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Answer
Please see the work below.
Work Step by Step
We know that $\Delta T_F=(\frac{9}{5})\Delta T_C$ We plug in the known values to obtain: $\Delta T_F=(\frac{9}{5})(10)$ $\Delta T_F=18F^{\circ}$
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Speaker: Affiliation:
Indian Institute of Technology
Department of Computer Science and Engineering Hauz Khas New Delhi 110016 Time: Venue: A-201 (STCS Seminar Room) Organisers:
We develop new techniques for rounding packing integer programs using iterative randomized rounding. It is based on a novel application of multidimensional Brownian motion in $\mathbb{R}^n$. Let $\overset{\sim}{x} \in {[0,1]}^n$ be a fractional feasible solution of a packing constraint $A x \leq 1,\ \ $ $A \in {\{0,1 \}}^{m\times n}$ that maximizes a linear objective function.
Our algorithm iteratively transforms $\overset{\sim}{x}$ to $\hat{x} \in {\{ 0,1\}}^{n}$ using a random walk, such that the expected values of $\hat{x}_i$'s are consistent with the Raghavan-Thompson rounding. In addition, it gives us intermediate values $x'$ which can then be used to bias the rounding towards a superior solution. Our algorithm gradually sparsifies $A$ to $A' \in {\{0,1 \}}^{m\times n}$ where each row in $A'$ has $\leq \log n$ non-zero coefficients with $A'\cdot x' \leq O(1)$. The reduced dependencies between the constraints of the sparser system can be exploited using {\it Lovasz Local Lemma}. Using the Moser-Tardos' constructive version, $x'$ converges to $\hat{x}$ in polynomial time to a distribution over the unit hypercube ${\cal H}_n = {\{0,1 \}}^n$ such that the expected value of any linear objective function over ${\cal H}_n$ equals the value at $\overset{\sim}{x}$.
We discuss application of these techniques when $A$ is a random matrix and also for a more general situation of a $k$-column sparse matrix (joint work with Dhiraj Madan).
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I have been recently working with generating functions in my discrete mathematics course, and I was interested in one particular generating function. I want to find the generating function for the number of ways one can split $n$ into odd parts. I can see that the first coefficients of the sequence are $$0,1,1,2,2,3,4,5,...$$ And so on. I've been trying to find a recursion, and while it seemed originally that the number of ways looked to be the ceiling of $\frac{n}{2}.$ But it seems like the $6$th coefficient suggests otherwise. Is there another possible recursion going on in this sequence? If so, how would I go about finding the generating function for it?
The generating function for this sequence is
$f(x)=\prod\limits_{n=1}^{\infty} \cfrac{1}{1-x^{2n-1}}$
since
$f(x)=(1+x+x^{1+1}+\cdots)(1+x^3+x^{3+3}+\cdots)\cdots (1+x^{2n-1}+x^{(2n-1)+(2n-1)})\cdots = (1+x+x^2+\cdots)(1+x^3+x^{2\cdot3}+\cdots)\cdots (1+x^{2n-1}+x^{2(2n-1)})\cdots=\cfrac{1}{1-x}\cfrac{1}{1-x^3}\cdots\cfrac{1}{1-x^{2n-1}}\dots=\prod\limits_{n=1}^{\infty} \cfrac{1}{1-x^{2n-1}}$
Moreover, you can express this function as follows,
$\prod\limits_{n=1}^{\infty} (1+x^n)=(1+x)(1+x^2)(1+x^3)\cdots=\cfrac{1-x^2}{1-x}\cfrac{1-x^4}{1-x^2}\cfrac{1-x^6}{1-x^3}\cdots=\cfrac{1}{1-x}\cfrac{1}{1-x^3}\cdots=f(x)$
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Basic Properties Regarding Ring Homomorphisms
Basic Properties Regarding Ring Homomorphisms
Recall from the Ring Homomorphisms page that if $(R, +_1, *_1)$ and $(S, +_2, *_2)$ are rings with multiplicative identities $1_R$ and $1_S$ respectively, then a ring homomorphism between them is a function $\phi : R \to S$ such that for all $a, b \in R$ we have that:
1) $\phi (a +_1 b) = \phi (a) +_2 \phi (b)$. 2) $\phi (a *_1 b) = \phi (a) *_2 \phi (b)$.
We now state some basic properties of ring homomorphisms.
(1)
Theorem 1: Let $(R, +_1, *_1)$ and $(S, +_2, *_2)$ be rings with additive identities $0_R$ and $0_S$ respectively. If $R$ and $S$ are homomorphic with homomorphism $\phi$ then $\phi (0_R) = 0_S$.
\begin{align} \quad \phi (0_R) = \phi(0_R + 0_R) = \phi (0_R) + \phi(0_R) \end{align}
Therefore $\phi (0_R) = 0_S$. $\blacksquare$
Lemma 2: Let $(R, +_1, *_1)$ and $(S, +_2, *_2)$ be rings with additive identities $0_R$ and $0_S$ respectively and multiplicative identities $1_R$ and $1_S$ respectively. If $R$ and $S$ are homomorphic with homomorphism $\phi$ then $\phi (-1_R) = -1_S$. (2) Proof: By the previous theorem we have that:
\begin{align} \quad 0_S = \phi (0_R) = \phi (1_R - 1_R) = \phi (1_R) + \phi(-1_R) = 1_S + \phi(-1_R) \end{align}
So $\phi (-1_R) = -1_S$. $\blacksquare$
Theorem 3: Let $(R, +_1, *_1)$ and $(S, +_2, *_2)$ be rings with additive identities $0_R$ and $0_S$ respectively. If $R$ and $S$ are homomorphic with homomorphism $\phi$ then for all $a \in R$, $\phi (-a) = -\phi(a)$. (3) Proof: By Lemma 2 we have that:
\begin{align} \quad \phi (-a) = \phi(-1 *_1 a) = \phi(-1_R) *_2 \phi (a) = -1_S \phi(a) = -\phi(a) \quad \blacksquare \end{align}
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The easiest explanation for why the maximum distance one can see is not simply the product of the speed of light with the age of the universe is because the universe is non-static.
Different things (i.e. matter vs. dark energy) have different effects on the coordinates of the universe, and their influence can change with time.
A good starting point in all of this is to analyze the Hubble parameter, which gives us the Hubble constant at any point in the past or in the future given that we can measure what the universe is
currently made of:
$$ H(a) = H_{0} \sqrt{\frac{\Omega_{m,0}}{a^{3}} + \frac{\Omega_{\gamma,0}}{a^{4}} + \frac{\Omega_{k,0}}{a^{2}} + \Omega_{\Lambda,0}} $$where the subscripts $m$, $\gamma$, $k$, and $\Lambda$ on $\Omega$ refer to the density parameters of matter (dark and baryonic), radiation (photons, and other relativistic particles), curvature (this only comes into play if the universe globally deviates from being spatially flat; evidence indicates that it is consistent with being flat), and lastly dark energy (which as you'll notice remains a
constant regardless of how the dynamics of the universe play out). I should also point out that the $0$ subscript notation means as measured today.
The $a$ in the above Hubble parameter is called the scale factor, which is equal to 1 today and zero at the beginning of the universe. Why do the various components scale differently with $a$? Well, it all depends upon what happens when you increase the size of a box containing the stuff inside. If you have a kilogram of matter inside of a cube 1 meter on a side, and you increase each side to 2 meters, what happens to the density of matter inside of this new cube? It decreases by a factor of 8 (or $2^{3}$). For radiation, you get a similar decrease of $a^{3}$ in number density of particles within it, and also an additional factor of $a$ because of the stretching of its wavelength with the size of the box, giving us $a^{4}$. The density of dark energy remains constant in this same type of thought experiment.
Because different components act differently as the coordinates of the universe change, there are corresponding eras in the universe's history where each component dominates the overall dynamics. It's quite simple to figure out, too. At small scale factor (very early on), the most important component was radiation. The Hubble parameter early on could be very closely approximated by the following expression:
$$H(a) = H_{0} \frac{\sqrt{\Omega_{\gamma,0}}}{a^{2}}$$
At around:
$$ \frac{\Omega_{m,0}}{a^{3}} = \frac{\Omega_{\gamma,0}}{a^{4}} $$$$ a = \frac{\Omega_{\gamma,0}}{\Omega_{m,0}} $$we have matter-radiation equality, and from this point onward we now have matter dominating the dynamics of the universe. This can be done once more for matter-dark energy, in which one would find that we are now living in the dark energy dominated phase of the universe. One prediction of living in a phase like this is an
acceleration of the coordinates of universe - something which has been confirmed (see: 2011 Nobel Prize in Physics).
So you see, it would a bit more complicating to find the distance to the cosmological horizon than just multiplying the speed of light by the age of the universe. In fact, if you'd like to find this distance (formally known as the comoving distance to the cosmic horizon), you would have to perform the following integral:
$$ D_{h} = \frac{c}{H_{0}} \int_{0}^{z_{e}} \frac{\mathrm{d}z}{\sqrt{\Omega_{m,0}(1+z)^{3} + \Omega_{\Lambda}}} $$
where the emission redshift $z_{e}$ is usually taken to be $\sim 1100$, the surface of last scatter. It turns out this is the true horizon we have as observers. Curvature is usually set to zero since our most successful model indicates a flat (or very nearly flat) universe, and radiation is unimportant here since it dominates at a higher redshift. I would also like to point out that this relationship is derived from the Friedmann–Lemaître–Robertson–Walker metric, a metric which includes curvature and expansion. This is something that the Minkowski metric lacks.
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There is a wide litterature for the classical Gauss sums. For $\chi$ a primitive Dirichlet character modulo $N$, it is given by $$\tau(\chi) = \sum_{n \text{ mod } N} \chi(n) \exp(2i\pi n/N).$$
An interesting fact is that primitive Dirichlet characters are essentially their own Fourier transform, up to this associated Gauss sum: $$\sum_{n \text{ mod } N} \chi(n) \exp(2i\pi nm/N) = \bar{\chi} (m) \tau(\chi). \qquad (\star)$$
I am interested in similar properties in general number fields $F$. Let $\omega$ be a nonzero element of $F$ and write $\mathfrak{a}$ for its denominator. Hecke defined some analogous Gauss sums by $$C(\omega) = \sum_{n \text{ mod } \mathfrak{a}} \exp(\mathrm{tr}(n^2 \omega)).$$
(here $n \text{ mod } \mathfrak{a}$ means that $n$ is an integer ideal in $\mathfrak{o}_F/\mathfrak{a}\mathfrak{o}_F$, where $\mathfrak{o}_F$ is the ring of integers of $F$). I have not found anything concerning the analogous Gauss sums as above, namely for a finite order character $\chi$ on integer ideals which is a character modulo $\mathfrak{a}$, $$\tau(\omega, \chi) = \sum_{n \text{ mod } \mathfrak{a}} \chi(n)\exp(\mathrm{tr}(n^2 \omega)).$$
Is there anything of this kind in the litterature? In particular, do we have an analogous result to $(\star)$?
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When reading some literatures on topological insulators, I've seen authors taking Brillouin zone(BZ) to be a sphere sometimes, especially when it comes to strong topological insulators. Also I've seen the usage of spherical BZ in these answers(1,2) by SE user Heidar. I can think of two possibilities:
(1)Some physical system has a spherical BZ. This is hard to imagine, since it seems to me that all lattice systems with translational symmetries will have a torodal BZ, by the periodicity of Bloch wavefunctions. The closest scenario I can imagine is a continuous system having $R^n$ as BZ, and somehow(in a way I cannot think of) acquires an one-point compactification.
(2)A trick that makes certain questions easier to deal with, while the true BZ is still a torus.
Can someone elaborate the idea behind a spherical BZ for me?
Update: I recently came across these notes(pdf) by J.Moore. In the beginning of page 9 he mentioned
We need to use one somewhat deep fact: under some assumptions, if $π_1(M)
= 0$ for some target space $M$, then maps from the torus $T^
2\to M$ are contractible to maps from the sphere $S^2
\to M$
I think this is a special case of the general math theorem I want to know, but unfortunately Moore did not give any reference so I'm not sure where to look.
EDIT: The above math theorem is intuitively acceptable to me although I'm not able to prove it. I can take this theorem as a working hypothesis for now, what I'm more interested in is, granted such theorem, what makes a $\pi_1(M)=0$ physical system candidate for strong topological insulators(robust under local perturbations), and why in $\pi_1(M)\neq0$ case we can only have weak topological insulators.
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I am working my way through LDA and I think I got they main idea of it. Please correct me if I am wrong. Given the Plate notation:
The variables $\alpha$ and $\beta$ are Dirichlet distribution parameters. The variable $Z_{d,n}$ assigns observed word $W_{d,n}$ to topic $\phi_k$, which is a distribution over words. Variable $\theta_d$ is the document-specific topic distribution. Both distributions $\theta_d$ and $\phi_k$ are drawn from Dirichlet distributions.
Now, only $W_{d,n}$ is observed and can be "directly" calculated. My question: What exactly is inferred/calculated with e.g. Gibbs sampling, variational Inference and so on?
For instance: For a Gaussian Naive Bayes classifier one assumes that the likelihood of each feature is Gaussian. In other words each feature has a Gaussian distribution:
$$ P(x) = \frac{1}{{\sigma \sqrt {2\pi } }} e^{ \frac{ - ( {x - \mu } )^2} {2\sigma ^2 } } $$
To find this distribution $\sigma$ and $\mu$ have to be determined which is pretty straight forward.
However, plainly said: What Numbers do I determine for LDA?
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$S$ cannot be a 3-dimensional real hyperboloid unless $\psi$ is of the same form as $\phi$, though it can be a 2-dimensional real hyperboloid*.Excluding the possibilities achievable when $\psi$is a real quadratic form, and working up to similarity of $\psi$ over $\mathcal{R}$,$S$ is one of the following:
$\quad\bullet\quad\emptyset$
$\quad\bullet\quad$ a point
$\quad\bullet\quad$ a 2-dimensional quadric surface in the form of a cone, an ellipsoid, a hyperboloid of two sheets*, or a hyperboloid of one sheet.
$\quad\bullet\quad$ the Cartesian product of a circle with two intersecting lines
$\quad\bullet\quad$ the Cartesian product of a hyperbola with two intersecting lines
PROOF:
By the inertia theorem, up to similarity over $\mathbb{R}$,there exist $a,b,c,d\in\{\pm 1,\pm i\}$such that $\psi(w,x,y,z)=aw^2+bx^2+cy^2+dz^2$.In the case where $a,b,c,d\in\{\pm 1\}$,we have the usual quadrics obtainable by real quadratic forms on $\mathbb{R}^4$,and in this case $S$ is a hyperboloid model if and only if $\psi$takes the form of $\phi$, contributing nothing new.So assume that not all of $a,b,c,d$ are $\pm 1$, and we will work up to permutation of coordinates.
Suppose $a,b,c,d\in\{\pm i\}$.Then $\psi=i\phi'$where $\phi'$ is a real quadratic form, therefore $i\phi'(w,x,y,z))$is purely imaginary, giving $\forall w,x,y,z\in\mathbb{R}:\Re(\psi(w,x,y,z))=0\neq-1$.So $S=\emptyset$.
Suppose $a\in\{\pm1\}$ and $b,c,d\in\{\pm i\}$.Then $\Re(\psi(w,x,y,z))=\pm w$ and $\Im(\psi)=i\phi'$where $\phi'$ now is a real quadratic form on $\mathbb{R}^3$.So we require $w=\pm 1$, and $\phi'(x,y,z)=b'x^2+c'y^2+d'z^2=0$(where $b'$ is the real number $\Im(b)$, etc.) which gives either $x=y=z=0$ (when the signs agree)or $(x,y,z)$ lie on a two-dimensional cone (when the signs do not agree).So $S$ is either the point $(\pm1,0,0,0)$, or a cone lying in the copy of $\mathbb{R}^3$ at $w=1$ or $w=-1$.
Suppose $a,b\in\{\pm1\}$ and $c,d\in\{\pm i\}$.If $a,b=1$ then $\Re(\psi(w,x,y,z))=w^2+x^2\neq-1$, making $S=\emptyset$.If $a,b=-1$ then $\Re(\psi(w,x,y,z))=-w^2-x^2=-1$ $\Longrightarrow(w,x)$ lie on a unit circle.If $a=1, b=-1$ then $\Re(\psi(w,x,y,z))=w^2-x^2=-1\Longrightarrow(w,x)$ lie on a regular hyperbola.But for $\phi(w,x,y,z)$ to be $-1$ we need $\Im(\phi(w,x,y,z))=0$.If $c=d=\pm i$ then this forces $y=z=0$.If $c=i$ and $d=-i$ then this forces $y^2=z^2$,i.e. $(y,z)$ lie on two perpendicular lines.Combining this with the observations about the real part, we have that $S$ is either $\emptyset$, or a cartesian product of a curve with a pair of perpendicular lines, where the curve is either a circle or a hyperbola.
Lastly, suppose $a,b,c\in\{\pm1\}$ and $d\in\{\pm i\}$. Then to get $\phi(w,x,y,z))=-1$ we require that $z=0$, and that the solution for $(w,x,y)$to $aw^2+bx^2+cy^2=-1$ is either empty, an ellipsoid, a hyperboloid of one sheet, or a hyperboloid of two sheets, as determined by the signs.
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The Annals of Statistics Ann. Statist. Volume 18, Number 2 (1990), 717-741. Large-Sample Inference for Log-Spline Models Abstract
Let $f$ be a continuous and positive unknown density on a known compact interval $\mathscr{Y}$. Let $F$ denote the distribution function of $f$ and let $Q = F^{-1}$ denote its quantile function. A finite-parameter exponential family model based on $B$-splines is constructed. Maximum-likelihood estimation of the parameters of the model based on a random sample of size $n$ from $f$ yields estimates $\hat{f, F}$ and $\hat{Q}$ of $f, F$ and $Q$, respectively. Under mild conditions, if the number of parameters tends to infinity in a suitable manner as $n \rightarrow \infty$, these estimates achieve the optimal rate of convergence. The asymptotic behavior of the corresponding confidence bounds is also investigated. In particular, it is shown that the standard errors of $\hat{F}$ and $\hat{Q}$ are asymptotically equal to those of the usual empirical distribution function and empirical quantile function.
Article information Source Ann. Statist., Volume 18, Number 2 (1990), 717-741. Dates First available in Project Euclid: 12 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aos/1176347622 Digital Object Identifier doi:10.1214/aos/1176347622 Mathematical Reviews number (MathSciNet) MR1056333 Zentralblatt MATH identifier 0712.62036 JSTOR links.jstor.org Citation
Stone, Charles J. Large-Sample Inference for Log-Spline Models. Ann. Statist. 18 (1990), no. 2, 717--741. doi:10.1214/aos/1176347622. https://projecteuclid.org/euclid.aos/1176347622
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The Internal Direct Product of Two Groups
Recall from The External Direct Product of Two Groups page that if $(G, *)$ and $(H, +)$ are two groups then the external direct product of these groups is the new group $(G \times H, \cdot)$ where $\cdot$ is the binary operation defined for all $(g_1, h_1), (g_2, h_2) \in G \times H$ by:(1)
We will now define another type of "direct product" known as the internal direct product of two groups.
Definition: Let $(G, *)$ be a group and let $(H, *)$ and $(K, *)$ be two subgroups of $(G, *)$. Then $(G, *)$ is said to be the Internal Direct Product of $(H, *)$ and $(K, *)$ if: 1) $G = \{ h * k : h \in H, k \in K \}$. 2) $H \cap K = \{ e \}$ where $e$ is the identity element in $G$. 3) $h * k = k * h$ for all $h \in H$ and for all $k \in K$.
For example, consider the group $(\mathbb{Z}_6, +)$ and the following subgroups:(2)
Note that $\{ h * k : h \in H, k \in K \} = \{ 0, 1, 2, 3, 4, 5 \} = G$, so the first condition is met. Also the identity for $\mathbb{Z}_6$ is $e = 0$ and $H \cap K = \{ 0 \}$ so the second condition is met. Lastly $\mathbb{Z}_6$ is an abelian group so the third condition is met.
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With the given \$X(e^{jw})\$ you won't get that answer.
Graphical proof:
If we try to plot \$X(e^{jw})\$, it will be a train of pulses with pulse width = \$\pi/2\$ and amplitude = \$\sqrt2\$ . And pulse will be centered at integer multiple of \$2\pi\$ as shown in the figure.
Calculating the value of \$Y\$ at \$w=0\$,
$$Y(e^{j\omega})|_{w=0} = \int_0^{2\pi}{X^2(e^{j\alpha})\frac{d\alpha}{2\pi}}$$
This is the \$\frac{1}{2\pi}\times\$ area under \$X^2(e^{j\omega})\$ from \$0\$ to \$2\pi\$.
$$Y(e^{j\omega})|_{w=0} = \frac{1}{2\pi}\times (2\times\frac{\pi}{4} + 2\times\frac{\pi}{4}) = \frac{1}{2}$$
Which does not satisfy \$1-\frac{|\omega|}{\pi}\$. But a slight change in \$X(e^{jw})\$ can give you the result.
Changing the question:
If \$X(e^{jw})\$ was defined as follows,$$X(e^{j\omega}) = \sqrt{2}\sum_{-\infty}^\infty rect(\frac{\omega + 2\pi k}{\pi})$$
Then, \$X(e^{jw})\$ will be a pulse train similar to previous one but the pulse width would be \$\pi\$. See the figure given below.
The black represents the \$X(e^{j\alpha})\$ and blue represents the \$X(e^{j\alpha+w})\$. (assume that 0 < w < \pi)
So the product
$${X(e^{j\alpha})\times X(e^{j\alpha+\omega})}$$
will be zero everywhere (0 to \$2\pi\$) except in the region marked by gray color. And the amplitude of this product will be \$\sqrt2\times \sqrt2 =2\$.
So the integral $$\int_0^{2\pi}{X(e^{j\alpha})*X(e^{j\alpha+\omega})d\alpha}$$
will be given by the area under this product curve. Which will be summation of area under two gray-rectangles:
$$\int_0^{2\pi}{X(e^{j\alpha})*X(e^{j\alpha+\omega})d\alpha} = (\frac{\pi}{2}-w)\times 2 + \frac{\pi}{2}\times 2 = 2\pi - 2w$$
If you do this for a right shift of signal, which corresponds to negative \$w\$, you will get the same result.
So we can write:
$$Y(e^{jw}) = \frac{1}{2\pi}\times (2\pi - |2w|) = 1-\frac{|w|}{\pi}$$
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Asteroidal triples¶
This module contains the following function:
is_asteroidal_triple_free()
Test if the input graph is asteroidal triple-free Definition¶
Three independent vertices of a graph form an
asteroidal triple if every twoof them are connected by a path avoiding the neighborhood of the third one. Agraph is asteroidal triple-free ( AT-free, for short) if it contains noasteroidal triple [LB1962].
Use
graph_classes.AT_free.description() to get some known properties ofAT-free graphs, or visit this page.
Algorithm¶
This module implements the
Straightforward algorithm recalled in [Koh2004] anddue to [LB1962] for testing if a graph is AT-free or not. This algorithm has timecomplexity in \(O(n^3)\) and space complexity in \(O(n^2)\).
This algorithm uses the
connected structure of the graph, stored into a\(n\times n\) matrix \(M\). This matrix is such that \(M[u][v]==0\) if \(v\in(\{u\}\cup N(u))\), and otherwise \(M[u][v]\) is the unique identifier (a strictlypositive integer) of the connected component of \(G\setminus(\{u\}\cup N(u))\) towhich \(v\) belongs. This connected structure can be computed in time \(O(n(n+m))\)using \(n\) BFS.
Now, a triple \(u, v, w\in V\) is an asteroidal triple if and only if it satisfies \(M[u][v]==M[u][w]\) and \(M[v][u]==M[v][w]\) and \(M[w][u]==M[w][v]\), assuming all these values are positive. Indeed, if \(M[u][v]==M[u][w]\), \(v\) and \(w\) are in the same connected component of \(G\setminus(\{u\}\cup N(u))\), and so there is a path between \(v\) and \(w\) avoiding the neighborhood of \(u\). The algorithm iterates over all triples.
Functions¶
sage.graphs.asteroidal_triples.
is_asteroidal_triple_free(
G, certificate=False)¶
Test if the input graph is asteroidal triple-free
An independent set of three vertices such that each pair is joined by a path that avoids the neighborhood of the third one is called an
asteroidal triple. A graph is asteroidal triple-free (AT-free) if it contains no asteroidal triples. See the
module's documentationfor more details.
This method returns
Trueis the graph is AT-free and
Falseotherwise.
INPUT:
G– a Graph
certificate– boolean (default:
False); by default, this method returns
Trueif the graph is asteroidal triple-free and
Falseotherwise. When
certificate==True, this method returns in addition a list of three vertices forming an asteroidal triple if such a triple is found, and the empty list otherwise.
EXAMPLES:
The complete graph is AT-free, as well as its line graph:
sage: G = graphs.CompleteGraph(5) sage: G.is_asteroidal_triple_free() True sage: G.is_asteroidal_triple_free(certificate=True) (True, []) sage: LG = G.line_graph() sage: LG.is_asteroidal_triple_free() True sage: LLG = LG.line_graph() sage: LLG.is_asteroidal_triple_free() False
The PetersenGraph is not AT-free:
sage: from sage.graphs.asteroidal_triples import * sage: G = graphs.PetersenGraph() sage: G.is_asteroidal_triple_free() False sage: G.is_asteroidal_triple_free(certificate=True) (False, [0, 2, 6])
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Main Page The Problem
Let [math][3]^n[/math] be the set of all length [math]n[/math] strings over the alphabet [math]1, 2, 3[/math]. A
combinatorial line is a set of three points in [math][3]^n[/math], formed by taking a string with one or more wildcards [math]x[/math] in it, e.g., [math]112x1xx3\ldots[/math], and replacing those wildcards by [math]1, 2[/math] and [math]3[/math], respectively. In the example given, the resulting combinatorial line is: [math]\{ 11211113\ldots, 11221223\ldots, 11231333\ldots \}[/math]. A subset of [math][3]^n[/math] is said to be line-free if it contains no lines. Let [math]c_n[/math] be the size of the largest line-free subset of [math][3]^n[/math]. [math]k=3[/math] Density Hales-Jewett (DHJ(3)) theorem: [math]\lim_{n \rightarrow \infty} c_n/3^n = 0[/math]
The original proof of DHJ(3) used arguments from ergodic theory. The basic problem to be considered by the Polymath project is to explore a particular combinatorial approach to DHJ, suggested by Tim Gowers.
Useful background materials
Some background to the project can be found here. General discussion on massively collaborative "polymath" projects can be found here. A cheatsheet for editing the wiki may be found here. Finally, here is the general Wiki user's guide
Threads (1-199) A combinatorial approach to density Hales-Jewett (inactive) (200-299) Upper and lower bounds for the density Hales-Jewett problem (final call) (300-399) The triangle-removal approach (inactive) (400-499) Quasirandomness and obstructions to uniformity (inactive) (500-599) Possible proof strategies (active) (600-699) A reading seminar on density Hales-Jewett (active) (700-799) Bounds for the first few density Hales-Jewett numbers, and related quantities (arriving at station)
We are also collecting bounds for Fujimura's problem.
Here are some unsolved problems arising from the above threads.
Here is a tidy problem page.
Bibliography M. Elkin, "An Improved Construction of Progression-Free Sets ", preprint. H. Furstenberg, Y. Katznelson, “A density version of the Hales-Jewett theorem for k=3“, Graph Theory and Combinatorics (Cambridge, 1988). Discrete Math. 75 (1989), no. 1-3, 227–241. H. Furstenberg, Y. Katznelson, “A density version of the Hales-Jewett theorem“, J. Anal. Math. 57 (1991), 64–119. B. Green, J. Wolf, "A note on Elkin's improvement of Behrend's construction", preprint. K. O'Bryant, "Sets of integers that do not contain long arithmetic progressions", preprint. R. McCutcheon, “The conclusion of the proof of the density Hales-Jewett theorem for k=3“, unpublished.
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Classical Mechanics Kinematics (Equations of Motion) Newton's Laws Vector physics Work Energy Power Momentum Conservation Statics (Friction) Circular Motion (Angular stuff) Kepler's Laws (Satellite Motion) Harmonic Motion (Springs) Problem 1
You may need to modify the Newtonian theory of gravitation at short ranges. Suppose the potential energy between 2 masses
m and m' is:
For short distances
r << π, calculate the force between m and m' Solution 1
Recall the Newtonian theory of gravitation
In this case,
M = m' and you only need to remember the middle part Problem 2
A musicall instrument called the "bugle" is dropped into a wishing well.
t is the time between releasing the bugle and hearing it clang on the bottom. If the speed of sound is 330 m/s and t = 2.059s what is the depth h of the well? Solution 2 Problem 3
Calculate the centripetal force required to keep a 4 kg mass moving in a horizontal circle of radius 0.8 m at a speed of 6 m/s. (
r is the radical vector with respect to the center) Solution 3
Centripetal force is toward the center of the circle.
Problem 4
A spring-mass system is suspected from the ceiling of an elevator. When the elevator is at rest, the system oscillates with period T. The elevateor now starts and moves upward with acceleration of a = 0.2 g. During this constant acceleration phase, what is the period T of the spring-mass system?
Solution 4
The spring mass system is independent of gravity, and given by
Hence, the acceleration of the elevator will not affect the period.
Problem 5
Each of the figures above shows blocks of mass 2m and m acted on by an external horizontal force F. For each figure, which of the following statements about the magnitude of the force that one block exerts on the other (F
12) is correct? (Assume that the surface on which the blocks move is frictionless.) Solution 5Answer Problem 6
A simple pendulum of length l is suspended fro the ceiling of an elevator that is accelerating upward with constant acceleration a. For small oscillations, what is the period T of the pendulum?
Solution 6Answer Problem 7
A uniform disk with a mass of m and a radius of r rolls without slipping along a horizontal surface and ramp. The disk has an initial velocity of u. What is the maximum height h to which the center of mass of the disk rises?
Solution 7Answer Problem 8
A mass m attached to the end of a massless rod of length L is free to swing below the plane of support, as shown in the figure above. The Hamiltonian for this system is given by
where q and f are defined as shown in the figure. On the basis of Hamilton's equations of motion, the generalized coordinate or momentum that is a constant in time is
Solution 8Answer Problem 9
A rod of length L and mass M is placed along the x-axis with one end at the origin, as shown in the figure above. The rod has linear mass density
where x is the distance from the origin. What is the x-coordinate of the rod's center of mass in relation to L?
Solution 9Answer Problem 10
Which of the following best illustrates the acceleration of a pendulumn bob at points a through e?
Solution 10C Problem 11
A stone is thrown at an angle of 45° above the horizontal x-axis in the +x-direction. Ignore air resistance. Which of the velocity versus time graphs shown above best represents v
x versus t and v y versus t,respectively? I, IV II, I II, III II, V IV, V Solution 11
If a stone is thrown at such an angle at an initial velocity, its horizontal v
x vs t graph should be constant and positive, like Graph II
Thus, choices (A) and (E) are out.
Recall the basic kinematics equation
Eliminate choice (D), since that shows a parabolic time dependence, when a linear one is required. Since the slope is negative, the v
y-graph should look like III. This leaves only one answer: (C)
If you forget the basic equations above, you can derive it all from summing up the net force
Integrate both sides to get velocity. Integrate again to get position.
Problem 12
Seven pennies are arranged in a hexagonal planar pattern so as to touch each neighbor as shown in the figure above. Each penny is a uniform disk of mass m and radius r. What is the moment of inertia of the system of seven pennies above an axis that passes through the center of the central penny and is normal to the plane of the pennies?
(7/2) m r 2 (13/2) m r 2 (29/2) m r 2 (49/2) m r 2 (55/2) m r 2 Solution 12Answer Problem 13
A thin uniform rod of mass M and length L is positioin vertically aboeve an anchored frictionless pivot point, and then allowed to fall to the ground. With what speed does the free end of the rod strike the ground?
√( g L/ 3) √( g L) √(3 g L) √(12 g L) 12 √( g L) Solution 13Answer Problem 14
The figure above shows a plot of the time-dependent force F
x(t) acting on a particle in motion along the x-axis. What is the total impulse delivered to the particle? 0 1 kg m / s 2 kg m / s 3 kg m / s 4 kg m / s Solution 14Answer Problem 15
A particle of mass m is moving along the x-axis with speed v when it collides with a particle of mass 2m initially at rest. After the collision the first particle has come to rest and the second particle has split into two equal-mass pieces that move at equal angles θ > 0 with the x-axis. Which of the following statements correcty describes the speeds of the 2 pieces?
Each piece moves with speed greater than v/2 One of the pieces moves with speed v, the other moves with speed less than v. One of the pieces moves with speed v/2, the other moves with speed greater than v/2 Each piece moves with speed v Each piece moves with speed v/2 Solution 15
A
Problem 16
Two identical blocks are connected by a spring. The combination is suspended at rest from a string attached to the ceiling as shown in the figure above. The string breakes suddenly. Immediately after the string breakes what is the downward acceleration of the upper block?
√(2) g g / 2 2 g 0 g Solution 16
C
Problem 17
The cylinder shown above, with mass M and radius R, has a radially dependent density. The cylinder starts from rest and rolls without slipping down an inclined plane of height H. At the bottom of the plane its translational speed is (8 g H / 7)
1/2. What is the rotational inertia of the cylinder? Solution 17Answer Problem 18
Two small equal masses m are connected by an ideal massless spring that has equilibrium length ℓ
0 and force constant k. The system is free to move without friction in the plane of the page. If p 1 and p 2 represent the magnitudes of the momenta of the 2 masses, what is a Hamiltonian for this system? Solution 18Answer Problem 19
A 2kg box hangs by a massless rope from a ceiling. A force slowly pulls the box horizontally to the side until the horizontal force is 10N. The box is then in equilibrium as shown above. The angle that the rope makes with the vertical is closes to:
arcsin 0.5 45 degrees arcsin 2.0 arctan 0.5 arctan 2.0 Solution 19 Problem 20
3 masses are connected by 2 springs as shown above. A longitudinal normal mode with frquency (below) is exhibited by what motion of A, B, and C?
Solution 20Answer Problem 21
Two wedges, each of mass m, are placed next to each other on a flat floor. A cube of mass M is balanced on the wedges as shown above. Assume no friction between the cube and the wedges, but a coefficient of static friction μ < 1 between the wedges and the floor. What is the largest M that can be balanced as shown without motion of the wedges?
\frac {m}{\sqrt{2}} \frac {\mu m}{1 - \mu} \frac {2 \mu m}{1 - \mu} \frac {\mu m}{\sqrt{2}} All M will balance Solution 21
C
Problem 22
A cylindrical tube of mass M can slide on a horizontal wire. Two identical pendulums, each of mass m and length l, hang from the ends of the tube, as shown above. For small oscillations of the pendulums in the planet of the paper, the eigenfrequencies of the normal modes of oscillation of this system are 0,
and?
\sqrt {\frac{g m}{l M}} \sqrt {\frac{g m}{l(M + 2m)}} \sqrt {\frac{g}{m}} \sqrt {\frac{g (M + m)}{l m}} \sqrt {\frac{g m}{l(M + m)}} Solution 22
C
Problem 23
A solid cone hangs from a frictionless pivot at the origin O, as shown above. If î, ĵ, k̂ are unit vectors, and a, b, and c are positive constants, which of the following forces F applied to the rim of the cone at a point P results in a torque τ on the cone with a negative component τ
z? F = ak̂, P is (0, b, -c) F =-ak̂, P is (-b, 0, -c) F = aĵ, P is (-b, 0, -c) F =-ak̂, P is (0, -b, -c) F = aĵ, P is (b , 0, -c) Solution 23
C
Problem 24
A car travels with constant speed on a circular road on level ground. In the diagram above, F
air is the force of air resistance on the car. Which of the other forces shown best represents the horizontal force of the road on the car's tires? F A F B F C F D F E Solution 24Answer Problem 25
A block of mass m sliding down an incline at constant speed is initially at heigh h above the ground, as shown in the figure above. The coefficient of kinetic friction between the mass and the include is μ. If the mass continues to slide down the incline at a constant speed, how much energy is dissipated by friction by the time the mass reasches the bottom of the incline?
0 m g h sinθ m g h μ m g h / sinθ m g h / μ Solution 25
C
Problem 26
Three equal masses m are rigidly connected to each other by massless rods of length ℓ forming an equilateral triangle. The assembly is to be given an angular velocity ω about an axis perpendicular to the triangle. For fixed ω the ratio of the kinetic energy of the assembly for an axis through B compared with that for an axis through A is equal to:
3 2 1 1/2 1/3 Solution 26Answer Problem 27
A bead is constrained to slide on a frictionless rod that is fixed at an angle θ with a vertical axis and is rotating with angular frequency ω about the axis. Taking the distance s along the rod as the variable, what is the Lagrangian for the bead?
Solution 27Answer Problem 28
A string consists of 2 parts attached at x = 0. The right part of the string (x > 0) has a mass μ
r per unit length and the left part of the string (x < 0) has mass μ ℓ per unit length. The string tension is T. If a wave of unit amplitude travels along the left part of the string, what is the amplitude of the wave that is transmitted to the right part of the string? Solution 28Answer Problem 29
Consider a particle moving without friction on a rippled surface, as shown above. Gravity acts down in the negative h direction. The elevation h(x) of the surface is given by h(x) = d cos(kx). If the particle starts at x=0 with a speed v in the x direction for what values of v will the particle stay on the surface at all times?
Solution 29Answer Problem 30
Two pendulums are attached to a massless spring, as shown above. The arms of the pendulums are of identical lengths ℓ, but the pendulum balls have unequal masses m
1and m 2. The initial distance between the masses is the equilibrium length of the spring, which has spring constant K. What is the highest normal mode frequency of the system? Solution 30 Problem 31
Small amplitude standing waves of wavelength λ occur on a string with tension T, mass per unit length μ and length L. One end of the string is fixed and the other end is attached to a ring of mass M that slides on a frictionless rod, as shown in the figure above. When gravity is neglected, which of the following conditions correctly determines the wavelength? (You might want to consider the limiting cases M goes to 0 and M goes to infinity)
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However there is already a cloud on the horizon: constraints. Here are two situations I've already encountered:
In SSP without recourse, the graph structure for a particular realization might not be fully connected. I want to ensure the underlying CSMC doesn't select an infeasible edge. In the Ad Selection Problem, the main challenge was ensuring that the same ad was not picked repeatedly.
I've been wondering what the right formulation of constrained CSMC would be, and have thought of two scenarios. The first ``average formulation'' would be appropriate for when the set of feasible choices is related to the problem instances in a stationary fashion. In this case, there is a distribution $D = D_x \times D_{\omega|x} \times D_{c|\omega,x}$, where $c: K \to \mathbf{R}$ takes values in the extended reals $\mathbf{R} = \mathbb{R} \cup \{ \infty \}$, and the components of $c$ which are $\infty$-valued for a particular instance are revealed as part of the problem instance via $\omega \in \mathcal{P} (K)$ (i.e., $\omega$ is a subset of $K$). The regret of a particular classifier $h: X \times \mathcal{P} (K) \to K$ is given by \[ r_{av} (h) = E_{(x, \omega) \sim D_x \times D_{\omega|x}} \left[ E_{c \sim D_{c|\omega,x}} \left[ c (h (x, \omega)) - \min_{k \in K}\; E_{c \sim D_{c|\omega,x}} \left[ c (k) \right] \right] \right]. \] The second ``minimax formulation'' would be appropriate for when the set of feasible choices will vary in an ad-hoc fashion which is unpredictable from past data. In this case, there is a distribution $D = D_x \times D_{\tilde c|x}$, where $\tilde c: K \to \mathbb{R}$ takes values in the regular reals $\mathbb{R}$. Then, an adversary comes in and manufactures a cost vector $c: K \to \mathbf{R}$ in the extended reals $\mathbf{R}$ by setting some of the components to $\infty$; these choices are revealed via $\omega$ prior to a decision being elicited. In this case the regret of a particular classifier is given by \[ r_{mm} (h) = E_{x \sim D_x} \left[ E_{\tilde c \sim D_{\tilde c|x}} \left[ \max_{\omega \in \mathcal{P} (K)}\; \left\{ c (h (x, \omega)) - \min_{k \in K}\; E_{c \sim D_{\tilde c|x}} \left[ c (k) \right] \right\} \right] \right]. \] Intuitively it feels like SSP without recourse would better fit the average formulation better, and ad serving would better fit the minimax formulation.
Now for some speculation. I suspect that the relative efficiency of the filter tree reduction of CSMC to binary is because it does not handle constraints well. In other words, reduction to regression is prepared to handle arbitrary constraints by limiting the scope of the argmax, but one pays for this additional flexibility with a worse regret bound. If there is a rule of thumb in machine learning, it's ``always solve the easiest version of your problem'', and there are lots of CSMC problems which are unconstrained (i.e., all decisions are allowed on all instances), so this additional flexibility is not warranted.
So the next step is to look at the various reductions of CSMC to regression or binary classification and analyze how well they perform on ``constrained CSMC''.
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Subgroups of Finite Groups with Unique Order are Normal Subgroups
Table of Contents
Subgroups of Finite Groups with Unique Order are Normal Subgroups
Proposition 1: Let $G$ be a finite group. If $G$ has only one subgroup $H$ of order $n \mid |G|$ then $H$ is a normal subgroup of $G$. Proof:For each $g \in G$ let $\phi_g : H \to gHg^{-1}$ be defined for all $h \in H$ by:
\begin{align} \quad \phi_g(h) = ghg^{-1} \end{align}
We claim that each $\phi_g$ is a bijection. Let $h_1, h_2 \in H$ and suppose that $\phi_g(h_1) = \phi_g(h_2)$. Then $gh_1g^{-1} = gh_2g^{-1}$. Multiplying this equation on the left by $g^{-1}$ and on the right by $g$ gives us that $h_1 = h_2$. So $\phi_g$ is injective. Furthermore, if $ghg^{-1} \in gHg^{-1}$ then $h \in H$ is such that $\phi_g(h) = ghg^{-1}$. So $\phi_g$ is surjective. Thus $\phi_g$ is a bijection. Since $G$ is a finite group and $H$ is a subgroup of $G$ we see that $|H| = n < \infty$. So, since $\phi_g : H \to gHg^{-1}$ is a bijection we have that $|H| = n = |gHg^{-1}|$. So $gHg^{-1}$ is a subgroup of $G$ of order $n$. Since $G$ has only one such subgroup, we must have that $gHg^{-1} = H$. And since this holds for all $g \in G$, we have that $H$ is a normal subgroup of $G$. $\blacksquare$
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This shows you the differences between two versions of the page.
Both sides previous revision Previous revision cs-401r:axioms-of-probability-theory [2014/09/05 17:02]
cs401rpml [Second Axiom]
cs-401r:axioms-of-probability-theory [2014/09/05 17:02] (current)
cs401rpml [Third Axiom]
Line 12: Line 12: == Third Axiom == == Third Axiom == If for all $0 \leq i, j$ such that $i \neq j$ the events $A_i$ and $A_j$ are disjoint (i.e., $A_i \cap A_j = \emptyset$), then If for all $0 \leq i, j$ such that $i \neq j$ the events $A_i$ and $A_j$ are disjoint (i.e., $A_i \cap A_j = \emptyset$), then - \begin{equation} + - P\left(\bigcup_{j=1}^{\infty}{A_j}\right) = \sum_{j=1}^{\infty}{P\left(A_j\right)} + $P\left(\bigcup_{j=1}^{\infty}{A_j}\right) = \sum_{j=1}^{\infty}{P\left(A_j\right)} $ - \end{equation} + In the simple case of two variables, if $A \cap B = \emptyset$ then: In the simple case of two variables, if $A \cap B = \emptyset$ then: - \begin{equation} + - P\left(A \cup B\right) = P\left(A\right) + P\left(B\right) + $P\left(A \cup B\right) = P\left(A\right) + P\left(B\right) $ - \end{equation} +
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I know I came across a 2 op-amp, 2 capacitor circuit that can be used for a single section of an Inverse Chebyshev (a.k.a. "Chebyshev Type II) or an Elliptic (Cauer) filter. It has a pair of zeros on the \$j\omega\$-axis at \$\pm j\omega_z\$, and with resonant frequency \$\omega_0<\omega_z\$ and the transfer function is:
$$ H(s) = \frac{1 + \left(\tfrac{s}{\omega_z}\right)^2 }{1+\tfrac{1}{Q}\tfrac{s}{\omega_0}+\left(\tfrac{s}{\omega_0}\right)^2} $$
I know how to brute force derive a circuit with a pair of integrators and the canonical form we learned in Linear Circuits class 4 decades ago. Such as this:
I just thought I saw a more elegant circuit, with one less op-amp and a couple fewer resistors, that pretty much guaranteed that the zeros lie on the \$j\omega\$-axis and at a higher frequency than the resonant frequency \$\omega_0\$. Anybody know how to save a couple of parts with this? Is there a single op-amp, two-capacitor, 4-resistor circuit that can do this?
A sorta Sallen-Key with zeros.
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In this answer, AccidentalFourierTransform explains that
Renormalisation has nothing to do with Classical vs Quantum. Any theory, classical or quantum mechanical, needs renomalisation if and only if it is non-linear.
Can anyone help me understand the idea of renormalization in classical field theory (CFT) and how is it similar/different from the idea of renormalization in QFT?
To be concrete, consider the Lagrangian of QED. But let us treat it classically i.e., the Dirac field and the electromagnetic fields as classical fields. How is the renormalization carried out in this CFT?
In QFT, renormalization has to do with integrating out high frequency Fourier modes in the partition function defined in terms the path-integral as $$Z[J]=\int D\phi \exp[i\int d^4x(\mathcal{L}+J\phi)].$$ The Wilsonian picture in QFT quantities get renormalized in the process of integrating out the short-distance physics. But no such path-integral exist in CFT, this picture of renormalization (and RG flow) is lost.
Is there a similar "Wilsonian-like" perspective to understand the necessity and meaning of renormalization in CFT? Or is it that the idea of renormalization is completely different in classical physics and has nothing to do with integrating out short-distance physics?
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Range of a Linear Map
Definition: If $T \in \mathcal L (V, W)$ then the Range of the linear transformation $T$ is the subset of $W$ defined as $\mathrm{range} (T) = \{ T(v) : v \in V \}$, that is, the set of all vectors $T(v) \in W$ to which are mapped to from vectors $v \in V$.
Before we look at some examples of ranges of vector spaces, we will first establish that the range of a linear transformation can never be equal to the empty set. This should intuitively make sense. If $T : V \to W$ is a linear transformation, then since $V \neq \emptyset$ (since $V$ is a linear transformation) then there must contain at least one element in $V$, and so this element is mapped to some vector in $W$. We will verify this with the following lemma.
Lemma 1: If $T \in \mathcal L (V, W)$ then the range of $T$ contains at least one element from $W$, that is $\mathrm{range} (T) \neq \emptyset$. Proof:Since $V$ and $W$ are vector spaces, then we have that both $V$ and $W$ are nonempty sets. Furthermore, both $V$ and $W$ contain their respective additive identities $0_V$ and $0_W$. From the Null Space of a Linear Map page, we know that $T(0_V) = 0_W$ and so $0_W \in \mathrm{range} (T)$, so $\mathrm{range} (T) \neq \emptyset$. $\blacksquare$
Now the following lemma will tell us that the $\mathrm{range} (T)$ is a subspace of $W$.
Lemma 2: If $T \in \mathcal L (V, W)$ then the subset $\mathrm{range} (T)$ is a subspace of $W$. Proof:Since $\mathrm{range} (T) \subseteq W$, all we must do is verify that $\mathrm{range} (T)$ is closed under addition, closed under scalar multiplication, and contains the zero vector of $W$. Let $w, y \in W$ and $a \in \mathbb{F}$. Since $w, y \in W$ we have that there exists vectors $u, v \in V$ such that $w = T(u)$ and $y = T(v)$. Therefore $w + y = T(u) + T(v)$ and since $T$ is a linear transformation then $T(u) + T(v) = T(u + v)$. Therefore $w + y = T(u + v)$ and so $(w + y) \in \mathrm{range} (T)$ so $\mathrm{range} (T)$ is closed under addition. Once again, since $w \in W$, there exists a vector $u \in V$ such that $w = T(u)$. So $aw = aT(u)$ and since $T$ is a linear tranformation then $aT(u) = T(au)$. Therefore $aw = T(au)$ and so $(au) \in \mathrm{range} (T)$ so $\mathrm{range} (T)$ is closed under scalar multiplication. From Lemma 1, we have that $0_W \in \mathrm{range} (T)$, and so $\mathrm{range} (T)$ contains the zero vector of $W$. Therefore $\mathrm{range} (T)$ is a subspace of $W$. $\blacksquare$ Notice that lemmas 1 and 2 above are analogous that of lemmas 1 and 2 from the Null Space of a Linear Map page. It is important to note that if $T : V \to W$ is a linear map from the vector spaces $V$ to $W$, then both the null space of $T$ and the range of $T$ are nonempty and the null space of $T$ is a subspace of $V$, while the range of $T$ is a subspace of $W$.
We will now look at some examples of ranges of linear transformations.
The Range of the Zero Map
If $0 \in \mathcal L (V, W)$ represents the zero map, then $\mathrm{range} (T) = \{ 0 \}$ since every vector $v \in V$ is mapped to $0_W \in W$.
The Range of the Identity Map
If $I \in \mathcal L (V, V)$ represents the identity map, then $\mathrm{range} (T) = V$ since every vector $v \in V$ is mapped to itself, so the range contains all vectors from $V$.
The Range of the Left Shift Operator
If $T \in \mathcal L (\mathbb{F}^{\infty}, \mathbb{F}^{\infty})$ represents the left shift operator, then $\mathrm{range} (T) = \mathbb{F}^{\infty}$ since any sequence $(x_2, x_3, ...) \in \mathbb{F}^{\infty}$ is mapped from the sequence $(x_1, x_2, ...) \in \mathbb{F}^{\infty}$.
The Range of the Right Shift Operator
If $T \in \mathcal L (\mathbb{F}^{\infty}, \mathbb{F}^{\infty})$ represents the right shift operator, then $\mathrm{range} (T) = \{ (0, x_1, x_2, ...) : x_1, x_2, ... \in \mathbb{F} \}$. We note that any sequence $(x_1, x_2, ...)$ where $x_1 \neq 0$ cannot be in the range of $T$ since the first term of any sequence under $T$ will be zero.
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My question is really as to whether I can consider the following results as proof of one another, since $(2a)$&$(2b)$ cannot be true unless $(0)$ is true, and vice versa.
So if for below I prove $(0)$ using the elementary means in Real Analysis, can I also then consider this to be proof for $(2a)$&$(2b)$?
I am looking to prove with the standard epsilon approach that:
$$x \in \mathbb R \Rightarrow\lim _{n\rightarrow \infty }\Bigl({\frac { \lfloor nx \rfloor }{n} }\Bigr)=x $$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(0)}$$
Which has become necessary in the following context:
It's well known that in the p-adic coefficient of a natural number: $$d_{n,p,j}= \Bigl\lfloor{n\,{p}^{\,j- {\lfloor \ln_p(n)\rfloor} -1}} \Bigr\rfloor-p\Bigl\lfloor{n\,{p}^{\,j- {\lfloor \ln_p(n)\rfloor} -2}} \Bigr\rfloor$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(1)}$$ as seen in the p-adic expansion of that number: $$n=\sum_{j=1}^{\lfloor \ln_p(n)\rfloor+1}d_{n,p,j}\,p^{\,\lfloor \ln_p(n)\rfloor+1-j}\quad\forall n \in \mathbb N$$
can likewise be used to separate any positive real number into it's integer and fractional parts:
$$\lfloor x\rfloor=\sum_{j=1}^{\lfloor \ln_p(x)\rfloor+1}d_{x,p,j}\,p^{\,\lfloor \ln_p(x)\rfloor+1-j}\quad\forall x \in \mathbb R^{+}$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(2a)}$$ $${\{x}\}=\sum_{j=\lfloor \ln_p(x)\rfloor+2}^{\infty}d_{x,p,j}\,p^{\,\lfloor \ln_p(x)\rfloor+1-j}\quad\forall x \in \mathbb R^{+}$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(2b)}$$
And because the corresponding finite sum for the above always computes a value much like the one I am trying to prove:$$\sum_{j=1}^{n}d_{x,p,j}\,p^{\,\lfloor\ln_p(x)\rfloor+1-j}=\frac{\lfloor Nx\rfloor}{N}$$Then if it is true that taking the limit $n \rightarrow \infty$ results in convergence to $x$
in the manner described above, this means that proving $(0)$ to be true also concurrently proves $(2a)$ & $(2b)$ to be true.
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Let $\operatorname{Klein}$ denote the category of principal homogeneous bundles. An object in this category is a tuple $\mathbf Q = (Q, P; G, H; q, a, \tilde a)$, where:
$G$ is a Lie group, and $H$ is a closed subgroup;
$P$ is a $G$-torsor and the manifold $Q$ is diffeomorphic to the quotient $G/H$;
The (left) actions $\tilde a : G \to \operatorname{Aut}(P)$ and $a : G \to \operatorname{Aut}(Q)$ are morphisms of topological groups; and
$q :P \to Q$ is a principal $H$-bundle.
A morphism $\Phi : \mathbf Q \to \mathbf Q'$ is described by a tuple $\Phi = (\varphi, \tilde \varphi, f)$, where $\varphi : Q \to Q'$ and $\tilde \varphi : P \to P'$ are diffeomorphisms which commute with the bundle maps $q : P \to Q$ and $q' : P' \to Q'$, and $f : G \to G'$ is a morphism of Lie groups mapping $H$ to $H'$, and which satisfies the identity $$\tilde a'_{f(g)} \circ \tilde \varphi = \tilde \varphi \circ \tilde a_g$$ for all $g \in G$.
I think that this is the correct categorical description of principal homogeneous bundles; please correct me if I'm wrong. I selected the name $\operatorname{Klein}$ in homage to Felix Klein and his Erlangen Program.
It seems that such a bundle $\mathbf Q$ contains all the data on its symmetries. Namely, I think that its automorphism group $\operatorname{Aut}(\mathbf Q)$ is isomorphic to its Lie group Lie group $G = G(\mathbf Q)$?
It is easy to see that there is a natural map $K : G \hookrightarrow \operatorname{Aut}(\mathbf Q)$, in that each $u \in G$ corresponds to a unique automorphism $K_u \in \operatorname{Aut}(\mathbf Q)$. The morphism $K_u = (k_u, \tilde k_u, c_u)$ is defined by $$k_u(q) = a_u(q), \quad \tilde k_u(p) = \tilde a_u(p), \quad \mathrm{and} \quad c_u(g) = ugu^{-1}.$$ That is, the morphism $K_u$ acts by left-multiplication on both $Q$ and $P$, but by left-conjugation on $G$.
Is this map $K$ surjective? i.e., is $\operatorname{Aut}(\mathbf Q)$ isomorphic to $G = G(\mathbf Q)$?
If the answer is yes, then I think that this captures the notion of "internal symmetries" of the bundle, since these are the transformations which preserve the bundle structure.
However, I know that groupoids also show up to describe symmetries in a categorical setting, and I would be interested to hear more on that point of view.
How can groupoids be used to describe symmetries in this category?
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Hey guys! I built the voltage multiplier with alternating square wave from a 555 timer as a source (which is measured 4.5V by my multimeter) but the voltage multiplier doesn't seem to work. I tried first making a voltage doubler and it showed 9V (which is correct I suppose) but when I try a quadrupler for example and the voltage starts from like 6V and starts to go down around 0.1V per second.
Oh! I found a mistake in my wiring and fixed it. Now it seems to show 12V and instantly starts to go down by 0.1V per sec.
But you really should ask the people in Electrical Engineering. I just had a quick peek, and there was a recent conversation about voltage multipliers. I assume there are people there who've made high voltage stuff, like rail guns, which need a lot of current, so a low current circuit like yours should be simple for them.
So what did the guys in the EE chat say...
The voltage multiplier should be ok on a capacitive load. It will drop the voltage on a resistive load, as mentioned in various Electrical Engineering links on the topic. I assume you have thoroughly explored the links I have been posting for you...
A multimeter is basically an ammeter. To measure voltage, it puts a stable resistor into the circuit and measures the current running through it.
Hi all! There is theorem that links the imaginary and the real part in a time dependent analytic function. I forgot its name. Its named after some dutch(?) scientist and is used in solid state physics, who can help?
The Kramers–Kronig relations are bidirectional mathematical relations, connecting the real and imaginary parts of any complex function that is analytic in the upper half-plane. These relations are often used to calculate the real part from the imaginary part (or vice versa) of response functions in physical systems, because for stable systems, causality implies the analyticity condition, and conversely, analyticity implies causality of the corresponding stable physical system. The relation is named in honor of Ralph Kronig and Hans Kramers. In mathematics these relations are known under the names...
I have a weird question: The output on an astable multivibrator will be shown on a multimeter as half the input voltage (for example we have 9V-0V-9V-0V...and the multimeter averages it out and displays 4.5V). But then if I put that output to a voltage doubler, the voltage should be 18V, not 9V right? Since the voltage doubler will output in DC.
I've tried hooking up a transformer (9V to 230V, 0.5A) to an astable multivibrator (which operates at 671Hz) but something starts to smell burnt and the components of the astable multivibrator get hot. How do I fix this? I check it after that and the astable multivibrator works.
I searched the whole god damn internet, asked every god damn forum and I can't find a single schematic that converts 9V DC to 1500V DC without using giant transformers and power stage devices that weight 1 billion tons....
something so "simple" turns out to be hard as duck
In peskin book of QFT the sum over zero point energy modes is an infinite c-number, fortunately, it's experimental evidence doesn't appear, since experimentalists measure the difference in energy from the ground state. According to my understanding the zro pt energy is the same as the ground state, isn't it?
If so, always it is possible to substract a finite number (higher exited state for e.g.) from this zero point enrgy (which is infinte), it follows that, experimentally we always obtain infinte spectrum.
@AaronStevens Yeah, I had a good laugh to myself when he responded back with "Yeah, maybe they considered it and it was just too complicated". I can't even be mad at people like that. They are clearly fairly new to physics and don't quite grasp yet that most "novel" ideas have been thought of to death by someone; likely 100+ years ago if it's classical physics
I have recently come up with a design of a conceptual electromagntic field propulsion system which should not violate any conservation laws, particularly the Law of Conservation of Momentum and the Law of Conservation of Energy. In fact, this system should work in conjunction with these two laws ...
I rememeber that Gordon Freeman's thesis was "Observation of Einstein-Podolsky-Rosen Entanglement on Supraquantum Structures by Induction Through Nonlinear Transuranic Crystal of Extremely Long Wavelength (ELW) Pulse from Mode-Locked Source Array "
In peskin book of QFT the sum over zero point energy modes is an infinite c-number, fortunately, it's experimental evidence doesn't appear, since experimentalists measure the difference in energy from the ground state. According to my understanding the zro pt energy is the same as the ground state, isn't it? If so, always it is possible to substract a finite number (higher exited state for e.g.) from this zero point enrgy (which is infinte), it follows that, experimentally we always obtain infinte spectrum.
@ACuriousMind What confuses me is the interpretation of Peskin to this infinite c-number and the experimental fact
He said, the second term is the sum over zero point energy modes which is infnite as you mentioned. He added," fortunately, this energy cannot be detected experm., since the experiments measure only the difference between from the ground state of H".
@ACuriousMind Thank you, I understood your explanations clearly. However, regarding what Peskin mentioned in his book, there is a contradiction between what he said about the infinity of the zero point energy/ground state energy, and the fact that this energy is not detectable experimentally because the measurable quantity is the difference in energy between the ground state (which is infinite and this is the confusion) and a higher level.
It's just the first encounter with something that needs to be renormalized. Renormalizable theories are not "incomplete", even though you can take the Wilsonian standpoint that renormalized QFTs are effective theories cut off at a scale.
according to the author, the energy differenc is always infinite according to two fact. the first is, the ground state energy is infnite, secondly, the energy differenc is defined by substituting a higher level energy from the ground state one.
@enumaris That is an unfairly pithy way of putting it. There are finite, rigorous frameworks for renormalized perturbation theories following the work of Epstein and Glaser (buzzword: Causal perturbation theory). Just like in many other areas, the physicist's math sweeps a lot of subtlety under the rug, but that is far from unique to QFT or renormalization
The classical electrostatics formula $H = \int \frac{\mathbf{E}^2}{8 \pi} dV = \frac{1}{2} \sum_a e_a \phi(\mathbf{r}_a)$ with $\phi_a = \sum_b \frac{e_b}{R_{ab}}$ allows for $R_{aa} = 0$ terms i.e. dividing by zero to get infinities also, the problem stems from the fact that $R_{aa}$ can be zero due to using point particles, overall it's an infinite constant added to the particle that we throw away just as in QFT
@bolbteppa I understand the idea that we need to drop such terms to be in consistency with experiments. But i cannot understand why the experiment didn't predict such infinities that arose in the theory?
These $e_a/R_{aa}$ terms in the big sum are called self-energy terms, and are infinite, which means a relativistic electron would also have to have infinite mass if taken seriously, and relativity forbids the notion of a rigid body so we have to model them as point particles and can't avoid these $R_{aa} = 0$ values.
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I consider a spin manifold $M$, with the Dirac operator $D$ and spinors $\psi$. I consider the differential equation:
$$ D \psi = \frac {1} {2} (\frac {d < \psi, \psi >}{<\psi, \psi>})^* .\psi $$
with the Clifford multiplication of the vectors.
Then it can be showed that the real gauge group ${\cal C }^{\infty} (M,R^*)$ acts over the solutions. Have I a nice moduli space (finite dimensional and compact)?
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The prime number theorem says that the density of prime numbers is inverse as the number of digits of $n$:
$$\displaystyle \frac{\{1 \leq k \leq n : \text{ prime } \}}{n} \approx \frac{1}{\log n}$$
Strategies to prove this result usually resort to estimating generating functions of various types. And it seems to be proven over and over again, e.g. in these notes on complex-analytic multiplicative number theory.
my issue with the above proofs is they are vary analytic and have estimates that I am not comfortable with.
In fact, instead of the prime number theorem let's try a much simpler estimate. Let $\Lambda(x)$ be the Van Mangoldt function.
$$ \Lambda(n) = \begin{cases} \log p & \text{ if }n = p^k \\ 0 & \text{otherwise} \end{cases} = -\sum_{d|n} \mu(d) \log(d)$$
Logarithms and fractions naturally come up with discussions in the hyperbolic metric. Is it possible to
lift classical proofs into hyperbolic geometry? In particular, I could try the result:
$$ \sum_{n \leq x} \Lambda(x) = x + o(x)$$
Does estimates like these of the Van Mangoldt function have a geometric interpretation, e.g. in terms of geodesics?
$$\pi_\Gamma(x) = \# \{ \gamma \in SL(2,\mathbb{Z}): N(\gamma) = e^{\ell(\gamma)} \leq x \} = \int_0^x \frac{dt}{\log t} + \text{ error } $$
In this case, there is no analogue of the Van Mangoldt function. These results involve very difficult spectral arguments and I am worse off than I started.
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NTS ABSTRACTSpring2019
Return to [1]
Contents Jan 23
Yunqing Tang Reductions of abelian surfaces over global function fields For a non-isotrivial ordinary abelian surface $A$ over a global function field, under mild assumptions, we prove that there are infinitely many places modulo which $A$ is geometrically isogenous to the product of two elliptic curves. This result can be viewed as a generalization of a theorem of Chai and Oort. This is joint work with Davesh Maulik and Ananth Shankar. Jan 24
Hassan-Mao-Smith--Zhu The diophantine exponent of the $\mathbb{Z}/q\mathbb{Z}$ points of $S^{d-2}\subset S^d$ Abstract: Assume a polynomial-time algorithm for factoring integers, Conjecture~\ref{conj}, $d\geq 3,$ and $q$ and $p$ prime numbers, where $p\leq q^A$ for some $A>0$. We develop a polynomial-time algorithm in $\log(q)$ that lifts every $\mathbb{Z}/q\mathbb{Z}$ point of $S^{d-2}\subset S^{d}$ to a $\mathbb{Z}[1/p]$ point of $S^d$ with the minimum height. We implement our algorithm for $d=3 \text{ and }4$. Based on our numerical results, we formulate a conjecture which can be checked in polynomial-time and gives the optimal bound on the diophantine exponent of the $\mathbb{Z}/q\mathbb{Z}$ points of $S^{d-2}\subset S^d$. Jan 31
Kyle Pratt Breaking the $\frac{1}{2}$-barrier for the twisted second moment of Dirichlet $L$-functions Abstract: I will discuss recent work, joint with Bui, Robles, and Zaharescu, on a moment problem for Dirichlet $L$-functions. By way of motivation I will spend some time discussing the Lindel\"of Hypothesis, and work of Bettin, Chandee, and Radziwi\l\l. The talk will be accessible, as I will give lots of background information and will not dwell on technicalities. Feb 7
Shamgar Gurevich Harmonic Analysis on $GL_n$ over finite fields Abstract: There are many formulas that express interesting properties of a group G in terms of sums over its characters.
For evaluating or estimating these sums, one of the most salient quantities to understand is the {\it character ratio}: $$trace (\rho(g))/dim (\rho),$$ for an irreducible representation $\rho$ of G and an element g of G. For example, Diaconis and Shahshahani stated a formula of this type for analyzing G-biinvariant random walks on G. It turns out that, for classical groups G over finite fields (which provide most examples of finite simple groups), there is a natural invariant of representations that provides strong information on the character ratio. We call this invariant {\it rank}. This talk will discuss the notion of rank for GLn over finite fields, and apply the results to random walks. This is joint work with Roger Howe (TAMU).
Feb 14
Tonghai Yang The Lambda invariant and its CM values Abstract: The Lambda invariant which parametrizes elliptic curves with two torsions (X_0(2)) has some interesting properties, some similar to that of the j-invariants, and some not. For example, $\lambda(\frac{d+\sqrt d}2)$ is a unit sometime. In this talk, I will briefly describe some of the properties. This is joint work with Hongbo Yin and Peng Yu. Feb 28
Brian Lawrence Diophantine problems and a p-adic period map. Abstract: I will outline a proof of Mordell's conjecture / Faltings's theorem using p-adic Hodge theory. Joint with Akshay Venkatesh. March 7
Masoud Zargar Sections of quadrics over the affine line Abstract: Abstract: Suppose we have a quadratic form Q(x) in d\geq 4 variables over F_q[t] and f(t) is a polynomial over F_q. We consider the affine variety X given by the equation Q(x)=f(t) as a family of varieties over the affine line A^1_{F_q}. Given finitely many closed points in distinct fibers of this family, we ask when there exists a section passing through these points. We study this problem using the circle method over F_q((1/t)). Time permitting, I will mention connections to Lubotzky-Phillips-Sarnak (LPS) Ramanujan graphs. Joint with Naser T. Sardari March 14
Elena Mantovan p-adic automorphic forms, differential operators and Galois representations A strategy pioneered by Serre and Katz in the 1970s yields a construction of p-adic families of modular forms via the study of Serre's weight-raising differential operator Theta. This construction is a key ingredient in Deligne-Serre's theorem associating Galois representations to modular forms of weight 1, and in the study of the weight part of Serre's conjecture. In this talk I will discuss recent progress towards generalizing this theory to automorphic forms on unitary and symplectic Shimura varieites. In particular, I will introduce certain p-adic analogues of Maass-Shimura weight-raising differential operators, and discuss their action on p-adic automorphic forms, and on the associated mod p Galois representations. In contrast with Serre's classical approach where q-expansions play a prominent role, our approach is geometric in nature and is inspired by earlier work of Katz and Gross.
This talk is based joint work with Eishen, and also with Fintzen--Varma, and with Flander--Ghitza--McAndrew.
March 28
Adebisi Agboola Relative K-groups and rings of integers Abstract: Suppose that F is a number field and G is a finite group. I shall discuss a conjecture in relative algebraic K-theory (in essence, a conjectural Hasse principle applied to certain relative algebraic K-groups) that implies an affirmative answer to both the inverse Galois problem for F and G and to an analogous problem concerning the Galois module structure of rings of integers in tame extensions of F. It also implies the weak Malle conjecture on counting tame G-extensions of F according to discriminant. The K-theoretic conjecture can be proved in many cases (subject to mild technical conditions), e.g. when G is of odd order, giving a partial analogue of a classical theorem of Shafarevich in this setting. While this approach does not, as yet, resolve any new cases of the inverse Galois problem, it does yield substantial new results concerning both the Galois module structure of rings of integers and the weak Malle conjecture. April 4
Wei-Lun Tsai Hecke L-functions and $\ell$ torsion in class groups Abstract: The canonical Hecke characters in the sense of Rohrlich form a
set of algebraic Hecke characters with important arithmetic properties. In this talk, we will explain how one can prove quantitative nonvanishing results for the central values of their corresponding L-functions using methods of an arithmetic statistical flavor. In particular, the methods used rely crucially on recent work of Ellenberg, Pierce, and Wood concerning bounds for $\ell$-torsion in class groups of number fields. This is joint work with Byoung Du Kim and Riad Masri.
April 11
Taylor Mcadam Almost-prime times in horospherical flows Abstract: Equidistribution results play an important role in dynamical systems and their applications in number theory. Often in such applications it is desirable for equidistribution to be effective (i.e. the rate of convergence is known). In this talk I will discuss some of the history of effective equidistribution results in homogeneous dynamics and give an effective result for horospherical flows on the space of lattices. I will then describe an application to studying the distribution of almost-prime times in horospherical orbits and discuss connections of this work to Sarnak’s Mobius disjointness conjecture. April 18
Ila Varma Malle's Conjecture for octic $D_4$-fields. Abstract: We consider the family of normal octic fields with Galois group $D_4$, ordered by their discriminant. In forthcoming joint work with Arul Shankar, we verify the strong Malle conjecture for this family of number fields, obtaining the order of growth as well as the constant of proportionality. In this talk, we will discuss and review the combination of techniques from analytic number theory and geometry-of-numbers methods used to prove these results. April 25
Michael Bush Interactions between group theory and number theory Abstract: I'll survey some of the ways in which group theory has helped us understand extensions of number fields with restricted ramification and why one might care about such things. Some of Nigel's contributions will be highlighted. A good portion of the talk should be accessible to those other than number theorists. April 25
Rafe Jones Eventually stable polynomials and arboreal Galois representations Abstract: Call a polynomial defined over a field K eventually stable if its nth iterate has a uniformly bounded number of irreducible factors (over K) as n grows. I’ll discuss some far-reaching conjectures on eventual stability, and recent work on various special cases. I’ll also describe some natural connections between eventual stability and arboreal Galois representations, which Nigel Boston introduced in the early 2000s. April 25 NTS
Jen Berg Rational points on conic bundles over elliptic curves with positive rank Abstract: Varieties that fail to have rational points despite having local points for each prime are said to fail the Hasse principle. A systematic tool accounting for these failures is called the Brauer-Manin obstruction, which uses the Brauer group, Br X, to preclude the existence of rational points on a variety X. In this talk, we'll explore the arithmetic of conic bundles over elliptic curves of positive rank over a number field k. We'll discuss the insufficiency of the known obstructions to explain the failures of the Hasse principle for such varieties over a number field. We'll further consider questions on the distribution of the rational points of X with respect to the image of X(k) inside of the rational points of the elliptic curve E. In the process, we'll discuss results on a local-to-global principle for torsion points on elliptic curves over Q. This is joint work in progress with Masahiro Nakahara. April 25
Judy Walker Derangements of Finite Groups Abstract: In the early 1990’s, Nigel Boston taught an innovative graduate-level group theory course at the University of Illinois that focused on derangements (fixed-point-free elements) of transitive permutation groups. The course culminated in the writing of a 7-authored paper that appeared in Communications in Algebra in 1993. This paper contained a conjecture that was eventually proven by Fulman and Guralnick, with that result appearing in the Transactions of the American Mathematical Society just last year. May 2
Melanie Matchett Wood Unramified extensions of random global fields Abstract: For any finite group Gamma, I will give a "non-abelian-Cohen-Martinet Conjecture," i.e. a conjectural distribution on the "good part" of the Galois group of the maximal unramified extension of a global field K, as K varies over all Galois Gamma extensions of the rationals or rational function field over a finite field. I will explain the motivation for this conjecture based on what we know about these maximal unramified extensions (very little), and how we prove, in the function field case, as the size of the finite field goes to infinity, that the moments of the Galois groups of these maximal unramified extensions match out conjecture. This talk covers work in progress with Yuan Liu and David Zureick-Brown May 9
David Zureick-Brown Arithmetic of stacks Abstract: I'll discuss several diophantine problems that naturally lead one to study algebraic stacks, and discuss a few results.
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The price of the stock XYZ follows a brownian motion pattern with starting price = 10, μ = 0 and σ = 20 (on annual basis). What's the probability that in 6 months the price is less or equal to 8? Also i must solve this with paper and pen (I can consult the Normal distribution tabel)
Let $(S_t)$ be the price process of your stock such that $S_t = S_0+ \mu t + \sigma B_t$ where $(B_t)$ is a standard Brownian motion. Then, since $B_t\sim N(0,t)$, we get $S_t\sim N(S_0+\mu t, \sigma^2 t)$. In six months, $t=\frac{1}{2}$, we have $S_{0.5}\sim N(10,200)$, i.e. $S_{0.5}=10+\sqrt{200}Z=10+10\sqrt{2}Z$ where $Z\sim N(0,1)$. Thus, \begin{align*} \mathbb{P}[\{S_{0.5}\leq8\}] &= \mathbb{P}[\{10+10\sqrt{2}Z\leq8\}] \\ &= \mathbb{P}\left[\left\{Z\leq-\frac{1}{5\sqrt{2}}\right\}\right] \\ &= \Phi\left(-\frac{1}{5\sqrt{2}}\right) \\ &= 1-\Phi\left(\frac{1}{5\sqrt{2}}\right) \\ &\approx 1-\Phi\left(0.141\right) \\ &\approx 0.444. \end{align*}
You get the last number from your normal table. Note that under your model, the stock price may be negative with positive probability. Furthermore, a model with a normal distributed stock price was originally proposed by Bachelier.
I think there is a typo in the previous answer- assuming arithmetic brownian is meant- here is my working:
$P\left[S_t \le 8\right]=P\left[S_0+\mu t+\sigma B_t \le 8\right]$
$=P\left[S_0+\mu t+\sigma \sqrt{t}Z \le 8\right]$
$=P\left[Z\le \frac{8-S_0-\mu t}{\sigma \sqrt{t}}\right]$
$=P\left[Z \le \frac{8-10}{20 \sqrt{0.5}}\right]$
$=P\left[Z \le \frac{-1}{10 \sqrt{0.5}}\right]$
$=P\left[Z\le -0.14\right]$
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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
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Table of Contents
The Riesz Representation Theorem for Hilbert Spaces
Let $H$ be a Hilbert space. For each $g \in H$ let $f_g : H \to \mathbb{R}$ be defined for all $h \in H$ by:(1)
Note that each $f_g$ is linear since if $h_1, h_2 \in H$ we have that:(2)
And for all $h \in H$ and all $a \in \mathbb{R}$ we have that:(3)
Furthermore, observe that $f_g$ is bounded since for all $h \in H$, since by The Cauchy-Schwarz Inequality for Inner Product Spaces:(4)
So $\| f_g \| \leq \| g \|$. Furthermore, we have that:(5)
This shows that $\| g \| \leq \| f_g \|$. So $\| f_g \| = \| g \|$.
The Riesz-Representation theorem tells us that every bounded linear functional on $H$ is of the form $f_g$.
Theorem (The Riesz Representation Theorem for Hilbert Spaces): Let $H$ be a Hilbert space. Then for every $f \in H^*$ there exists an $g \in H$ such that for every $h \in H$, $f(h) = \langle h, g \rangle$. Proof:Let $f \in H^*$. If $f = 0$ then take $g = 0$. Then $f(h) = \langle h, 0 \rangle = 0$ for all $h \in H$. So assume $f \neq 0$. Let $V = \ker f$. Then $V$ is a closed subspace of $H$. Since $f \neq 0$ we have that $V \neq H$. So $V^{\perp} \neq \{ 0 \}$. Let $t \in V^{\perp}$ be such that $\| t \| = 1$ and $f(t) \neq 0$. Note this is always doable. Just take any nonzero $t' \in V^{\perp}$ and let $\displaystyle{t = \frac{t'}{\| t' \|}}$. Then $\| t \| = 1$ and since $t' \in V^{\perp}$ we have that $f(t) \neq 0$. Let: Then for all $h \in H$ we have that: So for every $h \in H$ we have that: Since $g \in V^{\perp}$ we have that then for every $h \in H$: So for every $h \in H$ we have that $f(h) = \langle h, g \rangle$. $\blacksquare$
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Thesis BELLE2-MTHESIS-2019-006 Untagged Exclusive Analysis of the Semileptonic Decay $B\rightarrow\pi\ell\nu$ from Belle II Data in Preparation for |$V_{ub}$| Extraction
Svenja Granderath ; Prof. Dr. Jochen Dingfelder ; Dr. Peter Lewis
2019 Physikalisches InstitutBonn, Germany
Abstract: In this thesis an untagged exclusive analysis of the charmless semileptonic decay $B\rightarrow\pi\ell\nu$ using an integrated luminosity of 5.6 $fb^{−1}$ of early Belle II data was discussed. The analysis was performed in 5 bins of momentum transfer, which makes it sensitive to form factor parametrisations. This method was chosen to prepare for later |$V_{ub}| extraction using a larger dataset. The signal selections were separately optimised in the momentum transfer bins using a figure of merit (FOM) based on the statistical significance of signal on top of background. First, selections based on detector properties and the kinematics of the signal decay were performed. After that, boosted decision trees (BDTs) were employed to separate signal from continuum background, as well as from BB background. After selecting on the BB BDT output classifier the shapes in signal and background of the signal extraction variables, $M_{bc}$ and $\Delta E$, were very similar and the correlations between the yields were high. Therefore, in the end, a selection was only placed on the continuum suppression BDT output classifier. Finally, we set up and validated a simultaneous extended binned maximum likelihood fit on the $M_{bc}$ and $\Delta E$ distributions to extract signal in each momentum transfer bin.bFitting to data was not possible, due to very low statistics and high correlations in the yields of the templates. We performed toy studies to predict the luminosities necessary to achieve statistical significance of 3$\sigma$ and 5$\sigma$ in the total branching fraction for each of the modes. In the $B^{0}$ modes a luminosity of $\sim$ 15 $fb^{−1}$ and $\sim$ 20 $fb^{−1}$ is needed for 3$\sigma$ and 5$\sigma$ significance, respectively. For the $B^{\pm}$ modes higher luminosities are needed: ∼ 30 $fb^{−1}$ for 3$\sigma$ significance and ∼ 50 $fb^{−1}$ for 5$\sigma$ significance. Note: Presented on 19 09 2019 Note: MSc
The record appears in these collections: Books, Theses & Reports > Theses > Masters Theses
Record created 2019-10-04, last modified 2019-10-04
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$\DeclareMathOperator{Hom}{Hom}$Let $R$ be a ring with $1$, $D$ a divisible $\mathbb{Z}$-module. Denote $X=\Hom_{\mathbb{Z}}(R,D)$ as the set of homomorphisms from $R$ to $D$ when viewing $R$ and $D$ as $\mathbb{Z}$-modules. We can associate an $R$-module structure on $X$ by defining $(rf)(x) = f(xr)$ for $r\in R, f\in X$. I want to prove that $X$ is an injective $R$-module.
To accomplish this, it suffices to prove:
Given an injective $g: M\to N$, with $M,N$ being $R$-modules, the induced map $$g^{\ast}: \Hom_R(N,\Hom_{\mathbb{Z}}(R,D)) \to \Hom_R(M,\Hom_{\mathbb{Z}}(R,D))$$ is surjective.
I have made some progress in proving the claim, but I don't know utilize the assumption that $g$ is injective and $D$ is divisible. There is also a hint, which I don't know how to use either: namely $$\Hom_{\mathbb{Z}}(Y,D) \cong \Hom_{R}(Y,\Hom_{\mathbb{Z}}(R,D))$$ for any $R$-module $Y$, the map is given by: $f\mapsto \hat{f}$, with $\hat{f}(y)(r) = f(ry)$. Proving this does not require divisibility of $D$.
Any hint or help is appreciated.
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I am trying to learn about Green's functions as part of my graduate studies and have a rather basic question about them:
In my maths textbooks and a lot of places online, the basic Greens function G for a linear differential operator L is stated as
$$ L G = \delta (x-x') $$
which is all well and good. I am now reading Economou's text on GF in Quantum Physics where he goes to define Green's functions as solutions of inhomogenous DE of the type:
$$ [z - L(r)]G(r,r';z) = \delta (r-r') $$
Where $z = \lambda + is$ and L is a time independent, linear, hermitian differential operator that has eigenfunctions $\phi_n (r)$
$$ L(r) \phi_n (r) = \lambda_n \phi_n (r)$$
Where these $\lambda_n$ are the eigenvalues of L. Where does this z come from in the second equation and what is the link between this and the first one?
Edit: see my post below for a new couple of questions.
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Table of Contents
Adherent, Accumulation, and Isolated Points in Metric Spaces
Recall from the Adherent Points of Subsets in Euclidean Space page that if $S \subseteq \mathbb{R}^n$ then a point $\mathbf{x} \in \mathbb{R}^n$ is an adherent point of $S$ if there exists an $\mathbf{s} \in S$ such that $\mathbf{s} \in B(\mathbf{x}, r)$ for all positive real numbers $r > 0$, i.e., every ball centered at $\mathbf{x}$ contains at least one element from $S$.
Also recall from the Accumulation Points of Subsets in Euclidean Space page that if $S \subseteq \mathbb{R}^n$ then a point $\mathbf{x} \in \mathbb{R}^n$ is an accumulation point of $S$ if there exists an $\mathbf{s} \in S \setminus \{ \mathbf{x} \}$ such that $\mathbf{s} \in B(\mathbf{x}, r)$ for all positive real numbers $r > 0$, i.e., every ball centered at $\mathbf{x}$ contains at least one element from $S$ different from $\mathbf{x}$.
Lastly, recall from the Isolated Points of Subsets in Euclidean Space page that if $S \subseteq \mathbb{R}^n$ then a point $\mathbf{x} \in \mathbb{R}^n$ is an isolated point of $S$ if there exists a positive real number $r_0 > 0$ such that $B(\mathbf{x}, r_0)$ contains no points from $S$ other than $\mathbf{x}$.
We will now define all of these points in terms of general metric spaces. The definitions below are analogous to the ones above with the only difference being the change from the Euclidean metric to any metric.
Definition: Let $(M, d)$ be a metric space and let $S \subseteq M$. A point $x \in M$ is an Adherent Point of $S$ if there exists an $s \in S$ such that $s \in B(x, r) = \{ y \in M : d(x, y) < r \}$ for all positive real numbers $r > 0$, i.e., every ball centered at $x$ contains at least one element from $S$. Equivalently, $x \in M$ is an adherent point of $S$ if for all $r > 0$, $B(x, r) \cap S \neq \emptyset$.
Definition: Let $(M, d)$ be a metric space and let $S \subseteq M$. A point $x \in M$ is an Accumulation Point (or Limit Point) of $S$ if there exists an $s \in S \setminus \{ x \}$ such that $s \in B(x, r) = \{ y \in M : d(x, y) < r \}$ for all positive real numbers $r > 0$, i.e., every ball cenetered at $x$ contains at least one element from $S$ different from $x$. Equivalently, $x \in M$ is an accumulation point of $S$ if for all $r > 0$, $B(x, r) \cap (S \setminus \{ x \}) \neq \emptyset$.
Definition: Let $(M, d)$ be a metric space and let $S \subseteq M$. A point $x \in S$ is an Isolated Point of $S$ if there exists a positive real number $r_0 > 0$ such that $B(x, r) = \{ y \in M : d(x, y) < r_0 \}$ contains no points from $S$ other than $x$. Equivalently, $x \in S$ is an isolated point of $S$ if there exists an $r_0 > 0$ such that $B(x, r_0) \cap (S \setminus \{ x \}) = \emptyset$.
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I am trying to solve a set of DAEs.
\begin{equation} -4 \nu (\lambda(s))^{(-1 - 4 \nu)} \theta'(s) \lambda'(s) + (\lambda(s))^{(-4 \nu)} \theta''(s) = -\alpha_y \cos\theta(s) + \alpha_x \sin\theta(s) \end{equation}
\begin{equation} (\lambda(s))^{(-2 \nu)} \log(\lambda(s)) = f_s (\alpha_x \cos\theta(s) + \alpha_y \sin\theta(s)) \end{equation}
\begin{equation} \theta(0) = 0 \end{equation}
\begin{equation} \theta'(1) = \beta \end{equation}
where $\lambda$ and $\theta$ are two variables, varying over the range $s \in [0,1]$. $\alpha_x, \alpha_y, \beta, f_s, \nu$ are constants. When I try to solve them numerically using NDSolve, Mathematica gives me an error saying DAEs must be given as IVPs.
The code I use is given below
i[s] = (lambda[s])^(-4 nu)i'[s] = D[i[s],s]Eqn1 = theta''[s] i[s] + theta'[s] i'[s] == alphax Sin[theta[s]] - alphay Cos[theta[s]]Eqn2 = (lambda[s])^(-2 nu) Log[lambda[s]] == fs*(alphax Cos[theta[s]] + alphay Sin[theta[s]])BC1 = theta[0] == 0BC2 = theta'[1] == betaparam = {alphax->0.1, alphay->0.1, beta->0.1, nu->0.3, fs->10^-6}thetaSol = NDSolve[{Eqn1,Eqn2,BC1,BC2}/.param,{theta,lambda},{s,0,1}]
If I can solve the second equation to obtain $\lambda(s)$ as a function of $\theta(s)$, then I can eliminate the second equation and solve it as a second order ODE in $\theta(s)$. However, I believe this sort of equation has a solution using the Lambert W function.
Can I use Mathematica to solve this system of equations?
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The Nested Intervals Theorem
The Nested Intervals Theorem
We have just looked at what exactly a Nested Interval is, and we are about to look at a critically important theorem in Real Analysis. Before we look at the Nested Intervals Theorem let's first look at the following important lemma that will be used to prove the Nested Intervals Theorem.
Lemma 1: Let $a < b$ and let $c < d$ and let $I = [a, b]$ and $J = [c, d]$. Then $I \subseteq J$ if and only if $c ≤ a < b ≤ d$.
Now let's look at the Nested Intervals theorem.
Theorem 1: If the interval $I_n = [a_n, b_n]$ for $n \in \mathbb{N}$ is a sequence of closed bounded nested intervals then there exists a real number $\xi = \sup \{ a_n : n \in \mathbb{N} \}$ such that $\xi \in \bigcap_{n=1}^{\infty} I_n$. Proof of Theorem: We note that by the definition of nested intervals that $I_n \subseteq I_1$ for all $n \in \mathbb{N}$ so then $a_n ≤ b_1$. Now consider the nonempty set $A = \{ a_n : n \in \mathbb{N} \}$ that is bounded above by $b_1$. Thus this set has a supremum in the real numbers and denote it $\sup A = \xi$ so that $a_n ≤ \xi$ for all $n \in \mathbb{N}$. We now want to show that $\xi ≤ b_n$ for all $n \in \mathbb{N}$. We will do this by showing that $a_n ≤ b_k$ for all $n, k \in \mathbb{N}$. First consider the case where $n ≤ k$. We thus have that $I_n \supseteq I_k$ by the definition of nested intervals and so by lemma 1 we get that $a_n ≤ a_k ≤ b_k ≤ b_n$. Here we see that $a_n ≤ b_k$. Now consider the case where $n > k$. We thus have that $I_k \supseteq I_n$ by the definition of nested intervals and so by lemma 1 once again we have that $a_k ≤ a_n ≤ b_n ≤ b_k$. Once again we have that $a_n ≤ b_k$ So then $a_n ≤ b_k$ for all $n, k \in \mathbb{N}$, and so $b_k$ is an upper bound to the set $A$ and so $\sup A = \xi ≤ b_k$ for all $k \in \mathbb{N}$. Furthermore we have that $a_k ≤ \xi$ for all $k \in \mathbb{N}$, and so $\xi \in I_k$ for every $k \in \mathbb{N}$ and thus $\xi \in \bigcap_{n=1}^{\infty} I_n$ and so the set theoretic union is nonempty. $\blacksquare$
Theorem 2: If the interval $I_n = [a_n, b_n]$ for $n \in \mathbb{N}$ is a sequence of closed bounded nested intervals then there exists a real number $\eta = \inf \{ b_n : n \in \mathbb{N} \}$ such that $\eta \in \bigcap_{n=1}^{\infty} I_n$. Proof: We note that by the definition of nested intervals that $I_n \subseteq I_1$ for all $n \in \mathbb{N}$ so then $a_1 \leq b_n$. Now consider the nonempty set $B = \{ b_n : n \in \mathbb{N} \}$ that is bounded below by $a_1$. Thus this set has an infimum in the real numbers and denote it $\inf B = \eta$ so that $\eta \leq b_n$ for all $n \in \mathbb{N}$. We now want to show that $a_n \leq \eta$ for all $n \in \mathbb{N}$. We will do this by showing that $a_k ≤ b_n$ for all $n, k \in \mathbb{N}$. First consider the case where $n ≤ k$. We thus have that $I_n \supseteq I_k$ by the definition of nested intervals and so by lemma 1 we get that $a_n ≤ a_k ≤ b_k ≤ b_n$. Here we see that $a_k ≤ b_n$. Now consider the case where $n > k$. We thus have that $I_k \supseteq I_n$ by the definition of nested intervals and so by lemma 1 once again we have that $a_k ≤ a_n ≤ b_n ≤ b_k$. Once again we have that $a_k ≤ b_n$ So then $a_k ≤ b_n$ for all $n, k \in \mathbb{N}$, and so $a_k$ is an upper bound to the set $A$ and so $a_k \leq \eta = \inf B$ for all $k \in \mathbb{N}$. Furthermore we have that $\eta \leq b_k$ for all $k \in \mathbb{N}$, and so $\eta \in I_k$ for every $k \in \mathbb{N}$ and thus $\eta \in \bigcap_{n=1}^{\infty} I_n$. $\blacksquare$
Theorem 3: Let $A := \{ a_n : n \in \mathbb{N} \}$ and $B := \{ b_n : n \in \mathbb{N} \}$. If $\sup A = \xi$ and $\inf B = \eta$ then if the interval $I_n = [a_n, b_n]$ is a sequence of closed bounded nested intervals then $[\xi, \eta] = \bigcap_{n=1}^{\infty} I_n$. Proof: We first prove that $[\xi, \eta] \subseteq \bigcap_{n=1}^{\infty} I_n$. Now let $x \in [\xi, \eta] = \{ x \in \mathbb{R} : \xi \leq x \leq \eta$. Therefore $\xi \leq x \leq \eta$. But we know that $a_n \leq \xi$ for all $n \in \mathbb{N}$ and we know that $\eta \leq b_n$ for all $n \in \mathbb{N}$ and so $a_n \leq \xi \leq x \leq \eta \leq b_n$ for all $n \in \mathbb{N}$. Therefore $x \in [a_n, b_n]$ for all $n \in \mathbb{N}$ or in other words $x \in \bigcap_{n=1}^{\infty} I_n$. Therefore $[\xi, \eta] \subseteq \bigcap_{n=1}^{\infty} I_n$. We will now prove that $\bigcap_{n=1}^{\infty} I_n \subseteq [\xi, \eta]$. Let $x \in \bigcap_{n=1}^{\infty} I_n$. Then $a_n \leq x \leq b_n$ for all $n \in \mathbb{N}$. We also know that $a_n \leq \xi \leq \eta \leq b_n$ for all $n \in \mathbb{N}$. Suppose that $x$ is such that $a_n \leq x \leq \xi$. Since $\xi$ is the supremum of the set $A$ then there exists an element $a_x \in \{ a_n : n \in \mathbb{N} \}$ such that $x < a_x$ and so $x \not \in [a_x, b_x]$ and therefore $x \not \in \bigcap_{n=1}^{\infty} I_n$, a contradiction. Now suppose that $x$ is such that $\eta \leq x \leq b_n$ for all $n \in \mathbb{N}$. Since $\eta$ is the infimum of the set $B$ then there exists an element $b_x \in \{ b_n : n \in \mathbb{N} \}$ such that $b_x < x$ and so $x \not \in [a_x, b_x]$ and so $x \not \in \bigcap_{n=1}^{\infty} I_n$, once again, a contradiction. We must therefore have that $a_n \leq \xi \leq x \leq \eta \leq b_n$ and so $x \in [\xi, \eta]$ and so $\bigcap_{n=1}^{\infty} I_n \subseteq [\xi, \eta]$. Since $[\xi, \eta] \subseteq \bigcap_{n=1}^{\infty} I_n$ and $\bigcap_{n=1}^{\infty} I_n \subseteq [\xi, \eta]$ we have that $[\xi, \eta] = \bigcap_{n=1}^{\infty} I_n$. $\blacksquare$
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Abstract:
Let $\Omega$ be an open and connected subset of the complex plane. A real valued function $u: \omega \rightarrow \mathbb{r}$ is said to be harmonic if it has continuous first and second partial derivatives and satisfies Laplace's equation $\Delta u = \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} = 0$. We begin by investigating basic properties of harmonic functions and defining the harmonic conjugate gradient and harmonic conjugate functions. From this we prove the mean value property and maximum and minimum modulus principle for harmonic functions.
Suppose we have a two dimensional plate of homogeneous material whose boundaries are heated to a constant temperature. We investigate solving the Dirichlet problem, which involves Laplace's equation and specific boundary values. Along with examples, we show that we can solve the Dirichlet problem on a disk using the Poisson integral along with Fourier series.
We generalize the Dirichlet problem to any region by investigating conformal mappings. We find a solution to the Dirichlet problem on a general region in the plane by transforming it to a problem in the upper half plane and using the Poisson integral.
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Powers of a Matrix
Definition: Given a square matrix $A$, for $n$ being a nonnegative integer, $A^n$ is defined as the product matrix taking $A$ and multiplying it by itself $n$-times. If $A$ is invertible, then $A^{-n} = (A^{-1})^{n}$, or the product matrix taking $A^{-1}$ and multiplying it by itself $n$-times.
For example, consider the following matrix:(1)
To compute $A^2$, take $A$ and multiply it by itself:(2)
Additionally, if we wanted to computer $A^{-2}$, we would first need to find $A^{-1} = \begin{bmatrix} -2 & 1\\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}$, and then:(3)
Theorem 1: If $A$ is a square matrix and let $r$ and $s$ be integers and let $k \in \mathbb{R}$ be a scalar. Then: a) $A^r A^s = A^{r+s}$. b) $(A^r)^s = A^{rs}$. c) If $A^n$ is invertible, then $A^{-n} = (A^{-1})^n$ for $n > 0$. d) If $k ≠ 0$ and the matrix $A$ is invertible, then $(kA)^{-1} = \frac{1}{k}A^{-1}$. Proof of (a):The matrix $A$ multiplied by itself $r$ times and then multiplied by itself $s$ times is the same as $A$ multiplied by itself $r + s$ times. $\blacksquare$ Proof of (b):First take matrix $A$ and multiply it by itself $r$ times. Then take this product and raise it to the $s^{\mathrm{th}}$ power. The total number of times $A$ is multiplied by itself will be $r \cdot s$. $\blacksquare$ Proof of (c):Assume that $(A^n)^{-1} = (A^{-1})^n$ is true and multiply both sides on the left by $(A^n)$: Proof of (d):Suppose that $(kA)^{-1} = \frac{1}{k}A^{-1}$. It follows that $(kA)(\frac{1}{k}A^{-1} ) = k \cdot \frac{1}{k} AA^{-1} = I$. Therefore the inverse of $kA$ must be equal to $\frac{1}{k} A^{-1}$. $\blacksquare$
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Riemann-Stieltjes Integrals with the Greatest Integer Function as an Integrator
Recall from the Riemann-Stieltjes Integrals with Multiple-Discontinuity Step Functions as Integrators page that if $f$ is a function on the interval $[a, b]$ and $\alpha$ is a step function on $[a, b]$ with jump discontinuities at $x_1, x_2, ..., x_n$ in $[a, b]$, then at each $x_k$ for $k = 1, 2, ..., n$ if $f$ and $\alpha$ are both not discontinuous from the left at $x_k$ and are not both discontinuous at the right at $x_k$ then $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ and:(1)
Where $\displaystyle{\alpha (x_k^+) = \lim_{x \to x_k^+} \alpha(x)}$ and $\displaystyle{\alpha (x_k^-) = \lim_{x \to x_k^-} \alpha (x)}$ for each $k \in \{1, 2, ..., n \}$ and if $x_1 = a$ we define $\alpha (x_1^-) = \alpha(a)$ and if $x_n = b$ we define $\alpha (x_n^+) = \alpha(b)$.
Also recall from The Greatest Integer Function page that the greatest integer function output of a real number $x \in \mathbb{R}$ is denoted $[x]$ and is the greatest integer less than or equal to $x$, i.e., $[x] \leq x < [x] + 1$.
We will now look at a rather nice theorem which tells us that every finite sum can be expressed as a Riemann-Stieltjes integral.
Theorem 1: Let $(a_k)_{k=1}^{n}$ be a finite sequence of numbers. Then the sum $\sum_{k=1}^{n} a_k$ can be expressed as a Riemann-Stieltjes integral by defining $f(x) = \left\{\begin{matrix} 0 & \mathrm{if} \: x = 0\\ a_1 & \mathrm{if} \: 0 < x \leq 1 \\ a_2 & \mathrm{if} \: 1 < x \leq 2 \\ \vdots \\ a_n & \mathrm{if} \: n-1 < x \leq n \end{matrix}\right.$. Then $\int_0^n f(x) \: d [x] = \sum_{k=1}^{n} a_k$. Proof:We note that $f$ is continuous on the left at the discontinuties (integers) throughout $[0, n]$ and $\alpha$ is continuous on the right at the discontinuties throughout $[0, n]$. Therefore $f$ and $\alpha$ are not both discontinuous on the left at these discontinuities and not both discontinuous on the right at these discontinuities. Therefore $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[0, n]$ and: Example 1 Consider the finite sequence $(a_k)_{k=1}^{5} = (1, 2, 3, 4, 5)$. Express the sum of this sequence as a Riemann-Stieltjes integral.
This sequence has $n = 5$ terms and by Theorem 1 we have that:(3)
Where the function $f$ is given by:(4)
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Table of Contents
The Uniform Continuity of Composite Functions on Metric Spaces
Recall from the Uniform Continuity of Functions on Metric Spaces page that if $(S, d_S)$ and $(T, d_T)$ are two metric spaces, $A \subseteq S$, and $f : A \to T$ then $f$ is said to be uniformly continuous on $A$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x, y \in A$ with $d_S(x, y) < \delta$ we have that:(1)
We will now show that the composition of uniformly continuous functions is also uniformly continuous.
Theorem 1: Let $(S, d_S)$, $(T, d_T)$, and $(U, d_U)$ be metric spaces. Let $f : S \to T$ and $g : T \to U$. Then if $f$ is uniformly continuous on $S$ and $g$ is uniformly continuous on $f(T)$ then $g \circ f : S \to U$ is uniformly continuous on $S$. Proof:Let $\epsilon > 0$ be given. We want to find a $\delta > 0$ such that if $x, y \in S$ and $d_S(x, y) < \delta$ then: Since $g$ is uniformly continuous on $f(S)$ we have that there exists a $\delta_1 > 0$ such that if $f(x), f(y) \in S$ and $d_T(f(x), f(y)) < \delta_1$ then: Since $f$ is uniformly continuous on $S$ we have that there exists a $\delta_2 > 0$ such that if $x, y \in S$ and $d_S(x, y) < \delta_2$ then: Let $\delta = \delta_2$. Then for all $x, y \in S$ we have that $(**)$ holds, and since $(**)$ holds we have that $(*)$ holds. Thus $g \circ f$ is uniformly continuous on $S$. $\blacksquare$
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I am trying to reduce the following ODE to Bessel's ODE form and hence solve it: $$x^{2}y''(x)+x(4x^{3}-3)y'(x)+(4x^{8}-5x^{2}+3)y(x)=0\tag{1} \, .$$
I tried to solve it via the standard method, i.e., by comparing it with a generalised ODE form and finding the solution from then on. The general form (as given in Mary L. Boas-
Mathematical Methods in Physical Sciences) is:
$$y''(x)+\frac{1-2a}{x}y'(x)+\left((bcx^{c-1})^{2}+\frac{a^{2}-p^{2}c^{2}}{x^{2}}\right)y(x)=0\tag{2} \, ,$$ and the solution:$$y(x)=x^{a}Z_{p}(bx^{c})\tag{3} \, .$$
But I am unable to get the answer via this method. The solution which is as follows:
$$y(x)=x^{2}e^{-\frac{x^{4}}{2}}[AI_{1}(\sqrt{5}x)+BK_{1}(\sqrt{5}x)]\tag{4}$$ Is obtained using comparison with another standard form which is given as follows:
$$x^{2}y''(x)+x(a+2bx^{p})y'(x)+[c+dx^{2q}+b(a+p-1)x^{p}+b^{2}x^{2p})y(x)=0\tag{5} \, ,$$ and the solution as: $$y(x)=x^{\alpha}e^{-\beta x^{p}}[AJ_{\nu}(\lambda x^{q})+BY_{\nu}(\lambda x^{q})]\tag{6} \, .$$
Where: $\alpha=\frac{1-a}{2}$, $\beta=\frac{b}{p}$, $\lambda=\frac{\sqrt{d}}{q}$, $\nu=\frac{\sqrt{(1-a)^{2}-4c)}}{2q}$
If I divide through the ode by $x^{2}$, I would get the Fuchasian form: $$y''(x)+f(x)y'(x)+g(x)y(x)=0$$
The terms of $xf(x)$ and $x^{2}g(x)$ are expandable in convergent power series $\sum_{n=0}^{\infty}a_{n}x^{n}$, hence there exists a nonessential singularity at the origin. But I am unable to solve via the Frobenius method.
Hence, my question- How is the generalised form of equation $(5)$ arrived at and why can't I use $(2)$ instead? Rather than bringing this ODE to a non-standard form as given in equation $(5)$, is there a way to derive the equation itself (and deduce the general solution)? Any help is appreciated.
Edit:I found the following form in a book:
$$x^{2}y''(x)+x(a+2bx^{p})y'(x)+[c+dx^{2q}+fx^{q}+b(a+p-1)x^{p}+b^{2}x^{2p})y(x)=0\tag{7} \, .$$
The only difference between the above and equation $(6)$ is the extra term:$fx^{q}$ Now if I substitute $y=we^{-\frac{bx^{p}}{p}}$ in equation $(7)$, it simplifies to the following linear equation:
$$x^{2}w''(x)+axw'(x)+(dx^{2q}+fx^{q}+c)w(x)=0\tag{8}\,.$$
Now using the transformation $z=x^{q}$, and $y=wz^{k}$, where $k$ is the root of the following quadratic equation: $q^{2}k^{2}+q(a-1)k+c=0$; leads to a further simplified and linear form:
$$q^{2}zy''(z)+[qbz+2kq^{2}+q(q-1+a)]y'(z)+(dz+kqb+f)y(z)=0\tag{9}\,.$$ This equation has the solution: $y(x)=e^{kx}w(z)$, where $w(z)$ is the solution to the hypergeometric equation as given below
Now, let a function $\Omega(b,a;x)$ be an arbitrary solution to the degenerate hypergeometric equation:
$$xy''(x)+(a-x)y'(x)-by(x)=0\tag{10}\,.$$
And $Z_{\nu}(x)$ be an arbitrary solution of the Bessel equation. Now in equation $(10)$, if $b\neq0,-1,-2,-3,...$, the solution is given by the Kummer's series as: $$\Phi(b,a;x)=1+\sum_{k=1}^{\infty}\frac{(b)_{k}x^{k}}{(a)_{k}k!}$$ Where:$(b)_{k}=b(b+1)...(b+k-1)$ When $a$ is not an integer, the solution can be written as: $$y=C_{1}\Phi(b,a;x)+C_{2}x^{1-a}\Phi(b-a+1,2-a;x)$$ Make the following replacements: $b=2n$ and $a=n$ Now the series becomes:
$$\Phi(n,2n;x)=\Gamma\left(n+\frac{1}{2}\right)e^{\frac{x}{2}}\left(\frac{x}{4}\right)^{(-n+\frac{1}{2})}I_{n-\frac{1}{2}}(\frac{x}{2})$$ And $$\Phi(-n,-2n;x)=\frac{1}{\sqrt{\pi}}e^{\frac{x}{2}}\left(x\right)^{(-n+\frac{1}{2})}K_{n-\frac{1}{2}}(x)$$
Substituting the above in the solution of equation $(9)$, the general solution becomes:
$$y=e^{x(k+\frac{1}{2})}\left[C_{1}\Gamma\left(n+\frac{1}{2}\right)\left(\frac{x}{4}\right)^{(-n+\frac{1}{2})}I_{n+\frac{1}{2}}(x)+C_{2}\frac{1}{\sqrt{\pi}}\left(x\right)^{(-n+\frac{1}{2})}K_{n+\frac{1}{2}}(x)\right]$$ Which simplifies to:
$$y(x)=\left(x\right)^{(-n+\frac{1}{2})}e^{x(k+\frac{1}{2})}\left[C_{1}\Gamma\left(n+\frac{1}{2}\right)\left(\frac{1}{4}\right)^{(-n+\frac{1}{2})}I_{n+\frac{1}{2}}(x)+C_{2}\frac{1}{\sqrt{\pi}}K_{n+\frac{1}{2}}(x)\right]$$
Which is the final solution. I tried to do the same for equation $(6)$, but did not get the solution. Is it plausible to obtain the solution via the Kummer's series method? Any help is appreciated.
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Do you think, when you train a
Convolutional Neural Network (CNN) to classify between images it is exactly understanding the image as we humans perceive? It’s difficult to answer, as for most of the times Deep learning models are often considered to be a black box. We feed in the data and then we get the output. Whatever happens in between this flow is very difficult to debug. Though we get the accurate predictions, it may not be true that they are intelligent enough to perceive the same way as we do. Why does it happen?
Imagine a situation where you want to classify between
Elephants and Penguins (I know it’s fairly an easy task). Now you got the data and trained a CNN to classify the images and deployed it. The model might perform well with most of the data, but there are chances for misclassification to happen. One may ignore thinking it as a corner case. But what do you think when the object is clearly distinguishable?
Considering the above situation, it’s very obvious to find
plants/trees in the images of elephant and snow/water in the images of penguin. So, it is evident that the model has learned to distinguish between the colors/shades of green and white rather than actually learning to classify the objects.
In the above case, simple image processing techniques like
color channel statistics do on par with the trained model. Because there is no intelligence, the model is simply distinguishing between the colors. Now you may ask, how to know where exactly the CNN is looking? Well, the answer is Grad-CAM. Grad-CAM
Gradient weighted Class Activation Map is the technique that we implement in this blog post. At first, we need to know that this is not the only technique out there. The authors say,
Gradient-weighted Class Activation Mapping (Grad-CAM), uses the gradients of any target concept (say logits for ‘dog’ or even a caption), flowing into the final convolutional layer to produce a coarse localization map highlighting the important regions in the image for predicting the concept.
So, to explain in simple terms, we simply take the
final convolutional feature map and then we weigh every channel in that feature with the gradient of the class with respect to the channel. It’s just nothing but how intensely the input image activates different channels by how important each channel is with regard to the class. The best part is it doesn’t require any re-training or change in the existing architecture.
$$S^c = \frac{1}{Z}\sum_i\sum_j \sum_k \frac{\partial y^c}{\partial A_{ij}^k} A_{ij}^k$$
The spatial score of a specific class $S^c$ is nothing but the global average pooling over two dimensions $i$ and $j$ for the gradient of respective class output $y^c$ with respect to the feature map $A_{ij}^k$. Then, we multiply the resulting value with the feature map along with its channel axis $k$. We then average/pool the resultant over the channel dimension $k$. Thus, resulting in the spatial score map of size $i \times j$. The $\sum$ is used to describe the pooling and average operation.
The
ReLU activation is applied to the score map and then it is normalized to output positive region predictions. Implementation
For the purpose of this blog post, let’s take a
pre-trained VGG model and start implementing by importing necessary packages. from keras.applications.vgg16 import VGG16, preprocess_input, decode_predictions from keras.preprocessing import image import keras.backend as K import numpy as np import cv2 import sys
We use the
VGG16 model that’s shipped with Keras. And load certain helper functions required for loading and preprocessing our image. model = VGG16(weights="imagenet") img_path = sys.argv[1] img = image.load_img(img_path, target_size=(224, 224)) x = image.img_to_array(img) x = np.expand_dims(x, axis=0) x = preprocess_input(x)
Let’s initialize our model and load the image from the command line argument. The VGG network expects input size to be $(224 \times 224 \times 3)$ we resize our image the required size. Since we are passing only one image through the network, it’s required to expand the first dimension noting it as a batch of size $1$. We then
normalize our image by subtracting mean RGB values from the input image using a helper function
preprocess_input
preds = model.predict(x) class_idx = np.argmax(preds[0]) class_output = model.output[:, class_idx] last_conv_layer = model.get_layer("block5_conv3")
Here, in this case, let’s see the map for the
top prediction. So we get the predictions for the image and then we take the topmost class index. Remember that we can compute map for any class. Then, we take the output from the final convolutional layer in the VGG16 which is
block5_conv3. The resulting feature map will be of shape $14 \times 14 \times 512$.
grads = K.gradients(class_output, last_conv_layer.output)[0] pooled_grads = K.mean(grads, axis=(0, 1, 2)) iterate = K.function([model.input], [pooled_grads, last_conv_layer.output[0]]) pooled_grads_value, conv_layer_output_value = iterate([x]) for i in range(512): conv_layer_output_value[:, :, i] *= pooled_grads_value[i]
As explained above, we compute the gradient of the class output value with respect to the feature map. Then, we pool the
gradients over all the axes leaving out the channel dimension. Finally, we weigh the output feature map with the computed gradient values. heatmap = np.mean(conv_layer_output_value, axis=-1) heatmap = np.maximum(heatmap, 0) heatmap /= np.max(heatmap)
We then
average the weighed feature map along the channel dimension resulting in a heat map of size
14 \times 14. And, then we normalize the heat map to make the values in between 0 and 1.
img = cv2.imread(img_path) heatmap = cv2.resize(heatmap, (img.shape[1], img.shape[0])) heatmap = np.uint8(255 * heatmap) heatmap = cv2.applyColorMap(heatmap, cv2.COLORMAP_JET) superimposed_img = cv2.addWeighted(img, 0.6, heatmap, 0.4, 0) cv2.imshow("Original", img) cv2.imshow("GradCam", superimposed_img) cv2.waitKey(0)
Finally, we use OpenCV to read the image, resize the existing heatmap to the image size. We blend the original image and the heatmap to
superimpose the heatmap on to the image. An example is shown below.
From the above image, it is clearly visible where CNN is looking in the image to actually distinguish between the classes.
This technique is not only useful for localization but it also used for Visual Question and Answering, Image captioning etc., as mentioned in their paper itself.
Also, it is very much helpful in debugging about the data requirements for building an accurate model. Though hyperparameter tuning is not much associated with this, we can generalize the model better with extra data and data augmentation techniques.
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Astronomy Kepler's Laws of Planetary Motion The orbit of every planet is an ellipse with the sun at a focus A line joining a planet and the sun sweeps out equal areas during equal intervals of time The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. Hubble's Law The redshift in the light from a distant galaxy is proportional to its distance.
The light coming to us from distance galaxies has its wavelength increased due to the Doppler Effect. This phenomenon is called redshift. Virtually all the galaxies (except a few very close to us) show redshift. You can calculate the speed of a galaxy from the redshift
z in this equation:
Hubble analyzed this observation and concluded that those galaxies are moving away from us and also moving away from each other, i.e., the universe is expanding. He derived the following relationship between the velocity of the galaxy and its distance from us:
This is called Hubble's Constant
Problem 1
According to Hubble's Law, what is the relationship between the distance
D and speed v of a galaxy? Solution 1
According to Hubble's Law,
z is directly proportional to D, and v is directly proportional to z. Hence v is directly poroportional to D Problem 2 Solution 2Answer Problem 3
At the present time, the temperature of the universe (i.e. the microwave radiation background) is about 3K. When the temperature was 12K, typical objects in the universe, such as galaxies, were:
1/2 as distanct as they are today 4 times as distant as they are today 2 times as distant as they are today 1/4 as distanct as they are today separated by about the same distances as they are today Solution 3D Problem 4
If the Sun were suddenly replaced by a black hole of the same mass, it would have a Schwarzschild radius of 3,000 m. What effect, if any, would this change have on the orbits of the planets?
The orbits would remain unchanged. The planets would oscillate about their former elliptical orbits. The planets would move in spiral orbits. The orbits would precess much more rapidly. The planets would move directly toward the Sun Solution 4
Answer
Problem 5
A satellite of mass m orbits a planet of mass M in a circular orbit of radius R. What is the time required for one revolution?
Solution 5It's an intuitive question. The answer is an ellipse Problem 6
The primary source of the Sun's energy is a series of thermonuclear reactions in which the energy produced is c
2 times the mass difference between 2 hydrogen atoms plus 2 helium atoms and 1 carbon atom 3 helium atoms and 1 carbon atom 4 hydrogen atoms and 1 helium atom 2 hydrogen atoms and 1 helium atom 6 hydrogen atoms and 2 helium atoms Solution 6C Problem 7
A satellite orbits the Earth in a circular orbit. An astronaut on board perturbs the orbit slightly by briefly firing a control jet aimed toward the Earth's center. Afterward, what shape is the satellite's path? Is it an ellipse, a hyperbola, a bigger circle, a spiral, or does it show many radial oscillations per revolution?
Solution 7It's an intuitive question. The answer is an ellipse Problem 8
The magnitude of the Earth's gravitational force on a point mass is F(r), where r is the distance from the Earth's center to the point mass. Assume the Earth is a homogenous sphere of radius R.
Suppose there is a very small shaft in the Earth such that the point mass can be placed at a radius of R/2. What is
?
Solution 8It's an intuitive question. The answer is an ellipse Problem 9
Which of the following is most nearly the mass of the Earth? (the radius of the Earth is about 6.4E6 m.)
Solution 9
Answer
Problem 10
Suppose that the graivtational force law between 2 massive objects were
where ε is a small positive number. Which of the following statements would be FALSE?
A single planet could move in a stational circular orbit about the Sun The angular momentum of a single planet moving about the Sun would be conserved The periods of planets in circular orbits would be propotional to the (3 + ε)/2 power of their respective orbital radii A single planet could move in a stationary noncircular elliptical orbit around the Sun The total mechanical energy of the planet-Sun system would be conserved Solution 10
Stable non-circular orbits can only occur for the simple harmonic potential and the inverse-square law force. The answer is D
Problem 11
A black hole is an object whose gravitational field is so strong that even light cannot escape. To what approximate radius would Earth (mass 5.98E24 kg) have to be compressed in order to become a black hole?
Solution 11
The Schwarzschild radius (aka gravitational radius) is the radius of a sphere such that, if all the mass of an object is compressed within that sphere, the escape speed from the surface of the sphere would equal the speed of light. A black hole is an object that is smaller than its Schwarzchild radius. The radius is:
Kepler's Laws The orbit of every planet is an ellipse with the sun at a focus A line adjoining a planet and the sun sweeps out equal areas during equal intervals of time The square of the orbital period of a planet ( T 2) is directly proportional to the cube of the semi-major axis of its orbit ( a 3) Circular Orbits Gravitational pull is the centripetal force on the satellite F = G \frac{mM}{r^2} = \frac{mv^2}{r}because centripetal acceleration isa = \frac{v^2}{r} The speed of a satellite in circle orbit about a body of mass Mv = \sqrt{\frac{GM}{r}} The time period ( T) of the satellite is proportional to r 3/2(Kepler's Third Law)T^2 = \frac{4 \pi^2}{GM} r^3 Problem 12An astronomy observes a very small moon orbiting a planet and measure the moon's minimum and maximum distances from the planet's center and the moon's maximum orbital speed. Which of the following CANNOT be calculated from these measurements? Mass of the moon Mass of the planet Minimum speed of the moon Period of the orbit Semimajor axis of orbit Solution 12The mass of the moon Problem 13The period of a hypothetical earth satellite orbiting at sea level would be 80 minutes. In terms of the earth's radius R, what is the radius of a synchronus satellit orbit (period 24 hours)? e 3 R e 7 R e 18 R e 320 R e 5800 R e
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Alright, I have this group $\langle x_i, i\in\mathbb{Z}\mid x_i^2=x_{i-1}x_{i+1}\rangle$ and I'm trying to determine whether $x_ix_j=x_jx_i$ or not. I'm unsure there is enough information to decide this, to be honest.
Nah, I have a pretty garbage question. Let me spell it out.
I have a fiber bundle $p : E \to M$ where $\dim M = m$ and $\dim E = m+k$. Usually a normal person defines $J^r E$ as follows: for any point $x \in M$ look at local sections of $p$ over $x$.
For two local sections $s_1, s_2$ defined on some nbhd of $x$ with $s_1(x) = s_2(x) = y$, say $J^r_p s_1 = J^r_p s_2$ if with respect to some choice of coordinates $(x_1, \cdots, x_m)$ near $x$ and $(x_1, \cdots, x_{m+k})$ near $y$ such that $p$ is projection to first $m$ variables in these coordinates, $D^I s_1(0) = D^I s_2(0)$ for all $|I| \leq r$.
This is a coordinate-independent (chain rule) equivalence relation on local sections of $p$ defined near $x$. So let the set of equivalence classes be $J^r_x E$ which inherits a natural topology after identifying it with $J^r_0(\Bbb R^m, \Bbb R^k)$ which is space of $r$-order Taylor expansions at $0$ of functions $\Bbb R^m \to \Bbb R^k$ preserving origin.
Then declare $J^r p : J^r E \to M$ is the bundle whose fiber over $x$ is $J^r_x E$, and you can set up the transition functions etc no problem so all topology is set. This becomes an affine bundle.
Define the $r$-jet sheaf $\mathscr{J}^r_E$ to be the sheaf which assigns to every open set $U \subset M$ an $(r+1)$-tuple $(s = s_0, s_1, s_2, \cdots, s_r)$ where $s$ is a section of $p : E \to M$ over $U$, $s_1$ is a section of $dp : TE \to TU$ over $U$, $\cdots$, $s_r$ is a section of $d^r p : T^r E \to T^r U$ where $T^k X$ is the iterated $k$-fold tangent bundle of $X$, and the tuple satisfies the following commutation relation for all $0 \leq k < r$
$$\require{AMScd}\begin{CD} T^{k+1} E @>>> T^k E\\ @AAA @AAA \\ T^{k+1} U @>>> T^k U \end{CD}$$
@user193319 It converges uniformly on $[0,r]$ for any $r\in(0,1)$, but not on $[0,1)$, cause deleting a measure zero set won't prevent you from getting arbitrarily close to $1$ (for a non-degenerate interval has positive measure).
The top and bottom maps are tangent bundle projections, and the left and right maps are $s_{k+1}$ and $s_k$.
@RyanUnger Well I am going to dispense with the bundle altogether and work with the sheaf, is the idea.
The presheaf is $U \mapsto \mathscr{J}^r_E(U)$ where $\mathscr{J}^r_E(U) \subset \prod_{k = 0}^r \Gamma_{T^k E}(T^k U)$ consists of all the $(r+1)$-tuples of the sort I described
It's easy to check that this is a sheaf, because basically sections of a bundle form a sheaf, and when you glue two of those $(r+1)$-tuples of the sort I describe, you still get an $(r+1)$-tuple that preserves the commutation relation
The stalk of $\mathscr{J}^r_E$ over a point $x \in M$ is clearly the same as $J^r_x E$, consisting of all possible $r$-order Taylor series expansions of sections of $E$ defined near $x$ possible.
Let $M \subset \mathbb{R}^d$ be a compact smooth $k$-dimensional manifold embedded in $\mathbb{R}^d$. Let $\mathcal{N}(\varepsilon)$ denote the minimal cardinal of an $\varepsilon$-cover $P$ of $M$; that is for every point $x \in M$ there exists a $p \in P$ such that $\| x - p\|_{2}<\varepsilon$....
The same result should be true for abstract Riemannian manifolds. Do you know how to prove it in that case?
I think there you really do need some kind of PDEs to construct good charts.
I might be way overcomplicating this.
If we define $\tilde{\mathcal H}^k_\delta$ to be the $\delta$-Hausdorff "measure" but instead of $diam(U_i)\le\delta$ we set $diam(U_i)=\delta$, does this converge to the usual Hausdorff measure as $\delta\searrow 0$?
I think so by the squeeze theorem or something.
this is a larger "measure" than $\mathcal H^k_\delta$ and that increases to $\mathcal H^k$
but then we can replace all of those $U_i$'s with balls, incurring some fixed error
In fractal geometry, the Minkowski–Bouligand dimension, also known as Minkowski dimension or box-counting dimension, is a way of determining the fractal dimension of a set S in a Euclidean space Rn, or more generally in a metric space (X, d). It is named after the German mathematician Hermann Minkowski and the French mathematician Georges Bouligand.To calculate this dimension for a fractal S, imagine this fractal lying on an evenly spaced grid, and count how many boxes are required to cover the set. The box-counting dimension is calculated by seeing how this number changes as we make the grid...
@BalarkaSen what is this
ok but this does confirm that what I'm trying to do is wrong haha
In mathematics, Hausdorff dimension (a.k.a. fractal dimension) is a measure of roughness and/or chaos that was first introduced in 1918 by mathematician Felix Hausdorff. Applying the mathematical formula, the Hausdorff dimension of a single point is zero, of a line segment is 1, of a square is 2, and of a cube is 3. That is, for sets of points that define a smooth shape or a shape that has a small number of corners—the shapes of traditional geometry and science—the Hausdorff dimension is an integer agreeing with the usual sense of dimension, also known as the topological dimension. However, formulas...
Let $a,b \in \Bbb{R}$ be fixed, and let $n \in \Bbb{Z}$. If $[\cdot]$ denotes the greatest integer function, is it possible to bound $|[abn] - [a[bn]|$ by a constant that is independent of $n$? Are there any nice inequalities with the greatest integer function?
I am trying to show that $n \mapsto [abn]$ and $n \mapsto [a[bn]]$ are equivalent quasi-isometries of $\Bbb{Z}$...that's the motivation.
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UOV
The first thing you need to know about UOV is that the input variables are partitioned into vinegar variables $x_1, \ldots, x_v$ and oil variables $x_{v+1}, \ldots, x_{v+o}$. The number of vinegar variables is roughly twice the number of oil variables: $v = 2o$. The name oil and vinegar comes from the fact that oil variables do not mix with other oil variables in the secret polynomials. In fact, every polynomial of the secret key can be described as a quadratic form. For example, let $v=4$ and $o=2$ and $\mathbf{x}^\mathsf{T} = (x_1 \, x_2 \, x_3 \, x_4 \, x_5 \, x_6)$. Then every secret polynomial can be described as:$$ f_i(\mathbf{x}) = \mathbf{x}^\mathsf{T} \left( \begin{matrix}* & * & * & * & * & * \\* & * & * & * & * & * \\* & * & * & * & * & * \\* & * & * & * & * & * \\* & * & * & * & & \\* & * & * & * & & \\\end{matrix} \right) \mathbf{x} \enspace .$$Where the $*$ indicate randomly chosen elements, but the blank spaces indicate zeros. The oil variables, $x_5$ and $x_6$ are not mixed with themselves or each other.
Now, it's also important to know that there are exactly $o$ secret polynomials. The vectorial function whose components are $f_i$ is denoted by $\mathcal{F}$:$$ \mathcal{F}(\mathbf{x}) = \left( \begin{matrix}f_1(\mathbf{x}) \\\vdots \\f_o(\mathbf{x}) \end{matrix} \right) \enspace .$$(In our example there are just two components in $\mathcal{F}$.)
The public key $\mathcal{P}$ is generated from $\mathcal{F}$ by composing it with a random invertible linear transformation $S \in \mathsf{GL}_{o+v}(\mathbb{F}_q)$, which is really a square matrix whose inverse exists: $S \in \mathbb{F}_q^{(v+o)\times(v+o)}$. The public key is then$$ \mathcal{P} = \mathcal{F} \circ S \enspace .$$
In order to sign a document $d$ with hash $\mathbf{h} = \mathsf{H}(d)$, the signer does the following. First, choose a random assignment to the vinegar variables $x_1, \ldots, x_v$. Then the system of polynomial equations $\mathcal{F}(x_{v+1},\ldots,x_{v+o}) = \mathbf{h}$ is a linear $o \times o$ system. Solve it to obtain $x_{v+1}, \ldots, x_{v+o}$ and at this point the signer has the entire vector $\mathbf{x}$. Next, compute the signature $\mathbf{y}$ as $\mathbf{y} = S^{-1}\mathbf{x}$.
In order to verify a signature $\mathbf{y}$, the verifier evaluates the public key in $\mathbf{y}$, and checks if the result is equal to the hash of the document,
i.e.,$$ \mathcal{P}(\mathbf{y}) \stackrel{?}{=} \mathsf{H}(d) \enspace .$$
Notice how UOV uses a linear $o \times o$ system as a step in generating signatures, but how this step generalizes to any quadratic (!) system for which it is possible to find an inverse efficiently. For instance, you can level down recursively and put another UOV system there. This multi-layer approach is called Rainbow.
HFE, HFE-, and HFEv-
HFE-type systems are very different. The first thing you should know about them is that they use univariate polynomials over the extension field. So $\mathbb{F}_q$ is the base field (like in the UOV case). Then pick any irreducible polynomial $f(z) \in \mathbb{F}_q[z]$ of degree $n$, and then you can identify the set of polynomials $\mathbb{F}_q[z] / \langle f(z) \rangle$ of degree smaller than $n$ as a new field, the
extension field $\mathbb{F}_{q^n}$, whose operations are polynomial addition and polynomial multiplication modulo $f(z)$. Whenever you have a vector of $n$ values over the base field, say $\mathbf{a} \in \mathbb{F}_q^n$, you can embed this value in the extension field using the canonical embedding $\varphi : \mathbb{F}_q^n \rightarrow \mathbb{F}_{q^n} : \mathbf{a} \mapsto A = \sum_{i=1}^na_iz^{i-1}$. In particular, you can do this with the vector of variables: $\varphi(\mathbf{x}) = X$ and then you have a single variable $X$ over the extension field. Any list of $n$ multivariate polynomials in $\mathbf{x}$ over the base field maps to one univariate polynomial in $X$ over the extension field, and vice versa.
In order to generate a HFE secret key, choose a random univariate polynomial over the extension field $\mathcal{F}(X) \in \mathbb{F}_{q^n}[X]$ with degree at most $D$ and of the form$$ \mathcal{F}(X) = \sum_{i} \sum_{j} \alpha_{i,j} X^{q^i + q^j} \enspace .$$In other words, choose the coefficients $\alpha_{i,j}$ uniformly at random from $\mathbb{F}_{q^n}$ as long as $q^i+q^j \leq D$, and set $\alpha_{i,j}=0$ otherwise. Following this structure for $\mathcal{F}$ will guarantee that it maps to a list of multivariate quadratic equations over the base field. In the old days they recommended $D \approx 1000$ but now we understand how to make HFE secure with much smaller $D$,
e.g. $D \approx 10$ -- and that will only make the scheme faster! Next, choose two random invertible linear transformations $T, S \in \mathsf{GL}_n(\mathbb{F}_q)$ which are once again really invertible square matrices in $\mathbb{F}_q^{n \times n}$. The secret key consists of $(\varphi, \mathcal{F}, T, S)$. The public key is found by composing these transforms:$$ \mathcal{P} = T \circ \varphi^{-1} \circ \mathcal{F} \circ \varphi \circ S \enspace .$$
In order to sign a document $d$ with hash $\mathbf{h} = \mathsf{H}(d)$, the signer proceeds as follows. First, he computes $Y = \varphi(T^{-1}\mathbf{h})$. Then he uses Berlekamp's algorithm to factorize the polynomial $\mathcal{F}(X)-Y$ and find the roots $\{X_i\}$ that make it zero. He chooses one of these roots as $X$ and proceeds to compute the signature $\mathbf{z} = S^{-1}\varphi^{-1}(X)$.
In order to verify a signature, the verifier evaluates the public key $\mathcal{P}$ in $\mathbf{z}$ and checks whether he obtains the hash $\mathbf{h}$ of the document:$$ \mathcal{P}(\mathbf{z}) \stackrel{?}{=} \mathsf{H}(d) \enspace .$$
Unfortunately, HFE as described up until here turns out to be insecure. In particular, it is vulnerable to a direct algebraic attack involving efficient Gröbner basis algorithms. That's why you need at least one of the modifiers - (minus) or v (vinegar). It makes sense to use both. A good right-hand rule is $a + v + \mathsf{log}_2 D \geq 15$.
Minus means that some polynomials are dropped from the public key. In particular, the decomposition now involves a forgetful operator $\pi : \mathbb{F}_q^n \rightarrow \mathbb{F}_q^{n-a}$ that leaves the first $n-a$ components intact but drops the remaining $a$ ones:$$ \mathcal{P} = \pi \circ T \circ \varphi^{-1} \circ \mathcal{F} \circ \varphi \circ S \enspace .$$In order to generate a signature, the signer has to choose the value of the missing polynomials. This is no problem as a random choice will suffice.
Vinegar consists of adding $v$ vinegar variables. During signature generation, the value of these vinegar variables is also chosen at random, after which what's left to do is to invert the regular HFE system. Instead of finding a solution to the linear $o \times o$ system (as was the case for UOV), now one must find the preimage by factorizing the HFE polynomial.
Reading Materials
In terms of accessible reading materials, I can recommend two things. One, Ding and Yang have a chapter in the 2009 book "Post-Quantum Cryptography". Two, Ding, Gower and Schmidt have a book called "Multivariate Public Key Cryptosystems". Aside from that, the relevant papers good reads because they contain all the necessary information, for example to estimate security. On the other hand, they are a lot denser.
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Obviously (;-) @egreg’s answer is great and goes right to the point, but it does have an infinitesimal drawback: a pervert who tried
$\overline{\sin x}$ would get a surprise. Moreover, one could argue that the placement of the radical sign in
$\sqrt{\sin x}$ is suboptimal.
The positioning of superscripts in math formulas is a very low-level feature of TeX, which is described, along with many other “intimate” details of math typesetting, in Appendix G of
The TeXbook, to which you are referred for an explanation of the following solution.
\documentclass[a4paper]{article}
\usepackage[T1]{fontenc}
\usepackage{amsmath}
% egreg's recipe:
\makeatletter
\DeclareRobustCommand{\sinname}{%
\smash{\operator@font sin}\vphantom{s}%
}
% ... but let us use a different name for the operator:
\newcommand{\ssin}{\qopname\relax o{\sinname}}
\makeatother
% Our recipe:
\setbox0 = \hbox{$$} % load math fonts
\fontdimen18\scriptfont2 = 3.78970pt % turns out to be just enough
\begin{document}
The downside in egreg's answer:
\[\sqrt{\sin x}+\sqrt{\ssin x} + \overline{\sin x}+\overline{\ssin x}\]
That's not fine.
But, with our correction, the exponent of the ``usual'' \verb|\sin| will not be
rised that much:
\[ \frac{1}{\cos^{2}x} + \frac{1}{\sin^{2}x} + \frac{1}{x^{2}} \]
For an explanation, see \textsl{The \TeX book}, Appendix~G, Rule~18a; the
relevant passage is ``set \( u\gets h-q \)\,\ldots\ where $q$\,\ldots\ [is] the
[value] of~$\sigma_{18}$\,\ldots\ in the font corresponding to
[style]~$C{\uparrow}$''. What we are doing is to increase~$q$. The amended
value of~$u$ will be subsequently used in Rule~18c to position the superscript.
\end{document}
Here’s the output:
Let us also magnify the crucial portion:
Of course, this solution could have drawbacks too: changing in this way a parameter that operate at so low a level of TeX might cause side-effects which are not evident at first sight.
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Chokshi, Paresh and Kumaran, V (2007)
Stability of the viscous flow of a polymeric fluid past a flexible surface. In: Physics of Fluids, 19 . 034102-1-034102-15.
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Stability_of_the_viscous_flow_of.pdf
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Abstract
The instability in plane Couette flow of a viscoelastic fluid past a deformable surface is examined using the temporal linear stability theory in the zero Reynolds number limit. The polymeric fluid is described using the Oldroyd-B model and the flexible wall is modeled as a linear viscoelastic solid surface. The analysis shows that the wall flexibility tends to reduce the decay rate of the stable discrete modes for the polymeric flow past a rigid wall, and one of the discrete modes becomes unstable when the wall deformability parameter $\Gamma=V \eta /(GR)$ exceeds a certain critical value $\Gamma_c$. Here, V is the top-plate velocity, \eta is the zero shear viscosity of the polymeric fluid, G is the shear modulus of the wall, and R is the width of the fluid layer. The analysis reveals the presence of two classes of modes, the first of which becomes unstable for perturbations with wavelength comparable to the channel width (finite wavelength modes), and the second becomes unstable for perturbations with wavelength small compared to the channel width (short wave modes). The latter class of modes are found to be absent for the highly concentrated polymer solutions with \beta\leq 0.23, where \beta is the ratio of solvent-to-solution viscosity. We have mapped out the regions in the parameter space \bar{W}-H) where the finite wavelength and short wave modes are unstable, where $\bar{W}=(\lambda G/ \eta)$, and lambda is the relaxation time of the viscoelastic fluid. Fluid elasticity is found to have a stabilizing influence on the unstable mode, such that when the shortwave instability is absent for \beta \leq 0.23, the flow becomes stable for any Weissenberg number $\bar{W}>\bar{W}_ {max}$. Here, $\bar{W}_{max}$ increases proportional to H for H\gg1. However, when the shortwave instability is present, the instability persists for \bar{W}\gg 1. The behavior of both classes of modes with respect to the parameters, like \bar{W}, H, \beta, and the ratio of solid-to-fluid viscosity $\eta_o$ , is examined.
Item Type: Journal Article Additional Information: Copyright of this article belongs to American Institute of Physics. Department/Centre: Division of Mechanical Sciences > Chemical Engineering Depositing User: Satish MV Date Deposited: 11 Oct 2007 Last Modified: 19 Sep 2010 04:38 URI: http://eprints.iisc.ac.in/id/eprint/11073 Actions (login required)
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The Measure of Countable Subsets of Real Numbers
Recall from the Subsets of Real Numbers with Measure Zero that a subset $S \subset \mathbb{R}$ is said to have measure $0$ denoted $m(S) = 0$ if there exists a countable open interval covering $\{ I_k = (a_k, b_k) \}_{k \in K}$ (where $K$ is some countable indexing set) that covers $S$, i.e., $S \subseteq \bigcup_{k \in K} I_k$ and for $l(I_k) = b_k - a_k$ we have that:(1)
We noted that any singleton and any finite subset $S$ of $\mathbb{R}$ has measure $0$. We will now see further than any countable set also has measure $0$.
Theorem 1: Let $S \subset \mathbb{R}$ be countable. Then $m(S) = 0$. Proof:Let $\epsilon > 0$ be given. If $S$ is finite then we're done since we've already seen that a finite subset of $\mathbb{R}$ has measure $0$ so assume that $S$ is countably infinite, say: For each $x_k \in S$ let: Then $\{ I_k \}_{k=1}^{\infty}$ is countable and clearly covers $S$. We also see that: Taking the sum of $l(I_k)$ from $k = 1$ to $\infty$ gives us: (We use the fact that the series $\sum_{k=1}^{\infty} \frac{1}{2^k}$ converges to $1$). Therefore $m(S) = 0$. $\blacksquare$
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Preface: International Conference On Matrix Analysis And Its Applications -- Mattriad 2017, 2018 University of Tampere
Preface: International Conference On Matrix Analysis And Its Applications -- Mattriad 2017, Oskar Maria Baksalary, Natalia Bebiano, Heike Fassbender, Simo Puntanen Electronic Journal of Linear Algebra
No abstract provided.
Norm Inequalities Related To Clarkson Inequalities, 2018 University of Jordan
Norm Inequalities Related To Clarkson Inequalities, Fadi Alrimawi, Omar Hirzallah, Fuad Kittaneh Electronic Journal of Linear Algebra
Let $A$ and $B$ be $n\times n$ matrices. It is shown that if $p=2$, $4\leq p<\infty$, or $2
Bounds For The Completely Positive Rank Of A Symmetric Matrix Over A Tropical Semiring, 2018 University of Ljubljana
Bounds For The Completely Positive Rank Of A Symmetric Matrix Over A Tropical Semiring, David Dolžan, Polona Oblak Electronic Journal of Linear Algebra
In this paper, an upper bound for the CP-rank of a matrix over a tropical semiring is obtained, according to the vertex clique cover of the graph prescribed by the positions of zero entries in the matrix. The graphs that beget the matrices with the lowest possible CP-ranks are studied, and it is proved that any such graph must have its diameter equal to $2$.
Resolutions Of Finite Length Modules Over Complete Intersections, 2018 University of Nebraska-Lincoln
Resolutions Of Finite Length Modules Over Complete Intersections, Seth Lindokken Dissertations, Theses, and Student Research Papers in Mathematics
The structure of free resolutions of finite length modules over regular local rings has long been a topic of interest in commutative algebra. Conjectures by Buchsbaum-Eisenbud-Horrocks and Avramov-Buchweitz predict that in this setting the minimal free resolution of the residue field should give, in some sense, the smallest possible free resolution of a finite length module. Results of Tate and Shamash describing the minimal free resolution of the residue field over a local hypersurface ring, together with the theory of matrix factorizations developed by Eisenbud and Eisenbud-Peeva, suggest analogous lower bounds for the size of free resolutions of finite length ...
Supporting English Language Learners Inside The Mathematics Classroom: One Teacher’S Unique Perspective Working With Students During Their First Years In America, 2018 University of Nebraska - Lincoln
Supporting English Language Learners Inside The Mathematics Classroom: One Teacher’S Unique Perspective Working With Students During Their First Years In America, Amy Marie Fendrick Research and Evaluation in Literacy and Technology
Reflecting upon my personal experiences teaching mathematics to English Language Learners (ELL) in a public high school in Lincoln, Nebraska, this essay largely focuses on the time I spent as the only Accelerated Math teacher in my school building. From 2012 – 2017, I taught three different subjects at this high school: Advanced Algebra, Algebra, and Accelerated Math. This essay highlights why I chose to become a math and ELL teacher, as well as the challenges, issues, struggles, and successes I experienced during my time teaching. I focus on the challenges I faced teaching students who did not share my native ...
Tp Matrices And Tp Completability, 2018 William & Mary
Tp Matrices And Tp Completability, Duo Wang Undergraduate Honors Theses
A matrix is called totally nonnegative (TN) if the determinant of
every square submatrix is nonnegative and totally positive (TP)
if the determinant of every square submatrix is positive. The TP
(TN) completion problem asks which partial matrices have a TP
(TN) completion. In this paper, several new TP-completable pat-
terns in 3-by-n matrices are identied. The relationship between
expansion and completability is developed based on the prior re-
sults about single unspecied entry. These results extend our un-
derstanding of TP-completable patterns. A new Ratio Theorem
related to TP-completability is introduced in this paper, and it can
possibly be ...
Strongly Real Conjugacy Classes In Unitary Groups Over Fields Of Even Characteristic, 2018 College of William and Mary
Strongly Real Conjugacy Classes In Unitary Groups Over Fields Of Even Characteristic, Tanner N. Carawan Undergraduate Honors Theses
An element $g$ of a group $G$ is called strongly real if there is an $s$ in $G$ such that $s^2 = 1$ and $sgs^{-1} = g^{-1}$. It is a fact that if $g$ in $G$ is strongly real, then every element in its conjugacy class is strongly real. Thus we can classify each conjugacy class as strongly real or not strongly real. Gates, Singh, and Vinroot have classified the strongly real conjugacy classes of U$(n, q^2)$ in the case that $q$ is odd. Vinroot and Schaeffer Fry have classified some of the conjugacy classes of U ...
Counting Real Conjugacy Classes In Some Finite Classical Groups, 2018 William & Mary
Counting Real Conjugacy Classes In Some Finite Classical Groups, Elena Amparo Undergraduate Honors Theses
An element $g$ in a group $G$ is real if there exists $x\in G$ such that $xgx^{-1}=g^{-1}$. If $g$ is real then all elements in the conjugacy class of $g$ are real. In \cite{GS1} and \cite{GS2}, Gill and Singh showed that the number of real $\mathrm{GL}_n(q)$-conjugacy classes contained in $\mathrm{SL}_n(q)$ equals the number of real $\mathrm{PGL}_n(q)$-conjugacy classes when $q$ is even or $n$ is odd. In this paper, we use generating functions to show that the result is also true for odd $q ...
The Doubly Stochastic Single Eigenvalue Problem: An Empirical Approach, 2018 College of William & Mary
The Doubly Stochastic Single Eigenvalue Problem: An Empirical Approach, John Wilkes, Charles Royal Johnson Undergraduate Honors Theses
The doubly stochastic single eigenvalue problem asks what is the set DS
n of all complex numbers that occur as an eigenvalue of an n-by-n doubly stochastic matrix. For n < 5, this set is known and for the analogous set for (singly) stochastic matrices, the set is known for all n. For P k , the polygon formed by the k-th roots of unity, U n k=1 P k ⊆ DS n, as is easily shown. For n < 5, this containment is an equality, but for n = 5, the containment is strict (though it is close). Presented here is substantial, computational evidence that the containment is an equality for 6 ≤ n ≤ 10 and for what DS 5 actually is.
A New Upper Bound For The Diameter Of The Cayley Graph Of A Symmetric Group, 2018 William & Mary
A New Upper Bound For The Diameter Of The Cayley Graph Of A Symmetric Group, Hangwei Zhuang Undergraduate Honors Theses
Given a finite symmetric group S_n and a set S of generators, we can represent the group as a Cayley graph. The diameter of the Cayley graph is the largest distance from the identity to any other elements. We work on the conjecture that the diameter of the Cayley graph of a finite symmetric group S_n with S ={(12),(12...n)} is at most $ C(n,2). Our main result is to show that the diameter of the graph of S_n is at most (3n^2-4n)/2.
Nonassociative Right Hoops, 2018 Chapman University
Nonassociative Right Hoops, Peter Jipsen, Michael Kinyon Mathematics, Physics, and Computer Science Faculty Articles and Research
The class of nonassociative right hoops, or narhoops for short, is defined as a subclass of right-residuated magmas, and is shown to be a variety. These algebras generalize both right quasigroups and right hoops, and we characterize the subvarieties in which the operation x ^^ y = (x/y)y is associative and/or commutative. Narhoops with a left unit are proved to be integral if and only if ^ is commutative, and their congruences are determined by the equivalence class of the left unit. We also prove that the four identities defining narhoops are independent.
Putting Fürer's Algorithm Into Practice With The Bpas Library, 2018 The University of Western Ontario
Putting Fürer's Algorithm Into Practice With The Bpas Library, Linxiao Wang Electronic Thesis and Dissertation Repository
Fast algorithms for integer and polynomial multiplication play an important role in scientific computing as well as other disciplines. In 1971, Schönhage and Strassen designed an algorithm that improved the multiplication time for two integers of at most n bits to O(log n log log n). In 2007, Martin Fürer presented a new algorithm that runs in O (n log n · 2 ^O(log* n)) , where log*n is the iterated logarithm of n. We explain how we can put Fürer’s ideas into practice for multiplying polynomials over a prime field Z/pZ, which characteristic is a Generalized ...
Upper Bound For The Number Of Distinct Eigenvalues Of A Perturbed Matrix, 2018 Sookmyung Women's University,
Upper Bound For The Number Of Distinct Eigenvalues Of A Perturbed Matrix, Sunyo Moon, Seungkook Park Electronic Journal of Linear Algebra
In 2016, Farrell presented an upper bound for the number of distinct eigenvalues of a perturbed matrix. Xu (2017), and Wang and Wu (2016) introduced upper bounds which are sharper than Farrell's bound. In this paper, the upper bounds given by Xu, and Wang and Wu are improved.
Range-Compatible Homomorphisms Over The Field With Two Elements, 2018 Université de Versailles Saint-Quentin-en-Yvelines, Laboratoire de Mathématiques de Versailles
Range-Compatible Homomorphisms Over The Field With Two Elements, Clément De Seguins Pazzis Electronic Journal of Linear Algebra
Let U and V be finite-dimensional vector spaces over a field K, and S be a linear subspace of the space L(U, V ) of all linear operators from U to V. A map F : S → V is called range-compatible when F(s) ∈ Im s for all s ∈ S. Previous work has classified all the range-compatible group homomorphisms provided that codimL(U,V )S ≤ 2 dim V − 3, except in the special case when K has only two elements and codimL(U,V )S = 2 dim V − 3. This article gives a thorough treatment of that special case. The results ...
Implementation And Analysis Of The Nonlinear Decomposition Attack On Polycyclic Groups, Yoongbok Lee Undergraduate Honors Theses
Around two years ago, Roman'kov introduced a new type of attack called the nonlinear decomposition attack on groups with solvable membership search problem. To analyze the precise efficiency of the algorithm, we implemented the algorithm on two protocols: semidirect product protocol and Ko-Lee protocol. Because polycyclic groups were suggested as possible platform groups in the semidirect product protocol and polycyclic groups have a solvable membership search problem, we used poly- cyclic groups as the platform group to test the attack. While the complexity could vary regarding many different factors within the group, there was always at least one exponential ...
An Investigation Into The Properties Of Quaternions: Their Origin, Basic Properties, Functional Analysis, And Algebraic Characteristics, James Miller Masters Essays
No abstract provided.
Linear Algebra (Ung), 2018 University of North Georgia
Linear Algebra (Ung), Hashim Saber, Beata Hebda, Piotr Hebda, Benkam Bobga Mathematics Grants Collections
This Grants Collection for Linear Algebra was created under a Round Seven ALG Textbook Transformation Grant.
Affordable Learning Georgia Grants Collections are intended to provide faculty with the frameworks to quickly implement or revise the same materials as a Textbook Transformation Grants team, along with the aims and lessons learned from project teams during the implementation process.
Documents are in .pdf format, with a separate .docx (Word) version available for download. Each collection contains the following materials:
Linked Syllabus Initial Proposal Final Report
Cayley Graphs Of Psl(2) Over Finite Commutative Rings, 2018 Western Kentucky University
Cayley Graphs Of Psl(2) Over Finite Commutative Rings, Kathleen Bell Masters Theses & Specialist Projects
Hadwiger's conjecture is one of the deepest open questions in graph theory, and Cayley graphs are an applicable and useful subtopic of algebra.
Chapter 1 will introduce Hadwiger's conjecture and Cayley graphs, providing a summary of background information on those topics, and continuing by introducing our problem. Chapter 2 will provide necessary definitions. Chapter 3 will give a brief survey of background information and of the existing literature on Hadwiger's conjecture, Hamiltonicity, and the isoperimetric number; in this chapter we will explore what cases are already shown and what the most recent results are. Chapter 4 will ...
Potential Stability Of Matrix Sign Patterns, 2018 William & Mary
Potential Stability Of Matrix Sign Patterns, Christopher Hambric Undergraduate Honors Theses
The topic of matrix stability is very important for determining the stability of solutions to systems of differential equations. We examine several problems in the field of matrix stability, including minimal conditions for a $7\times7$ matrix sign pattern to be potentially stable, and applications of sign patterns to the study of Turing instability in the $3\times3$ case. Furthermore, some of our work serves as a model for a new method of approaching similar problems in the future.
Branching Matrices For The Automorphism Group Lattice Of A Riemann Surface, 2018 Rose-Hulman Institute of Technology
Branching Matrices For The Automorphism Group Lattice Of A Riemann Surface, Sean A. Broughton Mathematical Sciences Technical Reports (MSTR)
Let
S be a Riemann surface and G a large subgroup of Aut(S) ( Aut(S) may be unknown). We are particularly interested in regular n-gonal surfaces, i.e., the quotient surface S/G (and hence S/Aut(S)) has genus zero. For various H the ramification information of the branched coverings S/K -> S/H may be captured in a matrix. The ramification information, in particular strong branching, may be then be used in analyzing the structure of Aut(S). The ramification information is conjugation invariant so the matrix's rows and columns may be indexed by conjugacy ...
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Cauchy Product of Power Series
The following theorem will give us a way to in a sense, "multiply" two power series together.
Theorem 1 (The Cauchy Product of Power Series): Consider the power series $\sum_{n=0}^{\infty} a_nx^n$ with a radius of convergence $R_1$, and the power series $\sum_{n=0}^{\infty} b_nx^n$ with a radius of convergence $R_2$. Then whenever both of these power series convergent we have that $\left ( \sum_{n=0}^{\infty} a_nx^n \right ) \left ( \sum_{n=0}^{\infty} b_nx^n \right ) = \sum_{n=0}^{\infty} \left ( \left ( \sum_{j=0}^{n} (a_jb_{n-j}) \right ) x^n \right ) = \sum_{n=0}^{\infty} \left (\left (a_0b_n + a_1b_{n-1} + ... + a_{n+1}b_{1} + a_nb_0 \right)x^n \right )$. This power series has a radius of convergence $R$ such that $R ≥ \mathrm{min} \{ R_1, R_2 \}$.
It is common to reformulate the theorem for the Cauchy Product above as follows. Let $\sum_{n=0}^{\infty} a_nx^n$ and $\sum_{n=0}^{\infty} b_nx^n$ be two power series with radii of convergence $R_1$ and $R_2$ respectively. Then the Cauchy Product of these series can be defined as $\sum_{n=0}^{\infty} c_nx^n$ where:(1)
Furthermore, the Cauchy product $\sum_{n=0}^{\infty} c_nx^n$ has a radius of convergence $R$ at least larger or equal to the smaller of the two radii $R_1$, $R_2$, that is $R ≥ \mathrm{min} \{ R_1, R_2 \}$.
We will now look at some examples of applying the Cauchy product to some power series.
Example 1 Using the Cauchy Product determine a power series representation of the function $f(x) = \frac{1}{(1 - x)^2}$ for $-1 < x < 1$.
We know that for $\mid x \mid < 1$, $\frac{1}{1 - x} = 1 + x + x^2 + x^3 + ... = \sum_{n=0}^{\infty} x^n$. So to determine a power series representation of the function $f$, we will simply take the Cauchy product of $\sum_{n=0}^{\infty} x^n$ with itself.
First note that:(2)
Therefore it follows by the Cauchy product that $f(x) = \frac{1}{(1 - x)^2} = \sum_{n=0}^{\infty} (n + 1)x^n$ is a representation for $-1 < x < 1$.
Example 2 Using the Cauchy Product determine a power series representation of the function $f(x) = \frac{1}{(1 - x)^4}$.
From example 1 we have that for $\mid x \mid < 1$, $\frac{1}{(1 - x)^2} = \sum_{n=0}^{\infty} (n + 1)x^n$. We will take the Cauchy Product of this series with itself once again as follows:(3)
And so:(4)
Therefore it follows that $f(x) = \frac{1}{(1 - x)^4} = \sum_{n=0}^{\infty} \left ( \frac{n^3 + 6n^2 + 11n + 6}{6} \right) x^n$ for $-1 < x < 1$.
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De Bruijn-Newman constant
For each real number [math]t[/math], define the entire function [math]H_t: {\mathbf C} \to {\mathbf C}[/math] by the formula
[math]\displaystyle H_t(z) := \int_0^\infty e^{tu^2} \Phi(u) \cos(zu)\ du[/math]
where [math]\Phi[/math] is the super-exponentially decaying function
[math]\displaystyle \Phi(u) := \sum_{n=1}^\infty (2\pi^2 n^4 e^{9u} - 3 \pi n^2 e^{5u}) \exp(-\pi n^2 e^{4u}).[/math]
It is known that [math]\Phi[/math] is even, and that [math]H_t[/math] is even, real on the real axis, and obeys the functional equation [math]H_t(\overline{z}) = \overline{H_t(z)}[/math]. In particular, the zeroes of [math]H_t[/math] are symmetric about both the real and imaginary axes. One can also express [math]H_t[/math] in a number of different forms, such as
[math]\displaystyle H_t(z) = \frac{1}{2} \int_{\bf R} e^{tu^2} \Phi(u) e^{izu}\ du[/math]
or
[math]\displaystyle H_t(z) = \frac{1}{2} \int_0^\infty e^{t\log^2 x} \Phi(\log x) e^{iz \log x}\ \frac{dx}{x}.[/math]
In the notation of [KKL2009], one has
[math]\displaystyle H_t(z) = \frac{1}{8} \Xi_{t/4}(z/2).[/math]
De Bruijn [B1950] and Newman [N1976] showed that there existed a constant, the
de Bruijn-Newman constant [math]\Lambda[/math], such that [math]H_t[/math] has all zeroes real precisely when [math]t \geq \Lambda[/math]. The Riemann hypothesis is equivalent to the claim that [math]\Lambda \leq 0[/math]. Currently it is known that [math]0 \leq \Lambda \lt 1/2[/math] (lower bound in [RT2018], upper bound in [KKL2009]).
The
Polymath15 project seeks to improve the upper bound on [math]\Lambda[/math]. The current strategy is to combine the following three ingredients: Numerical zero-free regions for [math]H_t(x+iy)[/math] of the form [math]\{ x+iy: 0 \leq x \leq T; y \geq \varepsilon \}[/math] for explicit [math]T, \varepsilon, t \gt 0[/math]. Rigorous asymptotics that show that [math]H_t(x+iy)[/math] whenever [math]y \geq \varepsilon[/math] and [math]x \geq T[/math] for a sufficiently large [math]T[/math]. Dynamics of zeroes results that control [math]\Lambda[/math] in terms of the maximum imaginary part of a zero of [math]H_t[/math]. Contents [math]t=0[/math]
When [math]t=0[/math], one has
[math]\displaystyle H_0(z) = \frac{1}{8} \xi( \frac{1}{2} + \frac{iz}{2} ) [/math]
where
[math]\displaystyle \xi(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \zeta(s)[/math]
is the Riemann xi function. In particular, [math]z[/math] is a zero of [math]H_0[/math] if and only if [math]\frac{1}{2} + \frac{iz}{2}[/math] is a non-trivial zero of the Riemann zeta function. Thus, for instance, the Riemann hypothesis is equivalent to all the zeroes of [math]H_0[/math] being real, and Riemann-von Mangoldt formula (in the explicit form given by Backlund) gives
[math]\displaystyle \left|N_0(T) - (\frac{T}{4\pi} \log \frac{T}{4\pi} - \frac{T}{4\pi} - \frac{7}{8})\right| \lt 0.137 \log (T/2) + 0.443 \log\log(T/2) + 4.350 [/math]
for any [math]T \gt 4[/math], where [math]N_0(T)[/math] denotes the number of zeroes of [math]H_0[/math] with real part between 0 and T.
The first [math]10^{13}[/math] zeroes of [math]H_0[/math] (to the right of the origin) are real [G2004]. This numerical computation uses the Odlyzko-Schonhage algorithm. In [P2017] it was independently verified that all zeroes of [math]H_0[/math] between 0 and 61,220,092,000 were real.
[math]t\gt0[/math]
For any [math]t\gt0[/math], it is known that all but finitely many of the zeroes of [math]H_t[/math] are real and simple [KKL2009, Theorem 1.3]. In fact, assuming the Riemann hypothesis,
all of the zeroes of [math]H_t[/math] are real and simple [CSV1994, Corollary 2].
It is known that [math]\xi[/math] is an entire function of order one ([T1986, Theorem 2.12]). Hence by the fundamental solution for the heat equation, the [math]H_t[/math] are also entire functions of order one for any [math]t[/math].
Because [math]\Phi[/math] is positive, [math]H_t(iy)[/math] is positive for any [math]y[/math], and hence there are no zeroes on the imaginary axis.
Let [math]\sigma_{max}(t)[/math] denote the largest imaginary part of a zero of [math]H_t[/math], thus [math]\sigma_{max}(t)=0[/math] if and only if [math]t \geq \Lambda[/math]. It is known that the quantity [math]\frac{1}{2} \sigma_{max}(t)^2 + t[/math] is non-increasing in time whenever [math]\sigma_{max}(t)\gt0[/math] (see [KKL2009, Proposition A]. In particular we have
[math]\displaystyle \Lambda \leq t + \frac{1}{2} \sigma_{max}(t)^2[/math]
for any [math]t[/math].
The zeroes [math]z_j(t)[/math] of [math]H_t[/math] obey the system of ODE
[math]\partial_t z_j(t) = - \sum_{k \neq j} \frac{2}{z_k(t) - z_j(t)}[/math]
where the sum is interpreted in a principal value sense, and excluding those times in which [math]z_j(t)[/math] is a repeated zero. See dynamics of zeros for more details. Writing [math]z_j(t) = x_j(t) + i y_j(t)[/math], we can write the dynamics as
[math] \partial_t x_j = - \sum_{k \neq j} \frac{2 (x_k - x_j)}{(x_k-x_j)^2 + (y_k-y_j)^2} [/math] [math] \partial_t y_j = \sum_{k \neq j} \frac{2 (y_k - y_j)}{(x_k-x_j)^2 + (y_k-y_j)^2} [/math]
where the dependence on [math]t[/math] has been omitted for brevity.
In [KKL2009, Theorem 1.4], it is shown that for any fixed [math]t\gt0[/math], the number [math]N_t(T)[/math] of zeroes of [math]H_t[/math] with real part between 0 and T obeys the asymptotic
[math]N_t(T) = \frac{T}{4\pi} \log \frac{T}{4\pi} - \frac{T}{4\pi} + \frac{t}{16} \log T + O(1) [/math]
as [math]T \to \infty[/math] (caution: the error term here is not uniform in t). Also, the zeroes behave like an arithmetic progression in the sense that
[math] z_{k+1}(t) - z_k(t) = (1+o(1)) \frac{4\pi}{\log |z_k|(t)} = (1+o(1)) \frac{4\pi}{\log k} [/math]
as [math]k \to +\infty[/math].
Threads Polymath proposal: upper bounding the de Bruijn-Newman constant, Terence Tao, Jan 24, 2018. Polymath15, first thread: computing H_t, asymptotics, and dynamics of zeroes, Terence Tao, Jan 27, 2018. Polymath15, second thread: generalising the Riemann-Siegel approximate functional equation, Terence Tao and Sujit Nair, Feb 2, 2018. Polymath15, third thread: computing and approximating H_t, Terence Tao and Sujit Nair, Feb 12, 2018. Polymath 15, fourth thread: closing in on the test problem, Terence Tao, Feb 24, 2018. Polymath15, fifth thread: finishing off the test problem?, Terence Tao, Mar 2, 2018. Polymath15, sixth thread: the test problem and beyond, Terence Tao, Mar 18, 2018. Polymath15, seventh thread: going below 0.48, Terence Tao, Mar 28, 2018. Polymath15, eighth thread: going below 0.28, Terence Tao, Apr 17, 2018. Polymath15, ninth thread: going below 0.22?, Terence Tao, May 4, 2018. Other blog posts and online discussion Heat flow and zeroes of polynomials, Terence Tao, Oct 17, 2017. The de Bruijn-Newman constant is non-negative, Terence Tao, Jan 19, 2018. Lehmer pairs and GUE, Terence Tao, Jan 20, 2018. A new polymath proposal (related to the Riemann hypothesis) over Tao's blog, Gil Kalai, Jan 26, 2018. Code and data Writeup Test problem Zero-free regions
See Zero-free regions.
Wikipedia and other references Bibliography [A2011] J. Arias de Reyna, High-precision computation of Riemann's zeta function by the Riemann-Siegel asymptotic formula, I, Mathematics of Computation, Volume 80, Number 274, April 2011, Pages 995–1009. [B1994] W. G. C. Boyd, Gamma Function Asymptotics by an Extension of the Method of Steepest Descents, Proceedings: Mathematical and Physical Sciences, Vol. 447, No. 1931 (Dec. 8, 1994),pp. 609-630. [B1950] N. C. de Bruijn, The roots of trigonometric integrals, Duke J. Math. 17 (1950), 197–226. [CSV1994] G. Csordas, W. Smith, R. S. Varga, Lehmer pairs of zeros, the de Bruijn-Newman constant Λ, and the Riemann hypothesis, Constr. Approx. 10 (1994), no. 1, 107–129. [G2004] Gourdon, Xavier (2004), The [math]10^{13}[/math] first zeros of the Riemann Zeta function, and zeros computation at very large height [KKL2009] H. Ki, Y. O. Kim, and J. Lee, On the de Bruijn-Newman constant, Advances in Mathematics, 22 (2009), 281–306. Citeseer [N1976] C. M. Newman, Fourier transforms with only real zeroes, Proc. Amer. Math. Soc. 61 (1976), 246–251. [P2017] D. J. Platt, Isolating some non-trivial zeros of zeta, Math. Comp. 86 (2017), 2449-2467. [P1992] G. Pugh, The Riemann-Siegel formula and large scale computations of the Riemann zeta function, M.Sc. Thesis, U. British Columbia, 1992. [RT2018] B. Rodgers, T. Tao, The de Bruijn-Newman constant is non-negative, preprint. arXiv:1801.05914 [T1986] E. C. Titchmarsh, The theory of the Riemann zeta-function. Second edition. Edited and with a preface by D. R. Heath-Brown. The Clarendon Press, Oxford University Press, New York, 1986. pdf
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Absolute and Conditional Convergence of Double Series of Real Numbers
Recall from the Absolute and Conditional Convergence of Series of Real Numbers page that a series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is said to be absolutely convergent if $\displaystyle{\sum_{n=1}^{\infty} \mid a_n \mid}$ converges, and conditionally convergent if $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges and $\displaystyle{\sum_{n=1}^{\infty} \mid a_n \mid}$ diverges.
We will now impose an analogous definition to double series of real numbers.
Definition: A double series $\displaystyle{\sum_{m,n=1}^{\infty} a_{mn}}$ is said to be Absolutely Convergent if $\displaystyle{\sum_{m,n=1}^{\infty} \mid a_{mn} \mid}$ converges. If $\displaystyle{\sum_{m,n=1}^{\infty} a_{mn}}$ converges but $\displaystyle{\sum_{m,n=1}^{\infty} \mid a_{mn} \mid}$ diverges then we said that $\displaystyle{\sum_{m,n=1}^{\infty} a_{mn}}$ is Conditionally Convergent.
Like with regular series, any strictly positive double series that converges is absolutely convergent.
Theorem 1: If a double series $\displaystyle{\sum_{m,n=1}^{\infty} a_{mn}}$ converges absolutely then $\displaystyle{\sum_{m,n=1}^{\infty} a_{mn}}$ converges. Proof:Suppose that $\displaystyle{\sum_{m,n=1}^{\infty} a_{mn}}$ converges absolutely. We first note that for all $m, n \in \mathbb{N}$ that: Therefore we have that: Taking the double sum from $m, n = 1$ to $\infty$ of all sides gives us that: By comparison we must have that the double series $\displaystyle{\sum_{m,n=1}^{\infty} [a_{mn} + \mid a_{mn} \mid]}$ converges, and so: Therefore, $\displaystyle{\sum_{m,n}^{\infty} a_{mn}}$ is the sum of two convergent double series and is hence convergent. $\blacksquare$
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A simulation of 50,000 iterations gives the average distance after a 2-step (unit step) random walk on a 2 dimensional plane, which is around 1.27. But how can one mathematically prove this?
Any insight is highly appreciated!
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
You can rotate the coordinates so the first step is to $(1,0)$. If the second step is at angle $\theta$, the end point is then $(1+\cos \theta,\sin \theta)$. The distance from the origin is then $\sqrt{(1+\cos \theta)^2+\sin^2\theta}=\sqrt{2+2\cos \theta}$ The average of this over $\theta$ is $$\frac 1{2\pi}\int_0^{2\pi}\sqrt{2+2\cos \theta}d\theta=\frac 8{2\pi}\approx 1.2732$$
I get $\frac{0+2\sqrt 2 + 2} 4 \approx 1.207$ as the expected distance. The first step moves you to distance $1$ from the origin. The next step goes back to the origin with probability $1/4$, goes sideways with probability $1/2$, and goes in the same direction with probability $1/4$.
I suppose the discrepancy between what I calculate and what you report is due to Monte Carlo sampling error.
Assuming you are describing a walk on a two dimensional grid.
First step in any direction. Second step has four choices (assume equally probable), left, right, forward, and backward. The four eqiproable distances from start are $ \sqrt{2}, \sqrt{2},2,0$. The average is $1..2071$
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Let's say we have the following multiple regression model,
$$Y_i = \alpha + \beta_1 x_{i1} +\beta_2x_{i2} + ... + \beta_kx_{ik} + \varepsilon_i$$
with $ \varepsilon_i$ is $iid$ ~ $N(0,\sigma_{\varepsilon}^2) $ and least square estimators are $A, B_1,...,B_k $ for $\alpha, \beta_1,...,\beta_k $.
The sampling variance of a given slope coefficient $B_j$ is given by $$ Var(B_j) = \frac{1}{1-R_j^2}\times \frac{\sigma_\varepsilon ^2}{\sum_{i=1}^{n}(x_{ij}-\bar{x_j})^2} $$ where $R_j^2$ is the squared multiple correlation from the regression of $X_j$ on all the other $X$s. In the matrix notation of the above model, the variance of the parameter estimate vector $(\boldsymbol{B})$ is given by $$ Var({\boldsymbol{B}}) = \sigma_{\varepsilon}^2(X'X)^{-1} $$ My question is, how do we go from the $Var(\boldsymbol{B})$ equation to the $Var(B_j)$ equation where the multiple correlation is incorporated? Note that I took the $Var(B_j)$ equation from John Fox's Applied Regression Analysis & Generalized Linear Models (3rd Ed).
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I'm working through Griffith's Intro to Quantum Mechanics, attempting to solve problem 2.27.
Consider the double delta-function potential $$ V(x)= -\alpha [\delta(x+a)+\delta(x-a)] $$ where $\alpha$ and $a$ are constants.
b) How many bound states does it possess? Find the allowed energies, for $\alpha=\hbar ^2 /ma$ and for $\alpha=\hbar ^2 /4ma$, and sketch the wave function
I'm having trouble right off the bat with this problem. A cursory google search has informed me that I need to split the wave function up into cases of even and odd parities in order to reduce the amount of constants in the problem. I'm not sure what this means, or how to approach it mathematically.
Before searching for a hint, I split up the potential into three regimes: $$ x<-a, \, -a<x<a, \, x>a $$ and found the wave function for each case. I ended up with the following result: $$ \psi (x)= \begin{cases} Be^{kx} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (x<-a) \\ Ce^{-kx} + De^{kx} \ \ (-a<x<a) \\ Ee^{-kx} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (x>a) \end{cases} $$
From there, I used the continuity conditions for $\psi$ and $\psi'$ to solve for each constant. My general approach seems to be correct, barring my mistake with the parity.
To reiterate, I understand that the even and odd cases will reduce the amount of constant from 4 to 2, allowing us to solve for them using the continuity conditions for $\psi$ and $\psi'$. My confusion is regarding how to apply the parity of the wave function when starting the problem from scratch.
EDIT
Hi, all. So I've figured out my original question. With the hints given, I was able to to make some significant progress. Now I'm faced with solving the discontinuous derivative at $\pm a$.
I've found that: $$ \psi (x) = \begin{cases} Be^{kx} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (x<-a) \\ C(e^{-kx} + e^{kx}) \ \ (-a<x<a) \\ Be^{-kx} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (x>a) \end{cases} $$
When applying the continuity conditions, I found that $$ \psi_{-}(a) = \psi_{+}(a) = Be^{-ka}=C(e^{-ka} +e^{ka}) \\ \Rightarrow B=C[1+e^{2ka}] $$
I've been trying to figure out the derivative, but I'm unsure of how to approach this. If I could ask for the following hint, I think I can work the rest out. Why does the discontinuity of $d\psi/dx$ at $x=a$ imply the following? $$ -kBe^{-ka} - C(ke^{ka}-ke^{-ka}) = -\frac{2m\alpha}{\hbar ^2}Be^{-ka} $$
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Eigenvalues and Algebraic/Geometric Multiplicities of Matrix $A+cI$ Problem 378
Let $A$ be an $n \times n$ matrix and let $c$ be a complex number.
(a) For each eigenvalue $\lambda$ of $A$, prove that $\lambda+c$ is an eigenvalue of the matrix $A+cI$, where $I$ is the identity matrix. What can you say about the eigenvectors corresponding to $\lambda+c$? (b) Prove that the algebraic multiplicity of the eigenvalue $\lambda$ of $A$ is the same as the algebraic multiplicity of the eigenvalue $\lambda+c$ of $A+cI$ are equal. (c) How about geometric multiplicities?
Contents
Proof. (a) $\lambda+c$ is an eigenvalue of $A+cI$.
Let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$. Then we have
\[A\mathbf{x}=\lambda \mathbf{x}.\] It follows that we have \begin{align*} (A+cI)\mathbf{x}&=A\mathbf{x}+c\mathbf{x}\\ &=\lambda \mathbf{x}+c\mathbf{x}\\ &=(\lambda+c)\mathbf{x}. \end{align*}
Thus, we obtain
\[(A+cI)\mathbf{x}=(\lambda+c)\mathbf{x},\] where $\mathbf{x}$ is a nonzero vector. Hence $\lambda+c$ is an eigenvalue of the matrix $A+cI$, and $\mathbf{x}$ is an eigenvector corresponding to $\lambda-c$.
In summary, if $\lambda$ is an eigenvalue of $A$ and $\mathbf{x}$ is an associated eigenvector, then $\lambda+c$ is an eigenvalue of $A+cI$ and $\mathbf{x}$ is an associated eigenvector corresponding to $\lambda+c$.
(b) Algebraic multiplicities are the same
Let $p(t)=\det(A-tI)$ be the characteristic polynomial of $A$.
Let $q(t)$ be the characteristic polynomial of the matrix $A+cI$. Then we have \begin{align*} q(t)&=\det\left(\, (A+cI)-t \,\right)=\det\left(\, A-(t-c) \,\right)\\ &=p(t-c). \end{align*}
Let $\lambda_1, \dots, \lambda_k$ be distinct eigenvalues of $A$ with algebraic multiplicities $n_1, \dots, n_k$, respectively.
Then we have \[p(t)=\pm \prod_{i=1}^k (t-\lambda_i)^{n_i}.\] It follows that we have \begin{align*} q(t)&=p(t-c)\\ &=\pm \prod_{i=1}^k (t-c-\lambda_i)^{n_i}\\ &=\pm \prod_{i=1}^k \left(t-(\lambda_i+c)\right)^{n_i}.\\ \end{align*} From the last equation, we read that the eigenvalues of the matrix $A+cI$ are $\lambda_i+c$ with algebraic multiplicity $n_i$ for $i=1,\dots, k$. Thus, geometric multiplicities of $\lambda$ and $\lambda-c$ are the same. (c) How about geometric multiplicities?
From part (a), we know that eigenvectors of $\lambda$ are eigenvectors of $\lambda-c$.
Reversing the argument, the eigenvectors of $\lambda+c$ are eigenvectors of $\lambda$.
Thus, the eigenvectors correspond one to one, and the eigenspace of $\lambda$ is the same as the eigenspace of $\lambda+c$.
Hence their geometric multiplicities are the same. Applications
As applications of this problem, consider the following problems.
This is a problem of a Linear Algebra final exam at Harvard University.
For a solution of this problem, see the post “Eigenvalues and eigenvectors of matrix whose diagonal entries are 3 and 9 elsewhere“.
This is also a problem of a Linear Algebra final exam at Harvard University.
See the post “Determinant of matrix whose diagonal entries are 6 and 2 elsewhere” for a solution.
Add to solve later
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It would seem that far-away stars are at such a distance that I should be able to take a step to the side and not have the star's photons hit my eye. How do stars release so many photons to fill in such great angular distances?
The answer is simple: Yes, stars really do produce
that many photons. This calculation is a solid (though very rough) approximation that a star the size of the sun might emit about $10^{45}$ visible photons per second (1 followed by 45 zeros, a billion billion billion billion billion photons).
You can do the calculation: If you're 10 light-years away from that star, you are nevertheless getting bombarded by 1 million photons per square centimeter in each second.
$$\frac{10^{45}\ \text{photons}/\mathrm s}{4\pi (10 \ \text{lightyears})^2} \approx 10^6\ \text{photons}/(\mathrm{cm^2\ s)}$$
Although I agree with all three of the above answers let me present a slightly different perspective on the problem.
It's tempting to think of the light from the star as a flood of photons that behave like little bullets. However this is oversimplified because a photon is a localised object i.e. we observe a photon when something interacts with the light and localises it.
The light from the star is not a hail of photons but instead the star is transferring energy to the photon quantum field and this energy spreads out radially and evenly. If you were to describe the light as photons you'd have to say the photons were completely delocalised i.e. they are spread over the whole spherical wavefront and you could not say in which direction the photon was travelling.
As the energy reaches you it can interact with the rhodopsin molecules in your eye and transfer one photon's worth of energy. It's at this point, and only at this point, that the energy is localised into a photon. Even if the star were so dim that it only emitted a few photons worth of energy per second there would still be a finite probability that your eye could interact with it and detect a photon, though that probability would obviously be ludicrously small.
So stepping aside would make little difference because as long as your eye intersected the spherical wavefront somewhere there would still be a finite probability of detecting a photon and therefore seeing the star.
Have a look at my answer to Some doubts about photons for some related arguments.
The only stars you
can reliably see are ones that are spewing enough photons at your eyeballs to appear stable.
Any star which is so dim that photons entering your eye can literally be counted one by one, simply will not register in your vision, because your eye's retina is not sensitive enough.
So your question is basically embroiled in
observer bias; it assumes that the stars you see are all the stars there are, and it assumes that you could see a single photon if it hit your eye.
Allow me to channel something akin to the anthropic principle here. You can only see the stars that have a lot of photons reaching your eye. If a star were so far away that photons were reaching your eyes only occasionally then the star would be too dim for you to see in in the first place. Even if you could see the photons, the star would appear to blink.
So because you can see the star and it's relatively bright, that means there is enough of a continuous stream of photons reaching the Earth that stepping side to side doesn't change anything. Also, angular resolution isn't quantized so there is never a situation where stepping side to side (while maintaining the same radius from the star) ever changes the probability of receiving a photon.
A star radiates in all directions. You would still see the star regardless of the number of steps you take to any side, just not the same photons.
A laser radiates in only one direction (or in a very small cone). If you took a large enough step to the side (larger than the angular size of the emitted beam) so as to exit this cone, then you would no longer see the source.
A very non-technical answer, but in trying to get your head around this, have you thought about the speed of light?
The angle distended by the star on your eyeball (or by your eyeball on the star) is very small. So it seems like a very tiny region of space must be 'full of photons' for the star to be constantly visible, and since the point where you are standing is not special, all similar regions must be equally 'full'. But the region in question is actually a very narrow beam whose length is the speed of light multiplied by the time that images persist in our vision. If the latter is 50ms, the length of the column is 15,000 km - the diameter of the earth. In this there would need to be a few dozen photons for the star to be marginally visible iirc.
Not a rigorous explanation I know, but it might help reconcile your intuition with the science.
So starlight propagates spherically and each human eyeball creates localized photons just at the intersection of wavefront and retina. No matter where you are in relation to the star some part of this wavefront will reveal the photon stream. Some kind of sensor that could image the path of all the photons/wave functions as they were emitted would reveal a solid hemisphere of light expanding away from the star...
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D-meson nuclear modification factor and elliptic flow measurements in Pb–Pb collisions at $\sqrt {s_{NN}}$ = 5.02TeV with ALICE at the LHC
(Elsevier, 2017-11)
ALICE measured the nuclear modification factor ($R_{AA}$) and elliptic flow ($\nu_{2}$) of D mesons ($D^{0}$, $D^{+}$, $D^{⁎+}$ and $D^{s+}$) in semi-central Pb–Pb collisions at $\sqrt{s_{NN}} =5.02$ TeV. The increased ...
ALICE measurement of the $J/\psi$ nuclear modification factor at mid-rapidity in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV
(Elsevier, 2017-11)
ALICE at the LHC provides unique capabilities to study charmonium production at low transverse momenta ( p T ). At central rapidity, ( |y|<0.8 ), ALICE can reconstruct J/ ψ via their decay into two electrons down to zero ...
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I am using a PIC micro with a 10bit ADC to take readings from an analog signal with a frequency less than 300 hz. However that analog signal is in the range of -2 V and +2 V. How can I condition the signal to get it into a usable range (assuming the input to the ADC has to be positive) Also I do not have a positive and negative power supply.
important note: this answer was posted to solve the problem for -20V to +20Vinput, because that was what was asked. It's a clever method but doesn't work work if the input voltage limit stays between the rails.
You'll have to scale the voltage with a resistor divider so that you get a voltage between -2.5V and +2.5V, and add 2.5V. (I'm presuming a 5V power supply for your PIC).
The following calculation looks long, but that's only because I explain every step in detail. In reality it's so easy that you can do it in your head in no time.
First this:
R1 is the resistor between \$V_{IN}\$ and \$V_{OUT}\$,
R2 is the resistor between \$+5V\$ and \$V_{OUT}\$, and R3 is the resistor between \$V_{OUT}\$ and \$GND\$.
How many unknowns do we have? Three, R1, R2 and R3. Not quite, we can choose one value freely, and the other two are dependent on that one. Let's choose R3 = 1k. The mathematical way to find the other values is to create a set of two simultaneous equations from two (\$V_{IN}\$, \$V_{OUT}\$) pairs, and solve for the unknown resistor values. Any (\$V_{IN}\$, \$V_{OUT}\$) pairs will do, but we'll see that we can tremendously simplify things by carefully choosing those pairs, namely the extreme values: (\$+20V\$, \$+5V\$) and (\$-20V\$, \$0V\$).
First case: \$V_{IN} = +20V\$, \$V_{OUT}=+5V\$
Note that (and this is the key to the solution!) both ends of R2 see \$+5V\$, so there's no voltage drop, and therefore no current through R2. That means that \$I_{R1}\$ has to be the same as \$I_{R3}\$ (KCL). \$I_{R3}=\dfrac{+5V-0V}{1k\Omega}=5mA=I_{R1}\$. We know the current through R1, and also the voltage over it, so we can calculate its resistance: \$R1=\dfrac{+20V-5V}{5mA}=3k\Omega\$. Found our first unknown!
Second case: \$V_{IN} = -20V\$, \$V_{OUT}=0V\$
The same thing as with R2 happens now with R3: no voltage drop, so no current. Again according to KCL, now \$I_{R1}\$ = \$I_{R2}\$. \$I_{R1}=\dfrac{-20V-0V}{3k\Omega}=6.67mA=I_{R2}\$. We know the current through R2, and also the voltage over it, so we can calculate its resistance: \$R2=\dfrac{+5V-0V}{6.67mA}=0.75k\Omega\$. Found our second unknown!
So a solution is: \$R1 = 3k\Omega, R2 = 0.75k\Omega, R3 = 1k\Omega\$.
Like I said it's only the
ratio between these values which is important, so I might as well pick \$R1 = 12k\Omega, R2 = 3k\Omega, R3 = 4k\Omega\$. We can check this solution against another (\$V_{IN}\$, \$V_{OUT}\$) pair, e.g. (\$0V\$, \$2.5V\$). R1 and R3 are now parallel (they both have +2.5V-0V over them, so when we calculate their combined value we find \$0.75k\Omega\$, exactly the value of R2, and the value we needed to get \$+2.5V\$ from \$+5V\$! So our solution is indeed correct. [QC stamp goes here]
The last thing to do is to connect \$V_{OUT}\$ to the PIC's ADC. ADCs often have rather low input resistances, so this may disturb our carefully calculated equilibrium. Nothing to worry about, however, we simply have to increase R3 so that \$R3 // R_{ADC} = 1k\Omega\$. Suppose \$R_{ADC} = 5k\Omega\$, then \$\dfrac{1}{1k\Omega}=\dfrac{1}{R3}+\dfrac{1}{R_{ADC}}=\dfrac{1}{R3}+\dfrac{1}{5k\Omega}\$ From this we find \$R3=1.25k\Omega\$.
edit OK, that was clever and very simple, even if I say so myself. ;-) But why wouldn't this work if the input voltage stays between the rails? In the above situations we always had a resistor which had no current flowing through it, so that, following KCL, the current coming into the \$V_{OUT}\$ node via one resistor would leave via the other one. That meant that one voltage had to be higher than \$V_{OUT}\$, and the other lower. If both voltages are lower there would only flow current away from that node, and KCL forbids that.
The easiest way is to use a "resistor divider".
You didn't say what voltage this PIC is running at and therefore the A/D input range is, so let's use 5V for the example. Your input voltage range is 40V, and the output 5V, so you need something that attenuates by at least 8. You also need the result to be centered on 1/2 Vdd, which is 2.5V, whereas your input voltage is centered on 0V.
This can be accomplished with 3 resistors. One end of all three resistors are connected together and to the PIC A/D input pin. The other end of R1 goes to the input signal, R2 goes to Vdd, and R3 goes to ground. The resistor divider is formed by the R1 and the parallel combination of R2 and R3. You can adjust R2 and R3 to center the resulting range at 2.5V, but for simplicity explaining this we'll live with a little bit of assymetry and attenuate a little bit more to make sure both ends are limited to the Vss-Vdd range.
Let's say the PIC wants the analog signal to have a impedance of 10 kΩ or less. Again for simplicity, let's make R2 and R3 20 kΩ. The impedance feeding the PIC will be no more than the parallel combination of those, which is 10 kΩ. To get attenuation of 8, R1 needs to be 7 times R2//R3, which is 70 kΩ. However, since the result won't be exactly symmetric, we need to attenuate a little more to make sure -20V in won't result in less than 0V into the PIC. That actually requires attenuation of 9, so R1 must be at least 8 times R2//R3, which is 80 kΩ. The standard value of 82 kΩ will allow for some slop and margin but you still get most of the A/D range to measure the original signal.
Added:
Here is a example of finding the exact solution to a similar problem. This has no assymetry and has a particular specified output impedance. This form of solution can always be used when the A/D range is wholly within the input voltage range.
This is the standard circuit for that. You need to scale the resistor values for your required impedance.
If the signal is not DC, or if a DC reference isn't important, the signal can be capacitively coupled to the input of the ADC.
Alternatively, if your ground for the PIC is floating, you could tie your signal ground to 1/2 VDD of the PIC.
The following circuit should do the job:
3.3V + | \ / 1k \ | +-- ADC input | \ / 1k \ | +-- Signal input (-2V to +2V)
It's a potential divider. At -2V, the output will be 0.65V; at +2V, 2.65V.
All noise on the 3.3V rail will get transferred to the input, so use a good voltage reference to reduce this problem.
This will work with other supplies too, but the offset will shift.
Thomas' voltage adder with two identical resistors is indeed simple, but has the disadvantage that the input range to the ADC is reduced, which means that noise will have a bigger influence. Also the lower limit is at 0.65V. If your microcontroller doesn't have a \$V_{ADCREF-}\$ input (most controllers don't) that part of the input range will remain unused.
This is easy to fix: choose the resistor ratio so that \$V_{ADC}\$ will be 0V if the input is -2V. For a \$V_{DD}\$ of 5V this means the input resistor should be 2/5 of the pull-up resistor. At 2V input \$V_{ADC}\$ will be 2.86V. Set \$V_{ADCREF+}\$ to this level, and the -2V to +2V will cover the full ADC range.
If your \$V_{DD}\$ = 3.3V the input resistor should be 61% (\$\frac{2V}{3.3V}\$) of the pull-up. At +2V in \$V_{ADC}\$ will be 2.49V.
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The most frequently used evaluation metric of survival models is the concordance index (c index, c statistic). It is a measure of rank correlation between predicted risk scores $\hat{f}$ and observed time points $y$ that is closely related to Kendall’s τ. It is defined as the ratio of correctly ordered (concordant) pairs to comparable pairs. Two samples $i$ and $j$ are comparable if the sample with lower observed time $y$ experienced an event, i.e., if $y_j > y_i$ and $\delta_i = 1$, where $\delta_i$ is a binary event indicator. A comparable pair $(i, j)$ is concordant if the estimated risk $\hat{f}$ by a survival model is higher for subjects with lower survival time, i.e., $\hat{f}_i >\hat{f}_j \land y_j > y_i$, otherwise the pair is discordant. Harrell’s estimator of the c index is implemented in concordance_index_censored.
While Harrell’s concordance index is easy to interpret and compute, it has some shortcomings:
it has been shown that it is too optimistic with increasing amount of censoring [1], it is not a useful measure of performance if a specific time range is of primary interest (e.g. predicting death within 2 years).
Since version 0.8, scikit-survival supports an alternative estimator of the concordance index from right-censored survival data, implemented in concordance_index_ipcw, that addresses the first issue.
The second point can be addressed by extending the well known receiver operating characteristic curve (ROC curve) to possibly censored survival times. Given a time point $t$, we can estimate how well a predictive model can distinguishing subjects who will experience an event by time $t$ (sensitivity) from those who will not (specificity). The function cumulative_dynamic_auc implements an estimator of the cumulative/dynamic area under the ROC for a given list of time points.
The first part of this post will illustrate the first issue with simulated survival data, while the second part will focus on the time-dependent area under the ROC applied to data from a real study.
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Corollaries to the Open Mapping Theorem
Recall from The Open Mapping Theorem page that if $X$ and $Y$ are Banach spaces and if $T : X \to Y$ is a bounded linear operator then the range $T(X)$ is closed if and only if $T$ is an open map.
We will now present some corollaries to the open mapping theorem.
Corollary 1 tells us that if $T$ is a bijective bounded linear operator from a Banach space $X$ to a Banach space $Y$ then the inverse $T^{-1}$ is also bounded.
Corollary 1: Let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a bounded linear operator. If $T$ is bijective then $T^{-1}$ is a bounded linear operator. Proof:Since $T$ is bijective, $T(X) = Y$, which is closed. So by the open mapping theorem we have that $T$ is an open map. But again since $T$ is bijective, $T$ being open implies that $T^{-1}$ is a bounded linear operator. $\blacksquare$
Corollary 2 gives us a quicker way to determine if two norms on a space are equivalent or not - provided that $X$ equipped with both norms forms a Banach space.
Corollary 2: Let $(X, \| \cdot \|_1)$ and $(X, \| \cdot \|_2)$ be Banach spaces. If there is an $M > 0$ for which $\| x \|_2 \leq M \| x \|_1$ for every $x \in X$ then $\| \cdot \|_1$ and $\| \cdot \|_2$ are equivalent norms. Proof:Let $i : (X, \| \cdot \|_1) \to (X, \| \cdot \|_2)$ be the identity map defined for all $x \in X$ by $i(x) = x$. Then for all $x \in X$ we have that: So $i$ is bounded. Since $(X, \| \cdot \|_1)$ and $(X, \| \cdot \|_2)$ are Banach spaces and since $i$ is bijective, by corollary $1$ we have that $i^{-1} : (X, \| \cdot \|_2) \to (X, \| \cdot \|_1)$ is a bounded linear operator. So there exists an $m > 0$ such that: For all $x \in X$. Hence for all $x \in X$: So $\| \cdot \|_1$ and $\| \cdot \|_2$ are equivalent norms. $\blacksquare$
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Impacto
Junge, M. and Pérez García, David and Palazuelos Cabezón, Carlos and Villanueva, Ignacio and Wolf, Michael (2009)
Operator Space theory: a natural framework for Bell inequalities. Physical Review Letters, 104 . ISSN 0031-9007, ESSN: 1079-7114
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Abstract
In this letter we show that the field of Operator Space Theory provides a general and powerful mathematical framework for arbitrary Bell inequalities, in particular regarding the scaling of their violation within quantum mechanics. We illustrate the power of this connection by showing that bipartite quantum states with local Hilbert space dimension n can violate a Bell inequality by a factor of order $\frac{\sqrt{n}}{\log^2n}$ when observables with n possible outcomes are used. Applications to resistance to noise, Hilbert space dimension estimates and communication complexity are given.
Item Type: Article Additional Information:
Short (non-technical) version
Uncontrolled Keywords: Teoría cuántica Palabras clave (otros idiomas): Quantum Physics Subjects: Sciences > Physics > Mathematical physics
Sciences > Physics > Quantum theory
ID Code: 12160 Deposited On: 03 Feb 2011 09:12 Last Modified: 03 Dec 2014 12:45 Origin of downloads
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An integrator is a very important filter that proves useful in implementation of many blocks of a communication receiver. In continuous-time case, an integrator finds the area under the curve of a signal amplitude. A discrete-time system deals with just the signal samples and hence a discrete-time integrator serves the purpose of collecting a running sum of past samples for an input signal. Looking at an infinitesimally small scale, this is the same as computing the area under the curve of a signal sampled at an extremely high rate. For the following discussion, we assume a normalized sample time $T_S =1$.
For an input $s[n]$ and output $r[n]$, there are three main methods to implement a discrete-time integrator.
[Forward difference] The forward difference integrator is realized through the following equation.
\begin{align} r[n] &= \sum _{i=-\infty}^{n} s[i] \nonumber \\ &= \sum _{i=-\infty}^{n-1} s[i] + s[n] \nonumber \\ &= r[n-1] + s[n] \label{eqIntroductionForwardIntegrator} \end{align} For obvious reasons, the running sum $r[n-1]$ is a common component in all types of integrators; the differentiating factor is the term added to $r[n-1]$ as a replacement for area under the curve. For a forward difference integrator, this additional term is $s[n]$ — the current input. This is drawn in Figure below. Notice that the forward difference integrator computes its output after the arrival of the current sample, i.e., at time $n$.
The block diagram shown in the figure is used to implement a forward difference integrator in loop filter portion of a Phase Locked Loop (PLL) that is employed in carrier phase and symbol timing synchronization systems of a digital communication system.
[Backward difference] The backward difference integrator is realized through the following equation.
\begin{equation}\label{eqIntroductionBackwardIntegrator} r[n] = r[n-1] + s[n-1] \end{equation}
Here, the term added to $r[n-1]$ is the previous input $s[n-1]$. This is also illustrated in the Figure above. In contrast to the forward difference case, the backward difference integrator can compute its output after the last sample, i.e., at time $n-1$. This minor dissimilarity plays a huge role in analyzing the performance of the actual discrete-time system where the integrator is employed (refer to any DSP text to understand the role of poles and zeros of a system).
The block diagram shown in the figure is used to implement a backward difference integrator in a Numerically Controlled Oscillator (NCO) of a Phase Locked Loop (PLL).
[Average of a backward and forward difference] A third integrator can be implemented as
\begin{equation*} r[n] = r[n-1] + \frac{1}{2}\left(s[n-1]+s[n]\right) \end{equation*}
An integrator maintains a running sum of past samples. In time domain, the summation of a large number of values tends towards a mean value. When some numbers are small, some are large and some are in between, their running sum naturally pulls large variations towards the middle, thus smoothing them out. Large variations represent high frequencies and smoothing out such variations is the function of a lowpass filter.
To verify this fact in frequency domain, first consider that when an impulse is given as input to an integrator in time domain, it again forms a running sum of the impulse, which is a unit step signal. Hence, its impulse response is a unit step signal. The unit step signal is very wide in time domain, so it must be narrow in frequency domain. Consequently, it filters out the high frequency terms and passes only the low frequency content. We can say that it acts as a lowpass filter.
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You're asking a mathematical question. Given a function $f$, let $a_n = f(n)$, then does the limit $\epsilon \downarrow 0$ exist for
$$ \sum a_n e^{-n \epsilon} - \int f(t) e^{-t\epsilon}?$$The answer is no: the integral will generically only get ride of the 'hardest' divergence. The sum $\sum a_n e^{-n \epsilon}$ typically (if $\{a_n\}$ behaves reasonably well) has an expansion of the form$$ \alpha \, \epsilon^{-\nu} + \beta \, \epsilon^{-\nu + 1/2} + \ldots$$and similarly for the integral, but you can only show that the first terms of both small-$\epsilon$ expansions agree. An example if $f(x) = x^{3/2}$, $a_n = n^{3/2}$. The integral has a single pole in $\epsilon$ but the sum has 3 divergent terms.
There is also an ambiguity, i.e. what should the lower integration boundary be? $t=0$ is just a choice and changing to, say, $t=1/2$ changes the answer/divergent pieces. Look into Eucler-Maclaurin theory for a deeper understanding of this.
There is also a physical misunderstanding. You are "heat kernel" regulating the series $\sum a_n$, but it doesn't always make sense to tag on a factor of $\exp(-n \epsilon)$. Otherwise, why not pick $\exp(-\sqrt{n} \epsilon)$ or $\exp(-n^{2013} \epsilon)$? In practice you often need to pick something covariant, say an energy or eigenvalue $\lambda_n$ of an operator. It just happens that for bosonic strings, the spectrum consists of integers $n$, so that's why textbooks don't mention this.
The right thing to do then is to look for counterterms that can cancel the negative powers of $\epsilon$ in the small-epsilon expansion shown above.
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On the fractal nature of the set of all binary sequences with almost perfect linear complexity profile Abstract
Stream ciphers usually employ some sort of pseudo-randomly generated bit strings to be added to the plaintext. The cryptographic properties of such binary sequences can be stated in terms of the so-called linear complexity profile. This paper shows that the set of all sequences with an almost perfect linear complexity profile maps onto a fractal subset of [0, 1].
The space \(\mathbb{F}_2^\infty\) of all infinite binary sequences can be mapped onto [0, 1] by \(\iota :\left( {{a_i}} \right)_{i = 1}^\infty \mapsto \sum\nolimits_{i = 1}^\infty {{a_i}} {2^{ - 1}}\). Any such sequence admits a linear complexity profile (l.c.p.) , stating for each n that the initial string (a
l, …, a n L n L n L ≈ n/2, and so m( n):= 2 • L n n should vary around zero. A be the set of those sequences from \( \mathbb{F}_2^\infty \) whose l.c.p. is almost perfect in the sense of | d m( n)| < d∀ n(Niederreiter, 1988a). The subset of [0,1] obtained as ι( A ) is fractal and its Hausdorff dimension is bounded from above by d φ is the positive real root of \( {x^d} = \sum\nolimits_{i = 0}^{d - 1} {{x^i}} \), e.g. d φ 1= 1, φ 2= 1.618… (Fibonacci’s golden ratio). Thus, although all the A have Haar measure zero in FT, a sharper distinction can be made by looking at their Hausdorff dimension. As a by-product the paper gives explicit formulae for the number of sequences of length d nin A , for all n and d d. KeywordsLinear complexity Hausdorff dimension References Dai, Z.-D. (1989) Continued fractions and the Berlekamp–Massey algorithm, E.I.S.S. Report # 89/7, Europäisches Institut für Systemsicherheit, Karlsruhe. Lidl, R. and Niederreiter, H. (1994) Introduction to Finite Fields and Their Applications, revised ed., Cambridge University Press, Cambridge. Niederreiter, H. (1988a) Sequences with almost perfect linear complexity profile, in Advances in Cryptology–EUROCRYPT ‘87(eds. D. Chaum, W.L. Price), LNCS 304, 37–51, Springer, Berlin.Google Scholar Niederreiter, H. (1988b) The probabilistic theory of linear complexity, in Advances in Cryptology–EUROCRYPT ‘88(ed. C.G. Günther), LNCS 330, 191–209, Springer, Berlin.Google Scholar
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(Sorry was asleep at that time but forgot to log out, hence the apparent lack of response)
Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference
Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details
http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference
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Complex Numbers
We should already be familiar with the set of real numbers $\mathbb{R}$. We will now extend further and learn the basics about the set of complex numbers $\mathbb{C}$.
Definition: A number $z$ in the form $z = a + bi$ where $a, b \in \mathbb{R}$ and $i^2 = -1$ is said to be a Complex Number. The Real Part is denoted as $\Re (z) = a$, and the Imaginary Part is denoted $\Im (z) = b$. If $\Re (z) = 0$, that is $z = 0 + bi$, then we say that $z$ is an Imaginary Number, and if $\Im (z) = 0$, that is $z = a + 0i$, then we say $z$ is a Real Number.
For example, consider the complex number $z = 2 + 3i$. We note that the real part $\Re (z) = 2$ and the imaginary part $\Im (z) = 3$. Therefore we can represent the complex number $z$ as $z = \Re(z) + \Im(z) \cdot i$.
Commonly, an imaginary number can be represented in two dimensions as a vector along both a real axis and an imaginary axis. We can plot $z = a + bi$ as the unique vector $(\Re (z), \Im (z))$ as illustrated below:
We will now look at some basic properties of complex numbers. All of these properties also hold for real numbers.
Properties of Complex Numbers
Consider the complex numbers $z_1 = a_1 + b_1i$, $z_2 = a_2 + b_2i$ and $z_3 = a_3 + b_3 i$. The following properties hold:
Commutativity of Addition for Complex Numbers:$z_1 + z_2 = z_2 + z_1$. Associativity of Addition for Complex Numbers:$z_1 + (z_2 + z_3) = (z_1 + z_2) + z_3$. Existence of an Additive Identity for Complex Numbers:$z_1 + 0 = 0 + z_1 = z_1$. Existence of an Additive Inverse for Each Complex Number:$z_1 + (-z_1) = (-z_1) + z_1 = 0$. Commutativity of Multiplication for Complex Numbers:$z_1 \cdot z_2 = z_2 \cdot z_1$. Associativity of Multiplication for Complex Numbers:$z_1 \cdot (z_2 \cdot z_3) = (z_1 \cdot z_2) \cdot z_3$. Existence of a Multiplicative Identity for Complex Numbers:$z_1 \cdot 1 = 1 \cdot z_1 = z_1$. Existence of a Multiplicative Inverse for Each Complex Number:If $z \neq 0$ then $z \cdot z^{-1} = z^{-1} \cdot z = 1$. Distributive Property:$z_1 \cdot (z_2 + z_3) = z_1 \cdot z_2 + z_1 \cdot z_3$. Additive and Multiplicative Inverse of a Complex Number
We note that $-z$ is the additive inverse of a complex number $z = a + bi$ if $z + (-z) = 0$. The additive inverse of a complex number is easy in fact, it is $-z = -a - bi$ since:(1)
Finding the multiplicative inverse $z^{-1}$ is a little more complicated though, but can be calculated with the formula $z^{-1} =\frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2} \cdot i$ where both $a, b =\neq 0$ since:(2)
Example 1 If $z = 3 + 4i$ compute the additive inverse $-z$.
We note that $-z = -3 - 4i$ since $z + (-z) = (3 + 4i) + (-3 - 4i) = (3 - 3) + (4 - 4)i = 0 + 0i = 0$.
Example 2 If $z = 5 + 3i$ compute the multiplicative inverse $z^{-1}$.
Using the formula we obtain that: $z^{-1} = \frac{5}{34} - \frac{3i}{34} = \frac{5 - 3i}{34}$.
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The Divergence and Curl of a Vector Field In Two Dimensions
From The Divergence of a Vector Field and The Curl of a Vector Field pages we gave formulas for the divergence and for the curl of a vector field $\mathbf{F}(x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ on $\mathbb{R}^3$ given by the following formulas:(1)
Now suppose that $\mathbf{F}(x, y) = P(x, y) \vec{i} + Q(x, y) \vec{j}$ is a vector field in $\mathbb{R}^2$. Then we define the divergence and curl of $\mathbb{F}$ as follows:
Definition: If $\mathbf{F}(x, y) = P(x, y)\vec{i} + Q(x, y) \vec{j}$ and $\frac{\partial P}{\partial x}$ and $\frac{\partial Q}{\partial y}$ both exist then the Divergence of $\mathbf{F}$ is the scalar field given by $\mathrm{div} (\mathbf{F}) = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}$.
Definition: If $\mathbf{F}(x, y) = P(x,y)\vec{i} + Q(x, y) \vec{j}$ and $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$ both existence then the Curl of $\mathbf{F}$ is the vector field given by $\mathrm{curl} (\mathbf{F}) = \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \vec{k}$. It is important to note that the curl of $\mathbf{F}$ exists in three dimensional space despite $\mathbf{F}$ be a vector field on $\mathbb{R}^2$. Example 1 Find the divergence of the vector field $\mathbf{F}(x, y) = 2xy \vec{i} + 3 \cos y \vec{j}$.
We can apply the formula above directly to get that:(3)
Example 2 Find the divergence of the vector field $\mathbf{F}(x, y) = e^x y^2 \vec{i} + (x + 2y) \vec{j}$.
We can apply the formula above directly to get that:(4)
Example 3 Find the curl of the vector field $\mathbf{F}(x, y) = 2xy \vec{i} + 3 \cos y \vec{j}$.
We can apply the formula above directly to get that:(5)
Example 4 Find the curl of the vector field $\mathbf{F}(x, y) = e^x y^2 \vec{i} + (x + 2y) \vec{j}$.
We can apply the formula above directly to get that:(6)
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Main Page The Problem
Let [math][3]^n[/math] be the set of all length [math]n[/math] strings over the alphabet [math]1, 2, 3[/math]. A
combinatorial line is a set of three points in [math][3]^n[/math], formed by taking a string with one or more wildcards [math]x[/math] in it, e.g., [math]112x1xx3\ldots[/math], and replacing those wildcards by [math]1, 2[/math] and [math]3[/math], respectively. In the example given, the resulting combinatorial line is: [math]\{ 11211113\ldots, 11221223\ldots, 11231333\ldots \}[/math]. A subset of [math][3]^n[/math] is said to be line-free if it contains no lines. Let [math]c_n[/math] be the size of the largest line-free subset of [math][3]^n[/math]. [math]k=3[/math] Density Hales-Jewett (DHJ(3)) theorem: [math]\lim_{n \rightarrow \infty} c_n/3^n = 0[/math]
The original proof of DHJ(3) used arguments from ergodic theory. The basic problem to be considered by the Polymath project is to explore a particular combinatorial approach to DHJ, suggested by Tim Gowers.
Useful background materials
Some background to the project can be found here. General discussion on massively collaborative "polymath" projects can be found here. A cheatsheet for editing the wiki may be found here. Finally, here is the general Wiki user's guide
Threads (1-199) A combinatorial approach to density Hales-Jewett (inactive) (200-299) Upper and lower bounds for the density Hales-Jewett problem (final call) (300-399) The triangle-removal approach (inactive) (400-499) Quasirandomness and obstructions to uniformity (inactive) (500-599) Possible proof strategies (active) (600-699) A reading seminar on density Hales-Jewett (active) (700-799) Bounds for the first few density Hales-Jewett numbers, and related quantities (arriving at station)
There is also a chance that we will be able to improve the known bounds on Moser's cube problem.
Here are some unsolved problems arising from the above threads.
Here is a tidy problem page.
Bibliography
Density Hales-Jewett
H. Furstenberg, Y. Katznelson, “A density version of the Hales-Jewett theorem for k=3“, Graph Theory and Combinatorics (Cambridge, 1988). Discrete Math. 75 (1989), no. 1-3, 227–241. H. Furstenberg, Y. Katznelson, “A density version of the Hales-Jewett theorem“, J. Anal. Math. 57 (1991), 64–119. R. McCutcheon, “The conclusion of the proof of the density Hales-Jewett theorem for k=3“, unpublished.
Behrend-type constructions
M. Elkin, "An Improved Construction of Progression-Free Sets ", preprint. B. Green, J. Wolf, "A note on Elkin's improvement of Behrend's construction", preprint. K. O'Bryant, "Sets of integers that do not contain long arithmetic progressions", preprint.
Triangles and corners
M. Ajtai, E. Szemerédi, Sets of lattice points that form no squares, Stud. Sci. Math. Hungar. 9 (1974), 9--11 (1975). MR369299 I. Ruzsa, E. Szemerédi, Triple systems with no six points carrying three triangles. Combinatorics (Proc. Fifth Hungarian Colloq., Keszthely, 1976), Vol. II, pp. 939--945, Colloq. Math. Soc. János Bolyai, 18, North-Holland, Amsterdam-New York, 1978. MR519318 J. Solymosi, A note on a question of Erdős and Graham, Combin. Probab. Comput. 13 (2004), no. 2, 263--267. MR 2047239
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This post was inspired by this answer of Dave Penneys.
In the category of (irreducible hyperfinite II$_1$) subfactors, the morphisms of $(N \subset M)$ to $(N' \subset M')$ are usually defined as the $W^*$-morphisms $\phi: M \to M'$ with $\phi (N) \subset N'$.
Unfortunately, through this definition, the category of finite group is $\underline{not}$ a (natural) subcategory of the category of subfactors.
In fact, let $G$ and $G'$ be finite groups and $f: G \to G'$ be a surjective group-morphism, then in general, $f$ does $\underline{not}$ extend into a (usual) subfactor-morphism of $(R \subset R \rtimes G)$ to $(R \subset R \rtimes G')$. Here is the explanation in the answer of Dave: A II$_1$-factor is algebraically simple, so each morphism ofII$_1$-factors is either injective or zero. Thus every non-zero morphism is an isomorphism onto its image. I don't think the canonical surjection $G\to G'=G/\ker(f)$ actuallygives you a map of factors $R\rtimes G\to R\rtimes G'$. In particular,if we denote the implementing unitaries as $u_g$ for $g\in G$, the map$u_g\mapsto u_{g\ker(f)}$ does not extend to a non-zero map ofII$_1$-factors if $\ker(f)$ is non-trivial. The element $u_g-u_{g'}$ would map to zero if $g,g'\in \ker(f)$, and a non-trivial map of II$_1$-factors must be injective.
Question: Is there an $\underline{other}$ (natural) definition of subfactor-morphisms such that the category of finite groups is a (natural) subcategory of this "new" category of subfactors ?
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If we want to describe a static spherically symmetric star we can use a metric which matches the Schwarzschild solution with correct mass on the outside of the star but differs from Schwartzschild in the inside of the matter distribution.
Basically we solve the Einstein equations with a source $T_{\mu\nu}$, for instance $$T_{\mu\nu}=(\rho+p)u_{\mu}u_{\nu}+p\,g_{\mu\nu}$$ where $u_{\mu}$ has zero spatial components, meaning it is the velocity in a static fluid (this can also be seen as a consequence of Einstein equations).
Can we do something similar for a rotating star using the metric for a Kerr black hole?
I heard that it is a much more difficult problem and I would like to understand how difficult it is (Is it possible?) and what makes it so difficult.
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Galois Fields
We have already looked at some examples of fields on the Field Axioms page. We will now look at another specific type of field known as a Galois Field. Before we do that though, we must first get a little bit acquainted to a special type of relation known as the
modulus of an integer $a$ to a divisor $m$.
Definition: Let $a$ be an integer. Then we write $r = a \pmod m$ which says that $r$ is the remainder of $a$ when $a$ is divided $m$, that is for some positive integer $k$ we have that $a = km + r$.
For example, consider $7 \pmod 3$. We note that $1 = 7 \pmod 3$ since when $7$ is divided by $3$ we have a remainder of $1$, and so $7 = 2(3) + 1$. We are now ready to define what a Galois field is.
Definition: A Galois Field of the prime number $p$ denoted $GF(p)$ or $\mathbb{Z}_p$ is a set containing the nonnegative integers $0, 1, 2, ..., p - 1$ whose operation of addition between any two elements $a, b \in GF(p)$ is defined by $a + b \pmod p$ and whose multiplication is defined as $a \cdot b \pmod p$.
The simplest Galois Field is $GF(2)$ whose addition and multiplication tables are given below.
$+$ 0 1 0 0 1 1 1 0
$\cdot$ 0 1 0 0 0 1 0 1
We can verify that all eleven axioms of a field hold for $GF(2)$. For example, commutativity of addition and multiplication can be observed on both tables since the tables have symmetric values upon the top-left to top-right diagonal.
Example 1 Write the addition table for $GF(5)$ and verify that an additive identity $0$ exists such that $\forall a \in GF(5)$ that $a + 0 = 0 + a = a$
The addition table for $GF(5)$ is as follows where the red column and row verifies that $0$ is the identity element of $GF(5)$.
$+$ 0 1 2 3 4 0 0 1 2 3 4 1 1 2 3 4 0 2 2 3 4 0 1 3 3 4 0 1 2 4 4 0 1 2 3 Example 2 Write down the multiplication table for $GF(3)$ and verify that multiplication is commutative.
The multiplication table for $GF(3)$ is as follows. The commutativity of multiplication is represented by the symmetry across the diagonal of this multiplication table, that is for all $a, b \in GF(3)$, $a \cdot b = b \cdot a$.
$\cdot$ 0 1 2 0 0 0 0 1 0 1 2 2 0 2 1 Example 3 Explain why $GF(4)$ does not form a field.
In this case we note that $p = 4$ is not a prime number. When we write the addition table for $GF(4)$ we get that:
$+$ 0 1 2 3 0 0 1 2 3 1 1 2 3 0 2 2 3 0 1 3 3 0 1 2
So nothing is wrong with addition. We can verify that all of the addition axioms hold. The reason why $GF(4)$ does not form a field actually lies within the operation of multiplication.
$\cdot$ 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1
We first note that $1$ is the multiplicative identity. Suppose that we are given the element $2$ and want to find its multiplicative inverse $b$ such that $2b = 1$. If we look down the column/row containing $2$ we notice that no such element exists since $2b = 0$ or $2b = 2$ for all $b \in GF(4)$. Therefore not all elements in $GF(4)$ contain multiplicative inverses and so $GF(4)$ is not a field.
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A cylinder of radius $a$ is fixed in space. A hoop of mass $m$ and radius $b$ is in contact with the surface of the cylinder and it is able to rotate around of it without slipping. The force of gravity is acting on the system (figure below).
At $t=0$, the hoop is at rest and a tangential velocity $v_0$ is imposed in the lowest point of the hoop.
¿What must be the minimum magnitude of $v_0$ in order to make the hoop perform a complete loop around the cylinder?
I have no idea what condition must be imposed in order to describe the complete loop around the cylinder and then computing $v_0$. I think it has something to do with the rolling constraint and that the center of mass (COM) of the hoop must describe a complete arc lenght $2\pi(b-a)$.
The initial condition must be $\dot{\alpha}(t=0)=v_0/b$ (it comes from $\boldsymbol{v}_\perp=\boldsymbol{\omega}\times\boldsymbol{r}$), where $\alpha$ is the angle between a vertical axis fixed at the COM of the cylinder and a vertical axis at the COM of the cylinder.
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$X$ is multivariate normal distributed with $\mu = (2, 2)$ and $\Sigma = (1, 0 ; 0, 1)$ and I have 2 vectors $A = (1, 1)$ and $B = (1, -1)$. How can I than show that the linear combinations $AX$ and $BX$ are independent?
First of all independence follows from a lack of correlation when two random variables are jointly normal. Now it isn't hard to see that $AX$ and $BX$ are uncorrelated:
\begin{align} \text{Cov}(AX, BX) &= \text{Cov}(X_1 + X_2, X_1 - X_2) \\ &= \text{Var}(X_1) - \text{Var}(X_2) \\ &= 0 \end{align}
so we just need to show that $AX$ and $BX$ are jointly normal. To do that it's enough to show that all linear combinations of $AX$ and $BX$ are normally distributed. Can you finish the proof?
The
definition of joint normality of $W$ and $Z$ is that all linear combinations $aW+bZ$ of $W$ and $Z$ are normal random variables. Since $X$ and $Y$ are jointly normal, this shows that $W=X+Y$ and $Z=X-Y$ are normal random variables. Furthermore, $aW+bZ = (a+b)X+(a-b)Y$ is normal because it is a linear combination of jointly normal random variables $X$ and $Y$. We conclude that $X+Y$ and $X-Y$ are jointly normal random variables. Hence they are independent if their covariance is $0$. I leave it to you to use the bilinearity of the covariance function to determine whether they are indeed independent random variables.
Look, Ma! No explicit formulas for pdfs used anywhere!
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Let $G$ be a group. Let $a$ and $b$ be elements of $G$.If the order of $a, b$ are $m, n$ respectively, then is it true that the order of the product $ab$ divides $mn$? If so give a proof. If not, give a counterexample.
We claim that it is not true. As a counterexample, consider $G=S_3$, the symmetric group of three letters.Let $a=(1\, 2), b=(1 \,3)$ be transposition elements in $S_3$.The orders of $a$ and $b$ are both $2$.
Consider the product\[ab=(1\, 2)(1 \,3)=(1 \, 3 \, 2).\]Then it is straightforward to check that the order of $ab$ is $3$, which does not divide $4$ (the product of orders of $a$ and $b$).
Therefore, the group $G=S_3$ and elements $a=(1\, 2), b=(1 \,3)\in G$ serve as a counterexample.
Remark. (Abelian group case)
If we further assume that $G$ is an abelian group, then the statement is true.Here is the proof if $G$ is abelian.
Let $e$ be the identity element of $G$.\begin{align*}(ab)^{mn} &=a^{mn}b^{mn} && \text{ since $G$ is abelian}\\&=(a^m)^n(b^n)^m\\&=e^n e^m && \text{since the order of $a, b$ are $m, n$ respectively}\\&=e.\end{align*}Thus the order of $ab$ divides $mn$.
Related Question.
If the group is abelian, then the statement is true.
Problem. Let $G$ be an abelian group with the identity element $1$.Let $a, b$ be elements of $G$ with order $m$ and $n$, respectively.If $m$ and $n$ are relatively prime, then show that the order of the element $ab$ is $mn$.
Problem.Let $G$ be an abelian group.Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively.Prove that there exists an element $c$ in $G$ such that the order of $c$ is the least common multiple of $m$ and $n$.
Also determine whether the statement is true if $G$ is a non-abelian group.
Order of the Product of Two Elements in an Abelian GroupLet $G$ be an abelian group with the identity element $1$. Let $a, b$ be elements of $G$ with order $m$ and $n$, respectively.If $m$ and $n$ are relatively prime, then show that the order of the element $ab$ is $mn$.Proof.Let $r$ be the order of the element […]
The Order of $ab$ and $ba$ in a Group are the SameLet $G$ be a finite group. Let $a, b$ be elements of $G$.Prove that the order of $ab$ is equal to the order of $ba$.(Of course do not assume that $G$ is an abelian group.)Proof.Let $n$ and $m$ be the order of $ab$ and $ba$, respectively. That is,\[(ab)^n=e, […]
Non-Abelian Simple Group is Equal to its Commutator SubgroupLet $G$ be a non-abelian simple group. Let $D(G)=[G,G]$ be the commutator subgroup of $G$. Show that $G=D(G)$.Definitions/Hint.We first recall relevant definitions.A group is called simple if its normal subgroups are either the trivial subgroup or the group […]
Elements of Finite Order of an Abelian Group form a SubgroupLet $G$ be an abelian group and let $H$ be the subset of $G$ consisting of all elements of $G$ of finite order. That is,\[H=\{ a\in G \mid \text{the order of $a$ is finite}\}.\]Prove that $H$ is a subgroup of $G$.Proof.Note that the identity element $e$ of […]
Non-Abelian Group of Order $pq$ and its Sylow SubgroupsLet $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$.Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.Hint.Use Sylow's theorem. To review Sylow's theorem, check […]
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Problem 556
Let $\mathbf{v}$ be a nonzero vector in $\R^n$.
Then the dot product $\mathbf{v}\cdot \mathbf{v}=\mathbf{v}^{\trans}\mathbf{v}\neq 0$. Set $a:=\frac{2}{\mathbf{v}^{\trans}\mathbf{v}}$ and define the $n\times n$ matrix $A$ by \[A=I-a\mathbf{v}\mathbf{v}^{\trans},\] where $I$ is the $n\times n$ identity matrix.
Prove that $A$ is a symmetric matrix and $AA=I$.
Conclude that the inverse matrix is $A^{-1}=A$. Problem 555
Let $U$ and $V$ be vector spaces over a scalar field $\F$.
Define the map $T:U\to V$ by $T(\mathbf{u})=\mathbf{0}_V$ for each vector $\mathbf{u}\in U$. (a) Prove that $T:U\to V$ is a linear transformation. (Hence, $T$ is called the zero transformation.)
Add to solve later
(b) Determine the null space $\calN(T)$ and the range $\calR(T)$ of $T$. Problem 553
Let $T:\R^3 \to \R^3$ be the linear transformation defined by the formula
\[T\left(\, \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \,\right)=\begin{bmatrix} x_1+3x_2-2x_3 \\ 2x_1+3x_2 \\ x_2+x_3 \end{bmatrix}.\]
Determine whether $T$ is an isomorphism and if so find the formula for the inverse linear transformation $T^{-1}$.Add to solve later
Problem 552
For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A|I]$, where $I$ is the $3\times 3$ identity matrix.
Add to solve later
(a) $A=\begin{bmatrix} 1 & 3 & -2 \\ 2 &3 &0 \\ 0 & 1 & -1 \end{bmatrix}$ (b) $A=\begin{bmatrix} 1 & 0 & 2 \\ -1 &-3 &2 \\ 3 & 6 & -2 \end{bmatrix}$. Problem 551
Let $\mathbf{v}$ be a vector in an inner product space $V$ over $\R$.
Suppose that $\{\mathbf{u}_1, \dots, \mathbf{u}_n\}$ is an orthonormal basis of $V$. Let $\theta_i$ be the angle between $\mathbf{v}$ and $\mathbf{u}_i$ for $i=1,\dots, n$.
Prove that
\[\cos ^2\theta_1+\cdots+\cos^2 \theta_n=1.\] Problem 550
Consider the $2\times 2$ matrix
\[A=\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix},\] where $\theta$ is a real number $0\leq \theta < 2\pi$.
(a) Find the characteristic polynomial of the matrix $A$. (b) Find the eigenvalues of the matrix $A$.
Add to solve later
(c) Determine the eigenvectors corresponding to each of the eigenvalues of $A$. Problem 548
An $n\times n$ matrix $A$ is said to be
invertible if there exists an $n\times n$ matrix $B$ such that $AB=I$, and $BA=I$,
where $I$ is the $n\times n$ identity matrix.
If such a matrix $B$ exists, then it is known to be unique and called the
inverse matrix of $A$, denoted by $A^{-1}$.
In this problem, we prove that if $B$ satisfies the first condition, then it automatically satisfies the second condition.
So if we know $AB=I$, then we can conclude that $B=A^{-1}$.
Let $A$ and $B$ be $n\times n$ matrices.
Suppose that we have $AB=I$, where $I$ is the $n \times n$ identity matrix.
Prove that $BA=I$, and hence $A^{-1}=B$.Add to solve later
Problem 546
Let $A$ be an $n\times n$ matrix.
The $(i, j)$
cofactor $C_{ij}$ of $A$ is defined to be \[C_{ij}=(-1)^{ij}\det(M_{ij}),\] where $M_{ij}$ is the $(i,j)$ minor matrix obtained from $A$ removing the $i$-th row and $j$-th column.
Then consider the $n\times n$ matrix $C=(C_{ij})$, and define the $n\times n$ matrix $\Adj(A)=C^{\trans}$.
The matrix $\Adj(A)$ is called the adjoint matrix of $A$.
When $A$ is invertible, then its inverse can be obtained by the formula
For each of the following matrices, determine whether it is invertible, and if so, then find the invertible matrix using the above formula.
(a) $A=\begin{bmatrix} 1 & 5 & 2 \\ 0 &-1 &2 \\ 0 & 0 & 1 \end{bmatrix}$. (b) $B=\begin{bmatrix} 1 & 0 & 2 \\ 0 &1 &4 \\ 3 & 0 & 1 \end{bmatrix}$. Problem 543
Let $R$ be a ring with $1$.
Suppose that $a, b$ are elements in $R$ such that \[ab=1 \text{ and } ba\neq 1.\] (a) Prove that $1-ba$ is idempotent. (b) Prove that $b^n(1-ba)$ is nilpotent for each positive integer $n$.
Add to solve later
(c) Prove that the ring $R$ has infinitely many nilpotent elements. The Inner Product on $\R^2$ induced by a Positive Definite Matrix and Gram-Schmidt Orthogonalization Problem 539
Consider the $2\times 2$ real matrix
\[A=\begin{bmatrix} 1 & 1\\ 1& 3 \end{bmatrix}.\] (a) Prove that the matrix $A$ is positive definite. (b) Since $A$ is positive definite by part (a), the formula \[\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans} A \mathbf{y}\] for $\mathbf{x}, \mathbf{y} \in \R^2$ defines an inner product on $\R^n$. Consider $\R^2$ as an inner product space with this inner product.
Prove that the unit vectors
\[\mathbf{e}_1=\begin{bmatrix} 1 \\ 0 \end{bmatrix} \text{ and } \mathbf{e}_2=\begin{bmatrix} 0 \\ 1 \end{bmatrix}\] are not orthogonal in the inner product space $\R^2$.
Add to solve later
(c) Find an orthogonal basis $\{\mathbf{v}_1, \mathbf{v}_2\}$ of $\R^2$ from the basis $\{\mathbf{e}_1, \mathbf{e}_2\}$ using the Gram-Schmidt orthogonalization process. Problem 538 (a) Suppose that $A$ is an $n\times n$ real symmetric positive definite matrix. Prove that \[\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}\] defines an inner product on the vector space $\R^n$. (b) Let $A$ be an $n\times n$ real matrix. Suppose that \[\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}\] defines an inner product on the vector space $\R^n$.
Prove that $A$ is symmetric and positive definite.Add to solve later
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Exponential Functions Form a Basis of a Vector SpaceLet $C[-1, 1]$ be the vector space over $\R$ of all continuous functions defined on the interval $[-1, 1]$. Let\[V:=\{f(x)\in C[-1,1] \mid f(x)=a e^x+b e^{2x}+c e^{3x}, a, b, c\in \R\}\]be a subset in $C[-1, 1]$.(a) Prove that $V$ is a subspace of $C[-1, 1]$.(b) […]
Exponential Functions are Linearly IndependentLet $c_1, c_2,\dots, c_n$ be mutually distinct real numbers.Show that exponential functions\[e^{c_1x}, e^{c_2x}, \dots, e^{c_nx}\]are linearly independent over $\R$.Hint.Consider a linear combination \[a_1 e^{c_1 x}+a_2 e^{c_2x}+\cdots + a_ne^{c_nx}=0.\][…]
Subspace Spanned by Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$.(a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ […]
Every Basis of a Subspace Has the Same Number of VectorsLet $V$ be a subspace of $\R^n$.Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\}$ is a basis of the subspace $V$.Prove that every basis of $V$ consists of $k$ vectors in $V$.Hint.You may use the following fact:Fact.If […]
Linear Independent Vectors and the Vector Space Spanned By ThemLet $V$ be a vector space over a field $K$. Let $\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n$ be linearly independent vectors in $V$. Let $U$ be the subspace of $V$ spanned by these vectors, that is, $U=\Span \{\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n\}$.Let […]
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Original Question: What is the link between Gamma and the Volatility Risk?
It leads me to ask:
- What is the Volatility Risk definition and what are the good practices to measure it?
Thinking about that question, all I could figure out liking with it is this:
Consider a market $(S^0, S)$ composed of one non-risky asset $S^0$ and one risky asset $S$. The interest rate in this market is $r$ (supposed constant just to simplify. Also, consider an adapted square integrable process $\sigma$ and suppose that $S$ follows $$S_t= S_0 + \int_0^t r S_u ~du +\int_0^t \sigma_u S_u dW_u \quad , t \geq 0$$ under the risk-neutral probability. Let's suppose a trader evaluate the price $v$ of an option of maturity $T$ by fixing $\sigma_t=\bar{\sigma}(t,S_t)$, a function of local volatility. Then, I know that the coverage error is given by (one can show it by a simple aaplication of Feyman-Kac theorem and Itô's Lemma) \begin{align} \text{Err} = V_T- v(T, S_T) &= \int_0 ^T e^{r(T-t)} (\bar{\sigma}(t,S_t) - \sigma_t) \partial^2_x v(t,S_t)dt \\&=\int_0 ^T e^{r(T-t)} (\bar{\sigma}(t,S_t) - \sigma_t) \Gamma(t,S_t)dt,\end{align} where $V = (V_t)_{0 \leq t \leq T}$ is the replicating portfolio and $v(t,x)$ is the intrinsic value of the option at time $t \in [0,T]$ and spot $S_t=x$.
So, we conclude that if:
$\Gamma \geq 0$ (i.e. a convex price):an over-estimation of $\bar{\sigma}(t,S_t)$ secures a gain, and an under-estimation of $\bar{\sigma}(t,S_t)$ secures a loss. $\Gamma\leq 0$ (i.e. a concave price):an under-estimation of $\bar{\sigma}(t,S_t)$ secures a gain, and an over-estimation of $\bar{\sigma}(t,S_t)$ secures a loss. $\Gamma \approx 0$ (i.e. neutral Gamma hedging):the hedging is weakly sensitive to realized volatility. Am I going in the right direction to answer that?
I would appreciate any advice. Thanks in advance.
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I recently came across this in a textbook (NCERT class 12 , chapter: wave optics , pg:367 , example 10.4(d)) of mine while studying the Young's double slit experiment. It says a condition for the formation of interference pattern is$$\frac{s}{S} < \frac{\lambda}{d}$$Where $s$ is the size of ...
The accepted answer is clearly wrong. The OP's textbook referes to 's' as "size of source" and then gives a relation involving it. But the accepted answer conveniently assumes 's' to be "fringe-width" and proves the relation. One of the unaccepted answers is the correct one. I have flagged the answer for mod attention. This answer wastes time, because I naturally looked at it first ( it being an accepted answer) only to realise it proved something entirely different and trivial.
This question was considered a duplicate because of a previous question titled "Height of Water 'Splashing'". However, the previous question only considers the height of the splash, whereas answers to the later question may consider a lot of different effects on the body of water, such as height ...
I was trying to figure out the cross section $\frac{d\sigma}{d\Omega}$ for spinless $e^{-}\gamma\rightarrow e^{-}$ scattering. First I wrote the terms associated with each component.Vertex:$$ie(P_A+P_B)^{\mu}$$External Boson: $1$Photon: $\epsilon_{\mu}$Multiplying these will give the inv...
As I am now studying on the history of discovery of electricity so I am searching on each scientists on Google but I am not getting a good answers on some scientists.So I want to ask you to provide a good app for studying on the history of scientists?
I am working on correlation in quantum systems.Consider for an arbitrary finite dimensional bipartite system $A$ with elements $A_{1}$ and $A_{2}$ and a bipartite system $B$ with elements $B_{1}$ and $B_{2}$ under the assumption which fulfilled continuity.My question is that would it be possib...
@EmilioPisanty Sup. I finished Part I of Q is for Quantum. I'm a little confused why a black ball turns into a misty of white and minus black, and not into white and black? Is it like a little trick so the second PETE box can cancel out the contrary states? Also I really like that the book avoids words like quantum, superposition, etc.
Is this correct? "The closer you get hovering (as opposed to falling) to a black hole, the further away you see the black hole from you. You would need an impossible rope of an infinite length to reach the event horizon from a hovering ship". From physics.stackexchange.com/questions/480767/…
You can't make a system go to a lower state than its zero point, so you can't do work with ZPE. Similarly, to run a hydroelectric generator you not only need water, you need a height difference so you can make the water run downhill. — PM 2Ring3 hours ago
So in Q is for Quantum there's a box called PETE that has 50% chance of changing the color of a black or white ball. When two PETE boxes are connected, an input white ball will always come out white and the same with a black ball.
@ACuriousMind There is also a NOT box that changes the color of the ball. In the book it's described that each ball has a misty (possible outcomes I suppose). For example a white ball coming into a PETE box will have output misty of WB (it can come out as white or black). But the misty of a black ball is W-B or -WB. (the black ball comes out with a minus). I understand that with the minus the math works out, but what is that minus and why?
@AbhasKumarSinha intriguing/ impressive! would like to hear more! :) am very interested in using physics simulation systems for fluid dynamics vs particle dynamics experiments, alas very few in the world are thinking along the same lines right now, even as the technology improves substantially...
@vzn for physics/simulation, you may use Blender, that is very accurate. If you want to experiment lens and optics, the you may use Mistibushi Renderer, those are made for accurate scientific purposes.
@RyanUnger physics.stackexchange.com/q/27700/50583 is about QFT for mathematicians, which overlaps in the sense that you can't really do string theory without first doing QFT. I think the canonical recommendation is indeed Deligne et al's *Quantum Fields and Strings: A Course For Mathematicians *, but I haven't read it myself
@AbhasKumarSinha when you say you were there, did you work at some kind of Godot facilities/ headquarters? where? dont see something relevant on google yet on "mitsubishi renderer" do you have a link for that?
@ACuriousMind thats exactly how DZA presents it. understand the idea of "not tying it to any particular physical implementation" but that kind of gets stretched thin because the point is that there are "devices from our reality" that match the description and theyre all part of the mystery/ complexity/ inscrutability of QM. actually its QM experts that dont fully grasp the idea because (on deep research) it seems possible classical components exist that fulfill the descriptions...
When I say "the basics of string theory haven't changed", I basically mean the story of string theory up to (but excluding) compactifications, branes and what not. It is the latter that has rapidly evolved, not the former.
@RyanUnger Yes, it's where the actual model building happens. But there's a lot of things to work out independently of that
And that is what I mean by "the basics".
Yes, with mirror symmetry and all that jazz, there's been a lot of things happening in string theory, but I think that's still comparatively "fresh" research where the best you'll find are some survey papers
@RyanUnger trying to think of an adjective for it... nihilistic? :P ps have you seen this? think youll like it, thought of you when found it... Kurzgesagt optimistic nihilismyoutube.com/watch?v=MBRqu0YOH14
The knuckle mnemonic is a mnemonic device for remembering the number of days in the months of the Julian and Gregorian calendars.== Method ===== One handed ===One form of the mnemonic is done by counting on the knuckles of one's hand to remember the numbers of days of the months.Count knuckles as 31 days, depressions between knuckles as 30 (or 28/29) days. Start with the little finger knuckle as January, and count one finger or depression at a time towards the index finger knuckle (July), saying the months while doing so. Then return to the little finger knuckle (now August) and continue for...
@vzn I dont want to go to uni nor college. I prefer to dive into the depths of life early. I'm 16 (2 more years and I graduate). I'm interested in business, physics, neuroscience, philosophy, biology, engineering and other stuff and technologies. I just have constant hunger to widen my view on the world.
@Slereah It's like the brain has a limited capacity on math skills it can store.
@NovaliumCompany btw think either way is acceptable, relate to the feeling of low enthusiasm to submitting to "the higher establishment," but for many, universities are indeed "diving into the depths of life"
I think you should go if you want to learn, but I'd also argue that waiting a couple years could be a sensible option. I know a number of people who went to college because they were told that it was what they should do and ended up wasting a bunch of time/money
It does give you more of a sense of who actually knows what they're talking about and who doesn't though. While there's a lot of information available these days, it isn't all good information and it can be a very difficult thing to judge without some background knowledge
Hello people, does anyone have a suggestion for some good lecture notes on what surface codes are and how are they used for quantum error correction? I just want to have an overview as I might have the possibility of doing a master thesis on the subject. I looked around a bit and it sounds cool but "it sounds cool" doesn't sound like a good enough motivation for devoting 6 months of my life to it
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I'm currently researching 5G technology. Many videos/articles on this topic claim that 5G is designed for communicating on lower distances (than 4G), because of higher (than 4G) frequency. So to build a network we need more dense grid of access points. Is this some general rule, that the higher frequency means less range, or is it connected to this exact frequency range, specific to 5G?
In general, yes, higher frequencies attenuate more the further distance they travel. There are two effects that are responsible for this.
First, higher frequency radio waves tend to be absorbed more readily by objects (ie: the penetration depth in the material is shorter). You may have noticed this effect with your home WiFi network. If you have a dual-band router, you probably have two networks, one at 2.4 GHz and one at 5 GHz. You may have noticed that the 2.4 GHz network reaches some corners of your house that the 5 GHz network does not - this is because the 2.4 GHz network penetrates walls better than the 5 GHz network does.
In terms of fifth-generation cell networks, this is the principal effect that limits range: people prefer to live in houses and apartment buildings, and those houses absorb more RF energy at higher frequencies.
The second effect is less intuitive, and falls out the Friis equation for free-space path loss.
$$ FSPL= 20\log_{10}(d) + 20\log_{10}(f) + 20\log_{10}\left(\frac{4\pi}{c}\right)$$
Where \$FSPL\$ is the free space path loss in dB, \$d\$ is the distance in meters, and \$f\$ is the frequency in hertz, and \$c\$ is the speed of light. Note that as \$f\$ increases, the loss increases: a doubling in frequency results in a 6 dB increase in path loss.
This equation would lead you to believe that free space in some way attenuates higher frequencies more than lower frequencies, but that's not really the truth, even though it's a convenient lie we engineers tell ourselves. The "free space path loss" is actually a
geometric effect that captures the "spread" of the electromagnetic wave as it propagates in free space, the same as how the beam of a flashlight spreads out when it's shone out into a big dark room.
The reason it's frequency dependent is because it does not account for the gain of the transmitting and receiving antenna - they are assumed to be isotropic. An isotropic antenna at a higher frequency is physically smaller than one at a lower frequency, which means that geometrically they have a smaller
effective aperture.
If you keep the effective aperture of both antennas at both ends of the link the same physical size as you go up in frequency, you actually see that your path loss is independent of frequency. However, you would also find that the antennas are now directional: as you increase the aperture size, the transmitting antenna shoots a tighter and tighter beam, and the receiving antenna can only receive signals from a narrower and narrower cone. Since engineers usually think in terms of preserving a given antenna pattern and not in aperture size (since pointing antennas can be problematic), we commonly consider free space to have "loss" that increases with frequency when in reality that isn't quite the case.
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Blow-up for the 3-dimensional axially symmetric harmonic map flow into $ S^2 $
1.
Instituto de Matemáticas, Universidad de Antioquia, Calle 67, No. 53–108, Medellín, Colombia
2.
Departamento de Ingeniería Matemática-CMM, Universidad de Chile, Santiago 837-0456, Chile
3.
Department of Mathematical Sciences University of Bath, Bath BA2 7AY, United Kingdom
4.
Department of Mathematics, University of British Columbia, Vancouver, B.C., Canada, V6T 1Z2
$ S^2 $
$ \begin{align*} u_t & = \Delta u + |\nabla u|^2 u \quad \text{in } \Omega\times(0,T) \\ u & = u_b \quad \text{on } \partial \Omega\times(0,T) \\ u(\cdot,0) & = u_0 \quad \text{in } \Omega , \end{align*} $
$ u(x,t): \bar \Omega\times [0,T) \to S^2 $
$ \Omega $
$ \mathbb{R}^3 $
$ \Gamma \subset \Omega $
$ T>0 $
$ u(x,t) $
$ T $
$ \Gamma $
$ | {\nabla} u(\cdot ,t)|^2 \rightharpoonup | {\nabla} u_*|^2 + 8\pi \delta_\Gamma \quad\mbox{as}\quad t\to T . $
$ u_*(x) $
$ \delta_\Gamma $ Keywords:Blow-up, semilinear parabolic equation, harmonic map flow, codimension 2 singular set, finite time blow-up. Mathematics Subject Classification:Primary: 35K58, 35B44; Secondary: 35R01. Citation:Juan Dávila, Manuel Del Pino, Catalina Pesce, Juncheng Wei. Blow-up for the 3-dimensional axially symmetric harmonic map flow into $ S^2 $. Discrete & Continuous Dynamical Systems - A, 2019, 39 (12) : 6913-6943. doi: 10.3934/dcds.2019237
References:
[1]
J. B. van den Berg, J. Hulshof and J. R. King,
Formal asymptotics of bubbling in the harmonic map heat flow,
[2]
J. B. van den Berg and J. F. Williams,
(In-)stability of singular equivariant solutions to the Landau-Lifshitz-Gilbert equation,
[3]
K. C. Chang,
Heat flow and boundary value problem for harmonic maps,
[4]
K. C. Chang, W. Y. Ding and R. Ye,
Finite-time blow-up of the heat flow of harmonic maps from surfaces,
[5] [6] [7] [8] [9] [10] [11] [12]
F. H. Lin and C. Y. Wang,
Energy identity of harmonic map flows from surfaces at finite singular time,
[13] [14]
F. H. Lin and C. Y. Wang,
Harmonic and quasi-harmonic spheres. Ⅲ. Rectifiability of the parabolic defect measure and generalized varifold flows,
[15] [16]
J. Qing and G. Tian,
Bubbling of the heat flows for harmonic maps from surfaces,
[17]
P. Raphaël and R. Schweyer,
Stable blowup dynamics for the 1-corotational energy critical harmonic heat flow,
[18] [19]
P. M. Topping,
Repulsion and quantization in almost-harmonic maps, and asymptotics of the harmonic map flow,
show all references
References:
[1]
J. B. van den Berg, J. Hulshof and J. R. King,
Formal asymptotics of bubbling in the harmonic map heat flow,
[2]
J. B. van den Berg and J. F. Williams,
(In-)stability of singular equivariant solutions to the Landau-Lifshitz-Gilbert equation,
[3]
K. C. Chang,
Heat flow and boundary value problem for harmonic maps,
[4]
K. C. Chang, W. Y. Ding and R. Ye,
Finite-time blow-up of the heat flow of harmonic maps from surfaces,
[5] [6] [7] [8] [9] [10] [11] [12]
F. H. Lin and C. Y. Wang,
Energy identity of harmonic map flows from surfaces at finite singular time,
[13] [14]
F. H. Lin and C. Y. Wang,
Harmonic and quasi-harmonic spheres. Ⅲ. Rectifiability of the parabolic defect measure and generalized varifold flows,
[15] [16]
J. Qing and G. Tian,
Bubbling of the heat flows for harmonic maps from surfaces,
[17]
P. Raphaël and R. Schweyer,
Stable blowup dynamics for the 1-corotational energy critical harmonic heat flow,
[18] [19]
P. M. Topping,
Repulsion and quantization in almost-harmonic maps, and asymptotics of the harmonic map flow,
[1]
José M. Arrieta, Raúl Ferreira, Arturo de Pablo, Julio D. Rossi.
Stability of the blow-up time and the blow-up set under
perturbations.
[2] [3]
Yohei Fujishima.
On the effect of higher order derivatives of initial data on the blow-up set for a semilinear heat equation.
[4]
Asma Azaiez.
Refined regularity for the blow-up set at non characteristic points for the vector-valued semilinear wave equation.
[5]
Yohei Fujishima.
Blow-up set for a superlinear heat equation and pointedness of the initial data.
[6]
Sachiko Ishida, Tomomi Yokota.
Blow-up in finite or infinite time for
quasilinear degenerate Keller-Segel systems of
parabolic-parabolic type.
[7]
Xiumei Deng, Jun Zhou.
Global existence and blow-up of solutions to a semilinear heat equation with singular potential and logarithmic nonlinearity.
[8] [9]
Shota Sato.
Blow-up at space infinity of a solution with a moving singularity for a semilinear parabolic
equation.
[10]
Jong-Shenq Guo.
Blow-up behavior for a quasilinear parabolic equation with nonlinear boundary condition.
[11]
Yuta Wakasugi.
Blow-up of solutions to the one-dimensional semilinear wave equation
with damping depending on time and space variables.
[12] [13]
Zhiqing Liu, Zhong Bo Fang.
Blow-up phenomena for a nonlocal quasilinear parabolic equation with time-dependent coefficients under nonlinear boundary flux.
[14] [15]
Miaoqing Tian, Sining Zheng.
Global boundedness versus finite-time blow-up of solutions to a quasilinear fully parabolic Keller-Segel system of two species.
[16]
Hua Chen, Huiyang Xu.
Global existence and blow-up of solutions for infinitely degenerate semilinear pseudo-parabolic equations with logarithmic nonlinearity.
[17]
Mohamed-Ali Hamza, Hatem Zaag.
Blow-up results for semilinear wave equations in the
superconformal case.
[18]
Qiong Chen, Chunlai Mu, Zhaoyin Xiang.
Blow-up and asymptotic behavior of solutions to a semilinear integrodifferential system.
[19]
Van Tien Nguyen.
On the blow-up results for a class of strongly perturbed semilinear heat equations.
[20]
2018 Impact Factor: 1.143
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Yes, it is possible to extend the state space with respect to which $Y$ is a Feller process. Then, $X$ will be a dense open subset of the extension $\hat X$. Furthermore, for any initial distribution of $Y_0\in\hat X$, then $Y$ will have a continuous modification which necessarily satisfies $Y_t\not\in\hat X\setminus X$ for all positive times (almost surely). In your example with $Y$ being a Brownian motion and $\hat X=\mathbb{R}^3\setminus\{0\}$ then this process corresponds to adding back the origin.
This is actually a special case of a more general method called
Ray-Knight compactification, which applies to right-continuous Markov processes taking values in a Suslin space. Ray-Knight compactification does not always lead to processes which are Feller though, as they can have branch points in the extended state space. The type of processes obtained by this method are called Ray processes. See the 1975 paper by Getoor & Sharpe detailing Ray-Knight compactification, or, any reasonably comprehensive textbook on Markov processes should describe this method. In your question the conditions that $X$ is locally compact, $T_tC_0\subseteq C_b$ and $Y$ is continuous are enough to ensure that we obtain a Feller process so, in particular, there will be no branch points in the extended domain. One point before continuing; I'm assuming, as is standard in the definition of Feller processes, that the space $X$ has a countable base (locally compact, Hausdorff with a countable base, aka an lccb space).
Constructing the extension is a rather natural application of the Gelfand-Naimark theorem, although showing that the resulting process is Feller will still require some work. Let $\mathcal{A}$ be the smallest closed $C^*$-subalgebra of $C_b(X)$ containing $C_0(X)$ and closed under application of $T_t$. We can define $\hat X$ to be its spectrum (the set of nonzero $C^*$ homomorphisms $\mathcal{A}\to\mathbb{C}$), under the weak topology. Then, we identify $X$ as an (open, dense) subset of the (locally compact) space $\hat X$ in the same way as in the construction of the Stone-Chech compactification. By the Gelfand-Naimark theorem, every $f\in\mathcal{A}$ extends uniquely to an $\hat f\in C_0(\hat X)$, giving an isometry $\mathcal{A}\to C_0(\hat X)$, $f\mapsto\hat f$. Then, $T_t$ extends to a map $\hat T_t\colon C_0(\hat X)\to C_0(\hat X)$, $\hat T_t\hat f=\hat{T_tf}$. This is automatically a Markov transition function, but we can show that it is also Feller in your case. It is a bit of work, and I'll break it up into smaller statements.
1) $T_tC_b(X)\subseteq C_b(X)$.
Any nonnegative $f\in C_b(X)$ is a pointwise limit of an increasing sequence $f_n\in C_0(X)$ and, therefore $T_tf=\lim\_{n\to\infty}T_tf_n$ is also a limit of an increasing sequence in $C_0(X)$, so is lower semicontinuous. Applying the same statement to $\Vert f\Vert - f$ implies that $f$ is also upper semi-continuous, so is continuous.
2) If $f\in C_0(X)$ is nonnegative and $t_n\to0$ then $(T_{t_n}f-f)\_+\to0$ in the compact-open topology.
(Note: This doesn't use continuity of $Y$. And, in general, it is not necessary that $T_{t_n}f\to f$ in the compact open topology, even for $Y$ right-continuous.)It is enough to consider $f$ with compact support and, by right-continuity of the process $Y$, we have $T_{t_n}f(x)\to f(x)$ as $n\to\infty$ for all $x\in X$. Then, setting $g_m=m\int\_0^{1/m}T_sf\\,ds$, it can be seen that $\Vert T_{t_n}g_m-g_m\Vert\le 2mt_n\Vert f\Vert\to0$ as $n\to\infty$. Now, letting $S_m$ be the space of convex combinations of $\{g_m,g_{m+1},\ldots\}$, $f$ is a pointwise limit of a sequence in $S_m$, so is in its closure under the compact-open topology (this is a consequence of the Hahn-Banach theorem and the Riesz representation theorem). Therefore, there exists $f_m\in S_m$ converging to $f$ in the compact-open topology. For any $\epsilon > 0$ and compact $K\subseteq X$, take $m$ large enough such that $\vert f-f_m\vert\le\epsilon$ on the union of $K$ and the support of $f$. So,$$T_tf-f \le T_tf_m-f_m+T_t(f-f_m)_++(f_m-f)\le T_tf_m-f_m + 2\epsilon.$$on $K$. Letting $t$ decrease to zero gives $\limsup\_{t\to 0}(T_tf-f)\le2\epsilon$ uniformly on $K$.
3) For any $f\in\mathcal{A}$ and $t_n\to0$, $T_{t_n}f\to f$
uniformly as $n\to\infty$.
This is the point where continuity is required. It is not too hard to show that the space of $f\in C_b(X)$ satisfying the conclusion forms a $C^*$-algebra and is closed under $T_t$, so it is enough to prove it for nonnegative $f\in C_b(X)$ with compact support $K$. Also, the fact that $T_tC_0(X)\subseteq C_b(X)$ can be used to prove the strong-Markov property. If $Y_0\not\in K$ and, letting $\tau$ be the first time at which $Y$ hits $S$, continuity implies that $f(Y_{\tau})=0$ and $Y_\tau\in K$. So, for $x\not\in K$,$$\begin{align}T_tf(x)=\mathbb{E}\_x[f(Y_t)]&=\mathbb{E}\_x\left[T_{(t-\tau)\_+}f(Y_{\tau\wedge t})\right]\\\\&\le\sup\_{y\in K,s\le t}\left(T_sf(y)-f(y)\right)\_+.\end{align}$$The previous statement says that the right-hand-side tends to zero as $t\to0$ and, as it is independent of $x$, $T_tf\to0$ uniformly on $X\setminus K$.
Now let $f_n$ be the sequence converging uniformly on compacts to $f$ constructed in the proof of 2. Then, we have $f_n\to f$ uniformly on $X\setminus K$ as $n\to\infty$ so, in fact, $f_n$ tends uniformly to $f$. Taking the limits $t\to0$ and $n\to\infty$ in$$\Vert T_tf-f\Vert\le\Vert T_tf_n-f_n\Vert+2\Vert f_n-f\Vert$$shows that $\Vert T_tf-f\Vert\to0$.
This almost shows that $\hat T$ is Feller, just the following technical lemma is left.
4) $\hat X$ is an lccb space.
As $X$ is lccb, $C_0(X)$ has a countable dense subset $S$. The closure of $S$ under products, taking $\mathbb{Q}$-linear combinations and applying $T_t$ for $t\in\mathbb{Q}\_+$ is a countable dense subset of $\mathcal{A}$ (use 3). So $C_0(\hat X)\cong\mathcal{A}$ is separable. This implies that $\hat X$ has a countable base.
So, $\hat T_t$ is a Feller transition function on $\hat X$. Therefore, any Markov process $\hat Y$ w.r.t. this transition function has a right-continuous modification. We can also show that, regardless of the initial distribution, $\hat Y$ will be continuous and $\hat Y_t\not\in\hat X\setminus X$ for all times $t > 0$ (almost-surely). For any time $s > 0$ at which $Y_s\in X$ then the statement of the question implies that $\hat Y$ has a continuous modification lying in $X$ at all times $t\ge s$ (which is unique). So, it is enough to show that $\mathbb{P}(\hat Y_s\in X)=1$ for each time $s > 0$. But, this is standard (see Prop. 4.6 (iii) from the paper of Getoor & Sharpe).
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This question already has an answer here:
I want to calculate
$ \lim_{n \to \infty }{\frac{{[n(n+1)(n+2)...(2n-1)]}^\frac1n}n} $
using
$$\int_0^1 f(x)\,dx=\lim_{n\to\infty}\frac1n \sum_{k=1}^n f\left(\frac{k}n\right)$$
I know I have to convert $\frac nk$ to $x$, but I am confused since all the factors are multiplied together. Should I use $\log$?
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Flux, as I understand it, is the amount of substance passing through a particular surface over some time. So, from a simple perspective, considering photons that go through some virtual surface $A$ (or $S$, doesn't matter). They have a fixed speed in vacuum, $v=299,792,458$ $\text m/\text s$. To simplify even further, they're all hitting the surface head-on. So, if we wanted to figure out how many photons go through the surface, we conclude that at a constant velocity they will only pass through the surface if they are in the volume bounded by sweeping the surface area along the velocity vector (perpendicular to the surface, the opposite of its normal) a distance $d$ in the alloted time $t$: $d = vt$
So the flux volume is well-defined as $V(t) = Avt$. We could just look for a period of unit time and "drop" the dependency on time. But even then, it's useless if we cannot sample photon volume density $\rho_p$ to determine how many photons occupy a unit volume in order to determine the actual flux. That makes sense:
$$\Phi = \rho_pV(t)$$
And now comes along the electric flux and thwarts my understanding of the whole notion completely. An electric field is generated when a charge is dropped somewhere in space. Any other charge, especially idealized point charges, placed in its vicinity would experience a force exerted on them by the source charge, its magnitude modulated by the amount of charge. So, the electric field maps points in space with force vectors (ie. a vector field) whose direction and magnitude is parametrized by the interaction between the source charge and the point charge.
And this electric flux is defined as $\Phi = E \cdot S$ (I'll use $\Phi = E \cdot A$) and I just cannot interpret the semantics of this dot product, the product's dimensionality is not what I've come to expect from the notion of flux ($\text{Vm}^{-1}$ or any other).
How does this in any way show how much electric field flow goes through a surface? Furthermore, what is this vague thing called electric field flow? It seems like it is completely disconnected.
I've tried expressing it in different ways from the derivation of the expression of an electric field, which makes sense (non-vector form, dropped unit vector):
$$E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$$
I've intentionally separated the inverse square of the distance and the $4\pi$ which is, I presume, a part of the normalization factor (steradians of sphere) -- but I noticed that together they forge the area of a sphere $A_r = 4\pi r^2$. This way, I could see the expression as the uniform charge density distribution on the surface of a sphere, scaled by the vacuum permittivity.
$$E = \frac{Q}{A_s\varepsilon_0}$$
And then, due to presumed uniformity, by multiplying by an arbitrary area, I could get the flux, the amount of charge(?) flowing through a particular surface in unit time(?):
$$\Phi = E \cdot A = \frac{Q}{A_s\varepsilon_0} A $$
I could see that as flux, but I'm really not sure can I really reinterpret parts of the normalization factor and the inverse-square of the distance into the area of a sphere. From the perspective of voltage over distance it makes absolutely no sense to me.
Any help would be appreciated.
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Electronic Journal of Probability Electron. J. Probab. Volume 24 (2019), paper no. 15, 51 pp. Asymptotic properties of expansive Galton-Watson trees Abstract
We consider a super-critical Galton-Watson tree $\tau $ whose non-degenerate offspring distribution has finite mean. We consider the random trees $\tau _n$ distributed as $\tau $ conditioned on the $n$-th generation, $Z_n$, to be of size $a_n\in{\mathbb N} $. We identify the possible local limits of $\tau _n$ as $n$ goes to infinity according to the growth rate of $a_n$. In the low regime, the local limit $\tau ^0$ is the Kesten tree, in the moderate regime the family of local limits, $\tau ^\theta $ for $\theta \in (0, +\infty )$, is distributed as $\tau $ conditionally on $\{W=\theta \}$, where $W$ is the (non-trivial) limit of the renormalization of $Z_n$. In the high regime, we prove the local convergence towards $\tau ^\infty $ in the Harris case (finite support of the offspring distribution) and we give a conjecture for the possible limit when the offspring distribution has some exponential moments. When the offspring distribution has a fat tail, the problem is open. The proof relies on the strong ratio theorem for Galton-Watson processes. Those latter results are new in the low regime and high regime, and they can be used to complete the description of the (space-time) Martin boundary of Galton-Watson processes. Eventually, we consider the continuity in distribution of the local limits $(\tau ^\theta , \theta \in [0, \infty ])$.
Article information Source Electron. J. Probab., Volume 24 (2019), paper no. 15, 51 pp. Dates Received: 12 December 2017 Accepted: 30 January 2019 First available in Project Euclid: 22 February 2019 Permanent link to this document https://projecteuclid.org/euclid.ejp/1550826098 Digital Object Identifier doi:10.1214/19-EJP272 Citation
Abraham, Romain; Delmas, Jean-François. Asymptotic properties of expansive Galton-Watson trees. Electron. J. Probab. 24 (2019), paper no. 15, 51 pp. doi:10.1214/19-EJP272. https://projecteuclid.org/euclid.ejp/1550826098
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Earth scientists recently celebrated the fifty-year anniversary of the theory of plate tectonics. The notion that the Earth’s surface is composed of a small number of slowly but continually-moving ‘rigid’ plates floating on a convecting, fluid interior is now firmly established as the foundational concept of the discipline. And in the last decade, it has become routinely observable thanks to space geodesy (see Figure 1). If the plates can be considered rigid, then they are merely boundary conditions on a global circulation problem.
Figure 1. A map of surface velocities shows that the surface motions are organised into a strikingly simple pattern of surface “plates” that are almost rigid bodies. The surface strain rate–coloured contours—is concentrated at the plate boundaries. The map also displays the second invariant of the strain-rate, which is vanishingly small in the plate interiors and large at the plate boundaries. The arrows are colored by their angle to ‘North’ as a means of distinguishing the different plates. The scale is such that the longest vector represents 10cm/yr.
The theory of plate tectonics is actually a kinematic description of plate motion that describes how the plates move, but it does not tell us why they move as they do. Neither does it explain how and why new boundaries form or show what happens in the continental regions where deformation is more diffuse. We don’t know how or when plate tectonics began, or why plate motions are only observed on Earth and not on other planets or moons in the solar system. Addressing these questions requires a fully-dynamic theory of plate motion and constitutive behaviour.
The contours of the strain rate in Figure 1 show some regions where plate boundaries are relatively diffuse, meaning that the plates are not rigid there. Most of these regions occur in the continents; particularly obvious is the regional deformation associated with the collision of India and Eurasia which produced the Himalayan mountain belt (see Figure 2). The realization that the plates are not rigid is the first step towards addressing one of the most enduring and significant challenges facing the Earth sciences: the development of a coupled model of deep circulation, plate motions, and continental deformation.
Figure 2. One of the most dramatic departures from plate-like deformation on Earth occurred where the Indian subcontinent collided with the Eurasian continent. The map on the left is a satellite image with the flow lines from the plate motion vector field drawn in red. On the right is the same region showing 50 years of earthquake data for events larger than magnitude 4.5, colored by depth and superimposed on the strain rate.
Developing a global circulation model for the solid Earth is a grand challenge problem. Rock rheology, the manner in which rocks deform under tectonic stress, is very nonlinear and history-dependent because it represents processes that occur at a much smaller temporal and spatial scale than plate motions. Such processes include the mechanics of faults formation and the manner in which they slip via the accumulated effect of thousands of small earthquakes (see Figure 2), combined with the effects of grain-scale deformation, metamorphism, and melting.
Another aspect of the grand challenge is determining how to find and use observations which can constrain the many additional degrees of freedom required by large-scale, dynamic circulation models. Appreciating the scale of this challenge requires recognition of the fundamental difference between the geological record of the ocean floor (the plates) and the geological record of the continental crust. The oceanic crust is part of the 3D, deep circulation pattern of the solid Earth, with an overturn time of around 100 million years, comparable to the median age of the oceanic plates. The continental crust is buoyant enough to escape large-scale recycling and much more widely-distributed in age (ranging up to over three billion years).
A low-angle view of a numerical model of continental collision using the Underworld particle-in-cell finite element code. The (Indian) indentor heads towards the viewer and crumples the crust into a mountain belt in the foreground. In the background, the crust escapes away from the viewer pulled by the subduction zone in the background. Video credit: Louis Moresi.
Consider the Himalayan mountain belt, which is the result of subduction zone congestion by the buoyant and deformable continental crust. The continental collision results in the dramatic topography of the region as the crust is shortened and thickened (see video above and Figure 3 below). Examining the stratigraphy, cooling history, and metamorphism of the rocks allows one to infer the history of the uplifted surface, which is recorded by rocks in crumpled and overturned units. It is also indirectly recorded in the sedimentary record as rivers and glaciers carry material away from the mountain belts and into sedimentary basins.
Figure 3. A low-angle view of a numerical model of continental collision using the Underworld particle-in-cell finite element code. The map (1) interprets the model in terms of the India-Eurasia collision. In the simulation, the Indian indentor heads towards the viewer and crumples the crust into a mountain belt in the foreground. In the background, the crust escapes away from the viewer pulled by the subduction zone. Snapshots from the movie: (2) pre-collision and (3) late in the collision.
In sedimentary basins, accumulated detritus from the landscape stacks up, burying older material and creating a stratigraphic record of the topographic erosion. The difficulty in inverting this record to reveal the history of geological deformation is that the transport of eroded sediment can occur over long distances, is heavily dependent on the history of the topography, can be quite variable in time and space, is sensitive to very small-scale processes, and is coupled to the large-scale deformation. For example, the lengthy, tortuous pathways that the major rivers draining the Himalayas have to take to reach the ocean are clearly influenced by the topography rising and translating beneath the streams and rivers as they cut their channels.
While global circulation models are needed to bring the power of plate tectonic theory to the continental geological record, making these models meaningful will require embedding models which can reproduce observable aspects of the continental record.
Solving a simplified system of equations is necessary to model the flow in the mantle and the plates’ response. A typical template for this problem uses the Stokes equation for an incompressible fluid (or nearly so) driven by thermal buoyancy due to planetary cooling and deep radioactive decay:
\[ \nabla \cdot \boldsymbol{\tau} - \nabla p = g \rho \hat{z} \label{eq:stokes} \qquad (1) \] \[ \nabla \cdot \mathbf{u} = 0 \label{eq:incomp} \qquad (2) \] \[ \frac{D T}{D t} = \nabla( \kappa \nabla T ) + Q \label{eq:advdiff} \qquad (3) \]
Silicate rocks deform in the same way as extremely high-viscosity fluids on geological timescales, which eliminates the need to consider the effects of inertia and rotation. However, a significant complexity arises from the rheology, which is strongly dependent on deformation history and composition.
\[ \frac{{\tau} _ {ij}}{\eta} + \color{Red}{\frac{\dot{\tau} _ {ij} }{\mu}} + \color{Blue}{ \Lambda _ {ijkl} {\huge \tau}_ {kl} } = \frac{\partial u _ i}{\partial x _ j} + \frac{\partial u _ j}{\partial x _ i} \label{eq:constitutive} \qquad (4) \]
The second term (red) in the constitutive law \((4)\) is the effect of viscoelasticity, and the third term (blue) represents the effects of faulting and other forms of material failure as a non-isotropic and strain-dependent plasticity. Neglecting elastic stresses is common at the large length and time scale of these models, but plate boundaries are represented through the plasticity term and these are central to the formulation.
The most successful algorithms for dealing with the elliptic Stokes equation and the history-dependent rheology are hybrids with an Eulerian mesh suitable to build an efficient, parallel, multigrid solver for \((1)\) and Lagrangian particles embedded in the fluid which record the stress and strain history required for \((4)\). In the Underworld code [8], the particle properties map to the mesh via a finite element quadrature rule that must be constructed on-the-fly for the distribution of particles found in each element at every timestep.
\[ \frac{D h}{D t} = \color{Red} { \nabla\left( \kappa(h) \nabla h \right) } - \color{Blue}{ K A ^ m \left| h \right| ^n } + \dot h _ \mathrm{basement} \label{eq:surfaceprocess} \qquad (5)\]
\[ A = \! \! \! \! \int \limits _ { \xi, \mathrm{upstream}} \!\!\!\! \alpha ( \xi ) d\xi \label{eq:upstream} \qquad (6) \]
It is only necessary to consider those processes contributing to erosion, sediment transport, and deposition that are important at scales of hundreds of kilometers and hundreds of thousands to millions of years. Consider the changing height of a simple surface described by with the simplified description outlined in \((5)\). The time derivative of the surface height is computed as a balance between local, nonlinear, diffusion-like terms (red) and non-local terms related to the integrated upstream area for any point \((6)\). Such terms occur because, for example, the available energy of a river in eroding the basement is expected to relate to both the local slope and the total amount of water in the river. To know how much sediment a river is carrying, the net erosion rate upstream must be known for every point along the river. Figure 4 illustrates the network that can be constructed from downhill flow pathway for a discrete set of points representing the surface and how integrating \((5)\) produces a tree-like collection of pathways which cut into the surface. Such pathways form narrow, localised channels that can be very stable. However, where deposition occurs (when the energy of the flow decreases in a flatter region, for example) then the reverse is true: flows tend to diverge and multiple pathways across low-relief surfaces develop. It is much harder to apply the ordered traversal of a tree that is diverging in this way. A promising approach in this case, which is also efficient for domain-decomposed parallel computation, is to use repeated applications of the adjacency matrix of the downhill-directed graph corresponding to the flow network. Such approaches are needed when the surface is not modeled in isolation but as part of a large computation, such as the one shown in Figure 3.
Figure 4. Idealised erosion, transport and deposition models are based on computing height changes which depend on local slope, those dependent on long-range transport processes by rivers or glaciers, and motions due to large-scale tectonics. Long-range transport by rivers requires knowledge of information integrated from the entire upstream catchment of every point. Erosion tends to localise, whereas deposition has the opposite characteristic (1, 2). Upstream integrals can be calculated very efficiently with an appropriately-ordered traversal of the graph of downhill connectivity of the surface (3, 4). Parallel algorithms based on arbitrary spatial decompositions of the domain disrupt this efficient ordering.
The Association for Computing Machinery last month awarded the 2015 Gordon Bell prize to a team of researchers for a mantle convection model. The original research paper can be found here.
Acknowledgments: The maps were produced using the cartopy plotting package (Met Office, 2010) with data from the following sources: Global topography and bathymetry from ETOPO1 [1], plate motions and strain rates from the global strain rate project [5], and satellite images from Mapbox (mapbox.com). Seismic data was obtained using the obspy package [2].
Further Reading Many of the papers which first described the theory of plate tectonics are extremely accessible. Wilson’s 1963 paper [11] on continental drift and Heirtzler’s 1968 paper [4] on sea-floor spreading bracket the period of major discoveries. A review of computational plate tectonics [7] is written for a broad audience and describes the methods discussed here with examples. Applications of adaptive mesh refinement to a solid Earth global circulation model with emergent plate boundaries are described in [10]. More background on Figure 3 and the movie can be found in [9]. To learn more about the large-scale interaction between mantle circulation and the surface processes of erosion, sediment transport, and deposition, see [3].
References [1] Amante, C. & Eakins, B.W. (2009). ETOPO1 1 Arc-Minute Global Relief Model: Procedures, Data Sources and Analysis. NOAA Technical Memorandum NESDIS NGDC-24. National Geophysical Data Center, NOAA. doi:10.7289/V5C8276M.
[2] Beyreuther, M., Barsch, R., Krischer, L., Megies, T., Behr, Y., & Wassermann, J. (2010). ObsPy: A Python Toolbox for Seismology.
Seismological Research Letters, 81(3), 530–533. doi:10.1785/gssrl.81.3.530.
[3] Braun, J. (2010). The many surface expressions of mantle dynamics.
Nature Geoscience, 3(12), 825–833. doi:10.1038/ngeo1020.
[4] Heirtzler, J. R. (1968). Sea-Floor Spreading.
Scientific American, 219(6), 60–70.
[5] Kreemer, C., Holt, W. E., & Haines, A. J. (2003). An integrated global model of present-day plate motions and plate boundary deformation.
Geophysical Journal International, 154, 8–34.
[6] Met Office. (2010-2015). Cartopy: a cartographic python library with a matplotlib interface. Exeter, Devon. http://scitools.org.uk/cartopy.
[7] Moresi, L. N., Gurnis, M., Zhong, S., & Zhong, S. J. (2000). Plate tectonics and convection in the Earth’s mantle: Toward a numerical simulation.
Computing in Science & Engineering, 2(22). http://dx.doi.org/10.1109/5992.841793. [8] Moresi, L. N., Quenette, S., Lemiale, V., Mériaux, C., Appelbe, B., & Muhlhaus, H. B. (2007). Computational approaches to studying non-linear dynamics of the crust and mantle. Physics of the Earth and Planetary Interiors, 163(1-4), 69–82. doi:10.1016/j.pepi.2007.06.009. [9] Moresi, L. N., Betts, P. G., Miller, M. S., & Cayley R. A. (2014). Dynamics of continental accretion. Nature, 508(7495), 245–248. doi:10.1038/nature13033. [10] Stadler, G., Gurnis, M., Burstedde, C., Wilcox, L. C., Alisic, L., & Ghattas, O. (2010). The Dynamics of Plate Tectonics and Mantle Flow: From Local to Global Scales. Science, 329(5995), 1033–1038. [11] Wilson, J. T. (1963). Continental drift. Scientific American, 208(4), 86–100.
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Unit Normal and Unit Binormal Vectors to a Space Curve
We have already looked at an important class of vectors known as unit tangent vectors denoted $\hat{T}(t)$. If $\vec{r}(t) = (x(t), y(t), z(t))$ is a vector-valued function for $t \in [a, b]$ that is differentiable, then the unit tangent vector at $t$ denoted $\hat{T}(t)$ represents the vector tangent to the point corresponding to $t$ and with length/magnitude $1$. We are now going to look at two other important vectors for space curves in $\mathbb{R}^3$.
Unit Normal Vectors
Definition: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function that represents the smooth curve $C$ for $t \in [a, b]$. Then for $\kappa (t) \neq 0$, the Unit Normal Vector (also known as the Unit Principle Normal Vector) denoted $\hat{N}(t) = \frac{\hat{T'}(t)}{\| \hat{T'}(t) \|}$ is a vector that is perpendicular to $\hat{T}(t)$ and has unit length/magnitude $1$.
We note that $\hat{N}(t) \perp \hat{T}(t)$ since $\| \hat{T}(t) \| = 1$, and so by one of the theorems on the Derivative Rules for Vector-Valued Functions page we have that $\hat{T}(t) \perp \hat{T'}(t)$ which implies that $\hat{T}(t) \perp \hat{N}(t)$. Geometrically, the unit normal vector at $t$ is a vector that is perpendicular to $\hat{T}(t)$ and points in the direction to which the curve $C$ is bending as illustrated in the following image.
Often times it can be extremely tedious to calculate unit normal vectors due to the frequent appearance of large numbers of terms and a radicals in the denominators that need differentiation.
Unit Binormal Vectors
We will now look at another important set of vectors known as unit binormal vectors.
Definition: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function that represents the smooth curve $C$ for $t \in [a, b]$. Then for $\kappa (t) \neq 0$ Unit Binormal Vector denoted $\hat{B}(t) = \hat{T}(t) \times \hat{N}(t)$, that is $\hat{B}(t)$ is the cross product of $\hat{T}(t)$ and $\hat{N}(t)$ and is perpendicular to both and has length/magnitude $1$.
The following diagram represents the unit binormal vectors for arbitrary points on a curve.
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i have the following question: let $\Omega$ an open in $\mathbb{R}^n$, and let $T \in \mathcal{D}'(\Omega)$.
We suppose that there exists an positive constant $C$ such that $$ \forall \varphi \in \mathcal{D}(\Omega): |\langle T,\varphi\rangle| \leq C \sqrt{ \displaystyle\int_{\Omega} |\varphi(x)|^2 dx } $$
The question is to prouve that there exists $f \in L^2(\Omega)$ such that $$ \forall \varphi \in \mathcal{D}(\Omega): \langle T,\varphi\rangle = \displaystyle\int_{\Omega} f(x) \varphi(x) dx. $$
My problem is that i have no idea for an probable solution. Can you give me somes indocations please.
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Electronic Journal of Probability Electron. J. Probab. Volume 24 (2019), paper no. 16, 71 pp. Exceedingly large deviations of the totally asymmetric exclusion process Abstract
Consider the Totally Asymmetric Simple Exclusion Process (TASEP) on the integer lattice $ \mathbb{Z} $. We study the functional Large Deviations of the integrated current $ \mathsf{h} (t,x) $ under the hyperbolic scaling of space and time by $ N $, i.e., $ \mathsf{h} _{N}(t,\xi ) := \frac{1} {N}\mathsf{h} (Nt,N\xi ) $. As hinted by the asymmetry in the upper- and lower-tail large deviations of the exponential Last Passage Percolation, the TASEP exhibits two types of deviations. One type of deviations occur with probability $ \exp (-O(N)) $, referred to as speed-$ N $; while the other with probability $ \exp (-O(N^{2})) $, referred to as speed-$ N^2 $. In this work we study the speed-$ N^2 $ functional Large Deviation Principle (LDP) of the TASEP, and establish (non-matching) large deviation upper and lower bounds.
Article information Source Electron. J. Probab., Volume 24 (2019), paper no. 16, 71 pp. Dates Received: 29 June 2018 Accepted: 9 February 2019 First available in Project Euclid: 22 February 2019 Permanent link to this document https://projecteuclid.org/euclid.ejp/1550826099 Digital Object Identifier doi:10.1214/19-EJP278 Subjects Primary: 60F10: Large deviations Secondary: 82C22: Interacting particle systems [See also 60K35] Citation
Olla, Stefano; Tsai, Li-Cheng. Exceedingly large deviations of the totally asymmetric exclusion process. Electron. J. Probab. 24 (2019), paper no. 16, 71 pp. doi:10.1214/19-EJP278. https://projecteuclid.org/euclid.ejp/1550826099
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$\vec{s}=\vec{u}t+\frac{1}{2}\vec{a}t^2$
The above relation can be used for $x$-axis or y-axis separately, you just have to keep the vectors in your mind. This relation can be broken down into two components :
$s_x \hat{i} + s_y \hat{j}=(u\cos{\theta} \; \hat{i} + u\sin{\theta}\; \hat{j}) t + \frac{1}{2}(0\;\hat{i}+g\;(-\hat{j}))t^2$.
Things will be more clear if we separately compare the displacements along the two axes.
For $x$-axis :
$s_x =u\cos{\theta}\;t$ $\tag1$
The direction of all the vectors is the same and the projectile will only travel in a particular direction $(+\hat{i})$, so we can neglect the unit vectors and focus on the magnitude of the vectors. This is only for $x$-axis.
There will be no change in the horizontal component of the velocity, since there is no acceleration along $x$-axis to bring about that change. You increase the time, the displacement along $x$-axis will keep on increasing.
For $y$-axis :
$s_y.\hat{j}=u\sin{\theta}\;(\hat{j}) t-\frac{1}{2}g\;(\hat{j})t^2$. $\tag2$
In case of $y$-axis we have to consider the direction as well as the magnitude of the vectors, A projectile passes through the same height twice (except maximum height).
Range :
Range for a projectile is defined as the horizontal distance covered by the projectile in that time period (i.e., time of flight) in which the projectile remains in air (roughly speaking).
So, if you throw a projectile from the ground, the time of flight will be equal to the time that it takes for the projectile to attain the maximum height and again come back to the ground $(s_y=0)$.
And in this time, the horizontal distance covered by the projectile will be its range.
This time can only be found out by equation $(2)$, since we are considering displacement along $y$-axis to define the time of flight.
$s_y$ is $0$ at two instants, the initial instant when the projectile was thrown and the final instant when the projectile reaches the ground again.
We are interested only in the second case.
So, $T=\frac{2u\sin{\theta}}{g}$.
To find the range, you only have to put this value of time in equation $(1)$.
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Summary: Possibilities for the Solution Set of a System of Linear Equations Problem 288
In this post, we summarize theorems about the possibilities for the solution set of a system of linear equations and solve the following problems.
Determine all possibilities for the solution set of the system of linear equations described below.
(a) A homogeneous system of $3$ equations in $5$ unknowns. (b) A homogeneous system of $5$ equations in $4$ unknowns. (c) A system of $5$ equations in $4$ unknowns. (d) A system of $2$ equations in $3$ unknowns that has $x_1=1, x_2=-5, x_3=0$ as a solution. (e) A homogeneous system of $4$ equations in $4$ unknowns. (f) A homogeneous system of $3$ equations in $4$ unknowns. (g) A homogeneous system that has $x_1=3, x_2=-2, x_3=1$ as a solution. (h) A homogeneous system of $5$ equations in $3$ unknowns and the rank of the system is $3$. (i) A system of $3$ equations in $2$ unknowns and the rank of the system is $2$. (j) A homogeneous system of $4$ equations in $3$ unknowns and the rank of the system is $2$. Add to solve later
Contents
Problem 288 the possibilities for the solution set of a system of linear equations The case $m \geq n$? Solutions. (a) A homogeneous system of $3$ equations in $5$ unknowns. (b) A homogeneous system of $5$ equations in $4$ unknowns. (c) A system of $5$ equations in $4$ unknowns. (d) A system of $2$ equations in $3$ unknowns that has $x_1=1, x_2=-5, x_3=0$ as a solution. (e) A homogeneous system of $4$ equations in $4$ unknowns. (f) A homogeneous system of $3$ equations in $4$ unknowns. (g) A homogeneous system that has $x_1=3, x_2=-2, x_3=1$ as a solution. (h) A homogeneous system of $5$ equations in $3$ unknowns and the rank of the system is $3$. (i) A system of $3$ equations in $2$ unknowns and the rank of the system is $2$. (j) A homogeneous system of $4$ equations in $3$ unknowns and the rank of the system is $2$. the possibilities for the solution set of a system of linear equations
An $m\times n$
system of linear equations is \begin{align*} \tag{*} a_{1 1} x_1+a_{1 2}x_2+\cdots+a_{1 n}x_n& =b_1 \\ a_{2 1} x_1+a_{2 2}x_2+\cdots+a_{2 n}x_n& =b_2 \\ a_{3 1} x_1+a_{3 2}x_2+\cdots+a_{3 n}x_n& =b_3 \\ \vdots \qquad \qquad \cdots\qquad \qquad &\vdots \\ a_{m 1} x_1+a_{m 2}x_2+\cdots+a_{m n}x_n& =b_m, \end{align*} where $x_1, x_2, \dots, x_n$ are unknowns (variables) and $a_{i j}, b_k$ are numbers. Thus an $m\times n$ system of linear equations consists of $m$ equations and $n$ unknowns $x_1, x_2, \dots, x_n$. A system of linear equations is called homogeneous if the constants $b_1, b_2, \dots, b_m$ are all zero. Namely, a homogeneous system is \begin{align*} a_{1 1} x_1+a_{1 2}x_2+\cdots+a_{1 n}x_n& =0 \\ a_{2 1} x_1+a_{2 2}x_2+\cdots+a_{2 n}x_n& =0 \\ a_{3 1} x_1+a_{3 2}x_2+\cdots+a_{3 n}x_n& =0 \\ \vdots \qquad \qquad \cdots\qquad \qquad &\vdots \\ a_{m 1} x_1+a_{m 2}x_2+\cdots+a_{m n}x_n& =0. \end{align*} A solution of the system (*) is a sequence of numbers $s_1, s_2, \dots, s_n$ such that the substitution $x_1=s_1, x_2=s_2, \dots, x_n=s_n$ satisfies all the $m$ equations in the system (*). We sometimes use the vector notation and say \[\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}=\begin{bmatrix} s_1 \\ s_2 \\ \vdots \\ s_n \end{bmatrix}\] is a solution of the system. For example, every homogeneous system has the zero solution $x_1=0, x_2=0, \dots, x_n=0$, or \[\mathbf{x}=\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}.\] Here we summarize several theorems concerning with the possibilities for the number of solutions of a system of linear equations.
We say a system is
consistent if the system has at least one solution. A system is called inconsistent if the system has no solutions at all.
Thus, for example, if we find two distinct solutions for a system, then it follows from the theorem that there are infinitely many solutions for the system.
Next, since a homogeneous system has the zero solution, it is always consistent. Thus:
Let us refine these theorems. Suppose that an $m\times n$ system of linear equations is given. That is, there are $m$ linear equations and $n$ unknowns.
Thus, there are only two possibilities when $m < n$: No solution or infinitely many solutions.
This is obtained by noting that a homogeneous system always has the zero solution, hence consistent. By the previous theorem, the only possibility is infinitely many solutions.
Summary 1 The case $m \geq n$?
What happens when $m \geq n$?
In general, when the number of equations is greater than or equal to the number of unknowns, we cannot narrow down the possibilities. We need more information about the system. The key word is the rank of the system. For a given system (*), let $A$ be the coefficient matrix and let $\mathbf{b}$ be the constant term vector. Then we define the rank of the system to be the rank of the augmented matrix $[A\mid \mathbf{b}]$. Recall that the rank is defined as follows. We first reduce the matrix $[A\mid \mathbf{b}]$ to a matrix $[A’\mid \mathbf{b’}]$ in (reduced) row echelon form by elementary row operations. Then the rank of $[A\mid \mathbf{b}]$ is the number of nonzero rows in the matrix $[A’\mid \mathbf{b’}]$.
A free variable means an unknown that can be assigned arbitrary values. It follows that if a system has a free variable, then there are infinitely many solutions.
Caution: the theorem assumes that a given system is consistent. Summary 2 Problems and solutions
Determine all possibilities for the solution set of the system of linear equations described below.
(a) A homogeneous system of $3$ equations in $5$ unknowns. (b) A homogeneous system of $5$ equations in $4$ unknowns. (c) A system of $5$ equations in $4$ unknowns. (d) A system of $2$ equations in $3$ unknowns that has $x_1=1, x_2=-5, x_3=0$ as a solution. (e) A homogeneous system of $4$ equations in $4$ unknowns. (f) A homogeneous system of $3$ equations in $4$ unknowns. (g) A homogeneous system that has $x_1=3, x_2=-2, x_3=1$ as a solution. (h) A homogeneous system of $5$ equations in $3$ unknowns and the rank of the system is $3$. (i) A system of $3$ equations in $2$ unknowns and the rank of the system is $2$. (j) A homogeneous system of $4$ equations in $3$ unknowns and the rank of the system is $2$. Solutions.
In the solution, $m$ denotes the number of equations and $n$ denotes the number of unknowns in the given system.
(a) A homogeneous system of $3$ equations in $5$ unknowns.
Since the system is homogeneous, it has the zero solution, hence consistent. Since there are more unknowns than equations, there are infinitely many solutions.
(b) A homogeneous system of $5$ equations in $4$ unknowns.
Since the system is homogeneous, it has the zero solution. Since there are more equations than unknowns, we cannot determine further.
Thus the possibilities are either a unique solution or infinitely many solution. (If the rank $r$ of the system is $4$, then a unique solution. If $r<4$, then there are infinitely many solutions.) (c) A system of $5$ equations in $4$ unknowns.
Since $m > n$, we can only say that the possibilities are no solution, a unique solution, or infinitely many solution.
(d) A system of $2$ equations in $3$ unknowns that has $x_1=1, x_2=-5, x_3=0$ as a solution.
Since $m < n$, the system is either inconsistent or has infinitely many solutions. Since $x_1=1, x_2=-5, x_3=0$ is a solution of the system, the system is not inconsistent. Thus the only possibility is infinitely many solutions.
(e) A homogeneous system of $4$ equations in $4$ unknowns.
Since $m=n$, this tells nothing. But since the system is homogeneous it has the zero solution, hence consistent. The possibilities are either a unique solution or infinitely many solutions.
(f) A homogeneous system of $3$ equations in $4$ unknowns.
Since $m < n$, the system has no solution or infinitely many solutions. But a homogeneous system is always consistent. Thus, the only possibility is infinitely many solutions.
(g) A homogeneous system that has $x_1=3, x_2=-2, x_3=1$ as a solution.
The possibilities for the solution set for any homogeneous system is either a unique solution or infinitely many solutions. Since the homogeneous system has the zero solution and $x_1=3, x_2=-2, x_3=1$ is another solution, it has at least two distinct solution. Thus the only possibility is infinitely many solutions.
(h) A homogeneous system of $5$ equations in $3$ unknowns and the rank of the system is $3$.
A homogeneous system is always consistent. Since the rank $r$ of the system and the number $n$ of unknowns are equal, the only possibility is the zero solution (and the zero solution is a unique solution).
(i) A system of $3$ equations in $2$ unknowns and the rank of the system is $2$.
We don’t know whether the system is consistent or not.
If it is consistent, then since the rank $r$ and the number of unknowns are the same, the system has a unique solution. Thus the possibilities are either inconsistent or a unique solution.
Before talking about the rank, we need to discuss whether the system is inconsistent or not. For example, consider the following $3\times 2$ system
\begin{align*} x_1+x_2&=1\\ 2x_1+2x_2&=3\\ 3x_1+3x_2&=3. \end{align*} Then the augmented matrix is \[\left[\begin{array}{rr|r} 1 & 1 & 1 \\ 2 &2 &3 \\ 3 & 3 & 3 \end{array}\right].\] We reduce this by elementary row operations as follows. \begin{align*} \left[\begin{array}{rr|r} 1 & 1 & 1 \\ 2 &2 &3 \\ 3 & 3 & 3 \end{array}\right] \xrightarrow{\substack{R_2-2R_1\\R_3-3R_1}} \left[\begin{array}{rr|r} 1 & 1 & 1 \\ 0 & 0 &1 \\ 0 & 0 &0 \end{array}\right]. \end{align*} The last matrix is in echelon form and it has two nonzero rows. Thus, the rank of the system is $2$. However, the second row means that we have $0=1$. Hence the system is inconsistent.
(j) A homogeneous system of $4$ equations in $3$ unknowns and the rank of the system is $2$.
A homogeneous system is consistent. The rank is $r=2$ and the number of variables is $n=3$. Hence there is $n-r=1$ free variable. Thus there are infinitely many solutions.
Add to solve later
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The most elementary reason is that the Dirac field Hamiltonian is bounded below only when you use anticommutation relations on the creation/annihilation operators instead of commutators. A free quantum field theory with energy unbounded below has no stable vacuum.
It is easiest to demonstrate this in two dimensions, where there are no polarization issues.
Instructive 2d example
In two dimensions (one space one time), there is a nice dimensionally reduced analog, which is the right-moving (necessarily massless) Majorana-Weyl Fermion (the argument also works with 2d Dirac fermions with two components, but this is the simplest case). This is a single component field $\psi$ which obeys the equation of motion
$$ (\partial_t -\partial_x) \psi = 0 $$
This simple equation is derived from the 2d Dirac equation using the (real convention, explicitly real) 2d Dirac matrices (0,1;-1,0) an (0,1;1,0), which are $\gamma_0 = \sigma_x$ and $\gamma_1 = i\sigma_y$. They square to 1 and -1 respectively, and they anticommute, so they reproduce the 1+1 dimensional metric tensor. The $\gamma_5$ analog, which I'll call $\Gamma$ to accommodate different dimensions, is diagonal in this explicit representation, and $\Gamma=\sigma_z$.
The two eigenvectors of $\Gamma$ propagate independently by the 2d massless equation of motion
$$ \gamma_i \partial_i \psi = 0 $$
And further, because the $\gamma$ matrices are real, this is a Majorana representation (most physicists write the dirac equation with an i factor in front of the derivative, so that the Dirac matrices for a Majorana representation are purely imaginary. I'm using a mathematician's convention for this, because I like the equations of motion to be real. Others like the k-space propagator to not have factors of i in the k part. Unfortunately, physicists never settled on a unique sensible convention--- everyone has their own preferred way to write Dirac matrices). So it is sensible in the equation of motion to restrict $\psi$ to be Hermitian, since its Hermitian conjugate obeys the exact same equation.
So that the field has a k decomposition
$$ \psi(x) = \int a_k e^{ikx - ikt }dk$$
And the reality condition (Hermiticity) tells you that $a^{\dagger}(-k) = a(k)$ (one should say that the normalization of the $a$ operators expansion is not completely conceptually trivial--- the $a$'s are both relativistically and nonrelativistically normalized, because the spinor polarization $\sqrt{w}$ factor cancels the mass-shell hyperbola factor, so that the dk integration is not weighted by anything, it's just the normal calculus integral with uniform measure)
An operator with definite frequency, which (Heisenberg picture) evolves in time according to
$$ \partial_t O = i\omega O$$
Has the property that it is a raising operator--- acting with this operator adds $\omega$ to the energy. If $\omega$ is negative, $a$ is an annihilation operator. The condition that the vacuum is stable says that all the annihilation operators give 0 when acting on the vacuum state.
But notice that the frequency in the expansion of $\psi$ changes sign at $k=0$. This came from the linearity of the Dirac Hamiltonian in the momenta. It means that the operator $a_k$ acts to raise the energy for k>0, but acts to lower the energy for $k<0$. This means that the $k>0$ operators create, and the $k<0$ operators annihilate, so that the right way to $a^{\dagger}(-k)$ are creation operators, while the $k<0$ operators are annihilation operators.
The energy operator counts the number of particles of momentum k, and multiplies by their energy:
$$ H = \int_{k>0} k a^{\dagger}(k) a(k) dk $$
And this is manifestly not a local operator, it is defined only integrated over k>0. To make it a local operator, you need to extend the integration to all k, but then the negative k and positive k contributions have opposite sign, and they need to be equal. To arrange this, you must take anticommutation relations
$$ \{ a^{\dagger}(k),a(k)\} = i\delta(k-k') $$
And then
$$ H = {1\over 2} \int k a^{\dagger}(k) a(k) = \int \psi(x) i \partial_x \psi(x) dx $$
Note that this looks like it is a perfect derivative, and it would be if $\psi$ weren't anticommuting quantity. For anticommuting quantities,
$$ \partial_x \psi^2 = \psi \partial_x \psi + \partial_x \psi \psi $$
Which is zero, because of the anticommutation.
Deeper reasons
Although this looks like an accidental property, that the energy was negative without anticommutators, it is not. The deeper reason is explained with Euclidean field theory using a Feynman-Schwinger formalism, but this requires understanding of the Euclidean and path integral versions of anticommuting fields, which requires being comfortable with anticommuting quantities, which requires a motivation. So it is best to learn the shallow reason first.This post imported from StackExchange Physics at 2014-03-22 16:49 (UCT), posted by SE-user Ron Maimon
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Global weak solution to the quantum Navier-Stokes-Landau-Lifshitz equations with density-dependent viscosity
1.
School of Mathematics and Information Science, Guangzhou University, Guangzhou, 510006, China
2.
Institute of Applied Physics and Computational Mathematics, China Academy of Engineering Physics, Beijing, 100088, China
In this paper we investigate the global existence of the weak solutions to the quantum Navier-Stokes-Landau-Lifshitz equations with density dependent viscosity in two dimensional case. We research the model with singular pressure and the dispersive term. The main technique is using the uniform energy estimates and B-D entropy estimates to prove the convergence of the solutions to the approximate system. We also use some convergent theorems in Sobolev space.
Keywords:Navier-Stokes-Landau-Lifshitz equations, global weak solutions, energy estimate, B-D entropy estimate. Mathematics Subject Classification:Primary: 35A01, 35D30, 35M31, 35Q40; Secondary: 76N10. Citation:Guangwu Wang, Boling Guo. Global weak solution to the quantum Navier-Stokes-Landau-Lifshitz equations with density-dependent viscosity. Discrete & Continuous Dynamical Systems - B, 2019, 24 (11) : 6141-6166. doi: 10.3934/dcdsb.2019133
References:
[1]
R. A. Adams and J. F. Fournier,
[2]
A. I. Akhiezer, V. G. Yakhtar and S. V. Peletminskii,
[3]
F. Alouges and A. Soyeur,
On global weak solutions for Landau-Lifshitz equations: Existence and nonuniqueness,
[4]
P. Antonelli and S. Spirito,
Global existence of finite energy weak solutions of quantum Navier-Stokes equations,
[5]
P. Antonelli and S. Spirito, On the compactness of finite energy weak solutions to the quantum Navier-Stokes equations,
[6]
D. Bresch and B. Desjardins,
Existence of global weak solutions for 2D viscous shallow water equations and convergence to the quasi-geostrophic model,
[7]
D. Bresch and B. Desjardins,
On the construction of approximate solutions for the 2D viscous shallow water model and for compressible Navier-Stokes models,
[8]
D. Bresch, B. Desjardins and C. K. Lin,
On some compressible fluid models: Korteweg, lubrication, and shallow water systems,
[9]
D. Bresh, B. Desjardins and E. Zatorska,
Two-velosity hydrodynamics in fluid mechanics: Part Ⅱ Existence of global $\kappa$-entropy solutions to the compressible Navier-Stokes system with degenerate viscosities,
[10] [11] [12] [13] [14]
E. Feireisl,
[15] [16] [17] [18]
B. L. Guo and G. W. Wang,
Global finite energy weak solution to the viscous quantum Navier-Stokes-Landau-Lifshitz-Maxwell equation in 2-dimension,
[19]
B. Haspot, Existence of global strong solution for the compressible Navier-Stokes equations with degenerate viscosity coefficients in 1D,
[20] [21]
D. Hoff,
Global well-posedness of the cauchy problem for the Navier-Stokes equations of nonisentropic flow with discontinuous initial data,
[22]
D. Hoff,
Global solutions of the equations of one-dimensional, compressible flow with large data and forces, and with differing end states,
[23] [24] [25]
Z. Lei, D. Li and X. Y. Zhang,
Remarks of global wellposedness of liquid crystal flows and heat flow of harmonic maps in two dimensions,
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[28] [29]
A. Mellet and A. Vasseur,
Existence and uniqueness of global strong solutions for one-dimensional compressible Navier-Stokes equations,
[30] [31]
R. Moser, Partial regularity for the Landau-Lifshitz equation in small dimensions, MPI (Leipzig) preprint, 2002.Google Scholar
[32] [33]
A. Novotný and I. Straškraba,
[34] [35] [36] [37] [38] [39]
G. W. Wang and B. L. Guo,
Existence and uniqueness of the weak solution to the incompressible Navier-Stokes-Landau-Lifshitz model in 2-dimension,
[40] [41] [42] [43]
Y. L. Zhou, H. S. Sun and B. L. Guo,
On the solvability of the initial value problem for the quasilinear degenerate parabolic system: $\vec{Z}_t=\vec{Z}\times \vec{Z}_xx+\vec{f}(x, t, \vec{Z})$,
[44] [45] [46]
Y. L. Zhou, H. S. Sun and B. L. Guo,
The weak solution of homogeneous boundary value problem for the system of ferromagnetic chain with several variables,
[47]
Y. L. Zhou, H. S. Sun and B. L. Guo,
Some boundary problems of the spin system and the system of ferro magnetic chain Ⅰ: Nonlinear boundary problems,
[48]
Y. L. Zhou, H. S. Sun and B. L. Guo,
Some boundary problems of the spin system and the system of ferromagnetic chain Ⅱ: Mixed problems and others,
[49]
show all references
References:
[1]
R. A. Adams and J. F. Fournier,
[2]
A. I. Akhiezer, V. G. Yakhtar and S. V. Peletminskii,
[3]
F. Alouges and A. Soyeur,
On global weak solutions for Landau-Lifshitz equations: Existence and nonuniqueness,
[4]
P. Antonelli and S. Spirito,
Global existence of finite energy weak solutions of quantum Navier-Stokes equations,
[5]
P. Antonelli and S. Spirito, On the compactness of finite energy weak solutions to the quantum Navier-Stokes equations,
[6]
D. Bresch and B. Desjardins,
Existence of global weak solutions for 2D viscous shallow water equations and convergence to the quasi-geostrophic model,
[7]
D. Bresch and B. Desjardins,
On the construction of approximate solutions for the 2D viscous shallow water model and for compressible Navier-Stokes models,
[8]
D. Bresch, B. Desjardins and C. K. Lin,
On some compressible fluid models: Korteweg, lubrication, and shallow water systems,
[9]
D. Bresh, B. Desjardins and E. Zatorska,
Two-velosity hydrodynamics in fluid mechanics: Part Ⅱ Existence of global $\kappa$-entropy solutions to the compressible Navier-Stokes system with degenerate viscosities,
[10] [11] [12] [13] [14]
E. Feireisl,
[15] [16] [17] [18]
B. L. Guo and G. W. Wang,
Global finite energy weak solution to the viscous quantum Navier-Stokes-Landau-Lifshitz-Maxwell equation in 2-dimension,
[19]
B. Haspot, Existence of global strong solution for the compressible Navier-Stokes equations with degenerate viscosity coefficients in 1D,
[20] [21]
D. Hoff,
Global well-posedness of the cauchy problem for the Navier-Stokes equations of nonisentropic flow with discontinuous initial data,
[22]
D. Hoff,
Global solutions of the equations of one-dimensional, compressible flow with large data and forces, and with differing end states,
[23] [24] [25]
Z. Lei, D. Li and X. Y. Zhang,
Remarks of global wellposedness of liquid crystal flows and heat flow of harmonic maps in two dimensions,
[26] [27]
P. L. Lions,
[28] [29]
A. Mellet and A. Vasseur,
Existence and uniqueness of global strong solutions for one-dimensional compressible Navier-Stokes equations,
[30] [31]
R. Moser, Partial regularity for the Landau-Lifshitz equation in small dimensions, MPI (Leipzig) preprint, 2002.Google Scholar
[32] [33]
A. Novotný and I. Straškraba,
[34] [35] [36] [37] [38] [39]
G. W. Wang and B. L. Guo,
Existence and uniqueness of the weak solution to the incompressible Navier-Stokes-Landau-Lifshitz model in 2-dimension,
[40] [41] [42] [43]
Y. L. Zhou, H. S. Sun and B. L. Guo,
On the solvability of the initial value problem for the quasilinear degenerate parabolic system: $\vec{Z}_t=\vec{Z}\times \vec{Z}_xx+\vec{f}(x, t, \vec{Z})$,
[44] [45] [46]
Y. L. Zhou, H. S. Sun and B. L. Guo,
The weak solution of homogeneous boundary value problem for the system of ferromagnetic chain with several variables,
[47]
Y. L. Zhou, H. S. Sun and B. L. Guo,
Some boundary problems of the spin system and the system of ferro magnetic chain Ⅰ: Nonlinear boundary problems,
[48]
Y. L. Zhou, H. S. Sun and B. L. Guo,
Some boundary problems of the spin system and the system of ferromagnetic chain Ⅱ: Mixed problems and others,
[49]
[1]
Shijin Ding, Boling Guo, Junyu Lin, Ming Zeng.
Global existence of weak solutions for Landau-Lifshitz-Maxwell equations.
[2]
Xueke Pu, Boling Guo, Jingjun Zhang.
Global weak solutions to the 1-D fractional Landau-Lifshitz equation.
[3]
Yong Yang, Bingsheng Zhang.
On the Kolmogorov entropy of the weak global attractor of 3D Navier-Stokes equations:Ⅰ.
[4]
Jingrui Wang, Keyan Wang.
Almost sure existence of global weak solutions to the 3D incompressible Navier-Stokes equation.
[5] [6] [7]
Daniel Coutand, J. Peirce, Steve Shkoller.
Global well-posedness of weak solutions for the Lagrangian averaged Navier-Stokes equations on bounded domains.
[8]
Daniel Pardo, José Valero, Ángel Giménez.
Global attractors for weak solutions of the three-dimensional Navier-Stokes equations with damping.
[9]
Jun Zhou.
Global existence and energy decay estimate of solutions for a class of nonlinear higher-order wave equation with general nonlinear dissipation and source term.
[10]
Peter E. Kloeden, José Valero.
The Kneser property of the weak solutions of the three dimensional Navier-Stokes equations.
[11]
Peter Anthony, Sergey Zelik.
Infinite-energy solutions for the Navier-Stokes equations in a strip revisited.
[12]
John W. Barrett, Endre Süli.
Existence of global weak solutions to Fokker-Planck and Navier-Stokes-Fokker-Planck equations in kinetic models of dilute polymers.
[13]
Fei Jiang, Song Jiang, Junpin Yin.
Global weak solutions to the two-dimensional Navier-Stokes equations of
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On the global existence of classical solutions for compressible Navier-Stokes equations with vacuum.
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[20]
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I'm following Carroll's Spacetime and Geometry, chapter on Linearized Gravity, pag. 282. He splits up the metric perturbation in scalar, vector and tensor components, writes the Einstein tensor with respect to these components and plugs into the Einstein field equation. He claims that the vector part $w_i$ and the two scalars ($\Phi$ being a component of the metric by itself, $\Psi$ being the trace of the tensor part) are not "propagating degrees of freedom", because they appear in equations like $$ \text{space derivatives (field of interest)} = f \text{ (other fields and their derivatives of any kind)} $$
For example ($\nabla^2 = \delta^{ij}\partial_i \partial_j$):
$$ \nabla^2\Psi=k T_{00} -\frac{1}{2}\partial_i\partial_j s^{ij} $$
This is an equation for $\Psi$ with no time derivatives; if we know what $T_{00}$ and $s_{ij}$ are doing at any time, we can determine what $\Psi$ must be (up to boundary conditions at spatial infinity). Thus, $\Psi$ is not by itself a propagating degree of freedom; it is determined by the energy-momentum tensor and the gravitational strain $s_{ij}$.
I do not understand this. What does it mean to be a
propagating degree of freedom? What would be the difference if we had time derivatives acting on $\Psi$?
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In the diagram above, an incident light ray
PO strikes at point O the interface between two media of refractive indexes n 1 and n 2. Part of the ray is reflected as ray OQ and part refracted as ray OS. The angles that the incident, reflected and refracted rays make to the normal of the interface are given as θ i, θ r and θ t, respectively.
The relationship between these angles is given by the law of reflection and Snell's law.
The fraction of the incident light that is reflected from the interface is given by the
reflection coefficient R, and the fraction refracted by the transmission coefficient T. The Fresnel equations may be used to calculate R and T in a given situation.
The calculations of
R and T depend on polarisation of the incident ray. If the light is polarised with the electric field of the light perpendicular to the plane of the diagram above ( s-polarised), the reflection coefficient is given by:
<math>R_s = \left\{ \frac{\sin (\theta_i - \theta_t)}{\sin (\theta_i + \theta_t)} \right\}^2</math>
where θ
t can be derived from θ i by Snell's law.
If the incident light is polarised in the plane of the diagram (
p-polarised), the R is given by:
<math>R_p = \left\{ \frac{\tan (\theta_i - \theta_t)}{\tan (\theta_i + \theta_t)} \right\}^2</math>.
The transmission coefficient in each case is given by
T s = 1 - R s and T p = 1 - R p.
If the incident light is unpolarised (containing an equal mix of
s- and p-polarisations), the reflection coefficient is R = ( R s + R p ) / 2 .
At one particular angle for a given
n 1 and n 2, the value of R p goes to zero and an p-polarised incident ray is purely refracted. This is known as Brewster's angle.
When moving from a more dense medium into a less dense one (i.e.
n 1 > n 2), above an incidence angle known as the critical angle all light is reflected and R s= R p=1. This phenomenon is known as total internal reflection.
When the light is at near-normal incidence to the interface(θ
i ≈ θ t ≈ 0) , the reflection coefficient is given by:
<math>R = R_s = R_p = \left\{ \frac{n_1 - n_2}{n_1 + n_2} \right\}^2</math>.
Note that reflection by a window is from the front side as well as the back side, and that the latter also includes light that goes back and forth a number of times between the two sides. The total is 2R/(1+R).
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The minimum mass for the spontaneous collapse of a gas cloud is the Jeans mass. For low temperatures and high densities, this jeans mass is greater than the mass of a typical star. What are the implications of the fact that the Jeans masses of most clouds are larger than the masses of most stars?
The Jeans mass is given by:
$$ M_J = \left(\frac{5kT}{Gm}\right)^{3/2} \sqrt{\frac{3}{4\pi\rho}} $$
The important point to note if that the Jean's mass is proportional to $T^{3/2}$ and $\rho^{-1/2}$. As the cloud collapses its density increases, but the temperature stays roughly constant. It may seem odd that the temperature stays constant - the reason is that the cloud does not scatter IR light so it can cool efficiently.
The result is that as the cloud collapses the Jean's mass decreases, so the cloud fragments into smaller pieces. These in turn collapse and increase in density, so the Jean's mass falls further and the cloud fragments further. This process only stops when the cloud is thick enough to start scattering light and its temperature starts to rise. At this point the cloud will form a protostar.
I'm not sure the details of the process are well understood (they are not understood by me!) so I don't know at what mass the fragmentation stops. Presumably it is well above a typical stellar mass since only a fraction of the cloud will end up in the star.
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I noted that there are two natural ways to aggregate $c_A$ and $c_B$ for someone who adheres to Probabilism, the principle that says that credences should be coherent. You might first fix up Adila's and Benoit's credences so that they are coherent, and then aggregate them using linear pooling -- let's call that
fix- then-pool. Or you might aggregate Adila's and Benoit's credences using linear pooling, and then fix up the pooled credences so that they are coherent -- let's call that pool- then-fix. And I noted that, for some natural ways of fixing up incoherent credences, fix-then-pool gives a different result from pool-then-fix. This, I claimed, creates a dilemma for the person doing the aggregating, since there seems to be no principled reason to favour either method.
How do we fix up incoherent credences? Well, a natural idea is to find the coherent credences that are closest to them and adopt those in their place. This obviously requires a measure of distance between two credence functions. In last week's post, I considered two:
Squared Euclidean Distance (SED)For two credence functions $c$, $c'$ defined on a set of propositions $X_1$, $\ldots$, $X_n$,$$SED(c, c') = \sum^n_{i=1} (c(X_i) - c'(X_i))^2$$ Generalized Kullback-Leibler Divergence (GKL)For two credence functions $c$, $c'$ defined on a set of propositions $X_1$, $\ldots$, $X_n$,$$GKL(c, c') = \sum^n_{i=1} c(X_i) \mathrm{log}\frac{c(X_i)}{c'(X_i)} - \sum^n_{i=1} c(X_i) + \sum^n_{i=1} c'(X_i)$$
If we use $SED$ when we are fixing incoherent credences -- that is, if we fix an incoherent credence function $c$ by adopting the coherent credence function $c^*$ for which $SED(c^*, c)$ is minimal -- then fix-then-pool gives
the same resultsas pool-then-fix.
If we use GKL when we are fixing incoherent credences -- that is, if we fix an incoherent credence function $c$ by adopting the coherent credence function $c^*$ for which $GKL(c^*, c)$ is minimal -- then fix-then-pool gives
different resultsfrom pool-then-fix.
Since last week's post, I've been reading this paper by Joel Predd, Daniel Osherson, Sanjeev Kulkarni, and Vincent Poor. They suggest that we pool and fix incoherent credences in one go using a method called the Coherent Aggregation Principle (CAP), formulated in this paper by Daniel Osherson and Moshe Vardi. In its original version, CAP says that we should aggregate Adila's and Benoit's credences by taking the coherent credence function $c$ such that the sum of the distance of $c$ from $c_A$ and the distance of $c$ from $c_B$ is minimized. That is,
CAPGiven a measure of distance $D$ between credence functions, we should pick that coherent credence function $c$ such that minimizes $D(c, c_A) + D(c, c_B)$.
As they note, if we take $SED$ to be our measure of distance, then this method generalizes the aggregation procedure on coherent credences that just takes straight averages of credences. That is, CAP entails unweighted linear pooling:
Unweighted Linear PoolingIf $c_A$ and $c_B$ are coherent, then the aggregation of $c_A$ and $c_B$ is $$\frac{1}{2} c_A + \frac{1}{2}c_B$$
We can generalize this result a little by taking a weighted sum of the distances, rather than the straight sum.
Weighted CAPGiven a measure of distance $D$ between credence functions, and given $0 \leq \alpha leq 1$, we should pick the coherent credence function $c$ that minimizes $\alpha D(c, c_A) + (1-\alpha)D(c, c_B)$.
If we take $SED$ to measure the distance between credence functions, then this method generalizes linear pooling. That is, Weighted CAP entails linear pooling:
Linear PoolingIf $c_A$ and $c_B$ are coherent, then the aggregation of $c_A$ and $c_B$ is $$\alpha c_A + (1-\alpha)c_B$$ for some $0 \leq \alpha \leq 1$.
What's more, when distance is measured by $SED$, Weighted CAP agrees with fix-then-pool and with pool-then-fix (providing the fixing is done using $SED$ as well). Thus, when we use $SED$, all of the methods for aggregating incoherent credences that we've considered agree. In particular, they all recommend the following credence in $X$: $$\frac{1}{2} + \frac{\alpha(c_A(X)-c_A(\overline{X})) + (1-\alpha)(c_B(X) - c_B(\overline{X}))}{2}$$
However, the story is not nearly so neat and tidy if we measure the distance between two credence functions using $GKL$. Here's the credence in $X$ recommended by fix-then-pool:$$\alpha \frac{c_A(X)}{c_A(X) + c_A(\overline{X})} + (1-\alpha)\frac{c_B(X)}{c_B(X) + c_B(\overline{X})}$$ Here's the credence in $X$ recommended by pool-then-fix: $$\frac{\alpha c_A(X) + (1-\alpha)c_B(X)}{\alpha (c_A(X) + c_A(\overline{X})) + (1-\alpha)(c_B(X) + c_B(\overline{X}))}$$ And here's the credence in $X$ recommended by Weighted CAP: $$\frac{c_A(X)^\alpha c_B(X)^{1-\alpha}}{c_A(X)^\alpha c_B(X)^{1-\alpha} + c_A(\overline{X})^\alpha c_B(\overline{X})^{1-\alpha}}$$ For many values of $\alpha$, $c_A(X)$, $c_A(\overline{X})$, $c_B(X)$, $c_B(\overline{X})$ these will give three distinct results.
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The most elementary reason is that the Dirac field Hamiltonian is bounded below only when you use anticommutation relations on the creation/annihilation operators instead of commutators. A free quantum field theory with energy unbounded below has no stable vacuum.
It is easiest to demonstrate this in two dimensions, where there are no polarization issues.
Instructive 2d example
In two dimensions (one space one time), there is a nice dimensionally reduced analog, which is the right-moving (necessarily massless) Majorana-Weyl Fermion (the argument also works with 2d Dirac fermions with two components, but this is the simplest case). This is a single component field $\psi$ which obeys the equation of motion
$$ (\partial_t -\partial_x) \psi = 0 $$
This simple equation is derived from the 2d Dirac equation using the (real convention, explicitly real) 2d Dirac matrices (0,1;-1,0) an (0,1;1,0), which are $\gamma_0 = \sigma_x$ and $\gamma_1 = i\sigma_y$. They square to 1 and -1 respectively, and they anticommute, so they reproduce the 1+1 dimensional metric tensor. The $\gamma_5$ analog, which I'll call $\Gamma$ to accommodate different dimensions, is diagonal in this explicit representation, and $\Gamma=\sigma_z$.
The two eigenvectors of $\Gamma$ propagate independently by the 2d massless equation of motion
$$ \gamma_i \partial_i \psi = 0 $$
And further, because the $\gamma$ matrices are real, this is a Majorana representation (most physicists write the dirac equation with an i factor in front of the derivative, so that the Dirac matrices for a Majorana representation are purely imaginary. I'm using a mathematician's convention for this, because I like the equations of motion to be real. Others like the k-space propagator to not have factors of i in the k part. Unfortunately, physicists never settled on a unique sensible convention--- everyone has their own preferred way to write Dirac matrices). So it is sensible in the equation of motion to restrict $\psi$ to be Hermitian, since its Hermitian conjugate obeys the exact same equation.
So that the field has a k decomposition
$$ \psi(x) = \int a_k e^{ikx - ikt }dk$$
And the reality condition (Hermiticity) tells you that $a^{\dagger}(-k) = a(k)$ (one should say that the normalization of the $a$ operators expansion is not completely conceptually trivial--- the $a$'s are both relativistically and nonrelativistically normalized, because the spinor polarization $\sqrt{w}$ factor cancels the mass-shell hyperbola factor, so that the dk integration is not weighted by anything, it's just the normal calculus integral with uniform measure)
An operator with definite frequency, which (Heisenberg picture) evolves in time according to
$$ \partial_t O = i\omega O$$
Has the property that it is a raising operator--- acting with this operator adds $\omega$ to the energy. If $\omega$ is negative, $a$ is an annihilation operator. The condition that the vacuum is stable says that all the annihilation operators give 0 when acting on the vacuum state.
But notice that the frequency in the expansion of $\psi$ changes sign at $k=0$. This came from the linearity of the Dirac Hamiltonian in the momenta. It means that the operator $a_k$ acts to raise the energy for k>0, but acts to lower the energy for $k<0$. This means that the $k>0$ operators create, and the $k<0$ operators annihilate, so that the right way to $a^{\dagger}(-k)$ are creation operators, while the $k<0$ operators are annihilation operators.
The energy operator counts the number of particles of momentum k, and multiplies by their energy:
$$ H = \int_{k>0} k a^{\dagger}(k) a(k) dk $$
And this is manifestly not a local operator, it is defined only integrated over k>0. To make it a local operator, you need to extend the integration to all k, but then the negative k and positive k contributions have opposite sign, and they need to be equal. To arrange this, you must take anticommutation relations
$$ \{ a^{\dagger}(k),a(k)\} = i\delta(k-k') $$
And then
$$ H = {1\over 2} \int k a^{\dagger}(k) a(k) = \int \psi(x) i \partial_x \psi(x) dx $$
Note that this looks like it is a perfect derivative, and it would be if $\psi$ weren't anticommuting quantity. For anticommuting quantities,
$$ \partial_x \psi^2 = \psi \partial_x \psi + \partial_x \psi \psi $$
Which is zero, because of the anticommutation.
Deeper reasons
Although this looks like an accidental property, that the energy was negative without anticommutators, it is not. The deeper reason is explained with Euclidean field theory using a Feynman-Schwinger formalism, but this requires understanding of the Euclidean and path integral versions of anticommuting fields, which requires being comfortable with anticommuting quantities, which requires a motivation. So it is best to learn the shallow reason first.This post imported from StackExchange Physics at 2014-03-22 16:49 (UCT), posted by SE-user Ron Maimon
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Take the limit as x approaches 9 of $\frac{3}{\sqrt{x}-3} - \frac{18}{x-9}$ Multiplying by the conjugate $\sqrt{x}+3$ on the left side to get a common denominator of $x-9$, then using L'Hôpital's rule and solving gets the answer 1/2 Multiplying the denominators together to get a common denominator then using L'Hôpital's rule and solving get me 3 on the numerator and 0 on the denominator. I've done this problem a few times now but I cannot seem to find my mistake.
I am getting $1/2$ in both cases.
1st method: Applying L'Hôpital Rule (I have eaten the steps involved and shown only 1st equation and last equation)
$$\lim_{x\to9} \left(\frac{3}{\sqrt{x}-3}-\frac{18}{x-9}\right)=\lim_{x\to9} \frac{3}{2\sqrt{x}}=\frac{1}{2}.$$
2nd method: Applying L'Hôpital Rule (I have again eaten the steps involved and shown only 1st equation and last equation)
$$\lim_{x\to9} \frac{3x-18\sqrt{x}+27}{x^{3/2}-3x-9\sqrt{x}+27}=\lim_{x\to9}\frac{1}{\sqrt{x}-1}=\frac{1}{2}.$$
I made a mistake, my numerator was actually 0 not 3, which would result in "0/0" again. I would then have to apply l'hôpital's rule again then solve. My bad
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I get so frustrated with modular arithmetic. It seems like every example I look at leaves steps out. I am trying to solve this problem:
Solve the linear congruence equations for x:
$x \equiv 2 \mod 7$
$x \equiv 1 \mod 3$
Ok, so I start
We know that 1st equation has a solution when $7 \mid (x-2)$. So there exists an integer k where $x = 2 + 7k$.
Ok, great. So I substitute into the 2nd equation:
$ 2+7k \equiv 1 \mod 3 \implies \\ 7k \equiv -1 \mod 3 \implies \\ 7k \equiv 2 \mod 3 $
Now I need to find an inverse of this last congruence. How do I do that? I know there is one solution because gcd(7,3) = 1. This is the step I'm having problems on. If I can get the solution to $7k \equiv 2 \mod 3$ into the form $k = a + bj$ where $a,b \in \mathbb{N}$ then I know how to solve it.
Thank you.
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