48, latitude: 37, longitude: 45)>array([[[ 2061.799 , 2068.4844, ..., 9074.937 , 9794.214 ], [ 2068.3455, 2071.648 , ..., 12285.38 , 12331.378 ], ..., [ 2031.0612, 2031.0513, ..., 2022.7593, 2029.4838], [ 2029.6093, 2035.5366, ..., 2021.2681, 2032.0813]], [[ 2062.8289, 2070.104 , ..., 9076.426 , 9794.513 ], [ 2068.5178, 2072.9602, ..., 12283.5 , 12330.7705], ..., [ 2028.933 , 2029.8497, ..., 2023.5824, 2032.0865], [ 2025.4272, 2032.5043, ..., 2023.407 , 2032.3804]], ..., [[ 2045.6422, 2051.2817, ..., 9068.013 , 9790.583 ], [ 2053.2986, 2057.6807, ..., 12275.755 , 12324.499 ], ..., [ 1981.8191, 1980.5361, ..., 1987.5459, 1996.2969], [ 1977.6063, 1985.1229, ..., 1987.896 , 1998.8293]], [[ 2047.196 , 2053.632 , ..., 9075.75 , 9798.725 ], [ 2052.647 , 2056.1565, ..., 12282.7 , 12331.903 ], ..., [ 1985.239 , 1981.6675, ..., 1986.0985, 1996.3464], [ 1978.8458, 1985.9766, ..., 1985.4244, 1997.1885]]], dtype=float32)Coordinates: * longitude (longitude) float32 165.0 165.25 165.5 165.75 166.0 166.25 ... * latitude (latitude) float32 -39.0 -39.25 -39.5 -39.75 -40.0 -40.25 ... level int32 130 * time (time) datetime64[ns] 2015-02-01 2015-02-01T01:00:00 ...
So that’s essentially it. Most of it is already documented in IFS (2015), however, I found that actually finding this information was the hardest part. If there’s any errors or any questions arise from the above – don’t hesitate to contact me.
References
ECMWF, IFS. “Documentation–Cy40r1.”
Part III: Dynamics and numerical procedures, ECMWF (2015). [link]
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To show that the series $\sum_{k=1}^{\infty} {{1}\over {k}}$ diverges ; I have to prove that $${\sum_{k=1}^{n}}{{1}\over{k}} \ge {\log n}$$ The given hint is that $${{1}\over {k}}\ge \int_k^{k+1} {1\over x} dx$$
Now,evaluating the RHS, say $$L=\int_k^{k+1} {1\over x} dx\\=\log(k+1)-\log\ k\\=\log(1+{1\over k})\\={1\over k}-{1\over{2k^2}}+{1\over {3k^3}}-......$$ How can I tell this is $L \le {1\over k}?$
If this part is proved then I guess the following is like
:
$$\sum_{k=1}^n {1\over k}\ge\int_1^2{1\over x}dx +\int_2^3{1\over x}dx +....\int_n^{n+1} {1\over x} dx = \int_1^{n+1} {1\over x} dx=log(n+1)$$
How to reach the conclusion $?$ . This is going nowhere .
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School, from our house as the crow flies, is 5.73 km. If we neglect air resistance and deal strictly with ballistic flight then we can materialize a wonderful fantasy. Starting in the backyard, extending over the top of the house, is a
launch-o-rocket, a rail-like launcher that accelerates the school-bound student until he or she can cruise over the city and arrive without bother of traffic. Our charter is to find the acceleration of the student from the launch-o-rocket.
Finding the Initial Velocity
We rely on the well-known fact that the maximum distance in a throw occurs when the departure angle is 45°. The vertical speed and the horizontal speed are equal. We denote these two identical speeds as $s$. Since distance is time multiplied by speed, the distance from home to school $d$ is
$$ d = t\cdot s.$$
We know the distance $d = $ 5.73 km.
Turning to the vertical speed, the student departs the launch-o-rocket with vertical speed $s$, but is immediately subject to gravitational acceleration. Since the student’s upward flight is exactly matched by his or her downward flight. Because the flight is matched, the student spends $t/2$ time rising and $t/2$ time descending. Since the student has no vertical speed at the top, we know that his or her speed is
$$ s = g\frac{t}{2},$$
where $g$ is the gravitational acceleration 9.8 m/s
2.
Now, we have a system of equations
$$ d = t\cdot s $$
$$ s = g\frac{t}{2}.$$ The system looks like it has a many variables, but really there are only two, $s$ and $t$. We know $g$ and $d$. To solve the system we substitute for $s$ in the first equation with the second to get
$$ d = tg\frac{t}{2} = g\frac{t^2}{2}$$
Solve for $t$ $$ t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2\cdot 5.73\,\text{m}}{9.8\,\text{m/s}^2}} \approx 34.2\,\text{s}. $$ Not a bad commute, a little over half a minute.
With $t$ in hand, we can find the magnitude of the initial velocity. Remember that the initial velocity is $s$ in the horizontal direction and $s$ in the vertical direction, so the speed when leaving the launcher is
$$ \left| \mathbf{v}_0\right| = \sqrt{s^2 + s^2} = \sqrt{2s^2} = s\sqrt{2}. $$ The initial speed the student must attain is given by the very first equation, $d = s\cdot t$. Solving for $s$ with the value of $t$ we found, we get
$$ s = \frac{5.73\,\text{km}}{34.2\,\text{s}} = 168\,\text{m/s}. $$
Finding the Acceleration
The ramp lives on a footprint that is about 80 ft, or 24.4 m. It is also 24.4 m tall, so special zoning is surely required! The rail of the launch-o-rocket is the hypotenuse of a triangle, and that triangle has sides 24.4 m, and a total length of $\sqrt{2}\cdot 24.4\,\text{m} = 34.5\,\text{m}$.
The formula for position after a period of acceleration is
$$ p = \frac{1}{2}a\tau^2.$$ For our system, we also know that the acceleration is the change in speed divided by the change in time. Our speed goes from zero to 168 m/s in $\tau$. Again, we have a system of equations,
$$ 34.5\, \text{m} = \frac{1}{2}a\tau^2 $$
$$ a = \frac{168\,\text{m/s}}{\tau}.$$
Solve for $a$ by first solving the second equation for $\tau$, and then substituting that result into the first equation to get
$$ 34.5\, \text{m} = \frac{1}{2}a\left(\frac{168\,\text{m/s}}{a}\right)^2 $$
$$ a = \frac{\left( 168\, \text{m/s}\right)^2}{2 \cdot 34.5\,\text{m}}
= 407\, \text{m/s}^2 = 41.5\, g. $$
The typical onset of death occurs when acceleration exceeds about $10g$, so unfortunately,
the launch-o-rocket is a single try system.
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We are challenged to determine "how fast MMa can get" and, in so doing, to suggest rules "to choose different programming styles." The original solution takes 116 seconds (on my machine). At the time the question was posted, the solution time had been reduced by a factor of 1000 (10 doublings of speed) to 0.124 seconds by suggestions from users in chat.
This solution takes 1300 (0.0013 seconds) on the same machine, for a further 100-fold speedup (another 7 doublings): microseconds
euler39[p_] := Commonest @ Flatten[Table[ l,
{m, 2, Sqrt[p/2]},
{n, Select[Range[m - 1], GCD[m #, m + #] == 1 && OddQ[m - #] &]},
{l, 2 m (m + n), p, 2 m (m + n)}]];
Timing[Table[euler39[1000], {i, 1, 1000}];]
{1.311, Null}
It scales nicely (changing
Table to
ParallelTable to double the speed on larger problems):
AbsoluteTiming[euler39p[10^8]]
{120.8409117, {77597520}}
That is almost linear performance.
Note the simplicity of the basic operations: this program could be ported to any machine that can loop, add, multiply, and tally (the square root can be eliminated). I estimate that an efficient compiled implementation could perform the same calculation in just a few microseconds, using just 500 bytes of RAM, for up to another nine doublings in speed.
This solution was obtained through a process that, in my experience,
generalizes to almost all forms of scientific computation:
The problem was analyzed theoretically to identify an efficient algorithm. Resulting speed: 0.062 seconds.
A timing analysis identified an MMa post-processing bottleneck. Some tweaking of this improved the timing. Speed: 0.0036 seconds (3600 microseconds).
In comments, J.M. and Simon Woods each suggested better MMa constructs, together reducing the execution time to 2400 microseconds.
The MMa bottleneck was removed altogether by a re-examination of the
algorithm and the data structure, achieving a final reduction to 1300 microseconds (and considerably less RAM usage).
Ultimately a speedup factor of 90,000 was achieved, and this was done
solely by means of algorithmic improvements: none of it can be attributed to programming style. Better MMa programmers than me will doubtlessly be able to squeeze most of the next nine speed doublings by compiling the code and making other optimizations, but--short of obtaining a direct $O(1)$ formula for the answer (which more or less would circumvent the whole point of the exercise, which is to use the computer for investigating a problem rather than for mere implementation of a theory-derived solution)--no more real speedup is possible. Note that compilation would also take us out of the MMa way of doing computation and bring us down to the procedural level of C and other compiled code.
The important lesson of this experience is that algorithm design is paramount. Don't worry about programming style or tweaking code: use your mathematical and computer science knowledge to find a better algorithm; implement a prototype; profile it; and--always focusing on the algorithm--see what can be done to eliminate bottlenecks. In my experience, one rarely has to go beyond this stage.
Detail of the story, as amended several times during development of this solution, follow.
This problem invites us to learn a tiny bit of elementary number theory, in the expectation it can result in a substantial
change in the algorithm: that's how to really speed up a computation. With its help we learn that these Pythagorean triples can be parameterized by integers $\lambda \gt 0$ and $m \gt n \gt 0$ with $m$ relatively prime to $n$. We may take $x = \lambda(2 m n)$, $y = \lambda(m^2-n^2)$, and $z = \lambda(m^2+n^2)$, whence the perimeter is $p = 2 \lambda m (m+n)$.
The restrictions imposed by $p\le 1000$ and the obvious fact that $p$ is even give the limits for a triple loop over the parameters, implemented in a
Table command below. The rest can be done without much thought--inelegantly and slowly--with brute force post-processing to avoid double counting $(x,y,z)$ and $(y,x,z)$ as solutions, to gather and count the solutions for each $p$, and select the commonest one.
(Although a triple loop sounds awful--one's instinctive reaction is to recoil at what looks like a $O(p^3)$ algorithm--notice that $m$ cannot exceed $\sqrt{p/2}$ and $n$ must be smaller yet, leaving few options for $\lambda$ in general. This gives us something like a $O(p f(p))$ algorithm with $f$ slowly growing, which scales very well. This limitation in the loop lengths is the key to the speed of this approach.)
euler39[p_] := Module[{candidates, scores, best},
candidates =
Flatten[Table[{Through[{Min, Max}[2 l m n, l (m^2 - n^2)]], 2 l m (m + n)},
{m, 3, Floor[Sqrt[p/2]]},
{n, 1, m - 1},
{l, 1, If[GCD[m, n] > 1, 0, Floor[p / (2 m (m + n))]]}], 2];
scores = {Last[Last[#]], Length[#]} & /@
DeleteDuplicates /@
Gather[candidates[[Ordering[candidates[[;; , 2]]]]], Last[#1] == Last[#2] &];
best = Max[scores[[;; , 2]]];
Select[scores, Last[#] >= best &]
];
The amount of speedup is surprising. Accurate timing requires repetition because the calculation is so fast:
Timing[Table[euler39[1000], {i, 1, 1000}]]
{3.619, {{{840, 8}}, {{840, 8}}, ...
I.e., the time to solve the problem is $0.0036$ seconds or $1/17000$ minutes. This makes larger versions of the problem accessible (using
ParallelTable instead of
Table to exploit some extra cores in part of the algorithm):
euler39[5 10^6] // AbsoluteTiming
{55.1441541, {{4084080, 168}}}
Even accounting for the parallelization, the timing is scaling nicely: it appears to be acting like $O(p\log(p))$. The limiting factor in
MMA is RAM: the program needed about 4 GB for this last calculation and attempted to claim almost 20 GB for
euler39[10^7] (but failed due to lack of RAM on this machine). This, too, could be streamlined if necessary using a more compact data structure, and perhaps could allow arguments up to $10^8$ or so.
Perhaps a solution that is faster yet (for smaller values of $p$, anyway) can be devised by factoring $p$, looping over the factors $\lambda$, and memoizing values for smaller $p$. But, at $1/300$ of a second, we have already achieved a four order of magnitude speedup, so it doesn't seem worth the bother.
Remarkably, this is much faster than the built-in
PowersRepresentations solution found by Simon Woods.
Edit
At this point, J.M. and Simon Woods weighed in with better MMa code, together speeding up the solution by 50% (see the comments). In pondering their achievement, and wondering how much further one could go, it became apparent that the bottleneck lay in the post processing to remove duplicates. What if we could generate each solution exactly once? There would no longer be any need for a complicated data structure--we could just tally the number of times each perimeter was obtained--and no post-processing at all.
To assure no duplication, we need to check that when generating a triple $\{x,y,p\}$ with $x^2 + y^2 = z^2$ and $x+y+z=p$, we do not later also generate $\{y,x,p\}$: that's how the duplicates arise. The initial effort tracked possible duplicates by forcing $x\le y$. The improved idea is to look at
parity.
The parameter $\lambda$ is intended to be the greatest common divisor of $\{x,y,p\}$. When it is, $x=2 m n$ and $y = m^2-n^2$ must be relatively prime. Because $x$ is obviously even, $y$ must be odd: that uniquely determines which of these two numbers is $x$ and which is $y$. Therefore,
we do not need to check for duplicates if, in the looping, we guarantee that $2 m n$ and $m^2-n^2$ are relatively prime. A quick way to check is that (a) $m n$ and $m+n$ are relatively prime and (b) $m$ and $n$ have opposite parity. Making this check is essentially all the work performed by the algorithm: the rest is just looping and counting.
By eliminating the check for duplicates, the new solution doubled the speed once more, from 2400 microseconds to 1300 microseconds. Where does it spend its time? For an argument $p$ (such as $1000$),
Approximately $p/2$ calculations of a GCD (for the second loop over
n).
A loop of length $p/2 / (m(m+n))$ for each combination of $(m,n)$.
An easy upper bound for the total number of iterations is $\frac{p}{8}\log{p}$, demonstrating the $O(p\log{p})$ scaling. If we assume the GCD calculations take an average of $\log{p}$ arithmetic operations each, the total number of operations is less than $p\log{p}$ plus comparable loop overhead together with incrementing an array of counts. The post-processing would merely scan that array for the location of its maximum. At $3 \times 10^9$ operations per second and $p=10^3$, the timing for good compiled code would be 0.3 microseconds. Problems up to $p \approx 10^{10}$ could be handled in reasonable time (under a minute) and without extraordinary amounts of RAM.
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I wanted to read a bit about modular forms of half integral weight. There is the notion of a 'factor of automorphy' (as for example, R.Rankin uses it in the book 'Modular Forms and Functions'). Now I am confused about the following: In modern literature, people seem to follow Shimura, i.e. they define modular forms of half integral weight using some cover of GL$_2^+(\mathbb{R})$, that is the group $G$ consisting of all pairs $(\alpha, \phi)$ with $\alpha \in \text{GL}_2^+(\mathbb{R})$ and $\phi^2 = t \det(\alpha)^{-1/2} (c\tau + d)$ [where $t$ is some extra parameter in $S^1$] with the multiplication $$(\alpha, \phi)(\beta, \psi) = (\alpha \beta, \phi(\beta \tau) \psi(\tau))$$
My question is the following:
Why does one introduce this cover when there already exists a notion of modular forms of half-integral weight, namely in terms of these factors of automorphy?
Example: For the usual Theta Series $\sum_{n \in \mathbb{Z}} e^{2 \pi i n^2 \tau}$ one can define what it means that this is a modular form of weight $1/2$ without even mentioning the group $G$.
In the beginning I thought that this would help about the problem of 'choosing' branches of square roots in the following sense: I thought one takes a subgroup $\Delta \subset G$ and then defines the slash operator and demands that a modular form should satisfy $f|_\xi = f$ for all $\xi \in \Delta$. So all branches exist in 'friendly' coexistence and one does not have to choose branches anymore. What actually happens is the following: Shimura wants the natural projection $P : \Delta \to P(\Delta)$ to be a bijection but this
is precisely choosing a smart branch of the square root for every matrix in question. So this does not seem to be the important point.
Of course, there is Hecke operators. If one just goes with the notion of a factor of automorphy, then this map is no more defined on matrices that are needed for defining Hecke operators. Furthermore, there is vector valued modular forms, say, for the Weil repn. Here it would be unnatural to say that the forms should have a translation behavior for some fixed lift of SL$_2(\mathbb{Z})$, so in view of these points I can understand why one uses this cover of GL$_2^+(\mathbb{R})$ but I have the feeling that I am missing out a much more basic point here...
Thanks in advance,
Fabian Werner
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The Annals of Probability Ann. Probab. Volume 20, Number 2 (1992), 681-695. The A.S. Behavior of the Weighted Empirical Process and the LIL for the Weighted Tail Empirical Process Abstract
The tail empirical process is defined to be for each $n \in \mathbb{N}, w_n(t) = (n/k_n)^{1/2}\alpha_n(tk_n/n), 0 \leq t \leq 1$, where $\alpha_n$ is the empirical process based on the first $n$ of a sequence of independent uniform (0,1) random variables and $\{k_n\}^\infty_{n=1}$ is a sequence of positive numbers with $k_n/n \rightarrow 0$ and $k_n \rightarrow \infty$. In this paper a complete description of the almost sure behavior of the weighted empirical process $a_n\alpha_n/q$, where $q$ is a weight function and $\{a_n\}^\infty_{n=1}$ is a sequence of positive numbers, is established as well as a characterization of the law of the iterated logarithm behavior of the weighted tail empirical process $w_n/q$, provided $k_n/\log\log n \rightarrow \infty$. These results unify and generalize several results in the literature. Moreover, a characterization of the central limit theorem behavior of $w_n/q$ is presented. That result is applied to the construction of asymptotic confidence bands for intermediate quantiles from an arbitrary continuous distribution.
Article information Source Ann. Probab., Volume 20, Number 2 (1992), 681-695. Dates First available in Project Euclid: 19 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aop/1176989800 Digital Object Identifier doi:10.1214/aop/1176989800 Mathematical Reviews number (MathSciNet) MR1159568 Zentralblatt MATH identifier 0754.60028 JSTOR links.jstor.org Citation
Einmahl, John H. J. The A.S. Behavior of the Weighted Empirical Process and the LIL for the Weighted Tail Empirical Process. Ann. Probab. 20 (1992), no. 2, 681--695. doi:10.1214/aop/1176989800. https://projecteuclid.org/euclid.aop/1176989800
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June 5th, 2017, 09:00 AM
# 1
Newbie
Joined: Jun 2017
From: Bucharest
Posts: 5
Thanks: 0
Impossible limit!
Sorry for the title. There is a limit that's giving me hell. I cannot solve it, but I know the answer is 0.
lim x->0 with x<0 from f(x)=(e^(1/x))/(x^2)
I sure know the limit is 0, but I don't know why. It is a vertical asymptote and for x>0 the limit is +inf but for x<0, I don't know how to calculate it.
Last edited by skipjack; June 5th, 2017 at 11:12 AM.
June 5th, 2017, 11:11 AM
# 2
Math Team
Joined: Dec 2013
From: Colombia
Posts: 7,689
Thanks: 2669
Math Focus: Mainly analysis and algebra
Write $y=-\frac1x$ so that
$$\lim_{x \to 0^-} \frac{e^\frac1x}{x^2}=\lim_{y \to +\infty} \frac{y^2}{e^y}$$
There is a standard result that
$$\lim_{t \to +\infty} \frac{t^a}{e^{bt}}=0 \quad \text{for all positive $a$ and $b$}$$
June 5th, 2017, 11:17 AM
# 3
Global Moderator
Joined: Dec 2006
Posts: 21,015
Thanks: 2250
Let x = -1/u, then it's the limit of u²/e^u as u tends to positive infinity. How you prove that limit is zero depends on how much you know about the exponential function.
June 5th, 2017, 01:43 PM
# 4
Newbie
Joined: Jun 2017
From: Bucharest
Posts: 5
Thanks: 0
Thanks very much guys. So basically the exponential function grows faster than polynomial so that's why it tends to 0.
Last edited by skipjack; June 5th, 2017 at 09:41 PM.
June 5th, 2017, 03:23 PM
# 6
Math Team
Joined: Dec 2013
From: Colombia
Posts: 7,689
Thanks: 2669
Math Focus: Mainly analysis and algebra
For $c \gt 0$ and $t \gt 1$ we have $$\begin{array}{c l}
0 \lt \tfrac1t \lt t^{c-1} & \text{and for $x \gt 1$} \\
0 \lt \int_1^x \tfrac1t \mathrm dt \lt \int_1^x t^{c-1} \mathrm dt \\
0 \lt \ln{x} \lt \tfrac1c x^c - 1 \lt \tfrac1c x^c \\ \\
0 \lt \ln^a{x} \lt \tfrac1c x^{ac} & (a \gt 0) \\
0 \lt \frac{\ln^a{x}}{x^b} \lt \tfrac1c x^{ac-b} & (b \gt 0) \\
& \text{now, pick $c = \frac{b}{2a} \gt 0$} \\
0 \lt \frac{\ln^a{x}}{x^b} \lt \tfrac1c x^{-\frac{b}{2}} \\
\end{array}$$
and since $-\frac{b}{2} \lt 0$ the squeeze theorem gives $$\lim_{x \to \infty} \frac{\ln^a{x}}{x^b} = 0$$
Now with $y = \ln x$ we get $$\lim_{y \to \infty} \frac{y^a}{e^{by}} = 0$$
for any positive $a$ and $b$.
Tags impossible, limit
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Hi Oz./.. just wanted to clarify some things:
Originally Posted by
oz93666
I find it hard to read such articles .. I have problems with nearly every line , just take the first line ...
"The stars still have their secrets. We know why they shine .We know why they trinkle . But we ...."
Well no , I don't believe established science has an accurate explanation of how they shine . The model is breathtakingly elegant And sounds very reasonable and satisfactory , but many holes in it , also papered over.
Oz... we have decades worth of quantitative experimental evidence that verifies not only how many different kinds of stars burn their fuels, but also the transport of thermal energy from the interior to the exterior and the subsequent light (not just as a brightness, but also a frequency-dependent curve). I have direct research experience with this and can fill you in on the evidence if you wish.
As far as I remember from my PhD, the main unknowns with stellar physics being researched today are the following:
1. MHD phenomena (corona, mass ejections, loops, heliosphere) are very difficult to model
2. supernovae models (SN 1a and SNII models) are still very uncertain and very difficult to do
3. thermal pulses, novae, oscillations are still difficult to model and observables are sometimes difficult to obtain
4. Extreme environments (black holes, neutron star equation of state, binary accretion) are difficult to observe and model
5. triple-alpha, $\displaystyle ^{12}$C($\displaystyle \alpha, \gamma$)$\displaystyle ^{16}$O reactions are uncertain in the lab, introducing uncertainties to stellar models and galactic chemical evolution models
6. Low-mass regime: brown dwarfs, hot Jupiters, dwarfs are hard to observe because they are faint (lower part of Salpeter curve)
7. Mixing models are crude (convective-boundary mixing, thermohaline mixing and entrainment models often use diffusive approximations)
8. Interior state: direct observations are difficult, so helioseismology could be an important test of stellar models
9. Population studies of massive stars are difficult, because a lot of them have exploded already (upper part of Salpeter curve)
10. Origin of r-process elements
Among others.
We do know why stars twinkle ...dust in the atmosphere...
We know the twinkling effect is caused by intervening matter between the stars and telescopes, including the Earth's atmosphere and the interstellar medium. There are extensive reddening maps of the Milky Way and interstellar medium to predict the effect. We also have adaptive optics at observational facilities to help eliminate the effect of twinkling and other aberrations (such as coma).
But back to dark matter (and dark energy) .... bottomline .... observations didn't fit in with established theories of gravitation , so rather than examine the theory , people assumed the theory HAD to be correct, it was chiseled in stone , and the only answer was to hypothesise some magical , invisible , fairy mater ....sprinkled just where it was needed to make everything right again .... No experiments have ever found or detected evidence of this dark matter ... so it must remain just another unproven theory, and we have plenty of them ....
Since the original observations of galactic rotation profiles, scientists have explored both possibilities (additional mass and modified Newtonian dynamics) and it is still one of the longest unsolved problems in astrophysics. There's the occasional idiot who claims to "know" the answer as being in one camp or the other, but any honest scientist will admit that the solution could be either.
We have other evidence in the fossil record that the assumption that G is constant may not be true ... it could vary over time , hence be different in observations of very distant stars ....
Those papers were really, really shite though dude. I don't think there's any compelling evidence yet for any of the fundamental constants not being constant.
Imho, the dark matter article doesn't really shed light on anything new. It's probably just not well reported about the MOND efforts or the TEVES models, so they're just getting round to reporting on it.
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What's the effective key length of Two-Key Triple-DES, for some (possibly several) reasonably well-defined and sensible definitions of
effective key length, say assuming attack using ample chosen plaintext?
A sensible definition of effective key length (in bits) of a block cipher under some attack assumptions could be: one plus the base-2 logarithm of the number of evaluations of the cipher (for arbitrary key and plaintext) that could have been made for a cost equivalent to the expected cost of recovering the key using the best known attack applicable to said assumptions.
By this definition, for any secure block cipher, effective key length is at most about the number of key bits that can influence the output. Proof sketch: consider brute force key search, and that a block cipher having a cost of evaluation not much larger than a comparison would be insecure.
Two-Key Triple-DES (also known as TDEA Keying option 2) is a 64-bit block cipher with a 16-octet key, of which 112 bits can influence the output, defined by $$\operatorname{3DES}(K_1\|K_2,P)=\operatorname{DES}(K_1,\operatorname{DES}^{-1}(K_2,\operatorname{DES}(K_1,P)))$$ where $K_1$ and $K_2$ are 8-octet keys, of which 56 bits can influence the output.
NIST has stated that with $2^{40}$ plaintext/ciphertext pairs, the security is at least 80 bits, but gives no upper bound beyond the obvious 112 bits, which is quite a wide margin.
The effective key length of 3DES assuming two or three chosen plaintexts (per the definition proposed) is no more than about 109.5 bits (see proof below); that obviously can only be lower with more (chosen) plaintext/ciphertext pairs.
One of the best known attack is: Paul C. van Oorschot and Michael J. Wiener:
A Known-Plaintext Attack on Two-Key Triple Encryption (in proceedings of Eurocrypt 1990); but it is not easily translated into effective key length, at least for the definition that I propose, for the cost is dominated by memory accesses.
The 109.5 bits upper bound stated is 112, minus $\log_2(3)\approx 1.58$ because the core attack loop only performs a DES thus is about 3 times less costly than a 3DES, minus 1 thanks to the the complementation property, plus a little something for the two comparisons in the core loop.
More in details: 3DES inherit from DES the complementation property: $$\forall K, \forall P, \operatorname{3DES}(\overline K,\overline P)=\overline{\operatorname{3DES}(K,P)}$$ and that allows a brute force search with maximum cost quite close to $2^{111}$ DES encryptions, that is $\approx2^{111}/3$ 3DES encryptions (and expected cost half that), with two or three chosen plaintexts:
obtain, for some $P$, the ciphertexts $\operatorname{3DES}(K,P)\to C$ and $\operatorname{3DES}(K,\overline P)\to C'$ for each of $2^{55}$ value of $K_1$ with high bit zero compute $\operatorname{DES}(K_1,P)\to A$, $\operatorname{DES}^{-1}(K_1,C)\to B$, $\operatorname{DES}^{-1}(K_1,\overline{C'})\to B'$ for each of $2^{56}$ value of $K_2$ compute $\operatorname{DES}^{-1}(K_2,A)\to X$; this step dominates the cost if $X=B$ (implying $\operatorname{3DES}(K_1\|K_2,P)=C$) if $\operatorname{3DES}(K_1\|K_2,\overline P)=C'$, output $K_1\|K_2$ if $X=B'$ (implying $\operatorname{3DES}(\overline{K_1\|K_2},\overline P)=C'$) if $\operatorname{3DES}(\overline{K_1\|K_2},P)=C$, output $\overline{K_1\|K_2}$
This is guaranteed to output the right $K$, and (heuristically) will seldom output others for random choice of $K$; and then another plaintext/ciphertext pair allows to rule out a false positive with overwhelming odds. The cost is dominated by $2^{111}$ DES at step (1); $2^{112}$ relatively cheap comparisons at steps (2) and (3), with very occasional false positives costing an expected $3\cdot 2^{48}$ evaluations of DES for their resolution; and the $3\cdot 2^{55}$ DES for computation of $A$, $B$, $B'$.
This attack is impractical: the total effort devoted to bitcoin mining (which is comparable in nature) as of February 2016 would have explored much less than one millionth of the search space for a single attacked key.
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The mapping class group of the closed unit disk $D_n\subset\mathbb{C}$ with $n$ equally spaced points $P$ on the x-axis (0,1) removed is defined as the set of isotopy classes of self-homeomorphisms $D_n\rightarrow D_n$ which fix the boundary: $\phi\partial D_n=\partial D_n$, and permute the distinguished points: $\phi P=P$.
I am considering a representative $\phi:D_n\rightarrow D_n$ of an element of said mapping class group that induces the identity-automorphism of the fundamental group $\pi_1(D_n,x_0)$, i.e. $\phi_*=\text{id}_{\pi_1(D_n,x_0)}$, where $x_0\in\partial D_n$. I am convinced that $\phi$ has to be homotopic to the identity map $\text{id}_{D_n}$ but am unable to prove it. I know that since $\phi$ is a homeomorphism, $\phi^{-1}_*$ has to be the identity-automorphism as well, but I fail to see how I could use this.
Also, since $\phi$ fixes all loops, for every point $x$ I can choose some loop $\gamma$ on which $x$ lies, and get some homotopy $H_\gamma$ to a loop $\phi\circ\gamma$, on which $\phi(x)$ lies. But I don't see how I could combine those homotopies to get a well defined homotopy from $\text{id}_{D_n}$ to $\phi$, or whether that is even possible at all.
A hint would be appreciated.
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I'm trying to find the maximum of a function $f = a^T\mu$ where a is a vector when we have a constraint of the form:
$$g(\mu) = n\mu^T S^{-1}\mu - C = 0$$
where C is a fixed constant, $S^{-1}$ is a fixed matrix , $n$ is a number and $\mu$ is a vector of variables. In fact, my goal is to find a $\mu$ that satisfies $g(\mu)$ and at the same time, maximizes f.
I thought of using Lagrange multipliers but I'm stuck and I appreciate if you can help me solve this problem. Below is what I've done so far:
define: $L = f(\mu) - \lambda g(\mu)$. Then we have:
$\frac{\partial L}{\partial \mu} = a^T - 2 \lambda n \mu^TS^{-1} = 0 \rightarrow \mu^T = \frac{a^TS}{2\lambda n}$
$\frac{\partial L}{\partial \lambda} = g(\mu) = 0$
I don't know how to proceed. Like I don't know how to get rid of lambda in the expression $\mu^T = \frac{a^TS}{2\lambda n}$ I derived above.
Thanks very much for your help.
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So, since $V\in\mathbb R^{m\times n}$ has full column rank $n$, implying also $m\geq n$, we might obtain the pseudoinverse using$$V^+=(V^TV)^{-1}V^T$$since $V^TV$ is invertible and $V^*=V^T$ in $\mathbb R$.
Now, let's denote by $i$ the index of the row which is scaled and by $d_i$ - the scaling factor.Thus, a perturbed matrix $V^\prime$ can be expressed as$$V^\prime = D_i V= \text{diag}(\underbrace{1,\ldots,1}_{i-1},d_i,\underbrace{1,\ldots,1}_{m-i})V$$
Now, how can we express it as a low-rank update?
Consider$$u=e_i=\left[\begin{array}{c}0\\\vdots\\1 \\\vdots\\0\end{array}\right], \quad\quadv=(V_{i,*})^T$$
where $u$ is the $i$th elementary unit vector in $m$ dimensional space and $v$ is the transposed $i$th row of the original matrix $V$. Then:$$V^\prime = V+(d_i-1)uv^T$$
Now, let's express the update to the "correlation" matrix $V^T V$:$$\begin{aligned}{V^\prime}^T{V^\prime}&=(V+(d_i-1)uv^T)^T(V+(d_i-1)uv^T)\\&=V^TV+(d_i-1)v\underbrace{u^TV}_{v^T}+(d_i-1)\underbrace{V^Tu}_{v}v^T+(d_i-1)(d_i-1)v\underbrace{u^Tu}_{1}v^T\\&=V^TV+\left((d_i-1)+(d_i-1)+(d_i-1)^2\right)vv^T\\&=V^TV+(d_i^2-1)vv^T\end{aligned}$$
With that, we expressed a rank-1 update to the correlation matrix $V^T V$ for which we already know the inverse. For compactness, let's denote
$$\gamma=d_i^2-1$$.
Now, let's use Sherman-Morrison to find an update to the inverse of the correlation matrix:
$$\begin{aligned}({V^\prime}^T{V^\prime})^{-1}&=(V^TV+\gamma v v^T)^{-1}\\&=(V^TV)^{-1}-\frac{\gamma (V^TV)^{-1}vv^T(V^TV)^{-1}}{1+\gamma v^T (V^TV)^{-1}v}\end{aligned}$$
Let's denote the following matrix-vector product by$$w = (V^TV)^{-1}v$$Then, since $V^T V$ is symmetric
$$\begin{aligned}({V^\prime}^T{V^\prime})^{-1}&=(V^TV)^{-1}-\frac{\gamma ww^T}{1+\gamma v^T w}\\&=(V^TV)^{-1}-\tau ww^T, \quad \tau=\frac{d_i^2-1}{1+(d_i^2-1)v^Tw}\end{aligned}$$
So, in order to update the inverse of the correlation matrix (provided you have the computation for the original matrix done):
Compute matrix-vector product $w = (V^TV)^{-1}v$ in $n^2$ operations. Calculate dot-product of $w$ and original row $v$: $v^T w$ in $n$ operations. Calculate $\tau$ in constant time. Calculate the scaled by $\tau$ outer product $\tau ww^T$ in $n^2$ operations Calculate the update to the inverse of the correlation matrix by subtracting the outer product in $n^2$ operations.
So, you achieved $({V^\prime}^T{V^\prime})^{-1}$ in $3n^2+n$ operations. Probably, can be done in $2n^2+n$ if fused flops are used combining steps 4 and 5.
Now, you can calculate
$${V^\prime}^+ = ({V^\prime}^T{V^\prime})^{-1}V^TD_i$$
or leave it in the unassembled form depending on your future usage.
I am not sure if that will lead to significant speed-up for the requested $\mathbb R^{4\times 3}$ case but is worth trying.
Note, this particular technique relies on $V$ having full-column rank, otherwise, the pseudoinverse cannot be found this way and SVD decomposition has to be used -> where updates are also possible, but much trickier.
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Evan Schwab
1, Rene Vidal 2, and Nicolas Charon 3
1Electrical and Computer Engineering, Johns Hopkins University, Baltimore, MD, United States, 2Biomedical Engineering, Johns Hopkins University, Baltimore, MD, United States, 3Applied Mathematics and Statistics, Johns Hopkins University, Baltimore, MD, United States
Synopsis Sparse modeling of dMRI signals has become of major interest for advanced protocols such as HARDI which require a large number of q-space measurements. With few exceptions, prior work have developed bases to sparsely represent q-space signals per voxel with additional constraints of spatial regularity. In this work, we propose a single basis to represent an entire HARDI dataset by modeling spatial and angular domains jointly, achieving an unprecedented level of sparsity. With a globally compressed representation we can then redefine HARDI processing, diffusion estimation, feature extraction and segmentation, and drastically reduce acquisition time and data storage. Introduction
Diffusion magnetic resonance imaging (dMRI) is a 6D neuroimaging modality that measures 3D q-space signals at every voxel in a brain MRI volume to estimate the orientation of neuronal fiber bundles,
in vivo
. Most every dMRI reconstruction algorithm
1,2,3
finds a basis to represent a q-space signal and adds terms to enforce sparsity, orientation distribution non-negativity or spatial regularity based on similar basis representations within neighboring voxels. Some recent methods have considered sparse optimization over multiple voxels or an entire volume jointly to reduce redundancies and invoke implicit regularization but all of these methods still only operate on a q-space basis
4, 5,6,7,8,9,10
. In contrast, we propose a single basis representation that models q-space and the spatial domain jointly resulting in a drastically compressed model of an entire brain dMRI dataset. This sparse global representation has the potential to push the limits of dMRI acquisition time, data storage, diffusion estimation, fiber segmentation, and other processes. In this work we consider high angular resolution diffusion imaging (HARDI) which simplifies general positions in 3D q-space to 2D locations on the unit q-sphere, but our main idea can be used for any 6D dMRI protocol. We propose a single joint spatial-angular basis for sparse whole brain HARDI reconstruction and orientation distribution function (ODF) estimation.
Methods
A HARDI signal $$$S(v,\vec{g})\in \Omega \times \mathbb{S}^2$$$ is a function of two variables: $$$v$$$, and $$$\vec{g}$$$ the gradient direction, or angular location, on the unit sphere $$$\mathbb{S}^2$$$. Let $$$\Gamma$$$ be a basis on $$$\mathbb{S}^2$$$ then we can write $$$S(v,\vec{g}) = \sum_i a_i(v) \Gamma_i(\vec{g})$$$. We notice that each $$$a_i$$$ is a 3D volume so they can each be represented by a spatial basis for $$$\Omega$$$, say $$$\Psi$$$, as $$$a_i(v) = \sum_j c_{i,j} \Psi_j(v)$$$. Combining equations we have $$$S(v,\vec{g}) = \sum_i \sum_j c_{i,j} \Psi_j(v)\Gamma_i(\vec{g})$$$. Therefore we arrive at a joint spatial-angular basis representation of the HARDI signal. In matrix notation, $$$s = \Phi c$$$ where $$$s$$$ is the vectorized HARDI tensor $$$S$$$, $$$c$$$ is the vectorized $$$C = [c_{i,j}]$$$ and $$$\Phi = \Psi \times \Gamma$$$, the Kronecker product. To find a sparse global parameterization $$$c$$$ we aim to solve $$c = arg\min_c || s - \Phi c || \ \ s.t \ \ ||c||_0 < k.$$ We use an Orthogonal Matching Pursuit (OMP) algorithm designed to exploit the structure of $$$\Phi$$$. In particular, we choose $$$\Psi$$$ to be an orthonormal 3D wavelet basis since MRIs are known to be sparse in this domain. With $$$\Psi$$$ orthonormal we need not compute explicitly $$$P = \Phi^\top \Phi$$$ for OMP but can pre-compute the smaller $$$G = \Gamma^\top \Gamma$$$ and sparsely populate $$$P(i_{\Gamma},j_{\Psi}) = G(i_{\Gamma},j_{\Gamma})$$$ if the next chosen index $$$i’_{\Psi} = i_{\Psi}$$$ and zero otherwise.
For this work we explore two angular bases, $$$\Gamma$$$, the spherical harmonic (SH) basis and the over-complete spherical ridgelet (SR) basis
1 known to provide a sparse ODF representation in the spherical wavelet (SW) domain. Results
For our experiments we used the ISBI 2013 HARDI Challenge Phantom dataset (50x50x50 volume with 64 gradient directions, SNR=30). For simplicity we chose a single slice (z=25). As a full, non-sparse "ground truth" (GT) representation we used the traditional SH (order $$$L=4$$$) reconstruction per voxel by least squares, which exudes a total of 15x50x50 = 37,500 atoms. Figures 1 and 2 show a qualitative comparison between the GT and the proposed method using the SH and SR bases, respectively with a fraction of the total atoms. We notice an inherent denoising due to the spatial wavelets. In Figure 3 we present a quantitative comparison of the normalized mean squared error (NMSE) between the proposed SH and SR reconstructions and the GT as we increase the sparsity level. Then Figure 4 illustrates how spatial and angular redundancies are encoded by the global atoms over the entire dataset as the OMP algorithm progresses.
As an important note, using 2500 atoms for this dataset achieves a comparably accurate reconstruction and amounts to an average of only 1 atom per voxel, which is unachievable with a sparse angular basis alone.
By exploring new spatial-angular bases and dictionary learning techniques we believe we can achieve even more accurate reconstructions with much fewer atoms.
Conclusion
We have thus presented a framework for encoding an entire HARDI dataset with a single joint spatial-angular basis which achieves an unprecedented level of sparsity. This framework is important for globally subsampling HARDI signals to speed up acquisition, enforcing ODF non-negativity globally, spatial-angular feature extraction, fiber segmentation and data storage.
AcknowledgementsThis work funded by JHU start-up funds.
References
[1] Tristán-Vega, Antonio, and Carl-Fredrik Westin. "Probabilistic ODF estimation from reduced HARDI data with sparse regularization." Medical Image Computing and Computer-Assisted Intervention (MICCAI) 2011, pp. 182-190.
[2] Merlet, S. L. and Caruyer, E. and Ghosh, A. and Deriche, R. "A Computational Diffusion MRI and Parametric Dictionary Learning Framework for Modeling the Diffusion Signal and its Features." Medical Image Analysis. 2013, 17(7), pp. 830-843.
[3] Gramfort, A. and Poupon, C. and Descoteaux, M. "Denoising and Fast Diffusion Imaging with Physically Constrained Sparse Dictionary Learning." Medical Image Analysis. 2014, 18(1) 36-49.
[4] Ye, W. and Vemuri, B. C. and Entezari, A. "An over-complete dictionary based regularized reconstruction of a field of ensemble average propagators." International Symposium on Biomedical Imaging (ISBI) 2012, pp. 940-943.
[5] Gramfort, A. and Poupon, C, and Descoteaux, M. "Sparse DSI: Learning DSI structure for denoising and fast imaging." Medical Image Computing and Computer-Assisted Intervention (MICCAI) 2012. 288-296.
[6] Wu, Y. and Feng, Y. and Li, F. and Westin, C. F. "Global consistency spatial model for fiber orientation distribution estimation.” International Symposium on Biomedical Imaging (ISBI) 2015, pp. 1180-1183.
[7] Cheng, J. and Shen, D. and Basser, P. J. and Yap, P. T. "Joint 6D kq Space Compressed Sensing for Accelerated High Angular Resolution Diffusion MRI." Information Processing in Medical Imaging (IPMI) 2015, pp. 782-793
[8] Ourselin, S. and Alexander, D. C. and Westin, C.-F., and E Cardoso, M. J. "Leveraging EAP-Sparsity for Compressed Sensing of MS-HARDI in (k, q)-Space." Information Processing in Medical Imaging (IPMI) 2015, pp. 375-386.
[9] Ouyang, Y., Chen, Y., Wu, Y., & Zhou, H. M. "Total Variation and Wavelet Regularization of Orientation Distribution Functions in Diffusion MRI." Inverse Problems and Imaging, 2013, 7(2): 565-583.
[10] Ouyang, Y. and Chen, Y. and Wu, Y. "Vectorial total variation regularisation of orientation distribution functions in diffusion weighted MRI."International Journal of Bioinformatics Research and Applications, 2014, 10(1): 110-127.
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What can be said about $\sigma(T_1 \otimes T_2)$ and $\sigma(T_1) \otimes \sigma(T_2)$, when $T_i$ are topologies that aren't necessary second countable, and $\otimes$ denotes, at the left, the product topology, and at the right, the product $\sigma$-algebra ?
Do you know examples where neither of these is included in the other ?
EDIT : I got an answer of these two questions here : Is "product" of Borel sigma algebras the Borel sigma algebra of the "product" of underlying topologies? Sorry for the duplicate.
I was asking myself if there was a measurable space $(X,\Sigma)$, a topological vector space $(V,\mathbf{T})$ and $\forall i \in \{1,2\}$, $f_i : (X,\Sigma) \rightarrow (V,\mathbf{T})$ measurable, such that $f_1 + f_2$ was $\textbf{not}$ mesurable. This cannot happen if $V$ is second countable, but I don't know many not second countable topological vector spaces...
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This post contains several code blocks, you can copy them easily with the help of
importCode.
Analytic Solution
The analytic solution can be obtained with
LaplaceTransform and
FourierSinCoefficient. First, make a Laplace transform on the equation and b.c.s and plug in the i.c.s:
Clear[f];
f[x_] = x (1 - x);
eqn = D[u[t, x], {t, 2}] + D[u[t, x], {x, 4}] == 0;
ic = {u[0, x] == f@x, Derivative[1, 0][u][0, x] == 0};
bc = {u[t, 0] == 0, u[t, 1] == 0, Derivative[0, 2][u][t, 0] == 0,
Derivative[0, 2][u][t, 1] == 0};
teqn = LaplaceTransform[{eqn, bc}, t, s] /. Rule @@@ ic
Now we have an ODE, solve it with
DSolve:
tsol = u[t, x] /. First@DSolve[teqn/.
HoldPattern@LaplaceTransform[a_, __] :> a, u[t, x], x] // Simplify
Notice the replacement
HoldPattern@LaplaceTransform[a_, __] :> a is necessary because
DSolve has trouble in handling expression containing
LaplaceTransform. The last step is to transform the solution back, but sadly
InverseLaplaceTransform can't handle
tsol. At this point, one work-around is to turn to numeric inverse Laplace transform, you can use this or this package for this task. But for your specific problem, we can circumvent the issue by expanding
tsol with Fourier sine series:
easyFourierSinCoefficient[expr_, t_, {a_, b_}, n_] :=
FourierSinCoefficient[expr /. t -> t + a, t, n,
FourierParameters -> {1, Pi/(b - a)}] /. t -> t - a
easyTerm[t_, {a_, b_}, n_] := Sin[Pi/(b - a) n (t - a)]
term = easyTerm[x, {0, 1}, n];
coe = easyFourierSinCoefficient[tsol, x, {0, 1}, n]
$$-\left(i\left(\frac{(1+i) (-1)^n e^{i \sqrt{2} \sqrt{s}}}{(1+i) \pi n+i \sqrt{2} \sqrt{s}}\right.\right....$$
coe still looks complex, but inspired by those
(-1)^ns in it, we split it to odd and even part and simplify:
oddcoe =
Simplify[coe /. n -> 2 m - 1, m > 0 && m ∈ Integers] /. m -> (1 + n)/2
(* (8 s)/(n^3 π^3 (n^4 π^4 + s^2)) *)
evencoe = Simplify[coe /. n -> 2 m, m ∈ Integers] /. m -> n/2
(* 0 *)
InverseLaplaceTransform can handle the series form of the transformed solution without difficulty:
soloddterm = Function[n, #] &@InverseLaplaceTransform[oddcoe term, s, t]
(* Function[n, (8 Cos[n^2 π^2 t] Sin[n π x])/(n^3 π^3)] *)
To find the final solution, just summate:
solgenerator[n_] := Compile[{t, x}, #] &@Total@soloddterm@Range[1, n, 2];
sol = solgenerator[200];
Animate[Plot[sol[t, x], {x, 0, 1}, PlotRange -> .3], {t, 0, 1}]
The animation is similar to the one in the subsequent solution so I'd like to omit it.
Fully
NDSolve-based Numeric Solution
Go back to the old-fashioned
"TensorProductGrid", set
"DifferentiateBoundaryConditions" -> {True, "ScaleFactor" -> 100} (or
NDSolve will set
"ScaleFactor" to
0 so the inconsistent b.c.s will be severely ignored, for more information, check the obscure tutorial) and
DifferenceOrder -> 2:
mol[n_Integer, o_:"Pseudospectral"] := {"MethodOfLines",
"SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> n,
"MinPoints" -> n, "DifferenceOrder" -> o}}
mol[tf:False|True,sf_:Automatic]:={"MethodOfLines",
"DifferentiateBoundaryConditions"->{tf,"ScaleFactor"->sf}}
tu = NDSolveValue[{eqn, ic, bc}, u, {t, 0, 10}, {x, 0, 1},
Method -> Union[mol[27, 2], mol[True, 100]], MaxSteps -> Infinity];
Animate[Plot[tu[t, x], {x, 0, 1}, PlotRange -> .3], {t, 0, 10}]
NDSolve spits out the
NDSolveValue::eerr warning, but in many cases
NDSolveValue::eerr isn't a big deal, and the result indeed looks OK:
Partly
NDSolve-based Numeric Solution
Theoretically we can also set
"DifferentiateBoundaryConditions" -> False to avoid the inconsistent b.c.s being ignored, but strangely
NDSolve spits out the
icfail warning and fails. I'm not sure about reason, but found that we can manually discretize the spatial derivative and solve the obtained ODE set with
NDSolve to avoid the issue.
First, let's define a function
pdetoode that discretizes PDEs to ODEs (Additionally, though not related to OP's problem, I've also define a function
pdetoae that discretizes differential equations to algebraic equations based on
pdetoode. A
rebuild function is also created to combine the list of
InterpolatingFunctions to a single
InterpolatingFunction):
Clear[fdd, pdetoode, tooderule, pdetoae, rebuild]
fdd[{}, grid_, value_, order_, periodic_] := value;
fdd[a__] := NDSolve`FiniteDifferenceDerivative@a;
pdetoode[funcvalue_List, rest__] :=
pdetoode[(Alternatives @@ Head /@ funcvalue) @@ funcvalue[[1]], rest];
pdetoode[{func__}[var__], rest__] := pdetoode[Alternatives[func][var], rest];
pdetoode[front__, grid_?VectorQ, o_Integer, periodic_: False] :=
pdetoode[front, {grid}, o, periodic];
pdetoode[func_[var__], time_, {grid : {__} ..}, o_Integer,
periodic : True | False | {(True | False) ..} : False] :=
With[{pos = Position[{var}, time][[1, 1]]},
With[{bound = #[[{1, -1}]] & /@ {grid}, pat = Repeated[_, {pos - 1}],
spacevar = Alternatives @@ Delete[{var}, pos]},
With[{coordtoindex =
Function[coord,
MapThread[Piecewise[{{1, PossibleZeroQ[# - #2[[1]]]},
{-1, PossibleZeroQ[# - #2[[-1]]]}}, All] &, {coord, bound}]]},
tooderule@Flatten@{
((u : func) | Derivative[dx1 : pat, dt_, dx2___][(u : func)])[x1 : pat, t_,
x2___] :> (Sow@coordtoindex@{x1, x2};
fdd[{dx1, dx2}, {grid}, Outer[Derivative[dt][u@##]@t &, grid],
"DifferenceOrder" -> o, PeriodicInterpolation -> periodic]),
inde : spacevar :>
With[{i = Position[spacevar, inde][[1, 1]]}, Outer[Slot@i &, grid]]}]]];
tooderule[rule_][pde_List] := tooderule[rule] /@ pde;
tooderule[rule_]@Equal[a_, b_] :=
Equal[tooderule[rule][a - b], 0] //. eqn : HoldPattern@Equal[_, _] :> Thread@eqn;
tooderule[rule_][expr_] := #[[Sequence @@ #2[[1, 1]]]] & @@ Reap[expr /. rule]
pdetoae[funcvalue_List, rest__] :=
pdetoae[(Alternatives @@ Head /@ funcvalue) @@ funcvalue[[1]], rest];
pdetoae[{func__}[var__], rest__] := pdetoae[Alternatives[func][var], rest];
pdetoae[func_[var__], rest__] :=
Module[{t},
Function[pde, #[
pde /. {Derivative[d__][u : func][inde__] :>
Derivative[d, 0][u][inde, t], (u : func)[inde__] :> u[inde, t]}] /. (u : func)[
i__][t] :> u[i]] &@pdetoode[func[var, t], t, rest]]
rebuild[funcarray_, grid_?VectorQ, timeposition_: 1] :=
rebuild[funcarray, {grid}, timeposition]
rebuild[funcarray_, grid_, timeposition_?Negative] :=
rebuild[funcarray, grid, Range[Length@grid + 1][[timeposition]]]
rebuild[funcarray_, grid_, timeposition_: 1] /; Dimensions@funcarray === Length /@ grid :=
With[{depth = Length@grid},
ListInterpolation[
Transpose[Map[Developer`ToPackedArray@#["ValuesOnGrid"] &, #, {depth}],
Append[Delete[Range[depth + 1], timeposition], timeposition]],
Insert[grid, Flatten[#][[1]]["Coordinates"][[1]], timeposition]] &@funcarray]
The syntax of
pdetoode is as follows: the 1st argument is the function to be discretized, the 2nd argument is the independent variable in the resulting ODE system (usually it's the variable playing the role of "time" in the underlying model), the 3rd argument is the list of spatial grid, the 4th argument is difference order.
Notice
pdetoode is a general purpose function. You may feel some part of the source code confusing. To understand it, just notice the following truth:
a /. a | b[m_] :> {m} outputs
{}.
Derivative[][u] outputs
u.
Then discretize
eqn,
ic and
bc and remove redundant equations:
lb = 0; rb = 1;
torder = 2;
(* Difference order of x: *)
xdifforder = 2;
points = 25;
grid = Array[# &, points, {lb, rb}];
(* There're 4 b.c.s, so we need to remove 4 equations from every PDE/i.c.,
usually the difference equations that are the "closest" ones to the b.c.s
are to be removed: *)
removeredundant = #[[3 ;; -3]] &;
(* Use pdetoode to generate a "function" that discretizes the spatial derivatives of
PDE(s) and corresponding i.c.(s) and b.c.(s): *)
ptoofunc = pdetoode[u[t, x], t, grid, xdifforder];
odeqn = eqn // ptoofunc // removeredundant;
odeic = removeredundant/@ptoofunc@ic;
odebc = bc // ptoofunc;
(* Another possible treatment for i.c.s and b.c.s: *)
(* odeic = ptoofunc@ic; *)
(* odebc = With[{sf=100},
Map[D[#, {t,torder}]+sf#&,bc/.ptoofunc,{2}]];*)
sollst = NDSolveValue[{odebc, odeic, odeqn}, u /@ grid, {t, 0, 10}, MaxSteps -> Infinity];
(* Rebuild the solution for the PDE from the solution for the ODE set: *)
sol = rebuild[sollst, grid];
Animate[Plot[sol[t, x], {x, 0, 1}, PlotRange -> .3], {t, 0, 10}]
The animation is similar to the one in the aforementioned solution so I'd like to omit it. This approach seems to be more robust than the fully
NDSolve-based one, because even if the
xordereqn i.e. the difference order for spatial derivative is set to
4, it's still stable, while the fully
NDSolve-based one becomes wild when
t is large.
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Multicollinearity Harsha Achyuthuni December 23, 2018
I want to go thru some basic statistical concepts before continuing with the EDA and in-time problem. I want to explain what is multicollinearity and how and when to tackle it.
Collinearity
Collinearity is a linear association between two variables. Two variables are collinear if there is a linear relationship between them. For example, \({\displaystyle X_{1}}\) and \({\displaystyle X_{2}}\) are perfectly collinear if there exist parameters \({\displaystyle \lambda _{0}}\) and \({\displaystyle \lambda _{1}}\) such that, for all observations i, I have
\[X_{2i} = \lambda_0 + \lambda_1 X_{1i}\] Here \({\displaystyle \lambda _{1}}\) can be considered as the slope of the equation and \({\displaystyle \lambda _{0}}\) is the intercept.
Correlation
Correlation is a numerical measure of how close two variables are having a linear relationship with each other. The most popular form of the correlation coefficient is Pearson’s coefficient(\(r_{xy}\)).
Pearson’s product-moment coefficient
Pearson’s product-moment coefficient is the most common correlation coefficient. It is obtained by dividing the covariance of the two variables by the product of their standard deviations.
\[{\displaystyle r_{xy}=\mathrm {corr} (X,Y)={\mathrm {cov} (X,Y) \over \sigma _{X}\sigma _{Y}} = {\frac {\sum \limits _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})}{(n-1)s_{x}s_{y}}}={\frac {\sum \limits _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})}{\sqrt {\sum \limits _{i=1}^{n}(x_{i}-{\bar {x}})^{2}\sum \limits _{i=1}^{n}(y_{i}-{\bar {y}})^{2}}}},}\] where x and y are the means of X and Y, and \(s_{x}, s_{y}\) are the standard deviations of X and Y.
The Pearson correlation is \(+1\) in case of a perfect direct(increasing) linear relationship (correlation), \(-1\) in the case of a perfect decreasing (inverse) linear relationship(anti-correlation), and some value in the open interval \((-1, +1)\) in all other cases, indicating the degree of linear dependence between the variables. As it approaches zero there is less of a relationship(closer to uncorrelated). The closer the coefficient is to either \(-1\) or \(+1\), the stronger is the correlation between the variables.
If the variables are independent, Pearson’s correlation coefficient is 0, but the converse is not true because the correlation coefficient detects only linear dependencies between two variables. For example, suppose the random variable X is symmetrically distributed about zero, and \(Y = X^{2}\). Then Y is completely determined by X, so that X and Y are perfectly dependent, but their correlation is zero.
Multicollinearity
Multicollinearity occurs when independent variables in a regression model are correlated. This correlation is a problem because independent variables should be independent. If the degree of correlation between variables is high, it can cause problems when I fit the model and interpret the results.
A key goal of regression (or classification) analysis is to isolate the relationship between each independent variable and the dependent variable. The interpretation of a regression coefficient is that it represents the mean change in the dependent variable for each 1 unit change in an independent variable when I hold all of the other independent variables constant. That last portion is crucial for our discussion about multicollinearity.
The idea is that I can change the value of one independent variable and not the others. However, when independent variables are correlated, it indicates that changes in one variable are associated with shifts in another variable. The stronger the correlation, the more difficult it is to change one variable without changing another. It becomes difficult for the model to estimate the relationship between each independent variable and the dependent variable independently because the independent variables tend to change in unison.
What problems do multicollinearity cause? The coefficients become very sensitive to small changes in the model. I will not be able to trust the p-values to identify independent variables that are statistically significant.
That said, I need not always fix multicollinearity. Multicollinearity affects the coefficients and p-values, but it does not influence the predictions, precision of the predictions, and the goodness-of-fit statistics. If my primary goal is to make predictions, and I don’t need to understand the role of each independent variable, I don’t need to reduce severe multicollinearity.
In-time analysis problem
Now that I have discussed this much, let me continue with the in-time analysis problem. I want to analyze my entry time at office and understand how different factors affect it.
Collinearity checks are done after Univariate and multivariate analysis in EDA. One reason for doing so is that now I know how each factor looks like, and as collinearity is only a linear association between two explanatory variables, I can convert my factors to linear after applying some functions on them(log, square root etc).
After multivariate analysis, I currently have the factors
1. date (month / weekday / season etc) 2. main_activity (means of transport) 3. hours.worked (of the previous day) 4. travelling.time (time it took to travel from house to office) 5. home.addr (the place of residence) 6. diff.out.time (previous day out of office time)
Out of these factors, only travelling.time, hours.worked and diff.out.time are continuous variables, so, for now, I will restrict my analysis to these three factors.
The dependent variable is diff.in.time(the difference between my actual in time vs policy in-time)
diff.in.time date main_activity hours.worked travelling.time home.addr diff.out.time 75 34 2018-04-06 IN_VEHICLE 9.450000 774.642 Old House -7 26 22 2018-01-10 ON_FOOT 9.500000 308.578 Old House 8 88 17 2018-04-25 ON_FOOT 10.350000 1015.017 Old House 64 179 -7 2018-10-30 ON_BICYCLE 10.566667 1668.824 New House 101 123 -12 2018-07-11 IN_VEHICLE 9.533333 866.401 Old House 44
As a recap on univariate analysis, hours.worked and diff.in.out.time were exponential distributions(while checking for collinearity I will take their log for linear relationships),
while travelling.time is close to a normal distribution(with exponential tails, but I will ignore that for now)
Testing for collinearity
One of the most popular tests to check multicollinearity is VIF factor(which will be discussed and calculated in a different blog post). What I want to instead do is visually see how the independent factors influence one another. I can do it in two main ways:
Correlation matrix
A correlation matrix is a table showing correlation coefficients(\(r_{xy}\)) between variables. Each cell in the table shows the correlation between the two variables.
First I will make a reusable correlation matrix plotting function.
corr_matrix_plotting_fxn <- function(df){ library(reshape) library(ggplot2) # Making a correlation matrix cormat <- round(cor(df), 2) # Getting the upper triangular matrix cormat[upper.tri(cormat)]<- NA # Melt the correlation data and drop the rows with NA values melted_cormat <- reshape::melt(cormat, na.rm = TRUE) colnames(melted_cormat) <- c('Var1', 'Var2', 'value') melted_cormat <- melted_cormat %>% filter(!is.na(value)) # Plot the corelation matrix ggheatmap <- ggplot(melted_cormat, aes(Var2, Var1, fill = value))+ geom_tile(color = "black")+ scale_fill_gradient2(low = "red", high = "green", mid = "white", midpoint = 0, limit = c(-1,1), space = "Lab", name="Pearson\nCorrelation") + theme_minimal()+ # minimal theme theme(axis.text.x = element_text(angle = 45, vjust = 1, size = 12, hjust = 1))+ coord_fixed() + geom_text(aes(Var2, Var1, label = value), color = "black", size = 4) + theme( axis.title.x = element_blank(), axis.title.y = element_blank(), panel.grid.major = element_blank(), panel.border = element_blank(), panel.background = element_blank(), axis.ticks = element_blank(), legend.justification = c(1, 0), legend.position = c(0.6, 0.7), legend.direction = "horizontal")+ guides(fill = guide_colorbar(barwidth = 7, barheight = 1, title.position = "top", title.hjust = 0.5)) print(ggheatmap) }
In the above function, I will just input the data frame with the necessary columns(modified dependent variables) and I will get the plot.
# Selecting necessary columns after converting to linear data setscorr.variables <- travel %>% filter(diff.out.time > -15) %>% mutate(log.hours.worked = log(hours.worked), log.diff.out.time = log(diff.out.time + 15)) %>% dplyr::select(travelling.time, log.hours.worked, log.diff.out.time)# Plotting the matrixcorr_matrix_plotting_fxn(corr.variables)
I can see that Hours worked in the previous day and the out-time of the previous day are slightly correlated(Which conceptually seems likely).
Correlation Network
Using a correlation matrix I can only check if two variables are correlated. If multiple variables are correlated, I should use the Correlation Network. Even when only two variables are correlated, these plots tell me which variable among the two to keep and which to reject as a dependent variable in our model.
library("qgraph")# LegendNames <- c('Travelling time', 'Hours worked (previous day)', 'Previous day out time (diff)')# Renaming variables (not necessary)colnames(corr.variables) <- c('tr', 'hw', 'ot')cormat <- round(cor(corr.variables), 2)qgraph(cormat, graph = "pcor", layout = "spring", nodeNames = Names, legend.cex = 0.4)
From this plot, I can see that traveling time is not correlated to any other variable, and among the other two which are correlated, I will pick out time difference as my second variable and reject hours worked.
References: Basic statistical formulas
The formulas for some basic statistical terms used in this blog are given below.
The equation for
standard deviation. \[\sigma = \sqrt{\frac{\sum\limits_{i=1}^{n} \left(x_{i} - \bar{x}\right)^{2}} {n-1}}\]
The equation for
covariance \[cov_{x,y} = \frac{\sum\limits_{i=1}^{n}{(x_i-\overline{x}) \cdot (y_i-\overline{y})} }{n-1}\]
Created using RMarkdown
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Electronic Journal of Probability Electron. J. Probab. Volume 17 (2012), paper no. 5, 27 pp. Novel characteristics of split trees by use of renewal theory Abstract
We investigate characteristics of random split trees introduced by Devroye [SIAM J Comput 28, 409-432, 1998]; split trees include e.g., binary search trees, $m$-ary search trees, quadtrees, median of $(2k+1)$-trees, simplex trees, tries and digital search trees. More precisely: We use renewal theory in the studies of split trees, and use this theory to prove several results about split trees. A split tree of cardinality n is constructed by distributing n balls (which often represent data) to a subset of nodes of an infinite tree. One of our main results is a relation between the deterministic number of balls n and the random number of nodes N. In Devroye [SIAM J Comput 28, 409-432, 1998] there is a central limit law for the depth of the last inserted ball so that most nodes are close to depth $\ln n/\mu+O(\ln n)^{1/2})$, where $\mu$ is some constant depending on the type of split tree; we sharpen this result by finding an upper bound for the expected number of nodes with depths $\geq \mu^{-1}\ln n-(\ln n)^{1/2+\epsilon}$ or depths $\leq\mu^{-1}\ln n+(\ln n)^{1/2+\epsilon}$ for any choice of $\epsilon>0$. We also find the first asymptotic of the variances of the depths of the balls in the tree.
Article information Source Electron. J. Probab., Volume 17 (2012), paper no. 5, 27 pp. Dates Accepted: 16 January 2012 First available in Project Euclid: 4 June 2016 Permanent link to this document https://projecteuclid.org/euclid.ejp/1465062327 Digital Object Identifier doi:10.1214/EJP.v17-1723 Mathematical Reviews number (MathSciNet) MR2878784 Zentralblatt MATH identifier 1244.05058 Subjects Primary: 05C05: Trees Secondary: 05C80: Random graphs [See also 60B20] 68W40: Analysis of algorithms [See also 68Q25] 68P10: Searching and sorting 68R10: Graph theory (including graph drawing) [See also 05Cxx, 90B10, 90B35, 90C35] 60C05: Combinatorial probability 68P05: Data structures Rights This work is licensed under aCreative Commons Attribution 3.0 License. Citation
Holmgren, Cecilia. Novel characteristics of split trees by use of renewal theory. Electron. J. Probab. 17 (2012), paper no. 5, 27 pp. doi:10.1214/EJP.v17-1723. https://projecteuclid.org/euclid.ejp/1465062327
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Longitudinal relaxation times of
13 Macromolecular (MM) resonances are reported for a gray matter rich voxel at
9.4 T for the first time. In addition, a
sequence was optimized based on calculated magnetizations from Bloch
simulations for combinations of inversion times using a DIR MC-semiLASER. The
results from this work highlight the importance of accounting for specific peak
relaxations due to the ranging T
1 relaxation times of the MM peaks.
Bloch simulations (Figure
1) for DIR sequence with several combinations of TI
1
and TI 2 (assuming MMs and metabolites T 1 relaxation time
from Murali-Manohar et al., 3 and Deelchand et al., 4
respectively) were performed, and those combinations (Table 1) for which the
metabolites were considerably nulled were chosen. All data were acquired on a
9.4T Magnetom, Siemens. Four healthy volunteers participated in this study after
signed consent. MC-semiLASER 5 with DIR was used with TE = 24 ms, NEX
= 32, TR = 8000 ms 3. TI 1 and TI 2 were varied according
to the table in figure 1 to acquire a set of macromolecular spectra with
different degrees of saturation. Simulated Voigt lines were created for the MM
peaks and all the data were fitted in LCModel-v6.3 6.
Care was taken to fit the
residual metabolites in the spectra by inclusion of narrow lineshapes to
represent NAA
acetyl, -CH 3 tCr, -CH 2 tCr, and GPC
in varying combinations across the inversion series. A simulated metabolite
basis set, created with VeSPA 7, was used to account for the
metabolite signals in the minimum signal MM spectra acquired with TI 1/TI 2
= 1050/238 ms.
The signal $$$S$$$ acquired were fitted to a four-parameter bi-exponential model using the DIR signal equation:
$$S = a(1-2e^{\frac{-TI_{2}}{T_{1}}}+2e^{\frac{-(TI_{1}+TI_{2})}{T_{1}}}),\\ a\equiv\frac{\rho}{4kT\cdot R\cdot BW}$$
where $$$a$$$ is a constant with $$$\rho$$$ being the effective spin density, $$$k$$$ being the Boltzmann constant, $$$T$$$ being temperature, $$$R$$$ being the effective resistance of the coil while loaded, and $$$BW$$$ being the bandwidth of the receiver.
Typically, the MM influence in
metabolite spectra is handled as a whole ‘baseline’ and assumed to decay at the
same rate with regard to T
1 and T 2 relaxation. The
results from this work highlight the importance of accounting for peak specific
T 1 relaxation times of the MM peaks ranging from approximately 150
ms to 750 ms.
For certain inversion time combinations, metabolite residuals appeared as a contributor to the spectra. Reliable HSVD was typically not achievable, especially for the ppm range of 3.6-4.1ppm, potentially because of the broad linewidths encountered and low SNR. Therefore, these residuals were handled with peak fitting in LCModel by simulating narrow lineshapes to account for metabolite contributions to the MM spectra.
Correcting for the MM baseline in
a metabolite spectrum has always posed a challenge. The T
1
relaxation times of the individual MM peaks help us predict their behavior
better and hence, aid us to adopt the model of the MM contribution to specific
flip angle and repetition time combinations when fitting metabolite spectra.
1. Xin L, Schaller B, Mlynarik V, Lu H, Gruetter R. Proton T1 relaxation times of metabolites in human occipital white and gray matter at 7 T. Magnetic resonance in medicine. 2013 Apr;69(4):931-6.
2. Behar KL, Rothman DL, Spencer DD, Petroff OA. Analysis of macromolecule resonances in 1H NMR spectra of human brain. Magnetic Resonance in Medicine. 1994 Sep;32(3):294-302.
3. Murali-Manohar S, Wright AM and Henning A (October-2018): Challenges in estimating T1 Relaxation Times of Macromolecules in the Human Brain at 9.4T, MRS Workshop 2018 Metabolic Imaging, Utrecht, The Netherlands.
4. Deelchand DK, Van de Moortele PF, Adriany G, Iltis I, Andersen P, Strupp JP, Vaughan JT, Uğurbil K, Henry PG. In vivo 1H NMR spectroscopy of the human brain at 9.4 T: initial results. Journal of Magnetic Resonance. 2010 Sep 1;206(1):74-80.
5. Giapitzakis IA, Kreis R, Henning A. Characterization of the macromolecular baseline with a metabolite-cycled double-inversion recovery sequence in the human brain at 9.4 T.
6. Provencher SW. Estimation of metabolite concentrations from localized in vivo proton NMR spectra. Magnetic resonance in medicine. 1993 Dec;30(6):672-9.
7. Soher BJ, Semanchuk P, Todd D, Steinberg J, Young K. VeSPA: integrated applications for RF pulse design, spectral simulation and MRS data analysis. InProc Int Soc Magn Reson Med 2011 (Vol. 19, p. 1410).
8. Jones E, Oliphant E, Peterson P, et al. SciPy: Open Source Scientific Tools for Python, 2001-, https://www.scipy.org.
9. G. van Rossum, Python Tutorial, Technical Report CS-R9526, Centrum voor Wiskunde en Informatixa (CWI), Amsterdam, May 1995.
10. John D. Hunter. Matplotlib: A 2D Graphics Environment, Comuting in Science & Engineering, 9, 90-95 (2007), DOI:10.1109/MCSE.2007.53
11. Pfeuffer J, Tkáč I, Provencher SW, Gruetter R. Toward an in vivo neurochemical profile: quantification of 18 metabolites in short-echo-time 1H NMR spectra of the rat brain. Journal of magnetic resonance. 1999 Nov 1;141(1):104-20.
12. Giapitzakis IA, Avdievich N, Henning A. Characterization of macromolecular baseline of human brain using metabolite cycled semi‐LASER at 9.4 T. Magnetic resonance in medicine. 2018 Aug;80(2):462-73.
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Understanding and Specifying LIDT of Laser Components
This is
Section 3.1 of the Laser Optics Resource Guide.
Laser Induced Damage Threshold (LIDT) is defined within ISO 21254 as the “highest quantity of laser radiation incident upon the optical component for which the extrapolated probability of damage is zero”.
1 The purpose of LIDT is to specify the maximum laser fluence (for pulsed lasers, typically in J/cm 2) or the maximum laser intensity (for continuous wave lasers, typically in W/cm 2) that a laser optic can withstand before damage occurs. Because of the statistical nature of laser damage testing, LIDT cannot be considered as the fluence below which damage will never occur, but rather the fluence below which the damage probability is less than the critical risk level. The level of risk depends on several factors such as the beam diameter, the number of test sites per sample, and the number of samples tested in order to determine the specification.
Laser induced damage in optical components causes degradation in system performance that can even result in catastrophic failure. An incorrect understanding of LIDT may lead to significantly higher costs or to component failures. Especially when dealing with high power lasers, LIDT is an important specification for all types of laser optics including reflective, transmissive, and absorptive components. The lack of an industry consensus on how LIDT should be tested, how damage should be detected, and how the test data should be interpreted makes LIDT a complicated specification. An LIDT value on its own does not convey the diameter of the beam used for testing, how many shots per testing site were administered, or the way the test data was analyzed.
Introduction to LIDT
In order to determine whether a laser’s fluence may cause damage to an optic, the following specifications of the laser should be reviewed: power, beam diameter, and whether the laser is continuous wave or pulsed. For pulsed lasers, the pulse duration must also be considered.
Laser Intensity: Not as Straightforward as it Seems
The intensity of a laser beam is the optical power per unit area, typically measured in W/cm
2. The distribution of the intensity of the laser across a cross-section of the beam is the intensity profile. Some of the most common intensity profiles are flat top beams and Gaussian beams. Flat top beams, or top hat beams, have an intensity profile that is constant across a cross-section of the beam. Gaussian beams have an intensity profile that decreases as the distance from the center of the beam increases following a Gaussian function. The peak fluence of a Gaussian beam is twice as large as that of a flat top beam with the same optical power ( Figure 1). Figure 1: Comparison of Gaussian and flat top beam profiles with the same optical power 2
The effective beam diameter of a Gaussian beam also scales with fluence. As fluence increases, a larger portion of the beam’s width has sufficient fluence to initiate laser induced damage (
Figure 2). This can be avoided by using a flat top beam instead of a Gaussian beam (see our Gaussian Beam Propagation application note for more information about Gaussian beams). Figure 2: The effective diameter of a Gaussian beam increases as fluence increases, leading to a higher probability of laser induced damage as indicated by more damage sites falling under the width of the curves with the highest fluence
The intensity of a laser plays an important role in determining the required LIDT for optics used with it. Some lasers also contain unintentional regions of higher intensity called hot spots, which can contribute to laser induced damage.
Continuous Wave Lasers:
Damage from continuous wave (CW) lasers is typically a result of thermal effects caused by absorption in the optic’s coating or substrate.
3 Cemented optical components, such as achromats, tend to have lower CW damage thresholds because of absorption or scattering in the cement.
To understand a CW LIDT specification, it is necessary to know the laser’s wavelength, beam diameter, power density, and intensity profile (e.g., Gaussian or flat top). LIDT for CW lasers is specified in units of power per area, typically in W/cm
2. For example, if a 5mW, 532nm Nd:YAG laser with a flat top beam is used with a beam diameter of 1mm, then the power density is: (1)$$ \text{Power Density} = \frac{\text{Power}}{\text{Area}} = \frac{5 \text{mW}}{\pi \left( \frac{\text{Beam Diameter}}{2} \right)^2} = \frac{5 \text{mW}}{\pi \left( \frac{1 \text{mm}}{2} \right)^2} = 0.6366 \tfrac{\text{W}}{\text{cm}^2} $$
Therefore, if the LIDT specified for an optic is lower than 0.64W/cm
2 then the user risks optical damage at 532nm. An extra factor of 2 would need to be added if using a Gaussian beam. Pulsed Lasers:
Pulsed lasers emit discrete pulses of laser energy at a given repetition rate or frequency (
Figure 3). The energy per pulse is directly proportional to the average power and inversely proportional to the repetition rate of the laser ( Figure 4). (2)$$ \text{Pulse Energy} = \frac{\text{Average Power}}{\text{Repetition Rate}} $$
Damage from short nanosecond laser pulses is typically due to dielectric breakdown of the material resulting from exposure to the high electric fields in the laser beam.
2 Dielectric breakdown occurs when a current flows through an electrical insulator because the applied voltage exceeds the material’s breakdown voltage. For longer pulse widths or high repetition rate laser systems, laser induced damage may result from a combination of thermally induced damage and dielectric breakdown. This occurs because the pulse duration is still on the order of the time duration of electron-lattice dynamics, which is responsible for thermally induced damage. These thermal processes are negligible for ultrashort pulses of about 10ps or less. 4 In this case, nonlinear excitation of electrons from the valence band to the conduction band, through mechanisms such as multiphoton absorption, multiphoton ionization, tunnel ionization, and avalanche ionization, leads to damage. 5 Figure 3: The pulses of a pulsed laser are temporally separated by the inverse of the repetition rate Figure 4: Depiction of the pulse energy as a function of repetition rate for a given average power of a pulsed laser
LIDT for pulsed lasers is specified as a fluence with units of J/cm
2 as opposed to power density. It is important to recognize that while J/cm 2 does not contain a unit of time, the damage threshold is dependent on pulse duration. In most cases, the LIDT fluence value will increase as the pulse duration increases. To understand a pulsed LIDT specification, it is necessary to know the laser’s wavelength, beam diameter, pulse energy, pulse duration, repetition rate, and intensity profile (e.g., Gaussian or flat top). The relationship between the fluence of a pulsed laser, the pulse energy, and the beam diameter is defined by: (3)$$ \text{Fluence} = \frac{\text{Pulse Energy}}{\text{Area}} = \frac{\text{Pulse Energy}}{\pi \left( \frac{\text{Beam Diameter}}{2} \right)^2} $$
For example, a Q-switched (pulsed) laser with a pulse energy of 10mJ, pulse duration of 10ns, and a beam diameter of 10 microns will have the following fluence:
(4)$$ \text{Fluence} = \frac{10 \text{mJ}}{\pi \left( \frac{10 \text{μm}}{2} \right)^2 } = 12.7 \tfrac{\text{kJ}}{\text{cm}^2} $$
A fluence value in kilojoules is incredibly high and will almost certainly damage an optic, making factoring the beam diameter and not only laser energy in the calculations crucial.
Damage Mechanisms:
In addition to thermal buildup and dielectric breakdown, laser induced damage can be triggered by the interaction of the laser with some type of defect. Defects include subsurface damage left behind from grinding and polishing processes, microscopic particles of polishing abrasive left on the optic, or clusters of metallic elements left behind from coating. Each of these defect sources exhibits distinct absorption characteristics, as the nature and size of any given defect determines the laser fluence the optic can withstand without causing damage.
As previously mentioned, pulse duration has a large impact on which mechanisms lead to laser induced damage (
Figure 5). Pulse durations on the order of femtoseconds to picoseconds may excite charge carriers from the valence band of a material to the conduction band, leading to nonlinear effects including multiphoton absorption, multiphoton ionization, tunnel ionization, and avalanche ionization ( Table 1). Pulse durations on the order of picoseconds to nanoseconds may lead to damage by relaxing charge carriers from the conduction band back down to the valence band through carrier-carrier scattering and carrier-phonon scattering. Figure 5: Temporal dependence of various laser induced damage mechanisms 6
Damage Mechanicsm Description
Multiphoton Absorption
Absorption process where two or more photons with energies lower than the material’s bandgap energy are absorbed simultaneously, making absorption no longer linearly proportional to intensity.
Multiphoton Ionization Absorption of two or more photons whose combined energy leads to the photoionization of atoms in the material. Tunnel Ionization The strong electric field generated by ultra-short laser pulses allows electrons to “tunnel” through the potential barrier keeping the electrons bound to atoms, allowing them to escape. Avalanche Ionization The strong electric field generated by ultra-short laser pulses causes electrons to accelerate and collide with other atoms. This ionizes them and releases more electrons that continue to ionize other atoms. Carrier-Carrier Scattering Electrons accelerated by the electric field collide with other electrons, scattering them and causing them to collide with more electrons. Carrier-Phonon Scattering Electrons accelerated by the electric field excite phonons, or vibrations in the lattice of the material. Dielectric Breakdown A current flowing through an electrical insulator due to the applied voltage exceeding the material’s breakdown voltage. Thermal Effects Heat diffusion resulting from distortions and vibrations in the material caused by the energy of the laser pulses. Table 1: Descriptions of different damage mechanisms
Varying root causes of damage create different morphologies of laser induced damage (
Figure 6). Understanding these morphologies is important for coating and process development, but for laser optics applications, the morphology is only important in determining whether the damage significantly degrades the laser system’s performance. The amount of performance degradation a system can handle is application dependent. For example, in some situations a 10% reduction in transmission may be tolerable, while another system may fail if more than 1% of the incident light is scattered. According to the ISO 21254:2011 standard, any detectable change in an optic after exposure to a laser is considered damage. Figure 6: Various morphologies of laser induced damage resulting from different root causes Scaling LIDT:
It is important to keep in mind that damage threshold is dependent on wavelength and pulse duration. If the specified LIDT of an optic is at a different wavelength or pulse duration than that of the application, the LIDT must be evaluated at the application conditions. LIDT scaling should be avoided when possible; providing firm rules on scaling that are applicable in all situations is difficult, but general rules exist for scaling a LIDT value from the initial wavelength (λ
1) and pulse duration (τ 1) to a new wavelength (λ 2) and pulse duration (τ 2). 7 (5)$$ \text{LIDT} \! \left( \lambda_2, \tau_2, ∅_2 \right) \approx \text{LIDT} \! \left( \lambda_1, \tau_1, ∅_1 \right) \times \left( \frac{\lambda_2}{\lambda_1} \right) \times \sqrt{\frac{\tau_2}{\tau_1}} \times \left( \frac{∅_1}{∅_2} \right)^2 $$
This scaling should not be applied over large wavelength or pulse duration ranges. For example, Equation 5 would be adequate for a wavelength shift from 1064nm to 1030nm, but should not be applied for scaling an LIDT value at 1064nm to a drastically different wavelength, such as 355nm.
References International Organization for Standardization. (2011). Lasers and laser-related equipment -- Test methods for laser-induced damage threshold -- Part 1: Definitions and general principles (ISO 21254-1:2011). Paschotta, Rüdiger. Encyclopedia of Laser Physics and Technology, RP Photonics, October 2017, www.rp-photonics.com/encyclopedia.html. R. M. Wood, Optics and Laser Tech. 29, 517, 1998. Jing, X. et al., “Calculation of Femtosecond Pulse Laser Induced Damage Threshold for Broadband Antireflective Microstructure Arrays.” Opt. Exp. 2009, 17, 24137. Mao, S. S. et al., “Dynamics of Femtosecond Laser Interactions with Dielectrics.” Appl. Phys. A 2004, 79, 1695. Mazur, Eric, and Rafael R Gattass. “Femtosecond Laser Micromachining in Transparent Materials.” Nature Photonics, vol. 2, 2008, pp. 219–225. Carr, C. W., et al. “Wavelength Dependence of Laser-Induced Damage: Determining the Damage Initiation Mechanisms.” Physical Review Letters, 91, 12, 2003.
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AliPhysics 6f1d526 (6f1d526)
Scripts used in the analysis Algorithms Corrections Monte-carlo code Mid-rapidity tracklet code for dN/deta Tasks Topical
file AliAODForwardMult.h Per-event \( N_{ch}\) per \((\eta,\varphi)\) bin. file AliFMDEventPlaneFinder.h file AliForwardUtil.h Various utilities used in PWGLF/FORWARD. file AliLandauGaus.h Declaration and implementation of Landau-Gauss distributions. file AliLandauGausFitter.h Declaration and implementation of fitter of Landau-Gauss distributions to energy loss spectra.
class AliForwardCreateResponseMatrices class AliForwardTriggerBiasCorrection class AliForwardUtil struct AliForwardUtil::Histos struct AliForwardUtil::RingHistos class AliLandauGaus class AliLandauGausFitter
Code to do the multiplicity analysis in the forward psuedo-rapidity regions
The code in this section defines methods to measure the charged-particle pseudorapidity density over \(|\eta|<2\) using SPD tracklets.
The code in this module constitutes tools for analysing SPD tracklet data for the charged-particle pseudorapidity density. It is based on Ruben's original code (see pwglf_fwd_spd_tracklet_1), but differs in some important aspects.
This code also requires a pass of real-data (AliTrackletAODTask) and simulated (AliTrackletAODMCTask) ESDs, but the output is not near-final histograms but an array of data structures (AliAODTracklet) stored on the output AOD. The data structure contains basic information on each tracklet
Tracklet structures from simulations contain in addition
During the AOD production, no cuts, except those defined for the re-reconstruction are imposed. In this way, the AOD contains the minimum bias information on tracklets for all events,
And for AODs corresponding to simulated data, we can also
In this way, we do a single pass of ESDs for real and simulated data, and we can then process the generated AODs with various cuts imposed. The AODs are generally small enough that they can be processed locally and quickly (for example using ProofLite). This scheme allows for fast turn-around with the largest possible flexibility.
The final charged-particle pseudorapidity density is produced by an external class (AliTrackletdNdeta2).
Other differences to Ruben's code is that the output files are far more structured, allowing for fast browsing of the data and quality assurance.
The analysis requires real data and simulated data, anchored to the real data runs being processed. For both real and simulated data, the analysis progresses through two steps:
A pass over ESD plus clusters to generate an AOD branch containing a TClonesArray of AliAODTracklet objects. In this pass, there are no selections imposed on the events. In this pass, the SPD clusters are reprocesed and the tracklets are re-reconstructed.
In this pass, we also form so-called injection events. In these events, a real cluster is removed and a new cluster put in at some other location in the detector. The tracklets of the event is then reconstructed and stored. This procedure is repeated as many times as possible. The injection events are therefore superpositions of many events - each with a real cluster removed and replaced by a fake cluster. The injection events are used later for background estimates.
When processing simulated data, the tracklets are also inspected for their origin. A tracklet can have three distinct classes of origins:
The last class is the background from combinations of clusters that does not correspond to true particles. This background must be removed from the measurements.
The second class, tracklets from secondaries, also form a background, but these tracklets are suppressed by cuts on the sum-of-square residuals
\[ \Delta = \left[\frac{\Delta_{\theta}^2\sin^2\theta}{\sigma_{\theta}^2}+ \frac{(\Delta_{\phi}-\delta_{\phi})^2}{\sigma_{\phi}^2}\right] \]
\[ \frac{d^2N_X}{d\eta d\mathrm{IP}_z}\quad, \]
where \( X\) is \( M\) or \( I\) for real data, or \( M',I',P',S',C'\) or \( G'\) for simulated data.
For each of these tracklet samples, except \( G'\), we also form the 3-dimensional differential \(\Delta\) distributions
\[ \frac{d^3N_X}{d\eta d\mathrm{IP}_z d\Delta}\quad, \]
which are later used to estimate the background due to wrong combinations of clusters into tracks.
Once both the real and simulated data has passed these two steps, we combine the to data sets into the final measurement. The final measurement is given by
\[ R = \frac{G'}{(1-\beta')M'}(1-\beta)M, \]
where
\[ \beta' = \frac{C'}{M'}\quad\mathrm{and}\quad \beta = k\beta'\quad. \]
Here, \( k\) is some scaling derived from the 3-dimensional differential \(\Delta_M\) and \(\Delta_{M'}\) distributions .
There are classes for containing data, classes that represent analysis tasks, and classes that perform calculations, as well as specialized classes for analysis of simulation (MC) output.
The classes AliAODTracklet and AliAODMCTracklet stores individual tracklet parameters. The difference between the two are that AliAODMCTracklet also stores the PDG code(s) and transverse momentum (momenta) of the mother primary particle (which may be the particle it self).
The pass over the ESD is done by the classes AliTrackletAODTask and AliTrackletAODMCTask. These tasks generated the array of tracklets in the AOD events. The difference between the two is that AliTrackletAODMCTask inspects and groups each tracklet according to it's origin, and create pseudo-tracklets corresponding to the generated primary, charged particles.
\[ \frac{d^2N_X}{d\eta d\mathrm{IP}_z}\quad, \]
and
\[ \frac{d^3N_X}{d\eta d\mathrm{IP}_z d\Delta}\quad. \]
The first task does this for \( X=M\) and \( I\), while the second and thhird tasks does this for \( X=M',I',P',S',C'\) and \( G'\). The third task reweighs all tracklets according to the particle specie(s) and transverse momentum (momenta) of the mother primary particle(s). AliTrackletWeights defines the interface used for reqeighing the data.
To produce the AODs with the tracklet information in, one needs to run a train with a task of the class AliTrackletAODTask (or AliTrackletAODMCTask for simulated data) and a task of the class AliSimpleHeaderTask in it. This is most easily done using the TrainSetup (Using the TrainSetup facility) derived class TrackletAODTrain.
For example for real data from run 245064 of LHC15o using the first physics pass
runTrain --name=LHC15o_245064_fp_AOD \ --class=TrackletAODTrain.C \ --url="alien:///alice/data/2015/LHC15o?run=245064&pattern=pass_lowint_firstphys/*/AliESDs.root&aliphysics=last,regular#esdTree"
or for simulated data from the LHC15k1a1 production anchored to run 245064
runTrain --name=LHC15k1a1_245064_fp_AOD \ --class=TrackletAODTrain.C \ --url=alien:///alice/sim/2015/LHC15k1a1?run=245064&pattern=*/AliESDs.root&aliphysics=last,regular&mc#esdTree
(note the addition of the option "&mc" to the URL argument)
In both cases a sub-directory - named of the name argument - of the current directory is created. In that sub-directory there are scripts for merging the output, downloading results, and downloading the generated AODs.
It is highly recommended to download the generated AODs to your local work station to allow fast second step analysis. To download the AODs, go to the generated sub-directory an run the
DownloadAOD.C script. For example, for the real data analysis of run 245064 of LHC15o, one would do
(cd LHC15o_245064_fp_AOD && root -l -b -q DownloadAOD.C)
and similar for the analysis of the simulated data.
To produce the histograms for the final charged-particle pseudorapidity density , one needs to run a train with a task of the class AliTrackletAODdNdeta (or AliTrackletAODMCdNdeta for simulated data) in it. This is most easily done using the TrainSetup derived class TrackletAODdNdeta.
For example for real data from run 245064 of LHC15o where we store the AODs generated above on the grid
runTrain --name=LHC15o_245064_fp_dNdeta \ --class=TrackletAODdNdeta.C \ --url="alien:///alice/cern.ch/user/a/auser/LHC15o_245064_fp_dNdeta/output?run=245064&pattern=* /AliAOD.root&aliphysics=last,regular#aodTree"
or for simulated data from the LHC15k1a1 production anchored to run 245064
runTrain --name=LHC15k1a1_245064_fp_dNdeta \ --class=TrackletAODdNdeta.C \ --url=alien:///alice/cern.ch/user/a/auser/LHC15k1a1_245064_fp_AOD/output?run=245064&pattern=* /AliAOD.root&aliphysics=last,regular&mc#esdTree
(note the addition of the option "&mc" to the URL argument)
If we had downloaded the AODs, we can use ProofLite to do this step
runTrain --name=LHC15o_245064_fp_dNdeta \ --class=TrackletAODdNdeta.C \ --url="lite:///${PWD}/LHC15o_245064_fp_dNdeta?pattern=AliAOD_*.root#aodTree"
and similar for simulated data
runTrain --name=LHC15k1a1_245064_fp_dNdeta \ --class=TrackletAODdNdeta.C \ --url="lite:///${PWD}/LHC15k1a1_245064_fp_dNdeta?pattern=AliAOD_*.root&mc#aodTree"
(note the addition of the option "&mc" to the URL argument)
The final result is obtained by runnin the class AliTrackletdNdeta2 over the histograms from both real data and simulations. As an example, suppose we ran or histogram production on the Grid and have downloaded the merged results into
LHC15o_245064_fp_dNdeta/root_archive_000245064/AnalysisResult.root (using
LHC15o_245064_fp_dNdeta/Download.C) and LHC15k1a1_245064_fp_dNdeta/root_archive_245064/AnalysisResult.root (using
LHC15k1a1_245064_fp_dNdeta/Download.C). Then we should do
(see AliTrackletdNdeta2::Run for more information on arguments)
Alternatively one can use the script Post.C to do this. The script is used like.
where
simFile is the simulation data input file (or directory)
realFile is the real data input file (or directory)
outDir is the output directory (created)
process are the processing options
visualize are the visualisation options
nCentralities is the maximum number of centrality bins
Processing options:
0x0001 Do scaling by unity
0x0002 Do scaling by full average
0x0004 Do scaling by eta differential
0x0008 Do scaling by fully differential
0x0010 Correct for decay of strange to secondary
0x1000 MC closure test
Visualization options:
0x0001 Draw general information
0x0002 Draw parameters
0x0004 Draw weights
0x0008 Draw dNch/deta
0x0010 Draw alphas
0x0020 Draw delta information
0x0040 Draw backgrounds
0x0100 Whether to make a PDF
0x0200 Whether to pause after each plot
0x0400 Draw in landscape
0x0800 Alternative markers
By default, each plot will be made and the process paused. To advance, simple press the space-bar.
If we had made the histograms using ProofLite, we should do
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Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.
Dear Uncle Colin,
Is there an easy way to write series in sum notation? I have $1 + \frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}} + ...$ but no real clue about how to put it into a compressed form.
- Summing Is Giving Me Aneurysms
Hi, SIGMA, and thanks for your message!
I like to make a little table for these, especially if it's not obvious what's going on. Then I can ask myself what's changing and what's staying the same.
Here, I'd start:
Term Value 1 $1$ 2 $\frac{1}{2\sqrt{2}}$ 3 $\frac{1}{3\sqrt{3}}$ 4 $\frac{1}{4\sqrt{4}}$ ... ...
If you write the first term as $\frac{1}{1\sqrt{1}}$, you can see exactly what changes each time: the two numbers on the bottom of the fraction are each the same as the number of the term, suggesting the $n$th term is $\frac{1}{n\sqrt{n}}$. I'd probably write that as $n^{-\frac{3}{2}}$, giving the final answer of $\Sigma_1^\infty n^{-\frac{3}{2}}$.
Hope that helps!
-- Uncle Colin
* Updated 2017-01-11 to make a $\sigma$ a $\Sigma$. Thanks, @robjlow!
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Can I be a pedant and say that if the question states that $\langle \alpha \vert A \vert \alpha \rangle = 0$ for every vector $\lvert \alpha \rangle$, that means that $A$ is everywhere defined, so there are no domain issues?
Gravitational optics is very different from quantum optics, if by the latter you mean the quantum effects of interaction between light and matter. There are three crucial differences I can think of:We can always detect uniform motion with respect to a medium by a positive result to a Michelson...
Hmm, it seems we cannot just superimpose gravitational waves to create standing waves
The above search is inspired by last night dream, which took place in an alternate version of my 3rd year undergrad GR course. The lecturer talks about a weird equation in general relativity that has a huge summation symbol, and then talked about gravitational waves emitting from a body. After that lecture, I then asked the lecturer whether gravitational standing waves are possible, as a imagine the hypothetical scenario of placing a node at the end of the vertical white line
[The Cube] Regarding The Cube, I am thinking about an energy level diagram like this
where the infinitely degenerate level is the lowest energy level when the environment is also taken account of
The idea is that if the possible relaxations between energy levels is restricted so that to relax from an excited state, the bottleneck must be passed, then we have a very high entropy high energy system confined in a compact volume
Therefore, as energy is pumped into the system, the lack of direct relaxation pathways to the ground state plus the huge degeneracy at higher energy levels should result in a lot of possible configurations to give the same high energy, thus effectively create an entropy trap to minimise heat loss to surroundings
@Kaumudi.H there is also an addon that allows Office 2003 to read (but not save) files from later versions of Office, and you probably want this too. The installer for this should also be in \Stuff (but probably isn't if I forgot to include the SP3 installer).
Hi @EmilioPisanty, it's great that you want to help me clear out confusions. I think we have a misunderstanding here. When you say "if you really want to "understand"", I've thought you were mentioning at my questions directly to the close voter, not the question in meta. When you mention about my original post, you think that it's a hopeless mess of confusion? Why? Except being off-topic, it seems clear to understand, doesn't it?
Physics.stackexchange currently uses 2.7.1 with the config TeX-AMS_HTML-full which is affected by a visual glitch on both desktop and mobile version of Safari under latest OS, \vec{x} results in the arrow displayed too far to the right (issue #1737). This has been fixed in 2.7.2. Thanks.
I have never used the app for this site, but if you ask a question on a mobile phone, there is no homework guidance box, as there is on the full site, due to screen size limitations.I think it's a safe asssumption that many students are using their phone to place their homework questions, in wh...
@0ßelö7 I don't really care for the functional analytic technicalities in this case - of course this statement needs some additional assumption to hold rigorously in the infinite-dimensional case, but I'm 99% that that's not what the OP wants to know (and, judging from the comments and other failed attempts, the "simple" version of the statement seems to confuse enough people already :P)
Why were the SI unit prefixes, i.e.\begin{align}\mathrm{giga} && 10^9 \\\mathrm{mega} && 10^6 \\\mathrm{kilo} && 10^3 \\\mathrm{milli} && 10^{-3} \\\mathrm{micro} && 10^{-6} \\\mathrm{nano} && 10^{-9}\end{align}chosen to be a multiple power of 3?Edit: Although this questio...
the major challenge is how to restrict the possible relaxation pathways so that in order to relax back to the ground state, at least one lower rotational level has to be passed, thus creating the bottleneck shown above
If two vectors $\vec{A} =A_x\hat{i} + A_y \hat{j} + A_z \hat{k}$ and$\vec{B} =B_x\hat{i} + B_y \hat{j} + B_z \hat{k}$, have angle $\theta$ between them then the dot product (scalar product) of $\vec{A}$ and $\vec{B}$ is$$\vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos \theta$$$$\vec{A}\cdot\...
@ACuriousMind I want to give a talk on my GR work first. That can be hand-wavey. But I also want to present my program for Sobolev spaces and elliptic regularity, which is reasonably original. But the devil is in the details there.
@CooperCape I'm afraid not, you're still just asking us to check whether or not what you wrote there is correct - such questions are not a good fit for the site, since the potentially correct answer "Yes, that's right" is too short to even submit as an answer
|
Can I be a pedant and say that if the question states that $\langle \alpha \vert A \vert \alpha \rangle = 0$ for every vector $\lvert \alpha \rangle$, that means that $A$ is everywhere defined, so there are no domain issues?
Gravitational optics is very different from quantum optics, if by the latter you mean the quantum effects of interaction between light and matter. There are three crucial differences I can think of:We can always detect uniform motion with respect to a medium by a positive result to a Michelson...
Hmm, it seems we cannot just superimpose gravitational waves to create standing waves
The above search is inspired by last night dream, which took place in an alternate version of my 3rd year undergrad GR course. The lecturer talks about a weird equation in general relativity that has a huge summation symbol, and then talked about gravitational waves emitting from a body. After that lecture, I then asked the lecturer whether gravitational standing waves are possible, as a imagine the hypothetical scenario of placing a node at the end of the vertical white line
[The Cube] Regarding The Cube, I am thinking about an energy level diagram like this
where the infinitely degenerate level is the lowest energy level when the environment is also taken account of
The idea is that if the possible relaxations between energy levels is restricted so that to relax from an excited state, the bottleneck must be passed, then we have a very high entropy high energy system confined in a compact volume
Therefore, as energy is pumped into the system, the lack of direct relaxation pathways to the ground state plus the huge degeneracy at higher energy levels should result in a lot of possible configurations to give the same high energy, thus effectively create an entropy trap to minimise heat loss to surroundings
@Kaumudi.H there is also an addon that allows Office 2003 to read (but not save) files from later versions of Office, and you probably want this too. The installer for this should also be in \Stuff (but probably isn't if I forgot to include the SP3 installer).
Hi @EmilioPisanty, it's great that you want to help me clear out confusions. I think we have a misunderstanding here. When you say "if you really want to "understand"", I've thought you were mentioning at my questions directly to the close voter, not the question in meta. When you mention about my original post, you think that it's a hopeless mess of confusion? Why? Except being off-topic, it seems clear to understand, doesn't it?
Physics.stackexchange currently uses 2.7.1 with the config TeX-AMS_HTML-full which is affected by a visual glitch on both desktop and mobile version of Safari under latest OS, \vec{x} results in the arrow displayed too far to the right (issue #1737). This has been fixed in 2.7.2. Thanks.
I have never used the app for this site, but if you ask a question on a mobile phone, there is no homework guidance box, as there is on the full site, due to screen size limitations.I think it's a safe asssumption that many students are using their phone to place their homework questions, in wh...
@0ßelö7 I don't really care for the functional analytic technicalities in this case - of course this statement needs some additional assumption to hold rigorously in the infinite-dimensional case, but I'm 99% that that's not what the OP wants to know (and, judging from the comments and other failed attempts, the "simple" version of the statement seems to confuse enough people already :P)
Why were the SI unit prefixes, i.e.\begin{align}\mathrm{giga} && 10^9 \\\mathrm{mega} && 10^6 \\\mathrm{kilo} && 10^3 \\\mathrm{milli} && 10^{-3} \\\mathrm{micro} && 10^{-6} \\\mathrm{nano} && 10^{-9}\end{align}chosen to be a multiple power of 3?Edit: Although this questio...
the major challenge is how to restrict the possible relaxation pathways so that in order to relax back to the ground state, at least one lower rotational level has to be passed, thus creating the bottleneck shown above
If two vectors $\vec{A} =A_x\hat{i} + A_y \hat{j} + A_z \hat{k}$ and$\vec{B} =B_x\hat{i} + B_y \hat{j} + B_z \hat{k}$, have angle $\theta$ between them then the dot product (scalar product) of $\vec{A}$ and $\vec{B}$ is$$\vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos \theta$$$$\vec{A}\cdot\...
@ACuriousMind I want to give a talk on my GR work first. That can be hand-wavey. But I also want to present my program for Sobolev spaces and elliptic regularity, which is reasonably original. But the devil is in the details there.
@CooperCape I'm afraid not, you're still just asking us to check whether or not what you wrote there is correct - such questions are not a good fit for the site, since the potentially correct answer "Yes, that's right" is too short to even submit as an answer
|
For multi-rail shots, just expand the solution of Alex, above. Reflect the pool table left and right, above and below the original pool table. Include reflections of reflections. It might help to color the original four sides in four different colors, to keep track of which side is which in the reflected pool tables. Be sure to also reflect the target ball. Now draw a straight line from the cue ball to any
reflected target ball. That line will represent a possible direction for the cue ball to hit the target; the real and reflected sides that get crossed are the ones the real ball bounces off in its path to the target.
EDIT: Using
mathematics to analyze the theoretical solution:
Take the "billiard table" as being the region one ball radius in from the actual bumpers on the real table (Thank you, Johannes). Let the origin be at the lower left corner of this table, with width $W$ in the $x$ direction and height $H$ in the $y$ direction. Place the cue ball at $X_c,\,Y_c$, and the target at $X_t,\,Y_t$.
Assume you want to do a bounce off the left cushion and then the top cushion, before hitting the target. The reflection in the left cushion places the reflected target at $-X_t,\,Y_t$, while the reflection of this reflected target in the top cushion places the reflected reflected target at $-X_t,\,(2H-Y_t)$
The line from the cue ball to the doubly reflected target is at an angle $\theta$ where:$$\theta = \tan ^{-1} \left( \frac{(2H-Y_t)-Y_c}{-X_t-X_c} \right)$$
Note that the angle may need to have $\pi$ radians added, depending on the actual quadrant.
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I have some basic questions around the theorem giving quantum error correction conditions that give necessary & sufficient conditions to have an error correcting operation.
The theorem is stated this way (page 436 of Nielsen & Chuang)
Let C be a quantum code, and let P be the projector onto C.
Suppose $E$ is a quantum operation with operation elements ${E_i}$.
A necessary and sufficient condition for the existence of an error-correction operation $R$ correcting $E$ on C is that $PE_i^{\dagger}E_jP = \alpha_{ij} P$, for some Hermitian matrix $\alpha$ of complex numbers.
We call the operation elements ${E_i}$ for the noise $E$ errors,and if such an $R$ exists we say that ${E_i}$ constitutes a correctable set of errors.
So said differently, we have a noise map that act as:
$$E(\rho) = \sum_i E_i(\rho) E_i^{\dagger}$$
And the theorem given the projector $P$ on the code space gives us relations that the noise Kraus operator must follow to ensure us we are able to correct the error.
My questions
Question about the theorem itself:
If I take the example of a 3 dimensional code space (the one for 3qb code). If I take the noise map allowing only single bit-flips, then the theorem will tell me that I can find a recovery operation.
But if I considered double or 3 bits-flip then the theorem would tell me it is not possible ? Because 3qb code can only correct single errors. Am I correct with this statement (I would like to avoid to do big calculations I want to see if I understand well the things).
Question about the proof
He does it in two steps : sufficient condition check and then necessary check. My question is for the sufficient condition.
Results he uses:He started to show that actually we can rewrite:
$$N(\rho)=\sum_i F_i \rho F_i^{\dagger} $$
With: $F_i=\sum_k u_{i,k} F_k$ where $u$ is the unitary matrix that diagonalise the matrix $\alpha$ in the statement of the theorem: $u^{\dagger}.\alpha.u=d$
He also says that there exist a unitary $U_k$ such that $F_k P= \sqrt{d_kk} U_k P$
My question
And then he says that $F_k$ acting on the code space changes the code space. The resulting code space has a projector:
$$P_k = U_k P U_k^{\dagger}$$
How was he able to find this ? I agree with the result. We indeed have:
$$\forall |\psi\rangle \in C, ~ P_k \left( F_k |\psi\rangle \right) = F_k |\psi\rangle$$
But how was he able to find that the projector was this one ??
Other question
He assumes in the proof that $E$ is not necesseraly trace-preserving: why not. But that $R$ is trace preserving. Why would we have $R$ trace preserving but not $E$ ?
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Suppose I have a number $x$ represented in a residue number system, so $x = (x_1, \ldots, x_m)$, where $x_i \equiv x \pmod{p_i}$, and the $p_i$'s are all relatively prime (they can be distinct primes if it helps).
I would like to compute $x \bmod{p^*}$ for some $p^* \not\in \{p_1, \ldots, p_m\}$, but with low space.
Of course I could reconstruct $x$ itself, but this would require $\log (\prod_i p_i) = \sum_i \log p_i$ bits of working space. Rather, I would like the computation to use something more like $\max_i \log p_i$ bits of space. Is this possible?
As a concrete example, suppose $\{ p_1, \ldots, p_m\}$ are the first $m$ primes and I would like to compute $x \bmod 4$. Or $\{ p_1, \ldots, p_m\}$ are the first $m$
odd primes, and I would like to compute $x \bmod 2$.
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The domain of a function $f:X\to Y$ is normally defined as $\operatorname{dom}f\equiv X$, but I would like the
domain-function $\operatorname{dom}$ to be a funtion itself, i.e. I would like to define the function $\operatorname{Dom}$ defined by:
$$\forall X\,\forall Y\,\forall f\,(f:X\to Y\Rightarrow\operatorname{Dom} f\equiv X)$$
Now $\operatorname{Dom}(\operatorname{Dom}(\cdot))$ would be the set of all functions, and $\operatorname{Ran}(\operatorname{Dom}(\cdot))$ would be the set of all sets.
It is however impossible to construct the set of all sets, by Russel's paradox. My question is what I have done wrong in the above definition, and how to define $\operatorname{Dom}$ as a function.
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Well $e^i = \cos 1 + i \sin 1$ soo $e^{e^i} = e^{\cos 1 + i \sin 1} = e^{\cos 1}[\cos (\sin 1) + i \sin(\sin 1)] \approx e^{0.54030230586813971740093660744298}(\cos 0.8414709848078965066525023216303 + i \sin 0.8414709848078965066525023216303) = 1.7165256995489035179801866917897(0.66636674539288052633780454726237 +i0.55595536063449993982642571881368)$.
There's nothing particularly significant to this number.
I suppose one way thinking about it is $1 \approx \frac {\pi}3$ so $e^i \approx \frac 12 + i\frac {\sqrt{3}}2$ (give or take a rather
large degree of error).
So $e^{e^i} \approx e^{\frac 12}(\cos \frac {\sqrt {3}}2 + i\sin \frac {\sqrt{3}}2)$. $\frac{\sqrt {3}}2 \approx \frac 78 \approx \frac {\pi} 4$ (give or take a margin large enough to get you fired from any government job).
So $e^{e^i} \approx \sqrt{e}(\frac {\sqrt{2}}2+ i \frac {\sqrt{2}}2)$ (with a margin of error that if you were aiming for downtown tokyo you would still remain in the solar system).
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Iosif Pinelis has given an answer to question 1 and partial answers to 2 and 3. Since he advised me to turn my comments into an answer, here it is.
I will deal with the case where the axiom of choice holds, and mention the case where it does not only at the end.
I will start with some definitions. A measure $\mu$ on $(X,\mathcal{P}(X))$ is called $\kappa$-additive if for all pairwise disjoint families of sets $(S_i)_{i \in I}$ such that $|I| < \kappa$ (or, speaking loosely, such that the cardinality of the family is $< \kappa$), we have $\mu\left(\bigcup_{i \in I} S_i\right) = \sum_{i \in I}\mu(S_i)$. So "countably additive" is equivalent to $\aleph_1$-additive (Not $\aleph_0$-additive, notice the
strict inequality. Finite additivity is $\aleph_0$-additivity.).
A
real-valued measurable cardinal is a cardinal $\kappa$ such that there exists a $\kappa$-additive probability measure $\mu$ on $(\kappa, \mathcal{P}(\kappa))$ that vanishes on singletons, i.e. $\mu(\{x\}) = 0$ for all $x \in \kappa$. There are two kinds of real-valued measurable cardinals. Measurable cardinals are real-valued measurable cardinals such that there exists a probability measure taking only the values $0$ and $1$ (this can be defined equivalently in terms of the existence of an ultrafilter closed under intersections of cardinality $< \kappa$). Atomlessly measurable cardinals are those such that the measure $\mu$ is atomless, i.e. if $\mu(S) > 0$, there exists a $T \subseteq S$ such that $0 < \mu(T) < \mu(S)$.
The following facts were essentially proved by Ulam in his paper
Zur Masstheorie in der allgemeinen Mengenlehre, although with different terminology. They are proved in a modern way in Jech's Set Theory: Third Millennium Edition, Chapter 10. A real-valued measurable cardinal is never a successor cardinal, and never the union of a strictly smaller family of strictly smaller sets, i.e. a real-valued measurable cardinal is weakly inaccessible. A measurable cardinal is not embeddable in the powerset of a strictly smaller cardinal, i.e. measurable cardinals are strongly inaccessible. Atomlessly measurable cardinals embed in the continuum, so have cardinality less than or equal to it.
A key fact is that the smallest real-valued measurable cardinal is the smallest cardinal having a
countably-additive measure $\mu$ vanishing on singletons. This is how Ulam originally phrased things, so if $\kappa$ is the smallest real-valued measurable, he called cardinals $< \kappa$ unmeasurable and cardinals $\geq \kappa$ measurable. However, in the modern terminology, $\kappa^+$ is not real-valued measurable. Because the notion of "cardinals not admitting a countably-additive measure vanishing on singletons" is such a useful notion, and we can't use the term "unmeasurable" for it, Fremlin introduced the name measure-free for cardinals smaller than the first real-valued measurable cardinal.
Now we can approach the problem of whether non-separable measures exist. Any set $X$ can be given the discrete metric, and the Borel sets of the topology determined by this metric are exactly $\mathcal{P}(X)$. If $X$ has a countably additive measure $\mu$ vanishing on singletons, then $\mu$ does not have separable support, because separable subsets of a discrete space are countable sets, and $\mu$ vanishes on every countable set by countable additivity. So if there is a real-valued measurable cardinal, then there is a metric space without the property (*) in the question. As pointed out in Iosif's answer, Billingsley proves the converse as Theorem 2 of Appendix 3 in the first edition of
Convergence of Probability Measures and its Russian translation, making nontrivial use of the paracompactness of metric spaces. All together, a metric space has no inseparable measures iff all its discrete subspaces have measure-free cardinality.
The continuum hypothesis enters as follows. If there exists an atomlessly measurable cardinal $\kappa$, we have $\kappa \leq 2^{\aleph_0}$, and as real-valued measurable cardinals are weakly inaccessible, we not only have cardinals in between $\aleph_1$ and $2^{\aleph_0}$, but a weakly-inaccessible-cardinalful of them (in fact later results showed that there are even more than this). So if the continuum hypothesis holds, or even if the continuum is any smallish sort of cardinal, like $\aleph_3$ or $\aleph_{\epsilon_0 + 1}$ or $\aleph_{\omega_9}$, there are no atomlessly measurable cardinals. But there may still be measurable cardinals, which will necessarily be strongly inaccessible. So the continuum hypothesis implies that metric spaces that have cardinals smaller than the first inaccessible (in fact, again, we can actually go larger than the first inaccessible too, but it is hard to describe exactly how high we can go before reaching the first measurable cardinal), such as $2^{\aleph_0}$, $2^{2^{2^{\aleph_5}}}$ and even $\beth_{\omega_1}$, do not have inseparable measures.
However, Silver, Solovay and Kunen showed that if the existence of measurable cardinals is consistent with ZFC, then they are compatible with the generalized continuum hypothesis. Therefore the answer to question 2 is no. And for 3, of course, in the absence of the continuum hypothesis, the reals, equipped with the discrete metric, may be a counterexample (if atomlessly measurable cardinals exist).
Billingsley, like many other authors of his era (such as Schaefer in
Topological Vector Spaces and Gilman and Jerison in Rings of Continuous Functions), describes the question of whether real-valued measurable cardinals exist as an "unsolved problem". However, the existence of real-valued measurable cardinals is better understood as an axiom that exceeds the consistency strength of ZF(C), thanks to set-theoretic results that have been proven since then. That is to say, one can show that the nonexistence of real-valued measurable cardinals is consistent with ZFC (by taking V=L), but the consistency of real-valued measurable cardinals cannot be proven, assuming only the consistency of ZFC.
The reason for this is that if $\kappa$ is measurable, $V_\kappa$ is set-sized model of ZFC, so measurable cardinals prove the consistency of ZFC. Solovay showed in his article
Real-Valued Measurable Cardinals that the existence of atomlessly measurable cardinals and the existence of measurable cardinals are equiconsistent, so the existence of real-valued measurable cardinals also implies the consistency of ZFC.
I will now briefly discuss what can happen if choice does not hold. In Solovay's model (defined in an much earlier article) where the axiom of choice does not hold, but dependent choice does, and all subsets of $\mathbb{R}$ are Lebesgue-measurable, Lebesgue measure itself defines a probability measure on $([0,1], \mathcal{P}([0,1]))$ vanishing on singletons, which is therefore a non-separable measure on $[0,1]$, equipped with the discrete metric. So in some sense, the absence of choice makes things worse (we only need an inaccessible cardinal, not a measurable one to build this model).
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Current Transformer Current Transformers or CT's are Instrument Transformers that convert a generally high primary current I p to a k-times lower secondary current Ithat can be connected to standard measuring or protection devices. The primary and secondary windings are galvanically separated and can be on a different potential level. The transformation ratio k of a current transformer is the number of secondary turns s Nto the number of primary turns s Nand is equal to the primary current p Iover the secondary current p I. s
[math] k = \frac{N_s} {N_p} = \frac{I_p} {I_s} [/math]
[math] I_s = I_p . \frac {N_p} {N_s} [/math]
[math] I_s = \frac {I_p} {k} [/math]
Contents 1 Standards for current transformers 2 Functioning of a Current Transformer 3 General properties of a Current Transformer 4 The use and specification of Current Transformers Standards for current transformers According IEC
At the moment TC38 of the IEC is busy converting all the instrument transformers from the 60044-family to the new 61869 family with a general part and specific parts.
IEC 60044-1 Consolidated Edition 1.2 (incl. am1+am2) (2003-02) TC/SC 38 Instrument transformers - Part 1: Current
transformers
IEC 60044-3 Edition 2.0 (2002-12) TC/SC 38 Instrument transformers - Part 3: Combined transformers IEC 60044-6 Edition 1.0 (1992-03) TC/SC 38 Instrument transformers - Part 6: Requirements for protective current
transformers for transient performance
IEC 60044-8 Edition 1.0 (2002-07) TC/SC 38 Instrument transformers - Part 8: Electronic current transformers IEC 61869-1 Edition 1.0 (2007-10) TC/SC 38 Instrument transformers - Part 1: General requirements Other standard organisations IEEE Std C57.13-1993: IEEE Standard requirements for Instrument transformers Canada CAN3-C13-M83: Instrument transformers Australia AS 1675 Current transformers - Measurement and protection British Standard BS3938 Specifications for Current Transformers (Withdrawn and replaced by IEC 60044-1) Functioning of a Current Transformer
Just like a normal voltage transformer, a CT has a primary winding, a secondary winding and a magnetic core. In the window-type and bushing-type CT's, the primary winding is reduced to one wire passing trough the round or square shaped core, accounting for 1 turn. The primary current I
p will produce a magnetic field with induction B round the conductor. The magnetic induction B is amplified by the core material with very high magnetic permeability µ and will produce a primary flux that will magnetise the core with cross section A and induces a secondary voltage V s in the secondary winding with N turns.
[math]V_s = 4.44 \times 10^-8 f N A B [/math]
At the same time, a N times smaller voltage, opposed to the primary current will be induced in the primary wire creating a small extra resistance in the primary circuit. The induced secondary voltage will drive the secondary current I
s that will flow for the mayor part trough the connected load R b and for a small part (the error current) I e trough the internal resistance and induction. The internal resistance and induction represent the part of the current that is used to magnetise the core (Inductive part) and to heat-up the core material as iron-losses.Actually the magnetising current is taken from the primary side but that will only make the calculation model more difficult and does not form any additional value. The secondary current I s will also produce a secondary flux, opposite to the primary flux. The resulting flux in the CT core is a very small magnetising flux so that the core does not saturate at normal operation currents. The secondary current I s will be N times smaller than the primary current I p.
[math] I_s = \frac{I_p} {N} [/math]
The error current I
e exists for the major part of a purely inductive part; the magnetising current
I
m that can be seen on the magnetising curve of the CT and a small resistive part I g that represent the iron losses. The magnetising current I m is proportional to the field strength H
[math] I = \frac{l} {N} \times H[/math]
The copper losses are represented in the series resistor R
CT General properties of a Current Transformer The Primary current Ip
According to IEC 60044-1, the primary current I_p is standadise of the decadic series 1 - 1,25 - 1,5 - 2 - 2,5 - 3 - 4 - 5 - 6 - 7,5. When selecting a CT; the primary current of the CT must be at least the maximum current of the line in which the CT will operate. When the current is bigger than the rated primary current of the CT, the windings will overheat, age faster and finally the insulation will fail. According ANSI, the primary currents are fixed values; for single Ratio CT's Ip = 10; 15; 25; 40; 50; 75; 100; 200; 300; 400; 600; 800; 1200; 1500; 2000; 3000; 4000; 5000; 6000; 8000; 12000A.
The Secondary current Is
According IEC the secondary current can be 0.5, 1 , 2 or 5A. According ANSI the secondary current is allways 5A.
Dual or Multi-Ratio CT's
Dual ratio CT's exist in all standards but only according the ANSI standard, the ratio's are standardised. Note the for multi ratio CT's, many primary currents are mentionned and only one secondary current but in reality there is only one primary connection and 5 secondary terminals that allow 10 different ratings.
Ratio k
As already mentionned, the most important property of the current transformer is the
ratio k that is both the ratio of secundary turns to primary turns and the ratio of primary current to secondary current. Note the often the primary is only one turn and practically it's just the conductor passing trough the core. Since a small amount of energy is necessary to magnetising the core and to produce heat as iron loss in the core, the secondary output Ampere-turns is a bit less than the primary Ampere-turns. The difference in current is the error current or magnetising current. In case of very critical CT's, ratio-turn-correction is applied; remove some secondary turns so that the ratio is a bit higher and the output is thus a bit higher at rated current. Of course this can only be applied when the CT meets all accuracy requirements after ratio-turn correction. R CT The internal copper resistance
R
CT is often called the secondary DC resistance at 75°C. It's value depends on the length en cross section of the secondary winding wire Pouillet's law. So R CT also depends on the core dimensions; bigger core cross section implies a longer wire length per turn. The smaller R CT; the more the current transformer approaches the ideal current source. Accuracy
The accuracy of a CT is given by it's "class". The division into accuracy classes depends on the type of CT; we mainly distinguish measuring class CT's and Protection class CT's who are defined quite differently. We will discuss accuracy for both types further. Of course they both have a primary current I_p, a secondary current I_s and a ratio k. From these 3 parameters we can define some important property's related to accuracy.
The primary current vector Ip. The secondary current vector Is that is here represented k times larger to be able to compare them and to have an idea of the error current. In case the error would be 0, both vectors I_p and k.I_s would be identical. The total error vector (composite error) can be seen as the composition of: an amplitude error (ratio error), expressed in % and an angle error, expressed in radians or seconds.
Note that for protection CT's, the angle error is disregarded and only the total composite error is given in %. When examining the equivalent diagram, one would easily conclude that the error current can only be the magnetising current of the CT. Indeed, normally the magnetising current is very low but at the saturation point of the core, 50% increase in magnetising current produces only 10% extra secondary voltage so at saturation the error current rises quickly. Therefore, the property's accuracy and saturation of the core are closely linked. Hense the error vector is allways a reducion in secondary output current; negative error. Positive error is only possible by ratio-turn correction.
Load, Rated load and Burden
A Voltage transformer is unloaded when the secondary terminals are open; it behaves like a normal voltage source. A current transformer is just the opposite and is unloaded with the secondary terminals short-circuited. Stonger even, when the secondary terminals of a CT are open, there is no secondary flux to oppose the primary flux and the core goes to positive saturation on the positive current-sine and to negative saturation on the negative current sine. The induced seconday voltage is proportional to [math] -N.d\phi/dt[/math] and from -Vsat to +Vsat is a huge voltage. One might also conclude that the current transformen is raising the voltage in trying to drive the secondary current trough the open terminals. The insulation of the CT is not calculated for this situation and it will distroy the CT secondary winding and may cause fire at the terminals & high voltage injury.The nominal load of a CT is the rated resistive burden R
B; expressed in VA. The correct resistance can be calculated with below formula
[math] P = R_B.I_s^2 [/math]
[math] R_B = \frac{P} {I_s^2} [/math]
Example: A 50VA CT with rated secondary current of 5A is designed for a connected load of 50VA/5² = 2 Ohm. Measuring transformers are tested at rated load and at 1/4 of the rated load so this CT should be loaded within these limits to be sure the accuracy is within specification.
The use and specification of Current Transformers
Current transformers are used to measure high currents; higher than 5A. So the most important parameter in defining a CT is indeed the Ratio that gives us the Magnetude of primary current and the secondary current. But for the following specifications of the current transformer, the purpose of the CT is needed since measuring CT's and Protection CT's require different specifications. Indeed, there will be two mayor groups of Current Transformers:
Protection current transformers Measurement current transformers
Regarding specification, different standards have different ways in specifying CT's but it all comes down to specifying core property's (saturation point or knee-point) and secondary wire property's (R
CT) although it may look a totally different. Protection CT's
Protection CT's:
are meant to protect an elektrical installation in case of overcurrent or short circuit and their operating current range is above nominal current I nor more specific from I nto ALF times I n. It is important for the good functionning of the protection relays that the CT's are NOT saturated at ALF times rated current. Where ALF is the ratio of the expected maximum fault current over the rated current. It is thus important that the core material has a high saturation induction. their accuracy is not very high but most important is that the accuracy in fault conditions is high enough. This can only be the case when the core is not saturated in case of a fault current. Therefore their accuracy is best described with an Accuracy Limit and an Accuracy Limit Factor (ALF).E.g. a 5P20 CT has an Accuracy limit of 5% at 20 times rated current (Accuracy Limit Factor). The accuracy of this CT at rated current is 1%. They will be connected to one or more protection relays according the application, they can be defined in a few ways: The standard IEC protection class CT's are of class "P" that only takes the AC behaviour into account in IEC 60044-1 Class PX CT's are defined by the position of the knee-point (saturation point or knee-point voltage and magnetising current) and the secondary wire resistance R CT. Class PR CT's are defined like the PX CT's but they have a low remanence; less than 10%. Note that remanence in CT's can be 60-80% that may cause quick saturation in case of a fault-current DC offset in the remanent direction. A class PX CT can't have that problem. CT's for transient response class "TP" are defined by their connected load R B, time constant T Sand their overcurrent figure K SSC. These linearised CT's have air-gaps in the core to obtain extreme high saturation voltage and current.
Ex. A 5P10 CT at 10 times rated current has a maximum error of 5% and only 1% at nominal current. A 10P15 CT at 15 times rated current has a maximum error of 10% and 3% at nominal current.
Measurement CT's Are aimed to measure accurately within their normal operating range of 0 to I n. Therefore, the core material must have a high permeability (µ-metal) so that the magnetising current is low. Measurement CT's are often being used for billing of electrical power consumption and their accuracy is determinent for a lot of money. For the protection of the measuring instruments in case of a fault current, it is favorable that for currents far above rated current I n, the core is saturated and the output lowers so that the fault-current trough the meter is only a part of the expected current trough the meter. This is expressed by the Instrument Security Factor SF. Of course, the dilemma is that the CT must be accurate at I n(and 1,2 x I n) but at f.i. 5 times rated current ( FS 5) the CT may be saturated for at least 10%. The accuracy of a measurement CT is given by it's accuracy class that corresponds to the error% at rated current and at 1.2 times rated current I n. The standard accuracy classes according IEC are class 0.2, 0.5, 1, 3 en 5. For classes 3 and 5, no angle error is specified. The classes 0.2S and 0.5S have their accuracy shifted toward the lower currents. This means that they have 5 measuring points instead of 4 (or 2 for class 3 & 5). The accuracy of the CT must be within these limits at the given currents and with rated load and at 1/4 of the rated load. A measurement CT that is not loaded is therefore not necessary accurate! Ratio turn correction may have been applied to get the CT ratings witthin spec and then not loading gives a higher error.
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№ 9
All Issues $G$-convergence of periodic parabolic operators with a small parameter by the time derivative Abstract
In this paper, we consider a sequence $\mathcal{P}^k$ of divergent parabolic operators of the second order, which are periodic in time with period $T = \text{const}$, and a sequence $\mathcal{P}^k_{\psi}$ of shifts of these operators by an arbitrary periodic vector function $ \psi \in X = \{L^2((0, T) \times \Omega)\}^n$ where $\Omega$ is a bounded Lipschitz domain in the space $\mathbb{R}^n$. The compactness of the family $\{P_{Ψ^k} ¦ Ψ \in X, k \in ℕ\}$ in $k$ with respect to strong $G$-convergence, the convergence of arbitrary solutions of the equations with the operator $\mathcal{P}^k_{\psi}$, and the local character of the strong $G$-convergence in $Ω$ are proved under the assumptions that the matrix of coefficients of $L^2$ is uniformly elliptic and bounded and that their time derivatives are uniformly bounded in the space $L^2(Ω; L^2(0,T))$.
English version (Springer): Ukrainian Mathematical Journal 45 (1993), no. 4, pp 564-580. Citation Example: Sidenko N. R. $G$-convergence of periodic parabolic operators with a small parameter by the time derivative // Ukr. Mat. Zh. - 1993. - 45, № 4. - pp. 525–538. Full text
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Now showing items 1-10 of 15
A free-floating planet candidate from the OGLE and KMTNet surveys
(2017)
Current microlensing surveys are sensitive to free-floating planets down to Earth-mass objects. All published microlensing events attributed to unbound planets were identified based on their short timescale (below 2 d), ...
OGLE-2016-BLG-1190Lb: First Spitzer Bulge Planet Lies Near the Planet/Brown-Dwarf Boundary
(2017)
We report the discovery of OGLE-2016-BLG-1190Lb, which is likely to be the first Spitzer microlensing planet in the Galactic bulge/bar, an assignation that can be confirmed by two epochs of high-resolution imaging of the ...
OGLE-2015-BLG-1459L: The Challenges of Exo-Moon Microlensing
(2017)
We show that dense OGLE and KMTNet $I$-band survey data require four bodies (sources plus lenses) to explain the microlensing light curve of OGLE-2015-BLG-1459. However, these can equally well consist of three lenses ...
OGLE-2017-BLG-1130: The First Binary Gravitational Microlens Detected From Spitzer Only
(2018)
We analyze the binary gravitational microlensing event OGLE-2017-BLG-1130 (mass ratio q~0.45), the first published case in which the binary anomaly was only detected by the Spitzer Space Telescope. This event provides ...
OGLE-2017-BLG-1434Lb: Eighth q < 1 * 10^-4 Mass-Ratio Microlens Planet Confirms Turnover in Planet Mass-Ratio Function
(2018)
We report the discovery of a cold Super-Earth planet (m_p=4.4 +/- 0.5 M_Earth) orbiting a low-mass (M=0.23 +/- 0.03 M_Sun) M dwarf at projected separation a_perp = 1.18 +/- 0.10 AU, i.e., about 1.9 times the snow line. ...
OGLE-2017-BLG-0373Lb: A Jovian Mass-Ratio Planet Exposes A New Accidental Microlensing Degeneracy
(2018)
We report the discovery of microlensing planet OGLE-2017-BLG-0373Lb. We show that while the planet-host system has an unambiguous microlens topology, there are two geometries within this topology that fit the data equally ...
OGLE-2017-BLG-1522: A giant planet around a brown dwarf located in the Galactic bulge
(2018)
We report the discovery of a giant planet in the OGLE-2017-BLG-1522 microlensing event. The planetary perturbations were clearly identified by high-cadence survey experiments despite the relatively short event timescale ...
Spitzer Opens New Path to Break Classic Degeneracy for Jupiter-Mass Microlensing Planet OGLE-2017-BLG-1140Lb
(2018)
We analyze the combined Spitzer and ground-based data for OGLE-2017-BLG-1140 and show that the event was generated by a Jupiter-class (m_p\simeq 1.6 M_jup) planet orbiting a mid-late M dwarf (M\simeq 0.2 M_\odot) that ...
OGLE-2016-BLG-1266: A Probable Brown-Dwarf/Planet Binary at the Deuterium Fusion Limit
(2018)
We report the discovery, via the microlensing method, of a new very-low-mass binary system. By combining measurements from Earth and from the Spitzer telescope in Earth-trailing orbit, we are able to measure the ...
KMT-2016-BLG-0212: First KMTNet-Only Discovery of a Substellar Companion
(2018)
We present the analysis of KMT-2016-BLG-0212, a low flux-variation $(I_{\rm flux-var}\sim 20$) microlensing event, which is well-covered by high-cadence data from the three Korea Microlensing Telescope Network (KMTNet) ...
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I want to solve following system of ODEs:
$ \Bigg\{ \begin{array}{} \frac{\partial C}{\partial t}=\frac{2W_b}{Rsin(\theta)}(1-\frac{C}{\gamma})+\frac{\nu}{\pi R^2}\frac{\partial C}{\partial Z} \\ \frac{\partial R}{\partial t}=\frac{W_b}{\gamma sin(\theta)} \end{array} $
Where $W_b=k_bC_h(1-\frac{C}{C_h})^{\frac{4}{3}}$ and $\nu, k_b, C_h, h, \gamma, \theta$ are constants.
It is known that:
$ \frac{\partial C}{\partial Z}=\frac{C_i-C_{i-1}}{Z_i-Z_{i-1}} $, where $Z_i-Z_{i-1}=h$ and $h$ is a constant.
So, the first equation become this: $ \frac{\partial C}{\partial t}=\frac{2W_b}{Rsin(\theta)}(1-\frac{C}{\gamma})+\frac{\nu}{\pi R^2}\frac{C_i-C_{i-1}}{h} $
My question is how can I get the result of previous computation, namely $C_{i-1}$?
Here is the mathematica code I wrote:
Wb[C_] := kb*Ch*(1 - C/Ch)^(4/3);system := { c'[t] == (2*Wb[c[t]])/(r[t]*Sin[theta])*(1 - C/gamma) + v/(Pi*r[t]^2)*(c[t] - ?? )/h, r'[t] == Wb[c[t]]/(gamma*Sin[theta]), c[0] == 0, r[0] == 0.15};solution = First@NDSolve[system, {c, r}, {t, 0, 25900000}, Method -> "BDF"];
UPDATE: Simplified version of the problem.
This is a model based on a plug flow reactor.
$ \Bigg\{ \begin{array}{} \dot{x} = \frac{c_1 x f(x)}{y} + \frac{\partial x}{\partial z} \\ \dot{y} = c_2 f(x) \end{array} $
Where c1 and c2 are constants. This is a model of a physical process and it was shown that
z dimension is quantified by a chunks of a constant size
h and
x is monotonously increasing, so $\frac{\partial x}{\partial z}=\frac{\Delta x}{\Delta z}=\frac{x_i - x_{i-1}}{h}$
f[x_] := ...;system := { x'[t] == c1*x*f[x[t]]/y[t] + (x[t] - ?)/h, y'[t] == c2*f[x[t]], x[0] == 0, y[0] == 0.15};solution = First@NDSolve[system, {x, y}, {t, 0, 25900000}, Method -> "BDF"];
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[K] refers to Kontsevich's paper "Deformation quantization of Poisson manifolds, I".
Background
Let $X$ be a smooth affine variety (over $\mathbb{C}$ or maybe a field of characteristic zero) or resp. a smooth (compact?) real manifold. Let $A = \Gamma(X; \mathcal{O}_X)$ or resp. $C^\infty(X)$.
Denote the dg
Lie algebra of polyvector fields on $X$ (with Schouten-Nijenhuis bracket and zero differential) by $T$. Denote the dg Lie algebra of the shifted Hochschild cochain complex of $A$ (with Gerstenhaber bracket and Hochschild differential) by $D$.
Then the Hochschild-Konstant-Rosenberg theorem states that there is a quasi-isomorphism of dg
vector spaces from $T$ to $D$. However, the HKR map is not a map of dg Lie algebras. It is not a map of dg algebras, either (where the multiplication on $T$ is given by the wedge product and the multiplication on $D$ is given by the cup product of Hochschild cochains).
I believe "Kontsevich formality" refers to the statement that, while the HKR map is not a quasi-isomorphism --- or even a morphism --- of dg
Lie algebras, there is an $L_\infty$ quasi-isomorphism $U$ from $T$ to $D$, and therefore $D$ is in fact formal as a dg Lie algebra.
The first "Taylor coefficient" of the $L_\infty$ morphism $U$ is precisely the HKR map (see section 4.6.2 of [K]).
Moreover, this quasi-isomorphism $U$ is compatible with the dg
algebra structures on $T$ and $D$ (see section 8.2 of [K]), and it yields a "corrected HKR map" which is a dg algebra quasi-isomorphism. The "correction" comes from the square root of the $\hat{A}$ class of $X$. See this previous MO question. Questions
(0) Are all of my statements above correct?
(1) In what way is the $L_\infty$ morphism $U$ compatible with the dg
algebra structures? I don't understand what this means.
(2) When $X$ is a smooth (compact?) real manifold, I think that all of the statements above are proved in [K]. When $X$ is a smooth affine variety, I think that the statements should all still be true. Where can I find proofs?
(3) Moreover, the last section of [K] suggests that the statements are all still true when $X$ is a smooth
possibly non-affine variety. For a general smooth variety, though, instead of taking the Hochschild cochain complex of $A = \Gamma(X;\mathcal{O}_X)$, presumably we should take the Hochschild cochain complex of the (dg?) derived category of $X$. Is this correct? If so, where can I find proofs?
In the second-to-last sentence of [K], Kontsevich seems to claim that the statements for varieties are corollaries of the statements for real manifolds, but I don't see how this can possibly be true. In the last sentence of the paper, he says that he will prove these statements "in the next paper", but I'm not sure which paper "the next paper" is, nor am I even sure that it exists, since "Deformation quantization of Poisson manifolds, II" doesn't exist.
P.S. I am not sure how to tag this question. Feel free to tag it as you wish.
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I wonder
1 whether there is a known relativization barrier against proving $L\neq NP$. Hence I'm looking for a language $A$ for which $L^A=NP^A$.
My first idea was to try $A:=SAT$, but then I thought that $L^A\subset P^A = \Delta_2^P$ and $NP^A=\Sigma_2^P$. This seems to disqualify not only $A:=SAT$, but any complete problem $A$ from the polynomial hierarchy for my purpose.
My next idea was to try $A:=TQBF$, where $TQBF$ is the $PSPACE$-complete problem to decide true quantified Boolean formulas. But $P^A=NP^A$ is well known, and $L^A=NP^A$ is a stronger statement, so $L^A=NP^A$ would be well known, if anybody had proved it.
My question is just whether there is a known relativization barrier. We certainly can't prove that such a language $A$ can't exist, because otherwise we would get $L\neq NP$ as corollary.
1. I know that logarithmically space bounded TMs with stack recognize exactly the languages from $P$, independent of whether the TM is deterministic or not. I don't know whether this result relativizes, but I guess it does. So I wonder whether there can be a language $A$ with $NL^A=P^A$. But asking for $L^A=NP^A$ instead seems easier, because of the connection to the polynomial hierarchy.
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We use numbers all the time but rarely stop to think how they work. In this section we will consider decimal, binary and hexadecimal numbers, we will look at exponential numbers to see how the exponents work under different operations and we will look at scientific and engineering notation (also known as standard form and preferred standard form).
You've grown up with the decimal system. It is so widely used that most people do not realise that it is just one of an infinite number of number systems. The decimal system has ten symbols $0$ to $9$ (dec means ten in Latin). You can add $1$ to $0$ to get $1$, add $1$ to $1$ to get $2$ and so on. If you want to add $1$ to $9$ you have a problem because you have no more symbols left. What do you do?
A solution for this problem was found by Indian mathmaticians in about 650AD when they rediscovered a system used by the ancient Sumerians around 2500BC (more details here). They let the position of a character, as well as the actual character, be significant. So, to add $1$ to $9$ you set the $9$ back to $0$ and carry $1$ into the next column. That may seem obvious to you now, four and a half thousand years later, but it is so clever and it was ground breaking at the time.
Let’s see how that works in binary. In binary you only have two symbols, $0$ and $1$ (bi means two in Latin). Add $1$ to $0$ you get $1$. To add $1$ to $1$ you have the same problem as above. So, you set the first $1$ back to $0$ and carry $1$ into the next column. The result looks like $10$ but it has the value of $2$ in decimal.
How does it work in hexadecimal (hex means six in Latin so hexadecimal means six + ten = sixteen)? In hexadecimal, or hex for short, there are sixteen symbols, $0$ to $9$ followed by $A$ to $F$. $F$ in hex is equal to $15$ in decimal. To add $1$ to $F$ you set the $F$ back to $0$ and carry $1$ to the next column. The result looks like $10$ but this time it has a value of $16$ in decimal.
Subscripts are used to specify the base of a number. $10$, $10_d$ and $10_{10}$ all mean $10$ in decimal. If there is no subscript then you can assume the number is base $10$.
$10_b$ and $10_2$ mean $10$ in binary which is equal to $2$ in decimal.
$10_h$ and $10_{16}$ mean $10$ in hexadecimal which is equal to $16$ in decimal.
In most computers there are 8 binary digits, called bits, in a byte. Half a byte is called a nybble and a 4 bit nybble can have $2^4$ values from $0000$ to $1111$ (in binary) which is $0$ to $15$ (in decimal) or $0$ to $F$ (in hex).
Here is a table of the values:
Dec Binary Hex Dec Binary Hex 0 0000 0 8 1000 8 1 0001 1 9 1001 9 2 0010 2 10 1010 A 3 0011 3 11 1011 B 4 0100 4 12 1100 C 5 0101 5 13 1101 D 6 0110 6 14 1110 E 7 0111 7 15 1111 F
A two digit hexadecimal number ranges from $0_h$ to $FF_h$ which is $0$ to $1111\ 1111_b$ in binary and $0$ to $255$ in decimal. In a two digit hexadecimal number the right digit represents units and the left digit represents sixteens.
Converting between binary and hexadecimal is easy, particularly if you treat the binary number in nybbles.
Example 1.1.1
Convert $1010\ 1100$ to hex
In decimal the left digit is $8 + 2 = 10$ which is $A$ in hex.
The right digit is $8 + 4 = 12$ which is $C$ in hex.
So $1010\ 1100_b = AC_h$.
Converting from hex into binary is also easy.
Example 1.1.2
Convert $D6$ to binary
$D_h = 13$ in decimal which is $8 + 4 + 1 = 1101_b$.
$6_h = 6$ in decimal which is $4 + 2 = 0110_b$.
So $D6_h = 1101\ 0110_b$.
The easiest way to convert from decimal into binary is to divide by sixteen (divide by 2 four times) and find the remainder. You can then convert the 16s and the units.
Example 1.1.3
Convert $197$ to binary and hex
Divide by 2 four time ignoring the remainders
$197 \div 2 = 98, 98 \div 2 = 49, 49 \div 2 = 24$ and $24 \div 2 = 12$.
So there are $12$ sixteens in $197$. $12 \times 16 = 192$ so $197 = 12 \times 16 + 5$. That's the hard bit done.
$12 = 8 + 4 = 1100_b = C_h$.
$5 = 4 + 1 = 0101_b = 5_h$.
So $197 = 1100\ 0101_b = C5_h$
and not even a whiff of a calculator.
If you multiply a number by itself you get an exponential number. For example $2 \times 2 = 2^2$. The big $2$ is called the base and the little $2$ is called the exponent. If you multiply $2 \times 2 \times 2$ you get $2^3$. $2$ is the base and $3$ is the exponent.
Each time you multiply a number by itself you increase the exponent by 1 so $2 \times 2 = 2^2$, $2 \times 2 \times 2 = 2^3$ and so on.
$2 = 2^1$ $2 \times 2 = 2^2$ $2 \times 2 \times 2 = 2^3$ $2$ $4$ $8$
Each time we increment the exponent we multiply by the base. What happens if we decrement the exponent?
$2 \times 2 \times 2 = 2^3$ $2 \times 2 = 2^2$ $2 = 2^1$ $2^0 = 1$ $8$ $4$ $2$ $1$
You can see we are dividing by the base each time we decrement the exponent so it comes as no surprise that $2^0 = 1$. You can probably see that anything raised to the power of $0$ is equal to $1$ except for $0^0$.
What happens if we keep decrementing the exponent?
$2^1$ $2^0$ $2^{-1}$ $2^{-2}$ $2$ $1$ $1/2$ $1/4$
Which means $2^{-2} = 1/2^2$ or, more generally:
What is $3^2 \times 3^3$? Well, $3^2 = 9$ and $3^3 = 27$. Now $9 \times 27 = 243$ which is equal to $3^5$ so $3^2 \times 3^3 = 3^5$. To multiply exponential numbers we simply add the exponents. More generally we can write
Pssssttt - This is how logarithms and slide rules work.
What is $243 \div 9$?
From above we know $243 = 3^5$ and $9 = 3^2$. $243 \div 9 = 27$ and $27 = 3^3$
Writing this as exponential numbers $3^5 \div 3^2 = 3^3$. To divide number $A$ by number $B$ we simply subtract number $B$'s exponent from number $A$'s exponent. More generally we can write
We know that $x^a \times x^b = x^{a+b}$. What happens in the case of $(x^a)^b$?
First, let’s try this with some easy numbers. Try $(2^3)^2 = (8)^2 = 64 = 2^6$.
As $6 = 3 \times 2$ we can deduce
$2^3$ is one of the square roots of $2^6$. We could write $2^3 = \sqrt{2^6}$. Using our previous rule we could also write $2^3 = (2^6)^{1/2}$. So $(2^6)^{1/2} = \sqrt{2^6}$ or more generally:
Look at this table:
$1$ $10$ $100$ $1000$ $10000$ $10^0$ $10^1$ $10^2$ $10^3$ $10^4$
Now look at this table:
$log(1)$ $log(10)$ $log(100)$ $log(1000)$ $log(10000)$ $0$ $1$ $2$ $3$ $4$
Logarithms are exponents. Taking the logarithm of a number is the inverse of raising the number to a power.
All the logarithms above have the same base, base 10, so we could have written $log_{10}(1000)=3$ where $1000$ is the argument, $10$ is the base and $3$ is the exponent. Logarithms can have any positive number for the base but the two most common bases are $10$ and Euler's constant $e$. Logarithms to base $10$ are written $log_{10}()$ or, more normally, $log()$. Logarithms to base $e$are written $log_{e}()$ or, more normally, $ln()$.
Look at your calculator. If it has logs it will probably have functions for $10^x$ and $log(x)$ as well as $e^x$ and $ln(x)$.
We know we can write $100$ as $10^2$ and $1000$ as $10^3$. What about the numbers in between? Well, using a calculator or log tables we can see $125=10^{2.0969}$, $150=10^{2.1761}$ and $175=10^{2.2430}$.
If we multiply $100 \times 1000$ we get $100000$ which can be written as $10^5$. If we multiply $125 \times 100$ we get $12500$ which can be written as $10^{4.0969}$.
In the same way we can if we divide $100000$ by $100$ we get $1000$ so $10^5 \div 10^2 = 10^3$. If we divide $125$ by $100$ we get $1.25$ so $10^{2.0969} \div 10^2 = 10^{0.0969}$.
Working with the exponents ($2$, $3$ and $2.0969$) rather than the numbers ($100$, $1000$ and $125$) means we can find products by adding the exponents and we can find quotients by subtracting the exponents.
Example 1.3.1
Simplify $3log(2x) - 2log(3x)$
$3log(2x)$ $\ =\ $ $log(2x)+log(2x)+log(2x)$ $\ =\ $ $log((2x) \times(2x) \times(2x))$ $\ =\ $ $log(8x^3)$ and $2log(3x)$ $\ =\ $ $log(3x)+log(3x)$ $\ =\ $ $log((3x) \times(3x))$ $\ =\ $ $log(9x^2)$ so
$3log(2x) - 2log(3x)$ $\ =\ $ $log(8x^3)-log(9x^2)$ $\ =\ $ $log(8x^3/9x^2)$ $\ =\ $ $log(8x/9)$
Example 1.3.2
Simplify $ln(y)+2ln(4x) - ln(8x^5) = 0$
Start by putting the $y$ terms on the left and the $x$ terms on the right
$ln(y)$ $=$ $ln(8x^5)-2ln(4x)$ $=$ $ln(8x^5)-ln(16x^2)$ $=$ $ln(8x^5/16x^2)$ $=$ $ln(x^3/2)$ We have single logs to the same base on both sides of the equation so we can take inverse logs of both sides giving us: $y$ $=$ $x^3/2$
Avogadro's constant is quite a large number. It is $602214000000000000000000$ and represents the number of atoms or molecules in 1 mole of substance. It is not easy to see how many million millions there are.
Plank's constant, by comparison, is quite small. It is $0.000000000000000000000000000000000662607$. It is a a fundamental constant equal to the energy of a quantum of electromagnetic radiation divided by its frequency. How many million millionths are there?
Scientific notation, also known as
standard form, was devised to help manage very large and very small numbers. In standard form we have a single digit before a decimal point, as many digits after the decimal point as we need and the number is multiplied by $10$ raised to an integer power. The number is called the mantissa or the significand.
In scientific notation Avogadro's constant would be written $6.02214 \times 10^{23}$. Plank's constant would be written $6.62607 \times 10^{−34}$.
Multiplying numbers in standard form is easy. You multiply the significands and add the exponents. For division you divide the significands and subtract the exponents. For example:
Imagine you have an object that is $500000$ mm long. You would probably not say it was $500000$ mm long, you are more likely to say it was $500$ m or maybe half a kilometre long.
Engineering notation, also known as
preferred standard form, only uses powers that are factors of $3$.
Significant figures are the figures in a number that carry meaning There are four rules on how to determine how many significant figures there are in a number.
Rules on significant figures
Let's look at some examples.
Rule 1 $12.34$ has four non-zero digits so it has four significant figures. Rule 2 $102.304$ has four non-zero digits and two zeros each between significant figures. That makes both the zeros significant so there are six significant figures. Rule 3 $0.003$ has a leading zero before the decimal point and two leading zeros after. They are not significant. There is one non-zero digit so the number has one significant figure. Rule 4 $600$ has one non-zero digit and two trailing zeros so it has one significant figure. Rule 4 $6.2600$ has three non-zero digits and two trailing zeros after the decimal point so it has five significant figures.
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Question
Four fair six-sided dice are rolled. The probability that the sum of the results being $22$ is $$\frac{X}{1296}.$$ What is the value of $X$?
My Approach
I simplified it to the equation of the form:
$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\leq 6,\,\,1\,\,\leq i \,\,\leq 4 $
Solving this equation results in:
$x_{1}+x_{2}+x_{3}+x_{4}=22$
I removed restriction of $x_{i} \geq 1$ first as follows-:
$\Rightarrow x_{1}^{'}+1+x_{2}^{'}+1+x_{3}^{'}+1+x_{4}^{'}+1=22$
$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$
$\Rightarrow \binom{18+4-1}{18}=1330$
Now i removed restriction for $x_{i} \leq 6$ , by calculating the number of
bad cases and then subtracting it from $1330$:
calculating
bad combination i.e $x_{i} \geq 7$
$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$
We can distribute $7$ to $2$ of $x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$ i.e$\binom{4}{2}$
We can distribute $7$ to $1$ of $x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$ i.e$\binom{4}{1}$ and then among all others .
i.e
$$\binom{4}{1} \binom{14}{11}$$
Therefore, the number of bad combinations equals $$\binom{4}{1} \binom{14}{11} - \binom{4}{2}$$
Therefore, the solution should be:
$$1330-\left( \binom{4}{1} \binom{14}{11} - \binom{4}{2}\right)$$
However, I am getting a negative value. What am I doing wrong?
EDIT
I am asking for my approach, because if the question is for a larger number of dice and if the sum is higher, then predicting the value of dice will not work.
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Algebra is a way of generalising arithmetic by replacing numbers with letters. There are no rules about which letters to use but you will often find numbers replaced with $ a, b $ and $ c $ and variables represented by $ x, y $ and $ z $.
A term is a mathematical quantity that consists of a coefficient, a variable and an exponent. $ 3x^2 $ is a term. The coefficient is $ 3 $ the variable is $ x $ and the exponent is $ 2 $.
$ ax^n $ is a more generalised term. The coefficient is $ a $, the variable is $x$ and the exponent is $ n $.
The coefficient can have any numeric or algebraic value. If the value of the coefficient is 0 then the value of the term is zero. If the value of the coefficient is 1 then the coefficient is not normally written. $1x^2$ is usually written as $x^2$.
Two terms are like terms if each one has the same variable or variables and each variable is raised to the same power. $2x$ and $5x$ are like terms, $2x$ and $5xy$ are not. $4ax$ and $-3ax$ are like terms, $4ax^2$ and $-3ax$ are not.
Example 2.2: Collecting like terms
In the following expressions collect and simplify like terms. $2x+5x$ $=$ $7x$ $2x+5xy$ $=$ $2x+5xy$ $4ax-3ax$ $=$ $ax$ $4ax^2-3ax$ $=$ $4ax^2-3ax$
Expanding brackets means multiplying each term inside the brackets by the term outside the bracket. Take $ 5(2 + 4) = 5 \times 2 + 5 \times 4 = 10+20=30 $.
This can also be seen as $ 5(2 + 4) = 5 \times 6 = 30 $
In the same way the expression $ 5(x + y) $ implies $x+y$ multiplied by $5$ to produce $5x+5y$.
Example 2.3: Expanding brackets
Expand the brackets in the following expressions. $(2x-4y)$ $=$ $2x-4y$ $4(2x-4y)$ $=$ $8x-16y$ $-a(2x-4y)$ $=$ $-2ax+4ay$
You will often find brackets multiplied by other brackets like this $(x+3)(x-2)$. In cases like these every term in the second bracket is multiplied every term in the first bracket. If there are $m$ terms in the first bracket and $n$ terms in the second there will be $m \times n$ terns when you expand the brackets. In this example we will get $2 \times 2 = 4$ terms like this
$(x+3)(x-2)= x \times x + x \times (-2) + 3 \times x + 3 \times (-2)$.
Note the brackets round the $-2$. They are there because $\times$ is an operator and so is $-$. Operators act on terms not on other operators so $x \times -2$ doesn't mean anything.
To complete our example we can collect like terms giving $(x+3)(x-2)= x^2 + x -6$. Notice the expression is written in decending powers of $x$. This is the correct way to write polynomials.
$(x+3)(x-2) = x^2-2x+3x-6 = x^2+x-6$.
Some people find the acronym
FOIL helpful when expanding brackets. The letters stand for First, Outside, Inside and Last.
Factorisation is part of the process of simplification. Look at each term in an expression to see whether there are any common factors. If there are, collect them together.
Consider $2x-4y$. Both coefficients have the factor $2$ so we can take it outside a bracket giving:
$2x-4y = 2(x-2y)$
Now look at $8x-16y$. Both coefficients have the factor $8$ so we can take it outside a bracket giving:
$8x-16y = 8(x-2y)$
The final case from Example 2.3 is $-2ax+4ay$. Both coefficients have the factor $2a$ so we can take it outside a bracket giving:
$2ax-4ay = 2a(-x+2y)$ or $2ax-4ay = -2a(x-2y)$
The general form of a quadratic expression is $ax^2+bx+c$. Quadratic expressions can often be factorised in to two linear factors. Let's look at some examples. In section 2.3 we had the expression $x^2+x-6$
Example 2.4.1: Factorise $x^2+x-6$
$x^2+x-6$ = $x^2-2x+3x-6$ = $x(x-2)+3(x-2)$ = $(x+3)(x-2)$
If you have an expression like $x^2+bx+c$ see if you can find two numbers whose product is $c$ and whose sum is $b$. In Example 2.4.1 $c=-6$. This can be factorised as $3$ and $-2$. $3 \times -2 = -6$ and $3-2 = 1$. This is called
. factorisation by inspection
You can also factorise expressions like $ax^2+bx+c$ by inspection though the arithmetic is generally harder.
To complete the square for an expression like $x^2+bx+c$ we rearrange the expression to be $(x+b/2)^2 - (b/2)^2+c$. We set the expression equal to $0$ and subtract the numeric terms, $-(b/2)^2+c$, from both sides. Now we can take the square root of both sides and find the factors.
Imagine we have $x^2 + 8x +15$. $b=8$ so $b/2=4$.
$x^2 + 8x +15$ = $(x+8/2)^2-(8/2)^2+15$ Let $(x+4)^2-4^2+15$ = $0$ $(x+4)^2-16+15$ = $0$ $(x+4)^2-1$ = $0$ $(x+4)^2$ = $1$ $x+4$ = $\pm 1$ so $x$ = $-3$ or $-5$ If $x=-3$ then $x+3=0$ and if $x=-5$ then $x+5=0$ So $x^2 + 8x +15$ = $(x+3)(x+5)$
If you cannot factorise a quadratic expression by inspection or by completing the square you can use the quadratic formula. If we have $ax^2+bx+c$ then $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
Taking the example from section 2.4.1 $x^2 + x -6$ so $a=1$, $b=1$ and $c=-6$
$x$ = $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ = $\frac{-1 \pm \sqrt{(-1)^2-4 \times 1 \times (-6)}}{2 \times 1}$ = $\frac{-1 \pm \sqrt{1+24}}{2}$ = $\frac{-1 \pm 5}{2}$ so $x$ = $-3$ or $2$ If $x=-3$ then $x+3=0$ and if $x=2$ then $x-2=0$ So $x^2 + x - 6$ = $(x+3)(x-2)$
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Resonance Contents Classical Derivations Series Resonance
The classical circuit to demonstrate series resonance is the RLC circuit shown in the figure right, which shows a voltage source connected to R, L and C impedances in series. Given a fixed ac voltage source U operating at angular frequency [math] \omega [/math], the current in the circuit is given by the following:
[math]I = \frac{U}{Z} = \frac{U}{R + j \left(\omega L - \frac{1}{\omega C} \right)} [/math] [math]= \frac{U}{R + j \left(\frac{\omega^{2} LC - 1}{\omega C} \right)} [/math]
The current is at a maximum when the impedance is at a minimum. So given constant R, L and C, the minimum impedance occurs when:
[math]\omega^{2} LC - 1 = 0 \, [/math]
or
[math]\omega = \frac{1}{\sqrt{LC}} [/math]
This angular frequency is called the
resonant frequency of the circuit. At this frequency, the current in the series circuit is at a maximum and this is referred to as a point of series resonance. The significance of this in practice is when harmonic voltages at the resonant frequency cause high levels of current distortion. Parallel Resonance
The classical circuit to demonstrate series resonance is the RLC circuit shown in the figure right, which shows a current source connected to R, L and C impedances in parallel. Given a fixed ac current source I operating at angular frequency [math] \omega [/math], the voltage across the impedances is given by the following:
[math]V = IZ = \frac{I}{\frac{1}{R} + j \left(\omega C - \frac{1}{\omega L} \right)} [/math] [math]= \frac{I}{\frac{1}{R} + j \left(\frac{\omega^{2} LC - 1}{\omega L} \right)} [/math]
The voltage is at a maximum when the impedance is also at a maximum. So given constant R, L and C, the maximum impedance occurs when:
[math]\omega^{2} LC - 1 = 0 \, [/math]
or
[math]\omega = \frac{1}{\sqrt{LC}} [/math]
Notice that the resonant frequency is the same as that in the series resonance case. At this resonant frequency, the voltage in the parallel circuit is at a maximum and this is referred to as a point of parallel resonance. The significance of this in practice is when harmonic currents at the resonant frequency cause high levels of voltage distortion.
Resonance in Practical Circuits Series Resonance
Here a distorted voltage at the input of the transformer can cause high harmonic current distortion ([math]I_{h}[/math]) at the resonant frequency of the RLC circuit.
Parallel Resonance
In this more common scenario, a harmonic current source ([math]I_{h}[/math]) can cause high harmonic voltage distortion on the busbar at the resonant frequency of the RLC circuit. The harmonic current source could be any non-linear load, e.g. power electronics interfaces such as converters, switch-mode power supplies, etc.
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I'm reading Anton's Elementary Linear Algebra. I have come upon the rotation matrix.
$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$
They start the discussion with the fact that $T(e_1) = T(1,0) = (\cos \theta, \sin \theta)$ and $T(e_2) = T(0,1) = (-\sin \theta, \cos \theta)$.
This makes sense to me. But why do you need both $e_1$ and $e_2$ ? What about $e_3$ to rotate a vector in 3 dimensions?
EDIT
More specifically, I'm wondering about rotation in 3 dimensions.
About the z-axis:
$\begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{bmatrix}$
About the x-axis:
$\begin{bmatrix} 1 & 0 & 0\\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta\end{bmatrix}$
About the y-axis:
$\begin{bmatrix} \cos \theta & 0 & \sin \theta \\ 0 & 1 & 0 \\ -\sin \theta & 0 & \cos \theta\end{bmatrix}$
I see that both z-axis rotation and x-axis rotation follow the pattern of $\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$
Why is the rotation about the y-axis different?
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The ideal encryption scheme $E$ would be one that, for every ciphertext $C=E(K, M)$, if the key remains secret for the adversary, the probability of identifying $M$ is
negligible. Since that is not possible in practice, the second most reasonable approach is to define constraints strong enough to satisfy some definition of security. The $\operatorname{IND-}$ notation provides such definitions in terms of games, where a challenger keeps his key secret, and an adversary has certain capabilities and his target is to break the encryption system.
To keep it general, an encryption scheme will have a
key generation algorithm $KG$, which will generate a key pair $K_E$, $K_D$, an encryption algorithm $E$ and a decryption algorithm $D$. Encryption is always revertible, but the encryption and decryption key can be different (covering public key crypto): $D(K_D, E(K_E, M))=M$
IND-CPA: INDistinguishability under Chosen Plaintext Attack
In words: the adversary generates two messages of equal length. The challenger decides, randomly, to encrypt one of them. The adversary tries to guess which of the messages was encrypted.
Algorithm: Challenger: $K_E, K_D$ = KG(security parameter) Adversary: $m_0, m_1 = $ choose two messages of the same length. Send $m_0,m_1$ to the challenger. Perform additional operations in polynomial time including calls to the encryption oracle. Challenger: $b=$ randomly choose between 0 and 1 Challenger: $C:=E(K_E, m_b)$. Send $C$ to the adversary. Adversary: perform additional operations in polynomial time including calls to the encryption oracle. Output $guess$. If $guess=b$, the adversary wins
Further comment: the main concept introduced by this scenario is the polynomial bound. Now, our expectations from crypto are weakened from probability of winning is negligible to probability of winning . The restriction for the messages to be of the same length aims to prevent the adversary to trivially win the game by just comparing the length of the ciphertexts. However, this requirement is too weak, especially because it assumes only a single interaction between the adversary and the challenger. within a reasonable timeframe is negligible
IND-CCA1: INDistinguishability under Chosen Ciphertext Attack
In words: the target of the game is the same as in IND-CPA. The adversary has an additional capability: to call an encryption or decryption oracle. That means: the adversary can encrypt or decrypt arbitrary messages before obtaining the challenge ciphertext.
Algorithm: Challenger: $K_E, K_D$ = KG(security parameter) Adversary (a polynomially-bounded number of times): call the encryption or decryption oracle for arbitrary plaintexts or ciphertexts, respectively Adversary: $m_0, m_1 = $ choose two messages of the same length Challenger: $b=$ randomly choose between 0 and 1 Challenger: $C:=E(K_E, m_b)$Send $C$ to the adversary. Adversary: perform additional operations in polynomial time. Output $guess$ If $guess=b$, the adversary wins
Further comment: IND-CCA1 considers the possibility of repeated interaction, implying that security does not weaken with time.
IND-CCA2: INDistinguishability under adaptive Chosen Ciphertext Attack
In words: In addition to its capabilities under IND-CCA1, the adversary is now given access to the oracles after receiving $C$, but cannot send $C$ to the decryption oracle.
Algorithm: Challenger: $K_E, K_D$ = KG(security parameter) Adversary (as many times as he wants): call the encryption or decryption oracle for an arbitrary plaintext/ciphertext Adversary: $m_0, m_1 = $ choose two messages of the same length Challenger: $b=$ randomly choose between 0 and 1 Challenger: $C:=E(K_E, m_b)$Send $C$ to the adversary. Adversary: perform additional operations in polynomial time, including calls to the oracles, for ciphertexts different than $C$. Output $guess$. If $guess=b$, the adversary wins
Further comment: IND-CCA2 suggests that using the decryption oracle after knowing the ciphertext can give a reasonable advantage in some schemes, since the requests to the oracle could be customized depending on the specific ciphertext.
The notion of IND-CCA3 is added based on the reference provided by @SEJPM. I add it for completeness, but it seems important to point out that there are few resources about it, and my interpretation could be misleading.
IND-CCA3: (authenticated) INDistinguishability under adaptive Chosen Ciphertext Attack
In words: It is not possible to create a valid forgery with non-negligible probability. The adversary is given two pairs of encryption/decryption oracles. The first pair performs the intended encryption and decryption operations, while the second one is defined as follows: $\mathcal{E}_K$: returns encryptions of random strings. $\mathcal{D}_K:$ returns INVALID. Instead of being presented as a game, it is presented using the mathematical concept of advantage: the improvement of the probability of winning by using the valid oracle against the probability of success under the "bogus" oracle.
Formula: $\mathbf{Adv}^{ind-cca3}_{\pi}(A)=Pr\left[K\overset{\$}{\leftarrow}\mathcal{K}:A^{\mathcal{E}_K(\cdot),\mathcal{D}_K(\cdot)}\Rightarrow 1\right] - Pr\left[A^{\mathcal{E}_K(\$|\cdot|),\perp(\cdot)}\Rightarrow 1\right] $
Further comment: the paper where IND-CCA3 is introduced a focus on one fundamental idea. IND-CCA3 is equivalent to authenticated encryption.
Note that in the case of public-key cryptography the adversary is always given access to the public key $K_E$ as well as the encryption function $E(K_E, \cdot)$.
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I am looking for a counter example to the fact that a faithfully flat morphism isan effective descent morphism for the category of quasi-coherent sheaveswhen one forgets the quasi-compact hypothesis. If possible, I'd like a counter-exampleon the
descent of morphism aspects. In other words, I am looking fora morphism of scheme $X' \rightarrow X$, faithfully flat but not quasi-compact,and two quasi-coherent sheaves $F$ and $G$ on $X$, such that, calling $X''=X' \times_X X'$,$F'$, $F''$, $G'$, $G''$ thepull-back of $F$,$G$ to $X'$, $X''$, the sequence of modules$$0 \rightarrow Hom_X(F,G) \rightarrow Hom_{X'}(F',G') \rightarrow Hom_{X''}(F'',G'')$$is not exact (here the last morpheme is as it should the difference between the morphism induced by the two projections $X'' -> X'$. I wrote it this way because I am not sure how to do a double arrow in mathoverflow's tex:-)
I am looking for a counter example to the fact that a faithfully flat morphism isan effective descent morphism for the category of quasi-coherent sheaveswhen one forgets the quasi-compact hypothesis. If possible, I'd like a counter-exampleon the
I think the following should be a counter-example.
Let $X=\text{Spec}(\mathbb{C}[T])$ be the affine line and $X'=\bigsqcup_{x\in\mathbb{C}}\text{Spec}(\mathbb{C}[T]_{(T-x)})$ be its faithfully flat cover built from all local rings at closed points. Let $F=\mathcal{O}_{X}$ and $G=\bigoplus_{x\in\mathbb{C}}\mathcal{O}_X/(T-x)\simeq\bigoplus_{x\in\mathbb{C}}\mathbb{C}_x$ be two quasi-coherent sheaves on $X$.
Then, there is a natural quotient morphism $u':F'\to G'$ given by $\mathbb{C}[T]_{(T-x)}\to \mathbb{C}[T]/(T-x)$ on $\text{Spec}(\mathbb{C}[T]_{(T-x)})$. This morphism obviously satisfies the descent condition, as the only non-trivial conditions to check are on connected components of $X''$ that are isomorphic to the generic point of $X$, and $G''$ is zero on these components. However, this morphism does not come from a morphism $u:F\to G$ because $u(1)$ should have a non-zero component in every factor of the direct sum, which is impossible.
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Reading a paper about eta invariants I came across a zeta-like function.
I'm looking for the analytic continuation of $$\sum_{k=1}^\infty k(k+a)^{-s}$$ at $s=0$, where $a$ is positive.
In the paper he just says "The [...] term causes no problem and at $s=0$ has the value $\left[\frac{(4a^2-1)}{12}\right]$." Unfortunately, I really don't see the that.
My first approach so far; I tried a Taylor series at $s=3$: \begin{align} \sum_{k=1}^\infty k(k+a)^{-s} &= \sum_{k=1}^\infty k \sum_{l=0}^\infty (-1)^l \begin{pmatrix}2+l \\ l\end{pmatrix} (k+a)^{-3-l} (s-3)^{l} \end{align} and inserted $s=0$ $$\sum_{k=1}^\infty \frac{k}{(k+a)^3} \sum_{l=0}^\infty \begin{pmatrix}2+l \\ l\end{pmatrix} \left( \frac{3}{k+a} \right)^l$$ which is for $a\geq 3$ $$ \sum_{k=1}^\infty \frac{k}{(k+a-3)^3}~. $$ That seems to converge if I didn't miscalculate. But what are the further steps in order to get the result above? Or is there a more skilful approach?
Thanks in advance!
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Given a positive integer $n$ which is not a perfect square, it is well-known that Pell's equation $a^2 - nb^2 = 1$ is always solvable in non-zero integers $a$ and $b$.
Question:Let $n$ be a positive integer which is not a perfect square. Is there always a polynomial $D \in \mathbb{Z}[x]$ of degree $2$, an integer $k$ and nonzero polynomials $P, Q \in \mathbb{Z}[x]$ such that $D(k) = n$ and $P^2 - DQ^2 = 1$, where $a = P(k)$, $b = Q(k)$ is the fundamental solution of the equation $a^2 - nb^2 = 1$?
If yes, is there an upper bound on the degree of the polynomials $P$ and $Q$ -- and if so, is it even true that the degree of $P$ is always $\leq 6$?
Example: Consider $n := 13$.Putting $D_1 := 4x^2+4x+5$ and $D_2 := 25x^2-14x+2$, we have $D_1(1) = D_2(1) = 13$.Now the fundamental solutions of the equations$P_1^2 - D_1Q_1^2 = 1$ and $P_2^2 - D_2Q_2^2 = 1$ are given by
$P_1 := 32x^6+96x^5+168x^4+176x^3+120x^2+48x+9$,
$Q_1 := 16x^5+40x^4+56x^3+44x^2+20x+4$
and
$P_2 := 1250x^2-700x+99$,
$Q_2 := 250x-70$,
respectively. Therefore $n = 13$ belongs to at least $2$ different series whose solutions have ${\rm deg}(P) = 6$ and ${\rm deg}(P) = 2$, respectively.
Examples for all non-square $n \leq 150$ can be found here.
Added on Feb 3, 2015: All what remains to be done in order to turnLeonardo's answers into a complete answer to the question is to find out which valuesthe index of the group of units of $\mathbb{Z}[\sqrt{n}]$ in the group of unitsof the ring of integers of the quadratic field $\mathbb{Q}(\sqrt{n})$can take. This part is presumably not even really MO level, but it's justnot my field -- maybe someone knows the answer? Added on Feb 14, 2015: As nobody has completed the answer so far, it seemsthis may be less easy than I thought on a first glance. Added on Feb 17, 2015: Leonardo Zapponi has given now a complete answer tothe question in this note.
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Figure 7.1 shows a triangle with sides a, b and c and angles subtending the sides A, B and C respectively.
Figure 7.1: Triangle with side lengths a, b and c.
For the triangle in Figure 7.1 we can write
$\dfrac{a}{sin(A)}=\dfrac{b}{sin(B)}=\dfrac{c}{sin(C)}$
These equations are called the sine rule.
Note: The sine of an angle, $\theta$, is the same as the sine of the supplement of the angle, $\pi - \theta$ ($180 - \theta$). This means the sine rule fails for triangles with an obtuse angle.
As a precaution, always calculate all three angles and add the results together. If the sum is less than $\pi$ ($180$) then the angle opposite the longest side is probably obtuse.
For the same triangle we can also write
$a^2=b^2+c^2-2bc\ cos(A)$
This equation is called the cosine rule.
When angle $A=\pi/2$, so we have a right angled triangle and the cosine rule simplifies to Pythagoras's theorem.
$a^2=b^2+c^2$
Figure 7.2: Right angle triangle.
For right angled triangles we can write
$sin(C)= \dfrac{c}{a}$, $cos(C)= \dfrac{b}{a}$ and $tan(C)= \dfrac{c}{b}$.
For angle $C$ side $b$ is called the adjacent, side $c$ is called the opposite and side $a$ is called the hypotenuse.
A common mnemonic for right angled triangles is
SohCahToa which means $sin(C)=\frac{opposite}{hypotenuse}$, $cos(C)=\frac{adjacent}{hypotenuse}$ and $tan(C)=\frac{opposite}{adjacent}$
A quick way to work out whether to use $sin()$ or $cos()$ to find a length is to imagine standing at angle $C$. Look along the adjacent side. We could say we were looking in the central direction and the central direction gives us $cos(C) = c/a$.
The opposite side goes sideways and sideways gives us $sin(\theta)=c/a$.
Example 7.1 Find the angles for a triangle with sides $a=4$, $b=5$ and $c=6$. We need to use the cosine rule to find the first angle. $a^2$ $=$ $b^2+c^2-2bc\ cos(A)$ $2bc\ cos(A)$ $=$ $b^2+c^2-a^2$ $$cos(A)$$ $=$ $\dfrac{b^2+c^2-a^2}{2bc}$ $$A$$ $=$ $cos^{-1}(\dfrac{b^2+c^2-a^2}{2bc})$ $$A$$ $=$ $cos^{-1}(\dfrac{5^2+6^2-4^2}{2 \times 5 \times 4})$ $$A$$ $=$ $0.72$ radians (or $41.41^\circ$) Similarly we get $B$ $$B$$ $=$ $cos^{-1}(\dfrac{a^2+c^2-b^2}{2ac})$ $$B$$ $=$ $0.97$ radians (or $55.77^\circ$) and we get $C$ $$C$$ $=$ $cos^{-1}(\dfrac{a^2+b^2-c^2}{2ab})$ $$C$$ $=$ $1.45$ radians (or $82.82^\circ$) Sanity Check You might have been tempted to calculate two of the angles and subtract their sum from $\pi$ or $180^\circ$. If you did that you would have no easy way to check whether or not you had calculated three wrong angles. Better to calculate all three angles and add them together. $0.72+0.97+1.45=3.14$
(or $41.41+55.77+82.82=180$)
so we can be confident our answers are correct.
So far we have been sketching our diagrams on x-y axes. This system is attributed to René Descartes and is called the
Cartesian coordinate system. The system allows us to describe any point on a 2D plane.
An alternative way to describe a point on a 2D plane is by giving an angle from a fixed line and a radius. This system is calle the
polar coordinate system and is used a lot in navigation.
To convert from cartesian coordinates to polar coordinates we have:
$r^2=x^2+y^2$
$\theta = tan^{-1}(y/x)$
To convert from polar coordinates to cartesian coordinates we have:
$x=r\ cos(\theta)$
$y=r\ sin(\theta)$
Imagine a disc with a pen mounted near the edge. If the disc rotates about its centre the pen will draw a circle. Now, imagine there is a long strip of paper under the disc. If we pull the paper at a constant speed as the disc rotates, instead of a circle the pen will draw a wiggley line. The line is a sinusoidal curve and this is why sines, cosines and tangents are called circular functions.
Pythagoras's theorem states $x^2+y^2=r^2$. If we divide both sides by $r^2$ we get
$$(\dfrac{x}{r})^2+(\dfrac{y}{r})^2=1$$
Looking at figure 7.3 we can see $sin(\theta)= \dfrac{y}{r}$ and $cos(\theta)= \dfrac{x}{r}$ substituting these into our equation we get:
$$cos^2(\theta)+sin^2(\theta)=1$$
The angle $\theta$ was not used in our derivation which means $cos^2(\theta)+sin^2(\theta)=1$ is true for any value of $\theta$. This means the expression is more than an equation, it is an identity.
Here are some other trigonometric identities
$ \cos(\theta + \phi) = \cos(\theta)\cos(\phi)- \sin(\theta)\sin(\phi)$
$ \cos(\theta - \phi) = \cos(\theta)\cos(\phi)+ \sin(\theta)\sin(\phi)$
$ \sin(\theta + \phi) = \sin(\theta)\cos(\phi)+ \cos(\theta)\sin(\phi)$
$ \sin(\theta - \phi) = \sin(\theta)\cos(\phi)- \cos(\theta)\sin(\phi)$
$ \tan(\theta + \phi) = \dfrac{\tan(\theta)+ \tan(\phi)}{1-\tan(\theta) \tan(\phi)} $
$ \tan(\theta - \phi) = \dfrac{\tan(\theta)- \tan(\phi)}{1+\tan(\theta) \tan(\phi)} $
If we let $\theta = \phi$ we get
$ \cos(2\theta) = \cos^2(\theta)- \sin^2(\theta)$
$ \sin(2\theta) = 2 \sin(\theta)\cos(\theta)$
$ \tan(2\theta) = \dfrac{2\tan \theta}{1-\tan^2 \theta} $
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As its long ago since Erdős died and mathoverflow is the second best alternative to him (for discussing personal problems), I'd like to start a fruitful discussion about the following problem that I find very interesting.
Let $n \geq 3$ be an integer and let $\alpha(n)$ denote the least integer $k$ such that there exists a simple graph on $k$ vertices having precisely $n$ spanning trees. What is the asymptotic behaviour of $\alpha$ ?
Motivation.I was introduced to the question through this post on Dick Lipton's blog. As it turns out, the question was posed already in 1970 by the Czech graph theorist J. Sedlacek (On the minimal graph with a given number of spanning trees, Canad. Math. Bull. 13 (1970) 515–517) What is known?
Sedlacek was able to show that for every (not so) large $n$
$ \alpha(n) \leq \frac{n+6}{3}$ if $n \equiv 0 \pmod{3} $ and $\alpha(n) \leq \frac{n+4}{3}$ if $n \equiv 2 \pmod{3}. $
Following is a summary of what I was able to find out.
Since the equation $n = ab+ac+bc$ is solvable for integers $1 \leq a < b < c$ for all but a finite number of integers $n$ (see this post) it can be deduced (by considering the graph $\theta_{a,b,c}$ which has $ab+ac+bc$ spanning trees) that for large enough $n \not \equiv 2 \pmod{3}$
$$\alpha(n) \leq \frac{n+9}{4}.$$
Moreover, the only fixed points of $\alpha$ are 3, 4, 5, 6, 7, 10, 13 and 22.
By generalizing the approach and considering the graphs $\theta_{x_1,\ldots,x_k}$ one could try to lower the constant in the fraction of the inequality by an arbitrary amount. As it turns out it is not know weather every large $n$ is then expressible as $n = x_1\cdots x_k(\frac{1}{x_1} + \cdots + \frac{1}{x_k})$ for suitable integers $1 \leq x_1 < \cdots < x_k.$
Even if that method would work out, the bound would most probably still be suboptimal. According to the graph (created by randomly generating graphs and calculating the number of their spanning trees) it seems reasonable to conjecture that
Conjecture.
$$\alpha(n) = o(\log{n})$$
The conjecture is clearly justifiable for highly composite numbers $n$ (consider the graph obtained after identifying a common vertex of the cycles $C_{x_1},\ldots,C_{x_k}$ for suitable odd factors $x_1, \ldots,x_k$ of $n$) but It fails for $n$'s that are primes.
It is evident to me that I lack the tools necessary for attacking this conjecture so any kind of suggestions (where to look for a possible answer, what kind of tools should I learn..) related to it are very welcome!
Edit. If anyone is willing to work on this problem, I'd be glad to collaborate since I'd benefit much from it!
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A sine wave starts at zero when $\theta=0$, rises to its maximum, drops through zero to its minimum and then returns to zero. This cycle is repeated again and again. The vertical distance from zero to the peak value is called the
amplitude. The number of peaks per unit time is called the frequency. Frequency is generally given the symbol $\nu$ (called nu). The horizontal distance between two peaks is called the wavelength. Wavelength is usually given the symbol $\lambda$ (called lambda).
Figure 8.1: A sine wave $y=sin \theta$
A cosine wave is similar to a sine wave except it starts at its maximum, drops through zero to its minimum and then returns back through zero to its maximum again. This cycle, just like the sine wave, is repeated again and again.
Figure 8.2: A cosine wave $y=cos \theta$
Imagine a cosine wave and a sine wave adding together to create a combined wave. What would the combined wave be like? To make the question a bit more general let the sine wave have an amplitude of $a$ and the cosine wave have an amplitude of $b$.
Figure 8.3: A combined wave $y=asin \theta+bcos \theta$
We know from section 7.3 that $ \sin(\theta + \phi)$ $=$ $\sin \theta \cos \phi + \cos \theta \sin \phi $. so we can write $ r\sin(\theta + \phi)$ $=$ $r\sin \theta \cos \phi + r\cos \theta \sin \phi $. For the combined wave we have $y$ $=$ $a\sin \theta + b\cos \theta $. let $a$ $=$ $r\cos \phi $ and $b$ $=$ $r\sin \phi $ then $a^2 + b^2$ $=$ $r^2 cos^2 \phi + r^2\sin^2 \phi $ $=$ $r^2 (cos^2 \phi + sin^2 \phi) $ we know $cos^2 \phi + sin^2 \phi$ $=$ $ 1 $ which means $a^2 + b^2$ $=$ $ r^2 $ so $r$ $=$ $\sqrt{a^2 + b^2} $. If we divide $b$ by $a$ we get $\dfrac{b}{a}$ $=$ $ \dfrac{r sin \phi}{r cos \phi} $ $=$ $ tan \phi $ so $\phi$ $=$ $tan^{-1}(b/a)$.
Example 8.3b: Find the amplitude and phase of the combined wave $y=-4sin \theta+3cos \theta$ We know $ r\sin(\theta + \phi)$ $=$ $r\sin \theta \cos \phi + r\cos \theta \sin \phi $ and we are given $y$ $=$ $-4\sin \theta + 3\cos \theta$ Comparing the equations we can see: $rcos\phi$ $=$ $-4$ $rsin\phi$ $=$ $3$ $r^2 cos^2\phi + r^2 sin^2 \phi$ $=$ $(-4)^2+3^2$ $=$ $5$ If we divide $rsin\phi$ by $rcos\phi$ we get $tan\phi$ $=$ $3/-4$ so $\phi$ $=$ $tan^{-1}(3/-4)$ Your calculator gives an answer of $-0.64$ radians (-36.87°) so we need to add $\pi$ radians (180°) to get $\phi$ $=$ $2.50$ radians (143.13°) $y=-4sin \theta+3cos \theta$ $=$ $5sin(\theta+2.50)$
There is a set of problems that require us to solve equations of the form $a sin\theta+b cos\theta$ $=$ $c$ To solve these equations we transform $a sin\theta+b cos\theta$ into $rsin(\theta+\phi)$ and set the result equal to $c$. $rsin(\theta+\phi)$ $=$ $c$ $sin(\theta+\phi)$ $=$ $c/r$ $\theta+\phi$ $=$ $sin^{-1}(c/r)$ $\theta$ $=$ $sin^{-1}(c/r)-\phi$ If the result is negative you normally add $2\pi$ to get it into the range $0$ to $2\pi$. Look at the diagram below. You will see there is another solution. To find the other solution subtract $\theta$ and $\phi$ from $\pi$.
$\theta_1 = sin^{-1}(c/r)-\phi$As before, add $2\pi$ if the result is negative.
$\theta_2 = \pi-sin^{-1}(c/r)-\phi$
Example 8.4: Solve the following equation
$4sin \theta+3cos \theta = 2$ for values of $\theta$ between $0$ and $2\pi$
$r$ $=$ $\sqrt{3^2 + 4^2} $. $=$ $5$ and $tan \phi$ $=$ $3/4$ $\phi$ $=$ $0.64$ radians (36.87°) $5sin(\theta+0.64)$ $=$ $2$ $sin(\theta+0.64)$ $=$ $2/5$ $\theta+0.64$ $=$ $sin^{-1}(2/5)$ $\theta+0.64$ $=$ $0.41$ $\theta$ $=$ $0.41-0.64$ $\theta$ $=$ $-0.23$ Adding $2\pi$ we get $\theta_1$ $=$ $6.05$ radians (346.71°) $\theta_2$ $=$ $\pi-sin^{-1}(2/5)-\phi$ $=$ $\pi-0.41-0.64$ $=$ $2.09$ radians (119.6°) Sanity Check Look at the diagram above. You can see the smaller intersection of the purple sine wave and the green line is a bit after $\pi/2$ which is about $2$ so we can be confident with $2.09$. The second intersection is between $3\pi/2$ and $2\pi$ which is about $6$ so we can be confident with $6.05$.
A combined wave is made from two or more sine and/or cosine waves. Once a combined wave has been calculated we would like to know the position and magnitude of the turning points, that is the points at which the combined wave is at a maximum or a minimum.
Consider $ y= a sin \theta + b cos \theta + c $. Sines and cosines vary between $+1$ and $-1$ so the maximum value is equal to $\sqrt{a^2+b^2}+c$ and the minimum value is equal to $-\sqrt{a^2+b^2}+c$.
When the total angle equals $\pi / 2$ the sine is at its maximum value of $1$. This means a maximum is at $\theta+\phi = \pi / 2$ so $\theta=\pi / 2 - \phi$
The wave then drops through zero to its minimum value of $-1$ when the angle equals $3 \pi / 2$. This means a minimum is at $\theta+\phi = 3\pi / 2$ so $\theta=3\pi / 2-\phi$
There is an infinite number of maxima and minima. It is normal to give values in the range $0 ≤ \theta ≤ 2\pi$.
Example 8.5: Find the value and position of the first maximum and minimum of
$y= 4sin\theta+3cos \theta + 3$
$r$ $=$ $\sqrt{3^2 + 4^2}=5 $. and $\phi$ $=$ $tan^{-1}(3/4) = 0.64$ radians Maximum $=$ $5 + 3=8$ Minimum $=$ $-5 + 3=-2$ Maximum $=8$ and is at $\theta+0.64$ $=$ $\pi/2$ so $\theta_{max}$ $=$ $\pi/2 - 0.64=0.93$ radians Minimum $=-2$ and is at $\theta+0.64$ $=$ $3 \pi/2$ so $\theta_{min}$ $=$ $3 \pi/2-0.64=4.07$ radians
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It seems that commutative diagrams appeared sometime in the late 1940s -- for example, Eilenberg-McLane (1943) group cohomology paper does not have any, while the 1953 Hochschild-Serre paper does. Does anyone know who started using them (and how they convinced the printers to do this)?
I can muddy the waters...!
In a note of 3/20/1933 (
Nachlass, fasc. 449) and in a further undated note (fasc. 571), Hausdorff symbolized the functoriality property of homology (in our later terminology) with a commutative diagram of homomorphisms between the terms of two sequences of groups $(A_n)_{n\in\mathbf N}$, $(A'_n)_{n\in\mathbf N}$: ( Nachlass, fasc. 571, leaf 1).
Could this have, somehow, made its way out of Bonn (where Hausdorff lectured on combinatorial topology that year) and to Hurewicz, Eilenberg, Steenrod, et al.? This seems not impossible, as e.g. Tucker (1932, footnote 4) mentions discussion of Hausdorff letters in the Princeton seminar. Maybe also noteworthy: Hausdorff’s colleague and mentor in Bonn was Eduard Study, author of another early commutative diagram.
Addition: Six years earlier, Fritz London (son of yet another Bonn mathematician) already had, in Winkelvariable und kanonische Transformationen in der Undulationsmechanik, Z. f. Physik 40 (Dec. 1926) 193-210:
An excerpt from MacLane's "Categories for the working mathematician" (29-30)
The fundamental idea of representing a function by an arrow first appeared in topology about 1940, probably in papers or lectures by W. Hurewicz on relative homotopy groups; see [1941].
His initiative immediately attracted the attention of R.H. Fox (see Fox [1943]) and N.E. Steenrod, whose [1941] paper used arrows and (implicitly) functors; see also Hurewicz-Steenrod [1941]. The arrow $f: X\to Y$ rapidly displaced the occasional notation $f(X)\subset Y$ for a function. It expressed well a central interest of topology. Thus a notation (the arrow) led to a concept (category).
Commutative diagrams were probably also first used by Hurewicz.
Categories, functors and natural transformations themselves were discovered by Eilenberg-Mac Lane [1942a] in their study of limits (via natural transformations) for universal coefficient theorems for Čech cohomology. In this paper commutative diagrams appeared in print (probably for the first time). Thus Ext was one of the first functors considered. A direct treatment of categories in their own right appeared in Eilenberg- Mac Lane [1945]...
This last mentioned paper is the one referred to in KConrad's answer.
Here $g, g^*, g'$ are rays in space with polar planes $\gamma, \gamma^*, \gamma'$, $\mathfrak P$ is the “polarity” taking rays to planes and conversely, the $\mathfrak t_i$ are commuting collineations acting on both rays and planes, and Study uses $\mathfrak t_i\mathfrak P$ to denote the composition we would write $\mathfrak P\circ\mathfrak t_i$.
There's Russell's example from 1919, see here where conjugacy between relations is expressed diagrammatically.
Commutative diagrams really show their significance when dealing with categories, so I would guess they first appeared in that context. Look at the paper which first introduced categories: Eilenberg and Mac Lane's "General Theory of Natural Equivalences" (Trans. AMS) 58 (1945), 231--294. You can find that paper through JSTOR, but if you don't have access to JSTOR then get it at http://killingbuddha.altervista.org/FILOSOFIA/GToNe.pdf. In this paper, commutative diagrams are used only a few times and the authors don't say the diagrams
commute. Instead they use more elaborate descriptions like "the two paths around the diagram are equivalent". Even if this is not the first place a commutative diagram appeared, that they are used sparingly (considering the context) and described in an awkward way suggests it was one of the earliest times this concept jumped off the blackboard into print.
Weibel's paper on the history of homological algebra, which is available at his website, may provide some earlier sources if you track down references he gives.
EDIT: Perhaps the first book which made extensive use of commutative diagrams is Eilenberg and Steenrod's "Foundations of Algebraic Topology" (1952). On page xi, under the heading of a section they call New Methods, they introduce the term commutativity for certain diagrams of groups and homomorphisms.
Further to David Corfield's post: The diagram below appeared in Frege's 1893
Grundgesetze der Arithmetik (Vol. I, §§130, 172) and illustrates the construction of a mapping that Frege uses to prove that for any finite concept, the objectsfalling under it can be well-ordered.
Russell, of course, studied the
Grundgesetze closely; perhaps he found inspiration in Frege's figure (if not the almost childish arrows).
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15th century printers pioneered the use of movable type printing. Mirror images of each letter were cast in metal. These were arranged, letter by letter, in frames which printed one page at a time. A frame was called a matrix, several frames were called matrices. In mathematics, a matrix is a frame used to hold numbers or variables. Matrices were used before computers were invented but the advent of computers has made the use of matrices much more common.
A 2x2 matrix looks like this $\begin{bmatrix} 1 & 2 \\[0.3em] 4 & -3 \end{bmatrix}$
Matrices are always rectangular and can be any size. The size of an array is called its order.
Matrices are generally represented by capital letters and the individual numbers, called elements, are represented by lowercase letters. Annoyingly, the subscripts are written row then column.
$A=\begin{bmatrix} a_{11} & a_{12} \\[0.3em] a_{21} & a_{22} \end{bmatrix}$
Figure 9.0: Matrix $A$ with elements $a_{ij}$
You can add and subtract matrices of the same order. To add two matrices together you need to add the corresponding terms.
If $A=\begin{bmatrix} a_{11} & a_{12} \\[0.3em] a_{21} & a_{22} \end{bmatrix}$ and $B=\begin{bmatrix} b_{11} & b_{12} \\[0.3em] b_{21} & b_{22} \end{bmatrix}$ then $A+B$ $=$ $\begin{bmatrix} a_{11} & a_{12} \\[0.3em] a_{21} & a_{22} \end{bmatrix}+ \begin{bmatrix} b_{11} & b_{12} \\[0.3em] b_{21} & b_{22} \end{bmatrix}$ $=$ $\begin{bmatrix} a_{11}+b_{11} & a_{12}+b_{12} \\[0.3em] a_{21}+b_{11} & a_{22}+b_{22} \end{bmatrix}$ Example 9.1: Find $A+B$ given $A=\begin{bmatrix} 1 & 3 \\[0.3em] -5 & 2 \end{bmatrix}$ and $B=\begin{bmatrix} -2 & 4 \\[0.3em] 3 & 1 \end{bmatrix}$
$A+B$ $=$ $\begin{bmatrix} 1+-2 & 3+4 \\[0.3em] -5+3 & 2+1 \end{bmatrix}$ $A+B$ $=$ $\begin{bmatrix} -1 & 7 \\[0.3em] -2 & 3 \end{bmatrix}$
To subtract one matrix from another you need to subtract the corresponding terms.
If $A=\begin{bmatrix} a_{11} & a_{12} \\[0.3em] a_{21} & a_{22} \end{bmatrix}$ and $B=\begin{bmatrix} b_{11} & b_{12} \\[0.3em] b_{21} & b_{22} \end{bmatrix}$ then $A-B$ $=$ $\begin{bmatrix} a_{11} & a_{12} \\[0.3em] a_{21} & a_{22} \end{bmatrix}- \begin{bmatrix} b_{11} & b_{12} \\[0.3em] b_{21} & b_{22} \end{bmatrix}$ $=$ $\begin{bmatrix} a_{11}-b_{11} & a_{12}-b_{12} \\[0.3em] a_{21}-b_{11} & a_{22}-b_{22} \end{bmatrix}$ Example 9.1a: Find $A-B$ given $A=\begin{bmatrix} 1 & 3 \\[0.3em] -5 & 2 \end{bmatrix}$ and $B=\begin{bmatrix} -2 & 4 \\[0.3em] 3 & 1 \end{bmatrix}$
$A-B$ $=$ $\begin{bmatrix} 1-(-2) & 3-4 \\[0.3em] -5-3 & 2-1 \end{bmatrix}$ $A-B$ $=$ $\begin{bmatrix} 3 & -1 \\[0.3em] -8 & 1 \end{bmatrix}$
To multiply a matrix by a scalar you multiply each of the elements by the scalar.
If $A=\begin{bmatrix} a_{11} & a_{12} \\[0.3em] a_{21} & a_{22} \end{bmatrix}$ and $c$ is a scalar then $cA=\begin{bmatrix} ca_{11} & ca_{12} \\[0.3em] ca_{21} & ca_{22} \end{bmatrix}$
Example 9.2: Given $A=\begin{bmatrix} 1 & 3 \\[0.3em] -5 & 2 \end{bmatrix}$ and $c=3$ find $cA$.
$cA$ $=$ $\begin{bmatrix} 3\times1 & 3\times3 \\[0.3em] 3\times(-5) & 3\times2 \end{bmatrix}$ $cA$ $=$ $\begin{bmatrix} 3 & 9 \\[0.3em] -15 & 6 \end{bmatrix}$
You can multiply two matrices
iff (iff means if and only if) the number of columns in the first matrix is equal to the number of rows in the second.
If $A=\begin{bmatrix} a_{11} & a_{12} \\[0.3em] a_{21} & a_{22} \end{bmatrix}$ and $B= \begin{bmatrix} b_{11} & b_{12} \\[0.3em] b_{21} & b_{22} \end{bmatrix}$ then the matrices can be multiplied together because the number of columns in $A$ is the same as the number of rows in $B$.
To multiply $A$ by $B$ we take the first column of $B$ and put it over $A$, multiply then sum the corresponding terms.
If $A=\begin{bmatrix} a_{11} & a_{12} \\[0.3em] a_{21} & a_{22} \end{bmatrix}$ and $B=\begin{bmatrix} b_{11} & b_{12} \\[0.3em] b_{21} & b_{22} \end{bmatrix}$ then $A \times B$ $=$ $\begin{bmatrix} a_{11} & a_{12} \\[0.3em] a_{21} & a_{22} \end{bmatrix} \times \begin{bmatrix} b_{11} & b_{12} \\[0.3em] b_{21} & b_{22} \end{bmatrix}$ $=$ $\begin{bmatrix} a_{11} \times b_{11} + a_{12} \times b_{21} & a_{11} \times b_{12} + a_{12} \times b_{22} \\[0.3em] a_{21} \times b_{11} + a_{22} \times b_{21} & a_{21} \times b_{12} + a_{22} \times b_{22} \end{bmatrix}$ Example 9.3: Find $A \times B$ given $A=\begin{bmatrix} 1 & 3 \\[0.3em] -5 & 2 \end{bmatrix}$ and $B=\begin{bmatrix} -2 & 4 \\[0.3em] 3 & 1 \end{bmatrix}$
$A \times B$ $=$ $\begin{bmatrix} 1 & 3 \\[0.3em] -5 & 2 \end{bmatrix} \times \begin{bmatrix} -2 & 4 \\[0.3em] 3 & 1 \end{bmatrix}$ $=$ $\begin{bmatrix} 1 \times (-2) + 3 \times 3 & 1 \times 4 + 3 \times 1 \\[0.3em] -5 \times (-2) + 2 \times 3 & -5 \times 4 + 2 \times 1 \\[0.3em] \end{bmatrix}$ $A \times B$ $=$ $\begin{bmatrix} 7 & 7 \\[0.3em] 16 & -18 \end{bmatrix}$
Matrix division does not exist. If you want to find $X$ in the matrix equation $AX=C$ where $A$, $X$ and $C$ are matrices you first need to find the inverse of matrix $A$. When you multiply $A$ by it's inverse, $A^{-1}$, the result is the identity matrix, that is a matrix with $1$s on the main diagonal and $0$s everywhere else.
The inverse of $A$ is $A^{-1}$ where $A \times A^{-1} = I$
If $A$ was a $3\times3$ matrix $I$ would be $\begin{bmatrix} 1&0&0\\[0.3em] 0&1&0\\[0.3em] 0&0&1 \end{bmatrix}$
Having found $A^{-1}$ we can write:
$A X$ $=$ $C$ $A A^{-1} X$ $=$ $CA^{-1}$ $A A^{-1}$ $=$ $I$ so $X$ $=$ $CA^{-1}$
The determinant of a matrix can be thought of as the magnitude of the matrix. You can only calculate determinants for square matrices. For a 2x2 matrix, $A=\begin{bmatrix} a_{11} & a_{12} \\[0.3em] a_{21} & a_{22} \end{bmatrix}$ the determinant is given by $a_{11} \times a_{22} - a_{12} \times a_{21}$.
Example 9.5.1: Find the determinant of $A=\begin{bmatrix} 1 & 3 \\[0.3em] -5 & 2 \end{bmatrix}$.
$|A|$ $=$ $ 1 \times 2 - 3 \times (-5)$ $|A|$ $=$ $17$
To find the determinant of a 3x3 matrix take each term in the top row one at a time. Ignore the row and column that contain the term.
What is left is a 2x2 matrix that we can evaluate as shown above.
At the end of this first step we have
$a_{11} \times (a_{22} \times a_{33} - a_{23} \times a_{32})$
For the next step we take $a_{12}$. We ignore the top row and middle column which gives us
$a_{12} \times (a_{21} \times a_{33} - a_{23} \times a_{31})$
There is an added complication. The signs on the top row alternate as shown below.
This means we need to change the sign of the middle term. If it is negative we make it positive. If it is positive . . .
We repeat the process for the last term so we giving us the determinant of $|A|$ as:
$a_{11} \times (a_{22} \times a_{33} - a_{23} \times a_{32})$ $|A|$ $=$ $-a_{12} \times (a_{21} \times a_{33} - a_{23} \times a_{31})$ +$a_{13} \times (a_{21} \times a_{32} - a_{22} \times a_{31})$ Example 9.5.3: Find the determinant of $A=\begin{bmatrix} 2 & 5 & 3\\[0.3em] -1 & 6 & 2\\[0.3em] 3 & -1 & 2 \end{bmatrix}$.
$2 \times (6 \times 2 - 2 \times (-1))$ $|A|$ $=$ $-5 \times (-1 \times 2 - 2 \times 3)$ $+3 \times (-1 \times (-1) - 6 \times 3)$ $=$ $2 \times 14 -5 \times (-8) +3 \times (-17)$ $=$ $28 + 40 - 51$ $|A|$ $=$ $17$
The question should really be 'Where do we not use matrices?' There was a time when matrices were only used in specialist applications like mathematics and quantum mechanics but since computers became commonplace matrices underpin pretty much every computer application.
The name of the popular program MatLab is a concatenation of Matrix Laboratory. Artificial Intelligence, neural networks, statistical, drafting, analysis and modelling software are all based on matrices.
Computer graphics would be virtually impossible without the use of matrices. You can easily imagine a computer display as a 2D matrix but computer graphics, using matrices, take it a lot further than that. Movement in 3D, reflection of light, the shadows are all calculated using matrices. In the still from the film Gravity below the only real items in the image are the faces of the actors. Everything else looks real, moves as if it is real but is computer generated.
Imagine you have the 2D coordinates of an object in an array $A$. To scale the object you would multiply $A$ by the scaling matrix $S=\begin{bmatrix}S_x&0\\[0.3em]0&S_y\end{bmatrix}$.
To rotate $A$ you would multiply by the rotation matrix $R=\begin{bmatrix}cos(\theta)&sin(\theta)\\[0.3em]-sin(\theta)&cos(\theta)\end{bmatrix}$
This video shows how characters are animated by the film company Pixar.
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The Annals of Probability Ann. Probab. Volume 42, Number 1 (2014), 354-397. Explicit rates of approximation in the CLT for quadratic forms Abstract
Let $X,X_{1},X_{2},\ldots$ be i.i.d. ${\mathbb{R}}^{d}$-valued real random vectors. Assume that ${\mathbf{E}X=0}$, $\operatorname{cov} X=\mathbb{C}$, $\mathbf{E}\Vert X\Vert^{2}=\sigma ^{2}$ and that $X$ is not concentrated in a proper subspace of $\mathbb{R}^{d}$. Let $G$ be a mean zero Gaussian random vector with the same covariance operator as that of $X$. We study the distributions of nondegenerate quadratic forms $\mathbb{Q}[S_{N}]$ of the normalized sums ${S_{N}=N^{-1/2}(X_{1}+\cdots+X_{N})}$ and show that, without any additional conditions,
\[\Delta_{N}\stackrel{\mathrm{def}}{=}\sup_{x}\bigl|\mathbf{P}\bigl\{\mathbb{Q}[S_{N}]\leq x\bigr\}-\mathbf{P}\bigl\{\mathbb{Q}[G]\leq x\bigr\}\bigr|={\mathcal{O}}\bigl(N^{-1}\bigr),\]
provided that $d\geq5$ and the fourth moment of $X$ exists. Furthermore, we provide explicit bounds of order ${\mathcal{O}}(N^{-1})$ for $\Delta_{N}$ for the rate of approximation by short asymptotic expansions and for the concentration functions of the random variables $\mathbb{Q}[S_{N}+a]$, $a\in{\mathbb{R}}^{d}$. The order of the bound is optimal. It extends previous results of Bentkus and Götze [
Probab. Theory Related Fields 109 (1997a) 367–416] (for ${d\ge9}$) to the case $d\ge5$, which is the smallest possible dimension for such a bound. Moreover, we show that, in the finite dimensional case and for isometric $\mathbb{Q}$, the implied constant in ${\mathcal{O}}(N^{-1})$ has the form $c_{d}\sigma ^{d}(\det\mathbb{C})^{-1/2}\mathbf{E} \|\mathbb{C}^{-1/2}X\|^{4}$ with some $c_{d}$ depending on $d$ only. This answers a long standing question about optimal rates in the central limit theorem for quadratic forms starting with a seminal paper by Esséen [ Acta Math. 77 (1945) 1–125]. Article information Source Ann. Probab., Volume 42, Number 1 (2014), 354-397. Dates First available in Project Euclid: 9 January 2014 Permanent link to this document https://projecteuclid.org/euclid.aop/1389278527 Digital Object Identifier doi:10.1214/13-AOP839 Mathematical Reviews number (MathSciNet) MR3161488 Zentralblatt MATH identifier 1290.60021 Citation
Götze, Friedrich; Zaitsev, Andrei Yu. Explicit rates of approximation in the CLT for quadratic forms. Ann. Probab. 42 (2014), no. 1, 354--397. doi:10.1214/13-AOP839. https://projecteuclid.org/euclid.aop/1389278527
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In [20], the existence of Coulomb gauge with respect to a “small” connection on a disk plays an important role in the existence of conservation law. It is quite natural to expect “good estimate” under “good gauge,” and then the existence of conservation law can be obtained from a fixed point argument of systems of elliptic partial differential equations.
One can also obtain the above existence of Coulomb gauge by heat flow method. In fact, as a result of the existence of generalized solution of (1.3) when applied to a special case, we will present a flow version proof of the existence (cf. ([23], Theorem 2.1; [5], Proposition 2)).
In what follows, we will take \(\Sigma \)
to be a small disk (dimension two), not necessarily with Euclidean metric. A connection of vector bundle
E
over \(\Sigma \)
, where
E
is with rank
m
and structure group \(\mathrm {SO}(m)\)
, can be written as \(d+\Omega \)
, where \(\Omega \in \mathfrak {so}(m)\otimes T^*\Sigma \)
is a 1-form with value in \(\mathfrak {so}(m)\)
, then for any gauge transformation
S
,
Thus, if one sets \(A_0=d\)
as a trivial connection and \(A=d+\Omega \)
, the existence of generalized solution of (1.3
) implies that there exists some \(S\in C^\infty (\Sigma ,\mathrm {SO}(m))\)
, such that
$$\begin{aligned} 0=\nabla ^*_{d+\tilde{\Omega }}\tilde{\Omega }=\nabla _d^*\tilde{\Omega }+\left\{ \tilde{\Omega },\tilde{\Omega }\right\} =d^*\tilde{\Omega }. \end{aligned}$$
By Poincaré Lemma, there exists some \(\xi \in C^\infty (\Sigma , \mathfrak {so}(m))\)
, such that
$$\begin{aligned} {\left\{ \begin{array}{ll} \nabla ^\perp \xi =S^{-1}dS+S^{-1}\Omega S, &{}\quad x\in \Sigma \\ \xi =0, &{}\quad x\in \partial \Sigma . \end{array}\right. } \end{aligned}$$
(A.1)
The boundary condition holds since the flow implies that on \(\partial \Sigma \)
,
which shows that \(\xi \)
is a constant along \(\partial \Sigma \)
, and one can make \(\xi \)
vanish on boundary since it is determined up to a constant.
To show the estimates hold in Corollary D
, we need to construct \(S_0\)
properly. Parameterize Open image in new window
as \(x=(y,r)\)
, where
r
is the distance to \(\partial \Sigma \)
and
y
is the parameter of \(\partial \Sigma \)
. Firstly, let us assume that \(\Omega \in C^\infty (\Sigma ,\mathfrak {so}(m)\otimes T^*\Sigma )\)
and set the initial value as
$$\begin{aligned} S_0(x)=\exp \left\{ (1-\eta )r\Omega _\nu (x)\right\} . \end{aligned}$$
where \(\Omega _\nu (x)=-\Omega (r,y)(\frac{\partial }{\partial {r}})\)
, and \(\eta \in C_0^\infty (\Sigma )\)
, with \(\eta =0\)
on \(\Sigma _\delta \)
and \(\eta =1\)
on \(\Sigma {\setminus }\Sigma _{2\delta }\)
; moreover, one can require \(|\nabla \eta |\le 2/\delta \)
. To see the compatibility at the corner, note that, on the corner \(\partial \Sigma \times \left\{ 0\right\} \)
, \(S=S_0=\mathrm{Id}\)
, where \(\mathrm{Id}\)
is the constant section over \(\partial \Sigma \)
, which maps each point to the identity of
G
,
The energy of \(S_0\)
can be controlled as
$$\begin{aligned} 2\mathcal {E}(S_0) =\int _\Sigma |S_0^{-1}\mathrm{d}S_0+S_0^{-1}\Omega S_0|^2 \le \int _\Sigma (|\mathrm{d}S_0|^2+|\Omega |^2) =\Vert \mathrm{d}S_0\Vert _2^2+\Vert \Omega \Vert _2^2. \end{aligned}$$
We claim that \(\mathcal {E}(S_0)\)
can be taken as small as wanted. In fact, note that for \(x\in \Sigma _{2\delta }\)
, the derivative of \(\exp \)
is uniformly bounded with respect to \(\delta \)
, thus
$$\begin{aligned} |dS_0|\le C\left( (1-\eta )|\Omega _\nu |+r\left( |\nabla \eta ||\Omega _\nu |+|d\Omega _\nu |\right) \right) , \end{aligned}$$
and
$$\begin{aligned} |dS_0|^2\le C|\Omega |_{1,2}^2\left( 1+r^2|\nabla \eta |^2+r^2\right) . \end{aligned}$$
Finally,
$$\begin{aligned} \int _\Sigma |dS_0|^2&=\int _{\Sigma _{2\delta }{\setminus }\Sigma _{\delta }}|dS_0|^2 \le C\int _{\Sigma _{2\delta }{\setminus }\Sigma _{\delta }} |\Omega |_{1,2}^2(1+\delta ^2\cdot 1/\delta ^2+\delta ^2), \end{aligned}$$
which tends to 0 as \(\delta \rightarrow 0\)
by the absolute continuity of integration.
Thus, for any \(\varepsilon >0\)
, one can construct \(S_0\)
, such that \(\Vert dS_0\Vert _2\le \varepsilon \)
. Since the energy is monotonically decreasing along the flow,
$$\begin{aligned} \Vert \tilde{\Omega }\Vert _2^2=2\mathcal {E}(S)\le 2\mathcal {E}(S_0) \le \Vert \Omega \Vert _2^2+\varepsilon ^2. \end{aligned}$$
Moreover, since \(\nabla ^\perp \xi =S^{-1}dS+S^{-1}\Omega S\)
,
$$\begin{aligned} \Vert dS\Vert _2\le \Vert \Omega \Vert _2+\Vert \nabla ^\perp \xi \Vert _2 =\Vert \Omega \Vert _2+\Vert \tilde{\Omega }\Vert _2 \le 2\Vert \Omega \Vert _2+\varepsilon . \end{aligned}$$
By Poincaré inequality, one concludes that
$$\begin{aligned} \Vert dS\Vert _2+\Vert \xi \Vert _{1,2}\le C(m)\Vert \Omega \Vert _2. \end{aligned}$$
To show the required estimate for \(\Omega \in L^2(\Sigma ,\mathfrak {so}(m)\otimes T^*\Sigma )\)
, let us approximate it by \(\Omega _k\in C^\infty (\Sigma ,\mathfrak {so}(m)\otimes T^*\Sigma )\)
in \(L^2\)
with \(\Vert \Omega _k\Vert _2\le \Vert \Omega \Vert _2\)
. For \(A_0=\mathrm{d}\)
, Open image in new window
, Open image in new window
, run the gauge transformation heat flow,
One obtains \(S_k\in W^{1,2}(D,\mathrm {SO}(m))\cap C^\infty \)
, \(\xi _k\in W^{1,2}(D,\mathrm {SO}(m))\cap C^\infty \)
, which solves (A.1
), with
$$\begin{aligned} \Vert dS_k\Vert _2+\Vert \xi _k\Vert _{1,2}\le C(m)\Vert \Omega _k\Vert _2. \end{aligned}$$
Therefore, \(S_k\)
, \(\xi _k\)
are \(W^{1,2}\)
bounded, the weak compactness implies that \(S_k\)
, \(\xi _k\)
converge weakly to some \(S_\infty \)
, \(\xi _\infty \)
in \(W^{1,2}\)
, respectively. Take limits in (A.1
), one finds the required gauge. The weakly lower semi-continuity implies the required estimate, and we finish the proof of Corollary D
.
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For simplicity, consider in $\mathbb{R}^3$, and the Fourier transform of the following function $$f=(x_1+x_2+x_3)(1+|x|^2+x_1^2(x_2^2+x_3^2)+x_2^2x_3^2)^{-t+is},~~ \frac12<t<1,~~s\in \mathbb{R}.$$
Note that the function is not in $L^1$, in fact $f$ behaves like $|x|^{-2t+1}$ at $\infty$. So the Fourier transform is taken in the sense of tempered distribution.
Note also that the function is not radial symmetric, however $f$ is symmetric in the sense that if any of the variables are interchanged, one obtains the same function. I'm interested in the quantitative behavior of $\hat{f}$ near the origin. Especially, for what values of $\sigma_i\ge 0$,$i=1,2,3,$ such that $$ \hat{f}(\xi)\leq C|\xi_1|^{-\sigma_1}|\xi_2|^{-\sigma_2}|\xi_3|^{-\sigma_3},~~|\xi|<1. $$ The following is an "unsuccessful" approach. First let's estimate the Fourier transform of $f(x)=1/{(1+|x|^2+x_1^2(x_2^2+x_3^2)+x_2^2x_3^2)^{t+is}}$. Then one can write $$ f(x)=\frac{1}{(B^2+A^2x_1^2)^{t+is}}, $$ where$ A^2(x_2, x_3)=1+x_2^2+x_3^2$, $B^2(x_2, x_3)=A^2+x_2^2 x_3^2$. Recall that \begin{align}\label{equ4.3} \mathcal{F}^{-1}(\frac{1}{(1+|x|^2)^z})(\xi)=\pi^{-\frac{n}{2}}2^{-(\frac{n}{2}+z-1)}\frac{1}{\Gamma(z)}|\xi|^{z-\frac{n}{2}}K_{\frac{n}{2}-z}(|\xi|), \end{align} where $K_v$ denotes the modified Bessel function of the second kind. $K_v$ satisfies \begin{align}\label{equ4.4} \frac{d}{dz}(z^{-v}K_v(z))=z^{-v}K_{v+1}(z), ~~~ \forall v\in \mathbb{C}. \end{align}
Combine the above two formula and by scaling, one has \begin{align}\label{equ4.6} I(\xi)&\triangleq \int_{\mathbb{R}^3}{e^{ix\cdot\xi}f(x)dx}\nonumber\\ &=\pi^{-\frac{1}{2}}2^{\frac12-z}\frac{1}{\Gamma(z)} \int_{\mathbb{R}^{2}}{e^{ix'\cdot\xi'}B^{1-2z}A^{-1}(\frac{B}{A}|\xi_1|)^{z-\frac12}K_{\frac12-z}(\frac{B}{A}|\xi_1|)dx'}\nonumber\\ &=\pi^{-\frac{1}{2}}2^{\frac12-z}\frac{|\xi_1|^{z-\frac12}}{\Gamma(z)} \int_{\mathbb{R}^{2}}{e^{ix'\cdot\xi'}A^{-2z}(x')(\frac{B}{A})^{\frac12-z}K_{\frac12-z}(\frac{B}{A}|\xi_1|)}dx', \end{align} In order to take out the factor $\frac{B}{A}$ from $K_{\frac12-z}(\frac{B}{A}|\xi_1|)$, we can use the following multiplication theorem of Bessel function \begin{align}\label{equ4.7} \lambda^{\nu}K_{\nu}(\lambda z)=\sum_{l=0}^{\infty}\frac{(-1)^l}{l!}(\frac{(\lambda^2-1)z}{2})^lK_{\nu-l}(z),~~~\lambda>0. \end{align} However, the estimate is fine with respect to $\xi_1$, but it get trouble with $\xi'$ since now it becomes \begin{align} I(\xi)&=c_z\sum_{l=0}^{\infty}\frac{(-1)^l}{2^ll!}(\int_{\mathbb{R}^{n-1}}{e^{ix'\cdot\xi'}\frac{x_2^{2l}x_3^{2l}}{A^{2(z+l)}}} ~~~dx')|\xi_1|^{z-\frac12+l}K_{\frac12-z-l}(|\xi_1|) \end{align} It seems that the integral involved in the series above is too singular that doesn't imply the convergence. I'm not sure how to overcome this issue.
Any reference or comment about this topic is welcome!
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I put together a series of demos for a group of epidemiology students who are studying causal mediation analysis. Since mediation analysis is not always so clear or intuitive, I thought, of course, that going through some examples of simulating data for this process could clarify things a bit.
Quite often we are interested in understanding the relationship between an exposure or intervention on an outcome. Does exposure \(A\) (could be randomized or not) have an effect on outcome \(Y\)?
But sometimes we are interested in understanding
more than whether or not \(A\) causes or influences \(B\); we might want to have some insight into the mechanisms underlying that influence. And this is where mediation analysis can be useful. (If you want to delve deeply into the topic, I recommend you check out this book by Tyler VanderWeele, or this nice website developed at Columbia University.)
In the example here, I am using the simplest possible scenario of an exposure \(A\), a mediator \(M\), and an outcome \(Y\), without any confounding:
A key challenge of understanding and conducting a mediation analysis is how we should
quantify this concept of mediation. Sure, \(A\) has an effect on \(M\), which in turn has an effect on \(Y\), and \(A\) also may have an effect on \(Y\) through other pathways. But how can we make sense of all of this? One approach, which is a relatively recent development, is to use the potential outcome framework of causal inference to define the various estimands (or quantities) that arise in a mediation analysis. (I draw on a paper by Imai, Keele and Yamamoto for the terminology, as there is not complete agreement on what to call various quantities. The estimation methods and software used here are also described in the paper.) Defining the potential outcomes
In an earlier post, I described the concept of potential outcomes. I extend that a bit here to define the quantities we are interested in. In this case, we have two effects of the possible exposure: \(M\) and \(Y\). Under this framework, each individual has a potential outcome for each level of \(A\) (I am assuming \(A\) is binary). So, for the mediator, \(M_{i0}\) and \(M_{i1}\) are the values of \(M\) we would observe for individual \(i\) without and with exposure, respectively. That is pretty straightforward. (From here on out, I will remove the subscript \(i\), because it gets a little unwieldy.)
The potential outcomes under \(Y\) are less intuitive, as there are four of them. First, there is \(Y_{0M_0}\), which is the potential outcome of \(Y\)
without exposure for \(A\) and whatever the potential outcome for \(M\) is without exposure for \(A\). This is what we observe when \(A=0\) for an individual. \(Y_{1M_1}\) is the potential outcome of \(Y\) with exposure for \(A\) and whatever the potential outcome for \(M\) is with exposure for \(A\). This is what we observe when \(A=1\) for an individual. That’s all fine.
But we also have \(Y_{0M_1}\), which can never be observed unless we can intervene on the mediator \(M\) somehow. This is the potential outcome of \(Y\)
without exposure for \(A\) and whatever the mediator would have been had the individual been exposed. This potential outcome is controversial, because it is defined across two different universes of exposure to \(A\). Finally, there is \(Y_{1M_0}\). It is analogously defined across two universes, but in reverse. Defining the causal mediation effects and direct effects
The estimands or quantities that we are interested are defined in terms of the potential outcomes. The
for an individual are causal mediation effects
\[
\begin{aligned} CME_0 &= Y_{0M_1} – Y_{0M_0} \\ CME_1 &= Y_{1M_1} – Y_{1M_0}, \end{aligned} \]
and the
are causal direct effects
\[
\begin{aligned} CDE_0 &= Y_{1M_0} – Y_{0M_0} \\ CDE_1 &= Y_{1M_1} – Y_{0M_1}. \end{aligned} \]
A few important points. (1) Since we are in the world of potential outcomes, we do not observe these quantities for everyone. In fact, we don’t observe these quantities for anyone, since some of the measures are across two universes. (2) The two causal mediation effects under do not need to be the same. The same goes for the two causal direct effects. (3) Under a set of pretty strong assumptions related to unmeasured confounding, independence, and consistency (see
Imai et al for the details), the average causal mediation effects and average causal direct effects can be estimated using observed data only. Before I simulate some data to demonstrate all of this, here is the definition for the (and its decomposition into mediation and direct effects): total causal effect
\[
\begin{aligned} TCE &= Y_{1M_1} – Y_{0M_0} \\ &= CME_1 + CDE_0 \\ &= CME_0 + CDE_1 \end{aligned} \] Generating the data
I’m using the
simstudy package to generate the data. I’ll start by generating the binary potential outcomes for the mediator, \(M_0\) and \(M_1\), which are correlated in this example. \(P(M_1=1) > P(M_0=1)\), implying that exposure to \(A\) does indeed have an effect on \(M\). Note that it is possible that for an individual \(M_0 = 1\) and \(M_1 = 0\), so that exposure to \(A\) has an effect contrary to what we see in the population generally. (We don’t need to make this assumption in the data generation process; we could force \(M_1\) to be 1 if \(M_0\) is 1.)
set.seed(3872672)dd <- genCorGen(n=5000, nvars = 2, params1 = c(.2, .6), dist = "binary", rho = .3, corstr = "cs", wide = TRUE, cnames = c("M0", "M1"))
Observe treatment:
dd <- trtObserve(dd, 0.6, grpName = "A")
Initial data set:
dd
## id A M0 M1## 1: 1 0 0 1## 2: 2 1 0 0## 3: 3 1 0 1## 4: 4 0 1 1## 5: 5 1 0 0## --- ## 4996: 4996 1 0 1## 4997: 4997 0 0 0## 4998: 4998 1 1 1## 4999: 4999 1 1 0## 5000: 5000 0 0 0
\(Y_{0M_0}\) is a function of \(M_0\) and some noise \(e_0\), and \(Y_{0M_1}\) is a function of \(M_1\) and the same noise (this is not a requirement). However, if \(M_0 = M_1\) (i.e. the mediator is not affected by exposure status), then I am setting \(Y_{0M_1} = Y_{0M_0}\). In this case, \(CME_0\) for an individual is \(2(M_1 – M_0)\), so \(CME_0 \in \{-2, 0, 2\}\), and the population average \(CME_0\) will depend on the mixture of potential outcomes \(M_0\) and \(M_1\).
def <- defDataAdd(varname = "e0", formula = 0, variance = 1, dist = "normal")def <- defDataAdd(def, varname = "Y0M0", formula = "2 + M0*2 + e0", dist = "nonrandom")def <- defDataAdd(def, varname = "Y0M1", formula = "2 + M1*2 + e0", variance = 1, dist = "nonrandom")
The same logic holds for \(Y_{1M_0}\) and \(Y_{1M_1}\), though at the individual level \(CME_1 \in \{-5, 0, 5\}\):
def <- defDataAdd(def, varname = "e1", formula = 0, variance = 1, dist = "normal")def <- defDataAdd(def, varname = "Y1M0", formula = "8 + M0*5 + e1", dist = "nonrandom")def <- defDataAdd(def, varname = "Y1M1", formula = "8 + M1*5 + e1", dist = "nonrandom")
The
observed mediator (\(M\)) and outcome (\(Y\)) are determined by the observed exposure (\(A\)).
def <- defDataAdd(def, varname = "M", formula = "(A==0) * M0 + (A==1) * M1", dist = "nonrandom")def <- defDataAdd(def, varname = "Y", formula = "(A==0) * Y0M0 + (A==1) * Y1M1", dist = "nonrandom")
Here is the entire data definitions table:
varname formula variance dist link e0 0 1 normal identity Y0M0 2 + M0*2 + e0 0 nonrandom identity Y0M1 2 + M1*2 + e0 1 nonrandom identity e1 0 1 normal identity Y1M0 8 + M0*5 + e1 0 nonrandom identity Y1M1 8 + M1*5 + e1 0 nonrandom identity M (A==0) * M0 + (A==1) * M1 0 nonrandom identity Y (A==0) * Y0M0 + (A==1) * Y1M1 0 nonrandom identity
Based on the parameters used to generate the data, we can calculate the expected causal mediation effects:
\[
\begin{aligned} E[CME_0] &= E[2 + 2M_1 + e_0] – E[2+2M_0+e_0] \\ &= E[2(M_1 – M_0)] \\ &= 2(E[M_1] – E[M_0]) \\ &= 2(0.6 – 0.2) \\ &= 0.8 \end{aligned} \]
\[
\begin{aligned} E[CME_1] &= E[8 + 5M_1 + e_1] – E[8+5M_0+e_1] \\ &= E[5(M_1 – M_0)] \\ &= 5(E[M_1] – E[M_0]) \\ &= 5(0.6 – 0.2) \\ &= 2.0 \end{aligned} \]
Likewise, the expected values of the causal direct effects can be calculated:
\[
\begin{aligned} E[CDE_0] &= E[8 + 5M_0 + e_1] – E[2+2M_0+e_0] \\ &= E[6 + 5M_0 – 2M_0)] \\ &= 6 + 3M_0 \\ &= 6 + 3*0.2 \\ &= 6.6 \end{aligned} \]
\[
\begin{aligned} E[CDE_1] &= E[8 + 5M_1 + e_1] – E[2+2M_1+e_0] \\ &= E[6 + 5M_1 – 2M_1)] \\ &= 6 + 3M_1 \\ &= 6 + 3*0.6 \\ &= 7.8 \end{aligned} \]
Finally, the expected total causal effect is:
\[
\begin{aligned} ATCE &= E[CDE_0] + E[CME_1] = 6.6 + 2.0 \\ &= E[CDE_1] + E[CME_0] = 7.8 + 0.8 \\ &= 8.6 \end{aligned} \] And now, the complete data set can be generated.
dd <- addColumns(def, dd)dd <- delColumns(dd, c("e0", "e1")) # these are not neededdd
## id A M0 M1 Y0M0 Y0M1 Y1M0 Y1M1 M Y## 1: 1 0 0 1 0.933 2.93 7.58 12.58 0 0.933## 2: 2 1 0 0 2.314 2.31 6.84 6.84 0 6.841## 3: 3 1 0 1 3.876 5.88 9.05 14.05 1 14.053## 4: 4 0 1 1 5.614 5.61 12.04 12.04 1 5.614## 5: 5 1 0 0 1.469 1.47 8.81 8.81 0 8.809## --- ## 4996: 4996 1 0 1 2.093 4.09 8.82 13.82 1 13.818## 4997: 4997 0 0 0 1.734 1.73 7.28 7.28 0 1.734## 4998: 4998 1 1 1 3.256 3.26 12.49 12.49 1 12.489## 4999: 4999 1 1 0 5.149 3.15 12.57 7.57 0 7.572## 5000: 5000 0 0 0 1.959 1.96 5.23 5.23 0 1.959
Looking at the “observed” potential outcomes
The advantage of simulating data is that we can see what the average causal effects are based on the potential outcomes. Here are the average potential outcomes in the generated data set:
dd[,.( Y0M0 = mean(Y0M0), Y0M1 = mean(Y0M1), Y1M0 = mean(Y1M0), Y1M1 = mean(Y1M1))]
## Y0M0 Y0M1 Y1M0 Y1M1## 1: 2.39 3.2 8.99 11
The four average causal effects based on the data are quite close to the expected values:
dd[, .(ACME0 = mean(Y0M1 - Y0M0), ACME1= mean(Y1M1 - Y1M0), ACDE0 = mean(Y1M0 - Y0M0), ACDE1= mean(Y1M1 - Y0M1))]
## ACME0 ACME1 ACDE0 ACDE1## 1: 0.81 2.03 6.6 7.81
And the here is the average total causal effect from the data set:
dd[, mean(Y1M1 - Y0M0)]
## [1] 8.62
All of these quantities can be visualized in this figure. The lengths of the solid vertical lines are the mediated effects. The lengths of the dotted vertical lines are the direct effects. And the sums of these vertical lines (by color) each represent the total effect:
Estimated causal mediation effect from observed data
Clearly, the real interest is in estimating the causal effects from data that we can actually observe. And that, of course, is where things start to get challenging. I will not go into the important details here (again,
Imai et al provide these), but here are formulas that have been derived to estimate the effects (simplified since there are no confounders in this example) and the calculations using the observed data:
\[\small
\hat{CME_0} =\sum_{m\in0,1}E[Y|A=0, M=m][P(M=m|A=1)-P(M=m|A=0)] \]
# Estimate CME0dd[M == 0 & A == 0, mean(Y)] * (dd[A == 1, mean(M == 0)] - dd[A == 0, mean(M == 0)]) +dd[M == 1 & A == 0, mean(Y)] * (dd[A == 1, mean(M == 1)] - dd[A == 0, mean(M == 1)])
## [1] 0.805
\[\small
\hat{CME_1} =\sum_{m\in0,1}E[Y|A=1, M=m][P(M=m|A=1)-P(M=m|A=0)] \]
# Estimate CME1dd[M == 0 & A == 1, mean(Y)] * (dd[A == 1, mean(M == 0)] - dd[A == 0, mean(M == 0)]) +dd[M == 1 & A == 1, mean(Y)] * (dd[A == 1, mean(M == 1)] - dd[A == 0, mean(M == 1)])
## [1] 2
\[\small
\hat{CDE_0} =\sum_{m\in0,1}(E[Y|A=1, M=m] – E[Y|A=0, M=m])P(M=m|A=0) \]
# Estimate CDE0(dd[M == 0 & A == 1, mean(Y)] - dd[M == 0 & A == 0, mean(Y)]) * dd[A == 0, mean(M == 0)] +(dd[M == 1 & A == 1, mean(Y)] - dd[M == 1 & A == 0, mean(Y)]) * dd[A == 0, mean(M == 1)]
## [1] 6.56
\[\small
\hat{CDE_1} =\sum_{m\in0,1}(E[Y|A=1, M=m] – E[Y|A=0, M=m])P(M=m|A=1) \]
# Estimate CDE1(dd[M == 0 & A == 1, mean(Y)] - dd[M == 0 & A == 0, mean(Y)]) * dd[A == 1, mean(M == 0)] +(dd[M == 1 & A == 1, mean(Y)] - dd[M == 1 & A == 0, mean(Y)]) * dd[A == 1, mean(M == 1)]
## [1] 7.76
Estimation with mediation package
Fortunately, there is software available to provide these estimates (and more importantly measures of uncertainty). In
R, one such package is
mediation, which is available on CRAN. This package implements the formulas derived in the
Imai et al paper.
Not surprisingly, the model estimates are in line with expected values, true underlying effects, and the previous estimates conducted by hand:
library(mediation)med.fit <- glm(M ~ A, data = dd, family = binomial("logit"))out.fit <- lm(Y ~ M*A, data = dd)med.out <- mediate(med.fit, out.fit, treat = "A", mediator = "M", robustSE = TRUE, sims = 1000)summary(med.out)
## ## Causal Mediation Analysis ## ## Quasi-Bayesian Confidence Intervals## ## Estimate 95% CI Lower 95% CI Upper p-value ## ACME (control) 0.8039 0.7346 0.88 <2e-16 ***## ACME (treated) 2.0033 1.8459 2.16 <2e-16 ***## ADE (control) 6.5569 6.4669 6.65 <2e-16 ***## ADE (treated) 7.7563 7.6555 7.86 <2e-16 ***## Total Effect 8.5602 8.4317 8.69 <2e-16 ***## Prop. Mediated (control) 0.0940 0.0862 0.10 <2e-16 ***## Prop. Mediated (treated) 0.2341 0.2179 0.25 <2e-16 ***## ACME (average) 1.4036 1.2917 1.52 <2e-16 ***## ADE (average) 7.1566 7.0776 7.24 <2e-16 ***## Prop. Mediated (average) 0.1640 0.1524 0.17 <2e-16 ***## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## ## Sample Size Used: 5000 ## ## ## Simulations: 1000
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Carter-Wegman polynomial authenticators
For a given finite field $(\mathbb F,+,\times)$ of $f$ elements, define a Carter-Wegman polynomial authenticator for a message $M=m_1\|m_2\|\dots\|m_l$ of $l$ symbols in $\mathbb F$ as $$H_{(r,s)}(M)=s+\sum_{i=1}^l m_{(l+1-i)}\cdot r^i$$ where $r$ and $s$ are uniformly random independent secrets over $\mathbb F$ (I'm restricting to neither $r$ nor $s$ reused, even though $r$ could).
Common fields $(\mathbb F,+,\times)$ are $GF(2^b)$ (equivalently: field $(\{0,1\}^b,\oplus,\times)$ where $\times$ is binary polynomial multiplication followed by reduction modulo a public irreducible polynomial of degree $b$ ), and $(\mathbb Z_f,+,\times)$ for prime $f$. AES-GCM uses $GF(2^{128})$ with polynomial $x^{128}+x^7+x^2+x+1$. Poly1305 uses $\mathbb Z_f$ with prime $f=2^{130}-5$ (and some restriction of the domain of $r$, and of the $m_i$'s).
Questions Tight security bound for elementary attack model: for fixed public $l$ and $(\mathbb F,+,\times)$ of $f\gg l$ elements, an adversary chooses a message $M$ of $l$ symbols, obtains $H_{(r,s)}(M)$ for fresh uniformly random secrets $r$ and $s$, and makes one attempt to produce $H_{(r,s)}(M')$ for $M'\ne M$ of his/her choice (also of $l$ symbols). What's a tight upper bound of probability $\epsilon$ of success (undetected forgery) as a function of $f$ and $l$? Do we reduce $\epsilon$ by excluding some $r$, e.g. requiring $r\ne 0$ ? Are some fields better than others ? Concatenation of authenticators using the same field: we consider $H_{(r,s,r',s')}(M)=H_{(r,s)}(M)\;\|\;H_{(r',s')}(M)$, with the $r$, $s$, $r'$, $s'$ uniformly random independent one-time secrets in $\mathbb F$. What's a new tight upper bound for probability $\epsilon$ of undetected forgery as a function of $f$ and $l$? Concatenation of authenticators using different fields: we consider $H_{(r,s,r',s')}(M)=H_{(r,s)}(M)\;\|\;H'_{(r',s')}(M)$, with $H$ (resp. $H'$) using field $\mathbb F$ of $f$ elements (resp. $\mathbb F'$ of $f'$ elements), with $f'\gtrsim f$ and some public way to plunge $m_i$ into $\mathbb F'$ , $r$ and $s$ uniformly random independent one-time secrets in $\mathbb F$, $r'$ and $s'$ uniformly random independent one-time secrets in $\mathbb F'$. What's a tight upper bound for probability $\epsilon$ of undetected forgery as a function of $f$, $f'$ and $l$? Is this lower than in 2. ? Motivation
Wide Carter-Wegman polynomial authenticators are non-trivial to implement both efficiently and portably: portable C has no semantic for carry-less multiplication, and no type wider than 64 bits. A generic Poly1305 uses 25 integer multiplications plus significant additions and shifts for 128 bits of message processed (or double arithmetic which a lot of low-end platforms do not have in hardware). It is thus tempting to concatenate narrower authenticators, which are easy and usually efficient to implement: C99 provides arithmetic in $\mathbb Z_f$ for $f<2^{32}$, using the
w = ( (uint64_t)u * v ) % f; semantic, and many hardware+compilers have good support for that (or perhaps
w = ( (int64_t)u * v ) % f; for $f<2^{31}$ ).
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Quasi-linear Schrödinger–Poisson system under an exponential critical nonlinearity: existence and asymptotic behaviour of solutions
Article
First Online:
74 Downloads Abstract
In this paper we consider the following quasilinear Schrödinger–Poisson system in a bounded domain in \({\mathbb {R}}^{2}\):
depending on the parameter \(\varepsilon >0\). The nonlinearity
$$\begin{aligned} \left\{ \begin{array}[c]{ll} - \Delta u +\phi u = f(u) &{}\ \text{ in } \Omega , \\ -\Delta \phi - \varepsilon ^{4}\Delta _4 \phi = u^{2} &{} \ \text{ in } \Omega ,\\ u=\phi =0 &{} \ \text{ on } \partial \Omega \end{array} \right. \end{aligned}$$
fis assumed to have critical exponential growth. We first prove existence of nontrivial solutions \((u_{\varepsilon }, \phi _{\varepsilon })\) and then we show that as \(\varepsilon \rightarrow 0^{+}\), these solutions converge to a nontrivial solution of the associated Schrödinger–Poisson system, that is, by making \(\varepsilon =0\) in the system above. KeywordsVariational methods Nonlocal problems Schrödinger–Poisson equation Exponential critical growth Mathematics Subject Classification35Q60 35J10 35J50 35J92 35J61 Preview
Unable to display preview. Download preview PDF.
Notes References 1. 2. 3. 4. 5. 6.Figueiredo, G.M., Siciliano, G.: Existence and asymptotic behaviour of solutions for a quasi-linear Schrödinger–Poisson system under a critical nonlinearity. arXiv:1707.05353 7. 8. 9. 10. 11. Copyright information
© Springer Nature Switzerland AG 2019
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We know by the Riemann Existence Theorem that any Riemann surface can arise holmorphically as the branched cover of a sphere:
Therefore, any branched cover of the sphere can be represented by a rational function
polynomial $\frac{p (z)}{q(z)}: X \to \mathbb{C}$.
Given arbitrary polynomial $p$, do there exist ways to compute the monodromy around each critical points?
For example, $z \mapsto z^2 + \frac{1}{z^2}$ (not a polynomial) is 4-sheeted cover of the Riemann sphere $\hat{\mathbb{C}}$. It has 3 branch points, where $$p'(z) = 2z - \frac{2}{z^3} = 0 $$ It seems to have 4 branch points at $z = 1, i, -1, -i$.
How do we calculate of which sheets were permuted given $p(z)$?
This is motivated by trying to understand a proof of the Riemann Existence Theorem in the case of genus
0branched covers of the sphere $S^2$.
This question is the reverse direction of my previous question on branched covers: polynomial branched cover of the sphere with specified monodromy
Our example has a symmetry - and I am trying to avoid techniques which exploit them too much.
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(Reposted from Math StackExchange because this may be more suited for professional mathematicians.)
The Mountain Pass Theorem roughly says the following. Consider a differentiable functional $I: H \rightarrow \mathbb{R}$ from a Hilbert space $H$ to the reals which satisfies an appropriate compactness/properness condition (the Palais-Smale condition). Assume the following are true:
$I[0] = 0$; There exist positive constants $r$ and $a$ such that $I[q] \geq a$ for all $\|q\| = r$; There exists a $p$ (with $\|p\| > r$) at which $I[p] \leq 0$.
Then $I$ has a critical point $c$ at which $I[c] > 0$ (in fact $I[c] \geq a$).
My question is the following: can an analogous statement be made for functionals whose domain is
not a vector space? For a concrete example (which is the one I care about), let $\mathcal{M}$ be the space of all differentiable Riemannian metrics on the (two-dimensional) sphere, and let $I: \mathcal{M} \rightarrow \mathbb{R}$ be a functional from $\mathcal{M}$ to the reals. Obviously $\mathcal{M}$ is not a vector space and so the Mountain Pass Theorem can't be applied. But say the following properties hold: $I$ obeys some appropriate properness condition; $I[g^\mathrm{round}] = 0$, where $g^\mathrm{round}$ is the metric of the round sphere; There exists a neighborhood $U$ of $g^\mathrm{round}$ in $\mathcal{M}$ such that $I[g] > 0$ for any $g \in U \setminus \{g^\mathrm{round}\}$; There exists a metric $\bar{g} \in \mathcal{M}$ at which $I[\bar{g}] < 0$.
Then it seems like one should still be able to apply the spirit of the proof of the Mountain Pass Theorem to conclude that $I$ must have some critical point $g^\mathrm{crit}$ at which $I[g^\mathrm{crit}] > 0$. Obviously making this statement precise would require understanding what the "appropriate properness condition" above is, but in principle it seems to me like the spirit of the theorem should still apply.
Does anyone know if this is indeed the case?
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An autonomous equation is autonomous. Therefore, its solutions are related by time shifts: if $z=f(t)$ is a solution, then so is $z=f(t+C)$ for any constant $C$. In particular, there is nothing special about $x\to 0^+$. If one solution blows up going backwards in time, then they all do, and in exactly the same way, but
at different times.
Let's see how they do it. Let $(x_0,z_0)$ be the initial condition, from which we move backwards in time. If the solution exists at time $x_1<x_0$, then $$\int_{z_0}^{z(x_1)} 2\exp(-z^{2\alpha}/2)\,dz = x_0-x_1 \tag1$$ The left side of (1) cannot exceed $\int_{z_0}^{\infty} 2\exp(-z^{2\alpha}/2)\,dz$. Therefore, a blow-up occurs at time $T<x_0$ such that $$x_0-T = \int_{z_0}^{\infty} 2\exp(-z^{2\alpha}/2)\,dz \tag2$$
So, depending on your initial condition, one of the following three things will happen as $x\to 0^+$:
If $T<0$, then $z(x)$ is continuous at $x=0$. The blow-up is yet to occur. If $T>0$, then the solution blows up before reaching $x=0$. The question of its behavior as $x\to 0^+$ is moot. If $T=0$, we should consider the asymptotics of the integral in formula (2), plugging an arbitrary $x>0$ instead of $x_0$. This yields $$x = \int_{z(x)}^\infty 2\exp(-z^{2\alpha}/2)\,dz \tag3$$For sufficiently large $z$ we have $2\exp(-z^{2\alpha}/2)\le z^{-10}$ (this is also true with other numbers in place of $10$). Therefore, for $x$ sufficiently close to $0$,$$x \le \int_{z(x)}^\infty z^{-10}\,dz = \frac{1}{9z(x)^9}\tag4$$Rearrange (4) as $z(x)\le (9x)^{-1/9}$ and you are done.
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
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R/Normal.R
normal.Rd
The Normal distribution is ubiquituous in statistics, partially because of the central limit theorem, which states that sums of i.i.d. random variables eventually become Normal. Linear transformations of Normal random variables result in new random variables that are also Normal. If you are taking an intro stats course, you'll likely use the Normal distribution for Z-tests and in simple linear regression. Under regularity conditions, maximum likelihood estimators are asymptotically Normal. The Normal distribution is also called the gaussian distribution.
Normal(mu = 0, sigma = 1)
mu
The location parameter, written \(\mu\) in textbooks,which is also the mean of the distribution. Can be any real number.Defaults to
sigma
The scale parameter, written \(\sigma\) in textbooks,which is also the
A
Normal object.
We recommend reading this documentation on https://alexpghayes.github.io/distributions3, where the math will render with additional detail and much greater clarity.
In the following, let \(X\) be a Normal random variable with mean
mu = \(\mu\) and standard deviation
sigma = \(\sigma\).
Support: \(R\), the set of all real numbers Mean: \(\mu\) Variance: \(\sigma^2\) Probability density function (p.d.f):
$$ f(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-(x - \mu)^2 / 2 \sigma^2} $$
Cumulative distribution function (c.d.f):
The cumulative distribution function has the form
$$ F(t) = \int_{-\infty}^t \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-(x - \mu)^2 / 2 \sigma^2} dx $$
but this integral does not have a closed form solution and must be approximated numerically. The c.d.f. of a standard Normal is sometimes called the "error function". The notation \(\Phi(t)\) also stands for the c.d.f. of a standard Normal evaluated at \(t\). Z-tables list the value of \(\Phi(t)\) for various \(t\).
Moment generating function (m.g.f):
$$ E(e^{tX}) = e^{\mu t + \sigma^2 t^2 / 2} $$
set.seed(27) X <- Normal(5, 2) X#> Normal distribution (mu = 5, sigma = 2)random(X, 10)#> [1] 8.814325 7.289754 3.470939 2.085135 2.813062 5.590482 5.013772 7.314822 #> [9] 9.269276 5.475689pdf(X, 2)#> [1] 0.0647588log_pdf(X, 2)#> [1] -2.737086cdf(X, 4)#> [1] 0.3085375quantile(X, 0.7)#> [1] 6.048801### example: calculating p-values for two-sided Z-test # here the null hypothesis is H_0: mu = 3 # and we assume sigma = 2 # exactly the same as: Z <- Normal(0, 1) Z <- Normal() # data to test x <- c(3, 7, 11, 0, 7, 0, 4, 5, 6, 2) nx <- length(x) # calculate the z-statistic z_stat <- (mean(x) - 3) / (2 / sqrt(nx)) z_stat#> [1] 2.371708#> [1] 0.01770607#> [1] 0.01770607# p-value for one-sided test # H_0: mu <= 3 vs H_A: mu > 3 1 - cdf(Z, z_stat)#> [1] 0.008853033# p-value for one-sided test # H_0: mu >= 3 vs H_A: mu < 3 cdf(Z, z_stat)#> [1] 0.991147### example: calculating a 88 percent Z CI for a mean # same `x` as before, still assume `sigma = 2` # lower-bound mean(x) - quantile(Z, 1 - 0.12 / 2) * 2 / sqrt(nx)#> [1] 3.516675#> [1] 5.483325#> [1] 3.516675 5.483325#> [1] 3.516675#> [1] 5.483325### generating random samples and plugging in ks.test() set.seed(27) # generate a random sample ns <- random(Normal(3, 7), 26) # test if sample is Normal(3, 7) ks.test(ns, pnorm, mean = 3, sd = 7)#> #> One-sample Kolmogorov-Smirnov test #> #> data: ns #> D = 0.20352, p-value = 0.2019 #> alternative hypothesis: two-sided #># test if sample is gamma(8, 3) using base R pgamma() ks.test(ns, pgamma, shape = 8, rate = 3)#> #> One-sample Kolmogorov-Smirnov test #> #> data: ns #> D = 0.46154, p-value = 1.37e-05 #> alternative hypothesis: two-sided #>#> [1] 0.7#> [1] 7
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The writing is certainly less clear than it ought to be.Your instincts are right; they just didn't explain it well.In particular, they're not saying that the term in bracketsonly vanishes for symmetry transformations. It alwaysvanishes for a field obeying the Euler-Lagrange equations.
What they're really saying is:
A continuous transformation can be written as equation (2.9) The equation of motion will be invariant under that continuous transformation if the Lagrangian changes as equation (2.10) Let's see how the Lagrangian actually changes under the transformation of (2.9) It changes as a term that looks like $\alpha \partial_\mu \mathcal{J}^\mu$ plus a term that goes to zero because of the Euler-Lagrange equations for the original, un-transformed system The actual change we get is the change we demanded for the Lagrangian to be invariant up to a 4-divergence Therefore, the continuous transformation is a symmetry transformation
The text immediately following what the OP included says
The second term [in equation (2.11)] vanishes by the
Euler-Lagrange equation. We set the remaining term equal to
$\alpha \partial_\mu \mathcal{J}^\mu$ and find
\begin{equation}
\partial_\mu j^\mu(x) = 0\,
\qquad \mathrm{for} \qquad
j^\mu(x) = \frac{\partial \mathcal{L}} {\partial
(\partial_\mu \phi)} \Delta \phi - \mathcal{J}^\mu.
\end{equation}
This really makes it look like they're saying that $j^\mu$ is identically zero; in fact, I would argue that that's the most reasonablereading of the text. What they
should have said is thatwe get the generic result\begin{equation} \Delta \mathcal{L} = \partial_\mu \left( \frac{\partial \mathcal{L}} {\partial (\partial_\mu \phi)} \Delta \phi \right) \tag{A}\end{equation}whenever $\phi$ satisfies the Euler-Lagrange equations, butto call $\Delta \phi$ a symmetry transformation we requirethe existence of some other field $\mathcal{J}^\mu$ suchthat\begin{equation} \Delta \mathcal{L} = \partial_\mu \mathcal{J}^\mu \tag{B}\end{equation}regardless of whether or not the Euler-Lagrange equationsare satisfied. Again, (A) is generically true for anydifferentiable Lagrangian and field satisfying the E-Lequations. On the other hand, you have to use theparticular form of $\mathcal{L}$ and the particular form of$\Delta \phi$ to decide if (B) is true in any particularcase. (Also note that $\mathcal{J}^\mu = 0$ is perfectlyacceptable.)
It probably would have been clearer if P&S had begun bymoving the explanation of what a symmetry transformation is[the four sentences up to equation (2.10)] to a previousparagraph and expanding on it. Then they could introduce acontinuous transformation and show how the Lagrangianchanges. But why bother editing a piece of text forclarity, when it's only explaining something as frivolousand useless as Noether's Theorem? ;)
These issues are made much clearer in @joshphysics's version of theproof — in particular with his introduction of separate$\mathcal{J}^\mu$ and $\mathcal{K}^\mu$ vector fields. Theylook like they perform the same function, and yet they'redifferent, as he explains in the "Important Notes"sectionat the bottom of his answer, and in some of thecomments.
I also highly recommend theexcellent "Quantum Field Theory for the Gifted Amateur", which doesa typically superb job on Noether's Theorem, and is really what I think should bethe standard text. The title is deceptive because it'saimed at professional physicists, just not currentprofessional quantum-field theorists. I encourage you to atleast read the otheranswer, oreven take up "Gifted Amateur" if you really want tounderstand QFT.
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Deepak Jain
Deen Dayal Upadhyaya College, New Delhi
Model independent constraints on cosmic curvature
We use two model-independent methods to constrain the curvature of the universe. In the first method, we study the evolution of the curvature parameter (Ω0k) with redshift by using the observations of the Hubble parameter and transverse comoving distances obtained from the age of galaxies. Secondly, we also use an indirect method based on the mean image separation statistics of gravitationally lensed quasars. The basis of this methodology is that the average image separation of lensed images will show a positive, negative or zero correlation with the source redshift in a closed, open or flat universe respectively. In order to smoothen the datasets used in both the methods, we use a non-parametric method namely, Gaussian Process (GP). Finally from first method we obtain Ω0k=0.025±0.57 for a presumed flat universe while the cosmic curvature remains constant throughout the redshift region 0
Mayukh Raj Gangopadhyay
Saha Institute Of Nuclear Physics, Kolkata
A Study in non-canonical domain of Goldstone inflaton
Inflationary paradigm offers a very attractive solution to resolve the hot big bang cosmology puzzles. One of the problems of the standard inflationary models is most of the textbook models are ruled out or disfavoured by the Planck observations. Another theoretical discomfort about many a models of inflation is related to the super-Planckian field excursion of the inflaton field. In early 90's one elegant solution was proposed by Freese et. al. from the idea of symmetry breaking to produce the inflation potential where the inflaton is a Goldstone boson. Due to the shift symmetry property embedded through the symmetry breaking, the flatness of the potential is maintained which is essential for the model building of inflation. But after recent observations from Planck, natural inflation model is almost ruled out in the standard Λ CDM model. Natural inflation is one class of a general form of inflation known as Goldstone inflation. To have a successful Goldstone inflation, all scales related to the theory has to be sub-Planckian thus keeping the inflaton guarded against the UV correction from the quantum gravity effects. In this work, we have shown that Goldstone inflation in case of a non-canonical dynamical realisation, can have symmetry breaking scales less than the Planck scale and can have observational predictions in perfect agreement with the latest data. Thus, a non-canonical realisation of generalised Goldstone inflation can be one of the prime candidates to explain the inflationary dynamics.
Sandeep Kumar Kataria
IIA, Bangalore
A study of Isolated halos distribution in Large-scale structures of universe
In this study we have explored the mass spectrum of isolated halos along with their local densities from high redshift z~ 12 to present universe z~0. For conducting this study we have used Horizon Run 4 (Kim & Changbom et al 2015) data products. Horizon Run 4 is N-body simulation designed to study evolutions of galaxies and large-scale structures of the Universe. In this simulation halo mass resolution scale down to M s = 2.7 × 10^{11} h^{-1} M sun; which is better for isolated halo study. We have found that isolated halos shows maximum in their probability distribution for over density regions which corresponds to values ranging from 0.1 to 1 at z~0. Apart from this non-isolated halos shows maximum in their probability distribution for over density regions which varies from 1 to 10. We have also shown that masses for which isolated galaxies shows maximum probability distribution is more than non-isolated ones at z~0. This implies that most of the satellites of progenitors of isolated halos have merged into progenitors during their evolution. Further study includes merger tree history of these isolated halos
Bikash Chandra Paul
North Bengal University, Siliguri
Emergent Universe with interacting fluids
Emergent Universe is studied in modified gravity. It has no bigbang singularity. To understand evaluation of the universe it is necessary to have a static Einstein universe in the infinitely past and in course of time the present universe will emerge from the state. A non-linear equation of state is required here, which is essentially describes a composition of three fluids. It is shown that interacting fluid gives a physically realistic solution to describe the present universe.
Priyank Parashari
PRL, Ahmedabad
Viscosity as a solution to σ8 tension and it's effect on neutrino mass
It has been reported that there are some discordances between cosmic microwave background (CMB) and large scale structure (LSS) surveys. In particular, the value of \sigma 8, the r.m.s. fluctuation of density perturbations at 8 h^-1Mpc scale and H0, the value of Hubble parameter observed today, inferred from CMB and LSS observations, are not in agreement with each other. Moreover, these discordances can be resolved by using effective viscous description of cold dark matter (CDM) on large scales. Effective field theory of dark matter fluid on large scales predicts the presence of viscosity of the order of 10^-6 H0 M P^2. The same order of viscosity have also been found to resolve the discordance between LSS observations and Planck CMB data. Since massive neutrinos suppresses the matter power spectrum on the small length scales similar to the viscosity, we, therefore, use the viscous dark matter description to study the neutrinos and demonstrate that one can indeed constrain the masses of the neutrinos in a stringent way. We found the bounds on sum of neutrino masses to be \sum m \nu < 0.266 eV for normal hierarchy and \sum m \nu < 0.146 eV for inverted hierarchy.
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1,764 69
Hi PF
Given some linear differential operator ##L##, I'm trying to solve the eigenvalue problem ##L(u) = \lambda u##. Given basis functions, call them ##\phi_i##, I use a variational procedure and the Ritz method to approximate ##\lambda## via the associated weak formulation $$\langle L(\phi_i),\phi_j\rangle = \lambda \langle \phi_i,\phi_j\rangle.$$ As you can see, this expression is now a matrix equation, solutions to which are straightforward. For my particular problem, the basis functions are $$\phi_j = \cos\left( \frac{\pi j}{2}(x+1) \right) \cosh\left( \frac{\pi j}{2}(y+h) \right).$$ However, this solution, when inputted into the weak formulation equation, does not output correct eigenvalues. However, ##\phi_j## can be split into even and odd components: $$ \phi_j^o = \sin \left( \pi(j-1/2)x \right)\cosh\left( \pi(p-1/2)(y+h) \right)\\ \phi_j^e = \cos \left( \pi j x \right)\cosh\left( \pi j(y+h) \right) $$ Now to obtain eigenvalues I solve two separate equations, one for even eigenvalues and one for odd: $$\langle L(\phi_i^e),\phi_j^e\rangle = \lambda \langle \phi_i^e,\phi_j^e\rangle\\ \langle L(\phi_i^o),\phi_j^o\rangle = \lambda \langle \phi_i^o,\phi_j^o\rangle.$$ This latter approach gives correct solutions: why? Any insight or direction is greatly appreciated.
Given some linear differential operator ##L##, I'm trying to solve the eigenvalue problem ##L(u) = \lambda u##. Given basis functions, call them ##\phi_i##, I use a variational procedure and the Ritz method to approximate ##\lambda## via the associated weak formulation
$$\langle L(\phi_i),\phi_j\rangle = \lambda \langle \phi_i,\phi_j\rangle.$$
As you can see, this expression is now a matrix equation, solutions to which are straightforward. For my particular problem, the basis functions are $$\phi_j = \cos\left( \frac{\pi j}{2}(x+1) \right) \cosh\left( \frac{\pi j}{2}(y+h) \right).$$
However, this solution, when inputted into the weak formulation equation, does not output correct eigenvalues. However, ##\phi_j## can be split into even and odd components:
$$ \phi_j^o = \sin \left( \pi(j-1/2)x \right)\cosh\left( \pi(p-1/2)(y+h) \right)\\
\phi_j^e = \cos \left( \pi j x \right)\cosh\left( \pi j(y+h) \right)
$$
Now to obtain eigenvalues I solve two separate equations, one for even eigenvalues and one for odd:
$$\langle L(\phi_i^e),\phi_j^e\rangle = \lambda \langle \phi_i^e,\phi_j^e\rangle\\
\langle L(\phi_i^o),\phi_j^o\rangle = \lambda \langle \phi_i^o,\phi_j^o\rangle.$$
This latter approach gives correct solutions: why? Any insight or direction is greatly appreciated.
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I am trying to find a single primitive root modulo $11$. The definition in my textbook says "Let $a$ and $n$ be relatively prime integers with ($a \neq 0$) and $n$ positive. Then the least positive integer $x$ such that $a^x\equiv1\pmod{\! n}$ is called the order of $a$ modulo $n$ and is denoted by $\text{ord}_{n}a$".
So what I don't understand is how I can find a single primitive root modulo $11$ if I am not also given $a$.
Or is it that maybe I understand things after all since $2$ is a primitive root modulo $11$ since $2^{10} \equiv \phi(11)\equiv 10\pmod{\! 11}$ and $2$ is a generator for the group $\mathbb{Z}/11\mathbb{Z}$?
In any case, I am confused since I need to find a second primitive root modulo $11$ and I'm not sure how to do that other than by guessing and checking $a^{10}$ for $a \in \{3,4,5,6,7,8,9,10\}$. Any help would be appreciated.
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I'm trying to solve a system of nonlinear, coupled ODEs, where the governing equation for the $n$-th ODE is of the form:
$$\sum_k^M Q_{nk} \ddot{a}_k -\sum_{\ell}^M\sum_k^M S_n(\ell,k) \dot{a}_{\ell} \dot{a}_k + \frac{1}{2}U_n = 0$$
where
$$Q_{nk}=\frac{1}{2}\sum_{j=1}^MP_{jn}P_{jk},$$
with
$$P_{jn}=\frac{1}{\sqrt{m}n}(a_{n-j}+a_{j+n}-a_ja_n),\quad \mathrm{and} \quad a_{n-j}=0 \quad \mathrm{for}\quad j>n,$$
$$S_n(\ell,k)=-\frac{1}{2}\left(\frac{\partial Q_{k\ell}}{\partial a_n}-\frac{\partial Q_{n\ell}}{\partial a_k}-\frac{\partial Q_{nk}}{\partial a_{\ell}}\right),$$
and
$$U_n=\frac{\partial}{\partial a_n} \frac{1}{2\pi} \int_0^{2\pi}\left\{ \left(-\sum_{j=1}^M \left(\frac{a_j^2}{2j}-\frac{a_j}{j}\cos j\theta \right)\right)^2\,\left(1+\sum_{j=1}^M a_j \cos j\theta \right) \right\}d\theta.$$
Note, $(Q_{k\ell}, S_N(k,\ell), U_n)$ are all polynomials (of maximum order 4) in the dependent variables $a_i(t)$ and $M$ is the total number of dependent variables under consideration (i.e. $a_j\equiv 0$ for $j>M$). I have been able to solve for low $M$ cases (i.e. $M$ less than 10) using
NDSolve. For fixed initial conditions, I have been increasing $M$ and finding a significant increase in computation time. I'm wondering if there are any obvious ways to decrease computation time that I have overlooked. (Ideally, for the system I'm trying to model, I'd like to increase the number of dependent variables to around 200.)
I'm new to
Mathematica, and I'm sure I'm making some 'rookie mistakes' that are costing me in computation time. Also, if I'm barking up the wrong tree, that is I should be using a different platform, I'd like to find out before I invest too much effort into this project.
Some questions I have on basic problem set-up:
Is it more efficient to isolate the dependent variables in question. That is, have equations of the form $\ddot{a}_i = \ldots$ versus $\sum_k^M Q_{nk} \ddot{a}_k = \ldots$?
Is it worth fully simplifying the governing equations (i.e. using
FullSimplify) before executing the
NDSolvecommand?
Should I rewrite each second order ODEs as a pair of first order ODEs?
Any advice on this would be greatly appreciated.
Here is the code:
M = 10; (*Number of modes*)f[r_, s_] := If[r > s, 0, 1]; (*Will ensure we have no negative modes*)α[i_, t_] := If[i <= M, A[i, t], 0]; (*Truncate series at Mth mode*)A[0, t_] := 1; (*By definition*)P[m_, n_, t_] := 1/Sqrt[m*n^2]*(f[m, n]*α[n - m, t] + α[n + m, t] - α[m, t]*α[n, t]) ; Q[n_, l_, t_] := 1/2 Sum[P[m, n, t]* P[m, l, t], {m,1,M}] ; S[n_, k_, l_, t_] := 1/2 (D[Q[k, l, t], A[n, t]] - D[Q[n, l, t], A[k, t]] - D[Q[n, k, t], A[l, t]]); F[t_] := 1/(2 π) Integrate[Sum[((A[j, t]*Cos[j*ξ])/j - A[j, t]^2/(2j)),{j,1,M}]^2* (1+Sum[1+A[j, t]* Cos[j*ξ],{j,1,M}]), {ξ, 0, 2π}]; U[n_, t_] := D[F[t], A[n, t]];Table[A[j, t_] = B[j][t], {j, 1, M}]; (*Redefine dep variables so that independent variable is distinguished from dep var index*)Gov[n_, t_] := Sum[B[l]''[t]*Q[n,l,t],{l,1,M}]-Sum[Sum[S[n,k,l,t]B[k]'[t]B[l]'[t], {k,1,M}],{l,1,M}]+1/2U[n,t];eqns = {Table[Gov[j, t] == 0, {j, 1, M}], B[1][0] == 0.3, B[2][0] == 0.1, Table[B[j][0] == 0, {j, 3, M}], Table[B[j]'[0] == 0, {j, 1, M}]};TenMode = NDSolve[eqns, Table[B[n][t], {n, 1, M}], {t, 0, 10}]
I apologize for any errors in the code since I had some issues parsing from the symbolic notation present in my editor to a code that would be easily copied into this forum.
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Need help! I was working on a project when I required to use a projection operator. For an example case, I have the Bell state, $$|\psi\rangle = \frac1{\sqrt2}\left(\color{blue}{|0}0\rangle+|11\rangle\right)$$ which now I want to take to the state, $$|\psi'\rangle = |\color{blue}{0}0\rangle$$ by weeding out the states with the leftmost qubit as $1$.
Another example would be, $$\frac1{2}\left(|\color{blue}{0}0\rangle+|\color{blue}{0}1\rangle+|10\rangle+|11\rangle\right)\rightarrow\frac1{\sqrt2}\left(|00\rangle+|01\rangle\right).$$
Edit: Just putting another example to make my question clear. I want to weed out states when the third (leftmost) qubit is $1$ in the following example.
Suppose we have a three qubit state, $$|\psi\rangle=\displaystyle\frac1{N}\left(|\color{blue}{0}\rangle\otimes\left[|00\rangle+|01\rangle+|10\rangle\right] + |1\rangle\otimes|11\rangle\right)$$ it should then get transformed to $$|\psi'\rangle = \frac1{N'}|\color{blue}{0}\rangle\otimes\left[|00\rangle+|01\rangle+|10\rangle\right].$$
Is this possible? If yes, how can I implement it in the IBM Quantum Experience?
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The question is: let $X_1, ..., X_n \sim N(\mu, \sigma^2)$. Let $\tau$ be the .95 percentile, i.e. $P(X<\tau)$ = .95. What is the MLE of $\tau$?_
What I have tried:
$P(X<\tau) = P(Z < \frac{\tau - \mu}{\sigma}) = \Phi(\frac{\tau - \mu}{\sigma})$ where Z denotes a Standard Normal random variable, and $\Phi$ is the CDF for the normal distribution.
I think I need to invoke the delta method for this, since $\tau$ is a function of the parameters for the normal distribution, but I don't quite see how I can use this method for finding a quantile. I also don't see how I need to invoke the maximum likelihood. I do know, that for a normal distribution the MLE's are:
$\hat{\mu} = \sum_{i=1}^n X_i/n$ ,the sample mean and $\hat{\sigma} = \sqrt{\sum_{i=1}^n(X_i - \mu)^2/n}$. How would I need those? Any help is appreciated!
Edit: the question is one from Larry Wasserman's All of Statistics that I'm working through, practising for an exam I have this week. And I suppose that $\tau$ is almost equal to two $\sigma$, I believe it is ~$1.96 \sigma$, looking at a normal table?
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Illinois Journal of Mathematics Illinois J. Math. Volume 45, Number 1 (2001), 25-39. Local properties of polynomials on a Banach space Abstract
We introduce the concept of a smooth point of order $n$ of the closed unit ball of a Banach space $E$ and characterize such points for $E = c_0$, $L_p(\mu)$ ($1\leq p \le\infty$), and $C(K)$. We show that every locally uniformly rotund multilinear form and homogeneous polynomial on a Banach space $E$ is generated by locally uniformly rotund linear functionals on $E$. We also classify such points for $E = c_0$, $L_p(\mu)(1\leq p \le\infty)$, and $C(K)$.
Article information Source Illinois J. Math., Volume 45, Number 1 (2001), 25-39. Dates First available in Project Euclid: 13 November 2009 Permanent link to this document https://projecteuclid.org/euclid.ijm/1258138253 Digital Object Identifier doi:10.1215/ijm/1258138253 Mathematical Reviews number (MathSciNet) MR1849984 Zentralblatt MATH identifier 1001.46029 Subjects Primary: 46G25: (Spaces of) multilinear mappings, polynomials [See also 46E50, 46G20, 47H60] Secondary: 46B20: Geometry and structure of normed linear spaces 46B28: Spaces of operators; tensor products; approximation properties [See also 46A32, 46M05, 47L05, 47L20] Citation
Aron, Richard M.; Choi, Yun Sung; Kim, Sung Guen; Maestre, Manuel. Local properties of polynomials on a Banach space. Illinois J. Math. 45 (2001), no. 1, 25--39. doi:10.1215/ijm/1258138253. https://projecteuclid.org/euclid.ijm/1258138253
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Anomalous physical transport in complex networks Abstract
The emergence of scaling in transport through interconnected systems is a consequence of the topological structure of the network and the physical mechanisms underlying the transport dynamics. We study here transport by advection and diffusion in scale-free and Erdös-Rényi networks. Velocity distributions derived from a flow potential exhibit power-law scaling with exponent $${\nu}{\approx}{\gamma}+1$$, where $${\gamma}$$ is the exponent of network connectivity. Using stochastic particle simulations, we find anomalous (nonlinear) scaling of the mean-square displacement with time. We show the connection with existing descriptions of anomalous transport in disordered systems, and explain the mean transport behavior from the coupled nature of particle jump lengths and transition times.
Authors: Massachusetts Inst. of Technology (MIT), Cambridge, MA (United States) Publication Date: Research Org.: Massachusetts Inst. of Technology (MIT), Cambridge, MA (United States) Sponsoring Org.: USDOE Office of Science (SC), Basic Energy Sciences (BES) (SC-22) OSTI Identifier: 1505821 Grant/Contract Number: SC0003907 Resource Type: Accepted Manuscript Journal Name: Physical Review. E, Statistical, Nonlinear, and Soft Matter Physics Additional Journal Information: Journal Volume: 82; Journal Issue: 5; Journal ID: ISSN 1539-3755 Publisher: American Physical Society (APS) Country of Publication: United States Language: English Subject: 71 CLASSICAL AND QUANTUM MECHANICS, GENERAL PHYSICS Citation Formats
Nicolaides, Christos, Cueto-Felgueroso, Luis, and Juanes, Ruben. Anomalous physical transport in complex networks. United States: N. p., 2010. Web. doi:10.1103/physreve.82.055101.
Nicolaides, Christos, Cueto-Felgueroso, Luis, & Juanes, Ruben. Anomalous physical transport in complex networks. United States. doi:10.1103/physreve.82.055101.
Nicolaides, Christos, Cueto-Felgueroso, Luis, and Juanes, Ruben. Wed . "Anomalous physical transport in complex networks". United States. doi:10.1103/physreve.82.055101. https://www.osti.gov/servlets/purl/1505821.
@article{osti_1505821,
title = {Anomalous physical transport in complex networks}, author = {Nicolaides, Christos and Cueto-Felgueroso, Luis and Juanes, Ruben}, abstractNote = {The emergence of scaling in transport through interconnected systems is a consequence of the topological structure of the network and the physical mechanisms underlying the transport dynamics. We study here transport by advection and diffusion in scale-free and Erdös-Rényi networks. Velocity distributions derived from a flow potential exhibit power-law scaling with exponent ${\nu}{\approx}{\gamma}+1$, where ${\gamma}$ is the exponent of network connectivity. Using stochastic particle simulations, we find anomalous (nonlinear) scaling of the mean-square displacement with time. We show the connection with existing descriptions of anomalous transport in disordered systems, and explain the mean transport behavior from the coupled nature of particle jump lengths and transition times.}, doi = {10.1103/physreve.82.055101}, journal = {Physical Review. E, Statistical, Nonlinear, and Soft Matter Physics}, number = 5, volume = 82, place = {United States}, year = {2010}, month = {11} } Citation information provided by Web of Science
Web of Science
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Monte Carlo is most useful when you lack analytic tractability or when you have a highly multidimensional problem.For example, even using simple lognormal and poisson models, there exist path-dependent payoffs or multi-asset computations such that no analytic solution exists and such that any PDE finite difference solution would require 3 or more ...
We assume that the short interest rate $r_t$ follows the Hull-White model, that is, the short rate $r$ and the stock price $S$ satisfies a system of SDEs of the form\begin{align*}dr_t &= (\theta_t -a\, r_t)dt + \sigma_0 dW_t^1,\\dS_t &= S_t\Big[r_t dt + \sigma \Big(\rho dW_t^1 + \sqrt{1-\rho^2} dW_t^2\Big)\Big],\end{align*}where $a$, $\sigma_0$, ...
You have the correct approach.(1) The simulation generates sampled portfolio values, $P_1,P_2, \dots, P_n$ at time $t=T$. VaR is specified as a left-tail percentile.Order the sample as$$P_{(1)} \leq P_{(2)} \leq \dots \leq P_{(n)}.$$If you are considering $VaR_\alpha$ at the $100(1-\alpha) \% $ confidence level , then choose the smallest integer $k$ ...
You are typically interested in evaluating $E\left[ f(X_T)-f(\bar{X}_T^{(n)}) \right]$ (refered as the weak convergence)$X_t$ the solution of the sde : $dX_t^x=b(X_t^x)dt+\sigma(X_t^x)dW_t$$\bar{X}_t^{(n)}=b(\underline{t},X_{\underline{t}}^{(n)})\cdot (t-\underline{t})+\sigma(\underline{t},X_{\underline{t}}^{(n)})\cdot (W_{\underline{t}}-W_t)$ is your ...
Okay just to wind things down here, I think an important clarification is needed if readers might come and seek to a similar solution.The Geometric Brownian Motion (GBM) is a model of asset prices dynamics which is usually given as follows:$$ dS_t = \mu S_t dt + \sigma S_t dB_t$$where $B_t$ is a standard brownian motion which has several important ...
You need to compute your greeks as finite differences, but the full procedure may be pretty tricky. I will use vega $\aleph$ as the example here. Let's begin by designating your Monte Carlo estimator as a function $V(\sigma,s,M)$ where $\sigma$ is the volatility as usual, $s$ is the seed to your random number generator, and $M$ is the sample count.To ...
I believe this is a nice paper for you to start with. Check out what references it cited and who cited it.Markov Chain Monte Carlo Analysis of Option Pricing Models"Use the Markov Chain Monte Carlo (MCMC) method to investigate a largeclass of continuous-time option pricing models. These include: constant-volatility,stochastic volatility, price jump-...
Tools from the field of stochastic optimization are best suited for these problems. In particular, attached is a paper on non-parametric density estimation for stochastic optimization that describes an algorithm if state variables can be associated with draws from the predictive distribution.Here's another approach by Kuhn. These are all one-period ...
In this scenario, the "joint dynamics" are trivially computed since the option value is a known deterministic function of stock price. For example, the mean of the option value for time $\tau$ is$$\mu_O = \int_0^\infty BSM( S_\tau ) p(S_\tau) dS_\tau$$which is best computed using quadrature as available in standard numerical libraries like scipy. The ...
By definition the fair value of an option is given by an expectation value of the payoff, $\mathbf{E}\left[\textrm{payoff}(\textit{paths})\right]$. The probability distribution of the paths is the risk neutral measure. This is just an integral expression of the form you wrote. This applies to all option prices. Many options are, of course, special in the ...
You have the right idea, but it seems you don't know $\mu$, so using it in your error check doesn't seem correct. Also, checking the result every 10,000 iterations may not be optimal for deciding when to stop.To be clear, let $E(X) = \mu$ and $Var(X) = \sigma$. We're invoking the CLT when we write$$P\left( \left|\frac{\bar{X}_n - \mu}{\sigma/\sqrt{n}}\...
Here is the general approach you can follow to generate two correlated random variables. Let's suppose, X and Y are two random variable, such that:$$X \sim N(\mu_1, \sigma_1^2)$$$$Y \sim N(\mu_2, \sigma_2^2)$$and $$cor(X,Y)=\rho$$Now consider: $y=bx + e_i$, where $x$ $(=\frac{X-\mu_1}{\sigma_1}$) and $y$ $(=\frac{Y-\mu_2}{\sigma_2}$) both follow ...
Yes, the term Brownian Bridge seems to be used loosely. I assume you are talking about continuously monitored barriers by the way, since you mention the probability of the barrier being crossed in between the path time points. If that's the case then "naive" Monte Carlo simulation will have what is called "simulation bias". That's exactly because the ...
Yes, it can work. However, keep in mind that:you'll be safer if you don't share any objects between threads; see my answer here, in particular the last point;even if you use different seeds, there's no guarantee that the generated sequences won't overlap. If you're willing to change the engine code so that you can pass the relevant parameters, a safer ...
You have the right intuition but the approach is not quite right.The issuer has the right to call back the bond at a pre-defined call price. So your decision criterion is "call when the value of the bond >= contractual call price". We are comparing prices in the decision rule, not the YTM of the callable bond with the coupon of the bond.Note that ...
Apart from numerical stability errors, Cholesky and PCA (without dim reduction) shall produce exactly the same distribution, they are two symmetric decomposition of the same covariance matrix and thus are equivalent for transforming a standard normal vector.Of course when doing different things with PCA components, such as in dim reduction or quasi Monte ...
LSM is very fiddly.The most important things in my view are1) don't believe anyone who says that the choice of basis functions doesn't matter.2) implement an upper bounder, eg Andersen--Broadie (2003) or Joshi-Tang (2014) so you can tell if your prices are good3) do two passes, one to build the strategy, one to price, if they give very different ...
Quasi Random Numbers are more tricky than it might seem, using them as a black box like with PRNGs is risky. E.g. an unscrambled Sobol' sequence is uniform only asymptotically, while for realistic sample sizes there are subvolumes with significantly different densities. You often do not realize that because the convergence graph looks good anyway, it gives ...
In general these are the two basic approaches to QuantFinance:Sell side (market maker, risk neutral): You use risk-neutral probabilities ("$\mathbb{Q}$") e.g. in option pricing (to e.g. calculate your greeks and hedge your portfolio), so that you live on the spread.Buy side (market/risk taker): You use real-world probabilites ("$\mathbb{P}$") for e.g. ...
the problem is that the pay-off has discontinuous first derivative. Try a contract with pay-off that is twice differentiable and it will probably work.The problem is that all the value comes from the tiny number of paths within $\Delta S$ of the strike, and these paths have huge value.This is a well-known problem. As the bump size goes to zero, the ...
First you need to correct the formula to:$$W_t^2 = \rho W_t^1 + \sqrt{1-\rho^2} Z_t,$$where $Z_t$ is a BM independent of $W_t^1$If you calculate the variance and the covariance, then you see that it is true:$$V[W_t^1] = t$$and$$V[W_t^2] = \rho^2 V[W_t^1] + (1-\rho^2) V[Z_t] = \rho^2 t + (1-\rho^2) t = t,$$which is the desired variance.For the ...
The estimation error is a random variable and not a simple scalar. As such, when performing one-shot assessments, you could always end up observing that using $6400$ paths provides a "better" price estimate than using $100$ of them. What matters is to investigate the variance of the estimator rather than looking at pointwise values it can take (*)To get a ...
If implemented properly, least-squares Monte Carlo as originally suggested by Longstaff-Schwartz should allow you to identify sub-optimal exercise dates and a lower bound of the true option price. There are many articles out there discussing this non trivial topic. @MarkJoshi can probably shed some more light, see this nice paper.You claim that your LSM ...
I'm guessing, and correct me if I'm wrong, you want to create a number of possible paths the stock price could follow with the local volatilty given by GARCH depending on the simulated history, or in pseudocode:N <- numberOfPathsT <- numberOfStepsfor (i in 1:N) {newSeries <- pastPricesfor (t in 1:T) {epsilon <- normrnd(0,1)...
In quantitative finance, we sometimes find ourselves choosing a new stochastic model for what market variables are random, and how. For example, someone might decide that they like the SDE\begin{equation}dS = \mu\ S\ dt + \left( \frac{S_0}{S} \right)^{\frac32} \sigma\ S\ dW\end{equation}because they want to capture a leverage effect.Now, this SDE ...
Check this document out: link to pdf fileAlso, if you are concerned with actual performance of your code and want to implement efficient code then gsl libraries would be the first place look at: link. It's got everything you need.
The best I have seen so far is William Wheaton's work in this area. I don't know how much is described in his papers but he and Torto created a system that combined factor models for things like local and national price indexes with specific economics of commercial real estate ventures (such as balloon payments on construction milestones and the like).The ...
For such high-dimensional path problems you will want to use the Morokov technique (you can find the paper online), which takes QR samples for the "important" dimensions and then reverts to pseudorandom for the less important dimensions in an interest rate problem remarkably similar to yours. (Similar principles apply to using QR sequences in factor model ...
For completeness, let's restate that the discrete case goes like this:$$\Delta S_t = S_{t+\Delta t}- S_t = \mu S_t \Delta t + \sigma S_t \sqrt{\Delta t} Z_t $$with $Z_t \sim \mathcal{N}(0,1)$What you are doing in your case is to use the exact solution of the SDE to model the movement between two points of $S$.Essentially, you are doing the same thing ...
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Answer
The given models overestimate the total dormitory charge by $2$ .
Work Step by Step
The total dormitory charge for undergraduate education is given by: $\left[ \begin{align} & \left( 4565+5249 \right)+\left( 4824+5413 \right) \\ & +\left( 5046+5629 \right)+\left( 5241+5837 \right) \\ \end{align} \right]=41,804$. The model for dormitory charges at public colleges is given by: ${{a}_{n}}=225n+4357$. Here, $ n=1$ for 2010, $ n=2$ for 2011, $ n=3$ for 2012, and $ n=4$ for 2013. We know that the total dormitory charge at a public college is calculated by; ${{S}_{n}}=\frac{n}{2}\left( {{a}_{1}}+{{a}_{n}} \right)$ For $ n=4$, ${{S}_{4}}=\frac{4}{2}\left( {{a}_{1}}+{{a}_{4}} \right)$ Now, value of ${{a}_{1}}$ is; $\begin{align} & {{a}_{1}}=225\times 1+4357 \\ & =4582 \end{align}$ And, value of ${{a}_{4}}$ is; $\begin{align} & {{a}_{4}}=225\times 4+4357 \\ & =5257 \end{align}$ So, ${{S}_{4}}$ is; $\begin{align} & {{S}_{4}}=\frac{4}{2}\left( {{a}_{1}}+{{a}_{4}} \right) \\ & =\frac{4}{2}\left( 4582+5257 \right) \\ & =19678 \end{align}$ The model for dormitory charges at private colleges is given by ${{a}_{n}}=225n+4357$. Here, $ n=1$ for 2010, $ n=2$ for 2011, $ n=3$ for 2012, and $ n=4$ for 2013. We know that total dormitory charge at private colleges is calculated by ${{S}_{n}}=\frac{n}{2}\left( {{a}_{1}}+{{a}_{n}} \right)$ For $ n=4$, ${{S}_{4}}=\frac{4}{2}\left( {{a}_{1}}+{{a}_{4}} \right)$ Now, the value of ${{a}_{1}}$ is $\begin{align} & {{a}_{1}}=198\times 1+5037 \\ & =5235 \end{align}$ And, the value of ${{a}_{4}}$ is $\begin{align} & {{a}_{4}}=198\times 4+5037 \\ & =5829 \end{align}$ So, ${{S}_{4}}$ is $\begin{align} & {{S}_{4}}=\frac{4}{2}\left( {{a}_{1}}+{{a}_{4}} \right) \\ & =\frac{4}{2}\left( 5235+5829 \right) \\ & =22128 \end{align}$ So, the total dormitory charge in a public and private college is $19,678+22,128=41,806$. So, from the bar graph, the total dormitory charge is 41,804 and from the given models, the total dormitory charge is 41,806 dollars. Hence, the given model overestimate the total dormitory charge by 2.
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Im trying to solve the following Poisson equation:
$$\nabla^2\phi = F(x,y)$$ $$for\ x\in(0,\infty)\ and\ y\in(0,L)$$
$$\frac{\partial\phi(x,0)}{\partial y}=0\ , \ \frac{\partial\phi(x,L)}{\partial y}=0\ , \ \frac{\partial\phi(0,y)}{\partial x}=0$$ $$\phi(x,y)\rightarrow0\ uniformly\ as \ x \rightarrow\infty$$ $$$$
I want to solve this using the Fourier Transform or the separation of variable or the Green function. But I'm having trouble understanding since the equation is not homogeneous. And the boundary conditions make lots of trouble too.
Any help and suggestions are appreciated!
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Let $G=\langle x,y,z\mid z^2=1\rangle\cong \mathbb{Z}*\mathbb{Z}*\mathbb{Z}/2$.
I am interested to compute by hand (or by any other means) the quotient groups $\gamma_n/\gamma_{n+1}$ where $\gamma_n$ is the $n$th term of the lower central series. That is $\gamma_1=G$, $\gamma_2=[G,G]$, $\gamma_3=[\gamma_2,G]$, etc.
For $\gamma_1/\gamma_2$, it is essentially the abelianization of $G$, hence it is $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}/2$.
I am facing some difficulties computing $\gamma_2/\gamma_3$.
I computed that $\gamma_2=[G,G]=\langle x^{-1}y^{-1}xy,x^{-1}z^{-1}xz,y^{-1}z^{-1}yz\mid z^2=1\rangle$. (Update: This is probably wrong.)
However things start to get complicated with $\gamma_3=[\gamma_2,G]$. Is there an "easy" way to find $\gamma_2/\gamma_3$ or is brute-force the way to go?
Thanks.
Update: I think my expression for $\gamma_2$ may be wrong, there should be way more generators than just the 3 commutators $x^{-1}y^{-1}xy,x^{-1}z^{-1}xz,y^{-1}z^{-1}yz$, in fact $\gamma_2$ may not even be finitely generated?
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The first and most sensible question:-
Do you like carrots?
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not really,tell me do you like carrots?
Hey Mehul , do you have quadro phobia?
Hello Everyone!!!
Hello!
Hey
Can you help me out with this?
Do you like carrots?
I too have great interest in quantum physics and relativity .Keep eating carrots like Nihar because you too have the same number of letters in your name as is in carrots.I hope that you might solve the conflict of gravity and quantum mechanics in future.
so you want to do career in pure physics or something else?
It is very difficult for me to answer this question because of my wide range of interest in subjects of Mathematics,Physics,Computer Science and Astronomy.
which subject do you like other than mathematics and physics.
Computer Science and Astronomy.
Bonne journee is not literally used as 'good day' in francais. It's more like have a good day. Sorry to ruin it. :3
And potato or potato?
Got it thanks for that bro .Actually i am learning German not french
(http://shivamani patil)
Chinmay Sangawadekar
Sorry I was practicing to make the link . The blue one...
I wanted to spam :P
Please as you are a moderator do check this problem.Chinmay Sangawadekar
@A Former Brilliant Member – Formatting is alright.
Still it is not getting the rating.
@A Former Brilliant Member – Be patient , it will get rated as more and more people view it. If you have a valid reason for urgency for rating it , please tell .
@Chinmay Sangawadekar, i see that you have interest in contributing to wikis like multiplicative functions and mobius function. talk to me at slack so we could make the wiki better.
Brother , I have no access to slack , when I try , it says you are already invited , but I had deleted my emails already :( is there another way to communicate ? On Facebook ? @Aareyan Manzoor
login here. i had the same problems....
@Aareyan Manzoor – How is it ? @Aareyan Manzoor
@A Former Brilliant Member – oh yes that is cool... back to the topic of sections now. lets decide that what section are required for the wiki?
@Aareyan Manzoor – First intro.then theorem(definition)then proof(if necessary)(examples)
@Aareyan Manzoor – i enter my email and pass and got Sorry, you entered anincorrect emailaddress orpassword.
@Dev Sharma – Yeah same here , @Dev Sharma
https://crypto.stanford.edu/pbc/notes/numbertheory/mult.html , do check this out , it has many useful things which we can use for wiki
cool, just went through it.... i was planning to include those sums....
@Aareyan Manzoor – Yeah those are really awesome , so i have a plan , I will provide you with the stuff ,you select it and then tell me , then we will add those on wiki
@A Former Brilliant Member – ok.. that is a good plan... it is going to become overcrowded your message board. try facebook. i have the same username as brilliant (aareyan manzoor).
@Aareyan Manzoor – Sure I will add you immediately
@Chinmay Sangawadekar, lets talk about multiplicative function wiki first.
okay ,
go here for the wiki. what level audience do we target?
@Aareyan Manzoor – I think we should aim for high level audience
@A Former Brilliant Member – same in my mind. how 'bout lvl 3-5?
@Aareyan Manzoor – Sure , first we will make it for level 3-4 then we will ask Calvin Lin for guidance , then we go for advance
@A Former Brilliant Member – ok. the sections of the wiki then... how are we going to divide it?
@A Former Brilliant Member – now what sections are we going to divide the wiki in?
@Otto Bretscher , I wanted ask you about this
Does this mean , that n^{2015}=1 , ?
Yes, "primitive root of unity" means that 2015 is the smallest positive integer kkk such that ωk=1\omega^k=1ωk=1.
@Otto Bretscher , can you tell me a general formula to find the product of totient of divisors of any number n
@Isaac Buckley why did you choose Computer Sci. topic for this problem of yours ?
Can you solve it using number theory?
I dont know but problem looks like NT problem.Can you solve it by NT any tries ?
@A Former Brilliant Member – I'm not going to say you can't... but if anyone could... they would be a God in the eyes of mathematicians.
@Isaac Buckley – Oh yeah there is no specific formats for primes iant it?@Isaac Buckley
I'd love to see anyone try to solve it with number theory, a great prize awaits them.
@Isaac Buckley – M primes are so hard to find using comp. sci, i dont think number theory is the right approach!
@Isaac Buckley – Really , well is there any generelization for primes,in such form ? @Isaac Buckley
@A Former Brilliant Member – this
@A Former Brilliant Member – Well I'm not sure what you mean by that exactly. 2n−12^n-12n−1 is a possible prime only if nnn is prime. That's all I can say about this form.
Something you might be interesting in reading up on is Mills' constant. It's a way to find primes. It has major short comings though, still interesting.
@Isaac Buckley – Yes, but mills constant is hard to use as we need to have a good degree of acuracy!
@Aareyan Manzoor – Exactly! But it's still amazing that such a constant exists. It's also not known if it rational or not.
@Isaac Buckley – It is suprising how many numbers are yet to be proven irrational. for example e+πe+\pie+π is not proven to be irrational.
@Aareyan Manzoor – Yeah, I love problems like that. I might actually post a problem relating to e+πe+\pie+π later tonight. It came up in a past exam paper I took. If mills' constant was rational the implications could be pretty big, you could possibly (by magic) find its exact fraction and thus generate as many primes as you want.
@Isaac Buckley – Cool, maybe then someone might use it to prove the riemenn hypothesis!
@Aareyan Manzoor – Is it like that ?
@Isaac Buckley – Oh yes I agree with Aareyan, and thanks for the link.btw dont forget to try my problems and sorry for self advertising .;)
@Isaac Buckley – You there ?@Isaac Buckley
@Isaac Buckley – Is 2n−12^n - 12n−1 a sure prime or is it not sure for prime nnn? @Isaac Buckley
@A Former Brilliant Member – Only 49 n's have been found such that 2n−12^n-12n−1 iss prime..... so no!
@Aareyan Manzoor – Oh thanks for that info ..
@A Former Brilliant Member – It's just a prime candidate. Try and prove that if n is composite then 2n−12^n-12n−1 is also composite.
@Isaac Buckley – sure ,
I didn't, I chose number theory which was then changed to computer science by a moderator. I do not contest to it, it fits well in computer science.
@Aareyan Manzoor , @Nihar Mahajan , dont you think this is level 5 problem ?Somebody set level 3 for it.
It is okay. it doesnt use any hard concepts.
see thishttps://brilliant.org/problems/double-polynomials/?group=3Bg9df1xnfS9&ref_id=1094417 shouldnt it be lvl 5 ???
I think it must be level 3 only. Also , you must mention what ϕ\phiϕ function is.
Hey @Nihar Mahajan take a look at this I strongly feel that this problem should get level 5 (though it has no advanced concepts)atleast 200 or 195 points.Please do update the seed level.I will be very thankfull to you
@A Former Brilliant Member – Done. I think this is harder then the previous ones.
@A Former Brilliant Member – @Chinmay Sangawadekar in the future please dont tag mods like this, if a mod comes across a problem he will change the level. By tagging you are bothering them, mods are usually busy.
@Aareyan Manzoor see the link of problem
@Nihar Mahajan , have you started learning limits ?
I have finished it.
Great What source ? M Prakash notes ? @Nihar Mahajan
@A Former Brilliant Member – Calculus will be done in MPA during June-July. But due to curiosity I am learning it on my own.
@Nihar Mahajan – Hmm , so no school today for you ? @Nihar Mahajan
@A Former Brilliant Member – Nope, its saturday.
@Nihar Mahajan – Well ,also see my level two limits problem ,this@Nihar Mahajan
@A Former Brilliant Member – Yes , sure.
@A Former Brilliant Member – Solved it. @Chinmay Sangawadekar
@Nihar Mahajan – grt :)
@A Former Brilliant Member – Post more limit problems please, thanks :)
@Nihar Mahajan – Hey why dont you try and solve my NT and Algebra lvl 4-5 problems ?@Nihar Mahajan
@A Former Brilliant Member – Why don't you solve my all previous problems? @Chinmay Sangawadekar
@Nihar Mahajan – I was doing them only I did many of your recent problems @Nihar Mahajan
@Nihar Mahajan what is the teamdomain of brilliant on slack ?
@Nihar Mahajan you may try this
Thanks , I got it :)
@Nihar Mahajan , can you update my solution number ? I have written 6 and deleted my previous one so , can you do that ? or tell any staff member to do so ?
Sorry , I can't do that , you may ask Calvin on his messageboard.
@Nihar Mahajan , you may also try this
@Chinmay Sangawadekar You may try this , this .
solved ,,sorry ,, saw your comment now ,, Nice problems ..
@Nihar Mahajan try these problems ;) 1 , and 2
@Abhishek Sinha , Sorry to mention you here , but I wanted to know more about the Mathematics Department of MIT Can I get in there ? What is the process to do so ? Please tell..Also , what is your PhD topic ?
You should apply. The application procedure is publicly available. I am working on Network Control and Information Theory. Good luck!
@Otto Bretscher : Please see if it is true : We have , a≡1mod ba\equiv 1\mod ba≡1modb where , (a,b)(a,b)(a,b) are consecutive natural numbers such that , a>ba> ba>b , then , an≡1mod bna^{n}\equiv 1\mod bnan≡1modbn , this congruence is true .
No, this is not correct (unless I misunderstand your statement). As a counterexample, consider n=2,b=3,a=4n=2, b=3, a=4n=2,b=3,a=4.
Yeah right,,,btw also try my recent problem More Modular in 2016
hi
@Otto Bretscher sorry to mention you here , but in this problem , I have added 1 to the lambda function , so deno. Won't be 0 ...
Are we talking about the Liouville function as usual? That one takes the vales 1 and -1.
@Otto Bretscher
There is no fun in doing Number Theory without you...:(
I really don't want to post any more problems... then I have to deal with some of these obnoxious guys when they post solutions. Maybe I will give you some ideas and you can post, Comrade!
I am really sorry , all this has happened because of me ...:( @Otto Bretscher
@A Former Brilliant Member – Not at all... you are my best Comrade here, @Chinmay Sangawadekar
@Otto Bretscher can you link me to some solutions like this, sorry if your experience in brilliant is harmed by them.
@Aareyan Manzoor – Solutions like what, Comrade?
@Otto Bretscher –
I really don't want to post any more problems... then I have to deal with some of these obnoxious guys when they post solutions.
these solutions posted by obnoxious guys.
@Aareyan Manzoor – I don't want to name names, but I think you know the type.
@Otto Bretscher – yes, i think i will go through your profile searching on my own for these guys.
@Aareyan Manzoor – Aareyan please take strict actions for people like him...
Humble Request: Please don't stop posting problems because of obnoxious guys. You are a crucial part of community and so are your problems. We enjoy your problems very much. If some guys are troubling you by any means , you may report it in private or on message board of any staff and the staff will take the appropriate action. Grooming math enthusiasts and comrades like Chinmay , Aareyan , me need your help on Brilliant , but if you stop posting problems like this for some reason you are not responsible for , we would be discouraged :(
@Nihar Mahajan – i agree; without his problem i never would have learned abt half the things iknow now, please dont stop otto sir; you are one of the best problem posters here.
@Nihar Mahajan – I agree with you aareyan...and Nihar...
Oh lord no! Your problems are really integral to the community, sir. Please don't stop posting problems. As Nihar rightly did point out, you may report any such action to the staff, so that appropriate action can be taken. But please don't stop posting problems! :(
@Otto Bretscher
Please look into my latest problem , Gauss won't help,, it is getting a lot of reports , but still I think problem may be right...help me comrade...
It's great problem and your answer is correct, Comrade. You should write the Legendre symbol more clearly as (n(n+1)10009)\left(\frac{n(n+1)}{10009}\right)(10009n(n+1)) though
@Otto Bretscher dsorry to disturb you ...https://brilliant.org/problems/more-problems-in-2016-part-11/?group=JgjXt4dJvxew&ref_id=113
@Aareyan Manzoor @Sharky Kesa see this problem......
this
404 is a palindrome in base: -31
Check my profile then...
thisand this
Kindly see into those problems...They are getting low response and also they are waiting for your awesome solutions...;)
Looks like everybody is busy at school or work, including myself ;)
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math.PR
Change to browse by: References & Citations Bookmark(what is this?) Mathematics > Probability Title: On non-uniqueness in mean field games
(Submitted on 16 Aug 2019)
Abstract: We analyze an $N+1$-player game and the corresponding mean field game with state space $\{0,1\}$. The transition rate of $j$-th player is the sum of his control $\alpha^j$ plus a minimum jumping rate $\eta$. Instead of working under monotonicity conditions, here we consider an anti-monotone running cost. We show that the mean field game equation may have multiple solutions if $\eta < \frac{1}{2}$. We also prove that that although multiple solutions exist, only the one coming from the entropy solution is charged (when $\eta=0$), and therefore resolve a conjecture of ArXiv: 1903.05788. Submission historyFrom: Erhan Bayraktar [view email] [v1]Fri, 16 Aug 2019 23:57:23 GMT (345kb)
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Epidemiology: The SIR model
Simulation of differential equations with turtle graphics using JSXGraph.
Contents SIR model without vital dynamics
The SIR model measures the number of susceptible, infected, and recovered individuals in a host population. Given a fixed population, let [math]S(t)[/math] be the fraction that is susceptible to an infectious, but not deadly, disease at time t; let [math]I(t)[/math] be the fraction that is infected at time [math]t[/math]; and let [math]R(t)[/math] be the fraction that has recovered. Let [math]\beta[/math] be the rate at which an infected person infects a susceptible person. Let [math]\gamma[/math] be the rate at which infected people recover from the disease.
A single epidemic outbreak is usually far more rapid than the vital dynamics of a population, thus, if the aim is to study the immediate consequences of a single epidemic, one may neglect birth-death processes. In this case the SIR system can be expressed by the following set of differential equations:
[math] \frac{dS}{dt} = - \beta I S [/math] [math] \frac{dR}{dt} = \gamma I [/math] [math] \frac{dI}{dt} = -(\frac{dS}{dt}+\frac{dR}{dt}) [/math] Example Hong Kong flu initially 7.9 million people, 10 infected, 0 recovered. estimated average period of infection: 3 days, so [math]\gamma = 1/3[/math] infection rate: one new person every other day, so [math]\beta = 1/2[/math]
Thus S(0) = 1, I(0) = 1.27E-6, R(0) = 0, see [1].
The lines in the JSXGraph-simulation below have the following meaning:
* Blue: Rate of susceptible population * Red: Rate of infected population * Green: Rate of recovered population (which means: immune, isolated or dead)
The underlying JavaScript code
var brd = JXG.JSXGraph.initBoard('box', {axis: true, boundingbox: [-5, 1.2, 100, -1.2]});var S = brd.create('turtle',[],{strokeColor:'blue',strokeWidth:3});var I = brd.create('turtle',[],{strokeColor:'red',strokeWidth:3});var R = brd.create('turtle',[],{strokeColor:'green',strokeWidth:3}); var s = brd.create('slider', [[0,-0.3], [30,-0.3],[0,1.27E-6,1]], {name:'s'});brd.create('text', [40,-0.3, "initially infected population rate (on load: I(0)=1.27E-6)"]);var beta = brd.create('slider', [[0,-0.4], [30,-0.4],[0,0.5,1]], {name:'β'});brd.create('text', [40,-0.4, "β: infection rate"]);var gamma = brd.create('slider', [[0,-0.5], [30,-0.5],[0,0.3,1]], {name:'γ'});brd.create('text', [40,-0.5, "γ: recovery rate = 1/(days of infection)"]);var t = 0; // globalbrd.create('text', [40,-0.2, function() {return "Day "+t+": infected="+(7900000*I.Y()).toFixed(1)+" recovered="+(7900000*R.Y()).toFixed(1);}]); S.hideTurtle();I.hideTurtle();R.hideTurtle();function clearturtle() { S.cs(); I.cs(); R.cs(); S.hideTurtle(); I.hideTurtle(); R.hideTurtle();} function run() { S.setPos(0,1.0-s.Value()); R.setPos(0,0); I.setPos(0,s.Value()); delta = 1; // global t = 0; // global loop();} function turtleMove(turtle,dx,dy) { turtle.moveTo([dx+turtle.X(),dy+turtle.Y()]);} function loop() { var dS = -beta.Value()*S.Y()*I.Y(); var dR = gamma.Value()*I.Y(); var dI = -(dS+dR); turtleMove(S,delta,dS); turtleMove(R,delta,dR); turtleMove(I,delta,dI); t += delta; if (t<100.0) { active = setTimeout(loop,10); }}function stop() { if (active) clearTimeout(active); active = null;}function goOn() { if (t>0) { if (active==null) { active = setTimeout(loop,10); } } else { run(); }}
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From Rogawski ET 2e section 10.7, exercise 4.
Find the Maclaurin series for $f(x) = \dfrac{x^2}{1-8x^8}$.
$$\dfrac{x^2}{1-8x^8} = \sum_{n=0}^{\infty} [\textrm{_______________}]$$
Hi! I am working on some online Calc2 homework problem and I am not quite sure how to go about solving this Taylor series. I know I should substitute $8x^8$ for $x$ in the Maclaurin series for $1 \over (1-x)$, but the $x^2$ in the numerator of the problem is throwing me off. If someone could help me find the Maclaurin series and on what interval the expansion is valid, I would greatly appreciate it!
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Jeff, your questions were in some sense the motivation for my thesis. Let me say a few things that you probably already know before I try and answer your questions.
The $E_\infty$ algebra structure on integral cochains of a topological space $X$ is a homotopy invariant of $X$. If $X$ is nilpotent and of finite type, then the quasi-isomorphism type of the cochain algebra is a complete homotopy invariant. This is a theorem of Mandell:http://arxiv.org/abs/math/0311016.
It is also true that the $C_\infty$ multiplication on cochains is a complete invariant of the rational homotopy type of a simply connected space, but I don't know of a place where this is written down explicitly. The problem is that Quillen and Sullivan wrote their papers before ideas like infinity algebras and Kozul duality were part of the general consciousness of topologists. However, Quillen shows in Rational Homotopy Theory that the (cocomutative) coalgebra on chains is a complete invariant of the rational homotopy type of a simply connected space. He goes to some trouble to construct a cocomutative coalgebra; nowadays we would say that he is just constructing a particular representative of the quasi-isomorphism type of cochain $C_\infty$ coalgebra which happens to be strictly associative.
Now, let me try and restate Jeff's question 1, which makes sense over Q or over Z. Fix a simply connected integral or rational PD space $X$. We know:
1) The cochain algebra $C^*(X)$, considered as an E or C infinity algebra on the integral or rational cochains, is a complete integral or rational homotopy invariant given restrictions on the fundamental group.
2) The homotopy class of the map $\mu_X: X \to BG$ which determines in the Spivak normal fibration is a homotopy invariant of $X$. (I don't actually know much about the rational version of this statement, but it looks like it's laid out in Su's thesis linked above.)
3) The topological (or smooth or PL) structure set of manifold structures in the homotopy type of $X$ is again a homotopy invariant of $X$.
I interpret Jeff's question 1 to be the following: the cochain algebra knows the all the homotopy invariant information about $X$, so how do we see the info of (2) and (3) as features of the cochain algebra? The problem is that the cochain algebra depends only on the homotopy type of $X$ as a $space$, not as a Poincare duality space. I don't think that that you can ever see, for example, the structure set from only the higher homotopies of the cup product. (Though I don't have a formal proof that it's impossible.)
If you want to detect manifold structures, you instead need to look at the Poincare duality map. In my thesis, I explain how you can write down Ranicki's total surgery obstruction -- which detects whether or not the structure set is empty -- as an obstruction to the existence of "local" inverse to the Poincare duality map. (This statement is over Z. I don't know of a rational version of Ranicki's total surgery obstruction, and Ranicki told me he doesn't either.)
Thus, as I understand it, the higher multiplications are not exactly the right place to look for obstructions to manifold structures; you need to look instead at the inverse of the Poincare duality map.
I know of a couple of different answers to Jeff's question 2 about the relation between the colagebra and the algebra structure on cochains.
1) There is the following paper of Tradler and Zeinalian:http://arxiv.org/pdf/math/0309455v2One result of this paper is that the rational chains of a PD space form an $A_\infty$ coalgebra with an "$\infty$ duality". Presumably there is a dual statement for the cochains.
2) David Chataur has a result that for any PD space X, the PD map determines an equivalence of the cochains and chains of X as $E_\infty-C^*(X)$ modules. He sent me a sketch of the proof of this statement but I don't have his permission to disseminate it.
Morally, the answer should be that the PD map is an equivalence of "infinity Frobenius" algebras. Unfortunately there are many different definitions of Frobenius algebra, and there are technical problems with writing down the infinity versions of some these algebraic structures. (The ones that have a unit and a counit.) However see this paper of Scott Wilson's:http://arxiv.org/abs/0710.3550
That ended up being a long answer! Please ask if something isn't clear!
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Are you planning to study engineering, physics or another courseinvolving lots of applied mathematics at university?
In your first year you will undoubtedly take maths courses to bringyou up to speed with many of the advanced mathematical techniquesrequired in the serious applications of mathematics to the physicalsciences.
Whilst courses and content do, of course, vary, there is a generalcore of content most physical scientists will encounter and makeuse of in their first year of study. Unfortunately, however, it issometimes not immediately clear how the maths you are taught willbe useful in the physical and practical applications.
This article lists some of the topics you are very likely to comeacross, and explains just some of the areas where they can beapplied to physical science problems. The topics below also link toarticles and problems where you can learn about them in moredetail. The more familiar you are with these concepts and newvocabulary the smoother your transition to university should be.Good luck!
1. Vectormanipulation
An important definition here is the cross product. This defines avector ${\bf a} \times {\bf b}$ which is perpendicular to both${\bf a}$ and ${\bf b}$, and has magnitude $\mid {\bf a} \mid \mid{\bf b} \mid \sin \theta$ . Suppose we have a spherical objectwhich rotates about a fixed axis at a constant speed, such as theearth. Individual points on the surface will move at differentspeeds, depending on whether they are closer to the pole or theequator. By using the cross product we can concisely define thespeed at every point, it turns out it is given by $\Omega \times{\bf x}$, where $\Omega$ is the angular velocity of the earth andis a constant vector pointing along its axis, and ${\bf x}$ is theposition vector from the centre of the earth to the point on thesurface we're interested in.
The cross product is crucial in the physical sciences when usingMoments. Moments allow us to calculate how an object, pivoted at acertain point, will move when a force is exerted upon it. Considerpushing a door. If the (shortest) vector from the hinge axis to theplace where the door is pushed is ${\bf r}$, and the force exertedis ${\bf F}$, then the moment is ${\bf r} \times {\bf F}$, and isequal to the moment of inertia of the door (a property of the door)multiplied by the acceleration of the door around its axis, thehinge. This is a form of Newton's second law, Force = Mass xAcceleration. The further away from the hinge the door is pushed,the larger moment is exerted, and the faster the door moves. New definitions and techniques:-
Triple products, ${\bf a} \times {\bf b} \times {\bf c}$ and ${\bfa} \cdot {\bf b} \times {\bf c}$.Vectors are useful for:
Findingthe shortest distance between two lines. Defining theElectromagnetic equations. And lots more.Suggested NRICH problems
: V-P Cycles
; Flexi Quads
; Flexi-Quad TanFurther information - NRICHarticles
: Vectors - What AreThey?
; Multiplication ofVectors
2. Matrices
A Matrix is a way ofconcisely setting up several linear relationships at once, such asa system of simultaneous equations. These relationships maytransform a vector $(x_0,y_0)$ to a new vector $(x_1,y_1)$ in aparticular way, for example rotating the original vector by$\theta$ about the origin. Or the relationships may be a set oflinear differential equations which we need to solvesimultaneously.
You will have come acrossproblems involving a single spring attached to a single mass (seeBuilding UpFriction
for an in-depth question) but now imagine a coupledsystem, such as a train with two carriages with connectors that canbe modelled as springs (spring constants $k_1$ and $k_2$,unstretched lengths $L_1$ and $L_2$)
Say the train comes to astop, and we are interested in the subsequent motion of thecarriages. There are two degrees of freedom, $x_1$ and $x_2$, thepositions of the two carriages. We find the followingequations
$m_1 \ddot{x}_1 = - k_1(x_1 - L_1) + k_2 (x_2 - x_1 -L_c - L_2)$
$m_2 \ddot{x}_2 = -k_2(x_2 - x_1 - L_c - L_2)$
We can put this intomatrix form
$ \left(\begin{array}{cc} m_1 & 0 \\ 0 & m_2 \end{array} \right)\left( \begin{array}{c} \ddot{x}_1 \\ \ddot{x}_2 \end{array}\right) + \left( \begin{array}{cc} k_1 + k_2 & -k_2 \\ -k_2& k_2 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2\end{array} \right) = \left( \begin{array}{c} k_1 L_1 - k_2(L_2 +L_c) \\ k_2 (L_2 + L_c) \end{array} \right)$
Or, more concisely
$M \ddot{X} + K X =F$
where $M$, $K$ and $F$could be described as the mass, spring constant and forcingmatrices respectively. We can now use matrix techniques to solvethe problem.New definitions and techniques:
-Eigenvectors and eigenvalues. Matrix multiplication. Inverse of aMatrix. Determinants.Matrices are useful for:
Physicalsystems with several degrees of freedom. Solving simultaneousequations. Writing mathematical expressions concisely.Suggested NRICH problems
: The Matrix
; Fix Me or CrushMe
; Transformations for10
3. Complex numbers
When faced with adifferential equation to solve, very often in the physical sciencesyou look for solutions of the form $e^{i \omega t}$, (where $t$ isthe variable we are differentiating with respect to) and then solvefor $\omega$. [Note that engineers often use $j$ for the squareroot of -1, rather than $i$, so as not to confuse it with thesymbol for current.] This simplifies things considerably, asdifferentiating becomes equivalent to multiplication by $i \omega$.It also means undergraduates need to be very comfortable withmanipulating complex numbers, and the
$e^{i \omega t} = \cos(\omega t) + i \sin (\omega t)$
formula.
So how do we know inadvance that complex exponentials are a good idea for the solutionto a given problem? In part, the answer is Fourier Theory (seebelow), which shows that any function (pretty much), can bere-expressed as a sum of exponentials. Basically, exponentials arethe fundamental solutions to Linear Differential Equations, and allother solutions are made up of these building blocks.
New definitions and techniques:-$r e^{i \theta}$ notation. Complex roots. De Moivre'sTheorem.
Complex numbers are useful for:
Solving Linear Differential Equations in Electronics, Vibrationsand Mechanics. See for example the article AC/DC Circuits
.Fourier Series.
4. Fourier Series
Fourier Series are aclever way of re-expressing a periodic function, $f(x)$, as a sumof sines and cosines. There's quite a nice graphical illustrationof this for a square wave in the Wikipedia article on FourierSeries.
It is important toremember that sines and cosines are made up of exponentials (manypeople forget these useful relationships when doing problems - canyou remember them?), and thus a Fourier Series is also a clever wayof re-expressing a periodic function in terms of exponentials.Since exponentials are particularly easy to differentiate andintegrate, by splitting a problem into lots of exponentialcomponents we can often solve differential equations which wecouldn't solve with the original $f(x)$. Or there may be standardsolutions for exponentials which someone else has previously workedout (this is where engineers' databooks come in handy).
Now the really clever bitis that usually in real life we are only interested in a limitedrange of $x$, and so we can 'pretend' a function is periodic whenactually it isn't, by defining it outside the range of interest.Say we are only interested in $0< x< L$, we simply define$f(x) = f(x-L)$ for $x> L$ and, ta dah!, we have a periodicfunction.
New definitions and techniques:-Rate of convergence - how many terms of the Series will you need toget a good estimate for the original function?
Fourier Series are useful for:Solving Linear Systems, that is one or more linear differentialequation. Re-expressing a function in terms of exponentials orsines/cosines.
5. Differentialequations
A lot of undergraduateactivity in the physical sciences is about constructing and solvingDifferential Equations (DEs). Often the hardest part of a questionis writing down the correct equation from the wordy description youare given. If you haven't got the correct equation to begin with,solving it can prove impossible! Sign errors, such as confusingwhich direction a force is acting in, can lead to big problems. Theway to eliminate these is practice and careful working, as well asthinking about whether the equation you have found makes sense(e.g. if you increase the force in the equation, will the systemrespond as you would expect from common sense?).
Techniques for solvingdifferential equations are an important part of the physicalscientist's tool box. You will probably have met in A-Level: Changeof variables; Integrating factors; Separation of variables. Newtechniques used at university include:
Differenceequations. These are often used in solving asystem of equations when stepping forward in time. Instead ofsolving a DE to find a continuous function $f(t)$, where $t$ (time)can be any real number greater than zero, we seek $f$ at specificpoints in time, $f_1 = f(t_1)$ etc. This means we can write out aset of simultaneous equations to solve algebraically, instead ofdifferential equations.
Convolution(also called Green's functions). Suppose we have a differentialequation with some function $f(x)$ on the right hand sidee.g.
$\frac{d^2y}{dx^2} +\alpha y = f(x)$
Convolution is a cunningway of finding $y(x)$ for any $f(x)$just by working out what $y(x)$ would be if $f(x) = \delta(x -x_0)$. This involves another new mathematical idea you will learnabout, delta functions. The delta function $\delta (x-x_0)$represents an infinite spike at $x = x_0$, and although it seemsrather an odd concept at first (a bit like imaginary numbers), itturns out to be incredibly useful.
LaplaceTransforms and FourierTransforms. As implied in Difference equations above,solving algebraic equations (e.g. $3 y^2 - 5 y = 0$) is usuallymuch easier than solving equations involving derivatives. Applyinga Transform to a DE means integrating the whole thing with respectto a completely new variable, (conventionally $k$ or $\omega$).This gives us a new equation to solve, where crucially anyderivative terms from the original equation are now just algebraicterms, e.g. $\frac{d^2 y}{d x^2} \rightarrow - k^2 {\hat y}$.(${\hat y}$ is the new function we are now solving for). Cleverstuff.Differential equations are goodfor:
Modelling physical systems.Suggested NRICH problems: DifferentialElectricity
; Euler's BucklingFormula
; Ramping it Up
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cs.IT
Change to browse by: References & Citations Bookmark(what is this?) Computer Science > Machine Learning Title: Finite Precision Stochastic Optimization -- Accounting for the Bias
(Submitted on 22 Aug 2019 (v1), last revised 26 Aug 2019 (this version, v2))
Abstract: We consider first order stochastic optimization where the oracle must quantize each subgradient estimate to $r$ bits. We treat two oracle models: the first where the Euclidean norm of the oracle output is almost surely bounded and the second where it is mean square bounded. Prior work in this setting assumes the availability of unbiased quantizers. While this assumption is valid in the case of almost surely bounded oracles, it does not hold true for the standard setting of mean square bounded oracles, and the bias can dramatically affect the convergence rate. We analyze the performance of standard quantizers from prior work in combination with projected stochastic gradient descent for both these oracle models and present two new adaptive quantizers that outperform the existing ones. Specifically, for almost surely bounded oracles, we establish first a lower bound for the precision needed to attain the standard convergence rate of $T^{-\frac 12}$ for optimizing convex functions over a $d$-dimentional domain. Our proposed Rotated Adaptive Tetra-iterated Quantizer (RATQ) is merely a factor $O(\log \log \log^\ast d)$ far from this lower bound. For mean square bounded oracles, we show that a state-of-the-art Rotated Uniform Quantizer (RUQ) from prior work would need atleast $\Omega(d\log T)$ bits to achieve the convergence rate of $T^{-\frac 12}$, using any optimization protocol. However, our proposed Rotated Adaptive Quantizer (RAQ) outperforms RUQ in this setting and attains a convergence rate of $T^{-\frac 12}$ using a precision of only $O(d\log\log T)$. For mean square bounded oracles, in the communication-starved regime where the precision $r$ is fixed to a constant independent of $T$, we show that RUQ cannot attain a convergence rate better than $T^{-\frac 14}$ for any $r$, while RAQ can attain convergence at rates arbitrarily close to $T^{-\frac 12}$ as $r$ increases. Submission historyFrom: Himanshu Tyagi [view email] [v1]Thu, 22 Aug 2019 04:57:22 GMT (49kb) [v2]Mon, 26 Aug 2019 04:56:31 GMT (49kb)
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An Introduction to LaTeX
LaTeX is a way of writing documents that is more like writing a program than using Word. You write the "source code" using a text editor (Notepad or Word will do) then you "compile" it. You can get "front-ends" of various complexity to help you, but I'll concentrate on the basics here without depending on particular operating systems or extra programs.
So why use LaTeX What are the pros and cons?
Free! On Macs, PCs, Linux, etc. Source files are small and easily transferable. You can see how effects are achieved. Tricks can be passed on. Typesetting's better, especially maths (used by CUP) - see Why TeX? Solid? (TeX yes, LaTeX no?) Many "add-ons" (in fact, LaTeX is an add-on to TeX) Font selection is difficult, and there are few fonts LaTeX was designed with technical reports very much in mind. It encourages (almost insists on) structured writing and the separation of style from content. This is not the way that many people (especially non-programmers) are used to working. Not good at graphics, layouts Without a WYSIWYG front end, it's not always easy to find out how to do things.
Here's a latex file
\documentclass{article} \usepackage{times} \begin{document} \section{Introduction} Hello! \end{document}
If I save it as
foo.tex and process it I get the following, which isn't as scary as it looks.
This is pdfTeX, Version 3.1415926-2.5-1.40.14 (TeX Live 2013) restricted \write18 enabled. entering extended mode (./foo.tex LaTeX2e <2011/06/27> Babel and hyphenation patterns for english, dumylang, nohyphenation, lo aded. (/usr/share/texlive/texmf-dist/tex/latex/base/article.cls Document Class: article 2007/10/19 v1.4h Standard LaTeX document class (/usr/share/texlive/texmf-dist/tex/latex/base/size10.clo)) (/usr/share/texlive/texmf-dist/tex/latex/psnfss/times.sty) No file foo.aux. (/usr/share/texlive/texmf-dist/tex/latex/psnfss/ot1ptm.fd) [1{/usr/share/texliv e/texmf/fonts/map/pdftex/updmap/pdftex.map}] (./foo.aux) ){/usr/share/texlive/t exmf-dist/fonts/enc/dvips/base/8r.enc} Output written on foo.pdf (1 page, 14173 bytes). Transcript written on foo.log.
You don't usually need to worry about these details, unless something goes wrong ...
" pdfTeX, Version 3.1415926 ..."- this tells you the program that's going to do the eventual processing. The version number converges to pi.
"LaTeX2e <2011/06/27> ..."- information about the LaTeX "add-on"
"/usr/share/texlive/texmf-dist/tex/latex/base/article.cls"- where the
articleclass file is.
"/usr/share/texlive/texmf-dist/tex/latex/base/size10.clo"- where the file to set things up for a 10pt font is (
clostands for 'class option')
"/usr/share/texlive/texmf-dist/tex/latex/psnfss/times.sty"- where the package for the times font is
"No file foo.aux."- LaTeX looks for a
*.auxfile when it runs
"/usr/share/texlive/texmf-dist/tex/latex/psnfss/ot1ptm.fd"- a
*.fdfile is a font description file. This one describes the times font.
"(./foo.aux)"- an extra file produced for the next latex run. Sometimes several files are produced.
As you write bigger files you'll need more commands. If you're writingan essay you might not need to know many more:
\textit{...} does italics,
\includegraphics{filename.jpg} includes graphics (you need
\usepackage{graphicx} to make it work),
\tableofcontents adds a table of contents. I suggest you look up features as you go along using the resources mentioned below.
Accept LaTeX's funny little ways. It has built-in aesthetics
The first paragraph of a section isn't indented! It will leave space at the end of a page rather than squeeze a section header there. It doesn't like squeezing a bit of text onto a page that's mostly graphics
You can change this behaviour if you want (see Squeezing Space in LaTeX), but latex probably knows best.
Some free and commercial programs make life easier
kileis installed on our linux terminals. You still need to type LaTeX code, but Kile has many facilities (templates, wizards, etc) to make it easier - see a screen dump. proTeXt (for Windows) and Scientific Workplace (for Windows - it isn't free) work the same way
lyx is a WYSIWYG front-end for LaTeX that's getting better all the time. It's installed on our linux servers - see a screen dump Overleaf lets you write LaTeX docs and work collaboratively without needing to install anything.
LaTeX is good at maths, automated numbering, and enforcing stylistic uniformity. The following example shows how various types of cross-referencing can be easily maintained.
\documentclass{article} % It's fairly common to use a sans-serif font for headings % The following line, if uncommented, does this using a package % \usepackage{sfheaders} % If you don't have the package on your machine, download it % from http://www.ctan.org/tex-archive/macros/latex/contrib/sfheaders/ % If you want table numbers to be reset in each section, use the next 2 lines \usepackage{chngcntr} \counterwithin{table}{section} \usepackage[font=small,format=plain,labelfont=bf,up,textfont=it,up]{caption} % The first time you process this document you'll need to process it % twice to make the references work. \usepackage{times} \begin{document} \tableofcontents \listoffigures \listoftables \section{One} This document has no tables or figures, but it has the captions for them. Later you might be interested in table \ref{TABLEB} or in a footnote\footnote{some small print} \begin{table}[htbp] \caption{\label{TABLEA}A table} \end{table} \section{Two} Now for some maths \begin{equation} x=y\sum_{i=0}^\infty\sin(i)\label{SIN} \end{equation} \subsection{Detail} We'd better go into detail now \begin{figure}[htbp] \caption{\label{FIGA}A figure} \end{figure} \subsubsection{More detail} \begin{table}[htbp] \caption{\label{TABLEB}Another table} \end{table} Earlier in equation \ref{SIN} we set x to a value. \end{document}
Running
pdflatex on this twice will produce this PDF file. Then if you have
latex2html installed you can type "
latex2html
filename" to produce a tree of WWW pages. Books - "The LaTeX Companion (2nd edition)" will tell you a lot, but it's not really a beginner's book. Handouts The Not So Short Introduction to LaTeX2e is a 141 page introduction by Tobias Oetiker et al. It's probably all you need. There are versions in many languages. Using Imported Graphics in LaTeX and PDFLaTeX (by Keith Reckdahl) explains all there is to know about putting graphics into LaTeX documents. Our rather old Advanced LaTeX and LaTeX Maths and Graphics documents are cook-books of features that students commonly use. Online Resources - start at our LaTeX page tex.stackexchange.com is a forum for questions and answers Detexify - LaTeX symbol classifier lets you draw a symbol and will give you the corresponding LaTeX What Next
Maybe try
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On any one flip, the probability that all three coins come up the same is $\frac{2}{8}=\frac14$ and the probability that they don't is $\frac34$. Hence, the probability that the three coins come up the same for the first time at exactly the $n$-th flip is ${\left(\frac34\right)}^{n-1}\frac14$.
The expectation is therefore given by
$$\sum_{n\geq 1}n{\left(\frac34\right)}^{n-1}\frac14=\frac14\cdot\sum_{n\geq 1}n{\left(\frac34\right)}^{n-1}.$$
Now, consider $f(x)=\sum_{n\geq0}x^n=\frac{1}{1-x}$, $|x|<1$. We have that
$$f'(x)=\sum_{n\geq1}nx^{n-1}=\frac{1}{{(1-x)}^2}.$$
It follows that the expectation equals
$$\frac14\cdot f'\left(\frac34\right)=\frac14\cdot\frac{1}{{\left(\frac14\right)}^2}=\frac14\cdot16=4$$
This was checking the whole thing via definitions. But you could have skipped it all noting that the chance of 'success' at each step is $1/4$, so the expected value, naturally, would be $4$.
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I'm working through James Stewart's
Precalculus, and I have some confusion regarding this question: "Find the exact value of the trigonometric function at the given real number: $\cot \frac{25\pi}2$."
So, easy enough... $\frac{25\pi}2$ is simply a multiple of $\pi\over2$. $\cot$ is
undefined at intervals of $n\pi$ where $n$ is any integer, and the value of $\cot$ at $\pi\over2$ is $0$, as evidenced by the graph below:
Now comes the part, that befuddles me: working it out algebraically, I do the following.
$$\begin{align} \cot \frac{25\pi}2 &= \cot \frac\pi2 \\[4pt] &= \frac{1}{\tan(\pi/2)} \\[4pt] &= \frac{1}{\text{undefined}} \end{align}$$
It is here that I get stuck:
How is that $\cot$ is defined for that value when $\tan$ itself is not?
I would imagine that a numerator divided by a denominator of an undefined value would be undefined? Could anyone explain how the value of 0 for cotangent would be worked out in terms of tangent. Is it incorrect to relate cotangent and tangent like this:
$$cot = \frac1{tan}$$
I hope that makes sense and is not too a silly a question.
Thanks in advance!
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I am unsure if this is even possible in Mathematica, but my optics professor assigned a project to me on circular apertures and Fourier transforms. I have found plenty on the Airy function, but I am wondering if there is any way to produce an image like the one below in Mathematica:
Physicist chiming in -
Hi!.
I believe there has been some confusion here. It seems to me that OP is meaning to plot an Airy disk which was studied by G.B. Airy but is
not given by the Airy function. It is given by the Fourier transform of the indicator function of the unit circle, which actually happens to be a Bessel function (see e.g. wikipedia).
If I understood this right, then the correct solution is as follows:
DensityPlot[BesselJ[1, Sqrt[x^2 + y^2]]/Sqrt[ x^2 + y^2], {x, -60, 60}, {y, -60, 60}, PlotPoints -> 100, PlotRange -> All]
You can play around with the options of
DensityPlot to increase the contrast, add a legend, or change the colour scheme into something more similar to your intended image. I leave this to you.
--
For the mathematically inclined: we are dealing with what we physicists call Fraunhofer diffraction. Given a profile $f(x_1,x_2)$, the corresponding diffraction pattern is proportional to $\tilde f(\xi_1,\xi_2)$. In our case, the profile is a solid disk, so $f(x_1,x_2)=1_{D(1)}=\theta(1-\sqrt{x_1^2+x_2^2})$, with $\theta$ the step-function.
Unfortunately,
FourierTransform[HeavisideTheta[1 - Sqrt[x^2 + y^2]], {x, y}, {a, b}]
is returned unevaluated. Fortunately, this Fourier integral is a well-known result, the quoted Bessel function. It would be nice to know why MMA is unable to evaluate this classic integral. Oh well.
I feel a bit bad about sliding in so lately. @Nasser's answer was already very good. However, the function is very oscillatory, so I thought it might be a better idea to use a
Texture to render it. I use
ParametricPlot for that along with a suitable
TextureCoordinateFunction. Since the function is radially symmetric, the texture needs only be produced for a very narrow strip. This is done by the function
tex.
R = 20;(*half box size*)n = 1000;(*resolution of texture*)x0 = N[Range[0, R, R/n]];tex = cut \[Function] Module[{c, cols}, c = Map[ r \[Function] If[Abs[r] <= 10^-12, 0., Max[cut, Log10[Abs[2.0 BesselJ[1, Pi r]/(Pi r)]]]], Sqrt[2.] x0 ]; cols = ColorData["SunsetColors"] /@ Rescale[c]; Image[ConstantArray[Developer`ToPackedArray[List @@@ (cols)], 10]]];texfun = {x, y, u, v} \[Function] Evaluate[{Sqrt[(u^2 + v^2)]/R/Sqrt[2], 0.5}];plot = cut \[Function] ParametricPlot[{x, y}, {x, -R, R}, {y, -R, R}, TextureCoordinateFunction -> texfun, TextureCoordinateScaling -> False, PlotStyle -> {Opacity[1], Texture[tex[cut]]}, BoundaryStyle -> None, PlotPoints -> 100, PlotRangePadding -> 0, Axes -> False];
Moreover,
tex produces a logarithmic texture where all values below the truncation argument
cut are replaced by
cut. The effect of different truncatations can be seen below:
g = GraphicsGrid[Partition[ Table[ plot[cut], {cut, {-1., -1.5, -2., -2.5, -3., -3.5}} ], 3], ImageSize -> Full]
The Fourier transforms of Airy function is given by what is called Sombrero function
So here is something to get you started. This finds the radius to each pixel over some square region and applies the definition of
Sombrero and plots the result using
ListDensityPlot
meshgrid[x_List,y_List]:={ConstantArray[x,Length[x]],Transpose@ConstantArray[y,Length[y]]};n = 50; (* grid size *)z = 1/2; (*adjust as needed*)x0 = Range[1,n,z]-n/2; (*shift center*){xx,yy} = meshgrid[x0,x0]; r = Sqrt[xx^2+yy^2]; (*radius*)sombro = Map[ If[#==0,0,2.0 BesselJ[1,Pi #]/(Pi #)]&,r,{2}];ListDensityPlot[sombro,ColorFunction->"SunsetColors"]
You can play with different plots or option to make it closer to what you showed. For example, using
z=1/4 or different
n or other scaling one can do.
I can't improve on the great plots already presented, but MMA can indeed solve the problem. Start with an energy wave from a circular aperture
Eeq=dE==A dx dy Sin[2Pi f t-(2\[Pi] r)/lambda]
Point P on the observation screen has coordinates X, Y, Z, with the origin at the center of the aperture.Z is the distance from the center of the aperture to the center of the screen.Distance from center of aperture to point P on screen
r0 is
r0=Sqrt[X^2+Y^2+Z^2]
Dist r, from a point x,y,0 on the aperture to Point P on the screen is
r=Sqrt[(X-x)^2+(Y-y)^2+Z^2]
Use the small aperture approximation to make simplifications
rhs=(r^2-r0^2//ExpandAll)/.{x^2->0,y^2->0}Clear[r,r0]r^2-r0^2//Factor(*(r-r0) (r+r0)*)lhs=(%/.r+r0->2r0)Solve[lhs==rhs,r]//FlattenRrule=r->r0-(x X+y Y)/r0Eeq=Eeq/.Rrule (* dE==A dx dy Sin[2 Pi f t-(2 Pi (r0-(x X+y Y)/r0))/lambda] *)
Convert from x, y on the aperture to rho, phi and X, Y on the screen to R Phi.
x=rho Cos[phi];y=rho Sin[phi];X=R Cos[Phi];Y=R Sin[Phi];Eeq(* dE==A dx dy Sin[2 \[Pi] f t-(2 \[Pi] (r0-(R rho Sin[phi] Sin[Phi]+R rho Cos[phi] Cos[Phi])/r0))/lambda] *)
Since the diffraction pattern is symmetric about
Phi, simplify by setting
Phi to 0.
Eeq=Eeq/.Phi->0//ExpandAll (*-A dx dy Sin[-(2 Pi f t)-(2 Pi R rho Cos[phi])/(lambda r0)+(2 Pi r0)/lambda]*) Eeq=A dx dy Sin[2 Pi f t+(2 Pi R rho Cos[phi])/(lambda r0)-(2 Pi r0)/lambda]//.{-2 f Pi t+(2 Pi r0)/lambda->alpha,-((2 Pi R rho Cos[phi])/(lambda r0))->beta}//TrigExpand;Eeq=%/.{alpha->-2 f Pi t+(2 Pi r0)/lambda,beta->-((2 Pi R rho Cos[phi])/(lambda r0))}
Integrate over the aperture of radius a in cylindrical coordinates, dx dy -> rho drho dphi.
Eeq=%/.dx dy->rho drho dphi(*A dphi drho rho Cos[(2 Pi r0)/lambda-2 \[Pi] f t] Sin[(2 Pi R rho Cos[phi])/(lambda r0)]-A dphi drho rho Sin[(2 Pi r0)/lambda-2 Pi f t] Cos[(2 Pi R rho Cos[phi])/(lambda r0)]*)$Assumptions=a>0&&R>0&&lambda>0&&r0>0
Works in the following form:
En=A Cos[(2 Pi r0)/lambda-2 Pi f t]Integrate[rho Sin[(2 Pi R rho Cos[phi])/(lambda r0)],{phi,0,2 Pi},{rho,0,a}]-A Sin[(2 Pi r0)/lambda-2 Pi f t] Integrate[rho Cos[(2 Pi R rho Cos[phi])/(lambda r0)],{phi,0,2 Pi},{rho,0,a}](*-((a A lambda r0 BesselJ[1,(2 a Pi R)/(lambda r0)] Sin[(2 Pi r0)/lambda-2 Pi f t])/R)*)
Remember that R is the radial distance from the center of the screen to the observation pt. Get to the familiar form by
En/.R->(lambda r0 xx)/(2 a Pi)(*-((2 Pi a^2 A BesselJ[1,xx] Sin[(2 Pi r0)/lambda-2 Pi f t])/xx)*)
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This question seems to have been specifically designed to make you realise that the homology groups of a space can sometimes pick out homotopical information about a space which is not contained in the fundamental group alone. In this case, we have $X$ which is homotopy equivalent to $S^2\vee S^1\vee S^1\vee S^1$ (as you rightly pointed out, although were one small step away from reaching) and then we have a space which I'll call $Y$ which is homotopy equivalent to $S^1\vee S^1\vee S^1$.
Now then, $X$ and $Y$ are both connected and path connected (the first place we'd look to see if two spaces were homotopy equivalent). Also, $X$ and $Y$ have the same fundamental group by a trivial application of Van-Kampen's theorem giving $$\pi_1(X)\cong\pi_1(Y)\cong F_3$$ where $F_3$ is the free group on three generators. This also tells us that $H_0$ and $H_1$ will be isomorphic as $H_0$ counts path components and $H_1\cong\pi_1^{ab}$ the abelianisation of the fundamental group.
Where is the next place to look then? There are a few approaches which I'd suggest initially. The first would be to note that $\pi_2(X)$ is non-trivial (generated by the inclusion of $S^2$ in to the wedge product) and $\pi_2(Y)$ is trivial as $Y$ is a graph. However, you may not have met the higher homotopy groups yet so disregard this approach if you haven't.
Next, I would suggest just calculating homology groups. This is rather easy if you've had enough practice calculating simplicial homology and you should find that $H_2(X)$ is non-trivial (again generated by the inclusion map) whereas $H_2(Y)$ is trivial (using a standard dimension argument). This is probably the approach your text/teacher expects.
The last approach I would suggest, which you may be comfortable with, is to find the universal cover of $X$ and $Y$. We know that $Y$ is a graph and so its universal cover $\tilde{Y}$ is contractible, however $X$ has a universal cover $\tilde{X}$ which is not contractible (essentially by the same argument that $S^2$ is not contractible, the identity map $S^2\rightarrow S^2$ is not homotopic to the constant map). This last approach is probably more contrived than the other two approaches though as the easiest way to show that the identity map $S^2\rightarrow S^2$ is not homotopic to the constant map is to use cellular homology.
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I am reading Rudin and I am very confused what a derivative is now. I used to think a derivative was just the process of taking the limit like this $$\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{(x+h)-x}$$
But between Apostol and Rudin, I am confused in what sense total derivatives are derivatives.
Partial derivatives much more resemble the usual derivatives taught in high school
$$f(x,y) = xy$$
$$\frac{\partial f}{\partial x} = y$$
But the Jacobian doesn't resemble this at all. And according to my books it is a linear map.
If derivatives are linear maps, can someone help me see more clearly how my intuitions about simpler derivatives relate to the more complicated forms? I just don't understand where the limits have gone, why its more complex, and why the simpler forms aren't described as linear maps.
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Differential and Integral Equations Differential Integral Equations Volume 22, Number 7/8 (2009), 617-636. Exponential stability for the $2$-D defocusing Schrödinger equation with locally distributed damping Abstract
This paper is concerned with the study of the unique continuation property associated with the defocusing Schrödinger equation \begin{eqnarray*} iu_{t} +\Delta u - |u|^2u =0 ~\hbox{ in }\Omega \times (0,\infty), \end{eqnarray*} subject to Dirichlet boundary conditions, where $\Omega \subset \mathbb{R}^2$ is a bounded domain with smooth boundary $\partial \Omega=\Gamma$. In addition, we prove exponential decay rates of the energy for the damped problem \begin{eqnarray*} iu_{t} +\Delta u - |u|^2u +i a(x) u =0 \hbox{ in } \mathbb{R}^2 \times (0,\infty), \end{eqnarray*} provided that $a(x) \geq a_0 >0$ almost everywhere in $\Omega_{R}:=\{x\in \mathbb{R}^2 : |x| \geq R\}$, where $R>0$.
Article information Source Differential Integral Equations, Volume 22, Number 7/8 (2009), 617-636. Dates First available in Project Euclid: 20 December 2012 Permanent link to this document https://projecteuclid.org/euclid.die/1356019541 Mathematical Reviews number (MathSciNet) MR2532114 Zentralblatt MATH identifier 1240.35509 Subjects Primary: 35Q55: NLS-like equations (nonlinear Schrödinger) [See also 37K10] Secondary: 35B35: Stability 35B40: Asymptotic behavior of solutions 35B60: Continuation and prolongation of solutions [See also 58A15, 58A17, 58Hxx] Citation
Cavalcanti, M. M.; Domingos Cavalcanti, V. N.; Fukuoka, R.; Natali, F. Exponential stability for the $2$-D defocusing Schrödinger equation with locally distributed damping. Differential Integral Equations 22 (2009), no. 7/8, 617--636. https://projecteuclid.org/euclid.die/1356019541
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