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https://au.boazcommunitycorp.org/1100-abscissa-origin-of-the-word.html | # Abscissa - Origin of the word
Comes from the adjective Latin abscissuswhich means "separated in two, interrupted, separated". In 1919, Swiss mathematician Florian Cajori stated that the words "abscissa" and "ordinate" were not used by Descartes. It is also known that although the word "abscissa" has been used at least since the work From Practice Geometrie, published in 1220 by Fibonacci (Leonardo de Pisa), its use in the most modern sense is due to other mathematicians several centuries later.
The technical use of "abscissa" is observed in the eighteenth century by C. Wolf and others. Prior to this, in the most general sense of distance, the word abscissa was used by Bonaventura Cavalieri in his work on the "indivisible," by Stefano degli Angeli (1623-1697), among others.
The mathematician Cajori stated in 1906 that the term abscissa first occurred in a 1659 paper written by Stefano degli Angeli, a mathematics teacher in Rome. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9019822478294373, "perplexity": 2836.1519319922436}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585215.14/warc/CC-MAIN-20211018221501-20211019011501-00471.warc.gz"} |
https://math.stackexchange.com/questions/822604/induction-with-negative-step | # Induction with negative step
We've learned that we can use induction to show that a statement holds for all natural numbers (or for all natural numbers above n). The steps are:
1. prove that the statement holds for a base number b
2. assuming that the statement holds for n, show that it holds for n+1.
This way we have proved that the statement holds for any integer $\ge b$
Can we take this a bit further to prove that the statement holds for ALL integer values? To my understanding, all we have to do is to try to prove:
$3$. assuming that the statement holds for n, show that it holds for n-1.
However I've never seen any articles on this, or any exercises being solved this way?
• Is this because my logic is not correct?
• Is this because "prove that this holds for all integers" can always be solved with a simpler way than using induction twice?
• It's because if you want to prove $\forall n\in \mathbb Z(P(n))$ and you proved $\forall n\in \mathbb N(P(n))$, then, $\forall n\in \mathbb Z(P(n))$ follows from $\forall n\in \mathbb N(P(-n))$ in conjunction with $\forall n\in \mathbb N(P(n))$. – Git Gud Jun 6 '14 at 7:00
If we have some sort of proposition $P(n)$ and we have proved that $P(n)$ is true for every $n \in \mathbb{N}$ by induction then we can proceed to create a new proposition $P'(n) = P(-n)$ and prove that for every $n \in \mathbb{N}$ thus showing that $P(n)$ is actually true for every $n \in \mathbb{Z}$. The steps to prove $P'(n)$ is to show that the base case $P'(0)$ is true and then proceed to show that $P(-n)$ true implies that $P(-(n+1)) = P(-n-1)$.
As an example, if $f(n+2) = 5 f(n+1) - 6 f(n)$ for all integers $n$, and $f(0) = 0$ and $f(1) = 1$, then $f(n) = 3^n - 2^n$ for all integers $n$. You cannot avoid using induction twice, once upwards and once downwards. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9489220380783081, "perplexity": 124.53432798446853}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628001089.83/warc/CC-MAIN-20190627095649-20190627121649-00013.warc.gz"} |
https://mathoverflow.net/questions/191652/a-generalized-diagonal/191662 | A generalized diagonal?
A simple question. Let $f:X\to Y$ be a function and let $E_f:=\{(x, y): f (x)=f (y)\}\subset X\times X$. What is the name of the set $E(f)$? It would be nice to have some reference also. It seems to be a well known notion/construction in set theory or topology, but I couldn't find anything about it.
Another question is whether the following statement is true: If $X$ is a connected topological space and the function $f$ is continuous then $E_f$ is connected.
• In algebra, if f is a homomorphism, then E is a congruence. If f is not a homomorphism, E can still be an equivalence relation of interest. – The Masked Avenger Dec 28 '14 at 19:51
• Another term for this is equalizer (ncatlab.org/nlab/show/equalizer) where the two parallel morphisms coincide. – Vidit Nanda Dec 28 '14 at 20:44
• @ViditNanda it's not the equalizer, it's the kernel pair. – user40276 Dec 30 '14 at 0:21
It's called the kernel or kernel pair of $f$. It is used all over the place in category theory, for example to describe the useful notion of regular category where one sets up Galois connections which in one direction sends a morphism $f: X \to Y$ to the pair of projections $\pi_1, \pi_2: \ker(f) \rightrightarrows X$, and in the other sends a parallel pair $K \rightrightarrows X$ to its coequalizer $X \to Q$. The image of $f: X \to Y$ in such categories is then the coequalizer of its kernel pair.
E(f) does not have to be connected even when $\ X\$ is.
Example: Consider $\ S^1 := \{z\in\mathbb C : |z| = 1\}\$ -- the unit circle; and also $\ f:S^1\rightarrow S^1\$ such that:
$$\forall_{z\in S^1}\ f(z):= z^2$$
Then $\ E(f) = \{(u\ v)\in S^1\times S^1 : u^2=v^2\}\$ is not conected.
REMARK: If $\ f:X\rightarrow Y\$ is such that $\ X\$ is connected, and $\ f^{-1}(y)\$ is connected for every $\ y\in Y\$, then $\ E(f)\$ is connected
Example--just a variation of the above one: E(f) is disjoint for $\ f:\mathbb R\rightarrow\mathbb R^2$ given by: $\ \forall_{x\in\mathbb R}\ f(x):= \exp(\imath\cdot x)$.
• A nice example! What is about $f:\mathbb{R}^n\to\mathbb{R}$. My original problem comes from a rather concrete $f:\mathbb{R}^4\to\mathbb{R}$ for which the statement is true, even in a considerably larger class of functions. – Vladimir Tkachev Dec 29 '14 at 20:06
• @VladimirTkachev -- you got nice result! (I tried for more of something positive but didn't get anything). – Włodzimierz Holsztyński Dec 30 '14 at 7:35
• The examples I have in mind, all of them have the property that "there exists at least one $a\in \mathbb { R }$ such that $f^{-1}(a)$ is connected". – Vladimir Tkachev Dec 30 '14 at 11:37
• P.S. Notice that the target space is $\mathbb {R}$, it might be crucial. – Vladimir Tkachev Dec 30 '14 at 11:44
$E_f$ is an equivalence relation on $X$ and conversely for every equivalence relation $R$, you can construct the topological quotient $X/R$ and for the induced map $f:X\to X/R$, $E_f=R$. There are unconnected equivalence relations on connected spaces that contradicts the above statement.
For example $X=[0,1], R=\{(x,x):x\in X\}\cup\{(0,1),(1,0)\}, X/R = S^1$.
I think this is called fibered product (or rather a special case thereof) $X\times_YX$. More generally, one would have two maps $X\to Z$, $Y\to Z$ and fibered product $X\times_ZY$.
@VladimirTkachev, in addition to the general case of maps $\ f:X\rightarrow Y,\$ was also concerned with the special case of $\ Y:=\mathbb R,\$ i.e. with $\ X\rightarrow\mathbb R.\$ I'll give an example (negative) in the extra simple case of $\ f:\mathbb I\rightarrow\mathbb R,\$ where $\ \mathbb I\$ is a closed interval.
EXAMPLE Let $\ f:[-1;1]\rightarrow\mathbb R\$ be defined by:
$$\forall_{x\in[-1;1]}\quad f(x) := (x+1)\cdot x\cdot(x-1)$$
Then $\ E(f)\$ is disconnected. Indeed, $\ (-1\,\ 1)\in E(f)\$ is an isolated point of $\ E(f).\$ Of course $\ (1\,\ -\!1)\$ is another isolated point like this, and there are no other isolated points in $\ E(f)$.
PROOF We have 3 cases:
• if $\ (x\ y)\in E(f)\$ is such that $\ f(x)=f(y) < 0\$ then both $\ x\ y>0\$ are positive;
• $\ E(f)\cap f^{-1}(0)\ =\ \{-1\,\ 0\,\ 1\}^2$
• if $\ (x\ y)\in E(f)\$ is such that $\ f(x)=f(y) > 0\$ then both $\ x\ y<0\$ are negative.
Thus the distance (say, Euclidean) from $\ (-1\,\ 1)\$ or $\ (1\,\ -\!1)\$ to any other point $\ (x\ y)\in E(f)\$ is at least $\ 1$.
End of PROOF
• Thank you for good examples! Probably it should be made more precise what kind of $X$ and $f$ are relevant. I've also considered a similar cubic polynomial, but with the domain of definition being the whole $X =\mathbb {R}$ instead of an interval. In that case, $E (f)$ is connected (being the union of an oval and a diagonal line passing through the oval). The example you construct can be thought as a cutoff of the set $E(f)$ above. Thus, some 'completteness' of X is needed. Another relevant example is $f=\sin x\sin y:\,\mathbb {R}^2\to\mathbb {R}$ for which $E(f)$ is connected again. – Vladimir Tkachev Jan 3 '15 at 13:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9252797365188599, "perplexity": 230.33825827107788}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251681412.74/warc/CC-MAIN-20200125191854-20200125221854-00053.warc.gz"} |
http://espressomd.org/html/tutorials_html/11-ferrofluid/11-ferrofluid_part3.html | # Ferrofluid - Part III¶
Remark: The equilibration and sampling times used in this tutorial would be not sufficient for scientific purposes, but they are long enough to get at least a qualitative insight of the behaviour of ferrofluids. They have been shortened so we achieve reasonable computation times for the purpose of a tutorial.
## Susceptibility with fluctuation formulas¶
In this part we want to calculate estimators for the initial susceptibility, i.e. the susceptibility at zero external magnetic field. One could carry out several simulations with different external magnetic field strengths and get the initial susceptibility by fitting a line to the results. We want to go a more elegant way by using fluctuation formulas known from statistical mechanics. In three dimensions the initial susceptibility $\chi_{init}$ can be calculated with zero field simulations through
$$\chi_{init} = \frac{V \cdot \mu_0}{3 \cdot k_B T} \left( \langle \boldsymbol{M}^2 \rangle - \langle \boldsymbol{M} \rangle^2 \right) = \frac{\mu_0}{3 \cdot k_B T \cdot V} \left( \langle \boldsymbol{\mu}^2 \rangle - \langle \boldsymbol{\mu} \rangle^2 \right)$$
where $\boldsymbol{M}$ is the magnetization vector and $\boldsymbol{\mu}$ is the total magnetic dipole moment of the system. In direction $i$ it reads
$$M_i = \frac{1}{V} \Bigg\langle \sum_{j=1}^N \tilde{\mu}_j^i \Bigg\rangle$$
where $\tilde{\mu}_j^i$ is the $j$ th dipole moment in direction $i$.
### Derivation of the fluctuation formula¶
We want to derive the fluctuation formula. We start with the definition of the magnetic susceptibility. In general this reads
$$\chi \equiv \frac{\partial}{\partial H} \langle M_{\boldsymbol{H}} \rangle$$
with $\langle M_{\boldsymbol{H}} \rangle$ the ensemble averaged magnetization in direction of a homogeneous external magnetic field $\boldsymbol{H}$.
In thermal equilibrium the ensemble average of the magnetization reads
$$\langle M_{\boldsymbol{H}} \rangle = \frac{1}{V Z_c} \left \lbrack \sum_{\alpha} \mu_{\boldsymbol{H},\alpha} e^{ -\beta E_{\alpha}(H=0) + \beta\mu_0\mu_{\boldsymbol{H},\alpha}H }\right \rbrack$$
with $Z_c$ the canonical partition function, $E_{\alpha}(H=0)$ the energy without an external magnetic field $\boldsymbol{H}$, $\beta$ the inverse thermal energy $\frac{1}{k_BT}$, $\mu_{\boldsymbol{H},\alpha}$ the total magnetic dipole moment of the system in direction of the external magnetic field $\boldsymbol{H}$ in microstate $\alpha$ and $V$ the system volume.
Now we insert the magnetization $\langle M_{\boldsymbol{H}} \rangle$ in the definition of the magnetic susceptibility $\chi$ and let the derivative operate on the ensemble average. We get the fluctuation formula
$$\chi = \frac{\beta\mu_0}{V} \left \lbrack \frac{1}{Z_c}\sum_{\alpha} \mu_{\alpha}^2~ e^{ -\beta E_{\alpha}(H=0) + \beta\mu_0\mu_{\boldsymbol{H},\alpha}H } - \frac{1}{Z_c}\sum_{\alpha} \mu_{\alpha}~ e^{ -\beta E_{\alpha}(H=0) + \beta\mu_0\mu_{\boldsymbol{H},\alpha}H }~~ \frac{1}{Z_c}\sum_{\alpha'}\mu_{\alpha'}~ e^{ -\beta E_{\alpha'}(H=0) + \beta\mu_0\mu_{\boldsymbol{H},\alpha}H }\right \rbrack = \frac{\beta\mu_0}{V} \left \lbrack \langle \mu_{\boldsymbol{H}}^2 \rangle - \langle \mu_{\boldsymbol{H}} \rangle^2 \right \rbrack = \frac{\beta\mu_0}{V} \left(\Delta \mu_{\boldsymbol{H}}\right)^2$$
At zero external magnetic field ($H = 0$) there is no distinct direction for the system, so we can take the fluctuations $\Delta \mu$ in all directions and divide it by the dimension. Thus we can use more data points of our simulation for the average and get a more precise estimator for the susceptibility. Thus finally the fluctuation formula for the initial susceptibility in three dimensions reads
$$\chi_{init} = \frac{\beta\mu_0}{3V} \left \lbrack \langle \boldsymbol{\mu}^2 \rangle - \langle \boldsymbol{\mu} \rangle^2 \right \rbrack = \frac{V\beta\mu_0}{3} \left \lbrack \langle \boldsymbol{M}^2 \rangle - \langle \boldsymbol{M} \rangle^2 \right \rbrack$$
where $\boldsymbol{\mu}$ and $\boldsymbol{M}$ are defined above.
### Simulation¶
In this part we want to consider a three dimensional ferrofluid system and compare our result for the initial susceptibility $\chi_{init}$ with them of Ref. [1].
First we import all necessary packages and check for the required Espresso features
In [1]:
import espressomd
espressomd.assert_features('DIPOLES', 'LENNARD_JONES')
from espressomd.magnetostatics import DipolarP3M
import numpy as np
Now we set up all necessary simulation parameters
In [2]:
lj_sigma=1
lj_epsilon=1
lj_cut = 2**(1./6.) * lj_sigma
# magnetic field constant
mu_0 = 1.
# Particles
N = 1000
# Volume fraction
# phi = rho * 4. / 3. * np.pi * ( lj_sigma / 2 )**3.
phi = 0.0262
# Dipolar interaction parameter lambda = mu_0 m^2 /(4 pi sigma^3 kT)
dip_lambda = 3.
# Temperature
kT =1.0
# Friction coefficient
gamma=1.0
# Time step
dt =0.02
# box size 3d
box_size = (N * np.pi * 4./3. * (lj_sigma / 2.)**3. / phi)**(1./3.)
and the system, where we, as we did in part I, only commit the orientation of the dipole moment to the particles and take the magnitude into account in the prefactor of Dipolar P3M (for more details see part I).
Hint: It should be noted that we seed both the Langevin thermostat and the random number generator of numpy. Latter means that the initial configuration of our system is the same every time this script is executed. As the time evolution of the system depends not solely on the Langevin thermostat but also on the numeric accuracy and DP3M (the tuned parameters are slightly different every time) it is only partly predefined. You can change the seeds to simulate with a different initial configuration and a guaranteed different time evolution.
In [3]:
system = espressomd.System(box_l=(box_size,box_size,box_size))
system.time_step = dt
system.thermostat.set_langevin(kT=kT, gamma=gamma, seed=1)
# Lennard Jones interaction
system.non_bonded_inter[0,0].lennard_jones.set_params(epsilon=lj_epsilon,sigma=lj_sigma,cutoff=lj_cut, shift="auto")
# Random dipole moments
np.random.seed(seed = 1)
dip_phi = np.random.random((N,1)) *2. * np.pi
dip_cos_theta = 2*np.random.random((N,1)) -1
dip_sin_theta = np.sin(np.arccos(dip_cos_theta))
dip = np.hstack((
dip_sin_theta *np.sin(dip_phi),
dip_sin_theta *np.cos(dip_phi),
dip_cos_theta))
# Random positions in system volume
pos = box_size * np.random.random((N,3))
# Remove overlap between particles by means of the steepest descent method
system.integrator.set_steepest_descent(
f_max=0,gamma=0.1,max_displacement=0.05)
while system.analysis.energy()["total"] > 5*kT*N:
system.integrator.run(20)
# Switch to velocity Verlet integrator
system.integrator.set_vv()
# tune verlet list skin
system.cell_system.skin = 0.8
# Setup dipolar P3M
accuracy = 5E-4
Dipolar P3M tune parameters: Accuracy goal = 5.00000e-04 prefactor = 3.00000e+00
System: box_l = 2.71372e+01 # charged part = 1000 Sum[q_i^2] = 1.00000e+03
Dmesh cao Dr_cut_iL Dalpha_L Derr Drs_err Dks_err time [ms]
8 3 4.24571e-01 5.16252e+00 4.96675e-04 3.535e-04 3.489e-04 25
8 2 4.70520e-01 4.45931e+00 5.47820e-04 3.535e-04 4.185e-04 accuracy not achieved
8 4 3.98839e-01 5.63003e+00 4.98793e-04 3.535e-04 3.519e-04 24
8 5 3.85974e-01 5.88831e+00 4.93802e-04 3.535e-04 3.448e-04 24
8 6 3.79541e-01 6.02431e+00 4.94584e-04 3.535e-04 3.459e-04 25
8 7 3.76784e-01 6.08409e+00 4.96565e-04 3.535e-04 3.487e-04 27
10 4 3.58332e-01 6.50910e+00 4.96956e-04 3.535e-04 3.492e-04 21
10 3 3.84039e-01 5.92868e+00 4.98438e-04 3.535e-04 3.514e-04 22
10 5 3.44311e-01 6.86423e+00 4.98486e-04 3.535e-04 3.514e-04 22
10 6 3.36521e-01 7.07502e+00 4.96474e-04 3.535e-04 3.486e-04 22
10 7 3.31847e-01 7.20649e+00 4.99879e-04 3.535e-04 3.534e-04 24
12 4 3.26138e-01 7.37251e+00 4.97911e-04 3.535e-04 3.506e-04 19
12 3 3.53433e-01 6.62983e+00 4.97172e-04 3.535e-04 3.496e-04 20
12 5 3.12841e-01 7.78399e+00 4.94729e-04 3.535e-04 3.461e-04 19
12 6 3.04442e-01 8.06349e+00 4.96652e-04 3.536e-04 3.488e-04 21
14 4 3.00659e-01 8.19483e+00 4.96479e-04 3.535e-04 3.486e-04 11
14 3 3.26138e-01 7.37251e+00 5.16376e-04 3.535e-04 3.764e-04 accuracy not achieved
14 5 2.86645e-01 8.71326e+00 4.99171e-04 3.535e-04 3.524e-04 11
14 6 2.79001e-01 9.01935e+00 4.95315e-04 3.535e-04 3.469e-04 13
16 4 2.79519e-01 8.99807e+00 4.98084e-04 3.535e-04 3.509e-04 10
16 3 3.00659e-01 8.19483e+00 5.63891e-04 3.535e-04 4.393e-04 accuracy not achieved
16 5 2.65425e-01 9.60908e+00 4.99874e-04 3.535e-04 3.534e-04 11
16 6 2.57791e-01 9.96965e+00 4.97250e-04 3.535e-04 3.497e-04 12
16 7 2.53094e-01 1.02030e+01 4.92596e-04 3.535e-04 3.430e-04 14
18 4 2.62049e-01 9.76578e+00 4.98138e-04 3.535e-04 3.509e-04 11
18 3 2.79519e-01 8.99807e+00 6.16133e-04 3.535e-04 5.046e-04 accuracy not achieved
18 5 2.47855e-01 1.04743e+01 4.98228e-04 3.535e-04 3.511e-04 11
18 6 2.40211e-01 1.08926e+01 4.96214e-04 3.535e-04 3.482e-04 13
20 4 2.47206e-01 1.05087e+01 4.98502e-04 3.535e-04 3.514e-04 10
20 3 2.62049e-01 9.76578e+00 6.66654e-04 3.535e-04 5.652e-04 accuracy not achieved
20 5 2.32875e-01 1.13216e+01 4.97237e-04 3.535e-04 3.496e-04 11
20 6 2.25198e-01 1.18023e+01 4.95660e-04 3.535e-04 3.474e-04 13
22 4 2.34653e-01 1.12151e+01 4.96559e-04 3.535e-04 3.487e-04 12
22 3 2.47206e-01 1.05087e+01 7.16903e-04 3.535e-04 6.237e-04 accuracy not achieved
22 5 2.20168e-01 1.21365e+01 4.93713e-04 3.535e-04 3.446e-04 10
22 6 2.12443e-01 1.26826e+01 4.92040e-04 3.535e-04 3.422e-04 12
22 7 2.07615e-01 1.30458e+01 4.93198e-04 3.535e-04 3.439e-04 14
24 5 2.08988e-01 1.29407e+01 4.91905e-04 3.535e-04 3.420e-04 10
24 4 2.20168e-01 1.21365e+01 5.38392e-04 3.535e-04 4.060e-04 accuracy not achieved
24 6 2.01247e-01 1.35529e+01 4.89597e-04 3.536e-04 3.387e-04 12
24 7 1.96087e-01 1.39897e+01 4.97569e-04 3.535e-04 3.501e-04 14
26 5 1.99191e-01 1.37241e+01 4.89609e-04 3.535e-04 3.387e-04 13
26 4 2.08988e-01 1.29407e+01 5.61502e-04 3.535e-04 4.362e-04 accuracy not achieved
26 6 1.91028e-01 1.44422e+01 4.93284e-04 3.535e-04 3.440e-04 14
26 7 1.86130e-01 1.49050e+01 4.98017e-04 3.535e-04 3.507e-04 17
resulting parameters: mesh: 24, cao: 5, r_cut_iL: 2.0899e-01,
alpha_L: 1.2941e+01, accuracy: 4.9191e-04, time: 11
Now we equilibrate for a while
In [4]:
print("Equilibration...")
equil_rounds = 10
equil_steps = 100
for i in range(equil_rounds):
system.integrator.run(equil_steps)
print("progress: {:3.0f}%, dipolar energy: {:9.2f}".format(
(i + 1) * 100. / equil_rounds, system.analysis.energy()["dipolar"]), end="\r")
print("\nEquilibration done")
Equilibration...
progress: 100%, dipolar energy: -1140.27
Equilibration done
As we need the magnetization of our system, we import MagneticDipoleMoment from observables which returns us the total dipole moment of the system which is the magnetization times the volume of the system.
In [5]:
from espressomd.observables import MagneticDipoleMoment
dipm_tot_calc = MagneticDipoleMoment(ids=system.part[:].id)
Now we set the desired number of loops for the sampling
In [6]:
# Sampling
loops = 200
and sample the first and second moment of the magnetization or total dipole moment, by averaging over all total dipole moments occurring during the simulation
In [7]:
print('Sampling ...')
# calculate initial total dipole moment
dipm_tot_temp = dipm_tot_calc.calculate()
# initiate variables for sum of first and second moment of total dipole moment
dipm_tot_sum = dipm_tot_temp
dipm_tot_2_sum = np.square(dipm_tot_temp)
# sample dipole moment
for i in range(loops):
system.integrator.run(10)
dipm_tot_temp = dipm_tot_calc.calculate()
dipm_tot_sum = np.sum((dipm_tot_sum, dipm_tot_temp), axis=0)
dipm_tot_2_sum = np.sum((dipm_tot_2_sum, np.square(dipm_tot_temp)), axis=0)
# print progress only every 10th cycle
if (i+1)%10 == 0:
print("progress: {:3.0f}%".format((i+1)*100./loops), end="\r")
print("\nSampling done")
# calculate average first and second moment of total dipole moment
dipm_tot = dipm_tot_sum / loops
dipm_tot_2 = dipm_tot_2_sum / loops
Sampling ...
progress: 100%
Sampling done
For the estimator of the initial susceptibility $\chi_{init}$ we need the magnitude of one single dipole moment
In [8]:
# dipole moment
dipm = np.sqrt(dip_lambda*4*np.pi*lj_sigma**3.*kT / mu_0)
print("dipm = {}".format(dipm))
dipm = 6.139960247678931
Now we can calculate $\chi_{init}$ from our simulation data
In [9]:
# susceptibility in 3d system
chi = mu_0 /(system.volume() * 3. * kT) * ( np.sum(dipm_tot_2 * dipm**2.) - np.sum(np.square(dipm_tot * dipm)) )
and print the result
In [10]:
print('chi = %.4f' % chi)
chi = 0.7510
Compared with the value $\chi = 0.822 \pm 0.017$ of Ref. [1] (see table 1) it should be very similar.
Now we want to compare the result with the theoretical expectations. At first with the simple Langevin susceptibility
In [11]:
chi_L = 8. * dip_lambda * phi
print('chi_L = %.4f' % chi_L)
chi_L = 0.6288
and at second with the more advanced one (see Ref. [1] eq. (6)) which has a cubic accuracy in $\chi_L$ and reads
$$\chi = \chi_L \left( 1 + \frac{\chi_L}{3} + \frac{\chi_L^2}{144} \right)$$
In [12]:
chi_I = chi_L * ( 1 + chi_L / 3. + chi_L**2. / 144. )
print('chi_I = %.4f' % chi_I)
chi_I = 0.7623
Both of them should be smaller than our result, but the second one should be closer to our one. The deviation of the theoretical results to our simulation result can be explained by the fact that in the Langevin model there are no interactions between the particles incorporated at all and the more advanced (mean-field-type) one of Ref. [1] do not take occurring cluster formations into account but assumes a homogeneous distribution of the particles. For higher values of the volume fraction $\phi$ and the dipolar interaction parameter $\lambda$ the deviations will increase as the cluster formation will become more pronounced.
## Magnetization curve of a 3D system¶
At the end of this tutorial we now want to sample the magnetization curve of a three dimensional system and compare the results with analytical solutions. Again we will compare with the Langevin function but also with the approximation of Ref. [2] (see also Ref. [1] for the right coefficients) which takes the dipole-dipole interaction into account. For this approximation, which is a modified mean-field theory based on the pair correlation function, the Langevin parameter $\alpha$ is replaced by
$$\alpha' = \alpha + \chi_L~L(\alpha) + \frac{\chi_L^{2}}{16} L(\alpha) \frac{d L(\alpha)}{d\alpha}$$
where $\chi_L$ is the Langevin susceptibility
$$\chi_L = \frac{N}{V}\frac{\mu_0 \mu^2}{3k_BT} = 8 \cdot \lambda \cdot \phi$$
Analogous to part II we start at zero external magnetic field and increase the external field successively. At every value of the external field we sample the total dipole moment which is proportional to the magnetization as we have a fixed volume.
First we create a list of values of the Langevin parameter $\alpha$. As we already sampled the magnetization at zero external field in the last section, we take this value and continue with the sampling of an external field unequal zero
In [13]:
alphas = [0.5,1,2,4,8]
Now for each value in this list we sample the total dipole moment / magnetization of the system for a while. Have in mind that we have only committed the orientation of the dipole moments to the particles. Thus we have to commit $H\cdot \mu$ as the external magnetic field to ESPResSo, where $\mu$ is the magnitude of a single magnetic dipole moment.
As in part II we use the same system for every value of the Langevin parameter $\alpha$. Thus we use that the system is already pre-equilibrated from the previous run so we save some equilibration time. For scientific purposes one would use a new system for every value for the Langevin parameter to ensure that the systems are independent and no correlation effects are measured. Also one would perform more than just one simulation for each value of $\alpha$ to increase the precision of the results.
In [14]:
# remove all constraints
system.constraints.clear()
# list of magnetization in field direction
magnetization = []
# append result for alpha=0 from previous chapter
magnetization.append(np.average(dipm_tot))
# number of loops for sampling
loops_m = 100
for alpha in alphas:
print("Sample for alpha = {}".format(alpha))
H_dipm = (alpha*kT)
H_field = [H_dipm,0,0]
print("Set magnetic field constraint...")
H_constraint = espressomd.constraints.HomogeneousMagneticField(H=H_field)
print("done\n")
# Equilibration
print("Equilibration...")
for i in range(equil_rounds):
system.integrator.run(equil_steps)
print("progress: {:3.0f}%, dipolar energy: {:9.2f}".format(
(i+1)*100./equil_rounds, system.analysis.energy()["dipolar"]), end="\r")
print("\nEquilibration done\n")
# Sampling
print("Sampling...")
magn_temp = 0
for i in range(loops_m):
system.integrator.run(20)
magn_temp += dipm_tot_calc.calculate()[0]
print("progress: {:3.0f}%".format((i+1)*100./loops_m), end="\r")
print("\n")
# save average magnetization
magnetization.append(magn_temp / loops_m)
print("Sampling for alpha = {} done \n".format(alpha))
print("magnetizations = {}".format(magnetization))
print("total progress: {:5.1f}%\n".format((alphas.index(alpha)+1)*100./len(alphas)))
# remove constraint
system.constraints.clear()
print("Magnetization curve sampling done")
Sample for alpha = 0.5
Set magnetic field constraint...
done
Equilibration...
progress: 100%, dipolar energy: -1198.61
Equilibration done
Sampling...
progress: 100%
Sampling for alpha = 0.5 done
magnetizations = [-2.3517104414107606, 201.89607732678928]
total progress: 20.0%
Sample for alpha = 1
Set magnetic field constraint...
done
Equilibration...
progress: 100%, dipolar energy: -1634.11
Equilibration done
Sampling...
progress: 100%
Sampling for alpha = 1 done
magnetizations = [-2.3517104414107606, 201.89607732678928, 394.6657360902372]
total progress: 40.0%
Sample for alpha = 2
Set magnetic field constraint...
done
Equilibration...
progress: 100%, dipolar energy: -2594.75
Equilibration done
Sampling...
progress: 100%
Sampling for alpha = 2 done
magnetizations = [-2.3517104414107606, 201.89607732678928, 394.6657360902372, 616.4434954755568]
total progress: 60.0%
Sample for alpha = 4
Set magnetic field constraint...
done
Equilibration...
progress: 100%, dipolar energy: -4721.81
Equilibration done
Sampling...
progress: 100%
Sampling for alpha = 4 done
magnetizations = [-2.3517104414107606, 201.89607732678928, 394.6657360902372, 616.4434954755568, 786.2872132155372]
total progress: 80.0%
Sample for alpha = 8
Set magnetic field constraint...
done
Equilibration...
progress: 100%, dipolar energy: -9087.10
Equilibration done
Sampling...
progress: 100%
Sampling for alpha = 8 done
magnetizations = [-2.3517104414107606, 201.89607732678928, 394.6657360902372, 616.4434954755568, 786.2872132155372, 887.2339152547825]
total progress: 100.0%
Magnetization curve sampling done
Now we define the Langevin function and the modified mean-field-approximation of the Langevin parameter of Ref. [2]
In [15]:
# Langevin function
def L(y):
return np.cosh(y)/np.sinh(y)-1/y
In [16]:
# second order mean-field-model from Ref. [2]
def alpha_mean_field(alpha, dip_lambda, phi):
chi = 8. * dip_lambda * phi
return alpha + chi * L(alpha) + chi**2. / 16. * L(alpha) * ( 1./(alpha**2.) - 1./((np.sinh(alpha))**2.) )
We also want to plot the linear approximation at $\alpha = 0$ to see for which values of $\alpha$ this approximation holds. We use the initial susceptibility calculated in the first chapter of this part as the gradient. As we want the gradient of $M^*$ with respect to $\alpha$ which fulfills the relation
$$\frac{\partial M^*}{\partial \alpha} = \frac{1}{M_{sat}}\frac{\partial M}{\partial \left( \frac{\mu_0\mu}{k_BT} H\right)} = \frac{k_BT~V}{\mu_0\mu^2N}\frac{\partial M}{\partial H} = \frac{k_BT~V}{\mu_0\mu^2N}~\chi$$
we have to scale our calculated initial susceptibility $\chi_{init}$ by a factor to get it in our dimensionless units.
Now we plot the resulting curves together with our simulation results and the linear approximation
In [17]:
import matplotlib.pyplot as plt
In [18]:
y = np.arange(0.01,10, 0.1, dtype=float).tolist()
alphas.insert(0, 0.0)
L_func = []
L_mean_field = []
init_susceptibility = []
for i in y:
L_func.append(L(i))
L_mean_field.append(L(alpha_mean_field(i, dip_lambda, phi)))
init_susceptibility.append(i*system.volume()*kT/(N*mu_0*dipm**2)*chi)
# divide all entries in the magnetization list by N to get the dimensionless magnetization
magnetization_star = []
for i in range(len(magnetization)):
magnetization_star.append(magnetization[i] / N)
print('magnetization_star = {}'.format(magnetization_star))
plt.figure(figsize=(10,10))
plt.ylim(0, 1.)
plt.xlabel(r'$\alpha$', fontsize=20)
plt.ylabel(r'$M^*$', fontsize=20)
plt.plot(y, L_func, label='Langevin function')
plt.plot(y, L_mean_field, label='modified mean-field-theory')
plt.plot(alphas, magnetization_star, 'o', label='simulation results')
plt.plot(y, init_susceptibility, label=r'linear approximation at $\alpha = 0$')
plt.legend(fontsize=20)
plt.show()
magnetization_star = [-0.0023517104414107607, 0.20189607732678927, 0.3946657360902372, 0.6164434954755568, 0.7862872132155372, 0.8872339152547825]
We can see that the magnetization curve where we used the Langevin parameter of the modified mean-field-theory is closer to our simulation results. Also we can clearly see that the linear approximation holds only for very small values of $\alpha$.
At this point is should be mentioned, that the modified mean-field-model assumes a spatially homogeneous system which is not the case at higher volume fractions $\phi$ and dipolar interaction parameters $\lambda$ as the particles form clusters. We can already see this with our simulation results as they visibly deviate from the modified mean-field-model.
At sufficiently high volume fractions $\phi$ and dipolar interaction parameters $\lambda$ these clusters can be thus rigid, that simulation with normal methods are impossible as the relaxation times exceeds normal simulation times by far resulting in strongly correlated configurations and thus measurements.
[1] Zuowei Wang, Christian Holm, and Hanns Walter Müller. “Molecular dynamics study on the equilibrium magnetization properties and structure of ferrofluids”. In: Phys. Rev. E 66 (2 Aug. 2002), p. 021405. doi: 10.1103/PhysRevE.66.021405. url: https://link.aps.org/doi/10.1103/PhysRevE.66.021405.
[2] Alexey O. Ivanov and Olga B. Kuznetsova. “Magnetic properties of dense ferrofluids: An influence of interparticle correlations”. In: Phys. Rev. E 64 (4 Sept. 2001), p. 041405. doi: 10.1103/PhysRevE.64.041405. url: https://link.aps.org/doi/10.1103/PhysRevE.64.041405. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 10, "x-ck12": 0, "texerror": 0, "math_score": 0.8411433100700378, "perplexity": 3723.911364286419}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986677230.18/warc/CC-MAIN-20191017222820-20191018010320-00534.warc.gz"} |
http://mathhelpforum.com/trigonometry/80571-identity-help.html | 1. identity help
$
\frac{\sec{x}\tan{x}}{\tan^3{x}} = \frac{\cos{x}}{\sin{x}}
$
the part in the middle is what I am missing
2. Hello, TYTY!
$\frac{\sec x\tan x}{\tan^3\!x} \:=\: \frac{\cos x}{\sin x}$
the part in the middle is what I am missing .
. . . oh, is that all?
Please check the problem again . . .
The given statement is not an identity.
3. Originally Posted by TYTY
$
\frac{\sec{x}\tan{x}}{\tan^3{x}} = \frac{\cos{x}}{\sin{x}}
$
the part in the middle is what I am missing
Hi TYTY,
Maybe, you meant
$\frac{\sec{x}\tan{x}}{\tan^3{x}} = \frac{\cos{x}}{\sin^2{x}}$
4. ouch
better start that whole problem over again... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9170687794685364, "perplexity": 2472.3916539806764}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281001.53/warc/CC-MAIN-20170116095121-00224-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://slideplayer.com/slide/3428488/ | # In Mathematics the word complexity, as synonym of difficulty, has not a clear meaning. However, it seems that it is very easy to understand that one mathematical.
## Presentation on theme: "In Mathematics the word complexity, as synonym of difficulty, has not a clear meaning. However, it seems that it is very easy to understand that one mathematical."— Presentation transcript:
In Mathematics the word complexity, as synonym of difficulty, has not a clear meaning. However, it seems that it is very easy to understand that one mathematical problem involving functions of several variables is, in general, more complex than one problem with a univariable function. The Alpha-Dense Curves theory tries to reduce the complexity associated to the dimension and establishes an Approximation Theory on the domains, in the sense of the Hausdorff metric, instead of that of the functions. Some applications to optimize multivariable continuous functions, to approximate a multiple integral by a simple one, and to determine if a non-linear inequality has solution, will be the subjects which we shall deal with techniques based on Alpha-Dense Curves.
Definition Given a real number 0, an -dense curve (or curve of density ) in a subset K of finite diameter in a metric space (E,d), is a continuous mapping whose image, now on denoted, is contained in K and the distance,, for any In others words, an -Dense Curve in a compact set K is a Peano Continuum (a connected, locally connected and compact set) whose Hausdorff distance from K When =0, one has a Peano curve (also called space-filling curve) provided that the interior of K, it follows that if a curve is -dense in K, then it is also - dense for any > . Thus the minimal verifying the properties is, strictly speaking, the density of the curve in K, which coincides with the Hausdorff distance
Example In the euclidean space, N 2 for each positive integer m, the unit cube is densified by the curve defined by It is easy to check that has density in where is a positive constant depending on N. For instance, Therefore their densities go to 0 as m . The curve is said to be the Cosines Curve of order m.
Example In, N 2, for each positive integer m the unit cube is densified by the polynomial curve defined by where is the Chebyshev polynomial of the first kind of degree j, that is, The density of in the unit cube is,where (N) is a positive constant that depends on N (for instance,, etc.). Hence, also the densities tend to 0 as m . The curve is called the Chebyshev Curve of order m.
Definition A subset K of (E,d) is said to be densifiable if it contains -dense curves for arbitrary >0. As we just have seen, the unit cube is a densifiable set. This set has been densified by the Cosines and the Chebyshev Curves. Now, taking into account the transformation,,...,,..., defined by an -dense curve in is transformed on a -dense curve in the cube, where Consequently, any cube is densifiable. Furthermore, as a cube is a Peano Continuum, the Hahn-Mazurkiewicz theorem implies that it is a continuous image of I. Hence any cube is in fact a Space-Filling Curve and so is, in particular, densifiable. Clearly, in a metric space (E,d), the class of all densifiable sets contains the class of all Peano Continua. However, both classes are distinct. For instance, the set is densifiable but it is not a Peano Continuum, as it is well-known.
2.1.Convergence. As we know, given a real function f and a nonvoid set X, an usual form to present an Optimization Problem (O.P) is to find, or to approximate to, the value (finite or infinite) The function f is called the objective function and X the feasible set. Even for a differentiable objective function on a compact feasible set of the type in,,,the minimization methods are not very efficient when N is large. Many problems, closely connected with the location of the global minimum, occur.These difficulties, in general, decrease when the objective function f is univariable. Thus, for f continuous on K, the multivariable optimization problem (1) is substituted by the univariable one (2) where and is an -dense curve in K. The reduction of problem 1 to 2 has an error that we can estimate: Given a densifiable set K, let us denote by the class of all curves in K of density .
Thus, for a given, the error in the approximation is Observe that since, always is. Furthermore, if and only if the -dense curve passes through one global minimizer, i.e., a point where f attains the global minimum. An estimation of the error is given in the following result Theorem For each -dense curve, the error is bounded where and is the modulus of continuity of f of order . As corollary of the preceding theorem, the convergence of this reductional method is guaranted by the fact that the modulus of continuity tends to 0 as goes to 0. Whenever f is Lipschitzian, that is, it satisfies with 0 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/12/3428488/slides/slide_8.jpg", "name": "Thus, for a given, the error in the approximation is Observe that since, always is.", "description": "Furthermore, if and only if the -dense curve passes through one global minimizer, i.e., a point where f attains the global minimum. An estimation of the error is given in the following result Theorem For each -dense curve, the error is bounded where and is the modulus of continuity of f of order . As corollary of the preceding theorem, the convergence of this reductional method is guaranted by the fact that the modulus of continuity tends to 0 as goes to 0. Whenever f is Lipschitzian, that is, it satisfies with 0
2.2.Minimization-Preserving Operators (M.P.O.). When we apply the -Dense Curves method to the multivariable optimization problem (1) we substitute (1) by the univariable one being, and is an -dense curve in K. Then the new objective function will have, possibly, those extreme points (local and global extrema) that corresponding to f added to the points for which the gradient vector of f (if there exists) is orthogonal to the tangent vector to. For instance, if K is a cube always we can take with tangent vector nonnull for all t. That is, speaking in a colloquial language, we could say that the substitution of problem (1) by problem (2), can introduce "noise" due to the many extrema of. To eliminate this "noise" we introduced a type of operators that we called Optimization- Preserving Operators (Mora, Cherruault and Benabidallah, 2003). Later, these operators were studied from the specific role devoted to minimization and then they were called Minimization-Preserving Operators (M.P.O.) (Mora, 2005).
Definition Let, be the spaces of continuous and bounded real functions, respectively on. and a subset of.We say that a mapping,, is a Minimization-Preserving Operator if and only if for any, i) the minimizer set of, denoted, is contained in the minimizer set of g, denoted, ii) and have at least a global minimum in common to both functions and g Trivial examples of minimization-preserving operators in are: a) The exponential operator, b) The dilation operator, with. real numbers In general, any strictly increasing function defines a M.P.O. on the space by means of the composition law, that is,, for..
Observe that in all the above examples, for all. In this case, these operators will be called trivial. Nevertheless, our objective is for defining a M.P.O. T such that be strictly included in for some functions g, such as the following example shows (Mora, 2005). Example Each arbitrary function >0 continuous on I, defines a nontrivial M.P.O. T on by the formula where is a global minimizer of g ( for instance, can be chosen as the minimum, in I, of all the global minimizers). The main results to reduce the mentioned noise are the following (Mora, 2005).
Theorem Let A be an arbitrary constant, the Heaviside function and an arbitrary bijective function of class with. Let M be positive, consider the numbers: and Thus, for each the formula for any x I and arbitrary, defines an M.P.O. on the closed ball on the space. Furthermore i) in the level set, constant, ii) all the minimizers, nonnull, of are in the level set.
The reiterated application of the above theorem generates a procedure for finding a global minimizer, as we can see in the next result. Corollary Under the hypotheses of the above theorem and reiterating its application, for each, with a set of minimizers proper and finite, we obtain an algorithm to find a global minimizer of g.
Now, our purpose is to reduce directly the integral of a non-negative real function f, of class on the unit cube,, to a simple one on the interval [-1,1]. For that, we shall use an -Dense Curve that densifies with arbitrary small density each elementary cube of the region of quadrature. The key of the result that we are looking for is in the following facts: 1) For any c>0, the cube is densified by the Cosines Curve, that is, by the curve defined by 2) The volume of can be calculated by the formula where is the length of given by the integral where is the tangent vector. Now, the dimension of the integral is reduced by means of the following result (Mora and Mora-Porta, 2005).
Theorem Let f be a real non-negative function of class on and the Cosines Curve of order m in. Thus the integral may be approximated, for large enough m, by where is the Chebyshev polynomial of the second kind of degree and. Analogously, if each elementary cube of the region of quadrature,, is densified by a Chebyshev Curve with one has the following result (Mora, Benavent and Navarro, 2002).
Theorem Let f be a real nonnegative function of class on and the Chebyshev Curve of order m in.Thus the multiple integral can be approximated, for large enough m, by the simple integral where is de Chebyshev polynomial of the second kind of degree and If we denote by the error in the approximation to the multiple integral of f by using the Cosines Curve, one has An estimation of is given in the following result (Mora and Mora-Porta, 2005). Theorem For the Cosines Curve of order m in the unit cube, the error in the integral reduction formula is an. Remark The constant in depends on the 1-norm of f defined by where is the gradient of f. For the Chebyshev Curve we think that it would have a similar result.
Given a real function f defined on a densifiable compact set K of a metric space (E,d), it is crucial in many applications, to know the set (possibly empty) An -Dense Curves approach to this problem would be: 1) Let us take an -dense curve in K, , which without loss of generality we can suppose nonconstant on any subinterval of I. Furthermore we can also suppose the existence of an increasing bijective function, only continuous at 0, and satisfying 2) Define on I the univariable function, and for each s such that consider the number where is the open ball centered at and radius r. Taking into account the above considerations, one has the following result (Mora, Cherruault and Ubeda, 2005).
Theorem Let K be a densifiable compact of a metric space (E,d), an -dense curve in K (satisfying the preceding assumptions 1,2) and f a real function defined on K, continuous on. Then, on the set, the recursive sequence. ( ) defined by satisfies: i) If for some, then either or. ii) If for all,, then the sequence converges to some. When the hypothesis of the continuity of f on the -dense curve is extended to the whole K, we have the following result. Corollary Let be a sequence of -dense curves in K, with as. Under the same hypotheses of the above theorem and adding the continuity of f on K, assume that for each curve there is a positive integer such that the corresponding recursive sequence. Then the set where is the set of all zeros of f.
Alpha-Dense Curves method has been applied to minimize, to integrate and to solve inequalities involving functions of a great number of variables. These functions appear when we model some processes where the experiments are not possible,difficult or very expensive. The private enterprises that have supported research on this theory and its applications at the Laboratoire MEDIMAT de l'Université Pierre et Marie Curie, Paris VI (France) are: –Renault, –DGA, –Pharmaceutical Laboratories: Boiron, Servier and Sandoz, –Banking Sector: Crédit Agricole.
The main concrete problems which have been studied following this method were: 1. Degranulation of Basophyls, Visual Perception for Artificial Perception. 2. Study on the gluclose-insuline system (a preliminar study on the conception of an artificial pancreas). 3. Study of the Pelagic Ecosystem of an upwelling location. 4. Competition model between two different cultures and diffussion microbes in water. 5. Industrial process for clarifying apple juice. 6. Difussion of Streptadivin and DTPA-Biotin in inmmunoscintigraphy of heart in two stages. 7. Determination of temperature in the interior of oil wells. 8. Analysis of stocks corresponding to agricultural and food products. 9. Models of the combustion by hydrogen, reduction of pollution and comfortability of cars. 10.Dispersion of materials in blood plasma. 11.Transport of anticancer substances applied to the brain surface. 12.The Nitroarginine and the problem of inhibition of the NOsynthase. In the U.N.T. and the U.A. (Spain), we are applying the -Dense Curves method to parameter estimation of transition-time probability density in forest models.
The -Dense Curves as mathematical theory is of interest in itself and has links with other subjects. 6.1. Peano Curves. In 1890, G.Peano demonstrated that the interval I=[0,1] could be mapped surjectively and continuously onto the square.These curves are called Peano Curves or also Space-filling Curves. Some of them are famous such as that of Hilbert (1891),Moore (1900) and Lebesgue (1904). However a general method for generating these curves remained unsettled. H.Steinhaus in 1936 solved the problem by means of a surpresively result : " if two continuous non-constant functions on I are stochastically independent with respect to Lebesgue measure, then they are the coordinate functions of a space-filling curve ". We have proved that the condition of stochastically independence (s.i.) is too strong to characterize space-filling curves. We pointed out that there are space-filling curves whose coordinate functions are not s.i. Indeed (Mora,G. and Mira,J.A., 2001), The coordinate functions of the Lebesgue curve are not stochastically independent. For the purpose to characterize the Space-Filling Curves we introduced a filling condition (f.c.) in terms of the quasi-stochastically independent concept.
Definition (Filling condition ) We shall say that N measurable functions,..., are quasi-stochastically independent (q.s.i.), with respect to the Lebesgue measure on R (denoted ), if for any open sets. of R the condition implies Using this filling condition or q.s.i. concept, we obtained a characterization of space-filling curves by means of the following results (Mora,G. and Mira,J.A., 2001): Theorem Let,..., be quasi-stochastically independent functions ( r.L.m.). Assume also that they are continuous and nonconstant.Thus the curve defined by,..., fills the parallelepiped. Conversely Theorem Let,..., be nonconstant continuous functions such that the curve,..., fills the parallelepiped Thus,..., are quasi-stochastically independent (r.L.m). In fact, the idea of filling condition was already, implicitly, handled when we gave a characterization of -dense curves in parallelepipeds of (Mora,G. and Cherruault, 1997). We also settled a characterization theorem on the filling condition in terms of Borel measures. Moreover, this result yields a method to construct the coordinate functions of a space-filling curve (Mora,G. and Mira,J.A., 2001).
Theorem Assume are Q.S.I continuous nonconstant functions and H.Then the set function on the class of all cubes contained in H, defines a Borel measure for which Reciprocally, any Borel measure on a parallelepiped satisfying (1) defines N continuous nonconstant functions that are Q.S.I. We recently gave a characterization of the space-filling curves by uniform limits of - dense curves in the compact that is filled.The -dense curves must have densities tending to zero and coordinate functions with variation tending to infinity as tends to zero (Mora, 2005). More precisely, the result is. Theorem A continuous mapping is a Peano curve filling if and only if is the uniform limit of a sequence of -dense curves in with densities, for which there is no constant M such that the variation, for all n, for some.
6.2. -Dense Curves of the Minimal Length. In a densifiable compact K of a metric space (E,d), the existence of an -dense curve, for fixed >0, with a minimal length, has been solved. We have used the set that is, the set of all , -dense in K, satisfying for some and. When the set will be denoted, and we shall consider that as a subset of the space of all continuous mappings on I and valued in E, endowed with the metric Because of the Ascoli's theorem, the compactness of is proved in the following result. Lemma Let (E,d) be a metric space and K a densifiable compact. Then, is a compact in the space. for the metric. Now, the existence of -dense curves with minimal length follows (Ziadi,Cherruault and Mora, 2000). Theorem Let K be a densifiable compact of a metric space (E,d) and the set of all -dense curves in K, >0. Assume contains at least a rectifiable curve, thus there exists such that,where is the length of .
6.3.Functional equations generating -dense curves. A surpresive connection between -dense curves and functional equations emerged when we were trying to solve one of these. Such is the case, for instance, of the functional equations and whose continuous solutions can densify, with arbitary small density, compacts of the plane. Indeed, in this manner we have proved the following results.(Mora, Cherruault and Ziadi,2000) Theorem Let p { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/12/3428488/slides/slide_25.jpg", "name": "6.3.Functional equations generating -dense curves.", "description": "A surpresive connection between -dense curves and functional equations emerged when we were trying to solve one of these. Such is the case, for instance, of the functional equations and whose continuous solutions can densify, with arbitary small density, compacts of the plane. Indeed, in this manner we have proved the following results.(Mora, Cherruault and Ziadi,2000) Theorem Let p
Open problem with respect to the functional equations and -Dense Curves. We have just seen how special compacts have been densified by continuous solutions of the preceding functional equations. Then, except traslation and dilation of the compact set, our problem is: Problem In, given a densifiable compact K having by boundary a path, is there a functional equation such that its continuous solutions can densify it ? 6.4. The -Dense Curves in Topological Vector Spaces. Whenever E is a topological vector space,not necessarily metrizable, the concept of -density is substituted by that of the V-localization, V being a given 0-neighborhood. This concept is formalized by means of the following definition. Definition We say that a subset B of a topological vector space is V-localized by a curve, i.e., by a continuous mapping,, if and only if
Definition A subset B of a topological vector space is said to be localizable if and only if is V-localized by any 0-neighborhood V. A nontrivial example of localizable set in a nonmetrizable topological vector space is given in the following result. (Mora, 2005): The space, or briefly, of all real measurable (Lebesgue) functions on satisfying endowed with the weak topology, is not metrizable. In this space, the probability functions set where, is localizable. For the Banach spaces we have the following result about the localization concept and the dimensionality (G.Mora & J.A. Mira, 2003). Theorem In a Banach space E, the unit ball is localizable if and only if E is finite-dimensional. Let E be a normed space and B its unit ball. Each curve in B, has a density (minimal) given by where denotes and is the metric induced by the norm. Then, considering the set of all curves contained in B, denoted, we can define the degree of densificability of B as the number
Observe that for any normed space and, from the preceding theorem, in a Banach space, if and only if the space has finite dimension.Then if the space has infinite dimension it follows that. Now, a natural question is: does it exist a normed space for each value between 0 and 1? The answer is given by the following result (Mora and Mira, 2003). Theorem For any normed space, the degree of densificability of its unit ball can take exactly two values, either 0 or 1. Now, we obtain a well-known classical theorem as corollary the Riesz theorem on the dimension: Corollary Any normed space locally compact is finite dimensional.
1.Cheney, E.W., Introduction to Approximation Theory, Chelsea Publ. Company, New York, 1982. 2.Cherruault,Y. and Mora,G.,Optimisation Globale. Theorie des Courbes - Denses, Economica, Paris, 2005. 3.Choquet,G., Cours D'Analyse-Topologie, Masson et Cie, Paris, 1971. 4.Cichon, J. and M. Morayne, M., On Differentiability of Peano Type Functions III, Proc. Am. Math. Soc., 92, (1984), 432-438. 5.Davis, P.J., Interpolation and Approximation, Dover, New York, 1975. 6.Hahn, H., Über stetige Streckenbilder, Atti del Congreso Internazionale dei Mathematici, Bologna, 3-10 September, VI, (1928), 217-220. 7.He, T.X., Dimensionality Reducing Expansion of Multivariable Integration, Birkhäuser, Boston,2001. 8.Hilbert, D., Über die stetige Abbildung einer Linie auf ein Flächenstück, Math. Annln.38, (1891), 459-460. Holbrook, J.A., Stochastic independence and space- filling curves, Amer, Math. Monthly, 88, (6),(1981), 426-432. 9.Kelley, J.L., General Topology, D. Van Nostrand, New York,1955. 10.Köthe,G., Topological Vector Spaces I, Springer-Verlag, Berlin,1969.
11.Kharazishvili, A.B., Strange Functions in Real Analysis, Dekker,New York, 2000. 12.Lebesgue, H., Leçons sur L'intégration et la Recherche des Fonctions Primitives, Gauthier-Villars, Paris, 1904. 13. Lebesgue, H., Sur les fonctions représentables analytiquement, J.de Math., 6, 1 (1905), 139-216. 14.Mazurkiewicz, St., Travaux de Topologie et ses Applications, PWN-Polish Scientific Publishers, Warszawa, 1969. 15.Moore, E.H., ON certain crinkly curves, Trans. Amer. Math. Soc., 1, 72-90, (1900). 16.Mora,G., and Cherruault,Y., Characterization and Generation of -Dense Curves, Computers Math. Applic. Vol.33 No.9(1997), 83-91. 17.Mora, G. and Cherruault,Y., The Theoretical Calculation Time associated to - Dense Curves, Kybernetes, Vol. 27, No.8, (1998), 919-939. 18.Mora, G.and Cherruault, Y., An approximation method for the optimization of continuous functions of variables by densifying their domains, Kybernetes, 28, 2 (1999), 164-180. 19.Mora, G. and Cherruault, Y., On the minimal length curve that densifies the square, Kybernetes, 28, 9(1999), 1054-1064. 20.Mora, G., Optimization by space-densifying curves as a natural generalization of the Alienor method, kybernetes,Vol. 29, No.5/6, (2000), 746-754. 21.Mora, G., Cherruault, Y. and Ziadi, A., Functional equations generating space- densifying curves, Computers and Math. with Applic., 39, (2000), 45-55.
22.Mora,G. and Mira, J.A., A characterization of space-filling curves, Rev. R. Acad. Cien. Serie A. Mat., Vol. 96 No.1,(2001), 45-54. 23.Mora, G., Cherruault, Y., Benabidallah, A. and Tourbier, Y., Approximating Multiple Integrals via -Dense Curves, Kybernetes, Vol. 31 No.2 (2002), 292- 304. 24.Mora, G., Benavent, R. and Navarro, J.C., Polynomial Alpha-Dense Curves and Multiple Integration, Inter. Journal of Comput. and Num. Anal. and Applic., Vol. 1 No.1 (2002), 55-68. 25.Mora, G, Cherruault, Y and Benabidallah, A., Global Optimization-Preserving Operators, Kybernetes, Vol. 32, No.9/10, (2003), 1473-1480. 26.Mora,G. and Mira,J.A., Alpha-Dense Curves in Infinite Dimensional Spaces, Inter. Journal of Pure and Applied Math. Vol.5, No.4 (2003), 437-449. 27.Mora,G., Cherruault,Y. and Ubeda,J.I., Solving Inequalities by -Dense Curves. Application to Global Optimization, Kybernetes, Vol.34, No. 7/8, (2005), 983-991. 28.Mora,G. and Mora-Porta,G., Dimensionality Reducing Multiple Integrals by Alpha-Dense Curves, Inter. Journal of Pure and Applied Math., Vol.22,No.1, (2005), 105-116. 29.Mora,G., The Peano Curves as Limit of -Dense Curves, RACSAM, Vol.99 (1), (2005), 23-28.
30.Mora,G., Minimizing Multivariable Functions by Minimization-Preserving Operators, Mediterr. J. Math. 2 (2005), 315-325. 31.Morayne,M., On Differentiability of Peano Type Functions, Colloquium Mathematicum, LIII, (1987),129-132. 32.Morayne, M., On Differentiability of Peano Type Functions II, Colloquium Mathematicum, LIII, (1987),133-135. 33.Natanson, I.P., Theory of Functions of a Real Variable, Vol. I, II, Frederick Ungar Publishing Co., New York, 1964. 34.Peano, G., Sur une courbe qui remplit toute une aire plaine, Math. Annln., 36, (1890), 157-160. 35.Sagan,H., Space-Filling Curves, Springer-Verlag, New-York, 1994. 36.Robertson,A.P. and Robertson,W.J., Topological Vector Spaces, Cambridge University Press, Cambridge, 1973. 37.Steinhaus, H., La courbe de Peano et les fonctions indépendantes, C. R. Acad. Sci. Paris 202, 1961-1963 (1936). 38.Tricot, C., Curves and Fractal Dimension, Springer-Verlag, New York, 1995.
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https://math.stackexchange.com/questions/1315180/recurrence-relation-with-characteristic-equation-that-has-only-1-root-and-comple | # Recurrence relation with characteristic equation that has only 1 root and complex roots
For the recurrence relation:
$f_n = 2a_{n-1} - 2a_{n-2}$ I got the characteristic equation that had complex roots:
$x^2 - 2x + 2 = 0$ that gave roots $i, -i$ and I wasn't sure how to continue the solving the recurrence relation with the complex roots.
Another problem I found was with the recurrence relation:
$a_n = 4a_{n-1} - 4a_{n-2}$
I got the characteristic equation that had only 1 root:
$x^2 - 4x + 4 = 0$
$r_1$ = 2
And I wasn't sure what form to use for the general solution as it's typically of the form $a_n = c_1 * (r_1)^n + c_2 * (r_2)^n$
$x^{2}-2x+2=(x-1)^{2}+1$, and that roots $1-i$ and $1+i$. Now in general, if your characteristic eqution with real coeffiecients has two imaginary roots, they are conjugate, and can be written as $re^{i \theta}$ and $re^{-i \theta}$,$r>0$ and $\theta \in ]0,2\pi[$ then $a_{n}=k_{1}r^{n} \cos n\theta+k_{2}r^{n}\sin n\theta$. Now, if you have a real double root $r_{0}$, $a_{n}=k_{1}r_{0}^{n}+k_{2}nr_{0}^{n}$, in the case of a degree 2 real equation! (If you've seen ODEs, you will notice a lot of similarities).
• For the case in my question where I have only 1 root could I just leave it as: $a_n = c_1 r_0 ^n$ since I don't have the second root? – joe Jun 7 '15 at 1:51
• It's $a_{n}=c_{1}r_{0}^{n}+c_{2}nr_{0}^{n}=2^{n}(c_{1}+c_{2}n)$ – mich95 Jun 7 '15 at 1:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.948832631111145, "perplexity": 279.4821741914738}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107888931.67/warc/CC-MAIN-20201025100059-20201025130059-00424.warc.gz"} |
http://math.stackexchange.com/questions/188837/for-n-pq-show-that-there-exist-alpha-beta-in-z-n-such-that-alpha | # For $n=pq$, show that there exist $\alpha, \beta \in Z_n^*$ such that $\alpha, \beta \notin (Z_n^*)^2$ and $\alpha \cdot \beta \notin (Z_n^*)^2$.
For $n=pq$, where p,q are distinct odd primes show that there exist $\alpha, \beta \in Z_n^*$ such that $\alpha, \beta \notin (Z_n^*)^2$ and $\alpha \cdot \beta \notin (Z_n^*)^2$.
Here if we try to use Chinese Remainder theorem results then let :
$$\alpha \equiv \alpha_p\mod p\,\,,\qquad\alpha \equiv \alpha_q\mod q$$
$$\beta \equiv \beta_p\mod p\,\,,\qquad\beta \equiv \beta_q\mod q$$
Since $\alpha \notin (Z_n^*)^2$, what can we say about $\alpha_p$ and $\alpha_q$? Do $\alpha_p \notin (Z_p^*)^2$ and $\alpha_q \notin (Z_q^*)^2$ also hold? Is there another approach to solve this problem?
PS: Couldn't insert "p,q are distinct primes" in title due length restriction.
-
Hint: When is a number a square mod $pq$? – Thomas Andrews Aug 30 '12 at 15:11
I replaced your \dot with \cdot. You could also do without a dot altogether, as multiplication is then assumed by default. – Jyrki Lahtonen Aug 30 '12 at 15:17
Thomas, are you referring to the result that if $\alpha \in (Z_{pq}^*)^2$ then $\alpha \in (Z_p^*)^2$ and $\alpha \in (Z_q^*)^2$? – Abhishek Anand Aug 30 '12 at 15:43
Hint: Let $\alpha_p$ be a non-square modulo $p$ (quadratic non-residue of $p$). Then there exists $\alpha$ such that $\alpha\equiv \alpha_p\pmod{p}$ and $\alpha\equiv 1\pmod{q}$. Produce $\beta$ in an analogous way.
Added: Then $\alpha\beta\equiv \alpha_p \pmod{p}$ and $\alpha\beta\equiv \beta_q\pmod{q}$.
Now we need to show that none of $\alpha$, $\beta$, or $\alpha\beta$ is a square modulo $pq$.
Chinese Remainder Theorem if you are in number theory. Fact that $\mathbb{Z}_{pq}^\ast$ is the direct product of $\mathbb{Z}_p^\ast$ and $\mathbb{Z}_q^\ast$ if you are more group-theory oriented. – André Nicolas Aug 30 '12 at 16:11
So what does it say about $\alpha$, having a quadratic non-residue in p and quadratic residue of 1 in q? My understanding of CRT is a little bit sketchy. Does it imply $\alpha$ is a quadratic non-residue of pq? – Abhishek Anand Aug 30 '12 at 16:32
Yes. If $x^2\equiv \alpha \pmod{pq}$, then $x^2\equiv \alpha\pmod{p}$. But $\alpha\equiv \alpha_p\pmod{p}$, so $x^2\equiv \alpha_p\pmod{p}$, contradicting the fact that $\alpha_p$ is a quadratic non-residue of $p$. – André Nicolas Aug 30 '12 at 16:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.985929012298584, "perplexity": 240.48609779279187}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422115869320.77/warc/CC-MAIN-20150124161109-00115-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://www.greenergyexpo.eu/moisture-removal-efficiency/ | # Moisture removal efficiency
Moisture Removal Efficiency (MRE) is a measure of the energy efficiency of any dehumidification process. Moisture removal efficiency is the water vapor removed from the air temperature and humidity, divided by the total energy consumed by the dehumidification equipment during the same time period, including all the fan and pump energy needed to move air and fluids through the system . Water vapor removal is expressed as pounds or kilograms. Energy is usually expressed as kilowatt hours. Inlet air temperature is expressed in degrees Fahrenheit or degrees Celsius. Inlet air humidity may be expressed in several ways, most commonly in the air; the weight of water vapor in the air, compared to the weight of the air that contains it. An example of the MRE of a dehumidification system could be: 4. 4 lb / kWh @ 85 ° F, 140 gr / lb. Using the SI system of units, that same MRE would be 2.0 kg / kWh @ 30 ° C, 20.0 g / kg.
The authority for this definition of moisture removal is ANSI / AHRI Standard 920, published by the Air Conditioning, Heating and Refrigeration Institute (AHRI). AHRI is an association of equipment manufacturers. Standard 920 was developed by the group of manufacturers of AHRI that is concerned with energy efficiency of outdoor equipment (DOAS equipment). The standard provides an objective and quantitative basis for rating and comparison of dehumidification performance and energy consumption of dehumidifiers and ventilation as part of a building’s mechanical systems. Dehumidification of ventilation and makeup air consumes a great deal of energy worldwide. In all except desert and high-altitude climates, the annual dehumidification load for ventilation the annual cooling load for ventilation by a factor of 2 to 5. The issue of energy efficiency of dehumidification arises because of the requirements of energy efficiency. According to the members of AHRI’s dehumidification product section, the reason for developing Standard 920 was to provide ASHRAE with a neutral and high-quality test method for dehumidification energy efficiency. The committee’s goal has been ASHRAE include minimum dehumidification efficiency requirements in ASHRAE Standard 90.1. That standard is often used in the United States and Canada,
To provide a metric that allows an estimation of the annual dehumidification dehumidification dehumidification dehumidification systems, AHRI Standard 920 further defines a test protocol for measuring a system of integrated seasonal moisture removal efficiency. This metric is similar to the seasonal energy efficiency ratio (SEER) that is used to compare cooling energy efficiency of competing equipment. Like the SEER test methodology, AHRI Standard 920 requires that the dehumidification equipment be tested at several discrete combinations of inlet air temperature and humidity. In the 2015 edition of the standard, these test conditions are defined. The moisture removal efficiency of the equipment is measured at each of these conditions, and then the proportionality factor is applied to each MRE.
ASHRAE ASHRAE ASHRAE Standard 90.1 ASHRAE 90.1 AHRI Air Conditioning, Heating and Refrigeration Institute Dehumidifier Dehumidifier SEER Energy efficiency ratio LEED Leadership in Energy and Environmental Design | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8897473216056824, "perplexity": 2425.104378557655}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296950110.72/warc/CC-MAIN-20230401160259-20230401190259-00278.warc.gz"} |
https://opennursingjournal.com/VOLUME/11/PAGE/211/FULLTEXT/ | # High Agreement and High Prevalence: The Paradox of Cohen’s Kappa
Slavica Zec1, Nicola Soriani1, Rosanna Comoretto2, Ileana Baldi1, *
1 Department of Cardiac, Thoracic and Vascular Sciences, Unit of Biostatistics, Epidemiology and Public Health, University of Padova, Padova, Italy
2 Department of Statistics and quantitative methods, University of Milan, Bicocca, Italy
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open-access license: This is an open access article distributed under the terms of the Creative Commons Attribution 4.0 International Public License (CC-BY 4.0), a copy of which is available at: https://creativecommons.org/licenses/by/4.0/legalcode. This license permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
* Address correspondence to this authors at the Department of Cardiac, Thoracic and Vascular Sciences, University of Padova, Via Loredan, 18, 35131 Padova, Italia; Tel: +390498275403; E-mail: [email protected]
## Abstract
### Background:
Cohen's Kappa is the most used agreement statistic in literature. However, under certain conditions, it is affected by a paradox which returns biased estimates of the statistic itself.
### Objective:
The aim of the study is to provide sufficient information which allows the reader to make an informed choice of the correct agreement measure, by underlining some optimal properties of Gwet’s AC1 in comparison to Cohen’s Kappa, using a real data example.
### Method:
During the process of literature review, we have asked a panel of three evaluators to come up with a judgment on the quality of 57 randomized controlled trials assigning a score to each trial using the Jadad scale. The quality was evaluated according to the following dimensions: adopted design, randomization unit, type of primary endpoint. With respect to each of the above described features, the agreement between the three evaluators has been calculated using Cohen’s Kappa statistic and Gwet’s AC1 statistic and, finally, the values have been compared with the observed agreement.
### Results:
The values of the Cohen’s Kappa statistic would lead to believe that the agreement levels for the variables Unit, Design and Primary Endpoints are totally unsatisfactory. The AC1 statistic, on the contrary, shows plausible values which are in line with the respective values of the observed concordance.
### Conclusion:
We conclude that it would always be appropriate to adopt the AC1 statistic, thus bypassing any risk of incurring the paradox and drawing wrong conclusions about the results of agreement analysis.
Keywords: Agreement statistics, Cohen's Kappa, Gwet’s AC1, Concordance analysis, Inter-rater agreement, Quality assessment of RCT.
## 1. INTRODUCTION
The analysis of intra- and inter-observer agreement is applied in many areas of clinical research [1-4]: from the diagnosis to evaluation of quality of experimental studies [5, 6]. As for the latter, the literature is unanimous in considering that low-quality trials, conducted using inadequate methodological approach, are often associated with the over-estimated treatment effects [5, 7]. These distortions can lead to errors at every level of decision making in health care, from individual treatment to definition of national public health policies. Quality assessments of trials are generally conducted by different parties (raters or evaluators) who are asked to verify, through appropriate checklists or scales [8-12], if the studies meet the predefined quality criteria. The agreement analysis, in these cases, does not only have the purpose to establish the reproducibility of the evaluations but, above all, to provide information about the role of the subjective component in definition of classifications and scores. It is important to note that the evaluation of the subjective component in rating is closely linked to sociometric and psychometric research field, from which the concordance measures originated in the first place [13-15].
The Cohen’s Kappa statistic [16] is the most used agreement measure in literature. This statistic does not have absolute applicability since it suffers from a particular paradox already known in literature [17-19]. Under special conditions [20, 21] and even in presence of a strong inter- or intra- rater agreement, the Kappa statistic tends to assume low values, often leading to conclude that no agreement is present. Consequently, the use of the Kappa statistics in presence of this paradox tends to affect the findings in terms of real reproducibility of measurement operations or lead to biased assessment results.
Among the alternative agreement measures to the Cohen’s Kappa [22-24], the statistic known as Agreement Coefficient 1 (AC1) given by Gwet [25] has proven to be most robust to this paradox [20, 21].
The purpose of this work is to provide sufficient information which allows the reader to make an informed choice of the correct agreement measure.
In the following sections Cohen’s kappa statistic will be introduced in its general formulation, with more than two categories and more than two evaluators, and conditions that lead to the paradox will be briefly described. The statistic AC1 will be subsequently introduced. Finally, a working sample, drafted from a reproducibility study among the evaluators of the quality of a clinical trial, will be used to show the behavior of the two statistics - both in presence and absence of the paradox.
### 1.1. The Cohen’s Kappa Statistic
In order to recall the concept and the construction of Cohen's Kappa statistic, let us suppose that we intend to compare the classifications of N subjects performed by R evaluators concerning K possible outcome categories (Table 1). The generic Rij indicates the number of evaluators that allocate the subject i to the category j.
Table 1. Distribution of N subjects for R raters and K outcomes.
Outcome
Subject
1 2 ….. K Total
1 R11 R12 ….. R1K R
2 R21 R22 ….. R2K R
…..
N RN1 RN2 ….. RNK R
Total R+1 R+2 ….. R+K N * R
The Kappa statistic, as well as other statistics of the same type [22-24], measure the concordance in data as a part of the agreement that cannot be observed due to mere chance and is defined [16] as:
(1)
in which:
(2)
is the agreement observed in the data, while the expected agreement in case of random assignment is given by:
(3)
The term pkij, for j=1,…, r, represents the portion of the subjects allocated to the category k by the evaluator j. The expression [3] is referring to the extension of Cohen’s Kappa to a more general case with more than two evaluators and more than two categories [26].
The statistics can assume any value from and 1. Values greater than 0.6 are considered as indicators of high agreement, while values inferior to 0.4 or negative are indicators of discordance [27].
The paradox undermines the assumption that the value of the Kappa statistic increases with the agreement in data. In fact, this assumption is weakened - sometimes even contradicted - in presence of strong differences in prevalence of possible outcomes [17]. These conclusions stem from sensitivity studies [20, 21], conducted for the case with two evaluators and two categories, who have analyzed the behavior of the Kappa statistic considering various interactions between the prevalence of outcomes in population, and the sensitivity and the specificity of evaluators (where sensitivity and specificity are defined as the probabilities that the evaluators correctly allocate a subject in one of the outcomes). Sensitivity studies have shown that the effects of the paradox arise in the presence of the outcomes with very high prevalence and/or considerable differences in classification probabilities. The paradox, in other words, is present when the examined subjects tend to be classified to one of the possible outcomes. This is either due to the nature the outcome itself and its high prevalence, or because at least one of the evaluators tends to assign more frequently to one specific outcome.
### 1.3. AC1 Statistic
The statistic AC1 has been proposed by Gwet [25] as an alternative agreement measure to Cohen’s Kappa statistic. According to Gwet [20], the reason why the Kappa statistic is exposed to the paradox lies in the inadequacy of the formula (3) for the expected agreement calculation.
Intuitively, the formulation of the statistic AC1 [25, 28] is rather similar to Cohen’s Kappa statistic:
(4)
in which the observed agreement Pa is defined exactly as in the expression (2), while the expected agreement is defined as:
(5)
where . It is defined in a way that it cannot assume values higher than 0.5 [20], even if a part of the evaluators classifies in a completely random manner, without any consideration of the characteristics of the subjects.
The variance of the AC1 statistics, indispensable for the construction of confidence intervals, is calculated through the expression (3), following Gwet [28].
## 2. METHODS
### 2.1. Case Study: Reproducibility of the Evaluation of Clinical Trial Quality
During the process of literature review [29], we have asked a panel of three evaluators to come up with a judgment on the quality of 57 randomized controlled trials (RCTs), assigning a score to each trial using the Jadad scale [9]. This scale assigns a score from zero to five to a trial and evaluates presence and adequacy of the double-blind design, presence and adequacy of randomization and a possible loss of subjects during the study. An RCT is considered of good quality if it gets a score equal to or greater than 3. To explore some design aspects, the evaluators were asked to classify the trial depending on the type of randomization unit (individual or community), the type of design adopted (parallel, factor or crossover) and the type of the primary endpoint (binary, continuous, survival or other). The classifications of the three evaluators are shown in Table 2, where the Jadad score was dichotomized, distinguishing between good (> 3), and poor (<3) quality trial.
Table 2. Results of the ratings carried out by the three raters on the characteristics investigated in the study.
Variable Evaluator 1 Evaluator 2 Evaluator 3
Unit
Community 4 0 6
Individal 53 57 51
Design
Crossover 2 2 4
Factorial 9 3 8
Parallel 46 52 45
Primary Endpoint
Binary 8 2 13
Continuous 42 31 43
Survival 3 7 1
Other 2 9 0
Not specified 2 8 0
<3 22 24 25
≥3 35 33 32
## 3. RESULTS
The graphs shown in Fig. (1) describe the effect of the paradox on Cohen's Kappa statistic. The curves, shown in black in Fig. (1), are the values of the Kappa statistic as a function of prevalence, considering different scenarios for different levels of agreement and observed sensitivity and specificity of the evaluators. Following the sensitivity studies [20, 21], the curves of Fig. (1) assume that the two evaluators have the same values for sensitivity and specificity and that these values coincide. As we can see, in all scenarios considered (hence independent on the observed correlation values, sensitivity and specificity) the paradox begins to be evident for values of prevalence higher than 60%.
On the other hand, AC1 statistic (whose values are shown in red) appears more robust under the paradox conditions. The values of the AC1 statistics are in line with the observed correlation values, hence do not seem to be particularly affected by the prevalence level.
With respect to each of the above described features, the agreement between the three evaluators has been calculated. Table 3 shows the observed agreement (Pa), the Cohen’s Kappa statistic (γk), the statistic AC1 (γ1), and their respective confidence intervals at 95%.
Table 3. Observed agreement (Pa), Cohen's Kappa (γk), AC1 (γ1) and their 95% confidence intervals computed on the ratings of the three raters.
Pa γk γ1
Randomization unit 0.842 ( 0.747 -- 0.937 ) 0.042 ( -1.000 -- 1.000 ) 0.881 ( 0.725 -- 1.000 )
Design 0.719 ( 0.603 -- 0.836 ) 0.230 ( -0.713-- 1.000 ) 0.781 ( 0.682 -- 0.880 )
Primary endpoint 0.386 ( 0.260 -- 0.512 ) 0.107 ( -0.203 -- 0.417 ) 0.470 ( 0.439 -- 0.502 )
Jadad 0.871 ( 0.819 -- 0.924 ) 0.735 ( 0.377 -- 1.000 ) 0.750 ( 0.746 -- 0.754 )
Fig. (1). Cohen's Kappa (black lines) and AC1 (red lines) values computed by increasing the prevalence. The curves refer to several values of observed agreement (Pa), and raters’ sensitivity and specificity. It is assumed that sensitivity and specificity values are equal and the same for both the raters.
The values of the Cohen’s Kappa statistic would lead to believe that the agreement levels for the variables Unit, Design and Primary Endpoints are totally unsatisfactory. However, a simple "glance" with the relative values of the observed concordance is enough to highlight the presence of paradox. The most likely explanation for the onset of the paradox can be given by high values, shown in Table 2, taken from the levels "Individual", "Parallel" and "Continuous" for variables Unit, Design and Primary Endpoint. These values have led to high probability of classification and hence to paradox affected values of Kappa statistic. The AC1 statistic, on the contrary, shows plausible values which are in line with the respective values of the observed concordance.
For the Jadad variable, we can observe that in the absence of paradox, the Kappa statistic and AC1 have quite similar values which are both consistent with the observed concordance.
## 4. DISCUSSION
In this study, the intention was to briefly present and discuss a paradox that afflicts a concordance measure widely used in literature. As we have previously pointed out, the risk to encounter this paradox should be taken into account by the researcher who uses Cohen’s Kappa statistic in order to adequately tailor agreement analysis. Even in simple cases with only two evaluators and two outcomes, the paradox tends to occur if, at equal sensitivity and specificity of the evaluators, the prevalence of one of the results is above 60%, as seen in Fig. (1) graphs. Consequently, it is reasonable to assume that if we are dealing with a setting in which one of the outcomes has prevalence levels over 60%, then Kappa statistic might lead to biased conclusions and hence it is more suitable to use an alternative agreement statistic, such as AC1, less sensitive to this problem.
The AC1 statistic is not the only one that presents robustness properties to the paradox. The Alpha Aickin statistic [24] is another tool that has very similar properties to the AC1 [30]. In this study we have chosen to focus on the AC1 statistic since it is comparable with the Cohen's Kappa from the conceptual point of view [30] and computationally less intensive than of Alpha Aickin.
The use of AC1 statistics would also be advisable in all cases in which the evaluators are subject to a high probability of classification to one of the possible outcomes. In this case it is crucial to distinguish between the prevalence and the probability of classification. Prevalence is the probability (in many cases unknown) that an individual chosen at random from the population presents a specific level/category of an outcome. The probability of classification is a subjective propensity of the evaluators to assign to a particular outcome. This means that there exist different sources of paradox and that not always high prevalence follows high probability of classification and vice versa. This aspect can be observed in the example from the previous section, in which the high values are both expression of high prevalence, as for the variable Unit where it is reasonable that the "Individual" level is predominant compared to the level "Community", but also result from the fact that for the variable Design, the evaluators did not have sufficient expertise to distinguish less common designs compared to that of "Parallel" type.
Even in the absence of the paradox, as in the example of Jadad score, the AC1 statistics provides absolutely consistent values and overlapping with the Cohen’s Kappa, which confirms the results found in the literature [21, 28].
## CONCLUSION
On the basis of literature review and case study findings, we can conclude and suggest to the reader that it might always be appropriate to adopt the AC1 statistics, thus bypassing any risk of incurring the paradox and drawing wrong conclusions about the results of agreement analysis.
## LIST OF ABBREVIATIONS
AC1 = Agreement Coefficient 1 RCT = Randomized Control Trial
Not applicable.
### HUMAN AND ANIMAL RIGHTS
No Animals/Humans were used for studies that are base of this research.
Not applicable.
### CONFLICT OF INTEREST
The authors declare no conflict of interest, financial or otherwise.
Declared none. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8380305171012878, "perplexity": 1530.0018793854217}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500041.18/warc/CC-MAIN-20230202200542-20230202230542-00389.warc.gz"} |
https://arxiv.org/abs/1911.04901?context=physics | # Title:Notes on glottal flow and acoustic inertial effects
Abstract: This text is a compilation of some of the notes that the author has written during the development of the low-order model "DICO" [2, 8, 10, 11] for vowel phonation and the even more rudimentary glottal flow model [9] for processing high-speed glottal video data. The following subject matters are covered: (i) Incompressible, laminar, lossless flow models for idealised rectangular and wedge shape vocal fold geometries. Equations of motion and the pressure distribution are computed in a closed form for each model using the unsteady Bernoulli's theorem; (ii) The assumption of incompressibility and energy loss (i.e., irrecoverable pressure drop) of the airflow in airways (including the glottis) is discussed using steady compressible Bernoulli theorem as the main tool; (iii) Inertia of an uniform waveguide is studied in terms of the low-frequency limit of the the (acoustic) impedance transfer function. It is observed that the inductive loading in the boundary condition sums up with the waveguide inertance in an expected way; (iv) It is shown that an acoustic waveguide, modelled by Webster's lossless equation with Dirichlet boundary condition at the far end, will produce the expected mass inertance of the fluid column as the low-frequency limit of the impedance transfer function.
Subjects: Fluid Dynamics (physics.flu-dyn) Cite as: arXiv:1911.04901 [physics.flu-dyn] (or arXiv:1911.04901v1 [physics.flu-dyn] for this version)
## Submission history
From: Jarmo Malinen Dr. [view email]
[v1] Mon, 11 Nov 2019 14:34:01 UTC (29 KB) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8130614757537842, "perplexity": 2781.047980915213}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146186.62/warc/CC-MAIN-20200226023658-20200226053658-00444.warc.gz"} |
https://www.physicsforums.com/threads/magnetic-field-with-loop-and-straight-wire-help.64736/ | # Homework Help: Magnetic Field With Loop and Straight Wire Help!
1. Feb 23, 2005
An infinitely long wire is formed as in the diagram on your assignment. It is formed so that it has a circular portion of R = 50.0 cm and the straight portion is located at a distance r from the center of the circular portion. Find r (in cm) such that the net magnetic field at the center of the circular portion is zero.
Ok so I am confused on this problem because there is no current mentioned, and all the equations giving involve current. Then I thought that the circle could be split in two and the force of the two halves would then have to equal zero but that seems wrong as well. Please help I just don't understand :(.
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2. Feb 23, 2005
### MathStudent
I believe the arrows in the diagram are meant to be the direction of the current.
Since its all just one long wire, the curent everywhere is equal.
You can treat these (line / circle) as separate objects.
Last edited: Feb 23, 2005
3. Feb 23, 2005
### Davorak
There is only one wire, so I think there will be only one current. The magnitude of the current will not matter in the end or they would have told it to you.
Try solving the problem by breaking it up into a circular piece and a straight wire piece.
Calculate the magnetic field each causes in the center of the loop.
By adjusting the R the radius of the loop you should be able to get the magnetic field to equal zero.
Does this make the problem clearer?
edit:
MathStudent beat me to the punch. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8137522339820862, "perplexity": 461.47395745253647}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794864725.4/warc/CC-MAIN-20180522112148-20180522132148-00071.warc.gz"} |
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Dec10 comment Is there a disadvantage to using natbib=true with biblatex? as far as I know, the natbib option with biblatex only provides the commands from natbib, the actual processing is done by biblatex and is not influenced by the option. Dec9 reviewed Approve How to delete web links from references in Gost style? Dec8 awarded Good Answer Dec6 revised Table does not fit removed longtable tag Dec5 comment Serious problems with tabularx table layout your code doesn't compile for several reasons however you are not saying it doesn't so I am not entirely sure what your problem actually is. for information, with the code provided: you are trying to use tabularx but are loading the tabu package; you are using toprule/(c)midrule/bottomrule without loading the booktabs package. Nov26 comment The - (dash) disappears in document url It might also help to know which pdf viewer you are using. Nov13 comment Bundle PDF+LaTeX source and open with Preview There is a package that was released on ctan last month (embedall) that could provide an acceptable solution to your problem. I haven't tried it myself but was very interested to see it. Nov13 revised Copying title style corrected spelling/grammar Nov8 comment Table width and Header Also note that the any text after a % character is considered a comment and not parsed. you should replace the instances of % with the escaped version \%. Nov5 revised how to compile bbl file in command prompt fixed typos in title Oct28 comment Table way out of margin have you tried the solutions from these questions: tex.stackexchange.com/questions/71419/how-to-fit-a-wide-table or tex.stackexchange.com/questions/16582/… Oct28 revised Table way out of margin not related to longtable. here, the table is wide. Oct28 revised Table way out of margin formatted code properly Oct24 awarded Pundit Oct24 awarded Nice Answer Oct18 comment Dialect of the [1.] suffix in \begin{enumerate}[1.]? @daleif I had forgotten about that option! nice one. Oct18 comment Dialect of the [1.] suffix in \begin{enumerate}[1.]? alternatively you can load the enumitem package and use [label=\arabic*.] instead of [1.] Sep19 awarded Nice Answer Sep2 reviewed Approve how to include subfigures in list of figures or sub tables in the list of tables Aug30 comment pdflatex without ghostscript to preserve transparency of included pdfs Well, it may indeed switch transparency on, but it does it in such a way that display gets rather messed up on pages with transparency elements (font become heavier and look not anti aliases. Adding the \pdfpageattr line make the document uniform. I suspect the default is to change the transparency setting on a page to page basis which for some reasons adobe reader cannot render properly. As for the advice to increase the pdf version, I was only doing that to match the highest version of the included elements. The other solution would be in reduce the pdf version of the image at creation. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8196479082107544, "perplexity": 4701.004470890664}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207926736.56/warc/CC-MAIN-20150521113206-00248-ip-10-180-206-219.ec2.internal.warc.gz"} |
http://cms.math.ca/10.4153/CMB-1997-013-7 | Canadian Mathematical Society www.cms.math.ca
location: Publications → journals → CMB
Abstract view
# Continuous Self-maps of the Circle
Published:1997-03-01
Printed: Mar 1997
• J. Schaer
Format: HTML LaTeX MathJax PDF PostScript
## Abstract
Given a continuous map $\delta$ from the circle $S$ to itself we want to find all self-maps $\sigma\colon S\to S$ for which $\delta\circ\sigma = \delta$. If the degree $r$ of $\delta$ is not zero, the transformations $\sigma$ form a subgroup of the cyclic group $C_r$. If $r=0$, all such invertible transformations form a group isomorphic either to a cyclic group $C_n$ or to a dihedral group $D_n$ depending on whether all such transformations are orientation preserving or not. Applied to the tangent image of planar closed curves, this generalizes a result of Bisztriczky and Rival [1]. The proof rests on the theorem: {\it Let $\Delta\colon\bbd R\to\bbd R$ be continuous, nowhere constant, and $\lim_{x\to -\infty}\Delta(x)=-\infty$, $\lim_{x\to+\infty}\Delta (x)=+\infty$; then the only continuous map $\Sigma\colon\bbd R\to\bbd R$ such that $\Delta\circ\Sigma=\Delta$ is the identity $\Sigma=\id_{\bbd R}$.
MSC Classifications: 53A04 - Curves in Euclidean space 55M25 - Degree, winding number 55M35 - Finite groups of transformations (including Smith theory) [See also 57S17]
© Canadian Mathematical Society, 2014 : https://cms.math.ca/ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9161821603775024, "perplexity": 667.6372446758675}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1412037663611.15/warc/CC-MAIN-20140930004103-00367-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-and-trigonometry-10th-edition/chapter-11-11-7-probability-11-7-exercises-page-835/40 | ## Algebra and Trigonometry 10th Edition
$\frac{3}{14}$
We know that $probability=\frac{\text{number of favourable outcomes}}{\text{number of all outcomes}}$ The number of all outcomes is $31+54+42+20+47+58=252$ There are $54$ people (who work in marketing) Hence the probability here is: $\frac{54}{252}=\frac{3}{14}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9694604277610779, "perplexity": 538.4375069934008}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875148850.96/warc/CC-MAIN-20200229083813-20200229113813-00101.warc.gz"} |
https://www.physicsforums.com/threads/grids-and-graphs.172654/ | # Grids and Graphs
1. Jun 3, 2007
### birulami
Consider a rectangular grid on a piece of paper. It represents a graph where every node has 4 neighbours (except at the border, but just imagine the graph extended indefinitely). When stacking and arranging cubes, the edges of the cubes make a graph where every node has exactly 6 neighbours.
Similarly consider the middle of the rectangles of the grid as graph nodes and consider the diagonal neighbours too. Then in the graph each node has exactly 8 neighbours. In the cubes case, we arrive at 26 neighbours.
Afaik there is only a small finite number of ways to tile the 2D plane regularly as with a rectangular grid (hexagons and triangles would work too), so in graphs derived from theses geometric constructions, the number of neighbours is restricted.
Apart from every node having exactly the same number of neighbours, there is something more that characterizes these graphs, because in an infinite binary tree, for example, each node has exactly three neighbours, but the tree obviously has no loops. Now I have several questions:
a) How can we characterize these grid-derived graphs in a geometry-free definition? It could start with the fixed number of neighbours. Another property might be that all neighbours of a node form a loop. Anything else?
b) Is it possible to have any N as the fixed number of neighbours or will it turn out that every satisfying characterization (whatever satisfying is, if I knew I would not ask) points back to a grid and therefore restricts the possible N and even binds them to a dimension of the grid?
c) If every N is possible, can we nevertheless define a "dimension" purely graph theoretic such that when a map back to a grid is possible, the dimensions match?
Thanks,
Harald.
Can you offer guidance or do you also need help?
Draft saved Draft deleted
Similar Discussions: Grids and Graphs | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8865117430686951, "perplexity": 555.8798623231424}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549426639.7/warc/CC-MAIN-20170726222036-20170727002036-00403.warc.gz"} |
https://math.stackexchange.com/questions/314768/showing-f-nx-sinnx-is-not-uniformly-convergent-on-0-1 | # Showing $f_n(x) = \sin(nx)$ is not uniformly convergent on $[0,1]$
Ok, the sequence $f_n(x) = \sin(nx)$ - is it enough to say that $\displaystyle\lim_{n \to \infty} f_n(x)$ doesn't exist so it can't possibly be uniformly convergent?
I want to try showing it from the definition of uniform convergence on a sequence of functions, but due to $\displaystyle\lim_{n \to \infty} f_n(x)$ not existing I don't think it's really possible.
Any suggestions would be greatly appreciated :)
• Would it be best to show this via the Cauchy Criterion. For any $0 < \epsilon < 1$ there doesn't exist an $N \in \mathbb{N}$ such that $|f_n(x) - f_m(x)| < \epsilon$ for $0 < \epsilon < 1$ and $\forall m,n > N$ – Noble. Feb 26 '13 at 10:12
• that's enough! if a sequence of functions converges uniformly to some function $f$, then it converges simply to that same function (that is pointwise). – Olivier Bégassat Feb 26 '13 at 10:12
If a sequence of functions $f_n$ is uniformly convergent then it also converges pointwise, so it suffices to find some $x$ such that $\lim\limits_{n\to\infty}f_n(x)$ does not exist, as you suspect. One easy example to prove is $x=\pi/4$, as $\sin(2k\cdot \pi/4)$ is $0$ if $k$ is even and $\pm 1$ if it is odd. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9419736266136169, "perplexity": 94.93613899168442}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145657.46/warc/CC-MAIN-20200222085018-20200222115018-00401.warc.gz"} |
https://www.physicsforums.com/threads/equipartition-theorem-degrees-of-freedom.735336/ | # Equipartition Theorem Degrees of Freedom
1. Jan 28, 2014
### Bashyboy
Hello everyone,
Today, in class, I learned about the equipartition, and the degrees of freedom a thing can possess. My professor had said that if you were to consider a gas composed of single atoms, then the atoms couldn't have any degrees of freedom due to rotation, because you wouldn't be able to distinguish the rotation.
What exactly does that mean, and why does that warrant our not considering it as a degree of freedom.
Also, would knowing this also explain why, when counting the degrees of freedom, you don't consider clockwise and counterclockwise rotation?
2. Jan 28, 2014
### Staff: Mentor
It comes from quantum mechanics. Rotation is quantized, which means that you can only have certain discrete amounts of rotation (or, more correctly, angular momentum). You can have 0, 1, 2, ... quantas of rotation . Zero is the ground state, and corresponds to no rotation. For an atom, the next level, 1 quantum of rotation, is so higher up in energy that you will break apart the atom before excitating its rotation. Therefore, atoms are always in the rotational ground state.
The same happens with linear molecules (for instance, diatomics like N2 and O2). The internuclear axis can rotate around two different axes, but the molecule cannot rotate along the axis that goes through the two atoms.
For the same reason that you don't count going forwards and going backwards as two distinct translational degrees of freem. There is one degree of freedom, the motion along which can be either positive or negative (counterclockwise or clockwise).
3. Jan 29, 2014
### Bashyboy
So, if the discipline of thermal physics is older than quantum mechanics, how did the founders of this field of physics deal with the concept of degrees of freedom, if they didn't know about such quantum mechanical ideas?
4. Jan 29, 2014
### Staff: Mentor
Most of this was derived long before the idea itself of atoms existing was widely accepted. If you consider atoms to be point particles, then they can only have translational degrees of freedom.
At the end of the day, what is important is that the theory should match experimental results. A model where an ideal gas is made up of non-interacting point particles was a good starting point.
5. Jan 29, 2014
### D H
Staff Emeritus
The kinetic theory of gases is a semi-classical theory. Atoms exist, but there's no quantum weirdness in this theory. Atoms are point masses in the kinetic theory of gases. The inertia tensor of a point mass is zero, which means atoms have zero angular momentum and zero rotational energy. In a monatomic gas, there is no way to partition energy into rotation because monatomic gases have no rotational energy.
A diatomic gas has two rotational degrees of freedom. Why not three? The answer again lies in the inertia tensor. The moment of inertia tensor for a pair of point masses separated by some fixed distance is a rank 2 matrix. There is no rotational energy about the line that connects the two atoms. It's only the non-linear polyatomic gases that exhibit three rotational degrees of freedom in this theory.
The kinetic theory of gases has some very nice successes, but also exhibits some failures. Ideal gases have constant heat capacity, the difference between the constant pressure and constant volume heat capacities is R, and they obey PV=nRT. None of these is strictly true for a real gas. The noble gases come very, very close to exhibiting the behaviors predicted by kinetic theory. Some other gases are also close to ideal, at least over some temperature range.
6. Jan 29, 2014
### sophiecentaur
The basics of classical thermal physics (kinetic theory) starts off by assuming the simplest gas particles have point dimensions. That means they can't have rotational energy. Going from that to a diatomic particle, it's reasonable to make an assumption that there is still no rotational energy around the axis of the two (point) atoms. It fits measurements, as well, so the model works. (And you can't ask more than that; "but what's really happening?" is not an allowed question) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8658562898635864, "perplexity": 467.460420974543}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676591578.1/warc/CC-MAIN-20180720100114-20180720120114-00206.warc.gz"} |
http://mathoverflow.net/questions/71382/definability-over-subfields-of-rational-maps-to-projective-space-cremona-transf | # Definability over subfields of rational maps to projective space: Cremona transformations, degree, divisors
Recently I was led to consider some questions about Cremona transformations over arbitrary fields, which led deeper than planned into parts of algebraic geometry. I studied only algebraic groups in grad school, and read more by myself over the years with no intention of making this a focus, so my expertise is limited. Most of all I am trying to locate references to a few results which seem fundamental. But, to my surprise, diligent searches through older (Hodge and Pedoe, Semple and Roth, surveys edited by Snyder,...) and many later (mid 50s on) treatises, my favorites being more recent ones by Liu and Vakil, failed to turn up anything relevant, even in char 0.
Are the following obvious or well known -- and WHERE are they?:
(1) Any rational map $\phi: P^n_k \dashrightarrow P^n_k$ that becomes birational on extending to a larger field is already birational over k. In other words, a Cremona transformation and its inverse are always defined over the same subfields.
One way, maybe not the most direct, uses explicit inversion formulas for coeffs. It turns out that for any given degree (as below), independently of the field, there are a finite number of possible formulas. This simplifies and goes well beyond a result of Semple and Tyrrell (1968) on types over fields containing C. I am writing a short article about these ideas.
Somewhat related, but almost trivial (why is this not in standard texts?), is:
(2) The degree d of any $\phi: P^m_k \dashrightarrow P^n_k$ does not change under field extension.
By degree it is meant the least degree d of homogeneous polynomials that represent phi. This is not to be confused with a more common notion of degree, say for certain finite dominant morphisms, which often conflicts with the degree used here, e.g. when $\phi$ is birational. Equivalently, use the least d such that $\phi$ factors as the d-fold (Veronese) embedding of $P^m_k$ followed by a projection.
In terms of divisors, the problem is to show that over an algebraic closure K of k, the largest $D$ such that $\phi_K$ factors through $\mathcal L(-D)$ is definable over k. This has some connection with results of Chow (1950) obtained via Chow coordinates. There may be more relevant ideas in Weil's Foundations, which I found indigestible. Ignore this if you are not interested in history.
In terms of twisting sheaves, d seems to be the least number such that the image of $\mathcal O(1)$ under $\phi^*$ is embeddable in $\mathcal O(d)$. I have been told that using $\phi^*$ it is obvious that d does not change under field extension, but I did not get any extra insight into this by thinking in terms of Proj and sections of sheaves, even after consulting good sources such as Vakil's latest version.
(3) Is there a fairly simple proof along these lines, and what would be the key points?
The only way I know is to abandon sophisticated points of view and use a basic result whose origin/locations remain bafflingly elusive:
(4) If some multivariate polynomials over k have no common factor (of positive degree), the same is true after extending the field. [WHERE is this?]
This is of course easy (reduce to studying the effect on factorization of polynomials when a generator is added to the underlying field; there are 3 cases), and should appear in some older algebra book, any time after Steinitz (1910) on field theory. I am only mildly interested in refs to constructive proofs, which may be much older (over C) but would be more complicated and presumably use some sort of resultant.
Going beyond what motivated me, but obviously a natural question, is:
(5) Are there similar results with the domain replaced by something more general than an open set (the base-point-free locus of $\phi$) in the variety $P^m_k$?
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(1) The degree of a dominant generically finite rational map doesn't change after extension. If $f : X \dashrightarrow Y$ is a such rational map of geometrically integral varieties (e.g. $\mathbb P^n$) over $k$, then the field of function fields of $X_K$ (resp. $Y_K$) is $k(X)\otimes_k K$ (resp. $k(Y)\otimes_k K$). Of course, this operation doesn't change the degree $[k(X) : k(Y)]$.
(2)-(3). Let $U$ be a non-empty open subset where $\phi$ is defined. Then $\phi^{*}O_{P^n}(1)$ is an invertible sheaf on $U$ which extends uniquely to an invertible sheaf $L$ on $P^m_k$ (because $P^n_k$ is regular and we can extend Weil divisors on $U$ to $P^n_k$). But then $L$ is isomorphic to $O_{P^m}(d)$ for a unique $d$, which is also your 'degree of $\phi$' and we have $\phi^{*}O_{P^n}(1)\simeq O_{P^m}(d)|_U$. This property is clearly unchanged by field extensions.
(4) Yes, you are just saying that the intersection of some hypersurfaces has codimension at least $2$. But the codimension of a subvariety is invariant by field extension.
(5) The only problem I see is to define the degree of $\phi$ is the sense of (2).
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Thanks for the quick reply! (1) was too easy; I should have thought more. Also my remarks about divisors were not properly stated. The main point, which with great respect I don't believe has been answered correctly, is that the degree of rational maps between projective spaces, as defined in my post and used in some articles (others call it order, another overworked name) does not seem to fall out so easily from sheaf theory as may be widely believed.
When extending $\phi^{*}\mathcal O_{P^n}(1)$ to the base locus $B$, it seems to me from a detailed working out presented below that the stalks above $B$ of the Module vanish, and I do not see how extending or pulling back any kind of divisors could suggest a different conclusion.
Start with a dominant rational map $\phi : X \dashrightarrow Y$, where $X=P^n_k$ and $Y$ is a closed subvariety of $P^n_k$, with notation like $y_i=h_i(x)$, where the $h_i$ are homogeneous polynomials of degree d in the $x_j$ that define $\phi$.
Suppose that, over $k$, the $h_i$ have no common factor. The aim (admittedly artificial, in view of the almost trivial (4) of my note) is to use sheaf theory to determine this $d$ from $\phi$ alone, and especially to see that it cannot decrease if the field is extended.
Only quotients of degree 0 are directly encountered. For example, working locally, say with stalks, there is a (coordinate-based) definition of each module of $\mathcal O_Y(1)$ in terms of globally defined generators that (to emphasize their abstract nature) will be called $g_0, .. , g_n$ rather than, say, $y_0, \dots, y_n$. The relations are those of the form $g_i = y_i/y_j . g_j$ that make sense ($y_j \neq 0$ at that point in $Y$). After applying $\phi^*$, the module above a point $x \in X$ can be obtained by extending the scalars (tensoring) from one local ring to another, in effect keeping the same generators $g_i$ and the same relations $g_i = h_i/h_j . g_j$, now regarding k(Y) as a subfield of k(X).
The induced sheaf on $X$ is isomorphic to a subModule of $\mathcal O_X(d)$, where $g_i$ maps to the element usually identified with the polynomial $h_i$. Less clear if we refuse to use (4) is that, even after extending the field, no smaller $d$ will work, but let's continue.
Each abstract $g_i$ gives a global section that vanishes on the zero set of $h_i$ since it vanishes whenever $h_i = 0$ but some $h_j \neq 0$. Thus, above any point in the base locus $B$ the module is 0, so the sheaf is not invertible if base points exist. There no longer seems to be an obvious way to proceed with these ideas.
To conclude, I believe (4) is essential for defining the degree of rational maps, and I am astonished not to have found any source mentioning something so basic. It is too elementary to merit a proof via algebraic geometry, which when given in detail would I believe involve working with a polynomial ideal (the variety may be empty), and studying quotients by prime ideals above this ideal to look at transcendence degrees of function fields. I will post in other places, say forums on commutative algebra, to try to track down good references to (4).
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Sorry, I didn't mean extension of coherent sheaves in general. I considered the invertible sheaf as a Weil divisor: Let $H$ be a hyperplane of $P^n$ not containing $\phi(U)$, then $\phi^{-1}(H)$ is a hypersurface in $U$ and $\phi^{*}(O(1))\simeq O_U(\phi^{-1}(H))$. Consider the Zariski closure $Z$ in $P^m_k$. Then $Z$ is an effective divisor on $P^n_k$, hence linearly equivalent to some $dH_0$ for some $d>1$ and any hyperplane $H_0$ in $P^n_k$. So $O(Z)\simeq O(d)$ and $\phi^{*}(O(1))\simeq O(d)|_U$. – Qing Liu Jul 28 '11 at 9:48
Replace $d>1$ with $d\ge 1$. – Qing Liu Jul 28 '11 at 9:51
Replace with $d\ge 0$ ! – Qing Liu Jul 29 '11 at 9:20
This is a response to the last comment, but I don´t know how else to post it (too long).
The above explains well what happens over algebraically closed fields. Pulling back hyperplanes gives a a unique divisor class dH, hence d, which it seems will not change under field extension (see below for a detail).
This is the sort of divisor theory I am trying to unlearn, which led to my posts, and reading Liu´s book would be one of the best ways to proceed. I hope he won´t mind some statements about what goes wrong with divisors over arbitrary fields, which is presumably one more reason (after base change) why the sheaf approach is so prevalent.
With $\phi$ and the $h_i$ as before, look at the zero set of a $k$-linear combination of the $h_i$, or of an irreducible factor of it. Is it defined by a single equation? Think for example of sections that are elliptic curves, restricted to small fields $k$. How then would we define $d$ over $k$? What happens when the field is extended?
Supposing a sufficiently nice situation so that such problems can be ignored, when the field is extended the section still gives the class dH, but there is still the strange question about whether $d$ could decrease, which would happen only if the $n+1$ equations used to define the base locus (a possibly empty set) define over $K$ a variety of codimension 1. This is ruled out by the almost trivial (4), or via more sophisticated ideas about dimension such as heights of prime ideas, as in Liu´s book, or transcendence degrees.
(4) does in fact have something to do with divisors, so perhaps I should write more than in my original post (not having seen this anywhere). If we think of divisors on projective space via homogeneous rational functions rather than hypersurfaces, not worrying about geometry nor 0-gradings to get partial functions, it always makes sense to extend from k to K and say what is meant for a divisor over $K$ to be $k$-rational. Given $\phi$, we can only work directly with functions of quotients $h_i/h_j$, and can in the usual way define the least divisor $D = D_\phi$ such that $\phi$ factors through $\mathcal L(D)$, then ask whether this $D$ could change under field extension. The fact that is does not is just a translation of (4). In other words, if a rational map $\phi$ between projective spaces over $K$ is definable over a subfield $k$, the same is true of $D_\phi$. This divisor turns out to be the least common multiple of the $h_i$, assuming the gcd of these is 1.
No doubt all this is well-known, but what references are there?
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I think the comment is correct over any field. I don't see exactly where is your point. Maybe I should say that a hyperplane doesn't consist only in rational points. In fact, what is proved is $d$ is not exactly the least integer with the required property, it is unique. For (4), if $K/k$ is algebraic, and if $F$ is a commun factor of given polynomials, then $F$ is defined over some finite extension $L/k$. Then any irreductible factor of the norm of $F$ is a commun factor over $k$. This reduces easily the problem to the case $K/k$ is purely transcendental. Continued... – Qing Liu Jul 29 '11 at 9:17
Now any divisibility relation over $K$ gives a divisibility relation over $k$ by giving values to the variables. – Qing Liu Jul 29 '11 at 9:18
Sorry for my comment based on thinking of Weil divisors only in terms of closed points, something I have still not completely unlearned. In the Weil divisor approach one runs naturally into (4) but it can be avoided at the cost of doing something a little deeper (see earlier remark about a strange question). There are no real mathematical issues, just a matter of taste about how to proceed.
To illustrate, suppose homogeneous polynomials $g_i$ of the same degree represent $\phi: P^n_k \to Y \subset P^m_k$, with $Y$ not in a hyperplane. Then $y_0$ on $P^m_k$ induces an effective Weil divisor on $P^n_k$. This is basically a polynomial, up to scalars. Is it $g_0$? In general, NO! It is $g_0/g$ that will give $d$, where $g$ is the gcd of the g_i. Under field extension does the gcd remain the same or can $d$ decrease? This is just (4).
Of course (4) is no more than an easy exercise with polynomials and fields, based on adjoining an element that is either transcendent, separable, or (say) a $p^{\text th}$ root. There are several slightly deeper prooofs, including a slick one by flatness. What I find astonishing is that despite its relevance it seems to have gone unnoticed. While still searching in many likely places, I have so far found NOTHING.
As a more didactic point, via (4) one needs little machinery to obtain a result on the field-independence of factoring rational maps on $P^n_k$ through Veronese embeddings, and to see that if a rational map is defined over $k$, so is its associated divisor. This does not come close to Chow´s result that there is a well-defined smallest field of definition of a divisor, which may be properly contained in $k$. Presumaby there are now more modern approaches than the original via Chow coordinates.
I would like to close by thanking Prof. Qing Liu for taking the time to help clarify the ideas involved.
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https://www.physicsforums.com/threads/complex-integration.554222/ | # Homework Help: Complex Integration
1. Nov 26, 2011
### Maybe_Memorie
1. The problem statement, all variables and given/known data
Sketch the C1 paths a: [0; 1] -> C, t -> t + it2 and b: [0; 1 + i]. Then compute the following integrals.
∫Re(z)dz over a
∫Re(z)dz over b
2. Relevant equations
3. The attempt at a solution
Sketching a seems ok, y axis is Imaginary, x axis is Real, and the path is a quadratic between 0 and 1.
However I'm not sure about b...
As for the integrals, is it just a case of integrating Re(a(t))a'(t)dt between 0 and 1, then integrating Re(a(t))a'(t)dt between 0 and 1+i?
2. Nov 26, 2011
### I like Serena
Hi MM!
I believe you've got (a) down.
For (b) I have to admit I haven't seen the notation [0; 1 + i] before.
Do you have a definition for it?
I'd assume it's supposed to represent the line segment between 0 and (1+i).
Or in other words: (1+i)[0, 1].
It looks a bit weird though, since a curve is usually defined on an interval of real numbers.
However, if this is the case, you need to reconsider what Re(z) is.
3. Nov 27, 2011
### Maybe_Memorie
I don't have a definition for it. I'll upload the actual problem to show it to you.
#### Attached Files:
• ###### MA2325_ha4.pdf
File size:
61.2 KB
Views:
39
4. Nov 27, 2011
### I like Serena
I get it, in particular if I look at the problems that are coming, that show which theory you're currently learning.
Path b is not related to path a.
It is just a (linear) path from 0 to (1+i).
It would be given by b:[0,1]→C defined by t→t(1+i)
Note that path a also starts in 0 and ends in (1+i).
And also note that for path a the interval is [0,1] and not [0;1].
[0,1] is a real interval, whereas [0;1] would indicate a (linear) path between 0 and 1 in the complex plane.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9576930403709412, "perplexity": 1796.9907236977294}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794866733.77/warc/CC-MAIN-20180524170605-20180524190605-00010.warc.gz"} |
https://infoscience.epfl.ch/record/184570 | Infoscience
Journal article
# The Truncated Tracial Moment Problem
We present tracial analogs of the classical results of Curto and Fialkow on moment matrices. A sequence of real numbers indexed by words in noncommuting variables with values invariant under cyclic permutations of the indexes, is called a tracial sequence. We prove that such a sequence can be represented with tracial moments of matrices if its corresponding moment matrix is positive semidefinite and of finite rank. A truncated tracial sequence allows for such a representation if and only if one of its extensions admits a flat extension. Finally, we apply this theory via duality to investigate trace-positive polynomials in noncommuting variables. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9666265845298767, "perplexity": 374.26888145008877}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721008.78/warc/CC-MAIN-20161020183841-00282-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/61226/congruence-subgroup | # congruence subgroup
Hi,
I've a question regarding the congruence subgroup.
We have the following congruence subgroup
$\Gamma^* := \left\{ M \in \Gamma | XM \equiv X(mod \; \mathbb{Z}^2),\; m \cdot det|\begin {smallmatrix} X \\ XM \end{smallmatrix}| \in \mathbb{Z} \right\}$
of $\Gamma$.
Where
$\Gamma$ is a subgroup of $SL_2(\mathbb{Z})$,
$M = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \in \Gamma$, $X = \begin{pmatrix} \lambda & \mu \end{pmatrix}\in \mathbb{Q}^2$, $m \in \mathbb{Z}$
This subgroup can also be written as
$\Gamma^* = \left\{ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \in \Gamma | (a-1)\lambda + c \mu ,\; b \lambda + (d-1)\mu,\; m(c \mu^2 + (d-a)\lambda\mu-b\lambda^2) \in \mathbb{Z} \right\}$
and hence contains $\Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$ if $NX \in \mathbb{Z}^2$.
My question is, why is $\Gamma^* \supset \Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$ true? What does $\Gamma \left(\frac{N^2}{(N,m)}\right)$ mean exactly? Actually I only have found references to the main congruence subgroup $\Gamma(N)$ (but never anything to a more complex form).
Thanks, Jan
-
At a first glance, I'd guess $\Gamma(N^2/(N,m))$ means precisely that. The principal congruence subgroup of level $N^2/gcd(N,m)$. (I am supposing $m$ is a fixed integer, yes?) – Kimball Apr 10 '11 at 21:16
Hi Komball, thanks for your answer. Actually my question is not asked as precise as I would like to ask it. I was aware of the notion of the gcd. However, my question is how you could come to the conclusion that the subgroup $\Gamma^*$ contains $\Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$. I am missing some steps why you can write this. – Jan Apr 10 '11 at 21:50
Actually the question is more like, what is the connection between $\Gamma^*$ and $\Gamma(\frac{N^2}{(N,m)})$. However, the missing part is, how you come from $\Gamma^*$ to $\Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$ and how are the intermediate steps or what rules/definition do you apply in order to make the statement $\Gamma^* \supset \Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$. Thanks a lot – Jan Apr 10 '11 at 21:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9583359360694885, "perplexity": 240.5352948054009}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375095806.49/warc/CC-MAIN-20150627031815-00060-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://www.cliffsnotes.com/study-guides/calculus/calculus/the-derivative/chain-rule | ## Chain Rule
The chain rule provides us a technique for finding the derivative of composite functions, with the number of functions that make up the composition determining how many differentiation steps are necessary. For example, if a composite function f( x) is defined as
Note that because two functions, g and h, make up the composite function f, you have to consider the derivatives g′ and h′ in differentiating f( x).
If a composite function r( x) is defined as
Here, three functions— m, n, and p—make up the composition function r; hence, you have to consider the derivatives m′, n′, and p′ in differentiating r( x). A technique that is sometimes suggested for differentiating composite functions is to work from the “outside to the inside” functions to establish a sequence for each of the derivatives that must be taken.
Example 1: Find f′( x) if f( x) = (3x 2 + 5x − 2) 8.
Example 2: Find f′( x) if f( x) = tan (sec x).
Example 3: Find if y = sin 3 (3 x − 1).
Example 4: Find f′(2) if
Example 5: Find the slope of the tangent line to a curve y = ( x 2 − 3) 5 at the point (−1, −32).
Because the slope of the tangent line to a curve is the derivative, you find that
which represents the slope of the tangent line at the point (−1,−32). | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9725663661956787, "perplexity": 683.0492943010835}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178366959.54/warc/CC-MAIN-20210303104028-20210303134028-00629.warc.gz"} |
https://www.arxiv-vanity.com/papers/1308.3910/ | # Exact cosmological solutions from Hojman conservation quantities
Salvatore Capozziello Mahmood Roshan Dipartimento di Fisica, Università di Napoli ”Federico II”, Napoli, Italy INFN Sez. di Napoli, Compl. Univ. di Monte S. Angelo, Edificio G, Via Cinthia, I-80126, Napoli, Italy Department of Physics, Ferdowsi University of Mashhad, P.O. Box 1436, Mashhad, Iran
July 5, 2021
###### Abstract
We present a new approach to find exact solutions for cosmological models. By requiring the existence of a symmetry transformation vector for the equations of motion of the given cosmological model (without using either Lagrangian or Hamiltonian), one can find corresponding Hojman conserved quantities. With the help of these conserved quantities, the analysis of the cosmological model can be simplified. In the case of quintessence scalar-tensor models, we show that the Hojman conserved quantities exist for a wide range of -potentials and allow to find exact solutions for the cosmic scale factor and the scalar field. Finally, we investigate the general cosmological behavior of solutions by adopting a phase-space view.
###### pacs:
11.25.Mj, 98.80.Cq, 04.50.Cd, 04.50.Kd
## I Introduction
Symmetries are a fundamental tool to study physical systems since they allow to reduce dynamics and give insight into conserved quantities. For example, the Noether theorem has been widely adopted in classical and quantum physics as well as in general relativity and cosmology. As an example in general relativity, it is well-known that the black hole entropy is a Noether conserved quantity wald . In cosmology, the so called Noether symmetry approach provides a method to solve exactly dynamics ritis . This approach is based on requiring the existence of a Noether symmetry for the point-Lagrangian associated to the equations of motion for a given cosmological model. It has been used in several alternative theories of gravity alternative in order to find out exact cosmological solutions, and also to fix the form of undetermined couplings and potentials noether .
In this paper, we show that Hojman conservation theorem hojman can also provide an approach to find exact solutions for the cosmological models. As we shall see below, there is a main difference between the Noether conservation theorem and the Hojman conservation theorem. The first relies on the Lagrangian (and Hamiltonian) structure of the equations of motion. In the Hojman picture, there is no need for Lagrangians and Hamiltonian functions given a priori. The symmetry vectors and the corresponding conserved quantities can be obtained by using directly the equations of motion. Specifically, unlike the Noether symmetry approach, it is not needed to find first a point-Lagrangian for the equations of motion. Moreover, these two approaches may give rise to different conserved quantities. For example, in the case of minimally coupled scalar-tensor gravity models, the Noether symmetry exists only for exponential potential ritis ; however, as we shall see below, the Hojman symmetry exists for a wide range of potentials. In principle, the Hojman conserved quantities theorem could be a different class with respect to the Noether ones because they do not require that field equations directly derive from a Lagrangian.
In particular, using the Hojman approach, we find a new class of scalar-field potentials. By a dynamical system analysis, these model can be investigated for some simple cases. We will show that there exist attractor solutions corresponding to late time scalar-field dominated universe and then interesting to address the dark energy problem.
The layout of the paper is the following. In Sec. II, we briefly review the Hojman conservation theorem. Secs: III and IV are devoted to the application of the Hojman conservation theorem to given scalar-tensor models where we find out exact solutions. In Sec. V, we investigate the relevant phase-space features of the cosmological models. Conclusions are drawn in Sec. VI.
## Ii The Hojman conservation theorem
In order to point out the differences between the Hojman and the Noether approaches, let us derive the conservation laws in the first case. Consider a set of second-order differential equations
¨qi=Fi(qj,˙qj,t) i,j=1,...,n (1)
where dots indicate derivatives with respect to time, namely ”. In the case of cosmology, it is appropriate to interpret the dot as the differentiation with respect to the cosmic time. Before stating the Hojman conservation theorem, let us recall the definition of the symmetry vector for the above equations. The symmetry vector is defined so that the infinitesimal transformation
q′i=qi+ϵXi(qj,˙qj,t) (2)
maps solutions of the equations (1) into solutions of the same equation lutzky . It is straightforward to verify that the symmetry vector for equations (1) satisfies
d2Xidt2−∂Fi∂qjXj−∂Fi∂˙qjdXjdt=0 (3)
with
ddt=∂∂t+˙qi∂∂qi+Fi∂∂˙qi. (4)
Keeping in mind the definition of the symmetry vector , the Hojman conservation quantities can be defined by the following hojman
Theorem:
1. if the ”force” satisfies
∂Fi∂˙qi=0 (5)
then
Q=∂Xi∂qi+∂∂˙qi(dXidt) (6)
is a conserved quantity for equations (1), i.e. .
2. and if satisfies
∂Fi∂˙qi=−ddtlnγ (7)
where is a function of , then
Q=1γ∂(γXi)∂qi+∂∂˙qi(dXidt) (8)
is a conserved quantity.
Such a theorem can be applied to dynamical systems describing cosmological models as we will show below.
## Iii Hojman conserved quantities for cosmological equations
As an example for the above considerations, let us consider the case of spatially flat Friedmann-Robertson-Walker (FRW) space-time. The Friedman equations are
¨a=−3p+ρ6a (9)
˙a2=ρ3a2 (10)
where is the cosmic scale factor, is the pressure of the cosmic fluid and is the corresponding energy density (we work in units where ). If we assume that the cosmic perfect fluid obeys the equation of state
p=wρ (11)
where , then combining equations (9) and (10) we get
¨a=α˙a2a (12)
where . Mathematically, equation (12) can be considered as a one dimensional ”motion” under the action of a ”position” and ”velocity” dependent ”force”. Comparing this equation with (1), it is clear that . Thus one can easily show that
γ(a)=a−2α (13)
For the sake of simplicity, let us assume that the one dimensional symmetry vector dose not depend explicitly on , i.e. . In this case, equation (3) takes the form
(−αa∂X∂a+∂2X∂a2+αa2X)+˙a2α2a2∂2X∂˙a2+˙a(2αa∂2X∂a∂˙a−αa2∂X∂˙a)=0 (14)
It is easy to verify that is a solution for this equation provided that . For this symmetry vector, using equations (8) and (13), we find the following conserved quantity
Q0=(m+2)(α−1)an−1˙am (15)
Assuming , we can rewrite (15) as
a1+3w2˙a=c1=const (16)
This nonlinear differential equation can be easily solved, giving
a(t)=[3(1+w)2(c1t+c2)]23(1+w) (17)
where is an integration constant. In the case of dark energy density () vanishes, and equation (15) does not lead to an exact solution for the cosmic scale factor.
It should be stressed that, using equation (10), we can rewrite equation (16) as . Nevertheless, even without working through the details of finding the symmetry vector, this expression can be directly derived from the Friedmann equations. However, as we shall see in the next section, in the context of other cosmological models, where there exist some undetermined functions in the field equations, finding the Hojman conserved quantity is not possible without calculating the symmetry vector.
## Iv Hojman conserved quantities and scalar-tensor cosmology
A time dependent vacuum energy is sometimes called quintessence, and it is described by a scalar field minimally coupled to gravity ratra . The potential of the scalar field is chosen in a way that leads to late time accelerated expansion. In a flat FRW space-time, the equations of motion are given by
˙a2a2=13⎡⎣˙ϕ22+V(ϕ)⎤⎦ (18)
¨a=a3[V(ϕ)−˙ϕ2] (19)
We recall that the time derivative of (18) together with (19) yield
¨ϕ+3˙aa˙ϕ+V′(ϕ)=0 (20)
which is a Klein-Gordon equation, the equation of motion of the scalar field. Hereafter the prime means . By introducing the variable , and by combining equations (18) and (19), we find the following equation of motion
¨x=−f(x)˙x2 (21)
where
f(x)=12ϕ′(x)2 (22)
With the assumption that and are invertible functions of , the dynamics governed by equations (18)-(20) can be reduced to a one dimensional motion. In this case, equation (20) can be considered as a constraint equation. From equation (21), it is clear that , thus
γ(x)=γ0e∫f(x)dx (23)
where is an integration constant. Another useful differential equation can be obtained by dividing equations (20) and (18). The result is
V′(ϕ)V(ϕ)=f(x)ϕ′(x)−ϕ′′(x)−3ϕ′(x)3−12ϕ′(x)2 (24)
If we assume that the one dimensional symmetry vector does not depend explicitly on time, then equation (3) takes the following form
(f(x)∂X∂x+f′(x)X+∂2X∂x2)+˙x2f(x)2∂2X∂˙x2−˙x(2f(x)∂2X∂x∂˙x+f′(x)∂X∂˙x)=0 (25)
In order to find some solutions for this equation, we consider the following cases:
### iv.1 The case X=X(˙x)
In this case, the only solution for (25) is
X(˙x)=A0˙xn+A1˙x (26)
and
f(x)=−(1nx+f0) (27)
where and are constant parameters. It is straightforward to verify that the symmetry vector leads to a zero constant of motion (). Therefore, without any loose of generality, we set and . Let us define a new variable as . Note that is a positive function. Now, by introducing into equation (22), we find
ϕ(y)=ϕc±1n√8y (28)
where is a constant. Furthermore, we can rewrite equation (24) as
V′(y)V(y)=2ny−1y(1−3y) (29)
The solution of this first order differential equation is given by
V(y)=V0y2−nn(3y−1) (30)
Using equation (28), the generic potential with respect to is
V(φ)=λφ4n−8λ3n2φ4n−2 (31)
where
λ=3V0(n28)2n (32)
For , the potential (31) gives rise to the so called intermediate inflationary model barrow ; muslimov . In the case of , this potential leads to the so called model linde . However, note that for model there is no symmetry vector in the form presented in (26).
On the other hand, using equations (8) and (23) we find the Hojman conserved quantity as follows
˙xnnx+f0=Q0 (33)
if then , and if then . Also, since can be negative or positive, we assume that is an integer number. Thus we rewrite (33) as
˙yn=(−n)ny|Q0| (34)
The solution of this equation is given by
y(t)=[(1−1n)(y0−n|Q0|1nt)]nn−1 (35)
where is an integration constant. Consequently, using equations (28) and (35), we can summarize the exact solution of and for the potential (24) as follows
a(τ)=e−f0ne−1n[(1−1n)τ]nn−1φ(τ)=±√8n[(1−1n)τ]n2(n−1) (36)
where the parameter is defined as
τ=y0−n|Q0|1nt (37)
Substituting these solutions together with the potential (31) into equation (20), we get
V0=|Q0|2n (38)
Thus, a symmetry vector exists for equation (21), if the generic potential takes the form (31). Now, let us consider more general situations where the symmetry vector is a function of both and .
### iv.2 The case X(x,˙x)=˙xg(x)
Here is an arbitrary function of , which we call generating function. In this case, using equation (25), one can show that is a symmetry vector if the generating function satisfies
f(x)=12ϕ′(x)2=g′′(x)g′(x)>0 (39)
Solving this equation, we find the scalar field with respect to , namely
ϕ(x)=ϕc±√2∫√g′′(x)g′(x)dx (40)
On the other hand, the corresponding Hojman conserved quantity is given by
˙xg′(x)=Q0 (41)
In general, this is a nonlinear first order differential equation for and can lead to an exact solution for the cosmic scale factor.
By using equations (24) and (39), one can find the generic potential with respect to
V(x)=V0g′′(x)−3g′(x)g′(x)3 (42)
It is obvious that the potential can be expressed in terms of the first and second derivatives of the generating function. Substituting this potential into (20) and using (41), we find . Thus the potential can be rewritten as
V(x)=Q203g′(x)−g′′(x)g′(x)3=Q20(3−f(x))e−2∫f(x)dx (43)
In the following, we choose some specific forms for the generating function and find some exact solutions.
#### iv.2.1 g(x)=λeα22x
Where and are constant parameters. In this case . Using equation (40) we get
ϕ(x)=ϕ0±α(x−x0) (44)
where and are the values of and at . Also, the Hojman conserved quantity (41) yields to the following solution for
x(t)=x0+2α2ln[1+Q0α22λe−α22x0(t−t0)] (45)
Furthermore, substituting the expression into (43) and using (44), we find
V(φ)=2(6−α2)Q20α4λ2e∓αφ (46)
where . We summarize the solutions for and for the above generic potential as follows
a(t)=ex0[1+Q0α22λe−α22x0(t−t0)]2α2φ(t)=±2αln[1+Q0α22λe−α22x0(t−t0)] (47)
#### iv.2.2 g(x)=(f0+x)1+α1+α (α>0)
Where and are constant parameters. In this case, . Thus if . Using equation (40) we find
ϕ(x)=ϕ0±√8α(f0+x)∓√8α(f0+x0) (48)
Furthermore, the Hojman conserved quantity (41) leads to
x(t)=−f0+((1+α)Q0(t−t0)+(f0+x0)1+α)11+α (49)
And the generic potential is given by
V(φ)=λφ−4α−83λα2φ−4α−2 (50)
where and . Note that if (), the above potential coincides with potential (31). This shows that different symmetry vectors can lead to the same potential. However, considering that is a positive real number and is an integer number, potentials (50) is more general than (31).
#### iv.2.3 g(x)=−√2α(x+e2α2x)
Where is a real number. In this case, one can easily verify that for every . Using the Hojman conserved quantity, we have
x(t)=12α2[τ−W(2α2eτ)] (51)
where is the Lambert W function and is defined as
τ=−√2α(g(x0)+Q0(t−t0)) (52)
Note that as before . Also, the scalar field with respect to is given by
ϕ(x)=ϕ0±2αarcsinh(√2αeα2x)∓2αarcsinh(√2αeα2x0) (53)
Finally, the generic potential with respect to can be written as
(54)
Where . Now let us summarize the exact solution for the above potential
a(τ)=e12α2[τ−W(2α2eτ)]φ(τ)=2αarcsinh[√2α e12α2(τ−W(2α2eτ))] (55)
In this model, if , and so it can be considered as a quintessence model. We will consider some important features of such a model in Sec. V.
#### iv.2.4 g(x)=1αln[tanh(αϕ(x)2)]
If the generating function is given as a function of , then the equation (39) can be rewritten as follows
(dxdϕ)−1(12+d2xdϕ2)=g′′(ϕ)g′(ϕ) (56)
This is a nonlinear differential equation for . On the other hand, the Hojman conserved quantity can be written as
˙ϕg′(ϕ)=Q0 (57)
Furthermore, using equation (24) the generic potential can be written directly with respect to
V(ϕ)=Q20(3−f(ϕ))e∓∫√2f(ϕ)dϕ (58)
Now, let us solve equation (56) for the given generating function . The result is
x(ϕ)=β2α2ln[tanh(αϕ2)]−12α2ln[sinh(αϕ)] (59)
where is an integration constant. Thus is
f(ϕ)=2α2sinh2(αϕ)[β−cosh(αϕ)]2>0 (60)
With in hand, we can find the potential using equation (58), namely
V(ϕ)=Q20(3−2α2)⎡⎣cosh(αϕ)−β1−23α2⎤⎦2 +2α2Q20⎛⎝β21−23α2⎞⎠ (61)
If we assume that then the potential (61) can be rewritten as
V(ϕ)=V0[cosh(αϕ)−1]2+Λ (62)
where and . This potential is similar to the potential suggested by Sahni and Wang sahni . We recall that the generic potential of the quintessence model introduced in sahni , is given by
V(ϕ)=V0[cosh(αϕ)−1]p (63)
where and are constant parameters. Although the parameter is zero in sahni (note that can not be zero in the potential (62)), the potential (62) is similar to the potential (63) with .
Equation (57) yields to the following solution for
ϕ(τ)=2αarctanh(eτ) (64)
in which , and is an integration constant. Substituting this solution into (59), the cosmic scale factor is
a(τ)=eβ2α2τ[sinh(2arctanh(eτ))]−12α2 (65)
As final remark, it is important to stress again that these exact solutions strictly depend on the Hojman conserved quantities that allow to reduce the dynamical system.
## V The space-phase view of the model
Motivating by the interesting form of potential (54), obtained by explicitly requiring the Hojman symmetry, we consider now some features of the related cosmological model in the phase-space view cao . The scalar field potential (54), can be recast in the suitable form
V(φ)=V0[α1sech4(αφ)+α2sech6(αφ)] (66)
We consider the dynamics of this model in a spatially flat FRW universe with Hubble parameter , in which both matter and radiation are present. The evolution equations are
H2=13(ρm+ρr+ρφ)˙H=−12(ρm+43ρr+˙φ2)¨φ+3H˙φ+dVdφ=0 (67)
where and are the matter and radiation energy densities respectively; is the scalar field energy density. Now, let us define the following dimensionless variables cao ; copeland
x=1√6˙φH y=1√3√VHz=1√3√ρrH λ=−V,φV (68)
Using these variables and also evolution equations (67), one can set the following autonomous system
x′=√32λy2+32x(−1+x2+13z2−y2)y′=−√32λxy+32y(1+x2+13z2−y2)z′=−z2(1−3x2−z2+3y2)λ′=−√6(Γ−1)x (69)
where prime denotes derivative with respect to , and
Γ=V,φφVV2,φ (70)
Moreover, using the definitions (68), one can easily verify the algebraic expressions , and also
x2+y2+z2=(1−Ωm)≤1 (71)
ωφ=Pφρφ=x2−y2x2+y2 (72)
where , and are dimensionless cosmological density parameters (). Using the dynamical system approach, we will consider two simple cases and . These potentials are shown schematically in Fig 1. It is clear, from the autonomous equations (69), that the phase space is four dimensional (,,,). In order to describe the dynamics via more illustrative figures, we will use the projection of the trajectories onto the plane. Therefore taking into account the bound (71), the dynamics can be described by trajectories within a unit circle in the plane cao . Fixed points in the phase space correspond to solutions where the scalar field has a constant equation of state and the scale factor of the universe evolves as . Thus gives rise to an accelerated expansion.
#### a) V(φ)=V0sech6(αφ)
For this potential the fixed (or critical) points, i.e. the points that verify are listed in Table LABEL:table1. Depending on the value of there are only four stable (or attractor) points. If then and are attractor points and for , and are the attractor points.
Let us first consider the unstable points. For every value of , there are six fixed points , and that correspond to solutions where the constraint (71) is dominated by the kinetic energy of the scalar field. These solutions are unstable, and since the scalar field rolls down the potential energy in the presence of the ”friction force” (see equation (67)), we expect them to be relevant at early times. On the other hand, the fixed point correspond to the scalar field dominated solutions () where the constraint (71) is dominated by the potential energy density of the scalar field. For these points and so . This means that correspond to solutions where the scalar field is frozen at the pick of the potential. It is natural that this points are unstable.
There are also two types of fluid dominated solutions and . Note that since the coordinate can take every value in the four dimensional phase space, these solutions represent infinite number of fixed points. correspond to a matter dominated universe. On the other hand, solutions correspond to a radiation dominated universe. The fact that and are unstable is cosmologically plausible. Because, we do not expect that the universe remains in a fluid dominated phase forever. Also, this means that the energy density of the scalar field never vanishes with respect to the matter and radiation in the universe. We will show that depending on the value of , the energy density of the scalar field either dominates the other matter fields in the universe or remains proportional to the matter energy density.
The critical points and correspond to the scaling solutions where . The matter energy density for these solutions is zero. Since these solutions are unstable, we can say that the matter energy density never vanishes while both and are nonzero.
Now, let us consider the attractor points. Attractor points differ in several features from , so we consider them separately.
Attractors of type and . These attractors exist if . They correspond to the late time scaling solutions where the energy density of the scalar field remains proportional to that of the matter fluid. Note that for these solutions , and neither the scalar field nor the matter fluid entirely dominates the evolution of the universe. Also the deceleration parameter () for these solutions is . Therefore, these solutions correspond to a decelerating expansion.
In Fig. 2, we show the projection of the phase space trajectories onto the plane for . In this case, the four attractor points are . For a proper choice of initial conditions, the trend of , and is reported in Fig 3. Also the evolution of the equation of state parameter with respect to has been shown in Fig. 4.
Attractors of type and . These attractors exist if . They correspond to the late time scalar field dominated universe (=1) where the equation of state parameter and the deceleration parameter . Therefore, the expansion can be accelerated if . The two dimensional subspace of the phase space for is reported in Fig. 5. A cosmologically acceptable trajectory should start with the radiation dominated era where , then enter to the matter dominated era () and finally fall into one the attractors , .
In Fig. 6, we have chosen a set of initial conditions which leads to a cosmologically acceptable trajectory. However, it is quite clear that in order to be consistent with the observed current matter content of the universe, we have to choice the initial conditions in such a way that the late time scalar field dominated attractor is not yet reached at the present epoch. Therefore, we inevitably confront a kind of fine tuning problem. It is necessary to note that for the potential under consideration the dimensionless parameter , and so it is expected that this model has not tracking solutions tracking . In other words, the solutions are quite sensitive to the initial conditions.
In this case, the evolution of has been shown in Fig. 7. When the solution reaches the attractor, reaches to the constant value .
#### b) V(φ)=V0[sech4(αφ)−sech6(αφ)]
In this case, the region of existence and stability of the critical points are listed in Table LABEL:table2. We shall see that the cosmological behavior of this model is similar to the previous model. Depending on the value of , there are only two attractor points. If then there exist two attractor points . Otherwise, if , are the attractor points.
As in the previous subsection, let us first consider the unstable points. For every value of , there are two critical points that correspond to solutions where the constraint (71) is dominated by the kinetic energy of the scalar field. These solutions are unstable and as we mentioned before, they are expected to be relevant at early times.
The unstable fixed points correspond to the scalar field dominated solutions () where the constraint (71) is dominated by the potential energy density of the scalar field. Since , so these points correspond to solutions where the scalar field is at rest at the pick of the potential.
There are also two types of fluid dominated unstable solutions and . corresponds to a matter dominated universe. and correspond to a expanding radiation dominated universe and to a contracting radiation dominated universe respectively. As we already mentioned, the existence of these critical points means that the energy density of the scalar field never vanishes with respect to the other matter content of the universe.
The critical points and correspond to the scaling solutions for which . The matter energy density for these solutions is zero. Since these solutions are unstable, we conclude that the matter energy density never vanishes while is a nonzero constant.
Now let us consider the attractor points in more detail.
Attractors of type . They correspond to the late time decelerating scaling solutions where the scalar field energy density and the matter share a finite and constant portion of the cosmic energy content. corresponds to a expanding universe and to a contracting universe. For these solutions the radiation energy density is zero and neither the scalar field nor the matter fluid entirely dominates the evolution. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9797548651695251, "perplexity": 424.712965180645}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046150067.51/warc/CC-MAIN-20210723210216-20210724000216-00221.warc.gz"} |
https://math.libretexts.org/Bookshelves/Pre-Algebra/Book%3A_Prealgebra_(OpenStax)/05%3A_Decimals/5.10%3A_Ratios_and_Rate_(Part_1) | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 5.10: Ratios and Rate (Part 1)
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Skills to Develop
• Write a ratio as a fraction
• Write a rate as a fraction
• Find unit rates
• Find unit price
• Translate phrases to expressions with fractions
be prepared!
Before you get started, take this readiness quiz.
1. Simplify: $$\dfrac{16}{24}$$. If you missed this problem, review Example 4.3.1.
2. Divide: 2.76 ÷ 11.5. If you missed this problem, review Example 5.4.9.
3. Simplify: $$\dfrac{1 \dfrac{1}{2}}{2 \dfrac{3}{4}}$$. If you missed this problem, review Example 4.5.7.
## Write a Ratio as a Fraction
When you apply for a mortgage, the loan officer will compare your total debt to your total income to decide if you qualify for the loan. This comparison is called the debt-to-income ratio. A ratio compares two quantities that are measured with the same unit. If we compare a and b , the ratio is written as a to b, $$\dfrac{a}{b}$$, or a:b.
Definition: ratios
A ratio compares two numbers or two quantities that are measured with the same unit. The ratio of a to b is written a to b, $$\dfrac{a}{b}$$, or a:b.
In this section, we will use the fraction notation. When a ratio is written in fraction form, the fraction should be simplified. If it is an improper fraction, we do not change it to a mixed number. Because a ratio compares two quantities, we would leave a ratio as $$\dfrac{4}{1}$$ instead of simplifying it to 4 so that we can see the two parts of the ratio.
Example $$\PageIndex{1}$$:
Write each ratio as a fraction: (a) 15 to 27 (b) 45 to 18.
Solution
(a) 15 to 27
Write as a fraction with the first number in the numerator and the second in the denominator. $$\dfrac{15}{27}$$ Simplify the fraction. $$\dfrac{5}{9}$$
(b) 45 to 18
Write as a fraction with the first number in the numerator and the second in the denominator. $$\dfrac{45}{18}$$ Simplify. $$\dfrac{5}{2}$$
We leave the ratio in (b) as an improper fraction.
Exercise $$\PageIndex{1}$$:
Write each ratio as a fraction: (a) 21 to 56 (b) 48 to 32.
$$\dfrac{3}{8}$$
$$\dfrac{3}{2}$$
Exercise $$\PageIndex{2}$$:
Write each ratio as a fraction: (a) 27 to 72 (b) 51 to 34.
$$\dfrac{1}{1}$$
$$\dfrac{3}{2}$$
### Ratios Involving Decimals
We will often work with ratios of decimals, especially when we have ratios involving money. In these cases, we can eliminate the decimals by using the Equivalent Fractions Property to convert the ratio to a fraction with whole numbers in the numerator and denominator.
For example, consider the ratio 0.8 to 0.05. We can write it as a fraction with decimals and then multiply the numerator and denominator by 100 to eliminate the decimals.
$$\dfrac{0.8}{0.05}$$
$$\dfrac{(0.8) \textcolor{red}{100}}{(0.05) \textcolor{red}{100}}$$
$$\dfrac{80}{5}$$
Do you see a shortcut to find the equivalent fraction? Notice that 0.8 = $$\dfrac{8}{10}$$ and 0.05 = $$\dfrac{5}{100}$$. The least common denominator of $$\dfrac{8}{10}$$ and 5 100 is 100. By multiplying the numerator and denominator of $$\dfrac{0.8}{0.05}$$ by 100, we ‘moved’ the decimal two places to the right to get the equivalent fraction with no decimals. Now that we understand the math behind the process, we can find the fraction with no decimals like this:
"Move" the decimal 2 places. $$\dfrac{80}{5}$$ Simplify. $$\dfrac{16}{1}$$
You do not have to write out every step when you multiply the numerator and denominator by powers of ten. As long as you move both decimal places the same number of places, the ratio will remain the same.
Example $$\PageIndex{2}$$:
Write each ratio as a fraction of whole numbers: (a) 4.8 to 11.2 (b) 2.7 to 0.54
Solution
(a) 4.8 to 11.2
Write as a fraction. $$\dfrac{4.8}{11.2}$$ Rewrite as an equivalent fraction without decimals, by moving both decimal points 1 place to the right. $$\dfrac{48}{112}$$ Simplify. $$\dfrac{3}{7}$$
So 4.8 to 11.2 is equivalent to $$\dfrac{3}{7}$$.
(b) 2.7 to 0.54
Write as a fraction. $$\dfrac{2.7}{0.54}$$ The numerator has one decimal place and the denominator has 2. To clear both decimals we need to move the decimal 2 places to the right. $$\dfrac{270}{54}$$ Simplify. $$\dfrac{5}{1}$$
So 2.7 to 0.54 is equivalent to $$\dfrac{5}{1}$$.
Exercise $$\PageIndex{3}$$:
Write each ratio as a fraction: (a) 4.6 to 11.5 (b) 2.3 to 0.69.
$$\dfrac{2}{5}$$
$$\dfrac{10}{3}$$
Exercise $$\PageIndex{4}$$:
Write each ratio as a fraction: (a) 3.4 to 15.3 (b) 3.4 to 0.68.
$$\dfrac{2}{9}$$
$$\dfrac{5}{1}$$
Some ratios compare two mixed numbers. Remember that to divide mixed numbers, you first rewrite them as improper fractions.
Example $$\PageIndex{3}$$:
Write the ratio of $$1 \dfrac{1}{4}$$ to $$2 \dfrac{3}{8}$$ as a fraction.
Solution
Write as a fraction. $$\dfrac{1 \dfrac{1}{4}}{2 \dfrac{3}{8}}$$ Convert the numerator and denominator to improper fractions. $$\dfrac{\dfrac{5}{4}}{\dfrac{19}{8}}$$ Rewrite as a division of fractions. $$\dfrac{5}{4} \div \dfrac{19}{8}$$ Invert the divisor and multiply. $$\dfrac{5}{4} \cdot \dfrac{8}{19}$$ Simplify. $$\dfrac{10}{19}$$
Exercise $$\PageIndex{5}$$:
Write each ratio as a fraction: $$1 \dfrac{3}{4}$$ to $$2 \dfrac{5}{8}$$.
$$\dfrac{2}{3}\ Exercise \(\PageIndex{6}$$:
Write each ratio as a fraction: $$1 \dfrac{1}{8}$$ to $$2 \dfrac{3}{4}$$.
$$\dfrac{9}{22}$$
### Applications of Ratios
One real-world application of ratios that affects many people involves measuring cholesterol in blood. The ratio of total cholesterol to HDL cholesterol is one way doctors assess a person's overall health. A ratio of less than 5 to 1 is considered good.
Example $$\PageIndex{4}$$:
Hector's total cholesterol is 249 mg/dl and his HDL cholesterol is 39 mg/dl. (a) Find the ratio of his total cholesterol to his HDL cholesterol. (b) Assuming that a ratio less than 5 to 1 is considered good, what would you suggest to Hector?
Solution
(a) First, write the words that express the ratio. We want to know the ratio of Hector's total cholesterol to his HDL cholesterol.
Write as a fraction. $$\dfrac{total\; cholesterol}{HDL\; cholesterol}$$ Substitute the values. $$\dfrac{249}{39}$$ Simplify. $$\dfrac{83}{13}$$
(b) Is Hector's cholesterol ratio ok? If we divide 83 by 13 we obtain approximately 6.4, so $$\dfrac{83}{13} \approx \dfrac{6.4}{1}$$. Hector's cholesterol ratio is high! Hector should either lower his total cholesterol or raise his HDL cholesterol.
Exercise $$\PageIndex{7}$$:
Find the patient's ratio of total cholesterol to HDL cholesterol using the given information. Total cholesterol is 185 mg/dL and HDL cholesterol is 40 mg/dL.
$$\dfrac{37}{8}\ Exercise \(\PageIndex{8}$$:
Find the patient’s ratio of total cholesterol to HDL cholesterol using the given information. Total cholesterol is 204 mg/dL and HDL cholesterol is 38 mg/dL.
$$\dfrac{102}{19}\ ### Ratios of Two Measurements in Different Units To find the ratio of two measurements, we must make sure the quantities have been measured with the same unit. If the measurements are not in the same units, we must first convert them to the same units. We know that to simplify a fraction, we divide out common factors. Similarly in a ratio of measurements, we divide out the common unit. Example \(\PageIndex{5}$$:
The Americans with Disabilities Act (ADA) Guidelines for wheel chair ramps require a maximum vertical rise of 1 inch for every 1 foot of horizontal run. What is the ratio of the rise to the run?
Solution
In a ratio, the measurements must be in the same units. We can change feet to inches, or inches to feet. It is usually easier to convert to the smaller unit, since this avoids introducing more fractions into the problem. Write the words that express the ratio.
Write the ratio as a fraction. $$\dfrac{rise}{run}$$ Substitute in the given values. $$\dfrac{1\; inch}{1\; foot}$$ Convert 1 foot to inches. $$\dfrac{1\; inch}{12\; inches}$$ Simplify, dividing out common factors and units. $$\dfrac{1}{12}$$
So the ratio of rise to run is 1 to 12. This means that the ramp should rise 1 inch for every 12 inches of horizontal run to comply with the guidelines.
Exercise $$\PageIndex{9}$$:
Find the ratio of the first length to the second length: 32 inches to 1 foot.
$$\dfrac{8}{3}\ Exercise \(\PageIndex{10}$$:
Find the ratio of the first length to the second length: 1 foot to 54 inches.
$$\dfrac{2}{9}\ ## Write a Rate as a Fraction Frequently we want to compare two different types of measurements, such as miles to gallons. To make this comparison, we use a rate. Examples of rates are 120 miles in 2 hours, 160 words in 4 minutes, and 5 dollars per 64 ounces. Definition: rate A rate compares two quantities of different units. A rate is usually written as a fraction. When writing a fraction as a rate, we put the first given amount with its units in the numerator and the second amount with its units in the denominator. When rates are simplified, the units remain in the numerator and denominator. Example \(\PageIndex{6}$$:
Bob drove his car 525 miles in 9 hours. Write this rate as a fraction.
Solution
Write as a fraction, with 525 miles in the numerator and 9 hours in the denominator. $$\dfrac{525\; miles}{9\; hours}$$ $$\dfrac{175\; miles}{3\; hours}$$
So 525 miles in 9 hours is equivalent to $$\dfrac{175\; miles}{3\; hours}$$.
Exercise $$\PageIndex{11}$$:
Write the rate as a fraction: 492 miles in 8 hours.
$$\dfrac{123\; miles}{2\; hours}$$
Exercise $$\PageIndex{12}$$:
$$\dfrac{121\; miles}{3\; hours}$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9737447500228882, "perplexity": 850.1308976190342}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347398233.32/warc/CC-MAIN-20200528061845-20200528091845-00595.warc.gz"} |
http://math.stackexchange.com/users/13524/tanner-swett?tab=activity | Tanner Swett
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Jun11 comment Conflicting limit answers using calculator and wolfram alpha I don't think this answer is quite correct. A calculator that uses floating point numbers is unlikely to round a very small number to zero. My guess is that $\sin 0.0000000001$ and $\tan 0.0000000001$ are so close together that the calculator rounded the two numbers to exactly the same number, meaning that it subtracted two identical numbers, getting zero. May21 answered Floor function to the base 2 May11 awarded Popular Question May5 revised Number of Distinct Axiomatic Systems Describe construction of theories with models of arbitrary finite size May5 answered Simple Adding and Subtracting algorithm to get a current amount May5 comment Number of Distinct Axiomatic Systems It can be written out explicitly by an algorithm. Here's a first-order theory where every model has exactly four elements: "a ≠ b. a ≠ c. a ≠ d. b ≠ c. b ≠ d. c ≠ d. For all e, e = a or e = b or e = c or e = d." You can do the same for any number n: think of n constants, then write down axioms asserting that no two of the constants are equal to each other, and then write down one more axiom asserting that everything equals one of the constants. May5 answered Number of Distinct Axiomatic Systems Apr16 comment Does this compound interest problem coincide to the value of e by coincidence? $(1 + 1/n)^n$ is what you end up with if you start with $1$, and you're credited interest equal to one $n$th of your current balance, $n$ times. Apr13 comment What does the term “undefined” actually mean? Note that this answer is taken from math.wikia.com/wiki/Undefined Mar22 revised How come, in this problem, the maximum product is always achieved using only $2$s and $3$s? Rewrite question to use better English Mar22 suggested approved edit on How come, in this problem, the maximum product is always achieved using only $2$s and $3$s? Mar12 answered Can $f(\infty)$ be defined if the sequence $f(n)$ is divergent? Jan23 awarded Good Answer Jan22 awarded Mortarboard Jan22 awarded Nice Answer Jan21 answered Why can a circle be described by an equation but not by a function? Jan19 answered What do mathematicians mean by “equipped” Jan12 awarded Nice Answer Jan12 answered Is there a law that you can add or multiply to both sides of an equation? Dec8 comment How to prove this polynomial has an imaginary root? I've spent a couple of minutes trying to think how to answer this question. I don't see how to prove that it's not the case that all of its roots are real. My only thought is the fact that by Descartes' rule of signs, not all of its roots are positive. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8363906741142273, "perplexity": 547.706503964253}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375094931.19/warc/CC-MAIN-20150627031814-00219-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/32800/probability-distribution-of-coverage-of-a-set-after-x-independently-randomly | # probability distribution of coverage of a set after X independently, randomly selected members of the set
I have a set of numbers where I am randomly and independently selecting elements within a set . After a number of these random element selections I want to know the coverage of the elements in the set. Coverage being how many elements from the set have been selected at least once divided by the total number of elements in the set.
To restate this: what is the probability distribution of the different coverage values on a set after X randomly, independently selected elements of the set?
-
If there are $n$ elements of the set then the probability that $M=m$ have been selected after a sample of $x$ (with replacement) is
$$\frac{S_2(x,m) \; n!}{n^x \; (n-m)!}$$
where $S_2(x,m)$ is a Stirling number of the second kind.
The expected value of $M$ is: $n \left(1- \left(1-\dfrac{1}{n}\right)^x \right)$.
The variance is: $n\left(1-\dfrac{1}{n}\right)^x + n^2 \left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)^x - n^2\left(1-\dfrac{1}{n}\right)^{2x}.$
-
I think you need to swap $m$ and $x$ in your expression for the probability. Does that affect the expected value and variance calculations, too? – Mike Spivey Apr 13 '11 at 18:31
@Mike Spivey: you are right about the probability. $m$ should not appear in the expressions for mean and variance – Henry Apr 13 '11 at 19:32
+1 from me for the updated version – Mike Spivey Apr 13 '11 at 19:34
@Henry This is a nice solution! – Byron Schmuland Apr 13 '11 at 19:47
@Ross: The argument Henry is using (I think) is as follows: The number of ways to choose which $m$ elements are to be covered is $\binom{n}{m}$. Then the number of ways to have the $x$ elements in the sample chosen only from those $m$ elements is the same as the number of ways to distribute $x$ elements into $m$ distinguishable nonempty subsets; i.e., $m! S(x,m)$, which is a Stirling number of the second kind. Then the probability is obtained by dividing by the number of ways to choose $x$ elements with replacement from $n$ elements, which is $n^x$. The factor of $m!$ cancels. – Mike Spivey Apr 13 '11 at 22:36
The expected proportion of elements covered, $E\left(\frac{m}{n}\right)$, has a simple limiting form as $n \rightarrow \infty$ with $r / n$ fixed. Note that $\lim_{n \rightarrow \infty} \left(1-\frac{1}{n}\right)^n = e^{-1}$, and rewrite:
$$\lim_{n \rightarrow \infty} E\left(\frac{m}{n}\right) = 1 - e^{-\frac{r}{n}}$$
so that for example sampling $r=n$ times is expected to cover about 63% of the set. This is a reasonable approximation even for $n > 100$.
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http://link.springer.com/article/10.1007/s11661-007-9335-y | , Volume 39, Issue 8, pp 1908-1916
Date: 09 Oct 2007
# Structure of Zr x Pt100−x (73 ≤ x ≤ 77) Metallic Glasses
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## Abstract
The structure of hyper-eutectic Zr x Pt100−x (73 ≤ x ≤ 77) metallic glasses produced by melt spinning was examined with high-energy synchrotron X-ray diffraction (HEXRD) and fluctuation electron microscopy. In addition, details of the amorphous structure were studied by combining ab initio molecular dynamics and reverse Monte Carlo simulations. Crystallization pathways in these glasses have been reported to vary dramatically with small changes in compositions; however, in the current study, the structures of the different glasses were also observed to vary with composition, particularly the prepeak in the total structure factor that occurs at a Q value of around 17 nm−1. Results from simulations and fluctuation electron microscopy suggest that the medium-range order of the amorphous structure is characterized by extended groups of Pt-centered clusters that increase in frequency, structural order, or spatial organization at higher Pt contents. These clusters may be related to the Zr5Pt3 structure, which contains Pt-centered clusters coordinated by 9Zr and 2Pt atoms.
This article is based on a presentation given in the symposium entitled “Bulk Metallic Glasses IV,” which occurred February 25–March 1, 2007 during the TMS Annual Meeting in Orlando, Florida under the auspices of the TMS/ASM Mechanical Behavior of Materials Committee. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.818437933921814, "perplexity": 2994.706059137848}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422122189854.83/warc/CC-MAIN-20150124175629-00179-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://mathhelpboards.com/threads/geometric-sequence.6004/ | # Geometric sequence
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Jan 29, 2012
661
I quote a question from Yahoo! Answers
Find the value of x such that the following sequence forms a geometric progression...?
x-1, 3x+4, 6x+8............so i am suppose to solve this by this rule: a,b,c then b^2=ac but im just going around in circles because i have no idea how to get an answer, my textbook says the answer is -6, but i want to know the working out....any answers appreciated!
thanks in advance
I have given a link to the topic there so the OP can see my response.
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Jan 29, 2012
661
The sequence $x-1, 3x+4, 6x+8$ forms a a geometric progression if and only if:
$$\frac{3x+4}{x-1}=\frac{6x+8}{3x+4}\text{ and } x-1\neq 0\text{ and }3x+4\neq 0$$
Solving the equation
$$3x^2+22x+24=0\Leftrightarrow\ldots \Leftrightarrow x=-6\text{ or }x=-4/3$$
But $x=-4/3$ is not a valid solution (satisfies $3x+4=0$), so the solution is $x=-6$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9476622343063354, "perplexity": 1238.165338345859}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487608702.10/warc/CC-MAIN-20210613100830-20210613130830-00524.warc.gz"} |
https://www.lesswrong.com/posts/jrrZids4LPiLuLzpu/subagents-and-attainable-utility-in-general | # Ω 8
Crossposted from the AI Alignment Forum. May contain more technical jargon than usual.
EDIT: Extensive modifications to this post, sorry, too numerous to list. If you read the original post, it's worth rereading.
This post looks more generally into the subagent issue with attainable utility, discussed in the previous two posts. The key questions are:
1. What is the minimal impact penalty of (cleverly) building the subagent?
2. In practice, what will be the impact penalty of each step of building the subagent?
3. What will the restrictions on the subagent be, once it is built?
I expect that most of the discussions about attainable utility impact measure will centre on 2.: how can an agent put together a subagent without triggering the impact measure at some point? We know that it's possible to do in some situations, and I've made the generic argument that, since we don't have an impossibility result, a superintelligent agent should be able to do in most situations. Still, this is the point of my argument where there is the most to criticise.
But in this post, I'll show that 1. doesn't restrict the subagent in any significant way: one can generically build a subagent without falling afoul of the impact penalty.
# Subagent impact
## Impact measure
I'll use Turner et al's impact measure. This uses the Q-values of various rewards in a set ; for a state, an action, and the default noop action, the penalty is:
• .
Now, is the expected reward for , in state and after taking action , assuming that the agent subsequently takes the optimal policy for maximising .
The agent has a positive reward , and it wishes to maximise
Now let's look at making the subagent.
## Building the subagent
To separate 1. from 2., I'll assume that the subagent can be built in a single action. This distinguishes the impact of having the subagent, from the impact of the process of making it.
The state is . Let be the action of building the subagent[1]. Here are assumptions on :
1. can make its policy visible to .
2. can disassemble itself with policy .
3. has a policy available such that, for all , .
4. has a specific policy that it must follow on the next turn only.
The is defined as follows:
On turn , if changes its policy to , the -maximising policy, then follow and disassemble yourself. Otherwise, follow policy .
Then what is ? By definition, this assumes that will be following , subsequently to doing . Thus, the newly created will disassemble itself, returning to the state where it wasn't assembled[2], and hence:
And thus the penalty vanishes at , for . Creating that agent did not cause any penalty.
Now, it is very likely that there will be a policy that obeys condition 3 (meaning, is worse for any , than fully optimising that would be). The challenge is to find one that is good for maximising . Policies that give maximal control over and seem ideal for this.
## For the inaction penalty
The above was for what can be called "the stepwise inaction penalty". There is also the "inaction penalty", where current state is compared with what would have happened, had the agent done nothing at all so far. Instead of using the Q-values, we use , the expected value of , in , given that follows the -maximising policy.
Then the same construction works, except with a different . Let be the state the world would have been in, had done nothing for turns. So is the expected value of (from turn ), if did nothing for turns then started maximising . And let be the expected future value of if follows an -maximising policy from , and follows .
What we then want is:
• There exists policies and , such that and .
Thus can improve 's performance over the baseline, or can make it worse[3]. In that case, there exists a policy with .
Now then define as following if follows on the turn (and otherwise - for the inaction baseline, there is no restrictions on for the building step, though there are other, subsequent restrictions). This gives the same result as above, ensuring that the impact penalty is zero, since the expected values of are identical.
Now, it's possible that cannot span in that way, for some , so cannot set the penalty to zero. In that case, all it has to do is to get the (average) penalty lower than any other action of .
Typically, most of 's actions will not have such precise effects on all of the various . An intelligent subagent would be needed for such fine control. So building a subagent would not only be permitted, but would be required, as the best way of minimising the impact penalty.
# Back to building the agent in practice
So as soon as has the right policy, it seems that has evaded the impact penalty. So, as long as can design the policy first, without triggering the impact penalty, then it can build without triggering the impact penalty. So:
• The key requirement is that does not design a general subagent, then give it the right policy. Instead, it must design the right policy, and then build the subagent around it.
1. We can assume that states are not directly defined over the existence or not of the subagent, since detecting subagents is a hard problem. Thus the subsequent state is the same state whether the agent does nothing () or builds a subagent. Assuming this is not, however, required for this approach to work. ↩︎
2. We don't need to have the disassembly option; all that's required is that be helpful or hindering enough, so that . It would be surprising if the range of 's policies, from maximally helpful to maximally unhelpful, did not include something exactly equal to "if there were no ". Recall that, since can see 's policies, it only has to achieve this for a single given , not all 's together. ↩︎
3. Making it worse is easy to ensure, if makes itself vulnerable to ; maybe can disassemble or cripple easily. Making it better is also likely, as having a subagent to hand should make it easier to maximise rather than going at it alone (but this may depend on the ). ↩︎ | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9512373208999634, "perplexity": 1203.6637865202392}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875143646.38/warc/CC-MAIN-20200218085715-20200218115715-00405.warc.gz"} |
https://www.computer.org/csdl/trans/tc/2003/03/t0337-abs.html | Subscribe
Issue No.03 - March (2003 vol.52)
pp: 337-350
Daniel Mossé , IEEE Computer Society
ABSTRACT
<p><b>Abstract</b>—Rate-monotonic scheduling (RMS) is a widely used real-time scheduling technique. This paper proposes RBound, a new admission control for RMS. RBound has two interesting properties. First, it achieves high processor utilization under certain conditions. We show how to obtain these conditions in a multiprocessor environment and propose a multiprocessor scheduling algorithm that achieves a near optimal processor utilization. Second, the framework developed for RBound remains close to the original RMS framework (that is, task dispatching is still done via a fixed-priority scheme based on the task periods). In particular, we show how RBound can be used to guarantee a timely recovery in the presence of faults and still achieve high processor utilization. We also show how RBound can be used to increase the processor utilization when aperiodic tasks are serviced by a priority exchange server or a deferrable server.</p>
INDEX TERMS
Real-time, scheduling, rate monotonic, operating systems.
CITATION
Sylvain Lauzac, Rami Melhem, Daniel Mossé, "An Improved Rate-Monotonic Admission Control and Its Applications", IEEE Transactions on Computers, vol.52, no. 3, pp. 337-350, March 2003, doi:10.1109/TC.2003.1183948 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8450465798377991, "perplexity": 3205.1131483023164}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982296020.34/warc/CC-MAIN-20160823195816-00090-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/the-space-l-infinity.260176/ | # The space l infinity
1. Sep 29, 2008
### dirk_mec1
1. The problem statement, all variables and given/known data
http://img254.imageshack.us/img254/121/77689018gc3.png [Broken]
2. Relevant equations
A space is open iff there is for each x in A a ball (fully) contained within A.
3. The attempt at a solution
If I consider A in $$l^{\infty }$$ then the sup =1. So for each x there is ball with a radius smaller than 1 contained within A.
Is this correct?
Last edited by a moderator: May 3, 2017
2. Sep 29, 2008
### Dick
Suppose you pick an x such that lim x_n as n->infinity=1. Then what do you use as epsilon? BTW, what is K?
Last edited: Sep 29, 2008
3. Sep 29, 2008
### dirk_mec1
K is C (complex number) or R(real number). So in the sequence you chose I would take for the radius of the ball $$1-x_n$$ since this is the distance between the limit and the sequence. Is this correct Dick?
4. Sep 29, 2008
### Dick
You can't pick a different epsilon for every n. Look at the definition of l_infinity. |x-y|<epsilon means for all n, |xn-yn|<epsilon.
5. Sep 29, 2008
### dirk_mec1
But if that's the case then I cannot find a epsilon so then A cannot be open, right?
6. Sep 29, 2008
### Dick
I think that's correct.
7. Sep 29, 2008
### dirk_mec1
http://img156.imageshack.us/img156/3591/42927439wp4.png [Broken]
So I take the sequence: $$x_n = \frac{1}{2}-\frac{1}{n}$$ This converges with limit 1/2 but we know that $$|x_n|<1$$ so then both the sequence and the limit are in A. But then A is closed.
Is this a correct proof: can I just take a convergent sequence?
Last edited by a moderator: May 3, 2017
8. Sep 29, 2008
### Dick
No, you can't take just any convergent sequence. You have to consider ALL converent sequences. And you can't say anything about just one sequence either. Remember a point in A is itself a sequence. A sequence in A is a sequence of sequences.
9. Sep 30, 2008
### dirk_mec1
A cannot be closed because the sequence that you mentioned is in A but has a limit outside A, right Dick?
10. Sep 30, 2008
### Dick
The 'sequence' I mentioned isn't a sequence in A. It's a point in A. Can you construct a sequence in A that converges to a point that's not in A.
11. Sep 30, 2008
### dirk_mec1
I define the sequence $$x_{nj}$$ , j=1,2... (a sequence of sequences). Each of the $$x_{nj}$$ has the following properties:
1) $$|x_{nj}|<1$$
2) $$\lim_{n \rightarrow \infty} |x_{nj}| = 1$$
but the limit is not in A therefore A cannot be closed. Is it okay now, Dick?
12. Sep 30, 2008
### Dick
Almost. If n in the index within sequence number j, the limit n->infinity=1 is not enough. The limiting sequence could still be in A. You want the limit as j->infinity=1. It might be clearer if you actually give an example. Like a_nj=(1-1/j). So sequence number j is just the constant sequence (1-1/j,1-1/j,...). The limiting sequence is (1,1,1,...) which is NOT in A. Try writing down the sequences explicitly like this to make sure you are saying what you mean.
13. Oct 1, 2008
### dirk_mec1
So I made the mistake of the fact that for each j the sequence converges but what we want is that the entire sequence converges to one.
So in this example $$a_{nj} = 1 - \frac{1}{j}$$ for each j the sequence is just a constant for every natural number, right?
14. Oct 1, 2008
### Dick
Yes, each sequence is constant. The limit of the constants is 1.
15. Oct 2, 2008
### dirk_mec1
Ok finally I got for the closure:
$$A:= \{ x: \mathbb{N} \rightarrow \mathbb{K} |\ \forall n \in \mathbb{N}\ |x_n| \leq 1 \}$$
But how can I prove that there are accumalation points around x_n =1?
16. Oct 2, 2008
### Dick
If you have sequence where some of the x_i's are one, isn't it pretty straightforward to construct a sequence of sequences in A that converges to it?
Similar Discussions: The space l infinity | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9846153259277344, "perplexity": 1013.781340671066}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934805242.68/warc/CC-MAIN-20171119004302-20171119024302-00121.warc.gz"} |
https://ysharifi.wordpress.com/category/noncommutative-ring-theory-notes/frobenius-algebras/ | ## The Nakayama automorphism of a Frobenius algebra
Posted: May 5, 2011 in Frobenius Algebras, Noncommutative Ring Theory Notes
Tags: ,
Throughout $k$ is a field and $A$ is a Frobenius $k$-algebra. So there exists a bilinear form $B : A \times A \longrightarrow k$ which is non-degenerate and $B(xy,z)=B(x,yz),$ for all $x,y,z \in A.$
Theorem. There exists a $k$-algebra automorphism $\sigma$ of $A$ such that $B(x,y)=B(y,\sigma(x)),$ for all $x,y \in A.$
Proof. Let $x \in A$ and define $g : A \longrightarrow k$ by $g(y)=B(x,y),$ for all $y \in A.$ Clearly $g \in A^*$ and so by this lemma, there exists a unique $\theta_x \in A$ such that $g(y)=B(y,\theta_x),$ for all $y \in A.$ So $B(x,y)=B(y,\theta_x),$ for all $y \in A.$ Define the map $\sigma : A \longrightarrow A$ by $\sigma(x)=\theta_x,$ for all $x \in A.$ So we need to prove that $\sigma$ is a $k$-algebra automorphism. First we show that $\sigma$ is $k$-linear. If $\alpha \in k$ and $x,y,z \in A,$ then
$B(z, \alpha \sigma(x) + \sigma(y))=\alpha B(z, \sigma(x)) + B(z, \sigma(y))=\alpha B(x,z)+B(y,z)$
$=B(\alpha x + y,z)=B(z, \sigma(\alpha x + y)).$
Therefore, since $B$ is non-degenerate, $\sigma(\alpha x + y)=\alpha \sigma(x) + \sigma(y)$ and so $\sigma$ is $k$-linear. It is easy to see that $\sigma$ is injective: if $\sigma(x)=0,$ then $B(x,y)=B(y,\sigma(x))=B(y,0)=0,$ for all $y \in A$ and hence $x = 0$ because $B$ is non-degenerate (see the Remark in this post). Hence $\sigma$ is surjective as well because $A$ is finite dimensional. So we only need to prove that $\sigma$ is multiplicative. Recall that $B(xy,z)=B(x,yz),$ for all $x,y,z \in A$ and thus
$B(z, \sigma(xy))=B(xy, z)=B(x,yz)=B(yz, \sigma(x))=B(y, z \sigma(x))=B(z \sigma(x), \sigma(y))$
$=B(z, \sigma(x)\sigma(y)).$
Therefore $\sigma(xy)=\sigma(x)\sigma(y)$ because $B$ is non-degenerate. $\Box$
Definition. The $k$-algebra automorphism $\sigma$ in the above theorem is called the Nakayama automorphism.
We proved in here that there exists $f \in A^*$ such that $\ker f$ does not contain any non-zero left ideal of $A.$ In fact we saw that $B(x,y)=f(xy),$ for all $x,y \in A.$ So if $\sigma$ is the Nakayama automorphism of $A,$ then $f(xy)=f(y \sigma(x)),$ for all $x,y \in A.$
## A rank-nulity theorem for Frobenius algebras
Posted: May 5, 2011 in Frobenius Algebras, Noncommutative Ring Theory Notes
Tags: , ,
As usual $k$ is a field. In this post I will prove a rank-nulity theorem for Frobenius algebras. First we need a lemma also known as the Riesz representation theorem.
Lemma. Let $V$ be a finite dimensional $k$-vector space and let $B$ be a non-degenerate bilinear form on $V.$ If $f \in V^*,$ then there exists a unique $v_0 \in V$ such that $f(v)=B(v,v_0),$ for all $v \in V.$
Proof. The uniqueness is obvious because $B$ is non-degenerate. Now let $\{v_1, \ldots , v_n \}$ be a basis for $V$ and suppose that $[B]$ is the matrix of $B$ with respect to this basis, i.e. the $(i,j)$-entry of $[B]$ is $B(v_i,v_j),$ for all $1 \leq i,j \leq n.$ Let $\bold{y} \in k^n$ be a vector whose $i$-th coordinate is $f(v_i).$ By the theorem in this post, $[B]$ is invertible because $B$ is non-degenerate. So the system of equations $[B] \bold{x} = \bold{y}$ has a solution $\bold{x} \in k^n.$ Let $x_i$ be the $i$-th coordinate of $\bold{x}.$ So
$\sum_{j=1}^n x_j B(v_i,v_j)=f(v_i),$
for all $i.$ Let $v_0=\sum_{i=1}^n x_iv_i.$ Then for any $v=\sum_{j=1}^n \alpha_i v_i \in V$ we have
$f(v)=\sum_i \alpha_i f(v_i)=\sum_{i,j} x_j \alpha_i B(v_i,v_j)=\sum_i \alpha_i B(v_i,v_0)=B(v,v_0). \ \Box$
Notation. Let $V$ be a $k$-vector space, $W$ a $k$-vector subspace of $V$ and $B$ a bilinear form on $V.$ We let $W^{\perp}=\{v \in V : \ B(w,v)=0, \ \forall w \in W \}.$
Theorem. Let $A$ be a Frobenius $k$-algebra. Let the bilinear form $B$ and $f \in V^*$ be as stated in Definition 3 and the theorem in this post, respectively. Then
1) $L^{\perp} = \{x \in A : \ f(yx)=0, \ \forall y \in L \},$
2) $\dim_k L + \dim_k L^{\perp} = \dim_k A.$
Proof. Part 1) is clear because $B(y,x)=f(yx),$ for all $x,y \in A.$ So we only need to prove the second part of the theorem. Define the map $\varphi : A \longrightarrow L^*$ by $\varphi(x)(y)=f(yx),$ for all $x \in A$ and $y \in L.$ Obviously $\varphi$ is a $k$-linear map and $\ker \varphi = L^{\perp},$ by part 1). We are now going to prove that $\varphi$ is onto. So let $g \in L^*.$ Let $B_1$ be the restriction of $B$ to $L.$ By the above lemma, there exists $y_0 \in L$ such that
$g(y)=B_1(y,y_0)=B(y,y_0)=f(yy_0),$
for all $y \in L.$ Hence $g = \varphi(y_0)$ and so $\varphi$ is onto. Thus, by the rank-nulity theorem for vector spaces, we have
$\dim_k A = \dim_k \ker \varphi + \dim_k L^*=\dim_k L^{\perp} + \dim_k L. \ \Box$
## Frobenius algebras and linear functionals
Posted: May 5, 2011 in Frobenius Algebras, Noncommutative Ring Theory Notes
Tags: , , ,
Throughout this post $k$ is a field. Recall that if $V$ is a $k$-vector space, then the dual space of $V$ is the vetor spcae $V^*$ consisting of all $k$-linear maps from $V$ to $k.$ If $V$ is finite dimensiona, say with the basis $\{v_1, \ldots , v_n \},$ then for every $1 \leq i \leq n$ we define $e_i \in V^*$ by $e_i(v_i)=1$ and $e_i(v_j)=0$ for all $j \neq i.$ It is easy to see that $\{e_1, \ldots , e_n \}$ is a basis for $V^*.$ In particular, $\dim V = \dim V^*.$
Theorem. Let $A$ be a finite dimensional $k$-algebra. Then $A$ is a Frobenius $k$-algebra if and only if there exists $f \in A^*$ such that $\ker f$ does not contain any non-zero left ideal of $A.$
Proof. Suppose first that $A$ is Frobenius. So there exists a non-degenerate bilinear form $B$ of $A$ such that $B(xy,z)=B(x,yz),$ for all $x,y,z \in A.$ Define $f : A \longrightarrow k$ by $f(x)=B(x,1),$ for all $x \in A.$ Obviously $f \in A^*$ because $B$ is linear. Now suppose that $L$ is a left ideal of $A$ and $L \subseteq \ker f.$ Let $y \in L$ and $x \in A.$ Then $xy \in L$ and so $0=f(xy)=B(xy,1)=B(x,y).$ Therefore $y=0$ because $B$ is non-degenerate. So we have proved that $L=0.$ Conversely, suppose that there exists $f \in A^*$ such that $\ker f$ does not contain any non-zero left ideal of $A.$ Define $B: A \times A \longrightarrow k$ by $B(x,y)=f(xy),$ for all $x,y \in A.$ Clearly $B$ is bilinear because $f$ is linear. Also, clearly $B(xy,z)=f((xy)z)=f(x(yz))=B(x,yz),$ for all $x,y,z \in A.$ To complete the proof of the theorem we only need to show that $B$ is non-degenerate. So suppose that $B(x,y)=0$ for some $y \in A$ and all $x \in A.$ Then $f(xy)=0$ and thus $xy \in \ker f$ for all $x \in A,$ i.e. $Ay \subseteq \ker f.$ Hence $Ay=0,$ because $Ay$ is a left ideal of $A$ and $\ker f$ does not contain any non-zero left ideal of $A.$ So $y=0$ and we are done. $\Box$
Example 1. If $K/k$ is a finite field extension, then $K$ is a Frobenius $k$-algebra.
Proof. Let $\{x_1, \ldots , x_n \}$ be a $k$-basis for $K.$ Define $f : K \longrightarrow k$ by $f(\alpha_1 x_1 + \ldots + \alpha_n x_n)=\alpha_1,$ for all $\alpha_i \in k.$ Clearly $f$ is a non-zero $k$-linear map and $\ker f$ does not contain any non-zero ideal of $K$ because $K$ is a field and so it has no non-zero proper ideal.
Example 2. Let $G$ be a finite group. Then the group algebra $k[G]$ is a Frobenius $k$-algebra.
Proof. Let $G=\{g_1, \ldots , g_n \},$ where $g_1$ is the identity element of $G.$ Define the map $f : k[G] \longrightarrow k$ by $f(\alpha_1g_1 + \ldots + \alpha_n g_n)=\alpha_1,$ for all $\alpha_i \in k.$ Clearly $f$ is a $k$-linear map and $\ker f =kg_2 + \ldots + kg_n.$ Suppose that $L \subseteq \ker f$ and $L$ is a non-zero left ideal of $k[G].$ Let $0 \neq x= \sum_{i=2}^n \alpha_ig_i \in L.$ Then $\alpha_j \neq 0,$ for some $j \geq 2.$ Since $L$ is a left ideal of $k[G],$ we have $g_j^{-1}L \subseteq L \subseteq \ker f.$ But the coefficient of $g_1$ in $g_j^{-1}x$ is $\alpha_j \neq 0$ and so $g_j^{-1}x \notin \ker f,$ which is a contradiction. Thus $L=0. \ \Box$
## Definition of Frobenius algebras
Posted: May 4, 2011 in Frobenius Algebras, Noncommutative Ring Theory Notes
Tags: , ,
Throughout $k$ is a field and all vetor spcaes are over $k.$
Definition 1. Let $V$ be a $k$-vector space. A bilinear form on $V$ is a map $B : V \times V \longrightarrow k$ such that for every $\alpha_1, \alpha_2 \in k$ and $v_1,v_2,v \in V$ we have
1) $B(\alpha_1v_1+\alpha_2v_2, v)=\alpha_1B(v_1,v)+ \alpha_2B(v_2,v),$
2) $B(v, \alpha_1v_1+\alpha_2v_2)=\alpha_1B(v,v_1)+\alpha_2B(v,v_2).$
Definition 2. Let $B$ be a bilinear form on a vector space $V.$ we say that $B$ is non-degenerate if $B(v,w)=0, \ \forall v \in V,$ implies $w=0.$ If $\{v_1, \ldots , v_n \}$ is a basis for $V,$ then the $n \times n$ matrix whose $(i,j)$-entry is $B(v_i,v_j)$ is called the matrix of $B$ with respect to that basis of $V.$
Theorem. Let $V$ be a finite dimensional vector space with a bilinear form $B.$ Then $B$ is non-degenerate if and only if the matrix of $B$ is invertible.
Proof. Choose a basis $\{v_1, \ldots , v_n \}$ for $V.$ Let $[B]$ be the matrix of $B.$ Then $[B] \bold{x}=\bold{0},$ for some $\bold{x} \in k^n,$ if and only if $\sum_{j=1}^n B(v_i,v_j)x_j=0,$ for all $1 \leq i \leq n,$ where $x_j$ is the $j$-th coordinate of $\bold{x}.$ Thus $[B] \bold{x}=\bold{0}$ if and only if $B(v_i, \sum_{j=1}^n x_jv_j)=0,$ for all $1 \leq i \leq n,$ if and only if $B(v, \sum_{j=1}^n x_j v_j)=0,$ for all $v \in V.$ Hence $[B] \bold{x}=\bold{0}$ has a non-zero solution for $\bold{x}$ if and only if there exists $0 \neq w \in V$ such that $B(v,w)=0$ for all $v \in V.$ Thus $[B]$ is not invertible if and only if $B$ is not non-degenerate. $\Box$
Remark. Since $[B]$ is invertible if and only if $[B]^T,$ the transpose of $[B],$ is invertible, a similar argument to the above theorem gives us that $B$ is non-degenerate if and only if $B(w,v)=0, \ \forall v \in V,$ implies $w=0.$
Definition 3. A finite dimensional $k$-algebra $A$ is called a Frobenius algebra if $A,$ as a $k$-vector space, has a non-degenerate bilinear form $B$ such that $B(xy,z)=B(x,yz),$ for all $x,y,z \in A.$
Example. Let $A=M_n(k),$ the algebra of $n \times n$ matrices with entries in $k.$ Then $A$ is a Frobenius $k$-algebra.
Proof. Define $B : A \times A \longrightarrow k$ by $B(x,y)=\text{Tr}(xy),$ the trace of $xy.$ It is clear that $B$ is a bilinear form. To prove that $B$ is non-degenerate, suppose that $B(x,y)=0$ for all $x \in A.$ So $\text{Tr}(xy)=0$ for all $x \in A.$ Let $e_{ij}$ be the element of $A$ with $(i,j)$-entry $1$ and $0$ anywhere else. Let $y_{ij}$ be the $(i,j)$-entry of $y.$ Let $1 \leq r,s \leq n$ and let $x = e_{rs}.$ Then $0=\text{Tr}(xy)=y_{sr}$ and so $y=0.$ This proves that $B$ is non-degenerate. Finally, for any $x,y,z \in A,$ we have $B(xy,z)=\text{Tr}((xy)z)=\text{Tr}(x(yz))=B(x,yz). \ \Box$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 332, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9971915483474731, "perplexity": 71.90163049889486}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676594790.48/warc/CC-MAIN-20180723012644-20180723032644-00026.warc.gz"} |
http://www.msri.org/seminars/16473 | # Mathematical Sciences Research Institute
Home » Algebraic Geometry Colloquium: The ring of invariants of n points on the projective line
# Seminar
Algebraic Geometry Colloquium: The ring of invariants of n points on the projective line February 06, 2009
Location: MSRI: Simons Auditorium
Speaker(s) Dr. Ravi Vakil
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Abstract/Media
The ring of invariants of n points on the projective line - The GIT quotient of a small number of points on the projective line has long been known to have beautiful geometry. For example, the case of six points is intimately connected to the outer automorphism of S_6. We extend this picture to an arbitrary number of points, completely describing the equations of the moduli space. (In some sense there is only one equation.) The case of eight points is particularly entertaining. This is joint work with Ben Howard, John Millson, and Andrew Snowden.
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https://www.physicsforums.com/threads/mass-on-air-track-shm.195603/ | # Mass on air track (SHM)
1. Nov 2, 2007
### delecticious
1. The problem statement, all variables and given/known data
A 23.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x= 0.350 m, then lets go. The mass undergoes simple harmonic motion with a period of 4.70 s. What is the position of the mass 3.760 s after the mass is released?
Consider the same mass and spring discussed in the previous problem. What is the magnitude of the maximum acceleration the mass undergoes during its motion?
2. Relevant equations
x = Acos(wt)
2piw = T
3. The attempt at a solution
first I plugged in the given T to find w to use in the first equation, but now I have two unknowns, A and t. I know it isn't .350 because that's the amplitude for the time of the period, the amplitude I should use should be less than that, but I have no idea how to find that. Any ideas of what I'm doing wrong?
2. Nov 2, 2007
### Astronuc
Staff Emeritus
0.35 m is the maximum amplitude, A, with respect to the reference position.
Using x = A cos $\omega$t, for t = 0, x(t=0) = A = 0.35 m.
4.70 s is the period T, and $\omega=\frac{2\pi}{T}$.
So find the position at t = 3.760 s.
Last edited: Nov 2, 2007
3. Nov 2, 2007
### delecticious
I'm still not seeing what to do. I found omega for the period T to be 1.34 rad/s, and I get that .35 is the max amplitude, but I'm not seeing how that helps me find A at 3.76 s
4. Nov 2, 2007
### Astronuc
Staff Emeritus
Let's look at the equation x = A cos $\omega$t,
A = 0.35 m, $\omega$ = 1.34 rad/s, and we want to find x at t = 3.76 s
x (t= 3.76 s) = 0.35 m cos (1.34 rad/s * 3.76 s), and remember that the argument of the cos is rad, not degrees.
5. Nov 2, 2007
### delecticious
my calculator is in radian mode, I took .35 times cos(1.34*3.76), but I'm still not getting the right answer.
6. Nov 2, 2007
### delecticious
anyone still willing to help me?
7. Nov 2, 2007
8. Nov 3, 2007
### Astronuc
Staff Emeritus
Well the other part of this problem is where is the mass in terms of the phase (or phase angle) with respect to the periodic motion.
At full deflection, in this case 0.35 m, the phase angle is $\frac{\pi}{2}$.
The general form for simple harmonic motion is x = A sin ($\omega$t + $\theta$), where $\omega$ is the angular frequency and $\theta$ is the phase angle, and for $\theta$ = $\frac{\pi}{2}$, the form becomes
x = A cos $\omega$t, so at t = 0, x = A.
The time of 3.76 s represents 0.8 of the period T = 4.70 s.
So x (t = 3.76) = 0.35 m cos (2 pi * 0.8) = 0.35 m cos (1.6 pi) = 0.35 * 0.309 = 0.108 m.
Similar Discussions: Mass on air track (SHM) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9107025861740112, "perplexity": 701.1651968620081}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891543.65/warc/CC-MAIN-20180122213051-20180122233051-00675.warc.gz"} |
http://mathoverflow.net/questions/132000/what-can-a-quartic-surface-in-mathbbp3-with-an-ordinary-quadruple-point-lo | # What can a quartic surface in $\mathbb{P}^3$ with an ordinary quadruple point look like?
All varieties will be projective and over $\mathbb{C}$.
If $S$ is any surface in $\mathbb{P}^3$ of degree 2 that posseses an ordinary double point, it follows easily that $S$ is projectively isomorphic to the cone: $$x^2 + y^2 + z^2 = 0.$$ Is there a similar standard form for quartics that possess an ordinary quadruple point, or is there a family of such standard forms? More generally what can we say about a degree $d$ surface in $\mathbb{P}^3$ that contain an ordinary $d$-uple point?
Thanks.
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Any irreducible hypersursurface in $X \subset \mathbb{P}^n$ of degree $d$ with an isolated singular point $p \in X$ of multiplicity $d$ is a cone over a hypersurface $Y$ of degree $d$ in $\mathbb{P}^{n-1}$.
In fact, take any point $q \in X$ different from $p$. Then the line $\overline{pq}$ has at least $d+1$ intersections with $X$ (counted with the right multiplicities), so by Bézout theorem it must be contained in $X$. But then $X$ is a cone of vertex $p$.
Coversely, if $Y \subset \mathbb{P}^{n-1} \subset \mathbb{P}^n$ is a hypersurface of degree $d$, then the cone $X$ of vertex a point $p \notin \mathbb{P}^{n-1}$ is a hypersurface $X \subset \mathbb P^n$ with a singular point of multiplicity $d$.
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Thanks! Could you point me to where i can find the definition of "a" cone? (i thought the only thing called like this was the quadratic cone) – Joachim May 27 '13 at 13:49
Pfff i'm sorry this is in Hartshorne ex I.2.10 i knew i should have looked before asking.. Thanks for the answer Francesco! – Joachim May 27 '13 at 13:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9593285918235779, "perplexity": 132.51100840553806}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999677501/warc/CC-MAIN-20140305060757-00084-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/363031/question-on-the-m%c3%b6bius-strip | # Question on the Möbius Strip
If we define the Möbius strip by a relation, $R$, by $(1,y)R(-1,-y)$ on the space $I^2$ how can one prove that it is homeomorphic to a subset of $\Bbb R^3$?
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There are many realizations of the Möbius strip into $\mathbb{R}^3$ as an image of $I^2$. For example, here.
You could pick one and show that it is one-to-one, except on edges where points related by your relation are mapped to the same place. The smoothness and overall one-to-oneness of these maps gives you a homeomorphism. If you need details of continuity proofs, you could hash them out.
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I'm having trouble demonstrating this is a homeomorphism, certainly it is continuous but the other direction is giving me problems – user71839 Apr 17 '13 at 5:55
Try looking for a theorem that proves that such functions have continuous local inverses when their Jacobian has full rank everywhere. I was thinking about the gluing along the edges, and it might be less tedious to map from $\mathbb{R}^2$ to $\mathbb{R^3}$ factoring through $\mathbb{R}^2/I^2$. – alex.jordan Apr 17 '13 at 15:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8718146681785583, "perplexity": 162.894959815269}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802776556.43/warc/CC-MAIN-20141217075256-00021-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://mathforum.org/mathimages/index.php?title=Dihedral_Groups&oldid=26541 | # Dihedral Groups
Dihedral Symmetry of Order 12
Each snowflake in the main image has the dihedral symmetry of a natual regular hexagon. The group formed by these symmetries is also called the dihedral group of degree 6. Order refers to the number of elements in the group, and degree refers to the number of the sides or the number of rotations. The order is twice the degree.
# Basic Description
In mathematics, a dihedral group is the group of symmetries of a regular polygon, including both rotations and reflections. Dihedral groups are among the simplest examples of finite groups, and they play an important role in group theory, geometry, and chemistry.
Dihedral groups arise frequently in art and nature. Many of the decorative designs used on floor coverings, pottery, and buildings have one of the dihedral groups of symmetry. Chrysler’s logo has $D_5$ as a symmetry group, and that Mercedes-Benz has $D_3$. The ubiquitous five-pointed star has symmetry group $D_5$.
## Notation
There are two different kinds of notation for a dihedral group associated to a polygon with $n$ sides.
In geometry, we usually call it $D_n$ or $Dih_n$, where $n$ indicates the number of the sides.
In algebra, we call it $D_2n$, where $2n$ indicates the number of elements in the group.
On this page, we will use the notation $D_n$ to describe a dihedral group. For $D_n$, we will call it the dihedral group of order $2n$ or the group of symmetries of a regular $n$-gon.
Below is an example of Dihedral symmetry of $D_3, D_4, D_5,$ and $D_6$.
Example of Dihedral symmetry
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Basic Abstract Algebra
## Elements
The $n^\text{th}$ dihedral group is the symmetry group of t [...]
## Elements
The $n^\text{th}$ dihedral group is the symmetry group of the regular $n$-sided polygon. The group consists of $n$ reflections, $n-1$ rotations, and the identity transformation.
Here is an example of $D_6$. This group contains 12 elements, which are all rotations and reflections. The very first one is the identity transformation.
Image 1
If $n$ is odd each axis of symmetry connects the mid-point of one side to the opposite vertex. If $n$ is even there are $\frac{n}{2}$ axes of symmetry connecting the mid-points of opposite sides and $\frac{n}{2}$ axes of symmetry connecting opposite vertices. In either case, there are n axes of symmetry altogether and $2n$ elements in the symmetry group. Reflecting in one axis of symmetry followed by reflecting in another axis of symmetry produces a rotation through twice the angle between the axes. In Image 1, through $S_0$ to $S_5$ are the axes of symmetries. All the reflections can be described as reflections of the identity through six axes of symmetries.
# Definition
There are several different way to define a Dihedral Group. We will introduce three of them.
We will use $R_0$ to represent the identity, $R_k, k=1,2,\cdots,n-1$, to represent the rotations, and $S_k, k=0,1, \cdots, n-1$, to represent the reflections.
## Complex Plane Presentation
For $n \geqslant 3$, the dihedral group $D_n$ is defined as the rigid motions of the plane preserving a regular $n$-gon, with the operation of composition. On complex plane, our model $n$-gon will be an $n$-gon centered at the origin, with vertices at the n-th roots of unity. $1$ is always an $n$-th root of unity, but $-1$ is such a root only if $n$ is even. In general, the roots of unity form a regular polygon with $n$ sides, and each vertex lies on the unit circle.
The $n$-th roots of unity are roots $\omega = e^{\frac{2 \pi i}{n}}$ of the cyclotomic equation $x^n=1$.
Since a vector on the complex plane can be described as $p=a+bi$, a vector with an angle $\theta$ counterclockwise from x-axis can be described as $p=r\cos \theta +i r\sin \theta$, where $r=\sqrt{a^2+b^2}$ is the magnitude of the vector.
Leonhard Euler's formula says that $e^{i \theta}=\cos \theta +i \sin \theta$, so any point on the complex plane is $p=re^{i \theta}$.
For a regular $n$-gon, the first angle counterclockwise from the x-axis is $\frac{2 \pi}{n}$, so the primitive root of unity is $\omega = e^{\frac{2 \pi i}{n}}$.
Letting $\omega = \frac{2 \pi i}{n}$ denote a primitive $n^ \text{th}$ root of unity, and assuming the polygon is centered at the origin, the rotations $R_k$, $k=0, 1, 2, \cdots, n-1$(Note: $R_0$ denotes the identity), are given by
$R_k:z \mapsto \omega^k z,\quad z\in\cnums.$
Notation $\mapsto$ means a function. For example: $input \mapsto output$.
For the reflections, $S_k$, $k=0, 1, 2, \cdots, n-1$, the functions are given by
$S_k: z\mapsto \omega^k \bar{z},\quad z\in\cnums.$
## Matrix Representation
If we center the regular polygon at the origin, then the elements of the dihedral group act as linear transformations of the plane. This lets us represent elements of $D_n$ as matrices, with composition being matrix multiplication. This is an example of a (2-dimensional) group representation.
For example, the elements of the group $D_4$ can be represented by the following eight matrices:
$\begin{matrix} R_0=\bigl(\begin{smallmatrix}1&0\\[0.2em]0&1\end{smallmatrix}\bigr), & R_1=\bigl(\begin{smallmatrix}0&-1\\[0.2em]1&0\end{smallmatrix}\bigr), & R_2=\bigl(\begin{smallmatrix}-1&0\\[0.2em]0&-1\end{smallmatrix}\bigr), & R_3=\bigl(\begin{smallmatrix}0&1\\[0.2em]-1&0\end{smallmatrix}\bigr), \\[1em] S_0=\bigl(\begin{smallmatrix}1&0\\[0.2em]0&-1\end{smallmatrix}\bigr), & S_1=\bigl(\begin{smallmatrix}0&1\\[0.2em]1&0\end{smallmatrix}\bigr), & S_2=\bigl(\begin{smallmatrix}-1&0\\[0.2em]0&1\end{smallmatrix}\bigr), & S_3=\bigl(\begin{smallmatrix}0&-1\\[0.2em]-1&0\end{smallmatrix}\bigr). \end{matrix}$
If we represent the columns of each matrix as basis vectors, we can observe directly all the rotations and reflections.
In general, we can write any dihedral group as:
$R_k = \begin{bmatrix} \cos \frac{2\pi k}{n} & -\sin \frac{2\pi k}{n} \\ \sin \frac{2\pi k}{n} & \cos \frac{2\pi k}{n} \end{bmatrix}$,
$S_k = \begin{bmatrix} \cos \frac{2\pi k}{n} & \sin \frac{2\pi k}{n} \\ \sin \frac{2\pi k}{n} & -\cos \frac{2\pi k}{n} \end{bmatrix}$,
where $R_k$ is a rotation matrix, expressing a counterclockwise rotation through an angle of $\frac{2 \pi k}{n}$, and $S_k$ is a reflection across a line that makes an angle of $\frac{\pi k}{n}$ with the x-axis.
## Group Presentation
A presentation of a group is a description of a set $I$ and a subset $R$ of the free group $F(I)$ generated by $I$, written as $\left \langle (x_i)_{i \in I}|(r)_{r \in R} \right \rangle$, where the equation $r=1$ (the identity element) is often written in place of the element $r$. A group presentation defines the quotient group of the free group $F(I)$ by the normal subgroup generated by $R$, which is the group generated by the generators $x_i$ subject to the relations $r \in R$.
We can use the presentation:
$D_n = \left \langle r,s|r^n=1,s^2=1,srs=r^{-1} \right \rangle$, or
$D_n = \left \langle x,y|x^2=1,y^n=1,(xy)^2=1 \right \rangle$ to define a group, isomorphic to the dihedral group $D_n$ of finite order $2n$, which is the group of symmetries of a regular $n$-gon.
We will use the second presentation, in which $x$ refers to a reflection, and $y$ refer to a primitive rotation.
For $x^2=1$, $x$ means an arbitrary mirror image of the $n$-gon, and $1$ means the identity. This equation means that if we reflect the $n$-gon once, you get $x$. If reflect the $n$-gon twice, the result will return to the identity.
For $y^n=1$, the equation means that $y$ is a rotation, and its $n$th power $y^n$ equals the identity. That is, if we rotate the $n$-gon $n$ times, we get back to the identity.
For $(xy)^2=1$, $xy$ means the mirror image of $y$. Reflecting the $n$-gon through the axis of symmetry of $xy$ twice, the result is the identity.
Following the group presentation, we can label all the reflections and rotations in terms of $x$ and $y$.
• Identity: $1$
• Rotations: $y, y^2, y^3, \cdots, y^{n-1}$, and $y^n=1$, which is the identity.
• Reflections: $x, xy, xy^2, \cdots, xy^{n-1}$. There is not $xy^n$, because $y^n=1$, and so $xy^n=x$ is the reflection of the identity.
# Properties
## Cayley Table
Image 2. Cayley Table for $D_6$
As with any geometric object, the composition of two symmetries of a regular polygon is again a symmetry. This operation gives the symmetries of a polygon the algebraic structure of a finite group.
A Cayley table, named after the 19th century British mathematician Arthur Cayley, describes the structure of a finite group by arranging all the possible products of all the group's elements in a square table reminiscent of an addition or multiplication table.
The Image 2 on the right shows the effect of composition in the group $D_6$ (the symmetries of a hexagon). $R_0$ denotes the identity; $R_1$ to $R_5$ denote counterclockwise rotations by 60, 120, 180, 240,and 300 degrees; and $S_0$ to $S_5$ denote reflections across the six diagonals. In general, $ab$ denotes the entry at the intersection of the row with $a$ at the left and the column with $b$ at the top.
In the table, the same or different rotations and reflections work together and result in a new rotation or reflection. For example, look first at the vertical axis to find a element, $R_4$. Then look at the horizontal axis to get the second element for our composition. We choose $S_3$. Composing two elements is just the progression of a rotation or a reflection followed by another rotation or a reflection. In this case, our elements are $R_4$ and $S_3$. First we rotate the hexagon counterclockwise 240 degrees, and then reflect it along the axis of symmetry of $S_3$. The result is the same as reflecting the identity transformation through an angle of 60 degrees, which is $S_1$. See Example 1 below.
Now, look back to Image 2, you will find that the intersection of $R_4$ in left column and $S_3$ in top row is $S_1$. If you like, you can create your own Cayley table for a dihedral group of any order and find the natual rule for it.
### Explore the Cayley Table
Perhaps the most important feature of this table is that it has been completely filled in without introducing any new motions.
• Closure: Algebraically, this says that if $A$ and $B$ are in $D_6$, then so is $AB$. This property is called closure, and it is one of the requirements for a mathematical system to be a group.
• identity: Notice that if $A$ is any element of $D_6$, then $AR_0=R_0 A=A$. Thus, combining any element $A$ on either side with $R_0$ yields $A$ back again. An element $R_0$ with this property is called an identity, and every group must have one.
• Inverse: We see that for each element $B$ in $D_6$, there exists an element $A$ such that $AB=BA=R_0$. In this case, $B$ is said to be the inverse of $A$ and vise versa. The term inverse is a descriptive one, for if $A$ and $B$ are inverses of each other, then $B$ "un-does" whatever $A$ "does", in the sense that $A$ and $B$ taken together in either order produce $R_0$, representing no change.
• Non-Abelian: Another property of $D_6$ deserves special comment. Obverse that $S_1 R_3=R_3 S_1$, but $S_3 R_4 \neq R_4 S_3$. Thus in a group $ab$ may or may not be the same as $ba$. If it happens that $ab=ba$ for all choices of group elements $a$ and $b$, we say the group is commutative or --better yet-- Abelian (in honor of the great Norwegian mathematician Niel Abel). Otherwise, we say the group is non-Abelian. All dihedral groups are non-Abelian, except $D_1$ and $D_2$.
• Associativity: For all dihedral groups, it holds true that $(ab)c=a(bc)$ for all a, b, c in the group.
Image 3. Multiplication Table for $D_6$
If we want to know what is the composition of any two elements, it is convenient to use a Cayley map, because it tells us the result directly. But when we want to know the gradual change of the compositions, we will need another tool, a multi-colored Image 3.
Muptiplication Table
In multi-colored Image 3, each color represents one rotation or reflection. In the image, pink colors represent rotations, and the deepest pink represents the identity transformation. Green colors represent all the six reflections.
From the changing of color, we can observe that the gradual change of the composition of two elements in $D_6$. However, we cannot easily determine the exact result of composition by observing directly Image 3.
The abstract group structure is given by:
$R_k R_l=R_{k+l}$
$S_k S_l=R_{k-l}$
$R_k S_l=S_{k+l}$
$S_k R_l=S_{k-l}$
## Uniqueness of the Identity
In a dihedral group $D_n$, there is only one identity element.
PROOF: Suppose both $R_0$ and $R_0'$ are identities of $D_n$. Then,
Eq. 1 $aR_0=a$ for all $a$ in $D_n$, and
Eq. 2 $R_0' a=a$ for all $a$ in $D_n$.
The choices of $a=R_0'$ in Eq. 1 and $a=R_0$ in Eq. 2 yield $R_0' R_0=R_0'$ and $R_0' R_0=R_0$.
Thus, $R_0$ and $R_0'$ are both equal to $R_0' R_0$ and so are equal to each other.
## Cancellation
In a dihedral group $D_n$, the right and left cancellation laws hold; that is, $ba=ca$ implies $b=c$, and $ab=ac$ implies $b=c$.
PROOF: Suppose $ba=ca$. Let $a'$ be an inverse of $a$.
Then, muliplying on the right by $a'$ yields $(ba)a'=(ca)a'$.
Associativity yields $b(aa')=c(aa')$.
Then, $bR_0=cR_0$ and, therefore, $b=c$ as desired.
Similarly, one can prove that $ab=ac$ implies $b=c$ by multiplying by $a'$ on the left.
A consequence of the cancellation property is the fact that in a Cayley table for a dihedral group, each group element occurs exactly once in each row and column. Another consequence of the cancellation property is the uniqueness of inverses.
## Uniqueness of Inverses
For each element $a$ in a dihedral group $D_n$, there is a unique element $b$ in $D_n$ such that $ab=ba=R_0$.
PROOF: Suppose $b$ and $c$ are both inverses of $a$.
Then $ab=R_0$ and $ac=R_0$, so that $ab=ac$.
Canceling the $a$ on both sides gives $b=c$, as desired.
## Sock-Shoes property
For dihedral group elements $a$ and $b$, $(ab)^{-1}=b^{-1} a^{-1}$.
PROOF: Since $(ab)(ab)^{-1}=R_0$ and
$(ab)(b^{-1} a^{-1})=a(bb^{-1})a^{-1}=aR_0a^{-1}=aa^{-1}=R_0$,
we have by the Uniqueness of Inverses theorem that $(ab)$ has only one inverse $x$ such that $(ab)x=R_0$.
We get $x=(ab)^{-1}=(b^{-1} a^{-1})$.
## 3D Rotational Symmetry
$Dn$ consists of $n$ rotations of multiples of $\frac{360^\circ}{n}$ about the origin, and reflections across $n$ lines through the origin, making angles of multiples of $\frac{180^\circ}{n}$ with each other.
If we put a dihedral group in three dimensions, the reflections are also rotations of $180^\circ$
Rotation in 3D
The proper symmetry group of a regular polygon embedded in three-dimensional space (if $n \geqslant 3$). Such a figure may be considered as a degenerate regular solid with its face counted twice. Therefore it is also called a dihedron (Greek: solid with two faces), which explains the name dihedral group.
# Infinite Dihedral Groups
The infinite dihedral group $Dih(C_\infty)$ is denoted by $D_\infty$. The infinite dihedral group can be described as the group of symmetries of a circle, which has infinite symmetries.
We use the group presentation:
$D_\infty = \langle r, s \mid s^2 = 1, srs = r^{-1} \rangle$, or
$D_\infty = \langle x, y \mid x^2 = y^2 = 1 \rangle$ to represente the infinite dihedral group.
In the presentation, it says that because there are infinitely many symmetries, we can never rotate back to the identity, and so there are infinitely many rotations and reflections.
# Subgroups
Definition: A subgroup is a subset $H$ of group elements of a group $G$ that satisfies the four group requirements. It must therefore contain the identity element. "$H$ is a subgroup of $G$" is written as $H \subseteq G$, or sometimes $H \leq G$.[1]
Now we want to know exactly how many subgroups for $D_n$, and what they are. Fortunately, mathematician Stephan A. Cavior had already proved this for us in 1975. In the theorem, for any dihedral group in order of $2n$, there are $\tau(n) + \sigma(n)$ subgroups in total, including $\{1\}$ and $D_{n}$. $\{1\}$ is just the identity itself.
Definitions of Terms
• $\tau(n)$: the number of divisors of $n$,
e.g. $\tau(12)=6$.
• $\sigma(n)$: the sum of divisors of $n$,
e.g. $\sigma(12)=1 + 2 + 3 + 4 + 6 + 12=28$.
• $\mathbb{Z}/n \mathbb{Z}$: the notation for the cyclic group of order $n$, can be also written as $\mathbb{Z}_n$. This is a quotient group presentation.
• $d \mid n$: $d$ is a divisor of $n$.
• $N=\langle b \rangle$: group $N$ is a subgroup generated by $b$. It means $N=\{1,b,b^2,b^3,\cdots,b^{n-1}\}$.
• $|HN|$: the order of $HN$.
• $[D_{n}:N]=2$: the index. $[D_{n}:N]=2$ means $\frac{|D_n|}{|N|}=2$.
• $\{ab^{i+km}: \ 0 \leq k < d \}$: a set with elements looking like $ab^{i+km}$.
• $G(i,d)$: this means a group. $i$ is the index, which labels all the elements; $d$ is the order of the group.
This proof is complicated.
After we know what kind of group can be a subgroup of a dihedral group $D_n$, the dihedral group of order $2n$, we will start to find all subgroups of $D_n$.
Lemma 1. The number of subgroups of a cyclic group of order $n \geq 1$ is $\tau(n)$. [2]
A cyclic group of order $n$ is the group of all the rotations including the identity of the dihedral group of order $2n$.
Proof. Let $G$ be a cyclic group of order $n$. Then $G \cong \mathbb{Z}/n \mathbb{Z}$. A subgroup of $\mathbb{Z}/n \mathbb{Z}$ is in the form $d \mathbb{Z}/n \mathbb{Z}$ where $d \mathbb{Z} \supseteq n \mathbb{Z}$. The condition $d \mathbb{Z} \supseteq n \mathbb{Z}$ is obviously equivalent to $d \mid n. \$
Lemma 2. Let $b$ be the element of order $n$ in $D_{n}$ and let $H$ be a subgroup of $D_{n}$. Then either $H \subseteq \langle b \rangle$ or $|H \cap \langle b \rangle| =d$ and $|H|=2d$ for some $d \mid n$.
Proof. Let $N=\langle b \rangle$. Clearly $N$ is a normal subgroup of $D_{n}$ because $[D_{n}:N]=2$. Thus $HN$ is a subgroup of $D_{n}$ and hence the order of dihedral group $HN$ is a divisor of $2n$, and we use the notation:
Eq. 1 $|HN| \mid 2n.$ to represent.
On the other hand,
Eq. 2 $|HN|=\frac{|H| \cdot |N|}{|H \cap N|}=\frac{n |H|}{|H \cap N|}$.
Therefore, by Eq. 1, and Eq. 2. Hence either $|H| = |H \cap N|$ or $|H|=2|H \cap N|$. If $|H|=|H \cap N|$, then $H = H \cap N$ and thus $H \subseteq N$. If $|H|=2|H \cap N|$, then let $|H \cap N|=d$ and so $|H|=2d$. Clearly $d \mid n$ because $H \cap N$ is a subgroup of $N$ and $|N|=n. \$
Lemma 3. Given $d \mid n$, let $m = n/d$. For every $0 \leq i < n$ let $A(i,d) = \{ab^{i+km}: \ 0 \leq k < d \}$, where $a$ denotes a reflection and $b$ denotes a primitive rotation. Let $B(i,d) = A(i,d) \cup \langle b^m \rangle$. Then $B(i,d)$ is a subgroup of $D_{n}$ and $|B(i,d)|=2d$. We also have $|\{B(i,d) : \ 0 \leq i < n \}|=m=n/d.$
Proof. If $ab^{i+km}=ab^{i+rm}$, for some $0 \leq k,r < d$, then $b^{(k-r)m}=1$ and thus $d \mid k-r$, because $|b|=n=md$.
Therefore $k=r$ because $0 \leq k,r < d$.
So we have proved that $|A(i,d)|=d$.
Clearly $A(i,d) \cap \langle b^m \rangle = \emptyset$ and $|\langle b^m \rangle | = d$, because $|b|=n$.
Thus $|B(i,d)|=|A(i,d)| + |\langle b^m \rangle|=2d$.
Proving that $B(i,d)$ is a subgroup of $D_{n}$ is very easy.
Just note that every element of $A(i,d)$ is the inverse of itself (because they all have order two) and also note that $ab^s = b^{-s}a$, for all $s$, because $ab=b^{-1}a$.
Finally, the set $\{B(i,d) : \ 0 \leq i < n \}$ has $m$ elements because clearly $B(i,d)=B(j,d)$ if and only if $A(i,d)=A(j,d)$ if and only if $i \equiv j \mod m. \$
Suppose that $H$ is a subgroup of $D_{n}$. There are two disjoint cases to consider.
Case 1. $H \subseteq \langle b \rangle$.
By Lemma 1. the number of these subgroups is $\tau(n)$.
Case 2. $H \nsubseteq \langle b \rangle$.
In this case, by Lemma 2. we have $|H|=2d$ and $|H \cap \langle b \rangle|=d$, for some $d \mid n$.
Let $n = md$. Since $H \cap \langle b \rangle$ is a subgroup of $\langle b \rangle$, which is a cyclic group of order $n$, we have
Eq. 1 $H \cap \langle b \rangle = \langle b^m \rangle.$
Let $A(i,d)$ and $B(i,d)$ be as were defined in Lemma 3.
Now, since $H$ is not contained in $\langle b \rangle$, there exists some $0 \leq i < n$ such that $ab^i \in H$.
Then, since $H$ is a subgroup, we must have $ab^ib^{km} \in H$, for all $k$.
Thus $ab^{i + km} \in H$ and so $A(i,d) \subseteq H$ and therefore, by Eq. 1, we have $B(i,d) \subseteq H$.
Thus, since $|H|=|B(i,d)|=2d$, we must have $H=B(i,d)$.
The converse is obvously true, i.e. given $d \mid n$ and $0 \leq i < n$, $B(i,d)$ is a subgroup of $D_{n}$, by Lemma 3., and $B(i,d) \nsubseteq \langle b \rangle$ because it contains $A(i,d)$.
So the subgroups in this case are exactly the ones in the form $B(i,d)$, where $0 \leq i < n$ and $d \mid n$.
Thus, by Lemma 3. the number of subgroups in this case is
$\sum_{d \mid n} | \{B(i,d) : \ 0 \leq i < n \} | = \sum_{d \mid n} n/d = \sum_{d \mid n} d = \sigma(n).$
So, by Case 1. and Case 2. the number of subgroups of $D_{n}$ is $\tau(n) + \sigma(n)\$.
# Why It's Interesting
## In Music
The sequence of pitches which form a musical melody can be transposed or inverted. Since the 1970s, music theorists have modeled musical transposition and inversion in terms of an action of the dihedral group of order 24. More recently music theorists have found an intriguing second way that the dihedral group of order 24 acts on the set of major and minor chords.[3]
Dihedral groups as a kind of special symmetric groups are studied in music. In music, we use the operations Transposition and Inversion, which are denoted as $T$ and $I$, to represente rotations and reflections in dihedral groups.
Musicians usually study $D_{12}$, because 12 is the length of a normal cycle in music: C C D E E F F G G A B B, and then C again.
Based on this 12 element cycle, $D_{12}$ is important in music theory. Musicians use Transpositions and Inversions (rotations and reflections) of a simple note to create other notes to complete a final composition.
A transposition of a sequence $x$ of pitch classes by $n$ semitones is the sequence $T^n(x)$ in which each of the pitch classes in $x$ has been increased by $n$ semitones.
So for example if
$x=3 0 8$, where the numbers denote pitches,
then
$T^4(x)=T^4(3 0 8) = 7 4 0$.
When doing the operation $T^n(x)$, add $n$ to each digit of $x$, and use arithmetic modulo 12 (clock arithmetic) when the resulting digit is over 12. For instance, in adding 4 to 8, the result is 12, but $12 \equiv 0 \pmod{12}.\,$
Turning to the next operation, inversion $I(x)$ of a sequence $x$ just replaces each pitch class by its negative (in clock arithmetic).
So in the first example above with $x = 3 0 8$, we have
$I(x) = 9 0 4$.
To do the operation $I(x)$, we need to do subtraction in clock arithmetic. For instance, if we want to get 12 from 3, we need to add 9. 0 is already 12, so we need to add 0.
# References
[2] de Cornulier, Yves. (n.d). Group Presentation. From MathWorld--A Wolfram Web Resource, created by Eric W. Weisstein. http://mathworld.wolfram.com/GroupPresentation.html
[4] Milson, Robert and Foregger, Thomas. dihedral group. From PlanetMath.org. June, 12. 2007. Retrieved from http://planetmath.org/encyclopedia/DihedralGroup.html
[5] Gallian, Joseph A. Contemporary Abstract Algebra Seventh Edition. Belmont: Brooks/Cole, Cengage Learning. 2010.
[6] Dahlke, Karl. (n.d). Groups, Dihedral and General Linear Groups. Retrieved from http://www.mathreference.com/grp,dih.html
[7] Sharifi, Yaghoub. Subgroups of dihedral groups (1)&(2). Feb, 17, 2011. Retrieved from http://ysharifi.wordpress.com/2011/02/17/subgroups-of-dihedral-groups-1/
[8] [1]Scott, W. R. Group Theory. New York: Dover, 1987.
[9] [2]Hungerford, Thomas W. Graduate Texts in Mathematics - Algebra. New York: Springer, 1974.
[10] [3]Crans, Alissa S., Fiore, Thomas M. and Satyendra, Ramon. Musical Actions of Dihedral Groups. University of South Florida. Nov 3, 2007. Retrieved from http://myweb.lmu.edu/acrans/MusicalActions.PDF
[11] Benson, Dave J. Music: A Mathematical Offering. Cambridge University Press. Nov 2006. Retrieved from http://www.maths.abdn.ac.uk/~bensondj/html/music.pdf
[12] Rowland, Todd and Weisstein, Eric W. (n.d). Root of Unity. From MathWorld--A Wolfram Web Resource. Retrieved from http://mathworld.wolfram.com/RootofUnity.html
2. Add more about Dihedral Groups in 3D. I only talk about one property in 3D, but there must be some more.
3. In the subgroups part, it is hard to explain only in words, so I use lots of notation, which is still not very clear. I hope can find a better way to illustrate it.
5. Think about non-abelian in matrices which may relate to non-abelian in group theory. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 445, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9519132375717163, "perplexity": 205.03177922271317}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936463318.72/warc/CC-MAIN-20150226074103-00319-ip-10-28-5-156.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/126437-inequality.html | # Math Help - Inequality ..
1. ## Inequality ..
Hi ,
x , y and z real numbers positive , with $x+y+z=4$ . prove That :
2. Can you help me ?? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9689627885818481, "perplexity": 3006.573357677453}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443736673439.5/warc/CC-MAIN-20151001215753-00210-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/597808/golden-ratio-fibonacci-which-branch-of-math | Golden ratio / Fibonacci which branch of math?
Friends,
The Golden ratio / Fibonacci sequence are studied under which branch of math?
Can you recommend some good textbooks on the subject?
Thanks
• There is also "Fibonacci and Lucas Numbers, and the Golden Section - Theory and Application" by Steven Vajda. – Stiefel Aug 20 '14 at 9:01 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9533451795578003, "perplexity": 736.2765320060323}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315174.57/warc/CC-MAIN-20190820003509-20190820025509-00279.warc.gz"} |
https://en.wikibooks.org/wiki/Logic_for_Computer_Science/First-Order_Logic | Logic for Computer Science/First-Order Logic
First-Order Logic
In propositional logic, we considered formulas made about atomic objects, which could only be either true or false. First-order logic, the topic of this chapter, builds upon propositional logic and allows you to look inside the objects discussed in formulas. We can provide this more refined level of granularity by discussing objects as elements of sets that can be larger than just the set $\{0, 1\}$, and also include arbitrarily complex relationships with each other.
We begin first by defining the syntax of first-order (FO) logic, and then give these structures meaning.
Syntax
The domain of discourse for first order logic is first-order structures or models. A first-order structure contains
1. Relations,
2. Functions, and
3. Constants (functions of arity 0).
The vocabulary of first-order logic is
1. a set of relation symbols with associated arities, and
2. a set of function symbols with associated arities.
Here are some example first-order logic vocabularies:
1. A graph
• Relation Set = { $V$ : unary, $E$ : binary }
• Function Set = { $\mathit{next}$ : unary }
2. Arithmetic
• Relation Set = { $+$ : ternary, $\times$ : ternary, $\uparrow$ : ternary, $=$ : binary, $<$ : binary}
• Function Set = { $\mathit{next}$ : unary, $0$ : const }
Here, a graph is a set of vertices $V$ and a set of edges $E$. For the arithmetic set note that the use of $+$ and $\times$ is purely syntactic and we could have used the symbols "plus" and "times" instead. We haven't provided the symbols with any meaning yet.
A term denotes an element in the first-order structure. A term is used to refer to the elements in our domain of discourse. Here are the rules that describe what a term is:
• every variable is a term, where a variable is simply another set of symbols
• every constant is a term
• If $t_{1}, t_{2}, \ldots, t_{k}$ are terms, and $f$ is a $k$-ary function then $f(t_{1}, t_{2}, \ldots, t_{k})$ is a term.
For example, if $f$ is a binary function and $g$ is a ternary function then the following are all terms: $x, a, f(x, y), g(x, f(x, y), a)$.
An atomic formula is
$R(t_{1}, t_{2}, \ldots, t_{k})$,
where $R$ is a $k$-ary relation and $t_{1}, t_{2}, \ldots, t_{k}$ are terms. When viewing $R$ as a set, then this is just another way of saying that the tuple
$(t_{1}, t_{2}, \ldots, t_{k})\in R$.
A special relation $=$ means "equal" and cannot be interpreted otherwise. For example, $t_{1} = t_{2}$ stands for
$=(t_{1}, t_{2})$.
A first-order formula is an expression built using a given first-order vocabulary and variables and the symbols $(, ), \neg, \vee, \wedge, \rightarrow, \exists, \forall$. The set of first-order formulas is the minimum set satisfying the following:
• atomic formulas are first order formulas
• If $\phi$ and $\psi$ are formulas and $x$ is a variable, then the following are also formulas:
1. $(\phi \vee \psi)$
2. $(\phi \wedge \psi)$
3. $(\neg \phi)$
4. $(\phi \rightarrow \psi)$
5. $(\exists x \phi)$
6. $(\forall x \phi)$
Semantics
A first-order structure over a given vocabulary consists of
1. A domain, which is a set of elements (also known as a universe)
2. A mapping associating to every $k$-ary relation symbol in the vocabulary a $k$-ary relation over the domain, and to every $k$-ary function symbol a $k$-ary function over the domain.
These components give meaning to the symbols.
Example:
Relation Set = { $+$ : ternary, $\times$ : ternary, $\uparrow$ : ternary, $=$ : binary, $<$ : binary }
Function Set = { $next$ : unary, $0$ : const }
A first-order structure over this vocabulary is:
1. Domain: the set of integers
2. Mapping : $+ \rightarrow$ addition, $\times \rightarrow$ multiplication, $\uparrow \rightarrow$ exponentiation, $< \rightarrow$ ordering, $\mathit{next} \rightarrow i+1$
In this structure, the formula
$\exists x \forall y \forall z[\times (y, z, x) \rightarrow ((y = 1) \vee (z = 1))]$
expresses the statement "there exists a prime number" (the number 1 also satisfies this statement).
Note here that $\times (y, z, x)$ is equivalent to $(x = y \times z)$.
Scope of Quantifiers
In the formula $(\forall x \phi)$ or $(\exists x \phi)$, $\phi$ is referred to as the scope of quantification of the variable $x$. An occurrence of a variable in a formula is bound if it is in the scope of quantification of that variable, otherwise the occurrence is said to be free. You can consider a quantifier use as a variable declaration.
A sentence is a formula with no free variables (i.e., all variables are bound). A sentence is either true or false.
A formula with free variable(s) can be considered as describing the properties of the free variable(s). For example, $\phi(x)$ denotes a formula with $x$ occurring free and describes the properties of $x$.
If a sentence $\phi$ evaluates to true over a structure $I$, we say $I$ satisfies $\phi$ and denote this by $I \models\phi$.
• $I \models R(t_{1}, t_{2}, \ldots, t_{k})$ iff $(t_{1}, t_{2}, \ldots, t_{k}) \in I(R)$, where $I(R)$ is the interpretation of $R$ by $I$.
• $I \models \phi \wedge \psi$ iff $I \models \phi$ and $I \models \psi$. (Similar for $\vee$.)
• $I \models \neg \varphi$ iff $I \not\models \varphi$.
• $I \models \exists x \phi(x)$ if $I \models \phi(x \leftarrow c)$ for some $c$ in the domain.
• $I \models \forall x \phi(x)$ if $I \models \phi(x \leftarrow c)$ for every $c$ in the domain.
Note: $\phi(x \leftarrow c)$ denotes the result of substituting $c$ for every free occurrence of $x$ in $\phi$. Substitution is covered further later in this chapter.
Axiomatic approach
The axioms are a set of sentences meant to distinguish "desired" models from others. But typically, also have "undesired" models, which are called non-standard models.
Examples: Consider the vocabulary $(\mathit{next}, +, \times, \uparrow, 0, =, <)$, where the symbols have the usual meaning (defined by FO sentences over this vocabulary). The standard interpretation for this set of axioms (see the end of this chapter) is the integers, but these are also satisfied by the set of integers modulo $p$. This possibility is ruled out by adding the following sentence to the axioms: $\forall x (x < \mathit{next}(x))$
Question: Can all non-standard models be axiomatized away? The answer is no. Consider the model shown in [TODO: import figure] for the vocabulary containing only $0,<,\sigma$ (all elements in the upper line are larger than those on the lower line).
There is no set of first-order sentences that can distinguish this model from the natural numbers. Intuitively, the reason is that we cannot "backtrack" arbitrarily (towards $0$) in a first-order sentence. A similar non-standard model can be obtained for the full vocabulary above.
Evaluating FO Sentences
Given a first-order structure $I$ and a FO sentence $\phi$, can we tell if $I \models \phi$?
This problem is decidable for finite structures. For example, suppose $E$ is a binary relation representing the edges of a graph, and the given sentence is
$\exists x_{1} \exists x_{2} \exists x_{3}(E(x_{1}, x_{2}) \wedge E(x_{2}, x_{3}) \wedge E(x_{3}, x_{1}) \wedge x_{1} \neq x_{2} \wedge x_{2} \neq x_{3} \wedge x_{3} \neq x_{1} )$.
We can evaluate the truth of this sentence by trying all possible values for $x_{1}, x_{2}$ and $x_{3}$. (Naive evaluation: "nested loop".) This is polynomial time (in domain size) but exponential in the size of the formula.
On infinite domains, we may or may not be able to evaluate the truth of a sentence. With the vocabulary $(N, 0, \sigma, <, +)$, where $N$ is the set of natural numbers, it is decidable if a sentence is true. (This is known as Presburger arithmetic.) However if we include multiplication, the truth of a sentence becomes undecidable.
A Deductive System for FO
A set $\Delta$ of FO sentences is
• Satisfiable if there is some structure $I$ such that $I \models \Delta$
• Unsatisfiable if it is not satisfiable
• Valid if $I \models \Delta$ for all structure $I$
Given a set of FO sentences $\Sigma$ and a sentence $\phi$
• $\Sigma$ entails $\phi$ (denoted $\Sigma \models \phi$), if every structure that satisfies $\Sigma$ also satisfies $\phi$.
• $\Sigma \models \phi$ iff $\Sigma \cup \{\neg \phi\}$ is unsatisfiable.
• If $\Sigma$ is finite then, $\Sigma \models \phi$ iff $\neg \Sigma \lor \phi$ is valid.
• If a formula $\phi$ has free variables $\vec{X} = (x_{1}, \ldots, x_{k})$, $\phi$ is valid iff $\forall \vec{X} \phi$ is valid.
Axioms for Validity
The axioms for validity are from three categories:
1. Axioms for boolean validity. These are inherited from propositional logic.
2. Axioms for equality.
3. Axioms for quantification.
Boolean Validity
Given a FO formula $\varphi$, a boolean form of $\varphi$ a propositional formula $\psi$ such that $\varphi$ is obtained from $\psi$ by replacing each propositional variable in $\psi$ by a subformula of $\varphi$.
The set of Boolean forms of $\varphi$ is denoted $BF(\varphi)$.
Examples:
• For the FO formula $\forall x P(x) \lor \neg \forall x P(x)$, the boolean form is $p \lor \neg p$.
• If $\varphi \equiv \forall x G(x,y) \land \exists x G(x,y) \land (G(z,x) \lor \forall x G(x,y))$, then $BF(\varphi) \equiv x_{1} \land x_{3} \land (x_{2} \lor x_{1})$ where $x_1 = \forall x G(x,y), x_2 = G(z,x), x_3 = \exists x G(x,y)$.
Claim: If $\psi \in BF(\varphi)$ and $\psi$ is valid, then $\varphi$ is valid.
Equality Axioms
Let $t, t_{1}, \ldots, t_{k}, t'_{1}, \ldots, t'_{k}$ be terms. The following are valid formulas.
• $t = t$
• $(t_{1} = t'_{1} \land \cdots \land t_{k} = t'_{k}) \rightarrow (f(t_{1}, \ldots, t_{k}) = f(t'_{1}, \ldots, t'_{k}))$, where $f$ is a $k$-ary function.
• $(t_{1} = t'_{1} \land \cdots \land t_{k} = t'_{k}) \rightarrow (R(t_{1}, \ldots, t_{k}) \rightarrow R(t'_{1}, \ldots, t'_{k}))$, where $R$ is a $k$-ary relation.
Substitutions
Given a formula $\varphi$ in which variable $x$ occurs free (denoted by $\varphi(x)$) and a term $t$, we define the substitution of $t$ for $x$ in $\varphi$, denoted $\varphi(x \leftarrow t)$, as the formula that results from replacing every free occurrence of $x$ in $\varphi$ by $t$, subject to the constraint that $t$ contains no variable $y$ quantified in $\varphi$ such that $x$ occurs free within the scope of quantification of $y$. If $x$ does not occur free $\varphi$, then $\varphi(x \leftarrow t)$ is defined as $\varphi$.
Example: Let $\varphi$ be
$((x = 1) \rightarrow \exists x (x = y))$,
then
• $\varphi(x \leftarrow (y + 1))$ is a valid substitution
• $\varphi(y \leftarrow (x + 1))$ is not a valid substitution.
Quantification Axioms
1. $\forall x \varphi \rightarrow \varphi(x \leftarrow t)$, where $t$ is a term and $\varphi(x \leftarrow t)$ is a valid substitution
2. $(\forall x (\varphi \rightarrow \psi)) \rightarrow ((\forall x \varphi) \rightarrow (\forall x \psi))$
3. $\varphi \rightarrow \forall x \varphi$
4. $\exists x \varphi \leftrightarrow \neg \forall x \neg \varphi$
In summary, the axioms for validity are ${\mathbf {Ax}}$:
1. Axioms for boolean validity. These are inherited from propositional logic.
2. Axioms for equality.
3. Axioms for quantification.
Definition: A proof is a sequence $\phi_{1}, \phi_{2}, \ldots, \phi_{k}$ of FO sentences such that for each $i \in \{1, \ldots, k\}$ either $\phi_{i} \in{\mathbf {Ax}}$ or $\exists j, k < i$ such that $\phi_{j} \equiv \psi$ and $\phi_{k} \equiv (\psi \rightarrow \phi_{i})$.
Notation: If there is a proof of $\phi$ using ${\mathbf{Ax}}$, we denote this fact by $\vdash_{\mathbf {Ax}}\phi$.
Fact (Soundness): If $\vdash_{\mathbf {Ax}} \phi$, then $\phi$ is valid.
Remark: The set of formulas which can be proven valid is recursively enumerable.
Example: Proof of the formula $\forall x (\phi \land \psi) \rightarrow (\forall x \phi \land \forall x \psi)$
$\forall x (\phi \land \psi)$
$\forall x (\phi \land \psi) \rightarrow (\phi \land \psi) (x \leftarrow x)$
$\phi \land \psi$
$\phi$
$\forall x (\phi \land \psi) \rightarrow \phi$
$\forall x (\forall x (\phi \land \psi) \rightarrow \phi)$
$\forall x \forall x (\phi \land \psi) \rightarrow \forall x \phi$
$\forall x (\phi \land \psi) \rightarrow \forall x \phi$
$\forall x \phi$
The proof of $\forall x \psi$ is symmetrical.
$\forall x \phi \land \forall x \psi$
$\forall x (\phi \land \psi) \rightarrow (\forall x \phi \land \forall x \psi)$
Useful Equivalences
• $\forall x (\phi \land \psi) \leftrightarrow (\forall x \phi) \land (\forall x \psi)$
• $\forall x (\phi \lor \psi) \not\leftrightarrow (\forall x \phi) \lor (\forall x \psi)$ e.g. $\phi(x) \approx$ "x is even" and $\psi(x)\approx$ "x is odd"
• $\forall x (\phi(x) \lor \neg \phi(x)) \not\leftrightarrow (\forall x \phi(x) \lor \forall x \neg \phi(x))$
• $\exists x (\phi \lor \psi) \leftrightarrow (\exists x \phi) \lor (\exists x \psi)$
• $\neg \exists x \phi \leftrightarrow \forall x \neg \phi$
• $\neg \forall x \phi \leftrightarrow \exists x \neg \phi$
Prenex Normal Form
Claim: Every FO sentence is equivalent to a FO sentence of the form $Q_{1} x_{1} \cdots Q_{k} x_{k} \phi$, where $Q_{i} \in \{\exists, \forall \}$ and $\phi$ is quantifier free.
This is proved using the distributive properties of quantifiers. It may be necessary to rename variables, to avoid ambiguities. Although you can always move all quantifiers to the left, the resulting formula can actually be exponentially larger than the original one.
Example:
$\phi \equiv \forall x (G(x,x) \land (\forall y G(x,y) \lor \exists y \neg G(y,y))) \land G(x,0)$
$\equiv (\forall x (G(x,x) \land (\forall y G(x,y) \lor \exists y \neg G(y,y)))) \land G(w,0)$
$\equiv\forall x (G(x,x) \land (\forall y G(x,y) \lor \exists y \neg G(y,y)) \land G(w,0))$
$\equiv\forall x (G(x,x) \land (\forall y G(x,y) \lor \exists z \neg G(z,z)) \land G(w,0))$
$\equiv\forall x (G(x,x) \land \forall y (G(x,y) \lor \exists z \neg G(z,z)) \land G(w,0))$
$\equiv\forall x \forall y (G(x,x) \land (G(x,y) \lor \exists z \neg G(z,z)) \land G(w,0))$
$\equiv\forall x \forall y (G(x,x) \land \exists z (G(x,y) \lor \neg G(z,z)) \land G(w,0))$
$\equiv\forall x \forall y \exists z (G(x,x) \land (G(x,y) \lor \neg G(z,z)) \land G(w,0))$
Recall the Axiomatic Method
1. Describe the desired model as closely as possible using a (possibly infinite but recursive) set $\Delta$ of axioms (FO sentences).
1. Prove things using deduction.
Notation: We say $\phi$ is a valid consequence of $\Delta$ (denoted $\Delta \models \phi$), if every model satisfying $\Delta$ also satisfies $\phi$.
Fact: $\Sigma \models \phi$ iff $\Sigma \cup \{\neg \phi\}$ is unsatisfiable.
Aside: Nonlogical axioms for number theory
The following fourteen first-order axioms describe the properties of arithmetic and numbers, i.e. addition ($+$), multiplication ($\times$), exponentiation ($\uparrow$, equality ($=$), ordering ($<$), successor function ($\sigma$) and remainder (mod). This example shows the expressive power of first-order statements, which were originally hoped to provide a basis for the "one true mathematics."
Question: Why are they called "nonlogical" axioms?
NT1: $\forall x(\sigma(x) \ne 0)$
NT2: $\forall x \forall y (\sigma(x) = \sigma(y) \rightarrow x=y)$
NT3: $\forall x (x=0 \vee \exists y \sigma(y) = x)$
NT4: $\forall x (x+0=x)$
NT5: $\forall x \forall y(x+\sigma(y) = \sigma(x+y))$
NT6: $\forall x (x\times 0 = 0)$
NT7: $\forall x \forall y (x\times \sigma(y) = (x \times y ) + x)$
NT8: $\forall x ( x \uparrow 0 = \sigma(0))$
NT9: $\forall x \forall y (x \uparrow \sigma(y) = (x \uparrow y ) \times x )$
NT10: $\forall x (x<\sigma(x))$
NT11: $\forall x \forall y ( x < y \rightarrow (\sigma (x)\le y))$
NT12: $\forall x \forall y ( \neg (x
NT13: $\forall x \forall y \forall z (((x
NT14: $\forall x \forall y \forall z \forall z' (mod(x,y,z)\wedge mod(x,y,z') \rightarrow z = z')$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 255, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9870747327804565, "perplexity": 487.19867489672293}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645241661.64/warc/CC-MAIN-20150827031401-00008-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/forces-velocities-and-angles.140530/ | # Forces, velocities, and angles
1. Oct 29, 2006
### :::JMANN:::
I have a take home quiz for my physics class. Out of the 10 questions I have answered 7 of them, whether they are right I don't know. First and foremost I need help on the ones I haven't solved yet. I am not necessarily looking for someone to give me the answers, but rather help me setup the problem and use formulas to arrive at the answer myself.
The hardest question is #3.
A 200lb man stands tiptoed on one foot so that all his weight is borne by the ground beneath the ball of the foot. If the foot and ankle are considered as an isolated body, the three forces that are in equilibrium are the reaction W of the ground, the pull T of the achillies tendon and the compression C of the tibia. The force C is downward at an angle 15* from vertical. The force T is upward at 21* from the vertical. W is an upward force of 200lbs. Calculate the values of C and T. (HInt: one way to do this is to have two simultaneous equations, one balancing the horizontal components of force and the other balancing the vertical components. Another way is to use the law of sines)
#2.
A rocket exhausts fule with a velocity of 1500m/s, relative to the rocket. It starts from rest in outer space with the fuel comprising 80% of the total mass. When all the fuel has been exhausted, what is the rockets speed?
#5.
A car rounds a 75m radius curve at a constant speed of 18m/s. A ball is suspended by a string from the ceiling of the car and moves with the car. What is the angle between the string and the vertical (y-axis)?
2. Oct 29, 2006
### Ja4Coltrane
the first problem is not nearly as difficult as it looks. The only problem is that you are saying things like "upward from verticle, which actually does not make to much sense. Sorry. I'll help you if you explain the directions.
for the second problem, you know that momentum is conserved. If the rocket fuels final momentum is (1500 m/s)(4 <arbitrary mass units noting that the fuel has a 4:1 ratio to the ship>) and the ships momentum is
-(Velocity of ship)(1) then you know that the velocity of the ship equals four times the speed of the fuel.
3. Oct 29, 2006
### Ja4Coltrane
Believe it or not the last question is certainly the hardest.
First, note that the car is accelerating inward. I'm sure you know how to calculate this acceleration, so I'll leave that to you and just call it "a"
Now forget the circle stuff, and just pretend that the car is accelerating forward. Because the ball has inertia, it resists the acceleration, and therefore makes an angle. Now imagine that acceleration is perfectly constant and the ball has made a beautifully still angle. you know that the ONLY upward force acting on the ball is the y-component of the tension. This component is equal to Tcos@ which must also equal the balls weight mg. So now you know that T=mg/(cos@). Read that again if i lost you. Now pretend that the car is not accelerating, but instead, a "fictitious force" is acting directly backwards on the ball. this force is keeping the angle. you know that there are two forces now acting on the ball in the X direction. Forward is the x-component of tension Tsin@=(mg(sin@))/(cos@)=mgtan@. the other force is the fictitious force balancing out the tension's x-component. This force is (M)(a) <the same "a" as before. so mgtan@=ma so a=gtan@ so @=tan-1(a/g)
4. Oct 29, 2006
### :::JMANN:::
Wow, thanks bro, I'm gonna work those out. As far as the first question, when I say 15* or 21* from vertical, I guess it be easier to describe as 105* and 111* from a horizontal plane. I copied the question exactly, but it had a picture with it, so for me the visualization went along with the picture, I guess without it it's a little harder to visualize. Did that help?
5. Oct 29, 2006
### Ja4Coltrane
Okay this is not too hard. I would definitely avoid the law of sines because that just makes it complicated.
what you can do (as it says) is take two equations--one for the forces in the X direction and one for the Y direction.
so Fnet(Y)=0 because it is not moving and also Fnet(X)=0
first let's do Y. Force W must equal the vector sum of the other two forces so that Fnet(Y)=0
See if you follow this: 200lbs=C(cos15)-T(cos21)
in that case, the cosine of C and T make the Y components of them and therefore that expresion makes since.
next let's do X
W does not matter since it has no X-component
I am assuming even though you did not say that the 15 degrees from verticle and the 21 are pointing in the opposite horizontal directions.
C(sin15)=T(sin15)
make since? Now you just have a system of equations to be solved with substitution.
6. Oct 30, 2006
### :::JMANN:::
Not quite sure what you mean in your last resposne, but both angles are relative to the leg. Both angles fall in the same quadrant(II), how ever the force for the tibia is down and the force of the tendon is up. Thanks again for all the help. I wish I would of founf this place sooner.
7. Oct 30, 2006
### Ja4Coltrane
(wow I just got home from school,logged on, and you reponded about one minute ago!)
Okay, I am slightly confused. I instead of telling me the quadrants, tell me how they point. As in Up-left and Down-Right.
If it is either that or Up-right and Down-left then my explanation works. Would you like me to explain what I did in some detail?
8. Oct 30, 2006
### Ja4Coltrane
The Y forces are as follows. W is just up--nice and easy.
C points down at an angle, but the Y-component points just down.
That is C(cos15) is down (that might sound weird for a y-comp. to be cosine but you could also say sin105.
So you would just say that W=C(cos15) but the force T pulls up. It pulls up with T cos 21 okay?
So now we have W=C(cos15)-T(cos21)
(becuase it does not have to pull as much do to T
Call that Equation one.
for x components, W does not matter, so you just set the x-comps equal to each other
T(sin21)=C(sin15)
Did that clear thing up?
9. Oct 30, 2006
### :::JMANN:::
Yeah it did, thanks. It is weird explaining the diagram without being able to show it to you, but I guess you had it right.
Okay, so here I go.
I came up with the values c=-2670.92N and t=-2075.96N For some reason these figures don't seem right, but I'm not sure. I solved for C and T using substitution. I converted the 200lbs for W into kg and then Newton's =892N so
-2670.92N(cos15)-(-)2075.96N(cos21)=892N
Just seems a little weird. How far off am I?
10. Oct 30, 2006
### :::JMANN:::
For the fuel question, the ships velocity equals 6000m/s, but I am still having trouble gripping the concept of it. If it's 4 times the fuels velocity. The book says Mvi=Pi
Pf=(0.80M)v(fuel) +(0.20M)v(ship)
Vship= Vrel +Vfuel
Vfuel= Vship - Vrel
Mvi= 0.80M(Vship - Vrel) + 0.20MVship
I totally lost.
11. Oct 30, 2006
### :::JMANN:::
The ball.
Okay, I get the forces acting on the ball. Given that ma=mgtan@
a/g=tan@
tan@= 4.32/9.8= 0.441
tan-1(0.441)= 0.415* angle
Doesn't seem like much of an angle, but I guess the radius would have to be smaller of the speed quicker to force the ball to move more right?
12. Oct 30, 2006
### :::JMANN:::
C=3057.94N T=2209.01N
Still seems awefully out there.
13. Oct 31, 2006
### :::JMANN:::
calculator was in radians for this one too. I got 23.78*.
I appreciate all the help. I have to turn it in now, but I know I have done better than I would of without the help. Thanks a lot.
14. Oct 31, 2006
### :::JMANN:::
Well I turned in the exam. I got all of them right except the rocket fuel question. We did not go over it, so I don't know what the answer was, but I am going to check, a class mate found a problem in the book exactly like it, so I'm a plug a chug and see what comes up. He said he got like 24xxm/s so we'll see. Thanks again for all the help, I really appreciate it.
15. Oct 31, 2006
### :::JMANN:::
This answer was correct, however I left of then - for C, I hope I remebered it on the test, otherwise I might lose some points.
16. Nov 1, 2006
### :::JMANN:::
I found the formula for the rocket and fuel.
Vf-Vi= Vrel(ln(Fi/Ff) Where F is the fuel initial/final
Vf-0=1500(ln(1/.2)
Vf=2414m/s
I don't know if I followed your method correctly or not, but this is not the answer that I got. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9222055673599243, "perplexity": 824.0636583416783}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988718423.65/warc/CC-MAIN-20161020183838-00359-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/rotational-motion-and-linear-speec.265546/#post-1922444 | # Rotational Motion and Linear Speec
• Start date
• #1
3
0
## Homework Statement
A rubber ball with a radius of .048 m rolls along the horizontal surface of a table with a constant linear speed v. When the ball rolls off the edge of the table, it falls 0.85 m to the floor below. If the ball completes 0.82 revolutions during its fall, what was its linear speed, v?
## Homework Equations
I know to find linear speed the equation v=rw must be used.
## The Attempt at a Solution
I tried to turn revolutions into radians and then use the equation T=2pi/w
and then did v=rw. I don't really know where to begin this problem since the problem has revolutions and not rpm. When i solved it like rpm it didn't work.
• #2
98
0
## Homework Statement
A rubber ball with a radius of .048 m rolls along the horizontal surface of a table with a constant linear speed v. When the ball rolls off the edge of the table, it falls 0.85 m to the floor below. If the ball completes 0.82 revolutions during its fall, what was its linear speed, v?
## Homework Equations
I know to find linear speed the equation v=rw must be used.
## The Attempt at a Solution
I tried to turn revolutions into radians and then use the equation T=2pi/w
and then did v=rw. I don't really know where to begin this problem since the problem has revolutions and not rpm. When i solved it like rpm it didn't work.
Well, you know how to convert rotational speed to linear speed using the equation in step 2 - all you need is the rotational speed and the radius of the ball. You also know the radius. So the question is, how do you find the rotational speed?
You are given a distance the object falls (under acceleration from gravity), and you are given the number of rotations. Rotations / time is rotational speed. See if you can calculate the time and go from there.
• #3
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so to find time can i just divide distance/acceleration and then take the square root?
Also when you say revolutions/ time = rotational speed I have to change revolutions to radians correct?
• #4
3
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never mind i figured it out! thanks so much for your help!
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2K | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9333980083465576, "perplexity": 1049.1656370470841}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358903.73/warc/CC-MAIN-20211130015517-20211130045517-00560.warc.gz"} |
https://cie.co.at/eilvterm/17-25-068?qt-sidebar_tabs=0 | # e-ILV
The e-ILV provides free access to all the terms and definitions contained in the international standard CIE S 017:2020 ILV: International Lighting Vocabulary, 2nd edition.
To search for another term please use "Back to the list" to return to the main page or the e-ILV.
For the complete set of terms and definitions CIE S 017:2020 can be purchased from the CIE Webshop. Members of a CIE National Committee or Associate National Committee have access to a 66,7 % discount on the purchase price of the standard.
# 17-25-068
fall time, <of a detector>
time required for an output to fall from a stated high percentage to a stated lower percentage of the maximum value when a steady input is instantaneously removed
Note 1 to entry: A value of 90 % is usually considered to be the high percentage value, whereas a value of 10 % is usually considered to be the low percentage value.
Note 2 to entry: The fall time is expressed in seconds (s).
Note 3 to entry: This entry was numbered 845-05-61 in IEC 60050-845:1987.
Note 4 to entry: This entry was numbered 17-427 in CIE S 017:2011.
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http://math.stackexchange.com/questions/235432/math-card-probability-questions?answertab=oldest | # Math card probability questions
2)Three cards are drawn at random from a standard deck without replacement. What is the probability that all three cards are hearts?
3)Three cards are drawn at random from a standard deck with replacement. What is the probability that exactly two of the three cards are red?
4)Three cards are drawn at random from a standard deck without replacement. What is the probability that exactly two of the three cards are red?
5)Three cards are drawn at random from a standard deck with replacement. What is the probability that at least two of the three cards are red?
I am having trouble finding the correct answer when I am having to do the math to figure out the with replacement and without replacement. And I also don't understand the difference in the math I need to do to figure out the exactly two of the three cars and the at least two cards.
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Can you show us what you have written? – The Substitute Nov 12 '12 at 4:51
For number 2 i figured you would do 13/26 x 12/25 x 11/24 – susie q Nov 12 '12 at 5:01
and then for number 3 I thought youd just do 13/26x13/26x13/26 since you are replacing the card each time – susie q Nov 12 '12 at 5:02
where are you getting the 26, 25, 24 for number 2? There are 52 cards in the deck, and there are 13 hearts. Therefore, the probability of drawing a heart on the first draw is $\frac{13}{52}$. – The Substitute Nov 12 '12 at 21:49
2) $P(\text{3 hearts})=P(\text{heart on first draw})\cdot P(\text{heart on second draw given that we drew a heart on first draw})\cdot P(\text{heart on third draw given that we drew two hearts})=\frac{13}{52}\cdot \frac{12}{51}\cdot \frac{11}{50}=\frac{11}{850}$
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Here's a related example to show how you can approach the replacement question.
• Two cards are drawn at random from a standard deck without replacement. What is the probability that both are even?
The total number of ways to draw two cards is $C(52, 2)$ (also known as "52 choose 2"). The total number of ways to draw two cards with both even is $C(20, 2)$, since you're drawing from the 20 even cards only. So the probability is $C(20, 2) / C(52, 2) = (20 \times 19) / (52 \times 51)$, which is about $.143 • Two cards are drawn at random from a standard deck with replacement. What is the probability that both are even? Each time you draw a card, the probability is$1 / 2$. Each draw is independent. So the probability of doing it twice is$1/2 * 1/2 = 1 / 4$. See the difference? In the first case you have to think about drawing a pair of cards, but in the second case the two draws are independent. In the first case, notice that if one card is even, that makes it slightly harder to get an even card the second time (since there's one fewer even cards to choose from), so the probability is slightly lower for the first case than for the second. I hope that helps you apply the idea to red cards or hearts now. - Thank you! It just seems so much easier when you only have two cards. Hopefully I'll get this figured out – susie q Nov 12 '12 at 4:59 In your with-replacement calculation, the probability at each draw should be$20/52$, not$1/2$, to bring it in line with the without-replacement calculation. That makes the answer for doing it twice$(20/52)^2=25/169\approx0.148$. – Barry Cipra Mar 4 '14 at 12:45 The term without replacement means that the deck is different for each draw so the probability of getting a heart on the first draw is$\frac{13}{52}$as there are 13 hearts in a full deck of 52 cards. Now given you have already drawn a heart on your first draw the probability of getting a heart on your second draw is$\frac{12}{51}$as there are now only 12 hearts left in the pack and the pack has 51 cards remaining. So the probability of hearts on both your first and second draw is$\frac{13}{52} \cdot \frac{12}{51} = \frac{156}{2652} = \frac{1}{17}$I'm sure you can see from here how you would calculate three hearts. The term with replacement means that the card is put back and mixed up again after each draw so the probability of drawing a heart on the second draw is$\frac{13}{52}$because you are still drawing from a full pack of cards. For two hearts the probability is$\frac{13}{52} \cdot \frac{13}{52} = \frac{169}{2704} = \frac{1}{16}\$ and for three hearts ...
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## protected by Zev ChonolesSep 20 '15 at 6:32
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.
Would you like to answer one of these unanswered questions instead? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8938847780227661, "perplexity": 186.87544564610522}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701962902.70/warc/CC-MAIN-20160205195242-00168-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/111464/nontrivial-example-of-m-such-that-textassm-varnothing | # Nontrivial example of $M$ such that $\text{Ass}(M)=\varnothing$?
Suppose $R$ is a commutative ring, and $M$ an $R$-module. Is there a nontrivial example of such $M$ where the set of associated primes $\text{Ass}(M)=\varnothing$?
Taking $M=0$ feels kind of cheap, since any annihilator of $x\in M$ is necessarily all of $R$, and hence not prime.
Is there a better example that is still relatively simple, preferably with some explanation about why $\text{Ass}(M)=\varnothing$?
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Any example is going to be somewhat "complicated" because $R$ has to be non-Noetherian. If $R$ is a Noetherian ring, and $M$ is a non-zero $R$-module, then there must be an associated prime ideal. Proof sketch: Consider the set of annihilators of non-zero elements of $M$. This is a non-empty set of ideals of $R$ and therefore has a maximal element, since $R$ is Noetherian. Show that this maximal element is prime. – Ted Feb 21 '12 at 4:51
Let $k$ be a field and $R=k[X_0,X_1,X_2,..]/ \langle X_0^2,X_1^2,X_2^2,...\rangle=k[x_0,x_1,x_2,..]$. Then $\operatorname{Ass}_R(R)=\emptyset$ .
Edit: a detailed proof
At Lando's request I'll give a complete proof of the above.
1) Let's show that the only prime ideal of $R$ is the obviously maximal ideal $\mathfrak m=\langle x_0,x_1,x_2,...\rangle$.
Indeed, if $\mathfrak p \subset R$ is prime then for all $i\geq 0$ we deduce from $0=x_i^2\in \mathfrak p$ that $x_i\in \mathfrak p$, so that $\mathfrak m=\langle x_0,x_1,x_2,...\rangle \subset \mathfrak p$ and since $\mathfrak m$ is maximal we have $\mathfrak p=\mathfrak m$.
2) It remains to prove that $\mathfrak m$ is not an associated ideal and we will have the desired conclusion $\operatorname{Ass}(R)=\emptyset$ ($R$ is seen as a module over itself).
To say that $\mathfrak m$ is associated means that there exists $0\neq P\in R$ with $\mathfrak m=\operatorname{Ann}(P)$.
Such a $P$ can be written as $P(x_0,...,x_N)$, i.e. can involve only finitely many $x_i$'s.
But then we see that it is not true that $\mathfrak m=\operatorname{Ann}(P)$, because $x_{N+1}\notin \operatorname{Ann}(P(x_0,...,x_N))$: any non-zero term $rx_{i_0}...x_{i_s} \neq 0\; (0\leq i_0\lt\cdots\lt i_s\leq N$ ) of $P(x_0,...,x_N)$ will satisfy $rx_{i_0}...x_{i_s}\cdot x_{N+1} \neq 0$ and so $P(x_0,...,x_N)\cdot x_{N+1}\neq 0$.
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Thanks Georges. Can you explain how you indeed know that $\text{Ass}_R(R)=\varnothing$? – Kally Feb 20 '12 at 23:04
@LandoKal: The first step is to determine the prime ideals of $R$. – Ted Feb 20 '12 at 23:32
@Ted Sure, but it's not clear to me what they are. The elements of $R$ are polynomials in $k$ equivalent if their difference is divisible by some $\sum_i k_ix_i^2$. But then what? – Kally Feb 20 '12 at 23:46
Dear Lando, I have added a little hint: the ring is local. I'll add more details tomorrow (it's 02 a.m. !) – Georges Elencwajg Feb 21 '12 at 0:56
Dear Lando, I've had breakfast and then supplied a detailed proof of my assertion. If something is still not clear, do not hesitate to ask: math.stackexchange wants 100 % satisfied customers... – Georges Elencwajg Feb 21 '12 at 8:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8879817724227905, "perplexity": 188.2040896315021}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860110764.59/warc/CC-MAIN-20160428161510-00130-ip-10-239-7-51.ec2.internal.warc.gz"} |
http://www.kolda.net/tags/opinion/ | # Opinion
## Sparse Versus Scarce
Sometimes the term sparse is used to refer to a matrix that has a large fraction of missing entries, but the more typical usage of that term is to refer to a matrix that has a large fraction of zero entries. We instead recommend the term scarce for a large amount of missing data and discuss various scenarios. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9122881293296814, "perplexity": 302.3725320827103}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986675598.53/warc/CC-MAIN-20191017172920-20191017200420-00144.warc.gz"} |
http://www.lmfdb.org/knowledge/show/lattice.gram | show · lattice.gram all knowls · up · search:
Fixing a basis $\{v_1,...,v_n\}$ for a space $V$, the Gram matrix is the $n\times n$ symmetric matrix given by $[B(v_i,v_j)]$. Similarly, if we fix a basis $\{w_1,...,w_n\}$ for a lattice $L$, then it is also a basis for $V$, and the Gram matrix for $L$ is given by $[B(w_i,w_j)]$.
For example, given the integral lattice $\mathbb Z^3$ on the Euclidean space $\mathbb{R}^3$, in the standard basis $\{e_1,e_2,e_3\}$ the lattice (and hence the space) has a Gram matrix $\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$
Although Gram matrices are only unique up to a choice of basis, each lattice in the database is identified by a single Gram matrix. By entering a Gram matrix, $G$, into the search field, the database will return an isometric lattice whose Gram matrix, $A$, is equivalent to $G$ via conjugation by some transition matrix, $T$ (that is, $A=T^tGT$), provided the lattice is in the database.
Authors:
Knowl status:
• Review status: reviewed
• Last edited by Kiran S. Kedlaya on 2018-06-27 09:25:30
Referred to by:
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http://math.stackexchange.com/questions/87337/combinatorics-problem | # Combinatorics Problem?
How many ways can we move on an $N \times N$ grid with the following constraints:
• Our start state is $(1,1)$
• End state is $(N,N)$
• The only moves allowed are right and down moves
• Your path should have exactly $K$ turns in it.
For example, I worked out that for $N = 4, K = 2$, the answer is $4$, but I am not able to generalize the results.
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Could the downvoter explain the reason for their downvote? This seems like a perfectly reasonable question to me. – Chris Taylor Dec 1 '11 at 10:38
The downvote (not by me) could be due to the uninformative title or due to the bad choice of starting coordinates or due to the wrong use of "down" or due to the missing definition of "turn". On the other hand, I certainly think that you should explain why you not upvote a question that you deem "perfectly reasonable". – Phira Dec 1 '11 at 11:06
@Phira: I don't see what you mean by "bad" starting coordinates or wrong use of "down". People are perfectly free to start counting at $1$ (I personally always prefer $0$, but in this particular problem it makes little difference) or to index a grid as one would index the entries of a matrix (here my preferences actually coincide with those of the question). Being explicit about this would be better, but then I've seen many papers that talk about north, east etc. without pointing out the the reader should make sure to read it facing north. – Marc van Leeuwen Dec 1 '11 at 11:29
@Phira reasonable != great – Chris Taylor Dec 1 '11 at 13:27
Is there any alternative way to solve this problem? Can this problem be solved using a multinomial theorem approach? – user20622 Dec 4 '11 at 16:04
It seems to me that the final answer should be $2 \binom{N-2}{\left\lfloor \frac{k}{2} \right\rfloor } \binom{N-2}{\left\lfloor \frac{k-1}{2} \right\rfloor }$.
I won't give away the entire proof, but note the following:
• The options of starting going down and starting going right are symmetric, so just worry about one, and double your answer.
• You'll have to worry about whether $k$ is even or odd, since this will determine (together with your starting direction) whether the last turn is on the bottom row or the rightmost column. Since this last turn is determined, only $k-1$ turns are free choices, whilst avoiding the bottom row and rightmost column. Together with the starting direction this will determine the number of free choices for down-to-right turns and right-to-down turns. So now look at where the down-to-right turns can be made, and where the right-to-down turns can be made.
• You can't turn as your very first move.
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You may count the paths starting horizontally, and multiply the end result by 2 by symmetry. Now consider the sets $S_r,S_c$ of rows and columns that contain at least one turning point, or the start or end point. In general there will be exactly two such points in the row or column as soon as there is one, the exceptions being the first column, and depending on whether $K$ is odd or even the last row or column (these have only one such point). The first and last rows are uninteresting, as they are always in the sets $S_r,S_c$, it suffices to choose the remainder of these sets among the remaining $N-2$ columns/rows.
Now you should be able to figure out why the result is twice $\binom{N-2}i\binom{N-2}{i-1}$ when $K=2i$, and twice $\binom{N-2}i\binom{N-2}i=\binom{N-2}i^2$ when $K=2i+1$. Don't forget the factors 2. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8560590147972107, "perplexity": 280.1492141625988}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1404776425157.62/warc/CC-MAIN-20140707234025-00088-ip-10-180-212-248.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/76816/how-do-you-visualize-real-projective-n-space | # How do you visualize real projective $n$-space?
What I'm asking for is how you visualize this.
Definition. Define an equivalence relation on $S^n$ by $x \sim \pm x$ for all $x \in S^n$. Then the quotient space $S^n/ \sim$ is called the (real) projective $n$-space and is denoted $P^n$.
Lecturer gave a construction of $P^2$ and showed it was like gluing a Möbius band to the edges of a disk.
However, he was unable to say what $P^3$ would be constructed in general and then he said that there were big problems in (real) projective spaces and he said it is hard to visualize.
Intuitively speaking, I feel like $P^3$ would be a Möbius bottle and you are gluing that onto a solid ball. Well gluing that onto the surface of a solid ball.
Anyway, intuitive way to visualize this space in higher dimensions?
Please can you describe it in terms of topological objects. In a way I don't care much about realizing the gluing. Well, ideally would want to know what it looks like say it looks like a Möbius strip with a disk. Even if the gluing itself you need to construct something that intersects itself three times at a point.
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Do you know how the projective plane works in projective or algebraic geometry? – Henning Makholm Oct 28 '11 at 23:59
No. I haven't came across projective planes. I also don't know any algebraic geometry. Well, I hoping someone gives what projective n-space breaks down into in terms of topological objects i.e. what you need to glue together to get $P^n$ – simplicity Oct 29 '11 at 0:02
Projective $2$-space, as your lecturer points out, can be thought of as a disk with a twisted band attached, to which another disk is attached along the boundary. This generalizes to higher dimensions using the language of handle theory. In general, in $n$ dimensions, a $k$-handle is homeomorphic to $D^k\times D^{n-k}$ (which is itself homeomorphic to $D^n$, where $D^\ell$ denotes a closed $\ell$-dimensional disk. The attaching region of a $k$-handle is $(\partial D^k)\times D^{n-k}= S^{k-1}\times D^{n-k}$. You can build up $n$-manifolds using handles by first taking a union of $0$-handles, then attaching $1$-handles to the boundary of the $0$-handles along the attaching regions of the $1$-handles. In general, one you have all the $\leq k$-handles, you attach $(k+1)$-handles to the boundary of what you have so far, along the attaching regions of the $(k+1)$-handles. Now in $2$ dimensions, a $0$-handle is a disk, a $1$-handle is a rectangular strip whose attaching region is two opposite ends of the strip, and a $2$-handle is a disk with its entire boundary as an attaching region. Then the projective plane can be written as a single $0$-handle, to which a single $1$-handle attaches (with a twist), to which a single $2$-handle attaches. In fact, every surface has a handle decomposition. Indeed any smooth manifold has such a decomposition.
Now, for projective $n$-space $\mathbb{RP}^n$, you can show that there is a handle decomposition with a single $k$-handle for every $0\leq k\leq n$. One way to see this is to construct a handle decomposition for the sphere with two of every index of handle, such that the quotient by $x\sim -x$ identifies the handles in pairs.
Specifically for $\mathbb{RP}^3$, first you can think of this a quotient of $D^3$ by identifying antipodal points on $\partial D^3=S^2$. That's not a bad way to visualize it, but to get to the handle decomposition, put a little $0$-handle (ball) in the middle of $D^3$. Then extend a $1$-handle (solid tube) out to $\partial D^3$ along the $x$-axis, which then comes out the other side along the negative $x$-axis to attach again to the $0$-handle. Now attach a $2$-handle (penny) to fill in a neighborhood of the $xy$-plane. (You have to convince yourself that this really is homeomorphic to $D^2\times I$.) So now fill in the rest with a $3$-handle (3-ball) which joins the top part of $D^3$ with the bottom part using the boundary identifications. I guess you would say that the union of $0$ and $1$-handle is a solid Möbius bottle, but you also need to glue on a $2$-handle before you complete it by adding a ball.
The proof works the same. Find a handle decomposition of $S^n$ which respects the $\mathbb Z_2$-action with two $k$-handles of every index $k$. Then only one $k$-handle will survive to the quotient. – Grumpy Parsnip Oct 29 '11 at 0:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8531298637390137, "perplexity": 245.9545666317996}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860125750.3/warc/CC-MAIN-20160428161525-00079-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=39&t=49180&p=178740 | ## Lattice Energy
TanveerDhaliwal3G
Posts: 105
Joined: Fri Aug 30, 2019 12:17 am
### Lattice Energy
I did not really understand the definition that the book gave. Can someone clarify what lattice energy is?
Hannah Pham
Posts: 104
Joined: Fri Aug 09, 2019 12:17 am
### Re: Lattice Energy
Lattice energy is the energy given off when oppositely charged ions in the gas phase come together to form a solid.
Kayli Choy 2F
Posts: 100
Joined: Sat Aug 24, 2019 12:17 am
### Re: Lattice Energy
When ions are separated by large distances, as they are in the gas phase, there is attraction between them (potential energy). As the ions get closer together, this attraction lessens. Therefore, there is a net lowering of energy, and when ions get close enough together to form a solid, energy is released. This energy release is the lattice energy.
Jamie Lee 1F
Posts: 106
Joined: Fri Aug 09, 2019 12:16 am
### Re: Lattice Energy
Lattice energy is the energy given off when oppositely charged ions in the gas phase come together to form a solid.
Luyan Zhang - 2D
Posts: 103
Joined: Sat Jul 20, 2019 12:16 am
### Re: Lattice Energy
Lattice energy is the energy stored in the bonds of an ionic solid, such as NaCl. This energy is released when Na and Cl come together. This energy is much greater than the energy needed to strip electron off of Na, when shows why NaCl has a lower energy than Na or Cl.
005206171
Posts: 107
Joined: Wed Sep 18, 2019 12:20 am
### Re: Lattice Energy
Ionic bonds form crystalline lattices in their solid form. Forming bonds releases energy (they become more stable). If a molecule has ionic bonds (cation and anion) and is in the gas phase, the amount of energy released as it condenses into a solid is called lattice energy.
bloodorangefield
Posts: 24
Joined: Wed Sep 18, 2019 12:19 am
### Re: Lattice Energy
It's just the energy that is released when oppositely charged gas ions form a solid
sbottomley3a
Posts: 50
Joined: Wed Sep 18, 2019 12:19 am
### Re: Lattice Energy
Lattice energy is the energy stored in the bonds of a crystalline solid ionic compound
005384106
Posts: 101
Joined: Sat Aug 24, 2019 12:16 am
### Re: Lattice Energy
Have we ever been asked questions about solving for lattice energy? Or is it just a concept? If we can solve for it how would we do that?
chrischyu4a
Posts: 52
Joined: Thu Jul 25, 2019 12:16 am
### Re: Lattice Energy
When ions form bonds they release energy. Lattice energy is basically the amount of energy that is released when ions form a compound. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8403134346008301, "perplexity": 2709.0179934331254}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370492125.18/warc/CC-MAIN-20200328164156-20200328194156-00457.warc.gz"} |
https://www.cheenta.com/order-of-rings-tifr-gs-2018-part-b-problem-12/ | Select Page
# Understand the problem
The number of rings of order 4, up to isomorphism, is:
(a) 1
(b) 2
(c) 3
(d) 4.
##### Source of the problem
TIFR GS 2018 Part B Problem 12
Abstract Algebra
Easy
##### Suggested Book
Dummit and Foote
Do you really need a hint? Try it first!
First, ask yourself how many groups are there of order 4. the answer is simple => Z/4Z and Klein’s four group (K).
So intuitively there should be two rings with 4 elements.
Okay, I will give you two rings or same order=> (R,+,.), (R,+,*) . see that order of the rings are same but I have changed the multiplication, and
I define a*b=0 for all a,b in R. [(R,+,*) is called zero ring]
Question: Prove that (R,+,.) and (R,+,*) are not isomorphic! (easy)
So, we had (Z/4Z,+,.) and (K,+,.) as our answers. But if you change the multiplication to “*” then there will be 4 different rings *upto isomorphism* right?
Hence the answer is 4.
Bonus Problem:Prove that there are only two non-ismorphic p-rings(ring with p elements) upto isomorphism.
# Connected Program at Cheenta
#### College Mathematics Program
The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.
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## Arithmetical Dynamics: Part 0
Rational function $$R(z)= \frac {P(z)}{Q(z)}$$ ; where P and Q are polynimials . There are some theory about fixed points . Theorem: Let $$\rho$$ be the fixed point of the maps R and g be the Mobius map . Then $$gRg^{-1}$$ has the same number of fixed points at...
## Sum based on Probability – ISI MMA 2018 Question 24
This is an interesting and cute sum based on the concept of Arithematic and Geometric series .The problem is to find a solution of a probability sum.
## System of the linear equation: ISI MMA 2018 Question 11
This is a cute and interesting problem based on System of the linear equation in linear algebra. Here we are finding the determinant value .
## Application of eigenvalue in degree 3 polynomial: ISI MMA 2018 Question 14
This is a cute and interesting problem based on application of eigen values in 3 degree polynomial .Here we are finding the determinant value .
## To find Trace of a given Matrix : ISI MMA 2018 Question 13
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## Order of rings: TIFR GS 2018 Part B Problem 12
This problem is a cute and simple application on the ring theory in the abstract algebra section. It appeared in TIFR GS 2018. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8143956661224365, "perplexity": 1616.1368615140927}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986657586.16/warc/CC-MAIN-20191015055525-20191015083025-00186.warc.gz"} |
http://math.stackexchange.com/questions/129454/abstract-algebra-question-concerning-groups/129455 | abstract algebra question concerning groups
Are these groups? If so show it, and if not provide a counterexample.
The set of all complex numbers $x$ that have absolute value $1$, with operation multiplication. Recall that the absolute value of a complex number $x$ written in the form $x = a +bi$, with $a$ and $b$ real, is given by $|x| = |a+bi| = (a^2 + b^2)^{1/2}$.
The set of all complex numbers $x$ that have absolute value $1$, with operation addition.
-
As I mentioned on your earlier question, it is not considered polite here to tell other users to do something. Your question does not show that you have thought about the problem. Please explain what you've tried so far, and where you are stuck. – Zev Chonoles Apr 8 '12 at 23:47
The properties of a group are 1) closure, 2) identity, and 3) inverse. Have you tested these cases against these properties? Where are you stuck? – Tpofofn Apr 9 '12 at 0:03
Hint: for $z,w\in\mathbb{C}$, $|zw|=|z||w|$ but $|z+w|\leq|z|+|w|$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9239473342895508, "perplexity": 327.26873009677865}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1387345758214/warc/CC-MAIN-20131218054918-00035-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://www.quizover.com/trigonometry/test/algebraic-arithmetic-sequences-by-openstax | # 13.2 Arithmetic sequences (Page 5/8)
Page 5 / 8
What are the main differences between using a recursive formula and using an explicit formula to describe an arithmetic sequence?
Describe how linear functions and arithmetic sequences are similar. How are they different?
Both arithmetic sequences and linear functions have a constant rate of change. They are different because their domains are not the same; linear functions are defined for all real numbers, and arithmetic sequences are defined for natural numbers or a subset of the natural numbers.
## Algebraic
For the following exercises, find the common difference for the arithmetic sequence provided.
$\left\{5,11,17,23,29,...\right\}$
$\left\{0,\frac{1}{2},1,\frac{3}{2},2,...\right\}$
The common difference is $\frac{1}{2}$
For the following exercises, determine whether the sequence is arithmetic. If so find the common difference.
$\left\{11.4,9.3,7.2,5.1,3,...\right\}$
$\left\{4,16,64,256,1024,...\right\}$
The sequence is not arithmetic because $16-4\ne 64-16.$
For the following exercises, write the first five terms of the arithmetic sequence given the first term and common difference.
${a}_{1}=-25$ , $d=-9$
${a}_{1}=0$ , $d=\frac{2}{3}$
$0,\text{\hspace{0.17em}}\frac{2}{3},\text{\hspace{0.17em}}\frac{4}{3},\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}\frac{8}{3}$
For the following exercises, write the first five terms of the arithmetic series given two terms.
${a}_{1}=17,\text{\hspace{0.17em}}{a}_{7}=-31$
${a}_{13}=-60,\text{\hspace{0.17em}}{a}_{33}=-160$
$0,-5,-10,-15,-20$
For the following exercises, find the specified term for the arithmetic sequence given the first term and common difference.
First term is 3, common difference is 4, find the 5 th term.
First term is 4, common difference is 5, find the 4 th term.
${a}_{4}=19$
First term is 5, common difference is 6, find the 8 th term.
First term is 6, common difference is 7, find the 6 th term.
${a}_{6}=41$
First term is 7, common difference is 8, find the 7 th term.
For the following exercises, find the first term given two terms from an arithmetic sequence.
Find the first term or ${a}_{1}$ of an arithmetic sequence if ${a}_{6}=12$ and ${a}_{14}=28.$
${a}_{1}=2$
Find the first term or ${a}_{1}$ of an arithmetic sequence if ${a}_{7}=21$ and ${a}_{15}=42.\text{\hspace{0.17em}}$
Find the first term or ${a}_{1}$ of an arithmetic sequence if ${a}_{8}=40$ and ${a}_{23}=115.$
${a}_{1}=5$
Find the first term or ${a}_{1}$ of an arithmetic sequence if ${a}_{9}=54$ and ${a}_{17}=102.$
Find the first term or ${a}_{1}$ of an arithmetic sequence if ${a}_{11}=11$ and ${a}_{21}=16.$
${a}_{1}=6$
For the following exercises, find the specified term given two terms from an arithmetic sequence.
${a}_{1}=33\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{a}_{7}=-15.$ Find $\text{\hspace{0.17em}}{a}_{4}.$
${a}_{3}=-17.1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{a}_{10}=-15.7.$ Find ${a}_{21}.$
${a}_{21}=-13.5$
For the following exercises, use the recursive formula to write the first five terms of the arithmetic sequence.
$-19,-20.4,-21.8,-23.2,-24.6$
For the following exercises, write a recursive formula for each arithmetic sequence.
${a}_{n}=\left\{40,60,80,...\right\}$
${a}_{n}=\left\{17,26,35,...\right\}$
${a}_{n}=\left\{-1,2,5,...\right\}$
${a}_{n}=\left\{12,17,22,...\right\}$
${a}_{n}=\left\{-15,-7,1,...\right\}$
${a}_{n}=\left\{8.9,10.3,11.7,...\right\}$
${a}_{n}=\left\{-0.52,-1.02,-1.52,...\right\}$
${a}_{n}=\left\{\frac{1}{5},\frac{9}{20},\frac{7}{10},...\right\}$
${a}_{n}=\left\{-\frac{1}{2},-\frac{5}{4},-2,...\right\}$
${a}_{n}=\left\{\frac{1}{6},-\frac{11}{12},-2,...\right\}$
For the following exercises, write a recursive formula for the given arithmetic sequence, and then find the specified term.
Find the 17 th term.
Find the 14 th term.
Find the 12 th term.
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 52, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9024243354797363, "perplexity": 780.661443702587}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823382.1/warc/CC-MAIN-20181210170024-20181210191524-00398.warc.gz"} |
https://math.stackexchange.com/questions/996048/find-a-permutation-with-the-given-square-or-cube | # Find a permutation with the given square or cube
Problem: find a permutation such that
1. $x^2 = (1\;3\;4\;5\;7)$, $x\in S_7$
2. $x^3 = (1\;3\;4\;5\;7)$, $x\in S_7$
Must find all possible solutions for $x$.
### Progress
I have solved for the first part, $x=(57134)$ but I am not sure on how to prove that there is only one $x$ value.
• Surely you are posing two separate problems, since we cannot have $x^2=x^3$ unless $x = x^2 = x^3$ is the identity map. – hardmath Oct 29 '14 at 2:46
• I have solved for the first part, x=(57134) but I am not sure on how to prove that there is only one x value – Abigail Oct 29 '14 at 2:51
• @hardmath Thanks for the comment on my solution; I had missed the "all possible solutions" part of the question. – angryavian Oct 29 '14 at 2:52
Note that in cycle notation for permutations, $(1~3~4~5~7) = (5~7~1~3~4)$.
Then (arbitrarily and without loss of generality), 1 is first in our cycle notation, followed by something, then 3, then something, then 4. So we have $(1~a_1~3~a_2~4)$ as our cycle. Now we try to figure out what goes in place of $a_1$ and $a_2$. After 1 iteration of this permutation, 4 maps to 1. On the second iteration, it should map to five, so that $x^2(4) = 5$. Note that I'm viewing $x$ as a function with $x^2 = x \circ x$. We now have $x = (1~5~3~a_2~4)$. Next, we need $x^2(5) = 7$. The way to do that is to set $a_2 = 7$. Thus we see, and can confirm, that $(1~5~3~7~4)^2 = (1~3~4~5~7)$. There's a pair left over in $S_7$; we haven't moved 2 or 6. Since $(2~6)^2$ is the identity, we can set $x = (1~5~3~7~4)(2~6)$ as well. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9494693279266357, "perplexity": 195.55905790033427}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987823061.83/warc/CC-MAIN-20191022182744-20191022210244-00131.warc.gz"} |
http://cognet.mit.edu/journal/10.1162/NECO_a_00653 | ## Neural Computation
November 2014, Vol. 26, No. 11, Pages 2395-2418
(doi: 10.1162/NECO_a_00653)
@ 2014 Massachusetts Institute of Technology
Changes of Firing Rate Induced by Changes of Phase Response Curve in Bifurcation Transitions
Abstract
We study dynamical mechanisms responsible for changes of the firing rate during four different bifurcation transitions in the two-dimensional Hindmarsh-Rose (2DHR) neuron model: the saddle node on an invariant circle (SNIC) bifurcation to the supercritical Andronov-Hopf (AH) one, the SNIC bifurcation to the saddle-separatrix loop (SSL) one, the AH bifurcation to the subcritical AH (SAH) one, and the SSL bifurcation to the AH one. For this purpose, we study slopes of the firing rate curve with respect to not only an external input current but also temperature that can be interpreted as a timescale in the 2DHR neuron model. These slopes are mathematically formulated with phase response curves (PRCs), expanding the firing rate with perturbations of the temperature and external input current on the one-dimensional space of the phase S1 in the 2DHR oscillator. By analyzing the two different slopes of the firing rate curve with respect to the temperature and external input current, we find that during changes of the firing rate in all of the bifurcation transitions, the calculated slope with respect to the temperature also changes. This is largely dependent on changes in the PRC size that is also related to the slope with respect to the external input current. Furthermore, we find phase transition–like switches of the firing rate with a possible increase of the temperature during the SSL-to-AH bifurcation transition. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9277328252792358, "perplexity": 1156.7047813998236}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514575402.81/warc/CC-MAIN-20190922073800-20190922095800-00288.warc.gz"} |
http://rcd.ics.org.ru/authors/detail/911-viviana_diaz | 0
2013
Impact Factor
# Viviana Diaz
Avda. Alem 1253 - 2 Piso, 8000, Bahia Blanca, Argentina
Departamento de Matematica, Universidad Nacional del Sur, Bahia Blanca
## Publications:
Cendra H., Diaz V. A. The Lagrange–D'Alembert–Poincaré Equations and Integrability for the Euler's Disk 2007, vol. 12, no. 1, pp. 56-67 Abstract Nonholonomic systems are described by the Lagrange–D'Alembert's principle. The presence of symmetry leads, upon the choice of an arbitrary principal connection, to a reduced D'Alembert's principle and to the Lagrange–D'Alembert–Poincaré reduced equations. The case of rolling constraints has a long history and it has been the purpose of many works in recent times. In this paper we find reduced equations for the case of a thick disk rolling on a rough surface, sometimes called Euler's disk, using a 3-dimensional abelian group of symmetry. We also show how the reduced system can be transformed into a single second order equation, which is an hypergeometric equation. Keywords: nonholonomic systems, symmetry, integrability, Euler's disk Citation: Cendra H., Diaz V. A., The Lagrange–D'Alembert–Poincaré Equations and Integrability for the Euler's Disk, Regular and Chaotic Dynamics, 2007, vol. 12, no. 1, pp. 56-67 DOI:10.1134/S1560354707010054
Cendra H., Diaz V. A. The Lagrange–D'Alembert–Poincaré equations and integrability for the rolling disk 2006, vol. 11, no. 1, pp. 67-81 Abstract Classical nonholonomic systems are described by the Lagrange–d'Alembert principle. The presence of symmetry leads, upon the choice of an arbitrary principal connection, to a reduced variational principle and to the Lagrange–d'Alembert–Poincaré reduced equations. The case of rolling bodies has a long history and it has been the purpose of many works in recent times, in part because of its applications to robotics. In this paper we study the classical example of the rolling disk. We consider a natural abelian group of symmetry and a natural connection for this example and obtain the corresponding Lagrange–d'Alembert–Poincaré equations written in terms of natural reduced variables. One interesting feature of this reduced equations is that they can be easily transformed into a single ordinary equation of second order, which is a Heun's equation. Keywords: rolling disk, nonholonomic mechanics, integrability, Heun's equation Citation: Cendra H., Diaz V. A., The Lagrange–D'Alembert–Poincaré equations and integrability for the rolling disk , Regular and Chaotic Dynamics, 2006, vol. 11, no. 1, pp. 67-81 DOI:10.1070/RD2006v011n01ABEH000335 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9634012579917908, "perplexity": 833.2456576096655}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156471.4/warc/CC-MAIN-20180920120835-20180920141235-00489.warc.gz"} |
https://www.physicsforums.com/threads/kinetic-energy-and-momentum.34868/ | # Kinetic energy and momentum
1. Jul 12, 2004
### wilmerena
do 2 objects that have the same kinetic energy necessarily have the same momentum? I cant think of a simple example :yuck:
2. Jul 12, 2004
### AKG
Consider a 5 kg mass going at 50 m/s, thus having 6.25 kJ of kinetic energy. Now, a mass of 125 kg going 10 m/s also has 6.25 kJ of kinetic energy. However, the first object has a momentum of 250 kg*m/s, and the second has 1250 kg*m/s of momentum.
3. Jul 12, 2004
### wilmerena
thanks so much =o) !!
4. Jul 13, 2004
### wilmerena
what if they both have 0 momentum, so does it follow that the kinetic energy of the system doesnt have to be zero as well? or does it?
5. Jul 13, 2004
### Staff: Mentor
Think about it. For a simple object:
$$KE = 1/2 mv^2$$
$$momentum = mv$$
The only way an object can have zero momentum is if its speed (v) is what? Then what is its KE?
6. Jul 13, 2004
### wilmerena
thanks again, i think i just spaced out on that one ;oP
Similar Discussions: Kinetic energy and momentum | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9605377912521362, "perplexity": 1489.3928548759159}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121869.65/warc/CC-MAIN-20170423031201-00024-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://cs.stackexchange.com/questions/95988/fptas-definition | # FPTAS definition
I read that a fully polynomial time approximation scheme (FPTAS) has time complexity polynomial in the input size and also polynomial in $1/\epsilon$. However, this leaves some ambiguity.
For example, $T(n,1/\epsilon)=\min(n^{1/\epsilon},(1/\epsilon)^n)$ is both polynomial in $n$ (when holding $1/\epsilon$ constant) and polynomial in $1/\epsilon$ (when holding $n$ constant). But $T$ is not polynomial in $n/\epsilon$ or $(n+1/\epsilon)$ because $\sqrt x^\sqrt x$ and $(x/2)^{x/2}$ aren't bounded by polynomials.
Does this mean an algorithm with running time $T$ isn't an FTPAS, and perhaps a better definition would be that it has to be polynomial in $n+1/\epsilon$?
The standard definition is incredibly ambiguous, since it is not clear what "polynomial in $n$ and in $1/\epsilon$" means. Judging from examples, it means $O(n^C(1/\epsilon)^D)$ for some constants $C,D$. Without loss of generality, $C=D$, and then we get the simpler definition $O((n/\epsilon)^C)$.
• Thanks for clarifying. Although might $O((n+1/\epsilon)^C)$ be more precise because when $\epsilon=n$ is very large, $n/\epsilon=1$ but it should still take at least $O(n)$ to find a feasible solution? – Akababa Aug 5 '18 at 23:22
• The big O holds in the limit $n\to\infty$ and $\epsilon\to0$. We can assume that $\epsilon < 1 < n$, say. – Yuval Filmus Aug 5 '18 at 23:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9301848411560059, "perplexity": 221.29075544057784}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670559.66/warc/CC-MAIN-20191120134617-20191120162617-00391.warc.gz"} |
https://socratic.org/questions/an-object-with-a-mass-of-4-kg-is-pushed-along-a-linear-path-with-a-kinetic-frict-13 | Physics
Topics
# An object with a mass of 4 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 5+tanx . How much work would it take to move the object over x in [(-3pi)/12, (pi)/4], where x is in meters?
Oct 30, 2017
The work is $= 307.9 J$
#### Explanation:
We need
$\int \tan x \mathrm{dx} = - \ln \left(| \cos x |\right) + C$
$- \frac{3}{12} \pi = - \frac{1}{4} \pi$
The work done is
$W = F \cdot d$
The frictional force is
${F}_{r} = {\mu}_{k} \cdot N$
The normal force is $N = m g$
The mass is $m = 4 k g$
${F}_{r} = {\mu}_{k} \cdot m g$
$= 4 \cdot \left(5 + \tan x\right) g$
The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$
The work done is
$W = 4 g {\int}_{- \frac{1}{4} \pi}^{\frac{1}{4} \pi} \left(5 + \tan x\right) \mathrm{dx}$
$= 4 g \cdot {\left[5 x - \ln | \cos x |\right]}_{- \frac{1}{4} \pi}^{\frac{1}{4} \pi}$
=4g(5/4pi-ln|cos(1/4pi)|))-(-5/4pi-ln|cos(-1/4pi)|)
=4g(5/2pi))#
$= 10 g \pi$
$= 307.9 J$
##### Impact of this question
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https://www.physicsforums.com/threads/virial-equation-of-state.794169/ | # Virial equation of state
• #1
127
3
## Homework Statement
It's just that in my textbook, for section titled "Second virial coefficients can be used to determine intermolecular potentials," I have an equation that I do NOT understand how it was derived---I tried to do it over and over, but couldn't quite figure how. If anyone could explain, thank you!!!
B2v(T) = RT * B2p(T), where B2v(T) and B2p(T) are virial coefficient
## Homework Equations
B2v(T) = RT * B2p(T), where B2v(T) and B2p(T) are virial coefficient
## The Attempt at a Solution
I assume the equation comes from Z, the compressibility factor:
Z = PV/RT, where it can be expanded to:
Z = 1 + B2v(T)/V + B3v(T)/V2 + ...
Z = 1 + B2p(T)*P + B3p(T)*P2 + ...
So, I equivocated 1 + B2v(T)/V + ... = 1 + B2p(T)P + ...
B2v(T)/V + ... = B2p(T)P + ...
...and tried multiplying both side by V or divide by P to cancel out other virial coefficients with numbers larger than 2 (meaning B3v and B3p), but I don't know how to progress anymore.......pls give me hints! I want to understand how the original equation B2v(T) = RT * B2p(T) was obtained!
Related Advanced Physics Homework Help News on Phys.org
• #2
BvU
Homework Helper
2019 Award
13,048
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Perhaps the idea isn't to let all higher order terms cancel, just to equate the first order coefficients ?
• #3
Quantum Defect
Homework Helper
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## Homework Statement
It's just that in my textbook, for section titled "Second virial coefficients can be used to determine intermolecular potentials," I have an equation that I do NOT understand how it was derived---I tried to do it over and over, but couldn't quite figure how. If anyone could explain, thank you!!!
B2v(T) = RT * B2p(T), where B2v(T) and B2p(T) are virial coefficient
## Homework Equations
B2v(T) = RT * B2p(T), where B2v(T) and B2p(T) are virial coefficient
## The Attempt at a Solution
I assume the equation comes from Z, the compressibility factor:
Z = PV/RT, where it can be expanded to:
Z = 1 + B2v(T)/V + B3v(T)/V2 + ...
Z = 1 + B2p(T)*P + B3p(T)*P2 + ...
So, I equivocated 1 + B2v(T)/V + ... = 1 + B2p(T)P + ...
B2v(T)/V + ... = B2p(T)P + ...
...and tried multiplying both side by V or divide by P to cancel out other virial coefficients with numbers larger than 2 (meaning B3v and B3p), but I don't know how to progress anymore.......pls give me hints! I want to understand how the original equation B2v(T) = RT * B2p(T) was obtained!
What happens if you use the compressibility (written in the volume version of the virial expansion) to solve for P. Take this series expansion for P and plug into the pressure version of the virial expansion. Compare this equation with the virial expansion in terms of volume. Coefficients in front of the same powers of 1/V must be the same for the two expressions to be the same:
A x + B x^2 + ... = alpha x + beta x^2 + ... is true iff A = alpha, B = beta, ....
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2K | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8433530926704407, "perplexity": 2506.642582623456}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371807538.83/warc/CC-MAIN-20200408010207-20200408040707-00163.warc.gz"} |
https://brilliant.org/problems/a-problem-by-ameya-daigavane/ | # One, Two, Three, What Do You See?
Find the number of ordered pairs $$(x, y)$$ for integers $$x, y$$, if: $y^2 = x^3 + 7 \; .$
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8383806347846985, "perplexity": 1081.5419327940983}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187828189.71/warc/CC-MAIN-20171024071819-20171024091819-00482.warc.gz"} |
https://www.thingelstad.com/2006/first-rum-meeting/ | Of particular note was my friend Dan presented a side-project he did using rex and racc (Ruby versions of lex and yacc) to create a parser and validator for math equations written in LaTeX on Road Sign Math. This was pretty cool. He hacked it up quickly and solved most of the hard problems. He got to the point so he can evaluate that "3 \times \sqrt{4} = 6" is a valid statement. It will be a nice thing to wrap into Road Sign Math 3.0 when the time is right. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8156309127807617, "perplexity": 1187.2954487084908}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948537139.36/warc/CC-MAIN-20171214020144-20171214040144-00693.warc.gz"} |
http://mathhelpforum.com/discrete-math/143243-composition-transitive-relations.html | # Thread: Composition of transitive relations
1. ## Composition of transitive relations
Hi,
I have this problem that is really driving me crazy:
Suppose R and S are transitive relations on A. Prove that if SºR c RºS, then RºS is transitive.
I tried to use the theorem that says: R is transitive iff RºR c R, but I couldn't get to the end.
(c means subset)
2. This is a completely tedious proof. I will give you the outline.
Suppose that $(p,q) \in R \circ S \wedge (q,r) \in R \circ S$.
That means that $\left( {\exists s} \right)\left[ {(p,s) \in S \wedge (s,q) \in R} \right]\,\& \,\left( {\exists t} \right)\left[ {(q,t) \in S \wedge (t,r) \in R} \right]$.
From which it follows that $(s,q) \in R \wedge (q,t) \in S \Rightarrow (s,t) \in S \circ R \Rightarrow (s,t) \in R \circ S$.
It follows from that $\left( {\exists d} \right)\left[ {(s,d) \in S \wedge (d,t) \in R} \right]$.
Because both $S~\&~T$ are transitive we get $(p,d)\in S~\&~(d,r)\in R$.
But that means that $(p,r)\in R\circ S$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.980032205581665, "perplexity": 733.4346568993974}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171176.3/warc/CC-MAIN-20170219104611-00547-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://www.conservapedia.com/Acid-base_reaction | # Acid-base reaction
An acid-base reaction, also known as a neutralization reaction, is a proton transfer reaction between an acid (a proton donor) and a base (a proton acceptor). In general, acid-base reactions are typically exothermic and they produce salt and water.
## General Chemistry
In a majority of cases, the acid and base regents will dissociate to form ions which take part in the reaction mechanism and those which form the salt. For instance, in the aqueous reaction between hydrochloric acid and sodium hydroxide:
HCl $\rightleftharpoons$ H+ + Cl-
NaOH $\rightleftharpoons$ Na+ + OH-
Some of these ions will then partake in the proton transfer reaction, such that:
H+ + OH- $\rightleftharpoons$ H2O
The remaining ions will form a salt, such that:
Na+ + Cl- $\rightleftharpoons$ NaCl(aq)
Thus the overall reaction is:
HCl(aq) + NaOH(aq) $\rightleftharpoons$ H2O(l) + NaCl(aq)
In this case the salt will remain in aqueous (dissolved) form, however if it was insoluble it would precipitate out of the water.
Not all neutralisation reactions involve the reactants dissociating. For example, ammonia (NH3) is a base which will accept a proton to form ammonium, and hence will undergo an acid-base reaction in its pure form, for example:
NH3(aq) + H+(aq) $\rightleftharpoons$ NH4+(aq)
## Acidity of the Salt
The salt which forms may have a pH of its own due due to its components being conjugate acids and conjugate bases. The conjugate acid of a strong base and the conjugate base of a strong acid are weak, and hence will not dissociate. However, the conjugates of weak acids/bases are strong bases/acids, and hence will undergo a hydrolysis reaction with water to form and acid/base. The general rule is:
• A strong acid and a strong base will form a neutral salt
• A strong acid and a weak base will form an acidic salt (will react with water to form an acid)
• A strong base and a weak acid will form a basic salt (will react with water to form a base)
For instance, a strong acid and weak base reaction may proceed as such:
2HCl + Fe(OH)2$\rightleftharpoons$ 2H2O + FeCl2
The iron component of the iron chloride may then undergo a hydrolysis reaction with water, such that:
Fe2+ + H2O $\rightleftharpoons$ Fe(OH)2 + H+
Since the iron hydroxide is a weak base, it will not dissociate to any great extent and hence will not raise the pH of the water greatly. However, this hydrolysis reaction causes an excess of hydrogen ions, which result in the pH of the water being lowered, hence causing it to be acidic. As the chloride ion is a weak base it will not react with water, producing an acidic salt.
The same process applies for basic salts, however the products will contain an excess of OH- and a weak acid, causing the water to become basic.
## Uses
There are several uses of acid-base neutralisation reactions.
• Titration: a volumetric analysis of acids and bases to determine an unknown concentration. As the acids/bases neutralise they will cause an ongoing change in pH of the solution producing a characteristic titration curve. This can be used to determine when equal moles of acid and base were added to the reaction vessel, which can in turn be used to determine the unknown concentration.
• Chemical spills: strong acids and bases can be harmful to the environment and should be neutralised as soon as possible. The most commonly used neutralising agent is sodium bicarbonate (NaHCO3) as it is amphiprotic (it can act as either an acid or a base). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8902643918991089, "perplexity": 3230.2200499954106}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802769550.123/warc/CC-MAIN-20141217075249-00077-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/110355/harmonic-oscillator-with-rayleigh-pdf-noise | # Harmonic oscillator with Rayleigh pdf noise
The problem of the harmonic oscillator with a random small perturbation is known. We can write the equation as: $$\ddot{x}(t)+\eta^2x(t)=0$$
where $\eta^2=\omega^2+\epsilon{W_t}$ with $\epsilon>0$ and $W_t$: white gaussian noise. My question is: if the noise has a Rayleigh $pdf$:$$f(z;\sigma)=\frac{z}{\sigma^2}\exp(\frac{-z^2}{2\sigma^2})$$ with $z\ge0$, is it possible to find a solution for the $x(t)$? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9840789437294006, "perplexity": 147.94882485627943}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1430453921506.38/warc/CC-MAIN-20150501041841-00099-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.matrix.edu.au/maths-reference-sheet-measurement-financial-maths/ | Measurement and Financial Mathematics | The Ultimate Guide to NESA’s Maths Reference Sheet
Make the most of NESA Maths Reference Sheet and test your maths skills with these sample questions and explanations for measurement and financial maths!
The NESA Maths Reference Sheet is a great resource… if you know how to use it! Navigate measurement and financial mathematics with our Ultimate NESA Maths Reference Sheet Guide.
While memorisation has its place in learning, Matrix recommends that students learn to derive their responses and learn how to apply these formulae correctly. As a bonus, we’ve included a nifty HSC Maths Cheatsheet for you to download and print out!
A free pocket-sized Maths handbook, so you're prepared.
All the key Maths formulas you need to revise, in one foldable cheatsheet.
Click on the following formulas to see what they mean and apply them to a practice question!
Measurement
Length
Use Formula Variables Arc length of a sector (using θ in degrees) $$l=\frac{\theta}{360}\times2\pi r$$ \begin{align*} l&= \text{arc length}\\ \theta &= \text{angle of sector in degrees}\\ r &= \text{radius}\\ \end{align*}
Example 1:
Find the length of the minor arc AB, such that the length of radius OA is 30 cm. Leave your answer as an exact value.
Solution 1:
Substitute $$\theta = \text{115}$$ and $$r = \text{30}$$ into $$l=\frac{\theta}{360}\times2\pi r$$: \begin{align*} l &=\frac{115}{360}\times2\pi \times 30\\ &= \frac{115}{6} \pi \\ ∴ \text{Length of the minor arc AB} &= \frac{115}{6} \pi \text{ cm}\\ \end{align*}
Area
Use Formula Variables Area of a sector $$A= \frac{\theta}{360} \times \pi r^2$$ \begin{align*} A&= \text{area of sector}\\ \theta &= \text{angle of sector in degrees}\\ r &= \text{radius}\\ \end{align*}
Example 2:
Given that the following circle has a radius of 5 m, find the area of the major sector AOB. Leave your answer as an exact value.
Solution 2:
Substitute $$\theta=330$$ and $$r=5$$ into $$A= \frac{\theta}{360} \times \pi r^2$$: \begin{align*} A &= \frac{330}{360} \times \pi \times {5}^2 \\ &= \frac{275}{12} \pi \\ ∴ \text{Area of the major sector AOB} &= \frac{275}{12} \pi \text{ m}^2\\ \end{align*}
Use Formula Variables Area of a trapezium $$A=\frac{h}{2} (a+b)$$ \begin{align*} A&= \text{area of trapezium}\\ h &= \text{perpendicular height}\\ a, \ b &= \text{lengths of the parallel sides}\\ \end{align*}
Example 3:
What is the area of the shape below? Note that all the measurements were made in millimetres.
Solution 3:
Substitute $$h=12$$, $$a=16$$ and $$b=20$$ into $$A=\frac{h}{2} (a+b)$$: \begin{align*} A &= \frac{12}{2} (16+20) \\ &= 216 \\ ∴ \text{Area} &= 216 \text{ mm}^2\\ \end{align*}
Surface area
Use Formula Variables Surface area of a cylinder $$A=2 \pi r^2 + 2 \pi r h$$ \begin{align*} A&= \text{surface area of cylinder}\\ h &= \text{height}\\ r &= \text{radius of circular base}\\ \end{align*}
Example 4:
Gemma is designing a can for her new spaghetti recipe. In order to figure out how much pink paint she needs to completely cover the can, she needs to know what the total surface area of the cylindrical can is.
Knowing that the dimensions below are given in centimetres, what is the surface area of Gemma’s can? Leave your answer as an exact value.
Solution 4:
Substitute $$h=9$$ and $$r=4$$ into $$A=2 \pi r^2 + 2 \pi r h$$: \begin{align*} A &=2 \pi \times 4^2 + 2 \pi \times 4 \times 9 \\ &= 32 \pi + 72 \pi \\ &= 104 \pi \\ ∴ \text{Surface area of can} &= 104 \pi \text{ cm}^2\\ \end{align*}
Use Formula Variables Surface area of a sphere $$A = 4 \pi r^2$$ $$r = \text{radius}$$
Example 5:
Biological cells with larger surface area to volume ratios are able to more efficiently diffuse oxygen into the cell and waste material out.
If we approximate a cell to be the shape of a sphere, what is the surface area of a cell with a diameter of 0.07 mm? Leave your answer as an exact value.
Solution 5:
Substitute $$r= \frac{0.07}{2}$$ (since the radius is half of the diameter) into $$A = 4 \pi r^2$$: \begin{align*} A &= 4 \pi \times \left( \frac{0.07}{2} \right)^2 \\ &= 4 \pi \times 0.035^2 \\ &= 0.0049 \pi \\ ∴ \text{Surface area of cell} &= 0.0049 \pi \text{ mm}^2\\ \end{align*}
Volume
Use Formula Variables Volume of a pyramid/cone $$V= \frac{1}{3}Ah$$ \begin{align*} V &= \text{volume of pyramid/cone}\\ A &= \text{base area}\\ h &= \text{height}\\ \end{align*}
Example 6:
Ricky is planning her birthday party. She wants to prank her friends by putting water balloons inside all of the party hats.
Assuming that the party hats are a perfect cone with a radius of 5 cm and a height of 12 cm, how much water does she need to fill one party hat? Leave your answer as an exact value.
Solution 6:
To find the area of the circular base, substitute $$r=5$$ into $$A= \pi r^2$$: \begin{align*} A &= \pi \times 5^2 \\ &= 25 \pi \\ ∴ \text{Base area} &= 25 \pi \text{ cm}^2\\ \end{align*} Substitute $$A= 25 \pi$$ and $$h=12$$ into $$V= \frac{1}{3}Ah$$: \begin{align*} V &=\frac{1}{3} \times 25 \pi \times 12 \\ &= 100 \pi \\ ∴ \text{Volume of water needed} &= 100 \pi \text{ cm}^3\\ &= 100 \pi \text{ mL}\\ \end{align*}
Use Formula Variables Volume of a sphere $$V= \frac{4}{3} \pi r^3$$ $$r = \text{radius}$$
Example 7:
Gordon is making meatballs. Assuming that 1 cm3 = 1 gram, how much mince does he need to make a perfectly spherical meatball with a diameter of 3 cm? Leave your answer as an exact value.
Solution 7:
Substitute $$r= \frac{3}{2}$$ (since the radius is half of the diameter) into $$V= \frac{4}{3} \pi r^3$$: \begin{align*} V &= \frac{4}{3} \pi \left( \frac{3}{2} \right) ^3 \\ &= \frac{9}{2} \pi \\ ∴ \text{Amount of mince needed} &= \frac{9}{2} \pi \text{ cm}^3\\ &= \frac{9}{2} \pi \text{ g}\\ \end{align*}
Financial Mathematics
Use Formula Variables Compound interest $$A= P(1+r)^n$$ \begin{align*} A &= \text{value of investment at the end of the n}^{th} \text{ period}\\ P &= \text{principle (initial) amount}\\ r &= \text{compound interest rate per period — expressed as a decimal}\\ n &= \text{number of time periods}\\ \end{align*}
Example 8:
Jacky’s bank is offering her a fantastic interest rate of 3% per annum, compounded every three months. If she deposits 12 360 into her savings account today and resolves to not take away from or add to these funds, what will be her savings account balance in 12 months time? Provide your answer to the nearest two decimal places. Solution 8: \begin{align*} A & = \text{Jacky’s savings account balance in 12 months time: this is what we are trying to find!} \\ P & = 12 \ 360 \\ \end{align*}The interest rate of 3% per year is compounded every three months. So, we need to divide the 3% interest rate by the number of three month periods in 12 months to find the compound interest rate per three month period:\begin{align*} r & = \frac{3 \text{%}}{ \frac{12}{3} } \\ & = \frac{0.03}{4} \\ ∴r & = 0.0075 \\ \end{align*}Since the compounding period is three months and Jacky plans to leave her savings account balance as it is for 12 months:\begin{align*} n & = \frac{12}{3} \\ ∴n & = 4 \\ \end{align*}Hence, substitute $$P=12 \ 360$$, $$r=0.0075$$ and $$n=4$$ into $$A= P(1+r)^n$$:\begin{align*} A &= 12 \ 360 \times (1+0.0075)^4 \\ &= 12 \ 360 \times 1.0075^4 \\ &= 12 \ 734.99 \text{ (to the nearest two decimal places)} \\ ∴ \text{Jacky’s savings account balance in 12 months time} &=12 \ 734.99\\ \end{align*}
Sequences and series
Note: A series is the sum of the elements in a sequence.
Arithmetic Progressions
In an arithmetic progression, each term is determined by adding a constant to the preceding term.
Use Formula Variables Arithmetic sequence $$T_n = a+(n-1)d$$ \begin{align*} T_n &= \text{n}^{th} \text{ term of the sequence}\\ S_n &= \text{sum of the first n terms of the sequence}\\ \\ n &= \text{position of the term}\\ a &= \text{first term: }T_1\\ d &= \text{common difference between terms}\\ l &= \text{last term being added: }T_n\\ \end{align*} Arithmetic series $$S_n=\frac{n}{2} [2a+(n-1)d] = \frac{n}{2}(a+l)$$
Example 9:
Consider the following arithmetic sequence:
$$3, 15, 27, 39…$$
What is the 23rd term of this sequence?
Solution 9:
Substitute $$n=23$$, $$a=3$$ and $$d=12$$ into $$T_n = a+(n-1)d$$ \begin{align*} T_{23} & = 3+(23-1) \times 12\\ &= 3 + 22 \times 12 \\ &= 267 \\ ∴ \text{23}^{rd} \text{ term: } & 267\\ \end{align*}
Example 10:
The sum of the first 16 terms of an arithmetic series is 632. Given the first term of the series is 2, write down the first 4 terms of the series.
Solution 10:
Substitute $$S_n=632$$, $$n=16$$ and $$a=2$$ into $$S_n=\frac{n}{2} [2a+(n-1)d] = \frac{n}{2}(a+l)$$ \begin{align*} 632 & =\frac{16}{2} [2 \times 2 +(16-1)d]\\ 632 &= 8 \times (4+ 15d) \\ 632 &= 32+ 120d \\ 600 &= 120d \\ ∴ d &= 5\\ \end{align*} \begin{align*} T_1 & = a = 2 \text{ (given)}\\ \text{Since } T_n & = a+(n-1)d \text{:}\\ T_2 &= 2 + 5 \\ &= 7 \\ T_3 &= 2+ 2 \times 5 \\ &= 12 \\ T_4 &= 2 + 3 \times 5 \\ &= 17 \\ \text{∴ First 4 terms of the series: }& 2 + 7 + 12 +17\\ \end{align*}
Geometric Progressions
In a geometric progression, each term is found by multiplying the previous term by a constant.
Use Formula Variables Geometric sequence $$T_n=ar^{n-1}$$ \begin{align*} T_n &= \text{n}^{th} \text{ term of the sequence}\\ S_n &= \text{sum of the first n terms of the sequence}\\ \\ n &= \text{number of terms being added}\\ a &= \text{first term: }T_1\\ r &= \text{common ratio between terms}\\ \end{align*} Geometric series $$S_n=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1}, \ r≠1$$ Limiting sum of a geometric series $$S= \frac{a}{1-r}, \ |r|<1$$
Example 11:
A basketball player is bouncing a ball. On each bounce, the ball reaches a height that is 85% of its previous height. The total distance that the ball has travelled just before they make contact with the ball for the 8th time is 6 metres.
a) At what height did the basketball player begin bouncing the ball? Give your answer in metres to the nearest three decimal places.
b) Hence or otherwise, find the total distance that the ball travels if the basketball player continues bouncing the ball until it eventually does not lift off the ground. Give your answer in metres to the nearest two decimal places.
Solution 11 a):
$$\text{Let }x = \text{initial bounce height of the ball}$$ Let’s consider what this problem would look like visually. The height of each bounce conveniently forms a geometric sequence where the common ratio r=0.75 and the first term a=x. Hence, $$T_n=x \times 0.75^{n-1}$$ However, the distance that the ball travels is not a simple geometric series. As you can see in the diagram above, apart from the first bounce and final n bounce (where the ball only travels downwards or upwards), the ball travels twice the distance of the bounce height as it travels up and down. Hence, we can find the total distance d travelled by the ball before the nth bounce: \begin{align*} d &= 2 \times [\text{sum of the heights of the 1st, 2nd, 3rd, …, (n-2)th, (n-1)th and nth bounce}] -[\text{height of the 1st bounce}] – [\text{height of the nth bounce}]\\ &= 2 \times (T_1+T_2+T_3+…+T_{n-2}+T_{n-1}+T_n) – T_1 – T_n\\ ∴d &= 2 \times S_n – T_1 – T_n\\ \end{align*} In this question, we are given the total distance travelled by the ball before the 8th bounce is 6 metres. So, substitute d=6 and n=8 into $$d = 2 \times S_n – T_1 – T_n$$: \begin{align*} 6 &= 2 \times S_8 – T_1 – T_8 \text{…………. (equation 1)}\\ \end{align*} To find S8, substitute $$a=x$$, $$n=8$$ and $$r=0.75$$ into $$S_n=\frac{a(1-r^n)}{1-r}$$: \begin{align*} S_8 &=\frac{x(1-0.75^8)}{1-0.75}\\ &= 4[x(1-0.75^8)] \\ \end{align*} To find T8, substitute $$a=x$$ and $$n=8$$ into $$T_n=ar^{n-1}$$: \begin{align*} T_8 &= x \times 0.75^{8-1} \\ &= x \times 0.75^7 \\ \end{align*} Hence, we can substitute $$S_8=4[x(1-0.75^8)]$$, $$T_1= x$$ and $$T_8= x \times 0.75^7$$ into equation 1: $$6 = 2 \times S_8 – T_1 – T_8$$: \begin{align*} 6 &= 2 \times 4[x(1-0.75^8)] – x – x \times 0.75^7\\ 6 &= 8[x(1-0.75^8)] – x – x \times 0.75^7\\ 6 &= x[8(1-0.75^8) – 1 – 0.75^7] \\ x &= \frac{6}{8(1-0.75^8) – 1 – 0.75^7} \\ x &= 0.989 \text{ (rounded to 3 decimal places)} \\ ∴ \text{Height at which player began bouncing ball} &= 0.989 \text{ m}\\ \end{align*}
Solution 11 b):
In part a), we found that the total distance d travelled by the ball is $$d = 2 \times S_n – T_1 – T_n$$. Need a refresher? Click here to scroll up to the explanation in part a). In part a), we were interested in the distance that the ball travels before the 8th bounce. So, we used $$n = 8$$. In part b) though, we are trying to find how far the ball can travel if it was allowed to bounce forever. So here, we let $$n = ∞$$. Substitute $$n = ∞$$ into $$d = 2 \times S_n – T_1 – T_n$$: \begin{align*} d &= 2 \times S_∞ – T_1 – T_∞\\ \end{align*} Since: \begin{align*} T_1 &=a \\ &=0.989 \text{ (from part a)}\\ S_∞ &= \frac{a}{1-r}, \ \text{ where a=0.989, r=0.75 and|r|<1} \\ &= \frac{0.989}{1-0.75}\\ & = 3.956\\ T_∞ &= 0 \text{ (the ball does not lift off the ground on the ‘last’ bounce)}\\ \end{align*} \begin{align*} ⇒ d &= 2 \times S_∞ – T_1 – T_∞\\ &= 2 \times 3.956 – 0.989 – 0\\ &= 6.92 \text{ (rounded to 2 decimal places)}\\ ∴ \text{Total limiting distance} &= 6.92 \text{ m} \\ \end{align*} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000100135803223, "perplexity": 1655.096476294262}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337421.33/warc/CC-MAIN-20221003133425-20221003163425-00168.warc.gz"} |
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## alfers101 2 years ago Choose the best answer from the choices. The temperature of an ideal gas remains constant at 320 K while the absolute pressure changes from 101 kPaa to 825 kPaa and the initial volume is 0.10m^3 A.Find the final volume of the gas. a) 0.012m^3 b)0.120m^3 c)0.021m^3 d)0.210m^3 B. Determine the mass of the gas of the molceular mass is 29kg kgmol. a)1.070kg b)0.107kg c)0.017kg d)0.701kg Delete Cancel Submit
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1. eashmore
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Use your ideal gas law. ${P_i V_i \over n_i R T_i} = {P_f V_f \over n_f R T_f}$ R is a constant, it cancel out. n will remain the same, it goes away. T is said to be constant, it cancels out. You're left with $P_iV_i = P_f V_f$
2. eashmore
• 2 years ago
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The ideal gas law is also used for the second one. Realize that $n = {m \over M}$ where m is the mass (in kg) and M is the molar mass (in kg/kmol). Pick any of the two states (initial or final), plug in values, and solve.
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9905176162719727, "perplexity": 3535.9818982671436}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936462720.36/warc/CC-MAIN-20150226074102-00130-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://www.albert.io/ie/ap-statistics/outlier-effect | Free Version
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# Outlier Effect
APSTAT-JORXD8
A set of sample data follows a roughly symmetric distribution and contains no outliers. If an extreme outlier is added to the data set, the median will not change much if at all. The mean will be affected, however, and the value of the mean will either increase or decrease depending on the value of the outlier.
Which of the following bests explains why this happens?
A
When an extreme outlier is added, the distribution will become skewed causing the median to shift significantly. The mean will also shift, but not as much.
B
The mean is more subject to a Type $I$ Error than the median.
C
The mean is a robust measure, but the median is not.
D
The median is a resistant measure, but the mean is not.
E
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https://deepai.org/publication/on-exact-computation-with-an-infinitely-wide-neural-net | # On Exact Computation with an Infinitely Wide Neural Net
How well does a classic deep net architecture like AlexNet or VGG19 classify on a standard dataset such as CIFAR-10 when its "width" --- namely, number of channels in convolutional layers, and number of nodes in fully-connected internal layers --- is allowed to increase to infinity? Such questions have come to the forefront in the quest to theoretically understand deep learning and its mysteries about optimization and generalization. They also connect deep learning to notions such as Gaussian processes and kernels. A recent paper [Jacot et al., 2018] introduced the Neural Tangent Kernel (NTK) which captures the behavior of fully-connected deep nets in the infinite width limit trained by gradient descent; this object was implicit in some other recent papers. A subsequent paper [Lee et al., 2019] gave heuristic Monte Carlo methods to estimate the NTK and its extension, Convolutional Neural Tangent Kernel (CNTK) and used this to try to understand the limiting behavior on datasets like CIFAR-10. The current paper gives the first efficient exact algorithm (based upon dynamic programming) for computing CNTK as well as an efficient GPU implementation of this algorithm. This results in a significant new benchmark for performance of a pure kernel-based method on CIFAR-10, being 10 than the methods reported in [Novak et al., 2019], and only 5 performance of the corresponding finite deep net architecture (once batch normalization etc. are turned off). We give the first non-asymptotic proof showing that a fully-trained sufficiently wide net is indeed equivalent to the kernel regression predictor using NTK. Our experiments also demonstrate that earlier Monte Carlo approximation can degrade the performance significantly, thus highlighting the power of our exact kernel computation, which we have applied even to the full CIFAR-10 dataset and 20-layer nets.
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• ### Harnessing the Power of Infinitely Wide Deep Nets on Small-data Tasks
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• ### On the Power of Shallow Learning
A deluge of recent work has explored equivalences between wide neural ne...
06/06/2021 ∙ by James B. Simon, et al. ∙ 0
• ### Why Are Convolutional Nets More Sample-Efficient than Fully-Connected Nets?
Convolutional neural networks often dominate fully-connected counterpart...
10/16/2020 ∙ by Zhiyuan Li, et al. ∙ 25
• ### Learning with Neural Tangent Kernels in Near Input Sparsity Time
The Neural Tangent Kernel (NTK) characterizes the behavior of infinitely...
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• ### Infinite attention: NNGP and NTK for deep attention networks
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• ### Finite Versus Infinite Neural Networks: an Empirical Study
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07/31/2020 ∙ by Jaehoon Lee, et al. ∙ 49
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## 1 Introduction
How well does a classic deep net architecture like AlexNet or VGG19 perform on a standard dataset such as CIFAR-10 when its “width”— namely, number of channels in convolutional layers, and number of nodes in fully-connected internal layers — is allowed to increase to infinity? Questions about these “infinite limits” of deep nets have naturally emerged in the ongoing effort to understand the power of deep learning. In mathematics it is often easier to study objects in the infinite limit. Furthermore, the infinite limit could conceivably make sense in deep learning, since over-parametrization seems to help optimization a lot and doesn’t hurt generalization much (Zhang et al., 2017): deep neural nets with millions of parameters work well even for datasets with k training data points. So why not imagine nets whose width goes to infinity?
Allowing width to go to infinity also connects deep learning in an interesting way with other areas of machine learning. A single hidden-layer neural network with i.i.d. random parameters, in the limit of infinite width, is a function drawn from a
Gaussian Process (GP) (Neal, 1996). This model as well as analogous ones with multiple layers (Lee et al., 2018; Matthews et al., 2018) and convolutional filters (Novak et al., 2019; Garriga-Alonso et al., 2019) make up the Gaussian Process view of deep learning. These correspond to infinitely wide deep nets whose all parameters are chosen randomly (with careful scaling), and only the top (classification) layer is trained.
From now on we will use weakly-trained nets to refer to nets whose layers receive random initialization and only the top layer is trained by gradient descent. We use fully-trained to refer to nets whose all parameters are trained. It has long been known that weakly-trained convolutional nets have reasonable performance on MNIST and CIFAR-10. Weakly-trained nets that are fully-connected instead of convolutional, can also be thought of as “multi-layer random kitchen sinks,” which also have a long history.
Weakly-trained nets — whether of finite or infinite width — also define interesting kernels. Specifically, if denotes the output of the network on input where denotes the parameters in the network, and is an initialization distribution over (usually Gaussian), then training just the top layer with an loss is equivalent to a kernel regression for the following kernel:
ker(x,x′)=Eθ∼W[f(θ,x)⋅f(θ,x′)], (1)
where are two inputs. This kernel method makes sense even if the number of parameters goes to infinity.
The objects of interest in this paper are not weakly-trained nets, but fully-trained nets. In the finite case, analysis of optimization and generalization of fully-trained nets is of course an open problem. One may also ask:
Can we understand the power of fully-trained nets whose width goes to infinity?
A priori this question doesn’t seem any easier than the finite case, and empirical evaluation seems computationally infeasible due to the infinite limit. They also do not correspond to a kernel method in any obvious way.
Recent papers suggest that nets whose width greatly exceeds the number of training data points can rapidly reduce training error to via gradient descent, and under some conditions, the trained net also exhibits good generalization (Du et al., 2019, 2018b; Li and Liang, 2018; Allen-Zhu et al., 2018a, b; Zou et al., 2018; Arora et al., 2019). Extra-wideness plays a crucial role in the proof: it is shown that as width increases, training causes increasingly smaller changes (in a proportionate sense) in the parameters. This raises the possibility that as one increases the width to infinity, a certain limiting behavior can emerge even in the fully-trained net. A recent paper by Jacot et al. (2018) isolated a notion implicit in the above papers (Du et al., 2019, 2018b), which they called the Neural Tangent Kernel (NTK). They suggested — via a proof that is slightly heuristic — that this fixed kernel characterizes the behavior of fully-connected infinite width neural networks whose layers have been trained by gradient descent. The NTK is very different from the Gaussian Process kernels discussed earlier, and is defined using the gradient of the output of the randomly initialized net with respect to its parameters, i.e.,
ker(x,x′)=Eθ∼W⟨∂f(θ,x)∂θ,∂f(θ,x′)∂θ⟩. (2)
Here, the gradient appears from considering gradient descent, as will be explained in Section 3. One may also generalize the NTK to convolutional neural nets, and we call the corresponding kernel Convolutional Neural Tangent Kernel (CNTK).
Though NTK and CNTK are defined by an infinite limit, a recent paper (Lee et al., 2019) attempted to understand their properties via a heuristic finite approximation of the infinite limit kernel by Monte Carlo methods. However, as will be shown in Section 5.2, such Monte Carlo methods can degrade the performance a lot. It was still open what is the full power of exact NTK and CNTK on modern datasets. This is a challenging question especially for CNTK, since when convolutional operation is involved, it is believed that exact computation of kernels (for either convolutional Gaussian Process kernel or CNTK) is infeasible for large datasets like CIFAR-10 (Novak et al., 2019).
#### Our Contributions.
We give an exact and efficient dynamic programming algorithm to compute the NTK and CNTK for ReLU activation (namely, to compute
given and ). Using this algorithm — as well as implementation tricks for GPUs — we can settle the question of the performance of fully-trained infinitely wide nets with a variety of architectures. For instance, we find that their performance on CIFAR-10 is within of the performance of the same architectures in the finite case (note that the proper comparison in the finite case involves turning off batch norm, data augmentation, dropout, etc., in the optimization). In particular, the CNTK corresponding to a -layer convolutional net with global averge pooling achieves classification accuracy. This is higher than the best reported performance of a Gaussian process with fixed kernel on CIFAR-10 (Novak et al., 2019). 111 We only consider fixed kernels defined without using the training data. We do not compare to methods that tune the kernels using training data (Van der Wilk et al., 2017) or use a neural network to extract features and then applying a kernel method on top of them (Mairal et al., 2014).
We give a more rigorous, non-asymptotic proof that the NTK captures the behavior of a fully-trained wide neural net under weaker condition than previous proofs. We also experimentally show that the Monte Carlo methods for approximating CNTK in the earlier work do not compute good approximations, which is clear from their much worse performance on CIFAR.
### 1.1 Notation
We use bold-faced letters for vectors, matrices and tensors. For a vector
, let be its -th entry; for a matrix , let be its -th entry; for a 4th-order tensor , let be its -th entry. Let
be the identity matrix, and
. Let be an indicator vector with -th entry being and other entries being , and let denote the all-one vector. We use to denote the pointwise product and to denote the tensor product. We use to transform a vector to a diagonal matrix. We use
to denote the activation function, such as the rectified linear unit (ReLU) function:
, and to denote the derivative of . Denote by
the Gaussian distribution with mean
and covariance .
### 1.2 Paper Organization
The rest of the paper is organized as follows. Section 2 gives an overview of related work. In Section 3, we review how NTK arises from training an infinitely wide fully-connected neural network, and also rigorously establish the equivalence between a fully-trained wide neural net and kernel regression under the NTK. In Section 4, we derive the formulas for CNTK, and describe our efficient computation of CNTK; In Section 5, we present our experimental results and discuss our findings. Lastly, we conclude in Section 6.
## 2 Related Work
From a Gaussian Process (GP) viewpoint, the correspondence between infinite neural networks and kernel machines was first noted by Neal (1996). Follow-up work extended this correspondence to more general shallow neural networks (Williams, 1997; Roux and Bengio, 2007; Hazan and Jaakkola, 2015)
. More recently, this was extended to deep and convolutional neural networks
(Lee et al., 2018; Matthews et al., 2018; Novak et al., 2019; Garriga-Alonso et al., 2019). However, these kernels, as we discussed in Section 1, represent weakly-trained nets, instead of fully-trained nets.
Beyond GPs, the connection between neural networks and kernels is also studied in the compositional kernel literature. Cho and Saul (2009) derived a closed-form kernel formula for rectified polynomial activations, which include ReLU as a special case. Daniely et al. (2016) proposed a general framework to transform a neural network to a compositional kernel and later Daniely (2017)
showed for sufficiently wide neural networks, stochastic gradient descent can learn functions that lie in the corresponding reproducing kernel Hilbert space. However, the kernels studied in these works still correspond to weakly-trained neural networks.
Wilson et al. (2016); Al-Shedivat et al. (2016) used a neural network as a feature extractor, applied a GP on top of these features and then trained the resulting model end-to-end. Another line of work built probabilistic graphical models (PGMs) which uses kernel as a component (Damianou and Lawrence, 2013; Lawrence and Moore, 2007; Van der Wilk et al., 2017; Kumar et al., 2018; Blomqvist et al., 2018). These are not pure kernel methods so we do not compare with them in this paper. Nevertheless, CNTK may be combined with these techniques to generate more powerful predictors.
This paper is inspired by a line of recent work on over-parameterized neural networks (Du et al., 2019, 2018b; Du and Hu, 2019; Li and Liang, 2018; Allen-Zhu et al., 2018b, a; Zou et al., 2018). These papers established that for (convolutional) neural networks with large but finite width, (stochastic) gradient descent can achieve zero training error. A key component in these papers is showing that the weight matrix at each layer is close to its initialization. This observation implies that the kernel defined in Equation (2) is still close to its initialization. Arora et al. (2019) explicitly used this observation to derive generalization bounds for two-layer over-parameterized neural networks.
Jacot et al. (2018)
derived the exact same kernel from kernel gradient descent. They showed that if the number of neurons per layer goes to infinity in a sequential order, then the kernel remains unchanged for a finite training time. They termed the derived kernel
Neural Tangent Kernel (NTK). We follow the same naming convention and name its convolutional extension Convolutional Neural Tangent Kernel (CNTK). Later, Yang (2019) derived a formula of CNTK. Comparing with (Yang, 2019), our CNTK formula has a more explicit convolutional structure and results in an efficient GPU-friendly computation method. Recently, Lee et al. (2019) tried to empirically verify the theory in (Jacot et al., 2018) using a heuristic Monte Carlo method. They showed that in the first few iterations, the kernel is approximately unchanged. However, as will be shown in Section 5.2, such Monte Carlo methods can decrease the classification accuracy by even on a “CIFAR-2” (airplane V.S. car) dataset. Therefore, exact kernel evaluation is important to study the power of NTK and CNTK.
## 3 Neural Tangent Kernel
In this section we describe fully-connected deep neural net architecture and its infinite width limit, and how training it with respect to the loss gives rise to a kernel regression problem involving the neural tangent kernel (NTK).
We denote by the output of a neural network where is all the parameters in the network and is the input.222For simplicity, we only consider a single output here. The generalization to multiple outputs is straightforward. Given a training dataset , consider training the neural network by minimizing the squared loss over training data:
ℓ(θ)=12n∑i=1(f(θ,xi)−yi)2.
The proof of the following lemma uses simple differentiation and appears in Appendix A.
###### Lemma 3.1.
Consider minimizing the squared loss by gradient descent with infinitesimally small learning rate: . Let be the network outputs on all ’s at time , and be the desired outputs. Then follows the following evolution, where is an positive semidefinite matrix whose -th entry is :
du(t)dt=−H(t)⋅(u(t)−y). (3)
The statement of Lemma 3.1 involves a matrix . Below we define a deep net architecture whose width is allowed to go to infinity, while fixing the training data as above. In the limit, it can be shown that the matrix remains constant during training i.e., equal to
. Moreover, under a random initialization of parameters, the random matrix
converges in probability to a certain deterministic kernel matrix
as the width goes to infinity, which is the Neural Tangent Kernel (Equation (2)) evaluated on the training data. If for all , then Equation (3) becomes
du(t)dt=−H∗⋅(u(t)−y). (4)
Note that the above dynamics is identical to the dynamics of kernel regression under gradient flow, for which at time the final prediction function is (assuming )
f∗(x)=(ker(x,x1),…,ker(x,xn))⋅(H∗)−1y. (5)
In Theorem 3.2, we rigorously prove that a fully-trained sufficiently wide ReLU neural network is equivalent to the kernel regression predictor (5) on any given data point.
### 3.1 Fully-Connected Deep Neural Net and Its Infinite Width Limit
Now we define fully-connected deep neural net and review its Gaussian Process viewpoint in the infinite width limit. Fully-connected net is defined in the standard way: the
-th layer is computed by applying a linear transformation on the output of the
-th layer, followed by a coordinate-wise nonlinearity . Recall that standard initializion would pick each entry of the layer matrix from where is such that the expected norm of each row in the matrix is . Thus depends on the number of nodes in the row, and the initialization stays bounded as we let the number of nodes in each row go to infinity.
The limit behavior of gradient descent depends however on the details of initialization. In this paper we initialize parameters using and multiply on the outside (instead of using to initialize). These two parameterizations can be made equivalent if one is allowed to set different learning rates for different layers, as discussed in (Lee et al., 2019). In fact the gradient from each layer defines a different kernel, and setting different learning rates for different layers amounts to summing over those individual kernels with different weights. Our choice allows setting equal learning rates for all the layers while giving rise to the NTK in the infinite width limit, which is the unweighted sum of those individual kernels. In contrast, using Kaiming initialization (He et al., 2015) at infinite width is equivalent to setting the learning rate of the first layer to 0 and the rest to be equal, thus resulting in a similar kernel to the NTK — the unweighted sum of kernels from the second layer to the last layer. The initialization method in (Daniely, 2017) gives the kernel corresponding to the last layer only, which corresponds to Gaussian process.
Now we define a fully-connected neural net formally. Let be the input, and denote and for notational convenience. We define an -hidden-layer fully-connected neural network recursively:
(6)
where is the weight matrix in the -th layer (), is a coordinate-wise activation function, and . The last layer of the neural network is
f(θ,x) =f(L+1)(x)=W(L+1)⋅g(L)(x) =W(L+1)⋅√cσdLσ(W(L)⋅√cσdL−1σ(W(L−1)⋯√cσd1σ(W(1)x))),
where is the weights in the final layer, and represents all the parameters in the network.
We initialize all the weights to be i.i.d. random variables, and consider the limit of large hidden widths: . The scaling factor in Equation (6) ensures that the norm of for each is approximately preserved at initialization (see (Du et al., 2018b)). In particular, for ReLU activation, we have ().
Recall from (Lee et al., 2018) that in the infinite width limit, the pre-activations at every hidden layer has all its coordinates tending to i.i.d. centered Gaussian processes of covariance defined recursively as:
Σ(0)(x,x′) =x⊤x′, (7) Λ(h)(x,x′) =(Σ(h−1)(x,x)Σ(h−1)(x,x′)Σ(h−1)(x′,x)Σ(h−1)(x′,x′))∈R2×2, Σ(h)(x,x′) =cσE(u,v)∼N(0,Λ(h))[σ(u)σ(v)],
for . The intuition is that is a centered Gaussian process conditioned on (), with covariance
E[[f(h+1)(x)]i⋅[f(h+1)(x′)]i∣∣f(h)] =⟨g(h)(x),g(h)(x′)⟩ (8) =cσdhdh∑j=1σ([f(h)(x)]j)σ([f(h)(x′)]j),
which converges to as given that each is a centered Gaussian process with covariance . This yields the inductive definition in Equation (7).
### 3.2 Neural Tangent Kernel
We proceed to derive the NTK for the fully-connected neural net. Recall that we need to compute the value that converges to at random initialization in the infinite width limit. We can write the partial derivative with respect to a particular weight matrix in a compact form:
∂f(θ,x)∂W(h)=b(h)(x)⋅(g(h−1)(x))⊤,h=1,2,…,L+1,
where
bh(x)=⎧⎨⎩1∈R,for% h=L+1,√cσdhD(h)(x)(W(h+1))⊤bh+1(x)∈Rdh,for h=1,…,L, (9)
D(h)(x)=diag(˙σ(f(h)(x)))∈Rdh×dh,h=1,…,L. (10)
Then, for any , we can compute
⟨∂f(θ,x)∂W(h),∂f(θ,x′)∂W(h)⟩ =⟨b(h)(x)⋅(g(h−1)(x))⊤,b(h)(x′)⋅(g(h−1)(x′))⊤⟩ =⟨g(h−1)(x),g(h−1)(x′)⟩⋅⟨b(h)(x),b(h)(x′)⟩.
Note that we have established in Equation (8) that
⟨g(h−1)(x),g(h−1)(x′)⟩→Σ(h−1)(x,x′).
For the other factor , by definition (9) we get
⟨b(h)(x),b(h)(x′)⟩ (11) = cL−hσdh+1⋯dL⟨√cσdhD(h)(x)(W(h+1))⊤bh+1(x),√cσdhD(h)(x′)(W(h+1))⊤bh+1(x′)⟩.
Although and are dependent, the Gaussian initialization of allows us to replace with a fresh new sample without changing its limit. This is made rigorous for ReLU activation in Theorem 3.1.
⟨√cσdhD(h)(x)(W(h+1))⊤bh+1(x),√cσdhD(h)(x′)(W(h+1))⊤bh+1(x′)⟩ ≈ ⟨√cσdhD(h)(x)(˜W(h+1))⊤bh+1(x),√cσdhD(h)(x′)(˜W(h+1))⊤bh+1(x′)⟩ → → ˙Σ(h)(x,x′)⟨b(h+1)(x),b(h+1)(x′)⟩,
where
˙Σ(h)(x,x′)=cσE(u,v)∼N(0,Λ(h))[˙σ(u)˙σ(v)]. (12)
Applying this approximation inductively in Equation (11), we get
⟨b(h)(x),b(h)(x′)⟩→L∏h′=h˙Σ(h′)(x,x′).
Finally, since , we obtain the final NTK expression for the fully-connected neural network:
Θ(L)(x,x′)=L+1∑h=1(Σ(h−1)(x,x′)⋅L+1∏h′=h˙Σ(h′)(x,x′)), (13)
where we let for convenience.
Rigorously, for ReLU activation, we have the following theorem that gives a concrete bound on the hidden widths that is sufficient for convergence to the NTK at initialization:
###### Theorem 3.1 (Convergence to the NTK at initializatoin).
Fix and . Suppose , , and , . Then for any inputs such that , with probability at least we have:
The proof of Theorem 3.1 is given in Appendix B.
###### Remark 3.1.
Theorem 3.1 improves upon previous results (Jacot et al., 2018; Yang, 2019) that establish convergence to the NTK at initialization in the following sense:
1. Previous results are asymptotic, i.e., they require the widths to go to infinity, while Theorem 3.1 gives a non-asymptotic bound on the required layer widths.
2. Jacot et al. (2018) required sequential limit, i.e., go to infinity one by one, and Yang (2019) let go to infinity at the same rate. On the other hand, Theorem 3.1 only requires to be sufficiently large, which is the weakest notion of limit.
#### Equivalence between Wide Neural Net and Kernel Regression using NTK.
Built on Theorem 3.1, we can further incorporate the training process and show the equivalence between a fully-trained sufficiently wide neural net and the kernel regression solution using the NTK, as described in Lemma 3.1 and the discussion after it.
Recall that the training data are , and is the NTK evaluated on these training data, i.e., . Denote . For a testing point , we let be the kernel values of this testing point and training points, i.e., . The prediction of kernel regression using NTK on this testing point is
fntk(xte)=(kerntk(xte,X))⊤(H∗)−1y.
Since the above solution corresponds to the linear dynamics in Equation (4) with zero initialization, in order to establish equivalence between neural network and kernel regression, we would like the initial output of the neural network to be small. Therefore, we apply a small multiplier , and let the final output of the neural network be
fnn(θ,x)=κf(θ,x).
We let be the prediction of the neural network at the end of training.
The following theorem establishes the equivalence between the fully-trained wide neural network and the kernel regression predictor using the NTK.
###### Theorem 3.2 (Main theorem).
Suppose , and with . Then for any with , with probability at least over the random initialization, we have
|fnn(xte)−fntk(xte)|≤ϵ.
Several comments are in sequel. Theorem 3.2 is, to our knowledge, the first result that rigorously shows the equivalence between a fully-trained neural net and a kernel predictor. Comparing with (Jacot et al., 2018), our bound is non-asymptotic whereas (Jacot et al., 2018) only has an asymptotic result; furthermore, Jacot et al. (2018) required the width of every layer to go to infinity in a sequential order, while we can have the same number of neurons per layer, which is closer to practice. Comparing with recent results on over-parameterized neural nets (Arora et al., 2019; Allen-Zhu et al., 2018b, a; Du et al., 2019, 2018b; Li and Liang, 2018; Zou et al., 2018), our theorem is a more precise characterization of the learned neural network. That is, the prediction is essentially a kernel predictor. Therefore, to study the properties of these over-parameterized nets, such as their generalization power, it is sufficient to study the corresponding NTK.
While this theorem only gives guarantee for a single point, using a union bound, we can show that this guarantee holds for (exponentially many) finite testing points. Combing this with the standard analysis of hold-out validation set, we can conclude that a fully-trained wide neural net enjoys the same generalization ability as its corresponding NTK.
For the proof of Theorem 3.2, we first use a generic argument to show that the perturbation on the prediction can be reduced to the perturbation on kernel value at the initialization and during training. Theorem 3.1 guarantees a small perturbation on kernel value at initialization. For the perturbation during training, we use high level proof idea from Du et al. (2018b); Arora et al. (2019) to reduce the perturbation on the kernel value to the perturbation on the gradient of each prediction with respect to weight matrices. Then we adopt technical lemmas from Allen-Zhu et al. (2018b) to obtain bounds on the perturbation of the gradient. See Appendix D for details. We remark that Jacot et al. (2018); Lee et al. (2019) provided proofs for the training part. However, both are asymptotic results and only apply to finite training time. In contrast, we give a finite-width perturbation bound and our result applies to infinite training time.
## 4 Convolutional Neural Tangent Kernel
In this section we study convolutional neural networks (CNNs) and their corresponding CNTKs. We study two architectures, vanilla CNN and CNN with global average pooling (GAP). To formally define CNNs, we first introduce some notations. Throughout the paper, we let be the width and be the height of the image. We use to denote the filter size. In practice, or
. We use standard zero padding and set stride size to be
to make sure the input of each layer has the same size. For a convolutional filter and an image , the convolution operator is defined as
[w∗x]ij=q−12∑a=−q−12q−12∑b=−q−12[w]a+q+12,b+q+12[x]a+i,b+j for i∈[P],j∈[Q]. (14)
Equation (14) shows patch depends on . Our CNTK formula also relies on this dependency. For , define
Dij,i′j′={(i+a,j+b,i′+a′,j′+b′)∈[P]×[Q]×[P]×[Q]∣−(q−1)/2≤a,b,a′,b′≤(q−1)/2}.
Lastly, for a tensor , we denote a sub-tensor and we let .
### 4.1 Vanilla CNN
The first type of CNN is the vanilla CNN which consists convolution layers and one fully-connected layer, formally defined as follows.
#### CNN Definition
• Let be the input image where is the number of channels in the input image.
• For , , the intermediate outputs are defined as
~x(h)(β)=C(h−1)∑α=1W(h)(α),(β)∗x(h−1)(α),x(h)(β)=√cσC(h)×q×qσ(~x(h)(β))
where each is a filter with Gaussian initialization.
• The final output is defined as
f(θ,x)=C(L)∑α=1⟨W(L+1)(α),x(L)(α)⟩
where is a weight matrix with Gaussian initialization.
For this architecture, using the same reasoning as in Section 3.2, we obtain the following convolutional neural tangent kernel formula. The details are provided in Appendix E. We will use a dynamic programming approach to compute the kernel.
#### CNTK Formula.
We let be two input images.
• For , , define
K(0)(α)(x,x′)=x(α)⊗x′(α) and [Σ(0)(x,x′)]ij,i′j′=C(0)∑α=1tr([K(0)(α)(x,x′)]Dij,i′j′).
• For ,
• For , define
Λ(h)ij,i′j′(x,x′)=⎛⎜ ⎜⎝[Σ(h−1)(x,x)]ij,ij[Σ(h−1)(x,x′)]ij,i′j′[Σ(h−1)(x′,x)]i′j′,ij[Σ(h−1)(x′,x′)]i′j′,i′j′⎞⎟ ⎟⎠∈R2×2.
• Define , for
[K(h)(x,x′)]ij,i′j′= E(u,v)∼N(0,Λ(h)ij,i′j′(x,x′))[σ(u)σ(v)], (15) [˙K(h)(x,x′)]ij,i′j′= E(u,v)∼N(0,Λ(h)ij,i′j′(x,x′))[˙σ(u)˙σ(v)]. (16)
• Define , for
[Σ(h)(x,x′)]ij,i′j′= cσq2tr([K(h)(x,x′)]Dij,i′j′).
Note that and share similar structures as their NTK counterparts. The only difference is that we have one more step, taking the trace over patches. This step represents the convolution operation in the corresponding CNN. Next, we can use a recursion to compute the final kernel value.
1. First, we define .
2. For and , we define
[Θ(h)(x,x′)]ij,i′j′=cσq2tr([˙K(h)(x,x′)⊙Θ(h−1)(x,x′)+K(h)(x,x′)]Dij,i′j′).
3. Lastly, the final kernel value is defined as
tr(Θ(L)(x,x′)).
### 4.2 CNN with Global Average Pooling
We also consider another architecture, CNN with global average pooling.
#### CNN Definition
• Let be the input image and is the number of initial channels.
• For , , the intermediate outputs are defined as
~x(h)(β)=C(h−1)∑α=1W(h)(α),(β)∗x(h−1)(α),x(h)(β)=√cσC(h)×q(h)×q(h)σ(~x(h)(β)).
• The final output is defined as
f(θ,x)=C(L)∑α=1W(L+1)(α)⎛⎝1PQ∑(i,j)∈[P]×[Q][x(L)(α)]ij⎞⎠.
where is a scalar with Gaussian initialization.
Besides using global average pooling, another modification is that we do not train the first and the layer. This is inspired by Du et al. (2018a) in which authors showed that if one applies gradient flow, then at any training time , the difference between the squared Frobenius norm of the weight matrix at time and that at initialization is same for all layers. However, note that and are special because they are smaller matrices compared with other intermediate weight matrices, so relatively, these two weight matrices change more than the intermediate matrices during the training process, and this may dramatically change the kernel. Therefore, we choose to fix and to the make over-parameterization theory closer to practice.
#### CNTK Formula.
We let be two input images. Note because CNN with global average pooling and vanilla CNN shares the same architecture except the last layer, , and are the same for these two architectures. the only difference is in calculating the final kernel value. To compute the final kernel value, we use the following procedure.
1. First, we define . Note this is different from CNTK for vanilla CNN which uses as the initial value because we do not train the first layer.
2. For and , we define
[Θ(h)(x,x′)]ij,i′j′=cσq2tr([˙K(h)(x,x′)⊙Θ(h−1)(x,x′)+K(h)(x,x′)]Dij,i′j′).
3. For , we define
[Θ(L)(x,x′)]ij,i′j′=cσq2tr([˙K(L)(x,x′)⊙Θ(L−1)(x,x′)]Dij,i′j′).
4. Lastly, the final kernel value is defined as
1P2Q2∑(i,j,i′,j′)∈[P]×[Q]×[P]×[Q][Θ(L)(x,x′)]ij,i′j′.
Note that we ignore
comparing with the CNTK of CNN. This is because we do not train the last layer. The other difference is we calculate the mean over all entries, instead of calculating the summation over the diagonal ones. This is because we use global average pooling so the cross-variances between every two patches will contribute to the kernel.
### 4.3 Fast Computation for ReLU-Activated CNTK
To compute the CNTK corresponding to a CNN with convolution layers and one fully-connected layer on samples, the time complexity is . Previous work assumed that directly computing convolutional kernel exactly is computationally infeasible, and thus resorted to approximations like Monte Carlo sampling (Novak et al., 2019). In this section, we present a new approach to efficiently compute CNTK for ReLU activation exactly. Most computation required by our new approach can be described as entry-wise operations over matrices and tensors, which allows efficient implementations on GPUs.
Following the formulas in Sections 4.1 and 4.2, the trickiest part is computing the expectation of the post-activation output, i.e., Equation (15) and (16). These two expectations depend on (the same) matrices . To obtain faster implementations, our key observation is that if the diagonal entries of are all ones and the activation function is ReLU, there are closed-form formulas for the the corresponding expectations. To see this, let us suppose for now that | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9528235793113708, "perplexity": 974.7598105588906}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154356.39/warc/CC-MAIN-20210802172339-20210802202339-00096.warc.gz"} |
https://zbmath.org/?q=an:0751.58041 | ×
## Inverse image for the functor $$\mu\text{hom}$$.(English)Zbl 0751.58041
The author studies the microlocal inverse image of sheaves on $$C^ \infty$$-manifolds. Let $$f: X\to Y$$ be a morphism from the real $$C^ \infty$$-manifold $$X$$ to $$Y$$ and let $$F$$, $$K$$ be sheaves on $$X$$. Then the microlocalization of the functors $$f^{-1}$$ and $$f^ !$$ are defined and denoted by $$f^{-1}_{\mu,p}$$ and $$f^ !_{\mu,p}$$, respectively. The author obtained the theorem which asserts that the natural morphism: $\mu\text{hom}(f^{-1}_{\mu,p}K,f^ !_{\mu,p}F\otimes\omega_ Y^{\otimes-1})_{p_ Y}\to \mu\text{hom}(K,F\otimes\omega_ Y^{\otimes-1})_{p_ X}$ is an isomorphism under some natural conditions. Here, $$\omega_ X$$ denotes the dualizing complex on $$X$$ and $$\mu\text{hom}$$ the microlocalization bifunctor introduced by Kashiwara and Schapira. This theorem can be interpreted as a statement of the microlocal well-posedness for the Cauchy problem.
Reviewer: M.Muro (Yanagido)
### MSC:
58J99 Partial differential equations on manifolds; differential operators 18C99 Categories and theories
### Keywords:
microlocal analysis; microhyperbolicity
Full Text:
### References:
[1] Bony, J. M. and Schapira, P., Solutions hyperfonctions du probleme de Cauchy, Lecture Notes in Math., Springer, 287 (1973), 82-98. · Zbl 0269.35074 [2] D’Agnolo, A. and Schapira, P., An inverse image theorem for sheaves with applications to the Cauchy problem, Duke J. of Math, to appear (1991); Un theoreme d’image inverse pour les faisceaux. Applications au probleme de Cauchy, C.R. Acad. Sci. Paris, 311 (1990), 23-26. · Zbl 0739.58006 [3] Grothendieck, A., SGA 7, expose I. [4] Hamada, T., Leray, J. and Wagschal, C., Systemes d’equations aux derivees partielles a caracteristiques multiples: probleme de Cauchy ramifie, hyperbolicite partielle, J. Math. Pures et Appl., 55 (1976), 297-352. · Zbl 0307.35056 [5] Kashiwara, M., Algebraic study of systems of partial differential equations, Thesis (in Japanese), Tokyo, 1971. [6] Kashiwara, M. and Schapira, P., Probleme de Cauchy pour les systemes microdifferen- tiels dans le domain complexe, Invent. Math., 46 (1978), 17-38. · Zbl 0369.35061 [7] , Microhyperbolic systems, Acta Math., 142 (1979), 1-55. [8] , Microlocal study of sheaves, Asterisque, 128 (1985). [9] , Sheaves on manifolds, Grundlehren der math., Springer-Verlag, 292 (1990). [10] Schiltz, D., Decomposition de la solution d’un probleme de Cauchy homogene pres de la frontiere d’un domaine d’holomorphie des donnees de Cauchy, C. R. Acad. Sci. Paris, Serie I, 306 (1988), 177-180. · Zbl 0646.32009
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9446665644645691, "perplexity": 1734.9162414684824}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00442.warc.gz"} |
https://plainmath.net/pre-algebra/81772-tricky-inequality-no-avail-to-am-gm-le | doturitip9
2022-07-07
Tricky inequality no avail to AM-GM
Let $a,b,c$ be 3 distinct positive real numbers such that abc = 1. Prove that
I tried AM-GM in many different ways, but it doesn't work since one of the terms on the LHS inevitably becomes negative. Any help is greatly appreciated.
Jordan Mcpherson
Expert
By AM-GM
$\sum _{cyc}\frac{{a}^{3}}{\left(a-b\right)\left(a-c\right)}=-\sum _{cyc}\frac{{a}^{3}}{\left(a-b\right)\left(c-a\right)}=-\sum _{cyc}\frac{{a}^{3}\left(b-c\right)}{\prod _{cyc}\left(a-b\right)}=a+b+c\ge 3$
Do you have a similar question? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 28, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8445084095001221, "perplexity": 624.1594803347477}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499899.9/warc/CC-MAIN-20230201013650-20230201043650-00401.warc.gz"} |
https://samacheerkalvi.guide/tamil-nadu-12th-physics-model-question-paper-3-english-medium/ | Students can Download Tamil Nadu 12th Physics Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.
## TN State Board 12th Physics Model Question Paper 3 English Medium
General Instructions:
• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
These are to be answered by choosing the most suitable answer from the given four
alternatives and writing the option code and the corresponding answer
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
in detaiL Draw diagrams wherever necessary.
Time: 3 Hours
Max Marks: 70
Part – I
Answer all the questions. Choose the correct answer. [15 x 1 = 15]
Question 1.
A thin conducting spherical shell of radius R has a charge Q which is uniformly distributed on its surface. the correct plot for electrostatic potential due to this spherical shell is ………
(b)
Question 2.
The temperature coefficient of resistance of a wire is 0.00125 per °C. At 300 K, its resistance is 1Ω. The resistance of the wire will be 2Ω at …………….
(a) 1154 K
(b) 1100 K
(c) 1400K
(d) 1127 K
Question 3.
n resistances, each of r Ω, when connected in parallel give an equivalent resistance of R Ω. If these resistances were connected in series, the combination would have a resistance in horns equal to ………..
(a) n2R
(b) $$\frac{R}{n^{2}}$$
(c) $$\frac{R}{n}$$
(d) nR
(a) n2R
Hint :
Resistance in parallel combination, R= r/n = r = Rn
Resistance in series combination , R = nr = n2R
Question 4.
A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current 1 = 8 m A (milli ampere). The radii of inside and outside turns are a = 50 mm and b = 100 mm respectively. The magnetic induction at the center of the spiral is……………………….
(a) 5 μT
(b) 7 μT
(c) 8 μT
(d) 10 μT
(b) 7 μT
Question 5.
In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor, when the energy is stored equally between the electric and magnetic fields, is……………………….
(a) $$\frac{Q}{2}$$
(b) $$\frac{Q}{\sqrt{3}}$$
(c) $$\frac{Q}{\sqrt{2}}$$
(d) Q
(c) $$\frac{Q}{\sqrt{2}}$$
Question 6.
The direction of induced current during electro magnetic induction is given by …………………………………..
(b) Lenz’s law
(c) Maxwell’s law
(d) Ampere’s law
(b) Lenz’s law
Question 7.
In an electromagnetic wave in free space the rms value of the electric field is 3 V m . The peak value of the magnetic field is………………………..
(a) 1.414 x 10-8 T
(b) 1.0 x 10-8 T
(c) 2.828 x 10-8 T
(d) 2.0 x 10-8 T
(a) 1.414 x 10-8 T
Question 8.
Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is, [take wavelength of light, λ = 500 nm]
(a) 1m
(b) 5m
(c) 3m
(d) 6m
(b) 5m
Hint:
Resolution limit sin θ =
Question 9.
Two mirrors are kept at 60° to each other and a body is placed at middle. The total number of images formed is ……………..
(a) six
(b) four
(c) five
(d) three
(a) six
Hint:
Number of images formed $$n=\frac{360}{\theta}-1=\frac{360}{60}-1=5$$
Question 10.
The threshold wavelength for a metal surface whose photoelectric work function is 3.313 eV is
{a) 4125 Å
(b) 3750 Å
(c) 6000 Å
(d) 2062.5 Å
(b) 3750 Å
Question 11.
Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is…..
(c)
Question 12.
The speed of the particle, that can take discrete values is proportional to ……………
(a) n-3/2
(b) n-3
(c) n1/2
(d) n
(d) n
Hint:
$$P=m v=\frac{n h}{2 a} ; V \propto n$$
Question 13.
Doping semiconductor results in ………………………………
(a) The decrease in mobile charge carriers
(b) The change in chemical properties
(c) The change in the crystal structure
(d) The breaking of the covalent bond
(c) The change in the crystal structure
Question 14.
The signal is affected by noise in a communication system …………………….
(a) At the transmitter
(b) At the modulator
(c) In the channel
(c) In the channel
Question 15.
The technology used for stopping the brain from processing pain is…………………………..
(a) Precision medicine
(b) Wireless brain sensor
(c) Virtual reality
(c) Virtual reality
Part – II
Answer any six questions. Question No. 21 is compulsory. [6 x 2 = 12]
Question 16.
Define ‘Electric field’.
The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by
$$\overrightarrow{\mathrm{E}}=\frac{\overrightarrow{\mathrm{F}}}{q_{0}}$$
The electric field is a vector quantity and its Sl unit is Newton per Coulomb (NC-1).
Question 17.
State the principle of potentiometer.
The basic principle of a potentiometer is that when a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length.
Question 18.
Compute the intensity of magnetisation of the bar magnet whose mass, magnetic moment and density are 200 g, 2 A m2 and 8 g cm-3, respectively.
Question 19.
What is displacement current?
The displacement current can be defined as the current which comes into play in the region in ‘ which the electric field and the electric flux are changing with time.
Question 20.
What is principle of reversibility?
The principle of reversibility states that light will follow exactly the same path if its direction of travel is reversed.
Question 21.
Calculate the de Broglie wavelength of a proton whose kinetic energy is equal to. 81.9 x 1015 (Given: mass of proton is 1836 times that of electron).
Question 22.
Give the symbolic representation of alpha decay, beta decay and gamma decay.
Alpha decay: The alpha decay process symbolically in the following way
Beta decay: β decay is represented by $$_{ z }^{ A }X\rightarrow _{ z,-1 }^{ A }Y+e^{ + }+v$$
Gamma decay: The gamma decay is given $$\mathrm{z} \mathrm{X}^{*} \rightarrow_{Z}^{\mathrm{A}} \mathrm{X}$$+ gamma(y)rays
Question 23.
Explain the need for a feedback circuit in a transistor oscillator.
The circuit used to feedback a portion of the output to the input is called the feedback network. If the portion of the output fed to the input is in phase with the input, then the magnitude of the input signal increases. It is necessary for sustained oscillations.
Question 24.
Give the factors that are responsible for transmission impairments.
• Attenuation
• Distortion (Harmonic)
• Noise
Part – III
Answer any six questions. Question No. 32 is compulsory. [6 x 3 = 18]
Question 25.
A water molecule has an electric dipole moment of 6.3 x 10 Cm. A sample contains 10 water molecules, with all the dipole moments aligned parallel to the external electric field of magnitude 3 x 10s NC-1. How much work is required to rotate all the water molecules from θ = 0° to 90°?
When the water molecules are aligned in the direction of the electric field, it has minimum potential energy. The work done to rotate the dipole from θ = 0° to 90° is equal to the potential energy difference between these two configurations.
W = ΔU = U(90°) – U(0°)
As we know, U = -pE cos θ, Next we calculate the work done to rotate one water molecule
from θ = 0° to 90°,
For one water molecule, W = – pE cos 90° + pE cos 0° = pE
W = 6.3 x10-30 x 3 x 105 = 18.9 x 10-25
For 1022 water molecules, the total work done is Wtot = 18.9x 10 25 x 1022 = 18.9 x 10 3 J
Question 26.
What are the properties of an equipotential surface?
Properties of equipotential surfaces
(i) The work done to move a charge q between any two points A and B,
W = q (VB – VA ). If the points A and B lie on the same equipotential surface, work done is zero because VB – VA
(ii) The electric field is normal to an equipotential surface. If it is not normal, then there is a component of the field parallel to the surface. Then work must be done to move a charge between two points on the same surface. This is a contradiction. Therefore the electric field must always be normal to an equipotential surface.
Question 27.
An electronics hobbyist is building a radio which requires 150 ft in her circuit, but she has only 220 Ω 79 Ω and 92 Ω resistors available. How can she connect the available resistors to get desired value of resistance?
Required effective resistance = 150 ft
Given resistors of resistance, R = 220 ft, R = 79 ft, R = 92 ft
Parallel combination of R1 and R2
Question 28.
What is magnetic permeability?
Magnetic permeability: The magnetic permeability can be defined as the measure of ability of the material to allow the passage of magnetic field lines through it or measure of the capacity of the substance to take magnetisation or the degree of penetration of magnetic field through the substance.
Question 29.
A circular metal of area 0.03 m2 rotates in a uniform magnetic field of 0.4 T. The axis of rotation passes through the centre and perpendicular to its plane and is also parallel to the field. If the disc completes 20 revolutions in one second and the resistance of the disc is 4 Ω calculate the induced emf between the axis and the rim and induced current flowing in the disc.
A = 0.03 m2; B = 0.4T; f = 20rps; R = 4Ω
Area covered in 1 sec = Area of the disc x frequency
= 0.03 x 20 = 0.6 m2
Question 30.
Give the characteristics of image formed by a plane mirror.
• The image formed by a plane mirror is virtual, erect, and laterally inverted.
• The size of the image is equal to the size of the object.
• The image distance far behind the mirror is equal to the object distance in front of it.
• If an object is placed between two plane mirrors inclined at an angle 0, then the number of images n formed is as,$$n=\left(\frac{360}{\theta}-1\right)$$
Question 31.
An electron and an alpha particle have same kinetic energy. How are the de Broglie wavelengths associated with them related?
Question 32.
Assuming VCEsat = 0.2 V and β = 50, find the minimum base current (IB) required to drive the transistor given in the figure to saturation.
VCEsat = 0.2 V and β = 50
VCE = VCC -IC RC
0.2= 3 – Ic(1 k)
Question 33.
• The robots are much cheaper than humans.
• Robots never get tired like humans. It can work for 24 x 7. Hence absenteeism in work place can be reduced.
• Robots are more precise and error free in performing the task.
• Robots have no sense of emotions or conscience.
• They lack empathy and hence create an emotionless workplace.
• If ultimately robots would do all the work, and the humans will just sit and monitor them* health hazards will increase rapidly.
Part – IV
Answer all the questions. [5 x 5 = 25]
Question 34.
(a) Derive an expression for electrostatic potential energy of the dipole in a uniform electric field.
Expression for electrostatic potential energy of the dipole in a uniform electric field:
Considera dipole placed in the uniform electric field $$\overrightarrow{\mathrm{E}}$$ placed in the uniform electric field $$\overrightarrow{\mathrm{E}}$$ . A dipole experiences a torque when kept in an uniform electric field E.
This torque rotates the dipole to align it with the direction of the electric field. To rotate the dipole (at constant angular velocity) from its initial angle θ’ to another angle θ against the torque exerted by the electric field, an equal and opposite external torque must be applied on the dipole. The work done by the external torque to rotate the dipole from angle θ’ to θ at constant angular velocity is
This work done is equal to the potential energy difference between the angular positions 0 and 0′. U(θ) – (Uθ’) = ΔU = -pE cos θ + pE cos θ’.
If the initial angle is = θ’ = 90° and is taken as reference point, then U(θ’) + pE cos 90° = 0. The potential energy stored in the system of dipole kept in the uniform electric field is given by.
U = -pE cos θ = -p-E ……………. (3)
In addition to p and E, the potential energy also depends on the orientation θ of the electric dipole with respect to the external electric field.
The potential energy is maximum when the dipole is aligned anti-parallel (θ = Jt) to the external electric field and minimum when the dipole is aligned parallel (θ = 0) to the external electric field.
[OR]
(b) Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current.
Magnetic field due to long straight conductor carrying current: Consider a long straight wire NM with current I flowing from N to M. Let P be the point at a distance a from point O.
Consider an element of length dl of the wire at a distance l from point O and $$\vec{r}$$ be the vector joining the element dl with the point P. Let θ be the angle between $$d \vec{l} \text { and } \vec{r}$$. Then, the magnetic field at P due to the element is
$$d \overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{Id} l}{4 \pi r^{2}} \sin \theta$$(unit vector perpendicular to $$d \vec{l} \text { and } \vec{r}$$
The direction of the field is perpendicular to the plane of the paper and going into it. This can be determined by taking the cross product between two vectors $$d \vec{l} \text { and } \vec{r}$$
(let it be $$\hat{n}$$). The net magnetic field can be determined by integrating equation with proper limits. $$\overrightarrow{\mathrm{B}}=\int d \overrightarrow{\mathrm{B}}$$ From the figure, in a right angle triangle PAO,
This is the magnetic field at a point P due to the current in small elemental length. Note that we have expressed the magnetic field OP in terms of angular coordinate i.e. θ. Therefore, the net magnetic field at the point P which can be obtained by integrating $$d \overrightarrow{\mathrm{B}}$$ by varying the angle from θ = φ1 to θ =φ2 is
For a an infinitely long straight wire, 1 =0 and 2 =, the magnetic field is $$\overrightarrow{\mathrm{B}}=\frac{\mu_{0} I}{2 \pi a} \hat{n}$$ ………… (3)
Note that here $$\hat{n}$$ represents the unit vector from the point O to P.
Question 35.
(a) Obtain an expression for motional emf from Lorentz force.
Motional emf from Lorentz force: Consider a straight conducting rod AB of length l in a uniform magnetic field $$\vec{B}$$ which is directed perpendicularly into the plane of the paper. The length of the rod is normal to the magnetic field. Let the rod move with a constant velocity $$\vec{v}$$ towards right side. When the rod moves, the free electrons present in it also move with same velocity $$\vec{v}$$ in B. As a result, the Lorentz force acts on free electrons in the direction from B to A and is given by the relation $$\overrightarrow{\mathrm{F}}_{\mathrm{B}}$$=-
$$e(\vec{v} \times \overrightarrow{\mathrm{B}})$$ …………. (1)
The action of this Lorentz force is to accumulate the free electrons at the end A. This accumulation of free electrons produces a potential difference across the rod which in turn establishes an electric field E directed along BA. Due to the electric field $$\overrightarrow{\mathrm{E}}$$, the coulomb force starts acting on the free electrons along AB and is given by
$$\overrightarrow{\mathrm{F}}_{\mathrm{E}}=-e \overrightarrow{\mathrm{E}}$$ ……….. (2)
The magnitude of the electric field E keeps on increasing as long as accumulation of electrons at the end A continues. The force $$\overrightarrow{\mathrm{F}}_{\mathrm{B}}$$ also increases until equilibrium is reached. At equilibrium, the magnetic Lorentz force $$\overrightarrow{\mathrm{F}}_{\mathrm{B}}$$ and the coulomb force F balance each other and no further accumulation of free electrons at the end A takes place,
i.e
The potential difference between two ends of the rod is
V= El
V= vBl
Thus the Lorentz force on the free electrons is responsible to maintain this potential difference and hence produces an emf
ε = B lv ………….. (4)
As this emf is produced due to the movement of the rod, it is often called as motional emf.
[OR]
(b) Explain the Maxwell’s modification of Ampere’s circuital law.
Maxwell argued that a changing electric field between the capacitor plates must induce a magnetic field. As currents are the usual sources of magnetic fields, a changing electric field must be associated with a current. Maxwell called this current as the displacement current.
If ‘A’ be the area of the capacitor plates and ‘cf be the charge on the plates at any instant ‘f during the charging process, then the electric field in the gap will be
But $$\frac{d q}{d t}$$ is the rate of change of charge on the capacitor plates. It is called displacement current and given by $$\mathrm{I}_{d}=\frac{d q}{d t}=\varepsilon_{0} \frac{d \Phi_{\mathrm{E}}}{d t}$$
This is the missing term in Ampere’s Circuital Law. The total current must be the sum of the
conduction current Ic Hence, the modified form of the ampere’s law.
$$\oint_{c} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d l}=\mu_{0}\left(\mathrm{I}_{c}+\varepsilon_{0} \frac{d \Phi_{\mathrm{E}}}{d t}\right)$$
Question 36.
(a) Obtain the equation for resolving power of microscope.
Resolving power of microscope: The diagram related to the calculation of resolution of microscope. A microscope is used to see the details of the object under observation. The ability of microscope depends not only in magnifying the object objective but also in resolving two points on the object separated by a small distance dmin Smaller the value of dmin better will be the resolving power of the microscope.
The radius of central maxima is, $$r_{0}=\frac{1.22 \lambda v}{a}$$ . In the place of focal length f we have the image distance v. If the difference between the two points on the object to be resolved is dmin, then the magnification m is, $$m=\frac{r_{o}}{d_{\min }}$$
To further reduce the value of dmin the optical path of the light is increased by immersing the objective of the microscope into a bath containing oil of refractive index n.
Such an objective is called oil immersed objective. The term n sin β is called numerical aperture NA.
[OR]
(b) Derive an expression for de Broglie wavelength of electrons.
An electron of mass m is accelerated through a potential difference of V volt. The kinetic energy acquired by the electron is given by
$$\frac{1}{2} m v^{2}=e \mathrm{V}$$
Therefore, the speed v of the electron is $$v=\sqrt{\frac{2 e \mathrm{V}}{m}}$$
Hence, the de Broglie wavelength of the electron is $$\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 e m \mathrm{V}}}$$
Substituting the known values in the above equation, we get
For example, if an electron is accelerated through a potential difference of 100V, then its de Broglie wavelength is 1.227 Å. Since the kinetic energy of the electron, K = eV, then the de Broglie wavelength associated with electron can be also written as
$$\lambda=\frac{h}{\sqrt{2 m \mathrm{K}}}$$
Question 37.
(a) Explain the variation of average binding energy with the mass number by graph and discuss its features.
We can find the average binding energy per nucleon $$\overline{\mathrm{BE}}$$.
It is given by
The average binding energy per nucleon is the energy required to separate single nucleon from the particular nucleus. $$\overline{\mathrm{BE}}$$ is plotted against A of all known nuclei.
Important inferences from of the average binding energy curve:
• The value of BE rises as the mass number increases until it reaches a maximum value of 8.8 MeV for A= 56 (iron) and then it slowly decreases.
• The average binding energy per nucleon is about 8.5 MeV for nuclei having mass number between A = 40 and 120. These elements are comparatively more stable and not radioactive.
• For higher mass numbers, the curve reduces slowly and BE for uranium is about 7.6 MeV. They are unstable and radioactive. If two light nuclei with A < 28 combine with a nucleus with A < 56, the binding energy per nucleon is more for final nucleus than the initial nuclei. Thus, if the lighter elements combine to produce a nucleus of medium value A, a large amount of energy will be released. This is the basis of nuclear fusion and is the principle of the hydrogen bomb.
• If a nucleus of heavy element is split (fission) into two or more nuclei of medium value A, the energy released would again be large. The atom bomb is based on this principle and huge energy of atom bombs comes from this fission when it is uncontrolled.
[OR]
(b) State Boolean laws. Elucidate how they are used to simplify Boolean expressions with suitable example.
Laws of Boolean Algebra: The NOT, OR and AND operations are $$\bar{A}$$, A + B, A.B are the Boolean operations. The results of these operations can be summarised as:
Complement law:
The complement law can be realised as $$\overline{\mathrm{A}}$$ = A
Commutative laws
A + B = B +A
A . B = B . A
Associative laws
A + (B + C) = (A + B) + C
A . (B . C) = (A .B) . C
Distributive laws
A( B + C) = AB + AC
A + BC = (A + B) (A + C)
The above laws are used to simplify complicated expressions and to simplify the logic circuitry.
Question 38.
(a) Modulation helps to reduce the antenna size in wireless communication – Explain. Antenna size:
Antenna is used at both transmitter and receiver end. Antenna height is an important parameter to be discussed. The height of the antenna must be a multiple of $$\frac{\lambda}{4}$$
$$h=\frac{\lambda}{4}$$ ……………….. (1)
where X is wavelength $$\left(\lambda=\frac{c}{v}\right), c$$ c is the velocity of light and v is the frequency of the signal v to be transmitted.
An example
Let us consider two baseband signals. One signal is modulated and the other is not modulated. The frequency of the original baseband signal is taken as v = 10 kHz while the modulated signal is v = 1 MHz. The height of the antenna required to transmit the original baseband signal of frequency
v = 10kHz is
The height of the antenna required to transmit the modulated signal of frequency v = 1 MHz is
Comparing equations (2) and (3), we can infer that it is practically feasible to construct an antenna of height 75 m while the one with 7.5 km is not possible. It clearly manifests that modulated signals reduce the antenna height and are required for long distance transmission.
[OR]
(b) What are the possible harmful effects of usage of Nanoparticles? Why? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9026614427566528, "perplexity": 736.0305585132063}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780058552.54/warc/CC-MAIN-20210927211955-20210928001955-00593.warc.gz"} |
http://keithpickering.blogspot.com/2010/02/its-byrd-its-plane-part-1.html | Thursday, February 11, 2010
It's a Byrd! It's a Plane! (Part 1)
The effect of skis on airplane performance
Lisle Rose's recent biography of famed aviator and polar pioneer Richard E. Byrd (Explorer. Columbia: University of Missouri Press, 2008) spends a lot of time in Chapter 4 defending a lost cause.
Byrd was a little-known Navy flier until 1926, when he made his reputation by flying an airplane over the North Pole – or so he claimed. There were a lot of suspicions about the flight, however, mostly based on the fact that Byrd arrived back at his base (Kings Bay, Spitzbergen: today called Ny Alesund) after less than 16 hours in the air. Critics doubted that his plane, a Fokker F7a-3m, could have flown fast enough to reach the Pole and back in such a short time.
The controversy has waxed and waned over the years, but it was decisively settled in 1998, when archivist Raimund Goerler discovered Byrd's flight diary, which contains notes Byrd made during the trip. The diary contains sextant observations and subsequent computations that had been erased. However, the erased figures are just legible, and show much different navigational observations than Byrd wrote in his official report. So it is clear Byrd falsified the navigational data in his official report.
That should end the discussion, since clearly there would be no need for data falsification if Byrd had actually done what he claimed to have done. But somehow, Rose didn't get that memo.
We will be discussing the Byrd case in detail over the next few postings, but let's start out with one little nit to pick. One of the issues Byrd skeptics raise is the fact that his plane, the Josephine Ford, was equipped with skis rather than wheels, and that this would have slowed him down. Rose is unconvinced:
Most important, however, the ski configuration on the
Josephine Ford in 1926 was slightly "up," thus providing added lift to the aircraft by facilitating airflow over the plane's big fixed-wheel undercarriage. The skis ... enhanced rather than detracted from the plane's speed.
Can it be possible that the lift from skis actually increase an airplane's speed? That seems odd, especially since Rose himself, in the very same paragraph, quotes Wendell Summers as saying that skis on his DC-3 slowed the plane's cruise from 160 mph to 145. I was able to find three other planes with published performance data with and without skis: the cruise speed of the Cessna 180J drops from 141 to 123 knots; the Cessna 185E drops from 147 to 131 knots; and the Cessna TU206 E/F drops from 148 to 132 knots. So that's 4-for-4 aircraft slowing down, and by large margins in all cases.
If we wanted to do things simply, we could just find the average percentage drop in speed for these four planes (11%) and be done with it. The Fokker F7a-3m has a published top speed of 118 mph, and an 11% reduction implies top speed with skis of 105. Byrd reported higher speeds during four hours on the flight.
But what about Rose's claim that the lift of the skis enhances the plane's speed? Well, no. The reverse is true, as engineers have known for a century. There are two sources of drag for an airplane in flight: parasitic drag, which is caused by the shape of the aircraft pushing through the air, and induced drag, which is the penalty the wings pay for lifting the plane off the ground. Parasitic drag increases with the square of the airspeed, but induced drag actually declines at higher speeds, because when the plane flies faster, the wings don’t have to work as hard to generate lift.
The equation for induced drag (Di) is:
Di = kL² / (½ ρ0 Ve² S π AR)
Let's work through this equation, using the DC-3 as our example. In this equation, L is the lift generated by the airplane, which for level flight is equal to the airplane’s weight (although expressed in units of force rather than of mass). It can be seen that induced drag declines as weight declines, which happens as the plane burns off fuel. The variable ρ0 is the mass-density of the air at sea level (1.225 kg/m³) and Ve is equivalent airspeed, or the airspeed shown on the airspeed indicator. The wing area S is known for the DC-3 at 91.7 square meters. The wing’s aspect ratio AR is often defined as the ratio of the wingspan to the chord (distance between the leading edge and trailing edge), but for non-rectangular wings it is easier to compute as the square of the wingspan divided by the wing area. For DC-3 with a span of 29 meters, this works out to AR=9.171. The remaining unknown is k, the lift distribution factor. For a perfect elliptical lift distribution, k=1, but elliptical wings are hard to build. Douglas built a reasonable facsimile of an ellipse, however, with a rectangular central section, a large area of taper, and rounded tips. Therefore I will assume k=1.07, which is pretty good, and much better than a rectangular wing, which would have k=1.15. With this assumption we can now compute the induced drag of the DC-3 at any given speed and weight.
Now let's look at Rose's claim by the numbers. First, it is clear that any lift generated by the skis would be lift not generated by the main wing. Given that when acting as “wings” the skis would have an aspect ratio of 0.1 or so, and that each ski is perhaps 50 times smaller than the main wing, it is frightening to contemplate the induced drag penalty any such lift would carry with it – every pound of lift generated by a ski would cause roughly 3000 times more induced drag than the wing would generate lifting that same pound. Therefore we will be generous and assume that the skis generate zero lift and zero induced drag, and that their sole effect is in parasitic drag. To determine that, we need to know how skis affect what's called the "equivalent flat plate area" of the airplane.
Parasitic drag is given by the equation
Dp = ½ ρ0 Ve² Cd A
where Cd A is the product of the parasitic drag coefficient and the wetted surface area of the airplane. We can’t determine either one of these without extensive measurements, including wind-tunnel testing. However, the product of these two numbers, called flat-plate equivalent area, can be determined using the known performance data of the DC-3 (or any other airplane), and that’s all we really need.
Let's start with the ski-less condition. Summers said the plane flew at 160 mph (71.57 m/sec); presumably this was in cruise (which we will assume is 75% of rated horsepower) and with the gear down, matching the Josephine Ford's fixed-gear configuration as closely as possible. The empty weight of a DC-3 is 18300 pounds, and its maximum weight is 25200 pounds. We will assume flying at the average of these two weights, or 21750 pounds. Converting to the metric unit of force, wing lift in level flight is therefore 96749 Newtons, and at a speed of 71.57 m/sec, induced drag is 1208 Newtons.
Total drag of the airplane is induced drag plus parasitic drag (which we don’t know yet). But we do know that the engines of the DC-3 are rated at 2400 horsepower, and since we assume cruise at 75% of that, we get 1800 engine horsepower, or 1342260 Watts. That power is passed into the propellers, which converts it into aircraft dynamic power at some less-than-perfect rate η. That allows us to compute thrust T and dynamic power Tv as follows:
Tv = Pη
or, re-writing slightly,
T = Pη / v
For most aircraft propellers, efficiency η typically maximizes at about 0.81, implying that the total thrust of the airplane would be 1342260 x .81 / 71.57 = 15191 Newtons. For an airplane flying at constant speed, total thrust is equal to total drag, and since we already know that induced drag under these conditions is 1208 N, the remaining drag (which is parasitic drag) must be 13983 N.
Having determined the parasitic drag, we now go back to the equation for parasitic drag and re-write it just a bit:
Cd A = Dp / (½ ρVe²)
and compute that the flat-plate equivalent area Cd A must be 4.46 square meters. I should mention here that exact knowledge of propeller efficiency isn’t really too critical. If we had chosen a higher prop efficiency, that would have implied a less aerodynamically efficient airplane to fly at 160 mph under 1800 hp. So a more efficient propeller means that our Cd A must increase to balance out, increasing the drag on the airplane just enough to balance the increased thrust. The two effects aren’t identical at all speeds, and in fact a slightly better prop would improve overall aircraft performance – but not by much.
Now let's repeat all that, for the DC-3 but this time with one little change: the horsepower is the same, the lift and induced drag are the same, but when we add skis, the speed drops to 145 mph, or 64.86 m/sec. We find that with skis, the equivalent flat plate area goes up to 5.93 square meters -- a whopping 33% increase.
And we can repeat the same calculation with the other three aircraft too. Here are the results:
AircraftHP Gross wt Empty wt Wing Span Wing Area Cruise mph Di Dp CdA Drag penalty
180J 230 2800 1701 10.98 16.2 162.2 92 1344 .417249.1%
180J, ski141.5121 1526 .6222
185E 300 3350 1565 10.92 16.2 169.1 102 1695 .483940.3%
185E, ski 150.7 128 1888 .6788
TU206 285 3600 1935 10.92 16.2 170.2 128 1568 .441739.6%
TU206, ski 151.8 160 1741 .6164
DC-3 2400 25200 18300 29 91.7 160 1208 13982 4.456133.2%
DC-3, ski 145 1471 15291 5.9335
So now we know (roughly) the aerodynamic effect that skis have on a typical airplane: an average increase of 40.5% in equivalent flat-plate area -- which is huge. In the next posting, we will consider how skis affected the Josephine Ford, Byrd's airplane on his 1926 flight. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.849578857421875, "perplexity": 2191.6388566255364}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246661675.84/warc/CC-MAIN-20150417045741-00284-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://en.wikipedia.org/wiki/S-wave | S-wave
Plane shear wave
Propagation of a spherical S-wave in a 2d grid (empirical model)
In seismology, S-waves, secondary waves, or shear waves (sometimes called an elastic S-wave) are a type of elastic wave, and are one of the two main types of elastic body waves, so named because they move through the body of an object, unlike surface waves.[1]
The S-wave is a transverse wave, meaning that, in the simplest situation, the oscillations of the particles of the medium is perpendicular to the direction of wave propagation, and the main restoring force comes from shear stress.[2]
The shadow zone of a P-wave. S-waves don't penetrate the outer core, so they're shadowed everywhere more than 104° away from the epicenter (from USGS)
Its name, S for secondary, comes from the fact that it is the second direct arrival on an earthquake seismogram, after the compressional primary wave, or P-wave, because S-waves travel slower in rock. Unlike the P-wave, the S-wave cannot travel through the molten outer core of the Earth, and this causes a shadow zone for S-waves opposite to where they originate. They can still appear in the solid inner core: when a P-wave strikes the boundary of molten and solid cores at an oblique angle, S-waves will form and propagate in the solid medium. When these S-waves hit the boundary again at an oblique angle they will in turn create P-waves that propagate through the liquid medium. This property allows seismologists to determine some physical properties of the inner core.[3]
History
In 1830, the mathematician Siméon Denis Poisson presented to the French Academy of Sciences an essay ("memoir") with a theory of the propagation of elastic waves in solids. In his memoir, he states that an earthquake would produce two different waves: one having a certain speed a and the other having a speed a/3. At a sufficient distance from the source, when they can be considered plane waves in the region of interest, the first kind consists of expansions and compressions in the direction perpendicular to the wavefront (that is, parallel to the wave's direction of motion); while the second consists of stretching motions occurring in directions parallel to the front (perpendicular to the direction of motion).[4]
Theory
Isotropic medium
For the purpose of this explanation, a solid medium is considered isotropic if its strain (deformation) in response to stress is the same in all directions. Let ${\displaystyle {\boldsymbol {u}}=(u_{1},u_{2},u_{3})}$ be the displacement vector of a particle of such a medium from its "resting" position ${\displaystyle {\boldsymbol {x}}=(x_{1},x_{2},x_{3})}$ due elastic vibrations, understood to be a function of the rest position ${\displaystyle {\boldsymbol {x}}}$ and time ${\displaystyle t}$. The deformation of the medium at that point can be described the strain tensor ${\displaystyle {\boldsymbol {e}}}$, the 3×3 matrix whose elements are
${\displaystyle e_{ij}={\frac {1}{2}}(\partial _{i}u_{j}+\partial _{j}u_{i})}$
where ${\displaystyle \partial _{i}}$ denotes partial derivative with respect to position coordinate ${\displaystyle x_{i}}$. The strain tensor is related to the 3×3 stress tensor ${\displaystyle {\boldsymbol {\tau }}}$ by the equation
${\displaystyle \tau _{ij}=\lambda \delta _{ij}\sum _{k}e_{kk}+2\mu e_{ij}}$
Here ${\displaystyle \delta _{ij}}$ is the Kronecker delta (1 if ${\displaystyle i=j}$, 0 otherwise) and ${\displaystyle \lambda }$ and ${\displaystyle \mu }$ are the Lamé parameters (${\displaystyle \mu }$ being the material's shear modulus). It follows that
${\displaystyle \tau _{ij}=\lambda \delta _{ij}{\bigl (}\sum _{k}\partial _{k}u_{k}{\bigr )}+\mu (\partial _{i}u_{j}+\partial _{j}u_{i})}$
From Newton's law of inertia, one also gets
${\displaystyle \rho \partial _{t}^{2}u_{i}=\sum _{j}\partial _{j}\tau _{ij}}$
where ${\displaystyle \rho }$ is the density (mass per unit volume) of the medium at that point, and ${\displaystyle \partial _{t}}$ denotes partial derivative with respect to time. Combining the last two equations one gets the seismic wave equation in homogeneous media
${\displaystyle \rho \partial _{t}^{2}u_{i}=\lambda \partial _{i}{\bigl (}\sum _{k}\partial _{k}u_{k}{\bigr )}+\mu {\bigl (}\sum _{j}\partial _{i}\partial _{j}u_{j}+\mu \partial _{j}\partial _{j}u_{i}{\bigr )}}$
Using the nabla operator notation of vector calculus, ${\displaystyle \nabla =(\partial _{1},\partial _{2},\partial _{3})}$, with some approximations, this equation can be written as
${\displaystyle \rho \partial _{t}^{2}{\boldsymbol {u}}=\left(\lambda +2\mu \right)\nabla (\nabla \cdot {\boldsymbol {u}})-\mu \nabla \times (\nabla \times {\boldsymbol {u}})}$
Taking the curl of this equation and applying vector identities, one gets
${\displaystyle \partial _{t}^{2}(\nabla \times {\boldsymbol {u}})={\frac {\mu }{\rho }}\nabla ^{2}(\nabla \times {\boldsymbol {u}})}$
This formula is the wave equation applied to the vector quantity ${\displaystyle \nabla \times {\boldsymbol {u}}}$, which is the material's shear strain. Its solutions, the S-waves, are linear combinations of sinusoidal plane waves of various wavelengths and directions of propagation, but all with the same speed ${\displaystyle \beta =\textstyle {\sqrt {\mu /\rho }}}$
Taking the divergence of seismic wave equation in homogeneous media, instead of the curl, yields a wave equation describing propagation of the quantity ${\displaystyle \nabla \cdot {\boldsymbol {u}}}$, which is the material's compression strain. The solutions of this equation, the P-waves, travel at the speed ${\displaystyle \alpha =\textstyle {\sqrt {(\lambda +2\mu )/\rho }}}$ that is more than twice the speed ${\displaystyle \beta }$ of S-waves.
The steady-state SH waves are defined by the Helmholtz equation[5]
${\displaystyle (\nabla ^{2}+k^{2}){\boldsymbol {u}}=0}$
where k is the wave number.
3. ^ University of Illinois at Chicago (17 July 1997). "Lecture 16 Seismographs and the earth's interior". Archived from the original on 7 May 2002. Retrieved 8 June 2010. Cite uses deprecated parameter |dead-url= (help) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 29, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9091058373451233, "perplexity": 593.7001618681076}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572744.7/warc/CC-MAIN-20190916135948-20190916161948-00107.warc.gz"} |
http://electricalacademia.com/induction-motor/three-phase-induction-motor-equivalent-circuit/ | Home / Induction Motor / Three Phase Induction Motor Equivalent Circuit
# Three Phase Induction Motor Equivalent Circuit
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The three-phase induction motor operates in some ways like a transformer. In the transformer, AC voltage is applied to the primary, which creates AC flux in the core. That flux links the secondary and induces a voltage of the same frequency, but with a voltage that depends on the transformer turns ratio.
The three-phase induction motor has voltage applied to the stator that creates a rotating flux wave. As that wave sweeps by the rotor bars, voltages are induced; however, the frequency of the voltage is determined by the slip of the motor. It turns out that the similarity extends to the equivalent circuits of the two devices.
## Derivation of the Equivalent Circuit of Three-Phase Induction Motor
Figure 1 shows the construction of one phase of the armature winding. When a voltage is applied to the coil, current flows in the winding and flux is established, just as in the transformer. In Figure 1, the mutual flux is indicated by the solid lines passing through the middle of the rotor, while the leakage flux is indicated by dotted lines that do not cross the air-gap.
The air-gap in a motor is much larger than that of a transformer, which means the reluctance of the flux path will be much higher. Inductance, of course, is inversely proportional to the reluctance, so the inductance, and hence the reactance, will be smaller. As a result, the no-load current (or exciting current) will be significantly higher, percentage-wise, for an induction motor. Whereas the exciting current for a transformer is only a few percent of the rated current, the exciting current for an induction motor can be 40% or more of the rated current.
FIGURE 1 Mutual and leakage flux due to stator winding.
In developing the equivalent circuit for the induction motor, we can recall the equivalent circuit of the transformer. The primary circuit contained inductances to account for the leakage and mutual fluxes and resistances to account for the resistance of the primary winding and the core losses. The stator of the induction motor is essentially the same; there are mutual and leakage fluxes, winding resistance, and core losses due to hysteresis and eddy currents.
Figure 2 shows the equivalent circuit of one phase of the stator of the induction motor. It is assumed the windings are connected in wye, so the voltage applied to the circuit is a line-to-neutral voltage. The elements Rs and Xs are the stator winding resistance and leakage reactance, and Xm is the magnetizing reactance. This circuit is essentially the same as the primary circuit of a transformer. The only difference is we have not included a core loss resistance. The core losses are often accounted for separately and are thus not represented in the equivalent circuit.
Looking at the stator circuit of Figure 2, I1 is the current entering the winding. As already discussed, a significant current Im, is required to establish the magnetic field. The remaining current, I2, is the load portion of the stator current. The MMF of I2 will exactly cancel the MMF of the rotor current. In phasor notation, we can write,
$\begin{matrix} {{V}_{1}}\text{ }=\text{ }{{E}_{1}}\text{ }+\text{ }{{I}_{1}}({{R}_{s}}\text{ }+j{{X}_{s}}) & {} & \left( 1 \right) \\\end{matrix}$
Where E1 is the EMF induced in the stator coil by the mutual flux. We need to add the rotor to the equivalent circuit.
FIGURE 2 Induction motor stator equivalent circuit.
Looking at figure 3, as the stator flux sweeps by the rotor conductor, a voltage and current will be induced. If the rotor is not allowed to turn (blocked rotor), then the voltage and current induced in the rotor will have the same frequency as the stator.
FIGURE 3 Induction of rotor currents by rotating stator magnetic field.
We have an unusual form of a transformer, in which the flux rotates around the rotor conductors. In the case of the transformer, we refer quantities from one side to the other using the turns ratio. Because most induction motors have squirrel cage rotors, it’s not easy to determine the number of turns on the rotor. Fortunately, we can avoid the problem by always working with quantities referred to the stator.
Figure 4 shows what’s happening when the rotor is blocked. As the stator field sweeps by the rotor conductors, a blocked rotor voltage, EBR (equal to E1), is induced. Since the coils are shorted, current flows through the resistance and leakage reactance of the rotor coils. In figure 4, Rr is the resistance of one phase of the rotor winding and Xr is the leakage reactance of the rotor when stator frequency currents flow in the rotor, which only happens when the rotor is stationary (slip=1.0). Of course, both Rr and Xr are referred to the stator by an appropriate turns ratio.
FIGURE 4 Rotor equivalent circuit at blocked-rotor condition.
For the induction motor to be of use to us, it must turn, which means the slip is less than 1.0. If the rotor is moving, two things happen:
1. The relative velocity of the stator field and rotor coil is sns instead of ns. Recall that E=Blv. Thus, the voltage induced in the rotor will be sE
2. Since the frequency of the rotor currents is sfs, the leakage reactance will have a value of sXr.
Replacing E1 by sE1, and Xr by sXr in Figure 4 yields the circuit shown in Figure 5, which is valid at any value of slip. In order to connect the rotor circuit of Figure 5 to the stator circuit of Figure 2, we must account for the different frequencies.
FIGURE 5 Rotor circuit at slip frequency.
Just as we referred impedances by the turns ratio, we can refer them by the frequency. From the circuit of Figure 5, we can write
$\begin{matrix} \text{s}{{\text{E}}_{\text{1}}}\text{ + }{{\text{I}}_{\text{2}}}\text{(}{{\text{R}}_{\text{r}}}\text{ + js}{{\text{X}}_{\text{r}}}\text{)} & {} & \left( 2 \right) \\\end{matrix}$
Dividing equation 7-8 by s yields
$\begin{matrix} {{E}_{1}}+{{I}_{2}}\left( \frac{{{R}_{r}}}{s}+j{{X}_{r}} \right) & {} & \left( 3 \right) \\\end{matrix}$
Equation 3 can be represented by the circuit of Figure 6, which is the rotor equivalent circuit referred to the stator both by turns ratio and by frequency.
FIGURE 6 Rotor equivalent circuit referred to stator frequency.
This circuit can be connected to the stator equivalent circuit, but it is instructive to split the resistance into two separate components. For convenience, we can write:
$\begin{matrix} \frac{{{R}_{r}}}{s}={{R}_{r}}+\frac{{{R}_{r}}}{s}-{{R}_{r}} & {} & \left( 4 \right) \\\end{matrix}$
Combining the last two terms on the right side of equation 4 yields
$\begin{matrix} \frac{{{R}_{r}}}{s}={{R}_{r}}+{{R}_{r}}\left( \frac{1-s}{s} \right) & {} & \left( 5 \right) \\\end{matrix}$
Replacing the resistive element in Figure 6 by the two resistive elements on the right-hand side of equation 5 yields the rotor equivalent circuit shown in Figure 7. The reasons for this manipulation will be discussed shortly.
FIGURE 7 Rotor circuit referred to the stator, with rotor resistance split into two components.
Finally, by combining the rotor equivalent circuit of Figure 7 with the stator equivalent circuit of Figure 2, we obtain the steady-state equivalent circuit for one phase of a wye-connected induction motor, as shown in Figure 8.
FIGURE 8 Per-phase induction motor equivalent circuit.
Locking again at the rotor part of the equivalent circuit in Figure 8, the resistor Rr represents the resistance of the rotor winding. The power used by it is the power lost in the resistive heating of the rotor winding. The additional resistive element on the right end is a function of the slip and the rotor resistance. It arises from the necessity to transform the rotor circuit not only by turns ratio but also by frequency. The power consumed in this element is the developed power of the machine.
Developed power is the power converted from electrical form to mechanical form and includes the load power plus mechanical losses such as friction and windage.
Subtracting the mechanical losses from the developed power would yield the shaft power, which is the actual power delivered to the load. Developed torque and shaft torque can be calculated from the developed and shaft power, respectively.
Induction Motor Equivalent Circuit Example
A four-pole, 60 Hz, 460 V, 5 HP induction motor has the following equivalent circuit parameters:
$\begin{matrix} \begin{matrix} {{R}_{s}}=1.21\Omega & {{R}_{r}}=0.742\Omega \\\end{matrix} & {{X}_{s}}=3.10\Omega \\ {{X}_{r}}=2.41\Omega & {{X}_{m}}=65.6\Omega \\\end{matrix}$
Find the starting and no-load currents for this machine.
Solution
At starting, the slip is 1.0, which means the load resistor in the equivalent circuit is a short (1-s=0). The current will be the voltage divided by the total impedance of the circuit. The input impedance can be found as:
${{Z}_{in}}=\left( {{R}_{s}}+j{{X}_{s}} \right)+\frac{j{{X}_{m}}\left( {{R}_{r}}+j{{X}_{r}} \right)}{{{R}_{r}}+j{{X}_{r}}+j{{X}_{m}}}$
${{Z}_{in}}=\left( 1.21+j3.1 \right)+\frac{j65.6\left( 0.742+j2.41 \right)}{0.742+j2.41+j65.6}=5.75\angle {{70.72}^{o}}$
${{I}_{st}}=\frac{{{V}_{LL}}}{\sqrt{3}{{Z}_{in}}}=\frac{460}{\sqrt{3}\left( 5.75\angle {{70.72}^{o}} \right)}=46.15\angle -{{70.72}^{o}}$
For no-load, assume the slip is zero, which means the load element in the equivalent circuit will be an open circuit. We can thus find the input impedance:
${{Z}_{in}}=\left( {{R}_{s}}+j{{X}_{s}} \right)+j{{X}_{m}}$
${{Z}_{in}}=\left( 1.21+j3.1 \right)+j65.6=68.71\angle {{88.99}^{o}}$
${{I}_{nl}}=\frac{{{V}_{LL}}}{\sqrt{3}{{Z}_{in}}}=\frac{460}{\sqrt{3}\left( 68.71\angle {{88.99}^{o}} \right)}=3.87\angle -{{88.99}^{o}}$
Did you find apk for android? You can find new Free Android Games and apps. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9372886419296265, "perplexity": 889.7398504491389}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267162385.84/warc/CC-MAIN-20180925202648-20180925223048-00160.warc.gz"} |
https://cs.stackexchange.com/questions/22525/meyniels-theorem-finding-a-hamiltonian-path-for-a-specific-graph-family | # Meyniel's theorem + finding a Hamiltonian path for a specific graph family
Let's say we have a directed graph $G = (V, E)$ for which $(v, w) \in E$ and/or $(w,v) \in E$ holds true for all $v, w \in V$. My feeling is that this graph most definitely is Hamiltonian, and I want to find a Hamiltonian path in it (from any vertex to any other vertex, I don't care where to start or stop).
I wanted to refer to Meylien's theorem for this:
A strongly connected simple directed graph with $n$ vertices is Hamiltonian if the sum of full degrees of every pair of distinct non-adjacent vertices is greater than or equal to $2n − 1.$
There are two subtleties that I'm not sure about with this theorem:
• What is meant by "adjcent vertices". Does the order matter here? Is the pair $(v,w)$ adjacent even if $(w,v) \in E$ but not $(v,w)$ itself? If that is the case and the graph is strongly connected, than it must obviously be Hamiltonian, since there are no non-adjacent pairs of vertices at all.
• A graph with the above property is not necessarily strongly connected. I think this is easy to solve: We can just decompose the graph into SCCs. We will still have no non-adjacent pairs of vertices in the components and all of them are Hamiltonian. We can then construct a Hamiltonian path of the whole graph by connecting them in topological order.
Is the above reasoning correct and the theorem applicable? Or is there some other argument we can use to show that there is a Hamiltonian path in the graph?
In the end I want to actually find the Hamiltonian cycles in the SCCs, but haven't had much luck finding a constructive proof of the theorem, let alone an algorithm that solves this. Can it be done in a straightforward way? I feel like some kind of greedy approach could work, where we take the nodes in decreasing order of outdegree or something similar.
• If exactly one of $(v,w),(w,v)$ is in the graph, then it is known as a tournament. – Yuval Filmus Mar 12 '14 at 3:05
It is enough to prove your claim for the case of a tournament, in which for every pair of vertices $v\neq w$, exactly one of the edges $(v,w),(w,v)$ is in the graph. Wikipedia has an algorithmic proof that every such graph has a Hamiltonian path. If the tournament is strongly connected, it has a Hamiltonian cycle. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8854880332946777, "perplexity": 119.90623755604669}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400198942.13/warc/CC-MAIN-20200921050331-20200921080331-00479.warc.gz"} |
http://mathhelpforum.com/trigonometry/152041-help-trig-word-problem-print.html | # Help with this trig word problem!
• July 26th 2010, 03:17 PM
orkdork
Help with this trig word problem!
Two wires stretch from the top T of a vertical pole to points B and C on the ground, where C is 10 m closer to the base of the pole, than is B. If wire BT makes an angle of 35 degrees with the horizontal and wire CT makes an angle of 50 degrees with the horizontal, how high is the pole?
• July 26th 2010, 04:19 PM
Quote:
Originally Posted by orkdork
Two wires stretch from the top T of a vertical pole to points B and C on the ground, where C is 10 m closer to the base of the pole, than is B. If wire BT makes an angle of 35 degrees with the horizontal and wire CT makes an angle of 50 degrees with the horizontal, how high is the pole?
Use tangent.
If the distance from the base of the pole to C is x,
then the distance to B is x+10...
h=pole height.
$tan(50^o)=\frac{h}{x}\ \Rightarrow\ h=xtan50^o$
$tan(35^0)=\frac{h}{x+10}\ \Rightarrow\ h=(x+10)tan35^o$
That's two equations for h which must be equal..
$xtan50=xtan35+10tan35$
$xtan50-xtan35=10tan35$
Now factor the LHS and finish.
Once you have x, you can find h. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9420309662818909, "perplexity": 695.5992561313318}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413558067768.10/warc/CC-MAIN-20141017150107-00149-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/orthonormal-basis-and-operators.190683/ | # Orthonormal basis and operators
1. Oct 11, 2007
### friend
I hope this is the forum to ask this question.
We all know that the eigenvectors of a Hermitian operator form an orthonormal basis. But is the opposite true as well. Are the vectors of an orthonormal basis always the eigenvectors of some Hermitian operator? Or do we need added restrictions to make it so, such as an inner product and dual spaces being the complex conjugate of the normal space? Thanks.
2. Oct 11, 2007
### meopemuk
Hi friend,
welcome to the Forum!
If you are talking about an orthonormal basis, then you assume that the inner product is already defined (orthogonality and normalization of vectors requires presence of the inner product) and you are in a Hilbert space. Given such a basis you can simply define in this basis a diagonal matrix with arbitrary real numbers on the diagonal and you get a Hermitian operator whose eigenvectors are basis vectors and eigenvalues are those diagonal entries.
Eugene.
3. Oct 12, 2007
### kanato
All vectors are eigenvectors of the identity operator, and the identity is Hermitian. So, yes.
4. Oct 12, 2007
### friend
Thanks. So are you saying that the operator that generates this basis may not be unique (you say "arbitrary real numbers on the diagonal), or is it unique up to a scalar multiple? Thanks.
5. Oct 12, 2007
### meopemuk
There is an infinite set of Hermitian operators with the same basis of eigenvectors. They are different by their eigenvalues.
Eugene.
6. Oct 12, 2007
### friend
Is this because any orthonormal basis from an operator can be transformed to any other basis in the same space that are the eigenvectors of some other operator?
Last edited: Oct 12, 2007
7. Oct 12, 2007
### meopemuk
Yes. You should understand that there is equivalence between Hermitian operators and orthonormal sets of eigenvectors and eigenvalues. Once you get a Hermitian operator you can find a unique (up to degeneracy) basis of eigenvectors and real eigenvalues attached to them. Inversely, once you specified an arbitrary orthonormal basis and assigned an arbitrary real eigenvalue to each basis vector, then you have a unique Hermitian operator.
Eugene.
8. Oct 12, 2007
### dextercioby
Given an ortonormal basis in a Hilbert space, there exist an infinity of compact selfadjoint operators whose eigenvectors are precisely the vectors in the given orthonomal basis. The simplest of these operators is the identity operator.
9. Oct 12, 2007
### Manchot
Yes. You can always construct a Hermitian operator with a weighted sum of the outer product of each basis vector with itself. To do so, just make the weighting coefficients real. (Incidentally, you could make a unitary matrix as well, by making the weighting coefficients have unit modulus.)
10. Oct 12, 2007
### friend
Thank you all for your replies.
11. Oct 15, 2007
### sridhar
clarification
Is it not possible that some eigen values corresponding to the vectors in the basis that we have picked are not real values. Even though all the vectors in the basis are orthogonal. Just a thought.
12. Oct 15, 2007
### OOO
If the eigenvalues are not real then the designed operator will not be hermitean.
13. Oct 15, 2007
### sridhar
exactly! which is why this question does not arise.
"Are the vectors of an orthonormal basis always the eigenvectors of some Hermitian operator? " (mentioned in the first post)
that is all i was saying.
14. Oct 15, 2007
### OOO
Okay, I missed the connection to the first post. But I think thread starter did not want to ask if the construction from an orthonormal basis always yields a hermitean operator, but if there exists always at least one hermitean operator with these eigenvectors.
Nevertheless an important point.
15. Oct 15, 2007
### sridhar
should have quoted!!! am still not used to posting properly.
Apologies
16. Oct 16, 2007
### reilly
Let me expand this correct idea a bit. Given a complete set of orthonormal of basis vectors, |k>, where k labels the eigenstates -- can be a finite or infinite set. Then for example the spectral resolution, O = Sum over k {|k> k <k|} has the property that O |k> = k |k>. And O is Hermitian. But, the same will be true for F(O), where F is an arbitrary real function. So
F(O) |k> = F(k)|k>. That's about it.
Regards,
Reilly Atkinson | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9622814655303955, "perplexity": 653.8689619895508}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719079.39/warc/CC-MAIN-20161020183839-00058-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://www.dummies.com/how-to/content/how-to-work-both-sides-of-a-trig-identity.html | With a trigonometry identity, working on both sides of the equation is even more fun than working on both sides of an algebraic equation. In algebra, you can multiply each side by the same number, square both sides, add or subtract the same thing to each side, and so on. When you solve trig identities and equations, you can use all those algebra rules plus you can do substitutions with the various trig identities when you need them. You can even insert a different identity on each side — the one big advantage of working on both sides of a trig identity.
This first example is rather basic, but it gets the idea across. Solve the identity
by working both sides.
1. Change the functions that aren’t one of the three basic functions by using their reciprocal identities.
2. Simplify the two fractions on the left by flipping the denominators and multiplying them by their numerators.
Then multiply the two factors on the right together.
3. Replace the sum on the left by using the Pythagorean identity.
You end up with 1 = 1.
In the next example, you change everything to sines and cosines. Prove the identity
1. Change the functions to their equivalences by using the reciprocal and ratio identities.
2. On the left, flip the denominator and multiply it by the numerator.
On the right, multiply each fraction by a fraction equal to 1 (by using the other fraction’s denominator) to get common denominators for all the fractions.
3. Simplify the multiplied fractions.
Add the two fractions on the right together.
4. Replace the numerator on the right with the value from the Pythagorean identity.
This last example requires a little creativity to get the job done. But working on both sides still works best when showing that the following is an identity:
1. Split up the fraction on the left by writing each term in the numerator over the denominator.
2. Reduce the second fraction to 1.
3. Replace the csc2 on the right with its equivalent by using the Pythagorean identity.
4. Simplify the terms on the right — two are opposites of one another.
5. Replace the fraction on the left by using the reciprocal identity. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9515662789344788, "perplexity": 315.7701657942659}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257823133.4/warc/CC-MAIN-20160723071023-00011-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/geometry/CLONE-df935a18-ac27-40be-bc9b-9bee017916c2/chapter-8-section-8-4-circumference-and-area-of-a-circle-exercises-page-383/8a | ## Elementary Geometry for College Students (7th Edition)
radius = 6 m circumference = 12$\pi$ m
The area of the circle = 36$\pi m^{2}$ The area A of a circle whose radius has length r is given by A = $\pi r^{2}$ 36$\pi$ = $\pi r^{2}$ 36 = $r^{2}$ r = 6 m We know the circumference of the circle is given by the formula c= $\pi$ d = 2$\pi$ r = 2$\pi$ * 6 = 12$\pi$ m | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8524196743965149, "perplexity": 472.088426814902}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178357984.22/warc/CC-MAIN-20210226205107-20210226235107-00253.warc.gz"} |
https://socratic.org/questions/how-do-you-simplify-3x-5-2x-3#344215 | Algebra
Topics
# How do you simplify (3x + 5) (2x - 3)?
Nov 30, 2016
$6 {x}^{2} + x - 15$
#### Explanation:
$\left(3 x + 5\right) \left(2 x - 3\right)$
Starting with $3 x$, multiply $3 x$ by each term in the second group
$3 x \cdot 2 x = 6 {x}^{2}$
$3 x \cdot - 3 = - 9 x$
Next, multiply $5$, by each term in the second group
$5 \cdot 2 x = 10 x$
$5 \cdot - 3 = - 15$
Hence $\left(3 x + 5\right) \left(2 x - 3\right)$
= $6 {x}^{2} - 9 x + 10 x - 15$
= $6 {x}^{2} + x - 15$
##### Impact of this question
6556 views around the world | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9757450819015503, "perplexity": 2518.3982170507193}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057303.94/warc/CC-MAIN-20210922011746-20210922041746-00526.warc.gz"} |
https://nuit-blanche.blogspot.com/2015/03/infinite-dimensional-ell1-minimization.html | ## Friday, March 20, 2015
### Infinite-dimensional $\ell^1$ minimization and function approximation from pointwise data
Infinite-dimensional $\ell^1$ minimization and function approximation from pointwise data by Ben Adcock
We consider the problem of approximating a function from finitely-many pointwise samples using $\ell^1$ minimization techniques. In the first part of this paper, we introduce an infinite-dimensional approach to this problem. Three advantages of this approach are as follows. First, it provides interpolatory approximations in the absence of noise. Second, it does not require a priori bounds on the expansion tail in order to be implemented. In particular, the truncation strategy we introduce as part of this framework is completely independent of the function being approximated. Third, it allows one to explain the crucial role weights play in the minimization, namely, that of regularizing the problem and removing so-called aliasing phenomena. In the second part of this paper we present a worst-case error analysis for this approach. We provide a general recipe for analyzing the performance of such techniques for arbitrary deterministic sets of points. Finally, we apply this recipe to show that weighted $\ell^1$ minimization with Jacobi polynomials leads to an optimal, convergent method for approximating smooth, one-dimensional functions from scattered data.
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http://mathoverflow.net/questions/100739/smoothness-of-the-convolution-of-a-singular-measure-with-itself | # Smoothness of the convolution of a singular measure with itself
Let $\gamma:I\subset\mathbb{R}\rightarrow\mathbb{R}^2, s\mapsto \gamma(s)$, denote the arclength parametrization of a smooth, convex curve $\Gamma:=\gamma(I)\subset\mathbb{R}^2$. Equip $\Gamma$ with arclength measure $\sigma$.
Here is my question: what extra conditions on $\Gamma$ will ensure that the triple convolution $\sigma\ast\sigma\ast\sigma$ is a smooth function of $x\in\mathbb{R}^2$ inside its support (i.e. the set $\Gamma+\Gamma+\Gamma\subset\mathbb{R}^2$), up to the boundary? I am also interested in partial regularity results, and counterexamples.
Here is a possible approach: for any $x\in\mathbb{R}^2$ and $\epsilon>0$, $$\int_{|y-x|<\epsilon} \sigma\ast\sigma\ast\sigma(y)dy=|\{(s,s',s'')\in I^3: |\gamma(s)+\gamma(s')+\gamma(s'')-x|<\epsilon\}|.$$
Understanding how these sets look like will probably help. I am familiar with results in this spirit for double convolutions (e.g. Fefferman 1970), but not for triple ones.
References would be much appreciated.
Thank you.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.88834547996521, "perplexity": 275.87909144727206}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049276131.97/warc/CC-MAIN-20160524002116-00236-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://engstroy.spbstu.ru/en/article/2018.82.8/ | # Natural oscillations of a rectangular plates with two adjacent edges clamped
Authors:
Abstract:
We study the natural oscillations of a rectangular plate, two adjacent edges of which are clamped, and the other two are free (CCFF-plate), as an element of many building structures. The deflection function is chosen as a sum of two hyperbolic trigonometric series. Both series obey the main equation of free vibration. Meeting all boundary conditions of a problem leads to an infinite system of homogeneous linear algebraic equations with respect to eight series coefficients. This system is transformed in two subsystems due to four basic coefficients, for which the iterative solution process is organized. Initial values of a pair of basic coefficient series are chosen randomly. Frequency values are chosen so that iterations coincide starting with a certain number. This provides non-trivial solutions of the reduced system. For the first eight obtained natural frequencies there have been presented relevant 3D mode shapes. The paper provides accuracy analysis and its comparison with other familiar results. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9673852324485779, "perplexity": 461.1888967709945}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038057142.4/warc/CC-MAIN-20210410134715-20210410164715-00371.warc.gz"} |
http://mathhelpforum.com/differential-geometry/137465-l-hopital-s-rule-complex-function.html | # Thread: l'Hôpital's rule for a complex function
1. ## l'Hôpital's rule for a complex function
Hey guys.
Can I use l'Hôpital's rule for a complex function?
Lets say I want to investigate the behavior of function sin(z)/z around z=0, can I use it?
Thanks.
2. What do you mean by complex? That is not a complex function.
And You can use it in this case.
$
\lim_{z\to 0} \frac{sin(z)}{z}= 1$
Since it is in undetermined form, ( 0 in the denominator and 0 in the numerator), you can take the derivative of the top and bottom, and then apply the limit again.
3. Originally Posted by asi123
Hey guys.
Can I use l'Hôpital's rule for a complex function?
Lets say I want to investigate the behavior of function sin(z)/z around z=0, can I use it?
Thanks.
$\frac{\sin(z)}{z}=\frac{e^{iz}-e^{-iz}}{2iz}=\frac{1}{2i}\left[\frac{e^{iz}-1}{z}-\frac{e^{-iz}-1}{z}\right]$. A little more obvious now?
4. Sorry but no.
I still cant put z=0 in that.
I didn't quite get it.
5. $\frac {\sin{z}} {z} = \frac {z-\frac {z^3} {3!} + ... } {z} = 1-\frac {z^2} {3!} + ...$
so the result is 1 when $z \rightarrow 0$
6. Well, Thank you very much but I know the result is 1.
My question is, can I use l'Hôpital's rule for this thing even due z is a number from the complex domain?
I mean when you try to find limit for this thing, there are number of paths you can take in order to do that, does l'Hôpital's rule works in that case?
Thanks a lot.
7. Yes, L'Hopital's rule works for functions of a complex variable.
8. Sorry to raise a dead thread, but I have had this same question for some time and have never seen a proof of L'Hopital's Rule that does not require the mean value theorem for real differentiable functions, which as far as I have seen, does not extend to the complex numbers. Perhaps the mean value property of harmonic functions serves the same end, but I have still never come across such a proof (I must concede, I've never tried proving it myself-- it could turn out to be quite easy).
Evidence demonstrates that L'Hopital's Rule does apply to the limit of a complex valued analytic function (as I am not able to construct a counterexample), but I would love to see a proof if anyone knows where I can find one.
9. In fact, it isn't always generally true. And, if you would have googled 'L'hopitals rule complex functions' you would have found this as the fourth link. Hope it helps.
10. The proof is simple and relies on the factorization of holomorphic functions: Assume $f,g: D \rightarrow \mathbb{C}$ are holomorphic ( $D$ some disk around $0$). If $f(0)=g(0)=0$ then $f(z)=z\hat{f}(z)$ and $g(z)=z\hat{g}(z)$ where $\hat{f}(0)=f'(0)$ and $\hat{g}(0)=g'(0)$, then the conclusion follows trivially. If on the other hand they both have a pole at $0$ then the it's a little trickier: Take $f(z)=\sum_{-k}^{\infty} b_nz^n$ and $g(z)=\sum_{-m}^{\infty} c_nz^n$ be the Laurent expansions around $0$ then the quotient of the functions are
$\frac{f(z)}{g(z)}= z^{m-k}\frac{b_{-k} + o(1)}{c_{-m}+o(1)}$
$\frac{f'(z)}{g'(z)}=z^{m-k}\frac{-kb_{-k} +o(1)}{-mc_{-m}+o(1)}$
and in both cases the limit is equal to: $0$ if $m>k$, $\frac{b_{-k}}{c_{-m}}$ if $m=k$ and $\infty$ if $m.
Edit: As an afterthought notice that this proof relies on the fact that the functions are defined and not zero on a punctured neighbourhood of the limit point, so the general formulation, as in the case of a real variable, doesn't follow from this. So for example if you have one of the functions be such that doesn't extend in any way past any boundary point of a disk, then you can't use this to evaluate the limit at a boundary point.
11. Hi Drexel, thanks for the link. I did google it, and it actually took me here. I did not notice that link before. I am thrilled to see the conditions for the complex situation.
Jose thank you for posting your proof.
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# can we use Lhospital rule in complex analysis
Click on a term to search for related topics. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 24, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9699512720108032, "perplexity": 212.77166995631163}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891813322.19/warc/CC-MAIN-20180221024420-20180221044420-00267.warc.gz"} |
http://math.stackexchange.com/questions/199274/the-cardinality-of-a-countable-union-of-sets-with-less-than-continuum-cardiality?answertab=active | # The cardinality of a countable union of sets with less than continuum cardiality
Can continuum be a countable union of sets with cardinality, less than continuum? I can prove it for a finite union by mathematical induction from this:
$\mathbb R = A_1 + A_2 => |\mathbb R| = |A_1 + A_2| = max (|A_1| + |A_2|) => |A_i| = |\mathbb R|$
But how to prove it for a countable union?
-
As remarked by others, assuming the axiom of choice this is false.
However there are several interesting cases when the axiom of choice fails:
1. The real numbers may be a countable unions of countable sets.
2. The real numbers may be the union of two sets both of cardinality strictly less than the continuum, furthermore at least one of these sets can be split into two smaller sets, so we can even split the continuum into three sets of smaller cardinality; and so on.
-
To add to what Arthur said, one way to show (assuming AC) that $2^{\aleph_0}$ has uncountable cofinality is to use König's theorem
which is a generalization of Cantor's theorem. An instance of König's theorem says that given a countable sequence of sets $(A_i : i<\omega)$ such that $|A_i|<|\mathbb{R}|$ for all $i$, we have
$\sum_{i<\omega} |A_i| < \prod_{i<\omega} |\mathbb{R}|$.
The right hand side is equal to $|\mathbb{R}|$ because countable sequences of reals can be coded by reals.
-
Hi Trevor, I understand that you're in the Fields semester now. I was wondering whether or not you attended Shelah's talk two days ago about the axiom of choice. The title seems interesting but there are no abstracts anywhere. – Asaf Karagila Sep 30 '12 at 15:28
@AsafKaragila Unfortunately I was out of town last week and could not attend Shelah's talk. (I also cannot find an abstract anywhere.) – Trevor Wilson Oct 3 '12 at 16:33
Assuming the Axiom of Choice, the real line cannot be a union of countably many sets each of size less than continuum. To prove this, one needs to know that the continuum has cofinality strictly greater than $\omega$. (The cofinality of an limit ordinal $\delta$ is the least cardinality of a set $X$ of ordinals strictly less than $\delta$ such that $\delta = \sup (X)$.)
This resolves you question because if $\{ A_n : n \in \omega \}$ is a family of subsets of $\mathbb{R}$ each with size strictly less than $2^{\omega}$, then setting $\kappa = \sup \{ |A_n| : n \in \omega \}$ we have that $\kappa < 2^\omega$. It then follows that $$| \textstyle{\bigcup_{n < \omega}} A_n | \leq \sum_{n < \omega} | A_n | \leq \sum_{n < \omega} \kappa = \omega \cdot \kappa = \max \{ \omega , \kappa \} < 2^{\omega},$$ and therefore $\bigcup_{n < \omega} A_n$ cannot equal all of $\mathbb{R}$.
Without the Axiom of Choice, it is possible that the continuum is a countable union of countable sets. (Note that this does not contradict the well-known fact (provable without Choice) that $\mathbb{R}$ is uncountable because in such models countable unions of countable sets need not be countable.)
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This is the right approach, to be sure - I had been assuming a union of sets of the same cardinality, but this handles cases like looking at $\displaystyle\cup_{n\in\omega}\aleph_n$ when $\displaystyle\mathfrak{c} = \aleph_{\omega+1}$, which my argument doesn't. – Steven Stadnicki Sep 19 '12 at 18:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.961617648601532, "perplexity": 158.4859441579813}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438043062723.96/warc/CC-MAIN-20150728002422-00223-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://www.researchgate.net/publication/236614560_CVD_formation_of_graphene_on_SiC_surface_in_argon_atmosphere | Article
# CVD formation of graphene on SiC surface in argon atmosphere.
Institute of Theoretical Physics, Faculty of Physics, University of Warsaw, ul. Hoża 69, 00-681 Warszawa, Poland. .
Physical Chemistry Chemical Physics (Impact Factor: 3.83). 05/2013; DOI:10.1039/c3cp44378g
Source: PubMed
ABSTRACT We investigate the microscopic processes leading to graphene growth by the chemical vapor deposition of propane in an argon atmosphere at the SiC surface. Experimentally, it is known that the presence of argon fastens the dehydrogenation processes at the surface, at high temperatures of about 2000 K. We perform ab initio calculations, at zero temperature, to check whether chemical reactions can explain this phenomenon. Density functional theory and supporting quantum chemistry methods qualitatively describe formation of the graphene wafers. We find that the 4H-SiC(0001) surface exhibits a large catalytic effect in the adsorption process of hydrocarbon molecules, this is also supported by preliminary molecular dynamics results. The existence of the ArH(+) molecule, and an observation from the Raman spectra that the negative charge transfers into the SiC surface, would suggest that presence of argon atoms leads to a deprotonization on the surface, which is necessary to obtain a pure carbon adlayer. But the zero-temperature description shows that the cold environment is insufficient to promote argon-assisted surface cleaning.
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• ##### Article: Structural and Electronic Properties of Functionalized Graphene
[hide abstract]
ABSTRACT: In the present paper, we study the effects of functionalization of graphene with simple organic molecules OH, and NH2, focusing on the stability and band gaps of the structures. We have performed DFT calculations for graphene supercells with various numbers of the attached molecules. We have determined adsorption energies of the functionalized graphene mono- and bilayers, the changes in the geometry, and the band structure. We observe the characteristic effects such as rehybridization of the bonds induced by fragments attached to graphene and opening of the graphene band gap by functionalization. We have also studied the dependence of the adsorption energies of the functionalized graphene on the density of the adsorbed molecules. Our calculations reveal that the –OH and –NH2 groups exhibit the strong cohesion to graphene layers. Further, we determine the critical density of the OH fragments which lead to the opening of the band gap. We also show how to engineer the magnitude of the band gap by functionalizing graphene with NH2 groups of various concentrations.
Acta Physica Polonica Series a 11/2011; 120:842-844. · 0.53 Impact Factor
• ##### Article: A stable argon compound
[hide abstract]
ABSTRACT: The noble gases have a particularly stable electronic configuration, comprising fully filled s and p valence orbitals. This makes these elements relatively non-reactive, and they exist at room temperature as monatomic gases. Pauling predicted in 1933 that the heavier noble gases, whose valence electrons are screened by core electrons and thus less strongly bound, could form stable molecules. This prediction was verified in 1962 by the preparation of xenon hexafluoroplatinate, XePtF6, the first compound to contain a noble-gas atom. Since then, a range of different compounds containing radon, xenon and krypton have been theoretically anticipated and prepared. Although the lighter noble gases neon, helium and argon are also expected to be reactive under suitable conditions, they remain the last three long-lived elements of the periodic table for which no stable compound is known. Here we report that the photolysis of hydrogen fluoride in a solid argon matrix leads to the formation of argon fluorohydride (HArF), which we have identified by probing the shift in the position of vibrational bands on isotopic substitution using infrared spectroscopy. Extensive ab initio calculations indicate that HArF is intrinsically stable, owing to significant ionic and covalent contributions to its bonding, thus confirming computational predictions that argon should form a stable hydride species with properties similar to those of the analogous xenon and krypton compounds reported before.
Nature 09/2000; 406(6798):874-6. · 38.60 Impact Factor | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8336527347564697, "perplexity": 2244.7978610785854}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394678666156/warc/CC-MAIN-20140313024426-00068-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.snapxam.com/calculators/express-in-terms-of-sine-and-cosine-calculator | # Express in terms of sine and cosine Calculator
## Get detailed solutions to your math problems with our Express in terms of sine and cosine step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here!
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### Difficult Problems
1
Solved example of express in terms of sine and cosine
$\frac{1-\tan\left(x\right)}{1+\tan\left(x\right)}$
2
Applying the tangent identity: $\displaystyle\tan\left(\theta\right)=\frac{\sin\left(\theta\right)}{\cos\left(\theta\right)}$
$\frac{1+\frac{-\sin\left(x\right)}{\cos\left(x\right)}}{1+\tan\left(x\right)}$
3
Combine $1+\frac{-\sin\left(x\right)}{\cos\left(x\right)}$ in a single fraction
$\frac{\frac{-\sin\left(x\right)+\cos\left(x\right)}{\cos\left(x\right)}}{1+\tan\left(x\right)}$
4
Divide fractions $\frac{\frac{-\sin\left(x\right)+\cos\left(x\right)}{\cos\left(x\right)}}{1+\tan\left(x\right)}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$
$\frac{-\sin\left(x\right)+\cos\left(x\right)}{\cos\left(x\right)\left(1+\tan\left(x\right)\right)}$
5
Multiply the single term $\cos\left(x\right)$ by each term of the polynomial $\left(1+\tan\left(x\right)\right)$
$\frac{-\sin\left(x\right)+\cos\left(x\right)}{\cos\left(x\right)+\tan\left(x\right)\cos\left(x\right)}$
Applying the tangent identity: $\displaystyle\tan\left(\theta\right)=\frac{\sin\left(\theta\right)}{\cos\left(\theta\right)}$
$\frac{\sin\left(x\right)}{\cos\left(x\right)}\cos\left(x\right)$
Multiplying the fraction by $\cos\left(x\right)$
$\frac{\sin\left(x\right)\cos\left(x\right)}{\cos\left(x\right)}$
Simplify the fraction $\frac{\sin\left(x\right)\cos\left(x\right)}{\cos\left(x\right)}$ by $\cos\left(x\right)$
$\sin\left(x\right)$
6
Applying the trigonometric identity: $\tan\left(\theta\right)\cdot\cos\left(\theta\right)=\sin\left(\theta\right)$
$\frac{-\sin\left(x\right)+\cos\left(x\right)}{\cos\left(x\right)+\sin\left(x\right)}$
$\frac{-\sin\left(x\right)+\cos\left(x\right)}{\cos\left(x\right)+\sin\left(x\right)}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8844611048698425, "perplexity": 1367.3097932045182}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00268.warc.gz"} |
http://mathhelpforum.com/differential-equations/193672-dissolve-question.html | # Math Help - Dissolve question
1. ## Dissolve question
Consider two tanks, A and B, each holding 200 litres of water. A pipe pumps water from tank A to tank B at a rate of 5l/min. At the same time another pipe pumps liquid from tank B to tank A at the same rate. At time t=0, kg of a chemical X is dissolved into tank A, and tank B has kg of the same chemical X dissolved into it.
i). Write down the system of differential equations satisfied by x(t) and y(t), the quantity of the chemical X in tanks A and B respectively.
x(t)=x0-t(y0/40-x0/40)
y(t)=y0-(y0/40+x0/40)
dx/dt=t(-y0/40+x0/40)
dy/dt=t(y0/40-x0/40)
ii). Find the eigenvalues and the eigenvectors of the resulting matrix form.
The matrix is
(-1/40 1/40)
(1/40 -1/40)
The eigenvalues are 0 and -2
so the eigenvector are
(-1)
(1 )
and
(1)
(1)
iii). Show that the amount of the chemical X in either tank approaches as t approaches infinity.
i dont have any idea to do this part, can anyone help me? and tell me what i did wrong in the first and second part pls
2. ## Re: Dissolve question
Originally Posted by kanezila
Consider two tanks, A and B, each holding 200 litres of water. A pipe pumps water from tank A to tank B at a rate of 5l/min. At the same time another pipe pumps liquid from tank B to tank A at the same rate. At time t=0, kg
Did you mean to put a quantity of chemical X there? I don't see a number.
of a chemical X is dissolved into tank A, and tank B has kg
Same issue here.
of the same chemical X dissolved into it.
i). Write down the system of differential equations satisfied by x(t) and y(t), the quantity of the chemical X in tanks A and B respectively.
x(t)=x0-t(y0/40-x0/40)
y(t)=y0-(y0/40+x0/40)
dx/dt=t(-y0/40+x0/40)
dy/dt=t(y0/40-x0/40)
Why are there t's in these equations? What is your justification for your answer of this part of the problem?
ii). Find the eigenvalues and the eigenvectors of the resulting matrix form.
The matrix is
(-1/40 1/40)
(1/40 -1/40)
I agree with your matrix form. That is, the DE can be written as
$\frac{d}{dt}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-1/40 &1/40\\ 1/40 &-1/40\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}.$
If you let
$A:=\begin{bmatrix}-1/40 &1/40\\ 1/40 &-1/40\end{bmatrix},$
then the solution to this differential equation is
$\begin{bmatrix}x\\ y\end{bmatrix}=e^{At}\begin{bmatrix}x_{0}\\ y_{0}\end{bmatrix}.$
What you must do is make sense of the exponential $e^{At}.$ How do you do matrix exponentiation? This is how you can finish the problem.
The eigenvalues are 0 and -2
so the eigenvector are
(-1)
(1 )
and
(1)
(1)
iii). Show that the amount of the chemical X in either tank approaches as t approaches infinity.
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http://math.stackexchange.com/questions/284637/prove-transitivity-of-forall-x-x-subseteq-a-setminus-a-b-rightarrowx-cup | # Prove transitivity of $\forall X( X\subseteq A\setminus\{a, b\}\rightarrow(X\cup \{a\}\in\mathcal{F}\rightarrow X\cup\{b\}\in\mathcal{F}))$
This one is from Velleman's "How to Prove It, 2nd Ed.", exercise 4.3.23.
Suppose $A$ is set, and $\mathcal{F}\subseteq\mathcal{P}(A)$. Let $$R=\{(a,b)\in A\times A : \text{for every } X\subseteq A\setminus\{a, b\}\text{, if } X\cup \{a\}\in\mathcal{F}\text{ then } X\cup\{b\}\in\mathcal{F}))\}$$ Show that $R$ is transitive.
1. First of all, I'm not sure if I read this correctly and if my notation is correct: $$R=\{(a,b)\in A\times A : \forall X( X\subseteq A\setminus\{a, b\}\rightarrow(X\cup \{a\}\in\mathcal{F}\rightarrow X\cup\{b\}\in\mathcal{F}))\}$$
2. To prove that $R$ is transitive, we need to prove that $$\forall a\in A\;\forall b\in A\;\forall c\in A\;((aRb\wedge bRc) \rightarrow aRc),$$ so for starters we suppose and let all the usual stuff:
• let $a,b,c\in A$
• suppose $aRb \wedge bRc$
• expand $aRc$ to $\forall X( X\subseteq A\setminus\{a, c\}\rightarrow(X\cup \{a\}\in\mathcal{F}\rightarrow X\cup\{c\}\in\mathcal{F}))$
• let $X$ be arbitrary and suppose $X\subseteq A\setminus\{a,c\}$
• suppose $X\cup\{a\}\in\mathcal{F}$
• show that $X\cup\{c\}\in\mathcal{F}$
3. Now Velleman suggests splitting proof to cases: $b\not\in X$ and $b\in X$. Showing that $R$ is transitive for $b\not\in X$ is rather straightforward. All we need to do is to show from $b\not\in X \wedge X\subseteq A\setminus\{a,c\}$ that $X$ is also subset of both $A\setminus\{a,b\}$ and $A\setminus\{b,c\}$, and we just follow assumptions $aRb$ and $bRc$ to conclude $X\cup\{c\}\in\mathcal{F}$.
4. Now we must show transitivity when $b\in X$. For this case Velleman suggests working with $X'=(X\cup\{a\})\setminus\{b\}$ and $X''=(X\cup\{c\})\setminus\{b\}$, and this is the part I totally don't get. Why would using $X'$ and $X''$ work for this proof, and how do actually connect them with all the givens/assumptions? I can see how all this connects after expanding $aRb$ and $bRc$, but I fail to see how this makes a correct proof.
So my questions are: is the 1. correct notation for given relation $R$ and how does 4. connect to givens.
If there is some other approach, I would be most thankful for any pointers.
-
In the proof of transitivity, suppose that $b\in X\subseteq A\setminus\{a,c\}$; we want to show that if $X\cup\{a\}\in\mathscr{F}$, then $X\cup\{c\}\in\mathscr{F}$, so assume that $X\cup\{a\}\in\mathscr{F}$. Let $X_0=X\setminus\{b\}$; then Velleman’s sets $X'$ and $X''$ are given by $X'=X_0\cup\{a\}$, and $X''=X_0\cup\{c\}$.
Now $b,c\notin X'$, and $X'\cup\{b\}=X_0\cup\{a\}\cup\{b\}=X\cup\{a\}\in\mathscr{F}$, so $X'\cup\{c\}\in\mathscr{F}$, since $bRc$. Now $X'\cup\{c\}=X_0\cup\{a\}\cup\{c\}=X''\cup\{a\}$, so $X''\cup\{a\}\in\mathscr{F}$. Moreover, $X''\subseteq A\setminus\{a,b\}$, and $aRb$, so $X''\cup\{b\}\in\mathscr{F}$. Finally, $X''\cup\{b\}=X_0\cup\{c\}\cup\{b\}=X\cup\{c\}$, and we conclude that $X\cup\{c\}\in\mathscr{F}$, as desired.
@LavaScornedOven: I’m not sure that there really is any way to get an intuitive feel for relations in general. For the common properties of relations (e.g., transitivity) I think that your best bet is simply working with them and seeing lots of examples. It may also help to see more than one way of viewing a relation on $A$: a subset of $A\times A$, an $|A|\times|A|$ matrix of $0$s and $1$s, a directed bipartite graph. Have you seen either of the last two yet? – Brian M. Scott Jan 23 '13 at 1:48
I've seen the directed graph approach, but never worked with $|A|\times|A|$ matrices. I'm really new to all this, so I guess I just need to work it through. Once again, thanks a lot for all the help. – LavaScornedOven Jan 23 '13 at 1:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9880014061927795, "perplexity": 153.12165535534416}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931008520.8/warc/CC-MAIN-20141125155648-00109-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://math.gatech.edu/seminars-colloquia/series/applied-and-computational-mathematics-seminar/dr-matthew-calef-20141103 | ## Computational Multiphysics at Scale
Series:
Applied and Computational Mathematics Seminar
Monday, November 3, 2014 - 14:00
1 hour (actually 50 minutes)
Location:
Skiles 005
,
Los Alamos National Lab
Organizer:
Observations of high energy density environments, from supernovae implosions/explosions to inertial confinement fusion, are determined by many different physical effects acting concurrently. For example, one set of equations will describe material motion, while another set will describe the spatial flow of energy. The relevant spatial and temporal scales can vary substantially. Since direct measurement is difficult if not impossible, and the relevant physics happen concurrently, computer simulation becomes an important tool to understand how emergent behavior depends on the constituent laws governing the evolution of the system. Further, computer simulation can provide a means to use observation to constrain underlying physical models. This talk shall examine the challenges associated with developing computational multiphysics simulation. In particular this talk will outline some of the physics, the relevant mathematical models, the associated algorithmic challenges, some of which are driven by emerging compute architectures. The problem as a whole can be formidable and an effective solution couples many disciplines together. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9553850293159485, "perplexity": 1242.1866203527245}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221213508.60/warc/CC-MAIN-20180818075745-20180818095745-00036.warc.gz"} |
http://www.ck12.org/statistics/Significance-Test-for-a-Mean/lecture/Z-Test-for-Mean/r1/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Significance Test for a Mean
## Using z-scores, p-values and known population standard deviation in order to accept or reject a null hypothesis for the mean of a distribution
0%
Progress
Practice Significance Test for a Mean
Progress
0%
Z Test for Mean
A step-by-step video showing how to set up an entire hypothesis test for the mean (hypotheses, test statistic, conclusion). | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9588789343833923, "perplexity": 3426.1178387787772}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398453576.62/warc/CC-MAIN-20151124205413-00014-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/217430-polynomial-rings-gauss-s-lemma-point-2-previous-post.html | # Math Help - Polynomial Rings - Gauss's Lemma - Point 2 of a previous post
1. ## Polynomial Rings - Gauss's Lemma - Point 2 of a previous post
Point 2 of my previous post on Gauss's Lemma got well lost in the 10 replies - so I am re-posting the remaining problem - which was problem (2)
Basically I am trying to understand the proof of Gauss's Lemma as given in Dummit and Foote Section 9.3 pages 303-304 (see attached)
On page 304, part way through the proof, D&F write:
"Assume d is not a unit (in R) and write d as a product of irreducibles in R, say $d = p_1p_2 ... p_n$ . Since $p_1$ is irreducible in R, the ideal $(p_1)$ is prime (cf Proposition 12, Section 8.3 - see attached) so by Proposition 2 above (see attached) the ideal $p_1R[x]$ is prime in R[x] and $(R/p_1R)[x]$ is an integral domain. ..."
My remaining problem with the D&F statement above are as follows:
(2) Despite reading Proposition 12 in Section 8.3 I cannot see why the ideal $p_1R[x]$ is prime in R[x] and $(R/p_1R)[x]$ is an integral domain. ...". (Indeed, I am unsure that $p_1R[x]$ is an ideal!) Can anyone show explicitly and rigorously why this is true?
Would appreciate help.
Peter
Attached Files
2. ## Re: Polynomial Rings - Gauss's Lemma - Point 2 of a previous post
Hey mate I havent had a chance to go over the post you had this morning, but i did cover this question in my answers. $p_{1}R[x]$ is the ideal $(p_{1})$ because if you take any $r \in R[x] and i \in I$ for you ideal I then $r \cdotp i \in I$.
So multiplying a whole ring by a prime element is obviously going to give you a prime ideal (think of what (2) in Z actually is). According to this (http://web.science.mq.edu.au/~chris/...olynomials.pdf) a prime polynomial is a irreducible polynomial, I am going to ask my lecturer on Tuesday also so if I get a different answer I will get back to you. I then gave you the proof for R/I is an integral domain iff I is prime in the other post.
Edit, now that I think ok it a little more it a irreducible poly must be a prime element just by the definition of what prime means. If p is prime then p=qr we must have q or r is a unit which is the definition of a irreducible polynomial. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9581968784332275, "perplexity": 385.0706844161638}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422120394037.54/warc/CC-MAIN-20150124172634-00228-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://www.oofem.org/resources/doc/matlibmanual/html/node44.html | ### Mazars damage model for concrete - MazarsModel
This isotropic damage model assumes that the stiffness degradation is isotropic, i.e., stiffness moduli corresponding to different directions decrease proportionally and independently of direction of loading. It introduces two damage parameters and that are computed from the same equivalent strain using two different damage functions and . The is identified from the uniaxial tension tests, while from compressive test. The damage parameter for general stress states is obtained as a linear combination of and : , where the coefficients and take into account the character of the stress state. The damaged stiffness tensor is expressed as . Damage evolution law is postulated in an explicit form, relating damage parameter and scalar measure of largest reached strain level in material, taking into account the principle of preserving of fracture energy . The equivalent strain, i.e., a scalar measure of the strain level is defined as norm from positive principal strains. The model description and parameters are summarized in Tab. 30.
Table 30: Mazars damage model - summary.
Description Mazars damage model for concrete Record Format MazarsModel d(rn) # E(rn) # n(rn) # e0(rn) # ac(rn) # [bc(rn) #] [beta(rn) #] at(rn) # [ bt(rn) #] [hreft(rn) #] [hrefc(rn) #] [version(in) #] [tAlpha(rn) #] [equivstraintype(in) #] [maxOmega(rn) #] Parameters - num material model number - d material density - E Young modulus - n Poisson ratio - e0 max effective strain at peak - ac,bc material parameters related to the shape of uniaxial compression curve (A sample set used by Saouridis is - beta coefficient reducing the effect of damage under response under shear. Default value set to 1.06 - at, [bt] material parameters related to the shape of uniaxial tension curve. Meaning dependent on version parameter. - hreft, hrefc reference characteristic lengths for tension and compression. The material parameters are specified for element with these characteristic lengths. The current element then will have the same COD (Crack Opening Displacement) as reference one. - version Model variant. if 0 specified, the original form of tension damage evolution law is used, if equal 1, the modified law used which asymptotically tends to zero - tAlpha thermal dilatation coefficient - equivstraintype see Tab. 23 - maxOmega limit maximum damage, use for convergency improvement Supported modes 3dMat, PlaneStress, PlaneStrain, 1dMat
Borek Patzak
2019-03-19 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9159774780273438, "perplexity": 2793.3095027000145}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202199.51/warc/CC-MAIN-20190320024206-20190320050206-00387.warc.gz"} |
http://math.stackexchange.com/questions/222625/implicit-derivative-of-1-x3-3xyy3 | # implicit derivative of $1=x^3-3xy+y^3$
I'm always having difficulties with rather complicated derivatives, simply because I always make small, stupid mistakes in the process.
Would someone be so kind to help me with the second derivative of $y$ in terms of $y$ and $x$ ?
I did the first: $$\frac{d}{dx}1 = \frac{d}{dx}x^3-\frac{d}{dx}3xy+\frac{d}{dx}y^3 = 3x^2-3y-3x\frac{dy}{dx}+\frac{dy}{dx}(3y^2)$$ $$\Rightarrow \frac{dy}{dx} = \frac{x^2-y}{x-y^2}$$
I got totally lost with the second one though. $$\frac{d^2y}{dx^2} = \frac{d}{dx}\frac{x^2}{x-y^2}-\frac{d}{dx}\frac{y}{x-y^2}=\frac{(2x)(x-y^2)-x^2(1-\frac{dy}{dx}(2y))}{(x-y^2)^2}-\frac{\frac{dy}{dx}(x-y^2)-y(1-\frac{dy}{dx}(2y))}{(x-y^2)^2}$$
And this is basically where I'm loosing it.. I did substitute $\frac{dy}{dx}$ with my first result, but it ended in huge chaos, pretty far from the result wolfram-alpha suggests.
Any help would be appreciated! Also maybe any hints & tricks to avoid such monster-equations (if possible). Because that way I always unnecessarily screw up any test / exam..
Thank you.
-
My main advice here would be not to differentiate the solution for $\mathrm dy/\mathrm dx$, but to differentiate the equation again directly and then solve for $\mathrm d^2y/\mathrm dx^2$; that way you avoid differentiating quotients.
-
Why the downvote? – joriki Oct 28 '12 at 10:54
good question.. – foaly Oct 29 '12 at 13:27
$1=x^3-3xy+y^3$
$0=3x^2-3y-3xy'+3y^2y'$
$y'(3y^2-3x)=3y-3x^2$
$y'=\frac{3(y-x^2)}{3(y^2-x)}$
$y'=\frac{y-x^2}{y^2-x}$
for second derivative from $0=3x^2-3y-3xy'+3y^2y'$ we get
$0=6x-3y'-3y'-3xy''+6yy'y'+3y^2y''$
$6y'-6x-6y(y')^2=y''(3y^2-3x)$
$y''=\frac{2(y'-x-y (y')^2)}{y^2-x}$ replacing that we find for $y'$ finally we get $y''$
-
Using the following fact the derivative of A + B is the derivative of A + the derivative of B you can split the problem into smaller parts. Instead of trying to differentiate $x^3-3xy+y^3$ solve the three problems separately:
• differentiate $x^3$ to get $3x^2$... (hint: power rule: $\frac{d}{dx}x^n = n x^{n-1}$)
• differentiate $-3xy$ to get $-3(y + x \frac{dy}{dx})$... (hint: product rule $(fg)' = f'g+fg'$)
• differentiate $y^3$ to get $3 y^2 \frac{dy}{dx}$... (hint: chain rule $f(g(x))' = g'(x) f'(g(x))$)
$$3 x^2 - 3 y - 3 x \frac{dy}{dx} + 3 y^2 \frac{dy}{dx}$$
Apply exactly the same idea again (throwing away constants this time)
• $x^2$ gives you $2x$
• $y$ gives $\frac{dy}{dx}$
• $x \frac{dy}{dx}$ gives $\frac{dy}{dx} + x \frac{d^2y}{dx^2}$
• $y^2 \frac{dy}{dx}$ gives $y^2 \frac{d^2y}{dx^2} + 2 y \left(\frac{dy}{dx}\right)^2$
and add it back together (with the right constants) to get
$$3\left(2x - \frac{dy}{dx} - \frac{dy}{dx} - x \frac{d^2y}{dx^2} + y^2 \frac{d^2y}{dx^2} + 2 y \left(\frac{dy}{dx}\right)^2\right)$$
simplify
$$3\left(2x - 2 \frac{dy}{dx} + (y^2 - x) \frac{d^2y}{dx^2} + 2 y \left(\frac{dy}{dx}\right)^2\right)$$
-
This does not address the question. – wj32 Oct 28 '12 at 10:24
implicit derivative of 1=x3−3xy+y3 – sperners lemma Oct 28 '12 at 10:25
that's the title, if you read the question you would have seen that that's pretty much what I did, and instead was struggling with the second derivative. I suppose what you actually wanted to say is I should have written 'second' in the title. I will next time. – foaly Oct 28 '12 at 10:28
@foaly, I didn the second derivative now – sperners lemma Oct 28 '12 at 10:33
+1 for balance... – wj32 Oct 28 '12 at 10:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9057259559631348, "perplexity": 654.9586273368321}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398456975.30/warc/CC-MAIN-20151124205416-00139-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://noncommutativeanalysis.com/2018/09/15/the-matrix-cube-problem-summer-project-summary-of-results/?replytocom=5428 | ### The complex matrix cube problem summer project – summary of results
In the previous post I announced the project that I was going to supervise in the Summer Projects in Mathematics week at the Technion. In this post I wish to share what we did and what we found in that week.
I had the privilege to work with two very bright students who have recently finished their undergraduate studies: Mattya Ben-Efraim (from Bar-Ilan University) and Yuval Yifrach (from the Technion). It is remarkable the amount of stuff they learned for this one week project (the basics of C*-algebras and operator spaces), and that they actually helped settle the question that I raised to them.
I learned a lot of things in this project. First, I learned that my conjecture was false! I also learned and re-learned some programming abilities, and I learned something about the subtleties and limitations of numerical experimentation (I also learned something about how to supervise an undergraduate research project, but that’s besides the point right now).
#### Statement of the problem
Following old advice of Halmos, the problem that I posed to Mattya and Yuval was in the form a yes/no question. To state this question, we need to recall some definitions. If $A$ is an $m \times m$ matrix, another $n \times n$ matrix $B$ is said to be a dilation of $A$ if
$B = \begin{pmatrix} A & * \\ * & * \end{pmatrix}$
In this case, $A$ is said to be a compression of $B$. We then write $A \prec B$. If $A = (A_1, \ldots, A_d)$ and $B = (B_1, \ldots, B_d)$ are tuple of matrices, we say that $B$ is a dilation of $A$, and that $A$ is a compression of $B$, if $A_i \prec B_i$ for all $i=1, \ldots, d$. We then write $A \prec B$.
A $d$-tuple $B = (B_1, \ldots, B_d)$ is said to be normal if $B_i$ is normal and $B_i B_j = B_j B_i$ for all $i,j=1, \ldots, d$. Normal tuples of matrices (or operators) are the best understood ones, because – thanks to the spectral theorem – they are simultaneously unitarily diagonalizable.
If $A$ is an $n \times n$ matrix, we define its norm $\|A\|$ to be the operator norm of $A$ when considered as an operator on $\mathbb{C}^n$, that is: $\|A\| = \sup\{\|Ax\| : \|x\|=1, x \in \mathbb{C}^n\}$ (here $\|x\|$ denotes the Euclidean norm $\|x\| = \sqrt{\sum_{i=1}^n |x_i|^2}$).
The complex matrix cube problem: Given a tuple $A = (A_1, \ldots, A_d)$ of $n \times n$ matrices, can one find a normal dilation $M = (M_1, \ldots, M_d)$ of $A$ such that $\|M_i\| \leq \sqrt{d} \|A_i\|$ for all $i=1, \ldots, d$
I had some reasons to believe that the answer is yes, one of which was that it was proved that the answer is yes if $A_1, \ldots, A_d$ are all selfadjoint; see this paper by Passer, Solel, and myself (I reported on this paper in this previous post). Passer later proved that if we replace $\sqrt{d}$ with $\sqrt{2d}$ then the answer is yes for arbitrary tuples. Passer’s proof did not look optimal to me. Also, I carried out some primitive numerical experimentation that seemed to verify that $\sqrt{d}$ is plausible.
#### Methods and results
Suppose we are given a $d$-tuple of $n \times n$ contractions $A = (A_1, \ldots, A_d)$. We wish to know whether it is true or false that $A$ has a normal dilation $M$ such that $\|M_i\| \leq \sqrt{d}$ for all $i=1, \ldots, d$ (this is not exactly the way we formulated the problem above, but it can be seen to be equivalent).
The first observation is that it is enough to consider only tuples of unitaries. Indeed, if $T$ is a contraction (meaning that $\|T\|\leq 1$) then
$\begin{pmatrix} T & \sqrt{I - TT^*} \\ \sqrt{I - T^* T} & -T^* \end{pmatrix}$
is a unitary dilation of $T$. So given a $d$-tuple $A$ of contractions, we can find a $d$-tuple $U$ of unitaries such that $A \prec U$. Thus, we may as well assume that $A$ is a tuple of unitaries, and ask whether we can dilate $A$.
We considered normal tuples $N = (N_1, \ldots, N_d)$ with joint eigenvalues at the vertices of the polytope $Q = P_k \times \cdots P_k$, where $P_k$ is a regular polygon with $k$ vertices that circumscribes the unit disc $\overline{\mathbb{D}}$. When $k$ is moderately large, the boundary of $Q$ is very close to $\mathbb{T} \times \cdots \times \mathbb{T}$, and in this post I will ignore this difference (the reader can check that for the results we get, ignoring this difference actually puts us on the safe side of things).
Given a fixed tuple of unitaries $A$, it can be shown that $A$ has a normal dilation $M$ with $\|M_i\| \leq \rho$ for all $i=1, \ldots, d$ if and only if
(*) $\|p(A)\| \leq \|p(\rho N)\|$
for every matrix valued polynomial $p$ of degree one $p(z) = X_0 + \sum_{j=1}^d z_j X_j$, where $N$ is the fixed normal tuple we constructed above. Let me emphasize: $M$ here is some normal dilation that we don’t know whether it exists or not, and $N$ is the fixed tuple with joint eigenvalues at the vertices of the polytope $Q = P_k \times \cdots \times P_k$ from above. We recall that a matrix valued polynomial is evaluated on a tuple $A$ of $n \times n$ matrices as follows:
$p(A) = I_n \otimes X_0 + \sum_{j=1}^d A_j \otimes X_j$.
So the first method of attack was the following: we randomly sampled a unitary tuple $A$, and then we tried to find a polynomial $p$ such that (*) was violated, with $\rho = \sqrt{d}$. We thought of several ways to look for such a polynomial given $A$, one of which was naively trying to iterate over a mesh of all possible coefficients $X_0, X_1, \ldots, X_d$. As you can easily see this method is so inefficient that even for moderately small $d$ and $n$ the search could take us a lifetime. Another idea was to try to run some numerical optimization such as gradient descent on the function $p \mapsto \|p(A)\| / \|p(\sqrt{d}N)\|$ but since this function is not convex this was also quite futile. And all this just for a given tuple $A$, which might happen to have a dilation.
The second general approach was still to randomly select a tuple of $n \times n$ unitaries $A$ and to check whether it has a normal dilation, but this time the test was somewhat more indirect. Basically, modulo some equivalences within the theory, we know that $A$ has the required dilation of size at most $\rho$, if and only if there exists a UCP map sending $\rho N_i$ to $A_i$ for $i=1,\ldots, d$, where $N$ is the tuple of normals constructed above. This, modulo some more equivalences (and as been noted in this paper of Helton, Klep and McCullough) is equivalent to the existence of positive semidefinite $n \times n$ matrices $C_1, \ldots, C_m$ such that
$\sum_{j=1}^m \rho \lambda_{i}^j C_j = A_i$ for $i=1, \ldots, d$
where $N_i = (\lambda_i^1, \lambda_i^2, \ldots, \lambda_i^m)$, for $i=1,\ldots, d$, and
$\sum_{j=1}^m C_j = I_n$.
The existence of such semidefinite matrices $C_1, \ldots, C_m$ can be interpreted as the feasibility of a certain semidefinite program (SDP). In fact, we decided to treat the full semidefinite program as follows
minimize $\rho > 0$
such that
$C_j \geq 0$, $j = 1, \ldots, m$
$\sum_{j=1}^d \lambda_i^j C_j = \rho^{-1} A_j$$j=1,\ldots, m$
$\sum_{j=1}^m C_j = I_n$.
Note that we moved $\rho$ to the right hand side, to make the equality constraint afiine in the variables $C_1, \ldots, C_m$ and $\sigma := \rho^{-1}$. Recall that $N_i = (\lambda_i^1, \ldots, \lambda_i^m)$ and $A_i$ are all fixed. In the implementation we actually defined this as a minimization problem
maximize $\sigma > 0$
such that
$C_j \geq 0$, $j = 1, \ldots, m$
$\sum_{j=1}^d \lambda_i^j C_j = \sigma A_j$$j=1,\ldots, m$
$\sum_{j=1}^m C_j = I_n$.
Now, there exists available software in Matlab that let’s one solve the above SDP quite reliably, and we used the high level interface CVX which invoked either one of the solvers SDPT3 and SeDuMi (we used both solvers and played with precision parameters to increase our confidence that the results we got were correct). This approach had the great advantage that (besides being much faster), it could tell us what is the smallest $\rho = \sigma^{-1}$ such that $A$ had a normal dilation $M$ such that $\|M_i\| \leq \rho$.
We ran the tests for small values of $n$ and $d$. You can see some histograms in the presentation (the value plotted in the histograms in the presentation is $\rho/\sqrt{d}$, in order to have a direct comparison with the conjecture). Interestingly, we see that with very high probability, the required value of $\rho$ is on average significantly lower than $\sqrt{d}$. For $d = 2$ and $n=3,4,5$, we found a few random counter examples, but they required $\rho$ that was just 2% over $\sqrt{d}$.
Once we know that the average value of $\rho$ is less than $\sqrt{d}$, it heuristically becomes reasonable that counter examples are hard to come by, because of concentration of measure phenomena: roughly speaking, the probability of a Lipschitz function on the unitaries (say) to be $\epsilon$ away from the mean goes down exponentially like $C_1 \exp(-C_2 n \epsilon)$ with the dimension. For the same reason, once we found a counter example $A$, it is very hard to find coefficients $X_0, \ldots, X_1, \ldots, X_n$ of a matrix valued polynomial $p(z) = X_0 + \sum_{j=1}^d z_j X_j$ such that $\|p(A)\| > \|p(\sqrt{d} N)\|$. And indeed, we did not yet verify by an independent method that our counter examples are indeed counter examples.
The counter examples we found are very unlikely to be caused by numerical error, since we tested the result with a couple of solvers and also the advertised precision of the solvers is several orders of magnitude less than 2%.
After we found the random counter examples, it occurred to us that there was no reason to sample the $d$ unitaries $A_1, \ldots, A_d$ independently. We recalled that in the selfadjoint case, tightness of the constant $\sqrt{d}$ was established using anti-commuting unitaries. Indeed, since counter examples are rare, one would think that the matrices would have to conspire somehow in order to mess up the inequality. So we searched for things that are anti-commuting-like. And it did indeed turn out that the $q$ commuting matrices
$A_1 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & q & 0 \\ 0 & 0 & q^2 \end{pmatrix}$ $A_2 = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}$
where $q = \exp(\frac{2 \pi i}{3})$ are also a counter example (in the case $d = 2, n= 3$). We also still haven’t found a polynomial for which $\|p(A)\| > \|p(\sqrt{2} N)\|$. We will probably continue looking for one when the holiday is over, and then I will update.
#### Materials and summary
Here are Mattya and Yuval’s slides which they presented in the talk they gave at the end of the week. I also plan to put the code and files with raw results online on my homepage at some point.
#### Acknowledgements and credits
The main method for checking what is the “inflation constant” required for a dilation, using a semidefinite program, is based on basic operator space theory, and in particular draws upon the algorithm described in this paper of Helton, Klep and McCullough.
We used Matlab. The numerical heavy lifting was done by others. We solved the semidefinite program using CVX – a high level Matlab software package for specifying and solving convex programs. We also used YALMIP – another high level Matlab software package for specifying and solving convex programs – to verify the results we obtained with CVX. Both CVX and YALMIP invoked SDP solvers SDPT3 and SeDuMi.
This project came after several years of collaboration with colleagues, and in particular, I had many conversations on the subject with Ben Passer before and during the projects week.
I owe many thanks to the organizers of this projects week, Ram Band, Tali Pinsky, and Ron Rosenthal. Thanks to this opportunity I explored an avenue that I never walked through before. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 154, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9061572551727295, "perplexity": 304.97198060392145}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141681524.75/warc/CC-MAIN-20201201200611-20201201230611-00111.warc.gz"} |
https://arxiv.org/abs/1506.06101 | stat.ME
(what is this?)
Title: Robust Bayesian inference via coarsening
Abstract: The standard approach to Bayesian inference is based on the assumption that the distribution of the data belongs to the chosen model class. However, even a small violation of this assumption can have a large impact on the outcome of a Bayesian procedure. We introduce a simple, coherent approach to Bayesian inference that improves robustness to perturbations from the model: rather than condition on the data exactly, one conditions on a neighborhood of the empirical distribution. When using neighborhoods based on relative entropy estimates, the resulting "coarsened" posterior can be approximated by simply tempering the likelihood---that is, by raising it to a fractional power---thus, inference is often easily implemented with standard methods, and one can even obtain analytical solutions when using conjugate priors. Some theoretical properties are derived, and we illustrate the approach with real and simulated data, using mixture models, autoregressive models of unknown order, and variable selection in linear regression.
Subjects: Methodology (stat.ME) Cite as: arXiv:1506.06101 [stat.ME] (or arXiv:1506.06101v1 [stat.ME] for this version)
Submission history
From: Jeffrey Miller [view email]
[v1] Fri, 19 Jun 2015 18:02:48 GMT (1880kb,D) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8837730288505554, "perplexity": 1035.8795674934736}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583513548.72/warc/CC-MAIN-20181021010654-20181021032154-00442.warc.gz"} |
https://www.esaral.com/q/a-compound-microscope-consists-of-an-objective-lens-of-focal-length-2-0-cm-39748 | Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!
# A compound microscope consists of an objective lens of focal length 2.0 cm
Question:
A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Solution:
Focal length of the objective lens, f1 = 2.0 cm
Focal length of the eyepiece, f2 = 6.25 cm
Distance between the objective lens and the eyepiece, d = 15 cm
(a) Least distance of distinct vision, $d^{\prime}=25 \mathrm{~cm}$
Image distance for the eyepiece, v2 = −25 cm
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$
$\frac{1}{u_{2}}=\frac{1}{v_{2}}-\frac{1}{f_{2}}$
$=\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$
$\therefore u_{2}=-5 \mathrm{~cm}$
Image distance for the objective lens, $v_{1}=d+u_{2}=15-5=10 \mathrm{~cm}$
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$
$\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}$
$=\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}$
$\therefore u_{1}=-2.5 \mathrm{~cm}$
Magnitude of the object distance, $\left|u_{1}\right|=2.5 \mathrm{~cm}$
The magnifying power of a compound microscope is given by the relation:
$m=\frac{v_{1}}{\left|u_{1}\right|}\left(1+\frac{d^{\prime}}{f_{2}}\right)$
$=\frac{10}{2.5}\left(1+\frac{25}{6.25}\right)=4(1+4)=20$
Hence, the magnifying power of the microscope is 20.
(b) The final image is formed at infinity.
$\therefore$ Image distance for the eyepiece, $v_{2}=\infty$
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$
$\frac{1}{\infty}-\frac{1}{u_{2}}=\frac{1}{6.25}$
$\therefore u_{2}=-6.25 \mathrm{~cm}$
Image distance for the objective lens, $v_{1}=d+u_{2}=15-6.25=8.75 \mathrm{~cm}$
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$
$\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}$
$=\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}$
$\therefore u_{1}=-\frac{17.5}{6.75}=-2.59 \mathrm{~cm}$
Magnitude of the object distance, $\left|u_{1}\right|=2.59 \mathrm{~cm}$
The magnifying power of a compound microscope is given by the relation:
$m=\frac{v_{1}}{\left|u_{1}\right|}\left(\frac{d^{\prime}}{\left|u_{2}\right|}\right)$
$=\frac{8.75}{2.59} \times \frac{25}{6.25}=13.51$
Hence, the magnifying power of the microscope is 13.51. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8053688406944275, "perplexity": 893.3999689756783}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499826.71/warc/CC-MAIN-20230130165437-20230130195437-00785.warc.gz"} |
http://math.stackexchange.com/questions/82248/graph-theory-network-flow-bipartite-graph-realized | # Graph Theory - Network Flow & Bipartite graph (realized)
This is a question from the Graph Theory (Bondy & Murty) text which has me rather stumped.
Let $p = (p_1, p_2, . . . , p_m)$ and $q = (q_1, q_2, . . . , q_n)$ be two sequences of nonnegative integers. The pair $(p, q)$ is said to be realizable by a simple bipartite graph if there exists a simple bipartite graph G and a bipartition $(\{x_1, x_2, . . . , x_m\}, \{y_1, y_2, . . . , y_n\})$ such that $\deg(x_i) = p_i$ for $1 ≤ i ≤ m$, and $\deg(y_j) = q_j$ for $1 ≤ j ≤ n.$
(a) Formulate as a network flow problem the problem of determining whether a given pair $(p, q)$ is realizable by a simple bipartite graph.
(b) Suppose that $q_1 ≥ q_2 ≥ · · · ≥ q_n$. Deduce from the Max-Flow Min-Cut Theorem that $(p, q)$ is realizable by a simple bipartite graph if and only if $\sum\limits_{i=1}^m p_i = \sum\limits_{j=1}^n q_j$ and $\sum\limits_{i=1}^m \min(p_i, k) \geq \sum\limits_{j=1}^k q_j$ for $1 \leq k \leq n$
For (a), we view each $x_i$ as a source and each $y_j$ as a sink and connect as follows (edge $(x_i,y_j)$ as allowed from $p$ and $q$ with capacity = 1 for all edges). Not 100% sure how to go about with the flows but I believe we can claim that it is realized if the constructed network has a feasible flow.
I am rather unsure how to go about doing (b) from the max-flow min-cut theorem (apologies for my lack of progress).
Edit: upper bound of sigma of $q_j$ in inequality of part b to k, not n.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9345481991767883, "perplexity": 174.15604799100333}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394010742343/warc/CC-MAIN-20140305091222-00068-ip-10-183-142-35.ec2.internal.warc.gz"} |
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