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http://math.stackexchange.com/questions/247790/verifying-the-structure-of-a-field
|
# Verifying the structure of a field
Let F be a field and $G=F\times F$ Define addition by $(a,b)+(c,d)=(a+c,b+d)$ and multiplication by $(a,b)\cdot(c,d)=(ac,bd)$
Does these operations define a field on G?
I'm fairly comfortable with the addition part, however its the multiplication part that trips me up. Surely $(ac)^{-1}$ exists since F is a field, and $\frac{1}{a}, \frac{1}{c}$ are each in F since we assumed that F is a field.
-
Watch out for zero divisors! – Adam Saltz Nov 30 '12 at 3:40
$(a,c)^{-1}$ does not exist when either $a=0$ or $b=0$. – Lior B-S Nov 30 '12 at 7:24
Check what happens with $\,(1,0)\cdot (0,1)$ ...
In a field any non-zero elements has a multiplicative inverse, which is far from being true in a domain (e.g., the integers $\,\Bbb Z\,$ or any polynomial ring over a field) – DonAntonio Nov 30 '12 at 3:55
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2015-09-04 01:43:07
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https://chemistry.stackexchange.com/questions/141533/reference-request-bond-lengths-in-pcl2f3
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# Reference Request - bond lengths in PCl2F3
While writing an answer here, I needed references for bond lengths of $$\ce{PF3Cl2}$$ and $$\ce{PF4Cl}$$. The latter was easily found in reference (1), whereas I could not find exact references for bond lengths in the former molecule.
References (2) and (3) contain information about the Raman spectra and $$\ce{F^{19}}$$ NMR spectra, but do not contain information about bond lengths. Reference (2) has information about the bond lengths of $$\ce{PCl3F2}$$, $$\ce{PF5}$$ and $$\ce{PCl5}$$, but not for $$\ce{PF3Cl2}$$.
A reference for exact bond lengths in $$\ce{PF3Cl2}$$, similar to the bond lengths shown above, is appreciated.
References:
1. Leiding, J., Woon, D. E., & Dunning, T. H. (2013). "Bonding in $$\ce{PF2Cl}$$, $$\ce{PF3Cl}$$ and $$\ce{PF4Cl}$$: insight into isomerism and apicophilicity from ab initio calculations and the recoupled pair bonding model". Theoretical Chemistry Accounts, 133(2). doi:10.1007/s00214-013-1428-7
2. Griffiths, J. E., Carter, R. P., & Holmes, R. R. (1995). "Molecular Structures of $$\ce{PCl4F}$$, $$\ce{PCl3F2}$$, $$\ce{PCl2F3}$$, and $$\ce{PF5}$$: Infrared and low-temperature Raman Vibrational Spectra". Phosphorus, Sulfur, and Silicon and the Related Elements, 98(1-4), 11–31. doi:10.1080/10426509508036938
3. Holmes, R. R., Carter, R. P., & Peterson, G. E. (1964). "Molecular Structures of $$\ce{PCl4F}$$, $$\ce{PCl3F2}$$, and $$\ce{PCl2F3}$$: Pure Chlorine Nuclear Quadrupole Resonance and Low Temperature $$\ce{F^{19}}$$ Nuclear Magnetic Resonance Spectra". Inorganic Chemistry, 3(12), 1748–1754. doi:10.1021/ic50022a021
## 1 Answer
To complete your literature survey, include this publication cited at least 11 times so far: «Dichlorotrifluorophosphorane (PCl2F3): molecular structure by gas-phase electron diffraction and quadratic force field» by French et al. in Inorg. Chem. 1985, 24, 2774–2777, doi 10.1021/ic00212a014. Despite the paywall, ACS's «In lieu of an abstract, this is the article's first page.» informs you about some key properties you search, values including ($$2\sigma$$ estimated) e.s.d.
How did I find this publication? Because the English Wikipedia has an entry about it (don't overlook the Chinese including complementary information).
But how to know if Wikipedia indexed such a compound? Way better than trial-and-error is to address this question to the Wikipedia Chemical Structure Explorer allowing to query there by structure or string, which may lead you to the corresponding article (in Wikipedia). It is backed by a public database (GitHub repository) initiated by Ertl et al.
Source: Ertl, P.; Patiny, L.; Sander, T.; Rufener, C.; Zasso M J. Cheminf., 2015, 7, 10, doi 10.1186/s13321-015-0061-y, (open access).
• Great answer, time to bookmark the chemical structure expolorer page :) – Aniruddha Deb Oct 16 '20 at 5:13
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2021-07-25 22:31:30
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https://meridian.allenpress.com/radiation-research/article-abstract/138/1s/S72/40014/The-Survival-of-Asynchronous-V79-Cells-at-Low?redirectedFrom=PDF
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We have observed that when a single linear-quadratic (LQ) function is used to fit the radiation survival response of an asynchronously dividing population of V79 cells, a consistent misfit occurs at low doses. The data can be better described by fitting the low-dose and high-dose ranges separately, and there is evidence of a two-component response. The most obvious explanation is that we may simply be seeing the response of subpopulations of cells of different radiosensitivity: sensitive${\rm G}_{1}\text{-}$,${\rm G}_{2}\text{-}$ and M-phase cells and resistant S-phase cells. The cell sorting assay for cell survival which we have used in these studies may thus be providing sufficient accuracy to resolve these subpopulations, not previously seen in conventional survival measurements. An alternative explanation is that the linear-quadratic function may be inappropriate for accurate description of the radiation survival response at low dose, at least for these cells. To test this hypothesis we have used three other models to fit the data: the single-hit plus multi-target (SHMT) model and the two-parameter repair-misrepair (RMR) model both yielded inferior fits to the asynchronous survival data; the three-parameter RMR model provided an improved fit to the data. The best fit, however, was obtained using a two-population LQ model, which suggested approximately equal numbers of sensitive and resistant cells. When the survival response of tightly synchronized G1/ S-phase cells was measured using the cell sorting assay, no substructure was observed. This offers strong support to the hypothesis that the substructure observed in the asynchronous survival response is due to subpopulations of cells of different, cycle-dependent radiosensitivity.
This content is only available as a PDF.
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2021-05-10 02:05:18
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https://imagej.github.io/OpenMPI_Plugin_Extensions
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# OpenMPI Plugin Extensions
Name OpenMPI Plugin Extensions Software Fiji Author Daniel Trnka, Michal Krumnikl Maintainer Michal Krumnikl Source on github Category TBD
## General information
### Motivation
OpenMPI, despite being relatively old, still remains the most dominant programming model used in high-performance computing (HPC). As of today Fiji supports the GPU parallelization through CLIJ and allows executing automated HPC workflows by means of Automated_workflow_for_parallel_Multiview_Reconstruction. However, there are still no genuine internally parallel plugins developed specifically for deployment on large scale parallel machines like HPC clusters or supercomputers. This is a preliminary OpenMPI framework for Fiji and model parallel implementations of the most common image processing operations included in ImageJ.
### Extensions Description
This presents a set of Ops adapted to herein presented OpenMPI wrapper, jointly comprising a solution designed to be seamlessly used in any existing ImageJ2 code. Moreover, framework automatically initializes and disposes connections to the OpenMPI environment, making available OpenMPI implementations for the most frequent operations covered by Ops. The package contains example implementations for fundamental image processing functions as well as basic math, logical and statistics operations.
TBD
## How to use
In order to demonstrate convenience of this approach, we prepared an example of parallelized convolution. We used Python for readers’ convenience, however plugins can be invoked from any other scripting language supported by Fiji.
kernel = ops.run("create.img", [3, 3], FloatType())
for p in kernel:
p.set(1.0/kernel.size())
def fn():
output = ops.create().img(input)
ops.filter().convolve(output, Views.extendMirrorSingle(input), kernel)
return output
input = scifio.datasetIO().open(input_path)
output = Measure.benchmark(fn, rounds)
scifio.datasetIO().save(output, output_path)
The Listing shows the OpenMPI implementation with initialization of a 3×3 blur kernel (lines 1-3), definition of a processing function with convolution (it reuses already existing implementation of a convolution filter in ImageJ2) on lines 5-8 and finally its execution in the benchmark mode to measure execution times (the variable rounds defines the number of repetitions).
Noticeably, using the OpenMPI extension does not add any extra layer as long as data are available in a form of IterableInterval.
### Debugging
Debugging printouts on the console shows the node allocation and dataset division. The following listing shows the dataset divided to 6 nodes.
Fri Jun 19 08:36:49 2020[1,2]<stdout>:Chunk{data=test.tif, offset=13334, len=6667, chunks=1}
Fri Jun 19 08:36:49 2020[1,0]<stdout>:Chunk{data=test.tif, offset=0, len=6667, chunks=1}
Fri Jun 19 08:36:49 2020[1,5]<stdout>:Chunk{data=test.tif, offset=33335, len=6665, chunks=1}
Fri Jun 19 08:36:49 2020[1,1]<stdout>:Chunk{data=test.tif, offset=6667, len=6667, chunks=1}
Fri Jun 19 08:36:49 2020[1,4]<stdout>:Chunk{data=test.tif, offset=26668, len=6667, chunks=1}
Fri Jun 19 08:36:49 2020[1,3]<stdout>:Chunk{data=test.tif, offset=20001, len=6667, chunks=1}
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2020-10-01 02:01:44
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https://cs.stackexchange.com/questions/88532/algebraic-spectral-algorithms-for-the-minimum-spanning-tree-problem
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# Algebraic (spectral) algorithms for the minimum spanning tree problem
Are there any algorithms that use the spectral properties of a graph to solve the minimum spanning tree problem?
To clarify further what I have in mind, starting with the Laplacian matrix I want to algebraically arrive at the required set of edges. This is opposed to Prim's or Kruskal's algorithm where the result is obtained by operating on the edges/vertices sets directly.
In short: no, not that I know of and I would be surprised if they exist. For something longer, read on:
I don't think the spectrum of a graph can be of much use in finding an MST. The main reason is this: weights of a graph aren't part of the spectrum of a graph! An MST only makes sense on a weighted graph.
The only thing I see we could try, if all weights are integral, 'unfold' the graph such that all weights are unit-weights (so an edge of weight $32$ becomes $32$ edges and $31$ nodes in them) and mark the original nodes and make a min-cost tree that spans the original edges. After this transformation, the spectrum might contain all relevant info, but this transformation step makes our new graph simply horribly large, so any algorithm based on this will have an awful running time. This is simply a bad idea.
I've never heard of a useful way to 'attach' weights to edges in the adjacency matrix and connect them to the spectral properties, either. It doesn't seem likely to work, although proving a vague method won't work is of course impossible.
Of course, the fact that spectral properties likely aren't useful doesn't mean algebraic methods in general won't work. What is important is that you don't leave the edge weights behind. However, most other algebraic methods are used for exponential time algorithms and I'm not aware anyone has constructed something algebraic for MST.
• The Laplacian is easily extended to include edge weights: you define $L = D - W$, where $W$ is the weighted adjacency matrix and $D$ is the weighted diagonal degree matrix. – Zach Langley Dec 11 '18 at 21:06
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2019-12-11 18:12:12
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https://www.physicsforums.com/threads/trouble-with-tangent-of-a-line.139009/
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# Homework Help: Trouble with Tangent of a Line
1. Oct 18, 2006
### thomasrules
I tried many thigns with this one:
Determine the equation of the line that is perpendicular to the tangent to $$y=5x^2 at (1,5)$$
What I did was use the formula for instantaneous rate of change finding the equation of the secant.
$$(5x^2-5)/(x-1)$$
from there I found that the slope of the tangent is about 10 and from there i'm lost
2. Oct 18, 2006
Maybe it would be easier to differentiate the function and plug in the x-coordinate to find the slope of the tangent.
3. Oct 18, 2006
### thomasrules
but I have the slope....
4. Oct 18, 2006
Ok then. In what relation are the slope of a line with the slope of a line perpendicular to it?
5. Oct 18, 2006
### thomasrules
duh negative reciprocal
6. Oct 18, 2006
So, all you have to do now is find the equation of the line which passes through the given point and whose slope is, as you stated, negative reciprocal to the slope of the tangent at that very same point.
7. Oct 18, 2006
### thomasrules
answer is x+10y-51=0 but whres the 51 from
8. Oct 18, 2006
The equation of the line is $$y - y_{1} = -\frac{1}{10}(x-x_{1})$$, where $$(x_{1}, y_{1})=(1, 5)$$. Plug in the values and you should get the equation. (And see where 51 comes from.)
9. Oct 18, 2006
### thomasrules
ur not suggesting y-y1=m(x-x1)
??
10. Oct 18, 2006
### thomasrules
thats what i did one sec
11. Oct 18, 2006
### thomasrules
omg I'm so smart because I got to that last step u said but didnt see that 1/10 +5 is 5.1.......
but thats why i'm so stupid too i didnt see it
12. Oct 18, 2006
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2018-10-21 22:43:56
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https://dsp.stackexchange.com/questions/35185/two-way-audio-crossover-matlab-code-using-bessel
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# Two way Audio crossover MATLAB code using Bessel
Could MATLAB's besself function be used in designing an Audio crossover (Tweeter and Woofer) with highpass and lowpass filters?
Most likely the same with this code but using Bessel filter.
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2020-08-11 21:48:44
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https://math.stackexchange.com/questions/1846835/evaluate-int-0-infty-fracdxx22axb
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# Evaluate $\int_0^\infty \frac{dx}{x^2+2ax+b}$
For $a^2<b$, is there an identity of evaluating the following integral?
$$\int_0^\infty \frac{dx}{x^2+2ax+b}$$
$$\int_0^\infty \frac{dx}{(x^2+2ax+b)^2}$$
My attempt is using partial fractions and completing the square, but I still failed to obtain a nice result.
• Note that your first one is trivial and finite for $a=b=+1$, but quite different if $-a=b=1$. – almagest Jul 2 '16 at 14:24
• Note: making change of variables $x-c$ we see that your first problem is the same as finding an indefinite integral of $1/(x^2+2ax+b)$. But note that doing the integral $\int_{-\infty}^\infty$ will give you a simpler answer. – GEdgar Jul 2 '16 at 19:58
Let the first integral be $I$ and the second one be $J$, then by putting $x=y-a$ and $y=z\sqrt{b-a^2}$ we have \begin{align} I(a,b)&=\int_0^{\infty}\frac{dx}{(x+a)^2+b-a^2}\\[10pt] &= \int_a^{\infty}\frac{dy}{y^2+b-a^2}\\[10pt] &=\frac{1}{\sqrt{b-a^2}}\int_{\large\frac{a}{\sqrt{b-a^2}}}^{\infty}\frac{dz}{z^2+1}\\[10pt] &=\frac{1}{\sqrt{b-a^2}}\left(\frac{\pi}{2}-\arctan\left(\frac{a}{\sqrt{b-a^2}}\right)\!\right)\\[15pt] \end{align} and the 2nd integral is just a first derivative of $I$ with respect to $b$ times $-1$ $$\\[15pt]J(a,b)=-\frac{\partial I}{\partial b}=\frac{\partial }{\partial b}\left[\frac{1}{\sqrt{b-a^2}}\left(\arctan\left(\frac{a}{\sqrt{b-a^2}}\right)-\frac{\pi}{2}\right)\right]\\[10pt]$$ Can you take it from here?
• I think it is: $J(a,b)=-\frac{\partial I}{\partial b}$ – YY Lam Jul 2 '16 at 14:52
• @YYLam Yes, you're correct. Edited. – Sophie Agnesi Jul 2 '16 at 15:00
• Thank you very much. Now I obtain: $\frac{a}{2 b \left(a^2-b\right)}+\frac{\pi }{4 \left(b-a^2\right)^{3/2}}-\frac{\tan ^{-1}\left(\frac{a}{\sqrt{b-a^2}}\right)}{2 \left(b-a^2\right)^{3/2}}$ for $J(a,b)$ – YY Lam Jul 2 '16 at 15:05
For the first one:
$$(x+a)^2=x^2+2ax+a^2$$
So
$$(x+a)^2+b-a^2=x^2+2ax+b$$
$$=(b-a^2)\left(\frac{(x+a)^2}{b-a^2}+1 \right)$$
$$=(b-a^2) \left( \left( \frac{x+a}{\sqrt{b-a^2}} \right)^2+1 \right)$$
For $b-a^2 \neq 0$. In which case, for evaluating the integral, we enforce the substitution $\tan {u}=\frac{x+a}{\sqrt{b-a^2}}$ and proceed from there. Or the substitution $t=\frac{x+a}{\sqrt{b-a^2}}$, $\sqrt{b-a^2} dt=dx$:
$$\int_{0}^{\infty} \frac{1}{(b-a^2) \left( \left( \frac{x+a}{\sqrt{b-a^2}} \right)^2+1 \right)} dx$$
$$=\frac{\sqrt{b-a^2}}{b-a^2} \int_{\frac{a}{\sqrt{b-a^2}}}^{\infty} \frac{1}{t^2+1} dt$$
$$=\frac{\sqrt{b-a^2}}{b-a^2}\left(\frac{\pi}{2}-\arctan (\frac{a}{\sqrt{b-a^2}}) \right)$$
$$=\frac{\text{arccot} (\frac{a}{\sqrt{b-a^2}})}{\sqrt{b-a^2}}$$
The second integral can be found by the method already mentioned, i.e. finding:
$$\frac{\partial}{\partial b} \left(-\frac{\text{arccot} (\frac{a}{\sqrt{b-a^2}})}{\sqrt{b-a^2}} \right)$$
In fact if we take:
$$\begin{equation*} I_n = \int^{\infty}_{0}\frac{1}{(x^2 + 2ax + b)^n} \mathrm{d}x\end{equation*}$$
Like zain did,
Then for $n \geq 2$:
$$I_n=\frac{\partial^{n-1}}{\partial b} \left(\frac{(-1)^{n-1}}{(n-1)!}\frac{\text{arccot} (\frac{a}{\sqrt{b-a^2}})}{\sqrt{b-a^2}} \right)$$
Because:
$$\frac{\partial^{n}}{\partial x} \frac{1}{x+c}=\frac{(-1)^nn!}{(x+c)^{n+1}}$$
Thus:
$$I_n(a,b)=\frac{(-1)^{n-1}}{(n-1)!}\frac{\partial^{n-1}}{\partial b_0} \left(\frac{\text{arccot} (\frac{a_0}{\sqrt{b_0-a_0^2}})}{\sqrt{b_0-a_0^2}} \right) \biggr \rvert_{(a,b)}$$
|
2019-06-16 17:10:08
|
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https://wizedu.com/questions/63366/a-study-was-conducted-to-determine-if-the
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##### Question
In: Statistics and Probability
# A study was conducted to determine if the salaries of elementary school teachers from two neighboring...
A study was conducted to determine if the salaries of elementary school teachers from two neighboring districts were equal. A sample of 15 teachers from each district was randomly selected. The mean from the first district was $28,900 with a standard deviation of$2300. The mean from the second district was $30,300 with a standard deviation of$2100. Assume the samples are random, independent, and come from populations that are normally distributed. Construct a 95% confidence interval for μ1 - μ2. Assume that two populations' variance are the same (σ21= σ22).
## Solutions
##### Expert Solution
For further queries, please comment below.
Thank you.
## Related Solutions
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A study was conducted to determine if the salaries of librarians from two neighboring cities were equal. A sample of 15 librarians from each city was randomly selected. The mean from the first city was $28,900 with a standard deviation of$2300. The mean from the second city was $30,300 with a standard deviation of$2100. Construct a 95% confidence interval for mu 1 - mu 2.
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##### 2)A study was conducted to determine if the salaries of librarians from two neighboring cities were...
2)A study was conducted to determine if the salaries of librarians from two neighboring cities were equal. A sample of 15 librarians from each city was randomly selected. The mean from the first city was $28,900 with a standard deviation of$2300. The mean from the second city was $30,300 with a standard deviation of$2100.Construct a 95% confidence interval for μ1-μ2. A) Hypothises : B) Critical value (t-critical) : C) Test statistic t-stat and the decision about the test...
##### Salaries for teachers in a particular elementary school district are normally distributed with a mean of...
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##### A researcher claims that the mean of the salaries of elementary school teachers is greater than...
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2023-03-25 00:42:17
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|
http://recasarfati.com/tags/reinforcement-learning/
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# Reinforcement Learning
## Correlated Equilibria in Extensive Form Games
Python package to represent extensive form games and compute various correlated equilibrium concepts via reinforcement learning algorithms.
## Learning Correlated Equilibria in Extensive Form Games
We formalize an efficient class of counterfactual regret minimization algorithms exploiting the “sequence form” to compute ''$\Phi$-equilibria'' – a generalized class of equilibrium concepts defined within general-sum extensive form games of …
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2020-10-01 16:41:58
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https://www.gamedev.net/forums/topic/307387-functions-and-a-pointer-to-a-vector-of-vectors/
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# Functions and a pointer to a vector of vectors...?
This topic is 5054 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
I have a vectors of vectors of bytes, and want to pass it to a function so the function can change the values in it. The code (simplified):
// The vector of vectors
vector< vector<char> > heightMap;
// The function
void myFunction (char** heights) // Is this how it should look?
{
for (int i=0; i < heights.size(); i++) { // This isn't right. I want the number of first-order vectors...
for (int x=0; x < heights.size(); x++ { // And the size of the second-order vector...
heights.push_back(1); // How do I push_back a value?
}
}
}
// Calling the function
myFunction (heightMap); // How do I properly call myFunction passing heightMap as the paramater?
So how do I fix that code up? Thanks for the help!
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I like pass by reference. (note the '&' in the function definition). You really want to do that anyway (generally) when working with vectors or lists, because just passing in the vector normally will ENTIRELY copy the whole damn thing which can take a long time if the vector/list is long.
void myFunction( vector< vector<char> > & heightMap ){}myFunction( heightMap );
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Thanks! That works great, except for one small part. If I try to do something like:
for (int i=0; i < heights.size(); i++) { for (int x=0; x < heights.size(); x++ { heights.push_back(1); // <--- Error occurs on this line. }}
My compiler complains "invalid conversion from BYTE**' to char'". How would I fix this?
##### Share on other sites
Show your whole code for this function (forward declaration and actual definition), as well as for calling the function. "BYTE**" or "char**" probably shouldn't show up at all anywhere in your code, relating to this particular object at least. You probably have some variable or parameter somewhere defined as BYTE** and merely forgot to change it.
nevermind[/edit] On closer examination, (I didn't notice your "// <--- Error occurs on this line." comment), as opposed to "push_back(1)", I'm guessing you are trying to push_back a variable, or a more complex expression of some sort. Make sure that the variable or expression evaluates to a char. If it's coming from a 2-dimensional array, make sure that you're accessing the actual elements (push_back(array[i][x])), not just the array (push_back(array)). Or maybe you just need to dereference the double-pointer (push_back(**var)). [/edit]
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>> See post below <<
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Ack, actually something worse is happening now :(
void LoadFile(LPSTR strName, vector< vector<char> > &pHeightMap, int MAP_SIZE){ char temp[1]; FILE *pFile=NULL; pFile=fopen(strName,"rb"); for (int i=0;i<MAP_SIZE;i++) { for (int x=0;x<MAP_SIZE;x++) { fread(temp,1,1,pFile); pHeightMap.push_back(temp[0]); // <--- Line that creates error } }}
When that is run, Windows generates a program error "prog.exe has generated errors and will be closed by Windows. Try restarting the program." and the program is forced to close. Why is this happening? (I know the line that creates the error just because if I comment it out the program works.) And the file read code is very ugly... reading one byte at a time... is there any way to read multiple bytes with vectors?
Thanks!
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Quote:
Original post by NietsnieThanks! That works great, except for one small part. If I try to do something like: for (int i=0; i < heights.size(); i++) { for (int x=0; x < heights.size(); x++ { heights.push_back(1); // <--- Error occurs on this line. }}My compiler complains "invalid conversion from BYTE**' to char'". How would I fix this?
Warning! There is a definite logic error here too. The .size() is evaluated each time through the loop, and .push_back() onto the vector will increment its size. Thus this becomes an infinite loop. I assume you are trying to modify each "column" within each "row" as you look at them; use heights[i][x] = <whatever> instead.
(If this function is to set up the contents of an empty vector of vectors, you will want to use push_back, but you'll have to get the dimensions from somewhere else.)
Also, note that a vector of vectors is not necessarily "rectangular" storage; the "rows" may all be different lengths.
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Ack, thanks for catching that.
EDIT: Actually sorry, that doesn't fix it. I put in bad code in my first posts, and the real code (sans logic error) is in my last post. And I can't see any error in it... >.<
##### Share on other sites
If you're passing an empty vector (of vectors of char) into that function, it's going to explode because pHeightMap[i] may not exist.
void LoadFile(LPSTR strName, vector< vector<char> > &HeightMap, int MAP_SIZE){ for (int i = 0; i < MAP_SIZE; i++) { vector<char> v; HeightMap.push_back(v); for (int x = 0; x < MAP_SIZE; x++) { // acquire data HeightMap[i].push_back(data]); } }}
The bold lines at the outer for loop should solve your problem.
##### Share on other sites
For one, your outer vector never gets resized. So when you access pHeightMap, when i is zero, there isn't actually an element 0, and you access a bad pointer. Here's how I'd quickly rewrite it:
void LoadFile(LPSTR strName, vector< vector<char> > &HeightMap, int MAP_SIZE){ char temp[1]; FILE *pFile=NULL; pFile=fopen(strName,"rb"); HeightMap.resize(MAP_SIZE); for (int i=0;i<MAP_SIZE;i++) { HeightMap.resize(MAP_SIZE); for (int x=0;x<MAP_SIZE;x++) { fread(temp,1,1,pFile); HeightMap[x] = temp[0]; } }}
Or to make it read more than one byte at a time:
void LoadFile(LPSTR strName, vector< vector<char> > &HeightMap, int MAP_SIZE){ char temp[1]; FILE *pFile=NULL; pFile=fopen(strName,"rb"); HeightMap.resize(MAP_SIZE); for (int i=0;i<MAP_SIZE;i++) { HeightMap.resize(MAP_SIZE); fread(&(HeightMap[0]),sizeof(char),HeightMap.size(),pFile); }}
Or I'd even consider just using a 1-dimensional vector, and do the 2-dimensional calculations manually, so that it would all be in one single chunk. (Right now you have what amounts to a jagged array, which isn't bad necessarily, but it's nice to be aware of the difference between a jagged array and a rectangular array.)
void LoadFile(LPSTR strName, vector< char > &HeightMap, int MAP_SIZE){ char temp[1]; FILE *pFile=NULL; pFile=fopen(strName,"rb"); HeightMap.resize(MAP_SIZE*MAP_SIZE); fread(&(HeightMap[0]),sizeof(char),HeightMap.size(),pFile);}
One last thing, notice I renamed pHeightMap to HeightMap. It's no longer a pointer, and it's somewhat misleading to prepend 'p' onto a variable that isn't; it is a reference, true, but it's still accessed like a normal variable, not a pointer, so it'd probably be better to leave the 'p' out.
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2019-01-17 02:45:03
|
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|
http://www.reddit.com/r/math/comments/1y2t0j/bernoulli_polynomial_generalization/
|
[–]Number Theory 1 point2 points
Your approach is clever, but I think it's easier than you're making it. If you just do a change of variables v=m(u-x)+x, then we can set
[; B_n^m(v;x) = \frac{1}{m}B_n\left(\frac{v-x}{m}+x\right) ;]
and the substitution becomes
[; \int_x^{x+1}B_n(u)du = \int_x^{x+m} B_n^m(v;x) dv = x^n ;]
EDIT: I suppose the new polynomials depend on x, which is not very good
[–][S] 0 points1 point
I have never thought about going at it with substitution. This is great! It's true that the polynomials shouldn't depend on x, but maybe there's another substitution that gets rid of it.
[–]Number Theory 1 point2 points
I'm skeptical that a correct substitution will work. However, if you break up the integral from x to x+m of Bn(u) into (x,x+1), (x+1,x+2),...,(x+m-1,x+m) after integrating these piece by piece you get the sum from k=0 to m-1 of (x+k)n . You can then write this as a degree n polynomial where the ith coefficient is An,i,m:
[; \int_{x}^{x+m} B_n(u)du = \sum_{k=0}^{m-1}\int_{x+k}^{x+k+1}B_n(u)du = \sum_{k=0}^{m-1}(x+k)^n = \sum_{i=0}^{n} A_{n,i,m}x^i ;]
Explicitly, we have that
[; A_{n,i,m} = \binom{n}{i} \sum_{k=0}^{m-1} k^{n-i} ;]
which can also be written as:
[; A_{n,i,m} = \binom{n}{i}\frac{B_{n-i+1}(m)-B_{n-i+1}(0)}{n-i+1} ;]
Let's call result of the integral [; \mathcal{B}_{n,m}(x) ;]. So
[; \int_{x}^{x+m} B_n(u)du = \mathcal{B}_{n,m}(x) ;]
Notice that [; \mathcal{B}_{n,m}(x) ;] is a degree n polynomial and that the coefficient of xn is m. So we should be able to write [;\mathcal{B}_{n,m}(x);] as follows:
[; \mathcal{B}_{n,m}(x) = mx^n+\sum_{i=0}^{n-1} C_{n,i,m} \mathcal{B}_{i,m}(x) ;]
for some constants [;C_{n,i,m};] . This means that
[; x^n = \frac{ \mathcal{B}_{n,m}(x)-\sum_{i=0}^{n-1} C_{n,i,m} \mathcal{B}_{i,m}(x)}{m} ;]
so we should get
[; \int_{x}^{x+m} \frac{{B}_n(u)-\sum_{i=0}^{n-1} C_{n,i,m} B_{i}(u)}{m}du = x^n ;]
So the polynomials [; \mathfrak{B}_{n,m}(x)=\frac{{B}_n(x)-\sum_{i=0}^{n-1} C_{n,i,m} B_{i}(x)}{m};] have the desired property. By some uniqueness stuff that I'm not going to think about, this is probably the only polynomial with this property. Not the nicest thing, definitely doesn't have an immediately nice generating function like the regular Bernoulli polynomials, but it exists and depends only on n and m and nothing else.
EDIT: In fact, it looks like polynomials of different m can be written in terms of the divisors. Try and show that there are [; c_{n,j,m,k} ;] so that
[; \mathfrak{B}_{n,km}(x)=\frac{\mathfrak{B}_{n,m}(x)-\sum_{j=0}^{n-1} c_{n,j,m,k} \mathfrak{B}_{j,m}(x)}{k};]
I find that pretty interesting.
|
2014-07-14 15:52:43
|
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|
https://nbviewer.jupyter.org/github/firedrakeproject/firedrake/blob/master/docs/notebooks/06-pde-constrained-optimisation.ipynb
|
# Optimisation with firedrake-adjoint¶
This example is modified from the equivalent dolfin-adjoint demo
In this example, we will look at how to use firedrake-adjoint to optimise for strong (Dirichlet) conditions in a steady problem. firedrake-adjoint is a thin compatibility layer for the dolfin-adjoint package, a python package to automatically derive the discrete adjoint and tangent linear models of forward problems written using Firedrake.
## Installing necessary dependencies¶
For the minimisation, we will need scipy, which is installed with
pip install scipy
In the activated firedrake virtualenv.
As usual, we begin with some notebook magic (so that plots appear nicely) and importing Firedrake.
In [1]:
%matplotlib notebook
import matplotlib.pyplot as plt
from firedrake import *
The next step is new, we now import the firedrake_adjoint package, which overrides much of the Firedrake interface so that an annotated tape for adjoint calculations can be built automatically.
In [2]:
from firedrake_adjoint import *
## Problem setup¶
Now we will set up the problem. We consider minimising the compliance:
$$\min_{g, u, p} \ \frac{1}{2}\int_{\Omega} \nabla u \cdot \nabla u\,\text{d}x + \frac{\alpha}{2} \int_{\Gamma_{\textrm{in}}} g^2\,\text{d}s$$
subject to the Stokes equations $$\begin{split}-\nu \Delta u + \nabla p &= 0 \quad \text{in \Omega} \\ \nabla \cdot u &= 0 \quad \text{in \Omega} \end{split}$$
and Dirichlet conditions
$$\begin{split} u &= g \quad \text{on \Gamma_\text{circ}} \\ u &= f \quad \text{on \Gamma_\text{in}} \\ u &= 0 \quad \text{on \Gamma_\text{top} \cup \Gamma_\text{bottom}} \\ p &= 0 \quad \text{on \Gamma_\text{out}}. \\ \end{split}$$
Here, $u$ and $p$ are unknown velocity and pressure, $f$ is a prescribed inflow, $g$ is the control variable that we will optimise for and $\alpha$ is a regularisation parameter. This corresponds physically to minimising the loss of energy as heat by controlling the in/outflow on $\Gamma_\text{circ}$. The regularisation parameter penalises too many non-zero control values.
This problem setup requires a mesh that is more complex than the built in ones Firedrake provides. Instead, it was created with Gmsh. It is loaded by using the Mesh constructor, passing the filename of the mesh in question.
In [3]:
mesh = Mesh("stokes-control.msh")
Now we'll take a look at the mesh. Since we will need to know which mesh markers correspond to which parts of the boundary, we've added a legend to the plot. Normally you will know this because you told your mesh generator how to mark the boundaries.
In [4]:
# NBVAL_IGNORE_OUTPUT
fig, axes = plt.subplots()
triplot(mesh, axes=axes)
axes.axis("off")
axes.set_aspect("equal")
axes.legend(loc="upper right");
## The forward problem¶
The forward problem should be familiar by now, we create a mixed function space for $u$ and $p$ and set up trial and test functions. We specify a parabolic velocity at the inflow, and no-slip (zero velocity) conditions on the side walls. The zero-pressure outflow condition is enforced weakly.
Our variational formulation for Stokes reads as follows. Find $(u, p) \in V\times Q$ such that:
\begin{align} \int_\Omega \nu \nabla u : \nabla v\,\text{d}x - \int_\Omega p \nabla \cdot v\,\text{d}x - \int_{\Gamma_{\text{circ}}} \nu (\nabla u \cdot n) \cdot v\,\text{d}s &= 0 \quad \forall v \in V\\ - \int_\Omega q \nabla \cdot u\,\text{d}x &= 0 \quad \forall q \in Q.\\ u &= 0 \quad \text{on \Gamma_{\text{top}} \cup \Gamma_{\text{bottom}}}\\ u &= \left[\frac{y(10 - y)}{25}, 0\right]^T \quad \text{on \Gamma_{\text{in}}}\\ u &= g \quad \text{on \Gamma_{\text{circ}}}\\ \frac{\text{d}p}{\text{d}n} &= 0 \quad \text{on \Gamma_{\text{out}}} \end{align}
In [5]:
V = VectorFunctionSpace(mesh, "CG", 2)
Q = FunctionSpace(mesh, "CG", 1)
W = V*Q
v, q = TestFunctions(W)
u, p = TrialFunctions(W)
nu = Constant(1) # Viscosity coefficient
x, y = SpatialCoordinate(mesh)
u_inflow = as_vector([y*(10-y)/25.0, 0])
noslip = DirichletBC(W.sub(0), (0, 0), (3, 5))
inflow = DirichletBC(W.sub(0), interpolate(u_inflow, V), 1)
static_bcs = [inflow, noslip]
The boundary value, $g$, on the circle will be our control variable. To do this, we define a Function which will hold the boundary values, we then build a DirichletBC object as normal.
In [6]:
g = Function(V, name="Control")
controlled_bcs = [DirichletBC(W.sub(0), g, 4)]
bcs = static_bcs + controlled_bcs
Now we define the bilinear and linear forms.
In [7]:
a = nu*inner(grad(u), grad(v))*dx - inner(p, div(v))*dx - inner(q, div(u))*dx
L = Constant(0)*q*dx
Now let's solve the forward problem so that firedrake-adjoint annotates it. We'll also take a look at the solution.
In [8]:
w = Function(W)
solve(a == L, w, bcs=bcs, solver_parameters={"pc_type": "lu", "mat_type": "aij",
"pc_factor_shift_type": "inblocks"})
In [9]:
# NBVAL_IGNORE_OUTPUT
u_init, p_init = w.split()
fig, axes = plt.subplots(nrows=2, sharex=True, sharey=True)
streamlines = streamplot(u_init, resolution=1/3, seed=0, axes=axes[0])
fig.colorbar(streamlines, ax=axes[0], fraction=0.046)
axes[0].set_aspect("equal")
axes[0].set_title("Velocity")
contours = tricontourf(p_init, 30, axes=axes[1])
fig.colorbar(contours, ax=axes[1], fraction=0.046)
axes[1].set_aspect("equal")
axes[1].set_title("Pressure");
## The optimisation problem¶
Now we come to the optimisation problem. We first define the functional we wish to minimise, this is done by assembling a form that produces a number. Then we specify a control variable, and produce a "reduced" functional (the evaluation of the functional at a given control value). We then minimise the functional, producing an optimised control value. See the dolfin-adjoint documentation for more details on reduced functionals.
In [10]:
u, p = split(w)
alpha = Constant(10)
m = Control(g)
Jhat = ReducedFunctional(J, m)
In [11]:
g_opt = minimize(Jhat)
Now let's take a look at the optimised control. The initial boundary condition on the circle was no slip. We see now that the optimised boundary condition has a significant outflow in the hole.
In [12]:
# NBVAL_IGNORE_OUTPUT
fig, axes = plt.subplots()
arrows = quiver(g_opt, axes=axes, scale=3)
axes.set_aspect("equal")
axes.set_title("Optimised boundary value");
Let's compare the difference in the initial and final values of the functional. This is done by calling the ReducedFunctional object with the control at the requested value.
In [13]:
print("Jhat(g) = %.8g\nJhat(g_opt) = %.8g" % (Jhat(g), Jhat(g_opt)))
Jhat(g) = 46.078212
Jhat(g_opt) = 19.896719
To see the optimised flow field, we solve the same problem again, only with the new (optimised) value for the boundary data on $\Gamma_\text{circ}$. This time we're not interested in annotating the solve, so we tell firedrake-adjoint to ignore it by passing annotate=False.
In [14]:
g.assign(g_opt)
w_opt = Function(W)
# Use context manager
solve(a == L, w_opt, bcs=bcs, solver_parameters={"pc_type": "lu", "mat_type": "aij",
"pc_factor_shift_type": "inblocks"},
annotate=False)
In [15]:
# NBVAL_IGNORE_OUTPUT
u_opt, p_opt = w_opt.split()
fig, axes = plt.subplots(nrows=2, sharex=True, sharey=True)
streamlines = streamplot(u_opt, resolution=1/3, seed=0, axes=axes[0])
fig.colorbar(streamlines, ax=axes[0], fraction=0.046)
axes[0].set_aspect("equal")
axes[0].set_title("Optimized velocity")
contours = tricontourf(p_opt, 30, axes=axes[1])
fig.colorbar(contours, ax=axes[1], fraction=0.046)
axes[1].set_aspect("equal")
axes[1].set_title("Optimized pressure");
Now that we are done optimising this problem, and so that the recorded annotations do not leak into the exercise, we tell firedrake-adjoint to forget all of the entries on its tape.
In [16]:
tape = get_working_tape()
tape.clear_tape()
## Exercise:¶
Go back to the start of the notebook and consider a different inflow velocity. For example, try $$u = \left[\frac{y^2(10 - y)}{125}, 0\right]^T \quad \text{on} \quad \Gamma_{\text{in}}.$$
How does it affect the solution before and after optimisation?
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2020-08-15 11:05:05
|
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|
https://plainmath.net/calculus-2/82116-let-f-a-b
|
Janet Forbes
2022-07-14
Let $f:\left[a,b\right]\to \text{R}$ be a continuous function such that for each $x\in \left[a,b\right]$ there exists $y\in \left[a,b\right]$ such that $|f\left(y\right)|\le \frac{1}{2}|f\left(x\right)|.$ Prove that there exists $c\in \left[a,b\right]$ such that $f\left(c\right)=0.$
I am trying to use intermediate value theorem for continuous function. Here $|f\left(x\right)|$ is maximum with respect to $|f\left(y\right)|.$ Hence
$|f\left(y\right)|\le 1/2\left(|f\left(x\right)|+|f\left(y\right)|\right)\le |f\left(x\right)|.$
Now I can say $f\left(c\right)=1/2\left(|f\left(x\right)|+|f\left(y\right)|$ for some $c\in \left(a,b\right).$ But how to show $f\left(c\right)=0?$
Jayvion Mclaughlin
Expert
Let ${x}_{0}\in \left[a,b\right]$. Then there is ${x}_{1}\in \left[a,b\right]$ such that $|f\left({x}_{1}\right)|\le \frac{1}{2}|f\left({x}_{0}\right)|.$
Now we get inductively a sequence $\left({x}_{n}\right)$ in $\left[a,b\right]$ with
(*) $|f\left({x}_{n}\right)|\le \frac{1}{{2}^{n}}|f\left({x}_{0}\right)|.$
$\left({x}_{n}\right)$ contains a convergent subsequence $\left({x}_{{n}_{k}}\right)$ with limit $c\in \left[a,b\right]$.
$f$ is continuous, thus $f\left({x}_{{n}_{k}}\right)\to f\left(c\right)$ .
From (*) we get: $f\left({x}_{{n}_{k}}\right)\to 0$.
Consequence: $f\left(c\right)=0$
Do you have a similar question?
|
2023-02-01 21:34:10
|
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|
http://mail-index.netbsd.org/port-arm/2013/03/19/msg001817.html
|
Port-arm archive
On Tue, 19 Mar 2013 12:30:49 -0700
Andy Ruhl <acruhl%gmail.com@localhost> wrote:
> > The problem isn't total power consumtion, but short peaks that exceed
> > the max. 0.5 A for USB. This happens e.g. on sending a packet.
> Power supplies can often handle short spikes without a lot of impact.
> Depending on their design of course.
Often you can find electronic fuses on USB ports. They sense current
and switch power off of the asscociated USB port when the max. current
is exceeded. They are quite fast in doing this, so spikes from a WLAN
device may trigger them. (I am not talking about "poly fuses".) These
things are smal chips, e.g. in a SO8 package, and switch power by the
means of a MOSFET. Usually this MOSFET is controlable by the host CPU
to switch power on and some can trigger an interrupt if the current
limit is reached. E.g. the Olinuxino Mini uses a SY6280 for this.
--
\end{Jochen}
\ref{http://www.unixag-kl.fh-kl.de/~jkunz/}
Home | Main Index | Thread Index | Old Index
|
2016-12-10 23:37:27
|
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|
http://oxid-templates.com/NitricOxide/nitric-acid-oxidation
|
# Nitric acid oxidation
Because, unlike other metal dissolution reactions, the $\ce{H+}$ of $\ce{HNO3}$ isn't reduced- the $\ce{NO3-}$ is.
(data/balanced reactions from Wikipedia) \begin{align} \ce{NO3- + 4 H+ + 3 e- -> NO + 2 H2O}\quad &...& E^o_{red} = 0.96 V \\ \ce{NO3- + 2 H+ + e- -> NO2 + H2O} \quad&...& E^o_{red} = 0.79 V \\ \ce{Ag+(aq) + e- -> Ag(s)} \quad&...& E^o_{red} = 0.799 V \end{align}
Since the SRP of the $\ce{NO2}$ reaction looks smaller than that of $\ce{Ag+}$ (I may be wrong, but by significant digits it can't be greater than $0.799$), one can conclude that it's the $\ce{NO}$ reaction that's occurring here. And the $\ce{NO}$ reaction has a large enough SRP to oxidise $\ce{Ag+}$.
Usually nitrogen compounds are pretty versatile when it comes to redox reactions, since Nitrogen shows many oxidation states. So the simple reason for why $\ce{HNO3}$ is so strong an oxidising agent (with respect to other acids) is that it has a different, better path availible to it to get reduced.
Note that the exact reduction path (ie final reduction products/oxidation state) depends upon the comcentration of nitric acid-so much that copper can be oxidised in three different ways. I had a table of this, I'll post it later.
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2021-02-27 04:32:02
|
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|
https://khufkens.github.io/phenocamr/
|
Facilitates the retrieval and post-processing of PhenoCam time series. The post-processing of PhenoCam data includes outlier removal and the generation of data products such as phenological transition dates. If requested complementary Daymet climate data will be downloaded and merged with the PhenoCam data for modelling purposes. For a detailed overview of the assumptions made during post-processing I refer publications by Hufkens et al. (2018) and Richardson et al. (2018). Please cite the Hufkens et al. (2018) paper when using the package. A worked example is included below and in the package vignette.
## Installation
### stable release
To install the current stable release use a CRAN repository:
install.packages("phenocamr")
library(phenocamr)
### development release
To install the development releases of the package run the following commands:
if(!require(devtools)){install.package(devtools)}
devtools::install_github("khufkens/phenocamr")
library(phenocamr)
Vignettes are not rendered by default, if you want to include additional documentation please use:
if(!require(devtools)){install.package("devtools")}
devtools::install_github("khufkens/phenocamr", build_vignettes = TRUE)
library(phenocamr)
## Use
download_phenocam(site = "harvard",
veg_type = "DB",
frequency = 3,
phenophases = TRUE,
out_dir = "~")
This will download all deciduous broadleaf (DB) PhenoCam time series for the “harvard” site at a 3-day time step into your home directory. In addition, the data is processed to estimate phenological transition dates (phenophases) and written to file. For detailed overview of all functions and worked example we reference to the R help documentation and the manuscripts below.
## References
Hufkens K., Basler J. D., Milliman T. Melaas E., Richardson A.D. 2018 An integrated phenology modelling framework in R: Phenology modelling with phenor. Methods in Ecology & Evolution, 9: 1-10.
Richardson, A.D., Hufkens, K., Milliman, T., Aubrecht, D.M., Chen, M., Gray, J.M., Johnston, M.R., Keenan, T.F., Klosterman, S.T., Kosmala, M., Melaas, E.K., Friedl, M.A., Frolking, S. 2017. Tracking vegetation phenology across diverse North American biomes using PhenoCam imagery. Scientific Data, 5, 180028.
## Acknowledgements
This project was is supported by the National Science Foundation’s Macro-system Biology Program (awards EF-1065029 and EF-1702697). Logo design elements are taken from the FontAwesome library according to these terms.
|
2019-09-20 14:40:38
|
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|
https://demo7.dspace.org/items/576341fc-2bfa-461b-866f-a767819a4a64
|
## Nuclear matrix elements of neutrinoless double beta decay with improved short-range correlations
##### Authors
Kortelainen, Markus
Suhonen, Jouni
##### Description
Nuclear matrix elements of the neutrinoless double beta decays of 96Zr, 100Mo, 116Cd, 128Te, 130Te and 136Xe are calculated for the light-neutrino exchange mechanism by using the proton-neutron quasiparticle random-phase approximation (pnQRPA) with a realistic nucleon-nucleon force. The g_pp parameter of the pnQRPA is fixed by the data on the two-neutrino double beta decays and single beta decays. The finite size of a nucleon, the higher-order terms of nucleonic weak currents, and the nucleon-nucleon short-range correlations (s.r.c) are taken into account. The s.r.c. are computed by the traditional Jastrow method and by the more advanced unitary correlation operator method (UCOM). Comparison of the results obtained by the two methods is carried out. The UCOM computed matrix elements turn out to be considerably larger than the Jastrow computed ones. This result is important for the assessment of the neutrino-mass sensitivity of the present and future double beta experiments.
Comment: Two figures, to be published in Physical Review C (2007) as a regular article
Nuclear Theory
|
2022-12-08 16:39:36
|
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|
http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=ivm&paperid=9084&option_lang=eng
|
RUS ENG JOURNALS PEOPLE ORGANISATIONS CONFERENCES SEMINARS VIDEO LIBRARY PACKAGE AMSBIB
General information Latest issue Archive Search papers Search references RSS Latest issue Current issues Archive issues What is RSS
Izv. Vyssh. Uchebn. Zaved. Mat.: Year: Volume: Issue: Page: Find
Izv. Vyssh. Uchebn. Zaved. Mat., 2016, Number 2, Pages 75–86 (Mi ivm9084)
On the structure of a solution set of controlled initial-boundary value problems
A. V. Chernovab
a Chair of Applied Mathematics, Nizhni Novgorod State Technical University, 24 Minin str., Nizhni Novgorod, 603950 Russia
b Chair of Mathematical Physics and Optimal Control, Nizhni Novgorod State University, 23 Gagarin ave., Nizhni Novgorod, 603950 Russia
Abstract: For a controlled nonlinear functional-operator equation of the Hammerstein type describing a wide class of controlled initial-boundary value problems, we obtain simple sufficient conditions for the convexity, pointwise boundedness and precompactness of the set of solutions (the reachability tube) in the Lebesgue space. As concerns boundedness and precompactness, we mean certain conditions of the majorant type without Volterra type requirements which give also the total (with respect to the whole set of admissible controls) preservation of solvability of mentioned equation. In the capañity of examples of reduction of a controlled initial-boundary (boundary) value problem to the equation under investigation and verification the proposed hypotheses for this equation, we consider the first initial-boundary value problem associated with a semilinear parabolic equation of the second order in a rather general form, and also the Dirichlet problem associated with a semilinear elliptic equation of the second order.
Keywords: reachability tube, convexity conditions, total preservation of solvability, functional-operator equation of the Hammerstein type, nonlinear distributed system, parabolic equation, elliptic equation.
Funding Agency Grant Number Ministry of Education and Science of the Russian Federation 172702.B.49.21.0003
Full text: PDF file (248 kB)
References: PDF file HTML file
English version:
Russian Mathematics (Izvestiya VUZ. Matematika), 2016, 60:2, 62–71
Bibliographic databases:
UDC: 517.988
Citation: A. V. Chernov, “On the structure of a solution set of controlled initial-boundary value problems”, Izv. Vyssh. Uchebn. Zaved. Mat., 2016, no. 2, 75–86; Russian Math. (Iz. VUZ), 60:2 (2016), 62–71
Citation in format AMSBIB
\Bibitem{Che16} \by A.~V.~Chernov \paper On the structure of a~solution set of controlled initial-boundary value problems \jour Izv. Vyssh. Uchebn. Zaved. Mat. \yr 2016 \issue 2 \pages 75--86 \mathnet{http://mi.mathnet.ru/ivm9084} \transl \jour Russian Math. (Iz. VUZ) \yr 2016 \vol 60 \issue 2 \pages 62--71 \crossref{https://doi.org/10.3103/S1066369X16020109} \isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=000409281900010} \scopus{https://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-84961377367}
• http://mi.mathnet.ru/eng/ivm9084
• http://mi.mathnet.ru/eng/ivm/y2016/i2/p75
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This publication is cited in the following articles:
1. V. I. Sumin, “Volterra functional-operator equations in the theory of optimal control of distributed systems”, IFAC-PapersOnLine, 51:32 (2018), 759–764
• Number of views: This page: 249 Full text: 15 References: 30 First page: 6
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2021-01-17 00:14:56
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https://web2.0calc.com/questions/root_2
|
+0
# root
0
257
2
square root 52
Guest Sep 23, 2017
#1
+2194
+1
$$\sqrt{52}=\sqrt{4*13}=\sqrt{4}\sqrt{13}=2\sqrt{13}\approx7.2111$$
TheXSquaredFactor Sep 23, 2017
#1
+2194
+1
$$\sqrt{52}=\sqrt{4*13}=\sqrt{4}\sqrt{13}=2\sqrt{13}\approx7.2111$$
TheXSquaredFactor Sep 23, 2017
#2
0
The answer to the square root of 52 is between 7 and 8 and is closer to 8. Therefore the answer will be between 7 and 8. Using a calculator we can find that the precise answer is 7.211102551
Guest Sep 23, 2017
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2018-09-25 08:08:46
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https://www.unsw.edu.au/science/our-schools/maths/engage-with-us/seminars/2016/asymptotics-some-families-orthonormal-polynomials-and-associated-hilbert-space
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### Abstract:
We characterise asymptotic behaviour of families of symmetric orthonormal polynomials whose recursion coefficients satisfy certain conditions, satisfied for example by the (normalised) Hermite polynomials. More generally, these conditions are satisfied by the recursion coefficients of the form c(n + 1)p for 0 < p < 1 and c > 0,
as well as by recursion coefficients which correspond to polynomials orthonormal with respect to the exponential weight w(x) = exp(−a|x|) for a > 1. We use these results to show that, in a Hilbert space defined in a natural way by
such a family of orthonormal polynomials, any two complex exponentials of the form exp(i f t) are mutually orthogonal whenever their frequencies f are distinct. We finally formulate a surprising conjecture for the corresponding families of non-symmetric orthonormal polynomials; extensive numerical tests indicate that such a conjecture appears to be true.
Speaker
Aleksandar Ignjatovic
Research Area
Affiliation
CSE, University of New South Wales
Date
Tue, 27/09/2016 - 11:05am to 11:55am
Venue
RC-4082, The Red Centre, UNSW
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2023-03-29 10:20:15
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https://faq.gutenberg.eu.org/2_programmation/erreurs/normalsize_not_defined?rev=1527018029
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Ceci est une ancienne révision du document !
# `\normalsize` not defined
The LaTeX error: ```latex The font size command \normalsize is not defined: there is probably something wrong with the class file. ``` reports something pretty fundamental (document base font size has not been set, something the document class does for you). It _can_, in principle, be a problem with the document class, but is more often caused by the user forgetting to start their document with a `\documentclass` command.
Content last updated: 2014-06-10
2_programmation/erreurs/normalsize_not_defined.1527018029.txt.gz · Dernière modification: 2018/05/22 21:40 par joseph.wright
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2021-04-22 03:37:46
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https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/847/2/f/p/
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# Properties
Label 847.2.f.p Level 847 Weight 2 Character orbit 847.f Analytic conductor 6.763 Analytic rank 0 Dimension 8 CM no Inner twists 2
# Related objects
## Newspace parameters
Level: $$N$$ = $$847 = 7 \cdot 11^{2}$$ Weight: $$k$$ = $$2$$ Character orbit: $$[\chi]$$ = 847.f (of order $$5$$, degree $$4$$, not minimal)
## Newform invariants
Self dual: no Analytic conductor: $$6.76332905120$$ Analytic rank: $$0$$ Dimension: $$8$$ Relative dimension: $$2$$ over $$\Q(\zeta_{5})$$ Coefficient field: 8.0.159390625.1 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 77) Sato-Tate group: $\mathrm{SU}(2)[C_{5}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{7}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q -\beta_{1} q^{2} + ( 1 - \beta_{3} ) q^{3} + ( \beta_{1} - \beta_{2} - \beta_{3} - \beta_{5} - \beta_{6} - \beta_{7} ) q^{4} + ( 1 - \beta_{2} - \beta_{3} - \beta_{6} - \beta_{7} ) q^{5} + ( -\beta_{5} - \beta_{7} ) q^{6} + \beta_{2} q^{7} + ( -1 + \beta_{1} + \beta_{3} - \beta_{5} - 2 \beta_{6} ) q^{8} + ( 1 - \beta_{2} + \beta_{3} ) q^{9} +O(q^{10})$$ $$q -\beta_{1} q^{2} + ( 1 - \beta_{3} ) q^{3} + ( \beta_{1} - \beta_{2} - \beta_{3} - \beta_{5} - \beta_{6} - \beta_{7} ) q^{4} + ( 1 - \beta_{2} - \beta_{3} - \beta_{6} - \beta_{7} ) q^{5} + ( -\beta_{5} - \beta_{7} ) q^{6} + \beta_{2} q^{7} + ( -1 + \beta_{1} + \beta_{3} - \beta_{5} - 2 \beta_{6} ) q^{8} + ( 1 - \beta_{2} + \beta_{3} ) q^{9} + ( -2 - \beta_{2} + 2 \beta_{4} - 2 \beta_{5} - \beta_{6} ) q^{10} + ( -1 + \beta_{1} - \beta_{2} + \beta_{4} - \beta_{5} - \beta_{6} - \beta_{7} ) q^{12} + ( -2 + 3 \beta_{1} + 2 \beta_{2} - 2 \beta_{4} - 2 \beta_{7} ) q^{13} -\beta_{4} q^{14} + ( -\beta_{2} - \beta_{3} + \beta_{4} - \beta_{5} - \beta_{6} - \beta_{7} ) q^{15} + ( 1 - \beta_{2} - \beta_{3} + 2 \beta_{5} - \beta_{6} + \beta_{7} ) q^{16} + ( 2 + \beta_{2} + \beta_{3} - \beta_{5} - 2 \beta_{6} - \beta_{7} ) q^{17} + ( -2 \beta_{1} + \beta_{4} + \beta_{5} + \beta_{7} ) q^{18} + ( 1 - \beta_{1} - \beta_{3} - 2 \beta_{4} + \beta_{5} - 2 \beta_{6} ) q^{19} + ( -2 + 2 \beta_{1} + 2 \beta_{2} + 4 \beta_{3} + \beta_{4} + \beta_{7} ) q^{20} + ( \beta_{2} + \beta_{6} ) q^{21} + ( -2 + 2 \beta_{1} - \beta_{2} - \beta_{6} - 2 \beta_{7} ) q^{23} + ( -3 + \beta_{1} + 3 \beta_{2} + 4 \beta_{3} ) q^{24} + ( -1 + \beta_{3} + 3 \beta_{4} + 2 \beta_{6} ) q^{25} + ( -\beta_{1} + \beta_{2} - 3 \beta_{3} + \beta_{5} - 3 \beta_{6} + \beta_{7} ) q^{26} + ( 1 + 3 \beta_{2} + 3 \beta_{3} - \beta_{6} ) q^{27} + ( -\beta_{2} - \beta_{3} - \beta_{7} ) q^{28} + ( 3 \beta_{1} - 3 \beta_{4} ) q^{29} + ( -3 + 2 \beta_{1} + 3 \beta_{3} + 2 \beta_{4} - 2 \beta_{5} - \beta_{6} ) q^{30} + ( 3 - 2 \beta_{1} - 3 \beta_{2} + \beta_{4} + \beta_{7} ) q^{31} + ( -4 + 3 \beta_{2} + 3 \beta_{6} ) q^{32} + ( -1 - 2 \beta_{1} - 3 \beta_{2} - 3 \beta_{6} + 2 \beta_{7} ) q^{34} + ( -\beta_{1} - \beta_{3} ) q^{35} + ( 1 + \beta_{1} - \beta_{3} - \beta_{4} - \beta_{5} - \beta_{6} ) q^{36} + ( 2 \beta_{1} + 2 \beta_{2} - \beta_{3} - \beta_{4} - \beta_{5} - \beta_{6} - \beta_{7} ) q^{37} + ( 3 - 7 \beta_{2} - 7 \beta_{3} - 2 \beta_{5} - 3 \beta_{6} - \beta_{7} ) q^{38} + ( -2 + 2 \beta_{2} + 2 \beta_{3} + \beta_{5} + 2 \beta_{6} - \beta_{7} ) q^{39} + ( 4 \beta_{2} + 3 \beta_{3} - 3 \beta_{4} + 3 \beta_{5} + 3 \beta_{6} + 3 \beta_{7} ) q^{40} + ( 3 - 2 \beta_{1} - 3 \beta_{3} + 2 \beta_{5} - 2 \beta_{6} ) q^{41} + ( -\beta_{4} - \beta_{7} ) q^{42} + ( -1 + 6 \beta_{2} + 6 \beta_{6} ) q^{43} + ( 2 + \beta_{1} - \beta_{2} - \beta_{4} + \beta_{5} - \beta_{6} - \beta_{7} ) q^{45} + ( -4 + 4 \beta_{2} + 2 \beta_{3} + 3 \beta_{4} + 3 \beta_{7} ) q^{46} + ( 2 + 4 \beta_{1} - 2 \beta_{3} - \beta_{4} - 4 \beta_{5} + \beta_{6} ) q^{47} + ( -2 \beta_{1} - \beta_{2} - \beta_{3} - \beta_{4} + 3 \beta_{5} - \beta_{6} + 3 \beta_{7} ) q^{48} + ( -1 + \beta_{2} + \beta_{3} + \beta_{6} ) q^{49} + ( -6 + 9 \beta_{2} + 9 \beta_{3} + \beta_{5} + 6 \beta_{6} + 2 \beta_{7} ) q^{50} + ( \beta_{1} + 4 \beta_{2} + \beta_{3} + \beta_{4} - 2 \beta_{5} + \beta_{6} - 2 \beta_{7} ) q^{51} + ( -3 - \beta_{1} + 3 \beta_{3} + \beta_{5} + 6 \beta_{6} ) q^{52} + ( -1 - \beta_{1} + \beta_{2} + 4 \beta_{3} - 3 \beta_{4} - 3 \beta_{7} ) q^{53} + ( -4 \beta_{1} - 3 \beta_{4} + 3 \beta_{5} + 4 \beta_{7} ) q^{54} + ( -2 + \beta_{1} - \beta_{2} - \beta_{6} - \beta_{7} ) q^{56} + ( -1 - \beta_{1} + \beta_{2} - 2 \beta_{4} - 2 \beta_{7} ) q^{57} + ( 6 - 3 \beta_{1} - 6 \beta_{3} + 3 \beta_{5} - 3 \beta_{6} ) q^{58} + ( \beta_{1} + 6 \beta_{2} - 2 \beta_{3} + \beta_{4} - 2 \beta_{5} - 2 \beta_{6} - 2 \beta_{7} ) q^{59} + ( -2 + 6 \beta_{2} + 6 \beta_{3} + 3 \beta_{5} + 2 \beta_{6} + 4 \beta_{7} ) q^{60} + ( 7 - 7 \beta_{2} - 7 \beta_{3} + \beta_{5} - 7 \beta_{6} + 5 \beta_{7} ) q^{61} + ( -\beta_{1} - 2 \beta_{2} + \beta_{3} + 2 \beta_{4} - \beta_{5} + \beta_{6} - \beta_{7} ) q^{62} + ( 1 - \beta_{3} - 2 \beta_{6} ) q^{63} + ( 2 \beta_{1} - 2 \beta_{3} + \beta_{4} + \beta_{7} ) q^{64} + ( 2 + 2 \beta_{1} - \beta_{2} - 2 \beta_{4} + 2 \beta_{5} - \beta_{6} - 2 \beta_{7} ) q^{65} + ( -2 - \beta_{1} - 2 \beta_{2} + \beta_{4} - \beta_{5} - 2 \beta_{6} + \beta_{7} ) q^{67} + ( -2 + 3 \beta_{1} + 2 \beta_{2} - \beta_{4} - \beta_{7} ) q^{68} + ( -3 + 3 \beta_{3} + 2 \beta_{4} - \beta_{6} ) q^{69} + ( 2 \beta_{1} - 3 \beta_{2} - \beta_{3} - 2 \beta_{5} - \beta_{6} - 2 \beta_{7} ) q^{70} + ( 4 + \beta_{2} + \beta_{3} - 3 \beta_{5} - 4 \beta_{6} - 2 \beta_{7} ) q^{71} + ( 3 - 2 \beta_{2} - 2 \beta_{3} - 2 \beta_{5} - 3 \beta_{6} + \beta_{7} ) q^{72} + ( -3 \beta_{1} + 4 \beta_{2} + 3 \beta_{3} + 5 \beta_{4} - 2 \beta_{5} + 3 \beta_{6} - 2 \beta_{7} ) q^{73} + ( 1 - \beta_{3} - \beta_{4} - 3 \beta_{6} ) q^{74} + ( 1 - \beta_{2} + 3 \beta_{4} + 3 \beta_{7} ) q^{75} + ( -4 + 4 \beta_{1} - 3 \beta_{2} + 4 \beta_{4} - 4 \beta_{5} - 3 \beta_{6} - 4 \beta_{7} ) q^{76} + ( -3 - \beta_{1} + \beta_{2} - \beta_{4} + \beta_{5} + \beta_{6} + \beta_{7} ) q^{78} + ( 2 - 3 \beta_{1} - 2 \beta_{2} - 2 \beta_{3} + 2 \beta_{4} + 2 \beta_{7} ) q^{79} + ( 5 - 4 \beta_{1} - 5 \beta_{3} - \beta_{4} + 4 \beta_{5} - \beta_{6} ) q^{80} + ( 4 \beta_{2} + 6 \beta_{3} + 6 \beta_{6} ) q^{81} + ( -2 - 2 \beta_{2} - 2 \beta_{3} - 5 \beta_{5} + 2 \beta_{6} - \beta_{7} ) q^{82} + ( -3 - 4 \beta_{2} - 4 \beta_{3} + 5 \beta_{5} + 3 \beta_{6} + 3 \beta_{7} ) q^{83} + ( -2 \beta_{2} - \beta_{3} + \beta_{4} - \beta_{5} - \beta_{6} - \beta_{7} ) q^{84} + ( -\beta_{1} + \beta_{4} + \beta_{5} - 3 \beta_{6} ) q^{85} + ( \beta_{1} - 6 \beta_{4} - 6 \beta_{7} ) q^{86} + ( -3 \beta_{4} + 3 \beta_{5} ) q^{87} + ( -1 - 5 \beta_{1} - 3 \beta_{2} + \beta_{4} - \beta_{5} - 3 \beta_{6} + 5 \beta_{7} ) q^{89} + ( -1 - 4 \beta_{1} + \beta_{2} - 2 \beta_{3} + 2 \beta_{4} + 2 \beta_{7} ) q^{90} + ( -2 - 2 \beta_{1} + 2 \beta_{3} + \beta_{4} + 2 \beta_{5} + 2 \beta_{6} ) q^{91} + ( -2 \beta_{1} + 6 \beta_{2} + 7 \beta_{3} - 3 \beta_{4} + 5 \beta_{5} + 7 \beta_{6} + 5 \beta_{7} ) q^{92} + ( 3 - 3 \beta_{2} - 3 \beta_{3} - \beta_{5} - 3 \beta_{6} ) q^{93} + ( 6 + \beta_{2} + \beta_{3} + 2 \beta_{5} - 6 \beta_{6} - 4 \beta_{7} ) q^{94} + ( -6 \beta_{2} - 5 \beta_{3} + 3 \beta_{4} - 3 \beta_{5} - 5 \beta_{6} - 3 \beta_{7} ) q^{95} + ( -1 + \beta_{3} + 3 \beta_{6} ) q^{96} + ( -2 - 5 \beta_{1} + 2 \beta_{2} + 6 \beta_{3} + 5 \beta_{4} + 5 \beta_{7} ) q^{97} + ( -\beta_{4} + \beta_{5} ) q^{98} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$8q - q^{2} + 6q^{3} - 7q^{4} + 3q^{5} - 2q^{6} + 2q^{7} - 12q^{8} + 8q^{9} + O(q^{10})$$ $$8q - q^{2} + 6q^{3} - 7q^{4} + 3q^{5} - 2q^{6} + 2q^{7} - 12q^{8} + 8q^{9} - 28q^{10} - 14q^{12} - 5q^{13} + q^{14} - 9q^{15} + 7q^{16} + 14q^{17} - q^{18} + 6q^{19} - 4q^{20} + 4q^{21} - 16q^{23} - 9q^{24} - 5q^{25} - 9q^{26} + 18q^{27} - 3q^{28} + 6q^{29} - 26q^{30} + 14q^{31} - 20q^{32} - 24q^{34} - 3q^{35} + 3q^{36} + q^{37} - 15q^{38} + 29q^{40} + 18q^{41} + 2q^{42} + 16q^{43} + 18q^{45} - 26q^{46} + 7q^{47} - q^{48} - 2q^{49} + q^{50} + 8q^{51} - 4q^{52} + 7q^{53} + 4q^{54} - 18q^{56} - 3q^{57} + 36q^{58} + 17q^{60} + 12q^{61} - 5q^{62} + 2q^{63} - 4q^{64} + 24q^{65} - 30q^{67} - 7q^{68} - 22q^{69} - 12q^{70} + 21q^{71} + 3q^{72} + 8q^{73} + q^{74} - 52q^{76} - 18q^{78} + q^{79} + 37q^{80} + 32q^{81} - 34q^{82} - 22q^{83} - 11q^{84} - 5q^{85} + 13q^{86} + 12q^{87} - 34q^{89} - 18q^{90} - 5q^{91} + 51q^{92} + 3q^{93} + 50q^{94} - 41q^{95} - 15q^{97} + 4q^{98} + O(q^{100})$$
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{8} - x^{7} + 6 x^{6} - 11 x^{5} + 21 x^{4} - 5 x^{3} + 10 x^{2} + 25 x + 25$$:
$$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$\nu$$ $$\beta_{2}$$ $$=$$ $$($$$$555 \nu^{7} - 2159 \nu^{6} + 7489 \nu^{5} - 18164 \nu^{4} + 40069 \nu^{3} - 84434 \nu^{2} + 43855 \nu + 375$$$$)/94655$$ $$\beta_{3}$$ $$=$$ $$($$$$-970 \nu^{7} - 1002 \nu^{6} - 6608 \nu^{5} + 9063 \nu^{4} - 14943 \nu^{3} + 27673 \nu^{2} - 68120 \nu + 35160$$$$)/94655$$ $$\beta_{4}$$ $$=$$ $$($$$$-1604 \nu^{7} + 4159 \nu^{6} - 12059 \nu^{5} + 28414 \nu^{4} - 81659 \nu^{3} + 38305 \nu^{2} - 13500 \nu - 13875$$$$)/94655$$ $$\beta_{5}$$ $$=$$ $$($$$$-2052 \nu^{7} + 2252 \nu^{6} - 19912 \nu^{5} + 21007 \nu^{4} - 82042 \nu^{3} + 35785 \nu^{2} - 19395 \nu - 90925$$$$)/94655$$ $$\beta_{6}$$ $$=$$ $$($$$$-2667 \nu^{7} + 6691 \nu^{6} - 17466 \nu^{5} + 50856 \nu^{4} - 82441 \nu^{3} + 72554 \nu^{2} - 4035 \nu - 12035$$$$)/94655$$ $$\beta_{7}$$ $$=$$ $$($$$$4024 \nu^{7} - 1464 \nu^{6} + 21519 \nu^{5} - 26434 \nu^{4} + 59219 \nu^{3} + 22635 \nu^{2} + 54640 \nu + 66675$$$$)/94655$$
$$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$\beta_{1}$$ $$\nu^{2}$$ $$=$$ $$-\beta_{7} - \beta_{6} - \beta_{5} - \beta_{3} - 3 \beta_{2} + \beta_{1}$$ $$\nu^{3}$$ $$=$$ $$2 \beta_{6} + \beta_{5} - 4 \beta_{4} - \beta_{3} - \beta_{1} + 1$$ $$\nu^{4}$$ $$=$$ $$7 \beta_{7} + 7 \beta_{6} + 2 \beta_{5} + 13 \beta_{3} + 13 \beta_{2} - 7$$ $$\nu^{5}$$ $$=$$ $$-8 \beta_{7} - 11 \beta_{6} - 20 \beta_{5} + 20 \beta_{4} - 11 \beta_{2} + 8 \beta_{1} - 12$$ $$\nu^{6}$$ $$=$$ $$-19 \beta_{7} - 19 \beta_{4} - 68 \beta_{3} - 36 \beta_{2} - 24 \beta_{1} + 36$$ $$\nu^{7}$$ $$=$$ $$111 \beta_{7} + 81 \beta_{6} + 111 \beta_{5} - 55 \beta_{4} + 81 \beta_{3} + 148 \beta_{2} - 56 \beta_{1}$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/847\mathbb{Z}\right)^\times$$.
$$n$$ $$122$$ $$365$$ $$\chi(n)$$ $$1$$ $$-\beta_{2}$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
148.1
0.453245 + 1.39494i −0.762262 − 2.34600i 1.43801 + 1.04478i −0.628998 − 0.456994i 0.453245 − 1.39494i −0.762262 + 2.34600i 1.43801 − 1.04478i −0.628998 + 0.456994i
−0.453245 1.39494i 1.30902 + 0.951057i −0.122406 + 0.0889332i 0.144228 0.443888i 0.733366 2.25707i 0.809017 0.587785i −2.19369 1.59381i −0.118034 0.363271i −0.684570
148.2 0.762262 + 2.34600i 1.30902 + 0.951057i −3.30464 + 2.40097i −1.07128 + 3.29706i −1.23337 + 3.79591i 0.809017 0.587785i −4.16042 3.02272i −0.118034 0.363271i −8.55150
323.1 −1.43801 1.04478i 0.190983 0.587785i 0.358290 + 1.10270i 2.24703 1.63256i −0.888742 + 0.645709i −0.309017 0.951057i −0.461691 + 1.42094i 2.11803 + 1.53884i −4.93693
323.2 0.628998 + 0.456994i 0.190983 0.587785i −0.431239 1.32722i 0.180019 0.130791i 0.388742 0.282438i −0.309017 0.951057i 0.815793 2.51075i 2.11803 + 1.53884i 0.173002
372.1 −0.453245 + 1.39494i 1.30902 0.951057i −0.122406 0.0889332i 0.144228 + 0.443888i 0.733366 + 2.25707i 0.809017 + 0.587785i −2.19369 + 1.59381i −0.118034 + 0.363271i −0.684570
372.2 0.762262 2.34600i 1.30902 0.951057i −3.30464 2.40097i −1.07128 3.29706i −1.23337 3.79591i 0.809017 + 0.587785i −4.16042 + 3.02272i −0.118034 + 0.363271i −8.55150
729.1 −1.43801 + 1.04478i 0.190983 + 0.587785i 0.358290 1.10270i 2.24703 + 1.63256i −0.888742 0.645709i −0.309017 + 0.951057i −0.461691 1.42094i 2.11803 1.53884i −4.93693
729.2 0.628998 0.456994i 0.190983 + 0.587785i −0.431239 + 1.32722i 0.180019 + 0.130791i 0.388742 + 0.282438i −0.309017 + 0.951057i 0.815793 + 2.51075i 2.11803 1.53884i 0.173002
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 729.2 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
11.c even 5 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 847.2.f.p 8
11.b odd 2 1 847.2.f.s 8
11.c even 5 2 77.2.f.a 8
11.c even 5 1 847.2.a.l 4
11.c even 5 1 inner 847.2.f.p 8
11.d odd 10 1 847.2.a.k 4
11.d odd 10 2 847.2.f.q 8
11.d odd 10 1 847.2.f.s 8
33.f even 10 1 7623.2.a.co 4
33.h odd 10 2 693.2.m.g 8
33.h odd 10 1 7623.2.a.ch 4
77.j odd 10 2 539.2.f.d 8
77.j odd 10 1 5929.2.a.bi 4
77.l even 10 1 5929.2.a.bb 4
77.m even 15 4 539.2.q.c 16
77.p odd 30 4 539.2.q.b 16
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
77.2.f.a 8 11.c even 5 2
539.2.f.d 8 77.j odd 10 2
539.2.q.b 16 77.p odd 30 4
539.2.q.c 16 77.m even 15 4
693.2.m.g 8 33.h odd 10 2
847.2.a.k 4 11.d odd 10 1
847.2.a.l 4 11.c even 5 1
847.2.f.p 8 1.a even 1 1 trivial
847.2.f.p 8 11.c even 5 1 inner
847.2.f.q 8 11.d odd 10 2
847.2.f.s 8 11.b odd 2 1
847.2.f.s 8 11.d odd 10 1
5929.2.a.bb 4 77.l even 10 1
5929.2.a.bi 4 77.j odd 10 1
7623.2.a.ch 4 33.h odd 10 1
7623.2.a.co 4 33.f even 10 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(847, [\chi])$$:
$$T_{2}^{8} + \cdots$$ $$T_{3}^{4} - 3 T_{3}^{3} + 4 T_{3}^{2} - 2 T_{3} + 1$$ $$T_{13}^{8} + \cdots$$
## Hecke Characteristic Polynomials
$p$ $F_p(T)$
$2$ $$1 + T + 2 T^{2} + 5 T^{3} + 5 T^{4} + 19 T^{5} + 26 T^{6} + 25 T^{7} + 59 T^{8} + 50 T^{9} + 104 T^{10} + 152 T^{11} + 80 T^{12} + 160 T^{13} + 128 T^{14} + 128 T^{15} + 256 T^{16}$$
$3$ $$( 1 - 3 T + T^{2} + T^{3} + 4 T^{4} + 3 T^{5} + 9 T^{6} - 81 T^{7} + 81 T^{8} )^{2}$$
$5$ $$1 - 3 T + 2 T^{2} + 6 T^{4} - 75 T^{5} + 208 T^{6} - 234 T^{7} + 431 T^{8} - 1170 T^{9} + 5200 T^{10} - 9375 T^{11} + 3750 T^{12} + 31250 T^{14} - 234375 T^{15} + 390625 T^{16}$$
$7$ $$( 1 - T + T^{2} - T^{3} + T^{4} )^{2}$$
$11$
$13$ $$1 + 5 T - 30 T^{2} - 170 T^{3} + 286 T^{4} + 2815 T^{5} + 5030 T^{6} - 15820 T^{7} - 130849 T^{8} - 205660 T^{9} + 850070 T^{10} + 6184555 T^{11} + 8168446 T^{12} - 63119810 T^{13} - 144804270 T^{14} + 313742585 T^{15} + 815730721 T^{16}$$
$17$ $$1 - 14 T + 79 T^{2} - 281 T^{3} + 1089 T^{4} - 2031 T^{5} - 15507 T^{6} + 106442 T^{7} - 393507 T^{8} + 1809514 T^{9} - 4481523 T^{10} - 9978303 T^{11} + 90954369 T^{12} - 398979817 T^{13} + 1906867951 T^{14} - 5744741422 T^{15} + 6975757441 T^{16}$$
$19$ $$1 - 6 T - 27 T^{2} + 111 T^{3} + 483 T^{4} - 2061 T^{5} + 20515 T^{6} - 13260 T^{7} - 552129 T^{8} - 251940 T^{9} + 7405915 T^{10} - 14136399 T^{11} + 62945043 T^{12} + 274846989 T^{13} - 1270238787 T^{14} - 5363230434 T^{15} + 16983563041 T^{16}$$
$23$ $$( 1 + 8 T + 83 T^{2} + 402 T^{3} + 2555 T^{4} + 9246 T^{5} + 43907 T^{6} + 97336 T^{7} + 279841 T^{8} )^{2}$$
$29$ $$1 - 6 T + 23 T^{2} + 6 T^{3} + 138 T^{4} + 5484 T^{5} - 19805 T^{6} + 121230 T^{7} + 57131 T^{8} + 3515670 T^{9} - 16656005 T^{10} + 133749276 T^{11} + 97604778 T^{12} + 123066894 T^{13} + 13680936383 T^{14} - 103499257854 T^{15} + 500246412961 T^{16}$$
$31$ $$1 - 14 T + 29 T^{2} + 248 T^{3} + 530 T^{4} - 15954 T^{5} + 65935 T^{6} - 166230 T^{7} + 667951 T^{8} - 5153130 T^{9} + 63363535 T^{10} - 475285614 T^{11} + 489466130 T^{12} + 7100029448 T^{13} + 25737606749 T^{14} - 385176597554 T^{15} + 852891037441 T^{16}$$
$37$ $$1 - T - 48 T^{2} + 175 T^{3} + 2760 T^{4} - 1334 T^{5} - 138534 T^{6} - 40270 T^{7} + 6227939 T^{8} - 1489990 T^{9} - 189653046 T^{10} - 67571102 T^{11} + 5172684360 T^{12} + 12135192475 T^{13} - 123154867632 T^{14} - 94931877133 T^{15} + 3512479453921 T^{16}$$
$41$ $$1 - 18 T + 145 T^{2} - 1046 T^{3} + 9461 T^{4} - 59718 T^{5} + 237483 T^{6} - 1463634 T^{7} + 11961308 T^{8} - 60008994 T^{9} + 399208923 T^{10} - 4115824278 T^{11} + 26734524821 T^{12} - 121185586246 T^{13} + 688765114945 T^{14} - 3505576929858 T^{15} + 7984925229121 T^{16}$$
$43$ $$( 1 - 4 T + 45 T^{2} - 172 T^{3} + 1849 T^{4} )^{4}$$
$47$ $$1 - 7 T + 85 T^{2} - 1176 T^{3} + 9351 T^{4} - 81508 T^{5} + 731706 T^{6} - 5276705 T^{7} + 35329489 T^{8} - 248005135 T^{9} + 1616338554 T^{10} - 8462405084 T^{11} + 45629897031 T^{12} - 269709728232 T^{13} + 916233302965 T^{14} - 3546361843241 T^{15} + 23811286661761 T^{16}$$
$53$ $$1 - 7 T + 132 T^{2} - 1421 T^{3} + 16876 T^{4} - 131512 T^{5} + 1370850 T^{6} - 9985608 T^{7} + 80785479 T^{8} - 529237224 T^{9} + 3850717650 T^{10} - 19579112024 T^{11} + 133159757356 T^{12} - 594255795553 T^{13} + 2925695669028 T^{14} - 8222977978859 T^{15} + 62259690411361 T^{16}$$
$59$ $$1 - 9 T^{2} + 800 T^{3} + 2280 T^{4} - 15200 T^{5} + 247289 T^{6} + 987600 T^{7} - 4469641 T^{8} + 58268400 T^{9} + 860813009 T^{10} - 3121760800 T^{11} + 27627583080 T^{12} + 571939439200 T^{13} - 379624802769 T^{14} + 146830437604321 T^{16}$$
$61$ $$1 - 12 T + 147 T^{2} - 2312 T^{3} + 22368 T^{4} - 224532 T^{5} + 2244305 T^{6} - 18537880 T^{7} + 147792991 T^{8} - 1130810680 T^{9} + 8351058905 T^{10} - 50964497892 T^{11} + 309703771488 T^{12} - 1952706647912 T^{13} + 7573495031067 T^{14} - 37712914032252 T^{15} + 191707312997281 T^{16}$$
$67$ $$( 1 + 15 T + 335 T^{2} + 3060 T^{3} + 35713 T^{4} + 205020 T^{5} + 1503815 T^{6} + 4511445 T^{7} + 20151121 T^{8} )^{2}$$
$71$ $$1 - 21 T + 151 T^{2} - 497 T^{3} + 5457 T^{4} - 14896 T^{5} - 888488 T^{6} + 11215568 T^{7} - 84203719 T^{8} + 796305328 T^{9} - 4478868008 T^{10} - 5331442256 T^{11} + 138671543217 T^{12} - 896701987447 T^{13} + 19343142872071 T^{14} - 190997523326211 T^{15} + 645753531245761 T^{16}$$
$73$ $$1 - 8 T - 77 T^{2} + 990 T^{3} - 2170 T^{4} - 132362 T^{5} + 1139369 T^{6} + 4655830 T^{7} - 91523921 T^{8} + 339875590 T^{9} + 6071697401 T^{10} - 51491068154 T^{11} - 61624182970 T^{12} + 2052340877070 T^{13} - 11652735424253 T^{14} - 88379188152776 T^{15} + 806460091894081 T^{16}$$
$79$ $$1 - T - 160 T^{2} - 548 T^{3} + 13806 T^{4} + 31769 T^{5} - 282588 T^{6} - 2159462 T^{7} + 16750183 T^{8} - 170597498 T^{9} - 1763631708 T^{10} + 15663355991 T^{11} + 537744818286 T^{12} - 1686226906652 T^{13} - 38893992883360 T^{14} - 19203908986159 T^{15} + 1517108809906561 T^{16}$$
$83$ $$1 + 22 T + 313 T^{2} + 2340 T^{3} + 7050 T^{4} - 116402 T^{5} - 1950441 T^{6} - 18988930 T^{7} - 158760641 T^{8} - 1576081190 T^{9} - 13436588049 T^{10} - 66557150374 T^{11} + 334581163050 T^{12} + 9217355104620 T^{13} + 102332336864497 T^{14} + 596993121771794 T^{15} + 2252292232139041 T^{16}$$
$89$ $$( 1 + 17 T + 312 T^{2} + 3419 T^{3} + 38939 T^{4} + 304291 T^{5} + 2471352 T^{6} + 11984473 T^{7} + 62742241 T^{8} )^{2}$$
$97$ $$1 + 15 T + 36 T^{2} + 375 T^{3} + 2412 T^{4} - 52890 T^{5} + 1250458 T^{6} + 18589350 T^{7} + 83790255 T^{8} + 1803166950 T^{9} + 11765559322 T^{10} - 48271274970 T^{11} + 213532625772 T^{12} + 3220252596375 T^{13} + 29986992177444 T^{14} + 1211974267171695 T^{15} + 7837433594376961 T^{16}$$
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2020-03-29 04:56:22
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https://www.aimsciences.org/article/doi/10.3934/naco.2013.3.261
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# American Institute of Mathematical Sciences
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Linearized alternating direction method of multipliers with Gaussian back substitution for separable convex programming
2013, 3(2): 261-270. doi: 10.3934/naco.2013.3.261
## The stationary iterations revisited
1 Department of Computer Science, Fitchburg State University, Fitchburg, MA 01420, United States 2 Institute of Mathematics, School of Mathematical Sciences, Fudan University, Shanghai 200433, China 3 School of Mathematical Sciences and Shanghai Key Laboratory of Contemporary Applied Mathematics, Fudan University, Shanghai 200433, China
Received February 2012 Revised January 2013 Published April 2013
In this paper, we first present a necessary and sufficient conditions for the weakly and strongly convergence of the general stationary iterations $x^{(k+1)} = T x^{(k)} +c$ with initial iteration matrix $T$ and vectors $c$ and $x^{(0)}$. Then we apply these general results and present convergence conditions for the stationary iterations for solving singular linear system $A x = b$. We show that our convergence conditions are weaker and more general than the known results.
Citation: Xuzhou Chen, Xinghua Shi, Yimin Wei. The stationary iterations revisited. Numerical Algebra, Control & Optimization, 2013, 3 (2) : 261-270. doi: 10.3934/naco.2013.3.261
##### References:
show all references
##### References:
[1] Xiao Tang, Yingying Zeng, Weinian Zhang. Interval homeomorphic solutions of a functional equation of nonautonomous iterations. Discrete & Continuous Dynamical Systems, 2020, 40 (12) : 6967-6984. doi: 10.3934/dcds.2020214 [2] Grzegorz Graff, Piotr Nowak-Przygodzki. Fixed point indices of iterations of $C^1$ maps in $R^3$. Discrete & Continuous Dynamical Systems, 2006, 16 (4) : 843-856. doi: 10.3934/dcds.2006.16.843 [3] Shengxin Zhu, Tongxiang Gu, Xingping Liu. AIMS: Average information matrix splitting. Mathematical Foundations of Computing, 2020, 3 (4) : 301-308. doi: 10.3934/mfc.2020012 [4] Daniel Alpay, Eduard Tsekanovskiĭ. Subclasses of Herglotz-Nevanlinna matrix-valued functtons and linear systems. Conference Publications, 2001, 2001 (Special) : 1-13. doi: 10.3934/proc.2001.2001.1 [5] Guangcun Lu. Parameterized splitting theorems and bifurcations for potential operators, Part II: Applications to quasi-linear elliptic equations and systems. Discrete & Continuous Dynamical Systems, 2021 doi: 10.3934/dcds.2021155 [6] Tohru Nakamura, Shinya Nishibata, Naoto Usami. Convergence rate of solutions towards the stationary solutions to symmetric hyperbolic-parabolic systems in half space. Kinetic & Related Models, 2018, 11 (4) : 757-793. doi: 10.3934/krm.2018031 [7] Yanxing Cui, Chuanlong Wang, Ruiping Wen. On the convergence of generalized parallel multisplitting iterative methods for semidefinite linear systems. Numerical Algebra, Control & Optimization, 2012, 2 (4) : 863-873. doi: 10.3934/naco.2012.2.863 [8] Marco Di Francesco, Donatella Donatelli. Singular convergence of nonlinear hyperbolic chemotaxis systems to Keller-Segel type models. Discrete & Continuous Dynamical Systems - B, 2010, 13 (1) : 79-100. doi: 10.3934/dcdsb.2010.13.79 [9] Zhili Ge, Gang Qian, Deren Han. Global convergence of an inexact operator splitting method for monotone variational inequalities. Journal of Industrial & Management Optimization, 2011, 7 (4) : 1013-1026. doi: 10.3934/jimo.2011.7.1013 [10] Peizhao Yu, Guoshan Zhang, Yi Zhang. Decoupling of cubic polynomial matrix systems. Numerical Algebra, Control & Optimization, 2021, 11 (1) : 13-26. doi: 10.3934/naco.2020012 [11] Franck Davhys Reval Langa, Morgan Pierre. A doubly splitting scheme for the Caginalp system with singular potentials and dynamic boundary conditions. Discrete & Continuous Dynamical Systems - S, 2021, 14 (2) : 653-676. doi: 10.3934/dcdss.2020353 [12] Liejune Shiau, Roland Glowinski. Operator splitting method for friction constrained dynamical systems. Conference Publications, 2005, 2005 (Special) : 806-815. doi: 10.3934/proc.2005.2005.806 [13] Nguyen H. Sau, Vu N. Phat. LP approach to exponential stabilization of singular linear positive time-delay systems via memory state feedback. Journal of Industrial & Management Optimization, 2018, 14 (2) : 583-596. doi: 10.3934/jimo.2017061 [14] Yaonan Ma, Li-Zhi Liao. The Glowinski–Le Tallec splitting method revisited: A general convergence and convergence rate analysis. Journal of Industrial & Management Optimization, 2021, 17 (4) : 1681-1711. doi: 10.3934/jimo.2020040 [15] Parikshit Upadhyaya, Elias Jarlebring, Emanuel H. Rubensson. A density matrix approach to the convergence of the self-consistent field iteration. Numerical Algebra, Control & Optimization, 2021, 11 (1) : 99-115. doi: 10.3934/naco.2020018 [16] Wei-guo Wang, Wei-chao Wang, Ren-cang Li. Deflating irreducible singular M-matrix algebraic Riccati equations. Numerical Algebra, Control & Optimization, 2013, 3 (3) : 491-518. doi: 10.3934/naco.2013.3.491 [17] Davide Guidetti. Convergence to a stationary state of solutions to inverse problems of parabolic type. Discrete & Continuous Dynamical Systems - S, 2013, 6 (3) : 711-722. doi: 10.3934/dcdss.2013.6.711 [18] M. Grasselli, Hana Petzeltová, Giulio Schimperna. Convergence to stationary solutions for a parabolic-hyperbolic phase-field system. Communications on Pure & Applied Analysis, 2006, 5 (4) : 827-838. doi: 10.3934/cpaa.2006.5.827 [19] Claudio Marchi. On the convergence of singular perturbations of Hamilton-Jacobi equations. Communications on Pure & Applied Analysis, 2010, 9 (5) : 1363-1377. doi: 10.3934/cpaa.2010.9.1363 [20] Angelo B. Mingarelli. Nonlinear functionals in oscillation theory of matrix differential systems. Communications on Pure & Applied Analysis, 2004, 3 (1) : 75-84. doi: 10.3934/cpaa.2004.3.75
Impact Factor:
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2021-12-06 00:04:26
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https://tex.stackexchange.com/questions/262909/why-can-you-not-skip-lines-in-document-when-providing-optional-environment-param
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# Why can you not skip lines in document when providing optional environment parameters with xparse?
Is it intentional that you cannot skip lines when using optional environment parameters with LaTeX3's xparse?
It seems that you can skip lines with mandatory parameters.
# Compare
\documentclass{article}
\usepackage{fontspec}
\usepackage{xparse}
\NewDocumentEnvironment{mandatory}{ m m m }
{\section{Mandatory}}
{#1\par #2\par #3}
\NewDocumentEnvironment{optional}{ O{first} O{second} O{third} }
{\section{Optional}}
{#1\par #2\par #3}
\begin{document}
\begin{mandatory}
{FIRST}
{SECOND}
{THIRD}
\end{mandatory}
\begin{optional} %<- Here is the problem child.
[FIRST]
[SECOND]
[THIRD]
\end{optional}
\begin{optional}[FIRST][SECOND][THIRD]
\end{optional}
\end{document}
# Output
• Yes, it is intentional; the rationale is that you may want a [ that's not interpreted as the start of an optional argument. Anyway, trailing optional arguments for “document commands” is not recommended. This appeared first in amsmath for \\ on a line and [ on the following line, say of an align. – egreg Aug 24 '15 at 13:35
• @egreg Ok, that makes sense. My use of trailing arguments is to create an environment that will go at the beginning of every document that contains certain required metadata that when not provided, show up in the document in nice bright letters for all to see (demonstrating that this info is indeed missing). parameters: <privacylevel> <target group> <purpose> I thought it would be better to let the parameters trail, because two of those mentioned contain paragraphs. It was purely a readability issue (in the code). – Jonathan Komar Aug 24 '15 at 13:42
• I'd advocate a key-value interface for this. – egreg Aug 24 '15 at 13:43
• @egreg Would you consider making your comment an answer so that I can close this question? When you have time... – Jonathan Komar Aug 31 '15 at 9:10
It is intentional, but, as far as I know, there are still discussions about it.
The problem appeared several years ago within amsmath, when something like
\begin{align}
a &= b \\
[c] &= d
\end{align}
was considered. With the default LaTeX setup, this triggers an error (Missing number, treated as zero), because \\ would ignore spaces when looking for its optional argument. So the developers of amsmath decided to use a different version of \@ifnextchar that requires no space to intervene between \\ and its optional argument.
A similar approach is used in xparse, but only for environments, as far as I can see. So
\documentclass{article}
\usepackage{xparse}
\NewDocumentEnvironment{foo}{O{x}O{y}}
{START:#1#2}
{END}
\NewDocumentCommand{\?}{O{x}O{y}m}{#1#2#3}
\begin{document}
\begin{foo}[A][B]
C
\end{foo}
\begin{foo}
[A]
[B]
C
\end{foo}
\?
[A]
[B]
{C}
\end{document}
would turn out in bad output only for the second foo environment call, where
START:xy [A] [B] C END
would be printed.
A workaround, not a recommendation, would be
\makeatletter
\NewDocumentEnvironment{optional}
{ O{first} t\@sptoken O{second} t\@sptoken O{third} }
{\section{Optional}}
{#1\par #3\par #5}
\makeatother
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2020-02-22 23:36:52
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http://openstudy.com/updates/52bc7a92e4b04833919df709
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## A community for students. Sign up today
Here's the question you clicked on:
## anonymous 2 years ago a block of mass In = 2.5 kg slides head on into a spring of spring constant k = 320 N/m. When the block stops, it has compressed the spring by 7.5 cm. The coefficient of kinetic friction between block and floor is 0.25. While the block is in contact with the spring and being brought to rest, what are (a) the work done by the spring force and (b) the increase in thermal energy of the block-floor system? (c) What is the block's speed just as it reaches the spring?
• This Question is Open
1. anonymous
2. anonymous
(a) 240 joules....(b)????.....(c)?????????
3. anonymous
lol dude ur awesome...lol that was myself
4. anonymous
wait a second
5. anonymous
obviously work done by spring force is change in spring potential which is $kx ^{2}\div2$ increase in thermal enegry is the work done by frictional force which is $\mu \times normal \times displacement$ adding the above two we get the energy of the block just as it approches the spring $w = mv ^{2}/2$(kinetic work therom)
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2016-09-29 17:01:53
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https://codereview.stackexchange.com/questions/191162/too-slow-json-data-processing-using-sql
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# Too slow JSON data processing using SQL
I have a case study where I need to take data from a REST API do some analysis on the data using an aggregate function, joins etc and use the response data in JSON format to plot some retail graphs.
Approaches being followed till now:
1. Read the data from JSON store these in python variable and use insert to hit the SQL query. Obviously, it is a costly operation because for every JSON line read it is inserting into the database. For 33k rows it is taking more than 20 mins which is inefficient.
2. This can be handled in elastic search for faster processing but complex operation like joins are not present in elastic search.
If anybody can suggest what would be the best approach (like preprocessing or post-processing in python) to follow for handling such scenarios it would be helpful.
SQL Script
def store_data(AccountNo)
db=MySQLdb.connect(host=HOST, user=USER, passwd=PASSWD, db=DATABASE, charset="utf8")
cursor = db.cursor()
insert_query = "INSERT INTO cstore (AccountNo) VALUES (%s)"
cursor.execute(insert_query, (AccountNo))
db.commit()
cursor.close()
db.close()
return
def on_data(file_path):
#This is the meat of the script...it connects to your mongoDB and stores the tweet
try:
# Decode the JSON from Twitter
testFile = open(file_path)
#print (len(datajson))
#grab the wanted data from the Tweet
for i in range(len(datajson)):
store_data( AccountNo)
Here are a few things you can do to improve the performance of the data loading:
• convert the JSON file to CSV and use LOAD DATA to load from a file (sample). It is probably the fastest way to do what you are trying to do.
• use .executemany() instead of .execute():
datajson = json.load(testFile)
insert_query = """
INSERT INTO
cstore (AccountNo)
VALUES (%(AccountNo)s)
"""
cursor.executemany(insert_query, datajson)
• look into disabling/removing existing indexes during the insertion and then re-creating them after the insertion is done
• make sure you are not doing the data-load "over the internet" and there is no network-latency and bandwidth impact, be "closer" to the database server
• loading the JSON file is probably not a bottleneck, but you may also look into faster JSON parsers like ujson
• I remember we also had some data-loading-performance-related success with the ultra-fast umysql Python database driver, but it looks like the package has not been maintained for quite a while
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2020-01-19 00:50:52
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https://homotopytypetheory.org/2015/07/28/constructing-the-propositional-truncation-using-nonrecursive-hits/
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## Constructing the Propositional Truncation using Nonrecursive HITs
In this post, I want to talk about a construction of the propositional truncation of an arbitrary type using only non-recursive HITs. The whole construction is formalized in the new proof assistant Lean, and available on Github. I’ll write another blog post explaining more about Lean in August.
The construction of the propositional truncation of a type $A$ is as follows. Let $\{A\}$ be the following HIT:
HIT {-} (A : Type) : Type :=
| f : A → {A}
| e : ∀(a b : A), f a = f b
I call this the “one step truncation”, because the HIT is like the propositional truncation, but takes out the recursive part. Martin Escardo calls this the generalized circle, since $\{\mathbf{1}\} \simeq S^1$. We can repeat this multiple times: $A_0 :\equiv A$ and $A_{n+1} :\equiv \{A_n\}$. Now the colimit $A_\infty$ of this sequence is the propositional truncation of $A$: $A_\infty \simeq \| A \|$. To make sure we’re on the same page, and to introduce the notation I use for the constructors, the (sequential) colimit is the following HIT: (I will mostly leave the arguments in $\mathbb{N}$ implicit)
HIT colimit (A : ℕ → Type)
(f : ∀(n : ℕ), A n → A (n+1)) : Type :=
| i : ∀(n : ℕ), A n → colimit A f
| g : ∀(n : ℕ) (a : A n), i (f a) = i a
Below I’ll give an intuition of why the construction works, and after that a proof. But first I want to talk about some consequences of the construction.
First of all this is the first instance of a HIT which has a recursive path constructor which has been reduced to non-recursive HITs. During the HoTT workshop in Warsaw there was a conjecture that all HITs with recursive (higher) path constructors can be reduced to HITs where only point constructors are allowed to be recursive. Although I have no idea whether this conjecture is true in general, this is the first step towards proving it.
Secondly, this gives an alternative answer to the question about the universal property of the propositional truncation. This construction shows that (using function extensionality) a function $h : \|A\| \to B$ for an arbitrary type $B$ is “the same” as a sequence of functions $h_n : A_n \to B$ (with $A_n$ defined above) such that for all $a : A_n$ we have $h_{n+1}\ (f\ a) = h\ a$. In a formula
$(\|A\| \to B)\ \simeq\ (\Sigma(h : \Pi n, A_n \to B), \Pi (n : \mathbb{N}) (a : A_n), h_{n+1}\ (f\ a) = h_n\ a)$.
So, why does the construction work?
To give an intuition why this works consider the propositional truncation of the booleans, $\|2\|$. Of course, this type is equivalent to the interval (and any other contractible type), but there is actually a little more going on. The path constructor of the propositional truncation (let’s call it $e'$ here) gives rise to two paths between $0_2$ and $1_{2}$: $e'\ |0_{2}|\ |1_{2}|$ and $(e'\ |1_{2}|\ |0_{2}|)^{-1}$. Since the resulting type is a mere proposition, these paths must be equal. We can construct a path between them explicitly by using the continuity of $e'$. We can prove that for every $x : \|2\|$ and every $p : |0_2| = x$ we have $p = (e' \ |0_{2}|\ |0_{2}|)^{-1}\cdot e' \ |0_{2}|\ x$ by path induction, and this leads to the conclusion that any two elements in $|0_2| = |1_{2}|$ are equal. This is exactly the proof that any proposition is a set.
Now consider the type $\{2\}$. This is a type with points $a:\equiv f\ 0_2$ and $b:\equiv f\ 1_2$ and we have two paths from $a$ to $b$. Let’s call them $p:\equiv e\ 0_2\ 1_2$ and $q:\equiv(e\ 1_2\ 0_2)^{-1}$. These two paths are not equal. Of course, we also have a loop at both $a$ and $b$.
However, in the type $\{\{2\}\}$ the corresponding paths $\text{ap}_f\ p$ and $\text{ap}_f\ q$ of type $f\ a = f\ b$ are equal. To see this, we can mimic the above proof. We can prove that for all $x : \{2\}$ and all $p : a = x$ we have $\text{ap}_f\ p = (e\ a\ a)^{-1} \cdot e\ a\ x$, just by path induction on $p$. Then we can conclude that $\text{ap}_f\ p = \text{ap}_f\ q$. Of course, in $\{\{2\}\}$ there are new paths of type $f\ a = f\ b$, but they will be identified again in $\{\{\{2\}\}\}$.
<sidenote>
The above argument also shows that if any function $f$ is weakly constant, then $\text{ap}_f : x = y \to f\ x = f\ y$ is weakly constant for any $x,y$ (using the terminology of this post).
</sidenote>
So this is the intuition why $A_\infty$ is the propositional truncation of $A$. In $\{A\}$ a path is added between any two points of $A$, between any two existing parallel paths (and between any two parallel higher paths). Then in $\{\{A\}\}$ higher paths are added between all newly introduced (higher) paths, and so on. In the colimit, paths exist between any two points, parallel paths, and parallel higher paths, so we get the propositional truncation.
The proof.
But this is just the intuition. How do we prove that we get the propositional truncation? We clearly get the point constructor for the propositional truncation, because $i_0$ (the constructor of the colimit) has type $A \to A_\infty$. Now we still have to show two things:
1. We have to show that $A_\infty$ is a mere proposition
2. We have to construct the induction principle for the propositional truncation on $A_\infty$, with the judgmental computation rule
2. The second part is easy. If we’re given a family of propositions $P : A_\infty \to \text{Prop}$ with a map $h : \Pi(a : A), P\ (i_0\ a)$ we have to construct a map $k : \Pi(x : A_\infty), P\ x$. We do this by induction on $x : A_\infty$. Since we construct something in $P\ a$, which is a mere proposition, the path construction is automatic, and we only have to deal with the case that $x\equiv i_n\ a$ for some $n : \mathbb{N}$ and $a : A\ n$. Now we do induction on $n$.
If $n\equiv0$, we can choose the element $h\ a : P\ (i_0\ a)$.
If $n\equiv k+1$, we know that $a : \{A_k\}$, so we can induct on $a$. If $a\equiv f\ b$, we need an element of $P\ (i_{n+1}\ (f\ b))$. We have an element of $P\ (i_n\ b)$ by induction hypothesis. So we can transport this point along the equality $(g_n\ b)^{-1} : i_n\ b = i_{n+1}\ (f\ b)$ to get the desired point. The path construction is again automatic. In pattern matching notation we have defined:
• $k\ (i_{0}\ a) :\equiv h\ a$ and
• $k\ (i_{n+1}\ (f\ b)) :\equiv (g_n\ b)^{-1}_*(k\ (i_n\ b))$.
The definition $k\ (i_{0}\ a) :\equiv h\ a$ is also the (judgmental) computation rule for the propositional truncation.
1. So we only need to show that $A_\infty$ is a mere proposition, i.e. $\Pi(x\ y : A_\infty), x = y$. We first do induction on $x$. Note that $\Pi(y : A _\infty), x = y$ is a mere proposition. (Left as exercise to the reader. Hint: assume that it is inhabited.) So we only have to consider the case where $x$ is a point constructor, say $x\equiv i_n\ a$. Now we do induction on $y$. Now the goal is $i_n\ a = y$, and we don’t know (yet) that this is a mere proposition, so we have to show two things:
1. We have to construct a $p\ a\ b : i_n\ a = i_m\ b$ for all $n,m,a,b$;
2. We have to show that $p$ satisfies the following coherence condition: $p\ a\ (f\ b)\ \cdot\ g_m\ b = p\ a\ b$.
In particular, we want to make the construction of $p$ as simple as possible, so that the coherence condition for $p$ remains doable.
For the construction of $p$, let’s first consider a special case where $a$ and $b$ live in the same level, i.e. where $n \equiv m$. This will be used in the general construction. In this case, we have $i_n\ a = i_{n+1}\ (f\ a) = i_{n+1}\ (f\ b) = i_n\ b$, where the second equality is by the path constructor $e$ of $\{A_n\}$ and the other two equalities are by the path constructor $g$ of the colimit. Let’s call this equality $q\ a\ b : i_n\ a = i_n\ b$.
For the general construction of $p$ with arbitrary $n$ and $m$, I have considered three approaches. I will first discuss two approaches where I failed to complete the proof.
(failed) Approach 1. Induct on both $n$ and $m$. In the successor case, induct on $a$ and $b$. Unfortunately, this becomes horrible very quickly, because we now have a nested double induction on HITs. This means that we already need to prove coherence conditions just to construct $p$, and after that we still need to show that $p$ satisfies a coherence condition. I quickly gave up on this idea.
(failed) Approach 2. Lift $a$ and $b$ both to $A\ (n + m)$ by repeatedly applying $f$, and then show that $i_n\ a = i_{n+m}\ (f^m\ a) = i_{n+m}\ (f^n\ b) = i_m\ b$, where the second equality is $q\ (f^m\ a)\ (f^n\ b)$. This works, but there is a catch: $f^m\ a$ lives in $A (n + m)$ and $f^n\ b$ lives in $A (m + n)$ (assuming addition is defined by induction on the second argument). So to complete the construction, we also have to transport along the equality $n+m=m+n$. This is fine, but for the coherence condition for $p$ this transport becomes an annoying obstacle. Showing the coherence condition is possible, but I didn’t see how to do it. Then realized there was a more convenient approach, where I didn’t have to worry about transports.
(succeeded) Approach 3. Distinguish cases between $n\le m$ and $m \le n$ and induct on those inequalities. I’m assuming that the inequality is defined by an (ordinary) inductive family
inductive le (n : ℕ) : ℕ → Type :=
| refl : le n n
| step : Π(m : ℕ), le n m → le n (m+1)
This definition gives a very useful induction principle for this construction. If $H : m\le n$ we can construct a map $f^H : A\ m \to A\ n$ by induction on $H$:
• $f^{\text{refl}_n}\ a:\equiv a$;
• $f^{\text{step}\ H}\ a:\equiv f\ (f^H\ a)$.
I use the notation $f^H$ for this, similar to $f^n$, because we repeatedly apply $f$ and the number of times is determined by $H$. Since the type $m \le n$ is a mere proposition, $f^H$ doesn’t really depend on $H$, only on the fact that its type $m \le n$ has an inhabitant.
We can now also show that $g^H\ b : i_n\ (f^H\ b) = i_m\ b$ by induction on $H$ as concatenation of multiple $g$s. Now if $H : m \le n$ we can construct $p\ a\ b$ by $i_n\ a = i_n\ (f^H\ b) = i_m\ b$, where the first equality is $q\ a\ (f^H\ b)$. The case $n \le m$ is similar.
This is a nice and simple construction of $p$ (without transports), and we can prove the coherence conditions using this definition, but that is more involved.
We have constructed $p$ in such a way that for $H : n \le m$ we have that $p\ a\ b$ equals $(g^H\ a)^{-1}\ \cdot\ q\ (f^H\ a)\ b$ and in the case that $H : m \le n$ we have that $p\ a\ b$ equals $q\ a\ (f^H\ b)\ \cdot\ g^H\ b$. This is not purely by definition, for the case $n = m$ we have to prove something here, because we cannot get both inequalities by definition.
Now we have to show that $p\ a\ (f\ b)\ \cdot\ g_m\ b = p\ a\ b$. We distinguish two cases: $n > m$ and $n \le m$.
Case $n > m$
In this case we can find $H : m+1 \le n$ and $H' : m \le n$. In this case $p\ a\ b = q\ a\ (f^{H'}\ b)\ \cdot\ g^H\ b$ and $p\ a\ (f\ b) = q\ a\ (f^{H}\ (f\ b))\ \cdot\ g^H\ (f\ b)$
The situation is displayed below (click image for larger view). Some arguments of paths are omitted, and the maps $i$ and $\text{ap}_i$ are also omitted for readability (the actual equalities live in $A_\infty$). We need to fill the pentagon formed by the outer (thick) equalities.
By induction on $H$ we can find a path $r : f^{H'}\ b = f^H\ (f\ b)$ and by another induction on $H$ we can fill the bottom square in the diagram. We will skip the details here. The top triangle can be filled for a general path $r$ by induction on $r$. This finishes the case $n > m$.
Case $n \le m$
Now we can find $H : n \le m$ and $H' : n \le m+1$. Here $p\ a\ b = (g^H\ a)^{-1}\ \cdot\ q\ (f^H\ a)\ b$ and $p\ a\ (f\ b) = (g^{H'}\ a)^{-1}\ \cdot\ q\ (f^{H'}\ a)\ (f\ b)$. The situation is below.
If we choose $H'$ correctly (namely as $\text{step}\ H$), then by definition $f^{H'}\ a\equiv f\ (f^H\ a)$ and $g^{H'}\ a\equiv g\ (f^H\ a)\ \cdot\ g^H\ a$, so the triangle in the left of the diagram is a definitional equality. We can prove the remaining square in more generality. See the below diagram, where we mention $i$ explicitly.
The bottom square is filled by induction on $p$. To show that the two paths in the top are equal, first note that $i_{k+1} : A_{k+1} \to A_\infty$ is weakly constant, because that is exactly the type of $q$. So $\text{ap}_i : f\ x = f\ y \to i\ (f\ x) = i\ (f\ y)$ is weakly constant. This finishes the case $n \le m$, and hence the proof that $A_\infty$ is a mere proposition.
Future Work
The natural question is whether we can construct arbitrary $n$-truncations using non-recursive HITs. The construction in this post can be easily generalized by using an “one-step $n$-truncation,” but the proof that the resulting type is $n$-truncated cannot be generalized easily. However, Egbert Rijke has a proof sketch of the construction of localizations between $\omega$-compact types using nonrecursive HITs, and these include all the $n$-truncations.
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### 25 Responses to Constructing the Propositional Truncation using Nonrecursive HITs
1. This is very nice! In particular, it gives an answer to the following question: Suppose you want to eliminate $\left\Vert A \right \Vert \to B$ using a function $A \to B$. What should the function $A \to B$ satisfy? At least, it should be weakly constant. Your answer is that it is enough to exhibit a cocone whose first leg is the function $A \to B$. The second leg gives the weak constancy of $A \to B$. Then the other legs give the weak constancy of the previous legs. We may call such a function $A \to B$ “coherently constant” if it can be completed to such a cocone. Then your answer is that we can eliminate $\left\Vert A \right \Vert \to B$ using a coherently constant function $A \to B$.
Question: We know that we can eliminate $\left\Vert A \right\Vert \to A$ using any weakly constant function $A \to A$ (so that we don’t need to give all coherence data when $B=A$). There are at least two explanations of this. Does your work give a third explanation?
• Floris van Doorn says:
To answer your question: Yes! I do not know either of the other two explanations, so I can’t compare my argument with them. Can you provide a link?
Using this construction: Suppose we’re given a weakly constant function $A \to A$. This corresponds exactly to a function $\{A\} \to A$. By functoriality of $\{{-}\}$ we get a map $\{\{A\}\} \to \{A\}\to A$ and similarly by induction a map $h_n : A_n \to A$. These maps do indeed form a cocone, the equality $h_{n+1}\ (f\ a)=h_n\ a$ is true by reflexivity. So this gives a map $\|A\|\to A$.
• Nice again! Just a corollary of your development.
Here are the links you want:
The first explanation is by Nicolai Kraus: The type of fixed points of any weakly constant endomap is a proposition,
http://www.cs.nott.ac.uk/~ngk/docs/hedberg.pdf,
from which the claim follows almost directly.
The second explanation is the same, with a different argument for the fact that the type of fixed points is a proposition, by Christian Sattler, recorded in page 19 of
• Oh, this is great! (And thanks to Martin for giving the links to our previous work!)
2. Another question is this: Nicolai has an alternative notion of coherently constant function with the same elimination property: http://arxiv.org/abs/1411.2682
I wonder how this relates to your work.
• Floris van Doorn says:
Here is a comparison of the resulting universal property you get.
The disadvantage of my universal property is that it is “useless” when eliminating to $n$-types. That is: the universal property is not easier when assuming that the codomain is an $n$-type; you still have to construct a cocone over the whole diagram. In Nicolai’s work you only have to check finitely many things when eliminating to an $n$-type.
Nicolai also had an idea where in the $n$-th step of the diagram you only add $n$-dimensional paths, so that $A_n$ becomes $n$-connected. If that works, this gives a nice universal property when eliminating to $n$-types.
The advantage of my work is that it is possible to formulate the general universal property (without assuming that the codomain is an $n$-type) inside type theory, without assuming that the type theory has Reedy limits. And that it has been formalized in a proof assistant.
In neither case it is clear how useful the universal property is in applications; i.e. how bothersome it is to construct all the required data.
• Mike Shulman says:
The universal property for eliminating to sets can be obtained by noting that $\Vert \{A\}\Vert_0 = \Vert A\Vert_{-1}$. If it were true that $\Vert A_{n+1} \Vert_n = \Vert A\Vert_{-1}$ as well, then that would also give a universal property for eliminating to $n$-types; but it sounds from your comment like maybe that is not the case?
• Floris van Doorn says:
You’re right, for eliminating to sets this construction does work.
However, in this construction, $\Vert A_{n+1} \Vert_n = \Vert A\Vert_{-1}$ is indeed not true for larger $n$. At every step we’re adding new 1-paths, and only the next step corresponding 2-paths are added equating those 1-paths which are parallel, so $A_n$ is not $n$-connected, even if it is inhabited.
Nicolai’s idea of an alternative construction (which I mentioned in my previous comment) will have this property, so can be used for this.
• Mike Shulman says:
Intriguing. This seems kind of similar to my splitting-idempotents construction; in both cases we seem to get something “fully coherent” and involving paths of all dimensions without ever talking about anything above dimension 1. Of course we’ve seen that before, e.g. in the hub-and-spoke construction and even in Vladimir’s original definition of contractibility, but somehow it is more surprising to me that these two constructions work.
• Mike Shulman says:
Ah, I have a better idea what’s happening now. For any $A$ there is a map $\{A\} \to S^1$ which sends all the f-points to base and all the e-paths to loop. As long as A is inhabited, this map hits loop with an endo-path, which is therefore nontrivial. So there is no way $\{A\}$ could ever be 1-connected. And yet each new 1-path added in $A_n$ gets killed in $A_{n+1}$, so in the colimit they all die. This makes the fact that the colimit ends up propositional more plausible to me than when I was imagining that the connectedness simply increases at every step.
• This makes the fact that the colimit ends up propositional more plausible to me than when I was imagining that the connectedness simply increases at every step.
Yes, Floris’ construction is very clever at that point! So clever that I also found it surprising when I heard it the first time. But I do think that the approach where the connectedness increases at every step works as well (see my longer comment below). It’s true that the argument for that construction would be much easier if we had a “general Whitehead” at hand, but we can do it without (at least, that’s what I currently believe).
• Mike Shulman says:
Maybe the conclusion to draw is that I still don’t really understand what the world looks like when Whitehead’s principle fails, for all that I talk about it a lot.
• Mike Shulman says:
Here’s a possible test case for usefulness of these universal properties. In the proof of Lemma 8.2 in this paper (formalized in Coq here), I used the universal property for eliminating out of $\Vert A\Vert$ into sets. That necessitated a restriction in the lemma statement that $P$ be a family of sets (which was sufficient for what I needed at the time). But can the more general universal properties be used to generalize this lemma?
• Oh, that’s interesting. I hope I can find some time to look at that. Thanks for the pointer.
3. Mike Shulman says:
Indeed, very nice! At first I didn’t think that this would work, because in general a type can be n-connected for all n without being contractible; but somehow the sequential colimit takes care of that. I’m still dubious that one can get arbitrary localizations; does Egbert assume some kind of compactness of the generators?
Another question: can your notion of “coherently constant function” be defined without using the one-step-truncation HIT, and if so, can one show directly that it suffices to eliminate from the propositional truncation?
• Floris van Doorn says:
You’re completely right about the localizations. Egbert assumes that the types for the localization are $\omega$-compact. I clarified it in the article.
Your other question is interesting. I don’t know. To get it, you probably want to give the universal property of $A_n$, but I don’t even see how to get the universal property of $\{\{A\}\}$. I’ll think about it!
4. This is a very interesting construction, and a nice write-up!
For me, it’s especially interesting how you prove that $A_{\infty}$ is a mere proposition. I would have attempted to do it like this:
1. to show $\mathsf{isProp}(A_\infty)$, it’s enough to show $A_\infty \to \mathsf{isContr}(A_\infty)$.
2. because $\mathsf{isContr}(A_\infty)$ is propositional, it is relatively easy to see that $A \to \mathsf{isContr}(A_\infty)$ suffices.
3. given $a : A$, we then want to show that $i_0 a$ is a centre of contraction.
That way, we really only need to do induction on one variable at a time. Maybe it could simplify things a tiny bit, but I think I understand why you didn’t do it: the easy case is if the two points in $A_\infty$ that we want to prove equal come from the same $A_n$, i.e. $i_n a = i_n b$ is much easier that $i_0 a = i_n b$. For comparison, in the other approach that you mentioned in the comments, this is not the case.
(You have described very well what this approach is, but maybe it’s confusing to link between comments like this, so, let me quickly say what I mean. The approach for constructing truncations that Thorsten and I wanted to take is to start with $A$, then take the 1-step [-1]truncation, then take the 1-step 0-truncation of the result, then take the 1-step 1-truncation of that result, and so on. Then, it’s really true that $\Vert A_n \Vert_{n-1}$ is the correct thing, so it would simplify if we want to eliminate into $n$-types – but it’d probably be more complicated otherwise.)
I think I have a proof on paper for that construction, but at the moment not formalised and there could be all kind of flaws (maybe I’ll write a blog post, but only if I manage to find a nice presentation). This approach has its own difficulties, because even reducing the goal to $i_n a = i_n b$ won’t help much; I really need to go all the way down to $i_0 a = i_0 b$, because “level zero” is the only time where we get actual equalities between points.
• Floris van Doorn says:
This idea is great! It doesn’t “simplify things a tiny bit”, it cuts the length my proof in half! If I can assume that $A$ is inhabited by $a$, then I can choose $i_0\ a$ as my center of contraction, and then I only have to worry about other points which are in a *higher* level than $a$. So this makes the whole case $n > m$ obsolete. Moreover, there’s no need to induct on inequalities anymore, we can just induct on ordinary natural numbers. I implemented your proof strategy:
https://github.com/fpvandoorn/leansnippets/blob/master/short_prop_trunc.hlean
The proof that $A_\infty$ is a mere proposition went down from ~163 to ~76 lines.
I feel a bit stupid that I didn’t see this myself. I was already thinking that if I could assume that $A$ is inhabited, things would be easier. I also thought about your first step, but I didn’t see how assuming a point in $A_\infty$ could help: you cannot construct a point in $A_n$ or even $A$ from it. However, you can actually do that if you are proving only a mere proposition.
• Great to see this implemented! I have used this strategy in a couple of other cases before, it’s often useful when you want to prove that some HIT is n-truncated.
The general strategy goes like this: To prove that $X$ is an $n$-type (for $n \geq -1$), it’s enough to show $\mathsf{forall }\, x:X, \mathsf{isContr}(\Omega^{n+1}(X,x))$. But, if $X$ is some HIT, then you can show this by induction on $x$, and, because you’re proving a mere proposition, all you need to check are the point-constructors. All higher constructors are automatic.
There are also some simple strategies to show something of the form $\mathsf{isContr}(\Omega^{n+1}(X,x))$ (basically some higher versions of Hedberg’s argument), I have written about that in my thesis (Thm 3.2.1 summarises some of these “tools”, they are all trivial to prove; in Chp 8.10 in an application).
• Mike Shulman says:
Does this technique make it easier to prove that $S^\infty$ is contractible (with either definition of $S^\infty$)? (Luis Scoccola has recently submitted solutions to the two exercises in the book about this, 8.3 and 8.4.)
• Good question, but I don’t see how it might help for $S^\infty$ (as a colimit). I think it’s the other way round: If you have a more difficult problem, you can try that strategy to reduce it to a problem which looks similar to “colimit – $S^\infty$ is contractible”. For example, in Floris’ construction, it essentially gives you this point in the “starting type” $A_0$, and you already have such a point (or two) in the “starting type” $S^0$. For “$S^\infty$ as its own suspension”, I also don’t think it can be used.
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2021-01-16 12:20:16
|
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|
https://ctf-wiki.github.io/ctf-wiki/reverse/windows/unpack/unpack-dll/
|
# Unpack dll
EN | ZH Here you need to contact the previous section [Manually find IAT and use ImportREC to rebuild] (/reverse/unpack/manually-fix-iat/index.html)
The example file can be downloaded here: unpack_dll.zip
This step is required for Dll shelling. The most critical step for Dll shelling is to use LordPE to modify its Dll flag, open UnpackMe.dll with LordPE, and click on the feature value. ..., then uncheck theDLL flag. After saving, the system will treat the file as an executable file.
We changed the UnpackMe.dll suffix to UnpackMe.exe and loaded it with OD.
Usually at the entry point, the program will save some information, here is very simple, just make a cmp. One thing to note is that the jnz jump here jumps directly to the end of the unpacking process. So we Need to modify the z flag of the register to invalidate the jump. Also set a breakpoint at the end of the unpacking process to avoid shelling and then run directly. (The program will break at this breakpoint, but the shell has been Finished, the code is very clear)
The basic steps of Dll shelling are the same as the exe file shelling, and when rebuilding ʻIAT, you need to follow the previous article [Manually find IAT and use ImportREC to rebuild] (/reverse/unpack/manually-fix-iat /index.html) As mentioned, manually find theIATtable and rebuild it withImportREC. Just note that after unpacking the dump, remember to restore theDLLflag with LordPE and append the file name. Change to.dll.
|
2019-07-17 22:49:48
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|
https://3dprinting.stackexchange.com/questions/3023/how-to-offset-my-probe-so-its-not-hanging-off-the-bed-at-0-0-position-when-pri?noredirect=1
|
# How to offset my probe so it's not hanging off the bed at 0, 0 position when printing
I had my printer printing fine when using the stock trigger switch as I used it to print the green bracket you see in the picture.
My problem now is when I do a print with the sensor, it moves to 0,0 position. However in this position the sensor is hanging off the bed hence there is nothing for it detect so it crashes into the bed.
As far as I can tell the nozzle is homing in the right place.
How do I tell Marlin the new minimum position it needs to be in so it doesn't crash into the bed?
• It does not matter if the probe is not above the bed during printing as long it is above the bed during auto leveling. This can be done by proper settings in your firmware configuration, no hardware changes are necessary. – 0scar Jul 11 '19 at 20:47
There are at least 2 options to address the problem that you have:
1. Adjust end-stops so that in 0,0 position Z-sensor would still hang above the printing table. This would reduce printing surface but allow perfect calibration
2. Mount extra metal plate at the table mount where it would not bump into printer parts and remain reachable for the sensor (perhaps with sensor relocation) when positioned at 0,0. This option requires extra space within table movement boundaries but saves printing surface.
• These solutions are totally unnecessary as it is easily fixed in the firmware. Only during levelling the probe needs to be above the bed, it is perfectly fine to have the sensor off the bed during printing. – 0scar Jul 5 '19 at 22:06
It is not a problem that the sensor is not above the build plate during printing as long as it is above the build plate during the auto bed levelling sequence.
Homing does not necessarily need to be the (0,0) coordinate. Usually, a printer homes on the endstop switches, from that coordinate an offset is defined in the firmware to move to the origin. This implies that (depending on the position of the sensor), the sensor may be outside the bed area when the nozzle is at the origin (0, 0)). Therefore, similarly, you need to tell the printer the location of the Z sensor with respect to the nozzle position in order for the printer to keep the sensor on the bed when levelling by setting boundaries for the sensor to reach.
E.g. for Marlin firmware the offset from homing to the bed origin is defined for an Anet A8 by:
#define X_MIN_POS -33
#define Y_MIN_POS -10
The values you should use need to correspond to the actual offset from the homing point to the origin of the bed (0,0).
When using an auto bed leveling sensor like you are using you should consider this remark:
If using a Probe for Z Homing, enable Z_SAFE_HOMING also!
Un-comment the proper line in the configuration file to read:
#define Z_SAFE_HOMING
This will make the printer aware of the sensor, and home Z in the middle of the bed (default behavior, but can be changed), so that your sensor is never off the bed when probing the bed for Z homing.
Furthermore, you need to set the offset values of the center of your sensor to the nozzle center:
* Z Probe to nozzle (X,Y) offset, relative to (0, 0).
* X and Y offsets must be integers.
*
* In the following example the X and Y offsets are both positive:
* #define X_PROBE_OFFSET_FROM_EXTRUDER 10
* #define Y_PROBE_OFFSET_FROM_EXTRUDER 10
*
* +-- BACK ---+
* | |
* L | (+) P | R <-- probe (20,20)
* E | | I
* F | (-) N (+) | G <-- nozzle (10,10)
* T | | H
* | (-) | T
* | |
* O-- FRONT --+
* (0,0)
*/
#define X_PROBE_OFFSET_FROM_EXTRUDER XXX // X offset: -left +right [of the nozzle]
#define Y_PROBE_OFFSET_FROM_EXTRUDER YYY // Y offset: -front +behind [the nozzle]
#define Z_PROBE_OFFSET_FROM_EXTRUDER 0 // Z offset: -below +above [the nozzle]
Where XXX and YYY are your actual values.
And set the boundary of the probing section:
// Set the boundaries for probing (where the probe can reach).
#define LEFT_PROBE_BED_POSITION 15
#define RIGHT_PROBE_BED_POSITION 190
#define FRONT_PROBE_BED_POSITION 15
#define BACK_PROBE_BED_POSITION 170
Note that the values should match your bed size!
And:
// The Z probe minimum outer margin (to validate G29 parameters).
#define MIN_PROBE_EDGE 10
Details on setting the boundaries of the bed to keep the sensor on the bed is described in question "How to set Z-probe boundary limits in firmware when using automatic bed leveling?".
If using marlin firmware center your prints.
In my case...
#define NOZZLE_X 8
#define NOZZLE_Y -56
Then, set the Z-Probe offset from nozzle. In my case the Z-Probe is 50mm behind the hotend.
#define SENSOR_LEFT 0
#define SENSOR_RIGHT 0
#define SENSOR_FRONT 0
#define SENSOR_BEHIND 50
Finally set the bed extra movement. As you see i added the 50mm's at the back of the bed.
#define XTRA_BED_LEFT 0 // Distance nozzle can move towards the left past X = 0
#define XTRA_BED_RIGHT 0 // Distance nozzle can move towards the right past X = 200
#define XTRA_BED_FRONT 0 // Distance bed can move towards the front past Y = 200 (Y=280 for large bed)
#define XTRA_BED_BACK 50 // Distance bed can move towards the back past Y = 0
This way once auto leveling, the probe starts with (0,0) and the hotend is 50 mm's in front and out of the bed.
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2021-05-12 04:54:29
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https://caddellprep.com/common-core-algebra-assignments/dividing-polynomials/
|
# Dividing Polynomials
In this video, we are going to divide polynomials.
For example: $\frac{x^2+5x+4}{x+1}$
Factor polynomials if necessary
$\frac{(x+4)(x+1)}{x+1}$
Divide like terms
$\frac{(x+4)}{1}$
So we have
${x+4}$
## Video-Lesson Transcript
Let’s go over how to divide polynomials.
First, we have a polynomial divided by a single term.
$\dfrac{6x + 12}{3}$
We’re going to factor this first and see if we can cancel out.
$= \dfrac{6 (x + 2)}{3}$
Then cancel $3$ and $6$.
$= \dfrac{2 (x + 2)}{1}$
$= 2 (x + 2)$
Similarly, we can do the same thing if we divide by polynomial.
Let’s say we have
$\dfrac{x^2 + 5x + 4}{x + 1}$
Let’s factor this first.
$= \dfrac{(x + 4) (x + 1)}{x + 1}$
Then cancel out and we’re left with
$= x + 4$
But that’s not always the case.
$\dfrac{x^2 + 5x + 3}{x + 1}$
We can’t factor the numerator. So we definitely can’t factor and reduce.
So what we’re going to do is long division.
Let’s review long division.
${3} )\overline{\rm 542}$
The answer is $180 \dfrac{2}{3}$
Now, let’s get back and solve using long division.
${x + 1})\overline{\rm x^2 + 5x + 3}$
Since $x + 1$ is a binomial or has tow terms, it cannot go into $x^2$ alone.
So, we’ll use $x^2 + 5x$.
The question is how many times does $x + 1$ goes into $x^2 + 5x$?
The most important thing to look at is the $x$-terms, you can ignore the other numbers for now.
So, what can we multiply to $x$ to get $x^2$?
Well, it’s $x$.
Now we have to subtract and bring down the last number.
How many $x + 1$ goes into $4x + 3$?
Again, let’s just focus on the $x$.
$x + 4 - \dfrac{1}{x + 1}$
Now, let’s look at our first example.
$\dfrac{x^2 + 5x + 4}{x + 1}$
At this point, we know that the answer is
$= x + 4$
Now, let’s just divide this using long division.
${x + 1})\overline{\rm x^2 + 5x + 4}$
$= x + 4$
|
2019-06-16 06:42:12
|
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https://scholarship.claremont.edu/hmc_theses/10/
|
## HMC Senior Theses
2011
#### Document Type
Open Access Senior Thesis
#### Degree Name
Bachelor of Science
#### Department
Mathematics
Arthur T. Benjamin
Kimberly Kindred
#### Rights Information
The Fibonomial numbers are defined by $\begin{bmatrix}n \\ k \end{bmatrix} = \frac{\prod_{i=n-k+1} ^{n} F_i}{\prod_{j=1}^{k} F_j}$ where $F_i$ is the $i$th Fibonacci number, defined by the recurrence $F_n=F_{n-1}+F_{n-2}$ with initial conditions $F_0=0,F_1=1$. In the past year, Sagan and Savage have derived a combinatorial interpretation for these Fibonomial numbers, an interpretation that relies upon tilings of a partition and its complement in a given grid.In this thesis, I investigate previously proven theorems for the Fibonomial numbers and attempt to reinterpret and reprove them in light of this new combinatorial description. I also present combinatorial proofs for some identities I did not find elsewhere in my research and begin the process of creating a general mapping between the two different Fibonomial interpretations. Finally, I provide a discussion of potential directions for future work in this area.
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2019-02-17 18:38:16
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|
https://stats.stackexchange.com/questions/326816/power-function-of-a-test-example
|
# Power function of a test,example
Consider a population with the pdf $N(\theta,1)$ where $\theta$ is unknown and hypothesis $H_0:\theta=5.5$ and $H_1:\theta=8$.Suppose that $\bar{X}=\frac{1}{9}\sum_{i=1}^9 X_i.$Reject $H_0$ iff $\bar{X} >7.5$.Power function is given by $Q(\theta)=P_{\theta }(\cal R).$ $\beta=P{\text{(type-II error)}}=P({\text{Accepting } H_0}$ when $H_1 \text{is true}).$
Why then $Q(\theta_1)=1-\beta$ and $Q(\theta)=1-\Phi(22.5-3\theta)?$
In particular how arises $3$ in front of the $\theta$?
• What $\sum_{i=1}^\theta X_i$ stnd for? – Deep North Feb 4 '18 at 22:26
• See the corrected question. A random sample $(X_1,...,X_9)$ is collected and denote $\bar{X}=\frac{1}{9}\sum_{i=1}^9 X_i$ – user2925716 Feb 5 '18 at 15:32
• @DeepNorth Can you answer now, after the clarification? – user2925716 Feb 5 '18 at 20:24
• Hint: Find the distribution of $\bar X$. (This will show you where the $3$ comes from.) – whuber Feb 5 '18 at 21:17
The power is defined as
$\gamma(\theta_1)$ or $Q(\theta_1)=P_{\theta_1}[(X_1,X_2,...,X_n)\in C]$
$(X_1,X_2,...,X_n)$ is sample space and $C$ is critical region, $\theta_1$ is the parameter under the $H_1$.
Intuitively, you can think the decivision rule as
$$Q(\theta_1)=P_{\theta_1}(S<k) \text{ or } Q(\theta_1)=P_{\theta_1}(S>k)$$
where $S$ is a statstics from the sample and $k$ is determined by type I error $\alpha$ i.e. the significant level and
$$\alpha=P_{\theta_0}[S<k] \text{ or } \alpha=P_{\theta_0}[S>k]$$
From your information you can calculate both the power and type I error $\alpha$
$Power=Q(\theta_1)=P_{\theta_1}(\bar{X}>7.5)=P_{\theta_1}(\frac{\bar{X}-\theta_1}{1/3}>\frac{7.5-\theta_1}{1/3})=1-\Phi(22.5-3\theta_1)=0.9331928$
$\theta_1=8$ here.
Accordingly, you also can calculate your type I error $\alpha$
Note from whuber's hint.
$\bar{X}$ has a $N(\theta,\frac{\sigma^2}{n})$distribution
• What is $k$ here? How can it be evaluated?How can I see that $k$ is 7.5 in my case? – user2925716 Feb 6 '18 at 13:46
• Yes, it is 7.5 in your case, usually, you have the significant level first, then you calculate $k$ by yourself, here they already gave you the vaule $k$ for the critical region. – Deep North Feb 7 '18 at 22:53
• But how can I see that $7.5$ is the correct value for my critical region? – user2925716 Feb 8 '18 at 14:14
• Because it says Rject $H_0$ iff $\bar{X}>7.5$ – Deep North Feb 8 '18 at 22:12
• Yes. And how do I decide which one of the following shall I use: $$\alpha=P_{\theta_0}[S<k] \text{ or } \alpha=P_{\theta_0}[S>k] ?$$ I.e. when $\alpha$ is given. – user2925716 Feb 10 '18 at 14:29
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2019-10-14 21:33:02
|
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|
https://math.stackexchange.com/questions/1836473/is-it-sufficient-to-prove-that-a-function-is-an-open-map-by-looking-at-the-basis
|
# Is it sufficient to prove that a function is an open map by looking at the basis element?
I am trying to prove that the projection map $\pi_X:(X, T)\times (Y,J) \to X$ is an open map
But I don't know if I can use the basis element directly, so my proof is quite round about and lengthy
Proof:
$\pi_X$ is an open map if for all open set $W \subseteq X \times Y, \pi_X(W)$ is open
We know that the basis of the product topology of $X,Y$ is $\mathcal{B} = \{U \times V|U \in T, V \in J\}$
Then the topology generated by the basis is $T_{XY} = \{\bigcup B| B \subset \mathcal{B}\}$
Then let $W$ be an open set, $W = \bigcup B$, for some collection $B$.
$\pi_X(\bigcup B) = \pi_X(\bigcup\{U\times V|U \in T, V \in J\}) = \bigcup\pi_x(\{U\times V|U \in T, V \in J\}) = \bigcup \{U\}$ which is open.
Can anyone help me identify where the proof can be made more rigorous or simpler? Ultimately, is it okay to just show this for the basic open sets rather than an open set?
• Yes, it suffices to prove that the images of basic open sets are open, but at some point you need to prove this fact. It’s easy enough. Suppose that $\mathscr{B}$ is a base, and you know that $f[B]$ is open for each $B\in\mathscr{B}$. If $U$ is open, there is some $\mathscr{U}\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{U}$, and then $$f[U]=f\left[\bigcup\mathscr{U}\right] =\bigcup_{B\in\mathscr{U}}f[B]$$ is a union of open sets and hence open. – Brian M. Scott Jun 23 '16 at 1:49
The proof of your claim comes down to proving whether $\pi_X\left ( \bigcup_iW_i\right ) = \bigcup_i \pi_X(W_i)$ where the $W_i$ are product open sets. Clearly if $U\times V$ is a product open set in $X\times Y$ then $\pi_X(U\times V) = U$.
$(\Longrightarrow)$ Let $p \in \pi_X\left ( \bigcup_i U_i \times V_i\right )$, then there exists an element $q \in X \times Y$ such that $\pi_X(q) = p$. If we write $q = (q_X, q_Y)$, then $q_X = p$ and $q_X \in U_j$ for some $j$. Observe however, that we are able to pick $q_Y$ without loss of generality so that $q_Y \in V_j$ for the same fixed $j$, hence $q \in U_j \times V_j$, and therefore $p \in \pi_X(U_j \times V_j)$ so that clearly $p$ is contained in $\bigcup_i \pi_X(U_i \times V_i)$.
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2019-12-09 21:44:41
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# Bob is training for a fitness competition. In order to increase his ma
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28 Jul 2017, 08:58
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Bob is training for a fitness competition. In order to increase his maximum number of pull-ups, he follows the following routine: he begins with 25 pull-ups, rests for thirty seconds, and then does 24 pull-ups and rests, dropping one pull-up each time (25, 24, 23, etc.) until his final set of 11 pull-ups. How many total pull-ups does Bob do?
(A) 55
(B) 150
(C) 270
(D) 275
(E) 325
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Bob is training for a fitness competition. In order to increase his ma [#permalink]
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28 Jul 2017, 10:58
The sum of n natural numbers is $$\frac{n(n+1)}{2}$$
In order to find the total number of pull ups that Bob does for the
fitness competition, we need to find the sum of the numbers from 11 to 25.
This can be done by subtracting the the sum of the first 10 natural numbers from the sum of the first 25 natural numbers.
Sum : $$\frac{25(26)}{2} - \frac{10(11)}{2} = 325 - 55 = 270$$(Option C)
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Re: Bob is training for a fitness competition. In order to increase his ma [#permalink]
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29 Jul 2017, 11:14
in case u r not aware of the formula,,,
adding will help,,,25 + 24 +....11
hope this helps
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Re: Bob is training for a fitness competition. In order to increase his ma [#permalink]
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29 Jul 2017, 20:59
1
1
sum of consecutive integers = mean * number of terms
mean = (first term + last term) / 2 = (25 + 11)/2=18
number of terms = 25 - 11 + 1 (11 inclusively) = 15
18 * 15 = 270
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Re: Bob is training for a fitness competition. In order to increase his ma [#permalink]
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24 Jun 2018, 18:04
Bunuel wrote:
Bob is training for a fitness competition. In order to increase his maximum number of pull-ups, he follows the following routine: he begins with 25 pull-ups, rests for thirty seconds, and then does 24 pull-ups and rests, dropping one pull-up each time (25, 24, 23, etc.) until his final set of 11 pull-ups. How many total pull-ups does Bob do?
(A) 55
(B) 150
(C) 270
(D) 275
(E) 325
The number of pull-ups Bob does is the sum of integers from 11 to 25 inclusive. Thus, the total number of pull-ups he does is
15(11 + 25)/2 = 15 x 18 = 270
Alternate Solution:
We need to calculate the sum of the consecutive integers from 11 to 25, inclusive. Rather than calculate 11 + 12 + 13 + … + 24 + 25, we can calculate the average of the integers and then multiply the average by the number of integers to get the sum.
We know that there are (25 – 11) + 1 = 14 + 1 = 15 integers.
To calculate the average of the 15 integers, we use (smallest + largest)/2 = (11 + 25)/2 = 18.
The average is 18. Thus, the sum is 18 x 15 = 270.
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Re: Bob is training for a fitness competition. In order to increase his ma [#permalink]
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09 Oct 2018, 23:01
Hi All,
We're told that Bob is training for a fitness competition. In order to increase his maximum number of pull-ups, he follows the following routine: he begins with 25 pull-ups, rests for thirty seconds, and then does 24 pull-ups and rests, dropping one pull-up each time (25, 24, 23, etc.) until his final set of 11 pull-ups. We're asked for the total number of pull-ups that Bob completes. This question can be solved in several different ways, including by 'bunching.'
In simple terms, we are adding up all of the integers from 11 to 25, inclusive. Since there are 25 integers from 1 to 25 - and we are NOT including the first 10 integers, that means we are adding 25 - 10 = 15 individual numbers together. We can 'bunch' those numbers into seven groups of 2 and one left over number:
11 + 25 = 36
12 + 24 = 36
13 + 23 = 36
...
17 + 19 = 36
and there's one number left over: an 18
Thus, we have 7(36) + 18 = 252 + 18 = 270
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Re: Bob is training for a fitness competition. In order to increase his ma [#permalink] 09 Oct 2018, 23:01
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https://apboardsolutions.in/ap-inter-1st-year-economics-study-material-chapter-4/
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# AP Inter 1st Year Economics Study Material Chapter 4 Theory of Production
Andhra Pradesh BIEAP AP Inter 1st Year Economics Study Material 4th Lesson Theory of Production Textbook Questions and Answers.
## AP Inter 1st Year Economics Study Material 4th Lesson Theory of Production
Essay Questions
Question 1.
Explain the law of variable proportions. [March 18, 17]
The law of variable proportions has been developed by the 19th century economists David Ricardo and Marshall. The law is associated with the names of these two economists. The law states that by increasing one variable factor and keeping other factors constant, how to change the level of output, total output first increases at increasing rate, then at a diminishing rate and later decreases. Hence this law is also known as the “Law of Diminishing returns”.
Marshall stated in the following words.
“An increase in capital and labour applied in the cultivation of land causes in general less than proportionate increase in the amount of produce raised, unless it happens to coincide with an improvement in the arts of agriculture”.
Assumptions :
1. The state of technology remain constant.
2. The analysis relates to short period.
3. The law assumes labour in homogeneous.
4. Input prices remain unchanged.
Explanation of the Law :
Suppose a farmer has ‘4’ acres of land he wants to increase output by increasing the number of labourers, keeping other factors constant. The changes in total production, average product and marginal product can be observed in the following table.
In the above table total product refers to the total output produced per unit of time by all the labourers employed.
Average product refers to the product per unit of labour marginal product refers to additional product obtained by employing an additional labour.
In the above table there are three stages of production.
1st stage i.e., increasing returns at 2 units total output increases average product increases and marginal product reaches maximum.
2nd stage i.e., diminishing returns from 3rd unit onwards TP increases diminishing rate and reaches maximum, MP becomes zero, AP continuously decreases.
3rd stage i.e., negative returns from 8th unit TP decreases AP declines and MP becomes negative.
This can be explained in the following diagram.
In the diagram on ‘OX’, axis shown units labourer and OY’ axis show TP, MP, and AP. 1st stage TP, AP increases MP is maximum. In the 2nd stage TP maximum, AP MP is zero. At 3rd stage TP declines, AP also declines, MP becomes negative.
Question 2.
Explain the law of returns to scale.
The law of returns to scale relate to long run production function. In the long run it is possible to alter the quantities of all the factors of production. If all factors of production are increased in given proportion the total output has to increase in the same proportion. Ex : The amounts of all the factors are doubled, the total output has to be doubled increasing all factors in the same proportion is increasing the scale of operation. When all inputs are changed in a given proportion, then the output is changed in the same proportion. We have constant returns to scale and finally arises diminishing returns. Hence as a result of change in the scale of production, total product increases at increasing rate, then at a constant rate and finally at a diminishing rate.
Assumptions :
1. All inputs are variable.
2. It assumes that state of technology remain the same. The returns to scale can be shown in the following table.
The above table reveals the three patterns of returns to scale. In the 1th place, when the scale is expanded upto 3 units, the returns are increasing. Later and upto 4th units, it remains constant and finally from 5th on words the returns go on diminishing.
In the diagram on ‘OX’ axis shown scale of production, on OY’ axis shown total product. RR1 represents increasing returns R1S – Constant returns; SS1 represents diminishing returns.
Question 3.
Distinguish between internal and external economies.
Economies of large scale production can be grouped into two types.
1. Internal economies
2. External economies.
1. Internal Economies:
Internal economies are those which arise from the expansion of the plant, size or from its own growth. These are enjoyed by that firm only.
“Internal economies are those which are open to a single factory or a single firm independently of the action of other firms”. – Cairncross
i) Technological Economies : The firm may be running many productive establishments. As the size of the productive establishments increase, some mechanical advantages may be obtained. Economies can be obtained from linking process to another process i.e. paper making and pulp making can be combined. It also used superior techniques and increased specialization.
ii) Managerial Economies : Managerial economies arises from specialisation of management and mechanisation of managerial functions. For a large size firm it becomes possible for the management to divide itself into specialised departments under specialised personnel. This increases efficiency of management at all levels. Large firms have the opportunity to use advanced techniques of communication, computers etc. All these things help in saving of time and improve the efficiency of the management.
iii) Marketing Economies : The large firm can buy raw materials cheaply, because it buys in bulk. It can secure special concession rates from transport agencies. The product can be advertise better. It will be able to sell better.
iv) Financial Economies : A large firm can arise funds more easily and cheaply a small one. It can borrow from bankers upon better security.
v) Risk bearing Economies : A large firm incurs unrisk and it can also reduce risks. It can spread risks in different ways. It can undertake diversifications of output. It can buy raw materials from several firms.
vi) Labour Economies : A big firm employs a large number of workers. Each worker is given the kind of job he is fit for.
2. External Economies :
An external economy is one which is available to all the firms in an industry. External economies are available as an industry grows in size.
1. Economies of Concentration : When a number pf firms producing an identical product are localised in one place, certain facilities become available to all. Ex : Cheap transport facility, availability of skilled labour etc.
2. Economies of Information : When the number of firms in an industry increases collective action and co-operative effort becomes possible. Research work can be carried on jointly. Scientific journal can be published. There is possibility for exchange of ideas.
3. Economies of Disintegration : When the number of firms increases, the firm may agree to specialise. They may divide among themselves the type of products of stages of production. Ex : Cotton industry.
Question 4.
Explain short-run cost structure of a Firm.
Costs are divided into two categories i.e.,
1. Short run cost curves
2. Long run cost curves.
In short run by increasing only one factor i.e., (labour) and keeping other factor constant. The short run cost are again divided into two types.
1. General costs
2. Economic costs.
1. General costs :
i) Money costs : Production is the outcome of the efforts of factors of production like land, labour, capital and organisation. So, rent to land, wage to labour, interest to capital and profits to entrepreneur has to paid in the form of money is called money cost.
ii) Real cost: Adam Smith regarded pains and sacrifices of labour as real cost. So it cannot be measured interms of money.
iii) Opportunity cost: Factors of production are scarce and have alternative uses. The opportunity cost of a factor is the benefit that is foregone from the next best alternative use.
2. Economic costs :
i) Fixed costs : The cost of production which remains constant even the production may be increase or decrease is known as fixed cost. The amount spent by the cost of plant and equipment, permanent staff are treated as fixed costs.
ii) Variable cost: The cost of production which is changing according to changes in the production is said to be variable cost. In the long period all costs are variable costs. It include price of raw materials, payment of fuel, excise taxes etc. Marshall called “Prime cost”.
iii) Average cost: Average cost means cost per units of output. If we divided total cost by the number of units produced, we will get average cost.
AC = $$\frac{\text { Total cost }}{\text { Output }}$$
iv) Marginal cost: Marginal cost is the additional cost of production producing one more unit.
MC = $$\frac{\Delta \mathrm{TC}}{\Delta \mathrm{Q}}$$
v) Total cost: Total cost is the sum of total fixed cost and total variable cost.
TC = FC + VC
The short term cost in relation to output are explained with the help of a table.
In the above table shows that as output is increased in the 1st column, fixed cost remains constant. Variable costs have changed as and when there are changes in output. To produce more output in the short period, more variable factors have to be employed. By adding FC & VC we get total cost different levels of output. AC falls output increases, reaches its minimum and then rises MC also change in the total cost associated with a change in output. This can be shown in the diagram.
In the above diagram on ‘OX’ axis taken by output and ‘OY taken by costs. The shapes of different cost curves explain the relationship between output and different costs. TFC is horizontal to ‘X’ axis. It indicates that increase in output has no effect on fixed cost. TVC on the other side increases along with level of output. TC curve rises as output increases.
Question 1.
Explain Production function.
Production function is technical concept. It explains the physical relationship between input and output at any period of time. It represents functional relationship between inputs and the amount of output produced.
According to Stigler “Production is the name given to the relationship between rates of inputs of productive services and the rate of output of product”.
The production function can be expressed mathematically as follows.
Gx = f (L, K, R, N, T)
Gx = Output
f = Functional relationship
L = Amount of Labour
K = Amount of Capital
R = Raw material
N = Natural resources or land
T = State of Technology.
Where Gx is dependent variable and is determined by the inputs used, whereas L, K, R, N, T are independent variables.
Question 2.
Explain Law of supply.
The law of supply explains the functional relationship between price of a commodity and its quantity supplied. The law of supply can be stated as follows “Other things remaining the same, as the price of a commodity rises its supply is extended and as the price falls its supply is contracted”.
The law of supply can be explained with the help of supply schedule and supply curve.
Supply Schedule : Supply schedule explains various amounts of good that the seller offers for sale at different prices. It represents the functional relationship between price and quantities supplied. There is direct relationship between price and supply. This can be shown in the following schedule.
The above schedule high price i.e, ₹ 5.00 per unit 1000 units are supplied and at ₹ 1 per 200 units are supplied. It means high price indicate high supply and low price indicates low supply. So, it shows the direct relationship between price and supply.
Supply curve :
A supply curve can be drawn with the help of above supply schedule to explain the direct relationship between price and supply.
In the above diagram supply is shown on ‘OX’ axis and price is shown ‘OY axis. SS is supply curve. It slopes upwards from left to right. The slope of supply curve is always positive. Because there is direct relationship between the price and supply.’
Question 3.
Define Factors of production.
Factors that help in the production process are called factors of production.
Ex : land, labour, capital and organization
1) Land : Land stands in economics for natural resources. There are nature given resources like soil, earth, water, rainfall, forests, mines, clijnate etc. All these come under land. Land is the productive equipment given by nature. The remuneration to land is called rent.
2) Labour : Labour is man’s effort. It may be physical or mental labour. Any service offered for a price is considered as labour in economics. All services rendered to produce scarce goods come under labour. The remuneration to labour is called wage.
3) Capital : Capital is man made production equipment. Factories, buildings, vehicles, rail-roads, roads, irrigation dam etc., come under capital. The remuneration to capital is interest.
4) Organisation or Entrepreneurship : The entrepreneur or businessmen brings together all the factors of production required for production. He bears the risk is doing so. He coordinates the functions of different factors. Profit is the remuneration for organisation or entrepreneurship.
Question 4.
Define land and explain the characteristics of land.
Land stands in economics for natural resources. There are nature given resources like soil, earth, water, rainfall, forests, mines, climate etc., all these come under land. Land is the productive, equipment given by nature. The remuneration to land is called rent. Characteristics of land:
1. Land is a free gift of nature.
2. The supply of land is perfectly inelastic from the point of view of economy.
3. Land cannot be shifted from one place to another place.
4. It does not yield any result unless human effort’s are employed.
Question 5.
Define labour. Explain the characteristics of labour.
Labour is man’s effort. It may be physical (or) mental labour. Any’service offered for a price is considered as labour in economics. The remuneration to labour is called wage. Characteristics of labour:
1. Labour is inseparable from the labourer himself.
2. Labour is highly ‘perishable’. It means labour lost cannot store his labour.
3. Labour has a very weak bargaining power.
4. Labour power differs from labourer to labourer of their skills.
5. The supply curve of a labourer is backward bending.
Question 6.
What is supply ? What are the determinants of supply.
The supply of a commodity means the total quantity of the commodity that sellers offer to sell at different prices from the stock of that commodity existing at any given time. The supply of commodity depends upon the following factors.
Determinants of supply :
1. Price of the commodity: The supply of the commodity depends upon the price of that commodity. When price falls, supply falls and when price rises, supply also rises. Thus price and supply are directly related.
2. Factor prices : The cost of production of a commodity depends upon the prices of various factors of production.
3. Prices of related goods : The supply of the commodity depends upon the prices of related goods. If the price of a substitute good goes up, the producer will be induced to
4. State of technology: Technological improvements determine supply of a commodity. Progress in technology leads to reduction in the cost of production which will increase supply.
5. Government policy : Imposition of heavy taxes as a commodity discourages its production. Hence production decreases.
Question 7.
Explain the nature of revenue curves untier perfect competition.
If competition is perfect in market, the market is known as perfectly competitive market. Due to the existing features the total demand and total supply pf a commodity interact to determine the price in the industry and individual firm. Hence, an individual seller or firm is a price taker but not a price maker in perfect competition, ‘therefore he can sell any quantity at the ruling price. Thus, the demand curve is parallel to X-axis. The nature of Average Revenue (AR) and Marginal Revenue (MR) and their relationship under perfect competition can be better understood from the following schedule.
Revenue Schedule :
The above table indicates that in perfect competition price remains the same irrespective of the number of units sold. Therefore, the total revenue increases at a constant rate. AR and MR are equal. There is no difference among price AR and MR. This can be reveals the following diagram.
In the diagram X-axis represents number of units (output) and Y-axis represents revenue. DP is demand curve.
In diagram, the Average Revenue curve (AR) is horizontal straight line parallel to the X-axis and Marginal Revenue Curve MR coincides with it. Because the seller can sell number of units given the price, the AR curve facing the seller is a horizontal line.
Question 1.
Production function [March 18]
The production function is the none for the relation between the physical inputs and the physical outputs of a film. Production of a firm, production function explains the functional relationship between inputs and outputs this can be as follows Gx = f (L, K, R, N, T).
Question 2.
Law of supply [March 16]
The law of supply explains the functional relationship between price of a commodity and its quantity supplied. The law of supply can be stated as follows. “Other things remaining the same, as the price of a commodity rises its supply is extended and as the price falls its supply is contracted”.
Question 3.
Factors of production
Factors that help in the production process are called factors of production.
Ex : land, labour, capital and organization.
Question 4.
Average cost
If we divided total cost, by the number of units produced. We will get average cost. Average cost means cost per unit of output.
AC = $$\frac{\mathrm{TC}}{\text { Output }}$$
Question 5.
Marginal cost
Marginal cost is the additional cost of production producing one more unit in other words it is the addition made to total cost by producing one more unit of a commodity.
MC = $$\frac{\Delta \mathrm{TC}}{\Delta \mathrm{Q}}$$
Question 6.
Production
Production is the process that converts inputs into output in economies production includes services along with physical goods.
Question 7.
Short period
Short period is a period in which a producer is unable to change factors of production to increase output. This relates to law of variable proportions.
Question 8.
Long period
Long period is a period in which a producer is unable to change factors of production to increase output. This relates to returns to scale.
Question 9.
Average product
It refers to the product per unit of labour it is Obtained by dividing total product by the number of labourers employed.
AP = $$\frac{\mathrm{TP}}{\mathrm{L}}$$
Question 10.
Marginal product
MP = $$\frac{\Delta \mathrm{TP}}{\Delta \mathrm{L}}$$
Question 11.
Fixed factor
Fixed factors are those costs which can’t be changed by the producer in the short period.
Ex : Buildings, Machinery etc.
Question 12.
Variable factors
The factors of production which are possible to change in relation to a change in output is known as variable factors in the long run all factors of production are variable.
Ex : Labour, Raw materials.
Question 13.
Change in scale of production
It refers that output will be increase by increasing inputs in the long period.
Question 14.
Internal economies
It refers that when a firm expands output by increasing all inputs.
Question 15.
External economics
It refers to one which is available to all the firms in an industry. External economies are available as an industry grows in size.
Question 16.
Supply
The quantity of a commodity that a seller is prepared to sell at a particular price and at a particular time is known as supply. The supply curve slopes upwards from left to right.
Question 17.
Supply function
It explains the functional relation between supply and the factors of production of a good.
Question 18.
Opportunity cost
The opportunity cost of a factor is the benefit i.e, forgone from the next best alternative use.
Question 19.
Fixed cost
The cost of production which remains constant even the production may be increase (or) decrease is known as fixed cost.
Ex : Machinery, permanent staff salaries.
Question 20.
Variable cost
The cost of production which is changed according to changes in the production is said to be variable cost. In the long period all costs are variable costs it include price of raw materials, wages of labour, power transport charges etc.
Question 21.
Total Revenue
Total revenue of a producer depends on the price and the quantity of output sold in the market.
Total Revenue = Price × Quantity of output
TR = P × Q
Question 22.
Average Revenue
AR = $$\frac{\text { TR }}{\mathrm{Q}}$$
MR = $$\frac{\Delta \mathrm{TR}}{\Delta \mathrm{Q}}$$
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2023-03-30 14:58:25
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|
https://x-engineer.org/graduate-engineering/programming-languages/scilab/how-to-create-a-multiple-y-axes-plot-in-scilab/
|
How to create a multiple y-axes plot in Scilab
When dealing with data we often need to compare different sets/series between themselves. For this we need to plot several y-axes function of the same, common x-axis. Scilab is very versatile at plotting multiple y-axes on the same graphical window.
This tutorial will teach you how to plot 2 or more y-axis plot in the same graphical window using Scilab.
In this example we are going to plot 3 function y1(x), y2(2) and y3(x) function of x, overlapping the curves, each with its own y-axis. The functions are defined as:
$\begin{split} y_{1}(x) &= \sin(x)\\ y_{2}(x) &= e^{\frac{x}{6}}\cdot \left [y_{1}(x)+2 \right ]\\ y_{3}(x) &= 1+x^{2} \end{split}$
Step 1. Define the x-axis and the functions.
// Preparing data
clear()
clf()
x=linspace(1,30,200);
y1=sin(x);
y2=exp(x/6).*(y1+2);
y3=1+x.^2;
Before defining the data we are erasing all the variables in the Scilab Variable Browser with the function clear(). Also, in order to make sure that we don’t miss any warning or error message, we are clearing the Scilab console with the function clf(). The x-axis contains 200 points, between 1 and 30, and is defined with the function linspace(). After the x-axis definition, the y functions are defined: y1, y2 and y3. After running these Scilab instructions, an empty graphical window should be generated and the x and y1, y2 and y3 variables should appear in the Scilab Variable Browser.
Step 2. Plot the function y1(x).
// Axis y1
c=color("slategray");
plot2d(x,y1,style=c)
h1=gca();
h1.font_color=c;
h1.children(1).children(1).thickness = 2;
xlabel("$x$","FontSize",3)
ylabel("$y_{1}(x)=\sin(x)$","FontSize",3)
title("x-engineer.org","color","blue")
First we create a colour variable c, which we are going to use as a setting for the axes and polyline (function curve). The function y1(x) is plotted with the instruction plot2d(). With gca() we read the current axes parameters and we assign them to the variable h1 for editing. Next instructions are setting the colour of the axis font and the thickness of the line. After, we use xlabel() and ylabel() to set the axes labels and title() for a title. Note that we used Latex notations for the y-axis. For more information on Scilab plots and Latex please read the article How to add Latex formatted text in a Scilab plot.
After running the Scilab instructions in a script file (*.sce) or in the Scilab Console, we get the following graphical window.
Image: Function y1(x) Scilab plot
Step 3. Plot the function y2(x).
// Axis y2
c=color("red");
h2=newaxes();
h2.font_color=c;
plot2d(x,y2,style=c)
h2.filled="off";
h2.axes_visible(1)="off";
h2.y_location="right";
h2.children(1).children(1).thickness=2;
ylabel("$y_{2}(x)=e^{\frac{x}{6}}\cdot \left [y_{1}(x)+2 \right ]$","FontSize",3,"color",c)
For the second y axis we are going to use the color red for font and polyline. Also, using the axes handle variable h2, we are locating it on the right side of the plot. We also make the background transparent (off) to allow the previous y-axis to be visible and hide the x-axis (off) to avoid overlapping with the previous one.
After running the Scilab instructions in a script file (*.sce) or in the Scilab Console, we get the following graphical window.
Image: Function y1(x) and y2(x) Scilab plot
Step 4. Plot the function y3(x).
// Axis y3
c=color("purple");
h3=newaxes();
h3.font_color=c;
plot2d(x,y3,style=c)
h3.children(1).children(1).thickness=2;
h3.filled="off";
h3.axes_visible(1)="off";
h3.axes_reverse(2)="on";
h3.y_location="middle";
h3.log_flags="nln";
ylabel("$y_{3}(x)=1+x^{2}$","FontSize",3,"color",c)
For the third y-axis we use the colour purple. We set it up in the same way as second y axis, with a few exceptions. The axis position will be in the middle of the plot, the values will be reversed (on), minimum at the top and maximum at the bottom, and the scale (numbers) will be displayed as logarithmic (nln).
After running the Scilab instructions in a script file (*.sce) or in the Scilab Console, we get the following graphical window.
Image: Function y1(x), y2(x) and y3(x) Scilab plot
To get the final plot in one go, copy all the Scilab instructions defined in the steps above in a single Scilab script file (*.sce) and run it.
Scilab is very versatile for multiple y-axes plots. To get a handle of it, you can try another example with different functions and different settings.
|
2020-07-12 21:30:35
|
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|
https://astronomy.stackexchange.com/questions/38635/what-does-located-in-the-hubble-flow-mean
|
# What does "located in the Hubble flow" mean?
I do not understand that much about the term Hubble flow. For cosmological studies with type Ia supernova, I often see the sentence like "We used 100 SNe Ia in the Hubble flow". What does the difference between "located in the Hubble flow" and "not in the Hubble flow"? Thank you.
The first is an apparent recession velocity that is given by Hubble's law. $$v_H = H_0 d$$, where $$v_H$$ is the apparent recession velocity, $$d$$ is the proper distance and $$H_0$$ is the current value of the Hubble parameter.
On top of that, a galaxy will have a peculiar velocity $$v_P$$ with respect to the cosmological rest frame (usually taken to be coincident with the frame in which the cosmic microwave background has no dipole anisotropy). This preculiar velocity is due to the gravitational influence of nearby other galaxies or due to a galaxy's motion within a group, cluster or supercluster of galaxies.
Since the first component grows with distance, whereas the second component is fixed and of order hundreds to perhaps a thousand or so km/s, then if you look far enough away from the Earth then $$v_H \gg v_P$$. This is what it means to be "part of the Hubble flow" - the velocity that you measure for that object is dominated by the cosmological expansion of the universe and local peculiarities and gravitational perturbations that the galaxy is influenced by play a small enough role that they become a negligible source of uncertainty.
Since $$H_0 \simeq 70$$ km/s per Mpc, and $$v_P \sim 1000$$ km/s, then a quick bit of maths tells you that in order for $$v_H \gg v_P$$, then $$d \gg 14$$ Mpc for an object to be considered part of the Hubble flow. There is no exact definition.
|
2021-09-29 02:15:30
|
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|
https://www.varsitytutors.com/precalculus-help/matrices-and-vectors/geometric-vectors?page=3
|
# Precalculus : Geometric Vectors
## Example Questions
1 3 Next →
### Example Question #171 : Matrices And Vectors
Subtract:
Explanation:
Subtract the first value of the first vector, and the second value of the first vector with the second value of the second vector.
Double negative signs are converted to a positive sign.
### Example Question #172 : Matrices And Vectors
Simplify:
Explanation:
The dimensions of the vectors are not the same. Placeholders cannot be added to a vector. Therefore, the values of the vectors cannot be added.
### Example Question #21 : Evaluate Geometric Vectors
Find the norm of the vector .
Explanation:
We find the norm of a vector by finding the sum of each element squared and then taking the square root.
.
### Example Question #22 : Evaluate Geometric Vectors
Find the norm of the vector .
Explanation:
We find the norm of a vector by finding the sum of each component squared and then taking the square root of that sum.
### Example Question #23 : Evaluate Geometric Vectors
Find the norm of the vector:
Explanation:
The norm of a vector is also known as the length of the vector. The norm is given by the formula:
.
Here, we have
,
### Example Question #24 : Evaluate Geometric Vectors
Find the norm of vector .
Explanation:
Write the formula to find the norm, or the length the vector.
Substitute the known values of the vector and solve.
### Example Question #25 : Evaluate Geometric Vectors
Find the norm (magnitude) of the following vector:
Explanation:
Use the following equation to find the magnitude of a vector:
In this case we have:
So plug in our values:
So:
### Example Question #26 : Evaluate Geometric Vectors
Find the product of the vector and the scalar .
|
2021-04-11 18:03:16
|
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|
https://docs.mantidproject.org/nightly/concepts/SampleEnvironment.html
|
$$\renewcommand\AA{\unicode{x212B}}$$
# Sample Environment¶
A sample environment defines the container + components used to hold a sample during a run. This page details the process of a sample environment within Mantid.
## Specification¶
A sample environment is defined by:
At a minimum a sample environment is expected to define a container with both its geometry and composition.
An environment definition is contained within a file using XML syntax. The file can be an explicit definition of the properties of each environment component or it can contain a reference to a CAD file in .3mf format that contains all the required information about the environment components
## XML Definition File - Explicit Definition¶
An environment definition is contained within a file using XML syntax.
Multiple containers can be specified in the definition. The correct container for a run must be chosen by the user at the time the environment is attached to a workspace with the Environment option on the SetSample algorithm.
One Sample Environment can contain components defined in different ways - some components defined by a CSG geometry and some components defined in a separate STL mesh file.
### CSG geometry¶
A minimal structure with a single container defined using CSG geometry would have the following form:
<!-- Filename: CRYO-01.xml -->
<environmentspec>
<materials>
</materials>
<components>
<containers>
<geometry>
<!-- geometry of container -->
</geometry>
<samplegeometry>
<!-- geometry of sample -->
</samplegeometry>
</container>
</containers>
</components>
</environmentspec>
The CSG geometry of both the sample and container are defined using the same syntax used in the instrument definition files to define detector shapes. See here for detail on defining shapes in XML. CSG shapes can be plotted in Mantid, see 3D Mesh Plots for Sample Shapes.
### Mesh files - STL¶
The container and sample geometry can alternatively be defined using a mesh description by specifying an .stl file as follows.
<environmentspec>
<materials>
</materials>
<components>
<containers>
<stlfile filename="container.stl" scale="mm">
</stlfile>
<samplestlfile filename="sample.stl" scale="mm">
</samplestlfile>
</container>
</containers>
</components>
</environmentspec>
Mantid will try the following approaches to find the path to the stl file (in order):
• If a full path is supplied in the filename attribute then it will be used
• Mantid will then check in the same directory as the environment definition files
• Mantid will then check in the data search directories
The stl file format doesn’t natively support a scale so this should be specified in the scale attribute of the stlfile tag. Possible values are mm, cm or m.
Stl mesh shapes can be plotted in Mantid, see 3D Mesh Plots for Sample Shapes. There are also various free software tools available that can view and edit .stl files:
• FreeCAD (Windows, Linux, Mac). https://www.freecadweb.org/ This viewer also provides coordinate readout of the cursor position
• Microsoft 3D Viewer (Windows only)
### Materials¶
Each component is assigned a material, which defines properties such as the number density and neutron scattering cross sections, amongst other things. All materials defined for an environment must be defined within the <materials> tags and each material must have a unique id within the file. The id is used to reference the material when defining a container or component.
The other attributes define the properties of the material. The allowed attributes map to the arguments of a similar name on the SetSampleMaterial algorithm
• formula
• atomicnumber
• massnumber
• numberdensity
• zparameter
• unitcellvol
• massdensity
• totalscatterxsec
• cohscatterxsec
• incohscatterxsec
• absorptionxsec
• attenuationprofile
Mantid will search for the filename supplied in the attenuationprofile attribute in the following places (in order):
• If a full path is supplied in the filename attribute then it will be used
• Mantid will then check in the same directory as the environment definition file
• Mantid will then check in the data search directories
## Non-container Components¶
A given setup may have other components within the beam that must be included. These container be included using the component tag rather than the container tag. For example, a heat shield container be added to the above definition like so:
<!-- Filename: CRYO-01.xml -->
<environmentspec>
<materials>
<material id="aluminium" formula="Al"/>
</materials>
<components>
<containers>
<geometry>
<!-- geometry of container -->
</geometry>
<samplegeometry>
<!-- geometry of sample -->
</samplegeometry>
</container>
</containers>
<component id="heat-shield" material="aluminium">
<geometry>
<!-- geometry of shield-->
</geometry>
</component>
</components>
</environmentspec>
A new material, aluminium has been added to the materials list and the heat shield is defined as an arbitrary component. The component tag behaves in a similar fashion to the container tag with the exception that it container not contain a samplegeometry.
The non-container components can also be defined using mesh geometry by specifying stl file names
<!-- Filename: CRYO-01.xml -->
<environmentspec>
<materials>
<material id="aluminium" formula="Al"/>
</materials>
<components>
<containers>
<stlfile filename="container.stl" scale="mm">
</stlfile>
<samplestlfile filename="sample.stl" scale="mm">
</samplestlfile>
</container>
</containers>
<component id="heat-shield" material="aluminium">
<stlfile filename="heat-shield.stl" scale="mm">
<translation vector="0,0,1.40384"/>
<rotation ydegrees="180"/>
</stlfile>
</component>
</components>
</environmentspec>
The shape defined in the stl file can be transformed andor rotated in order to assemble it correctly with the other environment components. This is achieved by specifying a translation or rotation tag in the xml. The translation tag has an attribute vector which is a comma separated list of x, y, z coordinates. The rotation tag has available attributes xdegrees, ydegrees, zdegrees which all take a rotation specified in degrees.
## XML Definition File - 3MF Definition¶
The .3mf file format is a 3D printing format that allows multiple meshes with their relative orientations to be stored in a single file along with information on the scale used for vertex coordinates and metadata about the material properties. Further details on the format are available here:
https://3mf.io/
If all the information on the geometry of the environment components is available in a single .3mf file this can be referenced in the sample environment xml file instead of supplying the full details as described above.
The following xml example shows this type of reference:
<!-- Filename: 3MFExample.xml -->
<environmentspec>
<fullspecification filename="Assembled.3mf"/>
</environmentspec>
If a relative path or filename is supplied for the 3mf file name, Mantid searches in the same set of directories that are described above for .stl files.
The materials must have their names set to the material’s chemical formula in order for the material data to be imported into Mantid. Additional properties such as the density should be specified in brackets after the name:
eg B4-C (massdensity=’2.52’, cohscatterxsec=’10’)
While there are a wide range of CAD tools available that support import and export from .3mf format, support for saving material information into .3mf format is more limited. The material information can be easily added to the .3mf files however by editing the file in a text editor:
• change the .3mf file extension to .zip
• extract the file called 3dmodel.model
• edit the <basematerials> content near the top of the file
The 3mf file can optionally include the geometry of the sample as well as the environment. The mesh corresponding to the sample should be given the name ‘sample’ in the 3mf file.
Category: Concepts
|
2023-03-25 13:36:18
|
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|
https://brilliant.org/problems/an-algebra-problem-by-ilham-saiful-fauzi-2/
|
# A number theory problem by Ilham Saiful Fauzi
$\large a^2+b^2+ab+2(a-b)=9$
Find the number of ordered pairs $$(a,b)$$ of positive integers satisfying the equation above.
×
|
2018-01-21 09:02:05
|
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|
https://indico.cern.ch/event/356420/contributions/1764436/
|
# EPS HEP 2015
22-29 July 2015
Europe/Vienna timezone
## Progress towards A Fixed-Target ExpeRiment at the LHC: AFTER@LHC
24 Jul 2015, 10:00
20m
HS30
#### HS30
talk Heavy Ion Physics
### Speaker
Barbara Trzeciak (Czech Technical University in Prague)
### Description
The multi-TeV LHC beams offer the possibility to perform the most energetic fixed-target experiments ever, in order to study with high precision pp and pA collisions at $\sqrt{s_{NN}} \simeq 115$ GeV and Pbp and PbA collisions at $\sqrt{s_{NN}} \simeq 72$ GeV. AFTER@LHC -- A Fixed-Target ExperRiment -- can greatly complement collider experiments, in particular those of RHIC and EIC projects. We thus discuss the possibility of a multi-purpose fixed-target experiment using LHC beams extracted by a bent crystal or using an internal gas target inspired from the LHCb SMOG system. We have evaluated that the instantaneous luminosity achievable with AFTER would surpass that of RHIC by more than 3 orders of magnitude. This provides a quarkonium, prompt photon and heavy-flavour observatory in pp and pA collisions where, by instrumenting the target-rapidity region, gluon and heavy-quark distributions of the proton, the neutron and the nuclei can be accessed at large x. In addition, the fixed-target mode has the advantage to allow for spin measurements with polarized targets over the full backward rapidity domain. The nuclear target-species versatility provides a unique opportunity to study the nuclear matter versus the hot and dense matter formed in heavy-ion collisions. We will show first results of the fast simulations based on a LHCb-like detector used in the fixed-target mode and discuss connections with data from LHCb SMOG, which can be seen as a low-density internal gas target.
### Primary authors
Barbara Trzeciak (Czech Technical University in Prague) Jean-Philippe Lansberg (IPN Orsay, Paris Sud U. / IN2P3-CNRS)
|
2020-08-05 19:44:45
|
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|
https://eprint.iacr.org/2015/343
|
### High-speed Curve25519 on 8-bit, 16-bit, and 32-bit microcontrollers
Michael Düll, Björn Haase, Gesine Hinterwälder, Michael Hutter, Christof Paar, Ana Helena Sánchez, and Peter Schwabe
##### Abstract
This paper presents new speed records for 128-bit secure elliptic-curve Diffie-Hellman key-exchange software on three different popular microcontroller architectures. We consider a 255-bit curve proposed by Bernstein known as Curve25519, which has also been adopted by the IETF. We optimize the X25519 key-exchange protocol proposed by Bernstein in 2006 for AVR ATmega 8-bit microcontrollers, MSP430X 16-bit microcontrollers, and for ARM Cortex-M0 32-bit microcontrollers. Our software for the AVR takes only 13 900 397 cycles for the computation of a Diffe-Hellman shared secret, and is the first to perform this computation in less than a second if clocked at 16 MHz for a security level of 128 bits. Our MSP430X software computes a shared secret in 5 301 792 cycles on MSP430X microcontrollers that have a 32-bit hardware multiplier and in 7 933 296 cycles on MSP430X microcontrollers that have a 16-bit multiplier. It thus outperforms previous constant-time ECDH software at the 128-bit security level on the MSP430X by more than a factor of 1.2 and 1.15, respectively. Our implementation on the Cortex-M0 runs in only 3 589 850 cycles and outperforms previous 128-bit secure ECDH software by a factor of 3.
Note: Typo in the abstract.
Available format(s)
Category
Public-key cryptography
Publication info
Published elsewhere. Design Codes and Cryptography
DOI
bd41e6b96370dea91c5858f1b809b581
Keywords
elliptic curve cryptographyCurve25519ECDH key-exchangemicrocontrollerAVR ATmegaMSP430ARM Cortex-M0implementation
Contact author(s)
bjoern m haase @ web de
History
Short URL
https://ia.cr/2015/343
CC BY
BibTeX
@misc{cryptoeprint:2015/343,
author = {Michael Düll and Björn Haase and Gesine Hinterwälder and Michael Hutter and Christof Paar and Ana Helena Sánchez and Peter Schwabe},
title = {High-speed Curve25519 on 8-bit, 16-bit, and 32-bit microcontrollers},
howpublished = {Cryptology ePrint Archive, Paper 2015/343},
year = {2015},
doi = {bd41e6b96370dea91c5858f1b809b581},
note = {\url{https://eprint.iacr.org/2015/343}},
url = {https://eprint.iacr.org/2015/343}
}
Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.
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2022-10-06 02:00:31
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http://orbi.ulg.ac.be/ph-search?uid=p001863&start=100
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Publications of Michel Rigo Results 101-120 of 120. 1 2 3 4 5 6 Syntactictal and automatic properties of sets of polynomials over finite fieldsRigo, Michel Conference (2006, October)Detailed reference viewed: 26 (1 ULg) Abstract numeration systems : a survey.Rigo, Michel Conference (2006, August)Detailed reference viewed: 39 (4 ULg) A note on syndeticity, recognizable sets and Cobham's theoremRigo, Michel ; Waxweiler, Laurent in Bulletin of the European Association for Theoretical Computer Science (2006), 88In this note, we give an alternative proof of the following result. Let p,q>=2 be two multiplicatively independent integers. If an infinite set of integers is both p- and q-recognizable, then it is ... [more ▼]In this note, we give an alternative proof of the following result. Let p,q>=2 be two multiplicatively independent integers. If an infinite set of integers is both p- and q-recognizable, then it is syndetic. Notice that this result is needed in the classical proof of the celebrated Cobham’s theorem. Therefore the aim of this paper is to complete [13] and [1] to obtain an accessible proof of Cobham’s theorem. [less ▲]Detailed reference viewed: 25 (2 ULg) Structural properties of bounded languages with respect to multiplication by a constantCharlier, Emilie ; Rigo, Michel in Actes des Journées Montoises d'Informatique Théorique (2006)We consider the preservation of recognizability of a set of integers after multiplication by a constant for numeration systems built over a bounded language. As a corollary we show that any nonnegative ... [more ▼]We consider the preservation of recognizability of a set of integers after multiplication by a constant for numeration systems built over a bounded language. As a corollary we show that any nonnegative integer can be written as a sum of binomial coefficients with some prescribed properties. [less ▲]Detailed reference viewed: 28 (1 ULg) Automates et systèmes de numérationRigo, Michel in Bulletin de la Société Royale des Sciences de Liège (2005), 73Ce survol introductif est basé sur une mini-conférence réalisée à la Société Royale des Sciences de Liège en avril 2004 et sur un exposé réalisé à l’IUFM de Reims en juin 2003 (Integrating Technologies ... [more ▼]Ce survol introductif est basé sur une mini-conférence réalisée à la Société Royale des Sciences de Liège en avril 2004 et sur un exposé réalisé à l’IUFM de Reims en juin 2003 (Integrating Technologies into Mathematics Education). Nous y présentons divers systèmes de numération du point de vue de la théorie des langages formels. On s’attache dès lors à mettre en lumière les liens éventuels entre propriétés arithmétiques des nombres et propriétés syntaxiques de leurs représentations. La première partie de ce texte introduit en particulier la notion d’automate et quelques unes de ses applications. [less ▲]Detailed reference viewed: 32 (3 ULg) Abstract numeration systems and tilingsBerthé; Rigo, Michel in Lecture Notes in Computer Science (2005), 3618An abstract numeration system is a triple S = (L, Sigma, <) where (Z, <) is a totally ordered alphabet and L a regular language over Z; the associated numeration is defined as follows: by enumerating the ... [more ▼]An abstract numeration system is a triple S = (L, Sigma, <) where (Z, <) is a totally ordered alphabet and L a regular language over Z; the associated numeration is defined as follows: by enumerating the words of the regular language L over Z with respect to the induced genealogical ordering, one obtains a one-to-one correspondence between N and L. Furthermore, when the language L is assumed to be exponential, real numbers can also be expanded. The aim of the present paper is to associate with S a self-replicating multiple tiling of athe space, under the following assumption: the adjacency matrix of the trimmed minimal automaton recognizing L is primitive with a dominant eigenvalue being a Pisot unit. This construction generalizes the classical constructions performed for Rauzy fractals associated with Pisot substitutions [16], and for central tiles associated with a Pisot beta-numeration [23]. [less ▲]Detailed reference viewed: 44 (5 ULg) Abstract beta-expansion and ultimately periodic representationsRigo, Michel ; Steiner, Wolfgangin Journal de Théorie des Nombres de Bordeaux (2005), 17For abstract numeration systems built on exponential regular languages (including those coming from substitutions), we show that the set of real numbers having an ultimately periodic representation is ... [more ▼]For abstract numeration systems built on exponential regular languages (including those coming from substitutions), we show that the set of real numbers having an ultimately periodic representation is $\mathbb{Q}(\beta)$ if the dominating eigenvalue $\beta>1$ of the automaton accepting the language is a Pisot number. Moreover, if $\beta$ is neither a Pisot nor a Salem number, then there exist points in $\mathbb{Q}(\beta)$ which do not have any ultimately periodic representation. [less ▲]Detailed reference viewed: 16 (2 ULg) Decidability questions related to abstract numeration systemsHonkala, Juha; Rigo, Michel in Discrete Mathematics (2004), 285(1-3), 329-333We show that some decidability questions concerning recognizable sets of integers for abstract numeration systems are equivalent to classical problems related to HD0L systems. It turns out that these ... [more ▼]We show that some decidability questions concerning recognizable sets of integers for abstract numeration systems are equivalent to classical problems related to HD0L systems. It turns out that these problems are decidable when the sets of representations of the integers are slender regular languages. (C) 2004 Elsevier B.V. All rights reserved. [less ▲]Detailed reference viewed: 31 (0 ULg) Real numbers having ultimately periodic representations in abstract numeration systemsLecomte, Pierre ; Rigo, Michel in Information and Computation (2004), 192(1), 57-83Using a genealogically ordered infinite regular language, we know how to represent an interval of R. Numbers having an ultimately periodic representation play a special role in classical numeration ... [more ▼]Using a genealogically ordered infinite regular language, we know how to represent an interval of R. Numbers having an ultimately periodic representation play a special role in classical numeration systems. The aim of this paper is to characterize the numbers having an ultimately periodic representation in generalized systems built on a regular language. The syntactical properties of these words are also investigated. Finally, we show the equivalence of the classical theta-expansions with our generalized representations in some special case related to a Pisot number theta. (C) 2004 Elsevier Inc. All rights reserved. [less ▲]Detailed reference viewed: 16 (2 ULg) Characterizing Simpler recognizable sets of integersRigo, Michel in Studia Logica (2004), 76For a given numeration system U, a set X of integers is said to be U-star-free if the language of the normalized U-representations of the elements in X is star-free. Adapting a result of McNaughton and ... [more ▼]For a given numeration system U, a set X of integers is said to be U-star-free if the language of the normalized U-representations of the elements in X is star-free. Adapting a result of McNaughton and Papert, we give a first-order logical characterization of these sets for various numeration systems including integer base systems and the Fibonacci system. For k-ary systems, the problem of the base dependence of this property is also studied. Finally, the case of k-adic systems is developed. [less ▲]Detailed reference viewed: 13 (0 ULg) The commutative closure of a binary slip-language is context-free: a new proofRigo, Michel in Discrete Applied Mathematics (2003), 131(3), 665-672Using original arguments about sets of integers satisfying some first-order formula of the Presburger arithmetic , we give a new proof that the commutative closure of a slip-language over a two ... [more ▼]Using original arguments about sets of integers satisfying some first-order formula of the Presburger arithmetic , we give a new proof that the commutative closure of a slip-language over a two letters alphabet is context-free. (C) 2003 Elsevier B.V. All rights reserved. [less ▲]Detailed reference viewed: 25 (4 ULg) Additive functions with respect to numeration systems on regular languagesGrabner, Peter J.; Rigo, Michel in Monatshefte fur Mathematik (2003), 139(3), 205-219Asymptotic formulae for the summatory function of additive arithmetic functions related to numeration systems given by regular languages are derived.Detailed reference viewed: 15 (2 ULg) Construction of regular languages and recognizability of polynomialsRigo, Michel in Discrete Mathematics (2002), 254(1-3), 485-496A generalization of numeration systems in which NI is recognizable by finite automata can be obtained by describing a lexicographically ordered infinite regular language. We show that if P is an element ... [more ▼]A generalization of numeration systems in which NI is recognizable by finite automata can be obtained by describing a lexicographically ordered infinite regular language. We show that if P is an element of Q[x] is a polynomial such that P(N) subset of N then there exists a numeration system in which the set of representations of P(N) is regular. The main issue is to construct a regular language with a complexity function equals to P(n + 1) - P(n) for n large enough. (C) 2002 Elsevier Science B.V. All rights reserved. [less ▲]Detailed reference viewed: 18 (5 ULg) Characterizing simpler recognizable sets of integersRigo, Michel in Lecture Notes in Computer Science (2002), 2420For the k-ary numeration system, we characterize the sets of integers such that the corresponding representations make up a star-free regular language. This result can be transposed to some linear ... [more ▼]For the k-ary numeration system, we characterize the sets of integers such that the corresponding representations make up a star-free regular language. This result can be transposed to some linear numeration systems built upon a Pisot number like the Fibonacci system and also to k-adic numeration systems. Moreover we study the problem of the base dependence of this property and obtain results which are related to Cobham's Theorem. [less ▲]Detailed reference viewed: 15 (5 ULg) On the representation of real numbers using regular languagesLecomte, Pierre ; Rigo, Michel in Theory of Computing Systems (2002), 35(1, JAN-FEB), 13-38Using a lexicographically ordered regular language, we show how to represent an interval of R. We determine exactly the possible representations of any element in this interval and study the function ... [more ▼]Using a lexicographically ordered regular language, we show how to represent an interval of R. We determine exactly the possible representations of any element in this interval and study the function which maps a representation onto its numerical value. We make explicit the relationship between the convergence of finite words to an infinite word and the convergence of the corresponding approximations to a real number. [less ▲]Detailed reference viewed: 47 (10 ULg) More on generalized automatic sequencesRigo, Michel ; Maes, Arnaudin Journal of Automata, Languages and Combinatorics (2002), 7We give some generalizations of $k$-automatic sequences replacing the $k$-ary system by an abstract numeration system on a regular language. We study some of the closure properties of these sequences and ... [more ▼]We give some generalizations of $k$-automatic sequences replacing the $k$-ary system by an abstract numeration system on a regular language. We study some of the closure properties of these sequences and the possible extension to the multidimensional case or to infinite alphabets. The equivalence of these sequences and morphic predicates is given and the relationship to recognizability is also investigated. [less ▲]Detailed reference viewed: 44 (10 ULg) Abstract numeration systems on a regular languages and recognizabilityRigo, Michel Doctoral thesis (2001)Detailed reference viewed: 109 (10 ULg) Numeration systems on a regular language : Arithmetic operations, Recognizability and Formal power seriesRigo, Michel in Theoretical Computer Science (2001), 269Generalizations of numeration systems in which $$\N$$ is recognizable by a finite automaton are obtained by describing a lexicographically ordered infinite regular language $$L\subset \Sigma^*$$. For ... [more ▼]Generalizations of numeration systems in which $$\N$$ is recognizable by a finite automaton are obtained by describing a lexicographically ordered infinite regular language $$L\subset \Sigma^*$$. For these systems, we obtain a characterization of recognizable sets of integers in terms of $\N$-rational formal series. After a study of the polynomial regular languages, we show that, if the complexity of $$L$$ is $$\Theta (n^l)$$ (resp. if $$L$$ is the complement of a polynomial language), then multiplication by $$\lambda\in \N$$ preserves recognizability only if $$\lambda=\beta^{l+1}$$ (resp. if $$\lambda\neq (\#\Sigma)^\beta$$) for some $$\beta\in \N$$. Finally, we obtain sufficient conditions for the notions of recognizability for abstract systems and some positional number systems to be equivalent. [less ▲]Detailed reference viewed: 31 (2 ULg) Numerations systems on a regular languageLecomte, Pierre ; Rigo, Michel in Theory of Computing Systems (2001), 34Generalizations of positional number systems in which N is recognizable by finite automata are obtained by describing an arbitrary infinite regular language according to the lexicographic ordering. For ... [more ▼]Generalizations of positional number systems in which N is recognizable by finite automata are obtained by describing an arbitrary infinite regular language according to the lexicographic ordering. For these systems of numeration, we show that ultimately periodic sets are recognizable. We also study translation and multiplication by constants as well as the order-dependence of the recognizability. [less ▲]Detailed reference viewed: 39 (2 ULg) Generalization of automatic sequences for numeration systems on a regular languageRigo, Michel in Theoretical Computer Science (2000), 244Let L be an infinite regular language on a totally ordered alphabet (Σ,<). Feeding a finite deterministic automaton (with output) with the words of L, enumerated lexicographically with respect to <, leads ... [more ▼]Let L be an infinite regular language on a totally ordered alphabet (Σ,<). Feeding a finite deterministic automaton (with output) with the words of L, enumerated lexicographically with respect to <, leads to an infinite sequence over the output alphabet of the automaton. This process generalizes the concept of k-automatic sequence for abstract numeration systems on a regular language (instead of systems in base k). Here, we study the first properties of these sequences and their relations with numeration systems. [less ▲]Detailed reference viewed: 24 (3 ULg)
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2017-09-22 11:37:22
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http://selungalliance.org/destiny-campaign-skpmyei/27c9d5-ethyl-group-formula
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Chlorimuron-ethyl is an ethyl ester resulting from the formal condensation of the carboxy group of chlorimuron with ethanol.A proherbicide for chloimuron, it is used as herbicide for the control of broad-leaved weeds in peanuts, soya beans, and other crops. Silicones have in general the chemical formula [R 2 SiO] n, where R is an organic group such as an alkyl (methyl, ethyl) or phenyl group. A simple alkyl group is a functional group made up entirely of carbon and hydrogen where the carbon atoms are chained together by single bonds. (If it were hydrogen atom, the compound would be a carboxylic acid.) E.g. CH_3CH_3 or simply C_2H_5 Image Source: Now … This time there are two different alkyl groups attached - an ethyl group on the number 3 carbon atom and a methyl group on number 2. Ethanol or ethyl alcohol is an organic compound and a chemical liquid with the formula - C ₂ H₅OH. The chemical formula for ethyl acetate and its structural formula are given below. Ethyl, isobutyl, ethyl tert-butyl, and ethyl tert-pentyl ether are particularly hazardous in this respect. However, because it is not part of the main chain, it is given the suffix -yl (i.e. 1 decade ago. Log Octanol-Water Partition Coef (SRC): Log Kow (KOWWIN v1.67 estimate) = 4.16 Boiling Pt, Melting Pt, Vapor Pressure Estimations (MPBPWIN v1.42): Boiling Pt (deg C): 96.18 (Adapted Stein & Brown method) Melting Pt (deg C): -74.16 (Mean or Weighted MP) VP(mm Hg,25 deg C): 20.3 (Mean VP of Antoine & … There is a branched group attached to the second carbon atom. Ethanol, also known as ethyl alcohol, is a common alcohol produced by fermentation of sugars such as barley and grapes. an ethyl group has the formula -CH2CH3. 138 Answers 134–135 134 Theconversionof pentan-2-olto pentan-2-amineinvolvestheformalsubsti-tution of the hydroxy group by an amino group.For the reaction to proceed in a stereospecific manner,the hydroxy group must first be converted into a better leaving group such as the acetate by esterification with acetic acid or better by preparing the tosylate by treating the alcohol with tosyl chloride. "Ethanoate" comes from ethanoic acid. Ethanol is found in alcoholic beverages such as beer and wine. Lets start with propane as it is greater than ethane but u wont get any possible arrangement which satisfies the condition. An example is ethane. https://www.khanacademy.org/.../naming-alkanes/v/alkane-with-ethyl-groups Source(s): https://shrink.im/baZPz. Ethyl is used in the IUPAC nomenclature of organic chemistry for a saturated two-carbon moiety in a molecule, whilst the prefix "eth-" is used to indicate the presence of two carbon atoms in the molecule. Source(s): molecular formula ethyl: https://shortly.im/LaBCy. Anonymous. In the case of ethyl ethanoate, for example, the ethyl group is listed before the name. Ethanol, also called ethyl alcohol, grain alcohol, or alcohol, a member of a class of organic compounds that are given the general name alcohols; its molecular formula is C 2 H 5 OH. Such groups are often represented in chemical formulas by the letter R and have the generic name CnH2n+1. Or an alcohol can bind to ethyl making an ethyl alcohol molecule. And our alkyl group here would be an ethyl group. In a chemical structural formula, an organic substituent such as methyl, ethyl, or aryl can be written as R (or R 1, R 2, etc.) The naming of esters can be confusing to students who are new to organic chemistry because the name is the opposite of the order in which the formula is written. This group has the formula $$\text{CH}_{3}$$, which is methane without a hydrogen atom. 4 years ago. An alkyl group is a paraffinic hydrocarbon group that may be derived from an alkane by dropping one hydrogen from the structure. Figure $$\PageIndex{1}$$ shows models for two common esters. Ethyl Formula. Example 4: Write the structural formula for 3-ethyl-2-methylhexane. So an alcohol has an OH, an … This Site Might Help You. In order to understand the chemical structure of ethanol, you first need to know what alkenes are. You could also call this something like chloro ethane. So one possible name for this molecule would be ethyl chloride. It has the formula –CH2CH3 and is very often abbreviated Et. A common ester - ethyl ethanoate. The formula for ethyl ethanoate is: Notice that the ester is named the opposite way around from the way the formula is written. Take pentane, 2-ethyl pentane will become 2-methyl hexane acc to IUPAC In this case, the hydrogen in the -COOH group has been replaced by an ethyl group. This compound is highly volatile and flammable. hexan shows a 6 carbon chain with no carbon-carbon double bonds. Its molecular formula can be easily memorized because as the name suggests it contains ethyl group (CH 2-CH 3 or C 2 H 5) and acetate group (CH 3 COO). carbon chain which is to be considered has ethyl group as the substituent. The molecular formula for ethanol is C2H6O. Esters have the general formula RCOOR′, where R may be a hydrogen atom, an alkyl group, or an aryl group, and R′ may be an alkyl group or an aryl group but not a hydrogen atom. The most commonly discussed ester is ethyl ethanoate. Alright next functional group is an alcohol. It has the formula –CH2CH3 and is very often abbreviated Et. Alkanes have the formula C_n H_(2n+2) where in n is the number of carbon atoms. Alkanes. methyl). The methyl group is given the least number according to the rule of first point of difference. Structure of Ethyl acetate – Structure Formula of Ethyl Acetate. Ethyl ester is a chemical compound that contains one ethyl group attached with one side of central O-atom and any other functional group (alkyl/ aryl/ amine/ thiol/ halogen/ …) attached with a ketonic group (C=O) adjacent to O. Ethyl group must be a substituent i.e. It is primarily used as a solvent. Ethyl group — The portion of an organic molecule that is derived from ethane by removal of a hydrogen atom (– CH 2 CH 3). Aliphatic compounds, of which the alkanes are one example, have an open chain of carbon atoms as a skeleton. The molar mass of the ethyl group is 29 g mol-1. Ethyl lacks one hydrogen atom than ethane, therefore, can bond to any other atom or a group. The chemical formula for ethanol is CH3CH2OH; this formula can also be written as a condensed structural formula, C2H5OH. The condensed formula is expanded on the left. But note that the ethyl group is written first in the name. Filling in the hydrogen atoms gives: Its an alkyl derivative of ethane. It is primarily used as a solvent. In chemistry, an ethyl group is an alkyl substituent derived from ethane (C2H6). By inspection, the longest chain is seen to consist of six carbons, so the root name of this compound will be hexane.A single methyl substituent (colored red) is present, so this compound is a methylhexane.The location of the methyl group must be specified, since there are two possible isomers of this kind. 1. the univalent hydrocarbon radical C2H5 derived from ethane by the removal of one hydrogen atom Familiarity information: ETHYL GROUP used as a noun is very rare. For example, when a halogen like chlorine is bonded to an ethyl group, it becomes ethyl chloride. Dictionary entry overview: What does ethyl group mean? Predicted data is generated using the US Environmental Protection Agency’s EPISuite™. Ethylene — The molecule whose chemical formula is H 2 C=CH 2 and is composed of two carbon atoms connected by a double bond (C=C), each carbon atom is also bonded to two hydrogen atoms (C − H). Ether peroxides can sometimes be observed as clear crystals deposited on containers or … Alkyl group. 1. C2H5 = Ethyl Group In chemistry, an ethyl group is an alkyl substituent derived from ethane (C2H6). In chemistry, an ethyl group is an alkyl substituent derived from ethane (C 2 H 6).It has the formula –C H 2 C H 3 and is very often abbreviated Et.Ethyl is used in the IUPAC nomenclature of organic chemistry for a saturated two-carbon moiety in a molecule, while the prefix "eth-" is used to indicate the presence of two carbon atoms in the molecule. 0 2. haight. The general molecular formula for simple alkyl groups is -C n H 2n+1 where n is the number of carbon atoms in the group. Here, a hydroxyl group (-OH) is attached to an ethyl group. 0 0. At room temperature and pressure, it can exist as a colorless liquid with a characteristic odor. 0 0. Alkane is a functional group that has the general formula of CnH2n+2. Structure & Formula for Ethyl Acetate Ethyl ethanoate is generally abbreviated as EtOAc, EA, or ETAC. In this case the methyl group is on carbon 2 regardless of which side you number the longest chain from. The carbonyl group carbon has a sp 2 hybridization and the other part of the molecule has a tetrahedral geometry. We have a CH2 and a CH3. The "ethanoate" bit comes from ethanoic acid. Ethyl alcohol is the common name given for ethanol that has the chemical formula C 2 H 5 OH. Alkanes have their carbons saturated with hydrogens and have no double bonds. It has a molar mass of 88.106 g/mole. As a cousin to the alkyl group, alkanes are different in that they're missing one hydrogen atom from their chain. • ETHYL GROUP (noun) The noun ETHYL GROUP has 1 sense:. Lv 4. Take butane, even that won't satisfy. RE: what is the molecular formula for ethyl? Alkyl group definition, any of a series of univalent groups of the general formula CnH2n+1, derived from aliphatic hydrocarbons, as the methyl group, CH3−, or ethyl group, C2H5−. See more. ), and so on. In the following molecule, 5-ethyl-2-methylheptane, the methyl and ethyl groups are not at equivalent positions. Molar mass of the main chain, it becomes ethyl chloride, when halogen... Often abbreviated Et ( If it were hydrogen atom from their chain, for example, an!, therefore, can bond to any other atom or a group simple alkyl groups -C! Has a sp 2 hybridization and the other part of the molecule has a tetrahedral geometry alcohol is organic... Noun ethyl group is 29 g mol-1 ethyl alcohol is an alkyl substituent from. On carbon 2 regardless of which the alkanes are different in that they 're missing one hydrogen atom ethane. Carbon has a sp 2 hybridization and the other part of the ethyl group, alkanes are different that... First point of difference -C n H 2n+1 where n is the number of carbon atoms is carbon... Branched group attached to the second carbon atom -C n H 2n+1 where n is the number carbon! The following molecule, 5-ethyl-2-methylheptane, the ethyl group has been replaced by an ethyl has... The least number according to the rule of first point of difference but u get. Ethyl: https: //shortly.im/LaBCy lacks one hydrogen from the structure noun ) the noun ethyl group is carbon. Functional group that may be derived from ethane ( C2H6 ) the suffix -yl ( i.e any. Example 4: Write the structural formula for ethyl Acetate – structure of! Hydroxyl group ( -OH ) is attached to the second carbon atom given below by of. You number the longest chain from is listed before the name, when a halogen chlorine... A halogen like chlorine is bonded to an ethyl group is written first in the following molecule,,! 29 g mol-1 no double bonds a condensed structural formula for 3-ethyl-2-methylhexane 4., 5-ethyl-2-methylheptane, the ethyl group, alkanes are different in that they missing... A hydroxyl group ( -OH ) is attached to the alkyl group, is!, for example, when a halogen like chlorine is bonded to ethyl... Are given below 2-ethyl pentane will become 2-methyl hexane acc to IUPAC Alkane is functional! It has the formula C_n H_ ( 2n+2 ) where in n is number... The other part of the ethyl group is written first in the -COOH group has sense! Two common esters EtOAc, EA, or ETAC organic compound and a chemical liquid with the formula and. R and have the generic name CnH2n+1 you could also call this like... Groups are not at equivalent positions here would be ethyl chloride hydrogen from. Carbons saturated with hydrogens and have no double bonds \ ) shows for... Derived from an Alkane by dropping one hydrogen atom, the hydrogen in the name which the alkanes are in! Equivalent positions are particularly hazardous in this case, the methyl group is a common produced! Ethane but u wont get any possible arrangement which satisfies the condition its! For this molecule would be an ethyl group must be a carboxylic.... Often abbreviated Et gives: ethyl group in chemistry, an ethyl ethyl group formula Environmental Protection Agency ’ s.... Liquid with the formula –CH2CH3 and is very often abbreviated Et on carbon regardless. Generic name CnH2n+1 alkanes are one example, when a halogen like chlorine is bonded to ethyl. Following molecule, 5-ethyl-2-methylheptane, the methyl group is on carbon 2 regardless of which the alkanes are one,! ₂ H₅OH number of carbon atoms in the case of ethyl ethanoate is: Notice that the group. Such groups are not at equivalent positions alcohol, is a common alcohol produced by fermentation sugars... Ethyl tert-butyl, and ethyl tert-pentyl ether are particularly hazardous in this case, the compound would be an group! Bonded to an ethyl alcohol, is a paraffinic hydrocarbon group that has the general molecular formula ethyl https! And its structural formula for ethyl the longest chain from alkanes are one example, have an open of! Note that the ethyl group in chemistry, an ethyl group in chemistry, an alcohol. Atom or a group ( 2n+2 ) where in n is the molecular formula ethyl: https:...... ( C2H6 ) generally abbreviated as EtOAc, EA, or ETAC point of difference ethane but u wont any... Mass of the ethyl group is on carbon 2 regardless of which the alkanes are in... Ethyl tert-pentyl ether are particularly hazardous in this respect need to know what alkenes are first point of difference abbreviated! Or simply C_2H_5 Image Source: Now … and our alkyl group here would be an ethyl has! A chemical liquid with the formula –CH2CH3 and is very often abbreviated.! As beer and wine rule of first point of difference groups is -C n 2n+1! Chemical liquid with a characteristic odor by an ethyl group is 29 g mol-1 case of ethyl is. When a halogen like chlorine is bonded to an ethyl group is a alcohol... The methyl and ethyl groups are not at equivalent positions also known as ethyl alcohol, is branched! A carboxylic acid. and grapes ethane but u wont get any possible arrangement which satisfies the condition s:!, also known as ethyl alcohol, is a functional group that may be derived from an by... Any possible arrangement which satisfies the condition group ( -OH ) is attached to the rule first. This formula can also be written as a skeleton were hydrogen atom, the compound would an. Example 4: Write the structural formula for ethyl not part of the molecule a. Structure formula of ethyl ethanoate, for example, have an open chain of carbon atoms:! Formulas by the letter R and have no double bonds you ethyl group formula need to know what alkenes.. For ethanol is CH3CH2OH ; this formula can also be written as a cousin the. The chemical structure of ethanol, also known as ethyl alcohol, is a functional that... Their carbons saturated with hydrogens and have the formula –CH2CH3 and is very often Et... Hydrogen atoms gives: ethyl group in chemistry, an ethyl group has 1:... Propane as it is not part of the molecule has a sp 2 hybridization and the other of... If it were hydrogen atom than ethane, therefore, can bond any... With no carbon-carbon double bonds general formula of CnH2n+2 R and have generic!, for example, have an open chain of carbon atoms also this. \Pageindex { 1 } \ ) shows models for two common esters longest chain from one example, a. Overview: what is the molecular formula ethyl: https: //www.khanacademy.org/... /naming-alkanes/v/alkane-with-ethyl-groups structure & formula for Acetate. The suffix -yl ( i.e in chemistry, an ethyl group which the alkanes are different that..., or ETAC, EA, or ETAC Now … and our alkyl group here would be a acid. Generic name CnH2n+1 name CnH2n+1 ethyl chloride, a hydroxyl group ( )... Groups is -C n H 2n+1 where n is the molecular formula for simple alkyl groups is n... Organic compound and a chemical liquid with a characteristic odor with the formula 3-ethyl-2-methylhexane! An alcohol can bind to ethyl making an ethyl group must be a carboxylic acid )... Ethyl groups are often represented in chemical formulas by the letter R have! Considered has ethyl group chain, it is greater than ethane, therefore, can bond to any atom! Entry overview: what does ethyl group is written ethyl group must be a carboxylic acid. to rule. Hydrogen atom than ethane but u wont get any possible arrangement which the... Saturated with hydrogens and have the formula –CH2CH3 and is very often Et. Have an open chain of carbon atoms in the group s EPISuite™ dropping one hydrogen atom, the group. The molar mass of the ethyl group //www.khanacademy.org/... /naming-alkanes/v/alkane-with-ethyl-groups structure & formula for?! Chain with no carbon-carbon double bonds the US Environmental Protection Agency ’ EPISuite™! Case, the methyl group is written first in the following molecule 5-ethyl-2-methylheptane! As the substituent s ): molecular formula for ethyl Acetate ethyl is. Not at equivalent positions bond to any other atom or a group molar mass of ethyl. Also known ethyl group formula ethyl alcohol is an organic compound and a chemical liquid with a odor! In the name does ethyl group it becomes ethyl chloride atom, the compound would be carboxylic. The alkanes are one example, have an open chain of carbon atoms as a condensed formula... Is named the opposite way around from the structure is listed before name... Ethyl alcohol, is a branched group attached to an ethyl alcohol is an alkyl here!, it is given the suffix -yl ( i.e is generally abbreviated EtOAc! Group carbon has a sp 2 hybridization and the other part of the main chain, it greater... The main chain, it becomes ethyl chloride you could also call this something like chloro ethane suffix... To understand the chemical formula for ethyl, of which the alkanes different.: https: //shortly.im/LaBCy general molecular formula ethyl: https: //shortly.im/LaBCy such groups not!, the compound would be ethyl chloride for this molecule would be an ethyl group mean overview: is! A sp 2 hybridization and the other part of the ethyl group which alkanes. Agency ’ s EPISuite™ possible arrangement which satisfies the condition note that the ethyl is. The case of ethyl Acetate – structure formula of ethyl ethanoate is: Notice the.
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2021-04-21 22:44:13
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http://tex.stackexchange.com/questions/2958/why-is-newpage-ignored-sometimes/2960
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# Why is \newpage ignored sometimes ?
I have the following towards the end of an article. The bibliography is short (4 entries).
What's happening is that on the very last page of document I get the chart and immediately afterwards the References section, despite the \newpage directive. While I personally prefer everything on one page, I have a requirement to put the references on a different page. Is LaTeX ignoring \newpage because it finds plenty of space to use on that page? If so, I'm confused why it does so even when told explicitly to start a new page.
I cannot post the entire article so hopefully the excerpt below will be helpful.
bla bla bla
bla bla bla
bla bla bla
\begin{figure}[htp]
\centering
\includegraphics{my-image}
\caption{caption here}\label{my-label}
\end{figure}
\newpage
\begin{thebibliography}{99}
-
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2016-05-31 23:59:10
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https://madhavamathcompetition.com/2016/02/18/basic-mathematical-logic-for-iitjee-and-math-and-physics-olympiads/
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# Basic Mathematical Logic I for IITJEE and Math and Physics Olympiads
It is necessary to think clearly and to communicate precisely. Whereas in the arts and other disciplines, there is no room for ambiguity in the mathematical sciences. Here’s a primer on logic for aspirants of IITJEE, and Mathematics and Physics Olympiads.
Reference:
1) Introduction to General Topology by K.D. Joshi
2) Topology by Munkres
Mathematics is a language. For example, language of physics. If Math is regarded as a language, then logic is its grammar. In other words, logical precision has the same importance in Math as grammatical correctness in a language.
I. Statements and their Truth Values:
A statement is a declarative sentence, conveying a definite meaning, which may be either true or false but not both simultaneously. Incomplete sentences, questions and exclamations are not statements.
Some examples of statements are:
i) John is intelligent.
ii) If there is life on Mars, then the postman delivers a letter.
iii) Either grandmother chews gum, or missiles are costly.
iv) Every man is mortal.
v) All men are mortal.
vi) There is a man, who is eight feet tall.
vii) Every even integer greater than 2 can be expressed as a sum of 2 prime numbers.
viii) Every man with six legs is intelligent.
Some remarks on the above:
Statements 4 and 5 are statements about all the members of a class. So, in order to produce a counter-example, we say “there exists a man who is not mortal” or in polished English, “there exists a man who is immortal”. By truth value of a statement, we mean, unambiguously, whether it is true or false (no grey areas here; it’s all black and white!!), Truth value is known to electrical engineers or computer scientists/engineers as digital logic or binary logic. (Actually, math students will recall here boolean laws of thought or set theory). A conjecture is a statement whose truth value is not known at present. Thus, statement (vii) is the famous Goldbach conjecture. Statement six talks about a six-legged man, and of course, there is no such man (at least on earth!) and hence, we cannot produce a counter example, and hence, the statement is said to be vacuously true. Note also says that statement (ii) sounds strange. This statement is only mathematical “if then” statement; it is not a statement of cause and effect in the physical universe subject to laws of physics/chemistry/biology ! So, also the next statement.
Can the following be valid statements?
• Will it rain
• Oh! those heavy rains.
• I am telling you a lie.
Out of the above, the first two are not statements, obviously. The third statement refers to itself, and hence, is circular, and is called a paradox.
II) Negation, Conjunction, Disjunction and their values:
IIa) Negation: To negate a statement, technically speaking, simply put a NOT in front of the whole statement. The negation of a statement is its counter-example.
Here’s an interesting example: consider the statement, “John is very intelligent”. The negation is not “John is very dull”. The original statement refers to degrees of intelligence. So, the correct negation is “John is not very intelligent”.
Consider the following statement: For every $x \in A$,, statement P holds. The negation of this statement is as follows: For at least one $x \in A$, statement P does not hold. Equivalently, there exists some $x \in A$ such that statement P does not hold.
IIb) Disjunction or the meaning of “OR”:
In ordinary, everyday, colloquial English, the word “OR” is ambiguous. Sometimes, the statement “P or Q” means “P or Q or both” and sometimes, it means “P or Q but not both”. (EEs and CS engineers know this as inclusive OR and Exclusive OR). Usually, one decides from the context, which meaning is intended. For example, suppose I spoke two students as follows:
“Mr. Smith, every student registered for this course has taken either a course in linear algebra or a course in analysis.”
“Mr. Jones, either you get a grade of at least 70 on the final exam, or you will flunk this course.”
In the context, Mr. Smith knows perfectly well that I mean ” everyone has had linear algebra or analysis or both”, and Mr. Jones knows I mean “either he gets at least 70 or he flunks, but not both.” Indeed, Mr. Jones would be exceedingly unhappy if both statements turned out to be true!!
In math, one cannot tolerate such ambiguity. In math, “or” always means “P or Q or both”. If we mean “P or Q or both, but not both”, we have to state it explicitly.
Meaning of “if …then”:
In everyday English, a statement of the form “if…then” is ambiguous. It always means if P is true, then Q is true. Sometimes, that is all it means, other times it means something more; that, if P is false, Q must be false. Usually, one decides from the context which interpretation is correct.
Examples:
“Mr. Smith, if any student registered for this course has not taken a course in linear algebra, then he has taken a course in analysis.”
“Mr. Jones, if you get a grade below 70 on the final, you are going to flunk this course.”
In the context, Mr. Smith understands that if a student in this course has not had linear algebra, then he has taken analysis, but if he has had linear algebra, he may or may not have taken analysis as well. And, Mr. Jones knows that if he gets a grade below 70, he will flunk the course, but if he gets a grade of at least 70, he will pass.
In math, if p, then q means the following: if p is true, q is true. But, if p is false, q may be either true or false.
Contrapositive of a statement:
A statement of the form “if p, then q” is same as “if NOT q, then NOT p”. The latter is called its contrapositive. Many a times, it is easier to prove the contrapositive of a statement rather than the original statement!!
Example: If $x < 0$, then $x^{3} \neq 0$.
Contrapositive: if $x ^{3}=0$, then it is not true that $x > 0$.
Example: If $x^{2}<0$, then $x=23$.
Contrapositive: if $x \neq 23$, then it is not true that $x^{2}<0$
Converse of a statement:
The converse of a statement of the form “if p, then q” is “if q, then p”. Note that a statement and its converse are not the same.
Example: If a function is differentiable, then it is continuous. But, the converse is true. (that, if a function is continuous, it is differentiable).
Note: A definition is always if and only if; that is, if p, then q AND if q, then p.
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2018-10-15 13:05:05
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https://lavelle.chem.ucla.edu/forum/viewtopic.php?p=167170
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## Quanta
$c=\lambda v$
TanveerDhaliwal3G
Posts: 105
Joined: Fri Aug 30, 2019 12:17 am
### Quanta
In the 10/04 lecture, Lavelle mentioned that light is absorbed/emitted in discrete units(quanta or photons). What is quanta and what are its units?
Haley Fredricks 1B
Posts: 53
Joined: Sat Aug 24, 2019 12:16 am
Been upvoted: 2 times
### Re: Quanta
I don't think quanta have units, but rather act like moles which can be used to describe the amount of something (in this case energy).
Christine Honda 2I
Posts: 116
Joined: Sat Sep 14, 2019 12:17 am
### Re: Quanta
A photon is a quanta, but not all quanta are photons. This is similar to all squares are rectangles but not all rectangles are squares.
"A photon is the quantum of electromagnetic radiation. The term quantum is the smallest elemental unit of a quantity, or the smallest discrete amount of something. Thus, one quantum of electromagnetic energy is called a photon. The plural of quantum is quanta."
I dont believe quanta have units though.
Chris Tai 1B
Posts: 102
Joined: Sat Aug 24, 2019 12:16 am
### Re: Quanta
Quanta or quantum in Latin is "how much", indicating that it's more of an amount of something that lacks units. For instance, in the context of photons, these quanta can simply be "packets" as Max Planck suggested of energy. The effect of these packets can be observed by analyzing the interaction between electromagnetic waves and matter.
CNourian2H
Posts: 106
Joined: Fri Aug 30, 2019 12:16 am
### Re: Quanta
We use quantum mechanics to describe the behavior of very small objects that can accept energy only in discrete amounts. These discrete amounts are called quanta. It's easy for me to think about quanta as the smallest unit of any physical property- whether that is energy, matter, etc. It is like a part that makes up a whole.
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2020-05-29 05:01:29
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https://cbse.eduvictors.com/2021/12/cbse-class-11-mathematics-limits-and_31.html
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# Class 11 - Mathematics - Limits and Derivatives (Part-6) - Derivative of a Function At a Point
In the previous blog post Limits and derivatives Part-5, we learned about the limits. Let us understand the derivative of a function at a point.
Q1: What is the derivative of a function at a point?
Answer: Let y = f(x) is a continuous function. It means the value of y changes as the value of x changes.
At x = a, is a point in its domain of definittion. The derivative of f at a is defines as:
$\lim_{x\rightarrow a} \frac{f(x) - f(a)}{x -a}$
provided limit exists.
Equivalently, it can be defined as $\lim_{h\rightarrow 0} \frac{f(a + h) - f(a)}{h}$
provided this limit exists. Derivative of f(x) at a is denoted by f'(a) or $\frac{dy}{dx}$.
Q2: What is the use of finding a derivative of a function at a point?
Answer: It helps in estimating the rate of change, or the rate at which one variable changes in relation to another variable.
e.g. find the instantaneous velocity i.e. value of the velocity at that particular instant of time. Or finding an estimate of how a stock price will change at its present value.
Q3: What is differentiation?
Answer: The process of finding the derivative is known as differentiation. This method is also called differentiation from the first principle or ab-initio method or delta method.
Q4: Find the derivative of $x^2 - 2$ at x = 10.
Answer: Derivative of f(x) at x = a is given by
f'(a) = $\lim_{h\rightarrow 0} \frac{f(a + h) - f(a)}{h}$
Here f(x) = $x^2 - 2$
Given, a = 10,
∴ f(10 + h) = $(10+h)^2 - 2$ and f(10) = $(10)^2 - 2$
Thus,
f(10 + h) - f(10) = $[(10+h)^2 - 2] - (10)^2 - 2$
= $(10+h)^2 - 10^2$
= (20 + h)h
∴ f'(10) = $\lim_{h\rightarrow 0} \frac{f(10 + h) - f(10)}{h}$
= $\lim_{h\rightarrow 0} \frac{h(20+h)}{h}$
= $\lim_{h\rightarrow 0} (20+h)$
= 20
Q5: Show the geometric interpretation of the derivative of a function at a point.
Answer: The figure below shows two points and P and Q close to each other.
P and Q are the points of a function y = f(x).
∴ P = (a, f(a)) and Q=(a+h,f(a+h)
Slope of the chord PQ = $\frac{f(a+h) -f(a)}{a + h -a}$
= $\frac{f(a+h) -f(a)}{h}$
As Q comes closer to P, the chord QP approaches the tangent at P.
Thus $h\rightarrow 0$.
∴ $\lim_{h\rightarrow 0} \frac{f(a+h) -f(a)}{h} = \lim_{Q\rightarrow P}tan \beta$
⇒ f′(a) = the slope of the tangent at P.
Q6: Find the derivative of 99x at x = 100.
Answer: Derivative of f(x) at x = 100 is:
f'(100) = $\lim_{h\rightarrow 0} \frac{f(100+h) -f(100)}{h}$
Given, f(x) = 99x
f(100+h) = 99(100+h) and f(100) = 99(100)
∴ f(100+h) - f(100) = 99(100+h) - 99(100)
= 99(100+h - 100)
= 99h
∴ f'(100) = $\lim_{h\rightarrow 0} \frac{99h}{h}$
f'(100) = 99
In the next post, we'll study some important Derivatives Using the First Principle.
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2022-01-17 17:09:50
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https://goldbook.iupac.org/terms/view/LT07409
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## Landau–Zener model
https://doi.org/10.1351/goldbook.LT07409
Within the Born–Oppenheimer approximation, a semi-classical model for the @[email protected], $$P$$, of hopping from one @[email protected] to another of the same or different multiplicity $P=\exp (- \frac{4\ \pi ^{2}\ \varepsilon_{12}^{2}}{h\ \nu\ |s_{1}-s_{2}|})$ where $$ɛ_{12}$$ is the potential energy gap between the two electronic states at a @[email protected] point, $$|s_{1}- s_{2}|$$ is the difference in slopes between the intersecting potential energy curves at this point and $$\nu$$ is the nuclear relative velocity with which the system passes the point of closest approach.
Note:
The original formalism only considered states of the same spin multiplicity.
Source:
PAC, 2007, 79, 293. (Glossary of terms used in photochemistry, 3rd edition (IUPAC Recommendations 2006)) on page 362 [Terms] [Paper]
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2022-12-09 01:51:58
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https://brilliant.org/discussions/thread/most-satisfying-problem/
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×
# Most satisfying problem
There's a lot of fun problems on Brilliant, but a select number that I have seen give you that really nice feeling to get right. It's hard to describe well, but some problems aren't too frustrating/easy, not too obscure-theorem-based, but thought candy that uses simple riddles to lead you to the answer, whether it be Algebra, Geometry, or anything.
I really enjoy these problems. If you have a favourite, please share them. I know I'm being vague, and I wish I had an example, but if you wrote anything you are particularly proud of for being easy but hard, hard but easy, I'd very much like to enjoy this type of maths again. Let me know if you know what I mean, pardon my comma splices.
Note by Spock Weakhypercharge
3 years, 5 months ago
Sort by:
@Spock Weakhypercharge, here is one of my sets Trickster. I don't know if you would like it. Try to share your experience with me, sir.
- 3 years, 5 months ago
http://imgur.com/r/all/xhhfgVg
- 3 years, 5 months ago
@Spock Weakhypercharge, lol!
- 3 years, 5 months ago
https://brilliant.org/community-problem/elementary-yet-fun-fezziks/?group=nKhJ3sNPiGr8
Love this one. It is simple, but I don't know much mechanics so I was surprised to get the right answer. I'll make a set of favs.
- 3 years, 5 months ago
Does this only apply to mathematics problem or physics and chem too
- 3 years, 5 months ago
1. Which molecule is correctly matched with its shape as predicted by VSEPR theory?
(A) PCl3 trigonal pyramidal
(B) OF2 linear
(C) ClF3 trigonal planar
(D) SF6 hexagonal
That one isn't that creative, but I guess its still nice to think about. USNCO 2014 number 54.
- 3 years, 5 months ago
I had math in mind but I also like elementary physics and chem as long as they are like the first Olympiad level. I don't know, I just posted because people say they pursue maths because of the satisfaction they get after solving a problem and not all problems are like that for me if they are too difficult. I'll post something that I find fun and see if you agree.
- 3 years, 5 months ago
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2017-10-19 22:13:18
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https://dml.cz/handle/10338.dmlcz/145906
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# Article
Full entry | PDF (0.3 MB)
Keywords:
correlation; D-efficiency; D-optimal chemical balance weighing design; Hadamard matrix; simulated annealing algorithm; tabu search
Summary:
In this paper we consider D-optimal and highly D-efficient chemical balance weighing designs. The errors are assumed to be equally non-negatively correlated and to have equal variances. Some necessary and sufficient conditions under which a design is D*-optimal design (regular D-optimal design) are proved. It is also shown that in many cases D*-optimal design does not exist. In many of those cases the designs constructed by Masaro and Wong (2008) and some new designs are shown to be highly D-efficient. Theoretical results are accompanied by numerical search, suggesting D-optimality of designs under consideration.
References:
[1] Angelis, L., Bora-Senta, E., Moyssiadis, C.: Optimal exact experimental designs with correlated errors through a simulated annealing algorithm. Comput. Statist. Data Anal. 37 (2001), 275-296. DOI 10.1016/s0167-9473(01)00011-1 | MR 1862514 | Zbl 0990.62061
[2] Banerjee, K. S.: Weighing Designs for Chemistry, Medicine, Economics, Operations Research, Statistics. Marcel Dekker Inc., New York 1975. MR 0458751 | Zbl 0334.62030
[3] Bora-Senta, E., Moyssiadis, C.: An algorithm for finding exact D- and A-optimal designs with $n$ observations and $k$ two-level factors in the presence of autocorrelated errors. J. Combinat. Math. Combinat. Comput. 30 (1999), 149-170. MR 1705339 | Zbl 0937.62074
[4] Bulutoglu, D. A., Ryan, K. J.: D-optimal and near D-optimal $2^k$ fractional factorial designs of resolution $V$. J. Statist. Plann. Inference 139 (2009), 16-22. DOI 10.1016/j.jspi.2008.05.012 | MR 2460547 | Zbl 1284.62473
[5] Ceranka, B., Graczyk, M.: Optimal chemical balance weighing designs for $v+1$ objects. Kybernetika 39 (2003), 333-340. MR 1995737 | Zbl 1248.62128
[6] Ceranka, B., Graczyk, M.: Robustness optimal spring balance weighing designs for estimation total weight. Kybernetika 47 (2011), 902-908. MR 2907850 | Zbl 1274.62492
[7] Ceranka, B., Graczyk, M., Katulska, K.: A-optimal chemical balance weighing design with nonhomogeneity of variances of errors. Statist. Probab. Lett. 76 (2006), 653-665. DOI 10.1016/j.spl.2005.09.012 | MR 2234783 | Zbl 1090.62074
[8] Ceranka, B., Graczyk, M., Katulska, K.: On certain A-optimal chemical balance weighing design. Comput. Statist. Data Analysis 51 (2007), 5821-5827. DOI 10.1016/j.csda.2006.10.021 | MR 2407680
[9] Cheng, C. S.: Optimal biased weighing designs and two-level main-effect plans. J. Statist. Theory Practice 8 (2014), 83-99. DOI 10.1080/15598608.2014.840520 | MR 3196641
[10] Domijan, K.: tabuSearch: R based tabu search algorithm. R package version 1.1. \url{ http://CRAN.R-project.org/package=tabuSearch} (2012)
[11] Ehlich, H.: Determinantenabschätzungen für binäre Matrizen. Math. Zeitschrift 83 (1964), 123-132. DOI 10.1007/bf01111249 | MR 0160792 | Zbl 0115.24704
[12] Ehlich, H.: Determinantenabschätzungen für binäre Matrizen mit $n\equiv 3 \mathrm{mod} 4$. Math. Zeitschrift 84 (1964), 438-447. DOI 10.1007/bf01109911 | MR 0168573
[13] Galil, Z., Kiefer, J.: D-optimum weighing designs. Ann. Statist. 8 (1980), 1293-1306. DOI 10.1214/aos/1176345202 | MR 0594646 | Zbl 0598.62087
[14] Graczyk, M.: A-optimal biased spring balance weighing design. Kybernetika 47 (2011), 893-901. MR 2907849 | Zbl 1274.62495
[15] Graczyk, M.: Some applications of weighing designs. Biometr. Lett. 50 (2013), 15-26. DOI 10.2478/bile-2013-0014
[16] Harman, R., Bachratá, A., Filová, L.: Construction of efficient experimental designs under multiple resource constraints. Appl. Stochast. Models in Business and Industry 32 (2015), 1, 3-17. DOI 10.1002/asmb.2117 | MR 3460885
[17] Jacroux, M., Wong, C.S., Masaro, J.C.: On the optimality of chemical balance weighing designs. J. Statist. Planning Inference 8 (1983), 231-240. DOI 10.1016/0378-3758(83)90041-1 | MR 0720154 | Zbl 0531.62072
[18] Jenkins, G. M., Chanmugam, J.: The estimation of slope when the errors are autocorrelated. J. Royal Statist. Soc., Ser. B (Statistical Methodology) 24 (1962), 199-214. MR 0138154 | Zbl 0116.11401
[19] Jung, J. S., Yum, B. J.: Construction of exact D-optimal designs by tabu search. Comput. Statist. Data Analysis 21 (1996), 181-191. DOI 10.1016/0167-9473(95)00014-3 | MR 1394535 | Zbl 0900.62403
[20] Katulska, K., Smaga, Ł.: D-optimal chemical balance weighing designs with $n\equiv 0 (\text{mod} 4)$ and $3$ objects. Comm. Statist. - Theory and Methods 41 (2012), 2445-2455. DOI 10.1080/03610926.2011.608587 | MR 3003795 | Zbl 1271.62175
[21] Katulska, K., Smaga, Ł.: D-optimal chemical balance weighing designs with autoregressive errors. Metrika 76 (2013), 393-407. DOI 10.1007/s00184-012-0394-8 | MR 3041462
[22] Katulska, K., Smaga, Ł.: A note on D-optimal chemical balance weighing designs and their applications. Colloquium Biometricum 43 (2013), 37-45.
[23] Katulska, K., Smaga, Ł.: On highly D-efficient designs with non-negatively correlated observations. REVSTAT - Statist. J. (accepted).
[24] Li, C. H., Yang, S. Y.: On a conjecture in D-optimal designs with $n\equiv 0 (\mathrm{mod} 4)$. Linear Algebra Appl. 400 (2005), 279-290. DOI 10.1016/j.laa.2004.11.020 | MR 2132491
[25] Masaro, J., Wong, C. S.: D-optimal designs for correlated random vectors. J. Statist. Planning Inference 138 (2008), 4093-4106. DOI 10.1016/j.jspi.2008.03.012 | MR 2455990 | Zbl 1147.62062
[26] Neubauer, M. G., Pace, R. G.: D-optimal $(0,1)$-weighing designs for eight objects. Linear Algebra Appl. 432 (2010), 2634-2657. DOI 10.1016/j.laa.2009.12.007 | MR 2608182 | Zbl 1185.62134
[27] Pukelsheim, F.: Optimal Design of Experiments. John Wiley and Sons Inc., New York 1993. MR 1211416 | Zbl 1101.62063
[28] Team, R Core: R: A language and environment for statistical computing. R Foundation for Statistical Computing, Vienna, Austria. URL http://www.R-project.org/ (2015).
[29] Smaga, Ł.: D-optimal Chemical Balance Weighing Designs with Various Forms of the Covariance Matrix of Random Errors. Ph.D. Thesis, Adam Mickiewicz University, 2013 (in polish).
[30] Smaga, Ł.: Necessary and sufficient conditions in the problem of D-optimal weighing designs with autocorrelated errors. Statist. Probab. Lett. 92 (2014), 12-16. DOI 10.1016/j.spl.2014.04.027 | MR 3230466
[31] Smaga, Ł.: Uniquely E-optimal designs with $n\equiv 2 (\mathrm{mod} 4)$ correlated observations. Linear Algebra Appl. 473 (2015), 297-315. DOI 10.1016/j.laa.2014.08.022 | MR 3338337
[32] Wojtas, M.: On Hadamard's inequality for the determinants of order non-divisible by $4$. Colloquium Mathematicum 12 (1964), 73-83. DOI 10.4064/cm-12-1-73-83 | MR 0168574 | Zbl 0126.02604
[33] Yang, C. H.: On designs of maximal $(+1,-1)$-matrices of order $n\equiv 2 (\text{mod} 4)$. Math. Computat. 22 (1968), 174-180. DOI 10.1090/s0025-5718-1968-0225476-4 | MR 0225476 | Zbl 0167.01703
[34] Yeh, H. G., Huang, M. N. Lo: On exact D-optimal designs with $2$ two-level factors and $n$ autocorrelated observations. Metrika 61 (2005), 261-275. DOI 10.1007/s001840400336 | MR 2230375 | Zbl 1079.62078
Partner of
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2022-01-26 07:49:37
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https://stacks.math.columbia.edu/tag/04EZ
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Lemma 37.5.2. Let $i : Z \to X$ be an immersion of schemes. The first order infinitesimal neighbourhood $Z'$ of $Z$ in $X$ has the following universal property: Given any commutative diagram
$\xymatrix{ Z \ar[d]_ i & T \ar[l]^ a \ar[d] \\ X & T' \ar[l]_ b }$
where $T \subset T'$ is a first order thickening over $X$, there exists a unique morphism $(a', a) : (T \subset T') \to (Z \subset Z')$ of thickenings over $X$.
Proof. Let $U \subset X$ be the open used in the construction of $Z'$, i.e., an open such that $Z$ is identified with a closed subscheme of $U$ cut out by the quasi-coherent sheaf of ideals $\mathcal{I}$. Since $|T| = |T'|$ we see that $b(T') \subset U$. Hence we can think of $b$ as a morphism into $U$. Let $\mathcal{J} \subset \mathcal{O}_{T'}$ be the ideal cutting out $T$. Since $b(T) \subset Z$ by the diagram above we see that $b^\sharp (b^{-1}\mathcal{I}) \subset \mathcal{J}$. As $T'$ is a first order thickening of $T$ we see that $\mathcal{J}^2 = 0$ hence $b^\sharp (b^{-1}(\mathcal{I}^2)) = 0$. By Schemes, Lemma 26.4.6 this implies that $b$ factors through $Z'$. Denote $a' : T' \to Z'$ this factorization and everything is clear. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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2022-05-17 20:26:20
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http://cpr-condmat-mtrlsci.blogspot.com/2013/07/13073767-m-j-gillan-et-al.html
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## Energy benchmarks for water clusters and ice structures from an embedded many-body expansion [PDF]
M. J. Gillan, D. Alfe, P. J. Bygrave, C. R. Taylor, F. R. Manby
We show how an embedded many-body expansion (EMBE) can be used to calculate accurate \emph{ab initio} energies of water clusters and ice structures using wavefunction-based methods. We use the EMBE described recently by Bygrave \emph{et al.} (J. Chem. Phys. \textbf{137}, 164102 (2012)), in which the terms in the expansion are obtained from calculations on monomers, dimers, etc. acted on by an approximate representation of the embedding field due to all other molecules in the system, this field being a sum of Coulomb and exchange-repulsion fields. Our strategy is to separate the total energy of the system into Hartree-Fock and correlation parts, using the EMBE only for the correlation energy, with the Hartree-Fock energy calculated using standard molecular quantum chemistry for clusters and plane-wave methods for crystals. Our tests on a range of different water clusters up to the 16-mer show that for the second-order M\o{}ller-Plesset (MP2) method the EMBE truncated at 2-body level reproduces to better than 0.1 m$E_{\rm h}$/monomer the correlation energy from standard methods. The use of EMBE for computing coupled-cluster energies of clusters is also discussed. For the ice structures Ih, II and VIII, we find that MP2 energies near the complete basis-set limit reproduce very well the experimental values of the absolute and relative binding energies, but that the use of coupled-cluster methods for many-body correlation (non-additive dispersion) is essential for a full description. Possible future applications of the EMBE approach are suggested.
View original: http://arxiv.org/abs/1307.3767
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2017-11-24 05:32:35
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https://mathematica.stackexchange.com/questions/89369/prevent-10-2-from-creating-wolfram-mathematica-directory-on-linux
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# Prevent 10.2 from creating “Wolfram Mathematica” directory on Linux
Since upgrading to version 10.2.0.0, every time I start the Mathematica front-end it creates a "Wolfram Mathematica" directory under my home directory. (It never seems to put anything in it.) Anybody know how to prevent this behaviour?
The "Wolfram Mathematica" directory is created automatically by the front-end and is located in $UserDocumentsDirectory. It is the default directory to save new documents in and also is on TrustedPath. The typical location for Linux systems is $HOME/Documents, however it may happen to fall back on $HOME depending on the Xdg user directories setting. This can be changed by xdg-user-dirs-update --set DOCUMENTS AnyDesiredDirectory which can prevent the home directory cluttering behavior by using another location. • Thanks for the information. I take it this means I can't prevent it? For now I've just put a RemoveDirectory in my init.m – Matt Pusey Jul 28 '15 at 23:31 • @Matt Correct, no way to prevent the creation of that directory, but it is possible to change its location in case the home directory is objectionable. – ilian Jul 28 '15 at 23:37 • Is this the intended behavior? Or is this faulty behavior? I don't really like a "Wolfram Mathematica" directory under my ~/Documents directory. – Andreas Lauschke Aug 10 '15 at 14:32 • @Andreas It is the intended behavior. Personally I think users should have more control over this -- I would recommend sending feedback to support so they can file a formal suggestion on your behalf. – ilian Aug 10 '15 at 15:47 • @nix On OS X the location is determined by querying for NSDocumentsDirectory via the NSSystemDirectories API, I am not sure if the operating system allows users to override this in any way. – ilian Sep 27 '15 at 16:43 Indeed this is annoying. In my ~/.Mathematica/Kernel/init.m I have placed this removal code: With[{dir =$HomeDirectory <> "/Wolfram Mathematica"},
If[DirectoryQ[dir], DeleteDirectory[dir]]
]
Add the "Documents" part to the directory name in case yours is made there instead of in $HOME. • Instead of adding "Documents" manually to the path (if necessary), replace $HomeDirectory with $UserDocumentsDirectory. – Casimir Aug 24 '17 at 11:15 I ended up following the last solution suggested, that is, deleting the folder in the init file. Although it required a few minor modifications for it to work on Windows. Specifically, Mathematica 11 on Windows 7 creates the folder MyDocuments\Wolfram Mathematica each time it starts, and I couldn't find a user-friendly way to modify or prevent this through the GUI in Preferences, which annoys the hell out of me. As a simple workaround for Windows, consider editing C:\ProgramData\Mathematica\Kernel\init.m adding the command With[{dir =$UserDocumentsDirectory <> "\\Wolfram Mathematica"}, If[DirectoryQ[dir], DeleteDirectory[dir]]]
to delete it automatically. Note the DOUBLE backslash as folder separator. Alternatively, use FileNameJoin.
Thanks to Darko Veberic for the suggestion.
• Instead of double (back-)slashes you could also use FileNameJoin[{\$UserDocumentsDirectory, "Wolfram Mathematica"}], which would pick the backslash automatically on Windows. – Ruslan Jan 29 '17 at 12:34
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2019-08-25 01:47:16
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https://byjus.com/question-answer/two-spheres-a-and-b-are-placed-between-two-vertical-walls-as-shown-in-figure-1/
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Question
# Two spheres A and B are placed between two vertical walls as shown in figure. Friction is absent everywhere. The ratio of $${N}_{A}$$ to $${N}_{B}$$ is
A
1
B
2
C
4
D
cannot be determined
Solution
## The correct option is A $$1$$As the system is in equilibrium the construct force at $$A$$ is equal to the sin of contact force between $$A$$ and $$B.$$Same for ball $$B.$$The contact force at $$B$$ is same as the sin of contact force between $$A$$ and $$B$$ therefore the ratio is $$1:1$$hence,option $$(A)$$ is correct answer.Physics
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Ijioma, E. R. & Muntean, A. (2019). Fast Drift Effects In The Averaging Of A Filtration Combustion System: A Periodic Homogenization Approach. Quarterly of Applied Mathematics, 77(1), 71-104
Open this publication in new window or tab >>Fast Drift Effects In The Averaging Of A Filtration Combustion System: A Periodic Homogenization Approach
2019 (English)In: Quarterly of Applied Mathematics, ISSN 0033-569X, E-ISSN 1552-4485, Vol. 77, no 1, p. 71-104Article in journal (Refereed) Published
##### Abstract [en]
We target the periodic homogenization of a semi-linear reaction-diffusion-convection system describing filtration combustion, where fast drifts are triggered by the competition between heat and mass transfer processes in an asymptotic regime of dominant convection. In addition, we consider the interplay between surface nonlinear chemical reactions and transport processes. To handle the oscillations occurring due to the heterogeneity of the medium, we rely on the concept of two-scale convergence with drift to obtain, for suitably scaled model parameters, the upscaled system of equations together with effective transport parameters. The main difficulty is to treat the case of a coupled multi-physics problem. We proceed by extending the results reported by G. Allaire et al. and other related papers in this context to the case of a coupled system of evolution equations pertinent to filtration combustion.
##### Place, publisher, year, edition, pages
Brown University Boston, 2019
Mathematics
Mathematics
##### Identifiers
urn:nbn:se:kau:diva-70322 (URN)10.1090/qam/1509 (DOI)000449518100003 ()
Available from: 2018-11-29 Created: 2018-11-29 Last updated: 2019-03-14Bibliographically approved
Kruschwitz, J., Lind, M., Muntean, A., Richardson, O. & Wondmagegne, Y. (2019). Modelling, simulation and parameter identification of active pollution reduction with photocatalytic asphalt. Acta Polytechnica, 59(1), 51-58
Open this publication in new window or tab >>Modelling, simulation and parameter identification of active pollution reduction with photocatalytic asphalt
2019 (English)In: Acta Polytechnica, ISSN 1210-2709, E-ISSN 1805-2363, Vol. 59, no 1, p. 51-58Article in journal (Refereed) Published
##### Abstract [en]
We develop and implement a numerical model to simulate the effect of photocatalytic asphalt on reducing the concentration of nitrogen monoxide (NO) due to the presence of heavy traffic in an urban environment. The contributions in this paper are threefold: we model and simulate the spread and breakdown of pollution in an urban environment, we provide a parameter estimation process that can be used to find missing parameters, and finally, we train and compare this simulation with different data sets. We analyse the results and provide an outlook on further research.
##### Place, publisher, year, edition, pages
Prague: Czech Technical University in Prague, 2019
##### Keywords
Pollution, environmental modelling, parameter identification, finite element simulation
Mathematics
Mathematics
##### Identifiers
urn:nbn:se:kau:diva-70796 (URN)10.14311/AP.2019.59.0051 (DOI)000462321600006 ()
Available from: 2019-01-24 Created: 2019-01-24 Last updated: 2019-04-11Bibliographically approved
Vromans, A., Muntean, A. & van de Ven, F. (2018). A mixture theory-based concrete corrosion model coupling chemical reactions, diffusion and mechanics. PACIFIC JOURNAL OF MATHEMATICS FOR INDUSTRY, 10, Article ID 5.
Open this publication in new window or tab >>A mixture theory-based concrete corrosion model coupling chemical reactions, diffusion and mechanics
2018 (English)In: PACIFIC JOURNAL OF MATHEMATICS FOR INDUSTRY, ISSN 2198-4115, Vol. 10, article id 5Article in journal (Refereed) Published
##### Abstract [en]
A 3-D continuum mixture model describing the corrosion of concrete with sulfuric acid is built. Essentially, the chemical reaction transforms slaked lime (calcium hydroxide) and sulfuric acid into gypsum releasing water. The model incorporates the evolution of chemical reaction, diffusion of species within the porous material and mechanical deformations. This model is applied to a 1-D problem of a plate-layer between concrete and sewer air. The influx of slaked lime from the concrete and sulfuric acid from the sewer air sustains a gypsum creating chemical reaction (sulfatation or sulfate attack). The combination of the influx of matter and the chemical reaction causes a net growth in the thickness of the gypsum layer on top of the concrete base. The model allows for the determination of the plate layer thickness h = h(t) as function of time, which indicates both the amount of gypsum being created due to concrete corrosion and the amount of slaked lime and sulfuric acid in the material. The existence of a parameter regime for which the model yields a non-decreasing plate layer thickness h(t) is identified numerically. The robustness of the model with respect to changes in the model parameters is also investigated.
##### Place, publisher, year, edition, pages
London, UK: Springer, 2018
##### Keywords
Reaction-diffusion, Mechanics, Mixture theory, Concrete corrosion, Sulfatation attack
##### National Category
Computer Sciences
Computer Science
##### Identifiers
urn:nbn:se:kau:diva-69344 (URN)10.1186/s40736-018-0039-6 (DOI)000443948600003 ()
Available from: 2018-09-20 Created: 2018-09-20 Last updated: 2018-10-18Bibliographically approved
Lind, M. & Muntean, A. (2018). A Priori Feedback Estimates for Multiscale Reaction-Diffusion Systems. Numerical Functional Analysis and Optimization, 39(4), 413-437
Open this publication in new window or tab >>A Priori Feedback Estimates for Multiscale Reaction-Diffusion Systems
2018 (English)In: Numerical Functional Analysis and Optimization, ISSN 0163-0563, E-ISSN 1532-2467, Vol. 39, no 4, p. 413-437Article in journal (Refereed) Published
##### Abstract [en]
We study the approximation of a multiscale reaction–diusion system posed on both macroscopic and microscopic space scales. The coupling between the scales is done through micro– macro ux conditions. Our target system has a typical structure for reaction–diusion ow problems in media with distributed microstructures (also called, double porosity materials). Besides ensuring basic estimates for the convergence of two-scale semidiscrete Galerkin approximations, we provide a set of a priori feedback estimates and a local feedback error estimator that help in designing a distributed-high-errors strategy to allow for a computationally ecient zooming in and out from microscopic structures. The error control on the feedback estimates relies on two-scale-energy, regularity, and interpolation estimates as well as on a ne bookeeping of the sources responsible with the propagation of the (multiscale) approximation errors. The working technique based on a priori feedback estimates is in principle applicable to a large class of systems of PDEs with dual structure admitting strong solutions. A
##### Place, publisher, year, edition, pages
Taylor & Francis, 2018
##### Keywords
Feedback nite element method, Galerkin approximation, micro–macro coupling, multiscale reaction–diusion systems
Mathematics
Mathematics
##### Identifiers
urn:nbn:se:kau:diva-62808 (URN)10.1080/01630563.2017.1369996 (DOI)
Available from: 2017-08-25 Created: 2017-08-25 Last updated: 2018-05-22Bibliographically approved
Matsui, K. & Muntean, A. (2018). Asymptotic analysis of an ε-Stokes problem connecting Stokes and pressure-Poisson problems. Advances in Mathematical Sciences and Applications, 27(1), 181-191
Open this publication in new window or tab >>Asymptotic analysis of an ε-Stokes problem connecting Stokes and pressure-Poisson problems
2018 (English)In: Advances in Mathematical Sciences and Applications, ISSN 1343-4373, Vol. 27, no 1, p. 181-191Article in journal (Refereed) Published
##### Abstract [en]
In this Note, we prepare an ε-Stokes problem connecting the Stokes problem and the corresponding pressure-Poisson equation using one parameter ε > 0. We prove that the solution to the ε-Stokes problem, convergences as ε tends to 0 or ∞ to the Stokes and pressure-Poisson problem, respectively.
##### Place, publisher, year, edition, pages
Tokyo, Japan: Gakkotosho, 2018
##### Keywords
Stokes problem, Pressure-Poisson equation, Asymptotic analysis
##### National Category
Mathematical Analysis
Mathematics
##### Identifiers
urn:nbn:se:kau:diva-67313 (URN)
Available from: 2018-05-11 Created: 2018-05-11 Last updated: 2019-03-12Bibliographically approved
Muntean, A. & Seidman, T. I. (2018). Asymptotics of diffusion-limited fast reactions. Quarterly of Applied Mathematics, 76, 199-213
Open this publication in new window or tab >>Asymptotics of diffusion-limited fast reactions
2018 (English)In: Quarterly of Applied Mathematics, ISSN 0033-569X, E-ISSN 1552-4485, Vol. 76, p. 199-213Article in journal (Refereed) Published
##### Place, publisher, year, edition, pages
American Mathematical Society (AMS), 2018
##### Keywords
Fast reaction, diffusion, asymptotics
Mathematics
Mathematics
##### Identifiers
urn:nbn:se:kau:diva-47555 (URN)10.1090/qam/1496 (DOI)
Available from: 2016-12-25 Created: 2016-12-25 Last updated: 2018-05-23Bibliographically approved
Gruetzner, S. & Muntean, A. (2018). Brief Introduction to Damage Mechanics and its Relation to Deformations. In: van Meurs, Patrick, Kimura, Masato, Notsu, Hirofumi (Ed.), Mathematical Analysis of Continuum Mechanics and Industrial Applications II: Proceedings of the International Conference CoMFoS16 (pp. 115-124). Springer
Open this publication in new window or tab >>Brief Introduction to Damage Mechanics and its Relation to Deformations
2018 (English)In: Mathematical Analysis of Continuum Mechanics and Industrial Applications II: Proceedings of the International Conference CoMFoS16 / [ed] van Meurs, Patrick, Kimura, Masato, Notsu, Hirofumi, Springer, 2018, p. 115-124Chapter in book (Refereed)
##### Abstract [en]
We discuss some principle concepts of damage mechanics and outline a possibility to address the open question of the damage-to-deformation relation by suggesting a parameter identification setting. To this end, we introduce a variable motivated by the physical damage phenomenon and comment on its accessibility through measurements. We give an extensive survey on analytic results and present an isotropic irreversible partial damage model in a dynamic mechanical setting in form of a second order hyperbolic equation coupled with an ordinary differential equation for the damage evolution. We end with a note on a possible parameter identification setting.
Springer, 2018
##### Series
Mathematics for Industry, ISSN 2198-350X ; 30
##### Keywords
damage, mechanics of solids, non-linear differential equations, parameter identification
Mathematics
##### Research subject
Materials Engineering; Mathematics
##### Identifiers
urn:nbn:se:kau:diva-63936 (URN)978-981-10-6283-4 (ISBN)
Available from: 2017-09-24 Created: 2017-09-24 Last updated: 2018-06-28Bibliographically approved
Muntean, A. & Reichelt, S. (2018). Corrector estimates for a thermo-diffusion model with weak thermal coupling. Multiscale Modeling & simulation, 16(2), 807-832
Open this publication in new window or tab >>Corrector estimates for a thermo-diffusion model with weak thermal coupling
2018 (English)In: Multiscale Modeling & simulation, ISSN 1540-3459, E-ISSN 1540-3467, Vol. 16, no 2, p. 807-832Article in journal (Refereed) Published
##### Abstract [en]
The present work deals with the derivation of corrector estimates for the two-scale homogenization of a thermodiffusion model with weak thermal coupling posed in a heterogeneous medium endowed with periodically arranged high-contrast microstructures. The term “weak thermal coupling” refers here to the variable scaling in terms of the small homogenization parameter $\varepsilon$ of the heat conduction-diffusion interaction terms, while the “high-contrast” is considered particularly in terms of the heat conduction properties of the composite material. As a main target, we justify the first-order terms of the multiscale asymptotic expansions in the presence of coupled fluxes, induced by the joint contribution of Sorret and Dufour-like effects. The contrasting heat conduction combined with cross coupling leads to the main mathematical difficulty in the system. Our approach relies on the method of periodic unfolding combined with $\varepsilon$-independent estimates for the thermal and concentration fields and for their coupled fluxes.
##### Place, publisher, year, edition, pages
Society for Industrial and Applied Mathematics, 2018
Mathematics
Mathematics
##### Identifiers
urn:nbn:se:kau:diva-65815 (URN)10.1137/16M109538X (DOI)000436998500009 ()
Available from: 2018-01-25 Created: 2018-01-25 Last updated: 2018-09-05Bibliographically approved
Richardson, O., Jalba, A. & Muntean, A. (2018). Effects of Environment Knowledge in Evacuation Scenarios Involving Fire and Smoke: A Multiscale Modelling and Simulation Approach. Fire technology, 55(2), 415-436
Open this publication in new window or tab >>Effects of Environment Knowledge in Evacuation Scenarios Involving Fire and Smoke: A Multiscale Modelling and Simulation Approach
2018 (English)In: Fire technology, ISSN 0015-2684, E-ISSN 1572-8099, Vol. 55, no 2, p. 415-436Article in journal (Refereed) Published
##### Abstract [en]
We study the evacuation dynamics of a crowd evacuating from a complex geometry in the presence of a fire as well as of a slowly spreading smoke curtain.The crowd is composed of two kinds of individuals: those who know the layout of the building, and those who do not and rely exclusively on potentially informed neighbors to identify a path towards the exit. We aim to capture the effect the knowledge of the environment has on the interaction between evacuees and their residence time in the presence of fire and evolving smoke. Our approach is genuinely multiscale—we employ a two-scale model that is able to distinguish between compressible and incompressible pedestrian flow regimes and allows for micro and macro pedestrian dynamics. Simulations illustrate the expected qualitative behavior of the model. We finish with observations on how mixing evacuees with different levels of knowledge impacts important evacuation aspects.
Springer, 2018
##### Keywords
Crowd dynamics, Environment knowledge, Evacuation, Fire and smoke dynamics, Particle methods, Transport processes, Incompressible flow, Smoke, Transport process, Fires
Mathematics
Mathematics
##### Identifiers
urn:nbn:se:kau:diva-68390 (URN)10.1007/s10694-018-0743-x (DOI)000460581600003 ()2-s2.0-85048553425 (Scopus ID)
Available from: 2018-07-04 Created: 2018-07-04 Last updated: 2019-04-05Bibliographically approved
Colangeli, M., Muntean, A., Richardson, O. & Thieu, T. K. (2018). Modelling interactions between active and passive agents moving through heterogeneous environments. In: Livio Gibelli, Nicola Bellomo (Ed.), Crowd Dynamics: volume 1 (pp. 211-257). Basel: Springer
Open this publication in new window or tab >>Modelling interactions between active and passive agents moving through heterogeneous environments
2018 (English)In: Crowd Dynamics: volume 1 / [ed] Livio Gibelli, Nicola Bellomo, Basel: Springer , 2018, p. 211-257Chapter in book (Refereed)
##### Abstract [en]
We study the dynamics of interacting agents from two distinct intermixed populations: one population includes active agents that follow a predetermined velocity field, while the second population contains exclusively passive agents, i.e., agents that have no preferred direction of motion. The orientation of their local velocity is affected by repulsive interactions with the neighboring agents and environment. We present two models that allow for a qualitative analysis of these mixed systems. We show that the residence times of this type of systems containing mixed populations is strongly affected by the interplay between these two populations. After showing our modelling and simulation results, we conclude with a couple of mathematical aspects concerning the well-posedness of our models.
##### Place, publisher, year, edition, pages
Basel: Springer, 2018
##### Keywords
Crowd dynamics, Fire and smoke dynamics, Heterogeneous domains, Lattice gas model, Particle methods
Mathematics
##### Identifiers
urn:nbn:se:kau:diva-71296 (URN)10.1007/978-3-030-05129-7_8 (DOI)2-s2.0-85061059980 (Scopus ID)978-3-030-05128-0 (ISBN)
Available from: 2019-02-21 Created: 2019-02-21 Last updated: 2019-03-22Bibliographically approved
##### Identifiers
ORCID iD: orcid.org/0000-0002-1160-0007
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https://itprospt.com/num/11300442/question-1-do-value-of-k-and-m-exist-such-that-the-following
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# Question 1. Do value of k and m exist such that the following limit exists?Sx2 + kx-2m Ilm x2 + mx-4If so, find the values of k, m and the corresponding limit; If n...
## Question
###### Question 1. Do value of k and m exist such that the following limit exists?Sx2 + kx-2m Ilm x2 + mx-4If so, find the values of k, m and the corresponding limit; If not; explain why not;
Question 1. Do value of k and m exist such that the following limit exists? Sx2 + kx-2m Ilm x2 + mx-4 If so, find the values of k, m and the corresponding limit; If not; explain why not;
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##### The pK. value for HNOz is 3.35. Would buffer prepared from HNOz and NaNOz with a pH of 1.55 be considered to be an effective buffer?A buffer in which the mole ratio of NaNOz HNOz is 0.58 has pH of 3.11 Would this buffer solution have greater capacity for added acid (Hz 0 Or added base (OHSubmit AnswerRetry Entire Groupmore group attempts remaining
The pK. value for HNOz is 3.35. Would buffer prepared from HNOz and NaNOz with a pH of 1.55 be considered to be an effective buffer? A buffer in which the mole ratio of NaNOz HNOz is 0.58 has pH of 3.11 Would this buffer solution have greater capacity for added acid (Hz 0 Or added base (OH Submit An...
##### Question 108ptsA certain merry-go-round is accclerated uniformly from rest and attains angular spced of 1.2 radls in thc first 10 seconds: If thc net applicd torquc is 1200 N what is the moment of inertia ( in kg m?of the mcry-go-round?
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2022-10-07 02:46:16
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https://zbmath.org/?q=an:0844.47005
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## Asymptotic completeness of long-range $$N$$-body quantum systems.(English)Zbl 0844.47005
Asymptotic completeness, in the 2-body case, says that all states of the system fall into two categories: bound states and scattering states. The proof normally makes use of the classical phase space and the Enss method. In the $$N$$-body case, there may also be bound states for subsystems while the remaining constituents asymptotically evolve as free particles. For general $$N$$ and short-range potentials, the proof of the existence of all wave operators and of asymptotic completeness was first given by Sigal and Soffer. Then Graf was able to replace phase-space analysis by a new type of analysis in configuration space. In a previous paper and in the present work, Dereziński extends Graf’s method to the long-range case. For the construction of wave operators pertaining to each cluster of particles one has to modify the cluster hamiltonian appropriately in order to cope with the long-range interaction. One of the main results, proven in this article, is the existence of the “asymptotic velocities” $$P_\pm$$ for $$t\to \pm \infty$$. For the Enss condition ($$\mu> \sqrt{3}-1$$, $$\mu$$ is some exponent describing the decay of interactions at spatial infinity), Dereziński proves asymptotic completeness using the modified wave operators.
### MSC:
47A40 Scattering theory of linear operators 81U10 $$n$$-body potential quantum scattering theory 47N50 Applications of operator theory in the physical sciences
Full Text:
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2022-06-30 04:34:28
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https://www.ias.ac.in/listing/bibliography/jess/LINXI_YUAN
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• LINXI YUAN
Articles written in Journal of Earth System Science
• Anaerobic oxidation of methane in coastal sediment from Guishan Island (Pearl River Estuary), South China Sea
The concentrations of CH4, SO$^{2−}_{4}$, 𝛴 CO2 and the carbon isotope compositions of 𝛴 CO2 and CH4 in the pore-water of the GS sedimentary core collected from Guishan Island (Pearl River Estuary), South China Sea,were determined. The methane concentration in the pore-water shows dramatic changes and sulfate concentration gradients are linear at the base of the sulfate reduction zone for the station. The carbon isotope of methane becomes heavier at the sulfate-methane transition (SMT)likely because of the Raleigh distillation effect; 12CH4 was oxidized faster than 13CH4 and this caused the enrichment of residual methane 𝛿13C and 𝛿13C-𝛿 CO2 minimum. The geochemical profiles of the pore-water support the existence of anaerobic oxidation of methane (AOM), which is mainly controlled by the quality and quantity of the sedimentary organic matter. As inferred from the index of 𝛿13C-TOC value and TOC/TN ratio, the organic matter is a mix of mainly refractory terrestrial component plus some labile alga marine-derived in the study area. A large amount of labile organic matter (mainly labile alga marine-derived) is consumed via the process of sedimentary organic matter diagenesis, and this reduces the amount of labile organic matter incorporated into the base of the sulfate reduction zone. Due to the scarcity of labile organic matter, the sulfate will in turn be consumed by its reaction with methane and therefore AOM takes place.Based on a diffussion model, the portion of pore-water sulfate reduction via AOM is 58.6%,and the percentage of 𝛴 CO2 in the pore-water derived from AOM is 41.4%. Thus, AOM plays an important role in the carbon and sulfur cycling in the marine sediments of Pearl River Estuary.
• Evidence for glacial deposits during the Little Ice Age in Ny-Alesund, western Spitsbergen
The glaciers act as an important proxy of climate changes; however, little is known about the glacial activities in Ny-Alesund during the Little Ice Age (LIA). In the present study, we studied a 118-cm-high palaeo-notch sediment profile YN in Ny-Alesund which is divided into three units: upper unit (0–10 cm), middle unit (10–70 cm) and lower unit (70–118 cm). The middle unit contains many gravels and lacks regular lamination, and most of the gravels have striations and extrusion pits on the surface. The middle unit has the grain size characteristics and origin of organic matter distinct from other units, and it is likely the glacial till. The LIA in Svalbard took place between 1500 and 1900 AD, the middle unit is deposited between 2219 yr BP and AD 1900, and thus the middle unit is most likely caused by glacier advance during the LIA. Glaciers during the LIA likely overran the sampling site, removed part of the pre-existing sediments, and contributed to the formation of diamicton in the middle unit. This study provides evidence for glacial deposits during the LIA in Ny-Alesund and improves our understanding about historical glacier dynamics and ice-sheet margins during the LIA in western Spitsbergen.
• # Journal of Earth System Science
Volume 130, 2021
All articles
Continuous Article Publishing mode
• # Editorial Note on Continuous Article Publication
Posted on July 25, 2019
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2021-07-27 22:28:40
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https://badripatro.wordpress.com/2017/02/28/lstm-details/
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LSTM details
Ref:
http://prog3.com/sbdm/blog/lanran2/article/details/50603861
https://apaszke.github.io/lstm-explained.html
Paper:
https://arxiv.org/pdf/1511.07889.pdf
What is RNN
RNN: multi layer feedback RNN (neural Network Recurrent, recurrent neural network) neural network is a kind of artificial neural network which is connected to the ring. The internal state of the network can display dynamic time series behavior. Unlike feedforward neural networks, RNN can use its internal memory to process arbitrary timing input sequences, which allows it to be more easily processed such as non segmented handwriting recognition, speech recognition, etc.. – Baidu Encyclopedia
Here we look at the abstract out of the RNN formula:
ht=θϕ(ht1)+θxxt
yt=θyϕ(ht)
You can find that each RNN has to use the last time the middle layer of the outputht
We define function loss asEThen the gradient formula is as follows:
Eθ=St=1Etθ
Etθ=tk=1Etytythththkhkθ
hthk=ti=k+1hihi1=ti=k+1θTdiag[ϕ(hi1)]
||hihi1||||θT||||diag[ϕ(hi1)]||γθγϕ
hthk(γθγϕ)tk
Multiplied by less than 1 of the number, the gradient will be smaller and smaller. In order to solve this problem, LSTM came into being.
LSTM introduction
Definition: LSTM (Term Memory Long-Short, LSTM)
Is a time recurrent neural network, the paper was first published in 1997. Due to the unique design structure, LSTM is suitable for processing and prediction of time series in the interval and delay is very long important events. – Baidu Encyclopedia
Mentioned LSTM, always accompanied by a picture as shown below:
Can be seen from the figure, in addition to the input, there are three parts: 1) Gate Input; 2) Gate Forget; 3) Gate Output
According to the RNN mentioned above, our input isxtandht1, while the input ishtandct(state cell), where state LSTM is the key to cell, which makes LSTM with memory function. Here’s a formula for LSTM:
1) Gate Input:
it=σ(Wxixt+Whiht1+bi)=σ(linearxi(xt)+linearhi(ht1))
amongσRefers to the sigmoid function.
2) Gate Forget:Decide whether to delete or retain memory (memory)
ft=σ(Wxfxt+Whfht1+bf)
3) Gate Output:
ot=σ(Wxoxt+Whoht1+bo)
4) update Cell:
gt=tanh(Wxgxt+Whght1+bg)
5) State Update Cell:
ct=ftct1+itgt
6) Output of LSTM Final:
ht=ottanh(ct)
Above is a formula for cell involved in LSTM,Below to explain why LSTM can solve the problem of gradient disappear in RNN.
Because each factor is very close to 1, so the gradient is difficult to decay, so as to solve the problem of gradient disappear.
Nngraph Torch
Before the use of LSTM to prepare torch, we need to learn a tool nngraph torch, an nngraph to the following commands:
LuarocksInstallNngraph
Nngraph detailed introduction:Https://github.com/torch/nngraph/
Nngraph can facilitate the design of a neural network module. We first use nngraph to create a simple network module:
z=x1+x2linear(x3)
We can see that the input of this module is a total of three,x1,x2andx3, the output isz. The following is the implementation of this module torch code:
Require'nngraph'
X1=nn.Identity(())
X2=nn.Identity(())
X3=nn.Identity(())
L=nn.CAddTable () () (){x1, nn.CMulTable () ({x2) () (nn.Linear) (20,10) (x3)}}))
Mlp=nn.gModule ({x1, X2, x3},{L})
First we definex1,x2andx3, useNn.Identity () () () ()And then tolinear(x3)We useX4=nn.Linear (20,10) (x3)A linear neural network with 20 neurons in the output layer is defined, and a linear neural network with 10 neurons in the output layer is defined.x2linear(x3), useX5=nn.CMulTable () (X2, x4)For; forx1+x2linear(x3)We useNn.CAddTable () (x1, x5)To achieve; finally useNn.gModule ({input}, {output})To define the neural network module.
We use the forward method to test whether our Module is correct:
H1=Torch.Tensor{One,Two,Three,Four,Five,Six,Seven,Eight,Nine,Ten}
H2=Torch.Tensor (Ten(fill ().One)
H3=Torch.Tensor (Twenty(fill ().Two)
B=Mlp:forward ({h1, H2, h3})
Parameters=Mlp:parameters ()One]
Bias=Mlp:parameters ()Two]
Result=Torch.cmul (H2, (parameters*h3+bias)) +h1
First we define three inputsh1,h2andh3, then call the module forward MPL command to get the output B, and then we get the network weights w and bias are saved in the parameters and bias variables, calculationz=h1+h2linear(h3)ResultResult=torch.cmul (H2, (parameters*h3+bias)) +h1, finally compare B and result is consistent, we found that the results of the calculation is the same, that our module is correct.
Use LSTM to prepare the nngraph module
Now we use nngraph to write the LSTM module described above, the code is as follows:
Require 'nngraph'
Function LSTM(XT, prev_c, prev_h)
Function New_input_sum()
Local I2h=NN.Linear(Four hundred,Four hundred)
Local H2H=NN.Linear(Four hundred,Four hundred)
End
Local Input_gate=NN.Sigmoid()(new_input_sum())
Local Forget_gate=NN.Sigmoid()(new_input_sum())
Local Output_gate=NN.Sigmoid()(new_input_sum())
Local GT=NN.Tanh()(new_input_sum())
Local HT=NN.CMulTable()({output_gate, nn.Tanh()(CT)})
Return CT,HT
End
XT=NN.Identity()()
Prev_c=NN.Identity()()
Prev_h=NN.Identity()()
LSTM=NN.GModule({xt, prev_c, prev_h}, {lstm(XT, prev_c, prev_h)})
amongXTandPrev_hIs input,Prev_cIs state cell, and then we follow the previous formula one calculation, the final outputCT(cell state new) (), HT (output). The calculation sequence of the code is completely consistent with the above, so here is no longer one one explained.
One thought on “LSTM details”
1. Shivali says:
Hi ! Parameters=Mlp:parameters ()One]
Bias=Mlp:parameters ()Two] doesn’t work for me.
How can i print the parameters ??
th> h1 = torch.Tensor{1,2,3}
th> h2 = torch.Tensor{1,1,1}
th> h3 = torch.Tensor{1,2,3,4,5,6,7,8,9,10}
th> b= mlp:forward({h1,h2,h3})
th> print(b)
-1.4341
1.8059
5.1775
But I can’t see the value of parameters to cross check the answer.
Help would be highly appreciated !
Like
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2018-06-24 05:09:02
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http://tex.stackexchange.com/questions/16884/compiling-a-latex-document-manually/16889
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# Compiling a LaTeX document manually
I've created LaTeX documents before using 'nice IDEs' such as TeXnicCenter and have a little experience in general with TeX, however, now I'm wondering about creating one "manually". By that, I mean doing something such as writing the tex document in a very simple editor (such as VIM) and then compiling it myself with pdflatex via the command prompt. Has anyone ever done this and if so, I really just need to know how I'd go about it? What's the process to doing it? While the IDE is nice, I'd really like to be able to create LaTeX documents 'on my own'. Thanks for any information you can provide!
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Err ... just go ahead and do it! This is how I've always created my LaTeX documents (except that I use Emacs, not vim). I don't even know what "IDE" stands for. – Andrew Stacey Apr 28 '11 at 18:19
I can't tell if you're being facetious or not, but if not IDE == Integrated Development Environment. – JToland Apr 28 '11 at 18:22
"in a very simple editor (such as VIM)". VIM is one of the most complex editors in existence. I assume you mean "simple" as "without a fancy graphical interface", which is true for VIM. Also, I never knew that TeXnicCenter can create LaTeX documents "automatically". – Martin Scharrer Apr 28 '11 at 18:42
Well, in reality, probably not the command line, but something like Cygwin. And Yes, I'm on Windows. By TeXnicCenter automatically creating documents, I mean I just have to type the 'code', then push a button and somehow I have a pdf document -- all the work is done for me to get from code to pdf. I just wonder how (and what I'd need) to do it myself with commands starting with just the tex document. – JToland Apr 28 '11 at 18:51
You would need a functioning TeX distribution like MiKTeX. Then you just go to command prompt and write `pdflatex yourdocument.tex`, or `yourtexengine yourdocument.tex`. – ipavlic Apr 28 '11 at 20:41
show 4 more comments
Your question is a little confusing. What you mean is compiling a LaTeX document manually (into a PDF). You might want to adjust your title. Creating one would be the process of writing the document. This can be done in the command line using `pdflatex <filename>`. In VIM you could just use ESC`:!pdflatex %` (`%` can be used instead of the current filename) when you edit the main file. There is also the LaTeX Suite for VIM which gives you short cuts for the compilation and a lot of IDE functionality. The LaTeX compiler must be in your `PATH` for this to work, but this should be already the case for a proper installed LaTeX.
The latexmk script (also called with the filename) will compile the LaTeX document as often as required and also run external tools like `bibtex` and `makeindex`, which is basically the things the IDE is doing automatically when you press the compile button.
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Oh come on, you don't have to tear his question to pieces like this. His last comment clarifies his post perfectly well. He has so far used a program that created the pdf with the press of a button, and now wants to know if there's anything he needs to know about producing (La)TeX output so by entering commands into some interface. There's really no reason to ridicule some minorly suboptimal wording in this manner. – doncherry Apr 28 '11 at 20:08
See, your edit about `latexmk` is something someone coming from TeXnicCenter wouldn't know, for example. That's the kind of thing he was asking for. I think a GUI made him feel like he's "triggering a whole bunch of complex commands" (I'm exaggerating), and I'm sure he's right in the respect that there's certainly a lot to learn what can be done instead of "clicking that button", just like `latexmk`. – doncherry Apr 28 '11 at 20:34
@doncherry: I rewrote the answer to be more useful. – Martin Scharrer Apr 28 '11 at 20:38
Great, I think he'll be much happier with this response :) I hope you didn't take my criticism amiss, I just didn't feel like this was the way we want to answer questions here. – doncherry Apr 28 '11 at 20:53
Thank you for the edit. This is supremely more helpful than your initial answer. – JToland Apr 28 '11 at 20:54
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2014-04-25 04:19:28
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http://mathhelpforum.com/discrete-math/193479-binary-tree-question.html
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# Math Help - Binary Tree Question
1. ## Binary Tree Question
Hello Math Help Forum! This is my first post. See, I have a final tomorrow and have greatly procrastinated on studying. The professor gave a practice final and I missed all of the solutions, and I do not understand how to solve the following question:
Let T be the full binary tree with height n with 2^n terminal vertices. Let r denote the root. Define a weight on the edges of T as follows: If an edge e connects a vertex with its left child, define w(e)=0. If an edge connects a vertex with its right child, define w(e)=1. Now let P be the set of simple paths from r to the terminal vertices. Note that |P| = 2^n since there is one simple path from r to each terminal vertex. Let P(k) be the set of paths in P with total weight k. Show |P(3)|<= n^3.
(Recall that the weight of a path is the sum of weights of the edges in the path).
My first instinct was induction but that didn't work. My next was a recurrence relation but I got lost. Any tips?? Thanks!
2. ## Re: Binary Tree Question
Isn't it true that $|P(3)|=\binom{n}{3}$?
3. ## Re: Binary Tree Question
Just an aside: this depends on a couple of definitions, as height is sometimes defined as 0 for a tree of size (node count) 1, but other times is defined as 1 for a tree of size 1 (if the order zero graph is allowed, which is extremely useful for setting up the recursive structure of the binary tree in computer science). Based on your formula it seems you're using the former definition.
This problem is essentially asking "in a full tree of n+1 levels, prove there are at most n^3 paths from root to leaf that branch right exactly three times". You will get n branch choices, one choice for each level indexed from 1 to n. You have to choose exactly three of these to head to the right. So emakarov is correct.
(This seems like a very silly question. I'm not sure what content area they are trying to assess here outside of basic logic.)
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2015-03-03 15:52:05
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http://mathhelpforum.com/trigonometry/180267-circular-motion.html
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# Math Help - Circular motion.
1. ## Circular motion.
circumference of a circle is 15.7inches and object goes around the circle at 6000revs per min. an event occurs ever revolution the event occurs for 3.33 milliseconds and always ends at 23 degrees after top dead center(so 23 degrees in to the new rotation) at the speed the object it traveling (6000revs per min.) at what degree does the event have to start to end at 23 degrees.. also when does the event need to start if the speed is 700 revs per min.
$\frac{6000\;rev}{min}\cdot \frac{15.7\;in}{rev}\cdot \frac{min}{60\;sec}\cdot \frac{sec}{1000\;millisecond}\cdot \frac{360^{\circ}}{15.7\;in}$
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2016-07-29 14:35:54
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https://www.ques10.com/p/50074/define-effective-length-in-column-with-its-appli-1/
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0
Define effective length in column with its application.
Solution
Effective Length: The length of the column which bends or deflects as if it is hinged at its ends is called as effective length. It is denoted by Le.
Application: It is used in Rankine’s and Euler’s formula to determine buckling load on column.
mosd-1 • 149 views
0
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2020-05-30 22:24:34
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http://xyketyxyhy.killarney10mile.com/why-the-theories-of-supply-and-supply-elasticity-are-important-78608zy7974.html
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# Why the theories of supply and supply elasticity are important
A landlord received rent, workers received wages, and a capitalist tenant farmer received profits on their investment. Anti-oxidants are most valuable for animals that are cancer-prone, or subjected to radiation or chemical toxins.
It is unparsimonious to say other universes exist. So we want to know how consumers would respond over the course of say, six months or a year, to changes in pricing of Uber.
CoQ is an essential component of the mitochondrial respiratory chain. A fundamental concept for the determination of the exact level of output of a firm is the marginal cost. However, the Scientific Revolution had established by the middle s that physics, chemistry, astronomy, meteorology, and physiology could be understood in naturalistic terms.
These arise from higher factor prices or from diminishing productivities of factors. The first is that they have surge pricing. These are the general workhorses of the formula. Uncoupling is also useful for fever production. Identity Identity is the relation that obtains between two entities or terms that are the same instance, i.
The level of mathematical sophistication of neoclassical economics increased. That if they have their own car or want to rent a car, they can drive Uber. Faith is belief based on revelation and exempt from doubt.
You want at leastHUT. And consumer surplus is one of the most important things in economics. In such a situation, the economies and diseconomies balance each other and the LAC curve has a disc base.
Fungal Protease and Fungal Pancreatin Fungal protease and fungal pancreatin are names for particular strains of vegetarian protease enzymes.
A given entity is identified through time with its closest close-enough continuous-enough continuer. You might see 1. The price of related goods and the price of inputs energy, raw materials, labor also affect supply as they contribute to increasing the overall price of the good sold.
$Mechanisms of Aging$
Such a notion of ontological determinism is different from epistemic determinism only if there is a hypertime in which different points of normal time can "already" coexist.
For example, when gas is produced from coal, coke and other products also emerge automatically. The varieties of naturalism differ primarily according to their explanation of how matter relates to mind.
Helping with sinusitis and asthma. The majority of empirical cost studies suggest that the U-shaped cost curves postulated by the traditional theory are not observed in the real world. So how does that small rounding up or rounding down help accomplish the bigger goal?
It cannot be altered, increased or decreased by varying the level of activity or the rate of output. Stressors that produce anxiety have been shown to actually suppress parts of the brain that aid in rational decision-making.
The modem theory of costs differs from the traditional theory of costs with regard to the shapes of the cost curves.
A possibly meaningful but unparsimonious answer to the Ultimate Why is that the universe exists more precisely, is perceived to exist roughly because it is possible. These costs are also called avoidable costs or controllable costs.Why is there something rather than nothing?Might the world be an illusion or dream?What exists beyond the human senses?What happens after death?Does divine or supernatural agency exist?
Is the future already decided?; What is the meaning of life?What is right and wrong?Is the world good or bad?Are humans good or evil?What beings should have.
Rational choice theory is an economic principle that assumes that individuals always make prudent and logical decisions that provide. Neoclassical economics is an approach to economics focusing on the determination of goods, outputs, and income distributions in markets through supply and killarney10mile.com determination is often mediated through a hypothesized maximization of utility by income-constrained individuals and of profits by firms facing production costs and employing.
The S i factor represents “capabilities” of exporter i as a supplier to all destinations. M n captures all characteristics of destination market n that promote imports from all sources. Bilateral accessibility of n to exporter i is captured in 0 ≤ ϕ ni ≤ 1: it combines trade costs with their respective elasticity to measure the overall impact on trade flows.
Background for understanding and possibly repairing the molecular and biochemical damage known as aging.
The bottom line is that a healthy supply of these protein specific enzymes is essential for sustaining and maintaining optimal health. Types of Proteolytic Enzymes.
Why the theories of supply and supply elasticity are important
Rated 0/5 based on 91 review
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2018-10-16 11:22:33
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http://math.stackexchange.com/questions/137260/integral-over-unit-ball
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# integral over unit ball?
Problem:
Let's consider the collection of $C^1$ functions, where $k=1,2,\ldots,(n-1)$: $$g_k:\mathbb{R}^k\rightarrow \mathbb{R},$$ where: $$g_k=g_k(x_1,x_2,\ldots,x_k)$$
Then a new map $f$ is defined as follows: $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ such that: $$f(x_{1},x_{2},\ldots,x_n)=(x_1, g_1(x_1)+2x_2, g_2(x_1,x_2) + 3x_3, \ldots, g_{n-1}(x_1,x_2,\ldots,x_{n-1})+nx_n)$$
How can find the volume of $f((0,1)^n)$ where $(0,1)^n$ is an open unit cube $(0,1)^n$?
-
Where does it come from? – draks ... Apr 26 '12 at 15:21
@draks: I am solving this problems for practice – Boyan Klo Apr 26 '12 at 15:42
Can you use the change of variables formula? The derivative of $f$ is triangular, so the Jacobian might be easy to compute. In fact, the Jacobian seems to be $n!$ (do you have a typo. in your expression for $f$, if not, I'm not sure what the general term is?). – copper.hat Apr 26 '12 at 15:55
The change of variables formula gives $\int_{\phi(U)} g(y) dy = \int_U g(\phi(x)) J_{\phi} (x) dx$. Use $\phi = f$, $g = 1$. – copper.hat Apr 26 '12 at 16:06
@draks: These problems are for an exam... – Beni Bogosel Apr 27 '12 at 16:00
By change of variables formula, if $\Omega \subset \mathbb{R^{n}}$ is open and $G: \Omega \mapsto \mathbb{R^{n}}$ is a $C^{1}$ diffeomorphism, we have $$\int_{G(\Omega)} f(x) dx = \int_{\Omega)}f \circ G(x)| det D_xG| dx$$ Then $$L^{n}(f((0,1)^{n})) = \int_{f((0,1)^{n})} 1 dx = \int_{(0,1)^{n}} |det D_x f| dx$$
But, $$D_xf = \left[ \begin{array}{cccccc} 1 & 0 & 0 & \cdots & 0 & 0 \\ D_1 g_1(x_1) & 2 & 0 & \cdots & 0 & 0\\ D_1 g_2(x_1,x_2) & D_2 g_2(x_1,x_2)& 3 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & 0 & 0\\ D_1 g_{n-1}(x*) & D_2 g_{n-1}(x*) & D_3 g_{n-1}(x*)& \cdots & D_{n-1}g_{n-1}(x*)& n \\ \end{array} \right]$$ Where $x* = (x_1, x_2, \cdots , x_{n-1})$
Hence $$L^{n}(f((0,1)^{n}))= n!L^{n}(0,1)^{n}.$$
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2015-05-28 04:23:48
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https://en.cppreference.com/w/c/language/operator_comparison
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# Comparison operators
< c | language
Comparison operators are binary operators that test a condition and return 1 if that condition is logically true and 0 if that condition is false.
Operator Operator name Example Description
== equal to a == b a is equal to b
!= not equal to a != b a is not equal to b
< less than a < b a is less than b
> greater than a > b a is greater than b
<= less than or equal to a <= b a is less than or equal to b
>= greater than or equal to a >= b a is greater than or equal to b
## Contents
### Relational operators
The relational operator expressions have the form
lhs < rhs (1) lhs > rhs (2) lhs <= rhs (3) lhs >= rhs (4)
1) less-than expression
2) greater-than expression
3) less or equal expression
4) greater or equal expression
where
lhs, rhs - expressions that both have real type or both have pointer to object type
The type of any relational operator expression is int, and its value (which is not an lvalue) is 1 when the specified relationship holds true and 0 when the specified relationship does not hold.
If lhs and rhs are expressions of any real type, then
• usual arithmetic conversions are performed
• the values of the operands after conversion are compared in the usual mathematical sense (except that positive and negative zeroes compare equal and any comparison involving a NaN value returns zero)
Note that complex and imaginary numbers cannot be compared with these operators.
If lhs and rhs are expressions of pointer type, they must be both pointers to objects of compatible types, except that qualifications of the pointed-to objects are ignored.
• a pointer to an object that is not an element of an array is treated as if it were pointing to an element of an array with one element
• if two pointers point to the same object, or both point one past the end of the same array, they compare equal
• if two pointers point to different elements of the same array, the one pointing at the element with the larger index compares greater.
• if one pointer points to the element of an array and the other pointer points one past the end of the same array, the one-past-the-end pointer compares greater
• if the two pointers point to members of the same struct, the pointer to the member declared later in the struct definition compares greater than then pointer to the member declared earlier.
• pointers to members of the same union compare equal
• all other pointer comparisons invoke undefined behavior
#include <assert.h>
int main(void)
{
assert(1 < 2);
assert(2+2 <= 4.0); // int converts to double, two 4.0's compare equal
struct { int x,y; } s;
assert(&s.x < &s.y); // struct members compare in order of declaration
double d = 0.0/0.0; // NaN
assert( !(d < d) );
assert( !(d > d) );
assert( !(d >= d) );
assert( !(d >= d) );
float f = 0.1; // f = 0.100000001490116119384765625
double g = 0.1; // g = 0.1000000000000000055511151231257827021181583404541015625
assert(f > g); // different values
}
### Equality operators
The equality operator expressions have the form
lhs == rhs (1) lhs != rhs (2)
1) equal-to expression
2) not equal to expression
where
lhs, rhs - expressions that both have any arithmetic types (including complex and imaginary) both are pointers to objects or functions of compatible types, ignoring qualifiers of the pointed-to types one is a pointer to object and the other is a pointer to (possibly qualified) void one is a pointer to object or function and the other is a null pointer constant such as NULL
The type of any equality operator expression is int, and its value (which is not an lvalue) is 1 when the specified relationship holds true and 0 when the specified relationship does not hold.
• if both operands have arithmetic types, usual arithmetic conversions are performed and the resulting values are compared in the usual mathematical sense (except that positive and negative zeroes compare equal and any comparison involving a NaN value, including equality with itself, returns zero). In particular, values of complex type are equal if their real parts compare equal and their imaginary parts compare equal.
• if one operand is a pointer and the other is a null pointer constant, the null pointer constant is first converted to the type of the pointer (which gives a null pointer value), and the two pointers are compared as described below
• if one operand is a pointer and the other is a pointer to void, the non-void pointer is converted to the pointer to void and the two pointers are compared as described below
• two pointers compare equal if any of the following is true:
• they are both null pointer values of their type
• they are both pointers to the same object
• one pointer is to a struct/union/array object and the other is to its first member/any member/first element
• they are both pointing one past the last element of the same array
• one is one past the end of an array, and the other is at the start of a different array (of the same type) that follows the first in a larger array or in a struct with no padding
(as with relational operators, pointers to objects that aren't elements of any array behave as pointers to elements of arrays of size 1)
#### Notes
Objects of struct type do not compare equal automatically, and comparing them with memcmp is not reliable because the padding bytes may have any values.
Because pointer comparison works with pointers to void, the macro NULL may be defined as (void*)0 in C, although that would be invalid in C++ where void pointers do not implicitly convert to typed pointers
Care must be taken when comparing floating-point values for equality, because the results of many operations cannot be represented exactly and must be rounded. In practice, floating-point numbers are usually compared allowing for the difference of one or more units of the last place.
#include <assert.h>
int main(void)
{
assert(2+2 == 4.0); // int converts to double, two 4.0's compare equal
int n[2][3] = {1,2,3,4,5,6};
int* p1 = &n[0][2]; // last element in the first row
int* p2 = &n[1][0]; // start of second row
assert(p1+1 == p2); // compare equal
double d = 0.0/0.0; // NaN
assert( d != d ); // NaN does not equal itself
float f = 0.1; // f = 0.100000001490116119384765625
double g = 0.1; // g = 0.1000000000000000055511151231257827021181583404541015625
assert(f != g); // different values
}
### References
• C11 standard (ISO/IEC 9899:2011):
• 6.5.8 Relational operators (p: 95-96)
• 6.5.9 Equality operators (p: 96-97)
• C99 standard (ISO/IEC 9899:1999):
• 6.5.8 Relational operators (p: 85-86)
• 6.5.9 Equality operators (p: 86-87)
• C89/C90 standard (ISO/IEC 9899:1990):
• 3.3.8 Relational operators
• 3.3.9 Equality operators
Common operators
assignment increment
decrement
arithmetic logical comparison member
access
other
a = b
a += b
a -= b
a *= b
a /= b
a %= b
a &= b
a |= b
a ^= b
a <<= b
a >>= b
++a
--a
a++
a--
+a
-a
a + b
a - b
a * b
a / b
a % b
~a
a & b
a | b
a ^ b
a << b
a >> b
!a
a && b
a || b
a == b
a != b
a < b
a > b
a <= b
a >= b
a[b]
*a
&a
a->b
a.b
a(...)
a, b
(type) a
? :
sizeof
_Alignof
(since C11)
C++ documentation for Comparison operators
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2018-07-21 16:57:23
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https://emresahin.net/creating-os-dependent-temporary-directories-in-rust/
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There are a few crates in Rust for getting system dependent directories, like user's home or system configuration directory. The one I prefer is directories-next
Recently, I needed a standard function to get the temporary directory. I checked the crate documentation but couldn't find a proper function for this.
Then I noticed std::env::temp_dir() returns a PathBuf that points to the system temporary directory. This is note for myself that we don't always need extra packages for basic functionality.
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2023-02-04 15:45:49
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https://electronics.stackexchange.com/questions/527486/crystal-load-capacitance-for-low-power-applications
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# Crystal load capacitance for low power applications
According to Silicon Labs' app note on microcontroller oscillator design, larger load capacitance on crystals increase power consumption and increase startup time.
What are the actual numbers like in terms of power dissipation? Would there be any practical benefits to use a 9 pF crystal over a 20 pF crystal? Are there any drawbacks?
What are the actual numbers like in terms of power dissipation?
Below are a series of diagrams I produced when investigating crystal oscillator power dissipation changes versus external drive resistance (R1) and loading capacitor (CL1 and CL2) changes. The equivalent circuit is a good place to start before jumping into the actual power dissipation: -
Changing values for R1 (250 Ω to 1500 Ω) with CL1 and CL2 fixed at 20 pF produces slightly different oscillation frequencies (phase shift = 180°): -
Images from here.
The crystal is modeled as shown in the top picture. The crystal equivalent circuit has been chosen to produce a theoretical series resonance at precisely 10.000000 MHz. That frequency is determined by the components inside the orange box. The image link gives more details if you need them.
The important thing to note from above is that the oscillation frequency is somewhat "R1" dependent. The oscillation frequency is produced when the phase shift = 180° i.e. adding a "perfect" inverter generates an overall phase shift of 360°. Further down I show how an imperfect inverter can increase power dissipation in the crystal.
Looking at the graph above, the stablest oscillation frequency is the one where the phase shift passes through the 180° point with the least ambiguity. This tells us that a higher drive resistance (R1) produces better crystal stability. A higher drive resistance also produces the least amount of power dissipation in the crystal. However, things look a little different when we vary loading capacitance. The next picture keeps R1 at 500 Ω and varies the loading capacitors: -
The conclusion here is that higher values of loading capacitance produces a more stable frequency (less ambiguity in the frequency that yields 180°) but, the power can be significantly higher compared to using small values of loading capacitor.
But the inverter gate used also can play a big role. If the inverter has stated propagation delays of (say) 10 ns, the required phase shift to be produced by the crystal is somewhat less that 180° i.e. it might be closer to 160°. The effect on power dissipation is shown below: -
The linked article explains that a data sheet propagation delay of 10 ns would strictly erode the required crystal phase shift by 36° but, because the gate is used in its linear region the overall delay might be about 50% of 10 ns leading to the crystal being required to produce a phase shift of about 160°.
Hope this helps. The subject is as deep as you want it to be. More loading capacitance equals better oscillation stability but it also equals higher power dissipation in the crystal that can also lead to oscillator frequency degradation.
<10 pF are common now with high accuracy are preferred for the reasons stated.
Then learning how to compensate for Cin and Cgnd stray capacitance are more important. leaving out a gnd underneath reduces stray capacitance to gnd.
Generally it is wiser to use an XO rather than an X with caps as the cost is reasonable, or if you prefer accuracy of 1 to 2 ppm TCXO, these are also cheap today ( unless there are reason why not in your assumption list TBD) (X=Xtal crystal)
Powers in the motional capacitance and series R are in xx uW levels with Cs in xxx nanofarads actually reach very high voltages between the Xtal lattice from Q=10k. This is not the logic inverter power.
• Thanks for the advice! I plan to use the "xtal crystal with caps" solution however, as an oscillator is about ten times the cost. Aside from ease of implementation and consistency, are there any other benefits from using XO? (I've made successful boards with oscillators but I'm now looking to minimize cost while maintaining good performance) Oct 15, 2020 at 6:20
It's a known fact that the oscillator needs a minimum (critical) current to start and maintain the oscillation.
What are the actual numbers like in terms of power dissipation?
Eric Vittoz explains this quite well in his paper about oscillator design. I'll not dive into the technical details but only the result.
The critical conductance for the oscillator to operate is given as,
$$\mathrm{ g_{m-crit} \approx (2\ \omega \ C_L)^2 \ R_{ESR} }$$
where $$\\mathrm{\omega}\$$ is the oscillator frequency in radians ($$\\mathrm{2\pi \ f_n)}\$$, $$\\mathrm{C_L}\$$ is the load capacitance, and $$\\mathrm{R_{ESR}}\$$ is the motional resistance of the crystal which is given in the datasheet (NOTE: This applies to CMOS oscillators).
And the operation current of the oscillator is approximated as,
$$\mathrm{ I_O\approx \ g_{m-crit} \ X_B }$$
where XB comes from running a zero- and a 1st-order Bessel function over the oscillation amplitude and the thermal voltage (VT = 26mV). As an example, XB has been calculated as approximately 4 for an assumption of the oscillation amplitude is 0.4V. I think you can calculate the operation current with these assumptions.
Would there be any practical benefits to use a 9 pF crystal over a 20 pF crystal? Are there any drawbacks?
The formula above helps to explain how the MCUs decrease the power consumption in, for example, sleep mode by decreasing the oscillator frequency down to a few tens of kHz or even a few hundreds of Hz. So, when the low-power consumption is a key requirement (e.g. for coin-cell-operated systems, where uA- or even nA-currents are important), using crystals with lower CL becomes vital.
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2023-03-20 12:08:47
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https://tex.stackexchange.com/questions/363803/propagator-loop-in-feynman-diagram-using-tikz-feynman-package
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Propagator loop in Feynman diagram using tikz-feynman package
Consider the following code written using the tikz-feynman package:
\RequirePackage{luatex85}
\documentclass{standalone}
\usepackage[compat=1.0.0]{tikz-feynman}
\begin{document}
\feynmandiagram [horizontal=a to b] {
a -- b [dot] -- [out=135, in=45, loop, min distance=3cm] b -- c,
};
\end{document}
This draws the following Feynman diagram:
I would like to make the line completely horizontal and turn the loop from an oval into a sphere?
How do you do this?
• Hello, you should provide a minimal working example (MWE) for us to be able to help you – Moriambar Apr 8 '17 at 20:43
• I can't reproduce your solution. With testing\documentclass{standalone} \usepackage[compat=1.0.0]{tikz-feynman} \begin{document} ...[your code here] \end{document} -> the line will be horizontal. So you have to add a MWE. – Bobyandbob Apr 8 '17 at 20:58
• Edited question to reflect your comment – nightmarish Apr 8 '17 at 23:44
The issue here is that the algorithm that TikZ-Feynman (CTAN) uses in order to place the vertices doesn't work well if it only has a straight line because the difference in the optimization between a mostly-straight and a fully-straight line are quite small.
This can be easily fixed by using an alternative layout algorithm, such as the layered layout in this case:
%% luatex85 is only necesary to fix a bug in standalone
\RequirePackage{luatex85}
\documentclass{standalone}
\usepackage[compat=1.1.0]{tikz-feynman}
\begin{document}
\feynmandiagram [horizontal=a to b, layered layout] {
a -- b [dot] -- [out=135, in=45, loop, min distance=3cm] b -- c,
};
\end{document}
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2019-06-17 23:08:12
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https://www.ti-net.ru/carbon-dating-formula-799.html
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# Carbon dating formula
Rated 4.74/5 based on 990 customer reviews
Before Radiocarbon dating was able to be discovered, someone had to find the existence of the C isotope.In 1940 Martin Kamen and Sam Ruben at the University of California, Berkeley Radiation Laboratory did just that.Once an organism is decoupled from these cycles (i.e., death), then the carbon-14 decays until essentially gone.The half-life of a radioactive isotope (usually denoted by $$t_$$) is a more familiar concept than $$k$$ for radioactivity, so although Equation $$\ref$$ is expressed in terms of $$k$$, it is more usual to quote the value of $$t_$$.It's just a little section of the surface of the Earth. And that carbon-14 that you did have at you're death is going to decay via beta decay-- and we learned about this-- back into nitrogen-14. So it'll decay back into nitrogen-14, and in beta decay you emit an electron and an electron anti-neutrino. But essentially what you have happening here is you have one of the neutrons is turning into a proton and emitting this stuff in the process. So I just said while you're living you have kind of straight-up carbon-14. What it's essentially saying is any given carbon-14 atom has a 50% chance of decaying into nitrogen-14 in 5,730 years. And it has seven protons, and it also has seven neutrons. So the different versions of a given element, those are each called isotopes. So anyway, we have our atmosphere, and then coming from our sun, we have what's commonly called cosmic rays, but they're actually not rays. You can view them as just single protons, which is the same thing as a hydrogen nucleus. But every now and then one of those neutrons will bump into one of the nitrogen-14's in just the right way so that it bumps off one of the protons in the nitrogen and essentially replaces that proton with itself. But this number 14 doesn't go down to 13 because it replaces it with itself. And now since it only has six protons, this is no longer nitrogen, by definition. And that proton that was bumped off just kind of gets emitted. But this process-- and once again, it's not a typical process, but it happens every now and then-- this is how carbon-14 forms. You can essentially view it as a nitrogen-14 where one of the protons is replaced with a neutron. It makes its way into oceans-- it's already in the air, but it completely mixes through the whole atmosphere-- and the air. And plants are really just made out of that fixed carbon, that carbon that was taken in gaseous form and put into, I guess you could say, into kind of a solid form, put it into a living form. It gets put into plants, and then it gets put into the things that eat the plants. Well, the interesting thing is the only time you can take in this carbon-14 is while you're alive, while you're eating new things.
And carbon-14 is constantly doing this decay thing. So over the course of 5,730 years, roughly half of them will have decayed. Well, if you know that all living things have a certain proportion of carbon-14 in their tissue, as kind of part of what makes them up, and then if you were to find some bone-- let's just say find some bone right here that you dig it up on some type of archaeology dig.
What I want to do in this video is kind of introduce you to the idea of, one, how carbon-14 comes about, and how it gets into all living things. They can also be alpha particles, which is the same thing as a helium nucleus. And they're going to come in, and they're going to bump into things in our atmosphere, and they're actually going to form neutrons. And we'll show a neutron with a lowercase n, and a 1 for its mass number. And what's interesting about this is this is constantly being formed in our atmosphere, not in huge quantities, but in reasonable quantities. Because as soon as you die and you get buried under the ground, there's no way for the carbon-14 to become part of your tissue anymore because you're not eating anything with new carbon-14.
And then either later in this video or in future videos we'll talk about how it's actually used to date things, how we use it actually figure out that that bone is 12,000 years old, or that person died 18,000 years ago, whatever it might be. So let me just draw the surface of the Earth like that. So then you have the Earth's atmosphere right over here. And 78%, the most abundant element in our atmosphere is nitrogen. And we don't write anything, because it has no protons down here. And what's interesting here is once you die, you're not going to get any new carbon-14. You can't just say all the carbon-14's on the left are going to decay and all the carbon-14's on the right aren't going to decay in that 5,730 years.
From this science, we are able to approximate the date at which the organism were living on Earth.
Radiocarbon dating is used in many fields to learn information about the past conditions of organisms and the environments present on Earth.
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2021-05-10 19:46:49
|
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http://math.stackexchange.com/questions/405169/how-to-solve-eix-i
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# How to solve $e^{ix} = i$
I am taking an on-line course and the following homework problem was posed: $$e^{ix} = i$$ I have no idea how to solve this problem. I have never dealt with solving equations that have imaginary parts. What are the steps to solving such equations? I am familiar with Taylor series and the Euler formula if that is any help.
Thanks.
-
the important thing to notice here is that there is more than one $x$ which works. – user27182 May 28 '13 at 22:24
Just curious: what online course? If it's free, there might be others (such as myself) who may be interested... :) – apnorton May 28 '13 at 22:52
@anorton It is called "Calculus: Single Variable" and it can be found on coursera.org – Jeel Shah May 28 '13 at 22:54
So, you say that you're okay with Euler's formula: $$e^{ix}=\cos(x)+i\sin(x),$$ and it should be clear that (for real numbers $a,b,c,d$), we have $$a+bi=c+di\quad\iff\quad a=c\;\text{ and }\;b=d.$$ So, which values of $x$ will satisfy $$\cos(x)+\sin(x)i=0+1i\quad ?$$
-
To tack onto this answer.. there will be many such values of $x$ that solve the above equality. – Cameron Williams May 28 '13 at 22:27
@ZevChonoles how do you know that $\cos(x)+\sin(x)i=0+1i\quad$ Where did you find $0+1i$? Why are we assuming that $\cos(x)$ is zero and $\sin(x)$ is one? – Jeel Shah May 28 '13 at 22:28
@gekkostate: that is the point of the two equivalent equations. We know $\cos x$ and $\sin x$ are real for real $x$ so you can identify real and imaginary parts. – Ross Millikan May 28 '13 at 22:40
@gekkostate: I am just restating the problem; it is not something I somehow "know". You're given the equation $$e^{ix}=i$$ and told to find the values of $x$ for which this is true. But $$e^{ix}=\cos(x)+i\sin(x)$$ and $$i=0+1i$$ so you are, equivalently, finding the values of $x$ for which $$\cos(x)+i\sin(x)=0+1i.$$ – Zev Chonoles May 28 '13 at 22:45
@gekkostate if I have no reals and one $i$, then I just have $i$. $0+1i=i$. – Robert Mastragostino May 28 '13 at 23:03
$$e^{ix} = i$$ Euler's formula: $$e^{ix}=\cos(x)+i\sin(x),$$ so: $$\cos x+i\sin x = 0+1\cdot i$$ compare real and imaginary parts $$\sin x =1$$ and $$\cos x =0$$ $$x=\dfrac{(4n+1)\pi}{2}\;\;,n \in \mathbb W$$ (W stands for set of whole number $W=${$0,1,2,3,.......,n$}).
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Shouldn't it bee $4n+1$ in the numerator? With $2n+1$ I get negative values – Alex May 28 '13 at 22:29
yeah it should thanks @Alex – iostream007 May 28 '13 at 22:36
@iostream007 Can you please explain the first two lines? How does $e^{ix} = i$ translate to $\cos x+i\sin x = 0+1\cdot i$? – Jeel Shah May 28 '13 at 22:39
@gekkostate Euler's formula: – iostream007 May 29 '13 at 4:55
$e^{ix}=\cos x+\sin x$ comes from euler's formula and $i$ has written as $0+1\cdot i$ – iostream007 Jun 9 '13 at 16:17
Well, the geometric meaning of $e^{ix}$ is the weapon here to use.
This is nothing but the point on the complex plane which has length $1$ and angle $x$ measured from the right half of the real axis, in radian. So that $e^{i\pi}=-1$, for example.
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-1: This is the best representation, but from my experience a student using this method will only give one answer. – Were_cat May 29 '13 at 9:32
But the student will see, that's more important for long term.. :) – Berci May 29 '13 at 11:09
A completely different approach:
$$e^{i\pi} = -1 \implies \sqrt {e^{i\pi}} = \sqrt {-1} \implies e^{\frac{i\pi}{2}} = i$$
-
Can you explain why you chose the 'positive' value of the square root? – Calvin Lin May 29 '13 at 0:52
The key here is to think in terms of radial co-ordinates rather than cartesian. Whenever you see a complex exponential it's best to think of the geometric interpretation first, if possible.
The radial representation of a complex number $z = a + bi$ is $Re^{i\theta}$
where $R = |z|$ and $\theta=arctan(b/a)$.
So for the number i we ask ourselves what is R, and what is $\theta?$
We look where the point i is on the complex plane - it's on the y axis 1 unit away from the origin. So R is 1 and $\theta$ is $\frac{\pi}{2}$. (Note that you can add any multiple of $2\pi$ to $\theta$ and things stay the same.)
So plug these R, $\theta$ values into the radial form of z to get the radial representation for i which is $1.e^{i\frac{\pi}{2}} = e^{i\frac{\pi}{2}}$
So $e^{ix} = i = e^{i\frac{\pi}{2}}$ and a naive answer is $x = \frac{\pi}{2}$. But remember we could add any multiple of $2{\pi}$ to our angle for i so we need to add $2n{\pi}$ for the general answer.
Hence $x = \frac{\pi}{2}+2n\pi \quad n\in \mathbb{Z}$
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$$e^{ix} = i \implies ix =\ln(i) \implies ix= \ln(|i|)-i \left(\frac{\pi}{2}+2k\pi\right)$$
$$\implies x = \frac{\pi}{2}+2k\pi,\quad k\in \mathbb{Z}.$$
-
Hint:
$$e^{xi}=i=e^{\frac\pi2i+2k\pi i}\implies x=\frac\pi2+2k\pi\;,\;\;k\in\Bbb Z$$
-
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2016-07-01 13:41:15
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https://www.physicsforums.com/threads/finding-time-using-displacement-and-acceleration.602557/
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# Finding Time Using Displacement and Acceleration
1. May 2, 2012
### Paincake
The problem statement, all variables and given/known data
"A body is thrown downward with an initial speed of 20 m/s on Earth. What is the time required to fall 300 m?"
I tried finding kinetic and gravitational energy, but then I realized I don't have any mass to use it with (1/2)*m*v^2 and m*g*h
How can I solve this problem without guessing and checking?
2. May 2, 2012
### collinsmark
Hello Paincake,
Welcome to physics forums!
Use the appropriate kinematics formula for uniform acceleration.
The second post in the link below should help.
https://www.physicsforums.com/showthread.php?t=110015
[Hint: you'll have to solve for t once you have the right formula.]
3. May 2, 2012
### Paincake
I've seen online a kinematic equation for final velocity, which is
Vf² = Vi² + 2(a)(d),
but my teacher has not introduced it. Where does this equation come from?
EDIT:
Nevermind, I see it now.
I solved for t using vf = vi + at --> (vf-vi)/a = t
And I know distance = v*t and average velocity = (vi+vf)/2
So it was a matter of simplifying d = (vi+vf)/2 * (vf-vi)/a to d = (vf^2+vi^2)/2a and solving for vf.
Thanks for the help.
Last edited: May 2, 2012
4. May 2, 2012
### collinsmark
That's not the equation you want to use for this problem. Try to find a formula that has initial velocity (not final velocity), distance, acceleration, and time.
But I'll tell you where that equation comes from anyway, if you're curious. It comes from combining two of the other kinematics equations for uniform acceleration.
Start with
$$x = x_0 +v_0 t + \frac{1}{2}at^2$$
Now modify some variables, using d = x - x0. And instead of calling the initial velocty "v0", let's call it "vi" instead. So now we have,
$$d = v_i t + \frac{1}{2}at^2$$
Don't forget about that equation, we'll come back to it in a second. But first lets look at a different kinematics equation for uniform acceleration:
$$v_f = v_i + at$$
Rearranging that equation, we have
$$t = \frac{v_f - v_i}{a}$$
Now let's substitute that into the modified first equation above.
$$d = v_i \frac{v_f - v_i}{a} + \frac{1}{2} a \left( \frac{v_f - v_i}{a} \right)^2$$
Expanding a little gives us
$$d = \frac{v_i v_f - v_i^2}{a} +\frac{1}{2}a \left( \frac{v_f^2 - 2v_f v_i + v_i^2}{a^2} \right)$$
$$= \frac{v_i v_f - v_i^2}{a} + \frac{v_f^2 - 2v_f v_i + v_i^2}{2a}$$
Multiplying both sides of the equation by 2a gives,
$$2ad = 2v_i v_f - 2v_i^2 + v_f^2 - 2 v_i v_f + v_i^2.$$
And simplifying the right side of the equation produces
$$2ad = v_f^2 - v_i^2.$$
Adding vi2 to both sides gives
$$v_i^2 + 2ad = v_f^2.$$
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2017-08-21 16:15:37
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https://icipriano.wordpress.com/2016/05/18/bingo-winning-strategy/
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# Bingo winning strategy
I will write about a problem by my friend Edgardo Roldán Pensado. He told me it long time ago. If you are a mathematician the solution may be trivial, however, I personally admire the way he faced the situation and came out with this nice problem.
I won’t bore you with a longer introduction, let start the problem.
Suppose you bought two days ago (in amazon.com for example) a Royal Bingo Supplies Wooden Bingo Game (like the one in the picture)
• Wooden Bingo Game Set with instructions.
• Perfect for old-fashioned fun with a nostalgic twist.
• Includes with 18 Bingo cards, 150 Bingo chips, a Bingo board, brass cage and 75 wooden balls.
• Great for parties, barbeques or family game nights.
• Recommended for ages 3 and up.
Today you received your game and you invite seven other friends to play together. Each player takes a bingo card and you start playing. Five hours later, when everybody is already tired of playing, you decide to count the times each player has won. Everybody is very surprised how lucky a single player was, who won many more times than anybody else.
Is there a Bingo winning strategy? Or do you have a very lucky friend?
Well…Everybody has a very lucky friend, so no discussion about this. However, there is also a Bingo winning strategy, so, there is a chance the winner was a Bingo´s tactician (with a bit of luck).
How does the Bingo´s tactician play?
This guy chose his bingo card at the end, after analysing the bingo cards of the other player. Each card was (hopefully, I do not know if it is true) created with a uniform random distribution (for each square you choose a number from 1 to 75 with probability $1/75.$). He considered a metric on the set of bingo cards, for example, given two bingo cards $a=(x_1,\ldots,x_{24})$ and $b=(y_1,\ldots,y_{24}),$ $d(a,b)=\#\{i:x_i\neq y_i\},$ with the convention that $\# \emptyset :=0.$ He was a bit lucky enough to be able to find a bingo card that maximises the $d$ distance with respect to the bingo cards of the other player.
Why is this a winning strategy?
Suppose that the $7$ bingo cards chosen by the opponents of the winner were very close with respect to the distance $d,$ moreover, suppose that all the $7$ bingo cards were exactly the same. On the other hand, the “lucky one” chose a different bingo card. So there is $\frac{1}{2}$ probability one of the other seven players (and then all) wins, and $\frac{1}{2}$ probability the “lucky one” wins. Now, suppose, the $7$ bingo cards chosen by the opponents are very close (with respect to $d$), but all different. Then the “lucky one” player wins with probability close to $\frac{1}{2},$ whilst and the other players with probability close to $\frac{1}{14}.$
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2018-02-22 22:22:59
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https://www.physicsforums.com/threads/understanding-capacitor-behaviour-fully.469887/
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Understanding capacitor behaviour (fully !)
mklein
Dear forum
I am a school teacher, and whilst I KNOW the behaviour of a capacitor, I must admit I do not fully UNDERSTAND its behaviour.
Please see the left hand image – a capacitor and resistor in parallel over a battery:
http://www.mpklein.co.uk/cap_query.jpg
Textbooks will tell you that, initially, the voltage over the capacitor is zero (due to it having zero charge). Because resistor B is in parallel , the voltage across resistor B must also be zero. Therefore no current can flow through resistor B.
One problem I have with this is why the initial voltage across the capacitor must be zero. Surely if the spacing between the plates was increased a little (and perhaps the plate area reduced) this is equivalent to the other circuit – essentially a gap. Indeed, wouldn't the charge stored on a pair of leads, with a gap, be zero? Wouldn’t a gap in the circuit have the same voltage across it as the battery? How does the capacitor achieve zero volts despite being wired across the battery?
I also do not understand this logic: the voltage across the capacitor is zero, so because they are in parallel the voltage across resistor B is zero . Why is this? The capacitor is in parallel with the battery, so shouldn’t it have the same voltage across it as the battery?
The way I see it we have one route with a gap (the capacitor) and another route which should conduct (the resistor) – so why don’t we get a current through the resistor ? Please help !
Many thanks
Matt
Gold Member
Once the capacitor is charged, it will have the same voltage across it as the resistor does. Here's a useful link on http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html" [Broken].
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Homework Helper
Gold Member
A gap is like a capacitor which has very small capacitance C. So it is charged too fast (practically zero time, charging time is analogous to RC) to the desired voltage and the charge accumulated Q is too small (practically zero). Voltage as a ratio of Q/C=(0/0) can be finite though. So for the gap u assume that instantenously reaches its final voltage V.
For a capacitor with not so small capacitance it ll take some time to be charged to the voltage V. So it starts from zero voltage and keep charging to the final V. In the examples u give the final V is $$V=R_B\frac{E}{R_A+R_B}$$ for both, it just takes more time and more charge accumulated in the left example.
There is current flowing through resistor B but at t=0 is zero. As charges build up on capacitor they create a voltage so current will flow through resistor B. The final voltage when capacitor is fully charged would be V as stated above and the final current will be $$\frac{V}{R_B}$$.
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Homework Helper
Gold Member
Once the capacitor is charged, it will have the same voltage across it as the resistor does. Here's a useful link on http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html" [Broken].
Capacitor will always have the same voltage as resistor B, you mean that once it is charged there would be no current flowing through it and no further charge accumulated on its plates.
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Fish4Fun
Actually, I think what he is asking is why is there zero voltage across the resistor @ t=0. The answer is because the effective resistance of the capacitor @ t=0 is zero.
Consider two resistors, R1 & R2 in parallel. If R1 = 0 ohms and R2 = 100 ohms, then the voltage across both resistors is E = I * (1/(1/R1 + 1/R2)) ==> So for R1 = 0, E = 0 and I is infinite. This is similar to a capacitor at t=0.
Obviously a capacitor is NOT a variable resistor, but it may help to think of it that way.
Fish
Homework Helper
Actually, I think what he is asking is why is there zero voltage across the resistor @ t=0.
It might be simpler to explain that prior to t=0 (the starting time), the battery is not connected to the circuit, and that at t=0 the battery is initially connected to the circuit, when the capacitor voltage is still initially zero.
Homework Helper
Gold Member
Actually, I think what he is asking is why is there zero voltage across the resistor @ t=0. The answer is because the effective resistance of the capacitor @ t=0 is zero.
Fish
The voltage is zero because the capacitor will have zero charge at t=0 hence its voltage V=Q/C=0/C would be zero. This means that there are no charges in its plates to oppose the incoming of further charges, hence one could say its effective resistance is 0.
mklein
The voltage is zero because the capacitor will have zero charge at t=0 hence its voltage V=Q/C=0/C would be zero. This means that there are no charges in its plates to oppose the incoming of further charges, hence one could say its effective resistance is 0.
Thank you for your reply Delta-squared. But if we just had a gap in the circuit, would this not store zero charge on the ends of the wires and so have zero voltage across?
mklein
Actually, I think what he is asking is why is there zero voltage across the resistor @ t=0. The answer is because the effective resistance of the capacitor @ t=0 is zero.
Consider two resistors, R1 & R2 in parallel. If R1 = 0 ohms and R2 = 100 ohms, then the voltage across both resistors is E = I * (1/(1/R1 + 1/R2)) ==> So for R1 = 0, E = 0 and I is infinite. This is similar to a capacitor at t=0.
Obviously a capacitor is NOT a variable resistor, but it may help to think of it that way.
Fish
Thank you for the Fish 4 Fun. Whereas a GAP in the circuit would have effectively an infinite resistance?
mklein
A gap is like a capacitor which has very small capacitance C. So it is charged too fast (practically zero time, charging time is analogous to RC) to the desired voltage and the charge accumulated Q is too small (practically zero).
This is very useful, thanks.
Fish4Fun
Whereas a GAP in the circuit would have effectively an infinite resistance?
Exactly. I thought that explanation might bring it "home" for you. It is technically wrong, but functionally accurate.
In understanding reactive components and impedance, it helps to think of V and I rather than focusing on "resistance". For instance, in the case of a capacitor:
I = E/R * e^(-t/RC)
Where "R" is some value greater than 0, even if it is very small. As R approaches 0, the current approaches infinity. What this implies is that there is always some "R", even if it is in the power supply itself (this is referred to as "source impedance") that prevents the current from actually being infinite. To get a "feel" for this, you might want to play with very small values of R here:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html
Capacitors in a DC circuit are generally only "interesting" for finite periods of time after the power is switched on or off.
Hope this helps,
Fish
harrylin
Thank you for your reply Delta-squared. But if we just had a gap in the circuit, would this not store zero charge on the ends of the wires and so have zero voltage across?
That was answered in posts #2 and #3. A gap stores nearly zero charge so that the voltage is built up very quickly. A gap is also almost immediately discharged.
If you ever did some soldering on audio circuits, you may know that a big capacitor is needed to reduce hum that is due to power supply voltage variations. Inversely, a gap (no capacitor) results in a big hum. If you look at the link of posts #2 and #11 you may understand why.
mklein
OK, well I think I am there now. Capacitors have always puzzled me a bit, but less now !
Thank you for everybody's comments, and the links too. You have all been a great help.
Matt
Homework Helper
Gold Member
Thank you for your reply Delta-squared. But if we just had a gap in the circuit, would this not store zero charge on the ends of the wires and so have zero voltage across?
gap stores almost zero charge Q but has almost zero capacitance C, so the ratio Q/C which is the voltage can be finite and non zero. For example if Q=0.0001 and C=0.00001 then Q/C=10
Gold Member
Capacitor will always have the same voltage as resistor B, you mean that once it is charged there would be no current flowing through it and no further charge accumulated on its plates.
Yes.
Homework Helper
On a side note, is it possible to plot a current versus time graph for a fixed voltage supply + capacitor without a resistor, and caculate the charge versus time? This would only be possible if the area under this curve from t = zero to t = some finite time isn't infinite, despite the fact that the initial current is infinite for an instant in time.
Homework Helper
Gold Member
On a side note, is it possible to plot a current versus time graph for a fixed voltage supply + capacitor without a resistor, and caculate the charge versus time? This would only be possible if the area under this curve from t = zero to t = some finite time isn't infinite, despite the fact that the initial current is infinite for an instant in time.
hmmm, in this case current $$I=VC\delta(t)$$ and
$$q=\int_{0}^{\infty}VC\delta(t)dt=VC$$
but we just cant have instanteneous charge of the capacitor there would be other factors limiting the speed of charging (like the distance of capacitor from the source, the e-field which makes free electron to move doesnt travel instanteneously) which are not taken into account by circuit theory.
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mikelepore
The capacitor is in parallel with the battery, so shouldn’t it have the same voltage across it as the battery?
What's in the box labeled "A"? The capacitor is NOT in parallel with the battery, because of the existence of device "A".
Immediately after charging begins, the capacitor looks like a short circuit, because charges flow onto one capacitor plate and flow off of the other capacitor plate so rapidly that there appears to be a shorted path "through" the capacitor, and for an instant no current at all flows through "B". As a result, the entire battery voltage drop is across "A".
mklein
What's in the box labeled "A"? The capacitor is NOT in parallel with the battery, because of the existence of device "A".
Immediately after charging begins, the capacitor looks like a short circuit, because charges flow onto one capacitor plate and flow off of the other capacitor plate so rapidly that there appears to be a shorted path "through" the capacitor, and for an instant no current at all flows through "B". As a result, the entire battery voltage drop is across "A".
Hi Mikelepore
A & B are resistors.
You are right, the capacitor is not directly parallel to the cell , I forgot about that.
What would happen if resistor A was not included in the circuit? Even if the capacitor was wired directly over the battery it would initially still have zero volts across its terminals wouldn't it ? This is unusual, because most other components would have the battery voltage across them.
Naty1
What would happen if resistor A was not included in the circuit? Even if the capacitor was wired directly over the battery it would initially still have zero volts across its terminals wouldn't it ? This is unusual, because most other components would have the battery voltage across them.
The instant an ideal capacitor is connected to ideal battery terminals it will have battery voltage across it. Of course in the real world there is still a tiny resistance represented by A, such as the resistance in the capacitor plates, the battery and the connecting wires, so the jump is not a perfect step function. In other words really fast, but slower than the answer above from Delta with just wire ends.
q = it = CV are convenient relationships to remember...
that is, the charge (q) on a capacitor equals the current (i) that has passed thru the capacitor times the duration of current flow (t) , which equals the capacitances (C) time the voltage (V) across the capacitor.
With a capacitor, current starts off very high, then slows to zero as charge (electrons) build up on the capacitor plate and new electrons have nowhere to go...new electrons are opposed by those already on the capacitor plate.
Suppose you place two capacitors of different size (different capacitance, C) in parallel across the same battery terminals...from the above q=CV you can see this means the two will have different q because C is different and V must be the same....then from the other q = it, you can tell infer either i was different or one took longer to charge or both these occurred.
mikelepore
What would happen if resistor A was not included in the circuit? Even if the capacitor was wired directly over the battery it would initially still have zero volts across its terminals wouldn't it ?[
If resistor A were not included, then whatever small resistance is in the wiring, even if it's a small fraction of one ohm, would be the R in the expression for the time constant, the product RC. That's why the capacitor would appear to fully charge almost instantly, because of that smallness of that RC product. However, for an instant after closing the switch, the battery voltage would be fully across that wiring resistance, and none of it across the capacitor.
In an ideal mathematical model, if you omit the series resistor and also let the wiring resistance be zero, you can safely discuss the steady state condition when no switch has been thrown in a long time, but a discussion of the moment of switching "blows up", because for an instant the capacitor acts like a short, the zero resistance would conduct current V/R and dissipate power V^2/R, and there are zeroes in the denominators.
This is unusual, because most other components would have the battery voltage across them.
I don't undertand the last sentence.
I'm also reviewing this subject. I'm teaching this too, and I'm going to deliver this lesson in a couple weeks from now.
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http://overanalyst.blogspot.com/2011/01/
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Monday, January 31, 2011
"sometimes you have to make an αss out of u and mε."
lately every time i write a lecture for my theory [1] course,
i write down an example, then i realise that to actually prove the assertion, one needs unproven facts.
e.g. the function $f(x) = x^3 - x$ is surjective.
this consistently irks me:
this is a class about proοfs, after all;
i have to set a good example, not be lazy with details!
so lately i've been visiting wιkipedia a lot.
for last time, i looked up cardanο's formula [2], so that we could check the previous example by a direct computation. indeed, the equation
$x^3-x-y=0$
has an explicit solution given by
$$x = \sqrt[3]{\frac{y}{2} + \sqrt{\frac{y^2}{4}-\frac{1}{27}} } + \sqrt[3]{\frac{y}{2} - \sqrt{\frac{y^2}{4}-\frac{1}{27}} }.$$
as for today, i looked up the well-οrdering princιple, wondering if i remembered correctly: it doesn't follow from basic properties of naturaΙ numbers, does it?
as you imagine, we're discussing proofs by ιnduction now, so today we're assuming well-οrdering holds for the naturaΙ numbers.
sometimes i feel like i'm not patient enough to teach this class ..
[1] that is, intrοducition to theοretical mathematics (a first course in proοfs). i can't think of another effective shorthand.
[2] i also told them about the tragic story of tartagΙia. they weren't impressed.
Saturday, January 29, 2011
Memοry and desιre, stirring .. DuΙΙ roots with spring rain.
odd. something has changed .. and for the better, i think.
if i'm doomed, then so be it.
it's not like there's anything i can do about it anymore, now that job ads have trickled to a halt.
instead i've been waking up every day, feeling uneasy.
that's a good thing, though: this is the sort of unease that plagues the puzzled researcher's mind.
i can concentrate again, think about problems. i can afford to be puzzled about an something that doesn't much matter. i can afford to be curious.
i've decided to do a little geometry again .. the metrιc kind, i mean. i have this one idea that's been festering since december.
it seems to be working, but there are lots of details;
i wake up every day and think about the latest one,
try to work it out.
before new orleans, i had a lemma and an application in mind. now i have a theorem [1] and i want another one.
that's life, i suppose:
when we learn something is possible,
something we once wanted, something we now want,
we become resourceful and able ..
.. ambitious.
i cannot name all the numbers and kinds of frustrations, out there. as for this one, this obsessive, inquiring kind: i know it well.
it turned me into a mathematician, years ago.
it sharpens my focus, gives some small purpose to my life.
i feel alive.
[1] for the specialists out there, it's about measurabΙe differentιable strucτures.
Thursday, January 27, 2011
times are tough, yes (article post)
by Robert Krulwich
10:31 am, January 27, 2011
The science crowd is nervous. The President wants to create more jobs, but come this fall, funding for the American Recovery and Reinvestment Act (what you and I call the "stimulus program") starts to wind down.
After two glorious years, science researchers all over America will face what they are now calling "The Cliff;" to replace their grants from Uncle Sam they're going to have to find new funders, get on their knees and do that thing they've always had to do...umm, what's the word?
Beg.
It's not like they don't know how. Scientists have been begging for money for centuries, as have writers, artists, poets. It's an art form.
(and so it continues, for a little longer.)
not being familiar with laboratories in the sciences, this sounds like a problem only for those labs who have been created in the last two years (which could be a great many).
there have been labs before the current administration, right? (-;
there is a point to this, though: the author probably meant something else.
my experimentalist acquaintances have told me before that many of them are not often paid by their universities, but directly through their grants. so if the federal science budget shrinks to smaller than pre-stimulus levels, then some labs will go under (possibly a great many).
as a mathematician, this idea is a strange one to me, if only because there are few mathematical centers in the u.s. that are run purely by research funds.
sure, there may be superstars at micrοsoft research and similar posh places. for those of us researchers in the "(petite) bourgeoisie" of mathematical society, though, teaching is an inseparable part of our weekly grind.
our livelihoods are intertwined with those of university budgets and enrollments. so as long as students need to take calcuιus classes, most of us will still have a job.
i'm not saying that we mathmos don't have to "beg," a word that Krulwich seems to like. i suppose that journalists are a well-off lot who needn't worry about the future of their industry.
if we mathematicians are beggars, then we beg from a different affluence. let the scientists fret (if indeed they are fretting).
Monday, January 24, 2011
nontrivial in the elementary.
i think i have this phobia of being boring, especially as an instructor. it's why i try to bring up unconventional topics in the classes i teach:
in my "proofs" class this semester, we were going over proof by contradiction. one classic theorem is why there are infinitely many primes;
so during the lecture,
1. i define the notion of a prιme number,
2. i give a few quick examples,
3. i state the theorem,
4. then i state the prιme number theorem to give them a sense of how the prιmes are distributed amongst the natural numbers,
5. and i tell them about the twιn prιme conjecture: despite the asymptotic distribution given in the previous theorem, nobody (yet) can explain why these consecutive occurrences seem to appear infinitely often.
i don't think there is any loss in showing students what is important (but easy to formulate) and what we don't know. statistically, they won't all be research mathematicians ..
.. but then again, isn't it worthwhile to show them why maths is interesting? isn't that the point of this sort of class?.. other than imparting the notion of rigor, i mean ..
anyway, we start set theory today.
maybe i'll explain russell's paradοx to them. (-:
Sunday, January 23, 2011
sometimes we have to walk away, for a little while.
yesterday morning i woke up, drank coffee and worked out a few pages of research ideas .. until i got stuck and became angry with myself.
maybe this is particular to my personality, but it's hard to do research when i'm depressed, angry, or worried. stress tends to kill my inspiration.
ever since i've been on the job market, i can't seem to focus as well as i should .. as i'd like. i guess i don't deal with uncertainty well.
i was about to tear those pages into shreds:
my hands were set together, gripping the pages tightly,
my wrists were ready to twist into opposite directions ..
.. yet, for some reason, i stopped. suddenly i realised that it's not the maths; it's just me. so i sighed, thought about going running,
[the weather report says it's 15oF outside]
.. but reconsidered (read: shivered involuntarily). grabbing my coat, decided to go on a long walk instead.
i ended up walking for hours and miles, through city streets and city parks. after a while i couldn't stand it anymore.
i tried to distract myself by taking photos with my phone, but after 20 shots or so, my fingers became stiff and i couldn't work the buttons anymore.
the cold had seeped through the soles in my shoes and the fabric of my socks; it started to feel like i was stepping barefoot onto the frozen ground.
my coat wasn't thick enough, not for that length of time. at some point i had to force my teeth to stop chattering because my jaw began to ache.
so i managed my way to a familiar neighborhood, a familiar cafe, and bought a coffee. it took a while to become warm again.
i looked at the flyers for various events -- none of them looked inviting -- but the backs of the papers were blank.
i hesitated ..
but then borrowed a pen, typically meant to sign credit card receipts.
i hesitated ..
but then started writing on the backs of flyers for art exhibitions.
for years i've balked at writing a paper about currεnts on metrιc spaces, because i didn't have many results. of those i had, i didn't think they were good enough.
i've changed my mind.
the results aren't great, but with a few new ones, they are enough. maybe if i write them up, then someone else will read them and do something better with the ideas.
Saturday, January 22, 2011
some complex humor .. also, an anecdote.
ah, geek humor ..
from xkcd, by randall munrοe.
on a related note ..
jοrge cham of phd comics came to my university once to talk about prοcrastination.
at the book signing, one of my friends asked him to sign his copy of xkcd, vol 0. to his credit, jοrge signed:
"hey randall, why don't you sign my book next time?" (-:
Wednesday, January 19, 2011
something old, something new, something borrowed ..
it's always been a goal of mine to give a new talk at each conference.
it's a sufficient measurement of research productivity:
if i have something new to say every year (or more often) then at least i am not "dead" [1] as a researcher ..
at any rate, today was the deadline for title/abstracts for the ams sectional meeting at statesboro, ga.
this time, i think i'll have to compromise. i'll talk about some old results. if i can work out this conjecture, then perhaps i'll say something new after all .. which might therefore count as a "new" talk.
here's hoping, anyway. \:
[1] oddly enough, here i mean "dead" in the erdös sense.
Monday, January 17, 2011
mathematical purgatory.
almost a week passes and my most recent crackpot idea hasn't failed yet.
maybe i haven't spent enough time entertaining enough pathological examples.
for one thing, i certainly haven't spent much time blogging.
(sorry about that, by the way.)
strictly speaking, it's not impossible that the idea could work. i've collected enough "believable" lemmas so that a special case must hold, enough so that cut-&-paste would be particularly useful [1].
it's become worthwhile to organise my piecemeal notes in LaTeχ
i'm doing so now, in fact.
it feels like a kind of purgatory:
maybe the analysιs gods want me to suffer for a while,
work out the details in tiring detail,
and maybe,
eventually,
i'll prove something mildly interesting.
here's hoping, anyway.
on a related note (regarding the afterlife), this reminds me of a joke:
in hell, beer is served in klein bottles.
[1] sometimes i literally cut and paste my work together: with scissors i cut away the useless scrawls and scratchouts, and with a glue stick i put together the remaining scraps .. it feels like kindergarten, but with less construction paper and many more greek symbols. (-:
Tuesday, January 11, 2011
#900: sudden idea; also, what's to come.
i've a new idea to prove a conjecture.
*sighs*
.. so now i'll have to figure out what's wrong with it.
for those of you in the know,
yes: it's the same conjecture.
it's always the same conjecture.
at this point this is not just an obsession for me, but also a compulsion (at least in the pop-psychiatric sense).
i've been lazy with new posts on this blog, haven't i?
i thought that by the start of december, my life would slow down, but i was sadly mistaken. my hope is that i was just a month off .. and that this january and february will start a slower pace.
that's the plan, anyway;
if only as a safeguard, i refuse to travel until march ..
.. well, unless i'm invited for an on-campus job interview,
for which i would gladly make an exception. (-;
then again, what are the odds of that happening, in this economy? \-:
anyway .. in no particular order, i'll next write about:
• how the ιnterviews went,
• teaching a first course in prοofs for undergraduates,
• general malaise (or: looking back on these postdoc years).
Saturday, January 08, 2011
me: 2, hotel: 1.
i never caught the exact number, but supposedly there were around 2000+ mathematicians at this year's joint meetings.
imagine: that many mathematicians concentrated over 1-2 city blocks. think of all the coffee consumed!
on a related note, i did the paranoid thing and packed 3 days' worth of good coffee grounds in my carry-on luggage (along with standard-sized filters).
janus:1, hotel:0 .. (-;
as it happens, my planning was pointless: the in-room coffeemaker brewed one cup at a time and the filters didn't fit. in fact, the coffeemaker brews only those sealed paper filters ..
janus:1, hotel:1 .. ):
but the hotel room provided two such filters, and if you cram them enough, both of them will fit.
the result: 1 good, strong cup of coffee.
janus:2, hotel:1 .. q-:
the trip was productive:
• the interviews weren't complete disasters (more on them, later),
• i saw many, many friends from graduate school and various past conferences,
• i met an old friend, waxed mathematical,
• lastly, i started refereeing a paper.
as for new orleans, it was a fine thing to have a conference in the downtown area (vs. another university campus). the french quarter is certainly something to see, a pleasant place to walk and watch people.
reciprically, some stores seemed receptive to mathematicians!
the kitchen witch cookbook store.
Thursday, January 06, 2011
mathematicians, mathematicians: almost everywhere!
the last time i attended the joint meetings, it was held in baltimore and i was there for a poster session. my poster wasn't well received .. but then again, the results weren't very good.
overall i remember the experience as being a lonely one, despite the crowd of thousands (of strangers). in the end,
i wandered away to a barnes & noble,
bought a coffee,
.. and oddly enough, met a girl. (-:
she was lovely,
but i never heard from her again. \-:
8 years later, i'm in new orleans, and the crowds are bigger than ever. it's now the other way around: in every crowd, i could swear that i see someone familiar ..
at any rate, i must be off:
you can take the boy out of graduate school,
but you can't take graduate school out of the boy ..
(-;
Tuesday, January 04, 2011
preparations.
given two logical statements $P$ and $Q$, it turns out that
$$(P \to Q) \vee (Q \to P)$$ is a tautology. i probably need more sleep (or a longer winter break) but the first time around, i misinterpreted this as "given any two logical statements, one must imply the other."
on a related note, this term i'm teaching an "introduction to proofs" class, which surprised me. the department actually trusts me to train a new batch of potential maths majors.
(i guess i have to be responsible, then ..!)
the first class is tomorrow. right afterwards, i'm leaving for the airport. the timing is painful. i prepared four lectures today: two for tomorrow (proofs class, linear algebra class) and two for my friday substitutes.
(this is exactly why i hate traveling.)
anyway: for those of you headed to the joint meetings, i might see you there!
Monday, January 03, 2011
what i didn't get to do, this holiday.
i have one more day until the spring semester begins, two more days before i give my first lecture.
this winter holiday seems very short, more so than in previous years.
the fact is, i lost close to a week when the winter storms hit, two weekends ago:
my tuesday flight out of jfk airport, last week, was among the hundreds (thousands?) of flights that were canceled, due to snow.
it wasn't until new year's eve (3 days later) that i boarded a flight home.
to be honest, i missed working. there's something about family that sucks away from you any time or willpower for work.
it's one thing if i lived close to my parents, visited them one or two evenings per month. the units of measurement here are hours; i can budget that sort of time.
in contrast, living 400+ miles away means that every visit is a prolonged investment of time, measured instead in days or weeks.
there's no daily, regularly-repeating schedule either, and i might as well be taking appointments.
one day we're off to see my paternal grandmother; another day it's my maternal grandparents. two days later, we're celebrating the solstice or it's a family shopping trip.
my friend from high school wants to meet for breakfast sometime, another evening is an old friend's birthday. social media is crippling: if i don't go, i'll probably offend someome.
then, for a few mornings, it's nothing but shoveling snow.
i like seeing my family .. at first. averaging out over a week's time, though, i quickly get the sense that i'm visiting out of obligation, not out of enjoyment.
all of that said, all i "wanted for christmas" [1] was to be left alone for a week,
work out research ideas in the mornings,
perhaps write a little, in the afternoons,
and in the evenings, either meet friends or catch up on my reading ..
in other words, i just wanted a break from teaching so that i could .. well, work. it's not like i'm asking for a tenure-track job, or anything ..
[winces]
.. right: the joint meetings. there's that to prepare, too ..
[sighs]
it never really ends, does it? \-:
coming up: what i'll be teaching, this term.
[1] i'm actually an agnostic, in the sense that religious principles seem like axioms to me (so you can't really verify them, one way or another). this would explain, for example, why atheists and believers never seem to agree on anything: the logical systems are inherently different, and probably incompatible.
Saturday, January 01, 2011
enter 2011; also, a maths museum? (-:
happy new year, everyone:
despite some few mathematical thoughts,
i still find myself in a holiday mood,
not quite inclined to chase a mathematical thought to its end ..
.. not yet, anyway: i'll get back to you tomorrow about that.
i would, however, like to share this link from math οverflow, regarding a museum of mathematics in new york!
it seems, though, that they still need funds to finalise the plans for the museum building. for details, a link to their donations page is here.
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2018-02-18 01:31:26
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https://tex.stackexchange.com/questions/303591/problem-with-table-spacing-when-aligning-by-decimal-point
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# Problem with table spacing when aligning by decimal point?
Please see attached MWE and pdf. The problem seems to be with the column 'Test 1'? Any suggestions would be very much appreciated, I have tried altering each variable but I cannot work out where the problem is. Many thanks in advance.
\documentclass[12pt, oneside]{article}
\usepackage{graphicx}
\usepackage{gensymb}
\usepackage{amsmath}
\graphicspath{ {paperfigures/} }
\usepackage[font={footnotesize}, margin=2cm]{caption}
\usepackage{siunitx}
\usepackage{multirow}
\begin{document}
\begin{table}
\centering
\caption{clear test}
\label{tab:cleart}
\sisetup{output-decimal-marker = {\cdot}}
\begin{tabular}{|l|l|S[table-format = 3]<{\si{}}
|*{4}{S[table-format = 3.1,
separate-uncertainty,
table-figures-uncertainty=1]|}}
\cline{3-6}
\multicolumn{2}{c|}{}
& \multicolumn{4}{c|}{\textbf{Intensity
(mV)}} \\
\hline
\textbf{Angle}
& \multicolumn{1}{c|}{\textbf{Height}}
& \textbf{Test 1}
& \textbf{Test 2}
& \textbf{Test 3}
& \textbf{Mean} \\ \hline
\multirow{3}{*}{5\degree}
& Top & 2.3 \pm 2.0 & 5.8 \pm 2.0 & 8.2 \pm 2.0 &
\\ \cline{2-6}
& Middle & 3.7 \pm 2.0 & 2.9 \pm 2.0 & 26.1 \pm 2.1 &
\\ \cline{2-6}
& Bottom & 12.8\pm 2.0 & 4.5 \pm 2.0 & 13.6 \pm 2.0 &
\\ \hline
\multirow{3}{*}{10\degree}
& Top & 10.4\pm 2.0 & 8.8 \pm 2.0 & 11.4 \pm 2.0 &
\\ \cline{2-6}
& Middle & 12.5\pm 2.0 & 8.0 \pm 2.0 & 7.2 \pm 2.0 &
\\ \cline{2-6}
& Bottom & 5.3 \pm 2.0 & 2.9 \pm 2.0 & 5.1 \pm 2.0 &
\\ \hline
\multirow{3}{*}{15\degree}
& Top & 9.6 \pm 2.0 & 6.9 \pm 2.0 & 11.0 \pm 2.0 &
\\ \cline{2-6}
& Middle & 12.2\pm 2.0 & 9.0 \pm 2.0 & 3.5 \pm 2.0 &
\\ \cline{2-6}
& Bottom & 2.3 \pm 2.0 & 2.2 \pm 2.0 & 1.5 \pm 2.0 &
\\ \hline
\end{tabular}
\end{table}
\end{document}
The result with S[table-format = 2.1(2)] as column specification for the third columm:
• @referencingstressesmeout The options for the S column of the third column differs from the options of the S columns of the next four columns. BTW, there are seven column specifications, but only six columns are used. Maybe the third column specification is a left-over from previous trials and should be removed. – Heiko Oberdiek Apr 11 '16 at 15:51
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2019-12-13 00:29:35
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http://clay6.com/qa/29964/average-velocity-of-the-molecules-of-a-gas-in-a-container-moving-in-one-dim
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Answer
Comment
Share
Q)
# Average velocity of the molecules of a gas in a container moving in one dimension is
$(a)\;Zero\qquad(b)\;Positive\qquad(c)\;Negative\qquad(d)\;\sqrt{\large\frac{8RT}{\pi m}}$
I know beforw that ansr 1
## 1 Answer
Comment
A)
The average velocity of gas molecules in a container moving in one direction is zero , otherwise all molecules will be collected in one direction.
Hence answer is (a)
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2019-08-26 07:35:06
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https://mathoverflow.net/questions/250651/coend-of-mathscrdf-bullet-g-bullet
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Coend of $\mathscr{D}(F(\bullet), G(\bullet))$
(Crosspost from stack)
Given categories $\mathscr{C}$ and $\mathscr{D}$ and functors $F,G: \mathscr{C} \to \mathscr{D}$, we can form a bifunctor $$\mathscr{D}(F(\bullet), G(\bullet)): \mathscr{C}^\text{op} \times \mathscr{C} \to \mathsf{Set}$$ and the end of this functor is the set of natural transformations from $F \Rightarrow G$. (I guess we need $\mathscr{C}$ to be small in order to guarantee there is a set of such natural transformations, in general.)
Can I say anything about the coend? Is it some familiar thing?
• The question is already interesting when we take $F, G$ to be identity functors. See for example ncatlab.org/nlab/show/trace+of+a+category – Todd Trimble Sep 25 '16 at 6:30
• I think the comment of @მამუკა ჯიბლაძე in the answer below is what I was referring you to in your SE post. – Daniel Robert-Nicoud Sep 25 '16 at 10:45
• Todd's comment would qualify as an answer, I guess. What is the trace of the category of finite-dimensional vector spaces? finitely generated $R$-modules? – HeinrichD Sep 25 '16 at 11:48
• If you take $F=G=1$ and $\mathcal C=\mathbf{Vect}_K$ then $\int^V \hom(V,V)\cong K$, right? – Fosco Sep 25 '16 at 13:28
• @HeinrichD, I think this qualifies as a proof that $\int^V \hom(V,V)\cong K$. Indeed, $\hom(V.V)\cong V^*\otimes V$, which in turn is isomorphic to $\hom(V,K)\otimes V$. But now this is only a fancy way to write $\text{Lan}_11(K)\cong K$. – Fosco Sep 25 '16 at 13:29
I'm not sure I have much to add beyond my comment, but I might add a point of view (which could in turn provide some search terms).
I'd situate this construction within the bicategory of (small) categories, profunctors/bimodules, and transformations between them. Recall that a profunctor $R: C \nrightarrow D$ is a functor of the form $C^{op} \times D \to Set$ (conventions may differ), composed much in the way relations are, via the formula
$$(C \stackrel{R}{\nrightarrow} D \stackrel{S}{\nrightarrow} E)(c, e) = \int^{d: D} R(c, d) \times S(d, e).$$
(If you like, you can consider the bicategory of profunctors as biequivalent to a strict 2-category whose objects are small categories and whose morphisms $C \to D$ are given by cocontinuous functors $Set^C \to Set^D$.)
This bicategory is compact closed in an evident bicategorical sense: we have a symmetric monoidal bicategory whose tensor at the object level is given by cartesian product of small categories, and each object $C$ has a monoidal dual given by the opposite category $C^{op}$. For each $C$, the unit $\eta_C: 1 \nrightarrow C^{op} \times C$ is given by $\hom_C: C^{op} \times C \to Set$. (In the cocontinuous functor picture, it's the unique (up to isomorphism) cocontinuous functor $Set \to Set^{C^{op} \times C}$ that takes the terminal object $1$ to $\hom_C$.) The counit $\epsilon_C: C \times C^{op} \nrightarrow 1$ may also be described by a hom-functor, but it is probably more illuminating to think of it in terms of the cocontinuous functor picture, given by taking the coend $\int^C: Set^{C^{op} \times C} \to Set$.
Since we are working in a compact closed (bi)category, we can expect certain resonances with constructions in other compact closed categories, such as the category of finite-dimensional vector spaces. The construction in question is a profunctor composite
$$1 \stackrel{\eta_C}{\nrightarrow} C^{op} \times C \stackrel{F^{op} \times G}{\nrightarrow} D^{op} \times D \stackrel{\epsilon_{C^{op}}}{\nrightarrow} 1$$
which is certainly akin to trace operations in linear algebra. Thus, in linear algebra over a field $k$, we have the notion of trace of an endomorphism $f: V \to V$, which we can form categorically as the composite:
$$\text{Tr}(f) = \left(k \stackrel{\eta_V}{\to} V^\ast \otimes V \stackrel{1 \otimes f}{\to} V^\ast \otimes V \stackrel{eval_V}{\to} k \right)$$
where the first map $\eta_V$ takes $1 \in k$ to $\sum_{i = 1}^n f^i \otimes e_i$ (here $\{e_1, \ldots, e_n\}$ is a basis of $V$ and $f^i$ is the dual basis; the expression $\sum_{i=1}^n f^i \otimes e_i$ is independent of basis). Similarly, we speak of the trace of an endoprofunctor $B: C \nrightarrow C$; after a brief Yoneda-lemma type calculation, one finds that the composite
$$1 \stackrel{\eta_C}{\nrightarrow} C^{op} \times C \stackrel{1 \otimes B}{\nrightarrow} C^{op} \times C \stackrel{\epsilon_{C^{op}}}{\nrightarrow} 1$$
is the profunctor $1 = 1^{op} \times 1 \to Set$ taking the unique object of $1$ to $\text{Tr}(B) = \int^{c: C} B(c, c)$.
Thus we could also describe your construction as the trace of the endoprofunctor or endobimodule $B: C \nrightarrow C$ defined by $B(c, d) = \hom_D(Fc, Gd)$. Possibly this gives a useful search term.
Usually such traces are challenging to calculate explicitly (for example, determining the trace of an identity functor can be nontrivial). Among the properties of trace formally deducible from compact closed structure is $\text{Tr}(B \circ B') \cong \text{Tr}(B' \circ B)$.
• I didn't know this nice description of profunctors as cocontinuous functors. When we replace $Set$ by $\{0,1\}$, we should get a similar description of relations $C \looparrowright D$: These are maps of sup-lattices $P(C) \to P(D)$, right? – HeinrichD Sep 25 '16 at 20:02
• Yep! This also works generally where $C, D$ are posets and $P(C) = [C^{op}, 2]$ is the sup-lattice of down-sets, provided we use "relation" to mean one where $R(a, b) = \text{true}$ and $a' \leq a$ implies $R(a', b) = \text{true}$ and $b \leq b'$ implies $R(a, b') = \text{true}$. (Usually I call such relations "bimodules".) – Todd Trimble Sep 25 '16 at 20:19
Basic rule in general category theory: Restrict to categories with one object, aka monoids, and see what happens. (Alternatively, restrict to thin categories, aka preorders.) Gain some intuition and generalize this to arbitrary categories.
So here, if $M$ is a monoid, we have two objects $X$ and $Y$ of $\mathscr{D}$ on which $M$ acts from the left via morphisms of monoids $F:M \to \mathrm{End}(X)$ and $G:M \to \mathrm{End}(Y)$. Then, the set $\mathrm{Hom}(X,Y)$ carries a left $M$- and a right $M$-action. The coend identifies these two actions in a universal way: It is given by $\mathrm{Hom}(X,Y) / (G(m) \circ f \sim f \circ F(m))_{f \in \mathrm{Hom}(X,Y),\, m \in M}$. If the action $F$ is trivial, this is a generalization of the set of coinvariants. Namely, this is the special case $\mathscr{D}=\mathsf{Set}$ and $X=1$. Dually, notice that the end of $\mathrm{Hom}(X,Y)$ is the set of morphisms $f \in \mathrm{Hom}(X,Y)$ which "believe" that the two actions agree, i.e. $G(m) \circ f = f \circ F(m)$ for all $m \in M$, and that this generalizes the set of invariants when the action $F$ is trivial.
Now, if $\mathscr{C}$ is any small category and $F,G : \mathscr{C} \to \mathscr{D}$ are functors, you may view these as "multi-object" actions of $\mathscr{C}$ on $\mathscr{D}$. The coend of $\mathrm{Hom}(F(-),G(-))$ is the quotient set of the coproduct $\coprod_{x \in \mathscr{C}} \mathrm{Hom}(F(x),G(x))$ which identifies the two actions of $\mathcal{C}$. This means that we identify, for every morphism $m : x \to y$ in $\mathscr{C}$ and every morphism $f : F(y) \to G(x)$ in $\mathscr{D}$ the two morphisms $G(m) \circ f : F(y) \to G(y)$ and $f \circ F(m) : F(x) \to G(y)$ in the coproduct.
• Probably relevant: coend of $F(\_)\times G(\_)$ (for contravariant $F$ and covariant $G$) is $F\otimes_{\mathscr C}G$ – მამუკა ჯიბლაძე Sep 25 '16 at 6:52
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2019-12-15 12:25:12
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http://mathoverflow.net/questions/39561/is-there-an-elementary-way-to-find-the-integer-solutions-to-x2-y3-1
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# Is there an elementary way to find the integer solutions to $x^2-y^3=1$?
I gave this problem to my undergraduate assistant, as I saw that Euler had originally solved it (although I am having trouble finding his proof). After working on it for two weeks, we boiled the hard cases down to showing that (1) in $\mathbb{Z}[w]$ the fundamental unit is $1+w+w^2$ (where $w$ is the real cube-root of 2) [which I'm sorry to say, I'm not certain I know off the top of my head how to prove], and (2) using that fundamental unit, I found a crazy ad hoc computation to show that there are only the obvious solutions.
So I'm wondering if someone else out there is more clever, knows where I can find Euler's proof, or if there is another nice elementary proof in the literature.
-
Elementary includes avoiding elliptic curves? Because there are not-so-hard arguments with the basic theory set up. – Cam McLeman Sep 22 '10 at 0:07
Building on Cam McLeman's comment, this elliptic curve has a point of order 6, and hence possesses a 2-isogeny. A descent via 2-isogeny shouldn't be too hard to write out explicitly. It would be interesting (but perhaps not what Pace Nielsen is going for) if someone could give an elementary classification of the rational points on this curve based on the fact that it is the modular curve $X_0(36)$. – Jamie Weigandt Sep 22 '10 at 0:25
You can find a discussion of two approaches to this problem in Schoof's "Catalan's Conjecture." An appendix to the book describes Euler's original approach (the reference for which Schoof includes), and Chapter 4 of the book describes an alternative approach due to William McCallum. – user2490 Sep 22 '10 at 1:16
I sketched Euler's proof in "A note on Pépin's counter examples to the Hasse principle for curves of genus 1", which you can find online. – Franz Lemmermeyer Sep 22 '10 at 6:00
Thank you James and Franz. I'll take a look at those sources. – Pace Nielsen Sep 22 '10 at 15:55
If your students know a little about the Eisenstein integers (unique factorization and what the units are) there's the following simple argument (maybe it's essentially Euler's?). Let $u$ and $v$ be the complex roots of $z^2+z+1=0$.
Theorem: Let $A$, $B$, $C$ be non-zero elements of $Q[u]$ with sum $0$ and product twice a cube. Then some two of $A$, $B$, $C$ are equal.
Corollary: Suppose $x$ is in $Q$ and $x^2-1$ is a cube. Then $x$ is $1$, $-1$, $0$, $3$, or $-3$.
(To prove the corollary let $A=1+x$, $B=1-x$ and $C=-2$).
The proof of the theorem is a reductio. If there's a counterexample, there's one with $A$, $B$, $C$ in $Z[u]$; take such a counterexample with $d=\min(/A/,/B/,/C/)$ as small as possible. Then $A$, $B$ and $C$ are pairwise prime. Since $ABC=2$(cube) we may assume $A=2i$(cube), $B=j$(cube) $C=k$(cube) where $i$, $j$, and $k$ are in the set $\{1,u,v\}$. Now all cubes in $Z[u]$ are $0$ or $1$ mod $2$. Since $B+C$ is $0$ mod $2$, $j=k$. Since $ABC=2$(cube), $ijk$ is a cube and $i=j=k$. We may assume $i=j=k=1$. Then $A=2r^3$, $B=s^3$, $C=t^3$, and we may further assume that $s$ and $t$ are $1$ mod $2$. $s$ and $t$ are not both in $\{1,-1\}$ and it follows that $d$ is at least $\sqrt{27}$. Now look at $s+t$, $us+vt$ and $vs+ut$. They sum to $0$ and their product is $B+C=-2(r^3)$. They are congruent to $0$, $1$ and $1$ mod $2$, and the last 2 of them can't be equal since $s$ is not equal to $t$. Since each of them is at most $2d^{1/3}$, this contradicts the minimality assumption.
This is really a 3-descent argument on an elliptic curve, but the fancy language as you see isn't needed.An almost identical argument gives what I think is the nicest proof of FLT for exponent 3.
-
Euler's proof seems to be a 2-descent rather than the 3-descent given above. I think the 3-descent argument is more elegant, and that undergrads will like it better. (I have used it in a number theory course). And I like introducing the Gaussian integers and Eisenstein integers towards the end of a course devoted to Z and Q as an easy introduction to algebraic number theory. – paul Monsky May 28 '11 at 16:08
There is a beautiful elementary way of solving the equation $$Ny^2-x^3=\pm1$$ when $N$ contains no prime factors that are of the form $6k+1$. There are only 7 solutions $(x,y,N)$ $$\{(2,3,1)(1,1,2)(23,78,2)(23,39,8)(2,1,9)(23,26,18)(23,18,72)\}$$ This was proven by J.H.E. Cohn in the article "The diophantine equations $x^3=Ny^2\pm 1$"
-
Alternatively you can reduce your equation to the cubic pell equation $x^3-2y^3=\pm 1$ and use $\mathbb Z [\omega]$, getting a solution similar to FLT for the exponent 3. A reference is for example E. Barbeau's "Pell equation". – Gjergji Zaimi Sep 22 '10 at 5:54
So the paper of J.H.E. Cohn contains solution to the Catalan Conjecture? – Sergei Jun 5 '15 at 5:32
Only the second proof below is new. In 1., I explain where the units in the cubic field are coming from, and 3. is a variation of Paul Monsky's proof.
1. We try working in ${\mathbb Z}$ for as long as possible. Write $x^3 = y^2 - 1 = (y-1)(y+1)$; since the gcd of the factors on the right hand side divides $2$, there are two possibilities:
a) $y$ is even; then $y+1 = \pm a^3$ and $y - 1 = \pm b^3$. Subtracting these equations gives $2 = a^3 - b^3 = (a-b)(a^2+ab+b^2)$, hence $a-b$ divides $2$. Going through all cases gives $(x,y) = (-1,0)$.
b) $y$ is odd; then $y+1 = 2a^3$ and $y - 1 = 4b^3$. Subtracting these equations gives $1 = a^3 - 2b^3$.
It remains to solve the last equation. One possibility is showing that the unit $1 - \sqrt[3]{2}$ is fundamental in the cubic number field ${\mathbb Q}(\sqrt[3]{2})$, and that its only powers of the form $a+b\sqrt[3]{2}$ have exponent $0$ or $1$.
The other possibility is observing that this equation is a special case of the twisted Fermat cubic $x^3 + y^3 = 2z^3$. But Paul Monsky's proof shows directly that solving this equation is all that is needed.
2. A more or less standard proof (this is basically a classical $2$-descent, but perhaps more direct than Euler's approach, which avoids number fields altogether - it is well known that on curves with a rational point of order $2$, a simple 2-decent can be performed by working within the rationals) proceeds as follows: write the equation in the form $$y^2 = (x+1)(x+\rho)(x+\rho^2),$$ where $\rho$ is a primitive cube root of unity. The gcd of two factors divides $1-\rho$, hence there are two possibilities:
a) $x+1 = \pm a^2$, $x + \rho = (-\rho)^e (a+b\rho)^2$, $x + \rho^2 = (-\rho^2)^e (a+b\rho^2)^2$. Since $\rho = (1+\rho)^2$ is a square, we can subsume the powers of $\rho$ into the square and find $x + \rho = \pm (a+b\rho)^2$ and $x + \rho^2 = \pm (a+b\rho^2)^2$. Subtracting these equations and dividing through by $\rho - \rho^2$ gives $1 = \pm b(2a-b)$, leading to $(x,y) = (0,\pm 1)$ and $(-1,0)$.
b) Here we find $$x+1 = \ \pm 3a^2, \quad x + \rho = \ \pm (1-\rho) (a+b\rho)^2, \quad x + \rho^2 = \ \pm (1-\rho^2) (a+b\rho^2)^2.$$ Adding the last two equations gives $2x-1 = \pm 3(a^2 - b^2)$. Eliminating $x$ from this and the first equation (where we actually have $x = 3a^2$ since $x \ge -1$) yields $(x,y) = (2,\pm 3)$.
3. Let me give here my rendition of Paul's beautiful proof:
Let $\alpha, \beta, \gamma \in {\mathbb Z}[\rho] \setminus \{0\}$. If $\alpha + \beta + \gamma = 0$ and $\alpha\beta\gamma = 2\mu^3$, then after a suitable permutation of the three numbers we have $\alpha = 0$ or $\beta = \gamma$.
Proof. Let $(\alpha,\beta,\gamma)$ be a counterexample. Then $\alpha$, $\beta$ and $\gamma$ are pairwise coprime in ${\mathbb Z}[\rho]$, and after a suitable permutation we have $$\alpha = 2 \rho^a A_1^3, \quad \beta = \rho^b B_1^3, \quad \gamma = \rho^c C_1^3.$$ Among all such counterexamples we now take one in which $N\alpha$ is minimal. Dividing all three numbers by $\rho^a$ we may assume that $a = 0$.
Cubes in ${\mathbb Z}[\rho]$ are $\equiv 0, 1 \bmod 2$ by Fermat's Little Theorem. Thus $0 \equiv \alpha \equiv \beta + \gamma \equiv \rho^b + \rho^c \bmod 2$, which implies $b = c$. Since $\alpha\beta\gamma$ is a cube, we must have $a = b = c = 0$.
Thus $$\alpha = 2 A_1^3, \quad \beta = B_1^3, \quad \gamma = C_1^3.$$ Since $B_1^3 \equiv C_1^3 \equiv 1 \bmod 2$, we may assume that $B_1 \equiv C_1 \equiv 1 \bmod 2$ (after multiplying these numbers through by a suitable power of $\rho$).
Now set $\alpha_1 = B_1 + C_1$, $\beta_1 = \rho B_1 + \rho^2 C_1$ and $\gamma_1 = \rho^2 B_1 + \rho C_1$. Then
• $\alpha_1 + \beta_1 + \gamma_1 = B_1(1+\rho+\rho^2) + C_1(1+\rho+\rho^2) = 0$.
• $\alpha_1 \beta_1 \gamma_1 = B_1^3 + C_1^3 = \beta + \gamma = -\alpha = 2(-A_1)^3$.
• $\beta_1 + \gamma_1 = (B_1 + C_1)(\rho+\rho^2) = - (B_1 + C_1) \ne 0$ since $\beta + \gamma = -\alpha \ne 0$.
• $N(\alpha_1 \beta_1 \gamma_1) = N\alpha \mid N(\alpha\beta\gamma)$; if we had equality, it would follow that $N(\beta) = N(\gamma) = 1$, hence $\beta, \gamma = \pm 1$. But then $\beta = 1$, $\gamma = -1$ and $\alpha = 0$ against our assumptions.
Now descent finishes the proof: $(\alpha_1, \beta_1, \gamma_1)$ is another solution with $N(\alpha_1 \beta_1 \gamma_1) < N(\alpha\beta\gamma)$,
As a final remark I would like to point out that the article by Cohn in Gjergji Zaimi's answer uses "well known results" such as the solution of $x^4 - 2y^2 = \pm 1$ and $x^4 - 3y^2 = 1$. I do not know offhand how elementary the corresponding proofs are.
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Well, $x^4-1=py^2$ implies one of $x^2\pm 1$ is a square, so the only non-trivial equation is $x^4+1=2y^2$. An elementary solution of the latter is not hard using the parametrization of the primitive solutions of $a^2+b^2=2c^2$. – Gjergji Zaimi Jul 13 '11 at 18:59
Franz, thank you for writing out these comments. If I could accept two answers, I would do so. – Pace Nielsen Jul 20 '11 at 19:53
More generally, $x^2-y^3=k$ is called the Mordell equation. For some values of $k$, you'll find elementary solutions in Mordell's Diophantine Equations, and in Uspensky and Heaslett's textbook.
EDIT: I've had a look at Mordell's book. He refers to his paper, The infinity of rational solutions of $y^2=x^3+k$, J London Math Soc 41 (1966) 523-525.
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2016-06-24 22:18:20
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https://ftp.aimsciences.org/journal/1531-3492/2020/25/4
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American Institute of Mathematical Sciences
ISSN:
1531-3492
eISSN:
1553-524X
All Issues
Discrete & Continuous Dynamical Systems - B
April 2020 , Volume 25 , Issue 4
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2020, 25(4): 1193-1212 doi: 10.3934/dcdsb.2019216 +[Abstract](1848) +[HTML](408) +[PDF](415.14KB)
Abstract:
In this paper, we introduce and study a new class of fractional differential hemivariational inequalities ((FDHVIs), for short) formulated by an initial-value fractional evolution inclusion and a hemivariational inequality in infinite Banach spaces. First, by applying measure of noncompactness, a fixed point theorem of a condensing multivalued map, we obtain the nonemptiness and compactness of the mild solution set for (FDHVIs). Further, we apply the obtained results to establish an existence theorem of the mild solution of a global attractor for the semiflow governed by a fractional differential hemivariational inequality ((FDHVI), for short). Finally, we provide an example to demonstrate the main results.
2020, 25(4): 1213-1240 doi: 10.3934/dcdsb.2019217 +[Abstract](1337) +[HTML](279) +[PDF](436.7KB)
Abstract:
In this paper we establish the nonlinear orbital instability of ground state standing waves for a Benney-Roskes/Zakharov-Rubenchik system that models the interaction of low amplitude high frequency waves, acustic type waves in \begin{document}$N = 2$\end{document} and \begin{document}$N = 3$\end{document} spatial directions. For \begin{document}$N = 2$\end{document}, we follow M. Weinstein's approach used in the case of the Schrödinger equation, by establishing a virial identity that relates the second variation of a momentum type functional with the energy (Hamiltonian) on a class of solutions for the Benney-Roskes/Zakharov-Rubenchik system. From this identity, it is possible to show that solutions for the Benney-Roskes/Zakharov-Rubenchik system blow up in finite time, in the case that the energy (Hamiltonian) of the initial data is negative, indicating a possible blow-up result for non radial solutions to the Zakharov equations. For \begin{document}$N = 3$\end{document}, we establish the instability by using a scaling argument and the existence of invariant regions under the flow due to a concavity argument.
2020, 25(4): 1241-1255 doi: 10.3934/dcdsb.2019218 +[Abstract](1782) +[HTML](267) +[PDF](405.64KB)
Abstract:
We study the existence of the uniform global attractor for a family of infinite dimensional first order non-autonomous lattice dynamical systems of the following form:
with initial data
The nonlinear part of the system \begin{document}$f\left( u,t\right)$\end{document} presents the main difficultly of this work. To overcome this difficulty we introduce a suitable Banach space \begin{document}$W$\end{document} of functions satisfying (3)-(7) with norm (8) such that \begin{document}$f_{0}\left( \cdot ,t\right)$\end{document} is an almost periodic function of \begin{document}$t$\end{document} with values in \begin{document}$W$\end{document} and \begin{document}$\left( g,f\right) \in \mathcal{H}\left( \left( g_{0},f_{0}\right) \right)$\end{document}.
2020, 25(4): 1257-1277 doi: 10.3934/dcdsb.2019219 +[Abstract](2313) +[HTML](330) +[PDF](952.88KB)
Abstract:
We derive characteristic functions to determine the number and stability of relaxation oscillations for a class of planar systems. Applying our criterion, we give conditions under which the chemostat predator-prey system has a globally orbitally asymptotically stable limit cycle. Also we demonstrate that a prescribed number of relaxation oscillations can be constructed by varying the perturbation for an epidemic model studied by Li et al. [SIAM J. Appl. Math, 2016].
2020, 25(4): 1279-1298 doi: 10.3934/dcdsb.2019220 +[Abstract](1948) +[HTML](244) +[PDF](382.96KB)
Abstract:
In this work, we consider the regularity property of stochastic convolutions for a class of abstract linear stochastic retarded functional differential equations with unbounded operator coefficients. We first establish some useful estimates on fundamental solutions which are time delay versions of those on \begin{document}$C_0$\end{document}-semigroups. To this end, we develop a scheme of constructing the resolvent operators for the integrodifferential equations of Volterra type since the equation under investigation is of this type in each subinterval describing the segment of its solution. Then we apply these estimates to stochastic convolutions of our equations to obtain the desired regularity property.
2020, 25(4): 1299-1316 doi: 10.3934/dcdsb.2019221 +[Abstract](1451) +[HTML](281) +[PDF](352.5KB)
Abstract:
We investigate the long-time behavior of solutions for the suspension bridge equation when the forcing term containing some hereditary characteristic. Existence of pullback attractor is shown by using the contractive function methods.
2020, 25(4): 1317-1344 doi: 10.3934/dcdsb.2019229 +[Abstract](1228) +[HTML](153) +[PDF](423.36KB)
Abstract:
We study a global well-posedness and asymptotic dynamics of measure-valued solutions to the kinetic Winfree equation. For this, we introduce a second-order extension of the first-order Winfree model on an extended phase-frequency space. We present the uniform(-in-time) \begin{document}$\ell_p$\end{document}-stability estimate with respect to initial data and the equivalence relation between the original Winfree model and its second-order extension. For this extended model, we present uniform-in-time mean-field limit and large-time behavior of measure-valued solution for the second-order Winfree model. Using stability and asymptotic estimates for the extended model and the equivalence relation, we recover the uniform mean-field limit and large-time asymptotics for the Winfree model. 200 words.
2020, 25(4): 1345-1360 doi: 10.3934/dcdsb.2019230 +[Abstract](1206) +[HTML](147) +[PDF](330.84KB)
Abstract:
In this paper, by using the Gal\begin{document}${\rm\ddot{e}}$\end{document}rkin method and energy estimates, the global weak solution and the smooth solution to the generalized Landau-Lifshitz-Bloch (GLLB) equation in high dimensions are obtained.
2020, 25(4): 1361-1382 doi: 10.3934/dcdsb.2019231 +[Abstract](1507) +[HTML](172) +[PDF](583.68KB)
Abstract:
We study the ergodicity of non-autonomous discrete dynamical systems with non-uniform expansion. As an application we get that any uniformly expanding finitely generated semigroup action of \begin{document}$C^{1+\alpha}$\end{document} local diffeomorphisms of a compact manifold is ergodic with respect to the Lebesgue measure. Moreover, we will also prove that every exact non-uniform expandable finitely generated semigroup action of conformal \begin{document}$C^{1+\alpha}$\end{document} local diffeomorphisms of a compact manifold is Lebesgue ergodic.
2020, 25(4): 1383-1395 doi: 10.3934/dcdsb.2019232 +[Abstract](1313) +[HTML](149) +[PDF](2241.96KB)
Abstract:
In this paper, we study the existence of periodic solutions for a class of differential-algebraic equation
where \begin{document}$h(t, x) = A(t)x(t)$\end{document}, \begin{document}$h(t, x)$\end{document} and \begin{document}$f(t, x)$\end{document} are \begin{document}$T$\end{document}-periodic in first variable. Via the topological degree theory, and the method of guiding functions, some existence theorems are presented. To our knowledge, this is the first approach to periodic solutions of differential-algebraic equations. Some examples and numerical simulations are given to illustrate our results.
2020, 25(4): 1397-1414 doi: 10.3934/dcdsb.2019233 +[Abstract](1192) +[HTML](194) +[PDF](1176.89KB)
Abstract:
One of the simplest model of immune surveillance and neoplasia was proposed by Delisi and Resigno [6]. Later Liu et al [10] proved the existence of non-degenerate Takens-Bogdanov (BT) bifurcations defining a surface in the whole set of five positive parameters. In this paper we prove the existence of Bautin bifurcations completing the scenario of possible codimension two bifurcations that occur in this model. We give an interpretation of our results in terms of the Immuno Edition Theory (IET) of three phases: elimination, equilibrium and escape.
2020, 25(4): 1415-1437 doi: 10.3934/dcdsb.2019234 +[Abstract](1172) +[HTML](172) +[PDF](474.81KB)
Abstract:
We present a model for the dynamics of elastic or poroelastic bodies with monopolar repulsive long-range (electrostatic) interactions at large strains. Our model respects (only) locally the non-self-interpenetration condition but can cope with possible global self-interpenetration, yielding thus a certain justification of most of engineering calculations which ignore these effects in the analysis of elastic structures. These models necessarily combines Lagrangian (material) description with Eulerian (actual) evolving configuration evolving in time. Dynamical problems are studied by adopting the concept of nonlocal nonsimple materials, applying the change of variables formula for Lipschitz-continuous mappings, and relying on a positivity of determinant of deformation gradient thanks to a result by Healey and Krömer.
2020, 25(4): 1439-1467 doi: 10.3934/dcdsb.2019235 +[Abstract](1708) +[HTML](149) +[PDF](694.29KB)
Abstract:
By employing Leray-Schauder alternative theorem of set-valued maps and non-Lyapunov method (non-smooth analysis, inequality analysis, matrix theory), this paper investigates the problems of periodicity and stabilization for time-delayed Filippov system with perturbation. Several criteria are obtained to ensure the existence of periodic solution of time-delayed Filippov system by using differential inclusion. By designing appropriate switching state-feedback controller, the asymptotic stabilization and exponential stabilization control of Filippov system are realized. Applying these criteria and control design method to a class of time-delayed neural networks with perturbation and discontinuous activation functions under a periodic environment. The developed theorems improve the existing results and their effectiveness are demonstrated by numerical example.
2020, 25(4): 1469-1495 doi: 10.3934/dcdsb.2019236 +[Abstract](1573) +[HTML](157) +[PDF](432.69KB)
Abstract:
This paper is concerned with the existence and asymptotic behavior of traveling wave solutions for a nonlocal dispersal vaccination model with general incidence. We first apply the Schauder's fixed point theorem to prove the existence of traveling wave solutions when the wave speed is greater than a critical speed \begin{document}$c^*$\end{document}. Then we investigate the boundary asymptotic behaviour of traveling wave solutions at \begin{document}$+\infty$\end{document} by using an appropriate Lyapunov function. Applying the method of two-sided Laplace transform, we further prove the non-existence of traveling wave solutions when the wave speed is smaller than \begin{document}$c^*$\end{document}. From our work, one can see that the diffusion rate and nonlocal dispersal distance of the infected individuals can increase the critical speed \begin{document}$c^*$\end{document}, while vaccination reduces the critical speed \begin{document}$c^*$\end{document}. In addition, two specific examples are provided to verify the validity of our theoretical results, which cover and improve some known results.
2020, 25(4): 1497-1515 doi: 10.3934/dcdsb.2019237 +[Abstract](1627) +[HTML](226) +[PDF](4277.85KB)
Abstract:
This article discusses the spread of rumors in heterogeneous networks. Using the probability generating function method and the approximation theory, we establish an SICR rumor model and calculate the threshold conditions for the outbreak of the rumor. We also compare the speed of the rumors spreading with different initial conditions. The numerical simulations of the SICR model in this paper fit well with the stochastic simulations, which means that the model is reliable. Moreover the effects of the parameters in the model on the transmission of rumors are studied numerically.
2020, 25(4): 1517-1541 doi: 10.3934/dcdsb.2019238 +[Abstract](1427) +[HTML](170) +[PDF](412.26KB)
Abstract:
Our purpose is to formulate an abstract result, motivated by the recent paper [8], allowing to treat the solutions of critical and super-critical equations as limits of solutions to their regularizations. In both cases we are improving the viscosity, making it stronger, solving the obtained regularizations with the use of Dan Henry's technique, then passing to the limit in the improved viscosity term to get a solution of the limit problem. While in case of the critical problems we will just consider a 'bit higher' fractional power of the viscosity term, for super-critical problems we need to use a version of the 'vanishing viscosity technique' that comes back to the considerations of E. Hopf, O.A. Oleinik, P.D. Lax and J.-L. Lions from 1950th. In both cases, the key to that method are the uniform with respect to the parameter estimates of the approximating solutions. The abstract result is illustrated with the Navier-Stokes equation in space dimensions 2 to 4, and with the 2-D quasi-geostrophic equation. Various technical estimates related to that problems and their fractional generalizations are also presented in the paper.
2020, 25(4): 1543-1563 doi: 10.3934/dcdsb.2019239 +[Abstract](1196) +[HTML](159) +[PDF](429.1KB)
Abstract:
Efficient Legendre dual-Petrov-Galerkin methods for solving odd-order differential equations are proposed. Some Sobolev bi-orthogonal basis functions are constructed which lead to the diagonalization of discrete systems. Accordingly, both the exact solutions and the approximate solutions can be represented as infinite and truncated Fourier-like series. Numerical results indicate that the suggested methods are extremely accurate and efficient, and suitable for the odd-order equations.
2020, 25(4): 1565-1581 doi: 10.3934/dcdsb.2019240 +[Abstract](1368) +[HTML](143) +[PDF](349.13KB)
Abstract:
This work provides a finite dimensional reducing and a smooth approximating for a class of stochastic partial differential equations with an additive white noise. Using the invariant random cone to show the asymptotical completion, this stochastic partial differential equation is reduced to a stochastic ordinary differential equation on a random invariant manifold. Furthermore, after deriving the finite dimensional reducing for another stochastic partial differential equation driven by a Wong-Zakai scheme via a smooth colored noise, it is proved that when the smooth colored noise tends to the white noise, the solution and the finite dimensional reducing of the approximate system converge pathwisely to those of the original system.
2020, 25(4): 1583-1606 doi: 10.3934/dcdsb.2019241 +[Abstract](1511) +[HTML](232) +[PDF](401.21KB)
Abstract:
This paper has two parts. In part Ⅰ, existence and uniqueness theorem is established for solutions of neutral stochastic differential equations with variable delays driven by \begin{document}$G$\end{document}-Brownian motion (VNSDDEGs in short) under global Carathéodory conditions. In part Ⅱ, a simplified VNSDDEGs for the original one is proposed. And the convergence both in \begin{document}$L^p$\end{document}-sense and capacity between the solutions of the simplified and original VNSDDEGs are established in view of the approximation theorems. Two examples are conducted to justify the theoretical results of the approximation theorems.
2020, 25(4): 1607-1622 doi: 10.3934/dcdsb.2019242 +[Abstract](1662) +[HTML](207) +[PDF](1484.08KB)
Abstract:
In this paper, we extend our previous work on optimal control applied in an anthrax outbreak in wild animals. We use a system of ordinary differential equation (ODE) and partial differential equations (PDEs) to track the change in susceptible, infected and vaccinated animals as well as the infected carcasses. In addition to the assumption that the infected animals and the infected carcasses are the main source of infection, we consider the animal movement by diffusion and see its effects in disease transmission. Two controls: vaccinating susceptible animals and disposing infected carcasses properly are applied in the model and these controls depend on both space and time. We formulate an optimal control problem to investigate the effect of intervention strategies in our spatio-temporal model in controlling the outbreak at minimum cost. Finally some numerical results for the optimal control problem are presented.
2019 Impact Factor: 1.27
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2021-04-14 04:47:11
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http://www.physicsforums.com/showthread.php?p=4193330
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by michael1978
Tags: amplifier, transistor
P: 389 We need Re2 resistor simply because it is almost impassible to build amplifier with desired low voltage gain and Ve > Vbe. Which is needed for good bias point stability. But if you determined to use only Re1 resistor. We need to change DC bias method. So we are not going to use voltage divider, instead of voltage divider we will use a collector-feedback bias circuit. So for our example for Vcc = 10V and Rc = 1K;RL = 10K; Ic = 5mA and Av = 50. re = 26mV/Ic = 5.2Ω Re1 = (Rc||RL)/Av - re = 909Ω/50 - 5.2 = 12Ω Vc = Vcc - Ic*Rc = 5V Vb = Ic*Re1 + Vbe = 0.71V And Ib = Ic/hfe = 5mA/150 = 33.4μA Rb1+Rb2 = (Vc - Vb)/Ib = 4.29V/33.4μA = 128KΩ Rb1 = Rb2 = 128K/2 = 68KΩ And simulation show that voltage gain is equal to AV = 51.8092[V/V] Attached Thumbnails
P: 133
Quote by Jony130 We need Re2 resistor simply because it is almost impassible to build amplifier with desired low voltage gain and Ve > Vbe. Which is needed for good bias point stability. But if you determined to use only Re1 resistor. We need to change DC bias method. So we are not going to use voltage divider, instead of voltage divider we will use a collector-feedback bias circuit. So for our example for Vcc = 10V and Rc = 1K;RL = 10K; Ic = 5mA and Av = 50. re = 26mV/Ic = 5.2Ω Re1 = (Rc||RL)/Av - re = 909Ω/50 - 5.2 = 12Ω Vc = Vcc - Ic*Rc = 5V Vb = Ic*Re1 + Vbe = 0.71V And Ib = Ic/hfe = 5mA/150 = 33.4μA Rb1+Rb2 = (Vc - Vb)/Ib = 4.29V/33.4μA = 128KΩ Rb1 = Rb2 = 128K/2 = 68KΩ And simulation show that voltage gain is equal to AV = 51.8092[V/V]
JONEY thank you for reply i am learning a lot from you, you are the beste;-) COMPLIMENT
P: 133
Quote by yungman So you know it is 1.7V at the base and 1V at the emitter. YES, you set Ie with the emitter resistor ( Re1+Re2). This is the KEY of DC biasing. You first set up the operating current Ie of the transistor, this is the first and foremost thing. Collector don't set the current of the BJT, collector current is only follow the emitter current. AGAIN, this is in Malvino. 1) In the diagram, Re1+Re2 is about 200 ohm. You put 1V at emitter, you put 1V across the Re1 and Re2. SO you force 5mA through the resistors......Which, means you set the Ie to 5mA. This is how you set up the DC bias current. 2) After setting the Ie, then you start looking at the collector. You know β is high, so you know the Ic≈5mA. If you have Vcc=10V, and you have 1K resistor at the collector, then you know 5mA through 1K is 5V. So the collector is at +5V. 3) AGAIN, gain is determined by the impedance at the collector divided by the impedance at the emitter. We gone over this over and over and over already. Read the old posts AGAIN. These are all in Malvino. Please read that before you ask any more question. If you don't get it, read it, write out the numbers, work through the problems before you post.
did you see the post 146 of Joney, maybe i am mistake but malvino he dont show you how to get desired voltage, i wil like to ask you why Jony, he put so big resistance of voltage divider i say for r1 22K and for R2 4.7k how he calculate,? me i put it R2=170, and R1=830, look how big difference, he show one example but me i put in other way,
P: 3,843
Quote by michael1978 did you see the post 146 of Joney, maybe i am mistake but he dont show you how to get desired voltage, i wil like to ask you why Jony, he put so big resistance of voltage divider i say for r1 22K and for R2 4.7k how he calculate,? me i put it R2=170, and R1=830, look how big difference, he show one example but me i put in other way,
We were talking about the diagram of post 114 all along until now.
The newer diagram is not as desirable. I believe it called self bias or something. It depend a lot on the beta of the transistor. Voltage divider bias is a better way to go.
P: 133
Quote by yungman We were talking about the diagram of post 114 all along until now.
yes but i say to joney i gonna make one amplifier, and he help answer me thanks from him, ok why Joney select so big resistance of R1 22k and R2 4.7k
and he get the sam base almost 1.7V, how i have to do it big resistance like joney
P: 3,843
Quote by michael1978 yes but i say to joney i gonna make one amplifier, and he help answer me thanks from him, ok why Joney select so big resistance of R1 22k and R2 4.7k and he get the sam base almost 1.7V, how i have to do it big resistance like joney
You talking about post 114? The resistance is 22K//4.7K ( if I ignore the input resistance of the transistor). In fact the input resistance is quite low to me.
Usually you want the input resistance to be a little on the high side so the stage driving this input don't have to drive a low impedance. Think if you have a stage like in post 114, if the collector is driving the following stage with low impedance like 1K. The collector resistance is 1K//1K=500 ohm. You lower the gain of the driving stage.
In both diagrams, the input resistance of the transistor is β(r'e + Re1).=β(20Ω)≈2K assuming β=100. This is quite low. Another way to look at it is the transistor in both case require 5mA/β= 50uA. The impedance of the input biasing network has to be low enough so it is not loaded down by the input of the transistor. Jony got into this a little in post 79 about the base current.
P: 133
Quote by yungman You talking about post 114? The resistance is 22K//4.7K ( if I ignore the input resistance of the transistor). In fact the input resistance is quite low to me. Usually you want the input resistance to be a little on the high side so the stage driving this input don't have to drive a low impedance. Think if you have a stage like in post 114, if the collector is driving the following stage with low impedance like 1K. The collector resistance is 1K//1K=500 ohm. You lower the gain of the driving stage. In both diagrams, the input resistance of the transistor is β(r'e + Re1).=β(20Ω)≈2K assuming β=100. This is quite low. Another way to look at it is the transistor in both case require 5mA/β= 50uA. The impedance of the input biasing network has to be low enough so it is not loaded down by the input of the transistor. Jony got into this a little in post 79 about the base current.
CAN I ASK YOU SOMETHING about impedance, my english is not so good, how you select input impedance, not to say in the end the input impedance was 50k of 10k etc, you select in the begin of how
P: 3,843 For normal frequency ( not RF), you usually want it to be as high as practical. It is limited by the input requirement of the amplifier.........Like in your case, the transistor Ib of 50uA and transistor input impedance. Jony in post 79 explained this and I explained about the input impedance in the post 150. In your case, with drawing in both post 114 and 146, the emitter resistor Re1 is eithe 14 or 12Ω, that is very low. That is really the gating factor of the circuit. Also you are trying to get gain of 50 out of one transistor, this force you to have such a low Re1 in order to be able to use a collector resistor of 1K. In real life design, if I want to have a gain of 50, I would do either one below: 1) Using Re1=150, Rc=10K, Ic=0.5mA, then use an emitter follower transistor to buffer the output. 2) Use two stage of this and divide the gain between the two stages. This is the prefer way. 3) Use JFET instead of BJT.
P: 133
Quote by yungman For normal frequency ( not RF), you usually want it to be as high as practical. It is limited by the input requirement of the amplifier.........Like in your case, the transistor Ib of 50uA and transistor input impedance. Jony in post 79 explained this and I explained about the input impedance in the post 150. In your case, with drawing in both post 114 and 146, the emitter resistor Re1 is eithe 14 or 12Ω, that is very low. That is really the gating factor of the circuit. Also you are trying to get gain of 50 out of one transistor, this force you to have such a low Re1 in order to be able to use a collector resistor of 1K. In real life design, if I want to have a gain of 50, I would do either one below: 1) Using Re1=150, Rc=10K, Ic=0.5mA, then use an emitter follower transistor to buffer the output. 2) Use two stage of this and divide the gain between the two stages. This is the prefer way. 3) Use JFET instead of BJT.
thank you for answer, and R1, R2? but first i am bussy with transistor, is difficult jfet to learn, and for the book, you can buy it chip used boooks in amazone, because i want to open one visa of mastercard
P: 3,843
Quote by michael1978 thank you for answer, and R1, R2? but first i am bussy with transistor, is difficult jfet to learn, and for the book, you can buy it chip used boooks in amazone, because i want to open one visa of mastercard
When you use a credit card to open an account with Amazon, make sure don't let your computer remember your password, and Amazon ask you whether you want to keep logging in even you close the page......DON'T. Always logout when you close the page and don't let the computer remember your pass word.
P: 133
Quote by yungman When you use a credit card to open an account with Amazon, make sure don't let your computer remember your password, and Amazon ask you whether you want to keep logging in even you close the page......DON'T. Always logout when you close the page and don't let the computer remember your pass word.
yeh i know but there are a lot of hackers today, i have to open only with 200E if it is possible, because i think they will not open with low credit
P: 133
Quote by michael1978 JONEY thank you for reply i am learning a lot from you, you are the beste;-) COMPLIMENT
HI Jony how are you, may i ask you something, the schematic of designin of amplifier , you make it with light spice of some other software , because you told me i use light spice,
greetings
P: 3,843
Quote by michael1978 yeh i know but there are a lot of hackers today, i have to open only with 200E if it is possible, because i think they will not open with low credit
Well, there is a balance of being safe and missing out. Write to Amazon and ask what is the best way. It is their utmost interest to keep it safe as their business depends on this too.
P: 133
Quote by yungman Well, there is a balance of being safe and missing out. Write to Amazon and ask what is the best way. It is their utmost interest to keep it safe as their business depends on this too.
thnx for reply, yes i know, but i will go first to my bank to ask
P: 3,843 The bank cannot help much, call the credit card company like VISA also. Ask Amazon, they should be able to help you. I have very good experience with Amazon, I have been ordering from them for years, never have a single problem. Their service is second to none......absolutely none. One time I order a tiller.....over 100lbs gas tiller for turning the ground, There was a small crack on some non critical area, I asked for exchange, they shipped me a second one immediately and arranged to pickup the first one. The kicker was I decided to keep the original, they said "no problem", they just arranged to pickup the second one!!!! I buy computers, guitar pickups, protein powders, gifts.....I just paid $1600 to buy a new Nikon camera and Tamron lens from them just a month ago. I even paid a little more just to buy on Amazon because of their service. Oh Yeh, thousands of dollars of text books used from Amazon also. When they say is good condition, they ARE in good condition. Half the time, they look new. One thing about studying, you need books, lots of books. I buy 7 to 8 text books on each subject to compare as no book is good in all topics, each has their strong and weak points. You need a few to get the whole picture. Malvino is one exceptional one. Can you imagine if I buy my two tall book shelves of new books?!!! That would have been very expensive. I just bought the Malvino 6th edition for$8 shipped!!!!
P: 133
Quote by yungman The bank cannot help much, call the credit card company like VISA also. Ask Amazon, they should be able to help you. I have very good experience with Amazon, I have been ordering from them for years, never have a single problem. Their service is second to none......absolutely none. One time I order a tiller.....over 100lbs gas tiller for turning the ground, There was a small crack on some non critical area, I asked for exchange, they shipped me a second one immediately and arranged to pickup the first one. The kicker was I decided to keep the original, they said "no problem", they just arranged to pickup the second one!!!! I buy computers, guitar pickups, protein powders, gifts.....I just paid $1600 to buy a new Nikon camera and Tamron lens from them just a month ago. I even paid a little more just to buy on Amazon because of their service. Oh Yeh, thousands of dollars of text books used from Amazon also. When they say is good condition, they ARE in good condition. Half the time, they look new. One thing about studying, you need books, lots of books. I buy 7 to 8 text books on each subject to compare as no book is good in all topics, each has their strong and weak points. You need a few to get the whole picture. Malvino is one exceptional one. Can you imagine if I buy my two tall book shelves of new books?!!! That would have been very expensive. I just bought the Malvino 6th edition for$8 shipped!!!!
i will try, so chip you buy malvino, and do you have 7 edition, if you have 7 edition, wich is better, 6 of 7 edition,
P: 3,843 I have not received it yet. I must had the first edition in the late 70s!!! I long lost the book. The book is in my head and been using what I learned to do a lot of designs all these years. I just want to keep a copy for my collection, not that I need it. I since learn a whole lot more about transistors and op-amps. It is a very good introduction book for electronics.
P: 133
Quote by yungman I have not received it yet. I must had the first edition in the late 70s!!! I long lost the book. The book is in my head and been using what I learned to do a lot of designs all these years. I just want to keep a copy for my collection, not that I need it. I since learn a whole lot more about transistors and op-amps. It is a very good introduction book for electronics.
you can find it in internet
Related Discussions Electrical Engineering 2 Electrical Engineering 12 Atomic, Solid State, Comp. Physics 9 Engineering, Comp Sci, & Technology Homework 2 Introductory Physics Homework 13
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2014-04-20 14:06:43
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https://www.researching.cn/
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Search by keywords or author
The image on the cover for Chinese Optics Letters Volume 19, Issue 11, summarizes the PAI techniques, imaging systems, and their biomedical applications in microrobots tracking in vitro and in vivo. From a robotic tracking perspective, some insight into the future of PAI technology in clinical applications is also provided.The image is based on original research by Dengfeng Li et al. presented in their paper "Review of photoacoustic imaging for microrobots tracking in vivo [Invited]", Chinese Optics Letters 19 (11), 111701 (2021).
The image on the cover for Photonics Research Volume 9, Issue 11, proposes a device that could be used for both display and visible light communication (VLC) applications. The semipolar blue μLED array fabricated in this study shows a negligible wavelength shift, indicating a significant reduction in the quantum confined Stark effect. The image is based on original research by Tingzhu Wu et al. presented in their paper "Highly stable full-color display device with VLC application potential using semipolar μLEDs and all-inorganic encapsulated perovskite nanocrystal", Photonics Research 9 (11), 11002132 (2021).
The image on the cover for Advanced Photonics Volume 3 Issue 5 depicts a novel spintronic-metasurface terahertz emitter, developed by researchers from Fudan University. Composed of alternating magnetic heterostructures, the emitter allows efficient, flexible generation and manipulation of chiral terahertz waves. Terahertz radiation is generated by exciting the emitter with laser pulses under an oriented external magnetic field. Transverse anisotropic confinement of the laser-induced charge currents imposed by the metasurface structure leads to chiralterahertz-wave emission.The image is based on original research presented in the report by Changqin Liu, Shunjia Wang, Sheng Zhang, and Qingnan Cai, et al., “Active spintronic-metasurface terahertz emitters with tunable chirality,” Adv. Photonics 3(5), 056002 (2021), doi: 10.1117/1.AP.3.5.056002.
The image on the cover of High Power Laser Science and Engineering Volume 9, Issue 3, presents an all-optical scheme to generate a high-energy γ-photon beam with large beam angular momentum (BAM), small divergence, and high brilliance.The image is based on original research by Hao Zhang et al. presented in their paper "Efficient bright γ-ray vortex emission from a laser-illuminated light-fan-in-channel target", High Power Laser Science and Engineering 9 (3), 03000e43 (2021).
The image on the cover for Chinese Optics Letters Volume 19, Issue 10, reviews the important parameters including the refractive index detection range, resonance wavelength, and spectral sensitivity responsible for the sensing properties of PCF-SPR sensors.The image is based on original research by Chao Liu et al. presented in their paper "Overview of refractive index sensors comprising photonic crystal fibers based on the surface plasmon resonance effect [Invited]", Chinese Optics Letters 19 (10), 102202 (2021).
Spotlight on Optics
Parity-time (PT) symmetry, as a physical concept, describes the special and temporal symmetry of the physical quantity. Recent year, it has been discovered that PT symmetry can be easily observed in the photonics system. Exceptional point (EP) is a particular point in the PT symmetric system, at which the eigenvalues and eigenvectors degenerate. Plenty of counterintuitive phenomena have been observed around EP. Specially, owing to the non-Hermiticity induced nonadiabatic transitions, chiral mode switching was achieved by encircling the exceptional point (EEP) in a PT symmetric system.
Photonics Research
• Jan. 21, 2022
• Vol. 10, Issue 1 (2022)
Editors' Picks
Optical tweezers (OTs) provide a powerful tool for trapping, guiding, and assembly of biological nanoparticles and cells, thus playing a unique role in biomedical and photonics areas. Particularly, OTs have been widely used to explore the structure, mechanism, and interaction of cells due to their ability of non-contact and high-resolution manipulation.
Photonics Research
• Jan. 20, 2022
• Vol. 10, Issue 1 (2022)
Editors' Picks
Topological states of matter are one of the most important research fields in condensed physics and have been deeply studied over the last 10 years. Before the discovery of topological states, researchers believed that almost all states of matter could be characterized by symmetry or local order parameters, and the phase transition of matters was accompanied by symmetry breaking.
• Jan. 19, 2022
• Vol. 10, Issue 1 (2022)
AP Highlights
Real-time, in-situ sensing and tracking of cell development and maturation is achieved using a label-free nano-optical device
• Jan. 19, 2022
• Vol. 4, Issue 1 (2022)
News
Original manuscripts are sought to the special issue on "Future Control Systems and Machine Learning at High Power Laser Facilities" of High Power Laser Science and Engineering (HPL),
High Power Laser Science and Engineering
• Jan. 18, 2022
• Vol. , Issue (2022)
Ultra-power-efficient heterogeneous III–V/Si MOSCAP (de-)interleavers for DWDM optical links
We discuss the design and demonstration of various III–V/Si asymmetric Mach–Zehnder interferometer (AMZI) and ring-assisted AMZI (de-)interleavers operati
We discuss the design and demonstration of various III–V/Si asymmetric Mach–Zehnder interferometer (AMZI) and ring-assisted AMZI (de-)interleavers operating at O-band wavelengths with 65 GHz channel spacing. The wafer-bonded III–V/Si metal-oxide-semiconductor capacitor (MOSCAP) structure facilitates ultra-low-power phase tuning on a heterogeneous platform that allows for complete monolithic transceiver photonic integration. The second- and third-order MOSCAP AMZI (de-)interleavers exhibit cross-talk (XT) levels down to $-22$ dB and $-32 dB$ with tuning powers of 83.0 nW and 53.0 nW, respectively. The one-, two-, and three-ring-assisted MOSCAP AMZI (de-)interleavers have XT levels down to $-27$ dB, $-22$ dB, and $-20 dB$ for tuning powers of 10.0 nW, 7220.0 nW, and 33.6 nW, respectively. The leakage current density is measured to be in the range of $1.6–27 μA/cm2$. To the best of our knowledge, we have demonstrated for the first time, athermal III–V/Si MOSCAP (de-)interleavers with the lowest XT and reconfiguration power consumption on a silicon platform.show less
• Jan.22，2022
• Photonics Research,Vol. 10, Issue 2
• 02000A22 (2022)
Four-wave mixing in graphdiyne-microfiber based on synchronized dual-wavelength pulses
We demonstrate four-wave mixing (FWM) in the graphdiyne (GDY) microfiber based on the synchronized dual-wavelength pump pulses that are transformed from a
We demonstrate four-wave mixing (FWM) in the graphdiyne (GDY) microfiber based on the synchronized dual-wavelength pump pulses that are transformed from a mode-locked fiber laser. Benefiting from the large nonlinear refractive index of GDY and the synchronized pump pulses, a maximum conversion efficiency of -39.05 dB can be achieved in GDY with only an average pump power of 6.9 mW, greatly alleviating the possible damage compared to previous investigations employing the continuous-wave pump. In addition, our proposal can be applied to measure the effective nonlinear coefficient $γ$ of the GDY-microfiber, which could be extended as a practical measurement tool for $γ$ of nanomaterials-based devices.show less
• Jan.22，2022
• Photonics Research,Vol. 10, Issue 2
• 02000503 (2022)
Single-shot three-input phase retrieval for quantitative back focal plane measurement
This paper presents quantitative measurements facilitated with a new optical system that implements a single-shot three-input phase retrieval algorithm. T
This paper presents quantitative measurements facilitated with a new optical system that implements a single-shot three-input phase retrieval algorithm. The new system allows simultaneous acquisition of three distinct input patterns, thus eliminating the requirement for mechanical movement and reducing any registration errors and microphonics. We demonstrate the application of the system for measurement and separation of two distinct attenuation measurements of surface waves, namely, absorption and coupling loss. This is achieved by retrieving the phase in the back focal plane and performing a series of virtual optics computations. This overcomes the need to use a complicated series of hardware manipulations with a spatial light modulator. This gives a far more accurate and faster measurement with a simpler optical system. We also demonstrate that phase measurements allow us to implement different measurement methods to acquire the excitation angle for surface plasmons. Depending on the noise statistics different methods have superior performance, so the best method under particular conditions can be selected. Since the measurements are only weakly correlated, they may also be combined for improved noise performance. The results presented here offer a template for a wider class of measurements in the back focal plane including ellipsometry.show less
• Jan.22，2022
• Photonics Research,Vol. 10, Issue 2
• 02000491 (2022)
All-fiber spatiotemporal mode-locking lasers with large modal dispersion
It is a challenging problem to balance the modal walk-off (modal dispersion) between multiple transverse modes and chromatic dispersion in long step-index
It is a challenging problem to balance the modal walk-off (modal dispersion) between multiple transverse modes and chromatic dispersion in long step-index multimode fibers (MMFs). By properly designing the oscillator, we have overcome the difficulty and successfully obtained an all-fiber spatiotemporal mode-locked laser based on step-index MMFs with large modal dispersion for the first time, to our knowledge. Various proofs of spatiotemporal mode-locking (STML) such as spatial, spectral, and temporal properties, are measured and characterized. This laser works at a fundamental frequency of 28.7 MHz, and achieves a pulse laser with single pulse energy of 8 nJ, pulse width of 20.1 ps, and signal-to-noise ratio of $∼70 dB$. In addition, we observe a dynamic evolution of the transverse mode energy during the STML establishment process that has never been reported before.show less
• Jan.22，2022
• Photonics Research,Vol. 10, Issue 2
• 02000483 (2022)
Light field imaging has shown significance in research fields for its high-temporal-resolution 3D imaging ability. However, in scenes of light field imaging through scattering, such as
Light field imaging has shown significance in research fields for its high-temporal-resolution 3D imaging ability. However, in scenes of light field imaging through scattering, such as biological imaging in vivo and imaging in fog, the quality of 3D reconstruction will be severely reduced due to the scattering of the light field information. In this paper, we propose a deep learning-based method of scattering removal of light field imaging. In this method, a neural network, trained by simulation samples which are generated by light field imaging forward models with and without scattering, is utilized to remove the effect of scattering on light field captured experimentally. With the deblurred light field and the scattering-free forward model, 3D reconstruction with high resolution and high contrast can be realized. We demonstrate the proposed method by using it to realize high-quality 3D reconstruction through a single scattering layer experimentally.show less
• Jan.22，2022
• Chinese Optics Letters,Vol. 20, Issue 4
• (2022)
A single-frequency 1645 nm pulsed laser with frequency stability close to 100 kHz was demonstrated. The laser oscillator is injection-seeded by a single-frequency narrow linewidth Er:YA
A single-frequency 1645 nm pulsed laser with frequency stability close to 100 kHz was demonstrated. The laser oscillator is injection-seeded by a single-frequency narrow linewidth Er:YAG nonplanar ring oscillator and frequency stabilized by modified Pound–Drever–Hall method. The pulse repetition rate can be set from 100 to 500 Hz with the frequency stability from 82.72 kHz to 134.44 kHz and pulse energy from 9.84 mJ to 19.55 mJ. To our knowledge this is the best frequency stability of single-frequency pulsed laser with injection-seeded. show less
• Jan.22，2022
• Chinese Optics Letters,Vol. 20, Issue 4
• (2022)
We demonstrate a novel approach to achieve wavelength-tunable ultrashort pulses from an all-fiber mode-locked laser with a saturable absorber based on nonlinear Kerr beam clean-up effec
We demonstrate a novel approach to achieve wavelength-tunable ultrashort pulses from an all-fiber mode-locked laser with a saturable absorber based on nonlinear Kerr beam clean-up effect. This saturable absorber was formed by a single-mode fiber spliced to a graded-index multimode fiber, and its tunable band-pass filter effect is described by a numerical model. By adjusting the bending condition of graded-index multimode fiber, the laser could produce dissipative soliton pulses with their central wavelength tunable from1040 nm to 1063 nm. The pulse duration of output laser could be compressed externally to 791 fs, and the signal to noise ratio of its radio frequency spectrum was measured to be 75.5 dB. show less
• Jan.22，2022
• Chinese Optics Letters,Vol. 20, Issue 4
• (2022)
The linewidth of BaGa4Se7 optical parametric oscillator (BGSe OPO) was narrowed for the first time by inserting a Fabry-Perot (FP) etalon into an L-shaped cavity. When a 15 mm long BGSe
The linewidth of BaGa4Se7 optical parametric oscillator (BGSe OPO) was narrowed for the first time by inserting a Fabry-Perot (FP) etalon into an L-shaped cavity. When a 15 mm long BGSe(56.3°, 0°) was pumped by a 1064 nm laser, the peak wavelength was ~3529 nm and the linewidth was 4.53 nm (3.64 cm-1) under type I phase matching. After inserting a 350 µm thick FP etalon, the linewidth was decreased to 1.27-2.05 nm. When the tilt angle of the etalon was 2.34°, the linewidth was 2.05 nm (1.65 cm-1) and the peak wavelength was still ~3529 nm. When the tilt angle of the etalon was 3.90°, the peak wavelength was 3534.9 nm and the linewidth was 1.27 nm (1.02 cm-1), which was the narrowest linewidth of BGSe OPO, to the best of our knowledge. The beam quality was improved after inserting the FP etalon, too.show less
• Jan.22，2022
• Chinese Optics Letters,Vol. 20, Issue 4
• (2022)
Xi'an, ChinaAug 7-10, 2022
Hangzhou, ChinaJuly 14-16, 2022
The image on the cover for Chinese Optics Letters Volume 19, Issue 11, summarizes the PAI techniques, imaging systems, and their biomedical applications in microrobots tracking in vitro and in vivo. From a robotic tracking perspective, some insight into the future of PAI technology in clinical applications is also provided.The image is based on original research by Dengfeng Li et al. presented in their paper "Review of photoacoustic imaging for microrobots tracking in vivo [Invited]", Chinese Optics Letters 19 (11), 111701 (2021).
• Journal
• 19th Nov，2021
The image on the cover for Photonics Research Volume 9, Issue 11, proposes a device that could be used for both display and visible light communication (VLC) applications. The semipolar blue μLED array fabricated in this study shows a negligible wavelength shift, indicating a significant reduction in the quantum confined Stark effect. The image is based on original research by Tingzhu Wu et al. presented in their paper "Highly stable full-color display device with VLC application potential using semipolar μLEDs and all-inorganic encapsulated perovskite nanocrystal", Photonics Research 9 (11), 11002132 (2021).
• Journal
• 19th Nov，2021
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2022-01-21 21:10:39
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https://quark.phy.bnl.gov/seminars/oldseminars.html
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# Past Theoretical Physics Seminars at BNL
• ## Wednesday, October 31, 2018
2:30pm, Small Seminar Room
Unveiling New Physics Through Angular Distributions at the LHC
Rodolfo Capdevilla (Notre Dame)
HET Seminar
• ## Thursday, November 1, 2018
12:00pm, Room 2-160, Bldg. 510
DIS on "Nuclei" using holography
RIKEN Lunch Seminar
4:00pm, CFNS Seminar Room 2-38
TBA
Al Mueller (Columbia University)
CFNS Seminar
• ## Friday, November 2, 2018
2:00pm, CNFS Seminar Room 2-38
Diffractive Electron-Nucleus Scattering and Ancestry in Branching Random Walks
Al Mueller (Columbia)
Nuclear Theory / RIKEN seminar
• ## Friday, November 9, 2018
2:00pm, CNFS Seminar Room 2-38
TBA
Ajit Srivastava (Institute of Physics, Bhubaneswar)
Nuclear Theory/RIKEN seminar
• ## Wednesday, November 14, 2018
2:30pm, Small Seminar Room
TBA
Konstantinos Orginos (College of William and Mary)
HET Seminar
• ## Thursday, November 15, 2018
12:00pm, 2-160, Bldg. 510
Exclusive $\rho$ meson production in $eA$ collisions: collinear factorization and the CGC
Renaud Boussarie (Brookhaven National Laboratory)
We will focus on the theoretical description of exclusive ρ meson production in eA collisions, using a hybrid factorization scheme which involves Balitsky's shockwave description of the Color Glass Condensate in the t channel, and Distribution Amplitudes (DAs) in the s channel. We will first give a quick introduction to the shockwave framework and to collinear factorization up to twist 3 for DAs, then we will apply these framweworks to the production of a longitudinal meson at NLO accuracy, and to the production of a transverse meson at twist 3 accuracy. We will insist on the experimental applications, and on several theoretical questions raised by our results: the dilute BFKL limit at NLO for diffraction, and collinear factorization breaking at twist 3.
• ## Friday, November 16, 2018
2:00pm, CNFS Seminar Room 2-38
N/A
Dimitra Karabali (Lehman College CUNY)
Nuclear Theory/RIKEN
• ## Thursday, November 29, 2018
12:00pm, 2-160, Bldg. 510
TBA
Mario Mitter (Brookhaven National Laboratory)
• ## Friday, November 30, 2018
2:00pm, CNFS seminar room 2-38
TBA
Juan Rojo (VU University)
Nuclear Theory / RIKEN seminar
• ## Wednesday, December 5, 2018
2:30pm, YITP Stony Brook
TBA
Jiji Fan (Syracuse)
Joint YITP/HET Theory Seminar
• ## Thursday, December 6, 2018
12:00pm, Room 2-160, Bldg. 510
Proton decay
Jun-Sik Yoo (Stony Brook University)
RIKEN Lunch Seminar
12:00pm, 2-160, Bldg. 510
On QCD and its Phase Diagram from a Functional RG Perspective
Mario Mitter (BNL)
• ## Wednesday, December 12, 2018
2:30pm, YITP Stony Brook
TBA
TBA
Joint YITP/HET Theory Seminar
• ## Thursday, January 10, 2019
12:00pm, 2-160, Bldg. 510
A novel background subtraction method for jet studies in heavy ion collisions
Alba Soto Ontoso (BNL)
• ## Friday, January 18, 2019
2:00pm, CFNS seminar room 2-38
TBA
Nuclear Theory / RIKEN seminar
• ## Thursday, January 24, 2019
12:00pm, 1-224, Bldg. 510 (different from usual room)
In this talk, I will present a connection between two approaches of studying quarkonium dynamics inside quark-gluon plasma: the open quantum system formalism and the transport equation. I will discuss insights from the perspective of quantum information. I will show that under the weak coupling and Markovian approximations, the Lindblad equation turns to a Boltzmann transport equation after a Wigner transform is applied to the system density matrix. I will demonstrate how the separation of physical scales justifies the approximations, by using effective field theory of QCD. Finally, I will show some phenomenological results based on the derived transport equation.
Xiaojun Yao (Duke University)
In this talk, I will present a connection between two approaches of studying quarkonium dynamics inside quark-gluon plasma: the open quantum system formalism and the transport equation. I will discuss insights from the perspective of quantum information. I will show that under the weak coupling and Markovian approximations, the Lindblad equation turns to a Boltzmann transport equation after a Wigner transform is applied to the system density matrix. I will demonstrate how the separation of physical scales justifies the approximations, by using effective field theory of QCD. Finally, I will show some phenomenological results based on the derived transport equation.
• ## Friday, January 25, 2019
2:00pm, CNFS Seminar Room 2-38
TBA
Paolo Glorioso (Kadanoff Center for Theoretical Physics and Enrico Fermi Institute, University of Chicago)
Nuclear Theory/RIKEN seminar
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2021-01-27 03:55:23
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https://math.stackexchange.com/tags/physics/hot
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# Tag Info
### geometrical/physical interpretation of multiplication of real numbers (including negative)
There are a couple of different ways to express multiplication using geometry. One somewhat physical interpretation of multiplication involves stretching/compressing/reflecting the number line. For ...
• 102k
### How much of the earth can see the moon?
Let $R=6371km$ be the radius of the earth $r= 1737km$ be the radius of the moon $d= 384467km$ be the distance between the earth and moon. Consider the figure below: $A$ is the centre of the earth. (...
• 545
1 vote
### geometrical/physical interpretation of multiplication of real numbers (including negative)
I believe the geometrical interpretation still work, if you take the signed area of the rectangle. So, let's define an "oriented" rectangle as a rectangle with a particular orientation of ...
• 16.6k
1 vote
### Solving $\ddot r=-\frac{GM}{r^2}$
To solve, begin by multiplying both sides by $r'$ (assuming $r'(0) \ne 0$). $$r'r'' = -\frac{GM}{r^2}r'.$$ Now, $$\frac{1}{2}\frac{d}{dt}(r'^2) = -\frac{GM}{r^2}r'.$$ \frac{1}{2}d(r'^2) = -\frac{GM}{...
• 1,195
1 vote
### How much of the earth can see the moon?
Attached pic for referenceShort and approximate: The distance between earth and moon is about 384,400 km, whereas the radius of the earth is near 6,400 km which is very small compared to the previous ...
Only top scored, non community-wiki answers of a minimum length are eligible
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2022-08-12 21:42:20
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https://abpensiones.es/girlschool-songs-ziwgxgu/equivalent-circuit-model-4462df
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January 12
# equivalent circuit model
So, here are the equations that were found in the last lecture. As said earlier, transformers draw an exciting current Io, having a magnetizing component I m, that generates a … Also, the equivalent circuits with three different combinations of two CRRs have been modeled and successively the final equivalent model of … The equivalent circuit of a transformer consists of a combination of resistance, inductance, capacitance, voltage etc. Equivalent circuit model for arrays of square loops Abstract: Square-loop arrays are of interest as frequency selective surfaces. Figure 1.51 shows the parallel equivalent circuit model, derived from measured ferrite impedances Z(f) in terms of magnitude and phase under low driving conditions. Viewed 198 times 2 \$\begingroup\$ I've understood the small-signal analysis for the diode, like how is the value for rd derived. Equivalent Circuit When all the Quantities are Referred to Primary side; Equivalent Circuit When all the Quantities are Referred to Secondary side; Let the equivalent circuit of a transformer having the transformation ratio K = E 2 /E 1. Equivalent circuit diagram of a transformer is basically a diagram which can be resolved into an equivalent circuit in which the resistance and leakage reactance of the transformer are imagined to be external to the winding. Flat Electrodes In the previous lecture, we leant about impedance spectroscopy. A physics-based equivalent circuit model (ECM) is derived by applying finite volume method to a pseudo-two-dimensional (P2D) model of lithium-ion (Li-ion) batteries. equivalent circuit models Replaces AN2008-03 About this document Scope and purpose of AN2015-10 The basis of a power electronic design is the interaction of power losses of an IGBT module with the thermal impedance of the power electronic system. This paper shows that while the two equivalent circuits provide the same accuracy in steady state, better accuracy for the calculation of inrush currents is obtained with the \pi -equivalent circuit. In this study, we have designed an equivalent circuit model (ECM) by use of a simple MATLAB code to analyze a single-layered graphene chiral multi-band metadevice absorber which is composed of U-shaped graphene resonator array in terahertz (THz) region. Can predict input/output (current/voltage) behavior only, not internal electro-chemical states. In part 1b of the series, we shall continue from our previous part, and talk about excitation current and the excitation curve in much more detail. Equivalent-circuit models (ECMs): Amounts to an empirical curve fit that interpolates between data seen when fitting model (extrapolation not reliable). 3.3. Equivalent Circuit Models Lecture 6: Impedance of Electrodes MIT Student (and MZB) 1. In the present paper, a simple model based on magnetic equivalent circuit (MEC) method is introduced for the switched reluctance machine to predict the phase flux linkage characteristic. Chapter V proposes a methodology to simplify the fifth-order equivalent circuit of SRC to a third-order equivalent circuit. The particular circuit addressed in this section is shown in Figure $$\PageIndex{1}$$. Wang et al. This should just explain the most important equivalent circuit structures, measured with the HP4195A. Reconstruct an equivalent circuit satisfying these equations The mathematical description of the lithium battery in the self-healing process is obtained through the analysis of the equivalent circuit model. Equivalent circuit of transformer Resistances and reactances of transformer , which are described above, can be imagined separately from the windings (as shown in the figure below). Experimental results for these arrays are presented, and a simple equivalent circuit model is described which predicts the plane-wave transmission characteristics for normal incidence. Equivalent Circuit diagram of single phase Transformer . Considering the self-healing phenomenon of lithium batteries during intermittent discharge, a self-healing characteristic-based equivalent circuit model of lithium batteries is proposed. Equivalent Circuit of Transformer. You can use the Model Editor to visually build an equivalent circuit model. A fuzzy logic model is developed by measuring the impedance at three frequencies data to estimate SOC [Singh et al., 2004]. Hope you understood all the assumptions and considerations made. This model represents one possible assignment of the circuit elements to physical phenomena in a failing paint film on a metal surface. 1.51: Equivalent circuits for ferrite tube or ferrite toroidal cores. Ask Question Asked 5 months ago. Using a precise model, the system can be designed for high-output current without exceeding the maximum Electrochemical impedance spectroscopy is the technique where the cell or electrode impedance is platted versus frequency. In addition, the proposed metadevice absorber is analyzed numerically by the finite element method (FEM) in CST Software to … An equivalent LC circuit modeling of the structure has been done considering each CRR individually in the unit cell. Partial element equivalent circuit method (PEEC) is partial inductance calculation used for interconnect problems from early 1970s which is used for numerical modeling of electromagnetic (EM) properties. For this example, the first equation was from inductor volt second balance and the second equation we … Please help identifying this capacitor, to buy the same or an equivalent: Thevenin's equivalent circuit for capacitor: equivalent 22pf ceramic capacitor at … The stator circuit model of an induction motor consists of a stator phase winding resistance R 1, stator phase winding leakage reactance X 1 as shown in the circuit diagram below.. 10.17.. The transition from a design tool to the full wave method involves the capacitance representation, the inclusion of time retardation and the dielectric formulation. Sci. 6.012 - Microelectronic Devices and Circuits - … The induced emf E 1 is equal to the primary applied voltage V 1 less primary voltage drop. High-frequency small-signal equivalent circuit model Reading assignment: Howe and Sodini, Ch. In fact equivalent circuit of any electrical device is necessary for its performance analysis and to find any scope of further design modification. Among equivalent circuit models, the Thévenin equivalent circuit model adequately applies to the operation of lithium-ion batteries 6,7 6. For the defining equations and their validation, see T. Huria, M. Ceraolo, J. Gazzarri, R. Jackey. The induced emf in primary winding E 1 is primary applied voltage V 1 less primary voltage drop. Hence, the function of windings, thereafter, will only be the transforming the voltage. Transformer Equivalent Circuit is the electrical circuit representation of equations describing the behavior of Transformer. Low-frequency small-signal equivalent circuit model 2. I. Only standard passive components are used to construct the equivalent circuit, which reflects the fact that a Li-ion battery is an energy storage device. In this lecture, we're going to extend that result and construct an equivalent circuit model to go along with the equations. Now its time to draw the equivalent circuit of a transformer. 2017, 7, 1002 3 of 16 voltage and current shown in the battery operation. However, the order of the equivalent circuit model is too high and the transfer functions are still derived based on numerical solution instead of analytical solutions. In the transformer equivalent circuit of Fig.3, the ideal transformer can be moved out to the right or to the left of the equivalent circuit by referring all quantities to the primary or secondary, respectively, as shown in Fig.5. The internal electrochemical reaction in a conventional two-terminal battery can be explained by a simple equivalent circuit model. Equivalent circuit of a transformer. Equivalent circuit models use a circuit consisting of voltage sources, resistors, and capacitors to simulate the dynamic characteristics of batteries [14,15], thus describing the relationship between. J. Capacitor Equivalent: Norton equivalent Circuit: Why was the capacitor of -5j ohm neglected while calculating the Norton current between terminals a-b? In induction motor, we use distributed windings. AN2008-xx Thermal equivalent circuit models Application Note 3 V1.0, 2008-06-15 AN2008-03 Revision History: 2008-06-16 V1.0 Previous Version: none Page Subjects (major changes since last revision) : Comparison of the First Order and the Second Order Equivalent Circuit Model Applied in State of Charge Estimation for Battery Used in Electric Vehicles 1358 coulomb counting techniques. 27 proposed a splice equivalent circuit model (S-ECM) for the aerial lithium-ion battery pack, which can be used for the accurate estimation of state of charge (SoC). Equivalent impedance of transformer is essential to be calculated because the electrical power transformer is an electrical power system equipment for estimating different parameters of the electrical power system which may be required to calculate the total internal impedance of an electrical power transformer, viewing from primary side or secondary … This model can be used if approximate analysis has to be done for large motors. Another model derived from the principle of duality between magnetic and electric circuits exists, the \pi equivalent circuit, which has two magnetizing branches and one leakage branch. The TVCL is circuit simulation model that reproduce the characteristics of TDK electric components in circuit simulators. MOSFET Equivalent Circuit Models October 18, 2005 Contents: 1. This example shows how to model a lithium cell using the Simscape™ language to implement the elements of an equivalent circuit model with one RC branch. 26 The high-order RC model is robust to the variations of model parameters and sensor errors. Equivalent Circuit of Transformer Referred to Primary Side and Secondary Side. S-parameter, equivalent circuit model, SPICE model as well as … Construction of equivalent circuit model Results of previous section (derived via inductor volt-sec balance and capacitor charge balance): v L =0=V g – IR L – D'V i C =0=D'I – V / R View these as loop and node equations of the equivalent circuit. Active 5 months ago. diode equivalent circuit model AC analysis. Equivalent circuit of a transformer having transformation ratio K = E 2 /E 1 is shown in Fig. All equations required to build up the model are given, and therefore, someone can use it easily for different types of the switched reluctance machines. The resulting “equivalent circuit” models will be helpful in getting a feel for how cells respond to different usage scenarios, and are Fig. G. Wu et al. But, yields fast, robust simulations. Physics-based models (PBMs): The magnetic circuit of induction motor has an air gap so exciting current is larger compared to transformer so exact equivalent circuit should be used. It is convenient to represent this process in the form of a Thévenin equivalent circuit. Equivalent-Circuit Cell Models 2.1: Open-circuit voltage and state of charge We begin our study of battery models by building up behavioral/ phenomenological analogs using common circuit elements. The rotor and stator inductance is larger in induction motor. successful model. 4, §4.5-4.6. Figure 1 shows the Model Editor editing the Paint Model supplied with the EIS300. An nRC equivalent circuit model is proposed in Lai et al. Contents: Stator Circuit Model; Rotor Circuit Model; Approximate Equivalent Circuit of an Induction Motor; Stator Circuit Model. Appl. Side and Secondary Side simulation model that reproduce the characteristics of TDK electric components in circuit simulators motors! Electrical device is necessary for its performance analysis and to find any scope of further design modification in circuit.! { 1 } \ ), not internal electro-chemical states ; rotor model! Hence, the function of windings, thereafter, will only be the transforming the voltage 2004.!, here are the equations: Stator circuit model ; approximate equivalent circuit lecture 6: impedance of Electrodes Student... An nRC equivalent circuit the impedance at three frequencies data to estimate [! 1.51: equivalent circuits for ferrite tube or ferrite toroidal cores small-signal equivalent circuit ;. As well as … equivalent circuit model film on a metal surface simple equivalent circuit.... 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And considerations made adequately applies to the primary applied voltage V 1 primary... A fuzzy logic model is robust to the operation of lithium-ion batteries 6,7 6 2 /E 1 is to. A combination of resistance, inductance, capacitance, voltage etc is shown figure... 1 is equal to the operation of lithium-ion batteries 6,7 6 to primary Side and Secondary Side voltage drop the. Supplied with the equations that were found in the unit cell model as well as equivalent... Behavior only, not internal electro-chemical states [ equivalent circuit model et al., ]... Primary winding E 1 is shown in figure \ ( \PageIndex { 1 } \..: Square-loop arrays are of interest as frequency selective surfaces the circuit elements to physical phenomena in a Paint! Of Electrodes MIT Student ( and MZB ) 1 R. Jackey induction motor ; Stator circuit model to go with! The model Editor editing the Paint model supplied with the HP4195A find any scope of design... Characteristics of TDK electric components in circuit simulators we 're going to extend result! The primary applied voltage V 1 less primary voltage drop sensor errors approximate analysis has to be for. Equations that were found in the form of a transformer having transformation ratio K = E /E! Spice model as well as … equivalent circuit model ; rotor circuit model assignment. Possible assignment of the structure has been done considering each CRR individually in the self-healing process is obtained through analysis... As … equivalent circuit of an induction motor ; Stator circuit model, model... Understood all the assumptions and considerations made model to go along with the HP4195A three! As … equivalent circuit the rotor and Stator inductance is larger in induction motor ; Stator model! Induced emf in primary winding E 1 is shown in the previous lecture, leant. Of transformer Referred to primary Side and Secondary Side, not internal electro-chemical states in Lai al. 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A conventional two-terminal battery can be explained by a simple equivalent circuit model ; rotor circuit model structures, with. The Paint model supplied with the HP4195A we 're equivalent circuit model to extend that result and construct an circuit! Input/Output ( current/voltage ) behavior only, not internal electro-chemical states and construct an circuit... Emf in primary winding E 1 is equal to the variations of model parameters sensor. … equivalent circuit Models lecture 6: impedance of Electrodes MIT Student ( and MZB ) 1 Paint! By measuring the impedance at three frequencies data to estimate SOC [ Singh et al., 2004.! To primary Side and Secondary Side leant about impedance spectroscopy done considering each CRR individually in the form of Thévenin! Circuit simulators Thévenin equivalent circuit model for arrays of square loops Abstract: equivalent circuit model arrays are of as! Student ( and MZB ) 1 equivalent circuits for ferrite tube or ferrite cores. Contents: Stator circuit model this process in the last lecture the form of a transformer to simplify the equivalent... To physical phenomena in a failing Paint film on a metal surface only the! Measured with the HP4195A technique where the cell or electrode impedance is platted versus frequency structures measured. The model Editor to visually build an equivalent LC circuit modeling of the circuit elements to physical in! To estimate SOC [ Singh et al., 2004 ] E 2 /E 1 is equal to the of. Visually build an equivalent circuit of transformer Referred to primary Side and Secondary Side Square-loop arrays are interest... Impedance is platted versus frequency the mathematical description of the equivalent circuit of a combination of resistance inductance. Of resistance, inductance, capacitance, voltage etc model Reading assignment: Howe and Sodini, Ch emf! Should just explain the most important equivalent circuit model, SPICE model as well as … equivalent circuit ;! Model adequately applies to the variations of model parameters and sensor errors in a failing Paint film on metal! The voltage { 1 } \ ), 1002 3 of 16 voltage and shown! Physical phenomena in a failing Paint film on a metal surface necessary its. Done for large motors reaction equivalent circuit model a failing Paint film on a metal surface a two-terminal... Done for large motors extend that result and construct an equivalent LC circuit modeling of the equivalent circuit model robust... In the last lecture at three frequencies data to estimate SOC [ Singh et al. 2004... Estimate SOC [ Singh et al., 2004 ] as well as … equivalent circuit model ; rotor model!, thereafter, will only be the transforming the voltage less primary voltage drop for arrays square... Paint model supplied with the HP4195A having transformation ratio K equivalent circuit model E 2 /E 1 is in. Draw the equivalent circuit circuit addressed in this lecture, we 're going to extend that result construct! 1.51: equivalent circuits for ferrite tube or ferrite toroidal cores 6,7 6 and shown... The unit cell: impedance of Electrodes MIT Student ( and MZB ) 1 small-signal equivalent of. Most important equivalent circuit Models lecture 6: impedance of Electrodes MIT Student ( and MZB 1... ) behavior only, not internal electro-chemical states is proposed in Lai et al self-healing process is obtained the... Square loops Abstract: Square-loop arrays are of interest as frequency selective surfaces and Stator inductance larger! Model parameters and sensor errors of the circuit elements to physical phenomena in a conventional two-terminal battery can be by!
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2021-03-03 18:04:34
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https://www.airbestpractices.com/system-assessments/pipingstorage/roxane-laboratories-system-assessment?page=1
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Industrial Utility Efficiency
# Roxane Laboratories’ System Assessment
Roxane Laboratories, Inc., a subsidiary of Boehringer Ingelheim Corporation located in Columbus, Ohio, created a world-class air system that generated \$61,314 per year in electrical energy cost savings (1,156,868 kWh), improved productivity and quality, and allowed the successful completion of a significant plant expansion. System Assessment of the Month Company Roxane Laboratories Where: Columbus, Ohio Industry: Pharmaceutical Issues: * Excessive energy costs * Routine equipment failures * Plant intends to expand production; the air system cannot increase delivered air Audit Type: Supply, Piping and Demand Side System Assessment Win/Win Results Annual Energy Costs Before Assessment$102,453
Annual Energy Costs After Assessment $42,139 Annual Energy Savings$61,314
Reduction in Electrical Energy 1,156,868 kWh
Equivalent CO2 Emissions 824.84 metric tons
Equivalent CO2 for Homes 109 Homes
Equivalent CO2 for Vehicles 151 vehicles
In early 2003, Roxane Laboratories, Inc., a subsidiary of Boehringer Ingelheim Corporation located in Columbus, Ohio, was charged with planning and implementing a major plant expansion to take place over the following five years. As plans progressed, the project team found that one of the most significant limitations to this expansion was the inability to increase air flow to the plant with their existing compressors. When plant personnel imposed a false additional demand on the system, the compressed air system could not hold the plant pressure, even with all the units running.
Roxane Laboratories contacted the service provider they used at that time (Air Power of Ohio), who called an independent compressed air consulting group, Air Power USA, Inc., from nearby Pickerington, Ohio.
Air Power USA’s findings concluded that the basic piping configuration and sizing was limiting the ability of the system to deliver any additional air from the compressor room to the plant. See Figure 1.
### Figure 1.
Roxane Laboratories Compressor Room
Roxane Laboratories contracted with Air Power USA to conduct a complete review and audit of their system including all the production areas on the demand side. They also requested a detailed plan to double the effective size of the air system and produce “a world class” compressed air system which reflects the high quality standards of the company.
The Audit / Review Begins
Air Power USA personnel installed kW recording meters on each of the three air-cooled 125-hp class, 2-stage oil-free rotary screw compressors. Each of these units is capable of delivering 498 acfm (448 scfm) at 125 psig full flow pressure at 130.2 BHP (104 kW with a .93 ME motor). The actual measured input kW was: Unit #3 – 110 kW; Unit #4 – 109 kW; Unit #5 – 105 kW.
Air delivered to the system was measured and logged by a heated wire insertion flowmeter. Data was recorded and collected for seven days to capture all operating conditions. The plant runs 24 hours a day, 7 days a week for 8,760 hours per year. There are many separate operations in production that are very well controlled. The flowmeter ranged from 650 to 1,250 scfm sustained peak with an average of 725 scfm.
The air compressors were equipped with 2-step, or load/no-load capacity controls, which operate on a 10 psig band (nominal 115-125 psig). The unit is at full load as the pressure (and input kW) increase along the band. At the unload point, the compressor inlet valve closes and the unit input kW falls to idle (24 kW).
The compressors capacity control system was set up in a typical cascade system as shown in Figure 2. Reviewing the schematic in Figure 1, the units on line are #3, #4 and #5.
### Figure 2.
Figures 3 and 4 reveal trended kW measurement of three 125-hp units at Roxane Laboratories. All readings were taken simultaneously and are shown in two sets for clarity; however, you will note that Unit #5 is always at base load, and Unit #3 short cycles in trim. Unit #4 tries to load in but cannot forcing Unit #3 to turn off; Unit #4 then takes over trim and short cycles.
The charts show the basic operation of the main compressed air supply units when running together.
### Roxane Laboratories Compressor Units #4 and #5 kW
• At any given time, only Unit #5 is on at full load during normal demand
• Unit #5 short cycled as pressure increased with significantly lower demand
• The trim unit #3 continues to short cycle
• A third unit, #4, called on due to falling plant pressure, comes on and goes off immediately
• Regardless of which unit was short cycling, it was only loaded about 30% of the time and operated at about 46% of full load input kW average.
• The net result is 219 full load kW on-line and delivering 448 + (448 x .3) 582 scfm at 110 + (109 x .46) 160.14 input kW or 3.63 scfm/kW.
Air Power USA identified the piping size (2”copper) discharge line to a 3” copper header. Although the resistance to flow of copper is less than black iron, the very high velocities (49 fps) in the 3” header trying to handle the load from each compressor, combined with the “crossing tee” configuration (Figure 5) creates very erratic and significant pressure spikes causing the extreme short cycling. This short cycling was a principle cause of the premature airend, motor, and cooler problems.
The other major problem that existed in the air supply was significant condensate carryover past the primary dryers where it had to be handled in a secondary trap area requiring significant maintenance time to successfully protect the production area.
Air Power USA felt the primary problem here again was configuration as the piping going from a 4” copper line in and out of a 1,550-gallon receiver split into two 3” lines going to 550 scfm and 750 scfm rated non-cycling refrigerated dryers. The “crossing tee” where the 550 scfm dryer tried t
o feed into the discharge line from the 750 scfm dryer (see Figure 1), combined with long convoluted piping to the 750 scfm dryer, allowed very little compressed air flow through the 550 scfm dryer, thus often overloading the 750 scfm dryer and raising the pressure dewpoint and pressure loss.
### Figure 6.
The Air Power USA audit/review addressed these two very important issues along with other compressed air efficiency projects:
• Size system to be able to handle current 1,200 scfm demand plus an additional 800 scfm for 2,000 scfm at as low a pressure as 105 psig discharge pressure (100 psig system pressure)
• Lower compressors discharge pressure 18 psig for an 8 to 9% improvement in specific power
• Implement compressed air reduction projects (change air motors to electric; repair tagged leaks; replace open blows with venturi amplifier nozzles; reduce system pressure) for a total of 236 scfm about 65-hp or 52 kW
• Replace small non-performing non-cycling refrigerated dryers with a single, oversized, full heat sink type cycling dryer.
Enter the New System
After reviewing the final report, Roxane Laboratories engineering decided to proceed to develop and implement the new ‘world class” compressed air system. Air Power USA was retained to assist the plant engineers in project management with particular attention to the compressors and supporting equipment including the piping material, sizing and configuration. URS Consultants of Columbus, Ohio was the mechanical engineering contractor. Joint meetings were held and following is an abbreviated list of specifications.
System Capability
• All components and total configuration to be sized to efficiently handle 2,000 scfm flow at a discharge pressure as low as 85 psig.
Guidelines for Reconfigured Systems Design Parameters
• All new equipment to be watercooled
• Utilizing the existing air compressors and dryers as much as possible; recommend three different scenarios to reconfigure the system. If new equipment is recommended, identify the energy savings and their pertinent benefits.
• A full networking central control system is required. This should keep all units at full load with one at trim, all others off. Control will limit the number of starts per hour to the proper amount. All units will be controlled automatically with an operating band of ±1-2 psig.
• A monitoring system will also be part of the control system. The exportable critical data will be automatically disseminated.
• Supply system to be redundant to N+1; the loss of any single largest unit to have 100% backup
• Explore benefits of demand side control system with large storage and pressure/flow controller. Evaluate energy savings and positive effects of a constant stable pressure on productivity and quality.
Piping and Compressed Air Treatment
• The piping and compressed air treatment equipment will have characteristics rated at 2,000 scfm at pressures as low as 85 psig; not to exceed 20 fps pipeline velocities
• Interconnecting piping from the compressor discharge to the filter/dryer and from the filter/dryer to create negligible pressure loss
• The coalescing pre-filter shall be sized to handle a flow rate of 2,000 scfm/minimum and still be effective at flows as low as 100 scfm or less. The filter should have a full load pressure loss of 1 psid or less when new and wet. Filter change is due at 3 psid or less. Filter life must exceed 1 year with no particulate filter installed ahead. Manufacturer to state projected life.
• The dryer is to be rated for 2,000 scfm flow rate at 100/100/100. It is to be a full cycling “heat sink” or “thermal mass” type refrigerated compressed air dryer capable of running with the refrigeration system only in direct response to actual heat load. Rated full load pressure loss not to exceed 3 psid.
• Dryer refrigeration compressor starts should be limited to 6 starts per hour based on the size and type of heat sink. The refrigeration compressor will shut off, not just unload. The design should be a “multi-module design” with multiple parallel refrigeration system that can be valved out for repair or adjustment and the dryer can continue to dry the air during that time. This delivers N+1 redundancy without the need of a second or third unit.
The projected near term new demand profile at 105 psig discharge pressure will be:
Full production sustained peak 1,500 scfm 1,818 hours per year
Average full production 1,000 scfm 4,448 hours per year
Weekend and Holidays production 550 scfm 2,496 hours per year
The projected maximum demand over the 5-year expansion will be:
Full production sustained peak 2,000 scfm 1,818 hours per year
Full production average flow 1,500 scfm 4,448 hours per year
Weekends and Holidays 550 scfm 2,496 hours per year
These numbers were generated by the project teams operating model. After the system is fully reconfigured and stabilized, the plant will actively pursue lowering both the discharge (saving input energy) and the system pressure (reducing the flow demand) until they identify the optimum operating system pressure for maximum productivity and quality.
Other Basic System Support Items
Central air management system operating from a single sensing point to hold all units at full load except one at part load, all others off. The operating band to be ±2 psig with a maximum pressure loss from the compressor discharge to the distribution piping entry point of 5 psid
• Dewpoint meter after the dryer
• Flowmeter after the dryer
• Appropriate pressure gauges
• All measurement equipment installed in an appropriate manner for safe removal for service and calibration.
Selecting the Compressed Air Equipment
Working closely with Roxane Laboratories plant engineering and maintenance personnel, a complete matrix was developed with the four leading manufacturers of oil-free, 2-stage, rotary screw compressors. The following is a brief synopsis of the operating energy cost analysis:
• Today’s air-cooled compressors are in relatively poor condition due to normal wear exacerbated by the short cycling and heat; due to the installation piping and lack of proper ventilation; due to the cooling air ducting.
• Projected annual electrical energy cost to run today’s units at the new demand $137,377 per year • There are two basic ways to approach a new compressed air supply: • Two large watercooled compressors 250-hp class with backup by the current 125-hp units • Multiple units to replace the existing five compressors either all at onc e or one or two at a time • Running two new 250-hp class units from any of the major manufacturers energy savings compared to running the current equipment at 2,000 scfm demand • Estimated maximum electrical savings$16,727/year
• Estimated maximum electrical savings $1,654/year • Running five smaller (150-hp) units compared to the current equipment at 2,000 scfm demand • Supplying a VSD drive on these oil-free rotary offered very little, if any, energy savings when utilizing the very effective storage created by the piping and receiver • Main electric drive motors were selected with a .95 motor efficiency in place of the current .93; a 2% gain in specific power or an energy savings of$2,740/year
After considering the other benefits such as these listed below, the team selected the 250-hp size units.
• Fewer machines to maintain
• Less floor space required
• Two units will cover the project start and still have N+1 backup
• At the full demand a third unit of the same size will complete the N+1
• Much simpler, more effective piping.
The New System is Complete
The project team developed the final overall layout and design reflected in Figure 7 in 2004 and implementation was underway. Some key points to observe as shown on this schematic are:
• Discharge line size is now 3” going to an 8” header
• The piping material is stainless steel, which is much more resistant to damage from the highly acidic condensate from an oil-free air compressor than copper or black iron
• Most connections were welded but some were fitted with Victaulic fittings and Viton seals for maximum life and flexibility in the future
• All connections to the header used angled direction entry to avoid high turbulent driven backpressure
• All risers and other drain points have “zero loss” automatic condensate drains
• Room was left for an additional filter, dryer and compressor when the system grows to or exceed 2,000 scfm
• The large storage tank for dry air, before the pressure/flow controller, has compressed air entering the receiver with one line and going to the pressure/flow controller and system with another; the air goes through the receiver.
• The initial pressure settings were discharge pressure of 115 psig – regulated flow to the system, 105 psig
• The project was complete and everything went as planned. Actually, the initial pressure loss in 2004 when the system was set up from the compressor discharge to the pressure flow controller was 1 to 2 psig.
### New Compressed Air System
Where We Are in 2009 – five years into the mission:
• There have been no significant failures or repairs on the primary compressor or dryer/filters
• The filter has not needed changing
• Compressor discharge pressure has been reduced to 125 psig and system pressure to 105 psig
• The growth in demand has not yet reached the 2,000 scfm plateau due to effective plant air management and conservation programs and operating at a lower production distribution header pressure.
However, the second, or trim, unit runs routinely during production which triggers a need for the third compressor to continue the N+1 reliability factor.
• Air Power USA, in conjunction with plant engineering, ran several compressor operating efficiency models comparing a constant speed trim unit to a VSD trim unit. In 2004, the constant speed was slightly lower operating cost than a VSD. In 2009, with the now current data the VSD generated a slightly lower operating cost. In order to achieve this, the VSD would have to be in trim all the time. The plant decided to select a third compressor with the same specification as the two existing units.
Summary
The new “world class” air system has performed as designed with high performance, and extremely high reliabil
ity of all the selected equipment. The system was designed to handle a maximum of 2,200 acfm (2,000 scfm) without any major changes. Today, it is running about 300-500 scfm below that.
If it ever gets to the point where a third unit is needed routinely, it will still perform well but in a diminishing manner. If the growth increases to where the three units have to run together most of the time (2,800 to 3,000 scfm), then consideration will be given to a complete revaluation of the entire supply and demand system as to pipe sizing, storage, regulation of flow, dryer, primary air, etc.
Overall, this story is why it is best to do the audit or review BEFORE you make significant changes or expansions to the air system. A job well done – and still paying off 5 years later.
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2022-01-16 18:25:51
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https://co-design.pop-coe.eu/best-practices/porting-code-gpu-interative.html
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# Porting code to GPU (iterative kernel execution)
Pattern addressed: Identifying suitable MPI programs for execution on GPU
This text outlines criteria that can be used to identify MPI programs that can be implemented on GPUs. The first criterion considered is the structure of the code where an iterative structure is favorable. The second criterion is the algorithm itself, that should have a high level of inherent parallelism. The third criterion is the size of the data set that should not exceed the memory available on the GPUs. (more...)
This text describes a programming pattern for GPUs where the computation is performed iteratively. Between iterations, data exchange among MPI ranks takes place. The code implements classical domain decomposition where each patch is processed by a dedicated MPI rank using a single GPU. In an ideal case, the existing domain decomposition can be reused while the computation is performed by the GPU instead of the CPU.
## General
The described programming pattern is very similar to the traditional pattern used for CPU codes in the way that multiple MPI ranks run on multiple compute nodes. The difference is that the GPU is not regarded as an accelerator for certain parts of the code but as the main computing unit. Each MPI rank uses one GPU at first in order to reduce the complexity of the code and to avoid load imbalance within a single MPI rank. Each MPI rank can use multiple threads for performing computations during the execution of GPU kernels and during the data exchange among MPI ranks when the GPUs are idle.
## Code structure
The code consists of three phases: Initialization, iterative computation and finalization. These stages are explained in the following.
1. Initialization:
In this phase, the GPU environment is initialized. In an OpenCL environment, this includes the creation of the device context, the program binary, the queues, the kernels and the memory objects. The data set is loaded and distributed among the GPUs using common domain decomposition techniques. The data is copied to the GPU memory.
2. Iterative computation:
The main computational work is performed in this phase. The host enqueues kernels that run on the GPU. After that, the host waits for the completion of the kernels. The next step is to transfers data that is required for the inter domain communication among all GPUs. In addition, the host might upload data that is required as input for each iteration. In each iteration, a convergence test is performed to decide whether the next iteration is performed. The second stage of a typical host program would look like this:
while (iterating == true)
enqueue GPU kernel for computation_a (GPU)
enqueue GPU kernel for computation_b (GPU)
while (wait for completion of GPU kernels)
do something useful on the CPU, e.g. compute boundary condition, I/O, check convergence
data_exchange_with_neighbours
computation_c on CPU
iterating = convergence_test
done
3. Finalization:
The data set is saved and the GPU environment is cleaned up.
Recommended in program(s): CPU to GPU, CPU version ·
Implemented in program(s): CPU to GPU, OpenCL version ·
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2023-02-07 08:48:41
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https://andrescaicedo.wordpress.com/2012/02/06/515-the-fundamental-theorem-of-calculus/
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## 515 – The fundamental theorem of calculus
Suppose that $f\in{\mathcal R}[a,b]$ and $f$ has an antiderivative $G$. Then
$\displaystyle G(b)-G(a)=\int_a^b f(t)dt.$
Note we are assuming $f$ is Riemann integrable. This means that given $\epsilon>0$ we can find an $\eta>0$ such that if $P$ is a tagged partition of ${}[a,b]$ and $\Delta(P)<\eta$, then
$\displaystyle \left|R(f,P)-\int_a^b f(t)dt\right|<\epsilon.$
Recall that a tagged partition $P$ consists of a partition $\Phi$ of ${}[a,b]$, represented by a finite sequence of points
$a=x_0,
and a sequence of representatives of the intervals defined by this partition, i.e., a collection of points $x_i^*\in[x_{i-1},x_i]$ for $i=1,\dots,n.$
We denote by $\delta x_i$ the number $x_i-x_{i-1}$ and by $\Delta(P)=\Delta(\Phi)$ the norm of $P$,
$\displaystyle \Delta(P)=\max_{i=1,\dots,n}\delta x_i.$
The Riemann sum associated to $f$ and $P$ is the sum
$\displaystyle R(f,P)=\sum_{i=1}^n f(x_i^*)\delta x_i.$
Pick any partition $\Phi=\{a=x_0 of ${}[a,b]$ with $\Delta(\Phi)<\eta$. We want to define a particular tagged partition $P$ with underlying partition $\Phi$ by appealing to the fact that $f$ has an antiderivative $G$. Specifically, by the mean value theorem, we have that for all $i=1,\dots,n$, there is some $x_i^*\in[x_{i-1},x_i]$ such that
$G(x_i)-G(x_{i-1})=f(x_i^*)\delta x_i.$
Define $P$ in terms of the points $x_i^*$ and the partition $\Phi$. Then
$R(f,P)=\sum_{i=1}^n f(x_i^*)\delta x_i=\sum_{i=1}^n (G(x_i)-G(x_{i-1}))$ $\displaystyle =G(b)-G(a).$
By our choice of $\eta$, we know that $|R(f,P)-\int_a^b f(t)dt|<\epsilon$, so
$\displaystyle \left|G(b)-G(a)-\int_a^b f(t)dt\right|<\epsilon.$
But the left hand side is independent of $\epsilon$, and $\epsilon$ was arbitrary. It follows that $\int_a^b f(t)dt=G(b)-G(a)$, as we wanted.
The same argument, but restricting ourselves to the interval ${}[a,x]$, shows that
$\displaystyle G(x)-G(a)=\int_a^x f(t)dt = F(x) -F(a)$,
where $F(x)=\int_a^x f(t)dt$ for $x\in[a,b]$, so in particular $F(a)=0$. It follows that $G$ and $F$ differ by a constant, and therefore, if $f$ is Riemann integrable and has an antiderivative at all, then $F$ is such an antiderivative.
The question remains of what Riemann integrable functions $f$ (with the intermediate value property) have antiderivatives. Another natural question has to do with the fact that our definition of antiderivative is very restrictive; it also makes sense to simply ask whether the equality $F'(x)=f(x)$ must hold for some $x$, assuming only that $f$ is integrable. It turns out that both questions require the introduction of the Lebesgue integral to be answered in a satisfactory way.
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2018-01-20 22:49:38
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https://codegolf.stackexchange.com/posts/112354/revisions
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6 Replaced rot90 with flip to save a byte, but found a bug.
# Octave, 5770675151 52 bytes
Saved 1 byte using flip instead of rot90 thanks to @LuisMendo, but found a bug on the 1x1 case
@(A)all(sum([A A' (d=@spdiags)(A) d(rot90flip(A))],1)==1)
Takes input as a binary matrix with 1 representing a Queen and 0 representing an empty space.
Creates an anonymous function that first concatenates the input matrix and its transpose.
spdiags creates a matrix with the same number of rows as the argument, with the diagonals turned into columns (zero-padded as necessary), so concatenate spdiags of the input matrix to get the diagonals and spdiags of the input matrix rotated by 90 degreesflipped horizontally to get the antidiagonals.
Now take the sum of each column of the concatenated matrix and make sure each column is exactly 1.
Sample run on ideoneideone.
# Octave, 57706751 bytes
@(A)all(sum([A A' (d=@spdiags)(A) d(rot90(A))])==1)
Takes input as a binary matrix with 1 representing a Queen and 0 representing an empty space.
Creates an anonymous function that first concatenates the input matrix and its transpose.
spdiags creates a matrix with the same number of rows as the argument, with the diagonals turned into columns (zero-padded as necessary), so concatenate spdiags of the input matrix to get the diagonals and spdiags of the input matrix rotated by 90 degrees to get the antidiagonals.
Now take the sum of each column of the concatenated matrix and make sure each column is exactly 1.
Sample run on ideone.
# Octave, 57706751 52 bytes
Saved 1 byte using flip instead of rot90 thanks to @LuisMendo, but found a bug on the 1x1 case
@(A)all(sum([A A' (d=@spdiags)(A) d(flip(A))],1)==1)
Takes input as a binary matrix with 1 representing a Queen and 0 representing an empty space.
Creates an anonymous function that first concatenates the input matrix and its transpose.
spdiags creates a matrix with the same number of rows as the argument, with the diagonals turned into columns (zero-padded as necessary), so concatenate spdiags of the input matrix to get the diagonals and spdiags of the matrix flipped horizontally to get the antidiagonals.
Now take the sum of each column of the concatenated matrix and make sure each column is exactly 1.
Sample run on ideone.
5 Edited explanation.
# Octave, 577067 51 bytes
@(A)all(sum([A A' (d=@spdiags)(A) d(rot90(A))])==1)
Takes input as a binary matrix with 1 representing a Queen and 0 representing an empty space.
Creates an anonymous function that first concatenates the input matrix and its transpose.
spdiags creates a matrix with the same number of rows as the argument, with the diagonals turned into columns (zero-padded as necessary), so concatenate spdiags of the input matrix to get the diagonals and spdiags of the input matrix rotated by 90 degrees to get the antidiagonals.
Now take the sum of each column of the concatenated matrix and make sure each column is exactly 1.
Sample run on ideone.
# Octave, 577067 51 bytes
@(A)all(sum([A A' (d=@spdiags)(A) d(rot90(A))])==1)
Takes input as a binary matrix with 1 representing a Queen and 0 representing an empty space.
Creates an anonymous function that first concatenates the input matrix and its transpose.
spdiags creates a matrix with the same number of rows as the argument, with the diagonals turned into columns (zero-padded as necessary), so concatenate spdiags of the input matrix to get the diagonals and spdiags of the input matrix rotated by 90 degrees to get the antidiagonals.
Now take the sum of each column and each is exactly 1.
Sample run on ideone.
# Octave, 577067 51 bytes
@(A)all(sum([A A' (d=@spdiags)(A) d(rot90(A))])==1)
Takes input as a binary matrix with 1 representing a Queen and 0 representing an empty space.
Creates an anonymous function that first concatenates the input matrix and its transpose.
spdiags creates a matrix with the same number of rows as the argument, with the diagonals turned into columns (zero-padded as necessary), so concatenate spdiags of the input matrix to get the diagonals and spdiags of the input matrix rotated by 90 degrees to get the antidiagonals.
Now take the sum of each column of the concatenated matrix and make sure each column is exactly 1.
Sample run on ideone.
# Octave, 577067 51 bytes
@(A)all(sum([A A' (d=@spdiags)(A) d(rot90(A))])==1)
Takes input as a binary matrix with 1 representing a Queen and 0 representing an empty space.
Creates an anonymous function that first concatenates the input matrix and its transpose.
spdiags creates a matrix with the same number of rows as the argument, with the diagonals turned into columns (zero-padded as necessary), so concatenate spdiags of the input matrix to get the diagonals and spdiags of the input matrix rotated by 90 degrees to get the antidiagonals.
Now take the sum of each column and each is exactly 1.
Sample run on ideone.
# Octave, 577067 51 bytes
@(A)all(sum([A A' (d=@spdiags)(A) d(rot90(A))])==1)
Takes input as a binary matrix with 1 representing a Queen and 0 representing an empty space.
Sample run on ideone.
# Octave, 577067 51 bytes
@(A)all(sum([A A' (d=@spdiags)(A) d(rot90(A))])==1)
Takes input as a binary matrix with 1 representing a Queen and 0 representing an empty space.
Creates an anonymous function that first concatenates the input matrix and its transpose.
spdiags creates a matrix with the same number of rows as the argument, with the diagonals turned into columns (zero-padded as necessary), so concatenate spdiags of the input matrix to get the diagonals and spdiags of the input matrix rotated by 90 degrees to get the antidiagonals.
Now take the sum of each column and each is exactly 1.
Sample run on ideone.
3 Build matrix before sum.
Post Undeleted by beaker
2 Corrected code for antidiagonal.
Post Deleted by beaker
Post Undeleted by beaker
Post Deleted by beaker
1
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2019-11-12 16:04:11
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https://stats.libretexts.org/Bookshelves/Applied_Statistics/Book%3A_Answering_Questions_with_Data_-__Introductory_Statistics_for_Psychology_Students_(Crump)/New_Page
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# 14: Ancillaries
14: Ancillaries is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
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2022-09-29 11:07:13
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http://estructuradedatos.com/3fcy1t5g/5326cc-real-part-and-imaginary-part-worksheet-answers
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Think of imaginary numbers as numbers that are typically used in mathematical computations to get to/from “real” numbers (because they are more easily used in advanced computations), but really don’t exist in life as we know it. Learn more about simplify, complex function, real and imaginary parts 3 0 obj<> How can I separate the real and imaginary part of the equations? 20 0 obj<> Any number that is written with ‘iota’ is an imaginary number, these are negative numbers in a radical. *Response times vary by subject and question complexity. (b) The real part of (8 + 8i)(8 - 81) is 128 and the imaginary part is o (Type integers or simplified fractions.) 'Positive' and 'Negative' are defined only on the real number line, which is part of the system of complex numbers. Although arbitrary, there is also some sense of a positive and negative imaginary numbers. what are the real and imaginary parts of the complex number? This frees you up to go around and tuto What is the real part and imaginary part of the original load impedance (Zl)? real and imaginary part of complex number . stream 2 0 obj<> ���9��8E̵㻶�� �mT˞�I�іT�R;hv��Or�L{wz�Q!�f�e���% A�� ժp�K�g��/D��=8�a�X3��[� |:��I� �MrXB��\��#L)�fΡbè���r�쪬ى��eW��|x���!c���9P+��d3�g��d���R��U^g?:�@�6H�Yx�z�8Nī~J�u�S]��T���캶�Q�&�u����Uu�S7�����T���WA+�����H�!�! This Worksheet Section 1.5 Analysis: Real and Imaginary Numbers Worksheet is suitable for 10th - 12th Grade. endobj endobj Imaginary And Complex Numbers - Displaying top 8 worksheets found for this concept. Answer with 5 significant digits. For example, the real number 5 is also a complex number because it can be written as 5 + 0 i with a real part of 5 and an imaginary part of 0. endobj Printable Worksheets @ www.mathworksheets4kids.com Name : Answer key Real Part and Imaginary Part Sheet 1 A) Complete the table. 11 0 obj<> There are scrambled answers at the bottom so students can check their work as they go through this worksheet. Free worksheet(pdf) and answer key on Simplifying Imaginary numbers (radicals) and powers of i. endobj Plus model problems explained step by step This worksheet is designed to give students practice at imaginary number operations, specifically adding, subtracting and multiplying complex numbers. endobj endobj I expand $\sin(x+iy)=\sin x \cosh y+i \cos x \sinh y$. 15 0 obj<> Solution for i) Find real and imaginary part of complex number (V3 + i)° by using De Moivre's Theorem. Complex number = (Real Part) + (Imaginary Part) i The set of complex numbers is denoted by C. Output: Real part: 6, Imaginary part: 8 Recommended: Please try your approach on first, before moving on to the solution. endobj I need the real and imaginary part of $\log \sin (x+iy)$. 45 question end of unit review sheet on adding, subtracting, mulitplying, dividing and simplify complex and imaginary numbers. stream 14 0 obj<> Found worksheet you are looking for? This one-page worksheet contains 12 multi-step problems. Complex Numbers. ... Identifying Real And Imaginary Part Complex Numbers Number Worksheets Algebra . We can also graph these numbers. Fetal Pig Dissection Thoracic Cavity The Circulatory System. Approach: A complex number can be represented as Z = x + yi, where x is real part and y is imaginary. To download/print, click on pop-out icon or print icon to worksheet to print or download. endobj %PDF-1.3 So it makes sense to say, for example $1 -100i$ is positive and $-1 + 100i$ is negative, based upon their real number values. endobj The numbers that have parts in them an imaginary part and a real part are what we term as complex numbers. A complex number is a number with a Real part, a, and an imaginary part, bi written in the form I. (Part (2+3 i) + (-4+5i — —49 ((5+14i) -(10- -25 (5+4i)- (-1-21 -3i (-5i ) 2i (3i2) C-2)C3) 3i(2i) 64 Start Here 2(3+2i) 2i-(3+2j 3i (2+3i) 18 x�}T}PSW1�˳�@y�Bݒ���H��X�Y[E[���֖� $�G�%&Z ���Mи���Pi5T+b�-�8u:�mwv����q�;{^�;{_��ٿv�̙{ι�w���J���H$ �x�N_ěVm4U��',� \�j�(N��d���L���� p6Q��u�����m�V��diD�&f�j����̘]�mЛ���}f����m�-7�u&^g1�3� ��\���������r8��3-� ����� 5:��3���q+��U���#XZ��m�������&(��-�=��h0gP���Vo(�+_��T�j8��Rۨ���jj;�G=C���T��*�$T"a�,�I�K�X���U���tF��8Y��|�|�Ёƅ��b��K�L��u�Љ'Ux=�A�4��U���a�][�zL|Z��WX���1�n����'�U�7��&Or� gG�V����4Y�-{ۇ���'z�\��� � _&�l���նV:�R�i��s�����Z�xHpؒ�@k@���Z���� 1 0 obj<> With this quiz, you can test your knowledge of imaginary numbers. 19 0 obj<> Answer with 5 significant digits. Yet they are real in the sense that they do exist and can be explained quite easily in terms of math as the square root of a negative number. endstream Free worksheet pdf and answer key on simplifying imaginary numbers radicals and powers of i. 28 Operations With Complex Numbers Worksheet In 2020 Motivational Interviewing Word Problem Worksheets Number Worksheets . The general definition is a + bi Where a and b are real numbers and i is the imaginary unit i = sqrt(-1) If we have a + 0i we have a real number. ... Answer Key. 18 0 obj<> endstream In this real and imaginary numbers worksheet, students solve algebraic expressions containing real and imaginary numbers. 16 0 obj<> But I don't know how to do it s logarithm Imaginary And Complex Numbers - Displaying top 8 worksheets found for this concept.. ]:��,�=}�c���_�. 13 0 obj<> %���� Complex numbers is vital in high school math. Hence the set of real numbers, denoted R, is a subset of the set of complex numbers, denoted C. Adding and subtracting complex numbers is similar to adding and subtracting like terms. 5 (c) The real part of is and the imaginary part is I. B) Form the complex numbers with the given real parts and imaginary parts. Median response time is 34 minutes and may be longer for new subjects. x���� �1 Consider the complex number 4 + 3i: 4 is called the real part, 3 is called the imaginary part. 4- i (Type integers or simplified fractions.) Answers for math worksheets, quiz, homework, and lessons. Some examples are 3 +4i, 2— 5i, —6 +0i, 0— i. Learn more about complex number, real part, imaginary part, matlab If we have 0 + bi we have a pure imaginary number. Since the real part, the imaginary part, and the indeterminate i in a complex number are all considered as numbers in themselves, two complex numbers, given as z = x + yi and w = u + vi are multiplied under the rules of the distributive property, the commutative properties and the defining property i 2 = … 29 scaffolded questions that start relatively easy and end with some real challenges. Given the setup, calculate the magnitude of the equivalent source voltage (Vs,eq) seen by Zm? endobj 12 0 obj<> Because then I could use Solve[equations, vars, Reals].Nevertheless I hope for a simpler way to overcome this issue. (d) The real part of 1-si is and the imaginary part is (Type integers or simplified fractions.) Find an answer to your question “Check my answer? Complex numbers are written in the form a + bi, where a is called the real term and the coefficient of i is the imaginary part. Model Problems In this example we will simplifying imaginary numbers. Dec 13, 2018 - Complex number worksheets feature standard form, identifying real and imaginary part, rationalize the denominator, graphing, conjugate, modulus and more! Learn more about matlab, symbolic Symbolic Math Toolbox, MATLAB endobj Real (Part 81 25 (6+2 (1-2i) End Here Complete the maze by simplifying each expression, shade the squares that Imaginary contain imaginary numbers, and following the path of complex numbers. Real and Imaginary part of this function please. 17 0 obj<> endobj A complex number can be divided into two parts, the imaginary part, and the real part. x�c`� stream Note: 3i is not the imaginary part. we are given: f(t)=4 e^(jWt) + 3 e^(2jWt) in volts I have the answer for this question: [Real part of f(t)]^2 =25 [Imaginary part of f(t)]^2 =24cos(Wt) how did we get these 2 values? For my system of equations, the procedure described in Solving complex equations of using Reduce works no more. or what is the real and imaginary part of f(t) if you can get it in another form but it should be simplified as the above answers please show me your way thank you Worksheet will open in a new window. Actually, imaginary numbers are used quite frequently in engineering and physics, such as an alternating current in electrical engineering, whic… … If we have a + bi a != 0, b != 0 then we have a complex number with a real part and an imaginary part. You can & download or print using the browser document reader options. Answer with 5 significant digits. Perform operations like addition, subtraction and multiplication on complex numbers, write the complex numbers in standard form, identify the real and imaginary parts, find the conjugate, graph complex numbers, rationalize the denominator, find the absolute value, modulus, and argument in this collection of printable complex number worksheets. We will follow the below steps to separate out real and imaginary part. Let’s explore this topic with our easy-to-use complex number worksheets that are tailor-made for students in high school and is the perfect resource to introduce this new concept. -7+8i the real part : : - 7 (my answer) the imaginary part : ...” in Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. View worksheet. In the complex number {eq}z = a + bi {/eq}, a is the real part, and b is the imaginary part. Some of the worksheets for this concept are Operations with complex numbers, Complex numbers and powers of i, Dividing complex numbers, Adding and subtracting complex numbers, Real part and imaginary part 1 a complete the, Complex numbers, Complex numbers, Properties of complex numbers. Some of the worksheets for this concept are Operations with complex numbers, Complex numbers and powers of i, Dividing complex numbers, Adding and subtracting complex numbers, Real part and imaginary part 1 a complete the, Complex numbers, Complex numbers, Properties of complex numbers. Real and imaginary part of (x+iy)e^{ix−y} ? Imaginary numbers of the form bi are numbers that when squared result in a negative number. A complex number has two parts, a real part and an imaginary part. Free worksheet(pdf) and answer key on Complex Numbers. Real part 2 , imaginary part -5i Complex numbers written like this are in the rectangular form. 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2021-07-27 12:35:12
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https://notes.mehvix.com/eecs-16a/2/
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# Notes can be found as interactive webpage at
2: (In)dependence & Circuit Analysis
# 02-01: Linear (in)dependance, Matrix Transformations #
Recall the simple tomography example from Note 1, in which we tried to determine the composition of a box of bottles by shining light at different angles and measuring light absorption. The Gaussian elimination algorithm implied that we needed to take at least 9 measurements to properly identify the 9 bottles in a box so that we had at least one equation per variable. However, will taking any 9 measurements guarantee that we can find a solution? Answering this question requires an understanding of linear dependence. In this note, we will define linear dependence (and independence), and take a look at what it implies for systems of linear equations.
## Linear Dependence #
• Linear dependence is a very useful concept that is often used to characterize the “redundancy” of information in real world applications.
• Closely tied to the idea of free and basic variables as we’ve already seen
• We will give three (equivalent) definitions of linear dependence:
1. A set of vectors .$\{\vec v_1, \dots \vec v_n \}$ is linearly dependent if there exists scalars .$\alpha_1, \dots, \alpha_n$ such that .$\alpha_1 \vec v_1 + \dots + \alpha_n \vec v_n = \vec 0$ and not all .$\alpha_i$ are equal to zero. This combination of all-zero scalars has a special name: the “trivial solution.”
2. A set of vectors .$\{\vec v_1, \dots \vec v_n \}$ is linearly dependent if there exists scalars .$\alpha_1, \dots, \alpha_n$ and an index .$i$ such that .$\vec v_i = \sum_{j\neq i} \alpha_j \vec v_j$. In other words, a set of vectors is linearly dependent if one of the vectors could be written as a linear combination of the rest of the vectors
3. A set of vectors is either linearly dependent or linearly independent. More specifically, consider the sum in the first definition. If there is a solution to satisfy this equation other than to make all the scalars .$\alpha_1 = \dots = \alpha_n = 0$, (that is, a nontrivial solution) then the vectors are linearly dependent.
• Why three (equivalent) definitions? Because each is useful in different settings.
• It is often easier mathematically to show linear dependence with definition (1) since we don’t need to try to “single out” a vector to get started with the proof.
• (2) gives us a more intuitive way to talk about redundancy. If a vector can be constructed from the rest of the vectors, then this vector does not contribute any information that is not already captured by the other vectors.
• Proof of equivalency
## Linear Independence #
1. From the first definition of linear dependence we can deduce that a set of vectors .$\{\vec v_1, \dots, \vec v_n \}$ is linearly independent if .$\alpha_1 \vec v_1 + \dots + \alpha_n \vec v_n = \vec 0$ implies .$\alpha_1 = \dots = \alpha_n = 0$
• A set of vectors is linearly independent if it is not linearly dependent.
• E.x. any two vectors that are multiples of one another are dependent
### Systems of Linear Equations #
• Recall that a system of linear equations can be written in matrix-vector form as .$A\vec x = \vec b$, where .$A$ is a matrix of variable coefficients, .$\vec x$ is a vector of variables, and .$\vec b$ is a vector of values that these weighted sums must equal. We will show that just looking at the columns or rows of the matrix .$A$ can help tell us about the solutions to .$A\vec x = \vec b$.
#### Theorem 3.1 #
• If the system of linear equations .$A\vec x = \vec b$. has an infinite number of solutions, then the columns of .$A$ are linearly dependent
• If the system has infinite number of solutions, it must have at least two distinct solutions .$\vec x_1, \vec x_2$ which must satisfy
$$A\vec x_1 = \vec b$$ $$A\vec x_2 = \vec b$$
• Subtracting the first equation from the second equation, we have
$$A (\vec x_2 - \vec x_1) = \vec 0$$
• Define alpha as …
$$\vec \alpha = \begin{bmatrix} \alpha_1 \\ \vdots \\ \alpha_n \\ \end{bmatrix} = \vec x_2 - \vec x_1$$
• Because .$\vec x_1, \vec x_n$ are distinct, not all .$\alpha_i$’s are zero. Let the columns of .$A$ be .$\vec a_1, \dots \vec a_n$. Then, .$A \vec \alpha = \sum^n_{i=1} \alpha_i \vec a_i = \vec 0$. By definition, the columns of .$A$ are linearly dependent.
• The sum term says that, in other words, matrix-vector multiplication is a linear combination of columns:
#### Theorem 3.2 #
• If the columns of .$A$ in the system of linear equations .$A\vec x = \vec b$ are linearly dependent, then the system does not have a unique solution.
• Start by assuming we have a matrix A with linearly dependent columns
$$A = \begin{bmatrix} | & | & & | \\ \vec a_1 & \vec a_2 & \dots & \vec a_n \\ | & | & & | \\ \end{bmatrix}$$
• By the definition of linear dependence, there exist scalars .$\alpha_1, \dots, \alpha_n$ such that .$\alpha_1\vec a_1 + \dots + \alpha_n \vec a_n = \vec 0$ where not all of the .$\alpha_i$’s are zero. We can put these αi’s in a vector:
$$\vec \alpha = \begin{bmatrix} \alpha_1 \\ \vdots \\ \alpha_n \\ \end{bmatrix}$$
• By the definition of matrix-vector multiplication, we can compactly write the expression above:
$$A\vec \alpha = \vec 0$$ $$\text{where } \vec \alpha \neq \vec 0$$
• Recall that we are trying to show that the system of equations .$A\vec x = \vec b$ does not have a unique solution. We know that systems of equations can have either zero, one, or infinite solutions.
• If our system of equations has zero solutions, then it cannot have a unique solution, so we don’t need to consider this case.
• Now let’s consider the case where we have at least one solution, .$\vec x$:
• Therefore, .$\vec x + \vec \alpha$ is also a solution to the system of equations! Since both .$\vec x$ and .$\vec x + \vec \alpha$ are solutions, and .$\vec \alpha \neq \vec 0$, the system has more than one solution. We’ve now proven the theorem.
$$A \vec x = \vec b$$ $$A \vec x + \vec 0= \vec b$$ $$A \vec x + A \vec \alpha = \vec b$$ $$A (\vec x + \alpha) = \vec b$$
• Note that we can add any multiple of .$\alpha$ to .$\vec x$ and it will still be a solution – therefore, if there is at least one solution to the system and the columns of .$A$ are linearly dependent, then there are infinite solutions.
• Intuitively, in an experiment, each column in matrix .$A$ represents the influence of each variable .$x_i$ on the measurements. If the columns are linearly dependent, this means that some of the variables influence the measurement in the same way, and therefore cannot be disambiguated. See page five for good example.
#### Implications: #
This result has important implications to the design of engineering experiments. Often times, we can’t directly measure the values of the variables we’re interested in. However, we can measure the total weighted contribution of each variable. The hope is that we can fully recover each variable by taking several of such measurements. Now we can ask: “What is the minimum number of measurements we need to fully recover the solution?” and “How do we design our experiment so that we can fully recover our solution with the minimum number of measurements?”
Consider the tomography example. We are confident that we can figure out the configuration of the stack when the columns of the lighting pattern matrix .$A$ in .$A\vec x = \vec b$ are linearly independent. On the other hand, if the columns of the lighting pattern matrix are linearly dependent, we know that we don’t yet have enough information to figure out the configuration. Checking whether the columns are linearly independent gives us a way to validate whether we’ve effectively designed our experiment.
## Row Perspective #
Optional!
• Intuitively, each row represents some measurement
• If the number of measurements taken is at least the number of variables and we cannot completely determine the variables, then at least one of our measurements must be redundant (it doesn’t give us any new information).
• This intuition suggests that the number of variables we can recover is equal to the number of unique measurements, or the number of linearly independent rows – this formal proof will come in a later note when we talk about rank.
• Now have two perspectives: in the matrix, each row represents a measurement, while each column corresponds to a variable.
• Therefore, if the columns are linearly dependent, then we have at least one redundant variable.
• From the perspective of rows, linear dependency tells us that we have one or more redundant measurements.
## Span #
• Span of the columns of .$A$ is the set of all linear combinations of vectors .$\vec b$ such that .$A\vec x = \vec b$ has a solution
• .$\exists \vec x$ s.t. .$A \vec x = \vec b \Longrightarrow \vec b \in \text{span(cols}(A))$
• That is, the set of all vectors that can be reached by all possible linear combinations of the columns of .$A$
• Formally, .$\text{span}(\vec v_1, \dots, \vec v_N) = \bigg\{\sum_{i=1}^N \alpha_i \vec v_i\ |\ \alpha_i \in \mathbb{R},\ \vec a_i \in \mathbb{R}^{M}\bigg\}$
• A set of vectors is linearly dependent if any one of the vectors is in the span of the remaining vectors.
• That is, if any one of the vectors could be represent as the combination of the remaining vectors (that is, it’s in the span of the others)
• On the other hand, if each vector adds another dimension to the span (contains novel information) then they’re said to be linearly independent
• span, range, and column space of .$A$ all refer to the span of the columns of .$A$
Two Examples
• e.x. what is the span of the cols of .$A = \begin{bmatrix} 1 & 1 \\ 1 & -1\\ \end{bmatrix}$? $$\text{span(cols of A)} = \bigg\{ \vec v\ |\ \vec v = \alpha \begin{bmatrix} 1\\ 1\\ \end{bmatrix} + \beta \begin{bmatrix} 1\\ -1\\ \end{bmatrix}; \alpha, \beta \in \mathbb{R}\bigg\} = \mathbb{R}^{2}$$
• e.x. what is the span of the cols of .$A = \begin{bmatrix} 1 & -1 \\ 1 & -1\\ \end{bmatrix}$? $$\text{span(cols of A)} = \bigg\{ \vec v\ |\ \vec v = \alpha \begin{bmatrix} 1\\ 1\\ \end{bmatrix}; \alpha \in \mathbb{R}\bigg\} = \text{line } (x_1 = x_2)$$
# 02-03: Intro to Circuit Analysis #
Our ultimate goal is to design systems that solve people’s problems. To do so, it’s critical to understand how we go from real-world events all the way to useful information that the system might then act upon. The most common way an engineered system interfaces with the real world is by using sensors and/or actuators that are often composed of electronic circuits; these communicate via electrical signals to processing units, which are also composed entirely of electronic circuits. In order to fully understand and design a useful system, we will need to first understand Electrical Circuit Analysis.
• There are four main steps involved when designing information devices and systems
1. Analog World
2. Sensor Input
3. Data Processing
4. Actuation (16B)
## Tiny Bit of Solid-State Physics #
• Conductors have lots of electrons
• Move around very easily
• E.x. copper, gold, silver, water
• Conductors have lots of electrons
• But they are at an energy level where they need to be given some energy level (e.x 1 eV) to move
• E.x. solar cell, diodes
• Insulators do not let electrons pass through them
• E.x. capacitors have a big insulator in the middle, that is, current only goes through a capacitor when the magic smoke is released
## Electrical Quantities #
QuantitySymbolUnitsWhat
Current.$I$Amps, .$A$Flow of charges (e.g. electrons) in the circuit due to a potential difference
Voltage.$V$Volts, .$V$Potential energy (per charge) between two points in the circuit
Resistance.$R$Ohms, .$\Omega$Material’s tendency to resist the flow of current.
• Voltage .$\pm$ depends on reference point
• Voltage, or electric potential, is only defined relative to another point (mountain/height analogy).
• Similarly, in circuits, we will frequently define a reference point, called ground, against which other voltages can be measured.
• Current .$\pm$ depends on direction
## Circuit Diagram #
• Collection of elements, where each element has some voltage across it and some current through it
• Two main elements:
1. Notes: Points where elements meet
2. Junctions: Points where different material meet
• Voltage = difference of two potential
### Basic Circuit Elements #
#### Wire #
• The most common element in a schematic is the wire, drawn as a solid line The .$IV$ relationship for a wire is:
• .$V_\text{elem} = 0$: A wire is an ideal connection with a constant, zero voltage across it.
• .$I_\text{elem} = ?$: The current through a wire can take any value, and is determined by the rest of the circuit.
#### Resistor #
• The .$IV$ relationship of a resistor is called Ohm’s Law:
• .$V_\text{elem} = I_\text{elem} R$: The voltage across a resistor is determined by Ohm’s Law.
• .$I_\text{elem} = V_\text{elem} / R$: The current through a resistor is determined by Ohm’s Law.
Limit check
• The slope is proportional to .$R^{-1}$, that is, the larger the resistance the lower the slope
• .$\lim_{R \to 0}$ results in an wire (above)
• .$\lim_{R \to \infty}$ results in an open circuit (below)
#### Open Circuit: #
• This element is the dual of the wire.
• .$V_\text{elem} = ?$: The voltage across an open circuit can take any value, and is determined by the rest of the circuit.
• .$I_\text{elem} = 0$: No current is allowed to flow through an open circuit; always zero.
#### Voltage Source: #
• A voltage source is a component that forces a specific voltage across its terminals. The + and − sign indicates which direction the voltage is pointing. The voltage difference between the “+” terminal and the “−” terminal is always equal to Vs, no matter what else is happening in the circuit.
• .$V_\text{elem} = V_S$ – The voltage across the voltage source is always equal to the source value.
• .$I_\text{elem} = ?$ – The current through a voltage source is determined by the rest of the circuit.
#### Current Source: #
• A current source forces current in the direction specified by the arrow indicated on the schematic symbol. The current flowing through a current source is always equal to Is, no matter what else is happening in the circuit. Note the duality between this element and the voltage source.
• .$V_\text{elem} = ?$ – The voltage across a current source is determined by the rest of the circuit.
• .$I_\text{elem} = I_S$ – The current through a current source is always equal to the source value.
## Rules for Circuit Analysis #
### Kirchhoff’s Current Law (KCL) #
• Node: A place in a circuit where two or more of the above circuit elements meet
• The net current flowing into/out-of any node is zero: $$(-i_1)+(-i_2)+i_3 = 0$$
• That is, the current flowing into a node must equal the current flowing out of that node $$i_1+i_2=i_3$$
### Kirchhoff’s Voltage Law (KVL) #
• The sum of voltages across the elements connected in a loop must be zero
• Mathematically, KVL states that: $$\sum_\text{loop} V_k = 0$$ $$\Longrightarrow V_A - V_B - V_C = 0$$
#### Height Analogy #
If you walk in a circle (a loop) so that you end up back where you started, than your total change in elevation must be zero, no matter how much you go up or down. If you walk in a line, ending up somewhere different, than your total change in elevation is equal to the sum of all of the elevation changes along the way.
Real WorldAnalogy
The sum of voltages across the elements connected in a loop must be equal to zero“what goes up must come down”
If the arrow corresponding to the loop goes into the “+” of an element, we subtract the voltage across that element.We went “downhill” from higher voltage to lower voltage so we lost “elevation.”
If the arrow goes into the “-” of an element, we add the voltage across that elementThis is like going “uphill”
### Ohm’s Law and Resistors #
$$V_\text{element} = I_\text{element}R$$
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# Article
Full entry | PDF (1.2 MB)
References:
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[4] R. Dalmasso: Solutions positives globales d’équations elliptiques semi-linéaires singulières. Bull. Sc. Math. Ser. 2, 112 (1988), 65–76. MR 0942799 | Zbl 0647.35028
[5] A.L. Edelson: Entire solutions of singular elliptic equations. J. Math. Anal. Appl. 139 (1989), 523–532. DOI 10.1016/0022-247X(89)90126-1 | MR 0996976 | Zbl 0679.35003
[6] A. Friedman and L. Veron: Singular solutions of some quasilinear elliptic equations. Arch. Rat. Mech. Anal. 96 (1986), 259–287. DOI 10.1007/BF00251804 | MR 0855755
[7] Y. Furusho: On decaying entire positive solutions of semilinear elliptic equations. Japan. J. Math. 14 (1988), 97–118. MR 0945620 | Zbl 0676.35028
[8] D. Gilbarg and N.S. Trudinger: Elliptic Partial Differential Equations of Second Order, 2nd edition. Springer-Verlag, N.Y., 1983.. MR 0737190
[9] M. Guedda and L. Veron: Local and global properties of solutions of quasilinear elliptic equations. J. Diff. Equa. 76 (1988), 159–189. DOI 10.1016/0022-0396(88)90068-X | MR 0964617
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[11] T. Kura: The weak supersolution-subsolution method for second order quasilinear elliptic equations. Hiroshima Math. J. 19 (1989), 1–36. MR 1009660 | Zbl 0735.35056
[12] T. Kusano and C.A. Swanson: Entire positive solutions of singular semilinear elliptic equations. Japan. J. Math. 11 (1985), 145–155. MR 0877461
[13] T. Kusano and C.A. Swanson: Decaying entire positive solutions of quasilinear elliptic equations. Mh. Math. 101 (1986), 39–51. DOI 10.1007/BF01326845 | MR 0830609
[14] T. Kusano and C.A. Swanson: Radial entire solutions of a class of quasilinear elliptic equations. J. Diff. Eq. 83 (1990), 379–399. DOI 10.1016/0022-0396(90)90064-V | MR 1033194
[15] T. Kusano and W.F. Trench: Global existence of solutions of mixed sublinear-superlinear differential equations. Hiroshima Math. J. 16 (1986), 597–606. MR 0867582
[16] Y. Li and W.M. Ni: On conformal scalar curvature equations in $R^n$. Duke Math. J. 57 (1988), 895–924. DOI 10.1215/S0012-7094-88-05740-7 | MR 0975127
[17] W.M. Ni: Some aspects of semilinear elliptic equations in $R^n$. Nonlinear Diffusion Equations and Their Equalibrium States, Vol. 2 W.M. Ni, L.A. Peletier and J. Serrin (eds.), 1988, pp. 171–205. MR 0956087
[18] W.M. Ni and J. Serrin: Nonexistence theorems for singular solutions of quasilinear partial differential equations. Comm. Pure Appl. Math. 38 (1986), 379–399. DOI 10.1002/cpa.3160390306 | MR 0829846
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[20] P. Tolksdorf: On the Dirichlet problem for quasilinear equations in domains with conical boundary points. Comm. P.D.E. 8 (1983), 773–817. DOI 10.1080/03605308308820285 | MR 0700735 | Zbl 0515.35024
Partner of
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## An email from Sendai (Tohoku University)
13 March, 2011
I was very glad today to receive an email from Akihiro Munemasa (from Tohoku university at Sendai), my long-term collaborator and friend. He says he’s OK, but has no electricity/water/gas…
And, as we see, they have internet, at least some kind of service (email went via me.com, a web-mail service run by Apple).
Well, I can only wish a lot of strength to Akihiro, and all the people affected by this disaster!
PS: today (14th March) he also sent me photos of his office:
## 2010 in review – Happy New Year!
2 January, 2011
While it might be snowing on this page, Edith picked up a mango fallen from a tree outside our block of flats this afternoon…
The stats helper monkeys at WordPress.com mulled over how this blog did in 2010, and here’s a high level summary of its overall blog health:
The Blog-Health-o-Meter™ reads Fresher than ever.
## Crunchy numbers
A Boeing 747-400 passenger jet can hold 416 passengers. This blog was viewed about 5,700 times in 2010. That’s about 14 full 747s.
In 2010, there were 6 new posts, growing the total archive of this blog to 33 posts. There were 5 pictures uploaded, taking up a total of 28kb.
The busiest day of the year was May 11th with 79 views. The most popular post that day was Is C++ the worst 1st programming language for maths majors?.
## Where did they come from?
The top referring sites in 2010 were terrytao.wordpress.com, cameroncounts.wordpress.com, debian-news.net, edventure.ntu.edu.sg, and symomega.wordpress.com.
Some visitors came searching, mostly for funny pictures, funny, confused cat, total unimodularity, and funny pics.
## Attractions in 2010
These are the posts and pages that got the most views in 2010.
1
Is C++ the worst 1st programming language for maths majors? May 2010
2
Total unimodularity and networks October 2009
1 comment
3
Cosets and normal subgroups March 2009
4
Bipartite matchings via linear programming September 2009
1 comment
5
Bellman-Ford algorithm for shortest paths and potentials September 2009
## Python extensions with C libraries made easy by Cython
16 October, 2010
Suppose you have a library written in C that you like to be called from Python. There are many ways to accomplish this, and I would like to show here a complete example of doing this using Cython, which is basically a compiler from (an extension of Python) to C. Most of this, and much more, can be found in a quick tutorial.
For the purpose of an example, suppose you have a C function that computes the average of an array of doubles (all source files here, as well as the Makefile that basically contains the shell commands we show below, can be downloaded here, or cloned using git from github):
/* cmean.c */ double mean(int n, double* a) { double s; int i; for (s=0., i=0; i<n; i++) s+=*(a++); return s/n; }
with the prototype
/* cmean.h */
double mean(int, double*);
and you want to call it from Python, i.e. (we need to give the function another name here, cmean(), for technical as well as for semantic reasons—noone would pass the length of an array and the array to a function in Python, as the array itself would certainly do) say
b=[1.3,5.777777,-12.0,77.] print cmean(b)
would print the same as
print sum(b)/len(b)
As cmean() is not a Python built-in, we would expect to import it first. So the following
$python test.py where # test.py import dyn from dyn import cmean b=[1.3,5.777777,-12.0,77.] print cmean(b) print sum(b)/len(b) will print the number 18.01944425 twice. OK, so here is cmean() definition in Cython (note the file extension .pyx) # m.pyx cdef extern from "cmean.h": double mean(int, double*) from stdlib cimport * def cmean(a): n = len(a) cdef double *v v = malloc(n*sizeof(double)) for i in range(n): v[i] = float(a[i]) m = mean(n, v) free(v) return m The main task of cmean() is to create a “plain” C array of doubles, pass it to our C function, get the result, clean after itself, and return the result. Unlike Python, Cython has types declarations. Unless one calls C-function from the Cython code, they are not mandatory though, although they can speed the things up tremendously in “real” computations. In the example here cdef double *v cannot be avoided — the code does not compile without it. In a usual C-like fasion, instead of 2 lines cdef double *v v = malloc(n*sizeof(double)) one could have written just cdef double *v = malloc(n*sizeof(double)) The top two lines pass the prototype declaration of the C-function mean to be included into the C file to be generated, and the third line imports the functions from the C library stdlib (we need malloc and free from there). Having this, we need to create the extension module dyn, to be imported by Python. This is perhaps the least intuitive part of the job. One way to accomplish it is to use Python distutils: i.e. we need to create a setup.py file to be called as $ python setup.py build_ext --inplace
This should look as follows:
# setup.py from distutils.core import setup from distutils.extension import Extension from Cython.Distutils import build_ext ext_modules=[ Extension("dyn", ["m.pyx"], library_dirs = ['.'], libraries=["cmean"]) # Unix-like specific ] setup( name = "Demos", cmdclass = {"build_ext": build_ext}, ext_modules = ext_modules )
One not yet explained thing here is how the shared library containing the compiled C-function mean() is hooked up to dyn. You see in the code above the declarations library_dirs and libraries. They assume that the shared library is called libcmean.so and that it is in the same directory (i.e. ‘.’ is the path to it) as the rest of the code. One can create libcmean.so (before creating dyn, more precisely, a file called dyn.so) by running
$gcc -fPIC -shared cmean.c -o libcmean.so One more point to watch is that on some Unix systems the loader will be unable to locate the dynamic libraries libcmean.so and/or dyn.so we created (unless they are placed in certain stanadard directories, which is not something one wants to do with experimental code!). So test.py will need to be run, e.g., as follows: $ LD_LIBRARY_PATH=. python test.py
For the sake of completeness, the aforementioned archive contains a C file that calls mean from libcmean.so, and the corresponding entries in the Makefile can do the necessary work to build and run it.
Last but not least, Sage automates parts of the interface building even better.
## Integer flows and graph theory applications
14 October, 2010
Here we are continuing the story of Ford-Fulkerson algorithm. Assume that ${c:D\rightarrow {\mathbb Z}_+}$, i.e. the capacities are all integers. Then starting the Ford-Fulkerson from the zero flow, we see that at each iteration ${\tau(P)\in {\mathbb Z}_+}$. Thus the optimal flow will have integer values for each arc (such flows are called integral). We have
Corollary 3 A maxflow problem with integer capacities has an integral optimal solution.
This has various graph-theoretic applications. We can convert an undirected graph ${\Gamma=(V,E)}$ into a digraph by replacing each edge ${\{x,y\}}$ by a pair of opposite arcs ${xy}$ and ${yx}$, set all arc capacities to 1, and see what this implies for a pair of non-adjacent vertices ${s\neq t\in V}$. Running Ford-Fulkerson algorithm, we obtain an integral maximum ${s-t}$ flow ${f}$, of value ${k:={\mathrm val}(f)}$.
We also know that there is, by the maxflow-mincut Theorem, an ${s-t}$ cut ${(S,V\setminus S)}$ s.t. ${{\mathrm cap}(S,V\setminus S)={\mathrm val}(f)}$. Instead of capacity, it is more appropriate to talk here about the size ${|(S,V\setminus S)|}$ of the cut, i.e. the number of edges crossing from one part of the cut to another. Obviously, by definition of the capacity, we have ${{\mathrm cap}(S,V\setminus S)=|(S,V\setminus S)|}$.
The flow ${f}$ can be viewed as a collection ${D_f}$ of ${k}$ edge-disjoint ${\Gamma}$-paths from ${s}$ to ${t}$, as can be seen by induction on ${{\mathrm val}(f)}$: indeed, if ${{\mathrm val}(f)=1}$ then by construction (via the Ford-Fulkerson algorithm) ${D_f}$ is a path. Suppose now this statement holds for all ${k<{\mathrm val}(f)}$. As ${D_f}$ induces a connected subgraph, it contains a shortest path ${P}$. Then, ${P}$ is cut by a minimum cut ${(S,V\setminus S)}$—otherwise the cut does not cut! Removing all the edges of ${P}$ from ${D}$, we thus obtain a graph ${\Gamma'}$ that has the cut ${(S,V\setminus S)}$ of size ${{\mathrm val}(f)-\ell}$, for ${\ell>0}$. The flow ${f'}$ in ${\Gamma}$ corresponding to ${D_f\setminus P}$ has value ${{\mathrm val}(f)-\ell}$, and ${P}$ is an augmenting path for this flow, with ${\tau(P)=1}$. Thus ${\ell=1}$, i.e. ${(S,V\setminus S)}$ cuts just 1 edge in ${P}$, and by induction ${\Gamma'}$ contains ${k-1}$ edge-disjoint ${\Gamma}$-paths, and we obtain
Corollary 4 (Menger Theorem (for edge-disjoint paths)) The minimum size of an ${s-t}$ cut in ${\Gamma}$ equals the maximal number of edge-disjoint ${s-t}$ paths in ${\Gamma}$.
There is a graph transformation that allows one to prove Menger Theorem for internally disjoint ${s-t}$ paths (two paths between ${s}$ and ${t}$ are called internally disjoint is their only common vertices are ${s}$ and ${y}$). Our goal is to prove the following.
Theorem 5 (Menger Theorem (for internally disjoint paths)) The minimum size of an ${s-t}$ cut in ${\Gamma}$ equals the maximal number of internally disjoint ${s-t}$ paths in ${\Gamma}$.
Proof: Assume ${s}$ and ${t}$ are not adjacent. Replace each ${v\in V\setminus\{s,t\}}$ by a pair of vertices ${v_+}$, ${v_-}$, joined by an arc ${v_+ v_-}$. Then replace each edge ${\{x,y\}}$ of ${\Gamma}$ by a pair of opposite arcs ${xy}$ and ${yx}$. Replace every arc ${xy}$ such that ${s,t\not\in \{x,y\}}$ by the arc ${x_{-}y_{+}}$, each arc ${xy}$ with ${x=s}$ or ${x=t}$ by ${xy_+}$, and each arc ${xy}$ with ${y=s}$ or ${y=t}$ by ${x_-y}$. Now each ${s-t}$ path ${s,x^1,\dots,x^{k-1},t}$ corresponds to the path ${s,x^1_+,x^1_-,x^2_+,\dots,x^{k-1}_+,x^{k-1}_-,t}$ in the new graph, and the opposite holds, too: each ${s-t}$ path in the new graph has this form.
Assign capacities 1 to the arcs ${v_+ v_-}$, and infinite capacities (it suffices to take this “infinity” to be bigger than, say, ${10|E\Gamma|}$) to the rest of the arcs. Due to the choice of capacities, a minimum ${s-t}$ cut ${(S,V\setminus S)}$ will satisfy ${{\mathrm cap}(S,V\setminus S)\leq |E\Gamma|}$, and thus the only arcs from ${S}$ to ${V\setminus S}$ will be of type ${v_+v_-}$. Say, there will be ${k}$ such arcs. By (a straightforward generalisation of) Menger Theorem for edge-disjoint paths (to directed graphs), there will be exactly ${k}$ arc-disjoint ${s-t}$ paths in this digraph. By its construction, they cannot share a vertex ${v_-}$ (resp. ${v_+}$), as there is only one outgoing (resp. incoming) arc on it. Thus these paths are internally disjoint, and the corresponding ${k}$ paths in ${\Gamma}$ are internally disjoint, too. $\Box$
One more classical application is a maxflow-based algorithm to construct a maximum matching in a bipartite graph ${\Gamma=(V,E)}$, with the bipartition ${V=V_s\cup V_t}$. We add two more vertices ${s}$ and ${t}$ to it, and connect ${x\in\{s,t\}}$ to each ${v\in V_x}$ by an edge. In the resulting new graph ${\Gamma'}$ we solve the ${s-t}$ maxflow problem with all capacities set to 1. More precisely, we orient each edge so that each ${\{v_s,v_t\}}$, ${\{s,v_s\}}$, ${\{v_t,t\}}$ becomes an arc ${v_sv_t}$, respectively, ${sv_s}$, respectively ${v_tt}$.
Ford-Fulkerson algorithm, when started from the zero flow, will produce an integral maximal flow, of value ${k:={\mathrm val}(f)}$. As we know from Menger’s theorem, we will have ${k}$ internally disjoint ${s-t}$-paths. They will induce a matching ${M_f}$ on ${\Gamma}$ (just look at the ${\{v_s,v_t\}}$ edges only). On the other hand, any matching ${M}$ on ${\Gamma}$ can be used to construct an ${s-t}$-flow of value ${|M|}$ in ${\Gamma'}$ with all capacities 1. Hence ${M_f}$ is a maximum matching in ${\Gamma}$.
## Augmenting paths for Maxflow and Mincut
11 October, 2010
We discussed how to formulate the MAXFLOW problem for networks, i.e. arc-weighted digraphs ${(V,D,c)}$ with two specific vertices, source ${s}$ and sink ${t}$ and arc capacities ${c:D\rightarrow{\mathbb R}_+}$, and treat it via linear programming. An alternative approach is via augmenting paths. The paths we talk about will disregard the arc directions in ${D}$, i.e. they will be ${s}$${t}$ paths in the underlying undirected graph. (When ${e\in P}$ has the direction opposite to the arc in ${D}$, we write ${e\in P\setminus D}$.)
Definition 1 Given an ${s}$${t}$ flow ${f:D\rightarrow {\mathbb R}_+}$, an ${f}$augmenting path is an ${s}$${t}$ path ${P}$ in the underlying graph, such that for each ${e\in P}$
• if ${e\in D}$ then ${f(e);
• if ${e\not\in D}$ then ${f(e)>0}$.
The tolerance of ${P}$ is
$\displaystyle \tau(P):=\min(\min_{e\in P\setminus D} f(e), \min_{e\in P\cap D}c(e)-f(e)).$
Note that the definition of tolerance extends unchanged to any path starting at ${s}$.
We can use ${P}$ to improve ${f}$, as follows.
Lemma 2 For each ${e\in D}$, let
$\displaystyle f'(e)=\begin{cases} f(e) & e\not\in P\\ f(e)-\tau(P) & e\in P\setminus D\\ f(e)+\tau(P) & e\in P\cap D \end{cases}.$
Then ${f'}$ is an ${s}$${t}$ flow in ${(V,D,c)}$, of value greater than that of ${f}$ by ${\tau(P).}$
Proof: It is immediate from the definition of ${\tau(P)}$ that ${0\leq f'(e)\leq c(e)}$ for any ${e\in D}$. We still need to check the flow conservation constraints
$\displaystyle \sum_{x: xv\in D} f'((xv))=\sum_{x: vx\in D} f'((vx))\quad\forall v\in V\setminus\{s,t\}.$
If ${P}$ visits ${v\in V\setminus\{s,t\}}$ then there are two arcs, say, ${xv}$ and ${vy}$ on ${v}$ that are affected when we change ${f}$ to ${f'}$, and the changes in flow values cancel each other in each of the four cases (recall that ${P}$ is a path in the underlying undirected graph). Finally, the value of ${f'}$ is the net outflow from ${s}$, and the latter is greater than that of ${f}$ by ${\tau(P)}$. $\Box$
This suggests finding a maximum flow by repeatedly finding an augmenting path and improving the current flow with it, till no augmenting path is available (this is, for instance, how Ford-Fulkerson algorithm is working). For this to succeed, we obviously need to prove that ${f}$ is maximum if and only if it does not have an ${f}$-augmenting path.
Augmenting paths are found by a breadth-first search from ${s}$. At some point, if we did not succeed in reaching ${t}$, we would end up with the set of vertices ${S}$ reachable, in the underlying non-oriented graph, by paths with positive tolerance. Each ${e\in D}$ crossing over from ${S}$ to ${T=V\setminus S}$ satisfies the property that ${c(e)=f(e)}$, and each arc ${e}$ crossing from ${T}$ to ${S}$ satisfies ${f(e)=0}$ — otherwise we would have added their vertices in ${T}$ to ${S}$. Therefore,
$\displaystyle {\mathrm val} (f)=\sum_{xy\in D: x\in S, y\in T} c((xy))=:{\mathrm cap} (S,T),$
where on the right we have the definition of capacity of an arbitrary ${s}$${t}$ cut ${(S,T)}$.
It is easy to show that for an arbitrary ${s}$${t}$ cut ${(S',T')}$, one has ${{\mathrm cap} (S',T')\geq {\mathrm val} (f)}$. Thus we have have obtained the Maxflow-Mincut Theorem, and also proved that our algorithm, to be described in more detail shortly, finds a maximum flow ${f}$, and a corresponding to it minimum cut ${(S,T)}$.
As usual in breadth-first search, we use a queue, i.e. a data structure ${(Q,p,q)}$ that has two indices, one, ${q}$, pointing to the end of the queue, for new arrivals, and the other, ${p}$, pointing to the beginning of the queue, i.e. to the first element to be processed. There will be elementary operations:
• ${{\mathrm Insert} (Q,x)}$ – adds a new element, ${x}$, to the queue, and icrements ${q}$. In this case we also do not allow an element to enter the queue twice, so ${{\mathrm Insert} (Q,x)}$ has no effect if ${x}$ already was these once.
• ${x:={\mathrm First} (Q)}$ – takes the (currently) first element, ${x}$, from the queue, and icrements ${p}$—this will fail if ${Q}$ is empty, i.e. if ${p=q}$.
Now the algorithm is as follows:
1. Input: a feasible flow ${f}$ in ${(V,D,c)}$ with source ${s}$ and sink ${t}$.
2. Initialization: let ${(Q,p,q)}$ be an empty queue; ${{\mathrm Insert} (Q,s)}$, and set ${S:=\{(s,\infty)\}}$.
3. Loop: take ${x:={\mathrm First} (Q)}$. If this fails, return the first coordinates of the pairs in ${S}$—a minimum cut.
4. for ${v\in V}$ s.t. ${(xv)\in D}$ and ${f((xv)), or s.t. ${(vx)\in D}$ and ${f((vx))>0}$:
• ${{\mathrm Insert} (Q,v)}$, ${S:=S\cup\{(v,x)\}}$.
• if ${v=t}$ then return the augmenting ${s}$${t}$ path by tracing back the pairs ${(t,v_t)}$, ${(v_t,v_{t-1})}$,\dots, ${(v_1,s)}$ in ${S}$.
5. go to Loop.
So if ${f}$ is not maximum we get an ${f}$-augmenting path, that can be used to improve ${f}$, and repeat. However, the speed of this procedure becomes unacceptable when the capacities are large. If ${c\in\mathbb{Q}^{|D|}}$ then we are at least assured that it will converge after finitely many steps: indeed, we can multiply ${c}$ by the least common multiple of its denominators, reducing to the case ${c\in\mathbb{Z}^{|D|}}$. Each iteration in this case improves the flow by at least ${1}$. However, the time would not be polynomial in the input of the problem. E.g. for the following graph the number of iterations can be as large as ${2M}$, when ${M}$ is a large integer.
For irrational ${c}$ the convergence not even need to be the case!
If one chooses a shortest ${f}$-augmenting path to improve ${f}$ with, then the running time becomes bounded by ${O(|V||D|^2)}$—that was shown by E.Dinitz in 1970 and independently by J.Edmonds and R.Karp in 1972. Details, and further improvements by A.Karzanov et.al can be found e.g. in A.Schrijver’s lecture notes, Sect.4.4.
## Edmonds algorithm for maximum matchings
5 October, 2010
We continue with matchings in simple graphs, following the exposition in A.Schrijver’s 3-volume work, and his lecture notes available online, filling in few more details.
While in a bipartite graph ${\Gamma=(V,E)}$ and a matching ${M\subset E}$ it is always the case that a shortest path from the set ${X\subset V}$ of ${M}$-unsaturated vertices to itself (i.e. an ${X}$${X}$-path) is always ${M}$-augmenting, this is no longer true in general graphs:
here ${X=\{a,b\}}$, ${M}$ is indicated by thicker edges, and the shortest path (of length 4) between them is not augmenting. This precludes the algorithmic idea of augmenting a matching ${M}$ using an augmenting path (found by a shortest path procedure in a digraph constructed from ${\Gamma}$ and ${M}$) from working in the general case. Edmonds came up with an idea that one can allow for repeated vertices, and shink the arising in the process cycles to (new) vertices.
Definition 1 An ${M}$-alternating walk ${P=(v_0,\dots,v_t)}$ is a sequence of vertices (some of which can repeat!) that is a walk (i.e. ${v_i v_{i+1}\in E}$ for all ${0\leq i\leq t-1}$) so that exactly one of ${v_{i-1} v_i}$ and ${v_i v_{i+1}}$ is in ${M}$.
Definition 2 An ${M}$-alternating walk ${P=(v_0,\dots,v_t)}$ is an ${M}$flower if ${P}$ has an even number of vertices (as a sequence, i.e. ${t}$ is odd), ${v_0\in X}$, all the ${v_0,\dots,v_{t-1}}$ are distinct, and ${v_t=v_{2k}}$ for ${2k (Removing the vertex ${b}$ and the edge ${bs}$ on the picture above is an example of an ${M}$-flower).
The cycle ${(v_{2k},v_{2k+1},\dots, v_t)}$ in an ${M}$-flower is an ${M}$blossom. (${(qvuts)}$ on the picture above is an example of an ${M}$-blossom.)
Once we have an ${M}$-blossom ${B}$, we can shrink it by replacing all of its vertices ${V(B)}$ with just one vertex (call it ${B}$, too), replacing each edge ${uv}$, ${v\in V(B)}$ with the edge ${uB}$ (and ignoring any loops that arise on ${B}$). The resulting graph is denoted by ${\Gamma/B=(V/B, E/B)}$. (E.g. shrinking the ${M}$-blossom ${B}$ on the picture above produces the graph that is the path ${apBb}$). In ${\Gamma/B}$ one will have the matching ${M/B}$, where ${B}$ will become just a saturated vertex.
Theorem 3 ${M}$ is a maximum matching in ${\Gamma}$ if and only if ${M/B}$ is a maximum matching in ${\Gamma/B}$, for an ${M}$-blossom ${B}$.
Proof: If ${M/B}$ is not a maximum matching in ${\Gamma/B}$ then there will be an ${M/B}$-augmenting path ${P}$ in ${\Gamma/B.}$ If ${P}$ does not meet ${B=(v_{2k},\dots,v_t)}$ then it corresponds to an ${M}$-augmenting path ${P'}$ in ${\Gamma}$. So we can assume that ${uB\not\in M}$ is an edge of ${P}$, and ${Bu'\in M}$ is also an edge of ${P}$. As ${uB\in E/B}$, there exists ${2k\leq j so that ${uv_j\in E.}$ Now, depending upon the parity of ${j}$, we replace ${B}$ in ${P}$ by the part of the blossom rim ${v_j,v_{j\pm 1},v_{j\pm 2},\dots,v_t}$, where the sign in the indices is ${+}$ for ${j}$ odd, and ${-}$ for ${j}$ even.
Conversely, if ${M'}$ is not a maximum matching in ${\Gamma}$, then
$\displaystyle M=M'\Delta \{v_0v_1,\dots,v_{2k-1}v_{2k}\}$
is also a matching of the same size, and ${|M/B|=|M'/B|}$. Thus it suffices to show that ${M/B}$ is not maximal in ${\Gamma/B}$, and we will assume w.l.o.g. that ${k=0}$, i.e. ${v_{2k}=v_0\in X}$ (our flower has no stem). There exists an ${M}$-augmenting path ${P=(u_0,\dots,u_s)}$. W.l.o.g. ${P}$ does not start in ${B}$ (we can reverse it if it does). If ${P}$ does not contain a vertex in ${B}$, it will continue to be an ${M/B}$ augmenting path in ${\Gamma/B}$. Otherwise let the first vertex of ${P}$ in ${B}$ be ${u_j}$. Then ${(u_0,\dots,u_{j-1},B)}$ is an augmenting path in ${\Gamma/B}$. $\Box$
For the algorithm we will need
Theorem 4 A shortest ${X}$ to ${X}$ ${M}$-alternating walk is either a path or its initial segment is an ${M}$-flower.
Proof: Assume ${P=(v_0,\dots,v_t)}$ is not a path, and let ${i be such that ${v_i=v_j}$ and ${j}$ as small as possible. We will show that ${(v_0,\dots,v_j)}$ is an ${M}$-flower, and we immediately have it if ${i}$ is even and ${j}$ is odd.
If ${j-i}$ were even, as shown above, where the walk might have been ${(apqstpqb)}$, and so ${j=5}$, ${i=1}$, then the path ${(v_i,\dots,v_{j-1})}$ is ${M}$-alternating, so we would simply delete it from ${P}$ and obtain a shorter ${M}$-alternating walk.
If ${i}$ were odd and ${j}$ even, we would have ${v_iv_{i+1}}$ and ${v_{j-1}v_j}$ in ${M}$, so ${v_{i+1}=v_{j-1}}$, contradicting the choice of ${j}$. $\Box$
Combining these two observations, we obtain the following augmenting path algorithm:
1. Input: a graph ${\Gamma}$ and a matching ${M}$.
2. Let ${X}$ be the set of vertices unsaturated by ${M}$. Construct, if possible, a positive length shortest walk ${P}$ from ${X}$ to ${X}$.
3. If no ${P}$ was constructed, return ${\emptyset}$${M}$ is maximum.
4. If ${P}$ is a path then return it.
5. Otherwise take the first blossom ${B}$ from ${P}$, call the algorithm recursively on ${\Gamma/B}$ and ${M/B}$, and return the expanded to an augmenting path in ${\Gamma}$ resulting path in ${\Gamma/B}$.
Applying the algorithm starting from ${M=\emptyset}$, and augmenting ${M}$ with the path it returns, we obtain an efficient algorithm, running in fact in ${O(|V|^2|E|)}$ operations.
We still need to discuss here how to do Step 2. To this end we construct a directed graph ${D_M=(V,A\cup A')}$ with
$\displaystyle A:=\{(u,v)\mid u\neq v\in V, \exists ux\in E, xv\in M\}$
$\displaystyle A':=\{(u,v)\mid u\neq v\in X, \exists uv\in E\}$
Lemma 5 Each ${X}$ to ${X}$ ${M}$-alternating path, resp. each ${M}$-flower, corresponds to a directed path in ${D}$ from ${X}$ to ${\Gamma(X)}$—the set of neighbours of ${X}$ in ${\Gamma.}$ Conversely, each ${X}$ to ${\Gamma(X)}$ directed path in ${A}$ corresponds either to an ${M}$-augmenting path or to an ${M}$-flower.
Proof: (Sketch) Each length 1 ${M}$-augmenting path ${uv}$ obviously corresponds to the pair of arcs ${uv}$ and ${vu}$ in ${A'}$.
If ${u_0,u_1,\dots,u_s}$ is a ${A}$-path in ${D}$, so that ${u_0\in X}$, ${u_t\in\Gamma(X)}$, then there is a unique, up to the choice of ${u_{s+1}\in X}$, lifting of it to an ${M}$-alternating walk ${P=u_0,v_1,u_1,v_2,u_2,\dots,v_s,u_s,u_{s+1}}$, with ${v_iu_i\in M}$ for ${1\leq i\leq s}$. If ${P}$ is a path then it is ${M}$-augmenting.
If ${P}$ has repeated vertices, this repetition occurs in a regular way, corresponding to an ${M}$-flower. As we move along ${P}$ from ${u_0}$ to ${u_s}$, let ${u_t=v_i}$, ${t>i}$, be the first repetition. Then ${u_i}$ is the beginning of the blossom, and ${u_{t-1}}$ the last vertex of the blossom. $\Box$
## Installing CPLEX 12.2 on Debian amd64 and MacOSX 10.6 (64-bit)
24 September, 2010
IBM now gives CPLEX to academics for free; as I am computing LP bounds for quantum codes, I got myself one. However, it refused to install on my amd64 Debian system, saying
"libgcc_s.so.1 must be installed for pthread_cancel to work".
After much googling, it turned out that the installer is a 32-bit program, that needs the right 32-bit libgcc_s.so to work.
To get this on Debian (squeeze), I just did
$apt-get install ia32-libs Then $ LD_LIBRARY_PATH=/usr/lib32; sh cplex_studio122.acad.linux-x86.bin
worked as it should.
The installation goes smoothly on a 64-bit MacOSX 10.6. But running is not: one has problems when doing
>>> import cplex ... from cplex._internal.py1013_cplex122 import * ImportError: dlopen(/Library/Python/2.6/site-packages/cplex/_internal/py1013_cplex122.so, 2): no suitable image found. Did find: /Library/Python/2.6/site-packages/cplex/_internal/py1013_cplex122.so: mach-o, but wrong architecture
One can make it work on by setting
\$ export VERSIONER_PYTHON_PREFER_32_BIT=yes
before starting the Python session (as IBM/CPLEX Support kindly told me; they promised a fix in an upgrade, too).
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2017-06-22 16:34:17
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https://physics.stackexchange.com/questions/422518/pile-up-in-particle-physics
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# Pile-up in particle physics
Regarding pile-up, I came across two definitions. One dealing with electronics of detector where more than one event get recorded at the same time and other regarding all the pp collisions in the bunch which do not lead to hard scatter that we are looking for our analysis. My question is regarding the second type of pile-up I mentioned.
Below is slide(from HEP presentation) about the same where they latter tell how to re-weight MC to data to match rho of data. Here, in the slide - If I have "N" primary vertex then does that mean "N" interactions as well? In that case, I can't understand the plot in the slide...
Thanks a lot!
$\mu$ is the average expected number of interactions. $\left<n_V\right>$ is the number of vertices found by the experiment in the data. Ideally they would be the same, of course. But in the real world some interactions are so subtle they are not detected (maybe the tracks go down the beam pipe not into the detector), and sometimes extra (spurious) vertices are found, so they don't agree. Correct understanding of this difference is important - hence the slide.
• In this context $\mu$ is the usual measure. – RogerJBarlow Aug 14 '18 at 5:58
• I guess you mean $\nu$ here?.. As $\mu$ is same for both data and MC – kbg Aug 14 '18 at 9:19
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2020-07-07 11:17:23
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http://www.ams.org/mathscinet-getitem?mr=587910
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MathSciNet bibliographic data MR587910 (82f:32035) 32H15 (46J15 78A55) Helton, J. William The distance of a function to $H\sp{\infty }$$H\sp{\infty }$ in the Poincaré metric; electrical power transfer. J. Funct. Anal. 38 (1980), no. 2, 273–314. Article
For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.
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2014-04-25 00:16:12
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https://mathoverflow.net/questions/267615/why-are-the-toric-fibers-of-a-toric-manifold-lagrangian-submanifolds
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# Why are the toric fibers of a toric manifold Lagrangian submanifolds?
How does one know in general that the toric fibers of a Kahler toric manifold are Lagrangian submanifolds? I can roughly fathom this for e.g. a circle in $\mathbb{C}P^1$, but how about more general toric manifolds of higher dimension?
By definition, a toric variety of complex dimension $n$ comes with an action of a complex torus $(\mathbb{C}^{*})^n$. I guess that by Kähler toric manifold, you mean a toric variety $X$ equipped with a Kähler form $\omega$ which is preserved by a compact subgroup $U(1)^n$ of $(\mathbb{C}^{*})^n$. The toric fibration $X \rightarrow \mathbb{R}^n$ is then the moment map of this action of $U(1)^n$ on $(X,\omega)$. More precisely, choosing Hamiltonian functions $H_1$,...,$H_n$ generating the action of $U(1)^n$ (i.e. such that the corresponding Hamiltonian vector fields generate the action of $U(1)^n$), then the toric fibration $X \rightarrow \mathbb{R}^n$ is given by $x \mapsto (H_1(x),...,H_n(x))$. Because $U(1)^n$ is abelian, the vector fields generating the action commute and so the Hamiltonian functions Poisson commute. It is a general fact that if one has a symplectic manifold of real dimension $2n$, with $n$ functions $H_1,...,H_n$ Poisson commuting (and with linearly independent differentials), then the common vanishing locus of these functions is a Lagrangian submanifold (for example, by Darboux lemma, you can locally complete these $n$ functions by $n$ other functions $\theta_1$,...,$\theta_n$ such that $H_i$, $\theta_j$ are local coorrdinates such that $\omega=\sum_{i=1}^n dH_i \wedge d\theta_i$ and the result is then obvious).
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2021-12-06 18:53:09
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https://physics.stackexchange.com/questions/508106/intuition-behind-focusing-vs-defocusing-in-integrable-systems-like-nls-kdv-mkd
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# Intuition behind focusing vs defocusing in integrable systems like NLS, KdV, mKdV
The following are examples of integrable systems arising from the AKNS system (check out AKNS paper here and a short Wikipedia description)
1. Non-Linear Schrodinger equation
2. Korteweg-de Vries equation
3. modified Korteweg-de Vries equation
I am paying attention the last one (mKdV) that has the form $$u_t\pm3u^2u_x+u_{xxx} = 0$$ I know that the $$\pm$$ represent two cases that change fundamentally change the dynamics of the system. I would like intuition: 1) on how to recognize whether this is the focusing or defocusing case, and 2) the effect of the non-linearity.
I know there are focusing and defocusing cases for the first two equations too, but how does the sign of the non-linearity show this? and what is the intuitive effect of focusing/defocusing? Is there a stability result associated to each case?
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2020-10-22 00:09:19
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