text stringlengths 256 16.4k |
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Tagged: normal subgroup If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Problem 575
Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.Add to solve later
Problem 470
Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer.
Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$. Then prove that $H$ is a normal subgroup of $G$.
(
Michigan State University, Abstract Algebra Qualifying Exam) Problem 332
Let $G=\GL(n, \R)$ be the
general linear group of degree $n$, that is, the group of all $n\times n$ invertible matrices. Consider the subset of $G$ defined by \[\SL(n, \R)=\{X\in \GL(n,\R) \mid \det(X)=1\}.\] Prove that $\SL(n, \R)$ is a subgroup of $G$. Furthermore, prove that $\SL(n,\R)$ is a normal subgroup of $G$. The subgroup $\SL(n,\R)$ is called special linear group |
Suppose $3$ points are drawn uniformly at random from $[0,1]$. Call them $x_1,x_2,x_3$ with $x_1\leq x_2\leq x_3$. I am interested in the distances between them.
Fix $0<a<b<1$. I want to compute the probability $P$ that $$x_2-x_1 < a, \ \ \ \ x_3 -x_2 < a,\ \ \ \ x_3-x_1 > b$$ In other words, the two pairs of adjacent points are $a$-close to each other, but the two outside points are $b$-far apart. Clearly this can't happen unless $b<2a$.
Since we have a uniform distribution, we can integrate the constant density $\equiv 1$ over the appropriate region in $[0,1]^3$. Now, $x_2$ must be within $a$ of $x_1$, and $x_3$ has to be at least $b$ greater than $x_1$ but not more than $a$ above $x_2$. This gives the following:
$$P = \int_0^{1-2a}\int_{x_1}^{x_1+a}\int_{x_1+b}^{x_2+a}\ dx_3\ dx_2\ dx_1$$
The problem: $x_{1}$ doesn't have to be less than $1-2a$; it should be able to be as large as $1-b$ (well technically, arbitrarily close to it but still less than). For instance, if $a=.2, b=.3$ then an acceptable set of points would be $.69,.85,1$. Clearly there are $2$ cases to consider, but I'm having a hard time explicitly differentiating the cases in order to write down two integrals. It seems the second case should look like
$$\int_{1-2a}^{1-b}\int_{x_1}^{x_1+a}\int_{x_1+b}^1\ dx_3\ dx_2\ dx_1$$
However I don't feel that this is correct. Specifically, it doesn't look like the above guarantees that $x_3-x_2<a$. Are these the only two cases to consider? What are the appropriate bounds in the second integral? Is there a better way to approach this problem? |
I want to simply the following without use of a calculator: $\displaystyle \frac{2 \tan64°}{1-\tan^264°}$
At first I thought that this is tangent double angle identity = $\tan(2\times 64°) $
hence $=\tan(128°) $
Is that all? Really?
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
I want to simply the following without use of a calculator: $\displaystyle \frac{2 \tan64°}{1-\tan^264°}$
At first I thought that this is tangent double angle identity = $\tan(2\times 64°) $
hence $=\tan(128°) $
Is that all? Really?
$\displaystyle Yup,\ that's\ it.$
Yes, you are correct, $$ \frac{2 \tanx}{1-\tan^2x} = \tan 2x. $$ The given problem is absolutely in the given format. Therefore, you directly use this principle to get the required solution that you have already got. |
Suppose that $|f(z)|\leq A+B|z|^M$ and that $f$ is entire. Show that for all coefficients $c_j$ with $M<j$ in its power series extansion are $0$.
Attampt:
$$ |f(z)|=\left|\sum_{k=0}^\infty c_kz^k\right| \leq A+B|z|^M \Rightarrow \\ \left|\sum_{k=0}^\infty c_kz^{k-M}\right|\leq{A\over |z|^M}+B\underset{z \to \infty}{\rightarrow}B $$ $g(z):=\sum_{k=0}^\infty c_kz^{k-M}$ is continuous and thus $|g|$ is bounded by some $Q$.
How can I finish the exercise? |
Given a set S of
n
sensors in the plane we consider the problem of establishing an ad hoc network from these sensors using directional antennae. We prove that for each given integer 1 ≤
k
≤ 5 there is a strongly connected spanner on the set of points so that each sensor uses at most
k
such directional antennae whose range differs from the optimal range by a multiplicative factor of at most
$2 \cdot \sin(\frac{\pi}{k+1})$
. Moreover, given a minimum spanning tree on the set of points the spanner can be constructed in additional
O
(
n
) time. In addition, we prove NP completeness results for
k
= 2 antennae. |
Why is the reduction $\textbf{SAT} \leq_P \textbf{3SAT}$ possible, but $\textbf{SAT} \leq_P \textbf{2SAT}$ not possible, given, that $\textbf{SAT}$ is $\textbf{NP}$-complete, $\textbf{2SAT} \in \textbf{NP}$ and $\textbf{3SAT} \in \textbf{NP}$?
I believe the asker is wanting to know specifically why the standard approach to reducing $\text{SAT}$ to $\text{3SAT}$ does not continue to extend to the $\text{2SAT}$ case. Here is a walkthrough as to why.
Conversion of a SAT Instance to a 3-SAT Instance:
For convenience, let us assume everything is in $\text{CNF}$, or conjunctive normal form. The $\text{SAT} \le_p \text{3SAT}$ proof works by introducing dummy variables to spread the actual variables over multiple clauses.
Illustration: Consider a single clause in boolean formula $(x_1 \lor x_2 \lor x_3 \lor x_4)$. The $\text{3SAT}$ reduction works by introducing a dummy variable to represent a disjunction of two literals in hopes that it reduces the clause's length by one. For example, $y = x_3\lor x_4$. The clause can be rewritten as:
$(x_1 \lor x_2 \lor y) \land (\bar y \lor x_3 \lor x_4)$
Notice that these clauses are now in a $\text{CNF}$ for $\text{3SAT}$. If this is repeated multiple times for each clause, the final $\text{CNF}$ formula will be at most polynomial in length of the original. This is why a polynomial time reduction would suffice.
Attempting to Apply the Method to Convert to 2-SAT:
If we tried to use the trick above with a clause of length 3, $(a\lor b\lor c)$, then we would not get the desired improvement. Let our dummy variable here be $y=b \lor c$.
The result: $(a \lor y) \land (\bar y \lor b \lor c)$
As you can see, it still has a clause of size 3. If we used the method again on the second clause, we would end up with the same problem while also adding an additional clause of length 2.
It could be said that the simplest reason for this is because $\land$ and $\lor$ are binary operators. You still need to apply these operators on pairs of variables, then include an additional dummy variable (resulting in a clause of length 3).
There have been attempts at getting around this that are successful, but the resulting sequence of clauses become exponential in length. Handling such a formula is outside the scope of $\text{P}$, which is why such a reduction would not be a polynomial time reduction.
First of all, "not possible" is only under the assumption that $P\neq NP$.
While it is true that $2SAT\in NP$, we also know that $2SAT\in P$. Thus, if indeed $SAT\leq_p 2SAT$, then $P=NP$.
The reduction $SAT\le_p 3SAT$ is possible since we know that $3SAT$ is $NP$-complete, and therefore there is a reduction from every problem in $NP$ to $3SAT$.
one way to conceptualize this: this can be seen as a case of a more general phenomenon where various problems are "simpler" for "small" fixed parameters of the problem. this happens with many NP complete problems but also outside of NP (eg with undecidable problems becoming decidable for small fixed parameters). see also 2SAT, wikipedia for the proof that 2SAT is in P.
another way/angle to understand this phenomenon is that the Tseitin transformation allows conversion of arbitrary circuits (including SAT itself) to 3SAT but some of the conversions require 3 variables (but none require more than 3).
yet another way to visualize this concept is with the SAT transition point where it is (theoretically/empirically) found that formulas of particular structure move from "hard-easy-hard" based on alteration of basic form, in particular (with SAT) the clause/variable ratio, but the transition point phenomenon is known to apply to all NP complete problems.[1] in this case the (average) number of variables in clauses is also a source of a transition point from hard to easy.
[1] Phase transition behavior Toby Walsh |
MathModePlugin
Add math formulas to TWiki topics using LaTeX markup language
Description
This plugin allows you to include mathematics in a TWiki page, with a format very similar to LaTeX. The external program
latex2html
is used to generate
gif
(or
png
) images from the math markup, and the image is then included in the page. The first time a particular expression is rendered, you will notice a lag as
latex2html
is being run on the server. Once rendered, the image is saved as an attached file for the page, so subsequent viewings will not require re-renders. When you remove a math expression from a page, its image is deleted.
Note that this plugin is called MathModePlugin
, not LaTeXPlugin, because the only piece of LaTeX implemented is rendering of images of mathematics.
Syntax Rules <latex [attr="value"]* >
formula </latex>
generates an image from the contained
formula
. In addition attribute-value pairs may be specified that are passed to the resulting
img
html tag. The only exeptions are the following attributes which take effect in the latex rendering pipeline:
size: the latex font size; possible values are tiny, scriptsize, footnotesize, small, normalsize, large, Large, LARGE, huge or Huge; defaults to %LATEXFONTSIZE%
color: the foreground color of the formula; defaults to %LATEXFGCOLOR%
bgcolor: the background color; defaults to %LATEXBGCOLOR%
The formula will be displayed using a
math
latex environment by default. If the formula contains a latex linebreak (
\\
) then a
multline
environment of amsmath is used instead. If the formula contains an alignment sequence (
& = &
) then an
eqnarray
environment is used.
Note that the old notation using
%$formula$%
and
%\[formula\]%
is still supported but are deprecated.
If you might want to recompute the images cached for the current page then append
?refresh=on
to its url, e.g. click
here
to refresh the formulas in the examples below.
Examples
The following will only display correctly if this plugin is installed and configured correctly.
<latex title="this is an example">
\int_{-\infty}^\infty e^{-\alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}
</latex>
<latex>
{\cal P} & = & \{f_1, f_2, \ldots, f_m\} \\
{\cal C} & = & \{c_1, c_2, \ldots, c_m\} \\
{\cal N} & = & \{n_1, n_2, \ldots, n_m\}
</latex>
<latex title="Calligraphics" color="orange">
\cal
A, B, C, D, E, F, G, H, I, J, K, L, M, \\
\cal
N, O, P, Q, R, S, T, U, V, W, X, Y, Z
</latex>
<latex>
\sum_{i_1, i_2, \ldots, i_n} \pi * i + \sigma
</latex>
This is
new inline test.
Greek letters
\alpha
\theta
\beta
\iota
\gamma
\kappa
\delta
\lambda
\epsilon
\mu
\zeta
\nu
\eta
\xi
Plugin Installation Instructions Download the ZIP file Unzip
in your twiki installation directory. Content:
MathModePlugin.zip
File: Description:
data/TWiki/MathModePlugin.txt
lib/TWiki/Plugins/MathModePlugin/Core.pm
lib/TWiki/Plugins/MathModePlugin.pm
pub/TWiki/MathModePlugin/latex2img This plugin makes use of three additional tools that are used to convert latex formulas to images. These are Make sure they are installed and check the paths to the programs
latex,
dvipng and
convert in the latex2img shiped with this plugin
Edit the file
<path-to-twiki>/pub/TWiki/MathModePlugin/latex2img accordingly and set execute permission for your webserver on it
Visit
configure in your TWiki installation, and enable the plugin in the {Plugins} section.
Troubleshooting If you get error like
"fmtutil: [some-dir]/latex.fmt does not exist", run
fmtutil-sys --all on your server to recreate all latex formatstyles.
If your generated image of the latex formula does not show up, then you probably have encoding issues. Look into the source of the <img>-tag in your page's source code. Non-ASCII characters in file names might cause troubles. Check the localization in the TWiki configure page. Configuration
There are a set of configuration variables that an be set in different places. All of the below variables can be set in your
LocalSite.cfg
file like this:
$TWiki::cfg{MathModePlugin}{<Name>} = <value>;
Some of the below variables can
only
be set this way, some of the may be overridden by defining the respective prefrence variable.
Name Preference Variable Default
LatexBGColor
%LATEXBGCOLOR% white
LatexFGColor
%LATEXFGCOLOR% black
LatexFontSize
%LATEXFONTSIZE% normalsize
ImageType
%LATEXIMAGETYPE% 'png'
LatexPreamble
%LATEXPREAMBLE% '\usepackage{latexsym}'
ScaleFactor
%LATEXSCALEFACTOR% 1.2 default background color default font size default text color extension of the image type; possible values are 'gif' and 'png' factor to scale images latex preamble to include additional packages (e.g. \usepackage{mathptmx} to change the math font) ; note, that the packages
amsmath and
color are loaded too as they are obligatory
length of the hash code. If you switch to a different hash function, you will likely have to change this
HashCodeLength
32
ImagePrefix
'_MathModePlugin_'
Latex2Img
'.../TWiki/MathModePlugin/latex2img' string to be prepended to any auto-generated image the script to convert a latex formula to an image Plugin Info |
It turns out that I was a over-optimistic. I don’t see a way to do it with Zorn’s lemma that doesn’t get unbearably complicated, so I’m going to go ahead and use transfinite recursion to construct $S$. I’ll do my best to make the explanation self-contained; in particular, I’ll avoid transfinite ordinals entirely. Fortunately, this particular recursion is quite straightforward and greatly resembles more familiar recursive constructions of ordinary sequences of things. Also, I’ll write it up for the $\Bbb R^3$ version of the problem: the mathematics is basically the same, but you can visualize what’s going on much better.
There is a set $S\subseteq\Bbb R^3$ such that $H\cap P$ is dense in $P$ for each plane $P$ in $\Bbb R^3$, but $H$ does not contain $3$ collinear points.
The first key observation is that there are $|\Bbb R|=\mathfrak{c}$ planes in $\Bbb R^3$. Thus, we can enumerate them as $\{P_\xi:\xi\in\Bbb R\}$. We will also need the fact that (assuming the axiom of choice) there is a well-ordering, $\preceq$, of $\Bbb R$ such that $|\{\xi\in\Bbb R:\xi\preceq\eta\}|<\mathfrak{c}$ for each $\eta\in\Bbb R$. We may further assume that $0$ is the first element in this well-ordering, i.e., that $0\preceq\xi$ for all $\xi\in\Bbb R$. (This is just a convenience, to make it easy to talk about the first element.)
Next, for each $\xi\in\Bbb R$ let $\mathscr{B}_\xi=\{B_{\xi,n}:n\in\Bbb N\}$ be a countable base for the relative topology of the plane $P_\xi$.
Finally, for each $A\subseteq\Bbb R^3$ let $L(A)$ be the union of the lines determined by pairs of points of $A$.
Suppose that $\eta\in\Bbb R$, and for each $\xi\prec\eta$ we have constructed a set $S_\xi\subseteq\Bbb R$ such that:
$S_\xi\setminus\bigcup_{\zeta\prec\xi}S_\zeta$ is countable;
no three distinct points of $S_\xi$ are collinear;
$S_\zeta\subseteq S_\xi$ for all $\zeta\preceq\xi$; and
$S_\xi\cap P_\xi$ is dense in $P_\xi$.
This is the recursion hypothesis; we’ll use it to construct a set $S_\eta$ that satisfies all four conditions with $\xi$ replaced by $\eta$. Note that it is vacuously satisfied when $\eta=0$, so the basis step will be no different from any later step.
Now I’ll construct $S_\eta$.
Let $T_0=\bigcup_{\xi\prec\eta}S_\xi$. Let $x,y$, and $z$ be distinct points of $T_0$. Then there are $\zeta,\theta,\xi\prec\eta$ such that $x\in S_\zeta$, $y\in S_\theta$, and $z\in S_\xi$. Without loss of generality $\zeta\preceq\theta\preceq\xi$, so $x,y,z\in S_\xi$ and by the recursion hypothesis are not collinear. Thus, no three distinct points of $T_0$ are collinear. If $T_0\cap P_\eta$ is dense in $P_\eta$, let $S_\eta=T_0$; clearly this satisfies the recursion hypothesis with $\xi$ replaced by $\eta$.
Now assume that $T\cap P_\eta$ is not dense in $P_\eta$. $T_0$ is the union of fewer than $\mathfrak{c}$ countable sets, which means that it’s the union of countably many sets, each of cardinality less than $\mathfrak{c}$. $\Bbb R$ is not the union of countably many sets of smaller cardinality, so $|T_0|<\mathfrak{c}$. (Here I’m using a basic fact about infinite cardinal arithmetic; I don’t see a way to avoid it.) It follows that pairs of points of $T_0$ generate fewer than $\mathfrak{c}$ lines, and hence that $L(T_0)\cap P_\eta$ is the union of fewer than $\mathfrak{c}$ points and lines.
We’ll now do an ordinary recursion over $\Bbb N$ to expand $T_0$ to the desired set $S_\eta$.
Suppose that $n\in\Bbb N$, and for each $k<n$ we’ve chosen a point $x_{\eta,k}\in B_{\eta,k}$ in such a way that no three points of $T_n=T_0\cup\{x_{\eta,k}:k<n\}$ are collinear. If $T_n\cap B_{\eta,n}\ne\varnothing$, let $x_{\eta,n}$ be any point of $T_n\cap B_{\eta,n}$. Otherwise, fix any point $p\in B_{\eta,n}$. $L(T_n)\cap P_\eta$ is the union of fewer than $\mathfrak{c}$ points and lines, and there are $\mathfrak{c}$ lines through $p$, so there is a line $L_0$ in $P_\eta$ through $p$ that intersects each line in $L(T_n)$ in at most one point. Thus, $|L_0\cap L(T_n)|<\mathfrak{c}$, while $|L_0\cap B_{\eta,n}|=\mathfrak{c}$, so we may choose a point $x_{\eta,n}\in (L_0\cap B_{\eta,n})\setminus L(T_n)$. Clearly no three distinct points of $T_{n+1}=T_n\cup\{x_{\eta,n}\}$ are collinear, so the recursive construction goes through. Then we set $S_\eta=\bigcup_{n\in\Bbb N}T_n=T_0\cup\{x_{\eta,n}:n\in\Bbb N\}$ and see that all four conditions of the main recursion are satisfied with $\xi$ replaced by $\eta$.
It’s a theorem of set theory that we can now conclude that we have family of set $\{S_\eta:\eta\in\Bbb R\}$ that satisfy the recursion hypothesis for each $\eta\in\Bbb R$. This is entirely analogous to the ordinary recursive construction of the points $x_{\eta,n}$ for each $n\in\Bbb N$, at the end of which we concluded that we had the whole set of them, $\{x_{\eta,n}:n\in\Bbb N\}$, to be lumped in with $T_0$ to form the set $S_\eta$.
Now let $S=\bigcup_{\xi\in\Bbb R}S_\xi$. The verification that no three distinct points of $S$ are collinear is essentially the same as the verification of the corresponding fact for $T_0$ above, and the fourth clause of the recursion hypothesis ensures that $S$ has dense intersection with each plane in $\Bbb R^3$. |
I understand how the theorem works but how would you prove that a sequence $f_n$ of functions on set $S \subset \mathbb{R}$ converges uniformly iff
$$\lim_{n\to\infty} \sup\{|f(x) - f_n(x) | : x \in S \}=0$$
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
It's actually trivial if you write everything down.
Let $\epsilon > 0$
Let $N \in \mathbb N$ such that $n \geq N \Rightarrow \forall x \in S, |f_n(x)-f(x)|\leq \epsilon$
Then (by the very definition of $\sup$), for any $n \geq N$ we have that $\sup\{|f(x) - f_n(x) | : x \in S \} \leq \epsilon$
Equivalently said, $\lim_{n\to\infty} \sup\{|f(x) - f_n(x) | : x \in S \}=0$
Let $\epsilon >0$
Let $N \in \mathbb N$ such that $n \geq N \Rightarrow \sup\{|f(x) - f_n(x) | : x \in S \} \leq \epsilon$.
Then for any $n \geq N$ and any $x \in S$, $|f(x) - f_n(x) | \leq \sup\{|f(x) - f_n(x) | : x \in S \} \leq \epsilon$ , which is the definition of uniform convergence. |
Assemble a formula using the numbers $3$, $4$ and $6$ in any order to make
. 108
You may use the operations;
$x + y$
$x - y$
$x \times y$
$x \div y$
$x!$
$\sqrt{x}$
$\sqrt[\leftroot{-2}\uproot{2}x]{y}$
$x^y$
Brackets to clarify order of operations "(",")" Concatenate two or more of the three digits you start with (concatenation of numbers from calculations is not permitted)
as long as all operands are either $3$, $4$ and $6$.
Note that double, triple, etc. factorials (n-druple-factorials) are
not allowed though factorials of factorials are fine, such as $((6-3)!)! = 6!$. |
6 0
I'm trying to solve a problem from the book of qualifier problems by Cahn, but I don't see how he got his solution.
The problem is about a particle constrained to move on a smooth spherical surface with radius R. The particle starts at the equator of the sphere with an angular velocity of [tex] \omega [/tex], and the particle is fast in the sense that [tex]\omega ^2 R >> g[/tex].
The problem is to show that the depth z below the level of the equator is
[tex]z \approx \frac{2g}{\omega ^2} \sin ^2 \frac{\omega t}{2}[/tex].
I found that the energy of the particle can be used to write:
[tex] \frac{1}{2} [ \frac{(1+z^2)\dot{z}^2}{R^2-z^2} + \frac{\omega ^2 R^4}{R^2-z^2}]-gz= \frac{R^2 \omega ^2}{2}[/tex]
using cylindrical coordinates.
But the solutions claim instead that
[tex]\frac{1}{2} [ \frac{R^2 \dot{z}^2}{R^2-z^2} + \frac{\omega ^2 R^4}{R^2-z^2}]-gz= \frac{R^2 \omega ^2}{2}[/tex].
I don't see why I'm getting a different answer, or how these could be the same.
Also the solutions claim that the condition [tex]\omega ^2 R >> g[/tex] leads to [tex]z<<R[/tex]. I can see intuitively why that would be true, but I'm not sure how to show it formally.
Any help would be greatly appreciated.
Thanks.
The problem is about a particle constrained to move on a smooth spherical surface with radius R. The particle starts at the equator of the sphere with an angular velocity of [tex] \omega [/tex], and the particle is fast in the sense that [tex]\omega ^2 R >> g[/tex].
The problem is to show that the depth z below the level of the equator is
[tex]z \approx \frac{2g}{\omega ^2} \sin ^2 \frac{\omega t}{2}[/tex].
I found that the energy of the particle can be used to write:
[tex] \frac{1}{2} [ \frac{(1+z^2)\dot{z}^2}{R^2-z^2} + \frac{\omega ^2 R^4}{R^2-z^2}]-gz= \frac{R^2 \omega ^2}{2}[/tex]
using cylindrical coordinates.
But the solutions claim instead that
[tex]\frac{1}{2} [ \frac{R^2 \dot{z}^2}{R^2-z^2} + \frac{\omega ^2 R^4}{R^2-z^2}]-gz= \frac{R^2 \omega ^2}{2}[/tex].
I don't see why I'm getting a different answer, or how these could be the same.
Also the solutions claim that the condition [tex]\omega ^2 R >> g[/tex] leads to [tex]z<<R[/tex]. I can see intuitively why that would be true, but I'm not sure how to show it formally.
Any help would be greatly appreciated.
Thanks. |
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Now showing items 1-1 of 1
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... |
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Now showing items 1-5 of 5
Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV
(Springer, 2015-05-20)
The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ...
Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV
(Springer, 2015-06)
We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ...
Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV
(Springer, 2015-09)
Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ...
Coherent $\rho^0$ photoproduction in ultra-peripheral Pb-Pb collisions at $\mathbf{\sqrt{\textit{s}_{\rm NN}}} = 2.76$ TeV
(Springer, 2015-09)
We report the first measurement at the LHC of coherent photoproduction of $\rho^0$ mesons in ultra-peripheral Pb-Pb collisions. The invariant mass and transverse momentum distributions for $\rho^0$ production are studied ...
Inclusive, prompt and non-prompt J/ψ production at mid-rapidity in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2015-07-10)
The transverse momentum (p T) dependence of the nuclear modification factor R AA and the centrality dependence of the average transverse momentum 〈p T〉 for inclusive J/ψ have been measured with ALICE for Pb-Pb collisions ... |
I need help with this problem:
Decide if the next infinite series are convergent or divergent. The tools that you need to use are the Leibniz Theorem, the Comparison Test, Quotient Test and the Integral Test.
$\sum_{n=1}^\infty {{\sin\ (n\theta)}\over{n^2}}$ $\sum_{n=1}^\infty (-1)^n {{\log\ n}\over{n}}$
I don't know how to start. Can you please help me and explain me how to decide when to use one of the tests? Thanks. |
You do not need to know too much Fourier theory. Of course it helps, but it is not necessary. There are many different applications of Fourier methods e.g.
Carr Madan (1999): you damp the option price as a function of the log-strike price and compute the fourier transform of the entire option price Bakshi and Madan (2000), Duffie, Pan and Singleton (2000): You Fourier transform the density and obtain a formula similar to the Black-Scholes solution. A decomposition of the call option price in Delta and exercise probability. Lewis (2001): you Fourier transform the option payoff (these have typically polynomial growth, so you need a generalised Fourier transform which extends to the complex domain. This approach requires some complex analysis (residue theorem) and is in fact equivalent to Carr Madan (1999): Choosing a contour to integrate along or an optimal damping factor is the same question. CONV and COS method: These are stable numerical algorithms which, for instance, allow very fast computation of European, Bermudan and (as a limit) American options.
In general, keep in mind, that Fourier methods do not apply to strongly path dependent options (asians, look-backs etc.) Furthermore, the main idea is always to replace the integral (expectation) which occurs from risk-neutral pricing by another integral which contains the characteristic function.
Regarding the maths, there are two things I believe are particular helpful.
Gil-Paelz inversion theoremWhat is option pricing about? Assuming a model for the distribution of $S_T$ and compute the (discounted) expectation of the payoff. This is an integral of payoff times density. Many processes and models (e.g. SVJ, VG, NIG, CGMY etc) have complicated densities but easy characteristic functions. Note that\begin{align*}\varphi_{\ln(S_T)}(u) = \mathbb{E}^\mathbb{Q}\left[e^{iu\ln(S_T)}\right] = \int_\mathbb{R} e^{iux} f_{\ln(S_T)}(x) \mathrm{d}x,\end{align*}where $f_{\ln(S_T)}$ is the risk-neutral density of $\ln(S_T)$. Thus, the characterisitc function of $\ln(S_T)$ is the Fourier transform of its density $f_{\ln(S_T)}$. Hence, $\varphi_{\ln(S_T)}$ captures the distribution of $\ln(S_T)$ and we can show that\begin{align*}F_{\ln(S_T)}(x) &= \frac{1}{2}+\frac{1}{2\pi} \int_0^\infty \frac{e^{iux}\varphi_{\ln(S_T)}(-u)-e^{-iux}\varphi_{\ln(S_T)}(u)}{iu}\mathrm{d}u.\end{align*}This will help a lot in deriving different option pricing formulae. Note that we typically consider $\ln(S_T)$ rather than $S_T$ since the characteristic function already involves somehow the exponential function. Uncertainty principle. This is a concept from physics and states that if $f$ ''spreads out widely'', then its Fourier transform $\hat{f}$ is rather peaked and has a ''small'' support. This means that if you have an option with a short maturity, the density function will be peaked since there is no time for large movements. After all, a stock won't move a lot in a week or two. Thus, the chracterisitic function, $\hat{f}$, will spread out a lot and due to the oscillating behaviour, it may be challenging to integrate numerically.
Schmelzle wrote a nice survey paper but his work includes some errors in the section about the Lewis (2001) approach. If you're interested in the implementation of these models, have a look at Hirsa.
P.S. When I looked the first time at Fourier methods, my background was real analysis and probability theory. Just like yours. And I could follow it and understand it just fine. Don't you worry :) |
Let $V=\begin{pmatrix}a\\b\\c\end{pmatrix}$.
$[V]_{\times}$ denotes the antisymmetrical matrix, associated with the linear operation "taking the cross product with $V$", i.e.,
$$[V]_{\times}X:=V \times X=\begin{pmatrix} -cy+bz\\ \ \ \ cx-az\\-bx+ay \end{pmatrix},$$
which means that $$[V]_{\times}=\begin{pmatrix}\ \ \ 0&-c& \ \ \ b\\ \ \ \ c& \ \ \ 0&-a\\-b& \ \ \ a& \ \ \ 0\end{pmatrix}.$$
(this notation is described in (https://en.wikipedia.org/wiki/Cross_product)).
Remarks:
1) Very often $V$ is assumed to be unitary ($a^2+b^2+c^2=1$).
2) Here is a very interesting paper (https://arxiv.org/pdf/1312.0788v1.pdf) where this notation is used with many of its properties grouped in its Appendix A.
3) This notation doesn't look very old ; maybe one should post the question of its origin (robotics ?) on the "history of mathematics" section of SE.
Edit:
4) I have had a glance at the source paper you have indicated after I wrote my answer.
In the block-defined matrix, $\hat R^n_{b,k}f_k^b$ has to be a vector. There are two ways to consider this notation:
a) either as a single matrix $\hat R^n_{b,k}$ or $\left(\hat R^n_{b}\right)_k$applied to a vector $f_k^b$, yielding a vector.
b) or (much less likely) a linear combination of matrices applied to vectors, with a hidden summation sign according to Einstein notation (https://en.wikipedia.org/wiki/Einstein_notation), $\sum_{b=1}^m \hat R^n_{b,k}f_k^b$ where the summation is on the index that is present once in a lower, once in a upper position, here index $b$.
5) See also the recent document (https://people.eecs.berkeley.edu/~wkahan/MathH110/Cross.pdf) and the older one (http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.825.1726&rep=rep1&type=pdf) (1989) |
Existence of solutions to singular integral equations
1.
School of Electrical and Electronic Engineering, Nanyang Technological University, 50 Nanyang Avenue, Singapore 639798, Singapore
$u_i(t)=int_0^Tg_i(t,s)[a_i(s,u_1(s),u_2(s),...,u_n(s))+b_i(s,u_1(s),u_2(s),...,u_n(s))]ds,$ $t \in [0,T],$ $1<=i<=n,$
where $T>0$ is fixed
and the nonlinearities $a_i(t,u_1,u_2,\cdots,u_n)$ can be
singular at $t=0$ and $u_j=0$ where $j\in\{1,2,\cdots,n\}.$
Criteria are established for the existence of fixed-sign
solutions $(u_1^*,u_2^*,\cdots,u_n^*)$ to the above system, i.e.,
$\theta_iu_i^*(t)\geq 0$ for $t\in [0,T]$ and $1\leq i\leq n,$ where
$\theta_i\in \{1,-1\}$ is fixed. We also include an example to
illustrate the usefulness of the results obtained.
Mathematics Subject Classification:Primary: 45B05, 45G15; Secondary: 45M20. Citation:Patricia J.Y. Wong. Existence of solutions to singular integral equations. Conference Publications, 2009, 2009 (Special) : 818-827. doi: 10.3934/proc.2009.2009.818
[1]
Patricia J.Y. Wong.
On the existence of fixed-sign solutions for a system of generalized right focal problems with deviating arguments.
[2] [3] [4]
Dongyan Li, Yongzhong Wang.
Nonexistence of positive solutions for a system of integral equations on
$R^n_+$ and applications.
[5]
G. C. Yang, K. Q. Lan.
Systems of singular integral equations and applications to
existence of reversed flow solutions of Falkner-Skan
equations.
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Liang Ding, Rongrong Tian, Jinlong Wei.
Nonconstant periodic solutions with any fixed energy for singular Hamiltonian systems.
[8]
Daniela Giachetti, Francesco Petitta, Sergio Segura de León.
Elliptic equations having a singular quadratic gradient term and
a changing sign datum.
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Yohei Sato, Zhi-Qiang Wang.
On the least energy sign-changing solutions for a nonlinear elliptic system.
[11]
Yavdat Il'yasov, Nadir Sari.
Solutions of minimal period for a Hamiltonian system with a changing sign potential.
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Josep M. Olm, Xavier Ros-Oton.
Existence of periodic solutions with nonconstant sign in a class of generalized Abel equations.
[14]
Dumitru Motreanu.
Three solutions with precise sign properties for systems of quasilinear elliptic equations.
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Mingchun Wang, Jiankai Xu, Huoxiong Wu.
On Positive solutions of integral equations with the weighted Bessel potentials.
[20]
Lu Chen, Zhao Liu, Guozhen Lu.
Qualitative properties of solutions to an integral system associated with the Bessel potential.
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8.6 Estimation and order selection Maximum likelihood estimation
Once the model order has been identified (i.e., the values of \(p\), \(d\) and \(q\)), we need to estimate the parameters \(c\), \(\phi_1,\dots,\phi_p\), \(\theta_1,\dots,\theta_q\). When R estimates the ARIMA model, it uses
maximum likelihood estimation (MLE). This technique finds the values of the parameters which maximise the probability of obtaining the data that we have observed. For ARIMA models, MLE is similar to the least squares estimates that would be obtained by minimising\[ \sum_{t=1}^T\varepsilon_t^2.\](For the regression models considered in Chapter 5, MLE gives exactly the same parameter estimates as least squares estimation.) Note that ARIMA models are much more complicated to estimate than regression models, and different software will give slightly different answers as they use different methods of estimation, and different optimisation algorithms.
In practice, R will report the value of the
log likelihood of the data; that is, the logarithm of the probability of the observed data coming from the estimated model. For given values of \(p\), \(d\) and \(q\), R will try to maximise the log likelihood when finding parameter estimates. Information Criteria
Akaike’s Information Criterion (AIC), which was useful in selecting predictors for regression, is also useful for determining the order of an ARIMA model. It can be written as \[ \text{AIC} = -2 \log(L) + 2(p+q+k+1), \] where \(L\) is the likelihood of the data, \(k=1\) if \(c\ne0\) and \(k=0\) if \(c=0\). Note that the last term in parentheses is the number of parameters in the model (including \(\sigma^2\), the variance of the residuals).
For ARIMA models, the corrected AIC can be written as \[ \text{AICc} = \text{AIC} + \frac{2(p+q+k+1)(p+q+k+2)}{T-p-q-k-2}, \] and the Bayesian Information Criterion can be written as \[ \text{BIC} = \text{AIC} + [\log(T)-2](p+q+k+1). \] Good models are obtained by minimising the AIC, AICc or BIC. Our preference is to use the AICc.
It is important to note that these information criteria tend not to be good guides to selecting the appropriate order of differencing (\(d\)) of a model, but only for selecting the values of \(p\) and \(q\). This is because the differencing changes the data on which the likelihood is computed, making the AIC values between models with different orders of differencing not comparable. So we need to use some other approach to choose \(d\), and then we can use the AICc to select \(p\) and \(q\). |
383 13 Homework Statement The electric field outside and an infinitesimal distance away from a uniformly charged spherical shell, with radius R and surface charge density σ, is given by Eq. (1.42) as σ/0. Derive this in the following way. (a) Slice the shell into rings (symmetrically located with respect to the point in question), and then integrate the field contributions from all the rings. You should obtain the incorrect result of ##\frac{\sigma}{2\epsilon_0}##. (b) Why isn’t the result correct? Explain how to modify it to obtain the correct result of ##\frac{\sigma}{2\epsilon_0}##. Hint: You could very well have performed the above integral in an effort to obtain the electric field an infinitesimal distance inside the shell, where we know the field is zero. Does the above integration provide a good description of what’s going on for points on the shell that are very close to the point in question? Homework Equations Coulomb's Law
Hi! I need help with this problem. I tried to do it the way you can see in the picture. I then has this:
##dE_z=dE\cdot \cos\theta## thus ##dE_z=\frac{\sigma dA}{4\pi\epsilon_0}\cos\theta=\frac{\sigma 2\pi L^2\sin\theta d\theta}{4\pi\epsilon_0 L^2}\cos\theta##.
Then I integrated and ended up with ##E=\frac{\sigma}{2\epsilon_0}\int \sin\theta\cos\theta d\theta##. The problem is that I don't know what are the limits of integrations, I first tried with ##\pi##, but I got 0. What am I doing wrong? |
How can we derive the chi-squared probability density function (pdf) using the pdf of normal distribution?
I mean, I need to show that
$$f(x)=\frac{1}{2^{r/2}\Gamma(r/2)}x^{r/2-1}e^{-x/2} \>, \qquad x > 0\>.$$
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The way the question is expressed is a mess, but I'll assume it means this: if $X\sim N(0,1)$, how do you find the pdf of $X^2$? Here's one way. Remember that the pdf of $X$ is $$ \varphi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}. $$ Let $f$ be the pdf of $X^2$. Then $$ \begin{align} f(x) & = \frac{d}{dx} \Pr(X^2 \le x) = \frac{d}{dx} \Pr(-\sqrt{x}\le X\le\sqrt{x}) \\ \\ & = \frac{d}{dx} \frac{1}{\sqrt{2\pi}} \int_{-\sqrt{x}}^\sqrt{x} e^{-u^2/2} \;du = \frac{2}{\sqrt{2\pi}}\frac{d}{dx} \int_0^\sqrt{x} e^{-u^2/2} \;du \\ \\ & = \frac{2}{\sqrt{2\pi}} e^{-\sqrt{x}^2/2} \frac{d}{dx} \sqrt{x} = \frac{2}{\sqrt{2\pi}} e^{-x/2} \frac{1}{2\sqrt{x}} \\ \\ \\ & = \frac{e^{-x/2}}{\sqrt{2\pi x}}. \end{align} $$
Sometimes it might be written as $\dfrac{1}{\sqrt{2\pi}} x^{\frac12 - 1}e^{-x/2}$ so that you can see how it resembles the function involved in defining the Gamma function.
Your title said $1$ degree of freedom. But what you write seems to allow $r$ to be some number other than $1$. If you want to do that, then there's more work to do.
If $X \sim (\mu, \Sigma) \neq (0, I)$, the result you wish to provedoes not hold: even if the random variables are independent but havenonzero means, you get a
non-central $\chi^2$ pdf which is not what you are trying to show.
If $X_1, \ldots, X_n$ are independent standard normal random variables, then $X_i^2$ has a Gamma distribution with scale parameter $\frac{1}{2}$ and order parameter $\frac{1}{2}$. Then, $\sum_{i= 1}^n X_i^2 $ is a sum of $n$ independent Gamma random variables each with scale parameter $\frac{1}{2}$ and order parameter $\frac{1}{2}$ and is thus a Gamma random variable with scale parameter $\frac{1}{2}$ and order parameter $\frac{r}{2}$. |
Usually probability theory is taught with Kolgomorov's axioms. Do Bayesians also accept Kolmogorov's axioms?
In my opinion, Cox-Jaynes interpretation of probability provides a rigorous foundation for Bayesian probability:
Cox, Richard T. "Probability, frequency and reasonable expectation." American journal of physics 14.1 (1946): 1-13. Jaynes, Edwin T. Probability theory: the logic of science. Cambridge university press, 2003. Beck, James L. "Bayesian system identification based on probability logic." Structural Control and Health Monitoring 17.7 (2010): 825-847.
The axioms of probability logic derived by Cox are:
(P1): $\Pr[b|a]\ge0$ (by convention) (P2): $\Pr[\overline{b}|a]=1-\Pr[b|a]$ (negation function) (P3): $\Pr[b\cap c|a]=\Pr[c|b\cap a]\Pr[b|a]$ (conjunction function)
Axioms P1-P3 imply the following (Beck, James L. "Bayesian system identification based on probability logic." Structural Control and Health Monitoring 17.7 (2010): 825-847):
(P4): a) $\Pr[b|b\cap c] = 1$; b) $\Pr[\overline{b}|b\cap c] = 0$; c) $\Pr[b|c]\in[0,1]$ (P5): a) $\Pr[a|c \cap (a \Rightarrow b)]\le\Pr[b|c\cap(a \Rightarrow b)]$, b) $\Pr[a|c\cap(a \Leftrightarrow b)] = \Pr[b|c\cap(a \Leftrightarrow b)]$, where $a \Rightarrow b$ means that $a$ is contained in $c$, and $a \Leftrightarrow b$ means that $a$ is equivalent to $b$. (P6): $\Pr[a \cup b|c] = \Pr[a|c]+\Pr[b|c]-\Pr[a\cap b|c]$ (P7): Assuming that proposition $c$ states that one and only one of propositions $b_1,\ldots,b_N$ is true, then: a) Marginalization Theorem: $\Pr[a|c]=\sum_{n=1}^N P[a \cap b_n|c]$ b) Total Probability Theorem: $\Pr[a|c] = \sum_{n=1}^N \Pr[a|b_n\cap c]\Pr[b_n|c]$ c) Bayes' Theorem: For $k=1,\ldots,N$: $\Pr[b_k|a\cap c] = \frac{\Pr[a|b_k\cap c]\Pr[b_k|c]}{\sum_{n=1}^N \Pr[a|b_n\cap c]\Pr[b_n|c]}$
They imply Kolmogorov's statement of logic, which can be viewed as a special case.
In my interpretation of a Bayesian viewpoint, everything is always (implicitly) conditioned on our believes and on our knowledge.
The following comparison is taken from Beck (2010): Bayesian system identification based on probability logic
The Bayesian point of view
Probability is a measure of plausibility of a statement based on specified information.
Probability distributions represent states of plausible knowledgeabout systems and phenomena, not inherent properties of them. Probability of a model is a measure of its plausibilityrelative to other models in a set. Pragmatically quantifies uncertainty due to missing information without any claim that this is due to nature's inherent randomness. The Frequentist point of view
Probability is the relative frequency of occurrence of an
inherently random event in the long run. Probability distributions are inherentproperties of random phenomena. Limited scope, e.g. no meaning for the probability of a model. Inherent randomnessis assumed, but cannot be proven. How to derive Kolmogorov's axioms from the axioms above
In the following, section 2.2 of [Beck, James L. "Bayesian system identification based on probability logic." Structural Control and Health Monitoring 17.7 (2010): 825-847.] is summarized:
In the following we use: probability measure $\Pr(A)$ on subset $A$ of a finite set $X$:
[K1]: $\Pr(A)\ge 0, \forall A \subset X$ [K2]: $\Pr(X) = 1$ [K3]: $\Pr(A\cup B)=\Pr(A)+\Pr(B), \forall A,B \subset X$ if $A$ and $B$ are disjoint.
In order to derive (K1-K3) from the axioms of probability theory, [Beck, 2010] introduced propositon $\pi$ that states $x\in X$ and specifies the probability model for $x$. [Beck, 2010] furthermore introduces $\Pr(A) = \Pr[x\in A|\pi]$.
P1 implies K1 with $b=\{x\in A\}$ and $c=\pi$ K2 follows from $\Pr[x\in X|\pi]=1$; P4(a), and $\pi$ states that $x\in X$. K3 can be derived from P6: $A$ and $B$ are disjoint means that $x\in A$ and $x\in B$ are mutually exclusive. Therefore, K3:$\Pr(x\in A\cup B|\pi)=\Pr(x\in A|\pi)+\Pr(x\in B|\pi)$
After the development of Probability Theory it was necessary to show that looser concepts answering to the name of "probability" measured up to the rigorously defined concept they had inspired. "Subjective" Bayesian probabilities were considered by Ramsey and de Finetti, who independently showed that a quantification of degree of belief subject to the constraints of comparability & coherence (your beliefs are coherent if no-one can make a Dutch book against you) has to be a probability.
Differences between axiomatizations are largely a matter of taste concerning what should be what defined & what derived. But countable additivity is one of Kolmogorov's that isn't derivable from Cox's or Finetti's, & has been controversial. Some Bayesians (e.g. de Finetti & Savage) stop at finite additivity & so
don't accept all of Kolmogorov's axioms. They can put uniform probability distributions over infinite intervals without impropriety. Others follow Villegas in also assuming monotone continuity, & get countable additivity from that.
Ramsey (1926), "Truth and probability", in Ramsey (1931),
The Foundations of Mathematics and other Logical Essays
de Finetti (1931), "Sul significato soggettivo della probabilità",
Fundamenta Mathematicæ, 17, pp 298 – 329
Villegas (1964), "On qualitative probability $\sigma$-algebras",
Ann. Math. Statist., 35, 4. |
A new construction of rotation symmetric bent functions with maximal algebraic degree
School of Mathematics and Statistics, Henan University, Kaifeng 475004, China
$ n = 2m\ge4 $
$ n $
$ m $
$ f(x_0,x_1\cdots,x_{n-1}) = \bigoplus\limits_{i = 0}^{m-1}(x_ix_{m+i})\oplus \bigoplus\limits_{i = 0}^{n-1}(x_ix_{i+1}\cdots x_{i+m-2} \overline{x_{i+m}} ), $
$ \widetilde{f}(x_0,x_1\cdots,x_{n-1}) = \bigoplus\limits_{i = 0}^{m-1}(x_ix_{m+i})\oplus \bigoplus\limits_{i = 0}^{n-1}(x_ix_{i+1}\cdots x_{i+m-2} \overline{x_{i+n-2}} ), $
$ \overline{x_{i}} = x_{i}\oplus 1 $
$ x $
$ n $ Keywords:Orbit, rotation symmetric Boolean function, Walsh-Hadamard transform, bent function, algebraic degree. Mathematics Subject Classification:Primary: 58F15, 58F17; Secondary: 53C35. Citation:Sihong Su. A new construction of rotation symmetric bent functions with maximal algebraic degree. Advances in Mathematics of Communications, 2019, 13 (2) : 253-265. doi: 10.3934/amc.2019017
References:
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C. Carlet, Boolean functions for cryptography and error correcting codes, in
[3]
C. Carlet, G. Gao and W. Liu,
A secondary construction and a transformation on rotation symmetric functions, and their action on bent and semi-bent functions,
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C. Carlet, G. Gao and W. Liu, Results on constructions of rotation symmetric bent and semi-bent functions, in
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$\acute{E}$. Filiol and C. Fontaine, Highly nonlinear balanced Boolean functions with a good correlation-immunity, in
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G. Gao, X. Zhang, W. Liu and C. Carlet,
Constructions of quadratic and cubic rotation symmetric bent functions,
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S. Kavut, S. Maitra and M. Yücel,
Search for Boolean functions with excellent profiles in the rotation symmetric class,
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F. MacWilliams and N. Sloane,
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S. Su and X. Tang,
Systematic constructions of rotation symmetric bent functions, 2-rotation symmetric bent functions, and bent idempotent functions,
[19]
W. Zhang, Z. Xing and K. Feng, A construction of bent functions with optimal algebraic degree and large symmetric group, preprint, Cryptology ePrint Archive, : Submission 2017/229.Google Scholar
show all references
References:
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C. Carlet, Boolean functions for cryptography and error correcting codes, in
[3]
C. Carlet, G. Gao and W. Liu,
A secondary construction and a transformation on rotation symmetric functions, and their action on bent and semi-bent functions,
[4]
C. Carlet, G. Gao and W. Liu, Results on constructions of rotation symmetric bent and semi-bent functions, in
[5] [6]
$\acute{E}$. Filiol and C. Fontaine, Highly nonlinear balanced Boolean functions with a good correlation-immunity, in
[7] [8] [9]
G. Gao, X. Zhang, W. Liu and C. Carlet,
Constructions of quadratic and cubic rotation symmetric bent functions,
[10]
S. Kavut, S. Maitra and M. Yücel,
Search for Boolean functions with excellent profiles in the rotation symmetric class,
[11] [12]
F. MacWilliams and N. Sloane,
[13] [14] [15] [16] [17] [18]
S. Su and X. Tang,
Systematic constructions of rotation symmetric bent functions, 2-rotation symmetric bent functions, and bent idempotent functions,
[19]
W. Zhang, Z. Xing and K. Feng, A construction of bent functions with optimal algebraic degree and large symmetric group, preprint, Cryptology ePrint Archive, : Submission 2017/229.Google Scholar
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Wenying Zhang, Zhaohui Xing, Keqin Feng.
A construction of bent functions with optimal algebraic degree and large symmetric group.
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Behrouz Kheirfam.
A full Nesterov-Todd step infeasible interior-point algorithm for symmetric optimization based on a specific kernel function.
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Sihem Mesnager, Fengrong Zhang, Yong Zhou.
On construction of bent functions involving symmetric functions and their duals.
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Ken Ono.
Parity of the partition function.
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Jaume Llibre, Y. Paulina Martínez, Claudio Vidal.
Phase portraits of linear type centers of polynomial Hamiltonian systems with Hamiltonian function of degree 5 of the form $ H = H_1(x)+H_2(y)$.
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SelÇuk Kavut, Seher Tutdere.
Highly nonlinear (vectorial) Boolean functions that are symmetric under some permutations.
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Welington Cordeiro, Manfred Denker, Michiko Yuri.
A note on specification for iterated function systems.
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2018 Impact Factor: 0.879
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5.7 Matrix formulation Warning: this is a more advanced, optional section and assumes knowledge of matrix algebra.
Recall that multiple regression model can be written as \[ y_{t} = \beta_{0} + \beta_{1} x_{1,t} + \beta_{2} x_{2,t} + \cdots + \beta_{k} x_{k,t} + \varepsilon_{t} \] where \(\varepsilon_{t}\) has mean zero and variance \(\sigma^2\). This expresses the relationship between a single value of the forecast variable and the predictors.
It can be convenient to write this in matrix form where all the values of the forecast variable are given in a single equation. Let \(\bm{y} = (y_{1},\dots,y_{T})'\), \(\bm{\varepsilon} = (\varepsilon_{1},\dots,\varepsilon_{T})'\), \(\bm{\beta} = (\beta_{0},\dots,\beta_{k})'\) and \[ \bm{X} = \left[ \begin{matrix} 1 & x_{1,1} & x_{2,1} & \dots & x_{k,1}\\ 1 & x_{1,2} & x_{2,2} & \dots & x_{k,2}\\ \vdots& \vdots& \vdots&& \vdots\\ 1 & x_{1,T}& x_{2,T}& \dots& x_{k,T} \end{matrix}\right]. \] Then \[ \bm{y} = \bm{X}\bm{\beta} + \bm{\varepsilon}. \] where \(\bm{\varepsilon}\) has mean \(\bm{0}\) and variance \(\sigma^2\bm{I}\). Note that the \(\bm{X}\) matrix has \(T\) rows reflecting the number of observations and \(k+1\) columns reflecting the intercept which is represented by the column of ones plus the number of predictors.
Least squares estimation
Least squares estimation is performed by minimising the expression \(\bm{\varepsilon}'\bm{\varepsilon} = (\bm{y} - \bm{X}\bm{\beta})'(\bm{y} - \bm{X}\bm{\beta})\). It can be shown that this is minimised when \(\bm{\beta}\) takes the value \[ \hat{\bm{\beta}} = (\bm{X}'\bm{X})^{-1}\bm{X}'\bm{y} \] This is sometimes known as the “normal equation”. The estimated coefficients require the inversion of the matrix \(\bm{X}'\bm{X}\). If \(\bm{X}\) is not of full column rank then matrix \(\bm{X}'\bm{X}\) is singular and the model cannot be estimated. This will occur, for example, if you fall for the “dummy variable trap”, i.e., having the same number of dummy variables as there are categories of a categorical predictor, as discussed in Section 5.4.
The residual variance is estimated using \[ \hat{\sigma}_e^2 = \frac{1}{T-k-1}(\bm{y} - \bm{X}\hat{\bm{\beta}})' (\bm{y} - \bm{X}\hat{\bm{\beta}}). \]
Fitted values and cross-validation
The normal equation shows that the fitted values can be calculated using \[ \bm{\hat{y}} = \bm{X}\hat{\bm{\beta}} = \bm{X}(\bm{X}'\bm{X})^{-1}\bm{X}'\bm{y} = \bm{H}\bm{y}, \] where \(\bm{H} = \bm{X}(\bm{X}'\bm{X})^{-1}\bm{X}'\) is known as the “hat-matrix” because it is used to compute \(\bm{\hat{y}}\) (“y-hat”).
If the diagonal values of \(\bm{H}\) are denoted by \(h_{1},\dots,h_{T}\), then the cross-validation statistic can be computed using \[ \text{CV} = \frac{1}{T}\sum_{t=1}^T [e_{t}/(1-h_{t})]^2, \] where \(e_{t}\) is the residual obtained from fitting the model to all \(T\) observations. Thus, it is not necessary to actually fit \(T\) separate models when computing the CV statistic.
Forecasts and prediction intervals
Let \(\bm{x}^*\) be a row vector containing the values of the predictors (in the same format as \(\bm{X}\)) for which we want to generate a forecast . Then the forecast is given by \[ \hat{y} = \bm{x}^*\hat{\bm{\beta}}=\bm{x}^*(\bm{X}'\bm{X})^{-1}\bm{X}'\bm{Y} \] and its estimated variance is given by \[ \hat\sigma_e^2 \left[1 + \bm{x}^* (\bm{X}'\bm{X})^{-1} (\bm{x}^*)'\right]. \] A 95% prediction interval can be calculated (assuming normally distributed errors) as \[ \hat{y} \pm 1.96 \hat{\sigma}_e \sqrt{1 + \bm{x}^* (\bm{X}'\bm{X})^{-1} (\bm{x}^*)'}. \] This takes into account the uncertainty due to the error term \(\varepsilon\) and the uncertainty in the coefficient estimates. However, it ignores any errors in \(\bm{x}^*\). Thus, if the future values of the predictors are uncertain, then the prediction interval calculated using this expression will be too narrow. |
The given temperature of $T=50\ \mathrm{^\circ C}$ and the given pressure of $p=92.5\ \mathrm{mmHg}$ approximately correspond to equilibrium conditions for liquid water and steam. (Strictly speaking, the corresponding saturation pressure for a given temperature of $T=50.000\ \mathrm{^\circ C}$ is $p=92.647\ \mathrm{mmHg}$, which could be rounded to $p=92.6\ \mathrm{mmHg}$ and not $p=92.5\ \mathrm{mmHg}$; in the opposite direction, however, the corresponding saturation temperature for a given pressure of $p=92.500\ \mathrm{mmHg}$ is $T=49.968\ \mathrm{^\circ C}$, which could be rounded to $T=50\ \mathrm{^\circ C}$.) The question does not explain whether these values describe the initial or final state; however, since the values correspond to equilibrium conditions, we may assume that these values apply to the final state when equilibrium is established (e.g. this can be achieved by keeping the closed system at a constant temperature of $T=50\ \mathrm{^\circ C}$).
Strictly speaking, the question doesn’t even explain whether the added water initially is liquid or steam; however, this is not relevant for the final state. If the water is added as liquid, a part of it evaporates until the equilibrium state is reached in the closed container; if the water is introduced as steam, a part of it condenses until the equilibrium state is reached. Therefore, when the defined equilibrium is established, the container contains certain amounts of liquid water and steam irrespective of the initial conditions.
The question does not mention any air in the container. For simplicity’s sake, we may assume that the container has been evacuated before the experiment and contains only liquid water and vapour.
The available volume of the container is reduced by the volume of the liquid water. However, since the density of liquid water at the given temperature and pressure is about $\rho=988\ \mathrm{kg\ m^{-3}}$, the given mass of $m=1.00\ \mathrm g$ corresponds to a maximum volume of $V_\text{liquid}=1.01\ \mathrm{ml}=0.00101\ \mathrm l$, assuming that no water has evaporated. Considering that the total volume is given as $V=5.00\ \mathrm l$ (note the number of significant digits), the difference caused by the liquid water is not significant and may be neglected.
Note that the use of the non-SI unit “conventional millimetre of mercury” (unit symbol: mmHg) is deprecated; the use of SI units is to be preferred.$$1\ \mathrm{mmHg}\approx 133.3224\ \mathrm{Pa}$$You might want to convert the given values to coherent SI units, or find a value for the molar gas constant $R$ that is expressed in a suitable unit.
The following data are given:
Temperature $T=50\ \mathrm{^\circ C}=323.15\ \mathrm K$ Pressure $p=92.5\ \mathrm{mmHg}=12332.322\ \mathrm{Pa}$ Volume $V=5.00\ \mathrm l=0.00500\ \mathrm{m^3}$
The value of the molar gas constant is $R=8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}$.
[source]
The amount of vapour $n$ in the container may be estimated using the ideal gas law$$\begin{align}p\cdot V&=n\cdot R\cdot T\\[6pt]n&=\frac{p\cdot V}{R\cdot T}\\[6pt]&=\frac{12332.322\ \mathrm{Pa}\times0.00500\ \mathrm{m^3}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times323.15\ \mathrm K}\\[6pt]&=0.022949674\ \mathrm{mol}\end{align}$$(Note that this intermediate result is given with an excessive number of digits. It is deliberately not rounded in order to avoid round-off errors in subsequent calculations.)
Since the molar mass of water is $M=18.01528\ \mathrm{g\ mol^{-1}}$, the corresponding mass of vapour is$$\begin{align}m&=M\cdot n\\[6pt]&=18.01528\ \mathrm{g\ mol^{-1}}\times0.022949674\ \mathrm{mol}\\[6pt]&=0.413444803\ \mathrm g\\[6pt]&\approx0.41\ \mathrm g\end{align}$$(Note that this final result is rounded to two significant digits. The number of significant digits is estimated in view of the fact that the mass of water $m=1.00\ \mathrm g$, the volume $V=5.00\ \mathrm l$, and the pressure $p=92.5\ \mathrm{mmHg}$ are given with three significant digits, but the temperature $T=50\ \mathrm{^\circ C}$ is only given with two significant digits.)
This result is the mass of vapour $m_\text{vapour}$ in the container, whereas the question is about the mass of liquid water $m_\text{liquid}$, which can be calculated from the given total mass of water $m_\text{total}=1.00\ \mathrm{g}$.$$\begin{align}m_\text{total}&=m_\text{vapour}+m_\text{liquid}\\[6pt]m_\text{liquid}&=m_\text{total}-m_\text{vapour}\\[6pt]&=1.00\ \mathrm g-0.41\ \mathrm g\\[6pt]&=0.59\ \mathrm g\end{align}$$
Therefore, the correct answer is option 3) $0.59\ \mathrm g$.
By way of comparison, precise engineering calculations for water and steam usually do not rely on the ideal gas law but use so-called steam tables. We may use such steam tables to check the accuracy of our estimate that has been based on the ideal gas law.
For example, for a temperature of $T=50\ \mathrm{^\circ C}$ at equilibrium, we find the following values in REFPROP – NIST Standard Reference Database 23, Version 9.0:
Pressure $p=12352\ \mathrm{Pa}$ Liquid density $\rho_\text{liquid}=988.00\ \mathrm{kg\ m^{-3}}=988.00\ \mathrm{g\ l^{-1}}$ Vapour density $\rho_\text{vapour}=0.083147\ \mathrm{kg\ m^{-3}}=0.083147\ \mathrm{g\ l^{-1}}$
Furthermore, the database includes various other thermodynamic and transport properties, which are not required for the following calculations.
If you do not have access to professional steam tables, you may want to consider using the steam tables that are included in WolframAlpha. The corresponding results for a temperature of $T=50\ \mathrm{^\circ C}$ at equilibrium can be obtained using the input “water boiling at 50 °C”:
Pressure $p=12352\ \mathrm{Pa}$ Liquid density $\rho_\text{liquid}=988\ \mathrm{kg\ m^{-3}}=988\ \mathrm{g\ l^{-1}}$ Vapour density $\rho_\text{vapour}=0.08315\ \mathrm{kg\ m^{-3}}=0.08315\ \mathrm{g\ l^{-1}}$
Note: The indicated values for STP are not in accordance with IUPAC recommendations.
We know that the mass balance in the container is$$m_\text{total}=m_\text{vapour}+m_\text{liquid}\tag{1}$$where $m_\text{total}=1.0000\ \mathrm g$ is the total mass of water (liquid and vapour) in the container (in this example, the precision of this value is arbitrarily increased to five significant digits).
Since density is defined as$$\rho=\frac mV$$the corresponding volume balance can also be expressed in terms of mass:$$\begin{align}V_\text{total}&=V_\text{vapour}+V_\text{liquid}\\[6pt]&=\frac{m_\text{vapour}}{\rho_\text{vapour}}+\frac{m_\text{liquid}}{\rho_\text{liquid}}\tag{2}\end{align}$$where $V_\text{total}=5.0000\ \mathrm l$ is the total volume of the container (in this example, the precision of this value is arbitrarily increased to five significant digits).
Solving the system of the equations $\text(1)$ and $\text(2)$ yields the solutions
$$\begin{align}m_\text{vapour}&=\frac{\rho_\text{vapour}\cdot\left(V_\text{total}\cdot\rho_\text{liquid}-m_\text{total}\right)}{\rho_\text{liquid}-\rho_\text{vapour}}\\[6pt]&=\frac{0.083147\ \mathrm{g\ l^{-1}}\times\left(5.0000\ \mathrm l\times988.00\ \mathrm{g\ l^{-1}}-1.0000\ \mathrm g\right)}{988.00\ \mathrm{g\ l^{-1}}-0.083147\ \mathrm{g\ l^{-1}}}\\[6pt]&=0.41569\ \mathrm g\end{align}$$
and
$$\begin{align}m_\text{liquid}&=-\frac{\rho_\text{liquid}\cdot\left(V_\text{total}\cdot\rho_\text{vapour}-m_\text{total}\right)}{\rho_\text{liquid}-\rho_\text{vapour}}\\[6pt]&=-\frac{988.00\ \mathrm{g\ l^{-1}}\times\left(5.0000\ \mathrm l\times0.083147\ \mathrm{g\ l^{-1}}-1.0000\ \mathrm g\right)}{988.00\ \mathrm{g\ l^{-1}}-0.083147\ \mathrm{g\ l^{-1}}}\\[6pt]&=0.58431\ \mathrm g\end{align}$$
These results confirm that the above-mentioned estimates $m_\text{vapour}\approx0.41\ \mathrm g$ and $m_\text{liquid}\approx0.59\ \mathrm g$, which have been obtained using the ideal gas law and neglecting the volume of the liquid water, are reasonable and sufficient in order to answer the question. |
I just read "Boole's Algebra Isn't Boolean Algebra" by Theodore Halperin (behind a paywall here). I don't have a strong background in abstract algebra, so, frankly, the paper is a bit over my head but the gist of it is as follows: the algebra developed by Boole in the 19th century has some strange properties. Boole interprets every term in an expression as representing a set and restricts the domain of valid expressions on the basis of relationships between the underlying sets. In particular, he asserts that the expression $x + y$ is valid iff $x$ and $y$ represent disjoint sets, and $x - y$ is valid iff the intersection of $x$ and $y$ is nonempty. He also defines an operation $w = \frac x y$ such that $w$ has many solutions. Halperin goes to great lengths to construct a commutative ring that satisfies Boole's constraints.
I think that there might be a nice, intuitive, graph-theoretic interpretation of Boole's algebra where we can say something like "Given a complex expression in Boole's algebra, $\Phi = \{\phi_1, \phi_2\, \dots, \phi_n\}$, we can construct a (di)graph $G$ such that $\Phi$ is valid if and only if $G$ has some property $P$."
For example, to test if $X + Y$ is valid, we could do something like the following:
Let $G$ be a graph. Let $V(G) = X \cup Y \cup \{v_X, v_Y\}$, i.e. make a vertex for every element of the underlying set, plus a vertex for each term in the expression. Then define
$$E(G) = \{uv_X \mid u \in X \} \cup \{uv_Y \mid u \in Y \}$$
and let $\partial(G)$ be the set of all valid bonds formed by subsets of $V(G)$. It follows that $X + Y$ is valid if and only if $|\partial(G)| = 2$.
From there, I'm not sure how to go about forming graphs for complex expressions like $\frac {y(x + z)} {z^2 + 1}$ by composing simpler graphs. I imagine that there is literature about representing Boolean algebra on graphs, but I haven't been able to find it.
Does anyone have an elegant interpretation? Or a pointer to relevant literature that might get me started?
EDIT: I can't find the article for free anywhere, but wikipedia touches on the issues:
In places, Boole talks of terms being interpreted by sets, but he also recognises terms that cannot always be so interpreted, such as the term 2AB...Such terms he classes uninterpretable terms...uninterpretable terms cannot be the ultimate result of equational manipulations from meaningful starting formulae.
Halperin's paper shows that his algebra is isomorphic to "a commutative ring with unit having no additive or multiplicative nilpotents." I think, for that reason, that it will require heavier machinery to represent than normal Boolean logic operations. |
Modified 2019-04-14 by Liam Paull
Measuring performance in robotics is less clear cut and more multidimensional than traditionally encountered in machine learning settings. Nonetheless, to achieve reliable performance estimates we assess submitted code on several
episodes with different initial settings and compute statistics on the outcomes. We denote $\objective$ to be an objective or cost function to optimize, which we evaluate for every experiment. In the following formalization, objectives are assumed to be minimized.
In the following we summarize the objectives used to quantify how well an embodied task is completed. We will produce scores in three different categories:
Modified 2019-05-16 by Julian Zilly
Modified 2019-04-14 by Liam Paull
As a performance indicator for both the “lane following task” and the “lane following task with other dynamic vehicles”, we choose the integrated speed $v(t)$ along the road (not perpendicular to it) over time of the Duckiebot. This measures the moved distance along the road per episode, where we fix the time length of an episode. This encourages both faster driving as well as algorithms with lower latency. An
episode is used to mean running the code from a particular initial configuration.
$$ \objective_{P-LF(V)}(t) = \int_{0}^{t} - v(t) dt $$
The integral of speed is defined over the traveled distance of an episode up to time $t=T_{eps}$, where $T_{eps}$ is the length of an episode.
The way we measure this is in units of “tiles traveled”:
$$ \objective_{P-LF(V)}(t) = \text{# of tiles traveled} $$
Modified 2019-04-14 by Liam Paull
In an autonomous mobility-on-demand system a coordinated fleet of robotic taxis serves customers in an on-demand fashion. An operational policy for the system must optimize in three conflicting dimensions:
We consider robotic taxis that can carry one customer. To compare different AMoD system operational policies, we introduce the following variables:
\begin{align*} &d_E &= &\text{ empty distance driven by the fleet} \\ &d_C &= &\text{ occupied distance driven by the fleet} \\ &d_T = d_C + d_E &= &\text{ total distance driven by the fleet} \\ &N &= &\text{ fleet size} \\ &R &= &\text{ number of customer requests served} \\ &w_i &= &\text{ waiting time of request } i\in \{1,...,R\} \\ &W &= &\text{ total waiting time } W = \sum_{i=1}^{R} w_i \end{align*}
The provided simulation environment is designed in the standard reinforcement framework: Rewards are issued after each simulation step. The (undiscounted) sum of all rewards is the final score. The higher the score, the better the performance.
For the AMoD-Task, there are 3 different championships (sub-tasks) which constitute separate competitions. The simulation environment computes the reward value for each category and conatenates them into a vector of length 3, which is then communicated as feedback to the learning agent. The agent can ignore but the entry of the reward vector from the category that they wish to maximize.
The three championships are as follows:
In the
Service Quality Championship, the principal goal of the operator is to provide the highest possible service quality at bounded operational cost. Two negative scalar weights $\alpha_1\lt{}0$ and $\alpha_2\lt{}0$ are introduced. The performance metric to maximize is
\begin{align*} \mathcal{J}_{P-AMoD,1} = \alpha_1 W + \alpha_2 d_E \end{align*}
The values $\alpha_1$ and $\alpha_2$ are chosen such that the term $W$ dominantes the metric. The number of robotic taxis is fixed at some fleet size $\bar{N} \in \mathbb{N}_{\gt{}0}$.
In the
Efficiency Championship, the principal goal of the operator is to perform as efficiently as possible while maintaining the best possible service level. Two negative scalar weights $\alpha_3\lt{}0$ and $\alpha_4\lt{}0$ are introduced. The performance metric to maximize is
\begin{align*} \mathcal{J}_{P-AMoD,2} = \alpha_3 W + \alpha_4 d_E \end{align*}
$\alpha_3$ and $\alpha_4$ are chosen such that the term $d_E$ dominantes the metric. The number of robotic taxis is fixed at some fleet size $\bar{N} \in \mathbb{N}_{\gt{}0}$.
Modified 2019-04-14 by Liam Paull
In the
Fleet Size Championship, the goal is to reduce the fleet size as much as possible while keeping the total waiting time $W$ below a fixed level $\bar{W}\gt{}0$. The condition $W\leq\bar{W}$ is equivalent to guaranteeing average waiting times smaller than $\frac{\bar{W}}{R}$. Therefore, the performance score to maximize is
\begin{align*} \mathcal{J}_{P-AMoD,3} = \begin{cases} -N & \text{if }W\leq\bar{W} \\ -\infty & \text{else} \end{cases} \end{align*}
Modified 2019-05-16 by Julian Zilly
The following shows rule objectives the Duckiebots are supposed to abide by within Duckietown. These penalties hold for the embodied tasks (LF, LFV).
Modified 2019-05-16 by Julian Zilly
This objective means to penalize “illegal” driving behavior. As a cover for many undesired behaviors, we count the median time spent oustide of the drivable zones. This also covers the example of driving in the wrong lane.
Metric: The median of the time spent outside of the drivable zones.
\begin{align*} \mathcal{J}_{T-LF/LFV} = \text{median}(\{t_{outside}\}), \end{align*}
where $\{t_{outside}\}$ is the list of accumulated time outside of drivable zones per episode.
Modified 2019-05-16 by Julian Zilly
Modified 2019-04-14 by Liam Paull
In the single robot setting, we encourage “comfortable” driving solutions. We therefore penalize large angular deviations from the forward lane direction to achieve smoother driving. This is quantified through changes in Duckiebot angular orientation $\theta_{bot}(t)$ with respect to the lane driving direction.
For better driving behavior we measure the median per episode lateral deviation from the right lane center line.
$$ \objective_{C-LF/LFV}(t) = \text{median}(\{d_{outside}\}), $$
where $\{d_{outside}\}$ is the sequence of lateral distances from the center line.
No questions found. You can ask a question on the website. |
I'm trying to design a filter that allows through a 1kHz sine wave, based on my university lecture notes I have the following transfer function for a multiple feedback band pass filter:
$$A(s) = \frac{-H\omega_0s}{s^2+(1/Q)\omega_0 s+\omega_0^2}$$ where \$\omega_0\$ is the center frequency and \$Q\$ is the quality factor.
I have calculated \$Q\$ to be 16.6667 (bandwidth of 60Hz) and \$\omega_0 = 2\times \pi \times 1000\$.
My lecturer has informed me that I can treat \$H\$ in the above transfer function to be a specification for the passband gain, I wish a gain of 0dB at the center frequency so I set \$H = 1\$. The problem is when I calculate my capacitor and resistor values using the provided formulas in my lecture slides, my frequency response is centered at 1000Hz, however it has a gain of approx 25dB (my chosen cap values are 100nF and R1 = 1.59k, R2 = 41, R5 = 64k).
How do I appropriately choose \$H\$ so that I have a gain of 0dB at the passband (aka 1kHz)? |
A bunch of points:$\def\Spec{\mathrm{Spec}\ }$
• Let $A$ be a cluster algebra over a field $k$, let $(x_1, \ldots, x_n)$ be a cluster and let $L$ be the Laurent polynomial ring $k[x_1^{\pm}, \ldots, x_n^{\pm}]$. I imaging your intended question is whether the map $\Spec L \to \Spec A$ is an open immersion. (You ask about a map $\mathbb{A}^n \to X$, but there isn't a natural such map; $\Spec L$ is a torus, not affine space.)
The answer is yes. The question is equivalent to asking whether $L$ is a localization of $A$. I claim that $L = A[x_1^{-1}, \ldots, x_n^{-1}]$. Proof: On the one hand, $A \subset L$ (by the Laurent phenomenon) and $x_1^{-1}$, ..., $x_n^{-1} \in L$ (obviously), so $A[x_1^{-1}, \ldots, x_n^{-1}] \subseteq L$. On the other hand, $L$ is generated by the $x_j$ and their reciprocals, and these are all in $A[x_1^{-1}, \ldots, x_n^{-1}]$, so $L \subseteq A[x_1^{-1}, \ldots, x_n^{-1}]$. $\square$.
• $\Spec A$ is not generally the union of cluster tori. This is true even in the simplest case: The extended exchange matrix $\left( \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \right)$ gives the cluster algebra$$\mathbb{C}[x,x', y^{\pm 1}]/x x'-y-1.$$The two tori are $xy \neq 0$ and $x' y \neq 0$. The point $(x,x',y) = (0,0,-1)$ is in neither torus.
• Some sources define "the cluster variety" as the (quasi-affine) union of cluster tori. If you take that as the definition, it is obviously smooth.
• But I imagine what you care about is whether $\Spec A$ is smooth (or possibly $\Spec U$, where $U$ is the upper cluster algebra. No, that doesn't have to be smooth. (The singular points are not in any of the cluster tori, of course.) I think the simplest example is the Markov clsuter algebra, with $B$-matrix $\left( \begin{smallmatrix} 0 & 2 & -2 \\ -2 & 0 & 2 \\ 2 & -2 & 0 \end{smallmatrix} \right)$. This ring is not finitely generated, and there is a maximal ideal generated by all cluster variables where the Zariski tangent space is infinite dimensional. If you like at the upper cluster algebra instead, it is $k[\lambda, x_1, x_2, x_3]/x_1 x_2 x_3 \lambda - x_1^2- x_2^2 - x_3^3$, which is singular along the line $x_1 = x_2 = x_3 = 0$.
For another example, which doesn't have the $A$ versus $U$ issue, look at the $A_3$ cluster algebra with no frozen variables. From Corollary 1.17 in Cluster Algebras III, this is generated by $(x_1, x_2, x_3, x'_1, x'_2, x'_3)$ module the relations$$x_1 x'_1 = x_2+1,\ x_2 x'_2 = x_1 + x_3,\ x_3 x'_3 = x_2+1.$$
Look at the point $(x_1, x_2, x_3 , x'_1, x'_2, x'_3) = (0, -1, 0, 0,0,0)$. The Jacobian matrix of these $3$ equations with respect to the $6$ variables is$$\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \end{pmatrix}$$which has rank two, so this is a singular point.
• We do have Theorem 7.7 of Greg Muller's Locally Acyclic Cluster Algebras -- if the cluster algebra is locally acyclic, the $B$-matrix has full rank and our ground field has characteristic zero, then $\Spec A$ is smooth.
See Muller 1 and Benito-Muller-Rajchgot-Smith 2 for more on singularities of cluster varieties. |
Yes, they can. It's not great form, but they can.
For a simple example, consider the hamiltonian for a 1D particle in a potential $V(x)$,$$\hat H=\frac1{2m}\hat{ p}^2+V(\hat{x}).$$This holds equally well if the potential is a delta function at $x_0$, $V(x)=\kappa \delta(x-x_0)$, in which case$$\hat H=\frac1{2m}\hat{ p}^2+\kappa\delta(\hat{x}-x_0).$$Functions of an operator are usually defined by transplanting its power series, but the way it's really done is by acting on the eigenvalues, so for any potential $V(x)$ you have$$V(\hat x)|x⟩=V(x)|x⟩$$regardless of whether the function $V(x)$ is analytic or not - including such "ugly" beasts as discontinuous potentials. The same holds true for our delta function:$$\delta(\hat x-x_0)|x⟩=\delta(x-x_0)|x⟩,$$though of course this is still problematic because the delta function is not actually a function. To get to what this really means, you apply the definition of the delta function: when integrated against an arbitrary function $f(x)$ it returns a specific value. Thus, you do$$\int\mathrm d x\: f(x)\delta(\hat x-x_0)|x⟩=\int\mathrm d x\: f(x)\delta(x-x_0)|x⟩=f(x_0)|x_0⟩.$$That is more OK, but it might still worry you. If you look closely, our operator $\delta(\hat x-x_0)$ has turned what might have been a perfectly reasonable state $\int f(x)|x⟩\mathrm dx$ into the singular state $f(x_0)|x_0⟩$, which isn't really a state at all - it sits outside of the normal Hilbert space. The way to resolve that is not to look at the eigenkets but at the eigen
bras of the position operator, which is actually how functions of the position operator should work,$$⟨x|V(\hat x)=V(x)⟨x|$$(because then the wavefunction of $V(\hat x)|\psi⟩$ is $⟨x|V(\hat x)|\psi⟩=V(x)⟨x|\psi⟩=V(x)\psi(x)$). Applying this to the delta function, the real definition of $\delta(\hat x-x_0)$ is that operator for which$$\bbox[5px,border:1px solid black]{\int\mathrm d x\: f(x)⟨x|\:\delta(\hat x-x_0)=f(x_0)⟨x_0|}$$for arbitrary (sufficiently well-behaved) functions $f$.
(In mathematical terms, this is no longer worrisome, as it takes linear functionals over $\mathcal H$, like $\int\mathrm d x\: f(x)⟨x|$, into other linear functionals, like $f(x_0)⟨x_0|$. Our worry from before (that $|x_0⟩$ wasn't a state) has now been de-fanged: the argument says that $\delta(\hat x-x_0)$ can take a bounded functional and return a non-bounded functional. That's perfectly fine - $\hat x$ itself does the same.)
In the example you quote it is hard to know what is and is not an operator without further context, but nothing looks like it needs to be an operator for the calculation to work. Unless you provide further context, all quantities you mention are perfectly valid c-numbers (i.e. not operators).
To do the calculation, you simply substitute for $I$ and invert the order integration and expectation value to get$$G(\vec r,t) =\frac{1}{N}\sum_{i,j} \left <\frac{1}{(2\pi)^3}\int e^{i\vec Q·(\vec R_i(t)-\vec R_j(0)-\vec r)}\ d^3 \vec Q\right >.$$The integral evaluates to a single delta function in $\vec r$:$$G(\vec r,t) =\frac{1}{N}\sum_{i,j} \left <\delta\left(\vec R_i(t)-\vec R_j(0)-\vec r\right)\right >.$$This coincides with the other result you give once you cancel out the $\vec r'$ integral, using the $\delta(\vec r' - \vec R_j(t))$ term, to substitute $\vec R_j(t)$ for $\vec r'$. It's hard to know why the author did this; presumably he had reasons to choose the form you give. Without more context it's hard (impossible) to tell. |
Edit: I'm leaving the old post below, but before I want to write the proof as suggested by Bruce from his book, which uses the ideas in a more efficient way.
Assume that $\|p-q\|<1$, with $p,q\in A$, a unital C$^*$-algebra. Let $x=pq+(1-p)(1-q)$. Then, as $2p-1$ is a unitary, $$\|1-x\|=\|(2p-1)(p-q)\|=\|p-q\|<1.$$So $x$ is invertible. Now let $x=uz$ be the polar decomposition, $z=(x^*x)^{1/2}\in A$. Then $u=xz^{-1}\in A$. Also, $px=pq=xq$, and $qx^*x=qpq$, so $qx^*x=x^*xq$, and then $qz=zq$. Then$$pu=pxz^{-1}=xqz^{-1}=uzqz^{-1}=uqzz^{-1}=uq.$$So $q=u^*pu$.
=============================================
(the old post starts here)
(A good friend pointed me to the ideas in this answer, so I'm sharing them here)
The result holds in any unital C$^*$-algebra. So assume that $\|p-q\|<1$, with $p,q$ in a unital C$^*$-algebra $A\subset B(H)$.
Claim 1: There is a continuous path of projections joining $p$ and $q$.
Proof. Let $\delta\in(0,1)$ with $\|p-q\|<\delta$. For each $t\in[0,1]$, let $x_t=tp+(1-t)q$. Then$$\|x_t-p\|=\|(1-t)(p-q)\|<\delta(1-t),$$$$\|x_t-q\|=\|t(p-q)\|<\delta t.$$This, together with the fact that $x_t$ is selfadjoint, implies that $\sigma(x_t)\subset K=[-\delta/2,\delta/2]\cup[1-\delta/2,1+\delta/2]$ (since $\min\{t,1-t\}\leq1/2$). Now let $f$ be the continuous function on $K$ defined as $0$ on $[-\delta/2,\delta/2]$ and $1$ on $[1-\delta/2,1+\delta/2]$. Then, for all $t\in[0,1]$, $f(x_t)\in A$ is a projection. And$$t\to x_t\to f(x_t)$$is continuous, completing the proof of the claim. Edit: years later, I posted this answer to a question on MSE that proves the continuity.
Claim 2: We may assume without loss of generality that $\|p-q\|<1/2$.
This is simply a compacity argument, using that each projection in the path $f(x_t)$ is very near another projection in the path. The compacity allows us to make the number of steps finite, and so if we find projectons $p=p_0,p_1,\ldots,p_n=q$ and unitaries with $u_kp_ku_k^*=p_{k+1}$, we can multiply the unitaries to get the unitary that achieves $q=upu^*$.
Claim 3: If $\|p-q\|<1/2$, there exists a unitary $u\in A$ with $q=upu^*$.
Let $x=pq+(1-p)(1-q)$. Then $$\|x-1\|=\|2pq-p-q\|=\|p(q-p)+(p-q)q\|\leq2\|p-q\|<1,$$so $x$ is invertible. Let $x=uz$ be the polar decomposition. Then $u$ is a unitary. Note that$$qx^*x=q(qpq+(1-q)(1-p)(1-q))=qpq,$$so $q$ commutes with $x^*x$ and then with $z=(x^*x)^{1/2}$. Note also that $px=xq$, so $puz=uzq=uqz$. As $z$ is invertible, $pu=uq$, i.e.$$q=u^*pu.$$Note that $u=xz^{-1}\in A$. |
The subject of this paper is the mathematical theory of catastrophes. A catastrophe is a disaster. But there is, implicit in the word, the idea of a sudden change for the worse.
We all know and have used the approximation $\pi \simeq \frac{22}{7}$. It may have occurred to you to ask why this is a worthwhile value.
The geometrical proposition, named after the Greek mathematician and philosopher, Pythagoras, (~ 570-550 BC), deals with a unique property of right-angled triangles.
Q.817 Find all integers $x,y$ such that $x(3y - 5) = y^2 +1$. |
The Hammett plot is commonly invoked in organic chemistry to reason about the plausibility (or implausibility) of various reaction mechanisms. The vertical axis is essentially the logarithm of an equilibrium constant (or rate constant) measured relative to a hydrogen functional group, which I understand. However, I am at a complete loss for understanding what is being plotted on the horizontal axis, which conventionally are denoted $\sigma$ and $\sigma^\pm$. What are these parameters and how are their values determined for various functional groups? In practice it seems that people look them up in tables, but where do the values in these tables come from?
Hammett wanted to find a way to quantify the effects electron-withdrawing (EWG) and - donating (EDG) groups have on the transition state or intermediate during the course of a reaction. Initially he took the $pK_{\mathrm{a}}$-values of benzoic acids carrying the respective functional group in para or meta position (ortho acids and aliphatic acids weren't used because steric effects would overlay with the electronic effects) as a guide and plotted them against the logarithm of the reaction rates. So, he based his scale on the following reaction reaction:
Later Hammett decided not to use the $pK_{\mathrm{a}}$s themselves for his correlation but defined a new parameter, which he called $\sigma$. This $\sigma$ shows how electron-donating or -withdrawing a group is relative to $\ce{H}$ as a ratio of the $\log K_a$s or the difference of the $pK_{\mathrm{a}}$ between the substituted benzoate and benzoic acid itself. If the acid required to determine $\sigma$ for a new substituent was not available, $\sigma$ could be determined by correlation with other reactions. A different value of $\sigma$ for any given substituent was needed for the meta and para positions (sometimes called $\sigma_{\mathrm{m}}$ and $\sigma_{\mathrm{p}}$, respectively). The equation for $\sigma$ is
\begin{equation} \sigma_{\ce{X}} = \log \left( \frac{ K_{\mathrm{a}}(\ce{X-C6H4COOH}) }{ K_{\mathrm{a}}(\ce{H-C6H4COOH}) } \right) = pK_{\mathrm{a}}(\ce{X-C6H4COOH}) - pK_{\mathrm{a}}(\ce{H-C6H4COOH}) \end{equation}
or in short notation
\begin{equation} \sigma_{\ce{X}} = \log \left( \frac{ K_{a} (\ce{X}) }{ K_{a} (\ce{H}) } \right) = pK_{\mathrm{a}}(\ce{X}) - pK_{\mathrm{a}}(\ce{H}) \end{equation}
where $\ce{X}$ is the functional group, i.e. the EWG or EDG, whose effect on the reaction rate shall be evaluated. It is also possible to determine the $\sigma$ values via the logarithm of the ratio of the rate constants for the aforementioned reaction with substituent $\ce{X}$ (in short notation: $k_{\ce{X}}$) and with $\ce{X}=\ce{H}$ (in short notation: $k_{\ce{H}}$)
\begin{equation} \sigma_{\ce{X}} = \log \left( \frac{ k_{\ce{X}} }{ k_{\ce{H}} } \right) \end{equation}
Update
AcidFlask's comment reminded me that I forgot to say something about $\sigma^{+}$ and $\sigma^{-}$ values.As mentioned above the "normal" $\sigma$ values are based upon the difference of the $pK_{\mathrm{a}}$ between a substituted benzoate and benzoic acid itself.But for benzoates one cannot draw resonance structures that delocalize the negative charge onto the benzene ring via the $\pi$ electron system.Yet, many reations of interest create negative or positive charges that can be stabilized by delocalization via resonance with the substituent.For these reactions, one finds that Hammett plots using $\sigma$ values have considerable scatter.Therefore, two new substituent effect scales were produced, one for groups that stabilize negative charges via resonance ($\sigma^{-}$), and one for groups that stabilized positive charges via resonance ($\sigma^{+}$).The $\sigma^{-}$ scale is based upon the ionization of para-substituted phenols (I've found different accounts for this. According to the German Wikipedia the $\sigma^{-}$ scale is based upon the ionization of para-substituted anilins.),
for which groups like nitro can stabilize the negative charge via resonance. The $\sigma^{+}$ scale is based upon the heterolysis ($\mathrm{S}_{\mathrm{N}}1$) reaction of para-substituted cumyl chlorides (phenyldimethyl chloromethanes),
for which groups like amino can stabilize the positive charge via resonance. The $\sigma^{\pm}$ values can then be evaluated by the aforementioned equation
\begin{equation} \sigma_{\ce{X}}^{\pm} = \log \left( \frac{ K_{a}^{\pm} (\ce{X}) }{ K_{a}^{\pm} (\ce{H}) } \right) = \log \left( \frac{ k_{\ce{X}}^{\pm} }{ k_{\ce{H}}^{\pm} } \right) \end{equation}
but with the rate constants $k_{\ce{X}}^{\pm}$ and $k_{\ce{H}}^{\pm}$ (or the equilibrium constants $K_{a}^{\pm} (\ce{X})$ and $K_{a}^{\pm} (\ce{H})$) measured for their respective defining reactions. |
I am deriving the natural numbers with the Peano Axioms and have a question about the Axiom of mathematical induction. I am having trouble grasping the significance of Peanos formulation of this Axiom. For example I cannot see a difference between Peanos Axiom and the formulation:"If n is a natural number it can be reached by repeated incrementation starting from 0".
"If n is a natural number it can be reached by repeated incrementation starting from 0" is meaningfully different from the Peano Axiom of Induction. The Peano Axiom is not just about "is reachable" but is about any property. For example, "$x\geq 0$" is a property that we can use with the axiom of induction.
One issue is that the statement
A: "If n is a natural number it can be reached by repeated incrementation starting from 0"
seems to be trivially true: if $n>0$ is a natural number, then $n$ should be reachable by adding $1$ to itself $n$ times. This does not tell us about whether
other properties are maintained as we move upward through the set of naturals.
When we formalize the induction axioms in Peano arithmetic, the statement
B: "If $n>0$ is a natural number it can be reached by adding $1$ to itself $n-1$ times"
is just one consequence of the induction axioms. It is known that the entire set of induction axioms in Peano Arithmetic is not only infinite, but it is not implied by any of its finite subsets. So there are other, more complicated induction axioms that are not implied by $B$ alone; we need to include the entire set of induction axioms to get the full strength of PA.
On the other hand, the intuition that statement A should be true is one of the motivations for the other induction axioms: it helps us see why we might expect them to be true.
Statements $A$ and $B$ have interesting behavior in the context of nonstandard models. Although a nonstandard model of Peano Arithmetic will think that each nonstandard element $n$ can be reached by repeatedly adding $1$ to itself, the number of times that $1$ must be added to itself will also be nonstandard when $n$ is. Somewhat by definition, there is no way to reach a nonstandard number by adding 1 to itself a standard number of times.
The induction axiom can be formally stated as follows:
$\forall P\subset N: [0\in P \land \forall x\in P:[S(x) \in P] \implies P=N]$
Let $M$ be the subset of $N$ comprised of those and only those elements of $N$ that can be reached by repeated incrementation starting from $0$.
$M=\{0, S(0), S(S(0)), \cdots \}$
In other words, $M$ is smallest subset of $N$ such that $0\in M$ and for all $x\in M$, we also have $S(x)\in M$.
More formally:
$\forall a:[a\in M \iff a\in N \land \forall Q\subset N: [0\in Q \land \forall b\in Q: [S(b)\in Q] \implies a\in Q]]$
From the axioms of set theory, we know that this subset of $N$ will exist.
It is then a simple exercise to prove using the induction axiom that, for all $x\in N$ we also have $x\in M$.
EDIT 1: The converse is also true, so you are correct. They are indeed equivalent. Formal proof to follow.
EDIT 2: See machine-verified, formal proof (136 lines) as promised.
Theorem: Suppose we have a set $n$ (finte or otherwise), a function $s: n \to n$ and $n_0 \in n$. Then induction will hold on set $n$ with successor function $s$ and "first" element $n_0$ if and only if all elements of $n$ are reachable by repeated succession starting at $n_0$, i.e. the set $m = \{n_0, s(n_0), s(s(n_0)), ... \}$ contains every element of $n$.
More formally:
$\forall n,n_0,s:[n_0\in n \land s: n\to n\\\implies[\forall a\subset n:[n_0\in a \land \forall b\in a: [s(b)\in a] \implies n\subset a]\\\iff \exists m:[\forall a: [a\in m \iff a\in n \\\land \forall b\subset n:[n_0\in b \land \forall c\in b:[s(c)\in b]\implies a\in b]]]\\\land n\subset m]]$ |
Integrable Highest Weight Representations of Affine Lie algebras¶
In this section \(\mathfrak{g}\) can be an arbitrary Kac-Moody Lie Algebra made with a symmetrizable, indecomposable Cartan matrix.
Suppose that \(V\) is a representation with a weight space decompositionas in Roots and Weights. Let \(\alpha\) be a real root, and let\(\mathfrak{g}_\alpha\) be the corresponding root space, that is,the one-dimensional weight space for \(\alpha\) in the adjointrepresentation of \(\mathfrak{g}\) on itself. Then \(-\alpha\) is alsoa root. The two one-dimensional spaces \(\mathfrak{g}_\alpha\) and\(\mathfrak{g}_{-\alpha}\) generate a Lie algebra isomorphic to\(\mathfrak{sl}_2\). The module \(V\) is called
integrable if for eachsuch \(\alpha\) the representation of \(\mathfrak{sl}_2\) obtained thisway integrates to a representation of the Lie group \(\operatorname{SL}_2\).Since this group contains an element that stabilizes \(\mathfrak{h}\)and induces the corresponding simple reflection on the weight lattice,integrability implies that the weight multiplicities are invariantunder the action of the Weyl group.
If the Kac-Moody Lie algebra \(\mathfrak{g}\) is finite-dimensionalthen the integrable highest weight representations arejust the irreducible finite-dimensional ones. For a generalKac-Moody Lie algebra the integrable highest weight representationsare the analogs of the finite-dimensional ones, that is,with
WeylCharacterRing elements. Theirtheory has many aspects in common with the finite-dimensionalrepresentations of finite-dimensional simple Lie algebras,such as the parametrization by dominant weights, andgeneralizations of the Weyl denominator and characterformulas, due to Macdonald and Kac respectively.
If \(\Lambda\) is a dominant weight, then the irreducible highest weight module \(L(\Lambda)\) defined in Roots and Weights is integrable. Moreover every highest weight integrable representation arises this way, so these representations are in bijection with the cone \(P^+\) of dominant weights.
The affine case¶
Now we assume that \(\mathfrak{g}\) is affine. The integrablehighest weight representations (and their crystals) areextremely interesting. Integrable highest weight representations of\(\mathfrak{g}\) arise in a variety of contexts, from stringtheory to the modular representation theory of the symmetricgroups, and the theory of modular forms. The representation \(L(\Lambda_0)\)is particularly ubiquitous and is called the
basic representation.
Therefore in [KMPS] (published in 1990) tabulated data for many of these representations. They wrote
We present a vast quantity of numerical data in tabular form, this being the only source for such information. The computations are tedious and not particularly straightforward when it is necessary to carry them out individually. We hope the appearance of this book will spur interest in a field that has become, in barely 20 years, deeply rewarding and full of promise for the future. It would indeed be gratifying if these tables were to appear to the scientists of 2040 as obsolete as the dust-gathering compilations of transcendental functions appear for us today because of their availability on every pocket calculator.
As we will explain, Sage can reproduce the contents of these tables. Moreover the tables in [KMPS] are limited to the untwisted types, but Sage also implements the twisted types.
Although Sage can reproduce the tables in the second volume of [KMPS], the work remains very useful. The first volume is a down-to-earth and very helpful exposition of the theory of integrable representations of affine Lie algebras with explicit examples and explanations of the connections with mathematical physics and vertex operators.
The support of an integrable highest weight representation¶
Let \(\Lambda \in P^+\) and let \(V = L(\Lambda)\) be the integrablerepresentation with highest weight \(\Lambda\). If \(\mu\) is anotherweight, let \(\operatorname{mult}(\mu)\) denote the multiplicity ofthe weight \(\mu\) in \(L(\Lambda)\). Define the
support of therepresentation \(\operatorname{supp}(V)\) to be the set of \(\mu\)such that \(\operatorname{mult}(\mu) > 0\).
If \(\operatorname{mult}(\mu) > 0\) then \(\lambda-\mu\) is a linear combination of the simple roots with nonnegative integer coefficients. Moreover \(\operatorname{supp}(V)\) is contained in the paraboloid
where \((\, | \,)\) is the invariant inner product on the weight lattice and \(\rho\) is the Weyl vector (Untwisted Affine Kac-Moody Lie Algebras). Moreover if \(\mu \in \operatorname{supp}(V)\) then \(\Lambda - \mu\) is an element of the root lattice \(Q\) ([Kac], Propositions 11.3 and 11.4).
We organize the weight multiplicities into sequences called
string functions or strings as follows. By [Kac], Proposition 11.3or Corollary 11.9, for fixed \(\mu\) the function\(\operatorname{mult}(\mu - k\delta)\) of \(k\) is an increasing sequence.We adjust \(\mu\) by a multiple of \(\delta\) to the beginningof the positive part of the sequence. Thus we define \(\mu\) to be maximal if \(\operatorname{mult}(\mu) \neq 0\) but\(\operatorname{mult}(\mu + \delta) = 0\).
Since \(\delta\) is fixed under the action of the affine Weyl group, and since the weight multiplicities are Weyl group invariant, the function \(k \mapsto \operatorname{mult}(\mu - k \delta)\) is unchanged if \(\mu\) is replaced by \(w(\mu)\) for some Weyl group element \(w\). Now every Weyl orbit contains a dominant weight. Therefore in enumerating the string we may assume that the weight \(\mu\) is dominant. There are only a finite number of dominant maximal weights. Thus there are only a finite number of such strings to be computed.
Modular Forms¶
Remarkably, [KacPeterson] showed that each string is the set ofFourier coefficients of a weakly holomorphic modular form; see also[Kac] Chapters 12 and 13. Here
weakly holomorphic modular meansthat the form is allowed to have poles at cusps.
To this end we define the
modular characteristic
Here \(k = (\Lambda | \delta)\) is the
level of the representation and\(h^\vee\) is the dual Coxeter number (Labels and Coxeter Number).If \(\mu\) is a weight, define
Let \(\Lambda\) be a weight, which we may assume maximal. Then Kacand Peterson defined the
string function
Although these do arise as partition functions in string theory, the term “string” here does not refer to physical strings.
The string function \(c_\mu^\Lambda\) is a weakly holomorphic modular form, possibly of half-integral weight. See [Kac], Corollary 13.10, or [KacPeterson]. It can have poles at infinity, but multiplying \(c_\mu^\Lambda\) by \(\eta(\tau)^{\dim\,\mathfrak{g}^\circ}\) gives a holomorphic modular form (for some level). Here \(\eta\) is the Dedekind eta function:
The weight of this modular form \(\eta(\tau)^{\dim\,\mathfrak{g}^\circ} c^\Lambda_\lambda\) is the number of positive roots of \(\mathfrak{g}^\circ\).
Sage methods for integrable representations¶
In this section we will show how to use Sage to compute withintegrable highest weight representations of affine Lie algebras.For further documentation, see the reference manual
IntegrableRepresentation.
In the following example, we work with the integrable representationwith highest weight \(2 \Lambda_0\) for \(\widehat{\mathfrak{sl}}_2\),that is, \(A_1^{(1)}\). First we create a dominant weight inthe extended weight lattice, then create the
IntegrableRepresentationclass. We compute the strings. There are two, since there are twodominant maximal weights. One of them is the highest weight \(2\Lambda_0\),and the other is \(2\Lambda_1 - \delta\):
sage: L = RootSystem("A1~").weight_lattice(extended=True)sage: Lambda = L.fundamental_weights()sage: delta = L.null_root()sage: W = L.weyl_group(prefix="s")sage: s0, s1 = W.simple_reflections()sage: V = IntegrableRepresentation(2*Lambda[0])sage: V.strings(){2*Lambda[0]: [1, 1, 3, 5, 10, 16, 28, 43, 70, 105, 161, 236], 2*Lambda[1] - delta: [1, 2, 4, 7, 13, 21, 35, 55, 86, 130, 196, 287]}sage: mw1, mw2 = V.dominant_maximal_weights(); mw1, mw2(2*Lambda[0], 2*Lambda[1] - delta)
We see there are two dominant maximal weights, \(2 \Lambda_0\) and \(2 \Lambda_1 - \delta\). We obtain every maximal weight from these by applying Weyl group elements. These lie inside the paraboloid described in The support of an integrable highest weight representation. Here are a few more maximal weights:
sage: pairs = [(s0*s1*s0, mw1), (s0*s1, mw2), (s0, mw1), (W.one(), mw2),....: (W.one(), mw1), (s1, mw2), (s1*s0, mw1), (s1*s0*s1, mw2)]sage: [w.action(mw) for (w, mw) in pairs][-6*Lambda[0] + 8*Lambda[1] - 8*delta, -4*Lambda[0] + 6*Lambda[1] - 5*delta, -2*Lambda[0] + 4*Lambda[1] - 2*delta, 2*Lambda[1] - delta, 2*Lambda[0], 4*Lambda[0] - 2*Lambda[1] - delta, 6*Lambda[0] - 4*Lambda[1] - 2*delta, 8*Lambda[0] - 6*Lambda[1] - 5*delta]
We confirm that the string function for one in the Weyl orbitis the same as that for
mw2, calculated above:
sage: s1.action(mw2)4*Lambda[0] - 2*Lambda[1] - deltasage: [V.mult(s0.action(mw2)-k*delta) for k in [0..10]][1, 2, 4, 7, 13, 21, 35, 55, 86, 130, 196]
String functions of integrable representations often appear in the Online Encyclopedia of Integer Sequences:
sage: [oeis(x) for x in V.strings().values()] # optional - internet[0: A233758: Bisection of A006950 (the even part)., 0: A233759: Bisection of A006950 (the odd part).]
Reading what the OEIS tells us about the sequence OEIS sequence A006950, we learn that the two strings are the odd and even parts of the series
This is
not a modular form because of the factor \(q^{1/8}\) infront of the ratio of eta functions.
Let us confirm what the Online Encyclopedia tells us by computing the above product:
sage: PS.<q> = PowerSeriesRing(QQ)sage: prod([(1+q^(2*k-1))/(1-q^(2*k)) for k in [1..20]])1 + q + q^2 + 2*q^3 + 3*q^4 + 4*q^5 + 5*q^6 + 7*q^7 + 10*q^8 + 13*q^9 + 16*q^10 + 21*q^11 + 28*q^12 + 35*q^13 + 43*q^14 + 55*q^15 + 70*q^16 + 86*q^17 + 105*q^18 + 130*q^19 + O(q^20)
We see the values of the two strings interspersed in this product, with the \(2 \Lambda_0\) string values in the even positions and the \(2 \Lambda_1 - \delta\) values in the odd positions.
To compute \(c^{2\Lambda_0}_\lambda\), which is guaranteed to be a modular form, we must compute the modular characteristics. We are interested in the cases where \(\lambda\) is one of the two dominant maximal weights:
sage: [V.modular_characteristic(x) for x in [2*Lambda[0], 2*Lambda[1]-delta]][-1/16, 7/16]
This gives us the string functions
These are both weakly holomorphic modular forms. Any linear combination of these two is also a weakly holomorphic modular form. For example we may replace \(\tau\) by \(\tau/2\) in our previous identity and get
Let \(V\) be the integrable highest weight representation with highest weight \(\Lambda\). If \(\mu\) is in the support of \(V\) then \(\Lambda - \mu\) is of the form \(\sum_i n_i\alpha_i\) where \(\alpha_i\) are the simple roots. Sage employs an internal representation of the weights as tuples \((n_0, n_1, \ldots)\). You can convert weights to and from this notation as follows:
sage: L = RootSystem(['E',6,2]).weight_lattice(extended=True)sage: Lambda = L.fundamental_weights()sage: delta = L.null_root()sage: V = IntegrableRepresentation(Lambda[0])sage: V.strings(){Lambda[0]: [1, 2, 7, 14, 35, 66, 140, 252, 485, 840, 1512, 2534]}sage: V.to_weight((1,2,0,1,0))Lambda[0] - 3*Lambda[1] + 4*Lambda[2] - 2*Lambda[3] + Lambda[4] - deltasage: V.from_weight(Lambda[0] - 3*Lambda[1] + 4*Lambda[2] - 2*Lambda[3] + Lambda[4] - delta)(1, 2, 0, 1, 0)sage: V.from_weight(Lambda[0]-delta)(1, 2, 3, 2, 1)
In reporting the strings, one may set the optional parameter depth to get more or fewer values. In certain cases even the first coefficient of the string is significant. See [JayneMisra2014] and [KimLeeOh2017].
Catalan numbers (OEIS sequence A000108):
sage: P = RootSystem(['A',12,1]).weight_lattice(extended=true)sage: Lambda = P.fundamental_weights()sage: IntegrableRepresentation(2*Lambda[0]).strings(depth=1) # long time{2*Lambda[0]: [1], Lambda[1] + Lambda[12] - delta: [1], Lambda[2] + Lambda[11] - 2*delta: [2], Lambda[3] + Lambda[10] - 3*delta: [5], Lambda[4] + Lambda[9] - 4*delta: [14], Lambda[5] + Lambda[8] - 5*delta: [42], Lambda[6] + Lambda[7] - 6*delta: [132]}
Catalan triangle numbers (OEIS sequence A000245):
sage: IntegrableRepresentation(Lambda[0]+Lambda[2]).strings(depth=1) # py2 long time{Lambda[0] + Lambda[2]: [1], 2*Lambda[1] - delta: [12], Lambda[3] + Lambda[12] - delta: [3], Lambda[4] + Lambda[11] - 2*delta: [9], Lambda[5] + Lambda[10] - 3*delta: [28], Lambda[6] + Lambda[9] - 4*delta: [90], Lambda[7] + Lambda[8] - 5*delta: [297]}sage: sorted(IntegrableRepresentation(Lambda[0]+Lambda[2]).strings(depth=1).values()) # long time[[1], [3], [9], [12], [28], [90], [297]]
sage: P = RootSystem(['B',8,1]).weight_lattice(extended=true)sage: Lambda = P.fundamental_weights()sage: IntegrableRepresentation(Lambda[0]+Lambda[1]).strings(depth=1) # py2 long time{Lambda[0] + Lambda[1]: [1], 2*Lambda[0]: [1], 2*Lambda[1] - delta: [1], Lambda[2] - delta: [3], Lambda[3] - delta: [3], Lambda[4] - 2*delta: [10], Lambda[5] - 2*delta: [10], Lambda[6] - 3*delta: [35], Lambda[7] - 3*delta: [35], 2*Lambda[8] - 4*delta: [126]}sage: sorted(IntegrableRepresentation(Lambda[0]+Lambda[1]).strings(depth=1).values()) # long time[[1], [1], [1], [3], [3], [10], [10], [35], [35], [126]] |
Statistics - (F-Statistic|F-test|F-ratio) 1 - About
It's what you observe in the numerator relative to what you would expect just due to chance in the denominator.
The f statistic is the statistic that we'll obtain if we dropped out the predictors in the model.
A lot of people refer to it as an F-ratio because it's the variance between the groups relative to variants within the groups.
An ANOVA will tell with the F-ratio if:
there is an effect overall there is significant difference somewhere
The F-test has a family of F-distributions.
The F statistic tests the null hypothesis that none of the predictors has any effect. Rejecting that null means concluding that *some* predictor has an effect, not that *all* of them do.
2 - Articles Related 3 - Formulas
<MATH> \begin{array}{rrl} \text{F-ratio (F-test)} & = & \frac{\href{variance}{variance}\text{ between the groups}}{\href{variance}{variance}\text{ within the groups}} \\ & = & \frac{\text{systematic variance}}{\text{unsystematic variance}} \\ & = & \frac{\text{good variance}}{\text{bad variance}} \\ \end{array} </MATH>
The variance between the groups, or across the groups, is a good variance because it was created with our independent variable (within the experimental treatment).
The variants within the groups try to determine why in one group the score of individual differ. And we don't know because that variance is unsystematic, it's just what I would expect due to chance.
If you get a ratio of two or three, you're probably going to have a significant effect with the p-value less than 0.05.
<MATH> \begin{array}{rrl} \text{F-ratio (F-test)} & = & \frac{\href{variance}{variance}\text{ between the groups}}{\href{variance}{variance}\text{ within the groups}} \\ & = & \frac{{\href{variance}{\text{Mean Square (MS)}}}_{Betweeen}}{{\href{variance}{\text{Mean Square (MS)}}}_{Within}} \\ & = & \frac{{\href{variance}{\text{MS}}}_{A}}{{\href{variance}{\text{MS}}}_{S/A}} \\ \end{array} </MATH>
where:
A is the independent variable (the manipulation) S within A is the way to read that error term. So, it's subjects within groups
Mean square (MS) is variance
<MATH> \begin{array}{rrl} \text{F-ratio (F-test)} & = & \frac{{\href{variance}{\text{MS}}}_{A}}{{\href{variance}{\text{MS}}}_{S/A}} \\ & = & \frac{\text{Sum of the Squares (SS)}_A}{\href{degree_of_freedom}{df}_A}. \frac{\href{degree_of_freedom}{df}_{S/A}}{\text{Sum of the Squares (SS)}_{S/A}} \\ & = & \frac{SS_A}{\href{degree_of_freedom}{df}_A}. \frac{\href{degree_of_freedom}{df}_{S/A}}{SS_{S/A}} \\ \end{array} </MATH> where:
<math>SS_A</math> will compare each group mean to the grand mean to get the variance across groups. <math>SS_{S/A}</math> will look at each individual within a group and see how much they differ from their group mean. The <math>\href{degree_of_freedom}{df}_A</math> is the number of group minus one. The <math>\href{degree_of_freedom}{df}_{S/A}</math> is the number of subjects in a group minus one times the number of groups.
<MATH> \begin{array}{rrl} SS_A & = & n \sum_{j=1}^{N}(Y_j - Y_T)^2 \end{array} </MATH>
where:
<math>N</math> is the number of group <math>n</math> is the number of subjects in each group (because if they're very large groups then we have
to take that into account)
<math>Y_j</math> is a group mean <math>Y_T</math> is the grand mean
<MATH> \begin{array}{rrl} SS_{S/A} & = & \sum_{j=1}^{N}\sum_{i=1}^{n}(Y_{ij} - Y_j)^2 \end{array} </MATH>
where:
<math>N</math> is the number of group <math>n</math> is the number of subjects in each group <math>Y_{ij}</math> is an individual scores <math>Y_j</math> is a group mean |
let $(X, \leq_*)$ be a partially ordered set.
Assume there is an isomorphism $f: (X,\leq_*) \to (\mathbb Z, \leq)$
let $A \subseteq X$ be a well ordered subset of $X$ with an upper bound. Meaning there is an element $b \in X$ such that $\forall a \in A, a\leq_* b$.
Show that $A$ is a finite set.
I'm not sure how to approach this problem. And I'm not sure I understand what we are given too. on one hand, $(X,\leq_*)$ is partially ordered. But on the other hand, it is isomorphic to $(\mathbb Z,\leq)$ which is linearly ordered. |
I am trying to do a Molecular Dynamics Simulation of a complex fluid, confined between solid surfaces. I would like to find the flow rate as a function of fluid film thickness, $h$, for a plane poiseuille flow. (The flow rate is not expected to show an $h^3$ dependence, in general, if $h$ is small, in the order of a few nanometers).
The question I have is, how I can get an estimate for the simulation time required for the system to attain steady state, so the flow rate measurements can be made after such time. If we assumed continuum, we would have this equation:
$ \begin{align*} \frac{\partial M}{\partial t} = \nu \frac{\partial^2 M}{\partial y^2} - \frac{1}{\rho} \frac{\partial P}{ \partial x} \end{align*}$
to describe this unsteady Poiseuille flow($M$ being the Momentum, $\nu$ the kinematic viscosity and $P$ the pressure), with the applied pressure gradient being very high (since it is applied via adding high elementary body forces to the particles) to start with, when compared to the viscous diffusion term. The pressure gradient term can be assumed to be constant along the direction of flow $x$ and in time, $t$.
Now, if we didn't have the pressure term, we would have a momentum diffusion equation from where we could estimate the time required for the unsteadiness to decay. But, from the above equation (or from any other starting point), it is not obvious to me how one can get an estimate for the simulation time required for steady state to be reached.
Thank you very much for your patience and help! |
Question:
Find the Fourier transform of the signal
{eq}x(t)=e^{-t} u(t){/eq}, where {eq}u(t){/eq} is the unit step
function.
Fourier transform:
Fourier transform is a method, or we can say technique in which we transform a function of time say x(t) to a function of frequency. Fourier transform is related to Fourier series to some extent. So, we can say the Fourier transform is the special case of Fourier series because the derivation of the Fourier transform is come from here. Fourier series is widely used in many fields such as physics, mathematics, engineering, etc.
Answer and Explanation:
We have given that,
{eq}x\left( t \right) = e^{ - t} u\left( t \right), {/eq} here is the unit step.
We know that the Fourier transform of this given signal is,
{eq}\begin{align*} X\left( w \right) &= \int\limits_{ - \infty }^\infty {x\left( t \right)e^{ - jwt} dt} \\ &= \int\limits_{ - \infty }^\infty {e^{ - t} e^{ - jwt} u\left( t \right)dt} \\ \end{align*} {/eq}
Since, unit step is defined by {eq}\left( {0,\infty } \right) {/eq}
So,
{eq}\begin{align*} X\left( w \right) &= \int\limits_{ - \infty }^\infty {e^{ - t} \cdot e^{ - jwt} dt} \\ &= \int\limits_{ - \infty }^\infty {e^{ - t\left( {1 + jwt} \right)} dt} \\ &= \left[ {\dfrac{{e^{^{ - t\left( {jw + 1} \right)} } }}{{ - \left( {jw + 1} \right)}}} \right]_0^\infty \\ &= \left[ {\dfrac{{e^{ - \infty \left( {jw + 1} \right)} }}{{ - \left( {jw + 1} \right)}}} \right] - \left[ {\dfrac{{e^{ - 0\left( {jw + 1} \right)} }}{{ - \left( {jw + 1} \right)}}} \right] \\ &= 0 + \dfrac{1}{{1 + jw}} \\ X\left( w \right) &= \dfrac{1}{{1 + jw}} \\ \end{align*} {/eq}
Hence, Fourier transform of given signal is {eq}\dfrac{1}{{1 + jw}}. {/eq}
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from Calculus: Tutoring SolutionChapter 3 / Lesson 6 |
This blog post will outline how Novoda mines organisation data from our Github projects, talk about the data mining architecture we thought of, helping you understand the Github API and improve the visibility of your organisation data.
Here at Novoda, we are proud to say that all decisions we take are data-driven. Why choose one technology over another? Why prefer a generic architecture over a specific one? Why develop a custom made solution over a pre-existing library?
In combination with this line of questioning, we’ve been asking ourselves how can we extract and analyse relevant information from Github around: team dynamics, developer time spent on pull requests, details of cross-project collaboration and how much open source projects participation there is.
We can break down the above description into a list of goals for a report system that is able to give us metrics regarding:
Understanding this data may reveal fundamental clues in driving actions to improve team organisation, identify if and where bottlenecks are, then proactively solve them.
Github already has a statistics page for each repo, but that is a bit limited and cannot respond to all of our requirements. So we began exploring all possibilities for retrieving these statistics using the Github APIs.
At first we thought of requesting data on demand. For instance, getting statistics about a user in a certain period of the year would be solved by querying that user’s data from the API. But then we asked ourselves how we could
compare multiple users’ data, which led to the realisation that we should actually query data for all of organisation's (Novoda) developers at once.
\begin{align}
& (130~repos) * (1~issue~list + 5~issues) * (30~days) \simeq 23,400~requests \Rightarrow \ \Rightarrow~ & \frac{23,400~requests}{5000~requests~per~hour} \simeq 5~hours \end{align}
Should any of these requests fail, or should we want to change the date range even by a day, we would have to run the whole querying process all over again!
As good Computer Science scholars, we know that the next step would be to get all of Novoda Github data in our own data mart, then mine it to extract all the pieces of information we need. In regards to the type of data we want to extract out of the database, it would only contain the minimum subset of information we need, making it as small and efficient as possible.
Github provides us with two ways to access its information:
For requests using Basic Authentication or OAuth, you can make up to 5,000 requests per hour.
Currently set at 5000 request per hour, the generous Github rate limit allows for excellent integrations in minimum and medium-load clients, but it is definitely too low for us wanting to mine all of our historical data.
This issue is connected to another, more practical issue: where can we run the whole system? Downloading the history for the entire organisation would definitely take a few days at the very least (we’ve been active on Github since 2011), while the live one (using the WebHooks) would need to be always on and performant to save all live data.
After a lot of coffee and many Super Mario Kart 8 races, we came up with a multi-tiered solution that we believe is optimal for our use case.
Studying the Github API and data, we figured out that the data we needed to retrieve is tree-shaped:
By building and going through this tree we know for sure that all the information will be received and persisted in an orderly fashion, otherwise all our database constraints would be invalid (e.g., repositories must be retrieved and stored before moving on to issues).
The “live” system would run on the cloud, exposing an endpoint that listens to, converts and persists relevant information sent by Github.
The “historical/batch” system was a bit harder to figure out, since it is composed of simple operations that need to be repeated over time:
And what do you do when you have a complex, repeating sequence of steps? Easy, you make it recursive! Recursion helps you figure out the smaller parts of the problem and reduce them to a final state, that is: your solution.
We started calling our tree nodes “messages”, something that our system receives and knows how to handle, eventually
reducing it to:
When we have a local termination, it means that we have reached a leaf node in the tree. Our system will then analyse and process one message:
To simplify this, we linearised the tree to a queue employing a breadth-first search, so that the concept of “next position” is more explicit. A depth-first search would have worked just fine, since both ensure the same relative order between tree nodes.
What if the request fails because we have reached the API limit for the hour? Our solution is to schedule the same message on a queue for future analysis as soon as the API limit is reset by Github.
Let’s not worry about the actual implementation of the rescheduling system, keeping it abstract helps break all of the issues down and leave the implementation details for later.
There will be some kind of active mechanism that wakes up the queue and restarts the process and since our system will always pick up the first message in the queue (FIFO), we know for a fact that it will re-run the operation that failed before.
The same behaviour can be used for network/database errors, rescheduling the process to re-run immediately or after a short pause.
Something we have not considered yet is paging. To avoid exhausting the API limit in a short time, we should retrieve information in batches as large as possible. But the data we ask for isn’t stale: from the moment we ask for a page to the moment we ask for the following page new data might be added!
This may introduce two problems: data duplication and incomplete data.
As pages are ordered by descending date, if new data is added, old data might be retrieved again in following pages, then saved twice onto the database. The naive solution would be to
UPDATE OR INSERT all data received from the API… which is just what we did!
Since new data shifts old data towards later pages, we might not be able to retrieve the old information we need, as the number of messages generated from the previous node corresponds to the number of pages the Github API said we had at that particular moment in time. This means that we might be told we have 3 pages at the start, but the new data has shifted to a total of 5 pages, while our original count is still at 3!
Our solution consists in retrieving the page from the API, then analysing if Github says it is the last one or not: if it is not, it means that new data has been generated and we need to request these new pages, i.e. add new messages to the queue.
This leads to a change of spec: a tree node does not always generate new children, but also siblings! Having linearised our tree in a queue, this is a merely academic observation.
All of the information we mined from Github ended up in a database full of repositories, pull requests, merges, closes, comments and much more.
Our next blog posts will show you how we use all this data to find interesting correlations, patterns and outliers.
Running this system on a local machine means you have all the data only on your own local machine AND you can’t do much else whilst it is running! Our next blog posts will also detail the implementation of such architecture on Amazon Web Services, go through the challenges we faced and help you get set up if you want to achieve similar results.
We plan, design, and develop the world’s most desirable software products. Our team’s expertise helps brands like Sony, Motorola, Tesco, Channel4, BBC, and News Corp build fully customized Android devices or simply make their mobile experiences the best on the market. Since 2008, our full in-house teams work from London, Liverpool, Berlin, Barcelona, and NYC.
Let’s get in contact |
In order to decide whether the languages generated by two DFAs $A_1,A_2$ by the same, construct a DFA $A_\Delta$ for the symmetric difference $L(A_1) \Delta L(A_2) := (L(A_1) \setminus L(A_2)) \cup (L(A_2) \setminus L(A_1))$, and check whether $L(A_\Delta) = \emptyset$.
Here are some more details. You can construct $A_\Delta$ using the
product construction: construct a product automaton, and use $(F_1 \times \overline{F_2}) \cup (\overline{F_1} \times F_2)$ as the set of accepting states.
In order to check whether $L(A_\Delta)$ is empty or not, it suffices to check whether some accepting state is reachable from the initial state, and this can be done using BFS/DFS. |
11 0 1. Homework Statement
Consider a branch of [itex]\log{z}[/itex] analytic in the domain created with the branch cut [itex]x=−y, x≥0.[/itex] If, for this branch, [itex]\log{1}=-2\pi i[/itex], find the following.
[tex]\log{(\sqrt{3}+i)}[/tex]
2. Homework Equations
[tex]\log{z} = \ln{r} + i(\theta + 2k\pi)[/tex]
3. The Attempt at a Solution
This one is actually given in the textbook (odd numbered problem), but I'm having trouble understanding how the answer was arrived at.
The answer given: [itex]0.693 - i\frac{11\pi}{6}[/itex]
I can see easily that [itex]\log{\sqrt{(1)^2 + (\sqrt{3})^2}}=\ln{2} = 0.693...[/itex] The real part here makes sense since it's the (real) log of the modulus of the given complex number [itex]\sqrt{3}+i[/itex].
I can also understand that the branch cut is made along [itex]x=-y[/itex]. Where I'm getting confused is how the cut actually affects this log. So [itex]r = 2, \theta=\frac{\pi}{6}[/itex]. Winding around counterclockwise from 0, we reach [itex]\frac{\pi}{6}[/itex] easily, since it does not cross the branch cut at all.
Does the restriction [itex]\log{1}=-2\pi i[/itex] actually restrict this to moving around the circle clockwise from [itex]-\frac{\pi}{4}[/itex] such that [itex]-\frac{9\pi}{4} < \theta \le -\frac{\pi}{4}[/itex]? When using this log with principal values and restricted to a domain of analyticity of [itex]-\pi < \theta \le \pi[/itex] we traditionally wind around counterclockwise toward [itex]\pi[/itex] and clockwise toward [itex]-\pi[/itex]. This one, if I understand it correctly, winds around [itex]-2\pi[/itex] from the cut so that it's restricted to one set of values for an otherwise multi-valued log.
Why do both of these ways of figuring a single-valued log (the traditional principal valued log cut on the negative x axis and the one used in this problem) seem to involve winding around the axis different ways? What should I be understanding here that I'm not? |
Class of Quadrature Schemes
A class
"quad" to represent a quadrature scheme.
Details
A (finite) quadrature scheme is a list of quadrature points \(u_j\) and associated weights \(w_j\) which is used to approximate an integral by a finite sum: $$ \int f(x) dx \approx \sum_j f(u_j) w_j $$ Given a point pattern dataset, a
Berman-Turner quadrature scheme is one which includes all these data points, as well as a nonzero number of other (``dummy'') points.
These quadrature schemes are used to approximate the pseudolikelihood of a point process, in the method of Baddeley and Turner (2000) (see Berman and Turner (1992)). Accuracy and computation time both increase with the number of points in the quadrature scheme.
An object of class
"quad" represents a Berman-Turner quadrature scheme. It can be passed as an argument to the model-fitting function
ppm, which requires a quadrature scheme.
An object of this class contains at least the following elements:
data:
an object of class
"ppp"
giving the locations (and marks) of the data points.
dummy:
an object of class
"ppp"
giving the locations (and marks) of the dummy points.
w:
vector of nonnegative weights for the quadrature points
Users are strongly advised not to manipulate these entries directly.
The domain of quadrature is specified by
Window(dummy) while the observation window (if this needs to be specified separately) is taken to be
Window(data).
The weights vector
w may also have an attribute
attr(w, "zeroes") equivalent to the logical vector
(w == 0). If this is absent then all points are known to have positive weights.
To create an object of class
"quad", users would typically call the high level function
quadscheme. (They are actually created by the low level function
quad.)
Entries are extracted from a
"quad" object by the functions
x.quad,
y.quad,
w.quad and
marks.quad, which extract the \(x\) coordinates, \(y\) coordinates, weights, and marks, respectively. The function
n.quad returns the total number of quadrature points (dummy plus data).
An object of class
"quad" can be converted into an ordinary point pattern by the function
union.quad which simply takes the union of the data and dummy points.
See Also Aliases quad.object Documentation reproduced from package spatstat, version 1.55-1, License: GPL (>= 2) |
Here is a beautiful result from numerical analysis. Given any nonsingular $n\times n$ system of linear equations $Ax=b$, an optimal Krylov subspace method like GMRES must necessarily terminate with the exact solution $x=A^{-1}b$ in no more than $n$ iterations (assuming exact arithmetic).
The Cayley-Hamilton theorem provides a simple, elegant proof of this statement. To begin, recall that at the $k$-th iteration, minimum residual methods like GMRES solve the least-squares problem$$\underset{x_k\in\mathbb{R}^n}{\text{minimize }} \|Ax_k-b\|$$by picking a solution from the $k$-th Krylov subspace$$\text{subject to } x_k \in \mathrm{span}\{b,Ab,A^2b,\ldots,A^{k-1}b\}.$$If the objective $ \|Ax_k-b\|$ goes to zero, then we have found the exact solution at the $k$-th iteration (we have assumed that $A$ is full-rank).
Next, observe that $x_k=(c_0 + c_1 A + \cdots + c_{k-1}A^{k-1})b=p(A)b$, where $p(\cdot)$ is a polynomial of order $k-1$. Similarly, $\|Ax_k-b\|=\|q(A)b\|$, where $q(\cdot)$ is a polynomial of order $k$ satisfying $q(0)=-1$. So the least-squares problem from above for each fixed $k$ can be equivalently posed as a polynomial optimization problem with the same optimal objective
$$\text{minimize } \|q_k(A)b\| \text{ subject to } q_k(0)=-1,\; q_k(\cdot) \text{ is an order-} k \text{ polynomial.}$$Again, if the objective $\|q_k(A)b\|$ goes to zero, then GMRES has found the exact solution at the $k$-th iteration.
Finally, we ask: what is a bound on $k$ that guarantees that the objective goes to zero? Well, with $k=n$, and the optimal polynomial $q_n(\cdot)$ for our polynomial optimization problem is just the characteristic polynomial of $A$. According to Cayley-Hamilton, $q_n(A)=0$, so $\|q_n(A)b\|=0$. Hence we conclude that GMRES always terminate with the exact solution at the $n$-th iteration.
This same argument can be repeated (with very minor modifications) for other optimal Krylov methods like conjugate gradients, conjugate residual / MINRES, etc. In each case, the Cayley-Hamilton forms the crux of the argument. |
A chemical reaction may be a very complicated thing, involving many different paths and elementary reactions. Basically it always leads to the formation of an equilibrium, where the lowest possible energy will be the most favoured. Under certain conditions more than one state will be populated and hence there will be enough potential energy to interconvert molecules in certain states. We usually refer to that as an dynamic equilibrium.
Changes under equilibrium conditions will most likely be elementary reactions (or an infinitely complicated chain of these.)
The population of the various states seems to be static at equilibrium, since the overall change of enthalpy is zero. In a dynamic picture that only means, that forward and backward reaction cancel each other. In that sense they are the same reaction in different directions. Therefore they can be described as one trajectory on the potential energy surface. (If there is a chemical reaction involving many different elementary reactions and the equilibrium still occurs, then the mathematical description will be much more complicated as all elementary reactions will be correlated. However, the main conclusions should stay the same.)
\begin{aligned}\Delta G_r &= \Delta G_r^\circ + \mathcal{R}T\cdot \ln K = 0\\\Delta G_r^\circ &= − \mathcal{R}T \cdot \ln K\end{aligned}
The equilibrium constant may be derived from the mass-action law and should be defined as a product of activities (for the standard state). For $\ce{A + B <=> C}$ this will lead to$$K^\circ= \frac{a(\ce{C})}{a(\ce{A})\cdot a(\ce{B})}$$
The activities are proportional to concentrations (in first approximation) and are unit less ($c^\circ$ being the standard concentration $1 \:\mathrm{mol/L}$)
$$a=\gamma\frac{c}{c^{\circ}}$$
For reasonable dilutions ($c\to0\:\mathrm{mol/L}$) one can assume that the activity coefficient becomes one $\gamma\approx1$ and therefore rewrite the equilibrium constant with concentrations.$$K^\circ= \frac{c(\ce{C})/c^\circ}{c(\ce{A})\cdot c(\ce{B})/(c^\circ)^2}$$
At equilibrium the forward and backward reaction are coupled and therefore have to reflect the equilibrium constant.$$K^\circ= \frac{k_f}{k_b} = \frac{c_{\text{eq}}(\ce{C})/c^\circ}{c_{\text{eq}}(\ce{A})\cdot c_{\text{eq}}(\ce{B})/(c^\circ)^2}$$
As one can see, the constants also need to have the same unit. As they are at equilibrium, they are the same reaction with different populated states.
If you are moving away from the equilibrium, thing will start to get a little bit more messy. On reaction will be faster than the other and rates will not be comparable any more. |
According to quantum electrodynamics (QED), which encodes the properties of electrons and photons, electrons are excitations of an electron fiueld in the same way as photons are excitations of the electromagnetic field.
The fields wave, and the electrons (or photons), as far as they can be considered to be particles, are localized wave packets of excitations of these fields.
The particle interpretation is appropriate, however, only to the extent that the so-called geometrical optics approximation is valid. This means, in a particle interpretation, you shouldn't look too closely at the details, as then the particle properties become more and more fuzy and the wave properties become more and more pronounced.
But if you just look at quantum mechanics (QM) rather than QED, your question cannot be answered as the wave function is something unobservable, existing only in an abstract space,
What can be given an interpretation in QM are certain things one can compute from the wave function. The stuff of interest to chemists is the charge distribution, given by $\rho(x)=e|\psi(x)|^2$ for a single electron, by $\rho(x)=\int_{R^3} ~dy~ e|\psi(x,y)|^2$ for a 2-electron system, etc.; here $e$ is the electron charge. For electrons in a molecule, nothing is waving here anymore, as the wave aspect is eliminated by taking the absolute values.
Indeed, if this charge density is concentrated in a tiny region, one sees the particle aspect of electrons; if it is very spread out, one sees the wave aspect, revealed by high frequency oscillatory patterns in the charge density.
This is the chemist's interpretation. See Chapter A6:The structure of physical objects of A theoretical physics FAQ.
Physicists (especially if not well acquainted with the use of charge density information) are often brainwashed by the teaching tradition, and then think and express everything in terms of probabilities, giving QM an unnecessary flair of mystery. |
@DavidReed the notion of a "general polynomial" is a bit strange. The general polynomial over a field always has Galois group $S_n$ even if there is not polynomial over the field with Galois group $S_n$
Hey guys. Quick question. What would you call it when the period/amplitude of a cosine/sine function is given by another function? E.g. y=x^2*sin(e^x). I refer to them as variable amplitude and period but upon google search I don't see the correct sort of equation when I enter "variable period cosine"
@LucasHenrique I hate them, i tend to find algebraic proofs are more elegant than ones from analysis. They are tedious. Analysis is the art of showing you can make things as small as you please. The last two characters of every proof are $< \epsilon$
I enjoyed developing the lebesgue integral though. I thought that was cool
But since every singleton except 0 is open, and the union of open sets is open, it follows all intervals of the form $(a,b)$, $(0,c)$, $(d,0)$ are also open. thus we can use these 3 class of intervals as a base which then intersect to give the nonzero singletons?
uh wait a sec...
... I need arbitrary intersection to produce singletons from open intervals...
hmm... 0 does not even have a nbhd, since any set containing 0 is closed
I have no idea how to deal with points having empty nbhd
o wait a sec...
the open set of any topology must contain the whole set itself
so I guess the nbhd of 0 is $\Bbb{R}$
Btw, looking at this picture, I think the alternate name for these class of topologies called British rail topology is quite fitting (with the help of this WfSE to interpret of course mathematica.stackexchange.com/questions/3410/…)
Since as Leaky have noticed, every point is closest to 0 other than itself, therefore to get from A to B, go to 0. The null line is then like a railway line which connects all the points together in the shortest time
So going from a to b directly is no more efficient than go from a to 0 and then 0 to b
hmm...
$d(A \to B \to C) = d(A,B)+d(B,C) = |a|+|b|+|b|+|c|$
$d(A \to 0 \to C) = d(A,0)+d(0,C)=|a|+|c|$
so the distance of travel depends on where the starting point is. If the starting point is 0, then distance only increases linearly for every unit increase in the value of the destination
But if the starting point is nonzero, then the distance increases quadratically
Combining with the animation in the WfSE, it means that in such a space, if one attempt to travel directly to the destination, then say the travelling speed is 3 ms-1, then for every meter forward, the actual distance covered by 3 ms-1 decreases (as illustrated by the shrinking open ball of fixed radius)
only when travelling via the origin, will such qudratic penalty in travelling distance be not apply
More interesting things can be said about slight generalisations of this metric:
Hi, looking a graph isomorphism problem from perspective of eigenspaces of adjacency matrix, it gets geometrical interpretation: question if two sets of points differ only by rotation - e.g. 16 points in 6D, forming a very regular polyhedron ...
To test if two sets of points differ by rotation, I thought to describe them as intersection of ellipsoids, e.g. {x: x^T P x = 1} for P = P_0 + a P_1 ... then generalization of characteristic polynomial would allow to test if our sets differ by rotation ...
1D interpolation: finding a polynomial satisfying $\forall_i\ p(x_i)=y_i$ can be written as a system of linear equations, having well known Vandermonde determinant: $\det=\prod_{i<j} (x_i-x_j)$. Hence, the interpolation problem is well defined as long as the system of equations is determined ($\d...
Any alg geom guys on? I know zilch about alg geom to even start analysing this question
Manwhile I am going to analyse the SR metric later using open balls after the chat proceed a bit
To add to gj255's comment: The Minkowski metric is not a metric in the sense of metric spaces but in the sense of a metric of Semi-Riemannian manifolds. In particular, it can't induce a topology. Instead, the topology on Minkowski space as a manifold must be defined before one introduces the Minkowski metric on said space. — baluApr 13 at 18:24
grr, thought I can get some more intuition in SR by using open balls
tbf there’s actually a third equivalent statement which the author does make an argument about, but they say nothing about substantive about the first two.
The first two statements go like this : Let $a,b,c\in [0,\pi].$ Then the matrix $\begin{pmatrix} 1&\cos a&\cos b \\ \cos a & 1 & \cos c \\ \cos b & \cos c & 1\end{pmatrix}$ is positive semidefinite iff there are three unit vectors with pairwise angles $a,b,c$.
And all it has in the proof is the assertion that the above is clearly true.
I've a mesh specified as an half edge data structure, more specifically I've augmented the data structure in such a way that each vertex also stores a vector tangent to the surface. Essentially this set of vectors for each vertex approximates a vector field, I was wondering if there's some well k...
Consider $a,b$ both irrational and the interval $[a,b]$
Assuming axiom of choice and CH, I can define a $\aleph_1$ enumeration of the irrationals by label them with ordinals from 0 all the way to $\omega_1$
It would seemed we could have a cover $\bigcup_{\alpha < \omega_1} (r_{\alpha},r_{\alpha+1})$. However the rationals are countable, thus we cannot have uncountably many disjoint open intervals, which means this union is not disjoint
This means, we can only have countably many disjoint open intervals such that some irrationals were not in the union, but uncountably many of them will
If I consider an open cover of the rationals in [0,1], the sum of whose length is less than $\epsilon$, and then I now consider [0,1] with every set in that cover excluded, I now have a set with no rationals, and no intervals.One way for an irrational number $\alpha$ to be in this new set is b...
Suppose you take an open interval I of length 1, divide it into countable sub-intervals (I/2, I/4, etc.), and cover each rational with one of the sub-intervals.Since all the rationals are covered, then it seems that sub-intervals (if they don't overlap) are separated by at most a single irrat...
(For ease of construction of enumerations, WLOG, the interval [-1,1] will be used in the proofs) Let $\lambda^*$ be the Lebesgue outer measure We previously proved that $\lambda^*(\{x\})=0$ where $x \in [-1,1]$ by covering it with the open cover $(-a,a)$ for some $a \in [0,1]$ and then noting there are nested open intervals with infimum tends to zero.
We also knew that by using the union $[a,b] = \{a\} \cup (a,b) \cup \{b\}$ for some $a,b \in [-1,1]$ and countable subadditivity, we can prove $\lambda^*([a,b]) = b-a$. Alternately, by using the theorem that $[a,b]$ is compact, we can construct a finite cover consists of overlapping open intervals, then subtract away the overlapping open intervals to avoid double counting, or we can take the interval $(a,b)$ where $a<-1<1<b$ as an open cover and then consider the infimum of this interval such that $[-1,1]$ is still covered. Regardless of which route you take, the result is a finite sum whi…
W also knew that one way to compute $\lambda^*(\Bbb{Q}\cap [-1,1])$ is to take the union of all singletons that are rationals. Since there are only countably many of them, by countable subadditivity this give us $\lambda^*(\Bbb{Q}\cap [-1,1]) = 0$. We also knew that one way to compute $\lambda^*(\Bbb{I}\cap [-1,1])$ is to use $\lambda^*(\Bbb{Q}\cap [-1,1])+\lambda^*(\Bbb{I}\cap [-1,1]) = \lambda^*([-1,1])$ and thus deducing $\lambda^*(\Bbb{I}\cap [-1,1]) = 2$
However, what I am interested here is to compute $\lambda^*(\Bbb{Q}\cap [-1,1])$ and $\lambda^*(\Bbb{I}\cap [-1,1])$ directly using open covers of these two sets. This then becomes the focus of the investigation to be written out below:
We first attempt to construct an open cover $C$ for $\Bbb{I}\cap [-1,1]$ in stages:
First denote an enumeration of the rationals as follows:
$\frac{1}{2},-\frac{1}{2},\frac{1}{3},-\frac{1}{3},\frac{2}{3},-\frac{2}{3}, \frac{1}{4},-\frac{1}{4},\frac{3}{4},-\frac{3}{4},\frac{1}{5},-\frac{1}{5}, \frac{2}{5},-\frac{2}{5},\frac{3}{5},-\frac{3}{5},\frac{4}{5},-\frac{4}{5},...$ or in short:
Actually wait, since as the sequence grows, any rationals of the form $\frac{p}{q}$ where $|p-q| > 1$ will be somewhere in between two consecutive terms of the sequence $\{\frac{n+1}{n+2}-\frac{n}{n+1}\}$ and the latter does tends to zero as $n \to \aleph_0$, it follows all intervals will have an infimum of zero
However, any intervals must contain uncountably many irrationals, so (somehow) the infimum of the union of them all are nonzero. Need to figure out how this works...
Let's say that for $N$ clients, Lotta will take $d_N$ days to retire.
For $N+1$ clients, clearly Lotta will have to make sure all the first $N$ clients don't feel mistreated. Therefore, she'll take the $d_N$ days to make sure they are not mistreated. Then she visits client $N+1$. Obviously the client won't feel mistreated anymore. But all the first $N$ clients are mistreated and, therefore, she'll start her algorithm once again and take (by suposition) $d_N$ days to make sure all of them are not mistreated. And therefore we have the recurence $d_{N+1} = 2d_N + 1$
Where $d_1$ = 1.
Yet we have $1 \to 2 \to 1$, that has $3 = d_2 \neq 2^2$ steps. |
Electric Charges and Fields Coulomb's Law and Forces between Multiple Charges Coulombs inverse square law \tt F = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q_{1}Q_{2}}{r^{2}} \tt \frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} N/m 2& ε 0= 8.85 × 10 −12C 2/N-m 2 Coulombs law in vector form \tt \overline{F} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q_{1}Q_{2}}{r^{2}} \times \hat{r} Dielectric constant or relative permittivity K is given by \tt K = \varepsilon_{r} = \frac{\varepsilon}{\varepsilon_{0}} = \frac{F air}{F medium} Coulombs electrostatic force is a conservative force Coulombs electrostatic force is a central force Principle of super position \tt \overline{F} = \overline{F}_{12} + \overline{F}_{13} + \overline{F}_{14} + ---- View the Topic in this video From 27:31 To 40:23
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1. Electrostatic force of interaction acting between two stationary point charges is given by
F = \frac{1}{4 \pi \varepsilon_{0}}\frac{q_{1}q_{2}}{r^{2}}
2. Forces between Multiple Charges
F_{1} = \frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q_{1}q_{2}}{r_{12}^{2}} \hat{r}_{21} + \frac{q_{1}q_{2}}{r_{13}^{2}} \hat{r}_{31} + ...... + \frac{q_{1}q_{2}}{r_{n}^{2}} \hat{r}_{n 1}\right] |
It seems like integrating it is a big deal and everything, but I don't understand why. My teacher is making it seem really important in my calculus class for seemingly no reason.
Help would be appreciated.
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It seems like integrating it is a big deal and everything, but I don't understand why. My teacher is making it seem really important in my calculus class for seemingly no reason.
Help would be appreciated.
The importance of the integral isn't limited to the fact that there is no closed form anti-derivative. The integral is important, in my opinion, because it makes many calculations possible which would not be possible otherwise. These calculations provide us with insights into the nature of certain mathematical problems which is a hard thing to come by.
Here is a list of important instances of the Gaussian that I can think of off the top of my head,
The sequence of functions $\frac{n}{\sqrt{\pi}} exp{-x^2/n^2}$ defines the generalized function known as the "Dirac Delta Function". This is so important I can't even get into it here, look on wikipedia to learn more.
Gaussian integration can be used to prove the Fourier inversion theorem for a certain class of functions.
The Fourier transform of a Gaussian can actually be computed (because we know the integral) which allows us to construct a Gaussian wave packet in quantum mechanics. Wave packets are an invaluable tool for gaining intuition about a problem.
In probability theory the central limit theorem tells us that the sum of a large number of random outcomes has a Gaussian probability distribution regardless of the nature of the original distribution. Since averages and variances require integration of the probability distribution we couldn't solve many important problems in probability without being able to evaluate this integral.
You may have heard the phrase "When you have a hammer every problem looks like a nail". In this analogy the Guassian integral is a really good hammer and it allows us to turn normally intractable problems into nice easy nails. Hope this helps.
Are you familiar at all with probability and statistics? The famous "bell curve" of the normal distribution is given by $ke^{-x^2}$, for an appropriate constant $k$, or some translate of that curve, so integrating this function is essential to many calculations in statistics.
For more information, see https://en.wikipedia.org/wiki/Normal_distribution
It's also just interesting as an integral, because it's tricky to calculate, and you can learn some interesting and useful techniques by working on it. |
For an ideal fluid, if the vorticity is $\vec{\omega}=\nabla \times \vec{v}$, then Euler's equations can be rewritten as: $$\rho \dot{v}_i = \rho \epsilon_{ijk} v_j \omega_k - \frac{1}{2} \rho \partial_i v^2 - \partial_i p $$ Any textbook will then tell you that if you have a steady flow with zero vorticity: $$ \frac{1}{2} \rho \partial_i v^2 + \partial_i p = 0 $$ which is a differential form of Bernoulli's theorem. However as it is obvious from the previous equation the necessary and sufficient condition for this equation to hold is not a steady flow with $\vec{\omega}=0$ but a steady flow with $\vec{v} \times \vec{\omega}=0$, which is a more general condition. I am wondering if one can give a nice geometric interpretation of this condition. In other worlds, what is the geometric interpretation of a vector field having $\vec{v} \times (\nabla \times \vec{v}) = 0$?
The quantity $\vec{v}\times(\vec{\nabla}\times\vec{v})$ you are interested in here is known as the
Lamb vector in theoretical fluid dynamics. It's gone a bit out of fashion these days, but there's some useful material in the JFM paper by C. W. Hamman, J. C. Klewicki and R. M. Kirby, "On the Lamb vector divergence in Navier–Stokes flows". Perhaps more pertinently, your question is a near-duplicate of a question asked here in February with an answer that has some more detail also. A Google search on "Lamb vector" may turn up more material you find interesting. P.S.: Perhaps the main reason why people typically invoke the more restrictive condition of irrotationality to justify the use of the Bernoulli equation is that the space of irrotational flows is an invariant subspace of the Navier-Stokes equations (meaning, roughly, there's a large, and practically relevant, set of flows that satisfy this condition), whereas the subspace of flows with zero "vortex force" (Lamb vector) is not invariant, so non-trivial rotational flows with vanishing Lamb vector everywhere will not in general retain this state. I think...
I hasten to add that I'm not an expert on this; I'll readily admit that I don't even know what an interesting rotational flow with vanishing Lamb vector would look like.
The actual derivation of the Bernoulli equation comes from the vorticity form of the incompressible Navier-Stokes equation. In terms of vorticity, the Navier-Stokes equation take the form, $$ \frac{\partial \vec{V}}{\partial t} + \vec{\omega} \times \vec{V} = -\nabla\left(\frac{p}{\rho} + \frac{|\vec{V}|^2}{2} + k\right) + \nu \cdot \left(\nabla \times \vec{\omega}\right)$$ Now if we have steady flow, $\frac{\partial \vec{V}}{\partial t} = 0$, and if we further assume the flow is inviscid, then the equation reduces to, $$ \vec{\omega} \times \vec{V} = -\nabla\left(\frac{p}{\rho} + \frac{|\vec{V}|^2}{2} + k\right)$$ Obviously, if the flow is irrotational, namely $\vec{\omega} = \nabla \times \vec{V} = 0$, then we are left with, $$ \nabla\left(\frac{p}{\rho} + \frac{|\vec{V}|^2}{2} + k\right) = 0 $$ or equivalently, $$ \frac{p}{\rho} + \frac{|\vec{V}|^2}{2} + k = \textrm{constant}$$ This is the most famous form of the Bernoulli equation, which requires steady, incompressible, inviscid, and irrotational flow. Also, an important note on this relation, because the flow is irrotational, the Bernoulli equation can be applied across streamlines. Now for the case you specified, for instance, what if the flow is rotational? Well, you have to consider the direction of the vector quantity $\vec{\omega} \times \vec{V}$. The resulting vector of $\vec{\omega} \times \vec{V}$ is orthogonal to the velocity and vorticity vector. Therefore, along a streamline the quantity $\vec{\omega} \times \vec{V} = 0$. Hence, the resulting equation becomes, $$ \frac{p}{\rho} + \frac{|\vec{V}|^2}{2} + k = \mathrm{constant\big|_{streamline}}$$ Therefore, the conclusion is that the Bernoulli equation can be applied across streamlines if we have steady, incompressible, inviscid, and irrotational flow ($\vec{\omega} = \nabla \times \vec{V} = 0$). However, if the flow is rotational ($\vec{\omega} = \nabla \times \vec{V} \neq 0$), we can only apply the Bernoulli equation along a streamline. |
The dot notation means to write a sum of three terms by matching x, y, z matrices for the two particles, as if $\vec{\sigma}^1$ were a vector $(\sigma_x^1,\sigma_y^1,\sigma_z^1)$ and likewise for particle #2.$$\vec{\sigma}^1\cdot\vec{\sigma}^2 = \sigma_x^1\sigma_x^2 + \sigma_y^1\sigma_y^2 + \sigma_z^1\sigma_z^2$$
Note that $\sigma_i^1$ and $\sigma_i^2$ act separately as matrices; they don't multiply with each other.
We have $$\sigma_z^1 = \left(\begin{matrix}1&0\\0&-1\end{matrix}\right) $$but acting only on the particle #1 factor of the wavefunction, and$$\sigma_z^2 = \left(\begin{matrix}1&0\\0&-1\end{matrix}\right) $$action only on the particle #2 factor.
For example, if you have some wavefunction (not necessarily an eigenstate of anything) being particle #1 spin up and particle #2 down, $$\psi =\psi^1\psi^2 = {\left(\begin{matrix}1\\0\end{matrix}\right)}^1{\left(\begin{matrix}0\\1\end{matrix}\right)}^2$$and some arbitrary matrix acting on particle #1, $$U^1 = \left(\begin{matrix}a&b\\c&d\end{matrix}\right) $$and another arbitrary matrix for particle #2, $$V^2 = \left(\begin{matrix}e&f\\g&h\end{matrix}\right) $$then $$U^1 \psi = (U^1\psi^1 )\psi^2 = {\left(\begin{matrix}a\\c\end{matrix}\right)}^1{\left(\begin{matrix}0\\1\end{matrix}\right)}^2$$and$$V^2 \psi = \psi^1(V^2\psi^2 ) = {\left(\begin{matrix}0\\1\end{matrix}\right)}^1{\left(\begin{matrix}f\\h\end{matrix}\right)}^2$$
Note that tacking superscripts onto column spinors like that to indicate particle identity isn't normal practice outside of textbooks or places where things need to be explained in such detail. A common way to deal with this is to form a four dimensional space, the product of the two spin spaces of the particles. The basis vectors would be $\uparrow^1\uparrow^2, \uparrow^1\downarrow^2, \downarrow^1\uparrow^2, \downarrow^1\downarrow^2$ or some nice linear combination.
Think about it: we need an index to count along the spinor components (the 2-dimensional Hilbert space of complex numbers), an index to count spatial dimensions (x,y,z), and an index to count particles. We're dealing with a three-dimensional entity, a matrix with rows, columns and "another". Stick with theoretical physics, and we'll add chromodynamic charge, more spatial dimensions to deal with curved spacetime, and sprinkle in a dash of supersymmetry or technicolor or whatever is hot with the kids these days. And you'll refer to the Pauli matrices as "Clebsch-Gordan coefficients" but at least they're the simplest nontrivial case. Have fun! |
Conic Sections Hyperbola If the centre of the hyperbola lies at a point (h, k) and the axes are parallel to the co-ordinate axes, then the equation of the hyperbola is \tt \frac{\left(x-h\right)^{2}}{a^{2}}-\frac{\left(y-k\right)^{2}}{b^{2}}=1 The condition for the line y = mx + c to be a tangent to the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 is that c 2= a 2m 2– b 2and the co-ordinates of the points of contact are \tt \left(\pm\frac{a^{2}m}{\sqrt{a^{2}m^{2}-b^{2}}},\pm\frac{b^{2}}{\sqrt{a^{2}m^{2}-b^{2}}}\right) The equation of tangent to the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 interms of slope ‘m’ is \tt y = mx\ \pm\sqrt{a^{2}m^{2}-b^{2}} The equation of the normal to the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 at the point (x 1y 1) is \tt \frac{a^{2}x}{x_{1}}+\frac{b^{2}y}{y_{1}}=a^{2}+b^{2} The equation of the normal to the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 at the point (a sec θ, b tan θ) is \tt \frac{ax}{\sec\theta}+\frac{by}{\tan\theta}=a^{2}+b^{2} The equation of normal to the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 interms of slope ‘m’ is \tt y=mx\ \pm\ \frac{m\left(a^{2}+b^{2}\right)}{\sqrt{a^{2}-b^{2}m^{2}}} Equation of the pair of tangents drawn from a point P(x 1y 1) to the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 is SS 1= T 2where \tt S=\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}-1, S_{1}=\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}-1\ and\ T = \frac{xx_{1}}{a^{2}}-\frac{yy_{1}}{b^{2}}-1 The equation of the chord of the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 with P(x 1y 1) as its middle point is given by T = S 1 The equation of chord of contact of tangents drawn from a point P(x 1y 1) to the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 is T = 0 Eccentricity of the hyperbola \tt (e) = \sqrt{\frac{a^{2}+b^{2}}{a^{2}}}
Tricks: Two tangents can be drawn from a point to a hyperbola. The two tangents are real and distinct or coincident or imaginary according as the given point lies outside, on or inside the hyperbola. The equation of director circle of the hyperbola is x 2+y 2= a 2– b 2 Equation of Hyperbola Latus Rectum and Examples of Hyperbola
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1. If the centre of hyperbola is (h, k) and axes are parallel to the co-ordinate axes, then its equation is \tt \frac{\left(x-h\right)^{2}}{a^{2}}-\frac{\left(y-k\right)^{2}}{b^{2}}=1.
2.
Imp. terms \ Hyperbola
\tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
\tt -\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \ or \ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1 Centre (0, 0) (0, 0) Length of transverse axis 2a 2b Length of conjugate axis 2b 2a Foci (± ae, 0) (0, ± be) Equation of directrices x = ± a / e y = ± b / e Eccentricity \tt e=\sqrt{\left(\frac{a^{2}+b^{2}}{a^{2}}\right)} \tt e=\sqrt{\left(\frac{a^{2}+b^{2}}{b^{2}}\right)} Length of latus rectum 2b
2 / a 2a
2 / b Parametric co-ordinates (a sec Φ, b tan Φ)
0 ≤ Φ < 2π
(b sec Φ, a tan Φ)
0 ≤ Φ < 2π
Focal radii SP = ex
1 − a
S'P = ex
1 + a SP = ey
1 − b
S'P = ey
1 + b Difference of focal radii (S'P - SP) 2a 2b Tangents at the vertices x = − a, x = a y = − b, y = b Equation of the transverse axis y = 0 x = 0 Equation of the conjugate axis x = 0 y = 0
3. Equations of chord joining two points P(a sec θ
1, b tan θ 1) and Q (a sec θ 2 , b tan θ 2) on the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}= 1 is \tt y - b tan \theta = \frac{b \tan \theta_{2}-b \tan\theta _{1}}{a \sec \theta_{2}-a \sec\theta _{1}}.\left(x-a \sec \theta_{1}\right) or \frac{x}{a}\cos\left(\frac{\theta_{1}-\theta_{2}}{2}\right)-\frac{y}{b}\sin\left(\frac{\theta_{1}+\theta_{2}}{2}\right)=\cos\left(\frac{\theta_{1}+\theta_{2}}{2}\right)
4. Equation of chord of contact of tangents drawn from a point (x
1, y 1) to the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \ is \frac{xx_{1}}{a^{2}}-\frac{yy_{1}}{b^{2}}=1.
5. The equation of the chord of the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 bisected at point (x
1, y 1) is given by \tt \frac{xx_{1}}{a^{2}}-\frac{yy_{1}}{b^{2}}-1=\frac{x_1^2}{a^{2}}-\frac{y_1^2}{b^{2}}or \ T=S_{1} Equation of tangent Hyperbola a). Point Form The equation of the tangent to the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \ at \ \left(x_{1},y_{1}\right)is \ \frac{xx_{1}}{a^{2}}-\frac{yy_{1}}{b^{2}}=1. b). Parametric Form The equation of the tangent to the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \ at \ \left(a\sec \theta,b \tan \theta\right)is \ \frac{x}{a}\sec \theta -\frac{y}{b}\tan \theta =1. c). Slope Form The equation of the tangents of slope m to the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 are given by \tt y = mx\pm \sqrt{a^{2}m^{2}-b^{2}}. The coordinates of the point of contact are \tt \left(\pm\frac{a^{2}m}{\sqrt{a^{2}m^{2}-b^{2}}},\pm\frac{a^{2}}{\sqrt{a^{2}m^{2}-b^{2}}}\right). Normal Equation of Hyperbola a). Point Form The equation of the normal to the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \ is \frac{a^{2}x}{x_{1}}+\frac{b^{2}x}{y_{1}}= a^{2}+b^{2}. b). Parametric Form The equation of the normal at (a sec θ, b tan θ) to the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 is ax cos θ + by cot θ = a 2 + b 2. c). Slope Form The equations of the normal of slope m to the hyperbola \tt \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 are given by \tt y=mx\mp\frac{m\left(a^{2}+b^{2}\right)}{\sqrt{a^{2}-b^{2}m^{2}}} |
There are $n$ students that share the same apartment. At each evening, one of them must prepare dinner for everyone. There are $m$ evenings to schedule, with $m \geq n$, and you have to assign any singular evening to one student at a time.
In particular, student $i$ is disposed to accept cooking for $c_i$ evenings, with $1 \leq c_i \leq m$. Let $S$ be the global set of evenings (eg. $S=\{1,2,3,4,5\}$), for every student we define also $S_i \subseteq S$: you may assign evening $j$ to student $i$ only if $j \in S_i$.
So, given $m, n, c_i, S_i$, you have to define an algorithm that assigns each single evening to an available student.
I think a solution can be found using linear programming, although seems a little tricky. Let $x_{ij}$ be defined as follows:
$$ x_{ij} = \begin{cases} 1 & \hbox{if evening } j \hbox{ is assigned to student } i \\ 0 & \hbox{otherwise} \end{cases} $$
Then you have such constraints. Student $i$ can accept a max. of $c_i$ evenings, so: $$ \begin{array}{cr} \sum_j x_{ij} \leq c_i & \hbox{ for every student } i \cr \end{array} $$ You have to ensure that every evening is assigned to a student only, then: $$ \begin{array}{cr} \sum_i x_{ij} = 1 & \hbox{ for every evening } j \cr \end{array} $$ Now a way to include $S_i$ in the constraints is required. Before doing it we define: $$ p_{ij} = \begin{cases} 1 & \hbox{if evening } j \notin S_i \\ 0 & \hbox{otherwise} \end{cases} $$ So last constraints are: $$ \begin{array}{cr} \sum_j x_{ij} \cdot p_{ij} = 0 & \hbox{ for every student } i \cr \end{array} $$ Suppose you have $5$ evenings, with $S_i=\{2,4,5\}$. You'll have the constraint $x_{11} + x_{13} = 0$. Since it's a linear program, we have $x_{ij} \geq 0$ so the equivalence is satisfied only if terms are all zeros, i.e. those evenings ($1$ and $3$) are not assignable to that student.
I have some doubt with the objective function, that it should be the sum of every $x_{ij}$. Should we maximize or minimize, or are they the same for this model? We could also put a constraint to say that such sum must be equal to $m$ (the total amount of evenings to assign), but i think it's not mandatory since it's guaranteed by previous constraints. What do you think about?
Is there a better way to solve such problems? |
I duplicate here a question I asked on math.stackexchange.
Question:Which inequalities similar to the famous isoperimetric inequality is known? conjectured?
I recently learned about some inequalities which are all similar to the famous isoperimetric inequality. Each time we consider two size functionals $\Sigma$ and $\Sigma'$ and along all the
convex bodies (convex and compact) $K$ in $\mathbb{R}^d$ satisfying $\Sigma'(K)=1$, we give a bound for $\Sigma(K)$.For example in $\mathbb{R}^2$, with $\Sigma=\mathrm{Area}$ and $\Sigma'=\mathrm{Perimeter}$ we have an upper-bound given by the famous isoperimetric inequality.
If $\Sigma$ (resp. $\Sigma'$) is homogeneous of degree $k$ (resp. $k'$). The problem is equivalent to giving a bound to $$\frac{\Sigma(K)^{1/k}}{\Sigma'(K)^{1/k'}}$$ for all $K$ with $\Sigma'(K)\neq 0$. Below I list the inequalities I encountered and give a quite general definition of what I consider size functionals.
The classical isoperimetric inequality in higher dimensionsstates that for any convex body $K$ in $\mathbb{R}^d$ with positive $(d-1)$-intrinsic volume we have
$$0<\frac{V_d(K)^{1/d}}{V_{d-1}(K)^{1/(d-1)}}\leq \frac{V_d(\mathrm{Ball})^{1/d}}{V_{d-1}(\mathrm{Ball})^{1/(d-1)}}$$
where $V_d$ is the $d$-dimensional volume, $V_{d-1}$ the $(d-1)$-intrisic volume (twice the perimeter if $d=2$ and twice the surface area if $d=3$), and $\mathrm{Ball}$ is any $d$-dimensional ball.
The isodiametric inequalitystate that for any convex body $K$ in $\mathbb{R}^2$ with positive perimeter we have
$$\frac{\mathrm{Diameter}(\mathrm{Disk})}{\mathrm{Perimeter}(\mathrm{Disk})} \leq\frac{\mathrm{Diameter}(K)}{\mathrm{Perimeter}(K)} \leq\frac12$$
where $\mathrm{Diameter}(K)$ is the maximum distance between two points of $K$. It has been proved by Bieberbach in 1915 (in german), I found this reference in the introduction of the article Isodiametric Problems for Polygons by by Michael J. Mossinghoff. I guess this inequality is true in higher dimensions but I have no reference.
Jung's theoremstates that for any convex body $K$ in $\mathbb{R}^d$ with positive diameter we have the second of the following inequalities (the first is obvious)
$$\frac{\mathrm{Outradius}(\mathrm{Disk})}{\mathrm{Diameter(\mathrm{Disk})}}\leq \frac{\mathrm{Outradius}(K)}{\mathrm{Diameter(K)}}\leq \frac{\mathrm{Outradius}(\Delta_d)}{\mathrm{Diameter(\Delta_d)}}$$
where $\Delta_d$ is the $d$-dimensional regular simplex.
The hyperplane conjecturestates there exists a universal constant $C$ such that in any dimension, for any convex body $K$ in $\mathbb{R}^d$ with positive volume, we have
$$C\leq\frac{\mathrm{MaxSection}(K)^{1/(d-1)}}{\mathrm{Volume(K)}^{1/d}}<\infty$$
where $\mathrm{MaxSection}(K)=\max\left(V_{d-1}(K\cap H) : H \text{ any hyperplane of }\mathbb{R}^d\right)$ is the maximal hyperplane section of $K$.
More generally if we note $\mathcal{K}=\mathcal{K}_d$ the set of convex body of $\mathbb{R}^d$ we can consider any
size functional $\Sigma:\mathcal{K}\to\mathbb{R}_{\geq 0}$ satisfying the following natural axioms: $\Sigma$ is continuous, not identically zero, homogeneous of some degree $k$, that is: $\Sigma(\lambda K)=\lambda^k \Sigma(K)$. increasing under set inclusion, that is: $(K\subset M \Rightarrow \Sigma(K)\leq\Sigma(M)$ invariant under translation, that is: $\Sigma(K+x)=\Sigma(K)$.
This covers most of the size functionals we usually consider:
volume = area in dimension 2, surface area =perimeter in dimension 2, mean-width, min-width, max-width (=diameter), width with a given direction in-radius : the radius of the biggest ball include in $K$, out-radius : the radius of the smalles ball include in $K$, intrinsic volumes the maximal hyperplane section: $\max\left(V_{d-1}(K\cap H) : H \text{ any hyperplane of }\mathbb{R}^d\right)$ ...
Now for any choice of couple of size functionals $\Sigma$ and $\Sigma'$ of degree $k$ and $k'$, if $K$ is a convex body with $\Sigma'(K)\neq0$ the fraction $$\frac{\Sigma(K)^{1/k}}{\Sigma'(K)^{1/k'}}\in[0,\infty[$$ is invariant under translation or rescaling of $K$.
I am interested by lower or upper bound for such fraction once we have fixed the dimension $d$ and $\Sigma$ and $\Sigma'$. |
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... |
Haber-Bosch ammonia synthesis reaction: $$\ce{3H2 +N2 -> 2NH3}$$
According to the ideal gas law: $pV=nRT$ where constant volume implies $\displaystyle \frac nV=\frac p{RT}$
Then $\displaystyle K_\mathrm c=\frac{[\ce{NH3}]^{2}}{[\ce{H2}]^3\cdot [\ce{N2}]}$
When I substitute $\displaystyle \frac nV=c=\frac {p_{\text{gas}}}{RT}$ I get:
$$ K_\mathrm c=\frac{\left(\frac{p(NH_3)}{RT}\right)^2}{\left(\frac{p(H_2)}{RT}\right)^3\left(\frac{p(N_2)}{RT}\right)}=\frac{R^2T^2}{p^2}$$
But according to Le Chatelier's principle, increased pressure should result in an increased $K_\mathrm c$, but my $\displaystyle K_\mathrm c\approx \frac 1{p^2}$.
Where is the error? Is is impossible to combine the partial pressures like that because they scale differently with total pressure? Can one derive from the gas law the pressure dependency of this reaction?
Edit after solved:
First mistake was that the formula had $K_c$ (constant) instead of Q, $$ Q=\frac{\left(\frac{p(NH_3)}{RT}\right)^2}{\left(\frac{p(H_2)}{RT}\right)^3\left(\frac{p(N_2)}{RT}\right)}=\frac{R^2T^2}{p^2}$$ where Q is the current value/ratio of concentrations, so the reaction tries to get Q back to $K_c$ so for greater pressure p, Q is smaller so the reaction goes in the direction of NH3... |
I need help to prove the following exercise:
Let $\epsilon >0$. Show that there exists $\delta >0$ with the property: If $A$ is a unital $C^*$-algebra and $x\in A$ such that $\|x^*x-1\|<\delta,\;\|xx^*-1\|<\delta$, then there is a unitary element $u\in A$ with $\|x-u\|<\epsilon$.
To prove this, I use continuous functional calculus for the element $x^*x$ and $xx^*$ to define $u$. The assumptions $\|x^*x-1\|<\delta,\;\|xx^*-1\|<\delta$ imply $\sigma(x^*x), \sigma(xx^*)\subseteq [1-\delta , 1+\delta]$, where $\sigma(x^*x)$ denotes the spectrum of $x^*x$.
If $x$ were invertible I could try something like $u=x|x|^{-1}$, where $|x|=(x^*x)^{\frac{1}{2}}$(polar decomposition), but $x$ could be non-invertible I think. But $x^*x$ is invertible if $\delta <1$, but I also have to choose a number $\delta $ which depends on $\epsilon$ .
Here I'm stuck. Which continuous function is suitable for u or how to choose $u$? |
Entropy change is defined as the amount of energy dispersed
reversibly to or from the system at a specific temperature. Reversivility means that the temperature of the system must remain constant during the dispersal of energy to or from the system . But this criterion is only fulfilled during phase change & isothermal processes. But not all processes maintain constant temperature;temperature may change constantly during the dispersal of energy to or from the system . To measure entropy change ,say, from $300K$ & $310K$, the range is divided into infinitesimal ranges ,then entropy is measured in that ranges and then is integrated as said in this site . But I cannot understand how they have measured entropy change in that infinitesimal ranges as there will always be difference between the temperature however small the range might be . What is the intuition behind it? Change of entropy is measured at constant temperature,so how can it be measured in a range ? I know it is done by definite integration but can't getting the proper intuition . Also ,if by using definite integration to measure change, continuous graph must be there(like to measure change in velocity,area under the graph of acceleration is measured) . So what is the graph whose area gives change in entropy? Plz help me explaining these two questions.
Entropy change is defined as the amount of energy dispersed
For a reversible addition of heat, the entropy change is $\int \frac{dQ} {T}$, in other words the area under the graph of $\frac{1}{T}$ against $Q$ (heat added to system).
And yes, when a small amount of heat $\Delta Q$ is added, temperature T changes only a little, so $\frac{\Delta Q} {T}$ is well-defined. When added up this gives the integral.
Calorimetry is a nasty business. Whenever is possible entropy is measured by measuring mechanical parameters by taking advantage the Gibbs form of 2nd law. For example, if a thin rod is elastic then $dU = TdS+\sigma d\epsilon$ where $\sigma, \epsilon$ are the stress and strain. From the equality of derivatives you get $\frac {\partial S}{\partial \sigma}|_T = -\frac {\partial \epsilon}{\partial T}|_\sigma$ the right side of which you can measure directly and then you can integrate with respect to stress at constant $T$. Thus you only have to measure entropy as function of stress at one temperature, the rest you can get by measuring thermal expansion at constant stress. |
The comments by KF Gauss are on the right track: non-relativistic quantum electrodynamics (NRQED) is a good foundation. The question is then how to recover the "effective potential" description of X-ray diffraction from NRQED.This isn't my specialty, but I reviewed the NRQED derivation of elastic X-ray scattering from a crystal, using the references listed ...
When a photon interacts with an atom, three things can happen:elastic scattering, the photon keeps its energy and changes angle (mirror reflection)https://en.wikipedia.org/wiki/Elastic_scatteringinelastic scattering, the photon keeps part of its energy and changes angle (photon transferring vibrational and rotational energies to the molecules, heat up ...
The problem with fixed-target experiments it that you have to make the target, then install the target, then irradiate the target with your beam. If your target is short-lived, the timescale for each of these steps becomes more challenging.For example, in positron-emission tomography, the positron source is usually fluorine-18, which has a two-hour half-...
Angular momentum is $\mathbf{L} = \mathbf{r} \times \mathbf{p}$. Thus by definition, the impact parameter is a ratio of the magnitudes of angular momentum to linear momentum.$$b = \frac{L}{p}$$From the energy-momentum 4-vector (assuming units with $c = 1$),$$E^2 - p^2= m^2$$$$p = \sqrt{E^2-m^2}$$$$b = \frac{L}{\sqrt{E^2-m^2}}=\frac{L/m}{\sqrt{(E/m)^2-1}}...
In the beams we have in the laboratory, the particles do not superpose or interact with each other, due to the large space time distances for the individual particles in the beam. Within our experimental accuracies it is as if each electron is all alone in the universe when interacting with a positron. Whether polarized or not each incoming particle can be ... |
On Malinvestment Matthew Martin7/21/2014 01:48:00 PM
Tweetable
"Austrian economics per se doesn’t make empirical predictions–that’s either its virtue or its vice, depending on your methodological views."Ok, whatever. But the fact that Austrianism is a load of crap doesn't mean that everything any of them have said is totally useless. A lot of Austrianist writings talk about one concept in particular, called "malinvestment," that causes the business cycle. Although many mainstream models include risky investments (not exactly mainstream, but here's one!), few of them contain anything resembling "malinvestment." Yet, I suspect that most people, even most economists, secretly believe something like malinvestment is closer to the truth. Let's leave the models behind for a moment, and examine something that happened in the real world.
I live in a small-town suburb of a big city. In the mid-2000s we got a coffee shop, and it was always busy. A couple years later we got a second coffee shop, and it did decent business as well. Then we got a third coffee shop, a second location of the first coffee shop, a drive-through coffee shop stand, then a coffee-shop/book-store combo, then the national chains took notice and we got a starbucks, then a coffee-shop/church combo (you really can't make this stuff up). Now, the residents of this town are fairly wealthy, but remember this is a small town populated by people who commute down to the big city everyday, and who therefore consume most of their coffee outside of town. Needless to say, we simply couldn't sustain the coffee-shop boom. And because they competed against eachother, we didn't just watch the newcommers go out of business--they all went out of business, more or less at the same time. Not even the original coffee-shop, the one that had been thriving alone, could survive the debt it accumulated while competing against the entrants. But here's the thing: it's pretty clear that this town can support at least one permanent coffee shop.
This looks a lot more like malinvestment than any mainstream theory of business cycles I've ever seen. New Keynesian theory doesn't apply because there was no demand shock--aggregate demand for coffee remained constant the whole time. Real Business Cycle theory doesn't apply because there wasn't any collapse in our ability to produce coffee, we simply tried to produce more than people wanted. It looks superficially like a news-shock model in which hypothetical coffee-shop investors heard that coffee shops in my small-town suburb were the next facebook, but I think even the peddlers of these models secretly know that's not what's happening--no one was predicting that our-town's coffee consumption was booming, just that they would be able to steal enough of the market to make it as the One True Coffee Shop for this town.
If I had to coerce the reality into a neoclassical model, I'd say our coffee shop cycle was a case of failure to converge to equilibrium. Equilibrium models assume not that unprofitable business will be driven out of the market, but that there are no unprofitable businesses in the market. A problem with the standard equilibrium modelling is that they ignore the possibility that unprofitable entrants in a market can stay in business long enough to drive both themselves and profitable incumbents into bankruptcy. Consider a model (I promise, hardly any calculus!) of a coffee shop in a small town. It's a very small town actually--they only demand 10 cups of coffee every month. But that's ok because the shop earns 1 unit of net revenue after paying labor and materials on each cup, and their fixed costs come to 8 units a month. They're earning 2 units of profit every month and, like a good little corporation, remit this profit immediately to shareholders (so that the corporation has zero cash on hand). Now an upstart little cretin from the big city moves in and sees that this coffee shop model is profitable, and sets up an identical replica of it right across the street. Coffee drinkers are indifferent between these two identical shops, so they go to each half the time. This means that both shops are selling only 5 cups a month, accumulating debt at a rate of 3 units a month to cover fixed costs.
Of course, they cannot go into debt for ever. A bit of math reveals that at 5 percent interest the companies become permanently unprofitable after 10 months of competition: The coffee shop needs to borrow 3 units each period for T periods at 5 percent interest, so the real present value of the debts that the coffee shop will accumulate is a geometric series given by [$$]Debt=\sum_{t=o}^T\frac{3}{1.05^t}=63\left(1-\frac{1}{1.05^T}\right).[$$] After date T, the coffee shop will begin earning 2 units of profit for ever after, which it can use to repay the loans. The total real present value of all future profits--the maximum amount of debt the coffee shop could possibly repay--is [$$]Profits=\sum_{t=T}^\infty\frac{2}{1.05^t}=\frac{42}{1.05^T}.[$$] Thus, the loans cab be repaid only if [$$]\frac{42}{1.05^T} \geq 63\left(1-\frac{1}{1.05^T}\right).[$$] Solving reveals that the maximum financing possible is T=10 months. At the 11th month, the coffee shop will default because the size of it's debts exceeds the total sum it will earn over all future periods. At 11 months, both the incumbent and the entrant are rendered permanently, irretrievably unprofitable.
I only do that bit of math so that we can talk about monetary policy. Austrianists see a malinvestment as a story about optimal monetary policy. That's pretty hard to follow, because the connection is pretty ambiguous. Monetary policy affects the above only to the extent that changing the nominal interest rate target temporarily changes the real interest rate in the inequality above--as the real interest rate falls, the T date increases. This, of course, ignores all the demand-side effects that money supply, prices, and interest rates have on the demand for coffee, and even then it is unclear how a lower real rate would actually affect the coffee shop market--regardless of the T date, it is only profitable for at most one of the two companies to borrow at all.
Strictly speaking, the coffee shop situation above shouldn't happen in neoclassical models--such models assume we've already reached a situation where all unprofitable entrants have been driven out of business, without addressing the question of whether profitable businesses can remain in business long enough to outlast unprofitable new entrants. Yet we see it happening all the time in real life--an explosion of new entrants render formerly profitable markets unprofitable for both the new and incumbent firms, with the resulting bust leaving us with fewer firms than the market can profitably sustain. (side note: the last coffee-shop bust happened years ago; since then my small town has had several other boom-bust cycles, most recently involving dance studios.) My suspicion is that something like this coffee-shop malinvestment cycle is what causes macroeconomic business cycles. It's not that businesses are expecting demand to surge in particular markets and turn out to be wrong, but that misguided investors think they can make it in already saturated markets, driving both themselves and the incumbents out of business.
What are the policy implications of malinvestment? It's hard to say without an empirically grounded mathematical model. But my intuition suggests a few things: 1) counter-cyclical monetary policy, tightening access to finance when entrants are upsetting markets and loosening it when incumbents are collapsing, 2) macroprudential regulations that prevent stupid entrepreneurial decisions like entering an already saturated market with a lackluster business model, 3) bankruptcy protections that allow once-profitable incumbents to discharge debts accumulated while competing with insane market entrants. In otherwords, if you think markets are plagued by malinvestment that chases good investment out of business, the policy implications would seem to be much closer to standard Monetarist and Keynesian views than Austrianist free-market liquidationism. |
Data Mining - Entropy (Information Gain) Table of Contents 1 - About
Entropy is a function “Information” that satisfies:
where:
p1p2 is the probability of event 1 and event 2 p1 is the probability of an event 1 p1 is the probability of an event 2
where:
H stands for entropy E for Ensemble ???
The entropy of a distribution with finite domain is maximized when all points have equal probability.
Bigger is the entropy, more is the event unpredicatble
Higher entropy means there is more unpredictability in the events being measured .
Higher entropy mean that the events being measured are less predictable.
100% predictability = 0 entropy
Fifty Fifty is an entropy of one.
Why is picking the attribute with the most information gain beneficial? It reduces entropy, which increases predictability. Information gain is positive when there is a decrease in entropy from choosing classifier/representation. A decrease in entropy signifies an decrease in unpredictability, which also means an increase in predictability.
2 - Articles Related 3 - Example 3.1 - Flipping a Coin
A two class problem.
3.2 - Rolling a die
A six class problem.
<MATH> \begin{array}{rrl} Entropy & = & - \sum_{x}{ p_x log_2( p_x)} \\ & = & - 6 * (\frac{1}{6} log_2 (\frac{1}{6})) \\ & \approx & 2.58 \end{array} </MATH>
3.3 - Rolling a weighted die
The weighted die is more predictable than a fair die.
4 - How unpredictable is your data? 4.1 - Titanic
Titanic training set with a two class problem: survived or die
4.1.1 - 342 survivors
case: 342 survivors on a total of 891 passengers
4.1.2 - 50 survivors
case: 50 survivors on a total of 891 passengers
It's a more predictable data set.
5 - Document / Reference Bill Howe (UW) |
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I read that visibly pushdown languages are supposed to model the typical simple formal languages like XML better than deterministic context free languages. The visibly pushdown languages can be recognized in LOGSPACE. I wonder whether the well-formed formulas in predicate logic for a given signature are a visibly pushdown language, or can at least be recognized in LOGSPACE.
Here is a typical well-formed formula in predicate logic, for the given signature $(f(\cdot,\cdot), g(\cdot))$:
$\forall x (f(x,y)=z \land g(z)=x)$
A corresponding context free grammar would be
$P \to T=T$
$P \to (P \land P)$ $P \to (P \lor P)$ $P \to \lnot P$ $P \to \forall V P$ $P \to \exists V P$ $T \to V$
$V \to x$
$V \to y$ $V \to z$
$T \to f(T,T)$
$T \to g(T)$
Some people prefer to use reverse polish notation:
$xyfz=zgx=\land x\forall$
A corresponding context free grammar would be
$P \to TT=$
$P \to PP\land$ $P \to PP\lor$ $P \to P\lnot$ $P \to PV\forall$ $P \to PV\exists$ $T \to V$
$V \to x$
$V \to y$ $V \to z$
$T \to TTf$
$T \to Tg$
Because I have now written down two explicit grammars, let the question just be whether these two grammars generate visibly pushdown languages, or can at least be recognized in LOGSPACE. |
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Abstract
We investigated the pressure driven flow of concentrated colloidal dispersions in a converging channel geometry. Optical microscopy and image analysis were used to track tracer particles mixed into dispersions of sterically stabilized poly(methyl methacrylate) (PMMA) spheres. The dispersions were drawn into a round \unit[0.5]{mm} capillary at one of two pump speeds ($\equiv$ applied pressure): $v_1=\unit[0.245]{ml\,\, min^{-1}}$ and $v_2=\unit[0.612]{ml\,\, min^{-1}}$. We observed that the dispersions at particle volume fractions $\phi\leqslant0.50$ followed Hagen-Poiseuille flow for a simple fluid; i.e. the mean flow rate $\langle V\rangle$ is approximately proportional to pressure drop (pump speed) and inversely proportional viscosity $\eta$. Above this concentration ($\phi\geqslant0.505$), the dispersions exhibit granular-like jamming behavior with $\langle V\rangle$ becoming independent of the pressure drop. However, at the highest applied pressure ($v_2$), the dispersions are able to unjam and switch from granular-like behaviour back to a simple hard-sphere liquid like system, due to the formation of rotating vortices in the spatial flow pattern. This mechanism is consistent with computer simulations of granular systems and supports for example proposed explanations of anomalously low friction in earthquake faults.
Author(s): Campbell, A.I. and Haw, M.D. ORCID: https://orcid.org/0000-0003-3736-1857 Item type: Article ID code: 28467 Keywords: jamming, unjamming, concentrated, colloidal, dispersions, channel flows, Chemical technology, Physics, Chemistry(all), Condensed Matter Physics Subjects: Technology > Chemical technology
Science > Physics
Department: Faculty of Engineering > Chemical and Process Engineering Depositing user: Dr Mark Haw Date deposited: 23 Oct 2010 13:02 Last modified: 08 Sep 2019 01:20 Related URLs: URI: https://strathprints.strath.ac.uk/id/eprint/28467 Export data: |
Consider the GARCH(1,1) process\begin{align} r_{t+1} &= \sigma_{t+1} z_{t+1} \\\sigma^2_{t+1} &= \omega+\alpha r^2_t +\beta \sigma^2_{t}\end{align}for the returns $r_t$, with ${z_t} \sim N (0,1)$ IID.
In what follows, let us distinguish the conditional return variance$$ V [ r_{t+1} \vert \mathcal{F}_t ] = \sigma^2_{t+1} $$from the unconditional return variance$$ V [ r_{t+1} ]$$where $\mathcal{F}_t$ denotes the information available up to time $t$ (filtration), in our case all past values $\{\sigma_t,...,\sigma_0,r_t,...,r_0\}$.
First, notice that if both $\alpha$ and $\beta $ happen to be zero, then if $\omega$ is allowed to reach zero as well there is a possibility that the conditional variance will become zero.
Second, because $z_t$ are zero-mean, unit variance and i.i.d, we have that the unconditional returns' variance is strictly equal to (*)\begin{align}V[r_{t+1}] &= E[r_{t+1}^2]-E [r_{t+1}]^2 = E[r_{t+1}^2] \\&= E [ \sigma_{t+1}^2 z_{t+1}^2] = E [ \sigma_{t+1}^2] E [z_{t+1}^2] \\&= E[\sigma_{t+1}^2]\end{align}
We can use this relationship along with the GARCH definition to write\begin{align}V[r_{t+1}] &= E[\sigma_{t+1}^2] = E [ (\omega+\alpha r^2_t +\beta \sigma^2_{t}) ] \\&= \omega + \alpha V [r_t] + \beta V [r_t] \\&= \omega + V [r_t] (\alpha + \beta)\end{align}Now, assuming weak stationarity we should have that the unconditional variance $\sigma^2_\infty$ is such that$$ \sigma^2_\infty = V [r_{t+1}] = V [r_t] $$which using the above gives $$ \sigma^2_\infty = \frac {\omega}{1-\alpha-\beta} $$ From which we see why the condition $\alpha + \beta < 1$ is relevant but also that allowing $\omega $ to reach zero would also imply a possibility that the unconditional variance becomes zero.
At the end of the day, we therefore have that if $\omega=0$ while the other coefficients are positive, you will have (conditional) heteroskedasticity in the sense that conditional variance will evolve through time, but the unconditional stationary variance will be zero, with unrealistic consequence that returns become deterministic at some point.
(*) This could have been anticipated since \begin{align}V [r_{t+1}] &= V [r_{t+1} \vert \mathcal{F}_0] \\&= E [r_{t+1}^2 \vert \mathcal{F}_0] \\&= E [ E [ r_{t+1}^2 \vert \mathcal{F}_t] \vert \mathcal{F}_0] \\&= E [ V [ r_{t+1} \vert \mathcal{F}_t] \vert \mathcal{F}_0] \\&= E [ \sigma_{t+1}^2 \vert \mathcal{F}_0] \\ &= E [ \sigma_{t+1}^2 ] \end{align}where we have made use of the tower law (and the fact that $z_t $ are zero-mean and independently distributed so that $E [r_{t+1}]=0$). |
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's Theorem
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is a generalization of the [[Fundamental Theorem of Calculus]]states that if ''M'' is an oriented smooth [[manifold]] of [[dimension]] k and <math>\omega</math> is a smooth (''k''−1)-formwith compact support on ''M''∂''M'' denotes the boundary of ''M'' with its induced orientation, then
:<math>\int_M \mathrm{d}\omega = \oint_{\partial M} \omega\!\,</math>,
:<math>\int_M \mathrm{d}\omega = \oint_{\partial M} \omega\!\,</math>,
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where ''d'' is the [[exterior derivative]].
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Latest revision as of 15:18, 29 July 2016
Stokes' Theorem states that the line integral of a closed path is equal to the surface integral of any capping surface for that path, provided that the surface normal vectors point in the same general direction as the right-hand direction for the contour:
Intuitively, imagine a "capping surface" that is nearly flat with the contour. The curl is the microscopic circulation of the function on tiny loops within that surface, and their sum or integral results in canceling out all the internal circulation paths, leaving only the integration over the outer-most path.
[1] This remains true no matter how the capping surface is expanded, provided that the contour remains as its boundary.
Sometimes the circulation (the left side above) is easier to compute; other times the expresses the surface integral of the curl of vector field is easier to computer (particularly when it is zero).
[2]
Stated another way, Stokes' Theorem equates the line integral of a vector fields to a surface integral of the same vector field. For this identity to be true, the
direction of the vector normal n must obey the right-hand rule for the direction of the contour, i.e., when walking along the contour the surface must be on your left.
This is an extension of Green's Theorem to surface integrals, and is also the analog in two dimensions of the Divergence Theorem. The above formulation is also called as the "Curl Theorem," to distinguish it from the more general form of the Stokes' Theorem described below.
Stokes' Theorem is useful in calculating circulation in mechanical engineering. A conservative field has a circulation (line integral on a simple, closed curve) of zero, and application of the Stokes' Theorem to such a field proves that the curl of a conservative field over the enclosed surface must also be zero.
General Form
In its most general form, this theorem is the fundamental theorem of Exterior Calculus, and is a generalization of the Fundamental Theorem of Calculus. It states that if
M is an oriented piecewise smooth manifold of dimension k and is a smooth ( k−1)-form with compact support on M, and ∂ M denotes the boundary of M with its induced orientation, then ,
where
d is the exterior derivative.
There are a number of well-known special cases of Stokes' theorem, including one that is referred to simply as "Stokes' theorem" in less advanced treatments of mathematics, physics, and engineering:
When k=1, and the terms appearing in the theorem are translated into their simpler form, this is just the Fundamental Theorem of Calculus. When k=3, this is often called Gauss' Theorem or the Divergence Theorem and is useful in vector calculus:
Where R is some region of 3-space, S is the boundary surface of R, the triple integral denotes volume integration over R with dV as the volume element, and the double integral denotes surface integration over S with as the oriented normal of the surface element. The on the left side is the divergence operator, and the on the right side is the vector dot product.
When k=2, this is often just called Stokes' Theorem:
Here S is a surface, E is the boundary path of S, and the single integral denotes path integration around E with as the length element. The on the left side is the curl operator.
These last two examples (and Stokes' theorem in general) are the subject of vector calculus. They play important roles in electrodynamics. The divergence and curl operations are cornerstones of Maxwell's Equations.
Stokes' Theorem is a lower-dimension version of the Divergence Theorem, and a higher-dimension version of Green's Theorem. Green’s Theorem relates a line integral to a double integral over a region, while Stokes' Theorem relates a surface integral of the curl of a function to its line integral. Stokes' Theorem originated in 1850.
References ↑ Notes on Gauss' and Stokes' Theorem, see Section 7.6, p. 87 ↑ |
Why is it possible to do the following approximation?
$$\frac{\mathrm{d} x}{x+\mathrm{d} x} \approx \frac{\mathrm{d}x}{x} $$
Why does that $\mathrm{d} x$ in the denominator "count" less than the one in the numerator?
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The $dx$ in the denominator is dwarfed by the $x$. The difference between $\frac{0.000001}{1.000001}$ and $\frac{0.000001}{1}$ is miniscule.
It's $\displaystyle \frac{dx}{x+dx}=\sum\limits_{k=0}^\infty (-1)^k (\frac{dx}{x})^{k+1}$.
Because of $\displaystyle (\frac{dx}{x})^{k+1}<<\frac{dx}{x}$ for $k\in\mathbb{N}$ you can approximate $\displaystyle \frac{dx}{x+dx}\approx \frac{dx}{x}$.
This is only valid as long as $x\not\approx 0$.
We have $x+\mathrm dx\approx x$ because $\mathrm dx\approx 0$ and addition is continuous. Then provided $x\not\approx 0$, the claim follows by continutity of division.
You are allowed to approx when terms are added as the difference between the results will be negligible however you cannot do the same with multiplication or division $10000+0.0001=10000.0001$ which is quite close however $10000*0.0001$ is definitely nowhere near $10000 $ |
What is the reason that the color properties we call red, green and blue have become tied to quarks, while what we call anti-red, anti-green and anti-blue has become tied to anti-quarks? Do note that I am NOT refering to the names (you can call any color value anything you want, after all), but the actual properties! Whatever the color Red really is, could that property have been tied to anti-quarks instead of quarks during the formation of the big bang? Or is there something special about the property we call Red that means it could only, ever, have become tied to quarks, and not anti-quarks??
closed as unclear what you're asking by ACuriousMind♦, user36790, CuriousOne, Floris, Gert Jun 5 '16 at 2:17
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
(Do quarks have) color charges because of something?
Boris Struminsky is though to be one of the first to postulate an additional quantum number for quarks, allowing for the $\Omega^-$ hyperion (comprising three strange quarks) to exist without violating the Pauli exclusion principle. Color charge was, hence, introduced as solution to this problem.
Experimental evidence from $e^+e^-$ annihilation shows there to be three color charges.
Can quarks have anti-colors?
A basic tenet of QCD is that quarks must combine to be colorless. If quarks could gain anti-color we would expect to see quark-quark pairs but we don't.
If you take some quark flavors to have color charge, $\psi_f\to U\psi_f$ with $U\in SU(3)$, and some others to have anti-charge, $\psi_g\to U^*\psi_g$, then the lagrangian $$ {\cal L}=\sum_{h=f,g} \bar\psi_h i\gamma^\mu D_\mu \psi_h $$ will not be gauge invariant, because $D\to UDU^\dagger$. In other words, this theory violates conservation of color. |
The word "statistics " refers two meanings. In the singular sense, it deals with the collection, presentation, analysis and interpretation of numerical data and helps in making a decision. In the plural sense, it refers to the numerical facts and figures are sometimes known as statistical data.
A typical value which represents the characteristics of the entire mass of huge data is called the central value of the whole distribution. A measure of the central tendency is also known as a measure of location or an average.
The various measure of central tendency are as follow:
The arithmetic mean is defined as the total sum of observations divided by the total number of observations. Here we study about the arithmetic mean of continuous series.
Calculation of arithmetic means in continuous series
When the number of items is large, we have to divide them into groups. Such groups are known as simple classes. It should be noted that there is no gap between any two successive intervals. So the data is continuous and the series is called continuous series.
Arithmetic Mean can be calculated in three different methods.
Arithmetic Mean can be calculated in three different methods.
Solution:
Mean(\(\overline X\))= 5o
\(\sum fx = 750 \) N = ?
\begin{align*} Mean\:(\overline {X}) &= \frac{\sum fx}{N}\\ 50 &= \frac{750}{N}\\ or, N &= \frac{750}{50}\\ \therefore N &= 15 \: \: \: \: _{Ans} \end{align*}
Solution:
\(Mean \: (\overline{X}) =60 \\ \sum fx = 960 \\ Number \: of \: term \: (N) = ? \)
\begin{align*} Mean (\overline {X}) &= \frac{\sum fx}{N}\\ 60 &= \frac{960}{N}\\ or, N &= \frac{960}{60}\\ \therefore N &= 16 \: _{Ans} \end{align*}
Solution:
\(Mean \: (\overline{X}) =12 \\ \sum fx = 70 +10a\\ No. \: of \: term \: (N) = 5+a \)
\begin{align*}Mean \: (\overline{X}) &= \frac{\sum fx}{N}\\ 12 &= \frac{70 + 10a}{5 + a}\\ or, 60 +12a &=70 + 10a\\ or,12a -10a &= 70 - 60\\ or, 2a &= 10 \\ or, a &= \frac{10}{2}\\ \therefore a &= 5 \: \: _{Ans} \end{align*}
Solution:
\(Mean \: (\overline{X}) = 13 \\ No. \: of \: terms \: (N) = 5 + a \)
\begin{align*} Mean (\overline{X}) &= \frac{\sum X}{N} \\ or, 13 &= \frac{4 + 8+12+x+25}{5}\\ or, 65 &=49 + x\\ or, x &= 65 - 49 \\ x &= 16 \:\: _{Ans} \end{align*}
Solution:
First item mean \( (\overline {X_{1}}) = 7 \)
Number of first item \(n_{1} = 4 \) Second item mean \( (\overline {X_{2}}) = 12\) No. of Second item \(n_{2}\)= 3 Mean of 7 items \(\overline{X}= ?\)
\begin{align*} ( \overline {X}) &= \frac{n_1 \overline{X}_1 + n_2 \overline{X}_2}{n_1+n_2}\\ &= \frac{4 \times 7 + 3 \times 12}{4 + 3}\\ &= \frac{28 + 36}{7}\\ &= \frac{64}{7}\\ &= 9.14 \: \: \: _{ans} \end{align*}
\(Mean (\overline{X}) = 40 \\ No. \: of \: terms (N)= 7 \\ k = ? \)
\begin{align*} Mean(\overline{X}) &= \frac{\sum X}{N} \\ or, 40 &= \frac{10+20+30+40+50+ 60+30+k}{7}\\ or, 280 &= 240 + k\\ or, k &= 280 - 240\\ \therefore k &= 40 \: \: \: \: \: \: \: \: \: _{Ans} \end{align*}
Solution:
Expenditure (in Rs.)x Frequency(f) fx 24 2 48 25 4 100 30 3 90 35 4 140 40 2 80
\begin{align*} Mean(\overline{X}) &= \frac{\sum fx}{N}\\ &= \frac{458}{15}\\ &= 30.53 \\ \therefore Average \: expenditure &= Rs \: 30.53 \end{align*}
Solution:
Let, the age of remaining students be x.
\(average \: age (\overline{X}) = 9 \\ Number \: of \: terms (N) = 5 \)
\begin{align*} (\overline{X}) &= \frac{\sum X}{N}\\ or, 9 &= \frac{5+7+8+15+x}{5}\\ or, 45 &= 35 + x \\ \therefore x &=45 -35 = 10 \\ \: \\ \therefore & \text{The age of remaining student is 10} \end{align*}
Class-interval 10-20 10-30 10-40 10-50 10-60 10-70 10-80 10-90 Frequency 4 16 56 97 124 137 146 150
Solution:
Calculation of mean
Class interval cf. f mid-value (m) fm 10-20 4 4 15 60 20-30 16 12 25 300 30-40 56 40 35 1400 40-50 97 41 45 1845 50-60 124 27 55 1485 60-70 137 13 65 845 70-80 146 9 75 675 80-90 150 4 85 340 N = 150 \(\sum fm=6950\)
\begin{align*}Mean \: (\overline{X})&= \frac{\sum fm}{N}\\ &= \frac{6950}{150}\\ &= 46.33 \: _{Ans} \end{align*}
X 0-10 10-20 20-30 30-40 40-50 f 5 7 8 4 6
Solution:
Calculation of mean
class interval midvalue f fm 0-10 5 5 25 10-20 15 7 105 20-30 25 8 200 30-40 35 4 140 40-50 45 6 270 N= 30 \( \sum fm = 740 \)
\begin{align*} Mean (\overline{X}) &= \frac{\sum fm}{N}\\ &= \frac{740}{30}\\ &= 24.6 \: _{ans} \end{align*}
Solution:
To find the value of p.
Solution:
To find the value of p.
Age in year No. of teachers (f) m fm 10-20 3 15 45 20-30 8 25 200 30-40 15 35 525 40-50 p 45 45p 50-60 4 55 220 N=p+30 \( \sum fm = 45p + 990\)
We know that,
\begin{align*} Mean \: (\overline {X}) &= \frac{\sum fm}{N}\\ 36 &= \frac{45p + 990}{p+30}\\ or, 1080 + 36p &= 45p +990\\ or, 45p - 36p &= 1080 -990 \\ or, 9p &= 90 \\ \therefore p &= \frac{90}{9} = 10 \: _{Ans} \end{align*}
Solution:
To find the value of p.
Solution:
To find the value of p.
Age in year No. of teachers (f) m fm 10-20 3 15 45 20-30 8 25 200 30-40 15 35 525 40-50 p 45 45p 50-60 4 55 220 N=p+30 \( \sum fm = 45p + 990\)
We know that,
\begin{align*} Mean \: (\overline {X}) &= \frac{\sum fm}{N}\\ 36 &= \frac{45p + 990}{p+30}\\ or, 1080 + 36p &= 45p +990\\ or, 45p - 36p &= 1080 -990 \\ or, 9p &= 90 \\ \therefore p &= \frac{90}{9} = 10 \: _{Ans} \end{align*}
The mean of 50 observations was 250. It was detected on checking that the value of 165 was wrongly copied as 115 for computation of mean. Find the correct mean.
251
152
543
345
The mean of 40 observations was 160. It was detected on rechecking that the value of 165 was wrongly copied as 125 for computation of mean. Find the correct mean.
785
321
161
465
In a grouped data, the sum of the marks obtained by 50 students in mathematics is 3000.Find the average mark.
60
20
80
60
In a grouped data, the sum of the marks obtained by 50 students in mathematics is 3000.Find the average mark.
80
60
20
60
In a continuous series the value of a assumed mean is 35 and the sum of frequencies is 50.If mean of the data is 41 then find the sum of the product of deviation (d) and frequencies(f).
300
200
750
700
In a continuous series the value of a assumed mean is 25 and the sum of frequencies is 150.If mean of the data is 26.07 then find the sum of the product of deviation (d) and frequencies(f).
120
175
160
80
If the mean of a grouped data having ∑fm =306 is 18, find the value of N.
7
10
17
20
In a continuous frequency distribution data, the number of terms (N) = 40 and mean ((overline {X}) )= 75. If a class interval 40-50 having frequency 10 is included in the series, find the new mean.
92
69
54
22
In a continuous frequency distribution data, the number of terms (N) = 40 and mean (overline {X}) )= 75. If a class interval 70-80 having frequency 12 is excluded from the series, what will be the new mean?
In a continuous frequency distribution data, the number of terms (N) = 40 and mean ((overline {X}) )= 36. If a class interval 40-50 having frequency 10 is excluded from the series, what will be the new mean?
45
21
65
33
In a continuous series the mean of 50 observation was 40. It was detected on checking that the class interval 30-40 having frequency 10 was wrongly copied as class interval 30-50 having frequency 10 for computation of mean.Find the correct mean.
39
69
59
49
In a continuous series the mean of 50 observation was 40. It was detected on checking that the class interval 30-40 having frequency 10 was wrongly copied as class interval 30-50 having frequency 10 for computation of mean.Find the correct mean.
59
49
39
69
In a continuous series the mean of 40 observation was 60. It was detected on checking that the class interval 20-40 having frequency 10 was wrongly copied as class interval 20-40 having frequency 10 for computation of mean.Find the correct mean.
40.25
61.25
35.30
31 |
Groups¶ class
sage.categories.groups.
Groups(
base_category)¶
The category of (multiplicative) groups, i.e. monoids with inverses.
EXAMPLES:
sage: Groups() Category of groups sage: Groups().super_categories() [Category of monoids, Category of inverse unital magmas] class
CartesianProducts(
category, *args)¶
The category of groups constructed as Cartesian products of groups.
class
ElementMethods¶
multiplicative_order()¶
Return the multiplicative order of this element.
EXAMPLES:
sage: G1 = SymmetricGroup(3) sage: G2 = SL(2,3) sage: G = cartesian_product([G1,G2]) sage: G((G1.gen(0), G2.gen(1))).multiplicative_order() 12 class
ParentMethods¶
group_generators()¶
Return the group generators of
self.
EXAMPLES:
sage: C5 = CyclicPermutationGroup(5) sage: C4 = CyclicPermutationGroup(4) sage: S4 = SymmetricGroup(3) sage: C = cartesian_product([C5, C4, S4]) sage: C.group_generators() Family (((1,2,3,4,5), (), ()), ((), (1,2,3,4), ()), ((), (), (1,2)), ((), (), (2,3)))
We check the other portion of trac ticket #16718 is fixed:
sage: len(C.j_classes()) 1
An example with an infinitely generated group (a better output is needed):
sage: G = Groups.free([1,2]) sage: H = Groups.free(ZZ) sage: C = cartesian_product([G, H]) sage: C.monoid_generators() Lazy family (gen(i))_{i in The Cartesian product of (...)}
order()¶
Return the cardinality of self.
EXAMPLES:
sage: C = cartesian_product([SymmetricGroup(10), SL(2,GF(3))]) sage: C.order() 87091200
Todo
this method is just here to prevent
FiniteGroups.ParentMethodsto call
_cardinality_from_iterator.
extra_super_categories()¶
A Cartesian product of groups is endowed with a natural group structure.
EXAMPLES:
sage: C = Groups().CartesianProducts() sage: C.extra_super_categories() [Category of groups] sage: sorted(C.super_categories(), key=str) [Category of Cartesian products of inverse unital magmas, Category of Cartesian products of monoids, Category of groups] class
Commutative(
base_category)¶
Category of commutative (abelian) groups.
A group \(G\) is
commutativeif \(xy = yx\) for all \(x,y \in G\). static
free(
index_set=None, names=None, **kwds)¶
Return the free commutative group.
INPUT:
index_set– (optional) an index set for the generators; if an integer, then this represents \(\{0, 1, \ldots, n-1\}\)
names– a string or list/tuple/iterable of strings (default:
'x'); the generator names or name prefix
EXAMPLES:
sage: Groups.Commutative.free(index_set=ZZ) Free abelian group indexed by Integer Ring sage: Groups().Commutative().free(ZZ) Free abelian group indexed by Integer Ring sage: Groups().Commutative().free(5) Multiplicative Abelian group isomorphic to Z x Z x Z x Z x Z sage: F.<x,y,z> = Groups().Commutative().free(); F Multiplicative Abelian group isomorphic to Z x Z x Z class
ElementMethods¶
conjugacy_class()¶
Return the conjugacy class of
self.
EXAMPLES:
sage: D = DihedralGroup(5) sage: g = D((1,3,5,2,4)) sage: g.conjugacy_class() Conjugacy class of (1,3,5,2,4) in Dihedral group of order 10 as a permutation group sage: H = MatrixGroup([matrix(GF(5),2,[1,2, -1, 1]), matrix(GF(5),2, [1,1, 0,1])]) sage: h = H(matrix(GF(5),2,[1,2, -1, 1])) sage: h.conjugacy_class() Conjugacy class of [1 2] [4 1] in Matrix group over Finite Field of size 5 with 2 generators ( [1 2] [1 1] [4 1], [0 1] ) sage: G = SL(2, GF(2)) sage: g = G.gens()[0] sage: g.conjugacy_class() Conjugacy class of [1 1] [0 1] in Special Linear Group of degree 2 over Finite Field of size 2 sage: G = SL(2, QQ) sage: g = G([[1,1],[0,1]]) sage: g.conjugacy_class() Conjugacy class of [1 1] [0 1] in Special Linear Group of degree 2 over Rational Field class
ParentMethods¶
cayley_table(
names='letters', elements=None)¶
Returns the “multiplication” table of this multiplicative group, which is also known as the “Cayley table”.
Note
The order of the elements in the row and column headings is equal to the order given by the table’s
column_keys()method. The association between the actual elements and the names/symbols used in the table can also be retrieved as a dictionary with the
translation()method.
For groups, this routine should behave identically to the
multiplication_table()method for magmas, which applies in greater generality.
INPUT:
names- the type of names used, values are:
'letters'- lowercase ASCII letters are used for a base 26 representation of the elements’ positions in the list given by
list(), padded to a common width with leading ‘a’s.
'digits'- base 10 representation of the elements’ positions in the list given by
column_keys(), padded to a common width with leading zeros.
'elements'- the string representations of the elements themselves.
a list - a list of strings, where the length of the list equals the number of elements.
elements- default =
None. A list of elements of the group, in forms that can be coerced into the structure, eg. their string representations. This may be used to impose an alternate ordering on the elements, perhaps when this is used in the context of a particular structure. The default is to use whatever ordering is provided by the the group, which is reported by the
column_keys()method. Or the
elementscan be a subset which is closed under the operation. In particular, this can be used when the base set is infinite.
OUTPUT: An object representing the multiplication table. This is an
OperationTableobject and even more documentation can be found there.
EXAMPLES:
Permutation groups, matrix groups and abelian groups can all compute their multiplication tables.
sage: G = DiCyclicGroup(3) sage: T = G.cayley_table() sage: T.column_keys() ((), (5,6,7), ..., (1,4,2,3)(5,7)) sage: T * a b c d e f g h i j k l +------------------------ a| a b c d e f g h i j k l b| b c a e f d i g h l j k c| c a b f d e h i g k l j d| d e f a b c j k l g h i e| e f d b c a l j k i g h f| f d e c a b k l j h i g g| g h i j k l d e f a b c h| h i g k l j f d e c a b i| i g h l j k e f d b c a j| j k l g h i a b c d e f k| k l j h i g c a b f d e l| l j k i g h b c a e f d sage: M = SL(2, 2) sage: M.cayley_table() * a b c d e f +------------ a| c e a f b d b| d f b e a c c| a b c d e f d| b a d c f e e| f d e b c a f| e c f a d b sage: A = AbelianGroup([2, 3]) sage: A.cayley_table() * a b c d e f +------------ a| a b c d e f b| b c a e f d c| c a b f d e d| d e f a b c e| e f d b c a f| f d e c a b
Lowercase ASCII letters are the default symbols used for the table, but you can also specify the use of decimal digit strings, or provide your own strings (in the proper order if they have meaning). Also, if the elements themselves are not too complex, you can choose to just use the string representations of the elements themselves.
sage: C=CyclicPermutationGroup(11) sage: C.cayley_table(names='digits') * 00 01 02 03 04 05 06 07 08 09 10 +--------------------------------- 00| 00 01 02 03 04 05 06 07 08 09 10 01| 01 02 03 04 05 06 07 08 09 10 00 02| 02 03 04 05 06 07 08 09 10 00 01 03| 03 04 05 06 07 08 09 10 00 01 02 04| 04 05 06 07 08 09 10 00 01 02 03 05| 05 06 07 08 09 10 00 01 02 03 04 06| 06 07 08 09 10 00 01 02 03 04 05 07| 07 08 09 10 00 01 02 03 04 05 06 08| 08 09 10 00 01 02 03 04 05 06 07 09| 09 10 00 01 02 03 04 05 06 07 08 10| 10 00 01 02 03 04 05 06 07 08 09 sage: G=QuaternionGroup() sage: names=['1', 'I', '-1', '-I', 'J', '-K', '-J', 'K'] sage: G.cayley_table(names=names) * 1 I -1 -I J -K -J K +------------------------ 1| 1 I -1 -I J -K -J K I| I -1 -I 1 K J -K -J -1| -1 -I 1 I -J K J -K -I| -I 1 I -1 -K -J K J J| J -K -J K -1 -I 1 I -K| -K -J K J I -1 -I 1 -J| -J K J -K 1 I -1 -I K| K J -K -J -I 1 I -1 sage: A=AbelianGroup([2,2]) sage: A.cayley_table(names='elements') * 1 f1 f0 f0*f1 +------------------------ 1| 1 f1 f0 f0*f1 f1| f1 1 f0*f1 f0 f0| f0 f0*f1 1 f1 f0*f1| f0*f1 f0 f1 1
The
change_names()routine behaves similarly, but changes an existing table “in-place.”
sage: G=AlternatingGroup(3) sage: T=G.cayley_table() sage: T.change_names('digits') sage: T * 0 1 2 +------ 0| 0 1 2 1| 1 2 0 2| 2 0 1
For an infinite group, you can still work with finite sets of elements, provided the set is closed under multiplication. Elements will be coerced into the group as part of setting up the table.
sage: G=SL(2,ZZ) sage: G Special Linear Group of degree 2 over Integer Ring sage: identity = matrix(ZZ, [[1,0], [0,1]]) sage: G.cayley_table(elements=[identity, -identity]) * a b +---- a| a b b| b a
The
OperationTableclass provides even greater flexibility, including changing the operation. Here is one such example, illustrating the computation of commutators.
commutatoris defined as a function of two variables, before being used to build the table. From this, the commutator subgroup seems obvious, and creating a Cayley table with just these three elements confirms that they form a closed subset in the group.
sage: from sage.matrix.operation_table import OperationTable sage: G = DiCyclicGroup(3) sage: commutator = lambda x, y: x*y*x^-1*y^-1 sage: T = OperationTable(G, commutator) sage: T . a b c d e f g h i j k l +------------------------ a| a a a a a a a a a a a a b| a a a a a a c c c c c c c| a a a a a a b b b b b b d| a a a a a a a a a a a a e| a a a a a a c c c c c c f| a a a a a a b b b b b b g| a b c a b c a c b a c b h| a b c a b c b a c b a c i| a b c a b c c b a c b a j| a b c a b c a c b a c b k| a b c a b c b a c b a c l| a b c a b c c b a c b a sage: trans = T.translation() sage: comm = [trans['a'], trans['b'], trans['c']] sage: comm [(), (5,6,7), (5,7,6)] sage: P = G.cayley_table(elements=comm) sage: P * a b c +------ a| a b c b| b c a c| c a b
Todo
Arrange an ordering of elements into cosets of a normal subgroup close to size \(\sqrt{n}\). Then the quotient group structure is often apparent in the table. See comments on trac ticket #7555.
AUTHOR:
Rob Beezer (2010-03-15)
conjugacy_class(
g)¶
Return the conjugacy class of the element
g.
This is a fall-back method for groups not defined over GAP.
EXAMPLES:
sage: A = AbelianGroup([2,2]) sage: c = A.conjugacy_class(A.an_element()) sage: type(c) <class 'sage.groups.conjugacy_classes.ConjugacyClass_with_category'>
group_generators()¶
Return group generators for
self.
This default implementation calls
gens(), for backward compatibility.
EXAMPLES:
sage: A = AlternatingGroup(4) sage: A.group_generators() Family ((2,3,4), (1,2,3))
holomorph()¶
The holomorph of a group
The holomorph of a group \(G\) is the semidirect product \(G \rtimes_{id} Aut(G)\), where \(id\) is the identity function on \(Aut(G)\), the automorphism group of \(G\).
EXAMPLES:
sage: G = Groups().example() sage: G.holomorph() Traceback (most recent call last): ... NotImplementedError: holomorph of General Linear Group of degree 4 over Rational Field not yet implemented
monoid_generators()¶
Return the generators of
selfas a monoid.
Let \(G\) be a group with generating set \(X\). In general, the generating set of \(G\) as a monoid is given by \(X \cup X^{-1}\), where \(X^{-1}\) is the set of inverses of \(X\). If \(G\) is a finite group, then the generating set as a monoid is \(X\).
EXAMPLES:
sage: A = AlternatingGroup(4) sage: A.monoid_generators() Family ((2,3,4), (1,2,3)) sage: F.<x,y> = FreeGroup() sage: F.monoid_generators() Family (x, y, x^-1, y^-1)
semidirect_product(
N, mapping, check=True)¶
The semi-direct product of two groups
EXAMPLES:
sage: G = Groups().example() sage: G.semidirect_product(G,Morphism(G,G)) Traceback (most recent call last): ... NotImplementedError: semidirect product of General Linear Group of degree 4 over Rational Field and General Linear Group of degree 4 over Rational Field not yet implemented class
Topological(
category, *args)¶
Category of topological groups.
A topological group \(G\) is a group which has a topology such that multiplication and taking inverses are continuous functions.
REFERENCES:
example()¶
EXAMPLES:
sage: Groups().example() General Linear Group of degree 4 over Rational Field static
free(
index_set=None, names=None, **kwds)¶
Return the free group.
INPUT:
index_set– (optional) an index set for the generators; if an integer, then this represents \(\{0, 1, \ldots, n-1\}\)
names– a string or list/tuple/iterable of strings (default:
'x'); the generator names or name prefix
EXAMPLES:
sage: Groups.free(index_set=ZZ) Free group indexed by Integer Ring sage: Groups().free(ZZ) Free group indexed by Integer Ring sage: Groups().free(5) Free Group on generators {x0, x1, x2, x3, x4} sage: F.<x,y,z> = Groups().free(); F Free Group on generators {x, y, z} |
I will work over a field of characteristic $0$ so that reductive algebraic groups are linearly reductive; presumably there is a way to eliminate this hypothesis. In this case, for an integer $n>1$ and for a divisor $m$ such that $n>m>1$, there
does not exist $f$ as above. The point is to consider the critical locus of the determinant. The computation below proves that "cohomology with supports" has cohomological dimension equal to $2(n-1)$ for the pair of the quasi-affine scheme $\textbf{Mat}_{n\times n} \setminus \text{Crit}(\Delta_{n\times n})$ and its closed subset $\text{Zero}(\Delta_{n\times n})\setminus \text{Crit}(\Delta_{n\times n})$. Since pushforward under affine morphisms preserve cohomology of quasi-coherent sheaves, that leads to a contradiction.
Denote the $m\times m$ determinant polynomial by $$\Delta_{m\times m}:\textbf{Mat}_{m\times m} \to \mathbb{A}^1.$$ For integers $m$ and $n$ such that $m$ divides $n$, you ask whether there exists a homogeneous polynomial morphism $$f:\textbf{Mat}_{n\times n}\to \textbf{Mat}_{m\times m}$$ of degree $n/m$ such that $\Delta_{n\times n}$ equals $\Delta_{m\times m}\circ f$. Of course that is true if $m$ equals $1$: just define $f$ to be $\text{Delta}_{n\times n}$. Similarly, this is true if $m$ equals $n$: just define $f$ to be the identity. Thus, assume that $2\leq m < n$; this manifests below through the fact that the critical locus of $\Delta_{m\times m}$ is nonempty. By way of contradiction, assume that there exists $f$ with $\Delta_{n\times n}$ equal to $\Delta_{m\times m}\circ f$.
Lemma 1. The inverse image under $f$ of $\text{Zero}(\Delta_{m\times m})$ equals $\text{Zero}(\Delta_{n\times n})$. In other words, the inverse image under $f$ of the locus of matrices with nullity $\geq 1$ equals the locus of matrices with nullity $\geq 1$.
Proof. This is immediate. QED
Lemma 2. The inverse image under $f$ of the critical locus of $\Delta_{m\times m}$ equals the critical locus of $\Delta_{n\times n}$. In other words, the inverse image under $f$ of the locus of matrices with nullity $\geq 2$ equals the locus of matrices with nullity $\geq 2$.
Proof.By the Chain Rule, $$d_A\Delta_{n\times n} = d_{f(A)}\Delta_{m\times m}\circ d_Af.$$ Thus, the critical locus of $\Delta_{n\times n}$ contains the inverse image under $f$ of the critical locus of $\Delta_{m\times m}$. Since $m\geq 2$, in each case, the critical locus is the nonempty set of those matrices whose kernel has dimension $\geq 2$, this critical locus contains the origin, and this critical locus is irreducible of codimension $4$. Thus, the inverse image under $f$ of the critical locus of $\Delta_{m\times m}$ is nonempty (it contains the origin) and has codimension $\leq 4$ (since $\textbf{Mat}_{m\times m}$ is smooth). Since this is contained in the critical locus of $\Delta_{n\times n}$, and since the critical locus of $\Delta_{n\times n}$ is irreducible of codimension $4$, the inverse image of the critical locus of $\Delta_{m\times m}$ equals the critical locus of $\Delta_{n\times n}$. QED
Denote by $U_n\subset \text{Zero}(\Delta_{n\times n})$, resp. $U_m\subset \text{Zero}(\Delta_{m\times m})$, the open complement of the critical locus, i.e., the locus of matrices whose kernel has dimension precisely equal to $1$. By Lemma 1 and Lemma 2, $f$ restricts to an affine morphism $$f_U:U_n\to U_m.$$
Proposition. The cohomological dimension of sheaf cohomology for quasi-coherent sheaves on the quasi-affine scheme $U_n$ equals $2(n-1)$.
Proof.The quasi-affine scheme $U_n$ admits a morphism, $$\pi_n:U_n \to \mathbb{P}^{n-1}\times (\mathbb{P}^{n-1})^*,$$ sending every singular $n\times n$ matrix $A$ parameterized by $U_n$ to the ordered pair of the kernel of $A$ and the image of $A$. The morphism $\pi_n$ is Zariski locally projection from a product, where the fiber is the affine group scheme $\textbf{GL}_{n-1}$. In particular, since $\pi_n$ is affine, the cohomological dimension of $U_n$ is no greater than the cohomological dimension of $\mathbb{P}^{n-1}\times (\mathbb{P}^{n-1})^*$. This equals the dimension $2(n-1)$.
More precisely, $U_n$ is simultaneously a principal bundle for both group schemes over $\mathbb{P}^{n-1}\times(\mathbb{P}^{n-1})^*$ that are the pullbacks via the two projections of $\textbf{GL}$ of the tangent bundle. Concretely, for a fixed $1$-dimension subspace $K$ of the $n$-dimensional vector space $V$ -- the kernel -- and for a fixed codimension $1$ subspace $I$ -- the image -- the set of invertible linear maps from $V/K$ to $I$ is simultaneously a principal bundle under precomposition by $\textbf{GL}(V/K)$ and a principal bundle under postcomposition by $\textbf{GL}(I)$. In particular, the pushforward of the structure sheaf, $$\mathcal{E}_n:=(\pi_n)_*\mathcal{O}_{U_n},$$ is a quasi-coherent sheaf that has an induced action of each of these group schemes.
The invariants for each of these actions is just $$\pi_n^\#:\mathcal{O}_{\mathbb{P}^{n-1}\times (\mathbb{P}^{n-1})^*}\to \mathcal{E}_n.$$ Concretely, the only functions on an algebraic group that are invariant under pullback by every element of the group are the constant functions. The group schemes and the principal bundle are each
Zariski locally trivial. Consider the restriction of $\mathcal{E}_n$ on each open affine subset $U$ where the first group scheme is trivialized and the principal bundle is trivialized. The sections on this open affine give a $\mathcal{O}(U)$-linear representation of $\textbf{GL}_{n-1}$. Because $\textbf{GL}_{n-1}$ is linearly reductive, there is a unique splitting of this representation into its invariants, i.e., $\pi_n^\#\mathcal{O}(U)$, and a complementary representation (having trivial invariants and coinvariants). The uniqueness guarantees that these splittings glue together as we vary the trivializing opens. Thus, there is a splitting of $\pi_n^\#$ as a homomorphism of quasi-coherent sheaves, $$t_n:(\pi_n)_*\mathcal{O}_{U_n} \to \mathcal{O}_{\mathbb{P}^{n-1}\times (\mathbb{P}^{n-1})^*}.$$
For every invertible sheaf $\mathcal{L}$ on $\mathbb{P}^{n-1}\times (\mathbb{P}^{n-1})^*$, this splitting of $\mathcal{O}$-modules gives rise to a splitting, $$t_{n,\mathcal{L}}:(\pi_n)_*(\pi_n^*\mathcal{L})\to \mathcal{L}.$$ In particular, for every integer $q$, this gives rise to a surjective group homomorphism, $$H^q(t_{n,\mathcal{L}}):H^q(U_n,\pi_n^*\mathcal{L}) \to H^q(\mathbb{P}^{n-1}\times (\mathbb{P}^{n-1})^*,\mathcal{L}) .$$
Now let $\mathcal{L}$ be a dualizing invertible sheaf on $\mathbb{P}^{n-1}\times (\mathbb{P}^{n-1})^*.$ This has nonzero cohomology in degree $2(n-1)$. Thus, the cohomological dimension of sheaf cohomology for quasi-coherent $\mathcal{O}_{U_n}$-modules also equals $2(n-1)$.
QED
Since $f_U$ is affine, the cohomological dimension for $U_n$ is no greater than the cohomological dimension for $U_m$. However, by the proposition, the cohomological dimension for $U_m$ equals $2(m-1)$. Since $1<m<n$, this is a contradiction. |
A silly question that is bugging me. I am working my way through Baxter and Rennie (again) and I am getting my wires crossed on the short rate models in particular the straight forward Ho and Lee analysis.
So given the SDE (under the $\mathbb{Q}$ measure) $$ dr_t = \sigma dW_t +\theta_t dt $$ where $\theta_t$ is both deterministic and bounded and $\sigma$ is constant. This becomes $$ r_t = f(0,t) + \sigma W_t +\int_0^t \theta_s ds $$ (I hope).
How do I compute the integral $$ \int_t^T r_sds $$
Basically it comes down to computing $$ \int_t^T W_sds $$ and $$ \int_t^T \int_0^s \theta_k dk. $$ From first integral is just book work (though it be nice to see a derivation here other than by parts?) It is the later which I am not sure about, as the result is apparently $$ \int_t^T (T-s)\theta_s ds $$ Which I am puzzled by?
$\textbf{edit}$ Actually an important piece of information is that I am trying to compute $$ -\log\mathbb{E}_{\mathbb{Q}}\left(\mathrm{e}^{-\int_t^T r_sds}\vert r_t = x\right)=x(T-t) -\frac{1}{6}\sigma^2 (T-t)^3 + \int_0^T (T-s)\theta_sds $$ |
Is the union of finite and countably infinite sequence over alphabet $\Sigma=\{1\}$, countably infinite as well?
I understand this is similar a question to the one of finite and countably infinite strings over $\{0,1\}$, however I think that the previous argument is not valid in this case.
Now, countably infinite strings over $\{0,1\}$ can be of any kind, e.g.
$$e_i \in \Sigma^*, e_i=100000...$$ $$e_k \in \Sigma^*, e_k=110000...$$
However, I noticed that if I pose the same question with an alphabet of a character only, say $\{1\}$, there is only 1 infinite sequence that I can generate $11.....$, hence:
$$S = \Sigma^* \cup \{111..\}$$
From this assumption, I could map $0$ to the infinite $11...$ and assume that this set S is uncountable. Is this a valid argument? Can someone help me formalize this? |
Wondering if there are any documents, theories, or methodologies for dealing with mutable memory mathematically. Basically a formal algebraic model of how computers manipulate memory. Along the lines of, say I want to model an
add operation at a low level, and it writes its output to a register. Then we can model the computer state as a memory, and:
$$add : a \times b \times m \to m'$$
the function $add$ transforms the memory into a different state, holding the final value. Then the second time we call add, the memory is different than it was before. So it's always changing:
$$m \neq m' \neq m'' \neq \dots \neq m^{(n)}$$
Wondering if there are any formalisms out there that deal with this. I would specifically like to see it applied to abstract algebra or category theory. Whereas in abstract algebra typically, everything is immutable and you never worry about "state". I would like to explore the state / mutable memory from a math framework. |
You have two things to show: that if $D$ is closed, then $X$ is Hausdorff, and that if $X$ is Hausdorff, then $D$ is closed.
Suppose first that $D$ is closed in $X\times X$. To show that $X$ is Hausdorff, you must show that if $x$ and $y$ are any two points of $X$, then there are open sets $U$ and $V$ in $X$ such that $x\in U$, $y\in V$, and $U\cap V=\varnothing$. The trick is to look at the point $p=\langle x,y\rangle\in X\times X$. Because $x\ne y$, $p\notin D$. This means that $p$ is in the open set $(X\times X)\setminus D$. Thus, there must be a basic open set $B$ in the product topology such that $p\in B\subseteq(X\times X)\setminus D$. Basic open sets in the product topology are open boxes, i.e., sets of the form $U\times V$, where $U$ and $V$ are open in $X$, so let $B=U\times V$ for such $U,V\subseteq X$. Can you see how to use this to get the desired open neighborhoods of $x$ and $y$?
Now suppose that $X$ is Hausdorff. To show that $D$ is closed in $X\times X$, you need only show that $(X\times X)\setminus D$ is open. To do
this, just take any point $p\in(X\times X)\setminus D$ and show that it has an open neighborhood disjoint from $D$. I suggest that you try to reverse what I did above. First, $p=\langle x,y\rangle$ for some $x,y\in X$, and since $p\notin D$, $x\ne y$. Now use disjoint open neighborhoods of $x$ and $y$ to build a basic open box around $p$ that is disjoint from $D$. |
How many positive factorials are also perfect Squares. So for example $1!=1=1^2$. How many others exist other than 1?
Is there any way to prove this?
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Hint:
Bertrand's postulate(in actuality a theorem) states that for every prime $p$, there exists another prime number between $p$ and $2p$. This means that $\forall n>1, n!$ will always have a single power of some prime number(s).
For any positive integer $n\ge 2$, there exists a prime $p$ such that $\frac{n}{2}<p\le n$ This implies that $p\mid n!$. For $n\ge 5, p^2>n$. So if $n!$ has unique prime factorization $p_1^{k_1} \cdot p_2^{k_2} \cdots p_r^{k_r} $, then the exponent of $p$ must be $1$. Since perfect squares must have even exponents in their prime factorizations, we know $n!$ cannot be a perfect square for $n \ge 5$. |
Does a
symbolically elliptic sequence of differential operators have an analytic index? cohomology? For example, is there any concrete meaning of the Todd genus of an almost complex manifold in terms of the Dolbeault operator?
Consider a smooth manifold $M$ with vectors bundles $E_0$, $E_1$ and a linear differential operator $D\colon C^\infty(E_0)\to C^\infty(E_1)$ on their sections. The (principal) symbol $\sigma(D)$ of the degree $n$ operator is the leading term, ie, considered modulo operators of lower degrees. It can be interpreted as a vector bundle map from $E_0$ to $E_1$, not on $M$, but on its cotangent bundle, $\sigma(D)\in\Gamma(T^*M; Hom(E_0,E_1))$. Varying over the cotangent space at a particular point of $M$, it is homogeneous of degree $n$. Thus the symbol is zero along the zero section $M\subset T^*M$. The operator $D$ is
elliptic if the symbol is an isomorphism of bundles away from the zero section.
An elliptic operator on a closed manifold is Fredholm, with finite dimensional kernel and cokernel. The
index of a Fredholm operator is the difference between the dimensions of the kernel and cokernel. The index is a robust invariant. The index of a continuous family of Fredholm operators is constant. Two operators with the same symbol can be linearly interpolated, so they have the same index, so the index depends only on the symbol. Moreover, homotopy classes of elliptic symbols have constant indices (where a homotopy is a path of symbols all of which are elliptic, ie, isomorphisms). Determining the index from the homotopical data of the symbol is Gelfand’s index problem, solved by Atiyah—Singer, but that is not directly relevant to this question. For the purpose of this question, when I say analytic index I mean something concretely defined in terms of the operator, not produced by making arbitrary choices to introduce auxiliary operators.
Just as an exact chain complex generalizes an isomorphism, so does an elliptic complex generalize an elliptic operator. Let me define a composable sequence of differential operators $d_0,d_1\colon C^\infty(E_0)\to C^\infty(E_1)\to C^\infty(E_2)$ to be
symbolically elliptic if their symbols form a chain complex of vector bundles on the cotangent bundle, exact away from the zero section. Is this a good concept?
The usual concept is an
elliptic complex. While the ellipticity of an operator depends only on its symbol, for a sequence of composable operators to be an elliptic complex requires that it be symbolically elliptic and the non-symbolic condition that it be a chain complex, ie, that the operators compose to zero $d_1\circ d_0=0$. An elliptic complex, being a chain complex, has cohomology. Under the weaker condition of being symbolically elliptic, the sequence can be rolled up into a single elliptic operator $d_0+d_1^*\colon C^\infty(E_0\oplus E_2)\to C^\infty(E_2)$. Doing this uses the adjoint, defined by metrics on $M$ and metrics on the bundles. The sums of the even and odd cohomologies of an elliptic complex are naturally isomorphic to the kernel and cokernel of the single operator (eg, the Hodge theorem). In particular, they are finite dimensional. Moreover, the Euler characteristic of an elliptic complex is equal to the index of the associated elliptic operator.
A symbolically elliptic sequence of operators has an index that can be defined by making various choices. It can be perturbed to an elliptic complex. Or it can be folded up by metrics. But does the index have concrete analytic meaning about the original operators? In other words: can the kernel and cokernel of the rolled up operator be identified with something independent of choices? More weakly, if $D$ and $D’$ are the elliptic operators produced by different choices of metrics, are their kernels and cokernels canonically isomorphic? Are they even the same dimensions? (In the case of a length 2 complex, they have the same dimensions because the kernels are isomorphic to the kernel of $d_0$ plus the cokernel of $d_2$ and the dimension of the cokernels are fixed by the index.)
Similarly, an anti-self-adjoint elliptic operator $D=-D^*$ on a compact manifold has an (anti-self-adjoint) index in $\mathbb Z/2$, the parity of the dimension of the kernel. This index depends only on the symbol. An operator whose symbol is anti-self-adjoint can be deformed to a truly anti-self-adjoint operator with the same symbol, namely $\frac12(D-D^*)$. Can this index of this operator be interpreted in terms of the original operator? |
Relations and Functions Composition of Functions and Invertible Function Composite function: If f and g are two functions, then the composite function f and g is denoted by gof and it is defined only when codomain of f equal to domain of g. i.e. If f : A → B and g : B → C are two functions, then the composite function gof : A → C is defined as gof(x) = g[f(x)] ∀ x ∈ A. If f and g are two bijective functions then gof is also bijective. If gof is one one, then f is one one but g may not be one one. If gof is onto, then g is onto but f may not be onto. Inverse of a function: If the function f : x → y is bijective, then we defined inverse of f as f −1: y → x by the rule y = f(x) ⇔ x = f −1(y) ∀ x ∈ X, y ∈ Y. If f : A → B and g : B → C are two bijective functions then (gof) −1= f −1og −1 The graph of y = f(x) and y = f −1(x) intersects on the line y = x i.e the solution of f(x) = f −1(x) is solution of y = x Even and odd function: A function f(x) is said to be f(−x) = f(x) then it even function. A function f(x) is said to be f(−x) = −f(x), then it is odd function. Every function f(x) can be express as sum of even and odd functions. ie f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2} The graph of even function is symmetric about the y-axis. The graph of odd function is symmetric about the origin and is placed in first and third quadrant (or) second and fourth quadrants. If f(x) is even function, then f'(x) is odd function provided f'(x) is defined on R. If f(x) is odd function, then f'(x) is even function provided f'(x) is defined on R.
1.
f g fog gof Even Even Even Even Even Odd Even Even Odd Even Even Even Odd Odd Odd Odd
2. E = Even function, O = Odd function, N = Neither even nor odd function
E + E = E E − E = E O − O = O E E = E O + O = O E − O = O − E = N O O = E \frac{E}{E}=E E
n = E E O = O E = O \frac{O}{E}=\frac{E}{O}=O
\frac{O}{O}=E
O^{n}=\begin{cases} O &\tt n \ is \ odd\\E & \tt n \ \ is \ even\end{cases} All even functions are many one. Even extension: The even extension is obtained by defining a new function g(x) as g(x)=\begin{cases}f(x) & x \in [0, a]\\f(-x) & x \in [-a, 0]\end{cases} Odd extension: The odd entension is obtained by defining a new function g(x) as g(x)=\begin{cases}f(x) & x \in [0, a]\\-f(-x) & x \in [-a, 0]\end{cases} If f : A → B is a bijection, then f −1: B → A is also a bijective function. If f : A → B is a bijection, then (f −1) −1= f The inversion of a bijection is unique If f : A → B is a bijection then f −1of = I Aand fof −1= I B. If f : A → B is a bijection if and only if there exists a function g : B → A such that gof = I A. View the Topic in this video From 20:35 To 55:26
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If f : A → B and g : B → C are two function then the composite function of f and g, gof A → C will be defined as gof(x) = g[f(x)], ∀ x ∈ A Composite of functions is not commutative i.e., fog ≠ gof. Composite of functions is associative i.e., (fog)oh = fo(goh). If f : A → B and g : B → C are two bijections, then gof : A → C is bijection and (gof) −1= (f −1og −1). fog ≠ gof but if, fog = gof then either f −1= g or g −1= f also, (fog)(x) = (gof)(x) = (x). (f −1) −1= f If f and g are two bijections such that (gof) exists then (gof) −1= f −1og −1. |
Volume 75, Issue 10, October 2015 In this issue (52 articles)
Regular Article - Theoretical PhysicsH. Mohseni Sadjadi, Parviz Goodarzi Article:513
Regular Article - Experimental Physics
Measurements of the \(\mathrm{Z}\) \(\mathrm{Z}\) production cross sections in the \(2\mathrm{l} 2\nu \) channel in proton–proton collisions at \(\sqrt{s} = 7\) and \(8~\mathrm{TeV} \) and combined constraints on triple gauge couplingsV. Khachatryan, A. M. Sirunyan, A. Tumasyan, W. Adam… Article:511
Special Article - Tools for Experiment and Theory
Using wavelet analysis to compare the QCD prediction and experimental data on \(R_{e^+e^-}\) and to determine parameters of the charmonium states above the \(D\bar{D}\) thresholdV. K. Henner, C. L. Davis, T. S. Belozerova Article:509
Regular Article - Theoretical PhysicsSebastian Bahamonde, Mubasher Jamil Article:508
Regular Article - Theoretical PhysicsSeyed Hossein Hendi, Shahram Panahiyan… Article:507
Regular Article - Theoretical Physics
Transverse momentum dependent gluon fragmentation functions from \(J/\psi \ \pi \) production at \(e^+ e^-\) collidersGuang-Peng Zhang Article:503
Regular Article - Theoretical PhysicsBorzoo Nazari Article:501
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I forgot all about the fact that I asked this question here, but I answered it myself a while back.
Definitions:
A
surface $S$ is an oriented connected sum of $g \geq 0$ tori with $b \geq 0$ open disks removed, and $n \geq 0$ punctures in its interior. Homeomorphism classes of surfaces are in bijective correspondence with $\{(g,b,n): g,b,n \geq 0\}$.
Given a surface $S$, the group (under composition) of orientation preserving homeomorphisms that fix the boundary $\partial S$ is denoted by Aut$^+(S,\partial S)$. This is endowed with the compact-open topology. The path component of Aut$^+(S,\partial S)$ containing id$_S$ is denoted Aut$_0(S,\partial S)$.
Theorem:
Given a surface $S$, the following are three equivalent definitions of its
mapping class group, Mod$(S)$:
Mod$(S)_1 := \pi_0 (\mathrm{Aut}^+(S,\partial S), \mathrm{id}_S)$.
$\mathrm{Mod} (S)_2$ is the group of boundary-fixing isotopy classes of maps in $\mathrm{Aut}^+(S,\partial S)$.
$\mathrm{Mod}(S)_3 := \mathrm{Aut}^+(S,\partial S)/ \mathrm{Aut}_0(S,\partial S)$.
We first show that $1$ and $2$ are naturally isomorphic as sets. Explicitly, $\pi_0 (\mathrm{Aut}^+(S,\partial S), \mathrm{id}_S)$ is the collection of homotopy classes of continuous maps $\sigma: \boldsymbol 2 \to \mathrm{Aut}^+(S,\partial S)$ such that $\sigma (0) = \mathrm{id}_S$. Note that $\boldsymbol 2$ denotes the discrete space $\{0,1\}$. Suppose $\sigma, \tau : \boldsymbol 2 \to \mathrm{Aut}^+(S,\partial S)$ belong to the same homotopy class. Then there is a continuous map$$H: \boldsymbol 2 \times [0,1] \to \mathrm{Aut}^+(S,\partial S)$$ such that $H(1,0) = \sigma(1)$, $H(1,1) = \tau(1)$, and $H(0,t) = \mathrm{id}_S$ for each $t \in [0,1]$. Consider the map$$F: S \times [0,1] \to S$$defined by $F(s,t) := H(1,t)(s)$ for all $s \in S, t \in [0,1]$. Then $F(s,0) = \sigma(1)(s)$ and $F(s,1) = \tau(1)(s)$ for each $s \in S$. Given any $t \in [0,1]$, $F(-,t) \in \mathrm{Aut}^+(S,\partial S)$. Therefore $F$ is a boundary-fixing isotopy from $\sigma(1)$ to $\tau(1)$. In summary, given any homotopy class $[\sigma] \in \pi_0 (\mathrm{Aut}^+(S,\partial S), \mathrm{id}_S)$, evaluation at $1$ gives a boundary-fixing isotopy class of maps in $\mathrm{Aut}^+(S,\partial S)$.
Conversely, suppose $f,g \in \mathrm{Aut}^+(S,\partial S)$ are isotopic. One can similarly construct a homotopy between the maps $\sigma,\tau: \boldsymbol 2 \to \mathrm{Aut}^+(S,\partial S)$ defined by $\sigma(0)=\tau(0)= \mathrm{id}_S$, $\sigma(1) = f$, $\tau(1) = g$.
We now define the group structure on $\mathrm{Mod} (S)_2$. I claim that$$[f][g] = [f\circ g]$$for each $[f],[g]\in \mathrm{Mod}(S)_2$ defines a group structure on $\mathrm{Mod} (S)_2$. Let $f_0,f_1,g_0,g_1 \in \mathrm{Aut}^+(S,\partial S)$ such that $[f_0] = [f_1]$ and $[g_0] = [g_1]$. Then there are boundary-fixing isotopies $F^f, F^g: S \times [0,1] \to S$ such that $F^f(s,i) = f_i(s)$ and $F^g(s,i) = g_i(s)$ for all $s \in S$. The map$$F: S \times [0,1] \to S$$defined by$F(s,t) = F^f(F^g(s,t),t)$for all $s\in S, t\in [0,1]$ is then a boundary-fixing isotopy from $f_0 \circ g_0$ to $f_1 \circ g_1$. Thus $[f][g] = [f\circ g]$ is well defined.
It is now routine to see that this binary operation is a group operation, where$$[\mathrm{id}_S] = e,\ \text{and}\ [f]^{-1} = [f^{-1}].$$This also induces a group structure on $\mathrm{Mod}(S)_1$: For any $[\sigma],[\tau] \in \pi_0(\mathrm{Aut}^+(S,\partial S), \mathrm{id})$,$$[\sigma][\tau] = [\gamma],\ \text{where}\ \gamma(1) = \sigma(1) \circ \tau(1).$$
Finally, we determine how definitions $1$ and $2$ of $\mathrm{Mod}(S)$ correspond to $\mathrm{Mod}(S)_3$. $\mathrm{Aut}^+(S,\partial S)$ is naturally equipped with a group structure under composition. We first show that $\mathrm{Aut}_0(S,\partial S)$ is a normal subgroup of $\mathrm{Aut}^+(S,\partial S)$, and then that the quotient group $\mathrm{Aut}^+(S,\partial S)/\mathrm{Aut}_0(S,\partial S)$ is isomorphic to $\mathrm{Mod}(S)_2$.
Observe that there is a one-to-one correspondence between isotopies $F: S \times [0,1] \to S$ and paths $\gamma : [0,1] \to \mathrm{Aut}^+(S,\partial S)$. Given an isotopy $F$, simply define the path by$$t \mapsto (s \mapsto F(s,t)).$$ Thus $\mathrm{Aut}_0(S,\partial S)$ is the isotopy class of $\mathrm{id}_S$. From the above discussion about the group operation on $\mathrm{Mod}(S)_2$, it is now clear that $\mathrm{Aut}_0(S,\partial S)$ is a normal subgroup of $\mathrm{Aut}^+(S,\partial S)$. Thus the quotient is truly a group.
I now claim that $\varphi: \mathrm{Mod}(S)_3 \to \mathrm{Mod}(S)_2$ defined by $f + [\mathrm{id}] \mapsto [f]$ is a group isomorphism. Suppose $f,g \in \mathrm{Aut}^+(S,\partial S)$ such that $f+[\mathrm{id}] = g+[\mathrm{id}]$. Then there exists $h \in \mathrm{Aut}_0(S,\partial S)$ such that $f = g \circ h$. Then $[f] = [g\circ h] = [g][h] = [g][\mathrm{id}] = [g]$, so $\varphi$ is well defined. Clearly the map is surjective and injective. Finally, given any $f+[\mathrm{id}], g+[\mathrm{id}]$,$$\varphi(f+[\mathrm{id}])\varphi(g+[\mathrm{id}]) = [f][g] = [f\circ g] = \varphi (f\circ g + [\mathrm{id}]) = \varphi((f + [\mathrm{id}])(g + [\mathrm{id}])).$$ Therefore $\varphi$ is an isomorphism as required. This proves the equivalence of the three definitions for $\mathrm{Mod}(S)$. |
Average
Add $n$ numbers and divide the result by $n$.
Associative Property of Multiplication
When three real numbers are multiplied, it makes no difference which two are multiplied first.
Associative Property of Addition
When three real numbers are added, it makes no difference which two are added first.
Area formula
A formula used to find the area of geometric objects or figures. Area is measured in square units.
Approximately equal to
The symbol $\,\approx\,$ represents an answer that is rounded to a specified number of decimal places.
Alternative rule for subtracting two fractions
If $a$, $b$, $c$, and $d$ are integers with $b\ne 0$ and $d\ne 0$, then $\,\displaystyle{a\over b}-{c\over d}={{ad-bc}\over{bd}}$.
Alternative rule for adding two fractions
If $a$, $b$, $c$, and $d$ are integers with $b\ne 0$ and $d\ne 0$, then $\,\displaystyle{a\over b}+{c\over d}={{ad+bc}\over{bd}}$.
Algebraic inequality
An inequality that contains one or more variable terms.
Algebraic expression
A collection of variables and constants combined by using addition, subtraction, multiplication, or division.
Algebraic equation
Uses an equal sign to set two algebraic expressions equal to each other and is true for only certain values of the variable.
Additive Inverse Property
The sum of a real number and its opposite is zero.
Additive Inverse
Two numbers whose sum is zero.
Additive Identity Property
The sum of zero and a real number equals the number itself.
Addition Property of Inequalities
Add the same quantity to each side.
Adding fractions with unlike denominators
Rewrite the fractions so that they have like denominators. Then use the rule for adding fractions with like denominators.
Adding fractions with like denominators
Let $a$, $b$, and $c$ be integers with $c\ne0$. Then use the following rule: $\displaystyle{a\over c}+{b\over c}={{a+b}\over{c}}$
Add two integers with unlike signs
Find the absolute value of each integer. Subtract the lesser absolute value from the greater absolute value and attach the sign of the integer with the greater absolute value.
Add two integers with like signs
Add their absolute values and attach the common sign to the result.
Abundant number
A number where the sum of its proper factors is greater than the number.
Absolute value
The distance between any real number and 0 on the real number line. |
optimal lifted alignment - is a dynamic programming algorithm. Its input is a tree $T$ with $k$ strings assigned to its leaf nodes (their length is $n$).
The algorithm then assigns strings to the internal nodes s.t. the sum of distance (global alignment distance) on all the edges would be minimal, under the constraint that the string assigned to each node is one of the strings of its immediate children (hence the name "lifted").
The algorithm first computes distances $D$ between all pairs $O(k^2n^2)$. For each leaf node $v$ assigned $S_v$ we initialize: $d(S, v) = 0 | S = S_v$ and $d(S, v) = \infty | S \neq S_v$.
We then traverse post-order and for each internal node $v$, for each string $S$ assigned to one of its leaf descendants (all strings in its subtree) does $d(v,S) = \sum{\min(D(S,T) + d(w,T))}$ where $w$ are $v$'s immediate children. Overall = $O(k^2n^2 + k^3)$.
I've been told its possible to improve the complexity to $O(k^2n^2 + k^2)$ but I can't see how? Would appreciate any hint. |
I think the questions were about unconditional proofs or counter examples. I don't have an answer to any of those questions but I think it is still interesting to understand how the yoga of motives suggests natural answers to theses questions. Even though this may seem trivial to people familiar with the subject.
Let's work in the setting of Voevodsky's $\otimes$-triangulated categories $DM^{eff}(\mathbb{Q}):= DM_{gm}^{eff}(Spec(\mathbb{Q});\mathbb{Q}) \subset DM_{gm}(Spec(\mathbb{Q});\mathbb{Q}) =: DM(\mathbb{Q})$. Remember that the latter is obtained by formally inverting $\mathbb{Q}(1)$ and that it is a rigid $\otimes$-triangulated category.
Question 1: Is the ring of (effective) periods a field?
Following Beilinson's "Remarks on Grothendieck's standard conjecctures", let's assume
Motivic conjecture: There exists a non degenerate t-structure on Voevodsky's category $DM(\mathbb{Q}) := DM(Spec(\mathbb{Q});\mathbb{Q})$ and such that the Betti realization function $\omega_B: DM(\mathbb{Q}) \to D^bMod_f(\mathbb{Q})$ is a $t$-exact $\otimes$-functor.
This is an extremely strong conjecture as it implies the standard conjectures in characteristic 0.
Under this conjecture, the heart of the motvitic $t$-structure is a tannakian category $MM(\mathbb{Q})$. We have Betti and Rham realization functors $\omega_B,\omega_{dR}: MM(\mathbb{Q}) \rightrightarrows Mod_f(\mathbb{Q})$. And we can define $$ Per := Isom^\otimes(\omega_{dR},\omega_{B})$$This is a fpqc-torsor under the motivic Galois group $G_B := Aut^\otimes(\omega_B)$.
Define the algebra of motivic periods as the ring of regular functions on the Betti/de Rham torsor:$$ P_{mot} := \mathcal{O}(Per)$$ Integration of differential forms (or more generally the Riemann-Hilbert correspondance) defines an $\mathbb{C}$-point $$ Spec(\mathbb{C}) \longrightarrow Per$$The image of the corresponding morphism $P_{mot} \to \mathbb{C}$ is the ring of periods $P$.
Note: I think this whole part is actually known unconditionnally in the setting of Nori's motives (see arXiv:1105.0865v4).
Period conjecture: The morphism $P_{mot} \to P$ is an isomorphism.
Now based on these tiny little conjectures we can say
Prop: $P_{mot}$ is not a field so $P$ isn't either.
Proof: Indeed in his comment G-torsor whose ring of regular functions is a field. @quasi-coherent explained how faithfull flatness would imply that if $P_{mot}$ were a field then it would be algebraic over $\mathbb{Q}$ which contradicts the fact that $2\pi i$ belongs to the image of $P_{mot}\to \mathbb{C}$.
Question 2 Is it true that $(P_{mot}^{eff})^\times = \overline{\mathbb{Q}}^\times$?
This post is getting too long already so I'll try and write down the rest later but the basic idea is that invertible effective motives are Artin motives. This can be proved in terms of weights or niveau (level). |
Tagged: subspace Problem 709
Let $S=\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4},\mathbf{v}_{5}\}$ where
\[ \mathbf{v}_{1}= \begin{bmatrix} 1 \\ 2 \\ 2 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{2}= \begin{bmatrix} 1 \\ 3 \\ 1 \\ 1 \end{bmatrix} ,\;\mathbf{v}_{3}= \begin{bmatrix} 1 \\ 5 \\ -1 \\ 5 \end{bmatrix} ,\;\mathbf{v}_{4}= \begin{bmatrix} 1 \\ 1 \\ 4 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{5}= \begin{bmatrix} 2 \\ 7 \\ 0 \\ 2 \end{bmatrix} .\] Find a basis for the span $\Span(S)$. Problem 706
Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set
\[S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\] still a spanning set for $V$? If so, prove it. Otherwise, give a counterexample. Problem 663
Let $\R^2$ be the $x$-$y$-plane. Then $\R^2$ is a vector space. A line $\ell \subset \mathbb{R}^2$ with slope $m$ and $y$-intercept $b$ is defined by
\[ \ell = \{ (x, y) \in \mathbb{R}^2 \mid y = mx + b \} .\]
Prove that $\ell$ is a subspace of $\mathbb{R}^2$ if and only if $b = 0$.Add to solve later
Problem 659
Fix the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$, and let $\R^3$ be the vector space of $3 \times 1$ column vectors. Define
\[W = \{ \mathbf{v} \in \R^3 \mid \mathbf{b} \mathbf{v} = 0 \}.\] Prove that $W$ is a vector subspace of $\R^3$. Problem 658
Let $V$ be the vector space of $n \times n$ matrices with real coefficients, and define
\[ W = \{ \mathbf{v} \in V \mid \mathbf{v} \mathbf{w} = \mathbf{w} \mathbf{v} \mbox{ for all } \mathbf{w} \in V \}.\] The set $W$ is called the center of $V$.
Prove that $W$ is a subspace of $V$.Add to solve later
Problem 612
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 $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.
Add to solve later
(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$. Problem 611
An $n\times n$ matrix $A$ is called
orthogonal if $A^{\trans}A=I$. Let $V$ be the vector space of all real $2\times 2$ matrices.
Consider the subset
\[W:=\{A\in V \mid \text{$A$ is an orthogonal matrix}\}.\] Prove or disprove that $W$ is a subspace of $V$. Problem 604
Let
\[A=\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 &1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 2 & 2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) Find a basis for the null space $\calN(A)$. (b) Find a basis of the range $\calR(A)$. (c) Find a basis of the row space for $A$.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution Problem 601
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.
Let \[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix} a & b\\ c& -a \end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\] (a) Show that $W$ is a subspace of $V$. (b) Find a basis of $W$. (c) Find the dimension of $W$.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution |
A reversible melting of $32.0\ \mathrm g$ of ice at $0\ \mathrm{^\circ C}$ and $1\ \mathrm{atm}$. Solve for $\Delta G, \Delta S_\text{universe}$, and $\Delta A$
I'm also given latent heat of fusion/vaporization, heat capacity at constant pressure, ($C_p$) and the densities of water and ice at the appropriate temperature and pressure,
Solving for $\Delta S_\text{universe}$
from equation sheet: $$\Delta S_\text{vap} = \frac {q_{p,\text{vap}}}{T_\mathrm b}$$ Which is for vaporization and I assume I can transform that into $$\Delta S_\text{melt}= \frac {q_{p,\text{fus}}}{T_\mathrm m}$$ So solving for $\Delta S_\text{system}$ I get $$\Delta S_\text{system}=\Delta S_\text{melt}= \frac {q_{p,\text{fus}}}{T_\mathrm m}$$
But the question asked for $\Delta S_\text{universe}$ not $\Delta S_\text{system}$. I know that $$S_\text{universe}=S_\text{surroundings}+S_\text{system}$$, so $\Delta S_\text{universe}$ must be $$\Delta S_{universe}=\Delta S_\text{surroundings}+\Delta S_\text{system}$$ But I'm given no information on the surroundings. How can I solve this? Is it possible that the change in entropy in the surroundings constant such that $\Delta S_\text{universe}=\Delta S_\text{system}$?
Solving for $\Delta A$ $$A=U-TS$$ $$\Delta A = \Delta U -T\Delta S_\text{sys}$$ $$\Delta A = (q-P\Delta V) -T\Delta S_\text{sys}$$ $$\Delta A = (q-P[\rho_\text{water}m-\rho_\text{ice}m])-T\Delta S_\text{sys}$$
But I don't know how to solve $q$ for this problem. Additionally, I'm not sure if the ice melted completely, so perhaps the change in Volume is less that what I wrote in the equation.
Solving for $\Delta G$ $$\Delta G = \Delta H-T\Delta S_\text{sys}$$ $$\Delta G = \Delta U+P\Delta V-TS_\text{sys}$$ $$\Delta G = (q-P\Delta V)+P\Delta V-TS_\text{sys}$$ $$\Delta G = q-TS_\text{sys}$$
But same as before I don't know how to solve for $q$. |
16 0 Homework Statement A copper wire has an internal diameter ##d_1## of ##10## ##cm## and an external diameter ##d_2## of 20 ##cm##. If the wire is ##20m## long what is the resistance of the wire? Homework Equations ##R=\rho\frac{l}{\Sigma}##
What i did is: ##R=\rho\frac{l}{\Sigma}=\rho\frac{l}{d_2^2\pi-d_1^2\pi}##
The problem is that I don't have ##\rho##. Is there a way to find ##R## without knowing it? Many thanks.
The problem is that I don't have ##\rho##. Is there a way to find ##R## without knowing it?
Many thanks. |
Tagged: invertible matrix Problem 583
Consider the $2\times 2$ complex matrix
\[A=\begin{bmatrix} a & b-a\\ 0& b \end{bmatrix}.\] (a) Find the eigenvalues of $A$. (b) For each eigenvalue of $A$, determine the eigenvectors. (c) Diagonalize the matrix $A$.
Add to solve later
(d) Using the result of the diagonalization, compute and simplify $A^k$ for each positive integer $k$. Problem 582
A square matrix $A$ is called
nilpotent if some power of $A$ is the zero matrix. Namely, $A$ is nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix.
Suppose that $A$ is a nilpotent matrix and let $B$ be an invertible matrix of the same size as $A$.
Is the matrix $B-A$ invertible? If so prove it. Otherwise, give a counterexample. Problem 562
An $n\times n$ matrix $A$ is called
nonsingular if the only vector $\mathbf{x}\in \R^n$ satisfying the equation $A\mathbf{x}=\mathbf{0}$ is $\mathbf{x}=\mathbf{0}$. Using the definition of a nonsingular matrix, prove the following statements. (a) If $A$ and $B$ are $n\times n$ nonsingular matrix, then the product $AB$ is also nonsingular. (b) Let $A$ and $B$ be $n\times n$ matrices and suppose that the product $AB$ is nonsingular. Then: The matrix $B$ is nonsingular. The matrix $A$ is nonsingular. (You may use the fact that a nonsingular matrix is invertible.) 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 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 506
Let $A$ be an $n\times n$ invertible matrix. Then prove the transpose $A^{\trans}$ is also invertible and that the inverse matrix of the transpose $A^{\trans}$ is the transpose of the inverse matrix $A^{-1}$.
Namely, show that \[(A^{\trans})^{-1}=(A^{-1})^{\trans}.\] Problem 500
10 questions about nonsingular matrices, invertible matrices, and linearly independent vectors.
The quiz is designed to test your understanding of the basic properties of these topics.
You can take the quiz as many times as you like.
The solutions will be given after completing all the 10 problems.
Click the View question button to see the solutions. Problem 452
Let $A$ be an $n\times n$ complex matrix.
Let $S$ be an invertible matrix. (a) If $SAS^{-1}=\lambda A$ for some complex number $\lambda$, then prove that either $\lambda^n=1$ or $A$ is a singular matrix. (b) If $n$ is odd and $SAS^{-1}=-A$, then prove that $0$ is an eigenvalue of $A$.
Add to solve later
(c) Suppose that all the eigenvalues of $A$ are integers and $\det(A) > 0$. If $n$ is odd and $SAS^{-1}=A^{-1}$, then prove that $1$ is an eigenvalue of $A$. Problem 438
Determine whether each of the following statements is True or False.
(a) If $A$ and $B$ are $n \times n$ matrices, and $P$ is an invertible $n \times n$ matrix such that $A=PBP^{-1}$, then $\det(A)=\det(B)$. (b) If the characteristic polynomial of an $n \times n$ matrix $A$ is \[p(\lambda)=(\lambda-1)^n+2,\] then $A$ is invertible. (c) If $A^2$ is an invertible $n\times n$ matrix, then $A^3$ is also invertible. (d) If $A$ is a $3\times 3$ matrix such that $\det(A)=7$, then $\det(2A^{\trans}A^{-1})=2$. (e) If $\mathbf{v}$ is an eigenvector of an $n \times n$ matrix $A$ with corresponding eigenvalue $\lambda_1$, and if $\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_2$, then $\mathbf{v}+\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_1+\lambda_2$.
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Stanford University, Linear Algebra Exam Problem) Read solution |
As Kimball mentioned, this is all in Godement-Jacquet's monograph "Zeta Functions of Simple Algebras". For the case $n = 1$, this is just Tate's thesis. When the field is $\mathbb{Q}$, a good reference is Goldfeld-Hundley "Automorphic Representations and $L$-Functions for the General Linear Group".
The story is roughly the following. There is a functional equation of the following form:\[\Lambda(s,\pi) = \epsilon(s,\pi) \Lambda(1 - s,\widetilde{\pi}).\]Here $\pi$ is a unitary cuspidal automorphic representation of $\mathrm{GL}_n(\mathbb{A}_F)$, where $F$ is a number field, $\widetilde{\pi}$ denotes the contragredient, and $\Lambda(s,\pi) = \prod_v L_v(s,\pi_v)$ is the completed $L$-function of $\pi$ (including the archimedean factors). The epsilon factor $\epsilon(s,\pi)$ factorises as\[\epsilon(s,\pi) = \epsilon(s,\pi,\psi) = \prod_v \epsilon_v(s,\pi_v,\psi_v),\]where $\psi_v$ is an additive character of $F_v$ that is the local component of an additive character of $\mathbb{A}_F$. Each local epsilon factor may depend on $\psi_v$, but the global epsilon factor is independent of $\psi$.
Let $v$ be a nonarchimedean place of $F$ with associated local field $F_v$ having ring of integers $\mathcal{O}_v$, maximal ideal $\mathfrak{p}_v$, and whose residue field $\mathcal{O}_v / \mathfrak{p}_v$ has cardinality $q_v$. Let $\psi_v$ be an additive character of $F_v$. The conductor of $\psi_v$ is $\mathfrak{p}_v^{c(\psi_v)}$, where $c(\psi_v)$ is the least integer (possibly negative) for which $\psi_v$ is trivial on $\mathfrak{p}_v^{c(\psi_v)}$. The conductor of $\pi_v$ is $\mathfrak{p}_v^{c(\pi_v)}$, where $c(\pi_v)$ is the least integer (necessarily nonnegative) for which $\pi_v$ contains a nonzero vector fixed by the congruence subgroup$$ K_0(\mathfrak{p}^{c(\pi_v)}) = \left\{ \begin{pmatrix} a & b \\\ c & d \end{pmatrix} \in \mathrm{GL}_n(\mathcal{O}_v) : c \in \mathrm{Mat}_{1 \times (n - 1)}(\mathfrak{p}_v^{c(\pi_v)}), \ d - 1 \in \mathfrak{p}_v^{c(\pi_v)} \right\}.$$Then the local epsilon factor is\[\epsilon_v(s,\pi_v,\psi_v) = \epsilon_v\left(\frac{1}{2},\pi_v,\psi_v\right) q_v^{(n c(\psi_v) - c(\pi_v))\left(s - \frac{1}{2}\right)}.\]Suppose that $F_v$ is an extension of $\mathbb{Q}_p$. If we choose $\psi_v$ to be the composition of the standard unramified additive character of $\mathbb{Q}_p$ with the trace map $\mathrm{Tr}_{F_v/\mathbb{Q}_p}$, then $\epsilon_v(1/2,\pi_v,\psi_v)$ is a complex number of absolute value $1$. For example, if $n = 1$, so that $\pi_v$ is a character, $\epsilon_v(1/2,\pi_v,\psi_v)$ is a normalised Gauss sum; in general, it can be quite complicated. Moreover, $\mathfrak{p}_v^{-c(\psi_v)}$ is the different ideal $\mathfrak{D}_{F_v/\mathbb{Q}_p}$ and $q_v^{-c(\psi_v)}$ is the absolute discriminant $\Delta_{F_v/\mathbb{Q}_p} = N_{F_v/\mathbb{Q}_p}(\mathfrak{D}_{F_v/\mathbb{Q}_p})$ of the extension $F_v/\mathbb{Q}_p$.
Note that at all but finitely many places, the additive character $\psi_v$ is unramified ($c(\psi_v) = 0$) and the representation $\pi_v$ is unramified ($c(\pi_v) = 0$).
I haven't yet discussed the archimedean factors. The local $L$-function at an archimedean place $v$ of $F$ will be something of the form\[L_v(s,\pi_v) = \prod_{j = 1}^{n} \zeta_v(s + t_j),\]for some complex numbers $t_j$ (with some restrictions on the possible vertical lines that $t_j$ lies on, since I am assuming that $\pi$ is unitary). Here $\zeta_v(s) = \pi^{-s/2} \Gamma(s/2)$ for a real place $v$ and $\zeta_v(s) = 2(2\pi)^{-s} \Gamma(s)$ for a complex place $v$. Finally, the local epsilon factor at an archimedean place will just be something of the form $\epsilon_v(s,\pi_v,\psi_v) = i^k$ for some integer $k$ (in particular, this is independent of $s$). For more details on the archimedean $L$-functions and epsilon factors, see Knapp's paper "Local Langlands Correspondence: the Archimedean Case".
So the global epsilon factor is\[\prod_v \epsilon_v\left(\frac{1}{2},\pi_v, \psi_v\right) \prod_{v \text{ nonarchimedean}} q_v^{n c(\psi_v) \left(s - \frac{1}{2}\right)} \prod_{v \text{ nonarchimedean}} q_v^{- c(\pi_v) \left(s - \frac{1}{2}\right)}.\]The first term is $\epsilon(1/2,\pi)$, the global root number, which is some complex number of absolute value $1$. The second is\[\Delta_{F/\mathbb{Q}}^{-n\left(s - \frac{1}{2}\right)},\]where $\Delta_{F/\mathbb{Q}}$ is the absolute discriminant of the extension $F/\mathbb{Q}$; this is the norm $N_{F/\mathbb{Q}}(\mathfrak{D}_{F/\mathbb{Q}})$ of the different$$\mathfrak{D}_{F/\mathbb{Q}} = \prod_{v \text{ nonarchimedean}} \mathfrak{p}_v^{-c(\psi_v)}.$$The third is\[N_{F/\mathbb{Q}}(\mathfrak{q})^{-\left(s - \frac{1}{2}\right)},\]where\[\mathfrak{q} = \prod_{v \text{ nonarchimedean}} \mathfrak{p}_v^{c(\pi_v)}\]is the (relative) conductor of $\pi$ and $N_{K/\mathbb{Q}}$ denotes the norm. |
I've really confused myself here: $$ P(x) = ``\text{x has a tail}" $$ How would I write: $$ ``\text{Not everything has a tail}" $$
would it: $$ ¬∀x P(x) $$
be correct?
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I've really confused myself here: $$ P(x) = ``\text{x has a tail}" $$ How would I write: $$ ``\text{Not everything has a tail}" $$
would it: $$ ¬∀x P(x) $$
be correct?
"Everything doesn't have a tail" is not logically equivalent to ""Not everything has a tail."
Use instead $\lnot \forall xP(x)$, or equivalently $\exists x \lnot P(x)$, there is something that doesn't have a tail. |
Geoff's answer is exactly correct, but I wanted to give the specifics.
If you only want bounds that are easy to compute without being able to prove them yourself, then this answer should be just fine. The bounds are easy and reasonably tight. If you want to understand the proofs (which are not too bad, and involve many fun areas of finite groups), then read the book: Blackburn–Neumann–Venkataraman (2007).
Pyber (1993) showed that for $f(n)$ the number of isomorphism classes of groups of order $n$:
$$f(n) \leq n^{(2/27)\mu(n)^2+O(\mu(n)^{3/2})}$$
where $\mu(n) \leq \log_2(n)$ is the highest power of any prime dividing $n$. When $\mu(n)=1$ you have the square-free case you mentioned, and when $n=p^k$, then $k=\mu(n)$ and the bound is asymptotically sharp. For $p$-groups, pretty decent but slightly weaker bounds were first proven in Higman (1960), and improved in Sims (1965).
The best lower bounds that aren't ridiculously complex to compute and that I can think of just follow from $f(k) \leq f(n)$ if $k$ divides $n$. In other words, count the nilpotent groups of that order as a lower bound. For reference an explicit form of Higman's lower bound is:
$$f(p^k) \geq p^{\tfrac{2}{27} k^2(k-6)}$$
All of these results and more are explained very nicely in the book Blackburn–Neumann–Venkataraman (2007). I recommend it highly.
Blackburn, Simon R.; Neumann, Peter M.; Venkataraman, Geetha.“Enumeration of finite groups.”Cambridge Tracts in Mathematics, 173. Cambridge University Press, Cambridge, 2007. xii+281 pp. ISBN: 978-0-521-88217-0MR2382539DOI:10.1017/CBO9780511542756 Pyber, L.“Enumerating finite groups of given order.”Ann. of Math. (2) 137 (1993), no. 1, 203–220.MR1200081DOI:10.2307/2946623 Sims, Charles C.“Enumerating p-groups.”Proc. London Math. Soc. (3) 15 (1965) 151–166.MR169921DOI:10.1112/plms/s3-15.1.151 Higman, Graham.“Enumerating p-groups. I. Inequalities.”Proc. London Math. Soc. (3) 10 (1960) 24–30.MR113948DOI:10.1112/plms/s3-10.1.24 |
Answer
See the answer below.
Work Step by Step
From figure 6-10, the wavelength is 121.6 nm and the frequency: $\nu=c/\lambda$ $\nu=2.47\times10^{15}\ 1/s$ The associated transition is from $n=2$ to $n=1$.
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Stochastic Lotka-Volterra system with unbounded distributed delay
1.
School of Mathematics and Statistics, Huazhong University of Science and Technology, Wuhan, Hubei 430074, China, China
$\dot{x}(t)=\mbox{diag}(x_1(t), \cdots, x_n(t))(r+Ax(t)+B\int_{-\infty}^0x(t+\theta)d\mu(\theta))$
into the corresponding stochastic system
$dx(t)=\mbox{diag}(x_1(t), \cdots, x_n(t))[(r+Ax(t)+B\int_{-\infty}^0x(t+\theta)d\mu(\theta))dt+\beta dw(t)].$
This paper obtains one condition under which the above stochastic system has a global almost surely positive solution and gives the asymptotic pathwise estimation of this solution. This paper also shows that when the noise is sufficiently large, the solution of this stochastic system will converge to zero with probability one. This reveals that the sufficiently large noise may make the population extinct.
Mathematics Subject Classification:Primary: 34K50; 60H10; Secondary: 92D2. Citation:Fuke Wu, Yangzi Hu. Stochastic Lotka-Volterra system with unbounded distributed delay. Discrete & Continuous Dynamical Systems - B, 2010, 14 (1) : 275-288. doi: 10.3934/dcdsb.2010.14.275
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Traveling wave solutions for Lotka-Volterra system re-visited.
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A global stability result for an N-species Lotka-Volterra food chain system with distributed time delays.
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Bifurcation structures of positive stationary solutions for a Lotka-Volterra competition model with diffusion II: Global structure.
[6]
Li-Jun Du, Wan-Tong Li, Jia-Bing Wang.
Invasion entire solutions in a time periodic Lotka-Volterra competition system with diffusion.
[7]
Yang Wang, Xiong Li.
Uniqueness of traveling front solutions for the Lotka-Volterra system in the weak competition case.
[8]
Rui Wang, Xiaoyue Li, Denis S. Mukama.
On stochastic multi-group Lotka-Volterra ecosystems with regime switching.
[9]
Dejun Fan, Xiaoyu Yi, Ling Xia, Jingliang Lv.
Dynamical behaviors of stochastic type K monotone Lotka-Volterra systems.
[10]
Dan Li, Jing'an Cui, Yan Zhang.
Permanence and extinction of non-autonomous Lotka-Volterra facultative systems with jump-diffusion.
[11]
Francisco Montes de Oca, Liliana Pérez.
Balancing survival and extinction in nonautonomous competitive Lotka-Volterra systems with infinite delays.
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Ting-Hui Yang, Weinian Zhang, Kaijen Cheng.
Global dynamics of three species omnivory models with Lotka-Volterra interaction.
[13]
Guo-Bao Zhang, Ruyun Ma, Xue-Shi Li.
Traveling waves of a Lotka-Volterra strong competition system with nonlocal dispersal.
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Yuan Lou, Dongmei Xiao, Peng Zhou.
Qualitative analysis for a Lotka-Volterra competition system in advective homogeneous environment.
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Quadratic perturbations of a quadratic reversible Lotka-Volterra system with two centers.
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The sign of the wave speed for the Lotka-Volterra competition-diffusion system.
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Qualitative analysis of a Lotka-Volterra competition system with advection.
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Polymers Molecular Mass of Polymers and Biodegradable Polymers Molecular mass of polymer : Number average molecular mass : \overline{M_{n}} = \frac{\sum N_{i}M_{i}}{\sum N_{i}} N i → number of molecules of m 1 mass M i → molecular mass Weight average molecular mass : \overline{M_{w}} = \frac{\sum N_{i}M_{i}^{2}}{\sum N_{i}M_{i}} Poly Dispersity Index : PDI = \frac{\overline{M_{w}}}{\overline{M_{n}}} For natural polymers PDI = 1 Synthetic polymers PDI > 1 Biodegradable polymer : Aliphatic polyesters. eg : Poly β-hydroxybutyrate-co-β-hydroxyvalerate) (PHBV) Obtained by copolymerisation of 3 - hydroxy butanoic acid. It is used in packing Orthopedic devices and controlled release of drugs. LDPE (Low density polyethylene) Ethylene \xrightarrow[350 - 570 \ K]{1000 - 2000 \ atm} LDPE Free radical addition High branched chemically inert poor conductor of electricity. HDPE (High density polyethylene) Ethylene \xrightarrow[1 \ atm]{330 - 350 \ K} HDPE Zeigler Natta catalyst → (C 2H 5) 3Al + TiCl 4 High tensile strength relatively tough and hard, making containers, bottles, pipes, house ware etc Part1: View the Topic in this Video from 15:22 to 20:40 Part2: View the Topic in this Video from 52:11 to 58:15
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1.
Number-Average Molecular Mass \tt \left(\overline{M}_{n}\right) If N 1 molecules have molecular mass M 1 each, N 2 molecules have molecular mass M 2 each, N 3 molecules have molecular mass M 3 each and so on, then \overline{M}_{n}=\frac{\Sigma N_{i}M_{i}}{\Sigma N_{i}}
2.
Mass-Average Molecular Mass \tt \left(\overline{M}_{w}\right) Supposing, as before that N 1, N 2, N 3 etc., molecules have molecular mass M 1, M 2, M 3 etc., respectivley, then, \overline{M}_{w}=\frac{\Sigma N_{i}M_{i}^{2}}{\Sigma N_{i}M_{i}}
3.
Polydispersity Index It is the ratio of the mass average molecular mass to the number average molecular mass PDI=\frac{\overline{M}_{w}}{\overline{M_{n}}} |
MathRevolution wrote:
[Math Revolution
GMAT math practice question]
What is the sum of the solutions of the equation \((x-1)^2=|x-1|?\)
\(A. -1\)
\(B. 0\)
\(C. 1\)
\(D. 2\)
\(E. 3\)
\(?\,\,\,:\,\,\,{\text{sum}}\,\,{\text{of}}\,\,{\text{roots}}\)
\({\left( {x - 1} \right)^2} = \left| {x - 1} \right|\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,{\left( {x - 1} \right)^4} = {\left( {x - 1} \right)^2}\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered}
\,\,\,x = 1\,\,\,\left( {{\text{trivial}}\,\,{\text{inspection}}} \right) \hfill \\
\,\,{\text{for}}\,\,x \ne 1\,\,:\,\,\,\,\,\frac{{{{\left( {x - 1} \right)}^4}}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{{{\left( {x - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}}}\,\,\,\,\, \Rightarrow \,\,\,\,{\left( {x - 1} \right)^2} = 1\,\,\,\,\, \Rightarrow \,\,\,\,\,x = 0\,\,\,{\text{or}}\,\,\,x = 2 \hfill \\
\end{gathered} \right.\)
(*) When we "square" one equation (or, in general, put it to an EVEN positive power), we don´t loose original roots, but we eventually "add" new ones.
That´s why, in the end, we must check each POTENTIAL root in the original equation! All of them fit here. Therefore:
\(? = 1 + 0 + 2 = 3\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH
method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net |
Consider a finite sequence $x_i \in (0,1)$ for $i=1,\ldots, n$ and define $y_i=\dfrac{\Pi_{j=1}^n x_j }{x_i}$.
I solved this system for $x$ in terms of $y$ and got $$x_i=\dfrac{\left(\Pi_{j=1}^n y_j \right)^\frac{1}{n-1}}{y_i}.$$
Now pick some $m<n$. Is there a simple way of solving $(x_1,\ldots x_m, y_{m+1},\ldots, y_n)$ in terms of $(y_1,\ldots y_m, x_{m+1},\ldots, x_n)$?
If we take logs, this is a linear system. My only idea on how to solve is to write the system for $n=2,3$ and solve it by brute-force to see if some pattern emerges and if so, make a conjecture and prove it... But perhaps someone can easily see some clever trick I am missing. This is not homework. It is for some lemma I need in my paper. |
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...
Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...
Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
(The cotangent bundle is naturally a symplectic manifold)
Yeah
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say |
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Problem 9. (8 points) Let ana_nan be a sequence such that a1=2,a2=ka_1=2,a_2=ka1=2,a2=k for some kkk and for all n≥3n\ge3n≥3, an=an−1an−2a_n=a_{n-1}^{a_{n-2}}an=an−1an−2. Let p,q,r,sp,q,r,sp,q,r,s be the values of a1024a_{1024}a1024 where k=4,5,6,7k=4,5,6,7k=4,5,6,7, respectively. Find the last two digits of p+q+r+sp+q+r+sp+q+r+s.
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I think it's 46.
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I do it this way. I start with k=4k=4k=4. First we try small numbers first.
a1=2,a2=4,a3=42=16,a4=164=65536=‾36(mod100),a5=6553616...a_1 = 2, a_2 = 4, a_3 = 4^2 = 16, a_4 = 16^4 = 65536 \overline= 36 (mod 100), a_5 = 65536^{16} ... a1=2,a2=4,a3=42=16,a4=164=65536=36(mod100),a5=6553616...
It would be laborious for us to calculate term a5a_5a5, so we try to trace the pattern, by testing some small values.
361=3636^1 = 36361=36
362=1296=‾96(mod100)36^2 = 1296 \overline= 96 (mod 100)362=1296=96(mod100)
363=‾96×36(mod10000)=‾56(mod100)36^3 \overline= 96 \times 36 (mod 10000) \overline= 56 (mod 100)363=96×36(mod10000)=56(mod100)
364=‾56×36(mod10000)=‾16(mod100)36^4 \overline= 56 \times 36 (mod 10000) \overline = 16 (mod 100)364=56×36(mod10000)=16(mod100)
365=‾16×36(mod10000)=‾76(mod100)36^5 \overline= 16 \times 36 (mod 10000) \overline = 76 (mod 100)365=16×36(mod10000)=76(mod100)
366=‾76×36(mod10000)=‾36(mod100)36^6 \overline= 76 \times 36 (mod 10000) \overline = 36 (mod 100)366=76×36(mod10000)=36(mod100),
367=‾36×36(mod10000)=‾96(mod100)36^7 \overline= 36 \times 36 (mod 10000) \overline = 96 (mod 100)367=36×36(mod10000)=96(mod100),
and so on. Obviously, the pattern recurs every 5 terms. Since 16=‾1(mod5)16 \overline= 1 (mod 5)16=1(mod5), so last two digits of 655361665536^{16}6553616 is 363636. Thus, we have 363636 as the last two digits of ppp, since no matter what happens, we have 666 as the last digit of every term except term a1a_1a1 and a2a_2a2, and 6=‾1(mod5)6 \overline= 1 (mod 5) 6=1(mod5).
Then, when k=5k = 5k=5, we have
a1=2,a2=5,a3=52=25,a4=255...a_1 = 2, a_2 = 5, a_3 = 5^2 = 25, a_4 =25^5 ...a1=2,a2=5,a3=52=25,a4=255...
Since 25525^5255 is very hard to calculate, so we try to trace the pattern again,
51=55^1 = 5 51=5
52=255^2 = 2552=25
53=1255^3 = 12553=125
54=6255^4 =62554=625
and so on. Starting from 525^252, last two digits are 252525. Thus, 25525^5255 has 252525 as the last two digits, so as the term a5=2525a_5 = 25^{25}a5=2525. Therefore, qqq has 252525 as the last two digits.
Then, when k=6k = 6k=6, we have
a1=2,a2=6,a3=62=36,a4=366=‾36(mod100),a5=‾36(mod100)...a_1 = 2, a_2 = 6, a_3 = 6^2 = 36, a_4 = 36^6 \overline= 36(mod 100), a_5 \overline= 36 (mod 100) ...a1=2,a2=6,a3=62=36,a4=366=36(mod100),a5=36(mod100)....
Hey! This case is almost the same as when k=4k=4k=4! So, last digits of r=36r = 36r=36
Finally, when k=7k=7k=7, we have
a1=2,a2=7,a3=72=49,a4=497...a_1 = 2, a_2 = 7, a_3 = 7^2 = 49, a_4 = 49^7 ... a1=2,a2=7,a3=72=49,a4=497...
Again, for the last time in this problem, we trace the pattern.
491=4949^1 = 49491=49
492=240149^2 = 2401492=2401
493=‾01×49(mod100)=‾49(mod100)49^3 \overline= 01 \times 49 (mod 100) \overline = 49 (mod 100)493=01×49(mod100)=49(mod100)
494=‾49×49(mod10000)=‾01(mod100)49^4 \overline= 49 \times 49 (mod 10000) \overline = 01 (mod 100)494=49×49(mod10000)=01(mod100)
and so on. Obviously, the pattern recurs every two terms. Since 7=‾1(mod2)7 \overline= 1(mod 2)7=1(mod2), so 497=‾49(mod100)49^7 \overline= 49(mod 100)497=49(mod100), so as term a5=49343=‾49(mod100)a_5 = 49^{343} \overline= 49(mod 100)a5=49343=49(mod100). This implies that rrr has last two digits of 494949.
In the end, p+q+r+s=36+25+36+49=‾46(mod100)p+q+r+s = 36+25+36+49 \overline= \boxed {46}(mod 100)p+q+r+s=36+25+36+49=46(mod100).
Some Latex suggestions. Use:
\pmod n
\pmod n
which comes out as (modn) \pmod n (modn) which looks nicer.
@Anqi Li – Okay, I appreciate that and I will remember that
Correct! 8 points for you!
The last two digits of P=36,Q=25,R=36,S=49.Thus Last two digits of P+Q+R+S=46
Can you explain the second sentence? Do you mean that a1024a_{1024}a1024has 444 possible values or other thing else?
When k=4k = 4k=4 , a1024=pa_{1024} = pa1024=p and when k=5k = 5k=5 , a1024=qa_{1024} = qa1024=q and so on .
Okay, I got it. Thanks.
i dont know sir can u liease teach me i will be very glad to get teachings from you
Okay, first of all, I start with k=4k=4k=4. Then, we follow the rules to define the next term of the sequence, we have a1=2,a2=k=4,a3=a(3−1)a(3−2)=a2a1=42=16,and a4=164=65536a_1 = 2, a_2 = k = 4, a_3 = a_{(3-1)}^{a_{(3-2)}} = a_2^{a_1} = 4^2 = 16, \text{and } a_4 = 16^4 = 65536a1=2,a2=k=4,a3=a(3−1)a(3−2)=a2a1=42=16,and a4=164=65536. So, why I write 65536=‾36(mod100)65536 \overline= 36 \pmod {100}65536=36(mod100) is because 100∣65536−36100|65536 - 36100∣65536−36. So, you may ask me why I can't try 136,236136, 236136,236 or bigger, yes, you can. But remember we need last two digits. It is all about Modular Arithmetic. So, if you don't understand or don't know what is Modular Arithmetic, click on the link. (By the way, you have to understand the main property of modular, that is when a=‾b(modn)a\overline= b \pmod na=b(modn), that means n∣a−bn|a-bn∣a−b.)
Now, let's move on to the line 7. (I expect you understand line 5 and 6) It would be laborious for us to calculate 36336^3363 (In this competition, it is stated clearly that calculator or any calculating program is prohibited, so this way is faster and better), so I take the last two digits of 1296, which is 96, then since 36336^3363 is equal to 36 times of 1296, so I multiply 96 with 36, and this process goes on and on (But I stopped it when it reaches 36736^7367, since it is enough to prove that the pattern of the last two digits is recursive). So, you may ask me why I write (mod10000)\pmod {10000}(mod10000) instead of (mod100) or (mod1000)\pmod{100} \text{ or } \pmod{1000}(mod100) or (mod1000), from line 7 to line 12. Yes, you can, if you want to, there is no right and wrong about that, as long as your statement satisfy the main property I listed out. Okay, back to the solution, as we know, 6=‾1(mod5)6 \overline= 1\pmod56=1(mod5), so from here it indicates the first term in the pattern. Therefore, ppp has 36 as its last two digits, since the exponent ends with 6.
Then, we now move on to the case when k=5k=5k=5. Do you realize that except term a1a_1a1 and a2a_2a2, all of them ends with 25. This implies to every term and qqq execpt term a1a_1a1 and a2a_2a2 has 25 as the last two digits.
Alright, for the case when k=6k=6k=6 and k=7k=7k=7, I use the same way, even though the case when k=6k=6k=6, I skipped a lot, since the exponent ends with 6, just like when k=4k=4k=4. Finally, sum up p+q+r+sp+q+r+sp+q+r+s, we have 46 as the last two digits.
So, I had try my best to explain, I hope you had understand. This question tests you Pattern Recognition the most, since you have to check out the pattern what has happened to the last two digits when the exponents are different. Otherwise, if you choose to use brute force, then you will have to calculate for 1024 times, that would drive you nuts. (I can't imagine what is the real value of a1024a_{1024}a1024...)
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Binomial Theorem Binomial Theorem for Positive Integral Indices Tips : If ‘n’ is a positive integer, then (x + a) n= nC o. x n+ nC 1. x n-1. a + nC 2. x n-2. a 2+ ....... + nC n. a n nC o= nC n; nC r= nC n-r If ‘n’ is odd, there are two greatest binomial coefficients in the expansion which are \tt ^n C _\frac{n-1}{2}\ and\ ^n C _\frac{n+1}{2} If ‘n’ is even there is only one greatest binomial coefficient in the expansion (x + a) nwhich is \tt ^n C _\frac{n}{2} If n > 2, n ∈ N then (2n −1) n+ (2n) n< (2n+1) n If the coefficients of X r-1, X r, X r+1in (1 + x) nare in A.P then (n −2r) 2= n + 2 Tricks : Coefficient of X pY qZ rin (ax + by + cz) nis \tt \frac{n! a^{p}b^{q}c^{r}}{p! \times q! \times r!} where p+q+r = n and n, p, q, r ∈ N. (x + a) n– (x - a) n= 2 [ nC 1x n−1a + nC 3x n−3a 3+ nC 5x n−5a 5+ ...........] Part1: View the Topic in this video From 00:10 To 14:07 Part2: View the Topic in this video From 00:10 To 08:57 Part3: View the Topic in this video From 00:10 To 06:08 Part4: View the Topic in this video From 00:10 To 11:16 Part5: View the Topic in this video From 00:10 To 07:30
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Binomial Theorem for Positive Integer: If n is any positive integer, then (x + a) n = nC 0 x n + nC 1 x n−1 + a+ nC 2 x n −2 a 2 + ... + nC n a n. i.e., (x + a)^n \sum_{r=0}^n \ ^nC_r\ x^{n-r}a^{r}
2. (x − a)
n = nC 0 − nC 1 x n −1 a + nC 2 x n −2 a 2 − nC 3 x n −3 a 3 + ... + (−1) n. nC n a n i.e., (x-a)^n \sum_{r=0}^n \ \left(-1\right) \ ^nC_r \cdot x^{n-r}\cdot a^{r}
3. (1 + x)
n = nC 0 + nC 1 x + nC 2 x 2 + ... + nC n x n i.e., (1+x)^n=\sum_{r=0}^n \ ^nC_r \cdot x^{r}
4. (x − a)
n = nC 0 − nC 1 x + nC 2 x 2 − nC 3 x 3 + ... + \left(-1\right)^{r^{n}}C_{r}x^{r} + ...+ (-1) n nC n x n i.e.,(1-x)^n \sum_{r=0}^n \left(-1\right)^r \ {^{n}C_{r}}\cdot x^{r}
5. (x + a)
n + (x − a) n = 2 ( nC 0x n a 0 + nC 2 x n −2 a 2 + ...) (x + a) n − (x − a) n = 2 ( nC 1x n −1 a + nC 3 x n −3 a 2 + ...) |
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator |
It's my first plot and I tried to google for half an hour but couldn't figure out how to fix my attempt to replicate this graph. Could someone help me?
This is, what I have got:
Errors so far:
parabola (2,2)node[pos=0.8, right=10pt] {erfahren}; positioning doesn't work, anyways the parables are wrong I couldn't find how to draw these parables, something like -((x-2)^2)+2
Thank you in advance !
\documentclass{article}\usepackage{tikz}\usetikzlibrary{positioning}\begin{document}\begin{tikzpicture} \draw (0,0) -- (5,0) node[pos=0.5,below=14pt] {Zeit$\rightarrow$}; \draw (0,0) -- (0,3) node[pos=0.5, left=10pt][text width=2cm]{$\uparrow$\\Emotionale\\Insensität}; \draw (0,0) parabola (2,2)node[pos=0.8, right=10pt] {erfahren}; \draw[dashed] (0,0) parabola (3,2)node[pos=0.5, right=10pt] {vorhergesagt};\end{tikzpicture}\end{document} |
I've been wondering this ever since I started using internet for maths actively -- reading math books, this website and sites similar to this one. Sometimes I run into some notations that I'm not sure what they mean, or why are they used in a "weird" way. So I guessed that different notations are used in different parts of the world, and since I'm reading mostly american websites and blogs/books written in English, all of the notations are standard notations in America. So why does, for example, Serbia, have different notations for some things?
For example, in Serbia we use $\text{tg}$ and $\text{ctg}$ instead of $\tan$ and $\cot$. We mark derivatives as $f'(x)$, $f''(x)$, $f'''(x)$ and $f^{(n)}(x)$ for larger numbers instead of $\frac{\text{d}}{\text{dx}}f(x)$ (or however it actually works, I've never got the hang of the notation). In a question I posted not so long ago, I realized that $a=\overline{a_na_{n-1}a_{n-2}\dots a_0}$ doesn't represent a number made of stacking digits $a_i$ one to another for other mathematicians around the globe. In Serbia, we use overline to make sure the reader doesn't interpret it as $a=\prod_{i=0}^{n}a_i$, but rather as $a=\sum_{i=0}^{n}{10^{i}\cdot a_i}$. We use it for other purposes as well, such as repeating digits, conjugated complex numbers, of course (which I believe are standard everywhere).
I'm interested in why different notations are used around the world. Why are they not standardized? You could say that the language of mathematics is a language used by everyone around the world, but these are a few counterexamples I get on top of my head.
Please note that I'm not talking about usual marks for certain things such as $A$ for area. It's obviously used in English only because $A$ is the first letter of *
a*${}$rea (we use $P$ here for he same reason), although I believe $S$ is internationally accepted.
So where do certain notations come from, and why do (you think) they differ in different countries? |
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