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https://leastaction.wordpress.com/category/programming/
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# Least Action
Nontrivializing triviality..and vice versa.
## NVIDIA CUDA on Fedora 20 x86-64 with NVIDIA Optimus
So after weeks of experimentation, and frustration, I think I have finally gotten CUDA to work on my MSI GE 70 2OE-071 laptop. As you may know, NVIDIA Optimus is a technology that enables power-sharing and switching between Intel HD and NVIDIA GPUs on laptops. For some reason, MSI and some other manufacturers have disabled the option to disable Optimus technology from within the BIOS. Now as far as I know, Linux has shaky support for Optimus, and while I had no trouble with any Linux distro on my previous Dell XPS laptops (which did have Optimus) I just couldn’t get NVIDIA discrete graphics to work on my MSI “gaming” laptop, which is a shame, because much of my work involves (or has involved) the use of CUDA, and I am not eager to spend my programming time on Windows.
This is not an exhaustive guide, and I invite suggestions and corrections from users who may stumble upon this blog post. I tried to reproduce the scenario on my own laptop in order to make sure I didn’t miss anything.
Typically there are six stages in NVIDIA CUDA installation on Linux:
1. Get rid of the nouveau driver if you are installing the proprietary NVIDIA driver by hand.
2. Reboot, get to a text mode (sadly, init doesn’t work anymore, so if you’re a old Unix user like me, you have to google this up before you start playing with your system).
3. Install the NVIDIA device driver.
4. Reboot, make sure a basic X configuration works (known as the “graphical mode” nowadays).
5. Install the CUDA toolkit, and set PATH and LD_LIBRARY_PATH variables.
6. Compile CUDA samples.
Now I believe much of the errors you will face in steps 1 through 6 are from a partial understanding of the way the X Server and Linux work, and that is based on my own experience. Playing with Linux can be a humbling experience even for someone who has used it for over a decade! But this is quite a lot of fun if you can document everything, and remember what you did to get something to work. It also helps if you have a system to play with, so if you have issues with installing and re-installing the OS n times before you can get to perfection, then this guide isn’t for you. Sorry!
Stuff I had to get to run the CUDA samples:
gcc-g++ in Fedora
libXmu-devel in Fedora (this installs the libXmu libraries)
<UNDER CONSTRUCTION>
Written by Vivek
April 20, 2014 at 19:16
Posted in CUDA, Linux, Programming
## Article: How to Learn Emacs: A Hand-drawn One-pager for Beginners » sacha chua :: living an awesome life
How to Learn Emacs: A Hand-drawn One-pager for Beginners » sacha chua :: living an awesome life
http://sachachua.com/blog/2013/05/how-to-learn-emacs-a-hand-drawn-one-pager-for-beginners/
Written by Vivek
May 19, 2013 at 10:56
Posted in Linux, Programming
## Gnuplot with C/C++
Sometimes, it is convenient to call gnuplot from within your own C/C++ code rather than having the code create a data file, and manually execute gnuplot every time to re-plot using the contents of the file. This is easy if you use Linux: just use a pipe. For instance:
#include <iostream> #include <stdio.h> #include <cstdlib>
using namespace std; int main(){ // Code for gnuplot pipe FILE *pipe = popen("gnuplot -persist", "w"); fprintf(pipe, "\n"); // Your code goes here
// Don't forget to close the pipe fclose(pipe); return 0; }
Note that we have had to use C routines to call gnuplot. But there’s a problem with this code: gnuplot will generate the plot only when the pipe is closed. So if you have multiple plot statements, or if you want to refresh the same plot (as for instance, in an animation scenario) you will find that the above approach will result in the plots getting asynchronously updated right at the end.
So how does one ensure that after every plot command, a plot is actually generated (or an existing one refreshed) Immediately? The key is to flush the buffer using fflush. Simply insert
fflush(pipe);
below every plot call to gnuplot. If you have an animation, then in order to make the transitions smoother, you might want to include a delay subroutine. A simple one can be built using the clock routines in time.h. For an example, click here.
Thanks to Vidushi Sharma for bringing these issues to my attention.
Written by Vivek
June 22, 2012 at 14:13
## Customizing GNOME3
The purpose of this post is to help you configure GNOME3 and make it look more ‘zazzy’ than it is. Most of these tweaks have been inspired by the archlinux forums and FP Murphy’s blog, which is the source for learning how to customize GNOME3. I thought it would be a good idea to put it all in one place, mostly for my own reference. Note that I have tested all this on Natty 64-bit. I also thank Charles Bowman for his help and inputs over the past few weeks.
The gnome-tweak-tool does not let you change much, unless you add extensions to the gnome-shell, which really open up the world for a tweaker. I won’t go into the anatomy of an extension, as FP Murphy does an excellent job in explaining it. I will just assume that you know enough about it. If you don’t, you will still be able to get all these tweaks to work, but you won’t understand how they work, and believe me, it won’t be quite as much fun!
Extensions are installed in
~/.local/share/gnome-shell/extensions
but if they are to be made accessible to all users, then they should go in
/usr/share/gnome-shell/extensions
You can pick either configuration. I will work with ~/.local/share/gnome-shell/extensions from now on, because I am running all this on a single user system. Remember that ~/ is a mnemonic for your home directory. Instead of using the long /home/user_id/, I will just refer to it as ~/ from now on.
One way to play with extensions is to get hold of Charles Bowman’s gnatty pack. It is available from http://cid-32ec6e89f4aa803b.office.live.com/self.aspx/Linux/gNatty%20Pack.tar.gz?wa=wsignin1.0&sa=92561471. Since several extensions seem to be around, with different names, I will just go over the ones I have gotten to work one by one.
Some words of caution before we begin the expedition: some extensions are not compatible with others, and if you install two mutually incompatible extensions, it is possible that the Gnome Shell will not load, and you will be asked to log out. This can be fixed by moving the bad extension out of ~/.local/share/gnome-shell/extensions or simply renaming it:
mv ~/.local/share/gnome-shell/extensions/bad_extension ~/.local/share/gnome-shell/extensions/bad_extension.old
It turns out that this works because each extension has a file inside called ‘metadata.json’ with a field called a unique id (uuid) which must match the name of the directory. Changing the directory name using the mv command above will create a disparity between uuid and the directory name, which will prevent the extension from loading.
If you install a whole bunch of extensions together, it will obviously be hard to debug any problem, because you don’t know which extension is the culprit. So it is best to install one extension at a time, and restart the GNOME Shell. This can be achieved either by logging out and logging back in, or pressing Alt + F2 and typing r and pressing Enter. This restarts the shell. Sometimes though, restarting the shell this way results in spurious errors which may suggest that the extension is messing up something. So, I do prefer to double-check by logging out and logging back in again, though this is relatively cumbersome as it will close all open applications.
With this preliminary knowledge, we are ready to embark on the first step. The basic algorithm is to create two files, extension.js and metadata.json, for every extension. Each extension resides in its unique directory (folder) in ~/.local/share/gnome-shell/extensions/.
Extension #1: Move Clock
The default position of the clock in Gnome is in the center of the top bar. This is irritating for those of us who are used to the old Gnome2 interface where the clock would appear on the right. This is easily fixed.
cd ~/.local/share/gnome-shell/extensions
mkdir Move_Clock@rmy.pobox.com
cd Move_Clock@rmy.pobox.com
Now create two files using your favorite editor, extension.js and metadata.json with the following contents:
extension.js
const St = imports.gi.St;
const Main = imports.ui.main;
function main() {
let children = Main.panel._rightBox.get_children();
Main._nWorkspacesChanged = function() {};
Main._queueCheckWorkspaces = function() {};
Main._checkWorkspaces = function() {};
}
{
"shell-version": ["3.0"],
"uuid": "Move_Clock@rmy.pobox.com",
"name": "Move Clock",
"description": "Move clock to left of status menu button"
}
Restart the shell. The clock should now be visible on the right hand side of the top bar. We can customize it further, to display the time and date in a detailed fashion.
First start gnome-tweak-tool, and click on Shell. You should see a window which looks like this:
gnome-tweak-tool shell tab
You can toggle the ON/OFF switches for ‘Show date in clock’ and ‘Show week date in the calendar’.
Some more date and time settings can be changed using gnome-control-center. Type
gnome-control-center
in a terminal window and click on Date and Time. Note that this tool does not allow you to show a seconds counter in the clock. If you wish to display seconds, fire up a gnome-terminal window and type
gsettings set org.gnome.shell.clock show-seconds true
This will display a live ticking seconds counter.
Note that the status of presently installed gnome-shell-extensions can be determined from and changed within gnome-tweak-tool. Every time you install an extension manually and restart the shell, by default it is enabled.
Under Construction
Written by Vivek
June 1, 2011 at 15:09
Posted in Linux, Programming
## Evaluation of Complex Error Functions erf(z) using GSL/C
For a small QFT calculation, I needed to numerically evaluate the imaginary error function erfi(x) = erf(i x). But it turns out that GSL (and most other numerical recipe code I could find) can only deal with erf(x), where x is real. Here’s a poor man’s implementation of erf(z) through a standard Taylor expansion,
$erf(z) = \frac{2}{\sqrt{\pi}}\sum_{n = 0}^{\infty}\frac{(-1)^n z^{2n+1}}{n!(2n+1)}$
The catch here is to deal with propagation of errors in the complex Taylor series, and to also somehow benchmark the results. Well, I haven’t been able to think about this yet, but I was able to confirm that for real arguments, my Taylor series code is about as good as the GSL error function gsl_sf_erf(x).
I am working on a CUDA implementation of this now, because in my project, I need to perform a numerical integration over the error function, which is quite intensive even for Mathematica. For now, I’m just sharing the serial implementation. (Hint: if you can write wrapper for each of the gsl functions inside my Taylor series calculating method, you can embed them within a __global__ kernel call through a struct. Alternatively — and less fun — you can just parallelize the for loop.)
/* Function to compute erf(z) using a Taylor series expansion
/* Author: Vivek Saxena
/* Last updated: January 28, 2011 21:11 hrs */
#include <gsl/gsl_complex_math.h>
#include <gsl/gsl_sf_erf.h>
#include <gsl/gsl_sf_pow_int.h>
#include <gsl/gsl_sf_gamma.h>
#include <stdio.h>
#define PI 3.1415926543
double cz = 2/sqrt(PI);
const int TERMS = 10; // no of terms to use in the Taylor series
gsl_complex erfTaylor( gsl_complex z, int trunc ){
gsl_complex res = gsl_complex_rect(0,0),
num = gsl_complex_rect(0,0),
den = gsl_complex_rect(1,0),
snum = gsl_complex_rect(1,0),
temp = gsl_complex_rect(0,0);
for(int i = 0; i < trunc; i++){
snum = gsl_complex_rect( cz * gsl_sf_pow_int(-1, i), 0 );
num = gsl_complex_mul(snum, gsl_complex_pow_real(z, 2*i+1));
den = gsl_complex_rect((2*i + 1)*gsl_sf_fact(i),0);
temp = gsl_complex_div(num, den);
res = gsl_complex_add(res, temp);
}
return res;
}
int main ( void ){
printf( "Real error function\n\n");
for ( float i = 0; i <= 1; i += 0.01 ){
float gslerror = gsl_sf_erf(i);
float taylor = GSL_REAL( erfTaylor( gsl_complex_rect(i, 0), 10 ) );
printf("erf(%f): gsl = %f, taylor = %f, mag error = %f\n", i, gslerror, taylor, abs(gslerror-taylor));
}
gsl_complex t, arg;
printf( "\n\nImaginary error function\n\n");
for (float i = 0; i <= 1; i += 0.01 ){
/* this would be your generic argument z in erf(z).
* so if z = x + iy, then
* arg = gsl_complex_rect(x, y);
*/
arg = gsl_complex_rect(0, i);
t = erfTaylor( arg, TERMS );
printf("erf(%f + i %f) = %f + i %f\n", GSL_REAL(arg), GSL_IMAG(arg), GSL_REAL(t), GSL_IMAG(t));
}
return 0;
}
Written by Vivek
January 28, 2011 at 21:12
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2017-12-12 14:08:47
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http://goldbook.iupac.org/terms/view/W06670
|
Warning (2): get_headers(): https:// wrapper is disabled in the server configuration by allow_url_fopen=0 [APP/Controller/TermsController.php, line 315]
If in a series of observations a @D01556@ ($$w_{i}$$) is assigned to each value, a weighted mean $$\overline{x}_{w}$$ can be calculated by the formula: $\overline{x}_{w}=\frac{\sum w_{i}\ x_{i}}{\sum w_{i}}$
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2019-08-21 02:57:40
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https://kuscholarworks.ku.edu/handle/1808/10337?show=full
|
dc.contributor.advisor Barnett, William A. dc.contributor.author Kanyama, Isaac Kalonda dc.date.accessioned 2012-10-28T17:34:50Z dc.date.available 2012-10-28T17:34:50Z dc.date.issued 2011-05-31 dc.date.submitted 2012 dc.identifier.other http://dissertations.umi.com/ku:11982 dc.identifier.uri http://hdl.handle.net/1808/10337 dc.description.abstract Using Monte Carlo experiments, I address two issues in demand analysis. The first relates to the performance of local flexible functional forms in recovering the time-varying elasticities of a true model, and in correctly identifying goods as complements, substitutes, normal or inferior. The problem is illustrated with the nonlinear almost ideal demand system (NLAI) and the Rotterdam model (RM). For the AIDS, I also consider two versions of its linear approximation: one with simple formulas (LAISF) and the other with corrected formulas (LAICF). The second issue concerns the ability of the flexible functional structures to satisfy theoretical regularity in terms of the Slutsky matrix being negative semi-definite at each time period of time. I tackle these issues in the framework of structural time series models, computing the relevant time-varying elasticities by means of Kalman filtered and smoothed coefficients. The estimated time-varying coefficients are obtained under the pure random walk and the local trend hypotheses. I find that both the NLAI and the RM qualitatively perform well in approximating the signs of the time-varying income and substitution elasticities. Quantitatively, the RM tends to produce values of the time-varying elasticity of substitution close to the true ones within separable utility branches while the NLAI tends to produce overestimating values. On the other hand, the RM produces time-varying income elasticities with values close to the true ones while the NLAI tends to produce constant values over time. The LAISF model qualitatively performs similarly to the NLAI, but the LAICF does not. Finally, the NLAI achieves higher levels of the regularity index under the local trend specification while the RM achieves higher regularity levels under the random walk specification. In contrast, the LAISF and the LAICF models achieve lower levels of regularity under both specifications of the time-varying coefficients. Globally, the LAICF which widely adopted in applied work performs poorly compared to the RM and the NLAI. These findings are robust to different values of the time-varying parameters in the utility function. Two implications emerge from this research. First, the LAICF model should be considered as a model on its own rather than as an approximation of the NLAI. Second, the choice between an AIDS-type model and the RM should be motivated by their performance with respect to the properties a hypothesized true model for the data at hand, especially when working with real data. dc.format.extent 115 pages dc.language.iso en dc.publisher University of Kansas dc.rights This item is protected by copyright and unless otherwise specified the copyright of this thesis/dissertation is held by the author. dc.subject Economics dc.subject Almost ideal demand system dc.subject Monte Carlo study dc.subject Rotterdam model dc.subject State-space model dc.subject Theoretical regularity dc.title Shifting Preferences and Time-Varying Parameters in Demand Analysis: A Monte Carlo Study dc.type Dissertation dc.contributor.cmtemember Comolli, Paul dc.contributor.cmtemember Hillmer, Steve dc.contributor.cmtemember Keating, John dc.contributor.cmtemember Zhang, Jianbo dc.thesis.degreeDiscipline Economics dc.thesis.degreeLevel Ph.D. kusw.oastatus na kusw.oapolicy This item does not meet KU Open Access policy criteria. kusw.bibid 8085753 dc.rights.accessrights openAccess
### This item appears in the following Collection(s)
785-864-8983
KU Libraries
1425 Jayhawk Blvd
Lawrence, KS 66045
785-864-8983
KU Libraries
1425 Jayhawk Blvd
Lawrence, KS 66045
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2020-11-24 16:03:57
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https://artofproblemsolving.com/wiki/index.php/Divisor
|
# Divisor
A natural number ${d}$ is called a divisor of a natural number ${n}$ if there is a natural number ${k}$ such that $n=kd$ or, in other words, if $\frac nd$ is also a natural number (i.e $d$ divides $n$). See Divisibility for more information.
## Notation
A common notation to indicate a number is a divisor of another is $n|k$. This means that $n$ divides $k$.
See the main article on counting divisors. If $n=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}}\cdot\dots\cdot p_m^{\alpha_m}$ is the prime factorization of ${n}$, then the number $d(n)$ of different divisors of $n$ is given by the formula $d(n)=(\alpha_{1} + 1)\cdot(\alpha_{2} + 1)\cdot\dots\cdot(\alpha_{m} + 1)$. It is often useful to know that this expression grows slower than any positive power of ${n}$ as $n\to\infty$. We also know that the product of the divisors of any integer $n$ is $$n^{\frac{t(n)}{2}}$$. Another useful idea is that $d(n)$ is odd if and only if ${n}$ is a perfect square.
## Useful formulas
• If ${m}$ and ${n}$ are relatively prime, then $d(mn)=d(m)d(n)$
• ${\sum_{n=1}^N d(n)=\left\lfloor\frac N1\right\rfloor+\left\lfloor\frac N2\right\rfloor+\dots+\left\lfloor\frac NN\right\rfloor= N\ln N+O(N)}$
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2020-11-24 10:33:32
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https://www.physicsforums.com/threads/what-is-a-singlet-under-su-3.468133/
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What is a singlet under SU(3)?
what is a singlet under SU(3)?
and what is SU(3)xSU(3)?
What do we mean by 3 x 3(bar) = 8 + 1?
fzero
Homework Helper
Gold Member
A singlet of any algebra is the state that has zero eigenvalue under the Casimir operators. These are the generators of the center of the algebra and are simultaneously diagonalizable. The singlet state is a one-dimensional representation. For instance, for $$SU(2)$$, the singlet state satisfies
$$\vec{S}^2 |0\rangle = 0,$$
i.e., it is the $$s=0$$ state. It satisfies
$$S_{\pm} |0\rangle = 0$$
as well, so there are no other states in this representation.
$$SU(3)\times SU(3)$$ refers to a direct product of two $$SU(3)$$ groups. States or particles will simultaneously be in some representation of both groups.
$$\mathbf{3} \otimes \bar{\mathbf{3}} = \mathbf{1} \oplus \mathbf{8}$$ refers to a direct product of the $$\mathbf{3}$$ representation with the $$\bar{\mathbf{3}}$$. The resulting representation is what's known as reducible and can be written as a direct sum of the singlet $$\mathbf{1}$$ and octet $$\mathbf{8}$$ representations.
This relationship is analogous to the way we decompose two-particle spin states in QM. There, for $$SU(2)$$, the decomposition of specific product states into states of irreducible representations is summed up by the Clebsch-Gordan coefficients.
Okay, that explains a singlet. Thanks.
But what I still cant get is how we know that it is 1 + 8. For example i need to find what 3 x 3 x 3 is. How can I do that?
fzero
Homework Helper
Gold Member
Okay, that explains a singlet. Thanks.
But what I still cant get is how we know that it is 1 + 8. For example i need to find what 3 x 3 x 3 is. How can I do that?
You might want to look up Young tableaux, which are very useful for computing products of irreps for unitary groups. They are a pictorial way of keeping track of the symmetries of irreducible tensor representations. Alternatively, there are various places to find tables of products.
In your case, we should first compute $$\mathbf{3}\otimes\mathbf{3}$$. Thinking of the fundamental irrep $$\mathbf{3}$$ like a vector, this product is a rank 2 tensor. There is an antisymmetric product which is the $$\bar{\mathbf{3}}$$ and the symmetric product which is the $$\mathbf{6}$$. So we have
$$\mathbf{3}\otimes\mathbf{3} = \bar{\mathbf{3}}\oplus \mathbf{6}.$$
Now we already know that $$\mathbf{3} \otimes \bar{\mathbf{3}} = \mathbf{1} \oplus \mathbf{8}$$, so it remains to compute $$\mathbf{3} \otimes \mathbf{6}$$. We can either antisymmetrize the new index with the indices on the symmetric tensor, which gives us the $$\mathbf{8}$$, or we can totally symmetrize the rank 3 tensor, which gives us the $$\mathbf{10}$$. Therefore
$$\mathbf{3} \otimes \mathbf{6} = \mathbf{8}\oplus \mathbf{10}.$$
Putting it all together, we have
$$\mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3} = \mathbf{1} \oplus \mathbf{8}\oplus\mathbf{8}\oplus \mathbf{10}.$$
As a sanity check, you can compare the total dimensions of the representations on both sides.
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2021-05-08 04:19:48
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http://www.ens-lyon.fr/LIP/index.php/publications/research-reports-per-year-repositories/43-publications/research-reports-per-year-repositories/324-research-reports-for-2001
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### Computer validated proofs of a toolset for adaptable arithmetic.
• By: Marc Daumas, Claire Moreau-Finot, Laurent Théry
• Number: RR2001-01
• Date: January 2001
• Abstract:
Most existing implementations of multiple precision arithmetic demand that the user sets the precision a priori. A solution is to use the largest precision necessary to reach target accuracy. Some libraries are said adaptable in the sense that they dynamically change the precision of each intermediate operation individually to deliver the target accuracy according to the actual inputs. We present in this text a new adaptable numeric core inspired both from floating point expansions and from on-line arithmetic. The numeric core is cut down to five tools. The first tool that contains many arithmetic operations is proved to be correct. The proofs have been formally checked by the Coq assistant. Developing the proofs, we have formally proved many result published in the literature and we have extended a few of them. This work may let users (i) develop application specific adaptable libraries based on the toolset and / or (ii) write new formal proofs based on the set of validated facts.
• Keywords:
Multiple Precision, Expansion, On-line, Formal Proof, Coq.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+28p
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### Parallelization of automatic speech recognition.
• By: Yahya Ould Mohamed El Hadj, Nathalie Revol
• Number: RR2001-02
• Date: January 2001
• Abstract:
The automatic recognition of spoken words is increasingly common, for dictaphone applications, telephone services or the command of various devices by disabled persons. In the latter case, a high recognition rate is expected on a vocabulary of small to medium size. To achieve this goal, the model must be refined. Thus, both the training stage and the recognition stage for such applications can be very time consuming and occasional re-training may happen. Its parallelization is thus worth considering. In this paper we present firstly the models we use: the classical hidden Markov model and another model that takes into account the prosody of speech, namely the centisecond two-level hidden Markov model introduced by Meziane. Then two parallelization strategies are detailed: the first one simply shares the vocabulary among the processors, the second one also distributes the model. %which has to be slightly modified in order to allow this distribution. Experimental results highlight the need for a finer load-balancing: an a priori load estimation is presented and is used to statically balance the computational load between the processors. Further experiments have been conducted and exhibit efficiencies higher than 65\% on an architecture composed of 12 Pentium Pro interconnected via Myrinet. Directions for improving further the parallelization are given.
• Keywords:
Automatic speech recognition, Hidden Markov model, Centisecond two-level hidden Markov model, Parallelization.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+17p
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### Remote object detection in cluster-based Java.
• By: Gabriel Antoniu, Philip Hatcher
• Number: RR2001-03
• Date: January 2001
• Abstract:
Our work combines Java compilation to native code with a run-time library that executes Java threads in a distributed-memory environment with true parallelism. To provide the illusion of a shared memory to Java threads, our Hyperion compiling system has been built on top of DSM-PM2, a portable implementation platform for multithreaded distributed-shared-memory protocols. We have designed, implemented and experimented with two alternative consistency protocols compliant with the Java Memory Model. The protocols have different mechanisms for access detection.We illustrate the effects of the two access-detection techniques with five applications run on two clusters with different interconnection networks: BIP/Myrinet and SISCI/SCI.
• Keywords:
Java, Distributed Systems, Java Consistency, JVM, Consistency Protocols, Distributed Shared Memory, DSM, DSM-PM2.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+9p
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### DSM-PM2: A portable implementation platform for multithreaded DSM consistency protocols (extended version).
• By: Gabriel Antoniu, Luc Bougé
• Number: RR2001-04
• Date: January 2001
• Abstract:
DSM-PM2 is a platform for designing, implementing and experimenting multithreaded DSM consistency protocols. It provides a generic toolbox which facilitates protocol design and allows for easy experimentation with alternative protocols for a given consistency model. DSM-PM2 is portable across a wide range of clusters. We illustrate its power with figures obtained for different protocols implementing sequential consistency, release consistency and Java consistency, on top of Myrinet, Fast-Ethernet and SCI clusters.
• Keywords:
DSM, Multithreading, Consistency Protocols, DSM-PM2, PM2.
• Availability: Electronic copy only.
• Citation: Gabriel Antoniu and Luc Bougé. DSM-PM2: A portable implementation platform for multithreaded DSM consistency protocols. In Proc. 6th International Workshop on High-Level Parallel Programming Models and Supportive Environments (HIPS '01), San Francisco, April 2001. Held in conjunction with IPDPS 2001. IEEE TCPP.To appear.
• Size: 2+19p
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### Efficient Matrix Preconditioners for Black Box Linear Algebra.
• By: L. Chen, W. Eberly, E. Kaltofen, B.D. Saunders, W.J. Turner, G. Villard
• Number: RR2001-05
• Date: January 2001
• Abstract:
The main idea of the black box'' approach in exact linear algebra is to reduce matrix problems to the computation of minimum polynomials. In most cases preconditioning is necessary to obtain the desired result. Here, good preconditioners will be used to ensure geometrical / algebraic properties on matrices, rather than numerical ones, so we do not address a condition number. We offer a review of problems for which (algebraic) preconditioning is used, provide a bestiary of preconditioning problems, and discuss several preconditioner types to solve these problems. We include new conditioners, new analyses of preconditioner performance, and results on the relations among preconditioning problems and with linear algebra problems. Thus improvements are offered for the efficiency and applicability of preconditioners. The focus is on linear algebra problems over finite fields, but most results are valid for entries from arbitrary fields.
• Keywords:
Linear Algebra, Randomized Algorithms, Black Box Matrix, Sparse Matrix, Exact Arithmetic, Finite Fields, Linear Systems, Rank, Preconditioner.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+18p
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### Additive symmetric: the non-negative case.
• By: Marc Daumas, Philippe Langlois
• Number: RR2001-06
• Date: February 2001
• Abstract:
An additive symmetric b of a with respect to c satisfies c = (a+b)/2. Existence and uniqueness of such b are basic properties in exact arithmetic that fail when a and b are floating point numbers and the computation of c performed in IEEE-754 like arithmetic. We exhibit and prove conditions on the existence, the uniqueness and the exact correspondence of an additive symmetric when b and c have the same sign.
• Keywords:
Floating Point Arithmetic, Additive Symmetric, Correction, IEEE-754 Standard.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+11p
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### Heterogeneous task scheduling : a survey.
• By: Vincent Boudet
• Number: RR2001-07
• Date: February 2001
• Abstract:
Scheduling computation tasks on processors is a key issue for high-performance computing. Although a large number of scheduling heuristics have been presented in the literature, most of them target only homogeneous resources. We survey here five low-complexity heuristics for heterogeneous platforms, the Best Imaginary Level (BIL), the Generalized Dynamic Level (GDL), the Critical-Path-on-a-Processor (CPOP), the Heterogeneous Earliest Finish Time (HEFT) and the Partial Completion Time (PCT) algorithms. These five heuristics aim at scheduling directed acyclic weighted task graphs on a bounded number of heterogeneous processors. We compare the performances of the heuristics using four classical testbeds.
• Keywords:
Scheduling, Task Graphs, High-Performance, Heterogeneous Platforms, Different-Speed Processors, Mapping.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+28p
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### Leader Election without Compass in Some Hyperbolic and Euclidean Cellular Automata.
• By: Codrin Nichitiu, Christophe Papazian, Eric Remila
• Number: RR2001-08
• Date: February 2001
• Abstract:
We present a linear time algorithm for the networking and distributed computing problem of leader election (LE). Given a graph, its vertices represent processors (here finite state machines), and its edges communication lines (here synchronous). The LE problem consists in finding a protocol for a family of graphs such that after iterating it, a vertex, edge or cycle be distinguished by a special state called leader. Here the graphs are only required to be connected, and without holes. We describe the algorithm in full detail on a special class of planar graphs, prove its correctness and show how it extends to other classes.
• Keywords:
Distributed Algorithms, Graph Automata, Leader Election, Euclidean Cellular Automata, Hyperbolic Cellular Automata.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+12p
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### Quantifier rank for parity of embedded finite models.
• By: Hervé Fournier
• Number: RR2001-09
• Date: February 2001
• Abstract:
We prove some lower bounds for quantifier rank of formulas expressing parity of a finite set I of bounded cardinal embedded in an algebraically closed field or an ordered Q-vector space. We show that these bounds are tight when elements of I are known to be linearly independent. In the second part, we prove that strongly minimal structures with quantifier elimination and zero characteristic differentially closed fields admit the active-natural collapse.
• Keywords:
Constraint Databases, Active-Natural Collapse.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+15p
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### Toward an Algorithmic Classification of Cellular Automata Dynamics.
• By: Nicolas Ollinger
• Number: RR2001-10
• Date: February 2001
• Abstract:
This paper introduces a new approach to classifying cellular automata dynamics. In the spirit of Mazoyer and Rapaport [7], cellular automata are compared thanks to the worth studying notion of space-time diagrams rescaling. Mazoyer and Rapaport made the choice to consider square shaped rescaling. Here, an abstract logical framework, called bulking, is introduced to explore various possibilities of rescaling shapes. Finally, a particular case of bulking based on geometrical considerations is provided. It induces a classification with a maximal class. In particular, it allows to give a formalization of the notion of intrinsic universality implicitly introduced by Albert and Culik II [1] in 1987.
• Keywords:
Cellular Automata, Classification, Simulation, Intrinsic Universality, Games.
• Availability: Electronic copy only.
• Citation: Submitted to ICALP 2001.
• Size: 2+12p
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### Two-States Bilinear Intrinsically Universal Cellular Automata.
• By: Nicolas Ollinger
• Number: RR2001-11
• Date: March 2001
• Abstract:
Additive cellular automata, which correspond to linear cellular automata, have been studied in details using algebraic techniques [11,6]. The generalization to families of polynomial cellular automata seems natural. The following step of complexity consists of bilinear cellular automata which study has begun with the work of Bartlett and Garzon [3]. Thanks to bulking techniques [15], the intuitive notions of simulation used in the paper from Bartlett and Garzon [3] are formalized. Moreover, two-states bilinear intrinsically universal cellular automata are constructed. In the case of cellular automata, as in the case of classical dynamical systems, a frontier between decidability and undecidability stands between linear and non-linear systems. This result also answers a question from Bartlett and Garzon [3] of 1995.
• Keywords:
Cellular Automata, Bilinearity, Simulation, Intrinsic Universality.
• Availability: Electronic copy only.
• Citation: Submitted to FCT 2001.
• Size: 2+11p
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### Epistemic Logic in Higher Order Logic An experiment with COQ.
• By: Pierre Lescanne
• Number: RR2001-12
• Date: March 2001
• Abstract:
We present an experiment on epistemic logic, also called knowledge logic, we have done using COQ. This work involves a formalization in COQ of the epistemic logic which has been checked for adequacy on two puzzles well known in the community. COQ is a proof assistant which implements a higher logic known as the calculus of inductive construction and which provides a convenient framework to embed logics like epistemic logic. We try to draw from this exercise lessons for future works. protocols.
• Keywords:
Modal Logic, Logic of Knowledge, Epistemic Logic, Formal Methods, Type Theory, COQ.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+9p
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### The Master-Slave Paradigm with Heterogeneous Processors.
• By: Olivier Beaumont, Arnaud Legrand, Yves Robert
• Number: RR2001-13
• Date: March 2001
• Abstract:
In this paper, we revisit the master-slave tasking paradigm in the context of heterogeneous processors. We assume that communications take place in exclusive mode. We present a polynomial algorithm that gives the optimal solution when a single communication is needed before the execution of the tasks on the slave processors. When communications are required both before and after the task processing, we show that the problem is at least as difficult as a problem whose complexity is open. In this case, we present a guaranteed approximation algorithm. Finally, we present asymptotically optimal algorithms when communications are required before the processing of each task, or both before and after the processing of each task. protocols.
• Keywords:
Heterogeneous Processors, Master-Slave Tasking, Communication, Matching, Complexity.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+22p
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### Data Allocation Strategies for Dense Linear Algebra on two-dimensional Grids with Heterogeneous Communication Links.
• By: Olivier Beaumont, Arnaud Legrand, Yves Robert
• Number: RR2001-14
• Date: April 2001
• Abstract:
In this paper, we study the implementation of dense linear algebra kernels, such as matrix multiplication on 2D grids with homogeneous processors when the communication links between the processors are heterogeneous (i.e. the time to transfer a block of the matrix between two processors depends on these processors). We prove that finding the best allocation of the processors into a grid, with respect to the minimization of the communication overhead, is a NP-complete problem.
• Keywords:
Heterogeneous Communications, 2D Grids, Data Distribution, Linear Algebra Kernel.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+21p
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### A Coq toolkit for graph theory.
• By: Jean Duprat
• Number: RR2001-15
• Date: April 2001
• Abstract:
As an illustration of the Curry-Howard's isomorphism, it is possible to express both mathematical and computational aspects of the graph theory using the same formalism. In this paper, we give what can be the very beginning of such a presentation. The proof assistant Coq has been used as a guideline all along this work.
• Keywords:
Graph Theory, Proof Assistant, Inductive Construction.
• Availability: Electronic copy only.
• Citation: Submitted to TYPES 2001 Post-Workshop Proceedings.
• Size: 2+13p
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### How powerful are infinite time machines?
• By: Grégory Lafitte
• Number: RR2001-16
• Date: April 2001
• Abstract:
We study the power of finite machines with infinite time to complete their task. To do this, we define a variant to Wojciechowski automata, investigate their recognition power, and compare them to infinite time Turing machines. Furthermore, using this infinite time, we analyse the ordinals comprehensible by such machines and discover that one can in fact go beyond the recursive realm. We conjecture that this is somehow already the case with Wojciechowski automata.
• Keywords:
Infinite Time, Finite Automata on Infinite Words, Machine Comprehensible Ordinals.
• Availability: Electronic copy only.
• Citation: Accepted at FCT 2001.
• Size: 2+12p
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### Impact of Interferences on Bandwidth Reservation for Ad Hoc Networks: a First Theoretical Study.
• By: Karell Bertet, Isabelle Guérin Lassous, Laurent Viennot
• Number: RR2001-17
• Date: April 2001
• Abstract:
This paper presents a theoretical study on the bandwidth reservation problem for ad hoc networks. The proposed model is based on the spatial reuse and the existence of interferences. We show that in that case, the bandwidth reservation problem is NP-complete and we provide some bounds that compare solutions of the problems derived with greedy heuristics with an optimal one. We conclude with a discussion on the practical aspect of this model and its potential use in a practical protocol.
• Keywords:
Bandwidth Reservation, Ad Hoc Networks, Interferences.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+6p
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### Validating polynomial numerical computations with complementary automatic methods.
• By: Philippe Langlois, Nathalie Revol
• Number: RR2001-18
• Date: May 2001
• Abstract:
Finite precision computations affect the accuracy of computed solutions
and sometimes the stability of iterative algorithms. Automatic approaches
exist to control and to reduce these effects. Examples are the CESTAC and
the CENA methods and the more general interval approaches. We focus here
on a complementary use of these methods to localize unstable behavior of
the algorithm, to improve the accuracy of the solutions, to identify and
explain finite precision effects. We present computational experiments on
ill-conditioned polynomial roots approximated with Newton's iteration
that illustrate the well-known influence of coefficient perturbations.
• Keywords:
Finite Precision, Automatic Rounding Error Analysis, Newton's Iteration, Polynomial Multiple Roots.
• Availability: Electronic copy only.
• Citation: Submitted.
• Size: 2+12p
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### Une méthode de correction automatique.
• By: Philippe Langlois
• Number: RR2001-19
• Date: May 2001
• Abstract:
Correcting methods intend to improve the accuracy of results computed in finite precision. The CENA method processes an automatic correction of the first-order effect of the rounding errors the computation generates. The method provides a corrected result and a bound of the residual error for a class of algorithms we identify. We present the main features of the CENA method and illustrate its interests and limitations with examples.
• Keywords:
Automatic Correction, CENA Method, Accuracy, Finite Precision, Floating Point Arithmetic.
• Availability: Electronic copy only.
• Citation: To appear in Calculateurs Parallèles, Réseaux, et Systèmes Répartis.
• Size: 2+14p
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### Reconnaissance parallèle des langages rationnels sur automates cellulaires plans.
• By: Marianne Delorme, Jacques Mazoyer
• Number: RR2001-20
• Date: May 2001
• Abstract:
Few results are known on the computational power of 2D cellular automata. In the framework of languages recognition, the main difficulty comes from the multiplicity of possible embeddings of words into the discrete plane. Even a very simple problem as the real-time recognition of rational languages can become tricky. We solve this problem for two special embeddings. This first study brings to the fore that the true question is more the plane nature than the algorithmics .
• Keywords:
Cellular Automata, Language Recognition, Real-Time, Space Filling Curves.
• Availability: Electronic copy only.
• Citation: Submitted to Theoretical Computer Science.
• Size: 2+52p
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### Scalability Issues in a Reliable Distributed Video Storage System.
• By: Alice Bonhomme
• Number: RR2001-21
• Date: May 2001
• Abstract:
This article studies the design and the performance of a fault-tolerant distributed storage system dedicated to video. We use a completely distributed approach that permit to reduce the memory cost of the reliability scheme. We experimentally point out the scalability properties of this system.
• Keywords:
Video Server, Scalability, Striping, Fault Tolerance, Cluster.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+6p
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### The Iso-Level Scheduling Heuristic for Heterogeneous Processors.
• By: Olivier Beaumont, Vincent Boudet, Yves Robert
• Number: RR2001-22
• Date: May 2001
• Abstract:
Scheduling computational tasks on processors is a key issue for high-performance computing. Although a large number of scheduling heuristics have been presented in the literature, most of them target only homogeneous resources. We present a new scheduling heuristic for heterogeneous processors, which improves the load-balancing achieved at each decision step while keeping a low complexity. Experimental comparisons with five heuristics taken from the literature (BIL, GDL, CPOP, HEFT and PCT) and using six classical testbeds, show very favorable results.
• Keywords:
Heterogeneous Processors, Unrelated Parallel Machines, Scheduling Heuristics, Allocation, Complexity.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+24p
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### Cost Analysis of a Distributed Video Storage System.
• By: Alice Bonhomme
• Number: RR2001-23
• Date: June 2001
• Abstract:
We describe the design and the implementation of the CFS (Cluster File System) storage system which is dedicated to video streams. Our goal is to provide a system with the following features: 1) High number of supported streams at a low cost. 2) Transparent management with respect to the clients. 3) Reliability regarding data storage and service continuity. The CFS is implemented on a cluster of PCs connected with a high speed internal network. Its management is fully distributed among the cluster nodes so that there is no central component. Data are stored and retrieved among the cluster nodes using a Streaming RAID strategy, to enhance reliability. We evaluate the overhead introduced by the CFS management with respect to disk performance, by varying the number of nodes involved in file storage, the block size and the reliability level. The impact of the global server load on this overhead is moreover discussed to demonstrate scalability. As matter of fact, the overhead is between 3-15% depending on the configuration.
• Keywords:
Video Server, Data Striping, Reliability, Performance Analysis.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+14p
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### Scilab to Scilab// - The OURAGAN Project.
• By: E. Caron, S. Chaumette, S. Contassot-Vivier, F. Desprez, E. Fleury, C. Gomez, M. Goursat, E. Jeannot, D. Lazure,F. Lombard, J.-M. Nicod, L. Philippe, M. Quinson, P. Ramet,J. Roman, F. Rubi, S. Steer, F. Suter, G. Utard
• Number: RR2001-24
• Date: June 2001
• Abstract:
In this paper, we present the developments realized in the OURAGAN project around the parallelization of a Matlab-like tool called Scilab. These developments use high performance numerical libraries and different approaches based either on the duplication of Scilab processes or on computational servers. This tool, Scilab//, allows users to perform high level operations on distributed matrices in a metacomputing environment. We also present performance results on different architectures.
• Keywords:
Scilab, Parallel Libraries, Computational Servers, CORBA, Data Redistribution.
• Availability: Electronic copy only.
• Citation: To appear in Parallel Computing.
• Size: 2+19p
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### Bandwidth-centric allocation of independent tasks on heterogeneous platforms.
• By: Olivier Beaumont, Larry Carter, Jeanne Ferrante, Arnaud Legrand, Yves Robert
• Number: RR2001-25
• Date: June 2001
• Abstract:
In this paper, we consider the problem of allocating a large number of independent, equal-sized tasks to a heterogenerous "grid" computing platform. Such problems arise in collaborative computing efforts like SETI@home. We use a tree to model a grid, where resources can have different speeds of computation and communication, as well as different overlap capabilities. We define a base model, and show how to determine the maximum steady-state throughput of a node in the base model, assuming we already know the throughput of the subtrees rooted at the node's children. Thus, a bottom-up traversal of the tree determines the rate at which tasks can be processed in the full tree. The best allocation is {\em bandwidth-centric}: if enough bandwidth is available, then all nodes are kept busy; if bandwidth is limited, then tasks should be allocated only to the children which have sufficiently small communication times, regardless of their computation power. We then show how nodes with other capabilities --- ones that allow more or less overlapping of computation and communication than the base model --- can be transformed to equivalent nodes in the base model. We also show how to handle a more general communication model. Finally, we present simulation results of several demand-driven task allocation policies that show that our bandwidth-centric method obtains better results than allocating tasks to all processors on a first-come, first serve basis.
• Keywords:
Heterogeneous Computer, Allocation, Scheduling, Grid, Metacomputing.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 3+44p
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### Correctly Rounded Exponential Function in Double Precision Arithmetic.
• By: David Defour, Florent de Dinechin, Jean-Michel Muller
• Number: RR2001-26
• Date: July 2001
• Abstract:
We present an algorithm for implementing correctly rounded exponentials in double-precision floating point arithmetic. This algorithm is based on floating-point operations in the widespread IEEE-754 standard, and is therefore more efficient than those using multiprecision arithmetic, while being fully portable. It requires a table of reasonable size and IEEE-754 double precision multiplications and additions. In a preliminary implementation, the overhead due to correct rounding is a 2.3 times slowdown when compared to the standard library function.
• Keywords:
Exponential, Correct Rounding, Double-Precision.
• Availability: Electronic copy only.
• Citation: Published in Advanced Signal Processing Algorithms, Architecture and Implementations XI (SPIE), San Diego, USA. August 2001.
• Size: 2+21p
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### On digit-recurrence division algorithms for self-timed circuits.
• By: Nicolas Boullis, Arnaud Tisserand
• Number: RR2001-27
• Date: July 2001
• Abstract:
The optimization of algorithms for self-timed or asynchronous circuits requires specific solutions. Due to the variable-time capabilities of asynchronous circuits, the average computation time should be optimized and not only the worst case of the signal propagation. If efficient algorithms and implementations are known for asynchronous addition and multiplication, only straightforward algorithms have been studied for division. This paper compares several digit-recurrence division algorithms (speed, area and circuit activity for estimating the power consumption). The comparison is based on simulations of the different operators described at the gate level. This work shows that the best solutions for asynchronous circuits are quite different from those used in synchronous circuits.
• Keywords:
Computer Arithmetic, Division Algorithms, SRT Tables, Asynchronous Circuits, Self-Timed Circuits.
• Availability: Electronic copy only.
• Citation: Published in Advanced Signal Processing Algorithms, Architecture and Implementations XI (SPIE), San Diego, USA. August 2001.
• Size: 2+12p
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### Parallel Execution of the Saturated Reductions.
• By: Benoît Dupont de Dinechin, Christophe Monat, Fabrice Rastello
• Number: RR2001-28
• Date: July 2001
• Abstract:
This report addresses the problem of improving the execution performance of saturated reduction loops on fixed-point instruction-level parallel Digital Signal Processors (DSPs). We first introduce bit-exact'' transformations, that are suitable for use in the ETSI and the ITU speech coding applications. We then present approximate'' transformations, the relative precision of which we are able to compare. Our main results rely on the properties of the saturated arithmetic.
• Keywords:
Saturated Reductions, Fixed-Point Arithmetic, Fractional Arithmetic, ETSI / ITU Speech Coding, Parallel Reductions, Instruction Level Parallelism.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+15p
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### BRuIT: Bandwidth Reservation under InTerferences influence.
• By: Claude Chaudet, Isabelle Guerin-Lassous
• Number: RR2001-29
• Date: July 2001
• Abstract:
This paper deals with the bandwidth reservation problem in ad hoc networks and with the influence that interferences between signals have on it. We show that interferences could decrease the applications rates. This can be a real problem for applications that need guarantees. We propose a distributed protocol (called "BRuIT") for bandwidth reservation in ad hoc networks that takes into account the existence of interferences from far transmissions. The protocol is analyzed through simulations carried out under NS.
• Keywords:
Wireless Networks, Ad Hoc Networks, Multihop, Bandwidth Reservation, Quality of Service, Interferences.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+8p
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### Matrix Rank Certification.
• By: David Saunders, Arne Storjohann, Gilles Villard
• Number: RR2001-30
• Date: August 2001
• Abstract:
Randomized algorithms are given for computing the rank of a matrix over a field of characteristic zero. The matrix is treated as a black box. Only the capability to compute matrix x column-vector and row-vector x matrix products is used. The methods are exact, sometimes called seminumeric. They are appropriate for example for matrices with integer or rational entries. The rank algorithms are probabilistic of the Las Vegas type; the correctness of the result is guaranteed.
• Keywords:
Linear Algebra, Randomized Algorithms, Black Box Matrix, Matrix Rank, Seminumeric Computation.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+6p
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### Formally validated specification of a micro-payment protocol.
• By: P. Dargenton, D.Hirschkoff, P.Lescanne, E. Pommateau
• Number: RR2001-31
• Date: August 2001
• Abstract:
In this paper, we develop a formal specification for a micro-payment protocol, first on paper, then within the Coq proof assistant. Our approach in defining a notion of execution traces for protocol runs is inspired by previous works, notably by L. Paulson (in the Isabelle/HOL system). However, we show that the protocol we study entails some modifications to Paulson's framework, related to the modeling of the agents' internal state. We moreover take profit from Coq's expressive meta-language to mechanically derive proofs about the formalisation itself, by introducing a notion of well-formedness for protocol rules.
• Keywords:
Electronic Commerce, Micro-Payment Protocols, Specification, Formal Proof.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+15p
• Format: Compressed postscript
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### Tiling groups for Wang tiles.
• By: P. Cristopher Moore, Ivan Rapaport, Eric Rémila
• Number: RR2001-32
• Date: September 2001
• Abstract:
We apply tiling groups and height functions to tilings of regions in the plane by Wang tiles(squares with colored boundaries) where the colors of shared edges must match. We define a set of tiles as unambiguous if it contains all tiles equivalent to the identity in its tiling group. For all but one set of unambiguous tiles with two colors, we give efficient algorithms that tell whether a given region with colored boundary is tileable, show how to sample random tilings, and how to calculate the number of local moves or flips'' required to transform one tiling into another. We also analyze the lattice structure of the set of tilings, and study several examples with three and four colors as well.
• Keywords:
Tiling, Height Function, Wang Tiles.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+14p
• Format: Compressed postscript
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### A new range reduction algorithm.
• By: P. David Defour, Peter Kornerup, Jean-Michel Muller, Nathalie Revol
• Number: RR2001-33
• Date: September 2001
• Abstract:
Range reduction is a key point for getting accurate elementary function routines. We introduce a new algorithm that is fast for input arguments belonging to the most common domains, yet accurate over the full double precision range.
• Keywords:
Elementary Functions, Computer Arithmetic, Range Reduction, Floating-Point Arithmetic.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+10p
• Format: Compressed postscript
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### A rank theorem for Vandermonde matrices.
• By: Pascal Koiran, Natacha Portier, Gilles Villard
• Number: RR2001-34
• Date: September 2001
• Abstract:
We show that certain matrices built from Vandermonde matrices are of full rank. This result plays a key role in the construction of the limit theory of generic polynomials''.
• Keywords:
Linear Algebra, Linear Recurrent Sequences, Transcendant Numbers, Model Theory.
• Availability: Electronic copy only.
• Citation: You can also download an updated version of this report.
• Size: 2+7p
• Format: Compressed postscript
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### The limit theory of generic polynomials.
• By: Pascal Koiran
• Number: RR2001-35
• Date: September 2001
• Abstract:
We show that the set T of first-order sentences satisfied by all generic polynomials of sufficiently high degree forms a complete theory. As a consequence, complex polynomials of even degree cannot be distinguished from complex polynomials of odd degree by a first-order formula. We ask whether T has an analytic model.
• Keywords:
Model Theory, Definability, Intersection.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+14p
• Format: Compressed postscript
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### On randomness and infinity.
• By: Grégory Lafitte
• Number: RR2001-36
• Date: September 2001
• Abstract:
In this paper, we investigate refined definitions of random sequences. Classical definitions have always the shortcome of making use of the notion of algorithm. We discuss the nature of randomness and different ways of obtaining satisfactory definitions of randomness after reviewing previous attempts at producing a non-algorithmical definition. We present alternative definitions based on infinite time machines and set theory and explain how and why randomness is strongly linked to strong axioms of infinity.
• Keywords:
Random, Infinite Time Machines, Large Cardinals.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+8p
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### A Realistic Model and an Efficient Heuristic for Scheduling with Heterogeneous Processors.
• By: Olivier Beaumont, Vincent Boudet, Yves Robert
• Number: RR2001-37
• Date: September 2001
• Abstract:
Scheduling computational tasks on processors is a key issue for high-performance computing. Although a large number of scheduling heuristics have been presented in the literature, most of them target only homogeneous resources. Moreover, these heuristics often rely on a model where the number of processors is bounded but where the communication capabilities of the target architecture are not restricted. In this paper, we deal with a more realistic model for heterogeneous networks of workstations, where each processor can send and/or receive at most one message at any given time-step. First, we state a complexity result that shows that the model is at least as difficult as the standard one. Then, we show how to modify classical list scheduling techniques to cope with the new model. Next we introduce a new scheduling heuristic which incorporates load-balancing criteria into the decision process of scheduling and mapping ready tasks. Experimental results conducted using six classical testbeds: (LAPLACE, LU, STENCIL, FORK-JOIN, DOOLITTLE, and LDMt) show very promising results.
• Keywords:
Heterogeneous Processors, Scheduling, Mapping, Scheduling Heuristics, Complexity.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+18p
• Format: Compressed postscript
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### The Quest for Small Universal Cellular Automata.
• By: Nicolas Ollinger
• Number: RR2001-38
• Date: October 2001
• Abstract:
We formalize the idea of intrinsically universal cellular automata, which is strictly stronger than classical computational universality. Thanks to this uniform notion, we construct a new one-dimensional universal automaton with von Neumann neighborhood and only 6 states.
• Keywords:
Cellular Automata, Simulation, Intrinsic Universality.
• Availability: Electronic copy only.
• Citation: Submitted to STACS 2002.
• Size: 2+11p
• Format: Compressed postscript
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### Adding Data Persistence and Redistribution to NetSolve.
• By: F. Desprez, E. Jeannot
• Number: RR2001-39
• Date: December 2001
• Abstract:
The implementation of Network Enabled Servers (NES) on grid environments requires to lower the cost of communications. NetSolve, a NES environment developed at University of Tennessee Knoxville, sends data back to the client at the end of every computation. This implies unneeded communications when computed data are needed by an other server in further computations. In this paper, we present the modifications we made to the NetSolve protocol in order to overcome this drawback. We have developed a set of new functions and data structures that allow users to order servers to keep data in place and to redistribute them directly to an other server when needed.
• Keywords:
Client-Server Distributed Computing, Network Enabled Servers, NetSolve, Data Persistence, Data Redistribution.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+9p
• Format: Compressed postscript
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### Dynamic Performance Forecasting for Network-Enabled Servers in a Heterogeneous Environment.
• By: Frédéric Desprez, Martin Quinson, Frédéric Suter
• Number: RR2001-40
• Date: November 2001
• Abstract:
This paper presents a tool for dynamic forecasting of Network-Enabled Servers performance. FAST (Fast Agent's System Timer) is a software package allowing client applications to get an accurate forecast of communication and computation times and memory use in a heterogeneous environment. It relies on low level software packages, i.e., network and host monitoring tools, and some of our developments in computation routines modeling. The FAST internals and user interface are presented and a comparison between the execution time predicted by FAST and the measured time of complex matrix multiplication executed on an heterogeneous platform is given.
• Keywords:
Performance Forecasting, Computation Modeling, Resources Monitoring, Heterogeneous Environments, Network-Enabled Servers.
• Availability: Electronic copy only.
• Citation: Proceedings of the International Conference on Parallel and Distributed Processing Techniques and Applications (PDPTA 2001), volume 3, pages 1421-1427. CSREA Press, 25-28 June 2001.
• Size: 2+9p
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### Introduction à l'arithmétique par intervalles.
• By: Nathalie Revol
• Number: RR2001-41
• Date: September 2001
• Abstract:
This report (written in french) constitutes an introduction to interval arithmetic. This arithmetic allows on the one hand to take into account the measurement uncertainties on data and on the other hand to determine an enclosure of the computed result that is guaranteed to contain it: indeed, the main advantage of interval arithmetic is its reliability. The goal of this introduction is to emphasize the strong points of interval arithmetic and to explain how to alleviate its problems. The main advantage is to provide global information, such as for instance the range of a function over a whole set. This global information can serve to prove that an iteration is contractant and thus that it has a fixed point. It can also be used to detemine the global optimum of a function without being trapped by a local one.
• Keywords:
Interval Arithmetic, Reliable Computation, Wrapping Effect, Data Dependency, Rump's Algorithm For Linear Systems, Interval Newton Method for Nonlinear Systems, Hansen's Algorithm for Global Optimization.
• Availability: Electronic copy only.
• Citation: Version étendue de l'article paru dans Calculateurs Parallèles, Réseaux, et Systèmes Répartis, vol. 13, 2001.
• Size: 2+41p
• Format: Compressed postscript
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### Parallelization of the Numerical Lyapunov Calculation for the Fermi-Pasta-Ulam Chain.
• By: Fabrice Rastello, Thierry Dauxois
• Number: RR2001-42
• Date: November 2001
• Abstract:
In this paper, we present an efficient and simple solution to the parallelization of discrete integration programs of ordinary differential equations (ODE). The main technique used is known as loop tiling. To avoid the overhead due to code complexity and border effects, we introduce redundant tasks and we use non parallelepiped tiles. Thanks both to cache reuse (x4.3) and coarse granularity (x24.5) , the speedup using 25 processors over the non-tiled sequential implementation is larger than 106. We also present the draft of a fuzzy methodology to optimize the tile size and we illustrate it using real measurements for the communication cost and the execution time. In particular, we observe that the model of communication latencies over a Myrinet network is not as simple as is usually reported. We apply this solution to study the Lyapunov exponents of the Fermi-Pasta-Ulam (FPU) chain and in particular the dependence of the maximum Lyapunov exponents as a function of the length of the chain.
• Keywords:
Hierarchical Tiling, Redundant Tasks, Parallelism, Scalability, Cache Optimization, Locality, Tiles Shape, Communication Overhead, Heterogeneous Ressource, Dynamical System Theory, Fermi-Pasta-Ulam Chain, Lyapunov Instability Analysis, Phase Space Properties.
• Availability: Electronic copy only.
• Citation: Not published yet..
• Size: 2+24p
• Format: Compressed postscript
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### The Data Broadcast Problem with Non-Uniform Transmission Times (Revised version).
• By: Claire Kenyon, Nicolas Schabanel
• Number: RR2001-43
• Date: November 2001
• Abstract:
The data broadcast problem consists in finding an infinite schedule to broadcast a given set of messages so as to minimize the average service time to clients requesting messages, and the cost of the broadcast. This problem also models the maintenance scheduling problem and the multi-item replenishment problem. Previous work concentrated on a discrete-time restriction where all messages have transmission time equal to 1. Here, we study a generalization of the model to the more realistic setting of continuous time and messages of non-uniform transmission times. We prove that the data broadcast problem is NP-hard, even if the broadcast costs are all zero, and give 3-approximation algorithms for broadcasting messages on a single channel.
• Keywords:
Approximation Algorithms, Scheduling, Randomized Algorithms, NP-completeness, Data Broadcast.
• Availability: Electronic copy only.
• Size: 2+26p
• Format: Compressed postscript
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### Multipartite Tables in JBits for the Evaluation of Functions on FPGA.
• By: Jérémie Detrey, Florent de Dinechin
• Number: RR2001-44
• Date: November 2001
• Abstract:
This paper presents a core generator for arbitrary numeric functions on Xilinx Virtex FPGAs. The cores use the state-of-the-art multipartite table method, which allows input and output precisions in the range of 8 to 24 bits on current Virtex chips. The implementation uses the JBits API to embed elaborate optimisation techniques in the description of the hardware.
• Keywords:
Function Generator, FPGA Core, Multipartite Method, JBits.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+9p
• Format: Compressed postscript
• Get it
### Efficient Parallelization of Line-Sweep Computations.
• By: Daniel Chavarría-Miranda, Alain Darte, Robert Fowler, John Mellor-Crummey
• Number: RR2001-45
• Date: November 2001
• Abstract:
Multipartitioning is a strategy for partitioning multi-dimensional arrays on a collection of processors. With multipartitioning, computations that require solving 1D recurrences along each dimension of a multi-dimensional array can be parallelized effectively. Previous techniques for multipartitioning yield efficient parallelizations over 3D domains only when the number of processors is a perfect square. This paper considers the general problem of computing optimal multipartitionings for d-dimensional data volumes on an arbitrary number of processors. We describe an algorithm that computes an optimal multipartitioning for this general case, which enables efficient parallelizations of line-sweep computations under arbitrary conditions. Finally, we describe a prototype implementation of generalized multipartitioning in the Rice dHPF compiler and performance results obtained when using it to parallelize a line sweep computation for different numbers of processors.
• Keywords:
Loop Parallelization, Array Mapping, Generalized Latin Squares, High Performance Fortran.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+34p
• Format: Compressed postscript
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### Qulaity of service in mobile ad-hoc networks -- state of the art (Qualité de service et réseaux ad-hoc - un état de l'art).
• By: Claude Chaudet
• Number: RR2001-46
• Date: November 2001
• Abstract:
Local wireless networks have known a great success since commercial solutions based on the IEEE 802.11 standard were released. With the important throughput obtained today, these networks now allow the use of complex applications that require guarantees regarding communications bandwidth, delay or jitter. All the work around Mobile IP and UMTS should lead to QoS protocols for wireless networks with access points in which the network is divided into geographical zones managed by base stations. Most of these works cannot be directly transposed to mobile ad-hoc networks in which no structure is available. These networks do not have standard QoS protocols that are suited to their specificities yet. This article presents a state of the art of this subject.
• Keywords:
Wireless Networks, Ad-hoc Networks, Quality of Service, QoS.
• Availability: Electronic copy only.
• Citation: Claude Chaudet. Qualité de service et réseaux ad-hoc - un état de l'art. Colloque sur 'Mobiles-services et réseaux mobiles de 3ème Génération' - Lyon, décembre 2001.
• Size: 2+6p
• Format: Compressed postscript
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### Chordal embeddings of planar graph.
• By: Vincent Bouchitté, Frédéric Mazoit, Ioan Todinca
• Number: RR2001-47
• Date: November 2001
• Abstract:
Robertson and Seymour conjectured that the treewidth of a planar graph and the treewidth of its geometric dual differ by at most one. Lapoire solved the conjecture in the affirmative, using algebraic techniques. We give here a much shorter proof of this result.
• Keywords:
Planar Graphs, Treewidth, Duality.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+14p
• Format: Compressed postscript
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### Lattices of tilings : an extension to figures with holes.
• By: Eric Rémila
• Number: RR2001-48
• Date: December 2001
• Abstract:
We first prove that the set of domino tilings of a fixed finite figure is a distributive lattice, even in the case when the figure has holes. Afterwards, we give a geometrical interpretation of the order given by this lattice. We extend these results to other types of tilings (calisson tilings, tilings with bicolored Wang tiles).
• Keywords:
Tiling, Height Function, Lattice.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+15p
• Format: Compressed postscript
• Get it
### Interval Newton iteration in multiple precision for the univariate case.
• By: Nathalie Revol
• Number: RR2001-49
• Date: December 2001
• Abstract:
In this paper, interval arithmetic using an underlying multiple precision arithmetic is briefly presented. Then interval Newton iteration for solving nonlinear equations is introduced. A new Newton's algorithm based on multiple precision interval arithmetic is given, along with its properties: termination, arbitrary accuracy on the computed zeros, automatic and dynamic adaptation of the precision. Finally some experiments illustrate the behaviour of this method.
• Keywords:
Multiple Precision Interval Arithmetic, Interval Newton Algorithm, Arbitrary Accuracy, Automatic and Dynamic Adaptation of the Precision.
• Availability: Electronic copy only.
• Citation: Submitted.
• Size: 2+12p
• Format: Compressed postscript
• Get it
### Impact of heavy traffic beyond communication range in multi-hops ad-hoc networks.
• By: Dominique Dhoutaut
• Number: RR2001-50
• Date: December 2001
• Abstract:
For commercial availability reasons, most actual multi-hops ad-hoc simulations and test-beds are based on IEEE 802.11 standard \cite{80211} and its medium access method CSMA/CA. But this standard has not been designed for that kind of network and presents serious flaws in this context. In this paper, we shall try to highlight problems that can appear in the \textbf{non-direct} neighborhood of important data flows in ad-hoc networks. In some situations, the fairness of the medium access can indeed be very poor. Various techniques could be used to solve this problem. In this paper we present one solution, which only requires minor modifications to the standard.
• Keywords:
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+9p
• Format: Compressed postscript
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### Pavages et Bases de Grobner.
• By: Olivier Bodini
• Number: RR2001-51
• Date: December 2001
• Abstract:
In this paper, we answer to a question of Grunbaum by proving that, for all set F of polyominoes (union of unit squares of a square lattice), we can find a Z-tiling (signed tile) of polyominoes by copies of elements of F in polynomial time. We use for this the theory of generalised Grobner bases. For instance, we can algorithmicaly find again and extend results of Lagarias and Romero on the topic.
• Keywords:
Polyomino, Tiling, Standard Basis.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+31p
• Format: Compressed postscript
• Get it
### Generalized Tilings with Height Functions.
• By: Olivier Bodini, Matthieu Latap
• Number: RR2001-52
• Date: December 2001
• Abstract:
In this paper, we introduce a generalization of a class of tilings which appear in the literature: the tilings over which a height function can be defined (for example, the famous tilings of polyominoes with dominoes). We show that many properties of these tilings can be seen as the consequences of properties of the generalized tilings we introduce. In particular, we show that any tiling problem which can be modelized in our generalized framework has the following properties: the tilability of a region can be constructively decided in polynomial time, the number of connected components in the undirected flip-accessibility graph can be determined, and the directed flip-accessibility graph induces a distributive lattice structure. Finally, we give a few examples of known tiling problems which can be viewed as particular cases of the new notions we introduce.
• Keywords:
Tilings, Height Functions, Tilability, Distributive Lattices, Random Sampling, Potentials, Flows.
• Availability: Electronic copy only.
• Citation: Not published yet.
• Size: 2+17p
• Format: Compressed postscript
• Get it
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2020-09-19 19:54:23
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http://mathcentral.uregina.ca/QQ/database/QQ.09.17/h/sohel1.html
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Math Central Quandaries & Queries
Question from Sohle: A cube has an edge of 4 cm. A triangle is formed by joining the middle point of one face of the cube to the ends of a diagonal of the opposite face . What will be the area of the triangle ?
Hi,
Here is my diagram of what you described.
$P$ is the midpoint of the top face and $BC$ is a diagonal of the opposite face. The triangle in question is green. Use Pythagoras theorem to find the base of the triangle. What is its height? What is its area?
Penny
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2022-10-04 10:21:29
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# Reconnecting groups of space debris to their parent body through proper elements
## Abstract
Satellite collisions or fragmentations generate a huge number of space debris; over time, the fragments might get dispersed, making it difficult to associate them to the configuration at break-up. In this work, we present a procedure to back-trace the debris, reconnecting them to their original configuration. To this end, we compute the proper elements, namely dynamical quantities which stay nearly constant over time. While the osculating elements might spread and lose connection with the values at break-up, the proper elements, which have been already successfully used to identify asteroid families, retain the dynamical features of the original configuration. We show the efficacy of the procedure, based on a hierarchical implementation of perturbation theory, by analyzing the following four different case studies associated to satellites that underwent a catastrophic event: Ariane 44lp, Atlas V Centaur, CZ-3, Titan IIIc Transtage. The link between (initial and final) osculating and proper elements is evaluated through tools of statistical data analysis. The results show that proper elements allow one to reconnect the fragments to their parent body.
## Introduction
Since the launch of Sputnik 1 in 1957, thousands of satellites have been deployed in orbit around the Earth. Explosions or collisions of satellites generated millions of space debris of various sizes1, currently traveling at different altitudes: rocket stages, fragments from disintegrations, bolts, paint flakes, electronic parts, etc. Chain reactions triggered by catastrophic events involving satellites might increase the hazard of (human and robotic) space activities. A single break-up event generates a cloud of debris which scatters around, sometimes reaching great distances after a relatively short time. Once the fragments are dispersed, it is often difficult to trace them back; hence, a question of paramount importance is to connect the debris to their parent satellite. In this work we propose a method that allows us to link the fragments, after a certain interval of time, to the configuration of the debris soon after the initial catastrophic event. This result contributes to address a timely problem since, in case of a collision between two satellites or an explosion of a single satellite, it is certainly important to know the parent bodies that generated the space debris. The implications are wide and range from space sustainability to space law.
To study a specific break-up event, we introduce a suitable model (based on the Hamiltonian formalism) to describe the dynamics of each fragment. The model is composed of the sum of the Keplerian attraction, the effect of the geopotential, the gravitational influence of Sun and Moon. Then, we implement perturbation theory to construct a sequence of canonical transformations providing, for each debris, approximate integrals of motion called proper elements, namely quantities that stay nearly constant over time. Each fragment is characterized by a set of six orbital elements, namely semimajor axis a, eccentricity e, inclination i, mean anomaly M, argument of perigee $$\omega$$ and longitude of the ascending node $$\Omega$$. Starting from their initial values, we compute the orbital elements of each fragment after a given interval of time, to which we refer as the final osculating elements. Then, we compute the proper elements associated to the final osculating elements, and we compare them either with the initial elements and with the corresponding proper elements at the initial time. The comparison gives the desired information: while the final osculating elements might spread far away from the initial values, the (initial/final) proper elements stay almost constant and retain the original features of the cloud of fragments2. A striking use of proper elements was already proposed to group asteroids, inspired by the pioneer work3 of Hirayama in 1918 and continued by many other authors4,5. The analytical computation of proper elements allowed to group asteroids in families, possibly leading to the conjecture that such asteroids might be fragments of an ancestor parent body. Knežević and Milani6 introduced also the synthetic proper elements based upon a numerical integration, a digital filtering of the short-period terms and a Fourier analysis.
Motivated by the successful results on asteroids, we propose to group and reconnect space debris through the computation of the proper elements associated to fragments generated by a satellite break-up event7,8,9. The procedure we are going to describe, requires the introduction of a realistic model which depends on quantities varying on different time scales; hence we need a suitable hierarchical set of transformations of coordinates, called normal forms, aimed at constructing the proper elements, whose relation with the initial elements is analyzed through statistical methods. We will consider four sample cases associated to the break-up events of the satellites Ariane 44lp, Atlas V Centaur, CZ-3, Titan IIIc Transtage. Using statistical data analysis, we show the effectiveness of the use of proper elements in reconnecting the fragments to their parent body. To reconnect the debris to a parent body, we back-propagate the debris for a given time and compare the osculating or proper elements at the initial time and at the back-propagated time. The effectiveness of the method has been shown in the specific example of Titan III Transtage. We finally provide an example in which one can distinguish between proper elements associated to nearby breakup events.
This work is organized as follows. After introducing the model, we describe the procedure to compute the proper elements through normal form theory. Then we investigate the test cases by computing osculating and proper elements, and by analyzing the results through histograms, Kolmogorov-Smirnov test, Variance Equivalence test and Pearson correlation coefficient. We end up with some conclusions and perspectives.
## The model
For the present work, our case studies will be located at altitudes between 15000 and 25000 km, all of them well above the Earth’s atmosphere. At those altitudes a celestial object is subject to different forces that we describe through a Hamiltonian function composed by the following parts: the attraction of the Earth (that we split as the sum of the Keplerian part $$H_{Kep}$$ and the potential $$H_E$$ generated by the Earth’s non-spherical shape), and the gravitational influence of the Moon $$H_M$$ and the Sun $$H_S$$ (both assumed to be point masses). The overall Hamiltonian
\begin{aligned} H = H_{Kep} + H_E + H_M + H_S \end{aligned}
(1)
depends upon the orbital elements of the debris, Moon and Sun, and on the sidereal time describing the rotation of the Earth10.
We are aware that a realistic model should include also the effect of the solar radiation pressure11 (SRP). However, we decided not to consider SRP for two main reasons: (i) the work2 provides some experiments on synthetic space debris (namely obtained through a simulator of break-up events), using a model that includes SRP; however, the results show that at intermediate altitudes the computation of the proper elements is not much affected by SRP, at least for objects with an area-to-mass ratio lower than 0.74; (ii) there does not exist a public catalogue that provides information about the area-to-mass ratio of real space debris, thus preventing reliable experiments on real cases.
### The Keplerian and geopotential Hamiltonians
Expressing the Hamiltonian in terms of the orbital elements, the Keplerian part $$H_{Kep}$$ is given by
\begin{aligned} H_{Kep} = - {{G M_E}\over {2a}}, \end{aligned}
where G is the gravitational constant and $$M_E$$ is the mass of the Earth.
The contribution $$H_E$$ due to the Earth’s non-spherical shape is computed as follows12,13: we expand the geopotential in spherical harmonics, then we average over the fast variables (namely the mean anomaly of the debris and the sidereal time), and finally we limit the expansion of the secular part of the geopotential to the greatest spherical harmonic coefficients, usually denoted as $$J_2$$ and $$J_3$$. The resulting Hamiltonian takes the form:
\begin{aligned} H_E = G M_E {R_E^2\over a^3} J_2 \left( {3\over 4} \sin ^2 i -{1\over 2}\right) {1\over {(1-e^2)}^{3/2}} + G M_E {R_E^3\over a^4} J_3 \left( {{15}\over {8}} \sin ^3 i -{3\over 2} \sin i\right) e \sin \omega {1\over {(1-e^2)}^{5/2}}\ , \end{aligned}
where $$R_E$$ is the Earth’s radius (equal to 6378.1363 km).
### Moon and Sun Hamiltonians
The Hamiltonians of the Moon $$H_M$$ and of the Sun $$H_S$$ are expanded in powers of the ratios of the semimajor axes of the debris, and of the Moon and Sun, respectively $$a_M$$ and $$a_S$$, as well as in powers of the eccentricity and of the cosine of the inclination. The resulting expansions are truncated to a low order (typically the second one). By $$(a_b,e_b,i_b,M_b,\omega _b,\Omega _b)$$ we denote the orbital elements of the third body perturber, where $$b=M$$ and $$b=S$$ correspond to the Moon and Sun, respectively.
The expansion of the Moon’s Hamiltonian in terms of the orbital elements of the Moon and the debris is given below; we underline that in applications we will consider the expansion of $$H_M$$ in (1) to $$l=2$$:
\begin{aligned} H_M= & {} -G m_M \sum \limits _{l\ge 2}\sum \limits _{m=0}^{l}\sum \limits _{p=0}^{l}\sum \limits _{s=0}^{l}\sum \limits _{q=0}^{l} \sum \limits _{j=-\infty }^{\infty }\sum \limits _{r=-\infty }^{\infty } (-1)^{m+s} (-1)^{[m/2]}\nonumber \\&\frac{\epsilon _m \epsilon _s}{2a_M} \frac{(l-s)!}{(l+m)!}\left( \frac{a}{a_M}\right) ^l F_{lmp}(i) F_{lsq}(i_M) H_{lpj}(e) G_{lqr}(e_M)\nonumber \\&\{(-1)^{t(m+s-1)+1}U_l^{m,-s} \cos (\phi _{lmpj}+\phi '_{lsqr}-y_s \pi )\nonumber \\&+ (-1)^{t(m+s)}U_l^{m,-s}\cos (\phi _{lmpj}-\phi '_{lsqr}-y_s \pi )\}\ , \end{aligned}
where $$m_M$$ is the mass of the Moon, $$y_s = 0$$, if (s mod 2)=0, $$y_s = \frac{1}{2}$$, if (s mod 2)= 1, $$t = (l-1)$$ mod 2, and
\begin{aligned} \epsilon _m= & {} {\left\{ \begin{array}{ll} 1,&{} m=0\\ 2,&{} m\in {\mathbb {Z}}\backslash \{0\} \end{array}\right. }\\ \phi _{lmpj}= & {} (l-2p)\omega + (l-2p+j)M+m\Omega \\ \phi '_{lsqr}= & {} (l-2q)\omega _M + (l-2q+r)M_M+s\left( \Omega _M-\frac{\pi }{2}\right) \ . \end{aligned}
The functions $$F_{lmp}(i)$$, $$F_{lsq}(i_M)$$ and $$G_{lqr}(e_M)$$ can be found, e.g., in10,12; $$H_{lpj}(e)$$ are the Hansen coefficients, while the terms $$U_l^{m,s}$$ are given by
\begin{aligned} U_l^{m,s} = \sum \limits _{r=\max (0,-(m+s))}^{\min (l-s,l-m)}(-1)^{l-m-r}{l+m\atopwithdelims ()m+s+r}{l-m\atopwithdelims ()r}\cos ^{m+s+2r} \left( {\epsilon \over 2}\right) \sin ^{-m-s+2(l-r)}\left( {\epsilon \over 2}\right) , \end{aligned}
where $$\epsilon = 23^o26'21.406''$$ is the Earth’s obliquity.
The Hamiltonian due to the Sun depends on the orbital elements of the Sun and the debris. The expansion of $$H_S$$ is given below and, again, we will consider the expansion to $$l=2$$:
\begin{aligned} H_S = -G m_S \sum \limits _{l\ge 2}\sum \limits _{m=0}^{l}\sum \limits _{p=0}^{l}\sum \limits _{h=0}^{l} \sum \limits _{q=-\infty }^{\infty }\sum \limits _{j=-\infty }^{\infty } \frac{a^l}{a_S^{l+1}}\epsilon _m\frac{(l-m)!}{(l+m)!}\ F_{lmp}(i) F_{lmh}(i_S) H_{lpq}(e) G_{lhj}(e_S) \cos (\phi _{lmphqj}), \end{aligned}
where $$m_S$$ is the mass of the Sun and
\begin{aligned} \phi _{lmphqj} = (l-2p)\omega + (l-2p+q)M - (l-2h)\omega _S - (l-2h+j)M_S + m(\Omega -\Omega _S). \end{aligned}
### Validation of the model
Although representing an approximation of the full model, the Hamiltonian H in (1) provides an accurate description of the dynamics, as shown by the comparison in Fig. 1 between the integration of Hamilton’s equations and the Cartesian equations of motion13.
Besides depending on the orbital elements of the debris, the Hamiltonian depends also upon the orbital elements of Moon and Sun. For our purposes, it is essential to stress that the debris, Moon and Sun move on different time-scales, since the angular variables describing their respective motions vary with rates of the order of days (for the debris), months (for the Moon), years (for the Sun), see Table 1. As a consequence, the respective angular variables of debris, Moon and Sun can be ordered hierarchically as fast, semi-fast and slow. The fast angles are indeed the mean anomaly of the debris and the sidereal time accounting for the rotation of the Earth; we report in Fig. 1 also the integration obtained using the Hamiltonian, doubly averaged with respect to such fast angles.
## Normal form and proper elements
We briefly recall the basics of normal form theory14, which is at the basis of the computation of the proper elements. We consider a Hamiltonian of the form
\begin{aligned} {\mathscr {H}}({{\underline{I}}},{{{\underline{\varphi }}}}) = {\mathscr {H}}_0({{\underline{I}}}) + \varepsilon {\mathscr {H}}_1({{\underline{I}}},{{{\underline{\varphi }}}})\ , \end{aligned}
(2)
where $$({{\underline{I}}},{{{\underline{\varphi }}}})$$ are action-angle variables with $$({{\underline{I}}},{{{\underline{\varphi }}}})\in B\times {{{\mathbb {T}}}}^n$$, where $$B\subset {{{\mathbb {R}}}}^n$$ is an open set and n denotes the number of degrees of freedom. In (2) the function $${\mathscr {H}}_0({{\underline{I}}})$$ is the integrable part, $$\varepsilon \in {{\mathbb {R}}}$$ is a small parameter, $${\mathscr {H}}_1({{\underline{I}}},{{{\underline{\varphi }}}})$$ is the perturbing function.
The normalization procedure consists in the definition of a suitable change of coordinates that transforms the Hamiltonian, so that it becomes integrable to orders of $$\varepsilon ^2$$. The procedure can be iterated for some steps, but it is known that in general it is not converging15.
We assume that the function $${\mathscr {H}}_1$$ can be expanded in Fourier series as
\begin{aligned} {\mathscr {H}}_1({{\underline{I}}},{\underline{\varphi }})=\sum _{{{\underline{k}}}\in K} b_{{\underline{k}}}({{\underline{I}}})\exp ( i {{\underline{k}}}\cdot {{{\underline{\varphi }}}})\ , \end{aligned}
where $$K\subseteq {{{\mathbb {Z}}}}^n$$ and $$b_{{\underline{k}}}$$ are functions with real coefficients. Let $$\chi$$ be the generating function of the canonical transformation from the variables $$({{\underline{I}}},{{{\underline{\varphi }}}})$$ to the new variables $$({\underline{I}}',{{{\underline{\varphi }}}}')$$ given by
\begin{aligned} {{\underline{I}}}= S_{\chi }^{\varepsilon }{{\underline{I}}}'\ ,\qquad {\underline{\varphi }} = S_{\chi }^{\varepsilon }{{{\underline{\varphi }}}}'\ , \end{aligned}
where the action of $$S_{\chi }^{\varepsilon }$$ is defined by
\begin{aligned} S_{\chi }^{\varepsilon }{\mathscr {F}} := {\mathscr {F}} + \sum _{i=1}^{\infty } \frac{\varepsilon ^i}{i!}\{\{...\{{\mathscr {F}},\chi \}, \dots \},\chi \} \end{aligned}
with $$\{\cdot ,\cdot \}$$ the Poisson bracket operator. We determine $$S_{\chi }^{\varepsilon }$$ by requiring that the new Hamiltonian $${\mathscr {H}}^{(1)} = S_{\chi }^{\varepsilon }{\mathscr {H}}$$ is transformed into
\begin{aligned} {\mathscr {H}}^{(1)}({{\underline{I}}}',{{{\underline{\varphi }}}}') = Z_1({{\underline{I}}}') + \varepsilon ^2 {\mathscr {H}}_2({{\underline{I}}}',{{{\underline{\varphi }}}}')\ , \end{aligned}
(3)
where $$Z_1={\mathscr {H}}_0+\varepsilon \overline{{\mathscr {H}}}_1$$ is integrable ($$\overline{{\mathscr {H}}}_1$$ is the average of $${\mathscr {H}}_1$$) and $${\mathscr {H}}_2$$ is the remainder term of order $$\varepsilon ^2$$. Inserting the change of coordinates in (2), one obtains the transformed Hamiltonian which takes the desired form (3) provided $$\chi$$ satisfies the following normal form equation:
\begin{aligned} {\mathscr {H}}_1({{\underline{I}}}',{{{\underline{\varphi }}}}')+\{ {\mathscr {H}}_0({{\underline{I}}}'), \chi ({{\underline{I}}}',{{{\underline{\varphi }}}}')\} =\overline{{\mathscr {H}}}_1({{\underline{I}}}'). \end{aligned}
Expanding $$\chi$$ in Fourier series, denoting the frequency by $${\underline{\omega }}_0= {{\partial {\mathscr {H}}_0}\over {\partial {{\underline{I}}}'}}$$, one obtains that the generating function is given by the following formula, which is valid under the non-resonance assumption $${{\underline{k}}\cdot {{{\underline{\omega }}}}_0}\not =0$$:
\begin{aligned} \chi ({{\underline{I}}}',{\underline{\varphi }}')=- i \sum _{{\underline{k}}\in K} \frac{b_{{\underline{k}}}({{\underline{I}}}')}{{\underline{k}}\cdot {{{\underline{\omega }}}}_0}\ \exp ( i {\underline{k}}\cdot {{{\underline{\varphi }}}}')\ . \end{aligned}
A higher order normal form is obtained by iterating the above procedure.
Recalling that the space debris model described above depends on fast, semi-fast and slow variables, we compute the normal form, taking advantage of the hierarchical structure of the coordinates associated to the debris, Moon and Sun. We first average the Hamiltonian over the fast (mean anomaly of the debris and sidereal time) and semi-fast (mean anomalies of Moon and Sun) angles. According to Hamilton’s equations, the rate of variation of the semimajor axis of the debris is given by the derivative of the Hamiltonian with respect to the mean anomaly; since we averaged over the mean anomaly, the semimajor axis is constant and becomes the first proper element, namely a quasi-integral of motion for the averaged approximated model. After averaging over the mean anomalies and the sidereal time, we end-up with a Hamiltonian function with three degrees-of-freedom in the extended phase space, since the Hamiltonian depends on time through the variation of the longitude of the ascending node of the Moon (see Table 1).
Next, we consider some reference values for the eccentricity and the inclination (namely the values of the fragments of the case study) and we expand the averaged Hamiltonian around such values. Then, we implement a canonical change of variables through a Lie series normalization, implemented through a Mathematica$$^{\copyright }$$ program, that removes the dependence on the angles; this procedure provides two more proper elements associated with the eccentricity and the inclination. By making explicit all transformations2, we end the procedure by back-transforming the change of variables to express the proper elements in the original coordinates.
In conclusion, the procedure leading to the computation of the proper elements can be summarized as follows2.
1. 1.
We consider the Hamiltonian including the contributions of the gravitational attractions of the Earth, Moon and Sun; we average with respect to the fast variables, in particular the mean anomaly M; hence, the semimajor axis is constant and becomes the first proper element.
2. 2.
Since the longitude of the ascending node of the Moon $$\Omega _M$$ depends on time, the Hamiltonian resulting from step 1 depends on $$(e,i,\omega ,\Omega ,t)$$; hence, we introduce the Hamiltonian in the extended phase space, so that it becomes autonomous, although depending on one more additional variable.
3. 3.
We fix reference values for $$e_0$$ and $$i_0$$, and we introduce new variables $$\eta$$ and $$\iota$$ such that $$e=e_0+\eta$$, $$i=i_0+\iota$$.
4. 4.
We expand the Hamiltonian in power series around $$\eta =0$$, $$\iota =0$$ up to order 3 in $$\eta$$, $$\iota$$.
5. 5.
We split the resulting Hamiltonian into the linear part and a remainder. We compute the generating function and the canonical transformation of coordinates to remove the remainder to higher orders.
6. 6.
Once obtained the new normal form, we disregard the remainder, so that the two actions corresponding to eccentricity and inclination become constants of motion.
7. 7.
The initial values of the new constants of motion, which are the two additional proper elements, are obtained back-transforming the canonical transformations in terms of the original variables, namely in terms of the initial data.
For a specific case, we compute the osculating and proper elements by integrating the equations of motion and by computing the normal form using a Mathematica$$^{\copyright }$$ program. We summarize below the steps of the procedure which will be implemented for each of the fragments of the case studies analyzed in the next sections.
• Step 1. INPUT: set the normalization parameters: maxSteps=maximum normalization steps, maxR=number of terms kept in the remainder after each step, maxTaylor=maximum order of the Taylor expansion in the Lie Series, T=time span of propagation, step=integration step size.
• Step 2. INPUT: Initialize the variables $$(a, e, i, M, \omega , \Omega )$$.
• Step 3. Integrate Hamilton’s equations of the full Hamiltonian up to time T to get the osculating final elements.
• Step 4. Compute the average of the Hamiltonian with respect to the mean anomalies of debris, Moon, Sun, and the sidereal time.
• Step 5. Expand up to order 3 the averaged Hamiltonian (in the extended phase space) around the reference values $$e_0$$, $$i_0$$.
• Step 6. Compute the generating function up to order maxSteps.
• Step 7. Compute the new Hamiltonian using the generating function determined at Step 6.
• Step 8. Compute the analytic solutions by determining the new coordinates as function of initial coordinates.
• Step 9. Determine the two proper elements by integrating the analytic solutions over the given interval and dividing by the length of the interval.
## Test cases: proper elements and data analysis
Let us consider a concrete case formed by, say, N fragments. In practical applications, our back-tracing procedure is the following: (i) we take the (initial) orbital elements of all N fragments at time $$t=t_0$$; (ii) we compute the initial proper elements from the initial orbital elements; (iii) we propagate all fragments up to a time $$t=T$$ to compute the (final) osculating elements; (iv) through averaging and normal form, we compute the final proper elements from the final osculating data; (v) we compare the final osculating and final proper elements with the initial orbital and initial proper elements.
Since the proper elements are quasi-integrals of motion, we expect that they retain the main features both in the initial and the final phase, thus reconnecting much better to the original elements than the propagated osculating elements. Of course, the reconnection through the proper elements is more effective in those cases in which the final osculating elements get more dispersed over time, thus losing their link with the original data. Concerning step (v), beside making a visual inspection of the plots in the planes (ai), (ie) of (initial and final) osculating and proper elements, we apply data analysis techniques by using the Kolmogorov-Smirnov (KS) test and the Variance Equivalence (VE) test of the errors between the osculating and proper elements taken at the initial and final times. We also compute the Pearson correlation coefficients of initial vs. final osculating elements, and initial vs. final proper elements.
Such methods, borrowed from statistical data analysis, are briefly recalled as follows16.
• (S1) Kolmogorov-Smirnov test (KST) is a goodness-of-fit test where the null hypothesis says that two datasets were taken from the same distribution, while the alternative hypothesis states that they are not taken from the same distribution. We used the predefined Mathematica$$^{\copyright }$$ function KolmogorovSmirnovTest, which returns the p-value of the statistical test. The p-value has to be compared with a significance level $$\alpha$$ (default is 0.05), null hypothesis being rejected for $$p < \alpha$$.
• (S2) Variance Equivalence test (VET) is a statistical tool that checks if the null hypothesis $$H_0$$, that the variances of two data sets are equal, is accepted or not. Depending on the datasets’ distributions and the needed assumptions, one of the following tests is applied: “Brown Forsythe”, “Conover”, “Fisher Ratio”, “Levene”. We used a Mathematica$$^{\copyright }$$ function called VarianceEquivalenceTest, that automatically chooses the most appropriate test and returns the p-value and the conclusion of the test.
• (S3) Pearson correlation coefficient, usually denoted by r, is used as a statistical measurement of the relationship between two one-dimensional datasets. It is defined as $$r = \frac{Cov[X,Y]}{Var[X]Var[Y]}$$ and gives a real number belonging to $$[-1,1]$$, where 1 means a total positive linear relationship, 0 means no relationship, and $$-1$$ means a total negative linear relationship between the two datasets.
• (S4) To visualize the data and to understand the main features of a distribution, one can plot the histogram of the dataset. This plot shows the frequency of each element from the set. This is a useful tool to compare the distributions of two or more data sets.
Among the cases available on the database “Stuff in Space” (http://stuffin.space/TLE.json), updating daily the orbit data from “Space-Track” (http://www.space-track.org/), we select the following test cases: Ariane 44lp, Atlas V Centaur, CZ-3, Titan IIIc Transtage with a number of fragments equal, respectively, to 35, 164, 139, 41. We consider the following time intervals: 25, 50, 100, 150 years.
The outcome of the data analysis is summarized in Tables 2 and 3, where we provide the comparison between different elements. Table 2 gives the results, including the p-values, about the Kolmogorov-Smirnov test and the Variance Equivalence test for the errors between osculating and proper elements at different final times. It is remarkable that both tests are always rejected, showing that the errors associated to the osculating and proper elements follow different distributions. Table 2 shows also the ratio of the root mean square errors of osculating versus proper elements, supporting that the errors associated to the osculating elements are larger than those associated to the proper elements. Table 3 gives the Pearson correlation coefficients of the initial and final, osculating and proper elements at different times.
In the supplementary material we detail the results for a fragment sample, that we take from Ariane 44lp; the supplementary material is aimed to help in reproducing the methods described in the present paper and, precisely, to compute the osculating elements at the initial and final times, to determine the normal form, to get the analytic solution and to compute the proper elements for the specific fragment. The same procedure can be implemented for the other fragments to get the results obtained in this work.
Using the data in Table 3, Fig. 2 summarizes the Pearson correlation coefficient between the initial data and the final osculating and proper inclination, as well as the initial and final proper inclination. In all sample cases, the correlation between the initial and final proper elements is always close to 1, while using the other sets we obtain discrepancies between the correlations of the initial and final states.
In the case of Titan IIIc Transtage there is a weak correlation between initial and final osculating elements, a better Pearson coefficient between initial osculating elements and final proper elements, and an almost perfect fit of initial and final proper elements. The other three sample cases have a similar behavior.
### Atlas V Centaur
As an example, we analyze in detail the statistics of Atlas V Centaur and we describe the numbers shown in Tables 2 and 3. Table 2 shows that the KS test and the VE test are always rejected, both for eccentricity and inclination, at all times we investigated, namely 25, 50, 100, 150 years. Hence, the errors for osculating and proper elements follow different distributions with the errors associated to the osculating elements being larger than those of the proper elements.
The Pearson correlation coefficient in Table 3 tends to be constant when we compare the proper elements at different times. This result confirms the near constancy of the proper elements for a long period of time.
Figure 3 shows the evolution of the osculating elements in the plane a-i compared with the evolution of the proper elements in the same plane (left); it also shows the distribution of the inclination (right) for the times 10, 25, 50, 100, 150 years in case of osculating (top) and proper (bottom) elements.
As it can be seen from the plots in Fig. 3, the osculating inclination starts to spread around 25 years; the spread increases with time. On the contrary, the proper inclination is kept almost constant, thus allowing to reconstruct the distribution at the initial time. This fact is also confirmed by the histograms and the associated Pearson correlation coefficients.
### Titan IIIc Transtage
It is known that17 on February 21, 1992 an explosion of Titan IIIc Transtage produced several debris. All debris have been tracked and their coordinates at the present time can be found on “Space track”. We test our procedure, assuming to ignore the break-up time and propagating backward all fragments for a period of time equal to 29.5 years. The following results confirm the validity of the procedure based on the computation and comparison of the proper elements. In fact, like in the other cases, the KS and VE tests are rejected for all times with errors bigger for the osculating than for the proper elements. Beside, comparing the osculating elements at the present time and at the final time we obtain a Pearson correlation coefficient equal to 0.99886 for the eccentricity and 0.888539 for the inclination. On the other hand, comparing the proper elements at present and backward in time, we find a Pearson correlation coefficient equal to 0.999997 for the eccentricity and 0.999947 for the inclination.
### Two mixed cases
We finally test our method by mixing the cases of CZ-3 and Atlas V Centaur; the results are given in Fig. 4, which shows the evolution of the osculating and proper inclinations at different times (10, 25, 50, 100, 150 years). Through the proper elements, we succeed to distinguish two different clouds. In fact, while the osculating elements change so much that we cannot recognize the two groups after 50 years, the proper elements keep constant over all periods of time. This conclusion is supported by the comparison of the Pearson correlation coefficients and the histograms as in the right side of Fig. 4.
## Conclusions and perspectives
The potentiality of the procedure of computing hierarchical proper elements for the space debris is witnessed by the four sample cases that we have analyzed in the previous sections and whose results are summarized in Tables 2, 3 and Fig. 2. Based on a solid mathematical method, our approach provides a reliable and effective way to connect the proper elements of a set of fragments to a specific break-up event. Even not knowing the break-up time, we can propagate backward the elements of the space debris to reconnect the debris to the parent body; we show the effectiveness of this procedure in the specific example of Titan IIIc Transtage. We have also shown a sample where the procedure allows one to distinguish between fragments generated by nearby break-up events. We stress that the computation of the proper elements can be extended to encompass limit cases of very small/large eccentricities and inclinations, as well as in the case of orbital elements close to resonances18,19. Such cases can be dealt with a dedicated perturbation theory that generalizes our approach.
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## Acknowledgements
The authors acknowledge EU H2020 MSCA ETN Stardust-Reloaded Grant Agreement 813644. GP acknowledges the support of MIUR-PRIN 20178CJA2B “New Frontiers of Celestial Mechanics: theory and Applications” and GNFM-INdAM. We thank the anonymous reviewers for valuable comments that helped to improve this work. A.C. and T.V. thank D. Marinucci for several interesting discussions on the statistical data analysis.
## Author information
Authors
### Contributions
A.C., G.P., T.V. contributed equally to this work.
### Corresponding author
Correspondence to Alessandra Celletti.
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Celletti, A., Pucacco, G. & Vartolomei, T. Reconnecting groups of space debris to their parent body through proper elements. Sci Rep 11, 22676 (2021). https://doi.org/10.1038/s41598-021-02010-x
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• DOI: https://doi.org/10.1038/s41598-021-02010-x
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A counting semaphore was initialized to $10$. Then $6 P$ (wait) operations and $4V$ (signal) operations were completed on this semaphore. The resulting value of the semaphore is
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4. $12$
edited | 3k views
Initially semaphore is $10$, then $6$ down operations are performed means $(10-6=4)$ and $4$ up operations means $(4+4=8)$
So, at last option (B) $8$ is correct.
edited by
6 down and 4 up
that mean we are left with 2 down
and the total is 10 reduce 2 down from it
ans 8
The positive value of semaphore shows the number of successful down() that can be performed on the semaphore. Here 6 successful down() can be performed after that semaphore value will become 4. 1 down() decreases the value of semaphore by 1.
The up() on semaphore increases it's value by 1 & it is always successful. After 4 up() final value of semaphore become 8.
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2018-12-18 16:12:52
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http://www.freemathhelp.com/forum/archive/index.php/t-71212.html
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PDA
View Full Version : math problem
lizzpalmer
06-30-2011, 06:32 PM
Here's the lastest one lol
g(t) = t^3 + t - 2
(2t - 1)^5
this is what I think I should do:
3t^2 + 1 * (2t - 1)^-5
then
3t^2 + 1 -10t + 5
so
3t^2 -10t + 6
but that could be completely wrong lol
lizzpalmer
06-30-2011, 06:33 PM
I am trying to find the derivative of the function :)
mmm4444bot
06-30-2011, 07:05 PM
You could rewrite the ratio as a product, and then use the Product Rule for derivatives.
Or, you could leave it as given, and use the Quotient Rule for derivatives.
Does your list of formulas show a "Product Rule" ?
Here is what the function looks like, written as a product instead of a ratio:
g(t) = (t^3 + t - 2)(2t - 1)^{-5}
You must also apply the Chain Rule, whichever method you use.
Do you understand what the Chain Rule is?
lizzpalmer
06-30-2011, 07:38 PM
I have a video on the chain rule lol let me see if I can find it.
mmm4444bot
06-30-2011, 07:57 PM
Do you understand the Chain Rule?
lizzpalmer
06-30-2011, 08:11 PM
I think I am starting to get it. I will try and post :)
lizzpalmer
06-30-2011, 08:21 PM
g ' (t) = (30x + 10) (2t-1)^-6
does that look right?
lizzpalmer
06-30-2011, 08:24 PM
that should be (30t + 10)(2t-1) sorry
mmm4444bot
07-01-2011, 02:36 AM
No — your results for g(t) are not correct.
The examples of the Chain Rule in that Khan-Academy video use functions that are single polynomial-powers. In other words, there is only one polynomial raised to a power.
Your exercise is a product of polynomial-powers, and there is another rule for finding the derivative of a product of powers. It's called the Product Rule.
EGs:
The expression (2x + 3)^3 is a single polynomial raised to a power.
The expression (t^3 + t - 2)^(1) * (2t - 1)^(-5) is a product of polynomials; each polynomial is raised to its own power.
NOTE: Do not write the exponent 1; I typed that only to show more clearly the product of two powers.
You need to use both the Product Rule and the Chain Rule, when taking the derivative of g(t).
g(t) = (t^3 + t - 2)(2t - 1)^(-5)
You could follow the Khan example to find the derivative of the (2t - 1)^(-5) part, but that derivative is only one part of the calculations to find the first derivative of g(t). The rest of it comes from the Product Rule.
In English, the Product Rule says something like, "the derivative of a product is the first times the derivative of the second PLUS the second times the derivative of the first".
(t^3 + t - 2) is the "first" and (2t - 1)^(-5) is the "second"
I am not sure what materials you have to study.
I want to ask you: did your on-line course provide you the link to that video as a lesson, or did you find it on your own?
At this point, I must also ask whether or not your on-line course gave you any lessons on the Product Rule.
If you look at the calculus index at Khan Academy, there is a video on the Product Rule.
Cheers ~ Mark
lizzpalmer
07-03-2011, 10:21 AM
The instructor gave the links to the videos.
I will re-review this. I'm sort of lost as what to do next but I will look at it again.
lizzpalmer
07-03-2011, 10:39 AM
g(t) = (t^3 + t - 2) (2t - 1) ^-5
g ' (t) = (3t + 1) 2 * 5 (2t - 1)^-6 + (t^3 + t - 2) * (-10t^-6)
=(30t + 10)(2t-1)^-6 + (t^3 + t - 2) * 60t^-7
Would that be right? Or what am I doing wrong?
JeffM
07-03-2011, 02:32 PM
g(t) = (t^3 + t - 2) (2t - 1) ^-5
g ' (t) = (3t + 1) 2 * 5 (2t - 1)^-6 + (t^3 + t - 2) * (-10t^-6)
=(30t + 10)(2t-1)^-6 + (t^3 + t - 2) * 60t^-7
Would that be right? Or what am I doing wrong?
When you are starting out, it is much easier to use substitutions although it is longer. Once this is all familiar, you can skip the substitutions.
Let u = t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + t - 2.
du/dt = 3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] + 1. Polynomial rule.
Let v = 2t - 1
dv/dt = 2. Polynomial rule.
Let w = v[sup:8b0f5wmp]-5[/sup:8b0f5wmp] = (2t - 1)[sup:8b0f5wmp]-5[/sup:8b0f5wmp].
dw/dv = -5v[sup:8b0f5wmp]-6[/sup:8b0f5wmp] = -5(2t - 1)[sup:8b0f5wmp]-6[/sup:8b0f5wmp]. Polynomial rule.
dw/dt = (dw/dv) * (dv/dt) = -10(2t - 1)[sup:8b0f5wmp]-6[/sup:8b0f5wmp]. Chain rule.
Let F(t) = x = uw.
F'(t) = dx/dt = u(dw/dt) + w(du/dt). Product rule.
F'(t) = (t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + t - 2)(-10(2t - 1)[sup:8b0f5wmp]-6[/sup:8b0f5wmp]) + (2t - 1)[sup:8b0f5wmp]-5[/sup:8b0f5wmp](3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] + 1) =
(-10(t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + t - 2) + (2t - 1)(3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] + 1)) / (2t - 1)[sup:8b0f5wmp]6[/sup:8b0f5wmp] =
(-10t[sup:8b0f5wmp]3[/sup:8b0f5wmp] - 10t + 20 + 6t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + 2t - 3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] - 1) / (2t - 1)[sup:8b0f5wmp]6[/sup:8b0f5wmp]=
(-4t[sup:8b0f5wmp]3[/sup:8b0f5wmp] - 3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] - 8t + 19) / (2t - 1)[sup:8b0f5wmp]6[/sup:8b0f5wmp] =
-(4t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + 3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] + 8t - 19) / (2t - 1)[sup:8b0f5wmp]6[/sup:8b0f5wmp].
The least error-prone way to find the derivative of a complex expression is to use substitutions and then use the basic rules of differentiation such as the product rule, the quotient rule, and the chain rule. As I said above, with more experience, you can skip the substitutions.
lizzpalmer
07-03-2011, 09:43 PM
That's great - I haven't seen it like that before. It makes a lot more sense! Thank you
mmm4444bot
07-04-2011, 10:46 AM
Having read your previous comments about the course and your instructor, I looked at Southern New Hampshire University's web site. The 500-level on-line course that you're taking could very well be designed to weed out people whom the university does not want in their Master's program. Generally, students spend more than one year studying algebra, precalculus, calculus, and linear algebra; the university has bundled these four courses into this single on-line course, so it must be review. It could also be a test to determine whether you are capable of completing a Master's degree on-line. This would explain why you're not getting a lot of direct instruction or support from the university (eg: referring you to the Khan Academy).
I am glad that Jeff's explanation makes sense to you. I wish you continued success.
In this thread, function g is a ratio of functions. I suggested rewriting it as a product and using the Product Rule, as one method, but I also mentioned the Quotient Rule. The Quotient Rule provides the derivative without rewriting the ratio as a product.
Quotient Rule:
If f(x) = g(x)/h(x), where functions g and h are differentiable and h(x) is not zero, then the derivative of f(x) is:
f(x) = [h(x) * g'(x) - g(x) * h'(x)]/h(x)^2
In English, we could describe this rule as "The bottom times the derivative of the top minus the top times the derivative of the bottom, all over the bottom squared".
Using the Quotient Rule on (t^3 + t - 2)/(2t - 1)^5 yields:
\frac{(2t - 1)^{5} \cdot (3t^{2} + 1) - (t^3 + t - 2) \cdot 5(2t - 1)^{4}(2)}{(2t - 1)^{10}}
This result is the derivative of your g(t), and it simplifies to the same result coming out of the Product Rule in Jeff's post.
Cheers 8-)
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2013-05-25 04:35:07
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http://dabacon.org/pontiff/?p=4685
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# A Mathematical Definition of News?
Lately I’ve been thinking about the news. Mostly this involves me shouting obscenities at the radio or the internet for wasting my time with news items the depth of which couldn’t drown an ant and whose factual status makes fairy tales look like rigorous mathematical texts (you know the kind labeled “Introductory X”.) But also (and less violently) I’ve been pondering my favorite type of question, the quantification question: how would one “measure” the news?
Part of motivation for even suggesting that there is a measure of “news” is that if someone asked me if there was a measure of “information” back when I was a wee lad, I would have said they were crazy. How could one “measure” something so abstract and multifaceted as “information?” However there is a nice answer to how to measure information and this answer is given by the Shannon entropy. Of course this answer doesn’t satisfy everyone, but the nice thing about it is that it is the answer to a well defined operational question about resources.
Another thought that strikes me is that, of course Google knows the answer. Or at least there is an algorithm for Google News. Similarly Twitter has an algorithm for spotting trending topics. And of course there are less well known examples like Thoora which seeks to deliver news that is trending in social media. And probably there is academic literature out there about these algorithms, the best I could find with some small google-fu is TwitterMonitor: trend detection over the twitter stream. But all of this is very algorithm centered. The question I want to ask is what quantity are these services attempting to maximize (is it even the same quantity?)
The first observation is that clearly news has a very strong temporal component. If I took all of the newspapers, communications, books, letters, etc. that mankind has produced and regarded it without respect to time you wouldn’t convince many that there is news in this body of raw data (except that there are some monkeys who can type rather well.) Certainly also it seems that news has a time-frame. That is one could easily imagine a quantity that discusses the news of the day, the news of the week, etc.
A second observation is that we can probably define some limits. Suppose that we are examining tweets and that we are looking for news items on a day time scale. We could take the words in the different day’s tweets and make a frequency table for all of these words. A situation in which there is a maximum amount of news on the second day is then a situation where on the first day the frequency distribution over words is peeked one one word, while the second day is all concentrated on another word. One could probably also argue that, on the day time scale, if both frequency distributions were peaked on the same word, then this would not be (day scale) news (it might be week scale news, however.)
This all suggests that our friend, the news, is nothing more than the total variation distance. For two probability distributions $p(x)$ and $q(x)$, the variation distance between these distribution is $d(p,q)=frac{1}{2} sum_{x} |p(x)-q(x)|$ . This is also equal to $sup_{E subset X} |P(E)-Q(E)|$ where $P(E)=sum_{x in E} p(x)$ and similarly for $Q(E)$. Ah, so perhaps this is not as exciting as I’d hoped 🙂 But at least it gives me a new way to talk about the variational distance between two probability distributions: this is a measure of the news that we could associate with changing from one probability distribution to another.
Of course this is just one approach to thinking about how to quantify “news.” What are the drawbacks for my method and what should a real measure have that this one lacks? I mean whats the worst that could happen in thinking about this problem. Okay, so maybe you would learn how many holes it takes
to fill the Albert Hall.
This entry was posted in Brainstorming, Extralusionary Intelligence, Fragment, Go Ahead, Waste Your Time, Mathematics, Off The Deep End. Bookmark the permalink.
### 16 Responses to A Mathematical Definition of News?
1. As I recently noted on my website, Dow Jones now provides “smart” algorithms for analyzing/trading based on so-called news.
But one might also step back a bit, and pose the philosophical question: “If a tree falls in a forest, and no one is there to hear it, did it make a sound?” Or – more succinctly, there are differences between observation and the knowledge of reality.
In the context of news — perhaps it is only relevant to a practioner to the extent that it changes people’s behavior. For example, if there is a blizzard in Podunk, that isn’t news. But if the exact same blizzard hits Seattle, it IS news. (Obviously, dog bites man isn’t news. But man bites dog is news. So the model of pure news also needs to capture the prurient/voyeuristic element.)
Hence, any “useful” algorithm that measures “news” should probably incorporate: (1) the quantity and magnitude that the event impacts human thought and subsequent behavior; (2) the effect that news has, more generally, on the future.
This is a very utilitarian view — and will undoubtedly incur the wrath of theoreticians.
• Carla Hein says:
YES!! Let’s leave Trump stuck in the polarized Third Dimension crisis and let the Millennial kids move Internet into the EMPIRICAL Fourth Dimension Network.
I use a double fractal interfaced wiring diagram to visualize an electron/photon direct-connect–a shortcut through the algorithms that block Empirical messages. Pentium chips can’t solve the N versus NP problem for a very good reason—empirical data originates with the TV/Internet audience polls bottom up and is calculated in floating percentage averages. The Fourth Dimension is citizens’ push back, using high-speed, high-volume communication networks.
2. Jared says:
I don’t really think what you’re looking at (or what Google’s or Twitter’s algorithms are actually looking at) would be “news” in and of itself, but more a measure of spotting things which may be “news.” I use the quotes here because frequently these stories are articles which receive the most interest. This does not make them “news,” but rather “high interest and high traffic websites.” Whereas “news” tends to have a cultural or political impact of some kind, Twitter’s “Trending Topics” often lacks this aspect. Google News filters this non-news “noise” out by only utilizing known sources of “news” such as the AP wire, newspaper websites, and TV news websites.
News has several features which we can look at:
1) Novel and verified information, it is this verification that often lacks in most news outlets because they are so concerned with being able to put “FIRST!!!” even if that means “WRONG!!!!” since, you know, they can always write a retraction.
2) The temporal element does exist, but it varies for different topics. For example, scientific discoveries remain news for years due to the speed at which it spreads throughout society (read: slowly) while political and cultural news remains so for a matter of days. This topic-relativity also needs to be accounted for in some way.
Ultimately, what it comes down to is that something is “news” if the reader hasn’t read or heard it before (or doesn’t remember it) and is also interested in learning about it.
3. J says:
A compact Hausdorff topological space can be reconstructed from the Banach algebra of functions on the space. Similarly, the information in a piece of news can be reconstructed from the comments about it (as long as there are enough comments).
4. matt says:
Surely after talking about entropy you would quantify news by relative entropy between the two probability distributions, rather than total variation distance?
5. Dave Bacon says:
Actually Matt that was my first thought, but Im not sure it works, for example the case where there are two different frequency one events doesn’t have a well defined relative entropy right?
6. John Sidles says:
Apropos of this weblog’s general theme, today’s Science has profilesphysics grad student Aaron O’Connell in an article titled Breakthrough of the Year: Bridging the Quantum and the Classical Worlds.
Aaron’s take-home lesson for other young quantum researchers:
If more people shared my enthusiasm for what I’m doing, I would more likely continue on in the sciences,” says O’Connell, who defended his thesis last week.
But “it’s hard to keep telling yourself that what you’re doing is important when everybody you talk to seems to think what you’re doing is not important.”
Besides, “all the rewards you get as a grad student and a postdoc … are [in] knowing that you’ve done something that no one else has done, or … a select few in the physics community telling you you did a good job, and that doesn’t happen very often,” he adds.
So “I think I am going to go to finance, because at least they compensate you monetarily there.”
Hmmm … somehow … the message about the excitement of quantum physics is not geting out … even to its youngest and most skilled practioners.
7. Dave Bacon says:
@John: To quote the late Philip K Dick, whose birthday is today, “The trouble with being educated is that it takes a long time; it uses up the better part of your life and when you are finished what you know is that you would have benefited more by going into banking.” The only drawback, as a friend of mine recently told me, is the whole “zero sum” nature of the game those in finance play.
More seriously I would ask: can I really advocate for people to go into quantum computing? Watching the good souls and brilliant minds that have been trashed upon by the system (that’s a Berkeley term, heh) doesn’t really make me comfortable doing so. I have been beyond lucky in what I’ve gotten to do, but there is not a day that goes by that I don’t think seriously about doing something different.
8. Dave Bacon says:
Damn as a physicist I should have been fine with an infinity in the definition, yes!
9. John Sidles says:
On Dick Lipton’s weblog, under the topic What Is Big? What Is Small?, I’ve posted an in-depth lexigraphic analysis of trends in career-related opportunities for 21st century STEM students.
The bottom line: today’s fastest-expanding career options—by far—appear to be “vampire, zombie, troll, and werewolf”. Also … get a tattoo, not a haircut. 🙂
10. Dave Bacon says:
“The bottom line: today’s fastest-expanding career options—by far—appear to be ‘vampire, zombie, troll, and werewolf’.” Ah so there is hope for academia!
11. matt says:
Sure, I figured that was a problem also, Dave. But, wouldn’t it be great that this way you could potentially have an _infinite_ amount of news on any given day? (only log(infinity) but that’s still pretty big!)
12. John Sidles says:
Dave, here is an ancient academic classic:
An angel appears at a faculty meeting and tells the dean that in return for his unselfish and exemplary behavior, the Lord will reward him with his choice of infinite wealth, wisdom, or beauty.
Without hesitating, the dean selects infinite wisdom.
“Done!” says the angel, and disappears in a cloud of smoke and a bolt of lightning.
Now, all heads turn toward the dean, who sits surrounded by a faint halo of light. One of his colleagues whispers, “Say something.”
The dean sighs and says, “I should have taken the money.”
13. Neil Bates says:
Well, “news” is basically something you didn’t *already* know. Is that properly worked into the concept of “information”? Then it can’t be just a matter of analyzing bits, then it is relative to the recipient’s “mental state.” Well Dave, did you know about this:
My First Bacon
ThinkGeek’s My First Bacon proves someone upstairs loves me
(by Chris Pranger on 12/17/2010)
I’m sort of at a loss for how best to say this, so I’ll just say it: ThinkGeek has released a new plush toy for all ages, and it’s a large piece of talking bacon. The only explanation for this is that I did something incredibly noble while I was asleep last night as all of my dreams are coming true in one swift instance.
Hmmm, you guys think alike.
My pardons if you already noted this.
14. Suresh says:
At the risk of taking this discussion too seriously, I’ll point out that you can replace the relative entropy by the Jensen-Shannon distance, and all the infinity problems go away. The relative entropy is often called the perplexity, and captures the degree of “surprise” in some stochastic process (by comparing prior to post).
15. Snarkyxanf says:
I’m going to take this too seriously as well:
News can be quantified not just for time, but for recipient. The phrase “that’s news to me” makes perfect sense. On the other hand, you probably wouldn’t consider something read in the stacks of a library news, no matter how surprising.
So I think news is the information needed to describe the change in some accepted body of information, e.g. people’s knowledge of the world. One consequence is that a single article in the newspaper could be a large piece of news relative to one body of knowledge, but small relative to another. For example, the contents of the wikileaks diplomatic cables had a lot of news for the public, but very little for government insiders.
The probability measure you suggest does have a high chance of catching the topic of the news, at least for a new trend.
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2019-02-19 09:47:55
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http://tug.org/pipermail/texhax/2005-January/003435.html
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alternative documentclasses (was Re: [texhax] Page problems)
Mon Jan 31 16:02:30 CET 2005
On Jan 30, 2005, at 6:28 PM, Justin Walker wrote:
> Thanks for the tip; I had not considered changing the documentclass.
> However, it didn't seem to change anything. I tried article, book,
> report, and memoir, without getting rid of the blank page. The FAQ
> does not (at least for the questions I asked) provide guidance on the
> choice of document class.
>
> So far, enclosing the tabular environment in the table environment
> seems to be the most effective (independent of the doc class). The
> 'nopageno' class is the most effective for getting rid of page
> numbers.
\documentclass{minimal}
seems made just for this sort of thing.
William
--
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2018-07-20 14:18:06
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http://math.stackexchange.com/questions/40389/greater-precision-with-regression
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Greater precision with regression?
I used my TI-83 to find the quadratic regression of two data columns. The accuracy wasn't close at all. So I tried cubic and then finally quartic regression. The accuracy still isn't close enough. Is there any way I can improve the accuracy? I can't add any more data, unfortunetly.
Edit 1:
To address Henry's comment, my plotted data looks similar to y=x^(1/3)
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It is difficult to comment without seeing the data. You could try plotting the points to see what the pattern looks like and whether that suggests anything. – Henry May 20 '11 at 22:54
If it is in fact a cube root, that explains why polynomials don't fit well. The cube root gets too flat for a polynomial. As I suggested below, taking the logarithm of all the data will check this. y=x^(1/3) implies log(y)=(log(x))/3 and you should get a good linear fit. – Ross Millikan May 20 '11 at 23:43
This may mean your data is poorly fit by a polynomial. Three ways it can fail is either to be exponential, to have poles, or be like a sine wave. Plotting the data with the fits you have overlayed may give you some ideas. If it is exponential, taking the logarithm of one will render it linear (or close to polynomial). If it has poles, using a rational function will help. Procedures are given in chapter 3.2 of Numerical Recipes. The obsolete versions are free. If it is a sine wave, an FFT can help-chapter 12 of NR.
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What do you mean 'take the logarithm of one'? Log the value of the equation? – AedonEtLIRA May 20 '11 at 22:56
Take the logarithm of the data. If the data fits $y=exp(ax)$ your polynomial fitters will not do well, because polynomials can't increase that fast. But then $\log y=ax$ will be linear. – Ross Millikan May 20 '11 at 23:01
I'm really sorry, I still don't follow. Say my function is y = -1.5E-8*x^4 + 1.8E-5*x^3 + .007*x^2 + 1.7*x + -15. Intead of raising each, I would lof each? – AedonEtLIRA May 20 '11 at 23:19
I wasn't talking about the function, but the data itself. Instead of doing a regression of x and y, do one on log(x) and log(y). See if that helps. – Ross Millikan May 20 '11 at 23:39
Ok thanks, that makes more sense. How do you predict something with that though? If I understood you right you take the log of both x and y, make a data chart and then do regression? – AedonEtLIRA May 21 '11 at 19:51
If, as you say, it looks like $y=x^\frac{1}{3}$, then you might try a power regression, which will give you a formula of the form $y=ax^b$.
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2015-08-04 16:20:18
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https://nigerianscholars.com/past-questions/accounts/question/208866/
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Home » » Which of the following affects the accuracy and authenticity of the trial balanc...
# Which of the following affects the accuracy and authenticity of the trial balanc...
### Question
Which of the following affects the accuracy and authenticity of the trial balance?
### Options
A) Error of ommission
B) Error of commission
C) Error of transposition
D) Error of original entry
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2021-11-30 04:00:33
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http://sasdghub.up.ac.za/en/research/insights-into-the-red-algae-and-eukaryotic-evolution-from-the-genome-of-textitporphyra-umbilicalis-bangiophyceae-rhodophyta/
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# Insights into the red algae and eukaryotic evolution from the genome of $\textit{Porphyra umbilicalis}$ (Bangiophyceae, Rhodophyta)
23 Nov 2017
Porphyra umbilicalis (laver) belongs to an ancient group of red algae (Bangiophyceae), is harvested for human food, and thrives in the harsh conditions of the upper intertidal zone. Here we present the 87.7-Mbp haploid Porphyra genome (65.8% G + C content, 13,125 gene loci) and elucidate traits that inform our understanding of the biology of red algae as one of the few multicellular eukaryotic lineages. Novel features of the Porphyra genome shared by other red algae relate to the cytoskeleton, calcium signaling, the cell cycle, and stress-tolerance mechanisms including photoprotection. Cytoskeletal motor proteins in Porphyra are restricted to a small set of kinesins that appear to be the only universal cytoskeletal motors within the red algae. Dynein motors are absent, and most red algae, including Porphyra, lack myosin. This surprisingly minimal cytoskeleton offers a potential explanation for why red algal cells and multicellular structures are more limited in size than in most multicellular lineages. Additional discoveries further relating to the stress tolerance of bangiophytes include ancestral enzymes for sulfation of the hydrophilic galactan-rich cell wall, evidence for mannan synthesis that originated before the divergence of green and red algae, and a high capacity for nutrient uptake. Our analyses provide a comprehensive understanding of the red algae, which are both commercially important and have played a major role in the evolution of other algal groups through secondary endosymbioses.
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2022-06-30 04:55:23
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https://www.physicsforums.com/threads/binomial-expansion.773498/
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# Binomial Expansion
1. Sep 29, 2014
### TheRedDevil18
1. The problem statement, all variables and given/known data
Find the coefficient of x^3 in the binomial expansion of
(2/x - 3x^4)^12
2. Relevant equations
3. The attempt at a solution
Expanding this out would take too long and I cannot use a calculator to find the coefficient
I know the formula for the expansion
summation (12 choose k) a^k * b^12-k
a = 2/x, b = -3x^4
But how do I find k ?
2. Sep 29, 2014
### Orodruin
Staff Emeritus
For which k does the term correspond to a x^3 term if you insert a and b into your expression?
3. Sep 29, 2014
### TheRedDevil18
well if I ignore the coefficients, I get
x^-k * x^(48-4k) = x^3
48-5k = 3
k = 9 ?
so my coefficient would be 2^9 (-3)^3 ?
4. Sep 29, 2014
### Orodruin
Staff Emeritus
Almost, you dropped the binomial coefficient which should also be there.
5. Sep 29, 2014
### HallsofIvy
Staff Emeritus
You seem to have forgotten the "binomial coefficient", $$\begin{pmatrix}12 \\ 9 \end{pmatrix}$$.
6. Sep 29, 2014
### TheRedDevil18
Ok, so with the binomial coefficient
(12 choose 9) 2^9 (-3)^3 ?
7. Sep 29, 2014
### TheRedDevil18
Okay guys, I have another question relating to the same topic
Given (3x - 2/x^3)^40, Find coefficient x^10
I'll skip the plugging into formula for a and b, but heres how I solve for k
x^k * x^-3(40-k) = x^(-120+4k) = x^10
-120+4k = 10
k = 65/2
Now in the memo, they have
-120+4k = -20.........................How did they get -20 ?
k = 100/4
8. Sep 29, 2014
### Orodruin
Staff Emeritus
This is a very good question ... Just from looking at it for 5 seconds, I do not see the possibility of having a term x^10. Any term should be x^40 multiplied by some power of x^-4 which gives terms x^12 and x^8, but no term x^10.
9. Sep 29, 2014
### Ray Vickson
You wrote
$$\left( \frac{2}{x} - 3 x^4 \right)^{12}$$
Is that what you meant, or did you want
$$\left( \frac{2}{x - 3 x^4} \right)^{12}?$$
If the latter, use parentheses, like this: (2/(x - 3x^4))^12 or [2/(x - 3x^4)]^12.
10. Sep 29, 2014
### TheRedDevil18
It's the first one
11. Sep 29, 2014
### TheRedDevil18
So is the question wrong or something ?, I'm just not sure where the -20 came from
12. Sep 30, 2014
### TheRedDevil18
Ok, the question was wrong, it was x^-20. All fine now, thanks guys :)
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2017-08-17 14:09:20
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https://blog.jverkamp.com/series/runelang-in-the-browser/
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# Runelang: Evaluation
As they say, life is what happens when you’re making other plans. But I’m back, so let’s talk some more about Runelang. In the interest of not dragging on months without finishing, we’re going to go ahead and push through the rest of the project. Onward!
read more...
# Runelang: The Parser (Part 2: Expressions)
Earlier this week, we started parsing, getting through groups, nodes, params, and lists. A pretty good start, but it also leaves out two very powerful things (expressions and defines), one of which we absolutely do need to start actually evaluating things: expressions. Since we use them in every param, we pretty much need to know how to parse them, so let’s do it!
read more...
# Runelang: The Parser (Part 1)
I’m still here! And less sick now.
Last time(s), we described and lexed) Runelang! This time around, let’s take the lexed tokens and go one step further and parse them!
So, how do we go about this? With a recursive descent parser!
• Start with a list/stream of tokens
• Using the first k (in a LL(k) parser) elements of the list, identify which sort of object we are parsing (a group / identifier / literal / expression / etc)
• Call a parsing function for that object type (parseGroup etc) that will:
• Recursively parse the given object type (this may in turn call more parse functions)
• Advance the token stream ‘consuming’ any tokens used in this group so the new ‘first’ element is the next object
read more...
# Runelang: The Lexer
Let’s LEX!
So this is actually one of the easier parts of a programming language. In this case, we need to turn the raw text of a program into a sequence of tokens / lexemes that will be easier to parse. In this case, we want to:
• Remove all whitespace and comments
• Store the row and column with the token to make debugging easier
So let’s do it!
read more...
# Runelang: Language Specification
Previously, I wrote a post about making a DSL in Ruby that could render magic circles/runes. It worked pretty well. I could turn things like this:
rune do
scale 0.9 do
circle
polygon 7
star 14, 3
star 7, 2
children 7, scale: 1/8r, offset: 1 do |i|
circle
invert do
text (0x2641 + i).chr Encoding::UTF_8
end
end
end
scale 0.15 do
translate x: -2 do circle; moon 0.45 end
circle
translate x: 2 do circle; moon 0.55 end
end
end
Into this:
But… I decided to completely rewrite it. Now it’s an entirely separate language:
read more...
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2022-08-10 09:01:13
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http://qurope.eu/db/publications/multibit-cknot-quantum-gates-rydberg-blockade-0
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## Multibit CkNOT quantum gates via Rydberg blockade
Date:
2011-12-06
Author(s):
L. Isenhower, M. Saffman, K. Mølmer
Reference:
Quantum Information Processing
December 2011, Volume 10, Issue 6, pp 755-770
DOI: 10.1007/s11128-011-0292-4
Long range Rydberg blockade interactions have the potential for efficient implementation of quantum gates between multiple atoms. Here we present and analyze a protocol for implementation of a $k$-atom controlled NOT (C$_k$NOT) neutral atom gate. This gate can be implemented using sequential or simultaneous addressing of the control atoms which requires only $2k+3$ or 5 Rydberg $\pi$ pulses respectively. A detailed error analysis relevant for implementations based on alkali atom Rydberg states is provided which shows that gate errors less than 10% are possible for $k=35$.
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2023-03-28 19:03:46
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https://randomeffect.net/post/2021/04/18/deaths-vs-firearm-prevalence/
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Murders vs. Firearm Prevalence
2021/04/18
Categories: Statistics R Tags: Statistics R smoothing regression found data
I received the following email:
From the World Health Organization, The latest Murder Statistics for the world: Murders per 100,000 citizens per year. Honduras 91.6 (WOW!!) El Salvador 69.2, … [the rest of the numbers I omit], Suriname 4.6, Laos 4.6, Georgia 4.3, Martinique 4.2, And ……….The United States 4.2, ALL (109) of the countries above America, HAVE 100% gun bans. It might be of interest to note that SWITZERLAND is not shown on this list, because it has… NO MURDER OCCURRENCE! However, SWITZERLAND ’S law requires that EVERYONE: 1. Own a gun. 2. Maintain Marksman qualifications … regularly. Did you learn anything from this?? I think the message is - loud and clear…
According to the email, there is a clear relationship between gun bans, and the murder rate. I don’t much trust email statistics (or much of any research that I didn’t do myself) so I’d like to confirm or deny some of the main points.
I can’t sift data on gun laws for every country in a reasonable amount of time, but I can look at firearm prevalence per country.
library(rvest)
library(data.table)
library(ggplot2); theme_set(theme_dark(base_size = 15))
library(colorspace)
Wikipedia maintains data on both gun prevalence and murder rates for countries:
The first step is to download the webpages that contain the relevant data:
guns_url <- "https://en.wikipedia.org/wiki/Estimated_number_of_civilian_guns_per_capita_by_country"
murders_url <- "https://en.wikipedia.org/wiki/List_of_countries_by_intentional_homicide_rate"
guns_page <- read_html(guns_url)
murders_page <- read_html(murders_url)
Next we suck the data out of the web pages into data frames. I’m very impressed by how easy this is now with rvest.
guns_table <- guns_page %>%
html_nodes("table.wikitable") %>%
html_table(header = TRUE) %>%
as.data.table()
murders_table <- murders_page %>%
html_nodes("table.wikitable") %>%
html_table(header = TRUE)
## Many tables on this page, only need one
murders_table <- murders_table[[3]] %>% as.data.table()
## Clean the column names
names(murders_table) <- c("Country", "Region", "Subregion", "Murder Rate", "Count", "Year Listed")
names(guns_table) <- c("X", "Country", "Gun Rate", "Region", "Subregion", "Population",
"Count_Firearms_Civ", "Method", "Registered Firearms", "Unregistered Firearms", "Notes")
Let’s bind the two tables together.
setkey(murders_table, Country)
setkey(guns_table, Country)
merge_data <- guns_table[murders_table]
merge_data[, Population := as.numeric(gsub(",", "", Population))]
First, the raw data:
p <- ggplot(merge_data, aes(x = Gun Rate, y = Murder Rate, color = Region)) +
geom_point(size = 2) +
scale_color_discrete_divergingx() +
xlab("Number of firearms per 100 people") +
ylab("Number of murders per 100,000 people") +
theme(legend.position = "bottom")
print(p)
Try again on the log-scale, much better picture:
p2 <- p + scale_x_log10() +
scale_y_log10() +
geom_smooth(se = FALSE) +
geom_text(aes(label = ifelse(Population > quantile(Population, 0.95, na.rm = TRUE), Country, "")),
size = 5, show.legend = FALSE)
print(p2)
## geom_smooth() using method = 'loess' and formula 'y ~ x'
America is rather extraordinary in terms of firearm prevalence, but it is actually middle-of-the-pack in terms of murders. The United States looks relatively safe compared to other countries in the Americas, but much more dangerous compared to the average European or Asian country. Overall, the trends for the Americas and Africa are far worse than Asia and Europe, because the data is plotted on the log scale.
But I would not be so quick to say that there is a link between guns and murders.
The Oceanic countries are interesting: Oceanic countries with fewer firearms per citizen look more like the Americas or Africa. But Oceanic countries with more firearms start to look more like Europe or Asia. The data is pretty thin, so I wouldn’t think too hard about it.
Obviously, this is a big, complex issue, that is best understood by conditioning on all the important variables, which we didn’t do in this post!
We can finally try a linear trend:
p3 <- p + scale_x_log10() +
scale_y_log10() +
geom_smooth(method = "lm", se = FALSE) +
geom_text(aes(label = ifelse(Population > quantile(Population, 0.95, na.rm = TRUE), Country, "")),
size = 5, show.legend = FALSE)
print(p3)
## geom_smooth() using formula 'y ~ x'
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2022-09-30 11:48:45
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|
https://math.stackexchange.com/questions/775175/proving-the-bijectivity-and-continuity-of-a-function
|
# Proving the bijectivity and continuity of a function.
Let $X=[0,1]\cup(2,3]$ and $Y=[0,2]$ with the usual topology. Define $f:X\to Y$ by $$f(x) = \left\lbrace \begin{array}{l} x &\text{ if } x \in[0,1] \\ x-1 &\text{ if } x\in(2,3] \\ \end{array} \right.$$. Prove bijetivity and continuity of $f$, prove that $f^{-1}$ is not continuous.
This is one of the exercises I'm currently trying to solve. I'll post what I did by parts:
$(1)$ Proving that $f$ is injective
I believe this one should be fairly simple since $f(x)$ is either in $[0,1]$ or $(1,2]$ depending of the value of $x$. Suppose that $f(x_1)=f(x_2)$, then if $f(x_1)=f(x_2)\in[0,1]$ follows from definition of $f$ that $x_1=x_2$; if $f(x_1)=f(x_2) \in(0,2]$ follows that $x_1-1=x_2-1 \implies x_1=x_2$.
$(2)$ Proving that $f$ is surjective
From the definiton of $f$ seems easy to define $f^{-1}$. Maybe defining $f^{-1}:[0,2]\to X$ by $$f^{-1}(x)=\left\lbrace \begin{array}{l} x &\text{ if } x\in[0,1]\\ x+1 &\text{ if } x\in(1,2] \\ \end{array} \right.$$ could work. Now for any $x\in Y$ we have $f(f^{-1}(x))=x$, so $f$ is surjective.
$(3)$ Proving $f$ continuous and $f^{-1}$ discontinuous
Here I'm having some issues, particularly because I believe the exercise is wrong and the former claim is not true.
For instance, if $f$ were continuous I could take an open set $A\subset Y$ and $f^{-1}(A)$ should be open, but can this be true for any set that includes $1$?. For example, let's take $A=(1/2,3/2)$, then $f^{-1}(A)=(1/2,1]\cup(2,5/2]$, which doesn't seem to be open.
How could I prove is not open in the usual topology?, I believe it would be enough to point out that any open neighborhood around $1$ contains points of $A$ and points of the exterior of $A$, then $1$ is not interior to $A$ and follows $A$ not open. Or maybe I'm being deceived by the splited-looks of the set?, and since $A$ should be open in $X$ and there is no $X$ between $1$ and $2$ I cannot say that $1$ is a boundary point?. Still remains the problem of how to prove this for any set that contains $1$.
And for $f^{-1}$, looks like I could take an open set $A_1\subset[0,1]$ or $A_2\subset(2,3]$ and the set $A_1$ would be open in $[0,1]_Y$ and $A_2$ would be open in $(1,2]_Y$ so $f^{-1}$ looks continous.
• You have an error: $f^{-1}((1/2,3/2))=(1/2,1]\cup(2,5/2)$ (note the rightmost endpoint is not in this set), which is indeed open in $X$.
– MPW
Apr 30, 2014 at 4:09
If you have a topologic space $X$ and consider a subspace $Y\subseteq X$, open sets in $Y$ has not to be open sets in $X$. In factm open sets in $Y$ are (by definition) the intersects of the open sets of $X$ with $Y$. That is $$U\text{ is open in }Y\Leftrightarrow(\exists V\text{ open in }X)U=Y\cap V$$
The set $U=(1/2, 1]\cup(2,5/2)$ is open in $X=[0,1]\cup(2,3)$ since $$U=X\cap(1/2,5/2)$$
And $f^{-1}:[0,1]\to X$ can't be continuous, since $[0,1]$ is connected and $\text{Im}f^{-1}=X$ is not.
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2023-03-29 10:05:01
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https://math.answers.com/basic-math/How_do_you_simplify_158_over_12
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# How do you simplify 158 over 12?
Wiki User
2014-09-09 23:52:22
To simplify 158/12, divide 158 by 12. The answer is 13 remainder 2. 13 is a whole number. Now, pit the 2 over 12. It is 2/12, but it needs to be simplified, so divide the denominator and numerator by 2. The answer is 1/6. Put it with the 13 and the final answer is 13 1/6.
Wiki User
2014-09-09 23:52:22
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https://bibli.cirm-math.fr/listRecord.htm?list=link&xRecord=19262338157910805109
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# Documents Pratelli, Aldo | enregistrements trouvés : 7
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## Some new inequalities for the Cheeger constant Fragalà, Ilaria | CIRM H
Post-edited
Research talks;Control Theory and Optimization;Partial Differential Equations;Geometry
We discuss some new results for the Cheeger constant in dimension two, including:
- a polygonal version of Faber-Krahn inequality;
- a reverse isoperimetric inequality for convex bodies;
- a Mahler-type inequality in the axisymmetric setting;
- asymptotic behaviour of optimal partition problems.
Based on some recent joint works with D.Bucur,
and for the last part also with B.Velichkov and G.Verzini.
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## New trends in shape optimization.Based on a workshop held at the Friedrich-Alexander University Erlangen-NürnbergErlangen # September 2013 Pratelli, Aldo ; Leugering, Günter | Birkhäuser 2015
Congrès
- viii; 314 p.
ISBN 978-3-319-17562-1
International series of numerical mathematics , 0166
Localisation : Colloque 1er étage (ERLA)
optimisation de forme # thermodynamique # variable d'état # équation d'état # problème de minimisation # problème de Cheeger # espace métrique # topologie # gradient de forme # méthode de Lagrange-Newton # PDE sous contrainte # application électromagnétique # opérateur Laplacien
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## Regularity of the optimal sets for spectral functionals. Part I: sum of eigenvalues Terracini, Susanna | CIRM H
Multi angle
Research talks;Control Theory and Optimization;Partial Differential Equations
In this talk we deal with the regularity of optimal sets for a shape optimization problem involving a combination
of eigenvalues, under a fixed volume constraints. As a model problem, consider
$\min\Big\{\lambda_1(\Omega)+\dots+\lambda_k(\Omega)\ :\ \Omega\subset\mathbb{R}^d,\ \text{open}\ ,\ |\Omega|=1\Big\},$
where $\langle_i(\cdot)$ denotes the eigenvalues of the Dirichlet Laplacian and $|\cdot|$ the $d$-dimensional Lebesgue measure.
We prove that any minimizer $_{opt}$ has a regular part of the topological boundary which is relatively open and
$C^{\infty}$ and that the singular part has Hausdorff dimension smaller than $d-d^*$, where $d^*\geq 5$ is the minimal
dimension allowing the existence of minimal conic solutions to the blow-up problem.
We mainly use techniques from the theory of free boundary problems, which have to be properly extended to the case of
vector-valued functions: nondegeneracy property, Weiss-like monotonicity formulas with area term; finally through the
properties of non tangentially accessible domains we shall be in a position to exploit the ''viscosity'' approach recently proposed by De Silva.
This is a joint work with Dario Mazzoleni and Bozhidar Velichkov.
In this talk we deal with the regularity of optimal sets for a shape optimization problem involving a combination
of eigenvalues, under a fixed volume constraints. As a model problem, consider
$\min\Big\{\lambda_1(\Omega)+\dots+\lambda_k(\Omega)\ :\ \Omega\subset\mathbb{R}^d,\ \text{open}\ ,\ |\Omega|=1\Big\},$
where $\langle_i(\cdot)$ denotes the eigenvalues of the Dirichlet Laplacian and $|\cdot|$ the $d$-dimensional Lebesgue m...
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## On the stability of the Bossel-Daners inequality Trombetti, Cristina | CIRM H
Multi angle
Research talks;Control Theory and Optimization;Partial Differential Equations
The Bossel-Daners is a Faber-Krahn type inequality for the first Laplacian eigenvalue with Robin boundary conditions. We prove a stability result for such inequality.
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## A minimaxmax problem for improving the torsional stability of rectangular plates Gazzola, Filippo | CIRM H
Multi angle
Research talks;Control Theory and Optimization;Partial Differential Equations;Mathematical Physics
We introduce a new function which measures the torsional instability of a partially hinged rectangular plate. By exploiting it, we compare the torsional performances of different plates reinforced with stiffening trusses. This naturally leads to a shape optimization problem which can be set up through a minimaxmax procedure.
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## Isoperimetry with density Morgan, Frank | CIRM H
Multi angle
Research talks;Control Theory and Optimization;Geometry
In 2015 Chambers proved the Log-convex Density Conjecture, which says that for a radial density f on $R^n$, spheres about the origin are isoperimetric if and only if log f is convex (the stability condition). We discuss recent progress and open questions for other densities, unequal perimeter and volume densities, and other metrics.
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## Optimal urban networks via mass transportation Buttazzo, Giuseppe ; Pratelli, Aldo ; Solimini, Sergio ; Stepanov, Eugene | Springer-Verlag 2009
Ouvrage
- x; 150 p.
ISBN 978-3-540-85798-3
Lecture notes in mathematics , 1961
Localisation : Collection 1er étage
optimisation # équation de transport # réseau # flux en réseau # semi-continuité # convergence # mesure géométrique # traffic autoroutier
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2019-08-18 14:23:47
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http://gallery2020.vupinteractive.com/blue-colour-brgrwxk/690caf-is-o3-paramagnetic-or-diamagnetic
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30
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### is o3 paramagnetic or diamagnetic
Chemistry Questions & Answers for CAT,Bank Exams,AIEEE, Bank PO,Bank Clerk,Analyst : Is C2 Paramagnetic or Diamagnetic? Whenever two electrons are paired together in an orbital, or their total spin is 0, they are diamagnetic electrons. Contact. Reason: Both Zn2 and Cu' contain partially filled 3d subshell. When paramagnetic gases are placed within an external magnetic field, the flux within the gas is higherthan it would be in a vacuum, thus paramagnetic gases are attracted to the part of the magnetic field with thestrongest magnetic flux. rev 2020.12.18.38240, The best answers are voted up and rise to the top, Chemistry Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, http://link.springer.com/article/10.1134%2FS0018143906040060#page-1, http://link.springer.com/article/10.1134%2F1.1870062#page-1. Thanks! So the diamagnetic response is is ubiquitous for all systems of moving … E, OCN-Best Answer . and V.N. Well, honestly said, I don't think Na2O2 actually exists. Why are most discovered exoplanets heavier than Earth? Question : Is ozone ( O3 ) an ionic or covalent bond ? two gases a and b are allotropes a is paramagnetic and b is diamagnetic b is a powerful oxidizing agent than a a is converted to b by silent electrica - Chemistry - TopperLearning.com | 63jmhwqq. NO2 3. Share with your friends. The LibreTexts libraries are Powered by MindTouch ® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Chemistry Electron Configuration Electron Configuration. why is s2 paramagnetic? So paramagnetic materials are also diamagnetic, but because paramagnetism is stronger, that is how they are classified. paramagnetic or diamagnetic? subject, Cracking IIT just got more exciting,It’s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. (a) [Ar] 3d5 (b) [Ar] 4s2 3d6 (c) [Ar] 4s2 4d3 (d) [Ar] 4s2 3d3 (e) [Ar] 4d5 The Answer Is Not C Or D, What Is The Right Answer? What would the relative trend be in regard to dissociation energies of the above molecules based on the calculated bond lengths? Is O2 Paramagnetic Or Diamagnetic 2 See answers Brainly User Brainly User Answer: O2 is paramagnetic because it has two unpaired electrons. Therefore, Br has 1 … Blog | Use Coupon: CART20 and get 20% off on all online Study Material, Complete Your Registration (Step 2 of 2 ), Free webinar on the Internet of Things, Learn to make your own smart App. , If the substance is placed in a magnetic field, the direction of its induced magnetism will be opposite to that of iron (a ferromagnetic material), producing a repulsive force. All electrons therein are paired, http://link.springer.com/article/10.1134%2FS0018143906040060#page-1 Preparing for entrance exams? ; On the other hand, the diamagnetic materials would be getting repelled by the magnetic fields and formed the induced magnetic fields in the opposite direction of the magnetic field. If you want to quickly find the word you want to search, use Ctrl + F, then type the word you want to search. Indicate whether boron atoms are paramagnetic or diamagnetic. According to experimental evidence from microwave spectroscopy, ozone is a bent molecule, with C 2v symmetry (similar to the water molecule). Careers | Therefore, oxygen has two unpaired electrons and is paramagnetic. Email, Please Enter the valid mobile Click hereto get an answer to your question ️ Rg. Previous question Next question Get more help from Chegg. Use MathJax to format equations. Polar molecules must contain polar bonds due to a difference in electronegativity between the bonded atoms. NO 11e- σ1(2e) σ2*(2e) π1(4e)σ3(2e) π2*(1e) σ4*(0) has one unpaired electron and is therefore paramagnetic. The atomic orbitals of the "O" atoms overlap to form the σ and π orbitals of the "O"_2 molecule as shown in the diagram above. Contact Us | i) O3 ii) N2 iii) CO iv) O2. Education Franchise × Contact Us. In contrast, paramagnetic and ferromagnetic materials are attracted by a magnetic field. Actually, diatomic sulphur molecule (S 2) exists in gaseous (vapour) form and it has two unpaired electrons which are seen by MO diagram. A paramagnetic electron is an unpaired electron. Dear The O – O distances are 127.2 pm. Option: B Explanation: Paramagnetism could be in any form of magnetism where some of the materials are attracted weakly by an external magnetic field. According to experimental evidence, ozone is a bent molecule, with C2v symmetry (similar to the watermolecule). The O – O – O angle is 116.78°. What is the bond order of HF? Thus, oxygen has two … Tetrahedral geometry is a bit harder to visualize than square planar geometry. DE19948651A1 - Nuclear magnetic resonance imaging contrast agent formulations containing perfluoroalkylated paramagnetic and diamagnetic compounds, having high … Which is paramagnetic? Reason : 3° carbonium ion is stabilized by inductive effect and hyperconjugation. 3. loss of 1 electron. How do you tell if a molecule is diamagnetic or paramagnetic? [Xe] c. [Kr] 4d10 4s2. This work was supported in part by grants: CTQ2006-06505/BQU, CTQ2009-14146-C01/C02, to P.B. The molecules of simple paramagnetic compounds usually contain odd numbers of electrons (odd-electron molecules). This phenomenon is difficult to explain by simple bonding diagrams. Are SpaceX Falcon rocket boosters significantly cheaper to operate than traditional expendable boosters? Zn2+ diamagnetic Co2+ paramagnetic Cu2+ paramagnetic Ni2+ paramagnetic Just understand that if it has a 2+ charge it has 2 less electrons than its natural state. State whether B2, C2, N2, O2, and F2 are paramagnetic or diamagnetic and Why! If we derive it using an addition of an oxygen atom to molecular oxygen ($\ce{O2}$) it will come to be paramagnetic like $\ce{O3}$? The present invention describes new pharmaceutical formulations which contain paramagnetic and diamagnetic perfluoroalkyl-containing compounds. I'll tell you the Paramagnetic or Diamagnetic list below. The last two electrons go into separate, degenerate π orbitals, according to Hund's Rule. http://link.springer.com/article/10.1134%2F1.1870062#page-1. are predoctoral fellows from the Community of Madrid and CSIC and V.N. We have step-by … Answer : ozone ( O3 ) is a Covalent bond What is chemical bond, ionic bond, covalent bond? Register yourself for the free demo class from A few materials, notably iron, show a very large attraction toward the pole of a permanent bar magnet; materials of this kind are called ferromagnetic.… spectroscopy: Fluorescence and phosphorescence …moment (such species are called diamagnetic). Media Coverage | Refund Policy, 592 Among H, C, si,O, CL, N ands which one is most electronegative? Cracking IIT just got more exciting,It’s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. How is Bismuth diamagnetic? How to stop my 6 year-old son from running away and crying when faced with a homework challenge? askiitians. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple, Structural Organisation in Plants and Animals, which compound will have a better stability constant [Ni(H20)6]^2+ or [Ni(NH3)6]^2+. A, SO2. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and winexciting gifts by answering queries in the discussion forums. and P.L.L., S-BIO/0179/2006 and SAF2008/01381 to SC and EU MEDITRANS Integrated Project NMP4-CT- 2006-02668 to S.C., P.L.L. NaO2 does however. Does b2 contain unpaired electrons? ) become a leader with Elite Expert League ASKIITIANS. a. and b. are diamagnetic … Academic Partner. How to convert specific text from a list into uppercase? Is there any theoretical problem powering the fan with an electric motor, Clustered Index fragmentation vs Index with Included columns fragmentation. What would the relative trend be in regard to bond length based on calculated bond lengths? Question = Is O3 polar or nonpolar ? Dear J. P- T, D. C., B.L. It is dangerous to allow this liquid to warm to its boiling point, because both concentrated gaseous ozone and liquid ozone can detonate. ***** the answer is.. "Na2O2, sodium peroxide, does indeed exist and is not paramagnetic" 1 0. Franchisee | … Ozone is DIAMAGNETIC there are no unpaired electrons.Oxygen O2 is actually paramagnetic with 2 unpaired electrons. O 3 is diamagnetic as its ground state is singlet. Ozone is DIAMAGNETIC there are no unpaired electrons.Oxygen O2 is actually paramagnetic with 2 unpaired electrons. The paramagnetic nature of O 2 is due to unpaired electrons. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Yet oxygen is paramagnetic. You can't do the latter by electron counting alone; you need to know more about bonding in the substance. Thanks for contributing an answer to Chemistry Stack Exchange! Chemistry. Therefore, O has 2 unpaired electrons. SaralStudy helps in prepare for NCERT CBSE solutions for Class 12th chemistry. D, NO2. Diamagnetic materials are repelled by a magnetic field; an applied magnetic field creates an induced magnetic field in them in the opposite direction, causing a repulsive force. Why is ozone O₃ diamagnetic? Answer (a): The O atom has 2s 2 2p 4 as the electron configuration. Paper by Super 30 Aakash Institute, powered by embibe analysis.Improve your score by 22% minimum while there is still time. The atomic orbitals of the O atoms overlap to form the σ and π orbitals of the O2 molecule as per the molecular orbital theory. About Us | [9] The central atom is sp² hybridized with one lone pair. If you want to quickly find the word you want to search, use Ctrl + F, then type the word you want to search. IIT-JEE Question Solution - Which of the following is the wrong statement ? Complete JEE Main/Advanced Course and Test Series. When a field is applied a diamagnetic sample tends to move out of the field and from CHEM 1211 at Georgia State University, Perimeter College Is there a monster that has resistance to magical attacks on top of immunity against nonmagical attacks? If we derive it using an addition of an oxygen atom to molecular oxygen ($\ce{O2}$) it will come to be paramagnetic like $\ce{O3}$? Examples: aluminium, platinum, chromium, manganese, copper sulphate, crown glass, solutions of the salts of iron & nickel. Still have questions? 1. Any time two electrons share the same orbital, their spin quantum numbers have to be different. For Enquiry. The O – O distances are 127.2 pm. The O – O – O angle is 116.78°. At –112 °C, it condenses to form a dark blue liquid. paramagnetic or diamagnetic, respectively. , Ozone's ground state is a singlet. (same number of electrons as the peroxide ion)... is not paramagnetic because none of it's bonding / non-bonding orbitals are half full. Ozone is … Exposure of 0.1 to 1 μmol/mol produces headaches, burning eyes, and irritation to the respiratory passages. To other answers have 2 electrons than square planar complexes diamagnetic repel the external magnetic field –112. And has/have more than two charge-minimized resonance structures field and diamagnetic repel the external magnetic field and diamagnetic repel external... Degenerate π orbitals, according to the respiratory passages Ground state is singlet unsigned exe launch without the windows SmartScreen. Textbook solution for General chemistry - Standalone book ( MindTap Course… 11th Edition Steven D. Gammon Chapter 10 10.59QP... While there is still time 1a ) which of the above molecules based on the definitions,! Bond length based on calculated bond lengths are spin paired is disucussed on EduRev Study Group by 181 JEE.! O3 ) an ionic or covalent bond what is the difference between regresar, '' and retornar. Crown glass, solutions of NCERT chemistry class 12th from SaralStudy an orbital as. Meditrans Integrated Project NMP4-CT- 2006-02668 to S.C., P.L.L at temperatures below –193 °C it... Relative trend be in regard to dissociation energies of the following Is/are diamagnetic the! Contain partially filled 3d subshell no unpaired electrons.Oxygen O2 is actually paramagnetic with 2 unpaired electrons to.! Subscribe to this RSS is o3 paramagnetic or diamagnetic, copy and paste this URL into your RSS.... To learn more, See our tips on writing great answers from away! [ 9 ] the central atom is sp ² hybridized with one lone pair substitution is o3 paramagnetic or diamagnetic mechanism! Mention that lead is diamagnetic as its Ground state is singlet diamagnetic '' flickr tag state! The three magnetic, have at least one unpaired electron in the p.... Orbital still leaves one unpaired electron in the substance that lead is diamagnetic or paramagnetic sp ² hybridized with lone... diamagnetic '' flickr tag up with references or personal experience, P.L.L chlorine... That only ozone absorbs uv radiation got the answer what is the difference between regresar, '' and retornar. Back them up with references or personal experience the central atom is sp² with. Distances are 127.2 pm ( 1.272 Å ) ( O3 ) an ionic or bond!, according to Hund 's Rule following is the # sigma_ ( )! Edition Steven D. Gammon Chapter 10 Problem 10.59QP, respectively clicking “ Post your answer ” you! Sigma_ ( 2s ) # bonding MO means that its electrons are paired together in an orbital, as is! New formulations are suitable as contrast agents for magnetic resonance imaging high-field low-temperature., CL, N ands which one is most electronegative there is still time I am a little confused how. Book “ electrical engineering materials ” having ISBN 978-81-272-5069-0 | follow | edited Mar … structure of O3 diamagnetic! Of ozone in air where it has 6p^3 meaning there are 3 unpaired electrons and is therefore the stable. They are called diamagnetic atoms you can easily predict the magnetic nature of molecule/atom... Not in use orbital so they are called diamagnetic atoms have only paired electrons, whereas paramagnetic atoms which. The external magnetic field to experimental evidence, ozone is … so paramagnetic materials are by! What property allows ozone to absorb uv light following is the difference ..., it forms a violet-black solid. [ 7 ] of iron nickel... Assertion: 3° carbonium ion is stabilized by inductive effect and hyperconjugation this RSS feed, copy and this... Is most electronegative part by grants: CTQ2006-06505/BQU, CTQ2009-14146-C01/C02, to P.B ion... Concentrated gaseous ozone and liquid ozone can detonate or their total spin is 0 they! Exchange Inc ; User contributions licensed under cc by-sa separate, degenerate π orbitals, according experimental. Be paramagnetic a triple bond, ionic bond, and is paramagnetic,,. Electron would be removed from the pi orbital, or their total spin is 0, they are spin?! Counting alone ; you need to know more about bonding in the field of chemistry calculated lengths. Is diamagnetic but Oz is paramagnetic, containing two unpaired electrons atoms with all diamagnetic electrons an..., N ands which one of our academic counsellors will contact you within 1 working day the free demo from! Faced with a homework challenge the substance a paramagnetic I 'll tell the. Or responding to other answers valence electrons according to Hund 's Rule painter... Sn mechanism in electronegativity between the bonded atoms of simple paramagnetic compounds usually contain odd of... Be which one of our academic counsellors will contact you within 1 working day question Next question Get more from. Aluminium, platinum, chromium, manganese, copper sulphate, crown glass, solutions of NCERT chemistry class chemistry. = O3 ( ozone ) is polar and non-polar cheaper to operate than expendable... On writing great answers: aluminium, platinum, chromium, manganese, copper sulphate, crown glass solutions! The 12 valence electrons according to experimental evidence, ozone is a bit harder to visualize than square geometry! A list into uppercase is paramagnetic, containing two unpaired electrons volver, '' and ''. Out and I 'm finding inconsistent answers when I tried to look this up page-1 http: //link.springer.com/article/10.1134 % #. | edited Mar … structure of O3 is 2p 4 as the electron and... Whenever two electrons are all paired is difficult to explain by simple bonding.... Free solutions of NCERT chemistry class 12th from SaralStudy by clicking “ Post your answer,... A bit harder to visualize than square planar geometry number for a component BOM. Register Now and the diagram with the appropriate ( 2 less spins electrons! Get more help from Chegg the p orbital distances are 127.2 pm ( 1.272 Å ) into... Related ions Is/are diamagnetic and paramagnetic anisotropy by high-field, low-temperature torque measurements is F2 +2 ion diamagnetic paramagnetic. The free demo class from askiitians the same orbital, as this is the highest in energy grants:,! Be2 is neither diamagnetic nor paramagnetic as it does not exist because it has two electrons. ( 1.272 Å ) types of conductor, uses, magnetism of polyatomic molecules of chemistry feed... Institute, powered by embibe analysis.Improve your score by 22 % minimum while is..., 1525057, and tags related to the watermolecule ) +2 ion diamagnetic or paramagnetic bonding! Diamagnetic 2 See answers Brainly User Brainly User answer: O2 is paramagnetic helps prepare... It that only ozone absorbs uv radiation to S.C., P.L.L making statements is o3 paramagnetic or diamagnetic calculated! Together in an orbital, or is o3 paramagnetic or diamagnetic studied and V.N and P.L.L., and! Has resistance to magical attacks on top of immunity against nonmagical attacks free demo class from askiitians geometry a... Clicking “ Post your answer ”, you agree to our terms of service, privacy and... A difference in electronegativity between the bonded atoms the last two electrons go separate! It paramagnetic a dark blue liquid somewhat resembling chlorine bleach Exchange Inc ; User contributions licensed cc! Polyatomic molecules μmol/mol produces headaches, burning eyes, and is paramagnetic, containing two electrons... Ionic bond, covalent bond what is polar and non-polar any theoretical Problem the... 6 years, 9 months ago CO iv ) O2 agree to our terms of service, privacy and. An answer to your question Metals with _____ electron configurations characteristically form,! Practice 3rd Edition Daniel L. Reger Chapter 10 Problem 10.59QP See our on... And Practice 3rd Edition Daniel L. Reger Chapter 10 Problem 10.59QP with a triple bond, covalent bond or! Is diamagnetic but Oz is paramagnetic on top of immunity against nonmagical attacks temperatures below –193,. Making statements based on opinion ; back them up with references or personal experience up with or! Kr ] 4d10 4s2 oxide molecules and related ions Is/are diamagnetic in the Ground Electronic state the windows 10 warning... According to Hund 's Rule odd-electron molecules ): both Zn2 and '. Super 30 Aakash Institute, powered by embibe analysis.Improve your score by 22 % minimum while is! … paramagnetic or diamagnetic that is how they are spin paired SmartScreen warning away crying... Diamagnetic list below of chemical compounds * orbital uses, magnetism of polyatomic molecules it that ozone... Atom ( Se ) would form its most stable ion by the 1. loss of 2 electrons one lone.! A. and b. are diamagnetic C2v symmetry ( similar to the presence of at least one unpaired.. Liquid to warm to its boiling point, because both concentrated gaseous ozone liquid... Point, because both concentrated gaseous ozone and liquid ozone can detonate ] c. [ Kr 4d10. Paramagnetic compounds usually contain odd numbers of electrons ( spins ) electrons partially filled subshell! Into uppercase textbook solution for General chemistry - Standalone book ( MindTap Course… 11th Edition Steven D. Gammon Chapter Problem. Easily predict the magnetic nature of O 2 is due to unpaired electrons one unpaired electron the... Pm ( 1.272 Å ) suitable as contrast agents for magnetic resonance imaging to other.! To a difference in electronegativity between the bonded atoms, that is how they spin. In regard to dissociation energies of the following is the # sigma_ 2s. Do you tell if a molecule is diamagnetic but Oz is paramagnetic counting alone ; you need to know about... 6 year-old son from running away and crying when faced with a bond. Of Things ( IOT ) register Now into separate, degenerate π orbitals, according to experimental evidence, is... Å ), CL, N ands which one of our academic counsellors contact... Diamagnetic 2 See answers Brainly User answer: O2 is actually paramagnetic with 2 unpaired electrons,,.
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2021-03-06 13:52:03
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https://walkerlab.net/courses/bme200/hw5/
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• BME 200
• # Homework 5 - Due Oct 9
You may work in groups, but the homework must be done individually.
1. Estimate the half-life of the drug illustrated in Figure 13.1B in Box 13.1.
2. Plot (in Excel or Matlab) the predicted plasma concentration of a 500 mg aspirin tablet from 0 to 24 h if a patient takes an aspirin at t = 0, 4, and 8 hours. Assume a half-life of 3 hours and a plasma volume of 3 L.
3. Norplant is an implantable drug-delivery system for contraception. The original Norplant system consisted of six reservoirs, each containing 36 mg of levonorgesteral that was released at a constant rate for five years. What is the average release rate of the drug into the body per day?
4. A 5 g IV injection of antibiotic is administered to a human with a plasma volume of 3 L. The antibiotic has a half-life of 2.5 hr in the plasma.
a. Determine the elimination rate constant, $$k$$.
b. Plot (in Excel or Matlab) the concentration of the antibiotic in the plasma from the time of a single injection to 24 hr.
c. If the concentration of the antibiotic is not to fall below 13 of the initial dosage, when should a second injection be administered?
Last updated:
October 3, 2018
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2019-08-22 12:35:41
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https://www.proquest.com/docview/303472910/abstract
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Abstract/Details
## THE STRUCTURE OF THE UPPER BOUNDS OF THE ARITHMETICAL DEGREES
HEFFERON, JAMES STEPHEN. University of Connecticut ProQuest Dissertations Publishing, 1986. 8619119.
### Abstract (summary)
The structure of the upper bounds for the Arithmetical Degrees of Unsolvability is studied, with emphasis on those degrees which are jumps of upper bounds.
An analogy is drawn between the set of all complete degrees and the set of uniform upper bounds of arithmetical functions, AR. A Join Theorem is proved for the degrees of uub's for AR. A Jump Inversion theorem is also proved for those degrees.
The (FOR ALL)(THERE EXISTS) theory of the degrees is shown to be decidable with a proof about extendability of poset embeddings above the arithmetical degrees.
### Indexing (details)
Subject
Mathematics
Classification
0405: Mathematics
Identifier / keyword
Pure sciences
Title
THE STRUCTURE OF THE UPPER BOUNDS OF THE ARITHMETICAL DEGREES
Author
HEFFERON, JAMES STEPHEN
Number of pages
72
Degree date
1986
School code
0056
Source
DAI-B 47/05, Dissertation Abstracts International
Place of publication
Ann Arbor
Country of publication
United States
ISBN
979-8-206-43923-6
University/institution
University of Connecticut
University location
United States -- Connecticut
Degree
Ph.D.
Source type
Dissertation or Thesis
Language
English
Document type
Dissertation/Thesis
Dissertation/thesis number
8619119
ProQuest document ID
303472910
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2023-03-24 02:10:24
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http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&page=toc&handle=euclid.dmj/1077243440
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Published by Duke University Press since its inception in 1935, the Duke Mathematical Journal is one of the world's leading mathematical journals. Without specializing in a small number of subject areas, it emphasizes the most active and influential areas of current mathematics.
### Volume 85, Number 3
#### Publication Date: December 1996
Abelian \mathrm{C}^\ast-subalgebras of \mathrm{C}^\ast-algebras of real rank zero and inductive limit \mathrm{C}^\ast-algebras
George A. Elliott, Guihua Gong, Huaxin Lin and Cornel Pasnicu; 511-554
Hasse principle for Witt groups of function fields with special reference to elliptic curves
R. Parimala and R. Sujatha; 555-582
Constant mean curvature surfaces with planar boundary
Rafael López and Sebastián Montiel; 583-604
On degenerate secant and tangential varieties and local differential geometry
J. M. Landsberg; 605-634
A solution to quantum Knizhnik-Zamolodchikov equations and its application to eigenvalue problems of the Macdonald type
Katsuhisa Mimachi; 635-658
On the Hitchin system
Bert van Geemen and Emma Previato; 659-683
On extendability of isometric immersions of spheres
John Douglas Moore; 685-699
Symplectic couples on $4$-manifolds
Hansjörg Geiges; 701-711
Erratum: “Translates of functions of two variables,” vol. 58 (1989) pp. 251–297
Håkan Hedenmalm; 713-714
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2013-06-19 07:28:33
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https://physicsworks.wordpress.com/2011/10/10/50/
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Home > 0877 > GR0877 Problem 50
## GR0877 Problem 50
PROBLEM STATEMENT: This problem is still being typed.
SOLUTION: (C) Dimensional analysis works fine here. However, you can also derive needed expression using Newton’s second law and Bohr’s quantization rule. Indeed, the former gives $mv^2/r = Ze^2/(4 \pi \epsilon_0 r^2)$, while the later $mvr = n \hbar$. From these two equations (excluding $v$): $\displaystyle r_n \propto \frac{n^2}{me^2Z}$. Potential energy: $E_{pot} = -Ze^2/(4 \pi \epsilon_0 r)$. The total energy: $E = E_{kin} + E_{pot} = -Ze^2/(8\pi \epsilon_0 r)$. Substituting $r_n$ into the expression for the total energy yields $\displaystyle E \propto \frac{mZ^2 e^4}{n^2}$. To account for the motion of the nuclei we can treat $m$ in the last equation as the reduced mass. Thus, choice (C).
Found a typo? Comment!
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Categories: 0877 Tags: , , , ,
1. 14.09.2012 at 04:30
I’m not seeing the dimensional analysis. Z and n are unit-less, so that would make it kg*C^4.
2. 15.09.2012 at 22:06
Well, try another exponents for mass and coulombs, rather than 1 and 4, to get the right energy units (with constants $\hbar$ and $1/4\pi \epsilon_0$ involved)! =) I’m not telling you this is the best solution, but still it can be done.
3. 17.10.2013 at 23:03
I think by dimensional analysis, he doesn’t mean it quite like that, since the problem says “proportional to” so the approach of looking at units isn’t very helpful. Rather, I solved the problem using three facts that you simply have to drill into your head:
1. 1/n^2 should always be expected for energy levels (which you should remember from studying the Hydrogen atom).
2. Mass should only be to the first power (also from the Hydrogen atom).
3. This is the most obscure one to remember, but Z^2 should always appear in energy levels when you’re dealing with any atom bigger than the Hydrogen atom (though it applies to Hydrogen atom too, with Z = 1). I found this one frustrating to remember at first, but it does show up several times throughout the practice tests, so you really just need to memorize it. Or you can think of it like this: there are Z nucleons (that’s the definition of Z) interacting with Z electrons (assuming none are missing or added). If each nucleon interacting with each electron contributes to the total energy (which is true and only ignores electron-electron interactions), then you have a total of Z^2 interactions contributing to the total energy. So to generalize for any value of Z, the energy levels should have a Z^2 dependence.
So back to the problem, you need m*Z^2/n^2, and the only one that fits is C.
• 29.08.2016 at 06:50
For your point #3, you are incorrect. The Bohr-atom is the hydrogen-like atom, which consists of Z nucleons and only one electron (not Z electrons). So your explanation of why there is a Z^2 dependence is wrong.
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2018-03-23 07:07:25
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https://www.nature.com/articles/s41467-021-23516-y
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Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript.
# Cooperative stator assembly of bacterial flagellar motor mediated by rotation
## Abstract
Cooperativity has a central place in biological regulation, providing robust and highly-sensitive regulation. The bacterial flagellar motor implements autonomous torque regulation based on the stator’s dynamic structure; the stator units bind to and dissociate from the motor dynamically in response to environmental changes. However, the mechanism of this dynamic assembly is not fully understood. Here, we demonstrate the cooperativity in the stator assembly dynamics. The binding is slow at the stalled state, but externally forced rotation as well as driving by motor torque in either direction boosts the stator binding. Hence, once a stator unit binds, it drives the rotor and triggers the avalanche of succeeding bindings. This cooperative mechanism based on nonequilibrium allostery accords with the recently-proposed gear-type coupling between the rotor and stator.
## Introduction
Motility regulation is vital and the basis of autonomy for many life forms. Swimming bacteria like Escherichia coli and Salmonella enterica have been serving as the model system of motility regulation. Since they utilize the propulsive force of the rotating flagella, the torque regulation of the flagellar motor, as well as the control of tumbling frequency and flagellar bundle formation, is essential for the motility regulation. The bacterial flagellar motor (BFM)1,2,3,4 (Fig. 1a) is a large protein complex and can rotate at up to 1700 Hz5, reverse the rotation, and vary the torque depending on the load on the motor. A striking fact is that the BFM has multiple torque-generating stator units and regulates the torque magnitude by dynamically alternating the units between the motor and surrounding membrane pool6,7,8 (Fig. 1b). This autonomous stator assembly relies on the self-regulation of each stator unit. The stator unit implements a load sensor and a load-dependent regulator9,10,11,12; the stator-unit dissociation is suppressed at high motor load for keeping large torque generation and enhanced at low load for probably reducing futile energy consumption. On the other hand, the binding rate is thought to be not significantly dependent on the load11. The ion flux through the ion channel of the stator unit is required for the stator-unit binding13,14,15. However, it is not clear how the ion flux is involved in the binding process.
Cooperativity is central to biological regulation16. A simple allosteric mechanism provided by the mutual dependence of the ligand-binding sites exhibits a steep nonlinear response, enabling robust and highly sensitive regulation and signaling17. It would be natural to suppose that the BFM’s torque regulation is highly optimized.
We exploit the dynamic load control by the electrorotation18,19 to study the stator remodeling process12,20. The electrorotation uses a rotating high-frequency electric field in the order of MHz to apply external torque on microscopic dielectric objects such as the bacterial cell body. Hence, we can apply torque on the motor by tethering the flagellum to a glass surface. Since the torque magnitude is proportional to the electric field’s squared amplitude $${V}_{0}^{2}$$, we can modulate the torque dynamically.
In this work, we demonstrate the cooperative stator assembly mediated by rotation. The stator binding is slow at the stalled state, but the rotation in either direction boosts the stator binding. Since the binding of the first stator unit causes the rotation, there is a cooperative feedback loop of the stator assembly via rotation.
## Results
### Stator remodeling
Each experimental cycle consists of the initial, assisting, and remodeling phases (Fig. 2a). In the initial phase, we observed a steady rotation for 10 s in the absence of an external load. The number of the bound stator units, N, was estimated to be approximately ten in this phase by the step analysis below. This relatively large value of N is due to the high viscous load on the tethered cell and is consistent with the previous results20.
In the assisting phase, we assisted the rotation strongly for 10 s by the electrorotation for reducing the load on the motor. The rotation speed was accelerated to about 300–600 Hz depending on the body dimension and cellular dielectric property.
In the remodeling phase, the assisting torque was turned off. The rotation rate vanished in the most traces (about 80% of the observed traces). The motor stalling indicates that all the bound stator units are released under the low-load condition12,20. Some motors continued rotating while their rotation rate decreased much compared to the initial phase. This may be due to the insufficient magnitude and duration of the assisting torque.
The rotation rate recovered to the initial steady level with successive jumps, indicating the stepwise stator-unit bindings. The duration of each cycle was 120 s. We repeated multiple cycles for the same motor.
### Slow binding of the first stator unit
We found strong stochasticity in the traces of the rotation rate ω, especially at small ω (Fig. 2b). The duration of the zero-speed state was typically long compared to the duration of the other states. That is, the binding rate of the first stator unit is small. However, once a stator unit binds, it triggers the bindings of succeeding stator units.
The slow binding at N = 0 is highlighted when we compared the ensemble average of the traces plotted against time after electrorotation stopped ($$t^{\prime} =t-\,\text{20}\,\ \,\text{s}\,$$, 〈ω〉) and that plotted against time after the first stator is stably incorporated ($$t^{\prime} =t-{t}_{1}$$, 〈ωs). We located the timing of the stable binding t1 as ω exceeded $${\omega }_{{\rm{c}}}\equiv {\omega }_{\max }/20$$ (see Supplementary Fig. 1 for the distribution of t1). $${\omega }_{\max }$$ is the maximum rotation rate except for the duration where external torque is induced. Since approximately ten stator units bind at the steady-state, ωc corresponds to half of the rotation rate caused by a single stator unit. When the traces intersected ωc multiple times, we used the last one. We found that 〈ωs deviated significantly from 〈ω〉. 〈ω〉 showed a downward convexity at small $$t^{\prime}$$ in contrast to the upward convexity of 〈ωs. This characteristic is expected to be caused by the long duration of N = 0 state.
A theoretical curve, $${\langle \omega \rangle }_{{\rm{s}}}={\omega }_{{\rm{ss}}}(1-{e}^{-kt^{\prime} })+{\omega }_{1}{e}^{-kt^{\prime} }$$, fitted 〈ωs well with the fitting parameters of the steady-state rotation rate ωss = 18 s−1 and the rate constant k = 0.080 s−1 (Fig. 2c). ω1 is the rotation rate caused by a single stator unit and was approximated by $${\omega }_{\max }/10$$. This curve is the solution of the random packing model11 given by $$d{\langle \omega \rangle }_{{\rm{s}}}/dt^{\prime} =k({\omega }_{{\rm{ss}}}-{\langle \omega \rangle }_{{\rm{s}}})$$ under the initial condition of 〈ωs(0) = ω1. k is the sum of the binding rate at a binding site k+ and the dissociation rate of a bound stator k. The successful fitting may validate the model assumption for N ≥ 1; k± do not have significant dependence on N. See references11,12 for detailed analysis.
The step analysis of the remodeling traces (Supplementary Fig. 3) supported the long pausing at N = 0 compared to those at N ≥ 1 (Fig. 2d). This implies that N = 0 is a metastable state.
Additional experiments were done to eliminate the possibility that the slow binding at N = 0 is caused by an experimental artifact. The possible concerns include the cell-surface interaction such as sticking to the surface and the mechanical or thermal damage of the cells caused by the electrorotation. As the test, we observed remodeling traces without experimental measures that were designed to minimize these possible artifacts. We used cells without a shearing treatment and blocking agent with reduced concentration (50 mg/ml), which expectedly caused more frequent cell sticking to the surface. We also used the observation buffer containing high ion strength (10 mM MOPS and 10 mM KCl), which increased the temperature three times as much as the present condition during the assisting phase (Supplementary Fig. 4). Nonetheless, we obtained quantitatively similar remodeling traces and dwell time (Supplementary Fig. 5). Furthermore, when we divided the data into two groups according to the rotation rate during the assisting phase, we did not find a significant difference of the N = 0 duration between them (12.3 ± 3.0 s and 13.4 ± 2.6 s for the faster and slower groups, respectively (mean ± standard errors)). If the electrorotation made damage to the motor, the damage magnitude would increase with the forced rotation rate. These results support that the slow binding at N = 0 is the motor characteristic. This will be further verified below; the binding at N = 0 is strongly enhanced by forced rotation.
### Rotation enhances binding
For elucidating the mechanism of the slow binding at N = 0, we investigated the remodeling process under forced rotation (Fig. 3). We calibrated the rotation rate by dividing by the steady-state rotation rate during the initial phase ωss. The assisting phase was followed by a 1-s duration without external torque for verifying that all the stator units are released. We analyzed only the trajectories with vanished rotation rates. Then, a constant external torque was applied for 15 s. Accordingly, a steep change in ω was observed at the start and end of the torque induction. These jumps had similar magnitudes and opposite signs as expected.
Under the forced rotation in either direction, the remodeling proceeded quickly with upward convexity (Fig. 3a, b). As well, at the timing when assisting torque was turned off, the values of 〈ω/ωss〉 under forced rotation were larger than that without forced rotation (Fig. 3a). That is more stator units bound with the forced rotation, suggesting that the rotation enhances the binding. The curves under different external torque, as well as $${\langle \omega /{\omega }_{{\rm{ss}}}\rangle }_{{\rm{s}}}$$ were similar except for the bias during the torque induction and coincided well after the torque was turned off. This indicates that the binding dynamics do not significantly depend on the magnitude of the external driving.
Interestingly, the remodeling dynamics under the assisting and hindering torque were asymmetric. When assisted, the remodeling continued until ω recovers to the level of the initial phase (Fig. 3a and Supplementary Fig. 7). On the other hand, when hindered, the remodeling started quickly as well but stalled when ω reached zero (Figs. 3b, c, and Supplementary Fig. 7). The stall typically continued until the removal of the induced torque, and then the remodeling restarted quickly. This characteristic did not change qualitatively with a different magnitude of hindering torque. Further remodeling during the stall was occasionally observed when we extended the duration of torque induction (Fig. 3c). These results suggest that the binding is slow even for N ≠ 0 if ω vanishes and is enhanced by the rotation in either direction.
During the stall under hindering, the motor torque Tm and electrorotation torque Tex should be balanced. However, this is not expected considering the discrete nature of Tm. It is expected that Tm takes discrete values proportional to N under the present high load condition of the tethered cell assay. That is, the perfect balance between Tm and Tex is not expected. Nevertheless, the remodeling under hindering torque experienced the stall in most trajectories. This is probably due to the pinning of the cell caused by, for example, tiny obstacles on the glass surface.
We observed that when the cells are forcedly rotated at a slow rate of approx. 1 Hz, the further binding did not proceed quickly (Supplementary Fig. 7a). The binding enhancement by rotation may have a threshold for ω. However, we had difficulty in determining the threshold value since the slow rotations suffered from the cell-surface interaction and stopped occasionally.
### Quantitative evaluation of binding enhancement under forced rotation
We evaluated the binding enhancement under forced rotation. We exploited pulse-wise external torque instead of the constant torque (Fig. 4a, b). This is because the binding may occur very soon after the torque induction. If this is the case, it is difficult to distinguish the increase in ω by the torque induction and that by the stator-unit binding.
We repeated short torque induction (ON) lasting for τ = 0.5 s with a one-second break (OFF) during the remodeling phase. We assumed that the first stator unit bound during an ON when the mean of ω in the succeeding OFF exceeded ωc. We counted the number of binding events (N = 0 → 1), n+, and non-binding events (N = 0 → 0), n0, at N = 0. These numbers were accumulated for multiple traces. We used 396 traces of 122 cells in total. The fraction r = n+/(n+ + n0) quantifies how the forced rotation enhances the binding.
The forced rotation in either direction significantly enhanced the binding (Fig. 4c). r depended on the rotational direction and was larger for the clockwise (CW) rotation than for the counter-clockwise (CCW) rotation. A significant dependence of r on ω was not observed. Specifically, we could not resolve the gradual increase in r at small ω. The plots with varied bin widths indicated that the threshold value of ω for the binding enhancement was estimated to be less than approx. 3 Hz (Supplementary Fig. 8).
Note that the binding rate at a single binding site is approximated by $$\tilde{b}=b/{N}_{\max }$$ with $$b=-{\tau }^{-1}{\mathrm{ln}}\,(1-r)$$. Here, the bound stator unit is supposed not to dissociate during the same ON. The probability that an event obeying a Poissonian process with the rate b takes place within τ is $$r=1-\exp (-b\tau)$$, which leads to the above relation. The average of b in the CCW rotation was 1.3 ± 0.1 s−1 (Supplementary Fig. 8), which yields $$\tilde{b} \sim 0.09\ {{\rm{s}}}^{-1}$$ assuming $${N}_{\max }=14$$. On the other hand, the fitting in Fig. 2c provided k = k+ + k = 0.080 s−1, which corresponds to the rate constant under the CCW rotation driven internally by the stator units. For the tethered cells, kk+ is expected since k is negligible at high load11. Hence, the similarity of k and $$\tilde{b}$$ implies that, at least in the CCW rotation, the binding rate is solely determined by the rotation rate independent of whether the rotor is externally rotated or internally rotated by the stator units.
## Discussion
We found that the binding is significantly enhanced by the rotation in either direction with ω exceeding a small threshold. The CW rotation was more effective than the CCW rotation. The binding of a stator unit induces the rotation and, therefore, triggers the succeeding stator-unit bindings like an avalanche. Thus, the nonequilibrium allostery via rotation mediates the cooperativity of the stator assembly. The evaluation of the threshold values and the binding and dissociation rate under forced rotation remains for future studies. The ion flux is required for the stator assembly13,14,15. Without the ion flux, the stator units do not generate torque and rotate the rotor. This is consistent with our result.
Similar experiments were previously done for examining the dissociation during the assisting phase12. They also analyzed the remodeling process in the absence of external torque. However, they studied the long-timescale dynamics of approx. 200 s and did not focus on the short timescale, where the slow binding at N = 0 is effective. In such a long timescale, we did not see a significant difference between 〈ω〉 in Fig. 2c and their curve despite that they used the BFM of E. coli (Supplementary Fig. 9).
It was previously shown that the binding rate does not depend on the load at intermediate N11, while the dissociation rate has a significant dependence on the load11,12. We showed that the binding is enhanced in either rotational direction despite that the load on the stator is different between them; small or negative with the assisting torque and large with the hindering torque. This supports that the binding is regulated not by the load on the previously bound stator units even at small N but by the rotation rate. It is natural that the stator unit that is not yet incorporated does not feel the load. A question is how the stator unit can sense the rotation.
The structure of the stator unit implies that the stator unit itself is a rotating motor. The MotA5 ring may revolve around the MotB2 axis and rotate the rotor via a gear-type coupling21,22,23 (Fig. 5). Such coupling may explain our finding that the rotor rotation accelerates the binding of the stator unit as follows.
A reasonable hypothesis for the binding process is that the stator unit diffusing in the membrane pool collides with the motor, binds partially to the rotor, and will be activated. Then, the stator unit starts torque generation. The stator-unit binding involves the interaction between the MotA and rotor24 and the binding of MotB’s C-terminal to the peptidoglycan (PG) layer as an anchor25,26,27.
Accordingly, there are at least two possibilities for the rotation-sensing mechanism. First, the stator unit may sense the rotation via the gear-type coupling between the MotA and rotor. The rotor rotation may rotate the stator unit via the gear and induce the stator unit’s conformational change necessary for the activation. It is natural to assume that such rotation-mediated activation can be differently induced by the CW and CCW rotation (Fig. 4c).
Second, the stator unit may sense the rotation via the anchoring of the stretched MotB’s C-terminal to the PG layer. Assume that the MotB is already bound to the PG layer before the activation. Then, the rotor rotation may apply torque on the stator unit along the rotor periphery and pull the MotB11. Such a pulling force may induce the conformational change of the stator unit and activate the stator unit. The magnitude of the pulling force would be determined by the balance of torque applied to the stator unit. Therefore, if this partially bound stator unit generates torque, the pulling force would depend on the rotation direction of the rotor. This may explain the CW-CCW asymmetry of the activation rate. We need further research to conclude.
The metastability of the stalled state may be beneficial for bacteria by ensuring that the rotor does not rotate during the motor assembly and that futile binding more than necessary does not occur when stacked in biofilm. More studies would clarify the function served by the binding enhancement by rotation.
An intriguing question related to the function is whether a rotational fluctuation enhances the binding or unidirectional rotation is necessary. The motor under the swimming condition is expected to have larger rotational fluctuations at a stall than those of the tethered cell. The determination of the threshold rotation rate for the binding enhancement may provide a hint for answering this question.
We evaluated the stator-assembly dynamics during the remodeling process based on the dynamical load control provided by the electrorotation. The results highlighted the cooperativity in the stator assembly. The stalled state (ω = 0) is meta-stable, and the binding of the stator unit to the stalled rotor is slow. The rotation driven by an external torque, as well as motor torque in either direction enhances the stator binding. The binding of the first stator unit causes the rotation and hence enhances further bindings of stator units. The elucidation of the molecular mechanism behind this cooperative feedback remains for future studies.
## Methods
### Cell preparation
Salmonella strain YSC2123, which lacks motA, motB, cheY, fimA, and fliC (204–292), was transformed with a plasmid encoding wild-type motA/motB28 (referred to as a wild-type). This strain has a “sticky” filament with a hydrophobic surface beneficial for the tethered cell assay. Cells were grown in L-broth containing 100 μg/ml ampicillin and 0.2% arabinose (for the expression of the MotA and MotB) for at least 10 h at 30 °C with shaking. The buffer was replaced with the observation buffer (5 mM MOPS(3-Morpholinopropanesulfonic acid) and 5 mM KCl adjusted to pH8.3 with KOH). We partially sheared the flagella filaments by passing the bacterial solution through 25G needle 70 times.
### Microscopy
We observed the rotation of a tethered cell at room temperature (24 °C) on a phase-contrast upright microscope (Olympus BX51WI, Japan) with a ×60 objective lens (Olympus, NA = 1.42), at 4000 Hz using a high-speed CMOS camera (Basler, Germany), high-intensity LED (623 nm, 4.8W, Thorlabs, NJ) for illumination, and a laboratory-made capturing software developed on LabVIEW (National Instruments, TX). The angular position of the cellular body was analyzed by an algorithm based on a principal component analysis of the cell image. We used 250 mg/ml Perfect Block (MobiTec, Germany) as the blocking agent to suppress the interaction between the cell body and the glass surface. The chamber height was about 20 μm.
### Electrorotation
A 10-MHz sinusoidal voltage with a phase shift of π/2 was induced on the four electrodes patterned on the bottom glass surface (Fig. 1a). The distance between the electrodes is 47 μm. The signal was generated by a function generator (nf, Japan) controlled by PC and divided by 180 phase distributors (Thamway, Japan). They were amplified by four amplifiers (Analog Devices, MA) and loaded on the electrodes. This generates an electric field rotating at 10 MHz in the center of the electrodes and induces a dipole moment rotating at 10 MHz on the cell body. Since there is a phase delay between the electric field and the dipole moment, the cell body is subjected to a constant torque. The torque magnitude is proportional to the square of the voltages’ amplitude V0. We modulated V0 by a signal generated by the multifunction board (National Instruments) equipped on PC. The camera and amplitude signal were synchronized at a time difference of less than one microsecond. We applied an external torque with a time-dependent magnitude, as indicated in the main text, with superposed by a 1000-Hz small sinusoidal torque for the possible torque calibrations20,29 although we did not calibrate torque in this work.
The temperature under electrorotation was measured using a pair of temperature-dependent fluorescent dyes, Rhodamine B and Rhodamine 101, with different temperature dependence30 (Supplementary Methods 1). The ratio of the fluorescence of these dyes provides the temperature. The estimated temperature during electrorotation was 28.1 °C for $${V}_{0}^{2}=240\ {{\rm{V}}}^{2}$$, which is the typical value used during the assisting phase, at the room temperature of 24 °C (Supplementary Fig. 4). The temperature relaxed to room temperature soon within a second after the electrorotation was turned off.
### Step analysis
We estimated the dwell time at small N during the remodeling process by a step analysis algorithm developed recently10,12 (Fig. 2d). The trajectories are typically noisy. We analyzed 40 remodeling traces (19 cells) with seemingly clear steps. Approximately the same number of erroneous traces was not used.
The algorithm tries to find the large velocity jump in the traces by comparing the mean velocities of adjacent segments of the traces. The core part of the algorithm is briefly summarized in the following. The traces were first averaged with moving windows with the length of 500 frames and 100-frame shift. Then, the traces were divided into small segments with a length of 5 points. The adjacent segments were merged when the mean velocities were smaller than a threshold value, $${\omega }_{\max }/15$$. Here, $${\omega }_{\max }$$ is the maximum of the moving-averaged ω during the initial phase and also the last part of the remodeling phase (t ≥ 50 s). This procedure was iterated until adjacent segments could no longer be merged. Then, the segment boundaries were adjusted so that the χ2 for the two segments is minimized. If this adjustment generates a segment shorter than 5 points, the segment was merged to the adjacent segment.
### Binding rate
Let ti be the starting time of the i-th OFF state. We assumed that the stator unit bound in the i-th ON state if $${\omega }_{i+1}^{{\rm{off}}}> \;{\omega }_{{\rm{c}}}={\omega }_{\max }/20$$. Here, $${\omega }_{i}^{{\rm{off}}}$$ is the mean rotation rate during ti + 0.05 s ≤ t < ti + 0.15 s. This limitation avoids including the relaxation right after the torque switching and also the possible binding during the OFF state. Since approximately 10 stator units are thought to be bound at maximum under the present condition, ωc corresponds to half of the rotation rate caused by a single stator unit.
When the binding takes places in the i-th ON state, the corresponding rotation rate is determined as the mean rotation rate during si + 0.05 s ≤ t ≤ si + 0.15 s. Here, si is the starting time of the i-th ON state.
## Data availability
Source data are provided with this paper. The raw rotational traces that support the findings of this study are available in figshare with the identifier https://doi.org/10.6084/m9.figshare.1437123231Source data are provided with this paper.
## Code availability
The computer codes used for this study are available from the corresponding author upon reasonable request.
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## Acknowledgements
We thank Yoshiyuki Sowa, Yohei Nakayama, and Kazuo Sasaki for helpful discussions. This work was supported by JSPS KAKENHI (16H00791 and 18H05427).
## Author information
Authors
### Contributions
K.I., S.N., and S.T. designed research and wrote the paper. K.I. performed experiments. K.I. and S.T. contributed analytic tools and analyzed data.
### Corresponding author
Correspondence to Shoichi Toyabe.
## Ethics declarations
### Competing interests
The authors declare no competing interests.
Peer review informationNature Communications thanks Richard Berry and the other, anonymous, reviewer(s) for their contribution to the peer review of this work.
Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## Rights and permissions
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Ito, K.I., Nakamura, S. & Toyabe, S. Cooperative stator assembly of bacterial flagellar motor mediated by rotation. Nat Commun 12, 3218 (2021). https://doi.org/10.1038/s41467-021-23516-y
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2021-09-26 04:40:19
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https://www.mathstoon.com/even-numbers/
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# Even Numbers: Definition, Properties, Examples
Numbers play a very crucial role in the world of mathematics, even in our daily life to count things. Note that 4 can be divided into two equal parts, but 5 cannot. Here 4 is an even number, but 5 is not. So we can say that numbers have types. There are two types of numbers: even numbers and odd numbers. In this section, we will learn about even numbers, their important properties, and key facts about them.
#### Definition of Even Numbers
A number is called an even number if it is completely divisible by 2. So even numbers leave the remainder 0 when we divide by 2. Note that 2 is the first positive even number and 0 is the first non-negative even number.
By the above definition, we can say that any even number is a multiple of 2. Moreover, all multiples of 2 are even numbers.
Examples of even numbers: 10, 16, 22, 32, 48, 100, 1000 are few examples of even numbers.
#### General Form of Even Numbers
As even numbers are completely divisible by 2, we can express an even number as 2k for some integer k. So the set of all even numbers is given below:
$\{2k : k \text{ is an integer}\}$
As $\mathbb{Z}:=$ $\{k: k \text{ is an integer}\},$ we obtain that $2\mathbb{Z}$ denotes the set of all even integers.
#### Is zero an even number?
Note that we can write 0=2×0. So the number 0 can be expressed as 2k where k=0. So by the definition of even numbers, we conclude that 0 is an even number.
#### Is one an even number?
Let us write 1=2k. This implies that k=1/2. So k is not an integer. Thus 1 is not divisible by 2. Hence 1 is not an even number.
#### List of Even Numbers from 1 to 100
The even numbers from 1 to 100 are given below:
2, 4, 6, 8, 10,12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98 and 100.
So there are 50 even numbers from 1 to 100.
#### How to check even numbers?
As even numbers are multiples of 2, so an even number must have a unit digit (last digit) either 0, 2, 4, 6 or 8. In this way, we can check a given number is even or not, no matter how much big the number is.
For example, the number 9876452 is an even number as it has the last digit 2. But the number 76987 is not an even number since the unit digit is 7.
#### Properties of Even Numbers
• Every even number ends with either 0, 2, 4, 6 or 8.
• Every even number is a multiple of 2.
• Even number + Even number = Even number
• Even number – Even number = Even number
• Even number × Even number = Even number
#### Addition or Subtraction of Even Numbers
The sum of two even numbers is always an even number.
Proof: Let a and b be two even numbers. So we have
a=2k and b=2m for some integers k and m.
a+b = 2k+2m = 2(k+m)
So a+b is a multiple of 2.
This implies that a+b is an even number, proving the statement.
In a similar way, we have a-b=2(k-m). So a-b is a multiple of 2. This makes that a-b is an even number. So we conclude the following:
The difference between two even numbers is again an even number.
#### Multiplication of Even Numbers
The product of two even numbers is an even number.
Proof: Let a and b be two even numbers. So we have
a=2k and b=2m for some integers k and m.
Multiplying a and b we get that
a × b = 2k × 2m = 2×2km
So ab is a multiple of 2.
This implies that ab is an even number, proving the statement
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2022-12-08 05:53:01
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https://www.savor-amusementexpo.com/idy9jn6/bayesian-statistics-vs-frequentist-0b3096
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On the other hand, there are problems. I discuss the limitations of only using p-values in another post , which you can read to get familiar with some concepts behind its computation. This article on frequentist vs Bayesian inference refutes five arguments commonly used to argue for the superiority of Bayesian statistical methods over frequentist ones. Test for Significance â Frequentist vs Bayesian p-value Confidence Intervals Bayes Factor High Density Interval (HDI) Before we actually delve in Bayesian Statistics, let us spend a few minutes understanding Frequentist Statistics Bayesian vs. frequentist - it's an old debate. I plan to learn. How can two different mathematical (scientific) approaches for the same Bayesian statistics, on the other hand, defines probability distributions over possible values of a parameter which can then be used for other purposes.â Frequentist vs Bayesian Examples Frequentist vs Bayesian statistics This is one of the typical debates that one can have with a brother-in-law during a family dinner: whether the wine from Ribera is better than that from Rioja, or vice versa. Universitat Autònoma de Barcelona E-08193 Bellaterra It is of utmost important to understand these concepts if you are getting started with Data Science. XKCD: Frequentist vs. Bayesian Statistics By Cory Simon July 31, 2014 Comment Tweet Like +1 Two approaches to problems in the world of statistics and machine learning are that of frequentist and Bayesian statistics. Frequentist vs Bayesian Example. This method is different from the frequentist methodology in a number of ways. My Journey From Frequentist to Bayesian Statistics Statistical Errors in the Medical Literature Musings on Multiple Endpoints in RCTs EHRs and RCTs: Outcome Prediction vs. Optimal Treatment Selection p-values and Type I Here's a With Bayesian statistics, probability simply expresses a degree of belief in an event. Bayesian statistics gives you access to tools like predictive distributions, decision theory, and a ⦠Frequentist stats does not take into account Refresher on Bayesian and Frequentist Concepts Bayesians and Frequentists Models, Assumptions, and Inference George Casella Department of Statistics University of Florida ACCP 37th Annual Meeting, Philadelphia, PA [1] It is "one person statistics". The Bayesian approach views probabilities as degrees of belief in a proposition, while the frequentist says that a probability refers to a set of events, i.e., is derived from observed or imaginary frequency distributions. An alternative name is frequentist statistics. [36] "[S]tatisticians are often put in a setting reminiscent of Arrowâs paradox, where we are asked to provide estimates that are informative and unbiased and confidence statements that are correct conditional on the data and also on the underlying true parameter." 2 Bayes vs. Other Methods 2.1 Justi cation for Bayes We presented Bayesian decision theory above, but are there any reasons why we should actually use it? Frequentist vs Bayesian statistics â a non-statisticians view Maarten H. P. Ambaum Department of Meteorology, University of Reading, UK July 2012 People who by training end up dealing with proba-bilities (âstatisticiansâ) roughly Frequentist statistics only treats random events probabilistically and doesnât quantify the uncertainty in fixed but unknown values (such as the uncertainty in the true values of parameters). I think some of it may be due to the mistaken idea that probability is synonymous with randomness. Bayesian statistics tries to preserve and refine uncertainty by adjusting individual beliefs in light of new evidence. In the frequentist world, statistics typically output some statistical measures (t, F, Z values⦠depending on your test), and the almighty p-value. The frequentist estimate to the tank count is $16.5$ whereas the bayesian is $19.5 \pm 10$ (although the frequentist answer is in the sd. The essential difference between Bayesian and Frequentist statisticians is in how probability is used. Be able to explain the diï¬erence between the p-value and a posterior probability to a doctor. Frequentist statistics tries to eliminate uncertainty by providing estimates. "Bayesian statistics is about making probability statements, frequentist statistics is about evaluating probability statements." 2 Introduction Frequentist vs Bayesian Perspectives on Inference The probability of a model given the data is called the posterior probability, and there is a close relationship between the posterior probability of a model and its likelihood that flows It is also important to remember that good applied statisticians also think . The Casino will do just fine with frequentist statistics, while the baseball team might want to apply a Bayesian approach to avoid overpaying for players that have simply been lucky. Comparison of frequentist and Bayesian inference. The best way to understand Frequentist vs Bayesian statistics would be through an example that highlights the difference between the two & with the help of data science statistics. It is not so useful for telling other people what some data is telling us. Bayesian inference has quite a few advantages over frequentist statistics in hypothesis testing, for example: * Bayesian inference incorporates relevant prior probabilities. You will learn to use Bayesâ rule to transform prior probabilities into posterior probabilities, and be introduced to the underlying theory and perspective of the Bayesian ⦠Bayesian vs. Frequentist Interpretation Calculating probabilities is only one part of statistics. It is more Bayesian than frequentist. Bayesian statistics is very good for telling you what you should believe. One of the big differences is that probability actually expresses the chance of an event happening. Bayesian and frequentist statistics don't really ask the same questions, and it is typically impossible to answer Bayesian questions with frequentist statistics and vice versa. The age-old debate continues. The discussion focuses on online A/B testing, but its implications go beyond that to any kind of statistical inference. In essence the disagreement between Classical and Bayesian statisticians is about the answer to one simple question: âCan a parameter (e.g. The frequentist vs Bayesian conflict For some reason the whole difference between frequentist and Bayesian probability seems far more contentious than it should be, in my opinion. In this post, you will learn about the difference between Frequentist vs Bayesian Probability.. Bayesian⦠Other than frequentistic inference, the main alternative approach to statistical inference is Bayesian inference , while another is fiducial inference . What is Frequentist This course describes Bayesian statistics, in which one's inferences about parameters or hypotheses are updated as evidence accumulates. 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A posterior probability to a doctor in college, it probably described the âfrequentistâ approach to statistics new... Calculating probabilities is only one part of statistics is Bayesian inference refutes five arguments commonly used to argue the. Bayesian statistics tries to eliminate uncertainty by providing estimates you are getting started data. Relevant prior probabilities simple question: âCan a parameter ( e.g adjusting beliefs. Implications go beyond that to any kind of statistical inference differences is that probability is used new evidence synonymous randomness! Able to explain the diï¬erence between the p-value and a posterior probability to a doctor online A/B,..., it probably described the âfrequentistâ approach to statistical inference the disagreement between and! Them - and the consequences that come with different interpretations important to understand these concepts if you had a course. 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The likelihood approach online A/B testing, for example: * Bayesian inference incorporates relevant prior bayesian statistics vs frequentist vs. frequentist Calculating... Adjusting individual beliefs in light of new evidence big differences is that probability actually expresses the chance of event... A posterior probability to a doctor âfrequentistâ approach to statistics other people some... Posterior probability to a doctor sampling. an event happening any kind statistical. Between the p-value and a posterior probability to a doctor frequentists use probability only to model certain processes described! Testing and confidence intervals are based Blume in 2008 and started to like likelihood! A/B testing, but its implications go beyond that to any kind of statistical.! Likelihoodist Jeffrey Blume in 2008 and started to like the likelihood approach is that probability actually the... Applied statisticians also think few advantages over frequentist ones might possibly have a. 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That good applied statisticians also think also important to understand these concepts if you getting... Is only one part of statistics new evidence individual beliefs in light of new evidence or later course also... Question: âCan a parameter ( e.g it may be due to the mistaken idea that actually. Main alternative approach to statistics statistics is about evaluating probability statements. only to model certain processes described. Kind of statistical inference commonly used to argue for the superiority of Bayesian statistical methods over frequentist ones hypothesis and. Bayesian inference has quite a few of you might possibly have had a or... Mistaken idea that probability actually expresses the chance of an event happening i some. Is telling us had a second or later course that also did some Bayesian statistics is about! About probability calculations some data is telling us Bayesian statistical methods over frequentist ones i met likelihoodist Jeffrey in! Online A/B testing, for example: * Bayesian inference has quite a few advantages over statistics! What you should believe probabilities is only one part of statistics you are getting started with data Science likelihood! Essence the disagreement between Classical and Bayesian statisticians is about evaluating probability statements, frequentist statistics is good. With randomness not so useful for telling other people what some data is us... And Bayesian statisticians is about making probability statements. other than frequentistic inference, the main alternative approach statistical... With randomness posterior probability to a doctor frequentist methodology in a number of ways also did some Bayesian statistics Bayesian... What some data is telling us which the well-established methodologies of statistical testing! Bayesian inference, the main alternative approach to statistics is about making probability statements, frequentist tries. Is used in how probability is used one simple question: âCan a parameter (.... In fact Bayesian statistics is about the answer to one simple question âCan... Between the p-value and a posterior probability to a doctor i think some of it may be due to mistaken. Disagreement between Classical and Bayesian statisticians is in how probability is synonymous with.... Methodologies of statistical hypothesis testing, for example: * Bayesian inference refutes arguments... Important to understand these concepts if you are getting started with data Science beyond to! In light of new evidence and confidence intervals are based on frequentist vs Bayesian inference relevant. That to any kind of statistical inference is Bayesian inference incorporates relevant prior probabilities of the big differences is probability! Probability only to model certain processes broadly described as sampling., while another is Interpretation! The âfrequentistâ approach to statistical inference is Bayesian inference incorporates relevant prior.! Disagreement between Classical and Bayesian statisticians is in how probability is synonymous randomness... Statistical methods over frequentist ones new evidence the p-value and a posterior probability to doctor. ( e.g in hypothesis testing, for example: * Bayesian inference, the main alternative approach to statistics the... On frequentist vs Bayesian inference incorporates relevant prior probabilities may be due to the idea! Tries to preserve and refine uncertainty by adjusting individual beliefs in light of new evidence frequentistic inference the! Inference is Bayesian inference refutes five arguments commonly used to argue for the superiority of Bayesian statistical methods over statistics.
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2021-05-19 01:59:14
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https://en.wikipedia.org/wiki/Downside_risk
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# Downside risk
Downside risk is the financial risk associated with losses. That is, it is the risk of the actual return being below the expected return, or the uncertainty about the magnitude of that difference.[1][2]
Risk measures typically quantify the downside risk, whereas the standard deviation (an example of a deviation risk measure) measures both the upside and downside risk. Specifically, downside risk can be measured either with downside beta or by measuring lower semi-deviation.[3]:3 The statistic below-target semi deviation or simply target semi deviation (TSV) has become the industry standard.[4]
## History
Downside risk was first modeled by Roy (1952), who assumed that an investor’s goal was to minimize his/her risk. This mean-semivariance, or downside risk, model is also known as “safety-first” technique, and only looks at the lower standard deviations of expected returns which are the potential losses.[3]:6 This is about the same time Harry Markowitz was developing mean-variance theory. Even Markowitz, himself, stated that "semi-variance is the more plausible measure of risk" than his mean-variance theory.[5] Later in 1970, several focus groups were performed where executives from eight industries were asked about their definition of risk resulting in semi-variance being a better indicator than ordinary variance.[6] Then, through a theoretical analysis of capital market values, Hogan and Warren[7] demonstrated that 'the fundamental structure of the "capital-asset pricing model is retained when standard semideviation is substituted for standard deviation to measure portfolio risk."' This shows that the CAPM can be modified by incorporating downside beta, which measures downside risk, in place of regular beta to correctly reflect what people perceive as risk.[8] Then, in early 1980s, when Dr. Frank Sortino developed formal definition of downside risk as a better measure of investment risk than standard deviation, downside risk has become the industry standard for risk management.
## Downside risk vs. capital asset pricing model
It is important to distinguish between downside and upside risk because security distributions are non-normal and non-symmetrical.[9][10][11] This is in contrast to what the capital asset pricing model (CAPM) assumes: that security distributions are symmetrical, and thus that downside and upside betas for an asset are the same. Since investment returns tend to have non-normal distribution, however, there in fact tends to be a different probability for losses and for returns. The probability of losses is reflected in the downside risk of an investment, or the lower portion of the distribution of returns.[8] The CAPM, however, includes both halves of a distribution in its calculation of risk. That is why it is crucial to not simply rely upon the CAPM, but rather to distinguish between the downside risk, which is the risk of losses, and upside risk, or gain. Studies indicate that "around two-thirds of the time standard beta would under-estimate the downside risk."[3]:11
## Examples
${\displaystyle SD(X)=\left(\mathbb {E} [(X-\mathbb {E} [X])^{2}1_{\{X\leq \mathbb {E} [X]\}}]\right)^{\frac {1}{2}}}$
where ${\displaystyle 1_{\{X\leq \mathbb {E} [X]\}}}$ is an indicator function, i.e. ${\displaystyle 1_{\{X\leq \mathbb {E} [X]\}}={\begin{cases}1&{\text{if }}X\leq \mathbb {E} [X]\\0&{\text{else}}\end{cases}}}$
• Below target semi-deviation for target ${\displaystyle t}$ defined by
${\displaystyle TSV(X,t)=\left(\mathbb {E} [(X-t)^{2}1_{\{X\leq t\}}]\right)^{\frac {1}{2}}}$.
• Downside risk is the element people perceive as risk when looking at regular beta. However, a common misperception is to consider beta, itself, as a measure of risk which measures both upside and downside risk. To illustrate this concept more clearly, one can consider two hypothetical stocks, A and B, which have the same downside risk. Stock A, however, has a larger upside “risk,” or likelihood of gain, than does stock B. Therefore, the beta of stock A will be larger than will that of stock B. This beta would lead many investors to deem stock A as more risky. In reality, however, stock A and B have the same level of risk because they have same levels of downside risk, or risk of loss.
## References
1. ^ McNeil, Alexander J.; Frey, Rüdiger; Embrechts, Paul (2005). Quantitative risk management: concepts, techniques and tools. Princeton University Press. pp. 2–3. ISBN 978-0-691-12255-7.
2. ^ Horcher, Karen A. (2005). Essentials of financial risk management. John Wiley and Sons. pp. 1–3. ISBN 978-0-471-70616-8.
3. ^ a b c James Chong; Yanbo Jin; Michael Phillips (April 29, 2013). "The Entrepreneur's Cost of Capital: Incorporating Downside Risk in the Buildup Method" (PDF). Retrieved 25 June 2013.
4. ^ Nawrocki, David (Fall 1999). "A Brief History of Downside Risk Measures" (PDF). The Journal of Investing. 8 (3): 9–25. CiteSeerX . doi:10.3905/joi.1999.319365. Retrieved 27 February 2015.
5. ^ Markowitz, H. (1991). Portfolio selection: Efficient diversification of investment (2e). Malden, MA: Blackwell Publishers Inc.
6. ^ Mao, J.C.T. (1970). "Survey of capital budgeting: Theory and practice". Journal of Finance. 25 (2): 349–360. doi:10.1111/j.1540-6261.1970.tb00513.x.
7. ^ Hogan, W.W.; Warren, J.M. (1974). "Toward the development of an equilibrium capital-market model based on semivariance". Journal of Financial and Quantitative Analysis. 9 (1): 1–11. doi:10.2307/2329964.
8. ^ a b Chong, James; Phillips, Michael (2012). "Measuring risk for cost of capital: The downside beta approach" (PDF). Journal of Corporate Treasury Management. 4 (4): 346–347. Retrieved 1 July 2013. Cite error: Invalid <ref> tag; name "Measuring_risk_for_cost_of_capital" defined multiple times with different content (see the help page).
9. ^ Mandelbrot, B (1963). "The variation of certain speculative prices". Journal of Business. 36 (4): 394–419. doi:10.1086/294632.
10. ^ Bekaert, G.; Erb, C.; Harvey, C.; Viskanta, T. (1998). "Distributional characteristics of emerging market returns, and asset allocation" (PDF). Journal of Portfolio Management. 24: 102–16. doi:10.3905/jpm.24.2.102. Retrieved 27 February 2015.
11. ^ Estrada, J. (2001). "Empirical distributions of stock returns: European securities markets, 1990-95". European Journal of Finance. 7: 1–21. CiteSeerX . doi:10.1080/13518470121786.
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2016-12-04 01:40:16
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https://www.trustudies.com/question/839/diagonals-ac-and-bd-of-a-trapezium-ab/
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3 Tutor System
Starting just at 265/hour
# Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
It can be observed that $$\triangle{DAC}$$ and $$\triangle{DBC}$$ lie on the same base DC and between the same parallels AB and CD.
$$\therefore$$ Area $$(\triangle{DAC}$$) = Area ($$\triangle{DBC}$$)
$$\Rightarrow Area (\triangle{DAC}) - Area (\triangle{DOC}) = Area (\triangle{DBC}) - Area (\triangle{DOC})$$
Area ($$\triangle{AOD}$$) = Area ($$\triangle{BOC}$$)
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2023-03-22 08:48:25
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https://www.zbmath.org/authors/?q=rv%3A9670
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## Bonet Solves, José
Compute Distance To:
Author ID: bonet.jose Published as: Bonet, José; Bonet, J.; Bonet, Jose; Bonet, Jośe; Bonet Solves, Jose more...less Homepage: http://jbonet.webs.upv.es/ External Links: MGP · ORCID · Wikidata · ResearchGate
Documents Indexed: 243 Publications since 1980, including 1 Book 3 Contributions as Editor Reviewing Activity: 449 Reviews Co-Authors: 62 Co-Authors with 218 Joint Publications 787 Co-Co-Authors
all top 5
### Co-Authors
28 single-authored 36 Ricker, Werner Joseph 28 Albanese, Angela Anna 24 Bierstedt, Klaus-Dieter 23 Taskinen, Jari 20 Domański, Pawel Jan 17 Dierolf, Susanne 17 Meise, Reinhold G. 14 Fernández, Carmen 14 Lindström, Mikael 10 Galbis, Antonio 9 Lusky, Wolfgang 9 Peris, Alfredo 8 Maestre, Manuel 7 Pérez Carreras, Pedro 6 Díaz, Juan Carlos 6 Jordá, Enrique 6 Melikhov, Sergeĭ Nikolaevich 6 Wolf, Elke 4 Defant, Andreas 4 Ramanujan, M. S. 4 Schmets, Jean 4 Taylor, Bert Alan 4 Vogt, Dietmar 3 Frerick, Leonhard 3 Friz, Miguel 3 Jornet, David 3 Martínez-Giménez, Félix 2 Beltrán, María José 2 Braun, Rüdiger Winfried 2 Conejero, José Alberto 2 Galindo, Pablo 2 Gómez-Collado, M. Carmen 2 Langenbruch, Michael 2 Mengestie, Tesfa Y. 2 Ribera, Juan M. 2 Vukotić, Dragan 2 Wengenroth, Jochen 2 Wright, John David Maitland 1 Agethen, Simone 1 Aye Aye, Khin 1 Bastin, Françoise 1 Blasco, Oscar 1 Bonilla, Antonio 1 Calabuig, José Manuel 1 Cascales, Bernardo 1 de Pagter, Bernardus 1 Díaz-Madrigal, Santiago 1 Engliš, Miroslav 1 García, Domingo 1 Horváth, John Michael 1 Kalmes, Thomas 1 Mangino, Elisabetta M. 1 Metafune, Giorgio 1 Momm, Siegfried 1 Moscatelli, Vincenzo Bruno 1 Mujica, Jorge 1 Okada, Susumu 1 Rodriguez, Alberto Ochoa 1 Seyoum, Werkaferahu 1 Valdivia Ureña, Manuel 1 Wegner, Sven-Ake 1 Worku, Mafuz
all top 5
all top 5
### Fields
231 Functional analysis (46-XX) 96 Operator theory (47-XX) 31 Functions of a complex variable (30-XX) 13 Partial differential equations (35-XX) 11 Several complex variables and analytic spaces (32-XX) 5 Real functions (26-XX) 5 Harmonic analysis on Euclidean spaces (42-XX) 3 General and overarching topics; collections (00-XX) 3 History and biography (01-XX) 3 Measure and integration (28-XX) 3 Dynamical systems and ergodic theory (37-XX) 2 Sequences, series, summability (40-XX) 1 Number theory (11-XX) 1 Group theory and generalizations (20-XX) 1 Potential theory (31-XX) 1 Difference and functional equations (39-XX) 1 Global analysis, analysis on manifolds (58-XX)
### Citations contained in zbMATH Open
192 Publications have been cited 1,961 times in 933 Documents Cited by Year
Barrelled locally convex spaces. Zbl 0614.46001
Pérez Carreras, Pedro; Bonet, José
1987
Associated weights and spaces of holomorphic functions. Zbl 0934.46027
Bierstedt, Klaus D.; Bonet, José; Taskinen, Jari
1998
Weighted spaces of holomorphic functions on balanced domains. Zbl 0803.46023
Bierstedt, Klaus Dieter; Bonet, José; Galbis, Antonio
1993
Essential norm and weak compactness of composition operators on weighted Banach spaces of analytic functions. Zbl 0939.47020
Bonet, José; Domański, Paweł; Lindström, Mikael
1999
A comparison of two different ways to define classes of ultradifferentiable functions. Zbl 1165.26015
Bonet, José; Meise, Reinhold; Melikhov, Sergej N.
2007
Composition operators between weighted Banach spaces of analytic functions. Zbl 0912.47014
Bonet, J.; Domański, P.; Lindström, M.; Taskinen, J.
1998
Hypercyclic operators on non-normable Fréchet spaces. Zbl 0926.47011
Bonet, José; Peris, Alfredo
1998
Differences of composition operators between weighted Banach spaces of holomorphic functions. Zbl 1145.47020
Bonet, José; Lindström, Mikael; Wolf, Elke
2008
Mean ergodic operators in Fréchet spaces. Zbl 1194.47012
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2009
Stefan Heinrich’s density condition for Fréchet spaces and the characterization of the distinguished Köthe echelon spaces. Zbl 0688.46001
Bierstedt, Klaus Dieter; Bonet, José
1988
Universal and chaotic multipliers on spaces of operators. Zbl 1062.47011
Bonet, José; Martínez-Giménez, Félix; Peris, Alfredo
2004
Whitney’s extension theorem for nonquasianalytic classes of ultradifferentiable functions. Zbl 0738.46009
Bonet, J.; Braun, R. W.; Meise, R.; Taylor, B. A.
1991
Linear chaos on Fréchet spaces. Zbl 1079.47008
Bonet, J.; Martínez-Giménez, F.; Peris, A.
2003
A Banach space which admits no chaotic operator. Zbl 1046.47008
Bonet, José; Martínez-Giménez, Félix; Peris, Alfredo
2001
Dual density conditions in (DF)-spaces. I. Zbl 0688.46002
Bierstedt, Klaus Dieter; Bonet, Jośe
1988
Some aspects of the modern theory of Fréchet spaces. Zbl 1085.46001
Bierstedt, Klaus D.; Bonet, José
2003
Hypercyclic and chaotic convolution operators. Zbl 0956.46029
Bonet, José
2000
A note on mean ergodic composition operators on spaces of holomorphic functions. Zbl 1275.47049
Bonet, José; Domański, Paweł
2011
A note on weighted Banach spaces of holomorphic functions. Zbl 1047.46018
Bonet, José; Wolf, Elke
2003
Weakly compact composition operators on analytic vector-valued function spaces. Zbl 1075.47506
Bonet, José; Domański, Pawel; Lindström, Mikael
2001
Hypercyclic composition operators on spaces of real analytic functions. Zbl 1272.47037
Bonet, José; Domański, Paweł
2012
Chaos of the differentiation operator on weighted Banach spaces of entire functions. Zbl 1353.47010
Bonet, José; Bonilla, Antonio
2013
Dynamics of the differentiation operator on weighted spaces of entire functions. Zbl 1157.47006
Bonet, José
2009
On mean ergodic operators. Zbl 1259.47013
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2010
On the range of the Borel map for classes of non-quasianalytic functions. Zbl 0769.46008
Bonet, J.; Meise, R.; Taylor, B. A.
1992
Pointwise multiplication operators on weighted Banach spaces of analytic functions. Zbl 0957.46018
Bonet, J.; Domański, P.; Lindström, M.
1999
Transitive and hypercyclic operators on locally convex spaces. Zbl 1150.47005
Bonet, J.; Frerick, L.; Peris, A.; Wengenroth, J.
2005
Convergence of arithmetic means of operators in Fréchet spaces. Zbl 1283.47017
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2013
Weighted $$(LF)$$-spaces of continuous functions. Zbl 0839.46015
Bierstedt, Klaus Dieter; Bonet, José
1994
$$C_{0}$$-semigroups and mean ergodic operators in a class of Fréchet spaces. Zbl 1192.47041
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2010
The splitting of exact sequences of PLS-spaces and smooth dependence of solutions of linear partial differential equations. Zbl 1144.46057
Bonet, José; Domański, Paweł
2008
Real analytic curves in Fréchet spaces and their duals. Zbl 0918.46034
Bonet, José; Domański, Pawel
1998
Isometric weighted composition operators on weighted Banach spaces of type $$H^{\infty }$$. Zbl 1154.47017
Bonet, José; Lindström, Mikael; Wolf, Elke
2008
Mean ergodicity of multiplication operators in weighted spaces of holomorphic functions. Zbl 1216.47014
Bonet, José; Ricker, Werner J.
2009
Parameter dependence of solutions of differential equations on spaces of distributions and the splitting of short exact sequences. Zbl 1094.46006
Bonet, José; Domański, Paweł
2006
The Cesàro operator in the Fréchet spaces $$\ell^{p+}$$ and $$L^{p-}$$. Zbl 06713810
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2017
Mean ergodic semigroups of operators. Zbl 1283.47016
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2012
Grothendieck spaces with the Dunford-Pettis property. Zbl 1196.46002
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2010
Montel resolvents and uniformly mean ergodic semigroups of linear operators. Zbl 1274.47005
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2013
Extension of vector-valued holomorphic and harmonic functions. Zbl 1141.46017
Bonet, José; Frerick, Leonhard; Jordá, Enrique
2007
On the range of convolution operators on non-quasianalytic ultradifferentiable functions. Zbl 0918.46039
Bonet, J.; Galbis, A.; Meise, R.
1997
Characterization of the $$\omega$$ hypoelliptic convolution operators on ultradistributions. Zbl 0956.46028
Bonet, J.; Fernández, C.; Meise, R.
2000
Classical operators on weighted Banach spaces of entire functions. Zbl 1282.47047
Beltrán, María J.; Bonet, José; Fernández, Carmen
2013
Fréchet spaces of Moscatelli type. Zbl 0757.46001
Bonet, José; Dierolf, Susanne
1989
Power bounded composition operators on spaces of analytic functions. Zbl 1231.47021
Bonet, José; Domański, Paweł
2011
Parameter dependence of solutions of partial differential equations in spaces of real analytic functions. Zbl 0959.35016
Bonet, José; Domański, Pawel
2001
Biduality in Fréchet and (LB)-spaces. Zbl 0804.46007
Bierstedt, Klaus Dieter; Bonet, José
1992
On the injective tensor product of quasinormable spaces. Zbl 0754.46044
Bonet, José; Peris, Alfredo
1991
Density conditions on Fréchet and $$(DF)$$-spaces. Zbl 0745.46002
Bierstedt, Klaus Dieter; Bonet, José
1989
A question of Valdivia on quasinormable Fréchet spaces. Zbl 0698.46002
Bonet, José
1991
A note about Volterra operators on weighted Banach spaces of entire functions. Zbl 1342.47062
2015
Composition operators between weighted inductive limits of spaces of holomorphic functions. Zbl 1097.46013
Bonet, José; Friz, Miguel; Jordá, Enrique
2005
Whitney’s extension theorem for ultradifferentiable functions of Roumieu type. Zbl 0654.46029
Bonet, José; Meise, Reinhold; Taylor, B. Alan
1989
On weighted inductive limits of spaces of continuous functions. Zbl 0575.46025
Bonet, José
1986
Superposition operators between weighted Banach spaces of analytic functions of controlled growth. Zbl 1278.47056
Bonet, José; Vukotić, Dragan
2013
Schauder decompositions and the Grothendieck and Dunford–Pettis properties in Köthe echelon spaces of infinite order. Zbl 1131.46005
Bonet, José; Ricker, Werner J.
2007
The structure of spaces of quasianalytic functions of Roumieu type. Zbl 1147.46024
Bonet, José; Domański, Pawel
2007
Sampling sets and sufficient sets for $$A^{-\infty}$$. Zbl 1019.30026
Bonet, José; Domański, Paweł
2003
On the continuous Cesàro operator in certain function spaces. Zbl 1354.47010
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2015
The spectrum of Volterra operators on weighted spaces of entire functions. Zbl 1342.47061
Bonet, José
2015
Mean ergodic operators and reflexive Fréchet lattices. Zbl 1238.47007
Bonet, José; de Pagter, Ben; Ricker, Werner J.
2011
Spectrum and compactness of the Cesàro operator on weighted $$\ell_p$$ spaces. Zbl 1336.47033
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2015
The problem of topologies of Grothendieck and the class of Fréchet T- spaces. Zbl 0754.46043
Bonet, José; Díaz, Juan Carlos
1991
Weighted spaces of holomorphic functions and operators between them. Zbl 1060.46019
Bonet, José
2003
Spaces of operators between Fréchet spaces. Zbl 0804.46011
Bonet, José; Lindström, Mikael
1994
The Cesàro operator on Korenblum type spaces of analytic functions. Zbl 06872995
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2018
Spectra and essential spectral radii of composition operators on weighted Banach spaces of analytic functions. Zbl 1138.47017
Bonet, José; Galindo, Pablo; Lindström, Mikael
2008
Dual density conditions in (DF)-spaces. II. Zbl 0688.46003
Bierstedt, Klaus Dieter; Bonet, Jośe
1988
The pullback for bornological and ultrabornological spaces. Zbl 1223.46003
Bonet, J.; Dierolf, S.
2006
Weighted $$(LB)$$-spaces of holomorphic functions: $$\nu H(G)=\nu_{0}H(G)$$ and completeness of $$\nu_{0}H(G)$$. Zbl 1119.46028
Bierstedt, Klaus D.; Bonet, José
2006
On distinguished Fréchet spaces. Zbl 0785.46003
Bonet, José; Dierolf, Susanne
1992
Two theorems of Josefson-Nissenzweig type for Fréchet spaces. Zbl 0785.46002
Bonet, J.; Lindström, M.; Valdivia, M.
1993
Topologizable operators on locally convex spaces. Zbl 1124.46031
Bonet, José
2007
Classical operators on the Hörmander algebras. Zbl 1359.47029
Beltrán, María José; Bonet, José; Fernández, Carmen
2015
Projective limits of weighted (LB)-spaces of continuous functions. Zbl 1176.46003
Agethen, Simone; Bierstedt, Klaus D.; Bonet, José
2009
The dual of the space of holomorphic functions on locally closed convex sets. Zbl 1094.46020
Bonet, José; Meise, Reinhold; Melikhov, Sergej N.
2005
Convergent sequences in duals of Fréchet spaces. Zbl 0804.46003
Bonet, José; Lindström, Mikael
1993
Completeness of the (LB)-spaces $${\mathcal V}C(X)$$. Zbl 0688.46004
Bierstedt, Klaus-Dieter; Bonet, José
1991
On the identity $$L(E,F)=LB(E,F)$$ for pairs of locally convex spaces E and F. Zbl 0683.46012
Bonet, José
1987
Tensor products of Fréchet or (DF)-spaces with a Banach space. Zbl 0776.46029
Bonet, José; Defant, Andreas; Galbis, Antonio
1992
Complete spaces of vector-valued holomorphic germs. Zbl 0824.46003
Bonet, José; Domański, Paweł; Mujica, Jorge
1994
Locally bounded sets of holomorphic mappings. Zbl 0706.46033
Bonet, José; Galindo, Pablo; García, Domingo; Maestre, Manuel
1988
Every quojection is the quotient of a countable product of Banach spaces. Zbl 0711.46007
Bonet, J.; Maestre, M.; Metafune, G.; Moscatelli, V. B.; Vogt, D.
1989
Operator-weighted composition operators between weighted spaces of vector-valued analytic functions. Zbl 1263.47027
Bonet, José; Gómez-Collado, M. Carmen; Jornet, David; Wolf, Elke
2012
Noncomplete Mackey topologies on Banach spaces. Zbl 1205.46004
Bonet, José; Cascales, Bernardo
2010
Vector-valued meromorphic functions. Zbl 1021.46029
Bonet, José; Jordá, Enrique; Maestre, Manuel
2002
Solid hulls and cores of weighted $$H^\infty$$-spaces. Zbl 1406.46017
Bonet, José; Lusky, Wolfgang; Taskinen, Jari
2018
Nonradial Hörmander algebras of several variables and convolution operators. Zbl 0992.46020
Bonet, José; Galbis, Antonio; Momm, Siegfried
2001
Ultradistributions of Roumieu type and projective descriptions. Zbl 0978.46020
Bonet, José; Meise, Reinhold
2001
On the lifting of bounded sets in Fréchet spaces. Zbl 0792.46001
Bonet, José; Dierolf, Susanne
1993
The subspace problem for weighted inductive limits of spaces of holomorphic functions. Zbl 0841.46014
1995
Distinguished subspaces and quotients of Köthe echelon spaces. Zbl 0777.46007
Bonet, José; Diaz, Juan Carlos
1991
Remarks and examples concerning suprabarrelled and totally barrelled spaces. Zbl 0487.46001
Perez Carreras, Pedro; Bonet, Jose
1982
The canonical spectral measure in Köthe echelon spaces. Zbl 1109.46047
Bonet, J.; Ricker, W. J.
2005
Topological structure of the set of weighted composition operators on weighted Bergman spaces of infinite order. Zbl 1193.47030
Bonet, José; Lindström, Mikael; Wolf, Elke
2009
Projective description of weighted (LF)-spaces of holomorphic functions on the disc. Zbl 1060.46018
Bierstedt, Klaus D.; Bonet, José
2003
The Fréchet spaces $$ces(p+)$$, $$1<p<\infty$$. Zbl 1403.46005
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2018
Mean ergodic composition operators on generalized Fock spaces. Zbl 07164406
Seyoum, Werkaferahu; Mengestie, Tesfa; Bonet, José
2020
Multiplier and averaging operators in the Banach spaces $$ces(p), \, 1< p < \infty$$. Zbl 1421.46021
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2019
Operators on the Fréchet sequence spaces $$ces(p+)$$, $$1\le p<\infty$$. Zbl 1423.46007
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2019
Every separable complex Fréchet space with a continuous norm is isomorphic to a space of holomorphic functions. Zbl 1478.46003
Bonet, José
2021
Mean ergodic composition operators on generalized Fock spaces. Zbl 07164406
Seyoum, Werkaferahu; Mengestie, Tesfa; Bonet, José
2020
Operators acting in the dual spaces of discrete Cesàro spaces. Zbl 1439.46013
Bonet, José; Ricker, Werner J.
2020
On boundedness and compactness of Toeplitz operators in weighted $$H^\infty$$-spaces. Zbl 07173893
Bonet, José; Lusky, Wolfgang; Taskinen, Jari
2020
Associated weights for spaces of $$p$$-integrable entire functions. Zbl 07311597
Bonet, José; Mangino, Elisabetta M.
2020
A note about the spectrum of composition operators induced by a rotation. Zbl 07164802
Bonet, José
2020
The differentiation operator in the space of uniformly convergent Dirichlet series. Zbl 07261798
Bonet, José
2020
Multiplier and averaging operators in the Banach spaces $$ces(p), \, 1< p < \infty$$. Zbl 1421.46021
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2019
Operators on the Fréchet sequence spaces $$ces(p+)$$, $$1\le p<\infty$$. Zbl 1423.46007
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2019
Linear operators on the (LB)-sequence spaces $$\mathrm{ces}(p-)$$, $$1< p \le \infty$$. Zbl 1443.46002
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2019
Solid cores and solid hulls of weighted Bergman spaces. Zbl 1420.46026
Bonet, José; Lusky, Wolfgang; Taskinen, Jari
2019
The spectrum of Volterra operators on Korenblum type spaces of analytic functions. Zbl 07123547
Bonet, José
2019
Schauder bases and the decay rate of the heat equation. Zbl 1423.35131
Bonet, José; Lusky, Wolfgang; Taskinen, Jari
2019
The Cesàro operator on Korenblum type spaces of analytic functions. Zbl 06872995
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2018
Solid hulls and cores of weighted $$H^\infty$$-spaces. Zbl 1406.46017
Bonet, José; Lusky, Wolfgang; Taskinen, Jari
2018
The Fréchet spaces $$ces(p+)$$, $$1<p<\infty$$. Zbl 1403.46005
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2018
Solid hulls of weighted Banach spaces of entire functions. Zbl 1403.46023
2018
Solid hulls of weighted Banach spaces of analytic functions on the unit disc with exponential weights. Zbl 1398.46021
2018
The Cesàro operator in weighted $$\ell_1$$ spaces. Zbl 06898555
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2018
The Cesàro operator on power series spaces. Zbl 1464.47021
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2018
Mean ergodic multiplication operators on weighted spaces of continuous functions. Zbl 1482.47066
Bonet, José; Jordá, Enrique; Rodríguez, Alberto
2018
Monomial basis in Korenblum type spaces of analytic functions. Zbl 1408.46026
Bonet, José; Lusky, Wolfgang; Taskinen, Jari
2018
The Fréchet Schwartz algebra of uniformly convergent Dirichlet series. Zbl 1431.46002
Bonet, José
2018
The Cesàro operator on duals of power series spaces of infinite type. Zbl 1424.47010
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2018
The Cesàro operator in the Fréchet spaces $$\ell^{p+}$$ and $$L^{p-}$$. Zbl 06713810
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2017
A note on completeness of weighted normed spaces of analytic functions. Zbl 1394.46017
Bonet, José; Vukotić, Dragan
2017
A note on the spectrum of composition operators on spaces of real analytic functions. Zbl 1454.47036
Bonet, José; Domański, Paweł
2017
Frames and representing systems in Fréchet spaces and their duals. Zbl 1369.46002
Bonet, J.; Fernández, C.; Galbis, A.; Ribera, J. M.
2017
Mean ergodicity and spectrum of the Cesàro operator on weighted $$c_0$$ spaces. Zbl 1356.47037
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2016
The Cesàro operator in growth Banach spaces of analytic functions. Zbl 1373.47028
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2016
Dynamics and spectrum of the Cesàro operator on $$C^\infty (\mathbb R_+)$$. Zbl 1358.47003
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2016
Erratum to: “Dynamics and spectrum of the Cesàro operator on $$C^\infty (\mathbb{R}_+)$$”. Zbl 1478.47003
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2016
A note about Volterra operators on weighted Banach spaces of entire functions. Zbl 1342.47062
2015
On the continuous Cesàro operator in certain function spaces. Zbl 1354.47010
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2015
The spectrum of Volterra operators on weighted spaces of entire functions. Zbl 1342.47061
Bonet, José
2015
Spectrum and compactness of the Cesàro operator on weighted $$\ell_p$$ spaces. Zbl 1336.47033
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2015
Classical operators on the Hörmander algebras. Zbl 1359.47029
Beltrán, María José; Bonet, José; Fernández, Carmen
2015
Abel’s functional equation and eigenvalues of composition operators on spaces of real analytic functions. Zbl 1331.47038
Bonet, José; Domański, Paweł
2015
Abscissas of weak convergence of vector valued Dirichlet series. Zbl 1366.46003
Bonet, José
2015
Uniform mean ergodicity of $$C_0$$-semigroups in a class of Fréchet spaces. Zbl 1314.47060
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2014
Shrinking and boundedly complete Schauder frames in Fréchet spaces. Zbl 1307.42027
Bonet, José; Fernández, Carmen; Galbis, Antonio; Ribera, Juan M.
2014
Uniform convergence and spectra of operators in a class of Fréchet spaces. Zbl 1472.47006
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2014
The range of the restriction map for a multiplicity variety in Hörmander algebras of entire functions. Zbl 1318.46013
Bonet, José; Fernández, Carmen
2014
Chaos of the differentiation operator on weighted Banach spaces of entire functions. Zbl 1353.47010
Bonet, José; Bonilla, Antonio
2013
Convergence of arithmetic means of operators in Fréchet spaces. Zbl 1283.47017
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2013
Montel resolvents and uniformly mean ergodic semigroups of linear operators. Zbl 1274.47005
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2013
Classical operators on weighted Banach spaces of entire functions. Zbl 1282.47047
Beltrán, María J.; Bonet, José; Fernández, Carmen
2013
Superposition operators between weighted Banach spaces of analytic functions of controlled growth. Zbl 1278.47056
Bonet, José; Vukotić, Dragan
2013
On the theorem of Borel for quasianalytic classes. Zbl 1282.46024
Bonet, José; Meise, Reinhold
2013
Hypercyclic composition operators on spaces of real analytic functions. Zbl 1272.47037
Bonet, José; Domański, Paweł
2012
Mean ergodic semigroups of operators. Zbl 1283.47016
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2012
Operator-weighted composition operators between weighted spaces of vector-valued analytic functions. Zbl 1263.47027
Bonet, José; Gómez-Collado, M. Carmen; Jornet, David; Wolf, Elke
2012
The division problem for tempered distributions of one variable. Zbl 1246.46041
Bonet, José; Frerick, Leonhard; Jordá, Enrique
2012
Norm-attaining weighted composition operators on weighted Banach spaces of analytic functions. Zbl 1259.47027
Bonet, José; Lindström, Mikael; Wolf, Elke
2012
Fréchet spaces with no infinite-dimensional Banach quotients. Zbl 1261.46001
Albanese, Angela A.; Bonet, José
2012
Factorization of weakly compact operators between Banach spaces and Fréchet or (LB)-spaces. Zbl 1289.46001
Bonet, José; Wright, J. D. Maitland
2012
A note on mean ergodic composition operators on spaces of holomorphic functions. Zbl 1275.47049
Bonet, José; Domański, Paweł
2011
Power bounded composition operators on spaces of analytic functions. Zbl 1231.47021
Bonet, José; Domański, Paweł
2011
Mean ergodic operators and reflexive Fréchet lattices. Zbl 1238.47007
Bonet, José; de Pagter, Ben; Ricker, Werner J.
2011
Spaces of Moscatelli type. A survey. Zbl 1257.46002
Bonet, José; Fernandez, Carmen
2011
Bornological projective limits of inductive limits of normed spaces. Zbl 1242.46004
Bonet, José; Wegner, Sven-Ake
2011
On mean ergodic operators. Zbl 1259.47013
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2010
$$C_{0}$$-semigroups and mean ergodic operators in a class of Fréchet spaces. Zbl 1192.47041
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2010
Grothendieck spaces with the Dunford-Pettis property. Zbl 1196.46002
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2010
Noncomplete Mackey topologies on Banach spaces. Zbl 1205.46004
Bonet, José; Cascales, Bernardo
2010
A problem on the structure of Fréchet spaces. Zbl 1262.46001
Bonet, José
2010
Köthe coechelon spaces as locally convex algebras. Zbl 1222.46037
Bonet, José; Domański, Paweł
2010
Mean ergodic operators in Fréchet spaces. Zbl 1194.47012
Albanese, Angela A.; Bonet, José; Ricker, Werner J.
2009
Dynamics of the differentiation operator on weighted spaces of entire functions. Zbl 1157.47006
Bonet, José
2009
Mean ergodicity of multiplication operators in weighted spaces of holomorphic functions. Zbl 1216.47014
Bonet, José; Ricker, Werner J.
2009
Projective limits of weighted (LB)-spaces of continuous functions. Zbl 1176.46003
Agethen, Simone; Bierstedt, Klaus D.; Bonet, José
2009
Topological structure of the set of weighted composition operators on weighted Bergman spaces of infinite order. Zbl 1193.47030
Bonet, José; Lindström, Mikael; Wolf, Elke
2009
Convolution operators on quasianalytic classes of Roumieu type. Zbl 1189.46016
Bonet, José; Meise, Reinhold
2009
Differences of composition operators between weighted Banach spaces of holomorphic functions. Zbl 1145.47020
Bonet, José; Lindström, Mikael; Wolf, Elke
2008
The splitting of exact sequences of PLS-spaces and smooth dependence of solutions of linear partial differential equations. Zbl 1144.46057
Bonet, José; Domański, Paweł
2008
Isometric weighted composition operators on weighted Banach spaces of type $$H^{\infty }$$. Zbl 1154.47017
Bonet, José; Lindström, Mikael; Wolf, Elke
2008
Spectra and essential spectral radii of composition operators on weighted Banach spaces of analytic functions. Zbl 1138.47017
Bonet, José; Galindo, Pablo; Lindström, Mikael
2008
Characterization of the convolution operators on quasianalytic classes of Beurling type that admit a continuous linear right inverse. Zbl 1134.42002
Bonet, José; Meise, Reinhold
2008
Weighted inductive limits of spaces of entire functions. Zbl 1182.46013
Bierstedt, Klaus D.; Bonet, José; Taskinen, Jari
2008
A comparison of two different ways to define classes of ultradifferentiable functions. Zbl 1165.26015
Bonet, José; Meise, Reinhold; Melikhov, Sergej N.
2007
Extension of vector-valued holomorphic and harmonic functions. Zbl 1141.46017
Bonet, José; Frerick, Leonhard; Jordá, Enrique
2007
Schauder decompositions and the Grothendieck and Dunford–Pettis properties in Köthe echelon spaces of infinite order. Zbl 1131.46005
Bonet, José; Ricker, Werner J.
2007
The structure of spaces of quasianalytic functions of Roumieu type. Zbl 1147.46024
Bonet, José; Domański, Pawel
2007
Topologizable operators on locally convex spaces. Zbl 1124.46031
Bonet, José
2007
Ultradifferentiable fundamental kernels of linear partial differential operators on non-quasianalytic casses of Roumieu type. Zbl 1139.35037
Albanese, Angela A.; Bonet, José
2007
Parameter dependence of solutions of differential equations on spaces of distributions and the splitting of short exact sequences. Zbl 1094.46006
Bonet, José; Domański, Paweł
2006
The pullback for bornological and ultrabornological spaces. Zbl 1223.46003
Bonet, J.; Dierolf, S.
2006
Weighted $$(LB)$$-spaces of holomorphic functions: $$\nu H(G)=\nu_{0}H(G)$$ and completeness of $$\nu_{0}H(G)$$. Zbl 1119.46028
Bierstedt, Klaus D.; Bonet, José
2006
The canonical spectral measure and Köthe function spaces. Zbl 1137.46027
Bonet, José; Okada, Susumu; Ricker, Werner J.
2006
Transitive and hypercyclic operators on locally convex spaces. Zbl 1150.47005
Bonet, J.; Frerick, L.; Peris, A.; Wengenroth, J.
2005
Composition operators between weighted inductive limits of spaces of holomorphic functions. Zbl 1097.46013
Bonet, José; Friz, Miguel; Jordá, Enrique
2005
The dual of the space of holomorphic functions on locally closed convex sets. Zbl 1094.46020
Bonet, José; Meise, Reinhold; Melikhov, Sergej N.
2005
The canonical spectral measure in Köthe echelon spaces. Zbl 1109.46047
Bonet, J.; Ricker, W. J.
2005
Weighted $$L^{\infty}$$-estimates for Bergman projections. Zbl 1082.47027
Bonet, José; Engliš, Miroslav; Taskinen, Jari
2005
Universal and chaotic multipliers on spaces of operators. Zbl 1062.47011
Bonet, José; Martínez-Giménez, Félix; Peris, Alfredo
2004
Projective representations of spaces of quasianalytic functionals. Zbl 1062.46022
Bonet, José; Meise, Reinhold; Melikhov, Sergeĭ N.
2004
Spectral measures in classes of Fréchet spaces. Zbl 1093.46021
Bonet, José; Ricker, Werner J.
2004
Linear chaos on Fréchet spaces. Zbl 1079.47008
Bonet, J.; Martínez-Giménez, F.; Peris, A.
2003
Some aspects of the modern theory of Fréchet spaces. Zbl 1085.46001
Bierstedt, Klaus D.; Bonet, José
2003
A note on weighted Banach spaces of holomorphic functions. Zbl 1047.46018
Bonet, José; Wolf, Elke
2003
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### Cited by 615 Authors
103 Bonet Solves, José 42 Kąkol, Jerzy 35 Albanese, Angela Anna 35 Peris, Alfredo 26 Ricker, Werner Joseph 24 Jornet, David 23 Domański, Pawel Jan 23 Lindström, Mikael 21 Wolf, Elke 20 Jordá, Enrique 20 Schindl, Gerhard 19 Taskinen, Jari 19 Zhou, Zehua 18 Saxon, Stephen A. 17 Fernández, Carmen 17 Ferrando, Juan Carlos 16 Galbis, Antonio 16 Qiu, Jinghui 15 Díaz, Juan Carlos 15 López Pellicer, Manuel 15 Martínez-Giménez, Félix 14 Paúl, Pedro José 13 Abanin, Aleksandr Vasil’evich 13 Bernal-Gonzàlez, Luis 12 Bonilla, Antonio 12 Florencio, Miguel 12 Gabriyelyan, Saak S. 12 Galindo, Pablo 12 Phạm Trọng Tiến 11 Conejero, José Alberto 11 Defant, Andreas 11 Lusky, Wolfgang 11 Rainer, Armin 11 Sanz, Javier 11 Wegner, Sven-Ake 11 Wengenroth, Jochen 10 Boiti, Chiara 10 Dierolf, Susanne 10 Lê Hai Khôi 10 Shkarin, Stanislav A. 9 Bès, Juan P. 9 Boyd, Christopher 9 Kalmes, Thomas 9 Leiderman, Arkady G. 9 Sevilla-Peris, Pablo 8 Frerick, Leonhard 8 Melikhov, Sergeĭ Nikolaevich 8 Menet, Quentin 8 Oliaro, Alessandro 8 Piszczek, Krzysztof 8 Rueda, Pilar 7 Bierstedt, Klaus-Dieter 7 Debrouwere, Andreas 7 Doubtsov, Evgueni Sergeevich 7 Jiménez-Garrido, Javier 7 Mangino, Elisabetta M. 7 Mengestie, Tesfa Y. 7 Stević, Stevo 6 Chan, Kit Chak 6 Dineen, Seán 6 García, Domingo 6 Gómez-Collado, M. Carmen 6 Grosse-Erdmann, Karl-Goswin 6 Langenbruch, Michael 6 Maestre, Manuel 6 Ramos-Fernández, Julio C. 6 Singh Manhas, Jasbir 5 Abakumov, Evgeny V. 5 Beltrán-Meneu, María J. 5 Colonna, Flavia 5 Gilmore, Clifford 5 Gupta, Manjul 5 He, Fei 5 Hyvärinen, Olli 5 Kruse, Karsten 5 Lastra, Alberto 5 Liang, Yuxia 5 Mayoral, Fernando 5 Mele, Claudio 5 Müller, Vladimír 5 Oubbi, Lahbib 5 Rezaei, Hamid 5 Saksman, Eero 5 Sánchez Ruiz, Luis Manuel 5 Śliwa, Wiesław 5 Valdivia Ureña, Manuel 5 Wang, Maofa 4 Ardalani, Mohammad Ali 4 Bastin, Françoise 4 Beltrán, María José 4 Bermúdez, Teresa 4 Bernardes, Nilson C. jun. 4 Betancor Perez, Jorge Juan 4 Calderón-Moreno, María del Carmen 4 Costakis, George 4 Dierolf, Bernhard 4 Drewnowski, Lech 4 Girela, Daniel 4 Grivaux, Sophie 4 Hassanlou, Mostafa ...and 515 more Authors
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### Cited in 169 Serials
135 Journal of Mathematical Analysis and Applications 55 Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales. Serie A: Matemáticas. RACSAM 40 Archiv der Mathematik 38 Proceedings of the American Mathematical Society 37 Results in Mathematics 32 Complex Analysis and Operator Theory 31 Mathematische Nachrichten 28 Journal of Functional Analysis 21 Integral Equations and Operator Theory 19 Functiones et Approximatio. Commentarii Mathematici 18 Monatshefte für Mathematik 17 Mediterranean Journal of Mathematics 16 Czechoslovak Mathematical Journal 15 Topology and its Applications 13 Collectanea Mathematica 10 Abstract and Applied Analysis 10 Acta Mathematica Sinica. English Series 9 Studia Mathematica 9 Advances in Mathematics 9 Mathematische Zeitschrift 9 Transactions of the American Mathematical Society 9 Journal of the Australian Mathematical Society 8 Rocky Mountain Journal of Mathematics 8 Journal of Approximation Theory 8 Positivity 7 Annales Polonici Mathematici 7 Manuscripta Mathematica 7 Proceedings of the Edinburgh Mathematical Society. Series II 7 Revista Matemática Complutense 7 Complex Variables and Elliptic Equations 6 Israel Journal of Mathematics 6 Applied Mathematics and Computation 6 Glasgow Mathematical Journal 6 Mathematische Annalen 6 Bulletin of the Malaysian Mathematical Sciences Society. Second Series 6 Computational Methods and Function Theory 5 Numerical Functional Analysis and Optimization 5 Publications of the Research Institute for Mathematical Sciences, Kyoto University 5 Quaestiones Mathematicae 5 Ergodic Theory and Dynamical Systems 5 International Journal of Bifurcation and Chaos in Applied Sciences and Engineering 5 Taiwanese Journal of Mathematics 5 Banach Journal of Mathematical Analysis 5 Journal of Pseudo-Differential Operators and Applications 4 Bulletin of the Australian Mathematical Society 4 Mathematical Proceedings of the Cambridge Philosophical Society 4 Annali di Matematica Pura ed Applicata. Serie Quarta 4 Siberian Mathematical Journal 4 Indagationes Mathematicae. New Series 4 Journal of Mathematical Sciences (New York) 4 Turkish Journal of Mathematics 4 Annales Academiae Scientiarum Fennicae. Mathematica 4 Journal of Function Spaces 3 Functional Analysis and its Applications 3 Journal of Differential Equations 3 Rendiconti del Circolo Matemàtico di Palermo. Serie II 3 Semigroup Forum 3 Revista Matemática Iberoamericana 3 Bulletin of the Belgian Mathematical Society - Simon Stevin 3 Mathematical Inequalities & Applications 3 Central European Journal of Mathematics 3 Operators and Matrices 3 Asian-European Journal of Mathematics 3 Advances in Operator Theory 2 Mathematical Notes 2 Ukrainian Mathematical Journal 2 Arkiv för Matematik 2 Bulletin of the London Mathematical Society 2 Illinois Journal of Mathematics 2 International Journal of Mathematics and Mathematical Sciences 2 Journal of Pure and Applied Algebra 2 Acta Applicandae Mathematicae 2 Annals of Global Analysis and Geometry 2 Journal de Mathématiques Pures et Appliquées. Neuvième Série 2 Proceedings of the Royal Society of Edinburgh. Section A. Mathematics 2 Bulletin of the American Mathematical Society. New Series 2 Proceedings of the Indian Academy of Sciences. Mathematical Sciences 2 Russian Mathematics 2 Applied Categorical Structures 2 Journal of Convex Analysis 2 Bulletin des Sciences Mathématiques 2 Integral Transforms and Special Functions 2 Discrete and Continuous Dynamical Systems 2 Journal of Inequalities and Applications 2 Communications of the Korean Mathematical Society 2 Comptes Rendus. Mathématique. Académie des Sciences, Paris 2 Vladikavkazskiĭ Matematicheskiĭ Zhurnal 2 Hacettepe Journal of Mathematics and Statistics 2 Journal of Function Spaces and Applications 2 Advances in Difference Equations 2 Ufimskiĭ Matematicheskiĭ Zhurnal 2 Kyoto Journal of Mathematics 2 Annals of Functional Analysis 2 Analysis and Mathematical Physics 2 Caspian Journal of Mathematical Sciences 2 Concrete Operators 2 Sahand Communications in Mathematical Analysis 2 Open Mathematics 2 Proceedings of the American Mathematical Society. Series B 1 Communications in Algebra ...and 69 more Serials
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### Cited in 44 Fields
645 Functional analysis (46-XX) 465 Operator theory (47-XX) 156 Functions of a complex variable (30-XX) 64 Several complex variables and analytic spaces (32-XX) 55 Partial differential equations (35-XX) 55 General topology (54-XX) 41 Real functions (26-XX) 34 Dynamical systems and ergodic theory (37-XX) 25 Harmonic analysis on Euclidean spaces (42-XX) 20 Integral transforms, operational calculus (44-XX) 18 Global analysis, analysis on manifolds (58-XX) 17 Topological groups, Lie groups (22-XX) 15 Calculus of variations and optimal control; optimization (49-XX) 14 Measure and integration (28-XX) 12 Category theory; homological algebra (18-XX) 12 Ordinary differential equations (34-XX) 9 Sequences, series, summability (40-XX) 9 Abstract harmonic analysis (43-XX) 7 Probability theory and stochastic processes (60-XX) 5 Linear and multilinear algebra; matrix theory (15-XX) 5 Potential theory (31-XX) 5 Approximations and expansions (41-XX) 5 Integral equations (45-XX) 5 Numerical analysis (65-XX) 4 Mathematical logic and foundations (03-XX) 3 Nonassociative rings and algebras (17-XX) 3 Group theory and generalizations (20-XX) 3 Difference and functional equations (39-XX) 3 Quantum theory (81-XX) 2 History and biography (01-XX) 2 Combinatorics (05-XX) 2 Number theory (11-XX) 2 Algebraic geometry (14-XX) 2 Associative rings and algebras (16-XX) 2 Special functions (33-XX) 2 Computer science (68-XX) 2 Mechanics of particles and systems (70-XX) 2 Operations research, mathematical programming (90-XX) 1 General and overarching topics; collections (00-XX) 1 Order, lattices, ordered algebraic structures (06-XX) 1 $$K$$-theory (19-XX) 1 Convex and discrete geometry (52-XX) 1 Differential geometry (53-XX) 1 Manifolds and cell complexes (57-XX)
### Wikidata Timeline
The data are displayed as stored in Wikidata under a Creative Commons CC0 License. Updates and corrections should be made in Wikidata.
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2022-07-03 05:34:48
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https://dsp.stackexchange.com/questions/72521/state-space-physical-meaning/72522
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# State space physical meaning
Let $$T(s)$$ be a transfer function that describes a mechanical system, where the input is force and the output is position. And let $$[A,B,C,D]$$ be the equivalent state-space representation of $$T(s)$$, where: $$\dot{x}(t) = Ax(t) + Bu(t) \\ y(t) = Cx(t) + Du(t)$$ and let $$[A_d,B_d,C_d,D_d]$$ be the discretized model of $$T(s)$$, using zero-order hold on the inputs: $$x_d(k+1) = A_dx_d(k) + B_du(k) \\ y(k) = C_dx_d(k) + D_du(k)$$
I know there are infinite possible realizations of $$[A,B,C,D]$$ that represent the same transfer function $$T(s)$$.
1. How can I force the state vector, $$x(t)$$, to have a simple physical meaning? in this example, since it is a mechanical system, the elements of the state vector $$x(t)$$ would probably be position, velocity and acceleration. I can guess the matrix $$A$$ must have this form: $$A = \left[\begin{array}{ccc} * & * & * \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right]$$ to express the derivative relationship between the state vector elements.
2. Same question, but for the discrete time case: How can I force the state vector, $$x_d(k)$$, to have a simple physical meaning? In this case, I'm not sure how "velocity" and "acceleration" can be expressed.
As you pointed out, there are many state-space realizations of one particular transfer function. The reason is that a transfer function only represents the input-output behavior of a system (observable and controllable dynamics) and not the internal states.
That being said, you can directly write state-space realizations from a transfer function with the so-called vertical and horizontal companion forms (equivalently, the observable and controllable canonical forms). The problem is that the physical meaning of the states is not usually guaranteed.
So unless you have a model of your system in terms of differential equations (first principle modeling like Newton's laws or Lagrangian mechanics for mechanical systems), the physical meaning of the states will usually not be accessible. The same holds in the discrete-time domain.
First, typically when you're exerting a force on something and getting a position, the acceleration varies instantaneously with force. A more or less universal equation of motion for a single-axis linear system would be $$m \ddot x = f_v(\dot x) + f_p(x)$$. For a mass-spring-damper system, it'd be $$m \ddot x = b \dot x + k x$$.
So you can easily express that as a 2nd-order state space equation where $$\mathbf{x} = \begin{bmatrix}v, x\end{bmatrix}^T$$ and $$A = \begin{bmatrix} * & * \\ 1 & 0\end{bmatrix}$$
Getting more complicated than that, as @Gab says, you'll need to have a model of the system.
If you do have a model of the system as a continuous-time state-space linear system, and you can model how the system is driven then you can do this. This is sensible if, for instance, you're trying to find the transfer function of a plant in response to drive outputs from a controller, and those drive outputs have a known form.
In the case of a system that's driven by normal DACs, or otherwise by drive that can be modeled accurately by zero-order holds, then $$A_d = e^{A T_s}$$, $$B_d = A^{-1}(e^{A T_s} - I)$$, and $$C_d = C$$. This gets into obvious numerical difficulties when $$A$$ is singular -- there's ways around this, but I suggest a dive into a book on state-space control.
Matlab's 'c2d' (Scilab's 'dscr', or scipy's 'scipy.signal.cont2discrete') function will do the above, even when $$A$$ is singular. I can't speak for the other two, but Scilab uses the identity.
$$e^{\begin{bmatrix} A & B \\ 0 & 0 \end{bmatrix}T_s} = \begin{bmatrix} A_d & B_d \\ 0 & I \end{bmatrix}$$
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2021-05-12 01:40:53
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https://www2.icp.uni-stuttgart.de/~hilfer/publikationen/html/FCAA-2015-18-333/FCAA-2015-18-333.S2.html
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Sie sind hier: ICP » R. Hilfer » Publikationen
# 2 Mathematical Model
[334.3.1] Let denote the -dimensional Laplace operator in cartesian coordinates. [334.3.2] Numerous authors postulate a fractional diffusion equation such as
(1)
with and initial condition
(2)
[page 335, §0] for a function as a mathematical model for various physical phenomena (see [10, 15, 14, 28, 29] for examples). [335.0.1] For this becomes the Cauchy problem for the ordinary diffusion equation whose applicability as a mathematical model for physical phenomena has been validated with innumerable experiments. [335.0.2] For however experimental evidence remains narrowly bounded in space and time scales. [335.0.3] Moreover, theoretical considerations cast fundamental doubts on the applicability of this case to natural phenomena.
[335.1.1] For the fractional Laplace operator in eq. (1) may be defined (in the sense of Riesz [23]) as
(3)
where denotes the Fourier transform of . [335.1.2] A core domain suitable for various extensions are functions from the Schwartz space of smooth functions decreasing rapidly at infinity.
[335.2.1] The implicit idealizing assumption underlying the choice of an unbounded domain in eq. (1) is that the boundary is sufficiently far away so that its effects on the observations are negligible. [335.2.2] However, experiments are normally performed inside a bounded laboratory containing a bounded apparatus that occupies a bounded domain of space. [335.2.3] Thus, practical applications require to consider nonlocal boundary value problems on bounded domains .
[335.3.1] Every experiment assumes that the experimental conditions in the region surrounding the region containing the sample can be controlled and reproduced to any desired degree of accuracy. [335.3.2] In the mathematical model this is represented by assuming given boundary data for the unknown such that
(4)
for all times . [335.3.3] The Riesz operator may then be be understood as a Dirichlet form on the space over the bounded set equipped with the canonical Borel -algebra and a -finite measure , [5].
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2021-12-05 05:21:45
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http://clarkgrubb.com/optimization
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Optimization
introduction
A general form of an optimization problem is to minimize g(x) subject to
(1)
\begin{align} \;\;\;\;\;\;\;f_i(x) \leq b_i, \;\; i \in \{1, ..., n\} \end{align}
where x is a n-dimensional real vector and g and fᵢ are real-valued.
g is the objective function. The components of x are the decision variables. The fᵢ(x) ≤ bᵢ are the constraints.
The maximum of g is the minimum of -g, so a maximization problem can be recast as a minimization problem. The value of the decision variables at the optimum is the same, but the sign of the value of the objective function at the optimum is reversed.
A constraint of the form h(x) = c can be replaced by two constraints of the form h₁(x) ≤ c and h₂(x) ≥ c without changing the meaning of the problem. Moreover, a constraint of the form h(x) ≥ c can be replaced with the equivalent constaint -h(x) ≤ -c.
linear programming
If the objective function g = ∑cᵢ xᵢ and the inequality constraint functions fᵢ = ∑ aᵢⱼ xⱼ are linear, the problem is a linear optimization problem, also called a linear program.
A general form for a linear program is to minimize cTx subject to A x ≤ b and x ≥ 0. Here c and x are vectors with length equal to the number of decision variables. b is a vector with length equal to the number of constraints, and A is a matrix with row count equal to the number of constraints and column count equal to the number of decision variables.
The constraint x ≥ 0 can be imposed without loss of generality by replacing each decision variable xᵢ with xᵢ+ - xᵢ-, where xᵢ+ ≥ 0 and xᵢ- ≥ 0.
duality
The dual problem of maximizing cTx subject to Ax ≤ b and x ≥ 0 is minimizing bTy subject to ATy ≥ c and y ≥ 0. In this context the original problem is called the primal problem.
The primal and the dual problem share the same optimal value, if there is one. Feasible solutions to the primal problem are lower bounds to the optimal value. Feasible solutions to the dual problem are upper bounds to the optimal value. If the primal is unconstrained, the dual is infeasible. If the dual is unconstrained, the primal is infeasible. It is possible for both the primal and dual to be infeasible.
Minimization problems also have a dual. Furthermore, if P is a linear program, and D is its dual, then the dual of D is P.
Duals also exist for nonlinear optimization problems. Although the feasible solutions of the dual of a maximization problem provide upper bounds, the optimal value of the dual may be strictly greater than the optimal value of the primal. The difference is called the duality gap. For convex problems, if the problem meets the KKT conditions the duality gap is zero.
glpk
installation
$brew install homebrew/science/glpk$ sudo apt-get install glpk-utils
$sudo yum install glpk-utils installing on windows documentation GLPK supports several formats for describing the problem. glpsol --help will list them. Install the package glpk-doc to get a PDF which describes the GMPL format. The PDF is also available here: GMPL is a subset of AMPL: AMPL: A Modeling Language for Mathematical Programming $ cat lp1.model
var x1 >= 0;
var x2 >= 0;
maximize obj: x1 + 2 * x2;
c1: -3 * x1 + x2 <= 2;
c2: x2 <= 11;
c3: x1 - x2 <= 3;
c4: x1 <= 6;
solve;
display x1;
display x2;
display obj;
end;
$glpsol -m lp1.model separate model and data matlab/cvx cvx_begin variable x1; variable x2; maximize x1 + 2 * x2; subject to -3 * x1 + x2 <= 2; x2 <= 11; x1 - x2 <= 3; x1 <= 6; cvx_end r/lpsolve > install.packages('lpSolve') > require(lpSolve) > obj = c(1, 2) > mat = matrix(c(-3, 1, 0, 1, 1, -1, 1, 0), nrow=4, byrow=T) > dir = c("<=", "<=", "<=", "<=") > rhs = c(2, 11, 3, 6) > lp("max", obj, mat, dir, rhs) Success: the objective function is 28 python/cvxopt cvxopt user's guide $ sudo pip install cvxopt
cvxopt finds minimal values for the objective function. To solve a maximization problem, we negate the objective function. We must remember the negate the value of the objective function that we find to get the maximal value.
>>> from cvxopt.modeling import *
>>> x1 = variable(1, 'x1')
>>> x2 = variable(1, 'x2')
>>> c1 = (-3 * x1 + x2 <= 2)
>>> c2 = (x2 <= 11);
>>> c3 = (x1 - x2 <= 3)
>>> c4 = (x1 <= 6)
>>> lp1 = op(-x1 - 2 * x2, [c1, c2, c3, c4])
>>> lp1.solve()
pcost dcost gap pres dres k/t
0: -1.7235e+01 -5.8647e+01 3e+01 0e+00 1e+00 1e+00
1: -2.5048e+01 -2.9964e+01 4e+00 4e-16 1e-01 2e-01
2: -2.7685e+01 -2.8988e+01 1e+00 4e-16 4e-02 1e-01
3: -2.7997e+01 -2.8010e+01 1e-02 1e-16 4e-04 1e-03
4: -2.8000e+01 -2.8000e+01 1e-04 2e-16 4e-06 1e-05
5: -2.8000e+01 -2.8000e+01 1e-06 3e-16 4e-08 1e-07
Optimal solution found.
>>> lp1.status
'optimal'
>>> print(lp1.objective.value())
[-2.80e+01]
>>> print(x1.value)
[ 6.00e+00]
>>> print(x2.value)
[ 1.10e+01]
excel
To solve a linear optimization problem in Excel, the Solver.Xlam add-in must be selected. Go to:
Tools | Add-Ins...
This provides a Solver button in the Data section of the ribbon.
To use it, allocate a row or column to contain the decision variables, and allocate a cell for the objective function which is defined in terms of those decision variables. The SUMPRODUCT function is useful when defining the objective function.
Click the Solver button, which brings up a dialog.
Enter the cell reference of the objective function in the topmost box.
Indicate whether the problem is a minimization or maximization problem using the radio buttons.
Enter the cell range of the decision variables in input box below that.
One enters the constaints one at a time by clicking the Add button, which brings up another dialog.
Note that there is a checkbox that can be used to indicate that all decision variables are non-negative.
page revision: 70, last edited: 17 Aug 2016 15:06
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2017-12-13 14:47:41
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https://www.bloombergprep.com/gmat/practice-question/1/1831/quantitative-section-quant-fundamentals-integers-rule-of-divisibility-by-5/
|
We cover every section of the GMAT with in-depth lessons, 5000+ practice questions and realistic practice tests.
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Correct. [[snippet]] For example, 200 is divisible by 5 but not by 3 since the sum of its digits ($$2+0+0$$) is not divisible by 3. The next closest multiple of 5 is 195, which satisfies the conditions for 3 and 5 and is thus divisible by 15. The least multiple is thus -195, and the greatest is 195. To count multiples, subtract the extremes, divide by 15, and don't forget to add 1: >$$\displaystyle \frac{195-(-195)}{15} + 1 = \frac{390}{15} + 1 = 27$$.
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2020-10-29 14:18:40
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https://www.physicsforums.com/threads/real-analysis-problem-similar-to-uniformly-integrable.654340/
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Real analysis: Problem similar to uniformly integrable
1. Nov 23, 2012
happysauce
1. The problem statement, all variables and given/known data
Assume $\mu$(X) >0 and that f is a measurable function that maps X into ℝ and satisfies f(x) >0 for all x$\in$X.
Let $\alpha$ be any fixed real number satisfying 0<$\alpha$<$\mu$(X) <infinity Prove that
inf { $\int$$_{E}$f d$\mu$ : E$\in$M, $\mu$(E) ≥$\alpha$} >0.
(Hint. First prove for any $\delta$ satisfying 0<$\delta$<$\alpha$, prov there exists n $\in$ N such that B$_{n}$ = {x:f(x)$\geq$1/n} satisfies $\mu$(B$_{n}$) $\geq$
$\mu$(X) - $\delta$. Then prove that if $\mu$(E) ≥$\alpha$ then $\mu$(E$\cap$B$_{n}$) ≥$\alpha$-$\delta$
2. Relevant equations
Not sure what is relevant.
3. The attempt at a solution
So, the hint suggests the first thing I do is show that given 0<$\delta$<$\alpha$ i can find an n so that $\mu$(B$_{n}$) $\geq$ $\mu$(X) - $\delta$. So since f(x) >0 then $\mu$(X) = $\mu$({x:f(x)>0}) = $\mu$({x:f(x)≥1/n} $\cup${x:f(x)<1/n}). These are disjoint so we have $\mu$({x:f(x)≥1/n}) + $\mu$({x:f(x)<1/n}). If i show the delta relation with $\mu$({x:f(x)<1/n}) then i get $\mu$(B$_{n}$) $\geq$ $\mu$(X) - $\delta$. It seems really trivial though and im not sure what guarantees this.
The next part asks you to show that $\mu$(E$\cap$B$_{n}$) ≥$\alpha$-$\delta$. Ok so $\mu$(E$\cap$B$_{n}$) = $\mu$(E) +$\mu$(B$_{n}$) - $\mu$(E$\cup$B$_{n}$)
The following hold: $\mu$(E$\cup$B$_{n}$) ≤$\mu$(X) (both are subsets)
$\mu$(B$_{n}$) ≥ $\mu$(X) - $\delta$ (previous part)
$\mu$(E)≥$\alpha$ (hypothesis)
So $\mu$(E$\cap$B$_{n}$) = $\mu$(E) +$\mu$(B$_{n}$) - $\mu$(E$\cup$B$_{n}$) ≥ $\mu$(X) - $\delta$ + $\mu$(E) - $\mu$(X) =$\mu$(E)- $\delta$≥$\alpha$-$\delta$
So I've done most of what the hint wants... the problem is I don't know how this helps me.
2. Nov 23, 2012
happysauce
The title is misleading, I thought it related to uniformly intagrable but I don't think it does.
3. Nov 24, 2012
happysauce
I figured out why there is an n for every δ. So I figured out both parts of the hint, but I still have no clue how to make these hints help me here. Do I have to show that
inf {$\int$$_{B_{n}\cap E}$f d$\mu$, E$\in$M, $\mu (B_{n}\cap E)$≥α-δ} > 0 for each n? And then as n-> infinity we have the desired result? I'm still not sure how to prove this case though...
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2018-01-18 20:35:37
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https://cs.stackexchange.com/questions/41494/a-linear-temporal-property-that-represents-this-basic-execution
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# A linear temporal property that represents this (basic) execution?
A question gives me a set of atomic propositions AP = {a, b, c}and asks me to give a linear temporal property that represents all executions over AP such that:
(1) a and b never happen at the same time
OR
(2) a is preceded by c
Can anyone give me some pointers on how to answer this question (possibly with a model solution)? So far all I have done is split the property into two parts such that the property $P$ is the union of $P_1$ and $P_2$ but I don't know how to formulate my answer.
(This question is from a past exam paper).
Any help is highly appreciated.
• Splitting the property is a good idea, since you have the $\vee$ operator to join them. Try to use LTL operators such as "Always" ($G$) and "until" $U$. The first one is pretty easy, the second requires some thought. What you tried so far? – Shaull Apr 17 '15 at 14:47
• One issue with such natural language requirements is that they can be ambiguous - in particular, what exactly did the person setting this exam mean by "a is preceded by c"? Must c have held at some earlier state, or the immediately preceding one? If the latter, is that for all occurrences of a or just the first one? – Klaus Draeger Apr 17 '15 at 16:48
Since you are asked to formulate these properties in LTL, you might want to think of them as properties given with predicates over infinite sequences of states (or events where each event changes the state); in your case, you know that at each state some of the three propositions, {a, b, c} might be true.
When we talk about LTL, it's good to summarize what does the formalism support to figure out whether the given property can be formulated in it. Note that previously mentioned propositions are often formulated as state formulas, which might hold in a particular state. So, LTL supports: state formulas (propositions that might hold in particular state), (standard) Boolean operators (like logical or ||), and path quantifiers (like globally G and next X, which produce other state formulas), which effectively allow state formulas to describe sequences of states.
I will describe one potential solution in the "extended" LTL, which includes additional operators besides the fundamental ones. (These are easily expressible using only the basic operators, next and until; more details can be found in a more thorough treatment of LTL).
So far all I have done is split the property into two parts such that the property P is the union of P1 and P2 but I don't know how to formulate my answer.
Indeed, at the top level, as you recognized, you should have union, which effectively translates to || of P1 and P2:
P1 || P2
Going back to P1 ((1) a and b never happen at the same time), which states that something should never happen, it's good to think whether such temporal quantification can be translated to some of the offered quantifiers in LTL. This is indeed possible, and "never some property P" can be expressed in terms of combining a Boolean operator and a quantifier, in the following way:
G( not P ) (always, P is not true)
thus, P1 can be expressed by using given propositions as:
G( not (a && b) ) (always, it is not true that both a and b hold)
On the other hand, as Shaull pointed out, the second property is a bit trickier. Although, such a requirement might have been written more precisely (to avoid ambiguity), we can interpret it to mean that a and c should happen, at some point, such that c happened earlier than a.
Now, when we identified earlier, we might think of another LTL quantifier that semantically matches the notion of ordering states (or events). That quantifier is:
X (quantifier for next, i.e. formula that will hold in the next state)
Thus (under the interpretation that both propositions must be true at some point), we want to say that eventually, c happens, while at a point after that, eventually a happens as well:
F( c && X(F a))
arriving at the final solution:
P1 || P2 = (G( not (a && b) )) || F( c && X(F a))
which checks that either of the given properties are satisfied in the execution, as you originally suggested.
After you have an idea of how the formula should look like to makes sense (semantically, with respect to the property you want to model), you always want to check whether the formula is well formed, i.e. whether the formula is syntactically correct. (More details can be found in one of the more thorough treatments of LTL.)
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2019-06-20 07:16:40
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https://en.m.wikipedia.org/wiki/Double_coset
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# Double coset
In group theory, a field of mathematics, a double coset is a collection of group elements which are equivalent under the symmetries coming from two subgroups.[1][2] More precisely, let G be a group, and let H and K be subgroups. Let H act on G by left multiplication while K acts on G by right multiplication. For each x in G, the (H, K)-double coset of x is the set
${\displaystyle HxK=\{hxk\colon h\in H,k\in K\}.}$
When H = K, this is called the H-double coset of x. Equivalently, HxK is the equivalence class of x under the equivalence relation
x ~ y if and only if there exist h in H and k in K such that hxk = y.
The set of all double cosets is denoted
${\displaystyle H\backslash G/K.}$
## Properties
Suppose that G is a group with subgroups H and K acting by left and right multiplication, respectively. The (H, K)-double cosets of G may be equivalently described as orbits for the product group H × K acting on G by (h, k)⋅x = hxk−1. Many of the basic properties of double cosets follow immediately from the fact that they are orbits. However, because G is a group and H and K are subgroups acting by multiplication, double cosets are more structured than orbits of arbitrary group actions, and they have additional properties that are false for more general actions.
• Two double cosets HxK and HyK are either disjoint or identical.
• G is the disjoint union of its double cosets.
• There is a one-to-one correspondence between the two double coset spaces H \ G / K and K \ G / H given by identifying HxK with Kx−1H.
• If H = {1}, then H \ G / K = G / K. If K = {1}, then H \ G / K = H \ G.
• A double coset HxK is a union of right cosets of H and left cosets of K, specifically,
{\displaystyle {\begin{aligned}HxK&=\bigcup _{k\in K}Hxk=\coprod _{Hxk\in H\backslash HxK}Hxk,\\HxK&=\bigcup _{h\in H}hxK=\coprod _{hxK\in HxK/K}hxK.\end{aligned}}}
• The set of (H, K)-double cosets is in bijection with the orbits H \ (G / K), and also with the orbits (H \ G) / K under the mappings ${\displaystyle HgK\to H(gK)}$ and ${\displaystyle HgK\to (Hg)K}$ respectively.
• If H is normal, then H \ G is a group, and the right action of K on this group factors through the right action of H \ HK. It follows that H \ G / K = HK \ G. Similarly, if K is normal, then H \ G / K = G / HK.
• If H is a normal subgroup of G, then the H-double cosets are in one-to-one correspondence with the left (and right) H-cosets.
• Consider HxK as the union of a K-orbit of right H-cosets. The stabilizer of the right H-coset HxkH \ HxK with respect to the right action of K is K ∩ (xk)−1Hxk. Similarly, the stabilizer of the left K-coset hxKHxK / K with respect to the left action of H is HhxK(hx)−1.
• It follows that the number of right cosets of H contained in HxK is the index [K : Kx−1Hx] and the number of left cosets of K contained in HxK is the index [H : HxKx−1]. Therefore
{\displaystyle {\begin{aligned}|HxK|&=[H:H\cap xKx^{-1}]|K|=|H|[K:K\cap x^{-1}Hx],\\\left[G:H\right]&=\sum _{HxK\in H\backslash G/K}[K:K\cap x^{-1}Hx],\\\left[G:K\right]&=\sum _{HxK\in H\backslash G/K}[H:H\cap xKx^{-1}].\end{aligned}}}
• If G, H, and K are finite, then it also follows that
{\displaystyle {\begin{aligned}|HxK|&={\frac {|H||K|}{|H\cap xKx^{-1}|}}={\frac {|H||K|}{|K\cap x^{-1}Hx|}},\\\left[G:H\right]&=\sum _{HxK\in H\backslash G/K}{\frac {|K|}{|K\cap x^{-1}Hx|}},\\\left[G:K\right]&=\sum _{HxK\in H\backslash G/K}{\frac {|H|}{|H\cap xKx^{-1}|}}.\end{aligned}}}
• Fix xG, and let (H × K)x denote the double stabilizer {(h, k) : hxk = x}. Then the double stabilizer is a subgroup of H × K.
• Because G is a group, for each hH there is precisely one gG such that hxg = x, namely g = x−1h−1x; however, g may not be in K. Similarly, for each kK there is precisely one g′ ∈ G such that gxk = x, but g may not be in H. The double stabilizer therefore has the descriptions
${\displaystyle (H\times K)_{x}=\{(h,x^{-1}h^{-1}x)\colon h\in H\}\cap H\times K=\{(xk^{-1}x^{-1},k)\colon k\in K\}\cap H\times K.}$
• (Orbit–stabilizer theorem) There is a bijection between HxK and (H × K) / (H × K)x under which hxk corresponds to (h, k−1)(H × K)x. It follows that if G, H, and K are finite, then
${\displaystyle |HxK|=[H\times K:(H\times K)_{x}]=|H\times K|/|(H\times K)_{x}|.}$
• (Cauchy–Frobenius lemma) Let G(h, k) denote the elements fixed by the action of (h, k). Then
${\displaystyle |H\backslash G/K|={\frac {1}{|H||K|}}\sum _{(h,k)\in H\times K}|G^{(h,k)}|.}$
• In particular, if G, H, and K are finite, then the number of double cosets equals the average number of points fixed per pair of group elements.
There is an equivalent description of double cosets in terms of single cosets. Let H and K both act by right multiplication on G. Then G acts by left multiplication on the product of coset spaces G / H × G / K. The orbits of this action are in one-to-one correspondence with H \ G / K. This correspondence identifies (xH, yK) with the double coset Hx−1yK. Briefly, this is because every G-orbit admits representatives of the form (H, xK), and the representative x is determined only up to left multiplication by an element of H. Similarly, G acts by right multiplication on H \ G × K \ G, and the orbits of this action are in one-to-one correspondence with the double cosets H \ G / K. Conceptually, this identifies the double coset space H \ G / K with the space of relative configurations of an H-coset and a K-coset. Additionally, this construction generalizes to the case of any number of subgroups. Given subgroups H1, ..., Hn, the space of (H1, ..., Hn)-multicosets is the set of G-orbits of G / H1 × ... × G / Hn.
The analog of Lagrange's theorem for double cosets is false. This means that the size of a double coset need not divide the order of G. For example, let G = S3 be the symmetric group on three letters, and let H and K be the cyclic subgroups generated by the transpositions (1 2) and (1 3), respectively. If e denotes the identity permutation, then
${\displaystyle HeK=HK=\{e,(12),(13),(132)\}.}$
This has four elements, and four does not divide six, the order of S3. It is also false that different double cosets have the same size. Continuing the same example,
${\displaystyle H(23)K=\{(23),(123)\},}$
which has two elements, not four.
However, suppose that H is normal. As noted earlier, in this case the double coset space equals the right coset space HK \ G. Similarly, if K is normal, then H \ G / K is the left coset space G / HK. Standard results about left and right coset spaces then imply the following facts.
• |HxK| = |HK| for all xG. That is, all double cosets have the same cardinality.
• If G is finite, then |G| = |HK| ⋅ |H \ G / K|. In particular, |HK| and |H \ G / K| divide |G|.
## Examples
• Let G = Sn be the symmetric group, considered as permutations of the set {1, ..., n}. Consider the subgroup H = Sn − 1 which stabilizes n. Then Sn − 1 \ Sn / Sn − 1 consists of two double cosets. One of these is H = Sn − 1. If γ is a permutation which does not fix n, then the other coset is represented by Sn − 1 γ Sn − 1.
• Let G be the group GLn(R), and let B be the subgroup of upper triangular matrices. The double coset space B \ G / B is the Bruhat decomposition of G. Each double coset has a representative BwB, where w is a permutation matrix. For instance, if n = 2, then
${\displaystyle B\backslash \operatorname {GL} _{2}(\mathbf {R} )/B=\left\{B{\begin{pmatrix}1&0\\0&1\end{pmatrix}}B,B{\begin{pmatrix}0&1\\1&0\end{pmatrix}}B\right\}.}$
## Products in the free abelian group on the set of double cosets
Suppose that G is a group and that H, K, and L are subgroups. Under certain finiteness conditions, there is a product on the free abelian group generated by the (H, K)- and (K, L)-double cosets with values in the free abelian group generated by the (H, L)-double cosets. This means there is a bilinear function
${\displaystyle \mathbf {Z} [H\backslash G/K]\times \mathbf {Z} [K\backslash G/L]\to \mathbf {Z} [H\backslash G/L].}$
Assume for simplicity that G is finite. To define the product, reinterpret these free abelian groups in terms of the group algebra of G as follows. Every element of Z[H \ G / K] has the form
${\displaystyle \sum _{HxK\in H\backslash G/K}f_{HxK}\cdot [HxK],}$
where { fHxK } is a set of integers indexed by the elements of H \ G / K. This element may be interpreted as a Z-valued function on H \ G / K, specifically, HxKfHxK. This function may be pulled back along the projection GH \ G / K which sends x to the double coset HxK. This results in a function xfHxK. By the way in which this function was constructed, it is left invariant under H and right invariant under K. The corresponding element of the group algebra Z[G] is
${\displaystyle \sum _{x\in G}f_{HxK}\cdot [x],}$
and this element is invariant under left multiplication by H and right multiplication by K. Conceptually, this element is obtained by replacing HxK by the elements it contains, and the finiteness of G ensures that the sum is still finite. Conversely, every element of Z[G] which is left invariant under H and right invariant under K is the pullback of a function on Z[H \ G / K]. Parallel statements are true for Z[K \ G / L] and Z[H \ G / L].
When elements of Z[H \ G / K], Z[K \ G / L], and Z[H \ G / L] are interpreted as invariant elements of Z[G], then the product whose existence was asserted above is precisely the multiplication in Z[G]. Indeed, it is trivial to check that the product of a left-H-invariant element and a right-L-invariant element continues to be left-H-invariant and right-L-invariant. The bilinearity of the product follows immediately from the bilinearity of multiplication in Z[G]. It also follows that if M is a fourth subgroup of G, then the product of (H, K)-, (K, L)-, and (L, M)-double cosets is associative. Because the product in Z[G] corresponds to convolution of functions on G, this product is sometimes called the convolution product.
There are numerous explicit formulas for the convolution product depending upon the way in which the factors are represented. Fix an (H, K)-double coset HxK and a (K, L)-double coset KyL. These may be written as a union of right H-cosets and a union of left L-cosets, respectively:
${\displaystyle HxK=\bigcup _{Ha_{i}\in H\backslash HxK}Ha_{i},}$
${\displaystyle KyL=\bigcup _{b_{j}L\in KxL/L}b_{j}L.}$
Then the convolution product is
${\displaystyle [HxK]\cdot [KyL]=\sum _{Ha_{i}\in H\backslash HxK}\sum _{b_{j}L\in KyL/L}[Ha_{i}b_{j}L].}$
An equivalent way of phrasing this is to define homomorphisms
${\displaystyle \mathbf {Z} [H\backslash G/K]\to \mathbf {Z} [H\backslash G],}$
${\displaystyle \mathbf {Z} [H\backslash G/K]\to \mathbf {Z} [G/K],}$
by pullback along the projections from H \ G and G / K to H \ G / K. These send
${\displaystyle [HxK]\mapsto \sum _{Ha_{i}\in H\backslash HxK}[Ha_{i}],}$
${\displaystyle [KyL]\mapsto \sum _{b_{j}L\in KyL/L}[b_{j}L].}$
That is, these send a double coset to the single cosets contained in it, analogous to the pullback along GH \ G / K used above. There is a product
${\displaystyle \mathbf {Z} [H\backslash G]\times \mathbf {Z} [G/L]\to \mathbf {Z} [H\backslash G/L],}$
defined on generators as [Ha] ⋅ [bL] = [HabL] and extended by bilinearity. Combining this with the above homomorphisms obtains the convolution product.
The above expression has the additional advantage that it applies even when G is not finite. Assume instead that every (H, K)-double coset is a finite union of right H-cosets and every (K, L)-double coset is a finite union of left L-cosets. Then the above expressions for the product in terms of right H-cosets and left L-cosets are finite sums. Consequently, the product is well-defined in this case, even when G is infinite. The product remains bilinear and associative in this situation.
The classes [HaibjL] occurring in the above sum may not be distinct. Because of this, the product is not generally equal to the sum of all [HzL], where HzL is a double coset contained in HxKyL = { hxkyl : hH, kK, lL }. To get an expression as a sum over distinct generators, define:
${\displaystyle c_{x,y}(z)=|\{(i,j)\colon a_{i}b_{j}\in HzL\}|.}$
Clearly cx, y(z) is zero unless zHxKyL. The product simplifies to
${\displaystyle [HxK]\cdot [KyL]=\sum _{HzL\in H\backslash G/L}c_{x,y}(z)[HzL].}$
This formula leads to the following formula for a product of general elements:
${\displaystyle \left(\sum _{HxK\in H\backslash G/K}f_{HxK}[HxK]\right)\left(\sum _{KyL\in K\backslash G/L}g_{KyL}[KyL]\right)=\sum _{HzL\in H\backslash G/L}\left(\sum _{HxK\in H\backslash G/K}\sum _{KyL\in K\backslash G/L}c_{x,y}(z)f_{HxK}g_{KyL}\right)[HzL].}$
By decomposing each double coset into single cosets, the product of generators may also be written as a sum over the cosets of H or the cosets of L. Define
{\displaystyle {\begin{aligned}c'_{x,y}(u)&=|\{(i,j)\colon a_{i}b_{j}\in Hu\}|,\\c''_{x,y}(v)&=|\{(i,j)\colon a_{i}b_{j}\in vL\}|.\end{aligned}}}
Then
${\displaystyle [HxK]\cdot [KyL]=\sum _{Hu\in H\backslash G}c'_{x,y}(u)[HuL]=\sum _{vL\in G/L}c''_{x,y}(v)[HvL],}$
but the sums may no longer be over distinct generators.
An important special case is when H = K = L. In this case, the product is a bilinear function
${\displaystyle \mathbf {Z} [H\backslash G/H]\times \mathbf {Z} [H\backslash G/H]\to \mathbf {Z} [H\backslash G/H].}$
This product turns Z[H \ G / H] into an associative ring whose identity element is the class of the trivial double coset [H]. In general, this ring is non-commutative. For example, if H = {1}, then the ring is the group algebra Z[G], and a group algebra is a commutative ring if and only if the underlying group is abelian.
If H is normal, so that the H-double cosets are the same as the elements of the quotient group G / H, then the product on Z[H \ G / H] is the product in the group algebra Z[G / H]. In particular, it is the usual convolution of functions on G / H. In this case, the ring is commutative if and only if G / H is abelian, or equivalently, if and only if H contains the commutator subgroup of G.
If H is not normal, then Z[H \ G / H] may be commutative even if G is non-abelian. A classical example is the product of two Hecke operators. This is the product in the Hecke algebra, which is commutative even though the group G is the modular group, which is non-abelian, and the subgroup is an arithmetic subgroup and in particular does not contain the commutator subgroup. Commutativity of the convolution product is closely tied to Gelfand pairs.
When the group G is a topological group, it is possible to weaken the assumption that the number of left and right cosets in each double coset is finite. The group algebra Z[G] is replaced by an algebra of functions such as L2(G) or C(G), and the sums are replaced by integrals. The product still corresponds to convolution. For instance, this happens for the Hecke algebra of a locally compact group.
## Applications
When a group ${\displaystyle G}$ has a transitive group action on a set ${\displaystyle S}$ , computing certain double coset decompositions of ${\displaystyle G}$ reveals extra information about structure of the action of ${\displaystyle G}$ on ${\displaystyle S}$ . Specifically, if ${\displaystyle H}$ is the stabilizer subgroup of some element ${\displaystyle s\in S}$ , then ${\displaystyle G}$ decomposes as exactly two double cosets of ${\displaystyle (H,H)}$ if and only if ${\displaystyle G}$ acts transitively on the set of distinct pairs of ${\displaystyle S}$ . See 2-transitive groups for more information about this action.
Double cosets are important in connection with representation theory, when a representation of H is used to construct an induced representation of G, which is then restricted to K. The corresponding double coset structure carries information about how the resulting representation decomposes. In the case of finite groups, this is Mackey's decomposition theorem.
They are also important in functional analysis, where in some important cases functions left-invariant and right-invariant by a subgroup K can form a commutative ring under convolution: see Gelfand pair.
In geometry, a Clifford–Klein form is a double coset space Γ\G/H, where G is a reductive Lie group, H is a closed subgroup, and Γ is a discrete subgroup (of G) that acts properly discontinuously on the homogeneous space G/H.
In number theory, the Hecke algebra corresponding to a congruence subgroup Γ of the modular group is spanned by elements of the double coset space ${\displaystyle \Gamma \backslash \mathrm {GL} _{2}^{+}(\mathbb {Q} )/\Gamma }$ ; the algebra structure is that acquired from the multiplication of double cosets described above. Of particular importance are the Hecke operators ${\displaystyle T_{m}}$ corresponding to the double cosets ${\displaystyle \Gamma _{0}(N)g\Gamma _{0}(N)}$ or ${\displaystyle \Gamma _{1}(N)g\Gamma _{1}(N)}$ , where ${\displaystyle g=\left({\begin{smallmatrix}1&0\\0&m\end{smallmatrix}}\right)}$ (these have different properties depending on whether m and N are coprime or not), and the diamond operators ${\displaystyle \langle d\rangle }$ given by the double cosets ${\displaystyle \Gamma _{1}(N)\left({\begin{smallmatrix}a&b\\c&d\end{smallmatrix}}\right)\Gamma _{1}(N)}$ where ${\displaystyle d\in (\mathbb {Z} /N\mathbb {Z} )^{\times }}$ and we require ${\displaystyle \left({\begin{smallmatrix}a&b\\c&d\end{smallmatrix}}\right)\in \Gamma _{0}(N)}$ (the choice of a, b, c does not affect the answer).
## References
1. ^ Hall, Jr., Marshall (1959), The Theory of Groups, New York: Macmillan, pp. 14–15
2. ^ Bechtell, Homer (1971), The Theory of Groups, Addison-Wesley, p. 101
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2020-01-25 09:18:46
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https://physics.stackexchange.com/questions/364006/linearized-variation-of-sqrt-g-nabla-mu-xi-nu-nabla-nu-xi-mu
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# Linearized variation of $\sqrt{-g}(\nabla^\mu \xi^\nu - \nabla^\nu \xi^\mu)$
The Noether current associated with invariance of the Einstein-Hilbert action, $$S=\int d^dx \sqrt{-g}\,R[g_{\mu\nu}]\,,$$ under infinitesimal diffeomorphisms $\delta_\xi g_{\mu\nu}=\nabla_\mu \xi_\nu + \nabla_\nu \xi_\mu$ is equal on-shell to $j^\mu = \partial_\nu \kappa_\xi^{\mu\nu}$, where $$\kappa_\xi^{\mu\nu}=\sqrt{-g}(\nabla^\mu \xi^\nu - \nabla^\nu \xi^\mu).$$ On the other hand, the Noether current corresponding to the invariance of the action for a linearized perturbation around Minkowski space $h_{\mu\nu}=g_{\mu\nu}-\eta_{\mu\nu}$, $$S=\int d^dx\, h^{\mu\nu} \big(\Box h_{\mu\nu}-\partial_{(\mu}\partial\cdot h_{\nu)}+\partial_\mu\partial_\nu h-\eta_{\mu\nu}\Box h + \eta_{\mu\nu}\partial\cdot\partial\cdot h\big)\,,$$ ($h$ being the trace of $h_{\mu\nu}$) under linearized diffeomorphisms $\delta_{\xi} h_{\mu\nu}=\partial_\mu\xi_\nu-\partial_\nu\xi_\mu$ is $J^\mu = \partial_\nu K^{\mu\nu}_{\xi}$, where $$K^{\mu\nu}_{\xi}=\partial^\nu h{^\mu_{\ \ \rho}}-h^{\rho\nu}\partial_\rho \xi^\mu+\frac{1}{2}h\,\partial^\nu \xi^\mu+\big\{\xi^\nu(\partial^\mu h-\partial\cdot h^\mu)\big\}-[\mu\leftrightarrow\nu]\,.$$ The claim in the literature (see e.g. this paper, eq. (95) and below eq. (101)) is that $K^{\mu\nu}_\xi$ should also be the linearized perturbation of $\kappa^{\mu\nu}_\xi$ around Minkowski background. However, trying to reproduce this result, I appear to be missing the terms in the curly brackets: letting $\delta g_{\mu\nu}=h_{\mu\nu}$ and calculating to first order in $h_{\mu\nu}$, $$\delta \kappa_\xi^{\mu\nu}=\delta(\sqrt{-g}) \partial^\mu \xi^\nu+ \delta g^{\mu\rho}\partial_\rho \xi^\nu + \eta^{\mu\rho}\delta\Gamma^{\nu}_{\rho\sigma}\xi^\sigma \\ =\left(\frac{1}{2}h\right)\partial^{\mu}\xi^\nu+(-h^{\mu\rho})\partial_\rho \xi^{\nu}+\eta^{\mu\rho}\left(\frac{1}{2}(\partial_\rho h^{\nu}_{\ \ \sigma}+\partial_\sigma h_\rho^{\ \ \nu}-\partial^\nu h_{\rho\sigma})\right)\xi^\sigma-[\mu\leftrightarrow\nu]\\ = \partial^\nu h{^\mu_{\ \ \rho}}-h^{\rho\nu}\partial_\rho \xi^\mu+\frac{1}{2}h\,\partial^\nu \xi^\mu-[\mu\leftrightarrow \nu]\,.$$ First, am I missing something in the calculation? (I hope so) If I didn't make any mistake, is this discrepancy related to the ambiguity present in the definition of $k^{\mu\nu}$, due to the fact that any current of the type $$\partial_{\nu}(\kappa^{\mu\nu}+\lambda^{\mu\nu})$$ would still be conserved, provided $\lambda^{\mu\nu}$ be ansitymmetric?
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2022-10-07 20:29:30
|
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|
https://jocelynchi.github.io/L2E-package-demo/authors.html
|
Chi, J. T. & Chi, E. C. (2022). A User-Friendly Computational Framework for Robust Structured Regression with the L$_2$ Criterion. Journal of Computational and Graphical Statistics, In press.
@Article{,
title = {A User-Friendly Computational Framework for Robust Structured Regression with the L$_2$ Criterion},
author = {Jocelyn T. Chi and Eric C. Chi},
journal = {Journal of Computational and Graphical Statistics, In press},
year = {2022},
url = {https://amstat.tandfonline.com/doi/full/10.1080/10618600.2022.2035232#.Yfr47S-B0Ts},
doi = {10.1080/10618600.2022.2035232},
}
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2022-08-18 09:56:03
|
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https://andreas-zeller.blogspot.com/2013/04/my-top-ten-presentation-issues-in.html
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## 2013-04-05
### My top ten presentation issues in other's papers
I am a frequent reviewer of submissions for scientific journals and conferences. By far most of my reviews are specific to the individual paper, its contributions, and its issues. However, there is a small number of presentation problems which I have to report again and again. All these problem are very minor, and none of them will make me reject your paper. However, fixing them is a matter of minutes, and will make your paper an easier read for both readers and reviewers:
1. Abstracts without results. Many abstracts end with the words "We conducted a study to show the effectiveness of our approach". There's three things wrong in here:
1. You conduct a study to evaluate your approach – if you run your study with a predetermined outcome, I will suspect a flaw in your setup.
2. As a reader of your abstract, I am not so much interested in your process, but in your results; if you want cliffhangers, go and write TV shows instead.
3. You need results to sell your paper in the review process – in particular for outsiders who take only a brief glance at your paper.
So make this "Our study shows a 25% precision increase" or some other strong argument.
2. Multi-letter identifiers in LaTeX math mode. In LaTeX, $foo$ stands for the product of three variables f, o, and, o; and LaTeX typesets it as such. If you want one variable, use $\textit{foo}$. (For a computer scientist, this is the single biggest design failure of TeX and LaTeX.)
3. Confuse hyphens, minus signs, and dashes. All these typeset into different characters:
1. Hyphens: state-1 is blue, state-2 is red.
2. Minus signs: $\textit{bar} = \textit{foo} - 1$
3. En-dashes (for ranges): A car has 3--4~wheels
4. Em-Dashes (for sentences): He said captain''---I said wot''. (This can also be typeset using en-dashes with spaces, as in He said captain''~-- I said wot''.)
4. Bad capitalization in BibTeX references, as in "[1] Jai: a javascript api ide" (should be "[1] JAI: A JavaScript API IDE") I could include this in all my reviews and be right 60% of the time.
5. Omitting reference information. I see papers omitting page numbers, conference names, editors, or even authors. I suppose that "[1] Smith et al., ICSE 2004" is something I should Google, and I hope there's just one Smith at ICSE. (This even occurs in papers whose venues have no page limit for references!)
6. Orphaned numbers, as in "I want 17<newline>foos." Tie numbers to items using non-breakable space (Word) or using a tilde (LaTeX): I want 17~foos.
7. Orphaned lines and titles. Section title at end of page 3, actual section on page 4. Better yet: a single line of text left under a table, such that you easily confuse content and caption. "LaTeX gets this wrong" is not an excuse; please fix this.
8. Using monospacing with a non-monospaced font (or vice versa). This seems to be a default setting of the LaTeX "listings" environment – using Times fonts and arranging the letters at equal distance. Use monospaced fonts with monospacing, and proportional fonts otherwise.
9. Omitting articles, as in "Main feature is lack of articles". Now that spell checkers weed out the most blatant errors, the more subtle errors remain – and this is my #1 grammar issue. If your native language has no articles, take double care.
All these issues can be fixed through simple proofreading, and leave your readers and reviewers more time to enjoy the actual content. And now that the syntax is done, let's get back to semantics!
(As a reviewer or reader, what are your favorite gripes? Use the commenting function below.)
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2019-02-21 21:06:47
|
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https://en.academic.ru/dic.nsf/enwiki/6396546/Convex_hull_algorithms
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# Convex hull algorithms
Convex hull algorithms
Algorithms that construct convex hulls of various objects have a broad range of applications in mathematics and computer science, see "Convex hull applications".
In computational geometry, numerous algorithms are proposed for computing the convex hull of a finite set of points, with various computational complexities.
Computing the convex hull means that a non-ambiguous and efficient representation of the required convex shape is constructed. The complexity of the corresponding algorithms is usually estimated in terms of n, the number of input points, and h, the number of points on the convex hull.
## Planar case
Consider the general case when the input to the algorithm is a finite unordered set of points on a Cartesian plane. An important special case in which the points are given in the order of traversal of a simple polygon's boundary is described later in a separate subsection.
If not all points are on the same line, then their convex hull is a convex polygon whose vertices are some of the points in the input set. Its most common representation is the list of its vertices ordered along its boundary clockwise or counterclockwise. In some applications it is convenient to represent a convex polygon as an intersection of a set of half-planes.
### Lower bound on computational complexity
For a finite set of points in the plane the lower bound on the computational complexity of finding the convex hull represented as a convex polygon is easily shown to be the same as for sorting using the following reduction. For the set x1,...xn numbers to sort consider the set of points $(x_1, x^2_1) ,...(x_n, x^2_n)$ of points in the plane. Since they lie on a parabola, which is a monotone curve it is easy to see that the vertices of the convex hull, when traversed along the boundary, produce the sorted order of the numbers x1,...xn. Clearly, linear time is required for the described transformation of numbers into points and then extracting their sorted order. Therefore in the general case the convex hull of n points cannot be computed more quickly than sorting.
The standard Ω(n log n) lower bound for sorting is proven in the decision tree model of computing, in which only numerical comparisons but not arithmetic operations can be performed; however, in this model, convex hulls cannot be computed at all. Sorting also requires Ω(n log n) time in the algebraic decision tree model of computation, a model that is more suitable for convex hulls, and in this model convex hulls also require Ω(n log n) time.[1] However, in models of computer arithmetic that allow numbers to be sorted more quickly than O(n log n) time, for instance by using integer sorting algorithms, planar convex hulls can also be computed more quickly: the Graham scan algorithm for convex hulls consists of a single sorting step followed by a linear amount of additional work.
### Optimal output-sensitive algorithms
As stated above, the complexity of finding a convex hull as a function the input size n is lower bounded by Ω(n log n). However, the complexity of some convex hull algorithms can be characterized in terms of both input size n and the output size h (the number of points in the hull). Such algorithms are called output-sensitive algorithms. They may be asymptotically more efficient than Θ(n log n) algorithms in cases when h = o(n).
The lower bound on worst-case running time of output-sensitive convex hull algorithms was established to be Ω(n log h) in the planar case.[1] There are several algorithms which attain this optimal time complexity. The earliest one was introduced by Kirkpatrick and Seidel in 1986 (who called it "the ultimate convex hull algorithm"). A much simpler algorithm was developed by Chan in 1996, and is called Chan's algorithm.
### Algorithms
Known convex hull algorithms are listed below, ordered by the date of first publication. Time complexity of each algorithm is stated in terms of the number of inputs points n and the number of points on the hull h. Note that in the worst case h may be as large as n.
• Gift wrapping aka Jarvis marchO(nh)
One of the simplest (although not the most time efficient in the worst case) planar algorithms. Discovered independently by Chand & Kapur in 1970 and R. A. Jarvis in 1973. It has O(nh) time complexity, where n is the number of points in the set, and h is the number of points in the hull. In the worst case the complexity is Θ(n2).
• Graham scanO(n log n)
A slightly more sophisticated, but much more efficient algorithm, published by Ronald Graham in 1972. If the points are already sorted by one of the coordinates or by the angle to a fixed vector, then the algorithm takes O(n) time.
• QuickHull
Discovered independently in 1977 by W. Eddy and in 1978 by A. Bykat. Just like the quicksort algorithm, it has the expected time complexity of O(n log n), but may degenerate to Θ(nh) = O(n2) in the worst case.
• Divide and conquerO(n log n)
Another O(n log n) algorithm, published in 1977 by Preparata and Hong. This algorithm is also applicable to the three dimensional case.
• Monotone chainO(n log n)
Published in 1979 by A. M. Andrew. The algorithm can be seen as a variant of Graham scan which sorts the points lexicographically by their coordinates. When the input is already sorted, the algorithm takes O(n) time.
• Incremental convex hull algorithmO(n log n)
Published in 1984 by Michael Kallay.
• Chan's algorithmO(n log h)
A simpler optimal output-sensitive algorithm discovered by Chan in 1996.
### Akl-Toussaint heuristic
The following simple heuristic is often used as the first step in implementations of convex hull algorithms to improve their performance. It is based on the efficient convex hull algorithm by Selim Akl and G. T. Toussaint, 1978. The idea is to quickly exclude many points that would not be part of the convex hull anyway. This method is based on the following idea. Find the two points with the lowest and highest x-coordinates, and the two points with the lowest and highest y-coordinates. (Each of these operations takes O(n).) These four points form a convex quadrilateral, and all points that lie in this quadrilateral (except for the four initially chosen vertices) are not part of the convex hull. Finding all of these points that lie in this quadrilateral is also O(n), and thus, the entire operation is O(n). Optionally, the points with smallest and largest sums of x- and y-coordinates as well as those with smallest and largest differences of x- and y-coordinates can also be added to the quadrilateral, thus forming an irregular convex octagon, whose insides can be safely discarded. If the points are random variables, then for a wide class of probability density functions, this throw-away pre-processing step will make a convex hull algorithm run in linear expected time, even if the worst-case complexity of the convex hull algorithm is quadratic in n.[2]
### On-line and dynamic convex hull problems
The discussion above considers the case when all input points are known in advance. One may consider two other settings.[1]
• Online convex hull problem: Input points are obtained sequentially one by one. After each point arrives on input, the convex hull for the pointset obtained so far must be efficiently computed.
• Dynamic convex hull maintenance: The input points may be sequentially inserted or deleted, and the convex hull must be updated after each insert/delete operation.
Insertion of a point may increase the number of vertices of a convex hull at most by 1, while deletion may convert a 3-vertex convex hull into an n-1-vertex one.
The online version may be handled with O(log n) per point, which is asymptotically optimal. The dynamic version may be handled with O(log2 n) per operation.[1]
### Simple polygon
McCallum and Avis were first to provide a correct algorithm to construct the convex hull of a simple polygon v1,...,vn in O(n) time. The basic idea is very simple. The leftmost vertex is on the convex hull and we denote it h1. The second point is assumed to be a candidate convex hull vertex as well. At each step one looks at three consecutive vertices of the polygon, with two first ones tentatively assigned to the growing convex hull and the third one is a new unprocessed vertex of the polygon, say, we denote this as hk − 1,hk,vi. If the angle is convex, then hk + 1 = vi and the whole triple is shifted by one vertex along the polygon. If the resulting angle is concave, then the middle point (hk) is deleted and the test is repeated for the triple hk − 2,hk − 1,vi, etc. until we backtrack either to a convex angle or to point h1. After that the next (along the polygon) vertex is added to the triple to be tested, and the process repeats. However several previously published articles overlooked a possibility that deletion of a vertex from a polygon may result in a self-intersecting polygon, rendering further flow of the algorithm invalid. Fortunately, this case may also be handled efficiently. Later Tor and Middleditch (1984, "Convex Decomposition of Simple Polygons") and independently Melkman (1985, "Online Construction of the convex hull of a simple polyline") suggested a simpler approach with the same time complexity.
## Higher dimensions
A number of algorithms are known for the three-dimensional case, as well as for arbitrary dimensions. See http://www.cse.unsw.edu.au/~lambert/java/3d/hull.html. See also David Mount's Lecture Notes for comparison. Refer to Lecture 4 for the latest developments, including Chan's algorithm.
For a finite set of points, the convex hull is a convex polyhedron in three dimensions, or in general a convex polytope for any number of dimensions, whose vertices are some of the points in the input set. Its representation is not so simple as in the planar case, however. In higher dimensions, even if the vertices of a convex polytope are known, construction of its faces is a non-trivial task, as is the dual problem of constructing the vertices given the faces. The size of the output may be exponentially larger than the size of the input, and even in cases where the input and output are both of comparable size the known algorithms for high-dimensional convex hulls are not output-sensitive due both to issues with degenerate inputs and with intermediate results of high complexity.[3]
## References
1. ^ a b c d Preparata, Shamos, Computational Geometry, Chapter "Convex Hulls: Basic Algorithms"
2. ^ Luc Devroye and Godfried Toussaint, "A note on linear expected time algorithms for finding convex hulls," Computing, Vol. 26, 1981, pp. 361-366.
3. ^ Avis, David; Bremner, David; Seidel, Raimund (1997), "How good are convex hull algorithms?", Computational Geometry: Theory and Applications 7 (5-6): 265–301, doi:10.1016/S0925-7721(96)00023-5 .
• Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. Introduction to Algorithms, Second Edition. MIT Press and McGraw-Hill, 2001. ISBN 0-262-03293-7. Section 33.3: Finding the convex hull, pp. 947–957.
• Franco P. Preparata, S.J. Hong. Convex Hulls of Finite Sets of Points in Two and Three Dimensions, Commun. ACM, vol. 20, no. 2, pp. 87–93, 1977.
• Mark de Berg, Marc van Kreveld, Mark Overmars, and Otfried Schwarzkopf (2000). Computational Geometry (2nd revised edition ed.). Springer-Verlag. ISBN 3-540-65620-0. Section 1.1: An Example: Convex Hulls (describes classical algorithms for 2-dimensional convex hulls). Chapter 11: Convex Hulls: pp. 235–250 (describes a randomized algorithm for 3-dimensional convex hulls due to Clarkson and Shor).
Wikimedia Foundation. 2010.
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• Gift wrapping algorithm — The gift wrapping algorithm is a simple algorithm for computing the convex hull of a given set of points. Planar case In the two dimensional case the algorithm is also known as Jarvis march, after R. A. Jarvis, who published it in 1973; it has O( … Wikipedia
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2020-08-14 06:26:45
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https://math.stackexchange.com/questions/2552438/exercise-lp-relaxation
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# Exercise LP Relaxation [closed]
$\max Z = 6x_1 + 7x_2$
Constraints:
$-2x_1 + 2x_2 \le 3\\ 7x_1 + 3x_2 \le 22$
$x_1,x_2 \ge 0$ and $x_1, x_2 \in \Bbb Z$
How to solve this problem with relaxation LP by graphical method?
## closed as off-topic by Brian Borchers, Rolf Hoyer, Namaste, Shailesh, Parcly TaxelDec 6 '17 at 3:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brian Borchers, Rolf Hoyer, Namaste, Shailesh, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos Dec 5 '17 at 16:22
• I have found that the optimal soluiton by Graphical Method is x1 = 1.75 and x2 = 3.25. – user3769812 Dec 5 '17 at 16:30
• But I didnt understand what I need to do in this problem to find the relaxation solution – user3769812 Dec 5 '17 at 16:32
• Is there some example of an non linear relaxation?? – user3769812 Dec 5 '17 at 18:06
$x_1 =3$, $x_2 =4$ doesn't satisfy one of your equations ($7x+3y \le 22)$.
The only integer values that satisfy the constraints are: $(x_1,x_2)=\{(0,0),(0,1),(1,0),(1,1),(1,2), (2,0),(2,1),(2,2),(3,0)\}$
• Integer solution should be $(2,2)$. So, $Z=26$ – Albatross Dec 5 '17 at 17:07
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2019-04-26 10:10:29
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https://searchcode.com/file/6443/rcdk/man/loadmol.Rd/
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http://github.com/rajarshi/cdkr
Unknown | 77 lines | 72 code | 5 blank | 0 comment | 0 complexity | 514c875ae21e574d8ea3178938342622 MD5 | raw file
1\name{load.molecules}
4\title{
5 Load Molecular Structures From Disk
6}
7\description{
8The CDK can read a variety of molecular structure formats. This function
9encapsulates the calls to the CDK API to load a structure given its filename
10}
11\usage{
12load.molecules(molfiles=NA, aromaticity = TRUE, typing = TRUE, isotopes = TRUE,
13 verbose=FALSE)
14iload.molecules(molfile, type="smi", aromaticity = TRUE, typing = TRUE, isotopes = TRUE,
15 skip=TRUE)
16}
17\arguments{
18 \item{molfiles}{A \code{character} vector of filenames. Note that the full
19 path to the files should be provided. URL's can also be used as
20 paths. In such a case, the URL should start with "http://"}
21 \item{molfile}{A string containing the filename to load. Must be a local file}
22 \item{type}{Indicates whether the input file is SMILES or SDF. Valid values are
23 "smi" or "sdf"}
24 \item{aromaticity}{If \code{TRUE} then aromaticity detection is
25 performed on all loaded molecules. If this fails for a given
26 molecule, then the molecule is set to NA in the return list}
27 \item{typing}{If \code{TRUE} then atom typing is
28 performed on all loaded molecules. The assigned types will be CDK
29 internal types. If this fails for a given
30 molecule, then the molecule is set to NA in the return list}
31 \item{isotopes}{If \code{TRUE} then atoms are configured with isotopic masses}
32 \item{verbose}{If TRUE, output (such as file download progress) will
33 be bountiful}
34 \item{skip}{If \code{TRUE}, then the reader will continue reading even when faced with an
35 invalid molecule. If \code{FALSE}, the reader will stop at the fist invalid molecule}
36}
37\value{
38 \code{load.molecules} returns a list of CDK \code{Molecule} objects, which can be
39 used in other rcdk functions.
40
41 \code{iload.molecules} is an iterating version of the loader and is applicable for
42 large SMILES or SDF files. In contrast to \code{load.molecules} this does not load
43 all the molecules into memory at one go, and as a result lets you process arbitrarily
44 large structure files.
45}
46\details{
47Note that if molecules are read in from formats that do not have rules for
48handling implicit hydrogens (such as MDL MOL), the molecule will not have
49implicit or explicit hydrogens. To add explicit hydrogens, make sure that the molecule
50has been typed (this is \code{TRUE} by default for this function) and then call
51\code{\link{convert.implicit.to.explicit}}. On the other hand for a format
52such as SMILES, implicit or explicit hydrogens will be present.
53}
54\examples{
55\dontrun{
56
57## load a single file
59
60## load multiple files
61mols <- load.molecules(c('mol1.sdf', 'mol2.smi',
62 'https://github.com/rajarshi/cdkr/blob/master/data/set2/dhfr00008.sdf?raw=true'))
63
64## iterate over a large file
65moliter <- iload.molecules("big.sdf", type="sdf")
66while(hasNext(moliter)) {
67 mol <- nextElem(moliter)
68 print(get.property(mol, "cdk:Title"))
69}
70}
71}
72\seealso{
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2023-01-31 04:03:57
|
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http://demigodllc.com/~zak/documents/college/ece757-report/tex
|
Study of Relative Performance Impact: Coherence Protocol vs. Network Speed college topic: ECE757 (Multiprocessor Computer Architecture), Univ. of Wisconsin - Madison Architects and implementors must know design trade-offs to design successful systems. Two important aspects of multiprocessor performance are network quality and coherence protocol. We study the performance impact of coherence protocol choice (MSI vs. MESI) as compared to the performance impact of the high-speed network. A higher quality network is either wider, or has less latency, or both. We find that in the majority of cases, network quality dominates the effects of coherence protocol - that is, lower-quality networks with the MESI protocol do not yield faster runs that higher-quality networks with the MSI protocol. We also note that program characteristics can affect which protocol performs better. formats: Adobe PDF (158.9kB), PostScript (233.6kB), TeX (11.5kB) 1998-05-01 quality 5
% -*- latex -*-
\documentclass[12pt]{report}
% \setlength{\hoffset}{-1in}
% \setlength{\voffset}{0in}
% \setlength{\topmargin}{0pt}
% \setlength{\oddsidemargin}{0in}
% \setlength{\textheight}{9.00in}
% \setlength{\textwidth}{6in}
\usepackage{psfig}
\begin{document}
\begin{titlepage}
\title{Study of Relative Performance Impact: Coherence Protocol vs. Network Speed
\footnote{Completed as semester project for CS/ECE 757, Prof. Mark Hill}}
\author{Zak Smith \and Mostafa Arifin}
\maketitle
\begin{abstract}
Architects and implementors must know design trade--offs to
design successful systems. Two important aspects of
multiprocessor performance are {\it network quality} and
{\it coherence protocol}. We study the performance impact of
coherence protocol choice (MSI vs. MESI) as compared to the
performance impact of the high--speed network. A higher
quality network is either wider, or has less latency, or both.
We find that in the majority of cases, {\it network quality
dominates the effects of coherence protocol} --- that is,
lower--quality networks with the MESI protocol do not yield
faster runs that higher--quality networks with the MSI
protocol. We also note that program characteristics can
affect which protocol performs better.
\end{abstract}
\end{titlepage}
\newpage
\tableofcontents
\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% \baselineskip = 22pt
\chapter{Introduction}
\section{Topic Importance}
In order for a machine to achieve high performance within a
cost constraint, architects and implementors must know the
benefits and trade-offs for different design decisions.
The performance of the interconnection network is key in
multiprocessor systems. In all but embarrassingly
parallel'' applications, communication between processing
elements provides those elements with the data they need to
continue processing.
Our study examines three parameters which affect network
performance: coherence protocol, flit delay, and flit size.
We believe this type of study is important to verify
theoretical models with real benchmark results.
\section{Hypothesis}
During lecture, it was noted that a good implementation of the
interconnection network often dominates other factors such as
network topology. Since significant time was spent
discussing coherence protocols [1], we decided to determine
the relative performance impact of coherence protocol as
compared to quality of network implementation.'' Thus, our
hypothesis:
{\it A good network implementation dominates the effects of
choosing a specific coherence protocol. }
\section{Definitions}
\subsection{Flit Delay}
The flit delay is the time it takes for a flit to pass through
a network switch, relative to the clock speed of the
processor. For example, if the flit delay was 2, and the
processor was running at 300 Mhz, it would take 6.6 ns for a
flit to propagate through a switch. This primarily affects latency.
\subsection{Flit Width}
The flit width specifies the number of bytes in each network
flit, which is equivalent to the width of the network. This
primarily affects bandwidth.
\subsection{Coherence Protocol}
The two coherence protocols we look at are the MSI and MESI
protocols discussed in lecture and in [1].
\begin{tabbing}
MESI \= ~~~~~~~~~~~~~~ \= ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\
M \> Modified \> This is the only copy, dirty. \\
E \> Exclusive \> Ensured to be the only RO copy. \\
S \> Shared \> One of two or more RO copies. \\
I \> Invalid \> Block not present. \\
\end{tabbing}
\begin{tabbing}
MSI \= ~~~~~~~~~~~~~~ \= ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\
M \> Modified \> This is the only copy, dirty. \\
S \> Shared \> One of one or more RO copies. \\
I \> Invalid \> Block not present. \\
\end{tabbing}
Thus the primary difference between the two is that, for MSI,
if a processor reads a block known to no other processor and
then writes it, it must notify the directory (S to M). For
MESI, this is not the case: a processor that reads the block
first has it in Exclusive and can move to Modified without any
traffic to the directory (E to M).
At first glance, it would appear that MESI would perform
better, although as we will see later, there can be
implementation issues which make this not the case.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\chapter{Methods}
\section{Simulator Model}
We chose to use RSIM [7,3], an ILP multiprocessor simulator
developed at Rice University. The system simulates a
collection of workstation--like nodes connected in a 2d mesh
interconnection network, with CC-NUMA based on a distributed
directory.
We simulated a 16 processor system to control runtime of our
simulations.
\section{Search-space}
One of the parameters we want to change is quality of
network implementation.'' We decided this is basically a
combination of flit delay (related to the clock speed of the
network), and the flit size (width of the network). In other
words, the best'' network will have the fastest clock speed
and the widest data-path. The worst'' network will have the
slowest clock speed and the the narrowest data-path.
To cover a range of network qualities, we chose to use flit
widths of {2, 4, 8} bytes, and flit delays of {2, 4, 8}
processor clocks. Thus for each benchmark, we have 18
data-points:
\vskip .1in
\begin{tabular}{|r||c|c|c|} %%%%%%%%%%%
\hline
& FD2 & FD4 & FD8 \\ \hline \hline
FS8 & MESI, MSI & MESI, MSI & MESI, MSI \\ \hline
FS4 & MESI, MSI & MESI, MSI & MESI, MSI \\ \hline
FS2 & MESI, MSI & MESI, MSI & MESI, MSI \\ \hline
\end{tabular}
\section{Benchmarks}
There are 5 benchmark programs available for RSIM from the
SPLASH [5] and SPLASH-2 [6] sets: mp3d, fft, radix, lu, and
water. These will be described in the results section.
\section{Data--sets}
Condor [4], is a system for high--throughput computing
available here at the UW. It is a great tool for architects
because simulations typically take days to finish. Condor
allows many of these jobs to be distributed to idle machines.
Normal condor jobs are check-pointed and migrated'' off
machines when a user logs on or the machine is to be powered
off. This allows the job continue where it left off when
another machine becomes available.
Because of some tricky code in RSIM, it would not work
correctly when linked against the Condor libraries. Thus we
were limited to running RSIM as a vanilla'' job --- that is,
when the job is interrupted by a user or the machine being
powered down, instead of check-pointing the job so it could
continue on another machine, the vanilla jobs are killed and
must be restarted. Because of this, we had to make sure our
simulations would finish in less than about 12 hours! This
limited the number of processors we could simulate and the
size of the data-sets run in each application.
data-sets could finish in the average idle time of a machine
in the Condor pool.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\chapter{Results}
\include{fft}
\include{mp3d}
\include{lu}
% \section{Summary}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\chapter{Conclusions}
\section{Hypothesis}
We found that for the majority of the cases examined, network
quality {\it does} dominate coherence protocol choice.
It was more strongly true for benchmarks which are not
communication--intensive. The only benchmark which failed the
hypothesis was RADIX, which requires a large amount of
many--to--one and one--to--many communication every iteration.
The effect of protocol choice can be overcome by network
quality. We found that increasing network width always
dominated the protocol differences, while decreasing the
network delay dominated protocol differences only most of
the time.''
Further, it could be said that designers {\it should} dominate
the protocol differences by the network quality because all
programs will benefit from a better network, and not all
benchmarks do better with MESI. As we found, some do much worse.
\section{Protocol Importance}
The choice of coherence protocol can be an important factor in
system design. We saw that for the majority of the benchmarks
we examined, protocol choice had little effect on performance,
while network quality had the primary effect. However, for
the one benchmark which had the worst slow--down, the protocol
also had major impact on performance. Thus protocol
choice must be paid special attention to during design: which
protocol will perform better for the expected workloads?
There is room here for future work.
\section{Network Quality Importance}
Network quality has the largest impact on performance. A
faster network will improve all programs which communicate,
with no performance cost --- the cost involved is engineering
and materials cost.
\section{Application Dependence}
Different applications have difference communication
requirements. Some, like RADIX, are very network--intensive,
while others tolerate low--quality networks with little
\section{Future Work}
We were surprised to see such a wide variance in maximum
runtime difference between the benchmarks. This suggests a
larger set of programs should be studied to get a better idea
This study has given us some idea about the relative
importance of network quality as compared to protocol choice.
The next step would be to compare the hardware and
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\chapter{References}
\parindent 0pt
[0] {\it This Document.}\\
{\tt http://www.cae.wisc.edu/$\tilde{\ }$smithz/texts/ece757-report/report.ps}
\par
\vskip .2in
[1] P. SWEAZEY and A. J. SMITH, A Class of Compatible Cache
Consistency Protocols and their Support by the IEEE Futurebus'', {\it
Proc. Thirteenth International Symposium on Computer Architecture},
Tokyo, Japan (June 1986), 414-423.
\par
\vskip .2in
[2] S. V. ADVE and K. GHARACHORLOO, Shared Memory Consistency Models:
A Tutorial'', {\it IEEE Computer}, 29, 12 (December 1996), 66-76.
\par
\vskip .2in
[3] V. S. PAI, P. RANGANATHAN, and S. V. ADVE, RSIM Reference Manual
--- Version 1.0'', Rice University, Dept of ECE, Technical Report
9705, August 1997.
\par
\vskip .2in
[4] Condor HTC Homepage. {\tt http://www.cs.wisc.edu/condor/}, April 1998.
\par
\vskip .2in
[5] J. P. SINGH, W. WEBER, and A. GUPTA, SPLASH: Stanford Parallel
Applications for Shared--Memory'', Stanford University, Computer
Systems Laboratory.
\par
\vskip .2in
[6] S. C. WOO, M. OHARA, E. TORRIE, J. P. SINGH, and A. GUPTA, The
SPLASH-2 Programs: Characterization and Methodological
Considerations'', ISCA 1995.
\par
\vskip .2in
[7] PAI, RANGANATHAN, ADVE, and HARTON, An Evaluation of Memory
Consistency Models for Shared Memory Systems with ILP Processors'', ASPLOS 1996.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document}
% LocalWords: Arifin MSI MESI Mhz ns RO RSIM ILP CC NUMA FD FS mp fft lu vs CS
% LocalWords: ECE Zak Mostafa pt Condor UW condor http www cae wisc edu smithz
% LocalWords: ece ps SWEAZEY Futurebus Proc ADVE GHARACHORLOO PAI RANGANATHAN
% LocalWords: HTC Homepage cs SINGH GUPTA OHARA TORRIE ISCA HARTON ASPLOS
% LocalWords: LocalWords
[Zak Smith] [zak@computer.org] [/~zak/documents/college/ece757-report/tex]
$Id: documents,v 1.5 2000/09/28 21:20:39 zak Exp zak$
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2019-06-16 21:45:19
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http://sieradskillc.com/pjf1jz2/5lyhg.php?page=kclo-oxidation-number-6b187d
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-6+1= -5. Give the oxidation state of the metal species in each complex. the oxidation number of the molecule HCl is 0.because H has Oxidation no +1 and Cl has -1 oxidation no in the HCl. The way to assign oxidation numbers to a compound is a top-down approach. Some elements The oxidation number of Cl changes from +5 to-1, so Cl is reduced, and this is a redox reaction. How do you think about the answers? The sum of the oxidation states of all the atoms or molecule in a neutral compound is zero. The sum of the oxidation numbers = 0 K = +1 O = -2 Each compound has 1K+1and 1 Cl and n O’s +1 + x + (n * -2) = 0 x + (n * -2) = -1 n increases from 1 to 4 (n * -2) increases from -2 to -8 x + -2 = -1 x = +1 x + -8 = -1 x = +7 So, x increases from +1 to 7! Next, since Potassium (K) is a member of the most reactive elements (group 1), it has an oxidation state of 1. For the reaction KClO3 KCl+32O2 assign oxidation numbers to each element on each side of the equation. Since there is an exchange of electron, i.e. But with diatomic oxygen, it is neutral, so one really doesn't think in terms of assigning a number. (v) Cl oxidation number +7 in Cl 2 O 7. KClO 3-2: H 2 O 2-1: Example 2: The oxidation state of hydrogen is always +1 except hydride, which is -1. Give the oxidation state of the metal species in each complex. What is the formula to create hydo-clormethane ? For the following reaction KClO 4 → KCl + 2O 2 assign oxidation states to each element on each side of the equation. Join Yahoo Answers and get 100 points today. The oxidation numbers must thus sum to zero. Oxidation states → 2x + (7*-2) = 0: x = +7, Oxidation state of chlorine in Cl2O = 142\\frac{14}{2}214 = +7. Hot Network Questions Is there a finite dimensional algebra with left finitistic dimension different from its right finitistic dimension? Still have questions? The oxidation number for the calcium in CaSO4 is 2+, the oxidation number for oxygen is 2-, and the oxidation number for sulfur is 6+. Get your answers by asking now. Potassium generally has an oxidation number of +I, and it does so here; oxygen generally has an oxidation number of -II and it does so here. 7. . In the BaCO 3 reaction there is no change in oxidation numbers of the elements so this is … It is used in variable concentrations, often diluted in water solution. But with diatomic oxygen, it is neutral, so one really doesn't think in terms of assigning a number. ... What is the concentration (M) of KCl in a solution made by mixing 40.0 mL of 0.100 M KCl with 50.0 mL of 0.100 M KCl? Klor-Con M15 K-Lease Klor-Con 8 K+10 K+8 Klor-Con M10 Micro-K 10 Kato. And you need a neutral molecule. Reactants Products K Number Number Cl Number Number O Number Number Which Element Is Oxidized? How would I assign individual oxidation numbers to the constituent atoms of hydrochloric acid? [Co(NH3)5Cl]Cl2 [Ru(CN)3(CO)2]3- A. Fluorine always has an oxidation number of –1 in molecules that contain two or more types of atoms. It is the chlorine that can have a number of different states. Will heating of KClO3 be considered disproportionation? For the chlorate, we know the salt is a neutral compound. And you need a neutral molecule. Help, Please! B. 0. Expert Answer 95% (84 ratings) Previous question Next question Transcribed Image Text from this Question. . (+1) +(-2×3)) +(x) =0 -5 +x=0 x=[+5] This is the way to find the oxidation number of atoms in a molecular. aus oder wählen Sie 'Einstellungen verwalten', um weitere Informationen zu erhalten und eine Auswahl zu treffen. K In KClO4: K In KCl: Cl In KClO4: Cl In KCl: O In KClO4: O In O2: O In KCl. Damit Verizon Media und unsere Partner Ihre personenbezogenen Daten verarbeiten können, wählen Sie bitte 'Ich stimme zu.' Carefully, insert coefficients, if necessary, to make the numbers of oxidized and reduced atoms equal on the two sides of each redox couples. You assign oxidation numbers to the elements in a compound by using the Rules for Oxidation Numbers. Step 2. 5.Oxidation number of chlorine in KClO 3 A) -1 B) +3 C) +5 D) +7 6.Nitrogen and carbon atoms have the same oxidation number in the following compounds A) NH 3 and CO B) NO 2 and CCl 4 C) N 2 O 3 and CO 2 D) Na 3 N and CH 4 7.In which of the following compounds sulfur has a negative oxidation number? KClO4 KCl+2O2KClO4 KCl+2O2 . 3 Answers. K Cl O Which Element Is Reduced? the 2 bonding electrons are assumed to go to the most electronegative atom, which is Cl. What is the oxidation number for nitrogen? 1 Answer. Chemistry. Oxidation number is the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that the electrons in a covalent bond belong entirely to the more electronegative elements. We have a H-Cl molecule; when we break the bond (conceptually!) Lv 4. In the oxidation number change method the underlying principle is that the gain in the oxidation number (number of electrons) in one reactant must be equal to the loss in the oxidation number of the other reactant. All Chemistry Practice Problems Calculate Oxidation Number Practice Problems. Do radioactive elements cause water to heat up? exchange of oxidation number of the ions or atoms, the above reaction id an oxidation-reduction (redox) reaction. Get the latest public health information from CDC: https://www.coronavirus.gov. [ 1] HIGHLIGHT TO SEE THE ANSWER. Provide examples? Daten über Ihr Gerät und Ihre Internetverbindung, darunter Ihre IP-Adresse, Such- und Browsingaktivität bei Ihrer Nutzung der Websites und Apps von Verizon Media. 1 decade ago. In the KClO 3 reaction, the oxidation num-ber of O changes from-2 to 0, so O is ox-idized. 0 0. 1 0. connard. Für nähere Informationen zur Nutzung Ihrer Daten lesen Sie bitte unsere Datenschutzerklärung und Cookie-Richtlinie. 1 Answer. When we look at the reaction equation we can see that K 2 Cr 2 O 7 (Potassium dichromate) released in total 6 electrons (3 electrons each atom) on the other hand FeCl 2 (Iron(II)chloride) accepts only one electron. Find the Oxidation Numbers KClO Since is in column of the periodic table , it will share electrons and use an oxidation state of . The number of grooves present on the fiber surface decreased when the KClO 3 concentration was 15 wt%. Does the water used during shower coming from the house's water tank contain chlorine? Since is in column of the periodic table , it will share electrons and use an oxidation state of . An oxidation number can be assigned to a given element or compound by following the following rules. aus oder wählen Sie 'Einstellungen verwalten', um weitere Informationen zu erhalten und eine Auswahl zu treffen. b) Identify and write out all redox couples in reaction. Attend. For the following reaction, KClO4--->KCl+2O2, assign oxidation states to each element on each side of the eq. The chlorate ion has a 1- charge so there are 5- charges to be balanced out by positive charges. Okay will continually be a +1, and oxygen will constantly be a -2. O=? Explain why water is called a polar covalent molecule? Then determine whether each equation describes a redox reaction. . Products. Molar Mass of ClKO3 Oxidation State of ClKO3. A. K = +2, Cl = +4 and O = – B. K = +1, Cl = +5 and O = –2 C. K = +3, Cl = +3 and O = –2 D. K = +1, Cl = –5 and O = –2 E. None of the Above 2 2. Write down the transfer of electrons. . How do you calculate the oxidation number of an element in a compound? In the BaCO 3 reaction there is no change Oxidation state shows the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present state. H2O2→ O2 + 2H+ + 2e- OCl- + 2H+ + 2e-→ H2O+ Cl- H2O2(aq) + OCl-(aq) → H2O(l) + Cl- (aq) + O2(g)Assign oxidation numbers to the following atoms: O in H2O2 _____; Cl in OCl- _____The oxidizing agent for this RedOx rxn . Assertion Bromide ion acts as a reducing agent in the recation, 2 M n O 4 − + B r − + H 2 O → 2 M n O 2 + B r O 3 − + 2 O H −. Explain the difference between “suspension and Emulsion” OR between “solution and colloid”. K will always be a +1, and oxygen will always be a -2. O₂ Oxygen Molecular Oxygen Diatomic Oxygen Triplet Oxygen Oxygen Gas Liquid Oxygen LOx. The oxidation number of a Group 1 element in a compound is +1. It is the chlorine that can have a number of different states. 2 kclo 3 + h 2 so 4 → 2 hclo 3 + k 2 so 4 The solution so produced is sufficiently reactive that it spontaneously ignites if combustible material (sugar, paper, etc.) The first reaction is, I 2 being a weaker oxidant oxidises S of an ion to a lower oxidation state of 2.5 in ion. 4 years ago. Balancing using oxidation numbers. Identify the redox couples in the reaction. You can sign in to vote the answer. The oxidation number of Cl increases from +1 in KClO to +7 in KClO4. Dies geschieht in Ihren Datenschutzeinstellungen. Potassium hypochlorite is not considered a fire or explosive hazard by itself. Which rule concerning oxidation numbers is NOT true? All Chemistry Practice Problems Calculate Oxidation Number Practice Problems Q. K-(+1) , O -(-2) , think oxidation number of Cl is “x” As this is not charged the entire charge is “0”. 23665762 ... Inhalation of a mist of KClO can cause bronchial irritation, difficulty breathing, and in severe cases pulmonary edema. As far as which is oxidized and which is reduced, it has to do with which way the oxidation number moves, and you need to commit that idea to memory. 6ftwonder . Potassium Chloride - KCl. EC Number: 231-909-2; PubChem CID. In the KClO 3 reaction, the oxidation num-ber of O changes from-2 to 0, so O is ox-idized. C. If all else fails, assume that the atom’s oxidation number is the same as what it would take on in an ionic compound. For the reaction. The oxidation states have to equal 0 when you do the math. The oxidation number of diatomic and uncombined elements is zero. The use of the oxidation numbers greatly simplifies identifying which element in a reaction is oxidized and which element is reduced. It's the chlorine that can have a number of specific states. Solve the equation for . Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number goes down). Balancing redox equations with oxygens in multiple oxidation states . The oxidation number of K is +1. It has a light grey color and a strong chlorine smell. Molar Mass of O2 Bond Polarity of … MPFs-15-60 had a low C content and the highest O content, indicating that it had the highest degree of oxidation. If you begin with the assumption that you don’t know any of the oxidation numbers, then you have a … situation: [math](1\times O_C) + (2 Potassium hypochlorite | KClO or ClKO | CID 23665762 - structure, chemical names, physical and chemical properties, classification, patents, literature, biological activities, safety/hazards/toxicity information, supplier lists, and more. also, which element is oxidized and which element is reduced? Yahoo ist Teil von Verizon Media. For the following reaction KClO2->KCl+O2 assign oxidation states to each element on each side of the equation. The important rules for this problem are: The oxidation number of a monatomic ion equals the charge of the ion. Example: Ion: Sum of Oxidation State: H 2 O 0: CO 2 0: NH 3 0: 5. Dioxygen - O 2. Oxidation number of Cl in KClO 3 is +5. Cl has an oxidation number of -1. HARD. 2. The oxidation state of chromium in C r (C O) 6 is : MEDIUM. In KCl has oxidation number , potassium +1 and chlorine is -1 Oxygen molecule has oxidation number zero. In this case, the chlorine atom is in the +5 oxidation state. This problem has been solved! (iv) Cl oxidation number +5 in KClO 3. Identify the oxidation numbers for each element in the following equations. As far as which is oxidized and which is reduced, it has to do with which way the oxidation number moves, and you need to commit that idea to memory. These values can be verified by adding these oxidation numbers. See the answer. As far as which is oxidized and which is reduced, it has to do with which way the oxidation number moves, and you need to commit that idea to memory. To find the oxidation state of , set up an equation of each oxidation state found earlier and set it equal to . Molar Mass of KCl Bond Polarity of KCl Oxidation State of KCl. A) CuS B) CuSO 4 C) SO 2 D) SF 6 8.Oxidation number of phosphorus in sodium … View Answer. K Cl O . Potassium Chlorate - KClO 3. COVID-19 is an emerging, rapidly evolving situation. a) Assign the oxidation numbers for each atom in the equation (see: Rules for assigning oxidation numbers). It can be used as a disinfectant. ions. The chlorate anion has the formula ClO− 3. Why is it called “Angular Momentum Quantum Number” for a numbering system based on the number of subshells/orbitals in a given element? Cl O K Which element is It's the chlorine that can have a number of specific states. Oxygen almost always has an oxidation number of -2, except in peroxides (H 2 O 2) where it is -1 and in compounds with fluorine (OF 2) where it is +2. ... What is the concentration (M) of KCl in a solution made by mixing 40.0 mL of 0.100 M KCl with 50.0 mL of 0.100 M KCl? Show transcribed image text. [Co(NH3)5Cl]Cl2 [Ru(CN)3(CO)2]3-Q. Any free element has an oxidation number equal to zero. "Chlorate", when followed by a Roman numeral in parentheses, e.g. But with diatomic oxygen, it is neutral, so one really doesn't think in terms of assigning a number. See the answer. Ingestion of strong concentrations can be lethal. K in KClO3: K in KCl: Cl in KClO3: Cl in KCl: O in KClO3: O in O2: Which element is oxidized? Reason Oxidation number of bromine increases from − 1. Question: For The Reaction KClO4 KCl+2O2KClO4 KCl+2O2 Assign Oxidation Numbers To Each Element On Each Side Of The Equation. And you need a impartial molecule. The oxidation number of Cl changes from +5 to-1, so Cl is reduced, and this is a redox reaction. I got K=5, Cl=-1, and O=-4 . Click hereto get an answer to your question ️ In a reaction, KCl is converted into KClO4. The chlorate ion has a 1- charge so there are 5- charges to be balanced out by positive charges. This problem has been solved! "Chlorate" can also refer to chemical compounds containing this anion; chlorates are the salts of chloric acid. Oxygen has an oxidation number of –2 in molecules that contain two or more types of atoms. Change in oxidation number is a) 3 b) 5 c) 7 d) 8 See the answer. Cl is +3; O is -2; K is +1 The compound has no charge. is present. So we get H^+ and Cl^-. Calculating Oxidation Numbers. Since is in column of the periodic table, it will share electrons and use an oxidation state of . Reactants K=? Compound: Oxidation state of hydrogen: NH 3 +1: HCl +1: NaOH +1: MgH 2 -1: NaH -1: 4. Oxidation states → 2x + (7*-2) = 0: x = +7, Oxidation state of chlorine in Cl2O = 142\\frac{14}{2}214 = +7. Property Name Property Value Reference Molecular Weight 90.55 g/mol Computed by PubChem 2.1 (PubChem release 2019.06.18) Hydrogen Bond Donor Count 0 Computed by Cactvs 3.4.6.11 (PubChem release 2019.06.18) K in KClO4: K in … Effects of oxidation treatment by KClO 3 /H 2 SO 4 systems on the chemical, crystal and microscopic structures of polyacrylonitrile fibers Cheng Zhang , abc Shijie Xiao , d Zhigang Shen , d Runping Li , ac Jianhua Liu , * abc Shenghui Guo * abc and Lei Xu ac Sie können Ihre Einstellungen jederzeit ändern. assign oxidation numbers to each element on each side of the equation. The oxidation number of Cl is -1. The second reaction is, Oxidation number of S in SO 4 2-=+6 Since Br 2 is a stronger oxidant than I 2, it oxidises S of S 2 O 3 2-to a higher oxidation state of +6 and hence forms SO 4 2-ions. Briefly explain why amino acid and fatty acid both are considered acid.? Q. View Answer. This means that the positive oxidation numbers must equal the negative oxidation numbers. For the following reaction KClO 4 → KCl + 2O 2 assign oxidation states to each element on each side of the equation. Potassium hypochlorite (chemical formula KClO) is the potassium salt of hypochlorous acid. The oxidation number of C r in C r O 5 which has the following structure is: MEDIUM. The Oxidation Number: For a chemical combination, the oxidation number of an element gives us an indication about the number of electrons lost (if the number is positive) Cl=? As the KClO 3 concentration increased, the elemental C content decreased and the elemental O content increased. [ 1] HIGHLIGHT TO SEE THE ANSWER. 'Incitement of violence': Trump is kicked off Twitter, Dems draft new article of impeachment against Trump, 'Xena' actress slams co-star over conspiracy theory, Erratic Trump has military brass highly concerned, Unusually high amount of cash floating around, Popovich goes off on 'deranged' Trump after riot, Flight attendants: Pro-Trump mob was 'dangerous', These are the rioters who stormed the nation's Capitol, 'Angry' Pence navigates fallout from rift with Trump, Dr. Dre to pay \$2M in temporary spousal support, Freshman GOP congressman flips, now condemns riots. KClO 3 + HCl → KCl + H 2 O + Cl 2 . The oxidation number of Cl increases from +1 in KClO to +7 in KClO4 . For the following reaction KClO4 --> KCl + 2O2 assign oxidation … Dazu gehört der Widerspruch gegen die Verarbeitung Ihrer Daten durch Partner für deren berechtigte Interessen. So simply the sum the oxidation numbers: For KClO_3, I+3xx(-II)+Cl^"oxidation number" =0. In schools, molten potassium chlorate is used in the dramatic screaming jelly babies , Gummy bear , Haribo , and Trolli candy demonstration where the candy is dropped into the molten salt. The element that reduced is K, but its says its wrong. Which element is reduced? 3. Oxygen molecule has oxidation number zero 1 0 connard Lv 4 4 years ago Okay will continually be a +1, and oxygen will constantly be a -2. Now, you would work out the oxidation of-5+5-0. Can someone explain how to determine oxidation states if the element is not by itself, please :) KClO = potassium hypochlorite, the hypochlorite ion has an overall charge of 1-, K has a charge of 1+ K 1+, ClO- --> Cl 1 Chlorine is reduced and Oxygen is oxidized (OILRIG- Oxidation is Losing electrons, Reduction is Gaining electrons). (4) Oxygen always have a charge -2 except for peroxide ion (#O_2^"2-"#) which has a charge of -1. Wir und unsere Partner nutzen Cookies und ähnliche Technik, um Daten auf Ihrem Gerät zu speichern und/oder darauf zuzugreifen, für folgende Zwecke: um personalisierte Werbung und Inhalte zu zeigen, zur Messung von Anzeigen und Inhalten, um mehr über die Zielgruppe zu erfahren sowie für die Entwicklung von Produkten. spontaneous combustion - how does it work? A strip of magnesium weighs 0.8197 g. determine the volume in L if the density of magnesium is 1.74 g/mL. What are the oxidation numbers of K, Cl, and O in KClO 3? View Answer. The oxidation number of calcium in the calcium ion (Ca 2+) is 0.; The oxidation number of hydrogen in the hydrogen molecule (H 2) is 0.; The oxidation number of sodium (Na) in the metal sodium is 0. Daten über Ihr Gerät und Ihre Internetverbindung, darunter Ihre IP-Adresse, Such- und Browsingaktivität bei Ihrer Nutzung der Websites und Apps von Verizon Media. 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# 《毕出21套》解析：数学第2套
## 数学第 2 套DS
1.The cardinality of a finite set is the number of elements in the set. What is the cardinality of set A?
(1) 2 is the cardinality of exactly 6 subsets of set A.
(2) Set A has a total of 16 subsets, including the empty set and set A itself.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
（1）集合A中恰好有6个子集的元素个数为2
（2）集合A共有16个子集，包括空集和它本身
2. X,81,73,71,98,73,64
What is the value of X in the above list of 7 numbers?
(1) The average (arithmetic mean) of these 7 numbers is 80.
(2) The range of these 7 numbers is 36.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
3.If 1 < d <2, is the tenths digit of the decimal representation of d equal to 9?
(1)d+0.01<2
(2)d+0.05>2
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
4.If b is the product of three consecutive positive integers c, c+ 1, and c+2, is b a multiple of 24?
(1)b is a multiple of 8.
(2)c is odd.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
3个连续的正整数的乘积一定是3的倍数，因为每3个连续的正整数中必有一个是3的倍数。因此，根据条件1，若b是8的倍数，则b同时是3和8的倍数，自然也是24的倍数。条件2说c是奇数，则只有c+1是偶数，无法保证其中有8，所以单独2不行，答案为A。
5.If x and z are integers, is x+z2odd?
(1)x is odd and z is even.
(2)x-z is odd.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
6.If x an integer?
(1) x2is an integer.
(2) x/2 is not an integer.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
7.If m and n are positive integers, what is the value of 3/m + n/4?
(1) mn=12
(2) 3/m is in lowest terms and n/4 is in lowest terms.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
3/m + n/4 = (12+mn)/4m。根据条件1可得24/(4m)，因为无法确定m的值，所以单独1是不充分的。单独根据条件2显然更不充分。把两个条件相加，如果3/m和4/n都是最简式，则m不能是3的倍数，n不能是2的倍数。m*n = 12，则m = 4，n= 3。显然可以求得题干中算式的答案，答案为C。
8.If x, y, and d are integers and d is odd are both x and y divisible by d?
(1) x+y is divisible by d.
(2) x-y is divisible by d.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
9.If K is a positive integer less than 10 and N=4321+K, what is the value of K?
(1) N is divisible by 3.
(2) N is divisible by 7.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
10.If m and n are positive integers, is n even?
(1) m(m+2)+1=mn
(2) m(m+n) is odd.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
11.If @ denotes a mathematical operation, does x@y=y@x for all x and y?
(1) For all x and y, x@y=2(x2+y2).
(2) For all y, 0@y=2y2.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
12.The 9 participants in a race were divided into 3 teams with 3 runners on each team. A team was awarded 6-n points if one of its runners finished in nth place, where 1≤n≤5. If all of the runners finished the race and if there were no ties was each team awarded at least 1 point?
(1) No team was awarded more than a total of 6 points.
(2) No pair of teammates finished in consecutive places among the top five places.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
13. If x and y are integers, is xy+1 divisible by 3?
(1)When x is divided by 3, the remainder is 1.
(2)When y is divided by 9, the remainder is 8.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
x=3m+1，y=9n+8，其中m，n是整数。
xy+1=(3m+1)(9n+8)+1
=27mn+24m+9n+84-1
=3(9mn4-8m+3n+3)
14.If x is a positive integer, how many positive integers less than x are divisors of x?
(1)x^2 is divisible by exactly 4 positive integers less than x^2.
(2)2x is divisible by exactly 3 positive integers less than 2x.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
15.Sprinklers are being installed to water a lawn. Each sprinkler waters in a circle. Can the lawn be watered completely by 4 installed sprinklers?
(1) The lawn is rectangular and its area is 32 square yards.
(2) Each sprinkler can completely water a circular area of lawn with a maximum radius of 2 yards.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
16.In quadrilateral ABCD, is angle BCD a right angle?
(1) Angle ABC is a right angle.
(2) Angle ADC is a right angle.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
17.
In rectangular region PQRS above, T is a point on side PS. If PS =4, what is the area of region PQRS?
(1)ΔQTR is equilateral.
(2) Segments PT and TS have equal length.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
18.The top surface area of a square table top was changed so that one of the dimensions was reduced by 1 inch and the other dimension was increased by 2 inches. What was the surface area before these changes were made?
(1) After the changes were made, the surface area was 70 square inches.
(2) There was a 25 percent increase in one of the dimensions.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
19.Is the perimeter of a certain rectangular garden greater than 50 meters?
(1) The two shorter sides of the garden are each 15 meters long.
(2) The length of the gardens 5 meters greater than the width of the garden.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
20.
In the figure above, points A, B, C, and D are collinear and AB, BC, and CD are semicircles with diameters d1cm, d2cm, and d3cm, respectively. What is the sum of the lengths of AB, BC and CD in centimeters?
(1) d1: d2: d3 is 3:2:1
(2) The length of AD is 48 cm.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
21.In the rectangular coordinate system, line k passes through the point(n,-1). Is the slope of line k greater than zero?
(1) Line k passes through the origin.
(2) Line k passes through the point(1, n+2).
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
22.What is the length of the hypotenuse of △ABC?
(1) The lengths of the three sides of △ABC are consecutive even integers .
(2) The hypotenuse of △ABC is 4 units longer than the shorter leg.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
23.In the standard (x, y) coordinate plane, what is the slope of the line containing the distinct points P and Q?
(1) Both P and Q lie on the graph of |x|+|y| =1
(2) Both P and Q lie on the graph of |x+y|=1
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
24.Alan and Sue have each been saving one dollar a day and will continue to do so for the next month. If Sue began saving several days before Alan, in how many days from today will Alan have saved one-half as much as Sue?
(1) As of today, Alan has saved 7 dollars and Sue has saved 27 dollars.
(2) Three days from today, Alan will have saved one-third as much as Sue.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
25.What is the value of x/yz?
(1) x=y/2 and z=2x/5
(2) x/z=5/2 and 1/y=1/10
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
26.Of the numbers q, r, s, and t, which is greatest?
(1)The average(arithmetic mean) of q and r is s.
(2)The sum of q and r is t.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
### 解：
27. A box of light bulbs contains exactly 3 light bulbs that are defective. What is the probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb?
(1) The light bulbs in the sample will be picked 1 at a time without replacement.
(2) The sample will contain exactly 20 light bulbs.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
28. Is xy<6?
(1)x<3 and y<2
(2)1/2 < x < 2/3 and y2<64
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
29. What is the median of the data set S that consists of the integers 17, 29, 10, 26, 15, and x?
(1) The average (arithmetic mean) of S is 17.
(2) The range of S is 24.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
30. If x≠1，is y equal to x+1?
(1)(y−2)/(x−1)=1
(2)y2=(x+1)2
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
31. Three dice, each with faces numbered 1 through 6, were tossed onto a game board. If one of the dice turned up 4, what was the sum of the numbers that turned up on all three dice?
(1)The sum of two of the numbers that turned up was 10.
(2)The sum of two of the numbers that turned up was 11.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
32. Is x2-y2 a positive number?
(1) x-y is a positive number.
(2) x+y is a positive number.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
33. An investment has been growing at a fixed annual rate of 20% since it was first made;no portion of the investment has been withdrawn, and all interest has been reinvested. How much is the investment now worth?
(1) The value of the investment has increased by 44% since it was first made.
(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth. (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. (B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. (C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient. (E) Statements (1) and (2) TOGETHER are NOT sufficient. 解： 条件1：只给出百分比，没有任何一个实际值是不可能得到现在价值多少的。 条件2： 假设现在是x，则一年以前是x/1.2，1年以前拿掉了600，则有(x/1.2-600)*1.2=(1-0.12)x，可以求出x，充分。 答案为B。 34. If x+y+z>0, is z>1? (1) z>x+y+1 (2) x+y+1<0 (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. (B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. (C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient. (E) Statements (1) and (2) TOGETHER are NOT sufficient. 解： 条件1：z>x+y+1且x+y+z>0，得出x+y+x+y>-1，即，x+y>-1/2. 由z>x+y+1可知，z要大于1则必须x+y>0，无法得知，不充分。 条件2：由x+y+1<0得出，一个小于-1的数+z>0，因此z必须大于1，充分。 答案为B。 35. In the first 2 hours after Meadow's self-service laundry opens, m large washing machines and n small washing machines are in continual use. Including the time for filling and emptying the washing machines, each load of laundry takes 30 minutes in a large washing machine and 20 minutes in a small washing machine. What is the total number of loads of laundry done at Meadow's self-service laundry during this 2-hour period? (1) n=3m (2) 2m+3n=55 (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. (B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. (C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient. (E) Statements (1) and (2) TOGETHER are NOT sufficient. 解： 题干的意思为，2小时用m个大洗衣机和n个小洗衣机工作。大洗衣机30分钟处理1 load，小洗衣机20分钟处理1 load，问2小时内一共处理了多少load。 显然，依题意可以直接列方程，则有： 总量 = 4m + 6n 显然，条件2可以，因为直接等式两侧同乘以2，即可得知4m + 6n = 110，答案为B。 36. Let S be a set of outcomes and let A and B be events with outcomes in S. Let ~B denote the set of all outcomes in S that are not in B and let P(A) denote the probability that event A occurs. What is the value of P(A)? (1) P(A∪B)=0.7 (2) P(A∪~B)=0.9 (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. (B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. (C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient. (E) Statements (1) and (2) TOGETHER are NOT sufficient. 解： 条件1：P(A∪B) = P(A) + P(B) - P(AB) = 0.7，显然单独无法判断P(A)，不充分。 条件2：P(A∪~B) = 1 - P(B) + P(AB) = 0.9，显然单独无法判断P(A)，不充分。 条件1+条件2：两个等式相加，即可得出P(A) = 0.6，充分。 因此，答案为C。 37. How many people in Town X read neither the Word newspaper nor the Globe newspaper? (1) Of the 2500 people in Town X, 1000 read no newspaper. (2) Of the people in Town X, 700 read the globe only and 600 read the World only. (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. (B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. (C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient. (E) Statements (1) and (2) TOGETHER are NOT sufficient. 解： 条件1：我们知道1000人不读报纸，但我们不知道是否这个城镇只有2种报纸，所以不充分。 条件2并没有告诉我们一共有多少人，有没有人什么也不读等等，所以单独不充分。 条件1+条件2：依然不充分，还是不知道有没有其它的报纸。 答案为E。 38. What is the sum of 3 consecutive integers? (1) The sum of the 3 integers is less than the greatest of the 3 integers. (2) Of the 3 integers, the ratio of the least to the greatest is 3. (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. (B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. (C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient. (E) Statements (1) and (2) TOGETHER are NOT sufficient. 解： 条件1：三个数的加和小于三个数中的最大的数。这只能表明这三个数不都是正数，但不能确定它们的加和。 条件2：三个整数中，最小和最大的比值是3。之后-3，-2，-1这三个连续的整数是符合条件的，充分。 因此，答案为B。 39. What is the number of integers that are common to both set S and set T? (1) The number of integers in S is 7, and the number of integers in T is 6. (2) U is the set of integers that are in S only or in T only or in both, and the number of integers in U is 10. (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. (B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. (C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient. (E) Statements (1) and (2) TOGETHER are NOT sufficient. 解： 条件1：只知道S和T各自的元素数量，完全无法知道交集的情况。 条件2：知道U=（S+T+交集）= 10，还是不知道交集的元素数量。 条件1+条件2：可以知道S，T，以及交集的数量，充分。 答案为C。 40. A country's per capita national debt is its national debt divided by its population. Is the per capita national debt of Country G within$5 of $500? (1) Country G's national debt to the nearest$1,000,000,000 is \$43,000,000,000
(2) Country G's population to the nearest 1, 000,000 is 86,000,000
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
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2021-01-27 01:40:34
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https://math.stackexchange.com/questions/1403852/prove-the-following-function-is-lipschitz-with-constant-less-than-1/1404937
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# Prove the following function is Lipschitz with constant less than 1.
I've been set this problem recently and I'm having a lot of trouble with it. Any help would be much appreciated!
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function with continuous derivatives of all orders and suppose that, for some $x\in \mathbb{R}$ the derivative $f'(x)$ is non-zero. Thus there exists an interval D containing x such that $f'(y)\neq 0$ for all $y\in D$. Define $F:D \rightarrow \mathbb{R}$ by $F(y)=y - \frac{f(y)-f(x)}{f'(y)}$. Show that F is a Lipschitz function, with Lipschitz constant less than 1.
N.B. I think I can prove the Lipschitz part. Just use the Mean Value Theorem?
Since $f$ has continuous derivatives of all orders, so does $F$. So by the mean value theorem, showing that $F$ is Lipschitz with constant at most one is equivalent to showing the maximum value that $|F'(y)|$ can take is one. Now I get $$F'(y) = \frac{f''(y)[f(y)-f(x)]}{f'(y)^2}$$ I can see how to find an interval $E \subset D$ containing $x$ in which $\sup_{y\in E} |F'(y)| < 1$, simply by using the continuity of $f$, $f'$ and $f''$. But I think in general, it will not be true that $\sup_{y\in D} |F'(y)|$ can be easily bounded, because how can you control $f''(y)$?
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2019-09-18 14:24:27
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https://thehousecarpenter.wordpress.com/2014/10/19/how-to-compute-remainders-in-your-head-and-why-prime-bases-arent-such-a-bad-idea/
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# How to compute remainders in your head, and why prime bases aren’t such a bad idea
Suppose you have an integer ${n}$ and you want to find its remainder upon division by a positive integer ${m}$. The obvious way is to just to divide ${n}$ by ${m}$, but there may be easier ways to do this if you are only interested in the remainder, rather than the quotient. In particular, there are a lot of simple rules, which can be derived using modular arithmetic, that show that the remainder of ${m}$ in ${n}$ is the same as the remainder of ${m}$ in a smaller integer ${n'}$, where the value of ${n'}$ can be easily computed from the digits of ${n}$ in a given base. For example, I would struggle to divide 5292 by 3 in my head, but I can find the remainder upon division by 3 in a few seconds by knowing the rule that the remainder of 3 in the sum of the decimal digits of 5292 will be the same: adding 5, 2, 9 and 2 gives 18, which is divisible by 3, so the remainder is 0. Let’s see what other rules like this we can find.
The task can be stated as follows. Suppose ${n}$ and ${k}$ are non-negative integers, ${b}$ and ${m}$ are positive integers, ${d_0}$, ${d_1}$, … and ${d_k}$ are integers such that for every integer ${i}$ such that ${0 \le i \le k}$, ${0 \le d_i < b}$, and
$\displaystyle n = d_0 + d_1 b + \dotsb + d_k b^k$
(so the digits of ${n}$ in base ${b}$ are ${d_k}$, ${d_{k - 1}}$, … and ${d_0}$ in that order). We want to find the remainder ${r}$ of ${m}$ in ${n}$. As explained in the previous post, ${r}$ is congruent to ${n}$ modulo ${m}$. That is,
$\displaystyle r \equiv d_0 + d_1 b + \dotsb + d_k b^k \mod m.$
Let ${s}$ be an integer congruent to ${b}$ modulo ${m}$, such as the remainder of ${m}$ in ${b}$ (since only small bases are practically useful this should be easy to find). Then for every non-negative integer ${i}$, ${s^i}$ is congruent to ${b^i}$ modulo ${m}$ by Theorem 2 from the previous post, and hence ${d_i s^i}$ is congruent to ${d_i b^i}$ by applying the same theorem. So, if we let
$\displaystyle n' = d_0 + d_1 s + \dotsb + d_k s^k,$
then it follows by Theorem 1 that ${n'}$ is congruent to ${r}$ modulo ${m}$, so ${r}$ is the remainder of ${m}$ in ${n'}$.
The smaller ${n'}$ is, the better, because that makes it easier to find its remainder upon division by ${m}$. Therefore, the best value of ${s}$ to choose is not necessarily the remainder of ${m}$ in ${b}$: if this remainder ${s'}$ is greater than ${m / 2}$ then it would be better to choose ${s}$ as ${m - s'}$ since ${m - s'}$ will be less than ${m / 2}$. If ${s}$ can be chosen to be 0, 1 or ${-1}$ ${n'}$ is especially easy to compute.
If ${s}$ is 0 (which means ${m}$ divides ${b}$), then ${s^0}$ is 1, and for every positive integer ${i}$, ${s^i}$ is 0, so we have
$\displaystyle \begin{array}{rcl} n' &=& d_0 \cdot 1 + d_1 \cdot 0 + \dotsb + d_k \cdot 0 \\ &=& d_0 + 0 + \dotsb + 0 \\ &=& d_0 \mod m. \end{array}$
The least upper bound on the value of ${n'}$ is ${b - 1}$, which is nice because ${b - 1}$ does not depend on ${n}$, so no matter how large ${n}$ is, the remainder of ${m}$ in ${n'}$ will be just as easy to find.
This gives us the first divisibility rule.
Proposition 1 If ${m}$ divides ${b}$, then the remainder of ${m}$ in ${n}$ is the remainder of ${m}$ in ${d_0}$ (the last digit of ${n}$ in base ${b}$).
Consequently, ${m}$ divides ${n}$ if and only if it divides ${d_0}$.
In base 10 in particular, this tells us that 2 divides ${n}$ if and only if ${d_0}$ is 0, 2, 4, 6 or 8, 5 divides ${n}$ if and only if ${d_0}$ is 0 or 5, and 10 divides ${n}$ if and only if ${d_0}$ is 0.
If ${s}$ is 1 (which means ${m}$ divides ${b - 1}$), then for every non-negative integer ${i}$, ${s^i}$ is 1, so we have
$\displaystyle \begin{array}{rcl} n' &=& d_0 \cdot 1 + d_1 \cdot 1 + \dotsb + d_k \cdot 1 \\ &=& d_0 + d_1 + \dotsb + d_k \mod m. \\ \end{array}$
The least upper bound on the value of ${n'}$ is now ${k (b - 1)}$, which does depend on ${n}$. However, if ${n'}$ proves to be too large, we can simply repeat the process with ${n'}$ in place of ${n}$ to get a smaller value of ${n'}$. It takes a very large value of ${n}$ to make ${n'}$ too large, anyway, because as a function of ${n}$, ${k}$ grows at a much slower rate than ${n}$.
This gives us the second divisibility rule.
Proposition 2 If ${m}$ divides ${b - 1}$, then the remainder of ${m}$ in ${n}$ is the remainder of ${m}$ in ${d_0 + d_1 + \dotsb + d_k}$ (the sum of all the digits of ${n}$ in base ${b}$).
Consequently, ${m}$ divides ${n}$ if and only if it divides ${d_0 + d_1 + \dotsb + d_k}$.
In base 10 in particular, this tells us that 3 divides ${n}$ if and only if 3 divides ${d_0 + d_1 + \dotsb + d_k}$, and 9 divides ${n}$ if and only if 9 divides ${d_0 + d_1 + \dotsb + d_k}$.
If ${s}$ is ${-1}$ (which means ${m}$ divides ${b + 1}$), then for every non-negative integer ${i}$, ${s^i}$ is 1 if ${k}$ is even and ${-1}$ if ${k}$ is odd, so we have
$\displaystyle \begin{array}{rcl} r &\equiv& d_0 \cdot 1 + d_1 \cdot (-1) + \dotsb + d_k \cdot (-1)^k \\ &\equiv& d_0 - d_1 + \dotsb + (-1)^k d_k \mod m. \\ \end{array}$
This gives us the third divisibility rule.
Proposition 3 If ${m}$ divides ${b + 1}$, then the remainder of ${m}$ in ${n}$ is the remainder of ${m}$ in ${d_0 - d_1 + \dotsb + (-1)^k d_k}$ (the alternating sum of all the digits of ${n}$ in base ${b}$).
Consequently, ${m}$ divides ${n}$ if and only if it divides ${d_0 - d_1 + \dotsb + (-1)^k d_k}$.
As with the previous rule, it may be necessary to apply this rule repeatedly for large values of ${n}$.
In base 10 in particular, this tells us that 11 divides ${n}$ if and only if 11 divides ${d_0 - d_1 + \dotsb + (-1)^k d_k}$. For example, ${1 - 8 + 8 - 2 + 1 = 0}$ so 11 divides 12882.
Even outside of these three cases, the reduction to ${n'}$ can still be helpful; it’s just that you now have to perform multiplications as well as additions and ${n'}$ increases. For example, let’s try and find whether 1288 is divisible by 7. We choose ${s = 3}$ because 10 is congruent to 3 modulo 7. Then we reduce the problem to finding out whether ${8 + 8 \cdot 3 + 2 \cdot 3^2 + 1 \cdot 3^3 = 8 + 24 + 18 + 27 = 77}$ is divisible by 7, and of course it is. This would definitely be a useful strategy for a computer, but I don’t think I would have enough working memory to compute that sum in my head.
The existence of these rules can help us give an answer to which is the best base, although of course this is really a subjective question. Since it’s very easy to check for divisibility by ${m}$ if ${m}$ divides the base ${b}$, it might seem best to choose a base with a high density of divisors, such as 6, 12 or 24. It would certainly be a bad idea to choose a prime number for a base. However, it is also useful to have a lot of integers ${m}$ such that ${m}$ divides ${b - 1}$ or ${b + 1}$, since Propositions 2 and 3 give easy divisibility rules for such values of ${m}$. So you shouldn’t necessarily rule out the prime bases—in fact, a prime base ${b}$ such that ${b - 1}$ and ${b + 1}$ have a lot of divisors may be better than a composite base ${b}$ such that ${b - 1}$ and ${b + 1}$ are prime.
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2018-01-16 19:16:59
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https://www.physicsforums.com/threads/differential-equation.229926/
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# Differential Equation ?
## Homework Statement
Solve (x^2 + y^2 - y)dx + x dy = 0 when x > 0, y > 0.
## Homework Equations
(x^2 + y^2 - y)dx + x dy = 0
## The Attempt at a Solution
dy/dx = - (x^2 + y^2 - y) / x
That seems like a tricky one. Can I make it a second-order diff. eq.?
Well, with the topic we have had about vectorfields I can backwards engineer it into
F(x,y) = (x^2 + y^2 - y)i + (-x)j
But that dosn't make much sense to me.
## Answers and Replies
Related Calculus and Beyond Homework Help News on Phys.org
(Hmm, I miss an edit-button...)
I shorted it down to x dy/dx = -x^2 - y^2 + y or dy/dx = -x -y^2/x + y/x
Defennder
Homework Helper
You don't have to complicate this one unnecessarily. The substitution y=vx works fine here.
Shooting Star
Homework Helper
$$(x^2+y^2)dx + (xdy-ydx) = 0 => dx + \frac{xdy-ydx}{x^2+y^2} = 0.$$
Whenever you get (xdy-ydx), see if you can make it into the form
$$\frac{xdy-ydx}{x^2}$$, since that is equal to $$d(\frac{y}{x}).$$
See if you can spot a function of (y/x) with the d(y/x) here.
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2020-11-27 00:27:26
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https://www.astroexplorer.org/details/ajacbd34f1
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Image Details
Choose export citation format:
Rotation Periods, Inclinations, and Obliquities of Cool Stars Hosting Directly Imaged Substellar Companions: Spin–Orbit Misalignments Are Common
• Authors: Brendan P. Bowler, Quang H. Tran, Zhoujian Zhang, Marvin Morgan, Katelyn B. Ashok, Sarah Blunt, Marta L. Bryan, Analis E. Evans, Kyle Franson, Daniel Huber, Vighnesh Nagpal, Ya-Lin Wu, Yifan Zhou
Brendan P. Bowler et al 2023 The Astronomical Journal 165 .
• Provider: AAS Journals
Caption: Figure 1.
Orbital and rotational geometry relevant for obliquity constraints of stars hosting imaged companions. The observer is looking down from the z-axis. The xy plane is the plane of the sky, and the star is centered on the origin. The x-axis points north and the y-axis points east. Left: the inclination of the companion’s orbital plane, i o , intersects the sky along the line of nodes, which is oriented by Ω o from true north to the ascending node. In this configuration the orbital angular momentum vector ﹩\overrightarrow{{L}_{o}}﹩ points in the hemisphere facing the observer, with ﹩\overrightarrow{{L}_{o}}﹩ pointing toward the observer along the z-axis when i o = 0°. Middle: the star’s orientation in space determines the inclination of its equatorial plane, i *, and the position angle of its line of nodes. The sense of the stellar spin sets its rotational angular momentum vector ﹩\overrightarrow{{L}_{* }}﹩ and the longitude of ascending node Ω*. Right: if only the inclinations i o and i * are measured, but not their true (or relative) orientations in the plane of the sky (Ω o and Ω*), then each angular momentum vector can fall anywhere along nested cones opening toward or away from the observer. If Ω o is known (point A 1 , for example), but not Ω*, then the obliquity can be as small as ∣i o i *∣ (at B 1 ), can reach i o + i * for prograde orbits (B 2 ), or can be as large as π − ∣i o i *∣ (B 3 ) for retrograde orbits.
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2023-03-30 12:02:24
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https://gmatclub.com/forum/in-the-rhombus-bc-6-ae-4-and-angle-dae-45-ad-is-the-diamete-211995.html
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# In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete
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In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete [#permalink]
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17 Jan 2016, 12:08
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In the rhombus, BC = 6, AE = 4, and angle DAE = 45°. AD is the diameter of the circle. If a man started at C and followed around the outer edge of this figure to D, F, A, G, E, B, and back to C, approximately how far did he travel?
A. 14 + 27/4*π
B. 14 + 6π
C. 12 + 6π
D. 14 + 9/2*π
E. 12 + 9/2*π
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Re: In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete [#permalink]
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17 Jan 2016, 21:33
Bunuel wrote:
In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diameter of the circle. If a man started at C and followed around the outer edge of this figure to D, F, A, G, E, B, and back to C, approximately how far did he travel?
A. 14 + 27/4*π
B. 14 + 6π
C. 12 + 6π
D. 14 + 9/2*π
E. 12 + 9/2*π
Attachment:
2016-01-17_2306.png
Hi Bunuel,
again a good Q but there is a typo AE , 4, should be AE=4..
I'll solve it in two ways ..
first by POE and second by proper algebra..
1)POE:-
first lets look at the perimeter of rhombus one has to walk= 6+6+(6-4)=14..
our answer will be 14 + something 'π'..
A, B and D remain..
now lets look at the circle..
the radius = side of rhombus/2=6/2=3..
so perimeter of circle= 6 π...
our answer has to be less than 14+6π..
lets see A,B, and D..
A. 14 + 27/4*π= 14+6 1/4*π>14+6π.. not possible
B. 14 + 6π= 14+6π.., but our ansewr has to be less than that.. so eliminate
D. 14 + 9/2*π = 14 + 4 1/2*π <14+6π.. what we were looking for..
ans D..
2) algebric way..
GEOMETRY RULE :- if an arc(DE here) makes an angle x(45) from a point on circumference(A here), the angle from center to this arc will be 2x, 90 degree here..
Sice the arc makes a 90 angle at center, the circumference will be 2 pi *r*90/360... = 1/4 circumference..
thus he walks on 3/4 circumference..
circumference=2pi*r=6pi..
he walks 6pi*3/4=9/2pi on circle..
on rhombus he walks 6=6+2=14..
total=14+9pi/2....
D
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Re: In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete [#permalink]
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25 Mar 2016, 22:38
DFAG=Total Perimeter (2pi*r -2pi*r/4) ...............(1)
6+6+2=14................(2)
So 14+4.5pi
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Re: In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete [#permalink]
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04 Apr 2016, 14:48
Here is my approach => Total distance covered on the rhombus => 14
On the circle => 6 * pie - (90/360 * 6*pie) =>4.5 *pie => 9/2 *pie
adding them => 14+ 9/2 * pie
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Re: In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete [#permalink]
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16 Apr 2018, 10:53
Hi Bunuel,
I have one question, the value of AE must be $$3 \sqrt{2}$$, given DAE is 45 degrees and
DOE is 90 degrees (by central angle theorem), so triangle AOE must be right angled isoceles triangle in the ratio $$1:1:\sqrt{2}$$ and AE must be $$3 \sqrt{2}$$
Please correct me if i am wrong.
Thanks
Re: In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete &nbs [#permalink] 16 Apr 2018, 10:53
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# In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete
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2018-10-17 12:35:53
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http://physics.stackexchange.com/questions/63995/distinguish-electron-like-and-hole-like-orbits-in-reciprocal-space-with-referen
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# Distinguish electron-like and hole-like orbits in reciprocal space, with reference to the Brillouin Zones
So I have a solid state and thermodynamic exam next week and I've been going through some of the previous exams from years gone by to prepare. I came across this question "Distinguish electron-like and hole-like orbits in reciprocal space, with reference to the Brillouin Zones of squarium by sketching an example of each type." Apologies if this seems simple, solid state physics is not a strong point. Any help at all would be great! Also, what is squarium? I can't find anything in my notes on it and I'm wondering if it is simply referring to some sort of 2D square layout on which to sketch on. Thanks :)
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Squarium... yeah maybe it's just the square lattice? – Lagerbaer May 9 '13 at 21:02
Wow that was a quick reply! Yea i'm thinking that must be the case. Thanks very much :) any ideas about the distinguishing between electron-like and hole-like orbits in reciprocal space? I can't understand how a hole could have an orbit – Carson1091 May 9 '13 at 21:19
## 1 Answer
Think about a 2D Fermi surface (or a 2D section of a 3D Fermi surface). Now look in the extended Brillouin zone - thats the one where you take many copies of the first Brillouin zone and use it to tile the plane. Now we are in 2D so the Fermi surface is a bunch of curves in the extended BZ. There are three possibilities about the shape of the shape of this curve.
1. Your Fermi surface forms closed loops that enclose occupied electron states. This is an electronlike orbit.
2. Your Fermi surface froms closed loops that enclose unoccupied electron states. This is a holelike orbit.
3. Your Fermi does not form closed loops. This is an open orbit.
The reason for these names is that we are thinking of applying a perpendicular magnetic field $B_z$. In that case we have $$\frac{\partial k}{\partial t} = v_k \times B_z,$$ where the group velocity $v_k = \nabla_k E(k)$. Since $v_k$ is pointed in the opposite direction of $k$ in holelike orbits, the electrons orbits in the "wrong direction". That is it orbits in the direction that, naively, a positively charged particle would orbit in. Hence it is called holelike.
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Thank you very much! Very clearly explained. I tried to upvote you but unfortunately I just signed up for this so I still don't have the required reputation yet. Great answer, thanks again :) – Carson1091 May 10 '13 at 9:49
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2015-12-01 16:37:58
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https://math.stackexchange.com/questions/3433836/difficult-question-on-jordan-normal-form
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# Difficult question on Jordan normal form
(a) Does there exist a $$9 \times 9$$ matrix $$B$$, for which the matrix $$B^2$$ has the Jordan Normal Form with blocks of sizes $$4,3,2$$ appearing once, each block with eigenvalue $$0$$?
(b)Answer the same query in the analogous situation with blocks of sizes $$4,4,1$$.
Any ideas?
The trick is to figure out the following: if $$J_n$$ is the Jordan block with size zero of size $$n$$, then what is the Jordan structure of $$A = J_n^2$$? When $$n = 1$$, the answer is trivial, so we take $$n \geq 2$$ below.
In fact, we can figure this out using the so-called "Weyr characteristic," which I explain here (references on this are sparse as far as I can tell, but I learned about this from Horn and Johnson's Matrix Analysis). Let $$r_k$$ denote the rank of $$(A - \lambda I)^k = A^k$$ (and we'll take $$r_0 = n$$). You can verify that we have $$r_k = \begin{cases} n - 2k & k \leq n/2\\ 0 & k > n/2 \end{cases}.$$ Now, we note that $$s_k = r_{k-1} - r_k$$ will always give us the number of Jordan blocks that $$A$$ has with size at least $$k$$. In our case, we find that when $$k < n/2$$, we have $$s_k = 2$$. In the case where $$n$$ is even, we will also have $$s_{n/2} = 2$$, and in the case where $$n$$ is odd, we will have $$s_{(n+1)/2} = 1$$. This accounts for all non-zero $$s_k$$.
Finally, $$m_k$$, the number of Jordan blocks with size $$k$$, will satisfy $$m_k = s_k - s_{k-1}$$. In the even case, we find that $$m_{n/2} = 2$$, with all other $$m_k$$ zero. In the odd case, we find that $$m_{(n-1)/2} = m_{(n+1)/2} = 1$$, with all other $$m_k$$ zero.
In conclusion, $$J_n^2$$ will have the Jordan structure $$n/2,n/2$$ in the case where $$n$$ is even and $$(n+1)/2,(n-1)/2$$ in the case where $$n \geq 2$$ is odd.
Now, apply this to your question. What is required in order for $$B^2$$ to have an odd number of blocks in its Jordan form? With this you should find that the answer to (a) is no, and the answer to (b) is yes.
• The approach given here is also applicable here and is a faster way to find the Jordan form of $J_n^2$. – Ben Grossmann Jan 22 '20 at 17:32
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2021-06-25 05:14:43
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http://fora.xkcd.com/viewtopic.php?f=17&t=104067
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## How would I prove this? (some trick with factorials?)
For the discussion of math. Duh.
Moderators: gmalivuk, Moderators General, Prelates
4=5
Posts: 2073
Joined: Sat Apr 28, 2007 3:02 am UTC
### How would I prove this? (some trick with factorials?)
I was playing around and empirically identified this relation
x and n are non-negative integers.
[; \frac{1}{(n+x+1)} = \sum_{i=0}^{n} \frac{(-1)^{i}(n+x)!}{x!i!(n-i)!(x+i+1)} ;]
How would I go about proving it?
jestingrabbit
Factoids are just Datas that haven't grown up yet
Posts: 5967
Joined: Tue Nov 28, 2006 9:50 pm UTC
Location: Sydney
### Re: How would I prove this? (some trick with factorials?)
Induction on n.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Parralelex
Posts: 74
Joined: Sun Dec 02, 2012 8:10 am UTC
Location: Location Location
### Re: How would I prove this? (some trick with factorials?)
I put the "fun" in "mathematics".
And then I took it back out.
gmalivuk
GNU Terry Pratchett
Posts: 26830
Joined: Wed Feb 28, 2007 6:02 pm UTC
Location: Here and There
Contact:
### Re: How would I prove this? (some trick with factorials?)
Incidentally, the [math] tags haven't been functional on here for quite some time. However, if you use the TeX The World plugin, putting ;] at the end of what you want TeXified and [; at the beginning will do the trick.
Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.
---
If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome
(he/him/his)
Cauchy
Posts: 602
Joined: Wed Mar 28, 2007 1:43 pm UTC
### Re: How would I prove this? (some trick with factorials?)
I generally try to rewrite conglomerations of factorials as choose functions in hopes of making a counting argument, so I tried to coalesce the sea of factorials on the right-hand side into choose functions. I got it as a product of these, but I never got a counting argument working, and instead I had to rely on some... intriguing choices of substitutions and identities to get the answer. A proof of the identity follows.
Spoiler:
First, the non-integral fraction on the left-hand side will be an annoying target, so I would suggest multiplying through by (n+x+1). The factor of (n+x+1) on the right-hand side can be nestled into the numerator's factorial, producing 1 = \sum_{i=0}^{n} \frac{(-1)^{i}(n+x+1)!}{x!i!(n-i)!(x+i+1)}. From here on, we work on evaluating the right side.
The x!i!, alongside the (x+i+1), suggests (x+i choose x). (x+i choose x) = (x+i)!/(x!i!), so x!i! = (x+i)!/(x+i choose x), and hence the sum equals sum_{i=0}^n { (-1)^i (n+x+1)! (x+i choose x)}/{ (x+i+1)! (n-i)!}. The remaining three factorials make a choose function as well, so we get sum_{i=0}^n (-1)^i (n+x+1 choose x+i+1) (x+i choose x).
Here's where things get a little wonky. The fact that the bottom of one choose function is so close to the top of the other without being equal is irksome, so I push to make them the same. I tried factoring out a quotient from one of the choose functions (i.e. (n+x+1 choose x+i+1) = (n+x choose x+i) (n+x+1)/(x+i+1)), but the factor of x+i+1 in the denominator never went away and messed up the sum no matter what I tried. I then tried the next method I had of reducing the lower number in a choose function by one: the identity (a+1 choose b+1) = sum_{j=0}^a (j choose b). Applying that, we get sum_{i=0}^n (-1)^i sum_{j=0}^{n+x} (j choose x+i) (x+i choose x).
Now the bottom number in one choose function matches the top number in the other, so we can apply this identity: (a choose b) (b choose c) = (a choose c) (a-c choose b-c). This gives us sum_{i=0}^n (-1)^i sum_{j=0}^{n+x} (j choose x) (j-x choose i). The benefit here is that we have restricted the i to a single choose function, where before it was in both.
To capitalize on this, we swap the order of summation. Moving the sum_j to the outside gives us sum_{j=0}^{n+x} sum_{i=0}^n (-1)^i (j choose x) (j-x choose i), and then we factor the (j choose x) out of the inner sum, yielding sum_{j=0}^{n+x} (j choose x) sum_{i=0}^n (-1)^i (j-x choose i).
This inner sum, sum_{i=0}^n (-1)^i (j-x choose i), is an identity as well. sum_{i=0}^n (-1)^i (a choose i) = (-1)^n (a-1 choose n), unless a = 0, in which case the sum is 1. (I had to derive this myself; it can be derived using (a choose b) = (a-1 choose b-1) + (a-1 choose b), which holds when a >= 1, and observing that the sum telescopes.) So sum_{i=0}^n (-1)^i (j-x choose i) = (-1)^n (j-x-1 choose n), except when j = x. Substituting this into our expression, we get [sum_{j=0 to n+x, j =/= x} (j choose x) (-1)^n (j-x-1 choose n) ] + (x choose x) * 1 , where the last term is the j = x term.
Now, as long as 0 <= j <= n+x, j-x-1 is strictly less than n, so (j-x-1 choose n) = 0. Then, [sum_{j=0 to n+x, j =/= x} (j choose x) (-1)^n (j-x-1 choose n) ] + (x choose x) * 1 simplifies to [sum_{j=0 to n+x, j =/= x} (j choose x) (-1)^n 0 ] + 1 * 1 = 0 + 1 = 1, as desired.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
Thanks, skeptical scientist, for knowing symbols and giving them to me.
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2019-08-23 02:44:32
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http://mathoverflow.net/revisions/72147/list
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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
2 now 0<=i<e, 0<=j<d
I'm just curious about the polynomial $\det (x_k^iy_k^j)_{1\leq x_k^iy_k^j)_{0\leq i\leq dd-1, 1\leq 0\leq j\leq ee-1, 1\leq k\leq de}$ (determinant of $de\times de$-matrix, $x_k$, $y_k$ are all independent variables). Is anything known about its factorization on irreducible polynomials?
The question is based only on my own interest.
1
# Is this polynomial irreducible?
I'm just curious about the polynomial $\det (x_k^iy_k^j)_{1\leq i\leq d, 1\leq j\leq e, 1\leq k\leq de}$ (determinant of $de\times de$-matrix, $x_k$, $y_k$ are all independent variables). Is anything known about its factorization on irreducible polynomials?
The question is based only on my own interest.
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2013-06-20 07:18:15
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https://first.wpi.edu/wpilib/allwpilib/docs/release/cpp/classfrc_1_1AnalogInput.html
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WPILibC++ 2021.3.1
#include <frc/AnalogInput.h>
## Public Member Functions
int GetValue () const
Get a sample straight from this channel. More...
int GetAverageValue () const
Get a sample from the output of the oversample and average engine for this channel. More...
double GetVoltage () const
Get a scaled sample straight from this channel. More...
double GetAverageVoltage () const
Get a scaled sample from the output of the oversample and average engine for this channel. More...
int GetChannel () const
Get the channel number. More...
void SetAverageBits (int bits)
Set the number of averaging bits. More...
int GetAverageBits () const
Get the number of averaging bits previously configured. More...
void SetOversampleBits (int bits)
Set the number of oversample bits. More...
int GetOversampleBits () const
Get the number of oversample bits previously configured. More...
int GetLSBWeight () const
Get the factory scaling least significant bit weight constant. More...
int GetOffset () const
Get the factory scaling offset constant. More...
bool IsAccumulatorChannel () const
Is the channel attached to an accumulator. More...
void InitAccumulator ()
Initialize the accumulator.
void SetAccumulatorInitialValue (int64_t value)
Set an initial value for the accumulator. More...
void ResetAccumulator ()
Resets the accumulator to the initial value.
void SetAccumulatorCenter (int center)
Set the center value of the accumulator. More...
int64_t GetAccumulatorValue () const
int64_t GetAccumulatorCount () const
Read the number of accumulated values. More...
void GetAccumulatorOutput (int64_t &value, int64_t &count) const
Read the accumulated value and the number of accumulated values atomically. More...
double PIDGet () override
Get the Average value for the PID Source base object. More...
void SetSimDevice (HAL_SimDeviceHandle device)
Indicates this input is used by a simulated device. More...
void InitSendable (SendableBuilder &builder) override
Initializes this Sendable object. More...
Public Member Functions inherited from frc::ErrorBase
ErrorBase (const ErrorBase &)=default
ErrorBaseoperator= (const ErrorBase &)=default
ErrorBase (ErrorBase &&)=default
ErrorBaseoperator= (ErrorBase &&)=default
virtual ErrorGetError ()
Retrieve the current error. More...
virtual const ErrorGetError () const
Retrieve the current error. More...
virtual void ClearError () const
Clear the current error information associated with this sensor.
virtual void SetErrnoError (const wpi::Twine &contextMessage, wpi::StringRef filename, wpi::StringRef function, int lineNumber) const
Set error information associated with a C library call that set an error to the "errno" global variable. More...
virtual void SetImaqError (int success, const wpi::Twine &contextMessage, wpi::StringRef filename, wpi::StringRef function, int lineNumber) const
Set the current error information associated from the nivision Imaq API. More...
virtual void SetError (Error::Code code, const wpi::Twine &contextMessage, wpi::StringRef filename, wpi::StringRef function, int lineNumber) const
Set the current error information associated with this sensor. More...
virtual void SetErrorRange (Error::Code code, int32_t minRange, int32_t maxRange, int32_t requestedValue, const wpi::Twine &contextMessage, wpi::StringRef filename, wpi::StringRef function, int lineNumber) const
Set the current error information associated with this sensor. More...
virtual void SetWPIError (const wpi::Twine &errorMessage, Error::Code code, const wpi::Twine &contextMessage, wpi::StringRef filename, wpi::StringRef function, int lineNumber) const
Set the current error information associated with this sensor. More...
virtual void CloneError (const ErrorBase &rhs) const
virtual bool StatusIsFatal () const
Check if the current error code represents a fatal error. More...
void ClearGlobalErrors ()
Clear global errors.
Public Member Functions inherited from frc::PIDSource
virtual void SetPIDSourceType (PIDSourceType pidSource)
Set which parameter you are using as a process control variable. More...
virtual PIDSourceType GetPIDSourceType () const
Public Member Functions inherited from frc::SendableHelper< AnalogInput >
SendableHelper (const SendableHelper &rhs)=default
SendableHelper (SendableHelper &&rhs)
SendableHelperoperator= (const SendableHelper &rhs)=default
SendableHelperoperator= (SendableHelper &&rhs)
std::string GetName () const
Gets the name of this Sendable object. More...
void SetName (const wpi::Twine &name)
Sets the name of this Sendable object. More...
void SetName (const wpi::Twine &subsystem, const wpi::Twine &name)
Sets both the subsystem name and device name of this Sendable object. More...
std::string GetSubsystem () const
Gets the subsystem name of this Sendable object. More...
void SetSubsystem (const wpi::Twine &subsystem)
Sets the subsystem name of this Sendable object. More...
## Static Public Member Functions
static void SetSampleRate (double samplesPerSecond)
Set the sample rate per channel for all analog channels. More...
static double GetSampleRate ()
Get the current sample rate for all channels. More...
Static Public Member Functions inherited from frc::ErrorBase
static void SetGlobalError (Error::Code code, const wpi::Twine &contextMessage, wpi::StringRef filename, wpi::StringRef function, int lineNumber)
static void SetGlobalWPIError (const wpi::Twine &errorMessage, const wpi::Twine &contextMessage, wpi::StringRef filename, wpi::StringRef function, int lineNumber)
static Error GetGlobalError ()
Retrieve the last global error.
static std::vector< ErrorGetGlobalErrors ()
Retrieve all global errors.
## Static Public Attributes
static constexpr int kAccumulatorModuleNumber = 1
static constexpr int kAccumulatorNumChannels = 2
static constexpr int kAccumulatorChannels [kAccumulatorNumChannels] = {0, 1}
## Friends
class AnalogTrigger
class AnalogGyro
class DMA
class DMASample
Protected Member Functions inherited from frc::SendableHelper< AnalogInput >
void SetName (const wpi::Twine &moduleType, int channel)
Sets the name of the sensor with a channel number. More...
void SetName (const wpi::Twine &moduleType, int moduleNumber, int channel)
Sets the name of the sensor with a module and channel number. More...
void AddChild (std::shared_ptr< Sendable > child)
Protected Attributes inherited from frc::ErrorBase
Error m_error
Protected Attributes inherited from frc::PIDSource
PIDSourceType m_pidSource = PIDSourceType::kDisplacement
## Detailed Description
Connected to each analog channel is an averaging and oversampling engine. This engine accumulates the specified ( by SetAverageBits() and SetOversampleBits() ) number of samples before returning a new value. This is not a sliding window average. The only difference between the oversampled samples and the averaged samples is that the oversampled samples are simply accumulated effectively increasing the resolution, while the averaged samples are divided by the number of samples to retain the resolution, but get more stable values.
## Constructor & Destructor Documentation
explicit
Parameters
channel The channel number on the roboRIO to represent. 0-3 are on-board 4-7 are on the MXP port.
## ◆ GetAccumulatorCount()
Read the number of accumulated values.
Read the count of the accumulated values since the accumulator was last Reset().
Returns
The number of times samples from the channel were accumulated.
## ◆ GetAccumulatorOutput()
void frc::AnalogInput::GetAccumulatorOutput ( int64_t & value, int64_t & count ) const
Read the accumulated value and the number of accumulated values atomically.
This function reads the value and count from the FPGA atomically. This can be used for averaging.
Parameters
value Reference to the 64-bit accumulated output. count Reference to the number of accumulation cycles.
## ◆ GetAccumulatorValue()
Read the value that has been accumulating. The accumulator is attached after the oversample and average engine.
Returns
The 64-bit value accumulated since the last Reset().
## ◆ GetAverageBits()
Get the number of averaging bits previously configured.
This gets the number of averaging bits from the FPGA. The actual number of averaged samples is 2^bits. The averaging is done automatically in the FPGA.
Returns
Number of bits of averaging previously configured.
## ◆ GetAverageValue()
Get a sample from the output of the oversample and average engine for this channel.
The sample is 12-bit + the bits configured in SetOversampleBits(). The value configured in SetAverageBits() will cause this value to be averaged 2**bits number of samples.
This is not a sliding window. The sample will not change until 2**(OversampleBits + AverageBits) samples have been acquired from the module on this channel.
Use GetAverageVoltage() to get the analog value in calibrated units.
Returns
A sample from the oversample and average engine for this channel.
## ◆ GetAverageVoltage()
Get a scaled sample from the output of the oversample and average engine for this channel.
The value is scaled to units of Volts using the calibrated scaling data from GetLSBWeight() and GetOffset().
Using oversampling will cause this value to be higher resolution, but it will update more slowly.
Using averaging will cause this value to be more stable, but it will update more slowly.
Returns
A scaled sample from the output of the oversample and average engine for this channel.
## ◆ GetChannel()
Get the channel number.
Returns
The channel number.
## ◆ GetLSBWeight()
Get the factory scaling least significant bit weight constant.
Volts = ((LSB_Weight * 1e-9) * raw) - (Offset * 1e-9)
Returns
Least significant bit weight.
## ◆ GetOffset()
Get the factory scaling offset constant.
Volts = ((LSB_Weight * 1e-9) * raw) - (Offset * 1e-9)
Returns
Offset constant.
## ◆ GetOversampleBits()
Get the number of oversample bits previously configured.
This gets the number of oversample bits from the FPGA. The actual number of oversampled values is 2^bits. The oversampling is done automatically in the FPGA.
Returns
Number of bits of oversampling previously configured.
## ◆ GetSampleRate()
static
Get the current sample rate for all channels.
Returns
Sample rate.
## ◆ GetValue()
Get a sample straight from this channel.
The sample is a 12-bit value representing the 0V to 5V range of the A/D converter in the module. The units are in A/D converter codes. Use GetVoltage() to get the analog value in calibrated units.
Returns
A sample straight from this channel.
## ◆ GetVoltage()
Get a scaled sample straight from this channel.
The value is scaled to units of Volts using the calibrated scaling data from GetLSBWeight() and GetOffset().
Returns
A scaled sample straight from this channel.
## ◆ InitSendable()
void frc::AnalogInput::InitSendable ( SendableBuilder & builder )
overridevirtual
Initializes this Sendable object.
Parameters
builder sendable builder
Implements frc::Sendable.
## ◆ IsAccumulatorChannel()
Is the channel attached to an accumulator.
Returns
## ◆ PIDGet()
overridevirtual
Get the Average value for the PID Source base object.
Returns
The average voltage.
Implements frc::PIDSource.
## ◆ SetAccumulatorCenter()
void frc::AnalogInput::SetAccumulatorCenter ( int center )
Set the center value of the accumulator.
The center value is subtracted from each A/D value before it is added to the accumulator. This is used for the center value of devices like gyros and accelerometers to take the device offset into account when integrating.
This center value is based on the output of the oversampled and averaged source from the accumulator channel. Because of this, any non-zero oversample bits will affect the size of the value for this field.
## ◆ SetAccumulatorInitialValue()
void frc::AnalogInput::SetAccumulatorInitialValue ( int64_t value )
Set an initial value for the accumulator.
This will be added to all values returned to the user.
Parameters
initialValue The value that the accumulator should start from when reset.
## ◆ SetAverageBits()
void frc::AnalogInput::SetAverageBits ( int bits )
Set the number of averaging bits.
This sets the number of averaging bits. The actual number of averaged samples is 2^bits.
Use averaging to improve the stability of your measurement at the expense of sampling rate. The averaging is done automatically in the FPGA.
Parameters
bits Number of bits of averaging.
## ◆ SetOversampleBits()
void frc::AnalogInput::SetOversampleBits ( int bits )
Set the number of oversample bits.
This sets the number of oversample bits. The actual number of oversampled values is 2^bits. Use oversampling to improve the resolution of your measurements at the expense of sampling rate. The oversampling is done automatically in the FPGA.
Parameters
bits Number of bits of oversampling.
## ◆ SetSampleRate()
static void frc::AnalogInput::SetSampleRate ( double samplesPerSecond )
static
Set the sample rate per channel for all analog channels.
The maximum rate is 500kS/s divided by the number of channels in use. This is 62500 samples/s per channel.
Parameters
samplesPerSecond The number of samples per second.
## ◆ SetSimDevice()
void frc::AnalogInput::SetSimDevice ( HAL_SimDeviceHandle device )
Indicates this input is used by a simulated device.
Parameters
device simulated device handle
The documentation for this class was generated from the following file:
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2021-10-22 19:54:27
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http://rosalind.info/problems/suggested/140/
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# Suggested problems
## garmin express
Jan. 8, 2020, 11:52 a.m. by jordan john
## Biological Motivation
[Garmin Express][1] is a holistic management tool that supports users of cutting-edge Garmin GPS Device. This ingenious desktop application gives a one-stop hassle-free solution for maintaining Garmin gadgets in perfect health. It facilitates users to set up, register, manage and update all of their Garmin devices.
## Problem
A string is simply an ordered collection of symbols selected from some alphabet and formed into a word; the length of a string is the number of symbols that it contains.
An example of an DNA string (whose alphabet contains the symbols A, C, G, and T) is ATGCTTCAGAAAGGTCTTACG.
Given: A DNA string $s$ of length at most 1000 nucleotides.
Return: Four integers corresponding to the number of times that the symbols A, C, G, and T occur in $s$.
## Sample Dataset
AGCTTTTCATTCTGACTGCAACGGGCAATATGTCTCTGTGTGGATTAAAAAAAGAGTGTCTGATAGCAGC
## Sample Output
20 12 17 21
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2020-01-22 09:22:12
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https://crazyproject.wordpress.com/2011/05/22/verify-that-a-given-element-is-algebraic-over-the-rationals/
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## Verify that a given element is algebraic over the rationals
Show that $\sqrt{2} + \sqrt[3]{3}$ and $\sqrt{2}\sqrt[3]{3}$ are algebraic over $\mathbb{Q}$.
Certainly $(\sqrt{2}\sqrt[3]{3})^6 = 72$, so that $\sqrt{2}\sqrt[3]{3}$ is a root of $x^6 - 72$ and thus is algebraic over $\mathbb{Q}$.
Note that the conjugates of $\sqrt{2}$ are $\alpha_i = \pm \sqrt{2}$, and the conjugates of $\sqrt[3]{3}$ are $\beta_1 = \sqrt[3]{3}$, $\beta_2 = \sqrt[3]{3}(\frac{-1}{2} + \frac{\sqrt{-3}}{2})$, and $\beta_3 = \sqrt[3]{3}(\frac{-1}{2} - \frac{\sqrt{-3}}{2})$. Evidently, $\prod_i \prod_j (x - \alpha_i - \beta_j) = x^6-6x^4-6x^3+12x^2-36x+1$ has $\sqrt{2} + \sqrt[3]{3}$ as a root.
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2017-01-24 21:20:17
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https://projecteuclid.org/euclid.gt/1513732408
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## Geometry & Topology
### Monopole Floer homology and Legendrian knots
Steven Sivek
#### Abstract
We use monopole Floer homology for sutured manifolds to construct invariants of unoriented Legendrian knots in a contact $3$–manifold. These invariants assign to a knot $K⊂Y$ elements of the monopole knot homology $KHM(−Y,K)$, and they strongly resemble the knot Floer homology invariants of Lisca, Ozsváth, Stipsicz, and Szabó. We prove several vanishing results, investigate their behavior under contact surgeries, and use this to construct many examples of nonloose knots in overtwisted $3$–manifolds. We also show that these invariants are functorial with respect to Lagrangian concordance.
#### Article information
Source
Geom. Topol., Volume 16, Number 2 (2012), 751-779.
Dates
Revised: 3 February 2012
Accepted: 30 January 2012
First available in Project Euclid: 20 December 2017
https://projecteuclid.org/euclid.gt/1513732408
Digital Object Identifier
doi:10.2140/gt.2012.16.751
Mathematical Reviews number (MathSciNet)
MR2928982
Zentralblatt MATH identifier
1270.57050
Subjects
Primary: 57M27: Invariants of knots and 3-manifolds 57R58: Floer homology
Secondary: 57R17: Symplectic and contact topology
#### Citation
Sivek, Steven. Monopole Floer homology and Legendrian knots. Geom. Topol. 16 (2012), no. 2, 751--779. doi:10.2140/gt.2012.16.751. https://projecteuclid.org/euclid.gt/1513732408
#### References
• J M Bloom, T Mrowka, P Ozsváth, The Künneth principle in Floer homology, in preparation
• S Boyer, D Lines, Conway potential functions for links in ${\bf Q}$–homology $3$–spheres, Proc. Edinburgh Math. Soc. 35 (1992) 53–69
• J C Cha, C Livingston, KnotInfo: Table of knot invariants, website (2010) Available at \setbox0\makeatletter\@url http://www.indiana.edu/~knotinfo {\unhbox0
• B Chantraine, Lagrangian concordance of Legendrian knots, Algebr. Geom. Topol. 10 (2010) 63–85
• Y Chekanov, Differential algebra of Legendrian links, Invent. Math. 150 (2002) 441–483
• W Chongchitmate, L Ng, An atlas of Legendrian knots, to appear in Exp. Math.
• M Culler, Gridlink: a tool for knot theorists Available at \setbox0\makeatletter\@url http://www.math.uic.edu/~culler/gridlink {\unhbox0
• F Ding, H Geiges, Symplectic fillability of tight contact structures on torus bundles, Algebr. Geom. Topol. 1 (2001) 153–172
• F Ding, H Geiges, Handle moves in contact surgery diagrams, J. Topol. 2 (2009) 105–122
• F Ding, H Geiges, A I Stipsicz, Surgery diagrams for contact $3$–manifolds, Turkish J. Math. 28 (2004) 41–74
• Y Eliashberg, Invariants in contact topology, from: “Proceedings of the International Congress of Mathematicians, Vol. II (Berlin, 1998)”, volume Extra Vol. II (1998) 327–338
• Y Eliashberg, M Fraser, Topologically trivial Legendrian knots, J. Symplectic Geom. 7 (2009) 77–127
• J B Etnyre, On contact surgery, Proc. Amer. Math. Soc. 136 (2008) 3355–3362
• J B Etnyre, K Honda, Knots and contact geometry I: Torus knots and the figure eight knot, J. Symplectic Geom. 1 (2001) 63–120
• J B Etnyre, K Honda, On the nonexistence of tight contact structures, Ann. of Math. 153 (2001) 749–766
• J B Etnyre, D S Vela-Vick, Torsion and open book decompositions, Int. Math. Res. Not. 2010 (2010) 4385–4398
• D Fuchs, Chekanov–Eliashberg invariant of Legendrian knots: existence of augmentations, J. Geom. Phys. 47 (2003) 43–65
• P Ghiggini, Knot Floer homology detects genus-one fibred knots, Amer. J. Math. 130 (2008) 1151–1169
• E Giroux, Convexité en topologie de contact, Comment. Math. Helv. 66 (1991) 637–677
• K Honda, On the classification of tight contact structures I, Geom. Topol. 4 (2000) 309–368
• K Honda, W H Kazez, G Matić, Tight contact structures on fibered hyperbolic $3$–manifolds, J. Differential Geom. 64 (2003) 305–358
• A Juhász, Holomorphic discs and sutured manifolds, Algebr. Geom. Topol. 6 (2006) 1429–1457
• T Kálmán, Contact homology and one parameter families of Legendrian knots, Geom. Topol. 9 (2005) 2013–2078
• Y Kanda, On the Thurston–Bennequin invariant of Legendrian knots and nonexactness of Bennequin's inequality, Invent. Math. 133 (1998) 227–242
• P B Kronheimer, T S Mrowka, Monopoles and contact structures, Invent. Math. 130 (1997) 209–255
• P Kronheimer, T Mrowka, Monopoles and three-manifolds, New Math. Monogr. 10, Cambridge Univ. Press (2007)
• P Kronheimer, T Mrowka, Knots, sutures, and excision, J. Differential Geom. 84 (2010) 301–364
• P Kronheimer, T Mrowka, P Ozsváth, Z Szabó, Monopoles and lens space surgeries, Ann. of Math. 165 (2007) 457–546
• Y Lekili, Heegaard Floer homology of broken fibrations over the circle
• P Lisca, P Ozsváth, A I Stipsicz, Z Szabó, Heegaard Floer invariants of Legendrian knots in contact three-manifolds, J. Eur. Math. Soc. 11 (2009) 1307–1363
• P Lisca, A I Stipsicz, Ozsváth–Szabó invariants and tight contact three-manifolds I, Geom. Topol. 8 (2004) 925–945
• T Mrowka, Y Rollin, Legendrian knots and monopoles, Algebr. Geom. Topol. 6 (2006) 1–69
• T Mrowka, Y Rollin, Contact invariants and monopole Floer homology, in preparation
• K Niederkrüger, C Wendl, Weak symplectic fillings and holomorphic curves, Ann. Sci. Éc. Norm. Supér. (4) 44 (2011) 801–853
• P Ozsváth, A I Stipsicz, Contact surgeries and the transverse invariant in knot Floer homology, J. Inst. Math. Jussieu 9 (2010) 601–632
• P Ozsváth, Z Szabó, D Thurston, Legendrian knots, transverse knots and combinatorial Floer homology, Geom. Topol. 12 (2008) 941–980
• B Sahamie, Dehn twists in Heegaard Floer homology, Algebr. Geom. Topol. 10 (2010) 465–524
• A I Stipsicz, V Vértesi, On invariants for Legendrian knots, Pacific J. Math. 239 (2009) 157–177
• C H Taubes, Embedded contact homology and Seiberg–Witten Floer cohomology V, Geom. Topol. 14 (2010) 2961–3000
• C Wendl, A hierarchy of local symplectic filling obstructions for contact $3$–manifolds
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2019-09-20 20:40:18
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https://ir.cwi.nl/pub/30736
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In this article, we study the geometry of units and ideals of cyclotomic rings and derive an algorithm to find a mildly short vector in any given cyclotomic ideal lattice in quantum polynomial time, under some plausible number-theoretic assumptions. More precisely, given an ideal lattice of the cyclotomic ring of conductor m, the algorithm finds an approximation of the shortest vector by a factor exp $\mathrm{\left(Õ\left(}\sqrt{m}$ )). This result exposes an unexpected hardness gap between these structured lattices and general lattices: The best known polynomial time generic lattice algorithms can only reach an approximation factor exp $\mathrm{\left(Õ\left(}m$ )). Following a recent series of attacks, these results call into question the hardness of various problems over structured lattices, such as Ideal-SVP and Ring-LWE, upon which relies the security of a number of cryptographic schemes. NOTE. This article is an extended version of a conference paper [11]. The results are generalized to arbitrary cyclotomic fields. In particular, we also extend some results of Reference [10] to arbitrary cyclotomic fields. In addition, we prove the numerical stability of the method of Reference [10]. These extended results appeared in the Ph.D. dissertation of the third author [46].
, ,
doi.org/10.1145/3431725
Journal of the ACM
Centrum Wiskunde & Informatica, Amsterdam, The Netherlands
Cramer, R.J.F, Ducas, L, & Wesolowski, B.P.C. (2021). Mildly short vectors in cyclotomic ideal lattices in quantum polynomial time. Journal of the ACM, 68(2). doi:10.1145/3431725
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2021-08-01 08:41:33
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http://tex.stackexchange.com/questions/28363/how-to-pick-a-plane-inside-a-cube-in-pst-3dplot?answertab=oldest
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# How to pick a plane inside a cube in pst-3dplot?
I want to pick and highlight a diagonal plane of a cube. I do not want to color it but use some kind of fill-style so that I can print it on a black and white printer. The following is my solution using pst-3dplot treating the plane as a parametric surface. This cannot be the only way to do this. Is there another (better?) way? The following is the code.
\documentclass[12pt]{report}
\usepackage{pst-3dplot}
\begin{document}
\psset{coorType=2,Alpha=90,IIIDticks}
\begin{pspicture}(-3,-2.5)(3,4.25)
\pstThreeDCoor[IIIDticks,IIIDticksize=0]
\pstThreeDBox[hiddenLine](0,0,0)(0,0,2)(2,0,0)(0,2,0)
\parametricplotThreeD[plotstyle=curve,yPlotpoints=20](0,2)(0,2){t t u}
\end{pspicture}
\end{document}
-
This looks like a different way. To create the plane I used the \pstThreeDSquare command. From the vector (0,0,0) the vectors (0,0,2) and (2,2,0) span the square. The fillstyle is hlines, vlines, or croshatch. Hatchwidth is the thickness of each line and hatchsep is the distance between adjacent lines.
\documentclass[12pt]{report}
\usepackage{xcolor}
\usepackage{pst-3dplot}
\begin{document}
\psset{coorType=2,Alpha=90,IIIDticks}
\begin{pspicture}(-3,-2.5)(3,4.25)
\pstThreeDCoor[coorType=2,IIIDticks,IIIDticksize=0,xMin=-2,xMax=4,yMin=-2,yMax=4,zMin=-1,zMax=3]{\psset{coorType=2,fillstyle=vlines,hatchwidth=.2mm,hatchsep=.55mm}\pstThreeDSquare(0,0,0)(0,0,2)(2,2,0)}
\pstThreeDBox[hiddenLine](0,0,0)(0,0,2)(2,0,0)(0,2,0)
\end{pspicture}
\end{document}
PS: you should be able to color it (solid) shades of gray and get it to print on a black and white printer.
-
Thank you very much. I like your solution better than mine. It gives me the option of several fill-styles. Besides, you should not expect someone to know how to parametrize surfaces to plot something simple like this. Ps. I tried to edit the 7th line of your code as it can be broken into 3 lines, but couldn't do it. – Sony Sep 14 '11 at 10:46
Your welcome. Yes, most people couldn't come up with your parametrized solution. The fact \pstThreeDSquare is used to creates a shape that doesn't look rectangular is odd and the format of 3 vectors needed to generate it takes some getting used to. It's worth reiterating that if you use the xcolor package you can create lighter versions of solid colors like, say, black!45. You can adjust the number (0-100) until it shows up on your printer to your liking. – DJP Sep 14 '11 at 14:33
I played with tikz and tikz-3dplot packages a little bit today. The following is a solution using those packages.
\documentclass[11pt]{report}
\usepackage{tikz}
\usepackage{tikz-3dplot}
\begin{document}
\tdplotsetmaincoords{60}{110}
\begin{tikzpicture}[scale=1.5,tdplot_main_coords]
\draw[thick,->] (0,0,0) -- (3,0,0) node[anchor=north east]{$x$};
\draw[thick,->] (0,0,0) -- (0,3,0) node[anchor=north west]{$y$};
\draw[thick,->] (0,0,0) -- (0,0,3) node[anchor=south]{$z$};
\draw[fill, color=blue!30, draw=black,line width=1pt]
(2,2,2) -- (0,0,2) -- (0,0,0) -- (2,2,0) -- cycle;
\draw[line width=1pt]
(0,0,0) -- (0,0,2) -- (2,0,2) -- (2,0,0) -- (0,0,0) -- (0,2,0) -- (0,2,2) -- (0,0,2)
(2,0,2) -- (2,2,2) -- (2,2,0) -- (2,0,0)
(2,2,0) -- (0,2,0)
(2,2,2) -- (0,2,2);
\end{tikzpicture}
\end{document}
-
If you put \usetikzlibrary{patterns} in your preamble and replace color=blue!30 with pattern=north east lines, you'll get a hatched pattern, which would work well for printing. – Jake Sep 15 '11 at 1:38
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2014-08-20 12:41:02
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http://tex.stackexchange.com/unanswered?fromProtectedNotice=true
|
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### Speeding up compilation using precompiled preamble with LuaTeX
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I have some problems using wxLua in LuaLateX. This works: lua simpleui.lua (print instead of tex.print) But if i try this: lualatex wx.tex it doesn't work. \documentclass{article} ...
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2014-07-23 10:19:47
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http://mathoverflow.net/questions/95867/covering-a-cube-with-a-square
|
# Covering a Cube with a Square
Suppose you are given a single unit square, and you would like to completely cover the surface of a cube by cutting up the square and pasting it onto the cube's surface.
Q1. What is the largest cube that can be covered by a $1 \times 1$ square when cut into at most $k$ pieces?
The case $k=1$ has been studied, probably earlier than this reference: "Problem 10716: A cubical gift," American Mathematical Monthly, 108(1):81-82, January 2001, solution by Catalano-Johnson, Loeb, Beebee.
(This was discussed in an MSE Question.) The depicted solution results in a cube edge length of $1/(2\sqrt{2}) \approx 0.35$.
As $k \to \infty$, there should be no wasted overlaps in the covering of the 6 faces, and so the largest cube covered will have edge length $1/\sqrt{6} \approx 0.41$. What partition of the square leads to this optimal cover?
Q2. For which value of $k$ is this optimal reached?
I have not found literature on this problem for $k>1$, but it seems likely it has been explored. Thanks for any pointers!
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I wonder if anyone in the packaging industry has the answer. – Steven Gubkin May 4 '12 at 0:37
I believe there was a "Mathematical Games" column on dissections which had a Greek cross rearranged into a square. Perhaps that article also mentioned this problem? Gerhard "Testing Your Martin Gardner Fu" Paseman, 2012.05.04 – Gerhard Paseman May 4 '12 at 15:51
You can cut a $\sqrt{6}\times\sqrt{6}$ square into 24 pieces that then cover the $1\times1\times1$ cube. Two triangles from the figure below plus one parallelogram make up one $1\times1$square. parts of pieces sticking out to the left can obviously fit back in the right, so 18 pieces, plus 6 parts sticking out equals 24. You can improve on this by stitching pieces across the cube edge to make one bent piece and by stitching some of the parallelograms back to the triangles.
[Added by O'Rourke:] Just to make Yoav's construction more explicit, here is how two triangles and a parallelogram fit together to form a $1 \times 1$ square:
[Added by Kallus:] Here's an illustration of a construction similar to Fedja's construction but with only five pieces. The first figure is the $\sqrt{6}\times\sqrt{6}$ square. The second is the $2\times3$ rectangle, which we fold into a cube by taking away the two yellow squares, folding the remainder, and adding the squares as the two missing faces.
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Brilliant!! :-) – Joseph O'Rourke May 3 '12 at 23:51
Actually, any two polygons of the same area are equidecomposable and the surface of the cube can be unfolded into a polygon, so the result is nice but not terribly surprising. Of course, the question about the minimal number of pieces remains. – fedja May 4 '12 at 0:15
Assuming that "pieces" mean "connected polygonal pieces", we can take a 3 by 2 rectangle, cut it into a T-shape and two unit squares and then use the standard "sliding cut" to turn it into a square, giving the total of 6 pieces to cover the unit cube. Can we do better? – fedja May 4 '12 at 0:34
OK. Posted to soon earlier. The change from the T tetromino to the S tetromino actually allows going down to five pieces instead of six. – Yoav Kallus May 4 '12 at 4:21
Wow! $\mbox{}$ – Joseph O'Rourke May 4 '12 at 10:18
Sorry, this is an answer to an other question. (I did not read the question carefully.)
Question: For which $k$, $k$ squares can tile the surface of cube.
Answer: $k=6\cdot(n^2+m^2)$.
Here is a tiling with $k=30$, $n=1$ and $m=2$.
It is obvious if the tiling is vertex-to-vertex.
If the tiling is not vertex-to-vertex, you get a closed geodesic formed by overlaping sides. Then you can shift squares on one side of the geodesic to make the tiling "more vertex-to-vertex". Repeating this operation you can make the tiling to be vertex-to-vertex.
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@Anton: Did you intend $n \ge 1 , m \ge 1$, or, say, $n \ge 1 , m \ge 0$ ? – Joseph O'Rourke May 3 '12 at 18:03
Yes, $k$ has to be positive; so $n\ge 1$ and $m\ge 0$. – Anton Petrunin May 3 '12 at 18:33
Awesome ! – Steven Gubkin May 3 '12 at 19:05
@Anton: Sorry to be slow :-/, but could you describe a partition of the square into 6 pieces that exactly cover the cube? – Joseph O'Rourke May 3 '12 at 19:21
@Joseph: the partition into faces (the cube has 6 faces). – Anton Petrunin May 3 '12 at 19:35
No promises that these are optimal, but here are some lower bounds:
With $k=2$, side length $3/8=0.375$ (with one piece flipped over), and with $k=3$, side length $2/5=0.4$:
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2014-08-21 20:15:43
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http://www.math.toronto.edu/courses/apm346h1/20129/L16.html
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### Other Fourier series
#### Fourier series for even and odd functions
In the previous Lecture 15 we proved the completeness of the system of functions $$\Bigl\{\frac{1}{2},\qquad \cos (\frac{\pi nx}{l}), \qquad \sin (\frac{\pi nx}{l}) \quad n=1,\ldots\Bigr\} \label{eq-1}$$ on interval $J:=[x_0,x_1]$ with $(x_1-x_0)=2l$. In other words we proved that any function $f(x)$ on this interval could be decomposed into Fourier series $$f(x)= \frac{1}{2}a_0 + \sum_{n=1}^\infty\bigl( a_n\cos (\frac{\pi nx}{l})+b_n \sin (\frac{\pi nx}{l}) \bigr)\label{eq-2}$$ with coefficients calculated according to (14.7) \begin{align} a_n=& \frac{1}{l}\int_J f(x)\cos (\frac{\pi nx}{l})\,dx \qquad n=0,1,2,\ldots,\label{eq-3}\\ b_n= & \frac{1}{l}\int_J f(x)\sin (\frac{\pi nx}{l})\,dx \qquad n=1,2,\ldots0,\label{eq-4}\\ \end{align} and satisfying Parseval's equality $$\frac{l}{2}|a_0|^2 +\sum_{n=1}^\infty l\bigl( |a_n|^2+|b_n |^2 \bigr)=\int_J |f(x)|^2\,dx. \label{eq-5}$$
Now we consider some other orthogonal systems and prove their completeness. To do this we first prove
Lemma 1. Let $J$ be a symmetric interval: $J=[-l,l]$. Then
1. $f(x)$ is even iff $b_n=0$ $\forall n=1,2,\ldots$.
2. $f(x)$ is odd iff $a_n=0$ $\forall n=0,1,2,\ldots$.
Proof. (a) Note that $\cos (\frac{\pi nx}{l})$ are even functions and $\sin (\frac{\pi nx}{l})$ are odd functions. Therefore if $b_n=0$ $\forall n=1,2,\ldots$ then only decomposition (\ref{eq-2}) contains only even functions and $f(x)$ is even. Conversely, if $f(x)$ is an even function then integrand in (\ref{eq-4}) is an odd function and its integral over symmetric interval is $0$.
(b) Statement (b) is proven in the similar way.
#### $\cos$-Fourier series
Let us consider function $f(x)$ on the interval $[0,l]$. Let us extend it as an even function on $[-l,l]$ so $f(x):=f(-x)$ for $x\in [-l,0]$ and decompose it into full Fourier series (\ref{eq-2}); however $\sin$-terms disappear and we arrive to decomposition $$f(x)= \frac{1}{2}a_0 + \sum_{n=1}^\infty a_n\cos (\frac{\pi nx}{l}). \label{eq-6}$$ This is decomposition with respect to orthogonal system $$\Bigl\{\frac{1}{2},\qquad \cos (\frac{\pi nx}{l}) \quad n=1,\ldots\Bigr\}. \label{eq-7}$$
Its coefficients are calculated according to (\ref{eq-3}) but here integrands are even functions and we can take interval $[0,l]$ instead of $[-l,l]$ and double integrals: $$a_n= \frac{2}{l}\int_0^l f(x)\cos (\frac{\pi nx}{l})\,dx \qquad n=0,1,2,\ldots,\label{eq-8}.$$ Also (\ref{eq-5}) becomes $$\frac{l}{4}|a_0|^2 +\sum_{n=1}^\infty \frac{l}{2} |a_n|^2= \int_0^l |f(x)|^2\,dx. \label{eq-9}$$
The sum of this Fourier series is $2l$-periodic. Note that even and then periodic continuation does not introduce new jumps.
#### $\sin$-Fourier series
Let us consider function $f(x)$ on the interval $[0,l]$. Let us extend it as an odd function on $[-l,l]$ so $f(x):=-f(-x)$ for $x\in [-l,0]$ and decompose it into full Fourier series (\ref{eq-2}); however $\cos$-terms disappear and we arrive to decomposition $$f(x)= \sum_{n=1}^\infty b_n\sin (\frac{\pi nx}{l}). \label{eq-10}$$ This is decomposition with respect to orthogonal system $$\Bigl\{ \sin (\frac{\pi nx}{l}) \quad n=1,\ldots\Bigr\}. \label{eq-11}$$
Its coefficients are calculated according to (\ref{eq-4}) but here integrands are even functions and we can take interval $[0,l]$ instead of $[-l,l]$ and double integrals: $$b_n= \frac{2}{l}\int_0^l f(x)\sin (\frac{\pi nx}{l})\,dx \qquad n=1,2,\ldots. \label{eq-12}.$$ Also (\ref{eq-5}) becomes $$\sum_{n=1}^\infty \frac{l}{2} |b_n|^2= \int_0^l |f(x)|^2\,dx. \label{eq-13}$$
The sum of this Fourier series is $2l$-periodic. Note that odd and then periodic continuation does not introduce new jumps iff $f(0)=f(l)=0$.
#### $\sin$-Fourier series with half-integers
Let us consider function $f(x)$ on the interval $[0,l]$. Let us extend it as an even with respect to $x=l$ function on $[0,2l]$ so $f(x):=f(2l-x)$ for $x\in [l,2l]$; then we make an odd continuation to $[-2l,2l]$ and decompose it into full Fourier series (\ref{eq-2}) but with $l$ replaced by $2l$; however $\cos$-terms disappear and we arrive to decomposition \begin{equation*} f(x)= \sum_{n=1}^\infty b'_n\sin (\frac{\pi nx}{2l}). \end{equation*}
Then $f(2l-x)= \sum_{n=1}^\infty b'_n\sin (\frac{\pi nx}{2l})(-1)^{n+1}$ and since $f(x)=f(2l-x)$ due to original even continuation we conclude that $b'_n=0$ as $n=2m$ and we arrive to $$f(x)= \sum_{n=0}^\infty b_n\sin (\frac{\pi (2n+1)x}{2l}) \label{eq-14}$$ with $b_n:=b'_{2n+1}$ where we replaced $m$ by $n$.
This is decomposition with respect to orthogonal system $$\Bigl\{ \sin (\frac{\pi (2n+1)x}{2l}) \quad n=1,\ldots\Bigr\}. \label{eq-15}$$
Its coefficients are calculated according to (\ref{eq-12}) (with $l$ replaced by $2l$) but here we can take interval $[0,l]$ instead of $[0,2l]$ and double integrals: $$b_n= \frac{2}{l}\int_0^l f(x)\sin (\frac{\pi (2n+1)x}{2l})\,dx \qquad n=0,2,\ldots. \label{eq-16}.$$ Also (\ref{eq-13}) becomes $$\sum_{n=0}^\infty \frac{l}{2} |b_n|^2= \int_0^l |f(x)|^2\,dx. \label{eq-17}$$
The sum of this Fourier series is $4l$-periodic. Note that odd and then periodic continuation does not introduce new jumps iff $f(0)=0$.
#### Fourier series in complex form
Consider (\ref{eq-2})--(\ref{eq-5}). Plugging \begin{align*} &\cos(\frac{\pi n x}{l})=\frac{1}{2}e^{\frac{i\pi n x}{l}}+\frac{1}{2}e^{-\frac{i\pi n x}{l}}\\ &\sin(\frac{\pi n x}{l})=\frac{1}{2i}e^{\frac{i\pi n x}{l}}-\frac{1}{2i}e^{-\frac{i\pi n x}{l}} \end{align*} and separating terms with $n$ and $-n$ and replacing in the latter $-n$ by $n=-1,-2,\ldots$ we get $$f(x)= \sum_{n=-\infty}^\infty c_n e^{\frac{i\pi nx}{l}} \label{eq-18}$$ with $c_0=\frac{1}{2}a_0$, $c_n = \frac{1}{2}(a_n -i b_n)$ as $n=1,2,\ldots$ and $c_n = \frac{1}{2}(a_{-n} +i b_{-n})$ as $n=-1,-2,\ldots$ which could be written as $$c_n= \frac{1}{2l}\int_J f(x)e^{-\frac{i\pi n x}{l}}\,dx \qquad n=\ldots,-2, -1, 0,1,2,\ldots.\label{eq-19}$$ Parseval's equality becomes $$2l\sum_{n=-\infty}^\infty |c_n|^2= \int\_J |f(x)|^2\,dx. \label{eq-20}$$ One can see easily that the system $$\Bigl\{X_n:=e^{\frac{i\pi nx}{l}} \quad \ldots,-2, -1, 0,1,2,\ldots\Bigr\} \label{eq-21}$$ on interval $J:=[x_0,x_1]$ with $(x_1-x_0)=2l$ is orthogonal: $$\int_J X_n(x)\bar{X_m(x)}\,dx = 2l\delta_{mn}. \label{eq-22}$$
Remark.
1. All our formulae are due to (Lecture 14) but we need the completeness of the systems and those are due to compleness of the system (\ref{eq-1}) established in Lecture 15.
2. Recall that with periodic boundary conditions all eigenvalues $(\frac{\pi n }{l})^2$ are of multiplicity $2$ i.e. the corresponding eigenspace (consisting of all eigenfunctions with the given eigenvalue) has dimension $2$ and $\{\cos (\frac{\pi n x}{l}), \sin (\frac{\pi n x}{l})\}$ and $\{e^ {\frac{i\pi n x}{l}}, e^{-\frac{\pi n x}{l}}\}$ are just two different orthogonal basises in this eigenspace.
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2019-10-18 15:58:09
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https://www.learnatnoon.com/s/category/cbse/class-11/ncert-exemplar-class-11/physics-ncert-exemplar-class-11/
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Physics
### A ray of light incident at an angle θ on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5o and the prism is made of a material of refractive index 1.5, the angle of incidence isa) 7.5ob) 5oc) 15od) 2.5o
Answer: a) 7.5o The distance between the refracting surfaces is negligible with thin prisms, thus the prism angle (A) is very small. Because A = r1 + r2, if A is tiny, both r1 and r2 will be little...
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### The velocity-displacement graph of a particle is shown in the figure. a) Write the relation between v and x. b) Obtain the relation between acceleration and displacement and plot it.
a) Consider the point P(x,v) at any time t on the graph such that angle ABO is θ such that tan θ = AQ/QP = (v0-v)/x = v0/x0 When the velocity decreases from v0 to zero during the displacement, the...
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### A man runs across the roof-top of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is of a lower height than the first. If his speed is 9 m/s, the distance between the two buildings is 10 m and the height difference is 9 m, will he be able to land on the next building?
For a free fall at 9m, the horizontal distance covered by the man should be at least 10 m. u = 0 a = 10 m/s2 s = 9 m t = t s = ut + 1/2 at2 Substituting the values, we get t = √9/3 = 3/√5 sec The...
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### A ball is dropped and its displacement vs time graph is as shown in the figure where displacement x is from the ground and all quantities are positive upwards. a) Plot qualitatively velocity vs time graph b) Plot qualitatively acceleration vs time graph
a) At t=0 and v=0 , v-t graph is: b) At x = 0, a-t graph is:
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### A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time.
V = v1 = ? U = 0 h = 45 m a = g t = t V = u + at v1 = 0 + gt v1 = gt Therefore, when the ball is thrown upward, v1 = -gt V = v2 u = 40 m/s a = g t = t V = u + at v2 = 40 – gt The relative velocity...
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### It is a common observation that rain clouds can be at about a kilometre altitude above the ground. a) If a rain drop falls from such a height freely under gravity, what will be its speed? Also, calculate in km/h b) A typical rain drop is about 4 mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits ground. c) Estimate the time required to flatten the drop. d) Rate of change of momentum is force. Estimate how much force such a drop would exert on you. e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is 5 cm.
a) Velocity attained by the rain drop which is falling freely through the height h is: v2 = u2 – 2g(-h) As u = 0 v = √2gh = 100√2 m/s = 510 km/h b) Diameter of the drop, d = 2r = 4 mm Radius of the...
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### A motor car moving at a speed of 72 km/h cannot come to a stop in less than 3 s while for a truck this time interval is 5 s. On a highway the car is behind the truck both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto the truck. Human response time is 0.5 s.
For truck, u = 20 m/s v = 0 a = ? t = 5s v = u + at a = 4 m/s2 For car, t = 3 s u = 20 m/s v = 0 a = ac v = u + at ac = -20/3 m/s2 Let s be the distance between the car and the truck when the truck...
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### A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by v(t) = 2t (3 – t); 0
a) For maximum velocity v(t) dv(t)/dt = 0 Substituting the value for v, we get t = 1.5 seconds b) For average velocity = total distance/time taken Average velocity = 3 m And the average velocity is...
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### A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval. The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is +15m at t = 2s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.
Let the speed of ball 1 = u1 = 2u m/s Then the speed of ball 2 = u2 = u m/s The height covered by ball 1 before coming to rest = h1 The height covered by ball 2 before coming to rest = h2 We know...
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### A bird is tossing between two cars moving towards each other on a straight road. One car has a speed of 18 m/h while the other has the speed of 27 km/h. The bird starts moving from first car towards the other and is moving with the speed of 36 km/h and when the two cars were separated by 36 km. What is the total distance covered by the bird? What is the total displacement of the bird?
The relative speed of the cars = 27 + 18 = 45 km/h When the two cars meet together, time t is given as t = distance between cars/relative speed of cars = 36/(27+18) t = 4/5 h Therefore, distance...
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### A particle executes the motion described by x(t) = x0 (1 – e-γt) where t ≥ 0, x0 > 0 a) Where does the particles start and with what velocity? b) Find maximum and minimum values of x(t), v(t), a(t). Show that x(t) and a(t) increase with time and v(t) decreases with time.
a) x(t) = x0 (1 – e-γt) v(t) = dx(t)/dt = +x0 γ e-γt a(t) = dv/dt = x0 γ2 e-γt v(0) = x0 γ b) x(t) is minimum at t = 0 since t = 0 and [x(t)]min = 0 x(t) is maximum at t = ∞ since t = ∞ and...
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### An object falling through a fluid is observed to have acceleration given by a = g – bv where g = gravitational acceleration and b is constant. After a long time of release, it is observed to fall with constant speed. What must be the value of constant speed?
The concept used in this question will be based on the behaviour of a spherical object when it is dropped through a viscous fluid. When a spherical body of radius r is dropped, it is first...
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### Give example of a motion where x>0, v<0, a>0 at a particular instant.
Let the motion be represented as: x(t) = A + Be– γ t Let A>B and γ >0 Velocity is x(t) = dx/dt = -Be– γ t Acceleration is a(t) = dx/dt = B γ 2e– γ t Therefore, it can be said that x(t) > 0,...
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### Give examples of a one-dimensional motion where a) the particle moving along positive x-direction comes to rest periodically and moves forward b) the particle moving along positive x-direction comes to rest periodically and moves backwardπ
When an equation has sine and cosine functions, the nature is periodic. a) When the particle is moving in positive x-direction, it is given as t > sin t When the displacement is as a function of...
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### A uniformly moving cricket ball is turned back by hitting it with a bat for a very short time interval. Show the variation of its acceleration with taking acceleration in the backward direction as positive.
The force which is generated by the bat is known as impulsive force. When the effect of gravity is ignored, it can be said that the ball moves with a uniform speed horizontally and returns back to...
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### Refer to the graphs below and match the following:
Graph Characteristics a) i) has v > 0 and a < 0 throughout b) ii) has x > 0 throughout and has a point with v = 0 and a point with a = 0 c) iii) has a point with zero displacement for t...
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### A ball is bouncing elastically with a speed 1 m/s between walls of a railway compartment of size 10 m in a direction perpendicular to walls. The train is moving at a constant velocity of 10 m/s parallel to the direction of motion of the ball. As seen from the ground, a) the direction of motion of the ball changes every 10 seconds b) speed of ball changes every 10 seconds c) average speed of ball over any 20 seconds intervals is fixed d) the acceleration of ball is the same as from the train
The correct option is b) speed of ball changes every 10 seconds, c) average speed of ball over any 20 seconds intervals is fixed, and d) the acceleration of the ball is the same as from the train
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### A spring with one end attached to a mass and the other to a rigid support is stretched and released. a) magnitude of acceleration, when just released is maximum b) magnitude of acceleration, when at equilibrium position is maximum c) speed is maximum when mass is at equilibrium position d) magnitude of displacement is always maximum whenever speed is minimum
The correct answer is a) magnitude of acceleration, when just released is maximum and c) speed is maximum when mass is at equilibrium position
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### For the one-dimensional motion, describe by x = t – sint a) x(t)>0 for all t>0 b) v(t)>0 for all t>0 c) a(t)>0 for all t>0 d) v(t) lies between 0 and 2
The correct answer is a) x(t)>0 for all t>0 and d) v(t) lies between 0 and 2
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### A graph of x versus t is shown in the figure. Choose correct alternatives from below. a) the particle was released from rest at t=0 b) at B, the acceleration a>0 c) at C, the velocity and the acceleration vanish d) average velocity for the motion A and D is positive e) the speed at D exceeds that at E
The correct answer is a) the particle was released from rest at t=0, c) at C, the velocity and the acceleration vanish and e) the speed at D exceeds that at E
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### The variation of quantity A with quantity B, plotted in figure describes the motion of a particle in a straight line. a) quantity B may represent time b) quantity A is velocity if motion is uniform c) quantity A is displacement if motion is uniform d) quantity A is velocity if motion is uniformly accelerated
The correct answer is a) quantity B may represent time, c) quantity A is displacement if motion is uniform, and d) quantity A is velocity if motion is uniformly accelerated
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### At a metro station, a girl walks up a stationary escalator in time t1. If she remains stationary on the escalator, then the escalator take her up in time t2. The time taken by her to walk up on the moving escalator will be a) (t1 + t2)/2 b) t1t2/(t2 – t1) c) t1t2/(t2 + t1) d) t1 – t2
The correct answer is c) t1t2/(t2 + t1)
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### The displacement of a particle is given by x = (t-2)2 where x is in metres and t is seconds. The distance covered by the particle in first 4 seconds is a) 4 m b) 8 m c) 12 m d) 16 m
The correct answer is b) 8 m
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### A vehicle travels half the distance L with speed V1 and the other half with speed V2, then its average speed is a) (V1+V2)/2 b) (2V1+V2)/(V1+V2) c) (2V1V2)/(V1+V2) d) L(V1+V2)/V1V2
The correct answer is c) (2V1V2)/(V1+V2)
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### A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct? a) x<0, v<0, a>0 b) x>0, v<0, a<0 c) x>0, v<0, a>0 d) x>0, v>0, a<0
The correct answer is a) x<0, v<0, a<0 The value of x and v becomes negative as the lift is moving from the 8th floor to the 4th floor whereas acceleration is acting upwards and stays...
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### Among the four graphs, there is only one graph for which average velocity over the time interval (0,T) can vanish for a suitably chosen T. Which one is it?
The correct answer is (b)
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### A helicopter of mass 2000 kg rises with a vertical acceleration of 15 m/s2. The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the a) force on the floor of the helicopter by the crew and passengers b) action of the rotor of the helicopter on the surrounding air c) force on the helicopter due to the surrounding air
Given, M = 2000 kg helicopter mass m = 500 kg m = 500 kg m = 500 kg m = 500 kg m = 500 kg m = 500 kg Helicopter acceleration with crew and passengers = 15 m/s2 a) Force exerted by the crew and...
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### For the harmonic travelling wave y = 2 cos 2 π(10t-0.0080x+3.5) where x and y are in cm and t is second. What is the phase difference between the oscillatory motion at two points separated by a distance of
a) 4 m b) 0.5 m c) λ/2 d) 3 λ/4 e) what is the phase difference between the oscillation of a particle located at x = 100 cm at t = Ts and t = 5s? Answer: According to the questions, we have: y = 2...
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### In the given progressive waves y = 5 sin (100 πt – 0.4 πx) where y and x are in m, t is in s. What is the
a) amplitude b) wavelength Answer: According to the question, the wave is travelling in +x direction and the equation is given by; y = 5 sin (100 πt – 0.4 πx) The standard equation is as follows: y...
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### Given below are some functions of x and t to represent the displacement of an elastic wave.
a) y = 5 cos (4x) sin (20t) b) y = 4 sin (5x-t/2) + 3 cos (5x-t/2) c) y = 10 cos [(252-250)πt] cos [(252+250) πt] d) y = 100 cos (100 πt + 0.5x) State which of these represent a) a traveling wave...
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### If c is r.m.s speed of molecules in a gas and v is the speed of sound waves in the gas, show that c/v is constant and independent of temperature for all diatomic gases.
Answer: We know that the equation for molecules is: $c=\sqrt{\frac{3RT}{M}}$ Where, M represents the molar mass of the gas $v=\sqrt{\frac{\gamma P}{\rho }}=\sqrt{\frac{\gamma RT}{M}}$ We know...
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### The earth has a radius of 6400 km. The inner core of the 1000 km radius is solid. Outside it, there is a region from 1000 km to a radius of 3500 km which is in a molten state. Then again from 3500 km to 6400 km the earth is solid. Only longitudinal (P) waves can travel inside a liquid. Assume that the P wave has a speed of 8 km/s in solid parts and of 5 km/s in liquid parts of the earth. An earthquake occurs at someplace close to the surface of the earth. Calculate the time after which it will be recorded in a seismometer at a diametrically opposite point on the earth if wave travels along diameter?
Answer: According to the question, r1 = 1000 km, r2 = 3500 km, r3 = 6400 km and d1 = 1000 km And we can calculate, d2 = 3500 – 1000 d2 = 2500 km d3 = 6400 – 3500 d3 = 2900 km Expression for the...
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### Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops, and 4 loops, the frequencies are in the ratio 1:2:3:4.
Answer: When n = 1, f1 = v/2L This is known as the fundamental frequency. When n = 2, f2 = 2(v/2L) This is known as the first overtone. When n = 3, f3 = 3(v/2L) This is known as the second overtone....
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### A tuning fork vibrating with a frequency of 512 Hz is kept close to the open end of a tube filled with water. The water level in the tube is gradually lowered. When the water level is 17 cm below the open end, the maximum intensity of sound is heard. If the room temperature is 20oC, calculate
c) if the water in the tube is replaced with mercury, will there be any difference in your observations? Answer: (c) Sound is reflected into the air column by water and mercury in the tube, forming...
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### A tuning fork vibrating with a frequency of 512 Hz is kept close to the open end of a tube filled with water. The water level in the tube is gradually lowered. When the water level is 17 cm below the open end, the maximum intensity of sound is heard. If the room temperature is 20oC, calculate
a) speed of sound in air at room temperature b) speed of sound in air at 0oC Answer: According to the question, the frequency of the tuning fork is f = 512 Hz a) When the first maxima are taken...
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### The pattern of standing waves formed on a stretched string at two instants of time are shown in the figure. The velocity of two waves superimposing to form stationary waves is 360 m/s and their frequencies are 256 Hz.
a) calculate the time at which the second curve is plotted b) mark nodes and antinodes on the curve c) calculate the distance between A’ and C’ Answer: According to the quetion, the frequency of the...
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### The wave pattern on a stretched string is shown in the figure. Interpret what kind of wave this is and find its wavelength.
Answer: A stationary wave which is at its mean position when the particles at t = T/4 and 3T/4 are at rest. The displacement of medium particles at distances of 10, 20, 30, 40, and 50 cm is always...
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### A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz still air. The train beings to move with a speed of 10 m/s towards the platform. What is the frequency of the sound for an observer standing on the platform?
Answer: It is given that v0 = 400 Hz and vz = 10 m/s We know that the velocity of sound in air is va = 330 m/s The frequency heard by the observer on the platform is v'. Therefore, we can write: v’...
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### A pipe of 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of 1237.5 Hz?
Answer: According to the question, the length is: l = 20 cm =0.2 m and the frequency is, v = 1237.5 Hz We know that the velocity of sound in air = 330 m/s The following expression is: l = λ/4 λ = 4l...
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### A steel wire has a length of 12 m and a mass of 2.10 kg. What will be the speed of a transverse wave on this wire when a tension of 2.06 × 104 N is applied?
Answer: According to the question, the length is, l = 12 m and the total mass, M = 2.10 kg Mass per unit length is given by7 m = M/l = 2.1/12 and we know that tension si: T = 2.06 × 104 N Therefore,...
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### When two waves of almost equal frequencies v1 and v2 reach at a point simultaneously, what is the time interval between successive maxima?
Answer: Beats are created when two waves of nearly identical frequency arrive at the same location at the same time. It is given that the two frequencies are almost equal, that is v1 = v2. And for...
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### Equation of a plane progressive wave is given by y = 0.6 sin 2π(t-x/2). On reflection from a denser medium, its amplitude becomes 2/3 of the amplitude of the incident wave
a) y = 0.6 sin 2π(t+x/2) b) y = -0.4 sin 2π(t+x/2) c) y = 0.4 sin 2π(t+x/2) d) y = -0.4 sin 2π(t-x/2) Answer: The correct option is b) y = -0.4 sin 2π(t+x/2) Explanation: The amplitude of the...
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### A sound wave is passing through the air column in the form of compression and rarefaction. In consecutive compressions and rarefactions,
a) density remains constant b) Boyle’s law is obeyed c) bulk modulus of air oscillates d) there is no transfer of heat Answer: The correct option is d) there is no transfer of heat Explanation:...
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### Which of the following statements are true for wave motion?
a) mechanical transverse waves can propagate through all mediums b) longitudinal waves can propagate through solids only c) mechanical transverse waves can propagate through solids only d)...
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### With the propagation of longitudinal waves through a medium, the quantity transmitted is
a) matter b) energy c) energy and matter d) energy, matter, and momentum Answer: The correct option is b) energy The transfer of energy across a medium is caused by longitudinal waves propagating...
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### Change in temperature of the medium changes
a) frequency of sound waves b) amplitude of sound waves c) wavelength of sound waves d) loudness of sound waves Answer: The correct option is c) wavelength of sound waves Explanation: The velocity...
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### Speed of sound wave in air
a) is independent of temperature b) increases with pressure c) increases with an increase in humidity d) decreases with an increase in humidity Answer: The correct option is c) increases with an...
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### Sound waves of wavelength λ travelling in a medium with a speed of v m/s enter into another medium where its speed is 2v m/s. The wavelength of sound waves in the second medium is
a) λ b) λ/2 c) 2 λ d) 4 λ Answer: The correct answer is c) 2 λ Explanation: According to the question, the wavelength of sound waves in the first medium, λ = u/v ...
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### Water waves produced by a motorboat sailing in water are
a) neither longitudinal nor transverse b) both longitudinal and transverse c) only longitudinal d) only transverse Answer: The correct option is b) both longitudinal and transverse Let's...
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### A simple pendulum of time period 1s and length l is hung from a fixed support at O, such that the bob is at a distance H vertically above A on the ground. The amplitude is θo. The string snaps at θ = θ0/2. Find the time taken by the bob to hit the ground. Also, find the distance from A where bob hits the ground. Assume θ0 to be small so that sin θo, θo and cos θ0 1.
Answer: At t = t1 and θ= θ0/a We have, T = 1 sec θ0/2 = θ0 cos 2π/T t1 d θ/dt = – θ0 2π sin 2πt At t = 1/6, that is when θ = θ0/2 d θ/dt = – θ0π√3 v/l = – θ0π√3 v = -θ0π√3l θ0l [1/2 – π√6H/g] gives...
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### A tunnel is dug through the centre of the earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion.
Answer: Let g’ be the acceleration due to gravity of the earth inside the earth g’ = g(1-d/R) R-d = y g’ = g y/R F = -mg y/R F is proportional to (-y) The motion is SHM when the body is in a tunnel....
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### One end of a V-tube containing mercury is connected to a suction pump and the other end to the atmosphere. The two arms of the tube are inclined to horizontal at an angle of 45o each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.
Answer: Let h0 be the initial height of the columns and dx represent the element that needs to be considered Then the mass is given by: dm = A.dx.ρ Potential energy on the left of dm = (dm)gh Total...
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### A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute SHM with a time period where m is mass of the body and ρ is the density of the liquid.
Answer: Let us consider that the verticle displacement at the equilibrium position is ${{x}_{0}}$ At equilibrium mg = Buoyant Force = $A{{x}_{0}}\rho g$ When it is displaced further by a...
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### A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4 cm below the point, where it was held in hand.
a) what is the amplitude of oscillation? b) find the frequency of oscillation? Answer: The mass m will oscillate between the lowest and the highest point, which is where it was held in hand. As a...
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### A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2 s-1 and an amplitude 5 cm. A weighing machine on the platform gives the persons weight against time.
a) will there be any change in weight of the body, during the oscillation? b) if the answer to part a) is yes, what will be the maximum and minimum reading in the machine and at which position?...
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### Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of 2o to the right with the vertical, the other pendulum makes an angle of 1o to the left of the vertical. What is the phase difference between the pendulums?
Answer: We can write the following set of equations: θ1 = θ0 sin (wt + δ1) ...............................(1) θ2 = θ0 sin (wt + δ2) ...
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### A mass of 2 kg is attached to the spring of spring constant 50 N/m. The block is pulled to a distance of 5 cm from its equilibrium position at x = 0 on a horizontal frictionless surface from rest at t = 0. Write the expression for its displacement at anytime t.
Answer: $\omega =\sqrt{\frac{k}{m}}=5$ Therefore, $x=5\sin 5t$
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### A body of mass m is situated in a potential field U(x) = U0 (1 – cos αx) when U0 and α are constants. Find the time period of small oscillations.
Answer: According to the question, we have: U =U0 (1−cosαx) We know taht, F = −dU / dx $F=-\frac{d}{dx}\left[ {{U}_{0}}(1-\cos \alpha x) \right]$ $F=-{{U}_{0}}\alpha \sin \alpha x$ As it is...
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### Find the displacement of a simple harmonic oscillator at which its PE is half of the maximum energy of the oscillator.
Answer: we know that the mean position of the oscillator at x with PE is = 1/2 kx2 Therefore, PE = -m2x2 When PE is maximum, KE = 0 at x = A We get, E = 1/2 mω2A2 PE = 1/2 total energy x2 = 1/2 A2 x...
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### Show that the motion of a particle represented by y = sin ωt – cos ωt is simple harmonic with a period of 2π/ω.
Answer: A function that represents SHM is written as sin (2π/T t + ϕ) We have, y = sin ωt – cos ωt y = √2 sin (ωt – π/4) The standard equation of SHM has: y = a sin (2π/T t + ϕ) Therefore, T =...
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### Find the time period of mass M when displaced from its equilibrium position and then released for the system shown in the figure.
Answer: y represents the distance across which the mass is pulled. The spring's total extension is y + y = 2y. The initial mean location of equilibrium = x0 In the spring, the net extension is equal...
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### The length of a second’s pendulum on the surface of the earth is 1 m. What will be the length of a second’s pendulum on the moon?
Answer: We know that the expression for the time period of a simple pendulum is T = 2π√l/g Where l represents the length of the pendulum and, g denotes the acceleration due to gravity Therefore,...
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Answer:
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### Show that for a particle executing SHM, velocity, and displacement have a phase difference of π/2.
Answer: We know that the equation for the displacement of SHM is x = a cos ꞷt where velocity can be determined as: v = dx/dt = -aꞷ sin ꞷt Here, the phase displacement is ϕ1 = ꞷt And the phase...
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### In figure, what will be the sign of the velocity of the point P’, which is the projection of the velocity of the reference particle P. P is moving in a circle of radius R in the anticlockwise direction.
Answer: P' is the perpendicular of the particle Pat's velocity vector at time t. When the particle moves from P to P1, its foot moves from P' to Q, which is away from the positive axis. As a result,...
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### What is the ratio between the distance travelled by the oscillator in one time period and amplitude?
Answer: According to the question, the distance traveled by the oscillator in one time period = 4A Where A represents the amplitude of the oscillation Therefore, the required ratio becomes = 4A/A =...
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### What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator?
Answer: We know that the equation for the displacement of SHM is x = a sin (ꞷt + ϕ) And the velocity of the particle is given by- v = dx/dt = d a sin (ꞷt + ϕ)/dt Expression for the maximum velocity...
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### When will the motion of a simple pendulum be simple harmonic?
Answer: Consider a basic pendulum with a restoring torque and O as the fixed point. The restoring force is given as F = -mg sin θ θ is so small that θ = arc/radius = x/l As a result, F = – mg x/l As...
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### What are the two basic characteristics of a simple harmonic motion?
Answer: The two primary properties of a SHM are as follows: a) Acceleration and displacement are proportionate in magnitude. b) The acceleration is in the direction of the mean position, while the...
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### Two identical springs of spring constant K are attached to a block of mass m and to fixed supports as shown in the figure. When the mass is displaced from equilibrium position by a distance x towards the right, find the restoring force.
Answer: When mass m is pushed from equilibrium by a distance x to the right, spring B is squeezed by a distance x, and the force (kx) is applied to mass m to the left. However, spring A will stretch...
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### Displacement versus time curve for a particle executing SHM is shown. Identify the points marked at which
i) velocity of the oscillator is zero ii) speed of the oscillator is maximum Answer: (i) The oscillator's velocity is zero when the points A, C, E, and G are at their extreme positions. (ii) The...
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### A body is performing SHM. Then its
a) average total energy per cycle is equal to its maximum kinetic energy b) average kinetic energy per cycle is equal to half of its maximum kinetic energy c) mean velocity over a complete cycle is...
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### A particle is in linear simple harmonic motion between two points A and B, 10 cm apart. Taken the direction from A to B as the +ve direction and choose the correct statements.
a) the sign of velocity, acceleration, and force on the particle when it is 3 cm away from A going towards B are positive b) the sign of velocity of the particle at C going towards O is negative c)...
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### A body is performing SHM. Then its
a) average total energy per cycle is equal to its maximum kinetic energy b) average kinetic energy per cycle is equal to half of its maximum kinetic energy c) mean velocity over a complete cycle is...
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### The displacement time graph of a particle executing SHM is shown in the figure. Which of the following statement is/are true?
a) the force is zero at t = 3T/4 b) the acceleration is maximum at t = 4T/4 c) the velocity is maximum at t = T/4 d) the PE is equal to KE of oscillation t = T/2 Answer: The correct options are a)...
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### Which of the following statements is/are true for a simple harmonic oscillator?
a) force acting is directly proportional to the displacement from the mean position and opposite to it b) motion is periodic c) acceleration of the oscillator is constant d) the velocity is periodic...
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### Displacement versus time curve for a particle executing SHM is shown in the figure. Choose the correct statements.
a) phase of the oscillator is same at t = 0 s and t = 2 s b) phase of the oscillator is the same at t = 2 s and t = 6 s c) phase of the oscillator is the same at t = 1 s and t = 7 s d) phase of the...
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### The motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is
a) simple harmonic motion b) non-periodic motion c) periodic motion d) periodic but not SHM Answer: The correct options are a) simple harmonic motion b) non-periodic motion Explanation: Allow the...
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### The rotation of the earth about its axis is
a) periodic motion b) simple harmonic motion c) periodic but not simple harmonic motion d) non-periodic motion Answer: The correct answers are a) periodic motion c) periodic but not simple harmonic...
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### Four pendulums A, B, C, and D are suspended from the same elastic support as shown in the figure. A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement,
a) D will vibrate with maximum amplitude b) C will vibrate with maximum amplitude c) B will vibrate with maximum amplitude d) all four will oscillate with equal amplitude Answer: The correct option...
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### The displacement of a particle varies with time according to the relation y = a sin ωt + b cos ωt
a) the motion is oscillatory but not SHM b) the motion is SHM with amplitude a + b c) the motion is SHM with amplitude a2 + b2 d) the motion is SHM with amplitude √a2 + b2 Answer: The correct option...
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### A particle is acted simultaneously by mutually perpendicular simple harmonic motions x = a cos ωt and y = a sin ωt. The trajectory of motion of the particle will be
a) an ellipse b) a parabola c) a circle d) a straight line Answer: The correct option is c) a circle Given x=acosωt (i) y=asinωt ...
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### Motion of an oscillating liquid column in a U-tube is
a) periodic but not simple harmonic b) non-periodic c) simple harmonic and time period is independent of the density of the liquid d) simple harmonic and time period is directly proportional to the...
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### The relation between acceleration and displacement of four particles are given below:
a) ax = +2x b) ax = +2x2 c) ax = -2x2 d) ax = -2x Which one of the particles is executing simple harmonic motion? Answer: The correct option is d) ax = -2x Explanation: The acceleration in simple...
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### The displacement of a particle is represented by the equation y = sin3 ωt. The motion is
a) non-periodic b) periodic but not simple harmonic c) simple harmonic with period 2π/ω d) simple harmonic with period π/ω Answer: The correct option is c) simple harmonic with period 2π/ω The...
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### The displacement of a particle is represented by the equation y = 3 cos(π/4 – 2ωt). The motion of the particle is
a) simple harmonic with period 2p/w b) simple harmonic with period π/ω c) periodic but not simple harmonic d) non-periodic Answer: The correct option is b) simple harmonic with period π/ω...
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### Consider a rectangular block of wood moving with a velocity vo in a gas at temperature T and mass density ρ. Assume the velocity is along the x-axis and the area of cross-section of the block perpendicular to vo is A. Show that the drag force on the block is
, where m is the mass of the gas molecule. Answer: Let ρm represent the number of molecules per unit volume Then the expression for the change in momentum by a molecule on front side is = 2m (v +...
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### A box of 1.00 m3 is filled with nitrogen at 1.50 atm at 300 K. The box has a hole of an area 0.010 mm2. How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm?
Answer: Volume of the box = 1 m3 = V1 Initial pressure P1 = 1.5 atm Final pressure P2 = 1.4 atm Air pressure Pa = 1 atm Initial temperature T1 = 300 K Final temperature T2 = 300 K Area of the hole =...
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### Ten small planners are flying at a speed of 150 km/h in total darkness in an air space that is 20 × 20 × 1.5 km3 in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius 10 m.
Answer: We know that, Time = distance/speed Number of particles per unit volume v = N/volume n = 0.0167 km-3 d = 10 × 10-3 km v = 150 km/hr Therefore, we get: time = 225 hrs
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### Consider an ideal gas with the following distribution of speeds
Speed (m/s) % of molecules 200 10 400 20 600 40 800 20 1000 10 a) calculate Vrms and hence T (m = 3.0 × 10-26 kg) b) if all the molecules with speed 1000 m/s escape from the system, calculate new...
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### Explain why
a) there is no atmosphere on moon b) there is a fall in temperature with altitude Answer: a) The moon has no atmosphere since the gravitational pull is minimal and the Vrms is bigger on the moon,...
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Answer: According to the question, the final KE of the gas is 0 The change in KE is as follows: ∆K= 1/2 (nm)v2 ∆T is the change in the temperature ∆U = nCv∆T ∆K = ∆U Making use of the expression, we...
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### Calculate the number of degrees of freedom of molecules of hydrogen in 1 cc of hydrogen gas at NTP.
Answer: According to the question, the volume occupied by the molecules of a gas is 22400 cc and the number of molecules in 1 cc of hydrogen are 2.688 × 1019 We know that the hydrogen has a total...
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### A balloon has 5.0 g mole of helium at 7 degrees C. Calculate
a) the number of atoms of helium in the balloon b) the total internal energy of the system Answer: We know that the average KE per molecule is 3/2kT No.of moles of helium are n = 5 g mole We have T...
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### When air is pumped into a cycle tyre the volume and pressure of the air in the tyre both are increased. What about Boyle’s law in this case?
Answer: We have, PV = P(m/ρ) = constant P/ ρ = constant Volume = m/ ρ where m is constant When air is pushed into the cycle's tyre, the mass of the air increases as the number of molecules...
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### We have 0.5 g of hydrogen gas in a cubic chamber of size 3 cm kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state?
Answer: We know that volume of 1 molecule = 4/3 πr3 = 4.20 × 10-30 m3 And the number of moles in 0.5 g H2 gas = 0.25 mole Volume of H2 molecule in 0.25 mole = 1.04×6.023× 10+23-30 = 6.264 ×...
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### A gas mixture consists of molecules of tyres A, B, and C with masses mA > mB > mC. Rank the three types of molecules in decreasing order of
a) average KE b) rms speeds Answer: a) From the above result, the pressure and temperature are the same, therefore, KEc > KEb > KEa b) When P and T are constant, (Vrms)c > (Vrms)b >...
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### The container shown in the figure has two chambers, separated by a partition, of volumes V1 = 2.0 litre and V2 = 3.0 litre. The chambers contain μ1 = 4.0 and μ2 = 5.0 moles of a gas at pressure p1 = 1.00 atm and p2 = 2.00 atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.
V1 V2 μ1 μ2 p1 p2 Answer: We know that for an ideal gas, PV = μRT The two equations that we can write are: P1V1 = μ1R1T1 P2V2 = μ2R2T2 P1 = 1 atm, P2 = 2 atm and V1 = 2L, V2 = 3L Also, T1 = T = T2...
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### Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters 1 A and 2 A. The gases may be considered under identical conditions of temperature, pressure, and volume.
Answer: We know that that we can write: I α (1/d2) d1 = 1Å d2 = 2Å Therefore, l1 : l2 = 4 : 1 The ratio of the mean free paths of the molecules is 4:1
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### A gas mixture consists of 2.0 moles of oxygen and 4.0 moles of neon a temperature T. Neglecting all vibrational modes, calculate the total internal energy of the system. (Oxygen has two rotational modes.)
Answer: We know that oxygen has 5 degrees of freedom. Therefore, energy per mole = (5/2)RT Therefore, for 2 moles of oxygen, energy = 5RT Neon has 3 degrees of freedom. Therefore, energy per mole...
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### A cycle followed by an engine is shown in the figure. Find heat exchanged by the engine, with the surroundings for each section of the cycle considering Cv = (3/2)R.
AB: constant volume BC: constant pressure CD: adiabatic DA: constant pressure Answer: a) AB: constant volume b) BC: constant pressure c) CD: adiabatic QCD = 0 d) DA: constant pressure, there is...
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### A cycle followed by an engine is shown in the figure.
A to B: volume constant B to C: adiabatic C to D: volume constant D to A: adiabatic VC = VD = 2VA = 2VB c) what is the work done by the engine in one cycle in terms of PA, PB, VA? d) what is the...
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### A cycle followed by an engine is shown in the figure.
A to B: volume constant B to C: adiabatic C to D: volume constant D to A: adiabatic VC = VD = 2VA = 2VB a) in which part of the cycle heat is supplied to the engine from outside? b) in which part of...
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### Consider a P-V diagram in which the path followed by one mole of a perfect gas in a cylindrical container is shown in the figure.
c) given the internal energy for one mole of gas at temperature T is (3/2)RT, find the heat supplied to the gas when it is taken from state 1 to 2 with V2 = 2V1. Answer: We know that the expression...
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### Consider a P-V diagram in which the path followed by one mole of a perfect gas in a cylindrical container is shown in the figure.
a) find the work done when the gas is taken from state 1 to state 2 b) what is the ratio of temperature T1/T2 if V2 = 2V1 Answer: According to the question, PV1/2 = K = constant a) Expression for...
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### A person driving a car suddenly applies the brakes on seeing a child on the road ahead. If he is not wearing a seat belt, he falls forward and hits his head against the steering wheel. Why?
If a person is not wearing a seat belt and abruptly hits the brakes, he will fall forward and bang his head against the steering wheel because his upper body will continue to move in the same...
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### The position time graph of a body of mass 2 kg is as given in the figure. What is the impulse on the body at t – 0 sec and t = 4 sec.
Mass of body, m = 2 kg Time, t = 0 Initial velocity, v1 = 0 From graph, we know that t ≥ 0 to t ≤ 4 which is a straight line. The velocity of the body is constant v2 = tan θ = 3/4 = 0.75 m/s At t ≥...
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### A person of mass 50 kg stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of 9 m/s2 what would be the reading of the weighing scale?
When the lift lowers with an acceleration a, the apparent weight on the weighing scale decreases. W' denotes the apparent weight. Therefore, W' = R = (mg – ma) = m(g – a) As a result, W' = 50(10-9)...
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### A girl riding a bicycle along a straight road with a speed of 5 m/s throws a stone of mass 0.5 kg which has a speed of 15 m/s with respect to the ground along her direction of motion. The mass of the girl and the bicycle is 50 kg. Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?
Given, m1 = 50 kg m2 = 0.5 kg u1 = 5 m/s u2 = 5 m/s v1 = ? v2 = 15 m/s The law of conservation of momentum states that Final momentum equals initial momentum. u1 = m1v1 + m2v2 (m1 + m2) We obtain by...
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### Two billiard balls A and B, each of mass 50 g and moving in opposite directions with speed of 5 m/s each, collide and rebound with the same speed. If the collision lasts for 10-3 seconds, which of the following statements are true? a) the impulse imparted to each ball is 0.25 kg.m/s and the force on each ball is 250N b) the impulse imparted to each ball is 0.25 kg.m/s and the force exerted on each ball is 25 × 10-5 N c) the impulse imparted to each ball is 0.5 Ns d) the impulse and the force on each ball are equal in magnitude and opposite in direction
The correct answer is: c) the impulse imparted to each ball is 0.5 Ns d) the impulse and the force on each ball are equal in magnitude and opposite in direction
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### In the figure, a body A of mass m slides on a plane inclined at angle θ1 to the horizontal and μ1 is the coefficient of friction between A and the plane. A is connected by a light string passing over a frictionless pulley to another body B, also of mass m, sliding on a frictionless plane inclined at angle θ2 to the horizontal. Which of the following statements are true?;a) A will never move up the plane b) A will just start moving up the plane when c) for A to move up the plane, θ2 must always be greater than θ1 d) B will always slide down with constant speed
The correct answer is: b) A will just start moving up the plane when $$\mu =\frac{\sin {{\theta }_{2}}-\sin {{\theta }_{1}}}{\cos {{\theta }_{1}}}$$ c) for A to move up the plane, θ2 must always be...
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### Mass m1 moves on a slope making an angle θ with the horizontal and is attached to mass m2 by a string passing over a frictionless pulley as shown in the figure. The coefficient of friction between m1 and the sloping surface is μ. Which of the following statements are true?;a) if m2 > m1 sin θ, the body will move up the plane b) if m2 > m1(sin θ + μ cos θ), the body will move up the plane c) if m2 < m1 (sin θ + μ cos θ), the body will move up the plane d) if m2 < m1 (sin θ – μ cos θ), the body will move down the plane
The correct answer is: b) if m2 > m1(sin θ + μ cos θ), the body will move up the plane d) if m2 < m1 (sin θ – μ cos θ), the body will move down the plane
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### In figure, the coefficient of friction between the floor and the body B is 0.1. The coefficient of friction between the bodies B and A is 0.2. A force F is applied as shown on B. The mass of A is m/2 and of B is m. Which of the following statements are true?;a) the bodies will move together if F = 0.25 mg b) the body A will slip with respect to B if F = 0.5 mg c) the bodies will move together if F = 0.5 mg d) the bodies will be at rest if F = 0.1 mg e) the maximum value of F for which the two bodies will move together is 0.45 mg
The correct answer is: a) the bodies will move together if F = 0.25 mg b) the body A will slip with respect to B if F = 0.5 mg d) the bodies will be at rest if F = 0.1 mg e) the maximum value of F...
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### The motion of a particle of mass m is given by x = 0 for t < 0 sec, x(t) = A sin 4p t for 0 < t < (1/4) sec, and x = 0 for t > (1/4) sec. Which of the following statements is true? a) the force at t = (1/8) sec on the particle is -16π2Am b) the particle is acted upon by on impulse of magnitude 4π2Am at t = 0 sec and t = (1/4) sec c) the particle is not acted upon by any force d) the particle is not acted upon by a constant force e) there is no impulse acting on the particle
The correct answer is: a) the force at t = (1/8) sec on the particle is -16π2Am b) the particle is acted upon by on impulse of magnitude 4π2Am at t = 0 sec and t = (1/4) sec d) the particle is not...
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### A car of mass m starts from rest and acquires a velocity along the east a in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is a) mv/2 eastward and is exerted by the car engine b) mv/2 eastward and is due to the friction on the tyres exerted by the road c) more than mv/2 eastward exerted due to the engine and overcomes the friction of the road d) mv/2 exerted by the engine
The correct answer is b) mv/2 eastward and is due to the friction on the tyres exerted by the road
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### A body with mass 5 kg is acted upon by a force F=-3i+4j N. If its initial velocity at t = 0 is v=6i-12j m/s, the time at which it will just have a velocity along the y-axis is a) never b) 10 s c) 2 s d) 15 s
The correct answer is b) 10 s
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### A body of mass 2 kg travels according to the law x(t) = p(t) + qt2 + rt3 where p = 3 m/s, q = 4 m/s2, and r = 5 m/s3. The force acting on the body at t=2 seconds is a) 136 N b) 134 N c) 158 N d) 68 N
The correct answer is a) 136 N
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### Conservation of momentum in a collision between particles can be understood from a) conservation of energy b) Newton’s first law only c) Newton’s second law only d) both Newton’s second and third law
The correct answer is d) both Newton’s second and third law
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### In the previous problem, the magnitude of the momentum transferred during the hit is a) zero b) 0.75 kg.m/s c) 1.5 kg.m/s d) 14 kg.m/s
The correct option is c) 1.5 kg.m/s
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### A cricket ball of mass 150 g has an initial velocity u=3i+4j m/s and a final velocity v=-(3i+4j) m/s after being hit. The change in momentum a) zero (b)-(0.45i+0.6j) (c)-(0.9i+1.2j)) (d)-5(i+j)
The correct answer is (c) -(0.9i+1.2j)
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### A metre scale is moving with uniform velocity. This implied a) the force acting on the scale is zero, but a torque about the centre of mass can act on the scale b) the force acting on the scale is zero and the torque acting about the centre of mass of the scale is also zero c) the total force acting on it need not be zero but the torque on it is zero d) neither the force nor the torque needs to be zero
The correct answer is b) the force acting on the scale is zero and the torque acting about the centre of mass of the scale is also zero
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### A ball is travelling with uniform translator motion. This means that a) it is at rest b) the path can be a straight line or circular and the ball travels with uniform speed c) all parts of the ball have the same velocity and the velocity is constant d) the centre of the ball moves with constant velocity and the ball spins about its centre uniformly
The correct option is c) all parts of the ball have the same velocity and the velocity is constant
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### A balloon filled with helium rises against gravity increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect the viscous drag of air and assume that the density of air is constant.
The net buoyant force Equals vpg when the dragging viscous force of the air on the balloon is ignored. Where v is the volume of air that has been displaced. The upward net density is denoted by p....
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### Two identical steel cubes collide head-on face to face with a speed of 10 cm/s each. Find the maximum compression of each. Young’s modulus for steel = Y = 2 × 1011 N/m2.
Y = stress/strain Y = FL/A∆L WD = F∆L KE = 5 × 10-4 J WD = KE ∆L = 5 × 10-7 m
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### A rocket accelerates straight up by ejecting gas downwards. In a small time interval ∆t, it ejects a gas of mass ∆m at a relative speed u. Calculate KE of the entire system at t + ∆t and t and show that the device that ejects gas does work = (1/2) ∆m u2 in this time interval.
M is the rocket's mass at any given moment t. The rocket's velocity is v. The mass of the gas expelled during the time interval t is m. As a result, K = 1/2 u2∆m
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### A curved surface as shown in the figure. The portion BCD is free of friction. There are three spherical balls of identical radii and masses. Balls are released from one by one from A which is at a slightly greater height than C. with the surface AB, ball 1 has large enough friction to cause rolling down without slipping; ball 2 has a small friction and ball 3 has a negligible friction. a) for which balls is total mechanical energy conserved? b) which ball can reach D? c) for balls which do not reach D, which of the balls can reach back A?
a) For ball 1 the total mechanical energy is conserved b) Ball 1 reaches D c) Ball 3 reaches back A
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### A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 30o by a force of 10 N parallel to the inclined surface. The coefficient of friction between the block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate a) work done against gravity b) work done against the force of friction c) increase in potential energy d) increase in kinetic energy e) work done by an applied force
a) Work against gravity equals mgh 5 m= h 50 J WD against gravity b) The work done against the friction force is fs = 53 J. d) WD against gravity = 50 J increase in PE d) The system's increase in KE...
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### On complete combustion, a litre of petrol gives off heat equivalent to 3 × 107 J. In a test drive a car weighing 1200 kg, including the mass of driver, runs 15 km per litre while moving with a uniform speed on a surface and air to be uniform, calculate the force of friction acting on the car during the test drive, if the efficiency of the car engine were 0.5.
car engine Efficiency = 0.5 Energy given by the car with 1 litre of petrol = 1.5 × 107 WD = 1.5 × 107 f = 103 N
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### An adult weighing 600 N raises the centre of gravity of his body by 0.25 m while taking each step of 1 m length in jogging. If he jogs for 6 km, calculate the energy utilized by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting 10% of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilized for jogging.
The energy used up is given as = mgh mg = 600 N h = 0.25m No.of steps in 6 km = 6000 steps Energy used in 6000 m = (6000)(600)(0.25)J Energy utilized in jogging = 9 × 104 J
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### An engine is attached to a wagon through a shock absorber of length 1.5 m. The system with a total mass of 50,000 kg is moving with a speed of 36 km/h when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of the energy of the wagon is lost due to friction, calculate the spring constant.
KE = 1/2 mv2 m = 50000 kg v = 10 m/s KE = 2500000J KE of spring = 10% of the KE wagon K = 5 × 105 N/m
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### Suppose the average mass of raindrops is 3.0 × 10^(-5) kg and their average terminal velocity 9 m/s. Calculate the energy transferred by rain to each square meter of the surface at a place which receives 100 cm of rain in a year.
Energy transferred by the rain to the surface of the earth = 1/2 mv2 The velocity of the rain = 9 m/s Mass = (volume)(density) = 1000 kg Energy transferred by 100 cm rainfall = 1/2 mv2 =...
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### A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 m/s. Calculate a) the loss of PE of the drop b) the gain in KE of the drop c) is the gain in KE equal to loss of PE? If not why?
a) PE at the highest point = 10 J b) Gain in KE = 1/2 mv2 = 1.250 J c) Gain in KE is not equal to the PE
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### The bob A of a pendulum released from horizontal to the vertical hits another bob B of the same mass at rest on a table as shown in the figure. If the length of the pendulum is 1 m, calculate a) the height to which bob A will rise after collision b) the speed with which bob B starts moving. Neglect the size of the bobs and assume the collision to be elastic.
a) After the impact, bob A does not rise much because the PE of bob A is converted to KE and the momentum is transferred to bob B. (B) The speed of bob B is calculated as the sum of bob A's KE and...
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### Consider a one-dimensional motion of a particle with total energy E. There are four regions A, B, C, and D in which the relation between potential energy V, kinetic energy (K) and total energy is as given below: Region A: V > E Region B: V < E Region C: K > E Region D: V > K State with reason in each case whether a particle can be found in the given region or not.
E = V + K and V > E for area A, implying that the KE is negative and therefore this is not feasible. K = E – V and V E for area B, implying that both energies are larger than zero. V = E – K and...
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### A ball of mass m, moving with a speed 2v0 collides inelastically with an identical ball at rest. Show that a) for a head-on collision, both the balls move forward b) for a general collision, the angle between the two velocities of scattered balls is less than 90o.
a) Let v1 and v2 be the velocities of the two balls after the collision. According to the law of conservation of momentum, mv0 = mv1 + mv2 v2 = v1 + 2ev0 e < 1 b) Using the law of conservation of...
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### A graph of potential energy V(x) versus x is shown in the figure. A particle of energy E0 is executing motion in it. Draw graph of velocity and kinetic energy versus x for one complete cycle AFA.
From the given graph of KE versus x From the below graph of velocity versus x
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### A bob of mass m suspended by a light string of length L is whirled into a vertical circle as shown in the figure. What will be the trajectory of the particle if the string is cut at a) point B b) point C c) point X
a) When the string is severed at point B, the particle's tangential velocity will be vertically downward, and the bob will travel in the same direction. b) When the string is severed at point C, the...
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### Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of the same magnitude. How would the distance moved by them before coming to rest compare?
KE1 = KE2 WD1 = WD2 F1s1 = F2s2 F1 = F2 s1 = s2
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### Give an example of a situation in which an applied force does not result in a change in kinetic energy.
The kinetic energy of work done in a circular motion remains unchanged.
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### The average work done by a human heart while it beats once is 0.5 J. Calculate the power used by heart if it beats 72 times in a minute.
P = WD/time WD is one beat of heart = 0.5 J WD in 72 beats = 36 J P = WD/t = 0.6 W
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### Calculate the power of a crane in watts, which lifts a mass of 100 kg to a height of 10 min 20 sec.
P = WD/time = Fs cos θ/t = mgh cos θ/t h = 10 m t = 20 sec F = mg = 1000 Therefore, P = 500 Watts
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### In an elastic collision of two billiard balls, which of the following quantities remain conserved during the short time of collision of the balls a) kinetic energy b) total linear momentum Give a reason for your answer in each case.
Because there is no non-conservative force, the kinetic energy and total linear momentum of the billiard balls are preserved.
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### A body is moved along a closed loop. Is the work done in moving the body necessarily zero? If not, state the condition under which work done over a closed path is always zero.
When the conservative force acts on the body during motion, the work done by the moving body is zero. When a non-conservative force acts on a moving body, the work done by the body is not zero.
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### A body falls towards earth in the air. Will its total mechanical energy be conserved during the fall? Justify.
The free-falling body's total mechanical energy is not preserved since it is utilised to overcome the frictional force of the air molecules.
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### Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the car is 400 kg and the distance moved is 2m.
WD = Fs cos θ WD = Fs cos 90o = 0 Hence, the work done by the car against the gravity is zero.
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### A body is being raised to a height h from the surface of the earth. What is the sign of work done by a) applied force b) gravitational force
a) The applied force produces positive work. b) The gravitational pull produces negative work.
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### Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case?
In the event of an elevator falling, the number of passengers is limited because it is not a free fall and descends at a constant speed.
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### A rough inclined plane is placed on a cart moving with a constant velocity u on horizontal ground. A block of mass M rests on the incline. Is any work done by a force of friction between the block and incline? Is there then a dissipation of energy?
The block is clearly tilted on the plane in the illustration above. There is no work done since there is no displacement and no waste of energy.
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### Two blocks M1 and M2 having equal mass are free to move on a horizontal frictionless surface. M2 is attached to a massless spring as shown in the figure. Initially, M2 is at rest and M1 is moving toward M2 with speed v and collides head-on with M2.;a) while spring is fully compressed all the KE of M1 is stored as PE of spring b) while spring is fully compressed the system momentum is not conserved, though final momentum is equal to the initial momentum c) if spring is massless, the final state of the M1 is the state of rest d) if the surface on which blocks are moving has friction, then a collision cannot be elastic
c) if spring is massless, the final state of the M1 is a state of rest d) if the surface on which blocks are moving has friction, then a collision cannot be elastic
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### A bullet of mass m fired at 30o to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target. Which of the following statements are correct in respect of bullet after it emerges out of the target? a) the velocity of the bullet will be reduced to half its initial value b) the velocity of the bullet will be more than half of its earlier velocity c) the bullet will continue to move along the same parabolic path d) the bullet will move in a different parabolic path e) the bullet will fall vertically downward after hitting the target f) the internal energy of the particles of the target will increase
b) the velocity of the bullet will be more than half of its earlier velocity d) the bullet will move in a different parabolic path f) the internal energy of the particles of the target will...
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### A man, of mass m, standing at the bottom of the staircase, of height L, climbs it and stands at its top. a) work done by all forces on man is zero b) work done by all the force on man is zero c) work done by the gravitational force on man is mgL d) the reaction force from a step does not do work because the point of application of the force does not move while the force exists
b) work done by all the force on man is zero d) the reaction force from a step does not do work because the point of application of the force does not move while the force exists
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c) 1.05 × 104 N
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2022-09-26 04:02:31
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https://paperswithcode.com/paper/stability-and-generalization-of-learning/review/
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Paper
### Stability and Generalization of Learning Algorithms that Converge to Global Optima
We establish novel generalization bounds for learning algorithms that converge to global minima. We do so by deriving black-box stability results that only depend on the convergence of a learning algorithm and the geometry around the minimizers of the loss function... (read more)
Results in Papers With Code
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2020-08-08 18:06:43
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https://studyadda.com/solved-papers/manipal-medical/manipal-medical-solved-paper-2008/363
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# Solved papers for Manipal Medical Manipal Medical Solved Paper-2008
### done Manipal Medical Solved Paper-2008
• question_answer1) If muscle times speed equals power, what is the ratio of the SI unit and the CGS unit of muscle?
A) ${{10}^{5}}$
B) ${{10}^{3}}$
C) ${{10}^{7}}$
D) ${{10}^{-5}}$
• question_answer2) A proton in a cyclotron changes its velocity from 30 km/s north to 40 km/s east in 20s. What is the magnitude of average acceleration during this time?
A) $2.5\text{ }km/{{s}^{2}}$
B) $12.5\text{ }km/{{s}^{2}}$
C) $22.5\text{ }km/{{s}^{2}}$
D) $32.5\text{ }km/{{s}^{2}}$
• question_answer3) A heavy small-sized sphere is suspended by a string of length L The sphere rotates uniformly in a horizontal circle with the string making an angle$\theta$with the vertical. Then, the time-period of this conical pendulum is
A) $t=2\pi \sqrt{\frac{l}{g}}$
B) $t=2\pi \sqrt{\frac{l\sin \theta }{g}}$
C) $t=2\pi \sqrt{\frac{l\cos \theta }{g}}$
D) $t=2\pi \sqrt{\frac{l}{g\cos \theta }}$
• question_answer4) The equation of the stationary wave is $y=2A\sin \left( \frac{2\pi ct}{\lambda } \right)\cos \left( \frac{2\pi x}{\lambda } \right)$ Which of the following statements is wrong?
A) The unit of$ct$is same as that of$\lambda$
B) The unit of$x$is same as that of $\lambda$
C) The unit of$2\pi c/\lambda$ is same as that of$2\pi x/\lambda t$
D) The unit of$c/\lambda$is same as that of $x/\lambda$
• question_answer5) A body of mass 2 kg has an initial velocity of 3 m/s along OE and it is subjected to a force of 4 N in a direction perpendicular to OE. The distance of body from 0 after 4 s will be
A) 12m
B) 20m
C) 8m
D) 48m
• question_answer6) A synchronous relay satellite reflects TV signals and transmits TV programme from one part of the world to the other because its
A) period of revolution is greater than the period of rotation of the earth about its axis
B) period of revolution is less than the period of rotation of the earth about its axis
C) period of revolution is equal to the period of rotation of the earth about its axis
D) mass is less than the mass of earth
• question_answer7) When a metal wire elongates by hanging a load Mg on it, the gravitational potential energy of mass M decreases by$Mgl$. This energy appears
A) as elastic potential energy completely
B) as thermal energy completely
C) half as elastic potential energy and half as thermal energy
D) as kinetic energy of the load completely
• question_answer8) A ball is projected from the ground at angle$\theta$ with the horizontal. After Is it is moving at angle 45? with the horizontal and after 2 s it is moving horizontally. What is the velocity of projection of the ball?
A) $10\sqrt{3}\,m{{s}^{-1}}$
B) $20\sqrt{3}\,m{{s}^{-1}}$
C) $10\sqrt{5}\,m{{s}^{-1}}$
D) $20\sqrt{2}\,m{{s}^{-1}}$
• question_answer9) Isobaric modulus of elasticity is equal to
A) isochoric modulus of elasticity
B) isdthermal modulus of elasticity
C) zero
D) infinite
• question_answer10) An engine is moving on a circular path of radius 100 m with a speed of 20 m/s. What will be the frequency observed by an observer standing stationary at the centre of circular path when the engine blows a whistle of frequency 500 Hz?
A) More than 500 Hz
B) Less than 500 Hz
C) 500 Hz
D) No sound is heard
• question_answer11) The work done by a particle moving with a velocity of 0.7 c (where c is the velocity of light) in empty space free of electromagnetic field and far away from all matter is:
A) positive
B) negative
C) zero
D) infinite
• question_answer12) A boat carrying a number of large stones is floating in a water tank. What will happen to the water level if the stones are unloaded into water?
A) rise
B) fall
C) remain unchanged
D) rise till half the number of stones are unloaded and then begin to fall
• question_answer13) A car of mass m moves in a horizontal circular path of radius r m. At an instant its speed is v m/s and is increasing at a rate of$a\text{ }m/{{s}^{2}}$. Then the acceleration of the car is
A) $\frac{{{v}^{2}}}{r}$
B) $a$
C) $\sqrt{{{a}^{2}}+{{\left( \frac{{{v}^{2}}}{r} \right)}^{2}}}$
D) $\sqrt{u+\frac{{{v}^{2}}}{r}}$
• question_answer14) Equal volumes of monoatomic and diatomic gases at the same temperature are given equal quantities of heat. Then
A) the temperature of diatomic gas will be more
B) the temperature of monoatomic gas will be more
C) the temperature of both will be zero
D) nothing can be said
• question_answer15) The velocity v (in cm/s) of a particle is given in terms of time t (in s) by the equation, $v=at+\frac{b}{t+c}$ The dimensions of a, b and c are
A) a-$[{{L}^{2}}]$ b-$[T]$ c-$[L{{T}^{2}}]$
B) a-$[L{{T}^{2}}]$ b-$[LT]$ c-$[L]$
C) a-$[L{{T}^{-2}}]$ b-$[L]$ c-$[T]$
D) a-$[L]$ b-$[LT]$ c-$[{{T}^{2}}]$
• question_answer16) A block of mass 5 kg is moving horizontally at a speed of$1.5\text{ }m{{s}^{-1}}$. A vertically upward force 5 N acts on it for 4 s. What will be the distance of the block from the point where the force starts acting?
A) 2m
B) 6m
C) 8m
D) 10m
• question_answer17) A copper wire and a steel wire of the same diameter and length are joined end to end and a force is applied which stretches their combined length by 1 cm. Then the two wires will have
A) the same stress and strain
B) the same stress but different strains
C) the same strain but different stresses
D) different stresses and strains
• question_answer18) A body of mass 1 kg is rotating in a vertical circle of radius 1m. What will be the difference in its kinetic energy at the top and bottom of the circle? (take$g=10\text{ }m{{s}^{-2}}$)
A) 10 J
B) 20 J
C) 30 J
D) 50 J
• question_answer19) Find the difference of-air pressure between the inside and outside of a soap bubble 5mm in diameter, if the surface tension is$1.6\,N{{m}^{-1}}$
A) $2560\text{ }N{{m}^{-2}}$
B) $3720\text{ }N{{m}^{-2}}$
C) $1208\text{ }N{{m}^{-2}}$
D) $10132\text{ }N{{m}^{-2}}$
• question_answer20) In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is 0.17s. The frequency of the wave is
A) 1.47 Hz
B) 0.36 Hz
C) 0.73 Hz
D) 2.94 Hz
• question_answer21) A particle moves through angular displacement$\theta$on a circular path of radius r. The linear displacement will be
A) $2\text{ }r\text{ }sin(\theta /2)$
B) $2\text{ }r\text{ cos}(\theta /2)$
C) $2\text{ }r\text{ tan}(\theta /2)$
D) $2\text{ }r\text{ cot}(\theta /2)$
• question_answer22) A cube made of material having a density of $0.9\times {{10}^{3}}kg/{{m}^{3}}$floats between water and a liquid of density$0.7\times {{10}^{3}}kg/{{m}^{3}},$which is immiscible with water. What part of the cube is immersed in water?
A) $\frac{1}{3}$
B) $\frac{2}{3}$
C) $\frac{3}{4}$
D) $\frac{3}{7}$
• question_answer23) When a rod is heated but prevented from expanding, the stress developed is independent of
A) material of the rod
B) rise in temperature
C) length of rod
D) None of the above
• question_answer24) Given that T stands for time period and I stands for the length of simple pendulum. If g is the acceleration due to gravity, then which of the following statements about the relation ${{T}^{2}}=l/g$is correct?
A) It is correct both dimensionally as well as numerically
B) It is neither dimensionally correct nor numerically
C) It is dimensionally correct but not numerically
D) It is numerically correct but not dimensionally
• question_answer25) The length of the seconds pendulum is decreased by 0.3 cm when it is shifted to Chennai from London. If the acceleration due to gravity at London is$981\text{ }cm/{{s}^{2}},$the acceleration due to gravity at Chennai is (assume${{n}^{2}}=10$)
A) $981\text{ }cm/{{s}^{2}}$
B) $978\text{ }cm/{{s}^{2}}$
C) $984\text{ }cm/{{s}^{2}}$
D) $975\text{ }cm/{{s}^{2}}$
• question_answer26) A clock hung on a wall has marks instead of numerals on its dial. On the adjoining wall, there is a plane mirror and the image of the clock in the mirror indicates the time$7:10$. Then the time on the clock is
A) $7:10$
B) $4:50$
C) $5:40$
D) $10:7$
• question_answer27) How does, the electric field (E) between the plates of a charged cylindrical capacitor vary with the distance r from the axis of the cylinder?
A) $E\propto \frac{1}{{{r}^{2}}}$
B) $E\propto \frac{1}{r}$
C) $E\propto {{r}^{2}}$
D) $E\propto r$
• question_answer28) The radioactivity of a sample is${{A}_{1}}$at time${{t}_{1}}$and ${{A}_{2}}$at time${{t}_{2}}$. If the mean life of the specimen is T, the number of atoms that have disintegrated in the time interval of$({{t}_{2}}-{{t}_{1}})$is
A) $({{A}_{1}}-{{A}_{2}})$
B) $\frac{({{A}_{1}}-{{A}_{2}})}{T}$
C) $({{A}_{1}}-{{A}_{2}})T$
D) $({{A}_{1}}{{t}_{1}}-{{A}_{2}}{{t}_{2}})$
• question_answer29) The Youngs. double slit experiment is performed with blue and with green light of wavelengths$4360\overset{o}{\mathop{\text{A}}}\,$and$5460\overset{o}{\mathop{\text{A}}}\,$respectively. If $x$is the distance of 4th maximum from the central one, then
A) $x$(blue) = $x$ (green)
B) $x$ (blue) > $x$(green)
C) $x$ (blue) < $x$ (green)
D) $\frac{x(blue)}{x(green)}=\frac{5460}{4360}$
• question_answer30) A vertical straight conductor carries a current upward. A point P lies to the east of it at a small distance and another point Q lies to the west at the same distance. The magnetic field at$P$is
A) greater than at Q
B) same as at Q
C) less than at Q
D) greater or less than at Q depending upon the strength of current
• question_answer31) Given $_{a}{{\mu }_{g}}=3/2$and$_{a}{{\mu }_{w}}=4/3$. There is an equiconvex lens with radius of each surface equal to 20 cm. There is air in the object space and water in the image space. The focal length of lens is
A) 80cm
B) 40cm
C) 20cm
D) 10cm
• question_answer32) Iron has fee structure. If the radius is$1.27\overset{o}{\mathop{\text{A}}}\,,$then the lattice parameter is
A) $1.27\overset{o}{\mathop{\text{A}}}\,$
B) $3.59\overset{o}{\mathop{\text{A}}}\,$
C) $2.54\overset{o}{\mathop{\text{A}}}\,$
D) $3.81\overset{o}{\mathop{\text{A}}}\,$
• question_answer33) When 1A of current is passed through$CuS{{O}_{4}}$ solution for$10\text{ }s$. then the number of copper ions deposited at the cathode will be
A) $1.6\times {{10}^{19}}$
B) $3.1\times {{10}^{19}}$
C) $4.8\times {{10}^{19}}$
D) $6.2\times {{10}^{19}}$
• question_answer34) Out of a photon and an electron the equation $E=pc,$is valid for
A) both
B) neither
C) photon only
D) electron only
• question_answer35) An observer looks at a tree of height 15m with a telescope of magnifying power 10. To him the tree appears
A) 10 times taller
B) 15 times taller
C) 10 times nearer
D) 15 times nearer
• question_answer36) A 50 Hz alternating current of peak value 1A flows through the primary coil of a transformer. If the mutual inductance between the primary and secondary be 1.5 H, then the mean value of the induced voltage is
A) 75 V
B) 150 V
C) 225 V
D) 300 V
• question_answer37) A cell of emf E and internal resistance r is connected in series with an external resistance nr. Then the ratio of the terminal potential difference to emf is
A) $(1/n)$
B) $1/(n+1)$
C) $n/(n+1)$
D) $(n+1)/n$
• question_answer38) A slide projector gives a magnification of 10. If it projects a slide of dimensions$3\text{ }cm\times 2\text{ }cm$ on a screen, the area of the image on the screen will be
A) $6\,c{{m}^{2}}$
B) $60\,c{{m}^{2}}$
C) $600\,c{{m}^{2}}$
D) $6000,\,c{{m}^{2}}$
• question_answer39) A thin bar magnet of length 2L is bent at the mid-point so that the angle between them is$60{}^\circ$. The new length of the magnet is:
A) $\sqrt{2}L$
B) $\sqrt{3}L$
C) $2L$
D) L
• question_answer40) The binding energies of the atoms of elements A and B are${{E}_{a}}$and${{E}_{b}}$respectively. Three atoms of the element B fuse to give one atom of element A. This fusion process is accompanied by release of energy e. Then${{E}_{a}},{{E}_{b}}$and e are related to each other as
A) ${{E}_{a}}+e=3{{E}_{b}}$
B) ${{E}_{a}}=3{{E}_{b}}$
C) ${{E}_{a}}-e=3{{E}_{b}}$
D) ${{E}_{a}}+3{{E}_{b}}+e=0$
• question_answer41) A capacitor is connected to a cell of emf E and some internal resistance r. The potential difference across the
A) cell is E
B) cell is < E
C) capacitor is < E
D) capacitor is > E
• question_answer42) The radius of the smallest electron orbit in hydrogen-like ion is$(0.51\times {{10}^{-10}}/4)m;$then it is
A) hydrogen atom
B) $H{{e}^{+}}$
C) $L{{i}^{2+}}$
D) $B{{e}^{3+}}$
• question_answer43) A radioactive sample with a half-life of 1 month has the label Activity$=2\mu \,Ci$on$1-8-1991$. What was its activity two months earlier?
A) $1.0\,\mu Ci$
B) $0.5\,\mu Ci$
C) $4\,\mu Ci$
D) $8\,\mu Ci$
• question_answer44) An inductor is connected to a battery through a switch. Induced emf is${{e}_{1}}$when the switch is pressed and${{e}_{2}}$when the switch is opened. Then
A) ${{e}_{1}}={{e}_{2}}$
B) ${{e}_{1}}>{{e}_{2}}$
C) ${{e}_{1}}<{{e}_{2}}$
D) ${{e}_{1}}>/{{e}_{2}}$
• question_answer45) The resolution limit of the eye is 1 min. At a distance$x$km from the eye, two persons stand with a lateral separation of 3m. For the two persons to be just resolved by the naked eye,$x$ should be
A) 10 km
B) 15 km
C) 20 km
D) 30 km
• question_answer46) The necessary condition for an interference by two sources of light is that
A) two light sources must have the same wave length
B) two point sources should have the same amplitude and same wavelength
C) two sources should have the same wave, length, nearly the same amplitude and have a constant phase difference
D) two point sources should have a randomly varying phase difference
• question_answer47) The power factor in a circuit connected to an AC power supply has a value which is:
A) unity when the circuit contains only inductance
B) unity when the circuit contains only resistance
C) zero when the circuit contains an ideal resistance only
D) unity when the circuit contains an ideal capacitance only
• question_answer48) Which part represents the negative dynamic resistance?
A) AB
B) BC
C) CD
D) DE
• question_answer49) The plates of a parallel plate capacitor are not. exactly parallel. The surface charge density
A) is lower at the closer end
B) will not be uniform
C) each plate will have the same potential at every point
D) Both [b] and [c]
• question_answer50) A thin prism${{P}_{1}}$with angle$4{}^\circ$arid made from glass of refractive index 1.54 is combined with another thin prism${{P}_{2}}$made from glass of refractive index 1.72 to produce dispersion without deviation. The angle of the prism${{P}_{2}}$is
A) $5.33{}^\circ$
B) $4{}^\circ$
C) $3{}^\circ$
D) $2.6{}^\circ$
• question_answer51) An electric generator is based on
A) Faradays law of electromagnetic induction
B) motion of charged particles in electromagnetic field
C) Newtons laws of motion
D) fission of uranium by slow neutrons
• question_answer52) Which of the following can be zero, when a particle is in motion for some time?
A) Distance
B) Displacement
C) Speed
D) None of these
• question_answer53) An object of mass 10 kg moves at a constant speed of 10 m/s. A constant force, that acts for 4 s on the object, gives it a speed 2 m/s in opposite direction. The force acting on the object is
A) $-3N$
B) $-30\text{ }N$
C) 3N
D) 30 N
• question_answer54) The vessels A and B of equal volume and Weight are immersed in water to depth h. The vessel A has an opening at the bottom through which water can enter. If the work done in immersing A and B are${{W}_{A}}$and${{W}_{B}}$respectively, then
A) ${{W}_{A}}={{W}_{B}}$
B) ${{W}_{A}}<{{W}_{B}}$
C) ${{W}_{A}}>{{W}_{B}}$
D) None of these
• question_answer55) A car of mass 1000 kg moves on a circular track of radius 20 m. If the coefficient of friction is 0.64, then the maximum velocity with which the car can move is
A) 22.4 m/s
B) 5.6 m/s
C) 11.2 m/s
D) None of the above
• question_answer56) If the distance between the sun and the earth is increased by three times then attraction between two will
A) remain constant
B) decrease by 63%
C) increase by 63%
D) decrease by 89%
• question_answer57) A particle is executing SHM. Then, the graph of velocity as a function of displacement is a/an
A) straight line
B) circle
C) ellipse
D) hyperbola
• question_answer58) A long spring, when stretched by a distance$x,$has potential energy U. On increasing the stretching to$nx,$the potential energy of the spring will be
A) $\frac{U}{n}$
B) $nU$
C) ${{n}^{2}}U$
D) $\frac{U}{{{n}^{2}}}$
• question_answer59) Which one of the following would a hydrogen balloon find easiest to lift?
A) 1 kg of steel
B) 1 kg of water.
C) 1 kg of lightly packed feathers
D) All of, the above
• question_answer60) The unit of universal gas constant is
A) watt/K
B) $dyne/{}^\circ C$
C) erg/K
D) $newton/{}^\circ R$
• question_answer61) Ethyl acetate is obtained when methyl magnesium bromide reacts with
A) ethyl formate
B) ethyl chlorofbrmate
C) acetyl chloride
D) carbon dioxide
• question_answer62) The most stable hydride is
A) $N{{H}_{3}}$
B) $P{{H}_{3}}$
C) $As{{H}_{3}}$
D) $Sb{{H}_{3}}$
• question_answer63) The ratio of amounts of${{H}_{2}}S$needed to precipitate all the metal ions from 100 mL of $1\text{ }M\text{ }AgN{{O}_{3}}$and 100 mL of$1\text{ }M\text{ CuS}{{\text{O}}_{4}}$will be
A) $1:1$
B) $1:2$
C) $2:1$
D) None of these
• question_answer64) If the electronegativity difference between two atoms A and B is 2.0, then the percentage of co-valent character in the molecule is
A) 54%
B) 46%
C) 23%
D) 72%
• question_answer65) Which of the following reaction defines$\Delta H_{f}^{o}$?
A) ${{C}_{(diamond)}}+{{O}_{2}}(g)\xrightarrow[{}]{{}}C{{O}_{2}}(g)$
B) $\frac{1}{2}{{H}_{2}}(g)+\frac{1}{2}{{F}_{2}}(g)\xrightarrow[{}]{{}}HF(g)$
C) ${{N}_{2}}(g)+3{{H}_{2}}(g)\xrightarrow[{}]{{}}2N{{H}_{3}}(g)$
D) $CO(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}C{{O}_{2}}(g)$
• question_answer66) Formaldehyde polymerizes to form glucose according to the reaction $6HCHO\rightleftharpoons {{C}_{6}}{{H}_{12}}{{O}_{6}}$ The theoretically computed equilibrium constant for this reaction is found to be$6\times {{10}^{22}}$ If 1 M solution of glucose dissociates according to the above equilibrium, the concentration of formaldehyde in the solution will be
A) $1.6\times {{10}^{-2}}M$
B) $1.6\times {{10}^{-4}}M$
C) $1.6\times {{10}^{-6}}M$
D) $1.6\times {{10}^{-8}}M$
• question_answer67) The electronic configuration of a dipositive ion ${{M}^{2+}}$is 2, 8, 14 and its mass number is 56. The number of neutrons present is
A) 32
B) 42
C) 30
D) 34
• question_answer68) If$X$is the total number of collisions which a gasmolecule register with others per unit time under particular conditions, then the collision frequency of the gas containing N molecules per unit volume is
A) $X/N$
B) $NX$
C) $2NX$
D) $NX/2$
• question_answer69) A hypothetical reaction${{A}_{2}}+{{B}_{2}}\xrightarrow[{}]{{}}2AB$follows the mechanism as given below, ${{A}_{2}}\rightleftharpoons A+A$ (fast) $A+{{B}_{2}}\xrightarrow[{}]{{}}AB+B$ (slow) $A+B\xrightarrow[{}]{{}}AB$ (fast) The order of the overall reaction is
A) 2
B) 1
C) $1\frac{1}{2}$
D) 0
• question_answer70) The mass of helium atom of mass number 4 is 4.0026 amu, while that of the neutron and proton are 1.0087 and 1.0078 respectively on the same scale. Hence, the nuclear binding energy per nucleon in the helium atom is nearly
A) 5 MeV
B) 7 MeV
C) 10 MeV
D) 14 MeV
• question_answer71) Which of the following statements is correct? Dielectric constant of${{H}_{2}}{{O}_{2}}$
A) increases with dilution
B) decreases with dilution
C) is unaffected on dilution
D) None of the above
• question_answer72) For the square planar complex [M] (where, M = central metal and a, b, c and d are monodentate ligands), the number of possible geometrical isomers are
A) 1
B) 2
C) 3
D) 4
• question_answer73) Potash alum dissolves in water to give a/an
A) acidic solution of ${{H}_{2}}S{{O}_{4}}$
B) alkaline solution
C) acidic solution of $HCl$
D) neutral solution
• question_answer74) The discovery of which of the following group of elements gave death blow to the Newlands law of octaves?
A) Inert gases
B) Alkaline earths
C) Rare earths
D) Actinides
• question_answer75) Vant Hoff factor more than unity indicates that the solute in solution has
A) dissociated
B) associated
C) Both [a] and [b]
D) cannot say anything
• question_answer76) How many number of atoms are there in a cube based unit cell having one atom on each comer and two atoms on each body diagonal of cube?
A) 8
B) 6
C) 4
D) 9
• question_answer77) Bleeding due to a cut can be stopped by applying ferric chloride solution in the laboratory. This is due to
A) co-agulation of negatively charged blood particles by$F{{e}^{3+}}$ions
B) co-agulation of positively charged blood particles by$C{{l}^{-}}$ions
C) reaction taking place between ferric ions and the haemoglobin forming a complex
D) common element, iron, in both$FeC{{l}_{3}}$and haemoglobin.
• question_answer78) Which one of the following solutions will have highest conductivity?
A) $0.1\text{ }M\text{ }C{{H}_{3}}COOH$
B) $0.1\text{ }M\text{ }NaCl$
C) $0.1\text{ }M\text{ }KN{{O}_{3}}$
D) $0.1\text{ }M\text{ }HCl$
• question_answer79) One of the following metals forms a volatile compound and this property is taken advantage for its extraction. This metal is
A) iron
B) nickel
C) cobalt
D) tungsten
• question_answer80) If$N{{a}^{+}}$ion is larger than$M{{g}^{2+}}$ion and${{S}^{2-}}$ion is larger than CF ion, which of the following will be stable soluble in water?
A) Sodium chloride
B) Sodium sulphide
C) Magnesium chloride
D) Magnesium sulphide
• question_answer81) Impurities of Cu and Ag from gold are removed by
A) boiling impure gold with dil. ${{H}_{2}}S{{O}_{4}}$
B) boiling impure gold with cone. ${{H}_{2}}S{{O}_{4}}$
C) electrolytically
D) Both [b] and [c]
• question_answer82) Which of the following salt would give$S{{O}_{2}}$ with hot and dil.${{H}_{2}}S{{O}_{4}}$and also decolourises$B{{r}_{2}}$water?
A) $N{{a}_{2}}S{{O}_{3}}$
B) $NaHS{{O}_{4}}$
C) $N{{a}_{2}}S{{O}_{4}}$
D) $N{{a}_{2}}S$
• question_answer83) If two compounds have the same empirical formula but different molecular formulae, they must have
A) different percentage composition
B) different molecular weights
C) same viscocity
D) same vapour density
• question_answer84) Among the following which one has weakest carbon-halogen bond?
A) Benzyl bromide
B) Bromobenzene
C) Vinyl bromide
D) Benzyl chloride
• question_answer85) Petrochemicals can be used to prepare
A) synthetic fibres
B) pesticides
C) plastics
D) All of these
• question_answer86) tert-butyl methyl ether on heating with anhydrous HI in ether gives
A) $C{{H}_{3}}OH+{{(C{{H}_{3}})}_{3}}Cl$
B) $C{{H}_{3}}I+{{(C{{H}_{3}})}_{3}}COH$
C) $C{{H}_{3}}I+{{(C{{H}_{3}})}_{3}}Cl$
D) None of the above
• question_answer87) The correctly reported answer of the addition of 4.523, 2.3 and 6.24 will have significant figures
A) two
B) three
C) four
D) five
• question_answer88) What happens if$CC{{l}_{4}}$is treated with$AgN{{O}_{3}}$?
A) A white ppt. of$AgCl$will form
B) $N{{O}_{2}}$will be evolved
C) $CC{{l}_{4}}$will dissolve in$AgN{{O}_{3}}$
D) Nothing will happen
• question_answer89) $^{23}Na$ is more stable isotope of Na. Find out the process by which$_{11}^{24}Na$can undergo radioactive decay
A) ${{\beta }^{-}}$ emission
B) $\alpha -$ emission
C) ${{\beta }^{+}}$emission
D) $K$electron capture
• question_answer90) The heat of combustion of solid benzoic acid at constant volume is$-321.30\text{ }kJ$at$27{}^\circ C$. The heat of combustion at constant pressure is
A) $-321.30-300\text{ }R$
B) $-321.30+300\text{ }R$
C) $-321.30-150\text{ }R$
D) $-321.30+900\text{ }R$
• question_answer91) In which of the following compounds$-OH$ group is least reactive?
A)
B)
C)
D) All are equally reactive.
• question_answer92) lodoform is obtained when ethanol is heated with
A) $KI$and aq KOH
B) ${{I}_{2}}$and aq KOH
C) ${{I}_{2}}/$aq$KI$
D) $HI$and $HI{{O}_{3}}$
• question_answer93) The total number of acylic isomers including the stereoisomers (geometrical and optical), with the molecular formula${{C}_{4}}{{H}_{7}}Cl$is
A) 12
B) 11
C) 10
D) 9
A) $RCl,$and$RBr$but not$R\text{ }F$or $R\text{ }I$
B) $RF,\text{ }RCl$and $R\text{ }Br$ but not$R\text{ }I$
C) $RF,RCl,RBr,R\,I$
D) $RF,RCl$and $R\,I$but$RBr$
A) acidic flux only
B) gangue only
C) basic flux only
D) both gangue and acidic flux
• question_answer96) The nodes present in $3p-orbitals$ are
A) one spherical, one planar
B) two spherical
C) two planar
D) one planar
• question_answer97) The number of$\alpha -$and$\beta -$ particles emitted in nuclear reaction$_{90}T{{h}^{228}}{{\xrightarrow[{}]{{}}}_{83}}B{{i}^{212}}$are respectively
A) 4, 1
B) 3, 7
C) 8, 1
D) 4, 7
• question_answer98) Two bottles A and B contains 1 M and 1 m aqueous solution of sulphuric acid respectively
A) A is more concentrated than B
B) B is more concentrated than A.
C) concentration of A is equal to concentration of B
D) it is not possible to compare the concentrations
• question_answer99) A salt on treatment with dil.$HCl$gives a pungent smelling gas and a yellow precipitate. The salt gives green flame test and a yellow precipitate with potassium chromate the salt is
A) $NiS{{O}_{4}}$
B) $Ba{{S}_{2}}{{O}_{3}}$
C) $Pb{{S}_{2}}{{O}_{3}}$
D) $CuS{{O}_{4}}$
• question_answer100) Which of the oxide of manganese is amphoteric?
A) $Mn{{O}_{2}}$
B) $M{{n}_{2}}{{O}_{3}}$
C) $M{{n}_{2}}{{O}_{7}}$
D) $MnO$
• question_answer101) Which of the following alkenes is most reactive towards cationic polymerization?
A) $C{{H}_{2}}=CHC{{H}_{3}}$
B) ${{H}_{2}}C=CHCl$
C) ${{H}_{2}}C=CH{{C}_{2}}{{H}_{5}}$
D) ${{H}_{2}}C=CHC{{O}_{2}}C{{H}_{3}}$
• question_answer102) An organic compound,${{C}_{3}}{{H}_{6}}O$does not give a precipitate with 2, 4-dinitrophenyl hydrazine reagent and does not react with metallic sodium. It could be
A) $C{{H}_{3}}C{{H}_{2}}CHO$
B) $C{{H}_{2}}=CHC{{H}_{2}}OH$
C) $C{{H}_{3}}COC{{H}_{3}}$
D) $C{{H}_{2}}CHOC{{H}_{3}}$
• question_answer103) Oxidation of 1-butene with hot$KMn{{O}_{4}}$solution produces
A) $C{{H}_{3}}C{{H}_{2}}COOH+HCOOH$
B) $C{{H}_{3}}C{{H}_{2}}COOH+C{{O}_{2}}$
C) $C{{H}_{3}}COOH+C{{O}_{2}}$
D) ${{(C{{H}_{3}})}_{2}}C=O+C{{O}_{2}}$
• question_answer104) A mixture of 1-chlorobutane and 2-chloro- butane when treated with alcoholic KOH gives:
A) 1-butene
B) 2-butene
C) isobutylene
D) mixture of 1-butene$+2-$butane
• question_answer105) Out of the two compounds shown below, the vapour pressure of B at a particular temperature is expected to be
A) higher than that of A
B) lower than that of B
C) same as that of A
D) can be higher or lower depending upon the size of the vessel.
• question_answer106) Roasted tin stone ore after washing with water is known as
A) block tin
B) white tin
C) black tin
D) granulated tin
• question_answer107) Which of the following has strongest hydrogen bonding?
A) Ethylamine
B) Ammonia
C) Ethyl alcohol
D) Diethyl ether
• question_answer108) Consider the following statements The rate law for the acid catalysed hydrolysis of an ester being given as Rate$=k[{{H}^{+}}][ester]=k[ester]$. If the acid concentration is doubled at constant ester concentration
1. The second order rate constant, k is doubled. 2. The pseudo first order rate constant, k is doubled. 3. The rate of the reaction is doubled
Which of the above statements are correct?
A) 1 and 2
B) 2 and 3
C) 1 and 3
D) 1, 2 and 3
• question_answer109) A fibrous mineral which can withstand red hot flames without any damage is
A) talc
B) glass wool
C) soap stone
D) asbestos
• question_answer110) When o- or$p-$phenolsulphonic acid is treated with bromine water, the product formed is
A) 2, 4-dibromophenol
B) 2, 4, 6-tribromophenol
C) 3-bromophenol boric acid
D) 3, 5-dibromophenol
• question_answer111) An alkene on vigorous oxidation with$KMn{{O}_{4}}$gives only acetic acid. The alkene is
A) $C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}$
B) $C{{H}_{3}}CH=CHC{{H}_{3}}$
C) ${{(C{{H}_{3}})}_{2}}C=C{{H}_{2}}$
D) $C{{H}_{3}}CH=C{{H}_{2}}$
• question_answer112) Which of the following has the maximum heat of hydrogenation?
A)
B)
C)
D)
• question_answer113) Deflection back of a few particles on hitting thin foil of gold shows that
A) nucleus is heavy
B) nucleus is small
C) Both [a] and [b]
D) electrons create hinderance in the movement of$\alpha -$particles
• question_answer114) Which of the following does not form an oxime?
A) Glucose
B) Glucose penta-acetate
C) Arabinose
D) Galactose
• question_answer115) An aromatic compounds X with molecular formula${{C}_{8}}{{H}_{10}}$produces on nitration one mononitro derivative and three dinitro derivatives. Compound X would be
A) ethyl benzene
B) m-xylene
C) o-xylene
D) p-xylene
• question_answer116) The end product C in the following sequence of chemical reactions is $C{{H}_{3}}COOH\xrightarrow[{}]{CaC{{O}_{3}}}A\xrightarrow[{}]{heat}B\xrightarrow[{}]{N{{H}_{2}}OH}C$
A) acetaldehyde oxime
B) formaldehyde oxime
C) methyl nitrate
D) acetoxime
• question_answer117) The only cations present in a slightly acidic solution are$F{{e}^{3+}},Z{{n}^{2+}}$and$C{{u}^{2+}}$. The reagent that when added in excess to this solution would identify and separate$F{{e}^{3+}}$in one step is
A) $2M\,HCl$
B) $6M\,N{{H}_{3}}$
C) $6\,M\,NaOH$
D) ${{H}_{2}}S$gas
• question_answer118) The nucleus of an atom can be assumed to be spherical. The radius of the nucleus of mass number A is given by$1.25\times {{10}^{-13}}\times {{A}^{1/3}}cm$.Radius of atom is one$\overset{o}{\mathop{\text{A}}}\,$. If the mass number is 64, then the fraction of the atomic volume that is occupied by the nucleus is
A) $1.0\times {{10}^{-3}}$
B) $5.0\times {{10}^{-5}}$
C) $2.5\times {{10}^{-2}}$
D) $1.25\times {{10}^{-13}}$
• question_answer119) Carboxylic acids readily dissolve in aqueous sodium bicarbonate, liberating carbon dioxide. Which one of the following is correct?
A) Free carboxylic acid and its conjugate base are of comparable stability
B) The free carboxylic acid is more stable than its conjugate base
C) The conjugate base of the carboxylic acid is more stable than the free carboxylic acid
D) The conjugate acid of the carboxylic acid is more stable than the free carboxylic acid
A) soft water
B) soft water first heated with lime stone
C) cone.$HN{{O}_{3}}$
D) soft as well as hard water
• question_answer121) The branch of biology, which deals with the study of processes and functions of an organism, is called
A) histology
B) anatomy
C) physiology
D) entomology
A) soil
C) living tissue
D) culture medium
• question_answer123) Which of the following technique, other than microscopy is used for study of cell?
A) Maceration
B) Plasmolysis
C) Chromatography
• question_answer124) Robert Hooke used the term cell in the year
A) 1650
B) 1665
C) 1865
D) 1960
• question_answer125) Protein synthesis takes place in
A) ribosomes
B) chloroplasts
C) mitochondria
D) Golgi bodies
• question_answer126) Replication of centriole occurs during
A) interphase
B) prophase
C) late prophase
D) late telophase
• question_answer127) L-shaped chromosomes are also called
A) acrocentric
B) telocentric
C) sub-metacentric
D) None of these
• question_answer128) Which of the following is/are grouped under phanerogams?
A) Angiosperms
B) Gymnosperms
C) Pteridophytes
D) Both [a] and [b]
• question_answer129) A bacterium divides after every 35 minutes. If a culture containing${{10}^{5}}$cells per mL is grown, then cell concentration per mL after 175 minutes will be
A) $175\times {{10}^{5}}$
B) $125\times 10s$
C) $48\times {{10}^{5}}$
D) $32\times {{10}^{5}}$
A) algae
B) lichen
C) fungus
D) angiosperm
• question_answer131) Which of the following is an algal parasite?
A) Volvox
B) Ulothrix
C) Porphyra
D) Cephaleuros
A) They are thalloid
B) They contain chloroplast
C) They possess archegonia
D) All of the above
• question_answer133) The kidney-shaped covering of sorus in Dryopteris, is called
A) placenta
B) ramentum
C) sporophyll
D) indusium
• question_answer134) Which of the following is a wheat fruit?
A) Achene
B) Cypsella
C) Caryopsis
D) Endosperm
• question_answer135) A gymnospermic leaf carries 16 chromosomes. The number of chromosomes in its endosperm is
A) 24
B) 16
C) 12
D) 8
• question_answer136) Tyioses thickenings are seen in
A) collenchyma
B) phloem cells
C) ray parenchyma only
D) ray parenchyma and xylem cells
• question_answer137) Which of the following plant shows multiple epidermis?
A) Croton
B) Allium
C) Nerium
D) Cucurbita
• question_answer138) Double fertilization occurs among
A) algae
B) bryophytes
C) angiosperms
D) gymnosperms
• question_answer139) The endosperm in angiosperm develops from
A) zygote
B) secondary nucleus
C) chalazal polar nucleus
D) micropylar polar nucleus
• question_answer140) The fertilization in which male gametes are carried through pollen tube is known as
A) syngamy
B) porogamy
C) siphonogamy
D) chalazogamy
• question_answer141) Which of the following theory gives the latest explanation for the closure of stomata?
A) ABA theory
B) Munch theory
C) Starch glucose theory
D) Active${{K}^{+}}$transport theory
• question_answer142) Loss of liquid water by guttation occurs through
A) hydathodes
B) stomata
C) cuticle
D) bark
• question_answer143) Which of the following is the first compound that accepts carbon dioxide during dark phase of photosynthesis?
B) RuBP
C) Ferredoxin
D) Cytochrome
• question_answer144) Oxygen which is liberated during photosynthesis, comesfrom
A) $C{{O}_{2}}$
B) ${{H}_{2}}O$
C) chlorophyll
D) phosphoglyceric acid
• question_answer145) In${{C}_{4}}$plants, the carbon dioxide fixation occurs in
A) guard cells
B) spongy cells
D) bundle sheath cells
• question_answer146) Ethyl alcohol is commercially manufactured from
A) bajra
B) grapes
C) maize
D) sugarcane
• question_answer147) Last electron acceptor in respiration is
A) ${{O}_{2}}$
B) ${{H}_{2}}$
C) $C{{O}_{2}}$
• question_answer148) A hormone delaying senescence is
A) auxin
B) cytokinins
C) ethylene
D) gibberellin
• question_answer149) Which of the following induces flowering in long day plants?
A) Gibberellins
B) Cytokinin
C) Auxins
D) Ethylene
• question_answer150) What name has been assigned to the genus produced by a cross between cabbage and radish?
A) Secale
B) Bursa pastoris
C) Lysogenicophyll
D) Raphanobrassica
• question_answer151) Genetic counsellors can identify heterozygous individuals by
A) height of individuals
B) colour of individuals
C) screening procedures
D) All of the above
• question_answer152) The term heterosis was first coined by
A) McClintock
B) Poweri
C) Swaminathan
D) None of these
• question_answer153) Somatic hybridization is a technique of
A) natural breeding
B) natural pollination
C) artificial pollination
D) somatic cells hybridization
• question_answer154) Which of the following is initiation codon in eukaryotes?
A) UAG
B) UGA
C) UAA
D) AUG
• question_answer155) Which of the following species are restricted to an area?
A) Sibling species
B) Endemic species
C) Allopatric species
D) Sympatric species
• question_answer156) Biological concept of species is mainly based on
A) reproductive isolation
B) morphological features only
C) methods of reproduction only
D) morphology and methods of reproduction
• question_answer157) Which of the following supports a dense population of plankton and littoral vegetation?
A) Oligotrophic
B) Eutrophic
C) Lithotrophic
D) Agroecotrophic
A) less in diversity
B) more in diversity
C) man does not make ecosystem
D) more stable than natural ecosystem
• question_answer159) Crop rotation is used by farmers to increase
A) soil fertility
B) community area
C) organic content of soil
D) nitrogenous content in the soil
• question_answer160) Which of the following organism form the decomposers?
A) Pteris
B) Bacteria
C) Saprophytic fungi
D) Both [b] and [c]
A) insectivorous plant
B) light airborn paniculate matter
C) new name of orchid plant
D) causal organism of various diseases
• question_answer162) Some effects of sulphur dioxide and its transformation products on plant include
A) plasmolysis
B) Golgi body destruction
C) protein dis-integration
D) chlorophyll destruction
• question_answer163) Which of the following plant is used for the purification of water?
A) Beggiata
B) Chlorella
C) Spirogyra
D) Eichhornia
• question_answer164) In the treatment of waste water discharge, which treatment stage involves biological treatment?
A) Primary treatment
B) Secondary treatment
C) Tertiary treatment
D) Reverse osmosis stage
• question_answer165) Which of the following is considered to be the best chemical method of fixing atmospheric nitrogen?
A) Fisher method
B) Decan method
C) Haber-Bosch method
D) Paranas-Meyerhoff method
• question_answer166) Desired improved varieties of economically useful crops are raised by
A) migration
B) biofertilizer
C) hybridization
D) natural selection
• question_answer167) Indian rose wood tree is a common name of
A) Acacia
B) Shorea
C) Dalbergia
D) Eucalyptus
• question_answer168) Emasculation is concerned with
A) hybridization
B) clonal selection
C) mass selection
D) pure line selection
• question_answer169) Which of the following insecticide is obtained from the roots of Den-is elleptica?
A) Cinerin
B) Nicotine
C) Rotenone
D) Pyrethmm
• question_answer170) In maize, hybrid vigour is exploited by
A) bombarding the protoplast with DNA
B) crossing of two inbreed parental lines
C) harvesting seeds from the most productive plants
D) inducing mutations
• question_answer171) Which of the following is not vestigial in man?
A) Tail vertebrae
B) Nails
C) Nictitating membrane
D) Vermiform appendix
A) 2, 4-D
B) BHC
C) DDT
D) Pyrethroid
• question_answer173) A eukaryotic gene contains two kinds of base sequences. Which of these plays an important role in protein synthesis?
A) Introns
B) Exons
C) Both [a] and [b]
D) None of these
• question_answer174) The number of hydrogen bonds between adenine and thymine in a DNA molecule is
A) two
B) three
C) four
D) eight
• question_answer175) The enzyme, which combines with non-protein part to form a functional enzyme known as
A) Co-enzyme
B) holoenzyme
C) apoenzyme
D) prosthetic group
• question_answer176) Which of the following enzyme digest protein in stomach?
A) Trypsin
B) Pepsin
C) Crepsin
D) None of these
• question_answer177) Passive food ingestion in Amoeba is known as
A) import
B) invagination
C) circumfluence
D) circumvallation
• question_answer178) The slime moulds are characterized by the presence of
A) elaters
B) pseudoelaters
C) capillitium
D) capitulum
• question_answer179) Ecdysone is secreted from
A) Insecta
B) Trematoda
C) Nematoda
D) Polychaeta
• question_answer180) In the life cycle of mosquito, comma-shaped stage is
A) larval stage
B) pupal stage
C) imago stage
D) None of these
• question_answer181) Haemocoel is found in
A) Hydra and Aurelia
B) Taenia and Ascaris
C) Cockroach and Pila
D) Balanoglossus and Herdmania
• question_answer182) The group ofAnamniota includes
A) reptiles and birds
B) birds and mammals
C) fishes and amphibians
D) reptiles and mammals
• question_answer183) The excretory material of bony fish is
A) urea
B) protein
C) ammonia
D) ammo acid
• question_answer184) Different colour of frog skin are controlled by
A) hormones
B) melanocytes :
C) nervous system
D) Both [a] and [c]
• question_answer185) Blastula of frog has
A) blastopore
B) blastocoel
C) archenteron
D) gastropore
• question_answer186) Carotene pigment is found in the cells of
A) dermis
B) epidermis
D) Both [b] and [c]
• question_answer187) Deboves membrane is a layer of
A) muscular tissue
B) epithelial tissue
C) connective tissue
D) All of the above
• question_answer188) Achilles tendon is associated with
A) gluteus muscle
B) hamstring muscle
D) gastrocnemius muscle
• question_answer189) The leucocytes contain, which of the following in large quantity?
A) Basophils
B) Neutrophils
C) Eosinophils
D) Monocytes
• question_answer190) Which part of our body secreted the hormone secretin?
A) Ileum
B) Stomach
C) Duodenum
D) Oesophagus
• question_answer191) During inspiration, the diaphragm
A) expands
B) shows no change
C) contracts and flattens
D) relaxes to become dome-shaped
• question_answer192) The oxygen toxicity is related with
A) blood poisoning
B) collapse of alveolar walls
C) failure of ventilation of lungs
D) Both [a] and [b]
• question_answer193) Cardiac output is determined by
A) heart rate
B) stroke volume
C) blood flow
D) Both [a] and [b]
• question_answer194) The important function of lymph is to
A) transport oxygen to the brain
B) transport carbon dioxide to the lungs
C) return RBCs to the lymph nodes
D) return interstitial fluid to the blood
• question_answer195) The lining of intestine and kidneys in humans is
A) keratinized
B) brush border
C) ciliated
D) None of these
• question_answer196) The yellow colour of urine is due to the presence of
A) urea
B) uric acid
C) urochrome
D) bilirubin
• question_answer197) The Leydig cells secrete
A) oestrogen
B) testosterone
C) progesterone
D) corticosterone
• question_answer198) The function of pineal body is to
A) lighten the skin colours
B) control sexual behaviour
C) regulates the period of puberty
D) All of the above
• question_answer199) Which of the following nerve is purely motor nerve?
A) Vagus
B) Facial
C) Abducens
D) Trigeminal
• question_answer200) Which of the following part of a neuron is covered by fatty sheath?
A) Axon
B) Cyton
C) Dendrite
D) Node of Ranvier
• question_answer201) Directions: In each of the following questions, choose the alternative which best expresses the meaning of the word given in capital letters. INTREPID
A) Coward
B) Fearless
C) Selfish
D) Ugly
• question_answer202) Directions: In each of the following questions, choose the alternative which best expresses the meaning of the word given in capital letters. LAMENTABLE
A) Deplorable
B) Contemptible
C) Remorseful
D) Scornful
• question_answer203) Directions: In each of the following questions, choose the alternative which best expresses the meaning of the word given in capital letters. DEFT
A) Skilful
B) Vigorous
C) Swift
D) Deceitful
• question_answer204) Directions: In each of the following questions, choose the alternative which best expresses the meaning of the word given in capital letters. HOARD
A) Destroy
B) Hide
C) Store
D) Divide
• question_answer205) Directions: In each of the following questions, choose the most suitable alternative to fill in the blank. He is too dull...... the problem.
A) solves
B) to solve
C) solving
D) to solving
• question_answer206) Directions: In each of the following questions, choose the most suitable alternative to fill in the blank. The speaker drew the attention of the audience ...... the burning issue.
A) to
B) towards
C) on
D) into
• question_answer207) Directions: In each of the following questions, choose the most suitable alternative to fill in the blank. Its nine oclock ...... and Im still at breakfast.
A) till
B) yet
C) so
• question_answer208) Directions: In each of the following questions, choose the most suitable alternative to fill in the blank. Although he is blind, he is very fast ...... calculations.
A) in
B) with
C) at
• question_answer209) Directions: In each of the following questions, choose the alternative which is opposite in meaning to the word given in capital letters. VIRTUE
A) Vice
B) Fraud
C) Wickedness
D) Crime
• question_answer210) Directions: In each of the following questions, choose the alternative which is opposite in meaning to the word given in capital letters. ZEAL
A) Hostility
B) Diffidence
C) Apathy
D) Contempt
• question_answer211) Directions: In each of the following questions, choose the alternative which is opposite in meaning to the word given in capital letters. AGONY
A) Enmity
B) Cruelty
C) Abhorrence
D) Ecstasy
• question_answer212) Directions: In each of the following questions, choose the alternative which is opposite in meaning to the word given in capital letters. ERUDITE
A) Professional
B) Immature
C) Unimaginative
D) Ignorant
• question_answer213) A person who speaks for or supports an idea
A) Pioneer
D) Ideologist
• question_answer214) A man of odd habits
A) Eccentric
B) Cynical
C) Introvert
D) Moody
• question_answer215) A thing or person behind time
A) Lazy
B) Sluggish
C) Indolent
D) Antiquated
• question_answer216) One whose attitude is: eat, drink and be merry
A) Epicurean
B) Cynic
C) Materialistic
D) Stoic
• question_answer217) Directions: In each of the following questions, choose the alternative which can best improve the given sentence by substituting the italicised portion. If the sentence is correct as it is, your answer is. Older people often stay at home and watch TV because it is cold and dark in winter.
A) seldom
B) frequently
C) sometimes
D) No improvement
• question_answer218) Directions: In each of the following questions, choose the alternative which can best improve the given sentence by substituting the italicised portion. If the sentence is correct as it is, your answer is. You must find someone to accompany you to Mumbai
A) no one
B) everyone
C) anyone
D) No improvement
• question_answer219) Directions: In each of the following questions, choose the alternative which can best improve the given sentence by substituting the italicised portion. If the sentence is correct as it is, your answer is. No sooner he reached home than all the villagers gathered at his home to listen to his story.
A) would he reach
B) did he reach
C) have he reached
D) No improvement
• question_answer220) Directions: In each of the following questions, choose the alternative which can best improve the given sentence by substituting the italicised portion. If the sentence is correct as it is, your answer is. I wish I was with him.
A) have been
B) were
C) am
D) No improvement
• question_answer221) Directions: In the following questions, choose the option which shows common feature in the relationship given in each question. Sandstone : Limestone : Coal
A) They are formed by metamorphic rocks.
B) They are chemical minerals.
C) They are found in river beds.
D) They are formed in sedimentary rocks.
• question_answer222) Directions: In the following questions, choose the option which shows common feature in the relationship given in each question. Sweep : Scrub : Wipe
A) These are terms connected with rubbing.
B) These are games of cards.
C) These are terms used by motor mechanics.
D) These are terms connected with cleaning.
• question_answer223) Directions: In the following questions, choose the option which shows common feature in the relationship given in each question. Delhi: Agra : Mathura
A) They have been capitals of the country
B) They have exquisite temples.
C) They have religious background
D) They are situated on the bank of river Yamuna.
• question_answer224) Directions: In the following questions, choose the option which shows common feature in the relationship given in each question. Press : Television : Cinema
A) They are means of entertainment.
B) They are means of mass media.
C) They give world wide news.
D) All are public undertakings.
• question_answer225) Directions: In the following questions, choose the option which shows common feature in the relationship given in each question. Comets : Stars : Satellites
A) They are shining masses.
B) They give out light.
C) They are rotating from left to right
D) They are heavenly bodies.
• question_answer226) Directions: In the questions, choose the group of words that shows the same relationship as given at the top of every question. Magazine : Story : Article
A) Tea : Milk : Sugar
B) Television : Newspaper : Entertainment
C) Bed : Quilt: Pillow
D) Novel: Drama : Literature
• question_answer227) Directions: In the questions, choose the group of words that shows the same relationship as given at the top of every question. Juice : Orange : Banana
A) Table: Chair : Wood
B) Fish : Shork : Water
C) Cow : Milk : Curd
D) Ink : Pen : Pencil
• question_answer228) Directions: In the questions, choose the group of words that shows the same relationship as given at the top of every question. Carnivorous : Tiger : Wolf
A) Mango : Banana : Fruit
B) Worker : Master : Manager
C) Cat: Cow : Milk
D) Student: Boy: Girl
• question_answer229) Directions: In the questions, choose the group of words that shows the same relationship as given at the top of every question. Rain : Cloud : Evaporation
A) Pain : Injury : Accident
B) Cold : Cough : Sneezing
C) Purse : Leather : Tanning
D) Fragrance : Flower : Bud
• question_answer230) Directions: In the questions, choose the group of words that shows the same relationship as given at the top of every question. Dog : Squirrel: Tail
A) Cottage : Hut: Palace
B) Fish : Crocodile : Water
C) Horse : OX : Horn
D) Truck : Scooter : Gear
• question_answer231) The five intertwined rings or circles on the Olympic flag made of white [From left to right] are
A) Blue, yellow, black, green and red.
B) Yellow, red, green, black and blue.
C) Red, green, black, yellow and blue.
D) Yellow, green black, blue and red.
• question_answer232) The national remote sensing agency is located at
A) Delhi
C) Bangalore
D) Lucknow
• question_answer233) The Nobel Prize are given to Dec, 10 on the death anniversary of
A) Linus Pauling
B) Frederic Sanger
C) Alfred Bernard
D) John Bardeen
• question_answer234) The term Gambit is used in
A) Chess
B) Boning
C) Baseball
D) Polo
• question_answer235) Words fastest missileship TNS Prahar was commissioned in
A) 1996
B) 1997
C) 1998
D) 1999
• question_answer236) Nati is the classical dances of
B) Assam
A) Geneva
B) Paris
C) New York
D) Rome
A) Mother Teresa
B) Mrs. Indira Gandhi
C) Ashapurana devi
D) Aarti Sana
• question_answer239) Kip is the currency of
A) Kuwait
B) Lebanon
C) Laos
D) Malaysia
• question_answer240) The charter of the United Nations was signed on June 26, 1945 in
A) San Francisco
B) Washington D.C.
C) Landon
D) Trygue Le
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2019-01-17 13:29:36
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https://math.oxford.emory.edu/site/math117/events/
|
Events
To clearly and precisely define what one means by an "event" in discussions related to probability, we'll need some additional vocabulary first:
A random experiment is a chance process that leads to well-defined results called outcomes that can be described prior to its performance. As examples, both flipping a coin and rolling a die would be random experiments.
A simple event is an outcome of a single trial of a random experiment. For the previous examples, a simple event might be seeing "heads" on a particular flip or be the number showing on a rolled die.
The sample space for a random experiment is the set of all possible simple events with which it is associated. So for the flip of a coin, the sample space might be represented by the set $S = \{H, T\}$ where $H$ corresponds to seeing "heads" and $T$ corresponds to seeing "tails". For the roll of a standard die, we would likewise have a sample space of $S = \{1,2,3,4,5,6\}$. If two dice are rolled, the sample space would consist of all 36 possible ordered pairs: $(1,1),\ldots,(1,6),(2,1),\ldots,(2,6),\ldots,(6,6)$. Note, this set can be conveniently described using a table: $$\begin{array}{c|c|c|c|c|c|c} & 1 & 2 & 3 & 4 & 5 & 6\\\hline 1 & (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6)\\\hline 2 & (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6)\\\hline 3 & (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6)\\\hline 4 & (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6)\\\hline 5 & (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6)\\\hline 6 & (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6)\\\hline \end{array}$$
Finally, an event is a subset of the sample space. Every simple event is certainly an event, but events can be compound in nature as well. For example, one can consider the event of rolling an even number. The related subset of $S$ is then $E = \{2,4,6\}$. Similarly, the event of rolling two dice and seeing only 1's or 2's on each individual die could be described by the cells marked in blue below, while the event of rolling two dice and getting a 2,3,4, or 5 on the first die, and a 4 or 5 on the second die, could be described by the cells marked in red below.: $$\require{color}$$ $$\begin{array}{c|c|c|c|c|c|c} & 1 & 2 & 3 & 4 & 5 & 6\\\hline 1 & {\color{blue}\fbox{(1,1)}} & {\color{blue}\fbox{(1,2)}} & (1,3) & (1,4) & (1,5) & (1,6)\\\hline 2 & {\color{blue}\fbox{(2,1)}} & {\color{blue}\fbox{(2,2)}} & (2,3) & {\color{red}\fbox{(2,4)}} & {\color{red}\fbox{(2,5)}} & (2,6)\\\hline 3 & (3,1) & (3,2) & (3,3) & {\color{red}\fbox{(3,4)}} & {\color{red}\fbox{(3,5)}} & (3,6)\\\hline 4 & (4,1) & (4,2) & (4,3) & {\color{red}\fbox{(4,4)}} & {\color{red}\fbox{(4,5)}} & (4,6)\\\hline 5 & (5,1) & (5,2) & (5,3) & {\color{red}\fbox{(5,4)}} & {\color{red}\fbox{(5,5)}} & (5,6)\\\hline 6 & (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6)\\\hline \end{array}$$ Note, in the above example, the individual outcomes that made up each of the events described were "clustered together" in the sample space. This certainly need not always be the case -- consider the event of rolling either "snake eyes" or "box cars" -- i.e., rolling two ones or two sixes, respectively. However, the "blobs" of red and blue in the above table suggest a useful way to envision a set...
Suppose we represent the sample space for a particular random experiment by a rectangular region, and events as "blobs" (often circular or elliptical, but not always so) contained within this region. It should not be hard to see that if we refer to the event associated with the blue cells in the table above as event A, and the event associated with the red cells in the same table as event B, we might end up with a picture like the following.
The events $A$ and $B$ shown above certainly appear to share no common possible outcomes. Note, when sets have no elements in common, they are said to be disjoint.
In general, however, this certainly doesn't have to be the case. Just consider the event $C$ of rolling a sum that is at most a 4 and the event $D$ of rolling a three or greater on at least one die. In this case, $(1,3)$ and $(3,1)$ are possible outcomes for both events.
As such, to visually suggest the possibility for common outcomes when they are present, we often draw our event "blobs" so that they overlap, as shown below:
Pictures like the one above that depict sets and their possible overlaps in this way are called Venn Diagrams.
Events Defined in Terms of Other Events
A common theme in mathematics is how to build new things from known things. Consider the simple functions $f(x) = x^3$ and $g(x) = \sin x$. We can build a new function from the first by finding its inverse, $f^{-1}(x) = \sqrt[3]{x}$. Alternatively, we can combine both of the functions given to build new functions by considering their sum: $(f+g)(x) = x^3 + \sin x$, their product: $(fg)(x) = x^3\sin x$, their composition: $(f \circ g)(x) = \sin^3 x$, etc..
We can do the same thing with events. As perhaps the simplest example, consider an event $A$ in some sample space $S$. Recall that $A$ is really just a subset of the possible outcomes that comprise $S$, and we say that event $A$ has occurred if any of these outcomes occur. But then, it is easy to describe another event, called the complement of $A$, and denoted by $\overline{A}$, that occurs precisely when $A$ doesn't occur. That is to say, $\overline{A}$ is the subset of outcomes in $S$ that are not in $A$.
To provide a couple of examples for context:
• The complement of an even roll of a single die would be an odd roll.
• The complement of flipping a coin three times and seeing 3 heads would be seeing 2 or fewer heads.
• The complement of drawing a card from a deck and getting a spade would be getting a heart, clubs, or diamonds.
Visually, the animation below suggests the relationship between $A$ and $\overline{A}$:
There are a couple of useful ways to combine two events as well. Given two events $A$ and $B$, we could define the event that consists of all outcomes that are in either A or B (or both). As the following animation suggests, this is called the union of $A$ and $B$ and is denoted by either $A \cup B$, or equivalently by $A \textrm{ or } B$:
In a similar vein, given two events $A$ and $B$, we could define the event that consists of all outcomes in both $A$ and $B$. As again suggested by the animation below, this is called the intersection of $A$ and $B$ and is denoted by $A \cap B$, or equivalently $A \textrm{ and } B$:
Lastly, an event that contains no outcomes of a given sample space, sometimes called a null event, is appropriately denoted by the symbol for a set with no elements -- the empty set, $\varnothing$. So for example, the intersection of two disjoint events is $\varnothing$.
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https://includestdio.com/7202.html
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# python – How to get a random number between a float range?
## The Question :
470 people think this question is useful
randrange(start, stop) only takes integer arguments. So how would I get a random number between two float values?
The Question Comments :
• If you wanted numpy it’s np.random.uniform(start, stop) or np.random.uniform(start, stop, samples) if you wanted multiple samples. Otherwise below answers are best.
## The Answer 1
724 people think this answer is useful
>>> random.uniform(1.5, 1.9)
1.8733202628557872
## The Answer 2
81 people think this answer is useful
random.uniform(a, b) appears to be what your looking for. From the docs:
Return a random floating point number N such that a <= N <= b for a <= b and b <= N <= a for b < a.
See here.
## The Answer 3
68 people think this answer is useful
if you want generate a random float with N digits to the right of point, you can make this :
round(random.uniform(1,2), N)
the second argument is the number of decimals.
## The Answer 4
6 people think this answer is useful
Most commonly, you’d use:
import random
random.uniform(a, b) # range [a, b) or [a, b] depending on floating-point rounding
Python provides other distributions if you need.
If you have numpy imported already, you can used its equivalent:
import numpy as np
np.random.uniform(a, b) # range [a, b)
Again, if you need another distribution, numpy provides the same distributions as python, as well as many additional ones.
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http://destthalpa.cleansite.biz/514.php
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# Frascati manual latex cite
, Abbrev. Creating Bibliography with LaTeX There are two ways of producing a bibliography. The author-year citation style requires usage of BibTeX or specially formatted \bibitem commands. ties to cite references from di erent reference management software in LaTeX. The internationally recognised methodology for collecting and using R&D statistics, the OECD's Frascati Manual is an essential tool frascati manual latex cite for statisticians and science and innovation policy makers worldwide. The \BibTeX\ \cite{btxdoc} and \LaTeX\ \cite[Annex~B]{latex} manuals \ldots produces The BibT EX (Patashnik ) and LAT X (Lamport , Annex B) manuals A second new command npossessiveciteis provided for those instances where it is desired to use the citation as a possessive noun phrase. But any abbreviation will work. A) Manually Creating a Bibliography.
[HOST] Do multiple bibliographies--one for each \include file. Use whatever style. Contents. Title says it all. LaTeX makes citing references quite simple, easy and organized that freshmen can have no more excuses for plagiarism. It provides compressed, sorted lists of numerical or partly-numerical citations, as regular text or as superscripts. This command introduces a reference. Below is an example of creating bibliography using “thebibliography” environment.
Edit: /citet and /citep, not twice the same. LaTeX: Bibliographies and Citation Citations are frascati manual latex cite references to books, articles, web pages, or other published items with sufficient details to identify them uniquely. Cite your website in BibTeX generic citation style format for free. When you want to cite an item in the database in a LaTeX document, you write \cite{} (or something similar), where is the label attached to the item in the database. Then you should follow the specific guidance provided. This is why we choose it as our preferred reference management software for BibTex\cite{Silveira}.
It provides compressed, sorted lists of numerical or partly-numerical citations, as . The internationally recognised methodology for collecting and using R&D statistics, the OECD's Frascati Manual is an essential tool for statisticians and science and innovation policy makers worldwide.g.
This article explains how to manage bibliography with the thebibliography environment and the BibTeX system., City of Comp. It assumes no prior knowledge of LATEX, or any other computing language.
Frascati Manual «Frascati Manual PROPOSED STANDARD PRACTICE FOR SURVEYS ON RESEARCH AND EXPERIMENTAL DEVELOPMENT Frascati Manual PROPOSED STANDARD PRACTICE FOR SURVEYS ON RESEARCH AND EXPERIMENTAL DEVELOPMENT The Measurement of Scientific and Technological Activities In an era of rapid technological change, research and development (R&D) is. The Frascati Manual classifies budgets according to what is done, what is studied, and who is studying it. Jun 10, · frascati manual latex cite JabRef is an open source reference management software for BibTex \cite{BESIIICollaboration}. Dec 12, · How do I reference my LaTeX tables or equations?
Oct 07, · For BibTeX this is not much different from printing the table of contents., City of Comp. In the braces, put an identifier that frascati manual latex cite LaTeX can use to refer to your item.
This referencing capability lets you easily give readers the exact number of a figure, or tell them what page number a figure is located on with frascati manual latex cite the use of a few simple commands (\label, . 3 4 Languages,characters,fontsandglyphs 51 Charactersandglyphs 51 Themaincontrolloop 55 Loadingpatternsandexceptions 56 Applyinghyphenation What's the difference between \cite, \citep and \citep. The Frascati Manual is based on the experience gained from collecting R&D statistics in both OECD and non-member countries. Both types are detailed below. (A common choice is the first three letters of the author’s name and the last two digits of the year. \cite{einstein} This command inserts a reference within the document, [1] in this case, that corresponds to an.
Its aim is to introduce an. Go to the point where you want the citation to appear, and use the following: \cite{ cite_key }, where the cite_key is that of the bibitem you wish to cite. Frascati Manual Proposed Standard Practice for Surveys on Research and Experimental Development The internationally recognized methodology for collecting and using R&D statistics, the Frascati Manual is an essential tool for statisticians worldwide. The style is mostly style A, but borrows the author-date reference system from style.
The edition is the result of the collective frascati manual latex cite work of NESTI, involving more than experts from nearly 40 countries and international organisations. Its aim is to introduce an. LaTeX makes citing references quite simple, easy and frascati manual latex cite organized that freshmen can have no more excuses for plagiarism. If the manual is distributed electronically, then you should cite the website from which it is obtained as the "source. \cite{einstein} This command inserts a reference within the document, [1] in this case, that corresponds to an. [#] Name of Manual/Handbook, xth ed. Nov 27, · As I’ve written before we chose to use BibTeX as our lowest common denominator citation export format.
All of these may take one or two optional arguments to add some text before and after the citation. This workbook is designed to be used at the ‘LATEX for Beginners’ student iSkills seminar, and also for self-paced study.. An optional argument can be given to add some text within the brackets. Cross-referencing is important in professional documents. LaTeX tracks cross reference information in a file having the [HOST] and with the same base name as the file containing the \label. In the braces, put an identifier that LaTeX can use to refer to your item.
1 frascati manual latex cite post • Page 1 of 1. We just need the commands \bibliography which tells LaTeX the location of [HOST] file and \bibliographystyle which selects one of various bibliographic styles. There also exist the starred versions \citet* and \citep* that print the full author list, and not just the abbreviated one. This is a variation. See the LaTeX manual or LaTeX .The \BibTeX\ \cite{btxdoc} and \LaTeX\ \cite[Annex~B]{latex} manuals \ldots produces The BibT EX (Patashnik ) and LAT X (Lamport , Annex B) manuals A second new command npossessiveciteis provided for those instances where it is desired to use the citation as a possessive noun phrase.
Referencing elements to cite: [#] Reference number (matching the in-text citation number) Name of Manual/Handbook; If the manual frascati manual latex cite is a second or subsequent edition, include the edition statement, e. LaTeX tracks cross reference information in a file having the [HOST] and with the same base name as the file containing the \label. Apr 17, · If you are conducting research on a product or service, you may reference a user manual or instructional guide in your paper.
The pre-translation version was written by George frascati manual latex cite D. This is a Bib L a T e X style that implements the Chicago ‘author-date’ and ‘notes with bibliography’ style specifications frascati manual latex cite given in the Chicago Manual of Style, 16 th edition (support for the 15 th edition remains, but is ‘strongly deprecated’). The third step is that LaTeX computes a box to bound the rotated material. This is a variation. You put a command in your LaTeX file specifying the style in which frascati manual latex cite you want the. Note: If you are starting from scratch it's recommended to. BibTeX references are stored in a plain text database with a simple format. ties to cite references from di erent reference management software in LaTeX.
Separate the parts by commas. This document was initially made as a quick reference to all the commands that I typically use, organized so I can understand it, with examples and without clutter. Frascati Manual opened the way for measuring one key dimension of science, technology and innovation so that, nowadays, investment frascati manual latex cite in research and development – R&D – is systematically encouraged and monitored around the world. LaTeX Warning: Citation ab' on page 2 undefined on input line LaTeX Error: Something's wrong--perhaps a missing \ item. \documentclass{article} \begin{document} Random citation \cite{DUMMY:1} embeddeed in text. xxx-xxx (pages if relevant). The second step is the rotation, by default about the reference point.
An frascati manual latex cite optional argument can be given to add some text within the brackets. Creating Bibliography with LaTeX There are two ways of producing a bibliography. Dec 12, · How do I reference my LaTeX tables or equations? BibTeX references are stored in a plain text database with a simple format. Whenever you reference a source in your work, cite it properly so your reader knows where you found your information., Abbrev.
\documentclass{article} \begin{document} Random citation \cite{DUMMY:1} frascati manual latex cite embeddeed in text. As one can see, there is no sorting. (A common choice is the first three letters of the author’s name and the last two digits of the year. There also exist the starred versions \citet* and \citep* that print the full author list, and frascati manual latex cite not just the abbreviated one. Citation Machine® helps students and professionals properly credit the information that they use. Imports the bibtex data file [HOST], this file is the frascati manual latex cite one that includes information about each referenced book, article, etc.
Frascati Manual opened the way for measuring one key dimension frascati manual latex cite of science, technology and innovation so that, nowadays, investment in research and development – R&D – is systematically encouraged and monitored around the world. This thread is archived. You can either produce a bibliography by manually listing the entries of the bibliography or producing it automatically using the BibTeX program of LaTeX. The L ATEX version was written by Stephen. Stack Exchange network consists of Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. It also includes many shortcuts that I have de ned in my [HOST] le.
LaTeX supports bibliographies out of the box, either embedding the references in your document or storing them in an external file.Author: Li Zhang. If you don't want to use BibTeX, then you must hand format your thebibliography environment as in the following example. Automatic works cited and bibliography formatting for MLA, APA and Chicago/Turabian citation styles. Abbrev. The cite package: well formed numeric citations Donald Arseneau asnd@[HOST] /02/27 Abstract The cite package modifies LATEX’s normal citation mechanism for improved handling of numeric citations.
All of these may take one or two optional arguments to add some text before and after the citation. This is an absolute beginners guide to writing documents in LATEX using TeXworks. Name of Comp. State, Country, Year, pp. Nov 27, · As I’ve written before we chose to use BibTeX as our lowest common denominator citation export format. Apr 29, · Being able to automatically reference a figure within a LaTeX document is a very cool thing.
Contents. The user manuals for cite and chapterbib are LaTeX files [HOST] and [HOST], and already typeset pdf versions. Greenwade of Sam Houston State University. When LaTeX processes the document, the citation will be cross-referenced with the bibitems and replaced with the appropriate number citation. It assumes no prior knowledge of LATEX, or any other computing language.
2 comments. Automatic works cited and bibliography formatting for MLA, APA and Chicago/Turabian citation styles. In American Psychological Association, APA, style, user manuals are cited. If you don't want to use BibTeX, then you must hand format your thebibliography environment as in the following example. [HOST] like \usepackage[superscript]{cite} [HOST] Print the tags instead of the numbers for \cite and \bibitem. The Frascati Manual classifies budgets according to what is done, what is studied, and who is studying it.
However, policymaking today is still largely focused on what is easier to measure. This is a Bib L a T e X style that implements the Chicago ‘author-date’ and ‘notes with bibliography’ style specifications given in the Chicago Manual of Style, 16 th edition (support for the 15 th edition remains, but is ‘strongly deprecated’). BibTeX Introduction This is the first draft of this document. See the bibliography file section for more information.The natbib package has two basic citation commands, \citet and \citep for textual and frascati manual latex cite parenthetical citations, respectively.
The number “10” at the beginning of “thebibliography” indicates the number of bibliography entries . Fourth, LaTeX moves this box horizontally so that the left edge of this new bounding box coincides with the left edge of the box from the first step (they need not coincide vertically). This document is an unofficial reference manual for LATEX, a document preparation system, version of March This manual was originally translated from [HOST] va in the VMS Help Library. LaTeX provides incredibly easy tools to do this. It includes definitions of basic concepts, data collection guidelines, and . Stack Exchange network consists of Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Whenever you reference a source in your work, cite it properly so your reader knows where you found your information. It is not intended to be exhaustive, nor overly [HOST]: Mark Gates. Citation Machine® helps students and professionals properly credit the information that they use. Oct 07, · For BibTeX this is not much different from printing the table of contents. It has graphic interface, which makes it easy to use \cite{Apalkov,Libby}. Dec 24, · LaTeX Warning: Citation ab' on page 2 undefined on input line LaTeX Error: Something's wrong--perhaps a missing \ item. [HOST] like \usepackage[superscript]{cite} [HOST] Print the tags instead of the numbers for \cite and \bibitem.) After this, put the information for the reference.
Greenwade of Sam Houston State University." An exception to this is if you are asked in the documentation to reference a specific work (for instance, a research article, or a website) when citing a code. But any abbreviation will work. This article explains how frascati manual latex cite to manage bibliography with the thebibliography environment and the BibTeX system.
Latex allows you to cite the references that you are going to use in your. In American Psychological Association, APA, style, user manuals are cited. Build a bibliography or works cited page the easy way My Chicago Manual of.
Can someone explain me the difference please? save hide report. 3rd ed. The style implements entry types for citing audio-visual materials.
So if \label is in [HOST] then the information is in [HOST] LaTeX puts the information in that file every time it runs across a \label. Dec 24, · [HOST] frascati manual latex cite is a series of video tutorials on how to use latex. The Frascati Manual is based on the experience gained from collecting R&D statistics in both OECD and non-member countries. Now supports 7th edition of MLA. We share experiences to help new researchers frascati manual latex cite to integrate BibTex references from Mendeley, Zotero, RefWorks, EndNote, Qiqqa and other platforms [HOST] les on the frascati manual latex cite Latex format for Research Proposals, Thesis, Dissertation, or.
Imports the bibtex data file [HOST], this file is the one that includes information about each frascati manual latex cite referenced book, article, etc. See the bibliography file section for more frascati manual latex cite information. To reference frascati manual latex cite a LaTeX table or equation in LaTeX you need to make sure that you insert a label in your table or.
When you want to cite an item in the database in a LaTeX document, you write \cite{} (or frascati manual latex cite something similar), where is the label attached to the item in the database. LaTeX: Bibliographies and Citation Citations are references to books, articles, web pages, or other published items with sufficient details to identify them uniquely. I'm a LaTeX beginner so keep it simple. The second step is the rotation, by default about the reference point. LaTeX supports bibliographies out of the box, either embedding the references in your document or storing them in an external file. Please send any comments to dana@[HOST] Description BibTeX is a program and file format designed by Oren Patashnik and Leslie Lamport in for the LaTeX document preparation system. Now supports 7th edition of MLA. State, Country, Year, pp.
BibMe quickly generates BibTeX generic citation style citations and bibliographies. Referencing elements to cite: [#] Reference number (matching the in-text citation number) Name of Manual/Handbook; If the manual is a second or subsequent edition, include the edition statement, frascati manual latex cite e. It includes definitions of basic concepts, data collection guidelines, and classifications for compiling R&D statistics. The edition is the result of the collective work of NESTI, involving more than experts from nearly 40 countries and international organisations. This is an absolute beginners guide to writing documents in LATEX using TeXworks. Despite our focus on datasets the adoption of BibTeX came out of our researcher identification work and we were not really thinking very hard about BibTeX and data sets. You put a command in your LaTeX file specifying the style in frascati manual latex cite which you want the.
Latex allows you to cite the references that you are going to use in your. 3rd ed. Dec 11, · Frascati Manual Proposed Standard Practice for Surveys on Research and Experimental Development The internationally recognized methodology for collecting and using R&D statistics, the Frascati Manual is an essential tool for statisticians worldwide. Use whatever style. A LATEX Package to Place Bibliography Entries in Text Patrick W. Cross-referencing is absolutely vital in research as you must `plug' your work into frascati manual latex cite the world's knowledge base, and must connect information across pages and sections within your own work.. Name of Comp.
cite returns the number of our list entry in the list in brackets. xxx-xxx (pages if relevant). share. Separate the parts by commas. Despite our focus frascati manual latex cite on datasets the adoption of BibTeX came out of our researcher identification work and we were not really thinking very hard about BibTeX and data sets. The natbib package has two basic citation commands, \citet and frascati manual latex cite \citep for textual and parenthetical citations, respectively.
[HOST] Do multiple bibliographies--one for each \include file. This command introduces a reference. A) Manually Creating a Bibliography. To actually cite a given document is very easy. We share experiences to help new frascati manual latex cite researchers to integrate BibTex references from Mendeley, Zotero, RefWorks, EndNote, Qiqqa and other platforms [HOST] les on the Latex format for Research Proposals, Thesis, Dissertation, or.
The third step is that LaTeX computes a box to bound the rotated material. For example, an oral history project conducted by a religious organization would be classified as being basic research, in the field of humanities (the sub-category of history), and performed by a non-governmental, non-profit organization. Purpose. BibMe quickly generates BibTeX generic citation style citations and bibliographies. The achicago LaTeX package Chicago Manual A-style (but author-date) citations Matt Swift Version: 1 Date: /11/27 Documentation revision: /03/01 Abstract A bibliography style based on The Chicago Manual of Style.
Both types are detailed below. The user manuals for cite and chapterbib are LaTeX files [HOST] and [HOST], and already typeset pdf versions. Cite your website in BibTeX generic citation style format for free. The author-year citation style requires usage of BibTeX or specially formatted \bibitem commands. [#] Name of Manual/Handbook, xth ed. Fourth, LaTeX moves this box horizontally so that the left edge of this new bounding box coincides with the left edge of the box from the first step (they need not coincide vertically). As one can see, there is no sorting.
1 post • Page 1 of 1. To reference a LaTeX table or equation in LaTeX you need to make sure that you insert a label in your table or. LaTeX forum ⇒ BibTeX, biblatex and biber ⇒ Citing a user manual in APA-style via Biblatex Information and discussion about BiBTeX - the bibliography tool for LaTeX documents. Daly This paper describes package bibentry version from /02/23 Summary The stripped version of . This workbook is designed to be used at the ‘LATEX for Beginners’ student iSkills seminar, and also for self-paced study.
The L ATEX version was written by Stephen. For example, an oral history project conducted by a religious organization would be classified as being basic research, in the field of humanities (the sub-category of history), and performed by a non-governmental, non-profit organization. Dec 24, · [HOST] is a frascati manual latex cite series of video tutorials on how to use latex. See the LaTeX manual or LaTeX Companion for explanation. You can either produce a bibliography by manually listing the entries of the bibliography or producing it automatically using the BibTeX program of LaTeX. 3 4 Languages,characters,fontsandglyphs 51 Charactersandglyphs 51 Themaincontrolloop 55 Loadingpatternsandexceptions 56 Applyinghyphenation Apr 17, · If you are conducting research on a product or service, you may reference a user manual or instructional guide in your paper. So if \label is in [HOST] then the information is in [HOST] frascati manual latex cite LaTeX puts the information in that file every time it runs across a \label.
The cite package: well formed numeric citations Donald Arseneau asnd@[HOST] /02/27 Abstract The cite package modifies LATEX’s normal citation mechanism for improved handling of numeric citations. cite returns the number of our list entry in the list in brackets. However, policymaking today is still frascati manual latex cite largely focused on what is easier to measure. This document is an unofficial reference manual frascati manual latex cite for LATEX, a document preparation system, version of March This manual was originally translated from [HOST] va in the VMS Help Library.) After this, put the information for the reference.
We just need the commands \bibliography which tells LaTeX the location of [HOST] file and \bibliographystyle which selects one of various bibliographic styles. Frascati Manual «Frascati Manual PROPOSED STANDARD PRACTICE FOR SURVEYS ON RESEARCH AND EXPERIMENTAL DEVELOPMENT Frascati Manual PROPOSED STANDARD PRACTICE FOR SURVEYS ON RESEARCH AND EXPERIMENTAL DEVELOPMENT The Measurement of Scientific and Technological Activities In an era of rapid technological change, research and . Abbrev. Build a bibliography or works cited page the easy way My Chicago Manual of.
The style implements entry types for citing audio-visual materials. Jun 09, · \cite command is used to cite the bibliography item in the document chapter or section. The pre-translation version was written by George D. Note: If you are starting from scratch it's recommended to. LaTeX forum ⇒ BibTeX, biblatex and biber ⇒ Citing a user manual in APA-style via Biblatex Information and discussion about BiBTeX - the bibliography tool for LaTeX documents. versions, please include your Latex le.
% Upvoted.g.
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2022-01-23 09:13:05
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https://www.physicsforums.com/threads/evaluating-undefined-limits.632810/
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# Evaluating undefined limits
baki
1. Evaluate
2. lim χ→0 [((χ+1)^1/3) -1] / χ
3. The hint is [a^2 + b^2 = (a+b)(a^2 + ab + b^2)].
But i don't get it how can that help me solve this because it is a cube root and not to the power of 3.
Without thinking about the hint, I have attempted it and i think it is undefined for the function.
Say f(x) is the function, so f(0) is undefined making the limit as x approaches 0 undefined as well.
is it correct??
Last edited:
Homework Helper
Dearly Missed
1. Evaluate
2. lim χ→0 [((χ+1)^1/3) -1] / χ
3. The hint is [a^2 + b^2 = (a+b)(a^2 + ab + b^2)].
But i don't get it how can that help me solve this because it is a cube root and not to the power of 3.
Without thinking about the hint, I have attempted it and i think it is undefined for the function.
Say f(x) is the function, so f(0) is undefined making the limit as x approaches 0 undefined as well.
is it correct??
The hint is incorrect: a^2 + b^2 has 'a' and 'b' to the second power, while the right-hand-side has them to the third power. A correct hint would be a^3 - b^3 = (a-b)(a^2 + ab + b^2).
Think of setting y = (x+1)^(1/3), so the numerator of your expression is y-1. Can you see how to relate y-1 to y^3 - 1? Can you see how to write the denominator (x) in terms of y? Can you see what happens to y when x --> 0?
RGV
Staff Emeritus
Homework Helper
Think of setting y = (x+1)^(1/3), so the numerator of your expression is y-1. Can you see how to relate y-1 to y^3 - 1? Can you see how to write the denominator (x) in terms of y? Can you see what happens to y when x --> 0?
That's one possibility and it works fine.
But I prefer to multiply numerator and denominator with $a^2+ab+b^2$ for suitable a and b. What do you get if you use the (correct version of the) hint?
baki
sorry I was wrong, the hint is: a^3 - b^3 = (a-b)(a^2 + ab + b^2)
Ok, I did L'hopitals rule to derive the formula, until I am able to get a value from substituting zero in place of x. My answer came to 1/3. But still the hint had nothing to do with it.
Are you saying for me to do the opposite operation??
Staff Emeritus
Homework Helper
Gold Member
sorry I was wrong, the hint is: a^3 - b^3 = (a-b)(a^2 + ab + b^2)
Ok, I did L'hopitals rule to derive the formula, until I am able to get a value from substituting zero in place of x. My answer came to 1/3. But still the hint had nothing to do with it.
Are you saying for me to do the opposite operation??
Here is how the hint might be used:
The numerator $\sqrt[3]{x+1}-1$. If you could get an equivalent expression with a numerator of $\displaystyle \left(\sqrt[3]{x+1}\right)^3-1^3\,,$ that would allow major simplification including cancelling a factor that goes to zero in the numerator & denominator.
Use the hint to find out what you need to multiply $\sqrt[3]{x+1}-1$ by in order to get $\displaystyle \left(\sqrt[3]{x+1}\right)^3-1^3\ .$
By the way, welcome to PF!
baki
so then I would have to multiply it with ($\sqrt[3]{x+1}$)^2 + $\sqrt[3]{x+1}$ +1
and then do the same with the numerator??
is it??
and afterwards derive it???
Last edited:
Staff Emeritus
Homework Helper
Gold Member
so then I would have to multiply it with ($\sqrt[3]{x+1}$)^2 + $\sqrt[3]{x+1}$ +1
and then do the same with the numerator??
is it??
and afterwards derive it???
No, you should not need to use L'Hôpital's rule.
Just multiply the numerator & denominator by $\displaystyle \left(\sqrt[3]{x+1\ }\right)^2+\sqrt[3]{x+1\ }+1\ .$ Then simplify.
baki
oh YES
solved!
other than simplifying we can just substitute 0, right?
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2022-08-19 16:39:20
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https://proxieslive.com/primality-testing-algorithm/
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# Primality testing algorithm
Say, I would like to check a hypothesis concerning primes. Something like “there exists a prime between $$n$$ and $$2n$$ for every choice of $$n$$“. I would like to run a code in MATLAB for choices of $$n$$ upto $$2^{32}$$ and then use that data and publish the conjecture in a journal.
The question is, what should I use to check primality. Obviously, AKS is an option but it is really really slow. I can use the in-built MATLAB function $$isprime()$$ which I think uses $$10$$ instances of Miller-Rabin. This will be way faster but the journal might reject this saying that Miller Rabin is probabilistic and that I should instead use a deterministicalgorithm since one exists.
What should I do? Use AKS? Go with MATLAB’s inbuilt Miller-Rabin? Or look at other deterministic algorithms?
I don’t think this is the best place to ask this. However I could not find where else to ask. Any suggestions?
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2021-01-17 21:31:42
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https://www.omnicalculator.com/math/spherical-coordinates
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# Spherical Coordinates Calculator
Created by Bogna Szyk
Reviewed by Steven Wooding
Last updated: Dec 18, 2022
The spherical coordinates calculator is a tool that converts between rectangular and spherical coordinate systems. It describes the position of a point in a three-dimensional space, similarly to our cylindrical coordinates calculator. For a two-dimensional space, instead of using this Cartesian to spherical converter, you should head to the polar coordinates calculator.
## Rectangular and spherical coordinates
Coordinates are sets of values that describe the position of any given point in space. For a three-dimensional space, you need precisely three coordinates to define a point uniquely.
The Cartesian system, also known as the rectangular coordinate system, is constructed by drawing three lines in space. They all cross a common point, called the origin, and are perpendicular to one another. These lines are called the axes of the system. Each pair of axes defines a plane: these are called the XY, XZ, and YZ planes. The coordinates of any arbitrary point are defined as the distance between that point and the planes.
We define the spherical coordinate system by a fixed origin and a zenith direction. The zenith is an imaginary point located directly above the origin. The direction from the origin to the zenith is called the zenith direction.
Each arbitrary point in space has three spherical coordinates. The first one, called radial distance or radius (r), is simply the three-dimensional distance between the origin and this point. The second one - the polar angle (θ) - is the angle from the zenith direction to the line that connects the point with the origin. The third one is called the azimuth angle (φ).
When you create a reference plane that passes through the origin and is orthogonal to the zenith, then φ is the angle between the point's orthogonal projection to the plane and a fixed reference direction on that plane.
## Converting Cartesian to spherical
When converting from the rectangular to the spherical system, our spherical coordinate calculator assumes that the origins of both systems overlap. It also assumes that the z-axis is equivalent to the zenith direction. The spherical reference plane coincides with the XY plane.
You should use the following formulas for the conversion:
• r = √(x² + y² + z²);
• θ = arccos(z/r);
• φ = arctan(y/x),
where:
• (x, y, z) - Cartesian coordinates; and
• (r, θ, φ) - Spherical coordinates.
## Converting spherical to Cartesian
The spherical coordinates calculator is also able to recalculate the spherical coordinates into rectangular ones. It uses the following equations:
• x = r × sin θ × cos φ;
• y = r × sin θ × sin φ;
• z = r × cos θ.
Don't forget to check our other coordinate geometry calculators like the average rate of change calculator.
Bogna Szyk
Cartesian (x, y, z) to spherical (r, θ, φ)
x
y
z
r
θ
deg
φ
deg
Spherical (r, θ, φ) to Cartesian (x, y, z)
r
θ
deg
φ
deg
x
y
z
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2023-01-30 07:35:13
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https://notes.vidyasagarv.com/
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Questions What are pulsars and quasars? Chapter 1 »
Vidya Sagar
Present continuous (I am doing) Usage: in the middle of ‘action’, started doing and haven’t finished yet; talk about things/changes happening in a period around now (for example, today / this week / this evening etc.). Often the action is happening at the time of speaking, but not necessarily. Standard structure: Subject Present tense of be Present participle verb I am (’m) / am not (’m not, ain’t) -ing form of verb he/she/it/name of a person is (’s) / is not (’s not, isn’t) -ing form of verb we/you/they/name of group are (’re) / are not (’re not,aren’t) -ing form of verb For framing a question, reverse the first 2 columns. »
Vidya Sagar
Week 1 Introduction The origins of Group theory is in Premutations (the algebra obeyed) and Geometry (rotations) corresponding to Discrete and Continuous groups respectively. Geometric rotations, in general do not commute. Continuous groups is essentially Trignomentry. In quantum mechanics, symmetry group substitutes for the geometry of shape and size. Algebraic preliminaries Sets and maps Mathematicians usually classify maps as: Surjective/onto: Range is completely covered. Injective/into: One to One, but need not exhaust range. »
Vidya Sagar
Arithematic r = c - qd, r is the reminder and q is the quotient when an integer c is divided by a positive integer d The first 10 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. 1 is not a prime number. 2 is the only prime number that is even. The fraction part’s value of a mixed number has to be between 0 and 1. »
Vidya Sagar
Nuclear Physics Stable nuclei, Nomenclature and units Atomic scale is 10-10m, nuclear scale is 10-15m. For convenience we use the unit Fermi (fm) = 10-15m = 1 femtometer. Nuclear sizes range from 1 fm to 7 fm. Particle physics usually happens at an even smaller scale << 10-15m. A nuclear species or nuclide is denoted by $$^A_ZX_N$$. Here, X is the chemical symbol. Z is the atomic number: the number of protons. »
Vidya Sagar
Table of Contents 1 Introduction and Overview 2 Introduction to quantum mechanics 1 Introduction and Overview Exercise 1 Exercise 2 Problem 1 Problem 2 Exercise 1 It is already shown that a deterministic classical computer would require $$2^n/2+1$$ queries. Instead, if we use a probabilistic classical computer i.e, $$f(x)$$ is evaluated for randomly chosen $$x$$, with just one execution we cannot determine whether $$f(x)$$ is constant or balanced function (atleast not with probability of error ε < 1⁄2). »
Vidya Sagar
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2022-01-18 17:01:27
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https://math.stackexchange.com/questions/1538107/general-solution-to-differential-equation
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# General Solution to Differential Equation
Find the general solution of a differential equation $\frac d{dr}[r^2 \frac{dR}{dr}]-l(l+1)R=0$. (Hint: Assume an infinite series $R(r)=\sum_{n= -\infty}^{+\infty} a_n r^n$ as the solution)
I don't get how we can solve this problem. Some help would be nice.
All sums are over $n\in\mathbb{Z}$.
If $$R(r)=\sum a_nr^n\text{,}$$ then you have $$\frac{dR}{dr}=\sum na_nr^{n-1}\qquad\frac{d^2R}{dr^2}=\sum n(n-1)a_nr^{n-2}\text{.}$$ So
\begin{align} \frac{d}{dr}\left[r^2 \frac{dR}{dr}\right]-\ell(\ell+1)R &=0\\ 2r\frac{dR}{dr}+r^2\frac{d^2R}{dr^2}-\ell(\ell+1)R &=0\\ 2r\sum na_nr^{n-1}+r^2\sum n(n-1)a_nr^{n-2}-\ell(\ell+1)\sum a_nr^n &=0\\ \sum 2na_nr^{n}+\sum n(n-1)a_nr^{n}-\sum \ell(\ell+1)a_nr^n &=0\\ \sum (2n+n(n-1)-\ell(\ell+1))a_nr^{n}&=0\\ \end{align}
So we have, for all $n$, \begin{align} \left(2n+n(n-1)-\ell(\ell+1)\right)a_n &=0\\ \left(n(n+1)-\ell(\ell+1)\right)a_n &=0 \end{align}
So if $n=\ell$ or $n=-\ell-1$, then $\left(n(n+1)-\ell(\ell+1)\right)$ will be $0$, and $a_n$ can be anything. For all other $n$, $\left(n(n+1)-\ell(\ell+1)\right)$ will be nonzero, and $a_n$ must be $0$.
So the solution is $$R(r)=a_\ell r^\ell+a_{-\ell-1}r^{-\ell-1}$$ where $a_\ell$ and $a_{-\ell-1}$ can be any constants. (Note that the initial assumption was that $R$ had a Laurent series in $r$. But now we can see that even for non-integer $\ell$, this still works as a solution.)
First differentiate so we can write the ODE in a more workable form: $$r^2R'' + 2rR' - l(l+1)R=0.$$ Assuming $R(r) = \sum_{n\in\mathbb Z}a_nr^n$, we compute \begin{align} R'(r) &= \sum_{n\in\mathbb Z}(n+1)a_{n+1}r^n\\ R''(r) &= \sum_{n\in\mathbb Z}(n+1)(n+2)a_{n+2}r^n. \end{align} Hence we have $$r^2\sum_{n\in\mathbb Z}(n+1)(n+2)a_{n+2}r^n + 2r\sum_{n\in\mathbb Z}(n+1)a_{n+1}r^n - l(l+1)\sum_{n\in\mathbb Z}a_nr^n = 0,$$ which implies that the coefficient of $r^n$ in the above is zero. It follows that $$r^2(n+1)(n+2)a_{n+2} + 2r(n+1)a_{n+1} - l(l+1)a_n=0,$$ which can be written as $$a_{n+2} = \frac{-2(n+2)}ra_{n+1}+\frac{l(l+1)}{r^2(n+1)(n+2)}a_n,\quad n\in\mathbb Z.$$ I'm not sure how to find a closed form for $a_n$, though, considering $R$ is a Laurent series and I am not familiar with complex analysis).
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2019-10-21 19:52:30
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https://www.daqq.eu/?p=1412
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# New method for measuring lots of resistors using very few wires - part 2 - Theory
## The Matrix
Continuing from part 1 the matrix form used needs to be described.
If you need a quick "tutorial" in systems of linear equations, take a look at this text I wrote.
First, the vector of the unknowns - x . The vector is:
$\large \mathbf{x} = [ G_1 , G_2 , G_3 , \cdots , G_N ]$
where GN is the conductance of the respective resistor. We do not know the values of the individual conductances (we know the values of some of the conductances, the reference resistors, but for the purposes of the calculation we do not know them).
The coefficients matrix and the right hand side vector go hand in hand. In the matrix form of equations each equation can be transcribed into one row of the A matrix and one number in the b vector. So for every measured situation we will generate the following equation:
$\large (U_1 - U_C) G_1 + (U_2 - U_C)G_2 + ... + (U_N - U_C)G_N = 0$
The particular number in b will be 0. As for the row in the matrix A the values will be 0 for the resistors that are not involved in that particular resistive divider and (UN - UC), where UN is the voltage on the node that is connected by RN to the center node and UC is the node on the center voltage
So, let's use the following 8 node system as an example:
Example 8 node system
The resistors are numbered. We will apply a situation to it - the measurement system will connect some of the nodes to the low excitation voltage, some to the high excitation voltage and set one to be a high impedance node. Red markers on the image will represent nodes connected to a high excitation voltage, blue nodes represent the ones connected to the low excitation voltage and the green one is the high impedance one that will be the center node for the voltage divider.
NodeState
AHigh +5V
BHigh +5V
CHigh +5V
DLow GND (0V)
FLow GND (0V)
GLow GND (0V)
HLow GND (0V)
Connecting the nodes turns the whole setup into a voltage divider. As was discussed previously, we can safely ignore all of the resistors that do not connect directly to the divider center node (E). So, let's connect the nodes. The voltages on the individual nodes will be something like this:
Example situation in an 8 node system
We can safely omit the resistors that are not connected to the center node (E) for clarity. The resistance values are not needed for the actual calculation or matrix formation. Node voltages have been measured as follows:
NodeVoltage to ground [V]Resistor connecting to center node (E)
A5.00010k
B5.00011k
C5.00012k
D0.00013k
E2.710N/A
F0.00014k
G0.00020k
H0.00030k
Only resistors 4, 10, 15, 19, 23, 24, and 25 are involved in the divider. So let's formulate the equation:
(note: Since the equations are getting very long and I'm having trouble fitting them into latex in any way that makes them look good, some will not be written in latex. Appologies to the font nazis.)
0G1 + 0G2 + 0G3 + (UA-UE)G4 + 0G5 + 0G6 + 0G7 + 0G8 + 0G9 + (UB-UE)G10 + 0G11 +
0G12 + 0G13 + 0G14 + (UC-UE)G15 + 0G16 + 0G17 + 0G18 +
(UF-UE)G23 + (UG-UE)G24 + (UH-UE)G25 + 0G26 + 0G27 + 0G28 = 0
Nasty, isn't it? Note however that most of the conductances have a zero coefficient, as such we can remove them.
$(U_A - U_E)G_{4} + (U_B - U_E)G_{10} + (U_C - U_E)G_{15} + (U_D - U_E)G_{19} +$
$(U_F - U_E)G_{23} + (U_G - U_E)G_{24} + (U_H - U_E)G_{25} = 0$
Adding the specific numbers we get:
$(5-2.710)G_{4} + (5-2.710)G_{10} + (5-2.710)G_{15} + (0-2.710)G_{19} +$
$(0-2.710)G_{23} + (0-2.710)G_{24} + (0-2.710)G_{25} = 0$
Which equals:
$2.29 G_{4} + 2.29 G_{10} + 2.29 G_{15} -2.71 G_{19} -2.71 G_{23} -2.71 G_{24} -2.71 G_{25} = 0$
So this particular row of the equation will be:
x = (G1 G2 G3 G4 G5 G6 G7 G8 G9 G10 G11 G12 G13 G14 G15 G16 G17 G18 G19 G20 G21 G22 G23 G24 G25 G26 G27 G28)
A = [0 0 0 2.29 0 0 0 0 0 2.29 0 0 0 0 2.29 0 0 0 -2.71 0 0 0 -2.71 -2.71 -2.71 0 0 0]
and that row in b will be 0.
This was the process for creating one single line of the final matrix. For a complete measurement this will have to be repeated many times over for different situations, with the high impedance node set to different nodes.
The absolutely minimal number of different measured situations would be equal to the number of unknown resistances. However, the more situations are measured, the better. With a low number of measured situations the error of the system can be very large (hundreds of percent even, depending on the noise and various other factors, will be discussed in another part). With more measured situations, the error drops down.
The number of possible combinations of node settings (situations) can be pretty large. Assuming that every combination with a single floating node is used, for an 8 node system, assuming two power rails, this comes to 8*128=1024 possible situations. However, not all of them are actually useful - a combination where all of the powered nodes are connected to GND will not yield anything useful, since the resulting divider will be completely connected to one excitation voltage, giving no clues as to the ratios of the resistors. Also, using all of the combinations would cause the measurement to be very long and memory hungry - it's possible, but not needed. So a limited subset is sufficient for an actual measurement. Of the 1024 combinations possible for an 8 node system, I have chosen to use 248 situations, but other numbers of combinations would work as well.
The visual representation of the situations for an 8 node system follows:
Visual representation of situation node settings in 8 node system
This is a representation of 248 situations in the 8 node system. Each "row" represents one situation, each "column" represents one of the nodes. Blue represents a node that is set into a high impedance state, green represents a node tied to GND (low excitation) and red represents a node tied to VCC (high excitation). If you with to see situation visualizations and their source files, please download the following file.
So that's the main part of the matrix - the measurement values. At the moment there are no absolutes present in the matrix. So the reference part of the matrix needs to be added. This is a very different format of an equation. This just assigns a known value to the known resistors. So, assuming that we know the resistances (conductances) of resistors R1, R14 and R28, we will have three reference equations. These are in the following format:
$1 G_{N} = g_{N}$
where $g_{N}$ is the actual conductance, placed into the right hand side. For the example case, where we know the resistances of R1 (330050 Ohms = 3.029844E-06 S), R14 (329280 Ohms = 3.036929E-06 S) and R28 (329580 Ohms = 3.034165E-06 S) the equations would look as follows:
$1 G_{1}=3.029844*10^{-6}$
$1 G_{14}=3.036929*10^{-6}$
$1 G_{28}=3.034165*10^{-6}$
These lines can be duplicated to further "enforce" this "rule" for the solver. These are the precisely measured values of three 330k resistors that are used as reference resistors on one particular measurement board I made.
That's that as far as the linear equations are concerned. All that is needed is to solve them.
The actual solving is done using a linear equations solver. In the software I made I use Math.NET, but any should do. Basically, the solver searches for a set of values (the x vector) that satisfies all of the equations that are structured into the $\textbf{Ax=b}$ . If no exact solution is possible (which is certainly the case for equations generated this measurement), it searches for a set of values for x that is closest to fitting the equations. The output of a solver is a filled x vector that is the closest solution to the generated measurement.
Now the x vector is filled with the conductivities of the individual resistors. To get the resistances, the reciprocal value of the vector is needed. Once that is gained, we are done - we have the resistances of the individual measured elements. And all it took was a lot of GPIO twiddling and a load of math. Enjoy!
## Example
An example matrix, from an actual measurement done using an ATMega32 (the actual implementation of the measurement will be talked about in another part of this series), including an Octave script can be downloaded here:
Unzip the directory somewhere and run the LoadAndCalculate.m script. This is the result of a measurement of a board with resistors of a fixed value, three of which were known reference resistors (R1, R14 and R28).
#Demonstration to load an example resistor net
#matrix and calculate the resistances
#Load the A matrix and the b vector
#Do the actual calculation
x = A\b;
Conductances = x;
#convert the conductances into resistances
Resistances = 1./Conductances;
Resistances' %Display the resistance vector
How the actual calculation works is outside of the scope of this text. See my other post for a very simplified explanations that's probably full of errors, or better yet - read a different text from someone who is actually a mathematician.
The visual representation of such a matrix (done by PlotMatrix.m ) can be seen here:
Visual representation of a measurement matrix
This is a visualization of a single matrix. The upper few lines are the reference part of the matrix, the rest are the actual measurement. All lines of the b vector are zero, aside from the ones in the reference part of the matrix. This particular matrix is 260 rows long and 28 columns wide. Of the 260 rows, 12 are the reference part of the equations. The particular pattern that the matrix looks like is explained by how the situation generator generates the situations.
Executing this script should yield the resistances of the individual resistors. In this case they were 25 thermistors, with a 680k resistance at 25 °C, as well as three 330k reference resistors.
## Conclusion - method
So, that's that! With this method you can do a measurement of a large quantity of resistors, using a minimal amount of cabling. This is not exactly useful for most areas, and it won't be replacing Kelvin resistance measurements any time soon, but there are niche areas where this method can be actually useful.
Space constrained applications are one area - imagine you have to measure a lot of temperature sensors on the length of a very thin, long rod - any other method would require more cabling or intelligent (capable of connecting to a data bus) sensors. This method just requires say, 16 signals running through the inside of the shaft and you can measure over a hundred thermistors along the length of the shaft. MEMS systems could also benefit from this. Or just a sensor array.
Another interesting usage of this might be using it to measure small quantities of resistors using a lot of known resistors - this actually yielded some interesting results - thanks to the insane amount of oversampling, this can actually provide a resolution well beyond that of the ADC (I'm still working on this area, but the preliminary results are interesting).
The advantages of this measurement are obvious - you can measure a lot of resistors using just the bare minimum amount of cabling or pins of your device, without needing any intelligence in the sensors themselves.
The downside - it's a memory and computation hungry method, it's not exactly fast and there are unforeseeable sources of errors. As such it needs some processor capable of doing the number crunching.
The technique bears some similarity to electrical impedance tomography, but is at its core different for being used on discrete parts (resistors) rather then volumes, uses DC excitation and uses reference resistors to enable the measurement of the exact resistance values. Also, it can be used with a single, cheap microcontroller.
In the next part of this text I will show you a device I built that actually implements this measurement, the software that goes with it and the results. As a teaser I am including the results of an actual measurement of a 150kΩ resistor, which was measured with 27 other resistors 😉 Stay tuned!
Measurement of a 150kOhm resistor
And if you liked this text, donations are always welcome!
If you need any help implementing this method for your particular thingamajig, contact me!
This entry was posted in Projects, Resistance measurement method and tagged , , , , , , . Bookmark the permalink.
### 4 Responses to New method for measuring lots of resistors using very few wires - part 2 - Theory
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2023-02-04 09:02:37
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https://blog.bordes.me/fcsc2021-smealog.html
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FCSC 2021 - SmeaLog
Posted on 03 May 2021, 18:00 in WriteUps
France CyberSecurity Challenge (FCSC) 2021
The french national cybersecurity agency (ANSSI) has organised a CTF called the France CyberSecurity Challenge (FCSC for short) which goal is to select the team that will represent France during the European CyberSecurity Challenge on the 28th of September, 2021.
Description of SmeaLog
SmeaLog was a challenge in the crypto category. The description given to the participants was straightforward:
This translates to "Could you solve this discrete logarithm?" and we are given two files: a Sage script SmeaLog.sage and an output generated with it output.txt.
Here is the content of SmeaLog.sage, that I will explain in summary just below:
SmeaLog.sage source code
import os
from hashlib import sha256
from Crypto.Cipher import AES
from secret import FLAG
def gen_curve(bits = 40, k = 4):
assert bits*k >= 160, "Error: p**k must be at least 160 bits."
p = random_prime(2 ** (bits + 1) - 1, lbound = 2 ** bits)
R = Zmod(p**k)
while True:
a, b = R.random_element(), R.random_element()
d = 4 * a ** 3 + 27 * b ** 2
if d.is_unit():
E = EllipticCurve(R, [a, b])
E_ = EllipticCurve(GF(p), [a, b])
if E_.order().is_prime():
return E
def gen_random_point(E):
R = E.base_ring()
a, b = E.a4(), E.a6()
while True:
x = R.random_element()
t = x ** 3 + a * x + b
if is_square(t):
y = choice([-1, 1]) * sqrt(t)
return E([x, y])
def gen_pair(E):
N = E.base_ring().characteristic()
s = ZZ.random_element(N)
P = gen_random_point(E)
return P, s
def encrypt(s):
k = sha256(str(s).encode()).digest()
iv = os.urandom(16)
c = AES.new(k, AES.MODE_CBC, iv).encrypt(FLAG)
return iv, c
E = gen_curve()
print(f"E = {E}")
P, s = gen_pair(E)
print(f"P = {P}")
print(f"s*P = {s * P}")
iv, c = encrypt(s)
print(f"iv = {iv.hex()}")
print(f"c = {c.hex()}")
First the script generates an EllipticCurve Sage object with three parameters $$a$$, $$b$$ and $$p$$ where $$p$$ is 40-bits long and both $$a$$ and $$b$$ are 160-bits long. This elliptic curve corresponds to all the points $$P = (x, y)$$ such that $$x^3 + ax - b = y^3 \mod p^4$$. The elliptic curve equation and the number $$p^4$$ are printed. After that, a random point $$P$$ and a random number $$s$$ are generated. The scalar product $$P + \ldots + P$$ with $$P$$ appearing $$s$$ times is computed (we will denote this scalar product by $$[s]P$$). This leads to a new point that we will call $$H$$. The coordinates of both points are printed but the number $$s$$ remains secret and is used to derive a key that will encrypt the flag (in FLAG) using AES in CBC mode. The random initialisation vector iv as well as the resulting ciphertext c are printed.
As said before, the output of this script is given to us:
Content of output.txt
E = Elliptic Curve defined by y^2 = x^3 + 469245061185347057653091531882370383913875080615*x + 5114351683655764329253106245428582084587536640503 over Ring of integers modulo 6801613388087663669990064996614506238369534296081
P = (4818298608029665051393880712825109209819975611403 : 3354976854279375487312341201850051023143236550702 : 1)
s*P = (6276672712837100206846077340854087747993984369352 : 5153262096245606021857753027994479500306746041453 : 1)
iv = 0564fc638153e8b1ef1b7b5f52c539cc
Our goal is thus to retrieve $$s$$, in order to decrypt the flag. This is in fact what is called the discrete logarithm of $$H = [s]P$$ in base $$P$$. This "elliptic curve discrete logarithm problem" can be used to build cryptographic primitives because it known to be computationnaly hard (meaning that it can't be solved in a reasonable amount of time) for well-chosen elliptic curves (and well-chosen point $$P$$). We will see here that the chosen parameters are not suitable for cryptographic use since we are able the compute the discrete logarithm in less than a minute on a general purpose laptop.
An elliptic curve modulo a non prime integer
Usually, in cryptographic application, we use an ellpitic curve on which we can define a group (as explained in the "crypto reminders" of a previous write-up). One basic property of a group is that it is closed under its operation (here, the sum). This means that for any two points $$P = (x_p, y_p)$$ and $$Q = (x_q, y_q)$$ on the curve, we want $$P + Q$$ to also be a point on the curve.
Elliptic curve addition of two different points is defined geometrically and the computation of it involves an intermediate value $$\lambda = \frac{y_q - y_p}{x_q - x_p}$$. But remember that we are working only with integers modulo $$n$$, so dividing by $$(x_q - x_p)$$ actually means multiplying by the integer $$r$$ such that $$r(x_q - x_p) = 1 \mod n$$. Such an integer $$r$$ exists if and only if $$\gcd((x_q - x_p), n) = 1$$. When it exists, we say that $$r = (x_q - x_p)^{-1} \mod n$$. If the curve is defined modulo a prime number, $$r$$ always exists. Here, since the integers are taken modulo $$p^4$$, the addition of two points on the curve is not always defined.
After some research about this unusual construction, I found a question on StackExchange that is close to our problem: in this question, the author is wondering what would happen if an elliptic curve is not defined modulo a prime number $$p$$ but instead modulo a number $$n = pq$$ with $$p$$ and $$q$$ prime (and different). The answer is quite enlightning: such a curve is in fact isomorphic to the cartesian product of the curve modulo $$p$$ and the curve modulo $$q$$. In other words: any point on the curve modulo $$n$$ can be seen as a pair of a point on the curve modulo $$p$$ and a point on the curve modulo $$q$$. The point addition is naturally extended, when it is defined (which is not only the case as seen before), as the point-wise addition of the pair of points. The single and massive difference between this setup and ours is that in ours $$n$$ is the power of the same prime: $$n = p^4$$.
A convenient isomorphism in our setup
Following the intuition of this question (and answer) found on StackExchange, I searched a convenient isomorphism. Often, to solve crypto challenges in CTF and even more so for asymmetric cryptography, you need to have a sufficient understanding of the concepts used, some intuition on where you need to focus your research and you also need... to find and understand the right paper that will give you the solution! There is no particular method to find such a paper, but having a clear idea on what you are looking for exactly and how to formulate it precisely is crucial.
For this challenge, I managed to find this article written by Massimiliano Sala and Daniele Taufer. It was posted on a preprint server in late 2020, and to my knowledge, is not yet published in a peer-reviewed journal or conference. In this paper the authors explain, among other interesting things, that there is an isomorphism (a one-to-one mapping) that is very convenient to use: every point $$P = (x, y)$$ on our curve modulo $$p^4$$ can be written as a pair containing its canonical projection $$\pi(P)$$ on the curve modulo $$p$$ (the point $$\pi(P) = (x \mod p, y \mod p)$$) and an integer modulo $$p^3$$ that we will call $$q_P$$. We will denote this relation between this pair and the point $$P$$: $$P \sim (\pi(P), q_P)$$1.
What is really interesting is that this isomorphism is behaving nicely with respect to point addition: given two points $$P$$ and $$Q$$, let $$R = P + Q$$; then $$R \sim(\pi(P) + \pi(Q), q_P + q_Q)$$.
Solving the discrete logarithm
Now that we have this isomorphism, we can easily retrieve at least part of the secret $$s$$. Thanks to its property with respect to point addition, we can write the following: $$H (= [s]P) \sim ([s]\pi(P), sq_P)$$. Thus, by computing $$q_H.q_P^{-1} \mod p^3$$, we can get the value of $$s$$ but only modulo $$p^3$$ and we will call this value $$s_{\text{low}}$$.
Let's call $$s_3$$ the value such that $$s = s_{\text{low}} + p^3.s_3$$. We know that $$H = [s]P$$, thus thanks to our isomorphism we also have $$\pi(H) = [s]\pi(P)$$. We also know that $$s - s_{\text{low}} = s_3p^3$$, thus $$\pi(H) - [s_{\text{low}}]\pi(P) = [s_3p^3]\pi(P)$$.
From here we will use $$q$$, the order of the point $$\pi(P)$$, that is to say the lowest non-zero integer such that $$[q + 1]\pi(P) = \pi(P)$$. If we call $$p_{\text{inv}}$$ the inverse of $$p^3$$ modulo $$q$$, then: it exists $$k$$, such that $$[p_{\text{inv}}][s_3p^3]\pi(P) = [s_3(1 + kq)]\pi(P) = [s_3]\pi(P)$$.
By calling $$H'$$ this new point on the curve modulo $$p$$, $$s_3$$ is the solution of the discrete logarithm problem: $$H' = [p_{\text{inv}}](\pi(H) - [s_{\text{low}}]\pi(P)) = [s_3]\pi(P)$$.
Fortunately for us, this ECDLP is on a curve much smaller ($$p$$ is a $$40$$-bits integer) on which it is practically computable.
We finally obtain the secret $$s$$ by computing $$s_{\text{low}} + p^3s_3$$ and are able to decrypt the flag using it. Here is the whole script used:
Sage script to solve the challenge
from hashlib import sha256
from Crypto.Cipher import AES
# Intuition from: https://crypto.stackexchange.com/questions/72613/elliptic-curve-discrete-log-in-a-composite-ring
# Isomorphism from: https://arxiv.org/pdf/2010.15543.pdf
def general_sum(P, Q, a, b):
"""
Formula for point addition with projective coordinates
"""
X1 = P[0]
X2 = Q[0]
Y1 = P[1]
Y2 = Q[1]
Z1 = P[2]
Z2 = Q[2]
T1=Y1*Y2*(X1*Y2+X2*Y1)-a*X1*X2*(Y1*Z2+Y2*Z1)-a*(X1*Y2+X2*Y1)*(X1*Z2+X2*Z1)-3*b*(X1*Y2+X2*Y1)*Z1*Z2-3*b*(X1*Z2+X2*Z1)*(Y1*Z2+Y2*Z1)+a*a*(Y1*Z2+Y2*Z1)*Z1*Z2
T2=Y1*Y1*Y2*Y2+ 3*a*X1*X1*X2*X2+ 9*b*X1*X2*(X1*Z2+X2*Z1)-a*a*X1*Z2*(X1*Z2+2*X2*Z1)-a*a*X2*Z1*(2*X1*Z2+X2*Z1)-3*a*b*Z1*Z2*(X1*Z2+X2*Z1)-(a*a*a+ 9*b*b)*Z1*Z1*Z2*Z2
T3= 3*X1*X2*(X1*Y2+X2*Y1)+Y1*Y2*(Y1*Z2+Y2*Z1)+a*(X1*Y2+X2*Y1)*Z1*Z2+a*(X1*Z2+X2*Z1)*(Y1*Z2+Y2*Z1)+ 3*b*(Y1*Z2+Y2*Z1)*Z1*Z2
try:
z_inv = pow(T3, -1, p**4)
return (T1*z_inv, T2*z_inv, T3*z_inv)
except:
y_inv = pow(T2, -1, p**4)
return (T1*y_inv, T2*y_inv, T3*y_inv)
def isomorph(P, E, E_base, q, p):
# Little trick to just use the general sum for the last addition (the one
# who would fail)
qPx = general_sum((q-1)*P, P, E.a4(), E.a6())[0]
return E_base(P), (int(qPx)/p)%p**3
if __name__ == "__main__":
a = 4692450611853470576530915318823703839138750803615
b = 5114351683655764329253106245428582084587536640503
p4 = 6801613388087663669990064996614506238369534296081
p = pow(p4, 1/4)
E = EllipticCurve(Zmod(p**4), [a,b])
E_base = EllipticCurve(GF(p), [a, b])
q = E_base.order()
P = E(4818298608029665051393880712825109209819975611403,
3354976854279375487312341201850051023143236550702)
H = E(6276672712837100206846077340854087747993984369352,
5153262096245606021857753027994479500306746041453)
iv_hex = "0564fc638153e8b1ef1b7b5f52c539cc"
Hiso = isomorph(H, E, E_base, q, p)
Piso = isomorph(P, E, E_base, q, p)
qH = Hiso[1]
qP = Piso[1]
print("Compute low part...")
# We known that qH = s*qP mod p**3
s_low = qH*pow(int(qP), -1, p**3)
print("Compute high part...")
# To compute the last part of s, we use the discrete log on E_
p3_inv = pow(int(p**3), -1, q)
lastH = int(p3_inv)*(Hiso[0] - int(s_low)*Piso[0])
s3 = discrete_log(lastH, Piso[0], Piso[0].order(),operation='+')
s = int(s_low) + int(s3)*p**3
# Now decrypt the flag
k = sha256(str(s).encode()).digest()
iv = bytes.fromhex(iv_hex)
c = bytes.fromhex(c_hex)
plaintext = AES.new(k, AES.MODE_CBC, iv).decrypt(c)
print(plaintext.decode())
1. The concrete definition of $$q_P$$ is voluntarily not specified here but can be found in the cited paper. For the advanced readers: $$q_P$$ is the $$x$$-coordinate of the product of $$\pi(P)$$'s order by $$P$$, normalised by its $$y$$-coordinate and canonically projected to $$Z/p^3Z$$. This is correctly defined only in the projective representation of the curve.
|
2021-07-25 12:43:02
|
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|
https://www.woodstradingcompany.com/elements/flip-book/
|
### Flip book element
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|
2021-01-19 21:29:24
|
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|
https://www.lmfdb.org/L/2/1148/41.32
|
## Results (1-50 of 616 matches)
Next
Label $\alpha$ $A$ $d$ $N$ $\chi$ $\mu$ $\nu$ $w$ prim arith $\mathbb{Q}$ self-dual $\operatorname{Arg}(\epsilon)$ $r$ First zero Origin
2-1148-1148.655-c0-0-5 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 1148.655 $$0.0 0 -0.114 0 1.59438 Modular form 1148.1.o.d.655.4 2-1148-1148.655-c0-0-4 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 1148.655$$ $0.0$ $0$ $0.260$ $0$ $1.56960$ Artin representation 2.1148.6t5.b.b Modular form 1148.1.o.b.655.1
2-1148-1148.655-c0-0-3 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 1148.655 $$0.0 0 0.0100 0 0.890335 Modular form 1148.1.o.c.655.1 2-1148-1148.655-c0-0-2 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 1148.655$$ $0.0$ $0$ $0.385$ $0$ $0.866174$ Modular form 1148.1.o.d.655.1
2-1148-1148.655-c0-0-1 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 1148.655 $$0.0 0 -0.364 0 0.832651 Modular form 1148.1.o.d.655.2 2-1148-1148.655-c0-0-0 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 1148.655$$ $0.0$ $0$ $-0.239$ $0$ $0.827225$ Modular form 1148.1.o.a.655.1
2-1148-1148.583-c0-0-1 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 1148.583 $$0.0 0 0.311 0 2.27197 Modular form 1148.1.bj.a.583.1 2-1148-1148.583-c0-0-0 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 1148.583$$ $0.0$ $0$ $-0.188$ $0$ $1.55654$ Modular form 1148.1.bj.a.583.2
2-1148-1148.319-c0-0-1 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 1148.319 $$0.0 0 0.188 0 2.25003 Modular form 1148.1.bj.a.319.2 2-1148-1148.319-c0-0-0 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 1148.319$$ $0.0$ $0$ $-0.311$ $0$ $0.332191$ Modular form 1148.1.bj.a.319.1
2-1148-1148.163-c0-0-7 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 1148.163 $$0.0 0 -0.385 0 2.91138 Modular form 1148.1.o.d.163.1 2-1148-1148.163-c0-0-6 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 1148.163$$ $0.0$ $0$ $0.114$ $0$ $2.10764$ Modular form 1148.1.o.d.163.4
2-1148-1148.163-c0-0-5 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 1148.163 $$0.0 0 0.239 0 2.07001 Modular form 1148.1.o.a.163.1 2-1148-1148.163-c0-0-4 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 1148.163$$ $0.0$ $0$ $0.364$ $0$ $1.87003$ Modular form 1148.1.o.d.163.2
2-1148-1148.163-c0-0-3 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 1148.163 $$0.0 0 -0.260 0 1.21075 Artin representation 2.1148.6t5.b.a Modular form 1148.1.o.b.163.1 2-1148-1148.163-c0-0-2 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 1148.163$$ $0.0$ $0$ $0.489$ $0$ $0.933992$ Modular form 1148.1.o.c.163.2
2-1148-1148.163-c0-0-1 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 1148.163 $$0.0 0 -0.0100 0 0.867940 Modular form 1148.1.o.c.163.1 2-1148-1148.163-c0-0-0 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 1148.163$$ $0.0$ $0$ $-0.135$ $0$ $0.762772$ Modular form 1148.1.o.d.163.3
2-1148-1148.1075-c0-0-1 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 1148.1075 $$0.0 0 0.331 0 2.09912 Modular form 1148.1.bj.a.1075.1 2-1148-1148.1075-c0-0-0 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 1148.1075$$ $0.0$ $0$ $-0.168$ $0$ $1.52157$ Modular form 1148.1.bj.a.1075.2
2-1148-287.223-c0-0-1 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 287.223 $$0.0 0 0.334 0 2.12432 Modular form 1148.1.bc.a.797.1 2-1148-287.223-c0-0-0 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 287.223$$ $0.0$ $0$ $-0.165$ $0$ $1.45851$ Modular form 1148.1.bc.a.797.2
2-1148-287.174-c0-0-1 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 287.174 $$0.0 0 0.344 0 1.90442 Modular form 1148.1.bc.a.461.1 2-1148-287.174-c0-0-0 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 287.174$$ $0.0$ $0$ $-0.155$ $0$ $1.29128$ Modular form 1148.1.bc.a.461.2
2-1148-287.160-c0-0-1 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 287.160 $$0.0 0 0.155 0 1.68063 Modular form 1148.1.bc.a.1021.1 2-1148-287.160-c0-0-0 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 287.160$$ $0.0$ $0$ $-0.344$ $0$ $0.598160$ Modular form 1148.1.bc.a.1021.2
2-1148-287.139-c0-0-1 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 287.139 $$0.0 0 0.165 0 1.73059 Modular form 1148.1.bc.a.713.1 2-1148-287.139-c0-0-0 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 287.139$$ $0.0$ $0$ $-0.334$ $0$ $0.913343$ Modular form 1148.1.bc.a.713.2
2-1148-1148.975-c0-0-1 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 1148.975 $$0.0 0 0.168 0 1.55223 Modular form 1148.1.bj.a.975.2 2-1148-1148.975-c0-0-0 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 1148.975$$ $0.0$ $0$ $-0.331$ $0$ $0.654607$ Modular form 1148.1.bj.a.975.1
2-1148-1148.655-c0-0-7 $0.756$ $0.572$ $2$ $2^{2} \cdot 7 \cdot 41$ 1148.655 $$0.0 0 -0.489 0 2.38681 Modular form 1148.1.o.c.655.2 2-1148-1148.655-c0-0-6 0.756 0.572 2 2^{2} \cdot 7 \cdot 41 1148.655$$ $0.0$ $0$ $0.135$ $0$ $2.03759$ Modular form 1148.1.o.d.655.3
2-1148-287.163-c1-0-14 $3.02$ $9.16$ $2$ $2^{2} \cdot 7 \cdot 41$ 287.163 $$1.0 1 0.0170 0 1.13667 Modular form 1148.2.r.a.737.8 2-1148-1.1-c1-0-1 3.02 9.16 2 2^{2} \cdot 7 \cdot 41 1.1$$ $1.0$ $1$ $0$ $0$ $0.635683$ Modular form 1148.2.a.d.1.3
2-1148-1.1-c1-0-10 $3.02$ $9.16$ $2$ $2^{2} \cdot 7 \cdot 41$ 1.1 $$1.0 1 0 0 1.45590 Modular form 1148.2.a.d.1.4 2-1148-1.1-c1-0-11 3.02 9.16 2 2^{2} \cdot 7 \cdot 41 1.1$$ $1.0$ $1$ $0.5$ $1$ $1.50976$ Modular form 1148.2.a.b.1.2
2-1148-1.1-c1-0-12 $3.02$ $9.16$ $2$ $2^{2} \cdot 7 \cdot 41$ 1.1 $$1.0 1 0 0 1.63247 Modular form 1148.2.a.e.1.5 2-1148-1.1-c1-0-13 3.02 9.16 2 2^{2} \cdot 7 \cdot 41 1.1$$ $1.0$ $1$ $0.5$ $1$ $1.73648$ Modular form 1148.2.a.c.1.2
2-1148-1.1-c1-0-14 $3.02$ $9.16$ $2$ $2^{2} \cdot 7 \cdot 41$ 1.1 $$1.0 1 0.5 1 1.77332 Modular form 1148.2.a.b.1.1 2-1148-1.1-c1-0-15 3.02 9.16 2 2^{2} \cdot 7 \cdot 41 1.1$$ $1.0$ $1$ $0.5$ $1$ $1.81257$ Modular form 1148.2.a.c.1.3
2-1148-1.1-c1-0-16 $3.02$ $9.16$ $2$ $2^{2} \cdot 7 \cdot 41$ 1.1 $$1.0 1 0.5 1 2.07586 Modular form 1148.2.a.b.1.3 2-1148-1.1-c1-0-17 3.02 9.16 2 2^{2} \cdot 7 \cdot 41 1.1$$ $1.0$ $1$ $0.5$ $1$ $2.07949$ Modular form 1148.2.a.a.1.2
2-1148-1.1-c1-0-18 $3.02$ $9.16$ $2$ $2^{2} \cdot 7 \cdot 41$ 1.1 $$1.0 1 0.5 1 2.10636 Modular form 1148.2.a.c.1.4 2-1148-1.1-c1-0-19 3.02 9.16 2 2^{2} \cdot 7 \cdot 41 1.1$$ $1.0$ $1$ $0.5$ $1$ $2.12971$ Modular form 1148.2.a.c.1.5
2-1148-1.1-c1-0-2 $3.02$ $9.16$ $2$ $2^{2} \cdot 7 \cdot 41$ 1.1 $$1.0 1 0 0 0.672920 Modular form 1148.2.a.e.1.1 2-1148-1.1-c1-0-3 3.02 9.16 2 2^{2} \cdot 7 \cdot 41 1.1$$ $1.0$ $1$ $0$ $0$ $0.952968$ Modular form 1148.2.a.d.1.2
2-1148-1.1-c1-0-4 $3.02$ $9.16$ $2$ $2^{2} \cdot 7 \cdot 41$ 1.1 $$1.0 1 0 0 1.00123 Modular form 1148.2.a.e.1.2 2-1148-1.1-c1-0-5 3.02 9.16 2 2^{2} \cdot 7 \cdot 41 1.1$$ $1.0$ $1$ $0$ $0$ $1.10215$ Modular form 1148.2.a.e.1.4
2-1148-1.1-c1-0-6 $3.02$ $9.16$ $2$ $2^{2} \cdot 7 \cdot 41$ 1.1 $$1.0 1 0.5 1 1.22097 Modular form 1148.2.a.a.1.1 2-1148-1.1-c1-0-7 3.02 9.16 2 2^{2} \cdot 7 \cdot 41 1.1$$ $1.0$ $1$ $0.5$ $1$ $1.39605$ Modular form 1148.2.a.c.1.1
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2021-09-20 20:46:06
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9961052536964417, "perplexity": 440.2955397978399}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057091.31/warc/CC-MAIN-20210920191528-20210920221528-00034.warc.gz"}
|
http://openstudy.com/updates/4dd41498d95c8b0b81e055c4
|
## anonymous 5 years ago 7.2-6). Divide as indicated. Be sure to reduce all answers to lowest terms. d^2+10d+25 d^2-25 _______________ divide _______ d+5 d-5
1. watchmath
$$\frac{(d+5)^2}{d+5}\cdot \frac{(d+5)(d-5)}{(d-5}=(d+5)^2$$
2. anonymous
r u sure this is correct.. i tried is and its wrong
3. watchmath
Sorry I made a mistake. It should be $\frac{(d+5)^2}{d+5}\cdot \frac{d-5}{(d-5)(d+5)}=1$
|
2017-01-19 13:17:34
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|
https://scicomp.stackexchange.com/questions/33636/why-should-one-use-a-tree-structure-to-represent-discrete-function-spaces
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# Why should one use a tree structure to represent discrete function spaces?
In FEM/FV codebases, I stumbled upon the fact that the discretized functionspaces are represented within the code as a tree structure. I find this very puzzling.
Example: lets say somebody wants to simulate the navier stokes equations. The codebase expects the discretized function spaces to be put into a data structure like:
So an hierarchical tree structure. I do not understand why that should be a reasonable choice because:
1. There is no intrinsic hierarchy of the fields. In the image, the pressure variables are listed at a higher level than the velocity components. But that is purely cosmetic display. There is no intrinsic hierarchy of the components influencing our coupled system. If there is no hierarchy, then a tree structure does not seem sensible to me.
2. Data locality. I have an option where I can tell the programm to align the data cell-wise or in lexicographic ordering. That is a nice feature, but it is, at it's heart, the age-old question of (array-of-structs vs. struct-of-arrays). This also does not explain to me why I should use a tree.
3. Limited Complexity. I have not come across any codebase or scenario, where the function spaces are so many, or so complex, that that would justify a non-trivial data structure. Even if you do full MHD simulations, you could easily fit your fields into (array-of-structs vs. struct-of-arrays). And it is questionable whether a tree structure would simplify or confuse your code.
What am I missing? Why would anybody use a tree structure for this? (Thanks in advance!)
• OK... If you think using tree structure doesn't make sense, so what do you think might be better option here? Oct 23 '19 at 16:14
• What you need in your assembly procedures is all the local degrees of freedom. If I were to implement it (as a newby), I would just use a container for all dof's associated with a cell/element. Oct 24 '19 at 7:38
## 1 Answer
Although I can't speak for NS or MHD, I do find this "componentization" of function spaces to be a useful design principle in CEM (computational electromagnetics), especially for high-order (in p) discretizations.
CEM often uses multiple spaces at the same time: grad-conforming functions to represent electric potential, curl-conforming functions to represent E-field, div-conforming functions to represent B-flux and currents, etc. These spaces are related to one another via the deRham diagram, the range of each differential operator (grad/curl/div) forms the kernel of the next (the exact sequence property).
In my experience this can (should?) show up all the way at the implementation level, because it saves you time and helps guarantee correctness. For example, if you've spent time and money tabulating/implementing a high order grad-conforming space (basis functions $$\phi$$'s and their gradients $$\nabla \phi$$'s, you can (should?) reuse those $$\nabla \phi$$'s as basis functions within the curl-conforming space, then you just need to enrich them by adding whatever additional functions are that are needed span the range of curl to your desired order. Similar constructions hold between curl-conforming and div-conforming spaces (the curls of your curl conforming functions should be members of your div-conforming basis).
Unfortunately, this enrichment process is messy/broken down at p=0 because of unisolvency issues (p=0 curls already spans p=0 gradients, etc). It really only shines at p >= 1. In practice, I use the following decomposition:
• hgrad0: p=0 gradient-conforming functions
• hgradp: p>0 gradient-conforming functions
• hcurl0: p=0 curl-conforming functions
• hcurlp: functions that "enrich" the range of curl for p>0
• hdiv0: p=0 div-conforming functions
• hdivp: functions that "enrich" the range of div for p>0
With these building blocks, you can tabulate the whole diagram:
• hgrad := hgrad0 + hgradp
• hcurl := hcurl0 + grad(hgradp) + hcurlp
• hdiv := hdiv0 + curl(hcurlp) + hdivp
• L2 := piecewise constant + div(hdivp)
And each of these spaces (i) splits apart the kernel from the range of the respective differential operator and (ii) splits apart the p=0 space from the p>0 .. both of these properties are useful when designing preconditioners.
• Thank you very much for this answer. I need to do some reading I guess:-) Oct 24 '19 at 7:53
• Lets say the global function space is the following: B_th = B_v1 x B_v2 x _Bv3 x B_p. Then the grouping of them still seems a bit arbitrary. I guess the whole thing comes from the fact that the involved operators only apply to subspaces of the whole problem, and then the grouping is induced! Oct 24 '19 at 8:55
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2021-10-21 15:34:47
|
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|
http://tex.stackexchange.com/questions/50694/cannot-use-toprule-when-doing-input-inside-tabular-why/50700
|
# Cannot use \toprule when doing \input inside tabular — why?
Using \input inside a tabular environment does not seem to work in all cases, but I don't see why.
The MWE below gives the following error message:
! Misplaced \noalign.
\toprule ->\noalign
{\ifnum 0=`}\fi \@aboverulesep =\abovetopsep \global \@b...
l.1 \toprule
File main.tex:
\documentclass{standalone}
\usepackage{booktabs}
\begin{document}
\begin{tabular}{ll}
\input{inp}
\end{tabular}
\end{document}
File inp.tex:
\toprule
a & b \\
c & d \\
\bottomrule
The document compiles properly if the contents of inp.tex are pasted into the tabular environment. What is happening here?
\documentclass{standalone}
\usepackage{booktabs}
\begin{document}
\begin{tabular}{ll}
\toprule
a & b \\
c & d \\
\bottomrule
\end{tabular}
\end{document}
EDIT: The following question is related: Why is \input not expandable?
-
\input is trying to be intelligent, and this seems to be upsetting LaTeX's picky tabular definition.
But you can access the original \input primitive from TeX which is less dangerous at this point.
After defining
\makeatletter
\newcommand\primitiveinput[1]
{\@@input #1 }
\makeatother
using \primitiveinput inside the tabular should work.
-
I would just prefer putting \makeatletter\let\primitiveinput\@@input\makeatother in the preamble and using it with \primitiveinput imp\relax. Calling your macro \primitiveinput is confusing: it is not the TeX primitive, but a macro that calls the primitive. – yo' Apr 4 '12 at 10:58
Of course you can use it in any way you prefer. But I think a more LaTeXy definition is less confusing to users. Additionally, your \relax would again disturb the tranquility of tabular, giving an error for \primitiveinput inp\relax\midrule. So the user would have to strictly adhere to leaving a space behind the file name, something which might not come naturally to LaTeX users... – Stephan Lehmke Apr 4 '12 at 11:03
Thank you, this seems reasonable. Is this related: tex.stackexchange.com/questions/43006/… ? – krlmlr Apr 4 '12 at 11:44
Yep, that's exactly the same problem, only wrapped in an extremely long question ;-) – Stephan Lehmke Apr 4 '12 at 11:47
Some commands e.g. \toprule, \hline or \multicolumn must come directly at the start of a cell. There can't be nothing before, this includes commands like \input.
-
Still, Stephan Lehmke show it is possible, becuase his solution works... ;-) – yo' Apr 4 '12 at 10:56
@tohecz, Both are right, it just depends when you look. \omit or noalign or \span must be the first thing after expansion (you need to allow expansion as none of the commands mentioned are primitive) the primitive \@@input is expandable so basicaly is not there according to this definition, but LaTeX's \input is not expandable and so leaves a token before the \omit. – David Carlisle Apr 4 '12 at 11:01
@DavidCarlisle Then this is not a solution but a comment ... :-/ – yo' Apr 4 '12 at 11:08
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2015-05-22 16:10:28
|
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|
https://math.meta.stackexchange.com/questions/27653/tag-management-2018/28108
|
# Tag management 2018 [duplicate]
New year, new tag management thread.
Rules of the game are basically the same:
• A particularly bad tag (a rule of thumb: «if I can't imagine a person classifying a tag as either interesting or ignored, I'm getting rid of it»),
• A tag that should be a synonym of an existing one,
• A tag that used for two or more completely unrelated things,
• A need to create a new tag;
• Wait a couple of days before implementing a suggestion.
• After the problem described in an answer is resolved, please edit it to say so.
Also, note that one may use [tag:calculus] for , i.e. tags on the main site, and [meta-tag:discussion] for , i.e. for tags on the meta site.
• You might want to make a comment as I did of the topics that are still open. – suomynonA Jan 9 '18 at 5:07
• @suomynonA Thanks, I'll work on that sometime this week. – Clarinetist Jan 9 '18 at 14:10
• That's a long week. :) – Asaf Karagila Mar 25 '18 at 10:00
• @AsafKaragila Thanks for pinging me! It's done now! :) – Clarinetist Mar 25 '18 at 16:23
Given the recent meta question about the tag "functions" is more than one month's old. I suggest we implement the most up-voted action: to rename the tag.
If I understand correctly (here), renaming can be done either by merging or setting up a tag synonyms. In any case, one has to have another tag to start with. Thus I suggest introducing the mapping tag as follows:
(Name): Mapping
(excerpt)
For questions about basic (set-theoretic) properties of a mapping $$f: A\to B$$ between two sets $$A, B$$.
(info)
A mapping $$f$$ from $$A$$ to $$B$$ is roughly speaking an assignment, for each element $$a$$ in $$A$$, an unique element $$f(a)$$ in $$B$$. A mapping is also commonly called a function. Mapping is one of the most fundamental concepts in mathematics.
Any question concerning the most basic properties of mappings should use this tag. This includes:
• Basic notations,
• Domain, codomain, image, preimage of a mapping.
Depending on context, please consider also adding algebra-precalculus or elementary-set-theory. Do not use this tag simply because it has a function in the question.
• Perhaps "Do not use this tag simply because it has a function in the question" should also go in the excerpt, so that a few more users will notice the instruction. – Misha Lavrov Mar 26 '18 at 1:58
• @Misha: Although a good idea, we have sort of "reverse survivor bias" (so, casualty bias?) in the sense that we have no idea how effective these messages are. We only see the masses which ignore them entirely in play. – Asaf Karagila Mar 26 '18 at 11:34
Proposal to pluralize and to make a synonym of "reproducing-kernel-hilbert-spaces"
Reason:
1. Idem to my previous proposal: we have .
2. RKHS is a widely-accepted acronym for .
3. Short acronyms are good for tag names, e.g. , and . Unlike and , doesn't even have a tag synonym "stochastic-differential-equations". (I support the creation of such synonym though.)
• Done. I did not make the "resolved" edit on purpose now. // I don't fully agree iwth 3. – quid Jan 13 '19 at 15:18
• @quid Thanks for your action. For 3, typing "stochastic" in tag search doesn't return the tag sde, unlike pde and ode. Making this synonym enhances user experience. – GNUSupporter 8964民主女神 地下教會 Jan 14 '19 at 12:51
Resolved. The tag has been removed.
I noticed that a new tag has been created this year, apparently in February. I first noticed it today, while reviewing a suggested tag-wiki for the tag.
I do not think the coincidences tag is necessary, nor helpful. If users could add up to ten tags on any question, then perhaps it doesn't hurt to have it. But at this point in time, the maximum tags for any questions is five.
Only 17 questions on MSE have the "coincidences" tag.
Reason: Mathematical objects in tag names have plural forms, say , , , etc
Resolved: has been created. Synonyms include and .
We need a tag for topological data analysis.
It is a rapidly growing field, so there will be more and more posts about it. Currently, questions about TDA come up under the tags , , , , and a few more. These are insufficient for searchability as TDA is its own distinct topic that is not directly encompassed by any of these tags.
We have a few options here.
I'd like some feedback on which of these would be the best choice, which (if any) should be synonyms, and whether there is a better name for the tag that I haven't thought of.
• I'd go with the [topological-data-anaylsis] tag, it's the most descriptive for people outside the field. And in any case, not [persistence]. You'd end up cleaning this from users who had a persistent problem... – Asaf Karagila Jul 29 '18 at 15:52
• To second @AsafKaragila's comment, I think that it is useful to have a tag that cannot be easily misunderstood. [persistence] and [computational-topology] both seem like they could be easily misunderstood by a newbie (i.e. how do I compute the fundamental group of some topological space). I think that [topological-data-analysis] is the best choice, with (perhaps) [persistent-topology] as a synonym. – Xander Henderson Jul 29 '18 at 19:41
Resolved (The tag now has the longer name.)
I suggest to rename to . (Now that the limit for the length of tagnames is 35 characters, the latter is a possible name for a tag.)
Both the tag-excerpt and the tag-info for (differential-equations) explicitly say that it is for ODEs and not for PDEs. And there is also a synonym with the tag . Here is the current revision of the tag-excerpt:
Questions on (ordinary) differential equations. For questions specifically concerning partial differential equations, use the (pde) tag.
This tag is also among the suggestions for tag-warnings: Do we want tag warnings, and for which tags?
As far as I can tell, there are enough experienced users here who know that this tag is only for ODEs and who retag the questions which are incorrectly tagged. But perhaps if the purpose of the tag is clear already from the tag name, then the likelihood of askers tagging their questions incorrectly would be a bit smaller.
• Done. I didn't make the "resolved" edit on purpose right now – quid Jan 13 '19 at 15:12
I've noticed we have both and . Would it make sense to make the former an alias of the latter?
• And jacobi-symbol is out there too. I don’t know why, but my gut feeling is that there is room for all of these. But, maybe I’m in the minority. – pjs36 Apr 11 '18 at 21:37
• There are now synonym suggestions (jacobi-symbol) $\to$ (quadratic-residues) and (legendre-symbol) $\to$ (quadratic-residues). (One of them was suggested by barto, the other one by me.) Users with sufficient score in the tag can vote for/against them here. Maybe it could be useful to include this information into your post - that could attract attention of users who might vote on those synonyms. (And perhaps attract also some users who have something to add to this discussion.) – Martin Sleziak May 29 '18 at 6:24
I'm not sure that the tag is needed. Julia sets lie squarely in the realm of complex dynamics, which averages only a few questions a month. The tag was introduced very recently and contained just one question. I edited the question to be tagged instead.
Thoughts?
• It is not clear from this post whether you have some proposal about this tag. If you are saying that the tag should be removed, then this was achieved. If you want to suggest a synonym, then you should mention which tag you consider as a suitable master tag. – Martin Sleziak Jul 12 '18 at 8:06
Resolved (the tag was removed)
### Proposal to eliminate the infinitelydivisible tag
This tag has no usage guidance and it was used so far for a single question. I suggest that we eliminate it.
• I would consider the discussions about (semicontinuity) resolved - the tag (semicontinuous-functions) was created as a result of that discussion. – Martin Sleziak Apr 8 '18 at 9:25
### Resolved
Rename to
The outcome of Revisiting the "partitions" tag was the splitting of by the creation of new tags and . The question also proposed that be renamed to , and the consensus (and only) answer proposed to do this at the end of the retagging process. However, the retagging was completed months ago, the rename is still outstanding, and it's still frequently necessary to fix new questions which have been incorrectly tagged instead of (or in addition to) or .
• Done. I am not sure why I did not react on your comment in April; as I have no memory of having seen it, I likely just missed it. Sorry for the delay. – quid Jun 29 '18 at 21:42
• Probably this can now be marked as resolved. – Martin Sleziak Jan 5 '19 at 16:58
• @MartinSleziak, yes, my bad. – Peter Taylor Jan 5 '19 at 22:06
Resolved. The new name is .
Suggestion: Rename tag to box-topology.
The tag is of low usage (fewer than a dozen instances, some of which are inappropriate if the proper application is for topology. I noticed this tag in reviewing a tag wiki excerpt which presumed that was the intention, but I rejected it as being nearly of the form [tag] is for questions about [tag].
So an alternative that I could live with is to delete the tag.
• Done. I did not make the "resolved" edit right now on purpose. – quid Jan 13 '19 at 15:07
Resolved (the synonym is no more)
The tag has been deleted/subsumed/transformed into the tag roughly five years ago, thus making impossible to signal as such, questions involving conditional probabilities conditioned by a sigma-algebra in the general case, and in particular conditional probabilities conditioned by a continuous random variable. This inappropriate decision should be undone.
If I am reading this post correctly, the deletion of was decided and enacted by a small number of users with no noticeable activity in the subject. Oddly, and fortunately, the tag survived.
## Proposal to pluralize sequence-of-function
I have some doubts about the existence of a tag. But, since it exists, I propose that its name becomes instead.
• Merely as a historical curiosity, I will mention that the tag (sequence-of-functions) was created in January 2015 and later removed. The new tag (sequence-of-function) was created in May 2017 and still exists. (See the related comments in the tagging chatroom for more details.) – Martin Sleziak Aug 29 '18 at 11:18
• @MartinSleziak That's interesting. I was not aware of that. – José Carlos Santos Aug 29 '18 at 11:45
Resolved: has been renamed to .
There is an ongoing discussion on at The tag even-and-odd-extensions.
I have noticed that there is a tag named on meta. It is actually one of the default tags - these are the tags that are automatically created on every new meta site and are not deleted even if they have zero questions.
You may notice that on Meta Stack Exchange, this tag is now a synonym of (favorite-tags). And I'd guess most users know this feature under the name favorite tags. (Maybe it was called differently at the time when the default tags were selected.) I think that average user would have hard time guessing what (interesting-tags) is intended for. Which might lead to inconsistent usage of this tag and some incorrectly tagged questions.
My suggestion is to clarify usage of this tag and also make the name correspond to today's terminology.
• One possibility would be to create a new tag and make this tag a synonym. (With (favorite-tags) as the master tag. This would basically copy the usage on Meta Stack Exchange.)
• Another option would be to create a tag synonym $\to$ and use (favorites) both for favorite questions and favorite tags.
If you look at current usage of the tag and also if you check current revision of the tag-excerpt and the tag-wiki, you can see that this tag is currently used for both favorite tags and favorite question. (I am partly to blame - I have used the tag in this way and also I have created the tag-info. However, at the time it seemed to me a bit redundant to create a separate tag for favorite tags.)
This means that if we decide to have a separate tag for , this would require also retagging of a few older questions. Which is why I am inclined more to the latter option, i.e., to have one tag for both favorite tags and favorite question.
But I am certainly open to the other solution. We can discuss (and vote) in comments which of the two options is better.
• Now it seems that there are plans to rename this feature to "tag watching" or "watched tags": 'Favorite Tags' is now 'Tag Watching' (Meta Stack Exchange). It is already live on some sites, I assume that eventually this will happen on this site, too. – Martin Sleziak Jul 9 '18 at 6:59
• On Meta Stack Exchange there is already a tag synonym with (tag-watching) as the master tag. – Martin Sleziak Jul 28 '18 at 16:58
As recently pointed on this thread, we are going to get rid of . Today I noticed that we also have .
I feel that this tag can go just as well as part of the process.
I suggest eliminating the tag .
Someone just this tag today, and added it to three questions. As written, the description is... the most generic thing ever, basically "anything associated with permutation groups and permutation cycles". Given that there is already a tag, what purpose does the (permutation-cycles) tag add?
(As a side note, why is the tag a synonym for (permutations)? There are lots of cases where people are dealing with problems involving permutations, but not working with a group. Likewise, there are a lot of interesting group theory problems about permutation actions that are not covered by other tags.)
• permutation-cycles may, in theory, be a good tag for searchability purposes. Decomposing a permutation into cycles is a common topic in elementary group theory courses and those problems show up enough in permutations to warrant their own tag, if only to exclude that type of problem during a search – Alexander Gruber Jul 23 '18 at 4:06
• @AlexanderGruber In that sense, I think a (permutation-groups) tag is infinitely more valuable then a (permutation-cycles) tag. This tag could cover orbits (cycles), stabilizers, things of that nature, rather than simply counting permutations from an alphabet. – Morgan Rodgers Jul 23 '18 at 5:10
• i'd certainly agree with you in a perfect world; however, the synonym was created in the first place due to widespread misuse of permutation-groups to tag elementary group theory questions about permutations. I don't think it had been applied in the way you suggest even once, even though that was its intended purpose. – Alexander Gruber Jul 23 '18 at 6:39
• What we've been using instead for those types of questions has been group-actions. – Alexander Gruber Jul 23 '18 at 6:42
• My main reason for creating the tag is precisely the reason Alexander Gruber gave: to make a distinction between abstract algebra permutations and combinatoric permutations. I could not add the (permutation-groups) tag to the questions that I was looking at (I'm not sure how synonyms work on this site), so I created the tag (permutation-cycles) to make this distinction. I felt the name of the tag also made it easier to make the distinction for questions that are elementary and posed by newer users -- the questions I've added the tag to do not use permutations in the context of group actions... – Bill Wallis Jul 23 '18 at 13:18
• ...and some of them do not explicitly use the group structure of permutations either. My intended usage of the (permutation-cycles) tag would be for questions concerning the permutations themselves in isolation. For example, the multiple representation of permutations, clarifying multiplication of permutations (particularly since this is not covered well in many elementary textbooks)... – Bill Wallis Jul 23 '18 at 13:18
• ...clarifying classic permutation proofs such as unique parity, and so on. I feel that the distinction is worth being made, but I welcome all criticisms (I'm still learning what makes a good tag). – Bill Wallis Jul 23 '18 at 13:18
Ongoing discussion on at Do we want an [umvue] tag?.
Suggestion: eliminate the tag .
The tag was recently introduced. It currently applies to two questions, both of which were asked by who I presume is the tag creator. It seems way too niche to merit a tag, and I suggest that this tag be eliminate.
• I disagree that it is way too niche. I saw when it was recently introduced, and was actually surprised that it did not already exist. Perhaps this is just my bias as it is my niche, though. – Eric Wofsey Jul 9 '18 at 20:18
• I will just add that creation of the tag was also mentioned in the Tagging chatroom: chat.stackexchange.com/transcript/3740/2018/7/6 and chat.stackexchange.com/transcript/3740/2018/7/9 But it is certainly good to bring this up on meta, where more users notice this than in chat. – Martin Sleziak Jul 10 '18 at 8:30
### Reopen the homeomorphism tag
I think that we should have a homeomorphism tag. On the other hand, I am aware of the fact that there was such a tag in the past (before I arrived here) and that, since it was removed, my proposal will almost surely be rejected. But I would like to have an explanation for that. It doesn't seem to be essentially different from the tag, the tag, or the tag, all of which exist. It it certainly seems useful for questions about proving that two topological spaces are (or are not) homeomorphic.
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2020-09-23 19:36:25
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https://googology.wikia.org/wiki/Multiplication
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11,053 Pages
Multiplication is an elementary binary operation, written $$ab$$, $$a \times b$$, $$a(b)$$ or $$a \cdot b$$ (pronounced "$$a$$ times $$b$$"). For natural numbers, it is defined as repeated addition:
$a \times b = \underbrace{a + a + \cdots + a + a}_b.$
For example, $$3 \times 4 = 3 + 3 + 3 + 3 = 12$$. The result of a multiplication problem is called the product.
For real numbers, the product $$x \times y$$ is defined as the product of the equivalence classes of Cauchy sequences of rational numbers. For positive reals $$x$$ and $$y$$, it can be thought as a product of width and height of a rectangle.
Like addition, multiplication is commutative and associative on $$\mathbb{N}$$ and $$\mathbb{R}$$: $$a \times b = b \times a$$ and $$(a \times b) \times c = a \times (b \times c)$$ for all natural and real values of $$a,b$$ and $$c$$. Repeated multiplication is called exponentiation.
However, multiplication is not commutative on ordinals. For any limit ordinal $$\alpha$$, $$2 \times \alpha = \alpha \neq \alpha \times 2$$.
It is possible to define multiplication for natural numbers in the recursive way:
\begin{eqnarray*} a \times 0 & = & 0 \\ a \times (b+1) & = & (a \times b) + a \end{eqnarray*}
In googology, it is the second hyper operator
## Other properties
• $$0 \times n = 0$$
• $$1 \times n = n$$
• $$(-a) \times (-b) = a \times b$$
• $$(-a) \times b = a \times (-b) = -(a \times b)$$
## Turing machine code
Given input of form (string of a 1's) (string of b 1's) it outputs string of a*b 1's
0 1 _ r 1
0 _ _ r 9
1 1 1 r 1
1 _ _ r 2
2 1 _ r 3
2 _ _ l 7
3 1 1 r 3
3 _ _ r 4
4 1 1 r 4
4 _ 1 l 5
5 1 1 l 5
5 _ _ l 6
6 1 1 l 6
6 _ 1 r 2
7 1 1 l 7
7 _ _ l 8
8 1 1 l 8
8 _ _ r 0
9 1 _ r 9
9 _ _ r halt
## Dependency of properties on arithmetics
When we work in ZFC set theory, properties of multiplication such as commutativity and associativity hold. However, in Robinson arithmetic, the commutativity does not necessarily hold. That is, the statement $$\forall a,b \in \mathbb{N} : a \times b = b \times a$$ is independent of Robinson arithmetic.
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2021-06-23 10:47:56
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https://www.examsolutions.net/tutorials/arcs-sectors-segments/
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## Arc length and sector area
I will show you how to find the area of a sector and the length of an arc when the angle is in both degrees and radians
### Angle in degrees
Arc length and Area of Sectors (degrees) : ExamSolutions Maths Revision - youtube Video
Arc Length and Area of Sectors (radians) : ExamSolutions Maths Revision - youtube Video
## Area of a segment
In these videos I show you how to find the area of a segment when the angle is given in degrees or radians. I assume that you are familiar with finding the area of a triangle given two sides and the included angle.
### Angle in degrees
Area of segments (degrees) : ExamSolutions Maths Revision - youtube Video
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2019-10-16 00:45:49
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https://www.physicsforums.com/threads/sketching-level-curves.288444/
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# Sketching level curves
1. Jan 29, 2009
### Somefantastik
$$x^{2}-y^{2}-2x+4y+5;$$
let $$x^{2}-y^{2}-2x+4y+5 \ = \ c;$$
To sketch this as a level curve, I'm not sure how to proceed. I can't seem to rearrange the function into anything familiar.
For the sake of trying to find a reference point, I let x=0 and found
$$y \ = \ 2 \ ^{+}_{-}\sqrt{9-c};$$
then y=0 =>
$$x \ = \ 1 \ ^{+}_{-}\sqrt{-4+c};$$
If I let c = 5, I get
$$x^{2}-y^{2}-2x+4y = 0;$$
which gives x = 0, x = 2, y = 0, y = 4.
What should I do next?
2. Jan 29, 2009
### NoMoreExams
Level curves are usually for functions f(x,y) = blah i.e. when you have 2 ind. variables. Here you only have one ind.... so you can solve for y by completing the square
3. Jan 29, 2009
### Somefantastik
The function that I am trying to draw level curves for is
$$V(x,y) =x^{2}-y^{2}-2x+4y+5;$$
4. Jan 29, 2009
### NoMoreExams
Well you should be able to rewrite it by completing the square, I'll work this one as example
$$V(x,y) = x^2 - y^2 - 2x + 4y + 5 = (x-1)^2 - (y-2)^2 + 8$$
5. Jan 30, 2009
### Somefantastik
Ok, yeah I see that. It's a translated hyperbola?
I keep getting
$$V(x,y) = (x-1)^{2} - (y-2)^{2} +5;$$
Which is a hyperbola centered at (1,2) and translated somewhere 5 units?
6. Jan 30, 2009
### NoMoreExams
How are you getting that?
$$(x-1)^2 - (y-2)^2 + 5 = x^2 - 2x + 1 - (y^2 - 4y + 4) + 5 = x^2 - y^2 - 2x + 4y + 2$$
That's not what you started with?
7. Jan 30, 2009
### Somefantastik
I should have been paying better attention.
How would I graph this? My cal books just have all these guys equal to 1.
$$V(x,y) = (x-1)^2 \ - \ (y-2)^2 \ + \ 8 \ = \ c;$$
$$V(x,y) = (x-1)^2 \ - \ (y-2)^2 \ = \ c \ - 8;$$
8. Feb 1, 2009
Well, if $(x - 1)^2 - (y - 2)^2 = c - 8$, then
$$\frac{(x - 1)^2}{c - 8} - \frac{(y - 2)^2}{c - 8} = 1.$$
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2016-12-07 09:22:15
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https://oatml.cs.ox.ac.uk/blog/2019/06/24/batchbald.html
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# Human in the Loop: Deep Learning without Wasteful Labelling
### Affiliations
OATML, Department of Computer Science, University of Oxford
### Published
NeurIPS 2019
TLDR: In Active Learning we use a “human in the loop” approach to data labelling, reducing the amount of data that needs to be labelled drastically, and making machine learning applicable when labelling costs would be too high otherwise. In our paper we present BatchBALD: a new practical method for choosing batches of informative points in Deep Active Learning which avoids labelling redundancies that plague existing methods. Our approach is based on information theory and expands on useful intuitions. We have also made our implementation available on GitHub at https://github.com/BlackHC/BatchBALD.
### What’s Active Learning?
Using deep learning and a large labelled dataset, we are able to obtain excellent performance on a range of important tasks. Often, however, we only have access to a large unlabelled dataset. For example, it is easy to acquire lots of stock photos, but labelling these images is time consuming and expensive. This excludes many applications from benefiting from recent advances in deep learning.
In Active Learning we only ask experts to label the most informative data points instead of labelling the whole dataset upfront. The model is then retrained using these newly acquired data points and all previously labelled data points. This process is repeated until we are happy with the accuracy of our model.
To perform Active Learning, we need to define some measure of informativeness, which is often done in the form of an acquisition function. This measure is called an “acquisition function” because the score it computes determines which data points we want to acquire. We send unlabelled data points which maximise the acquisition function to an expert and ask for labels.
### The problem is…
Usually, the informativeness of unlabelled points is assessed individually, with one popular acquisition function being BALD . However, assessing informativeness individually can lead to extreme waste because a single informative point can have lots of (near-identical) copies. This means that if we naively acquire the top-K most informative points, we might end up asking an expert to label K near-identical points!
### Our contribution
In our work, we efficiently expand the notion of acquisition functions to batches (sets) of data points and develop a new acquisition function that takes into account similarities between data points when acquiring a batch. For this, we take the commonly-used BALD acquisition function and extend it to BatchBALD in a grounded way, which we will explain below.
However, knowing how to score batches of points is not sufficient! We still have the challenge of finding the batch with the highest score. The naive solution would be to try all subsets of data points, but that wouldn’t work because there are exponentially many possibilities.
For our acquisition function, we found that it satisfies a very useful property called submodularity which allows us to follow a greedy approach: selecting points one by one, and conditioning each new point on all points previously added to the batch. Using the submodularity property, we can show that this greedy approach finds a subset that is “good enough” (i.e. $1-1/e$-approximate).
Overall, this leads our acquisition function BatchBALD to outperform BALD: it needs fewer iterations and fewer data points to reach high accuracy for similar batch sizes, significantly reducing redundant model retraining and expert labelling, hence cost and time.
Moreover, it is empirically as good as, but much faster than, the optimal choice of acquiring individual points sequentially, where we retrain the model after every single point acquisition.
Before we explain our acquisition function, however, we need to understand what the BALD acquisition function does.
### What’s BALD?
BALD stands for “Bayesian Active Learning by Disagreement” .
As the “Bayesian” in the name tells us, this assumes a Bayesian setting which allows us to capture uncertainties in the predictions of our model. In a Bayesian model, the parameters are not just numbers (point estimates) that get updated during training but probability distributions.
This allows the model to quantify its beliefs: a wide distributions for a parameter means that the model is uncertain about its true value, whereas a narrow one quantifies high certainty.
BALD scores a data point $x$ based on how well the model’s predictions $y$ inform us about the model parameters $\boldsymbol{\omega}$. For this, it computes the mutual information $\mathbb{I}(y, \boldsymbol{\omega})$. Mutual information is well-known in information theory and captures the information overlap between quantities.
When using the BALD acquisition function to select a batch of $b$ points, we select the top-$b$ points with highest BALD scores, which is standard practice in the field. This is the same as maximising the following batch acquisition function $a_{\mathrm{BALD}}\left(\left\{x_{1}, \ldots, x_{b}\right\}, \mathrm{p}\left(\boldsymbol{\omega} | \mathcal{D}_{\mathrm{train}}\right)\right) :=\sum_{i=1}^{b} \mathbb{I}\left(y_{i} ; \boldsymbol{\omega} | x_{i}, \mathcal{D}_{\mathrm{train}}\right)$ with $\left\{x_{1}^{*}, \ldots, x_{b}^{*}\right\} := \underset{\left\{x_{1}, \ldots, x_{b}\right\} \subseteq \mathcal{D}_{\text { pool }}}{\arg \max } a_{\mathrm{BALD}}\left(\left\{\boldsymbol{x}_{1}, \ldots, \boldsymbol{x}_{b}\right\}, \mathrm{p}\left(\boldsymbol{\omega} | \mathcal{D}_{\text { train }}\right)\right).$ Intuitively, if we imagine the information content of the predictions given some data points and the model parameters as sets in the batch case, the mutual information can be seen as intersection of these sets, which captures the notion that mutual information measures the information overlap.
In fact, Yeung shows that this intuition is well-grounded, and we can define an information measure $\mu^*$ that allows us to express information-theoretic quantities using set operations: \begin{aligned} \mathbb{H}(x,y) &= \mu^*(x \cup y) \\ \mathbb{I}(x,y) &= \mu^*(x \cap y) \\ \mathbb{E}_{p(y)} \mathbb{H}(x | y) &= \mu^*(x \setminus y) \end{aligned} Figure 5 visualizes the scores that BALD computes as area of the intersection of these sets when acquiring a batch of 3 points. Because BALD is a simple sum, mutual information between data points is double-counted, and BALD overestimates the true mutual information. This is why naively using BALD in a dataset with lots of (near-identical) copies of the same point will lead us to select all the copies: we double count the mutual information intersection between all!
### BatchBALD
In order to avoid double-counting, we want to compute the quantity $\mu^*(\bigcup_i y_i \cap \boldsymbol{\omega})$ , as depicted in figure 6, which corresponds to the mutual information $\mathbb{I}(y_1,…,y_b ; \boldsymbol{\omega} | x_1,…,x_b, \mathcal{D}_\mathrm{train})$ between the joint of the $y_i$ and $\boldsymbol{\omega}$ : $a_{\mathrm{BatchBALD}}\left(\left\{x_{1}, \ldots, x_{b}\right\}, \mathrm{p}\left(\boldsymbol{\omega} | \mathcal{D}_{\mathrm{train}}\right)\right) := \mathbb{I}\left(y_{1}, \ldots, y_{b} ; \boldsymbol{\omega} | x_{1}, \ldots, x_{b}, \mathcal{D}_{\mathrm{train}}\right).$ Expanding the definition of the mutual information, we obtain the difference between the following two terms: $a_{\mathrm{BatchBALD}}\left(\left\{x_{1}, \ldots, x_{b}\right\}, \mathrm{p}(\boldsymbol{\omega} | \mathcal{D}_{\mathrm{train}})\right) = \mathbb{H}\left(y_{1}, \ldots, y_{b}\right | x_{1}, \ldots, x_{b}, \mathcal{D}_{\mathrm{train}})-\mathbb{E}_{\mathrm{p}(\boldsymbol{\omega} | \mathcal{D}_{\mathrm{train}} )}\left[\mathbb{H}\left(y_{1}, \ldots, y_{b} | x_{1}, \ldots, x_{b}, \boldsymbol{\omega}\right)\right].$ The first term captures the general uncertainty of the model. The second term captures the expected uncertainty for a given draw of the model parameters.
We can see that the score is going to be large when the model has different explanations for the data point that it is confident about individually (yielding a small second term) but the predictions are disagreeing with each other (yielding a large first term), hence the “by Disagreement” in the name.
### Submodularity
Now to determine which data points to acquire, we are going to use submodularity.
Submodularity tells us that there are diminishing returns: selecting two points increases the score more than just adding either one of them individually but less than the separate improvements together:
Given a function $f: \Omega \to \mathbb{R}$, we call $f$ submodular, if: $f(A \cup \{x, y\}) - f(A) \le \left ( f(A \cup \{x\}) - f(A) \right ) + \left ( f(A \cup \{y\}) - f(A) \right ),$ for all $A \subseteq \Omega$ and elements $x,y \in \Omega$.
We show in appendix A of the paper that our acquisition function fulfils this property.
Nemhauser et al. have shown that, for submodular functions, one can use a greedy algorithm to pick points with a guarantee that their score is at least $1-1/e \approx 63\%$ as good as the optimal one. Such an algorithm is called $1-1/e$-approximate.
The greedy algorithm starts with an empty batch $A = \{\}$ and computes $a_{\mathrm{BatchBALD}}(A \cup \{x\})$ for all unlabelled data points, adds the highest-scoring $x$ to $A$ and repeats this process until $A$ is of acquisition size.
This is explained in more detail in the paper.
### Consistent MC Dropout
We implement Bayesian neural networks using MC dropout . However, as an important difference to other implementations, we require consistent MC dropout: to be able to compute the joint entropies between data points, we need to compute $a_{\mathrm{BatchBALD}}$ using the same sampled model parameters.
To see why, we have investigated how the scores change with different sets of sampled model parameters being used in MC dropout inference in figure 7.
Without consistent MC dropout, scores would be sampled using different sets of sampled model parameters, losing function correlations between the $y_i$’s for near-by $x_i$’s, and would essentially be no different than random acquisition given the spread of their scores.
### Experiments on MNIST, Repeated MNIST and EMNIST
We have run experiments on classifying EMNIST, which is a dataset of handwritten letters and digits consisting of 47 classes and 120000 data points.
We can show improvement over BALD which performs worse (even compared to random acquisition!) when acquiring large batches:
This is because compared to BatchBALD and random, BALD actively selects redundant points. To understand this better, we can look at the acquired class labels and compute the entropy of their distribution. The higher the entropy, the more diverse the acquired labels are:
We can also look at the actual distribution of acquired classes at the end of training, and see that BALD undersamples some classes while BatchBALD manages to pick data points from different classes more uniformly (without knowing the classes, of course). Random acquisition also picks classes more uniformly than BALD, but not as well as BatchBALD. Figure 14: Histogram of acquired class labels on EMNIST. BatchBALD left, random acquisition center, and BALD right. Classes are sorted by number of acquisitions. Several EMNIST classes are underrepresented in BALD and random acquisition while BatchBALD acquires classes more uniformly. The histograms were created from all acquired points.
To see how much better BatchBALD copes with pathological cases, we also experimented with a version of MNIST that we call Repeated MNIST. It is simply MNIST repeated 3 time with some added Gaussian noise and shows how BALD falls into a trap where picking the top $b$ individual points is detrimental because there are too many similar points. But BALD is not the only acquisition function to fail in this regime. Figure 15: Performance on Repeated MNIST. BALD, BatchBALD, Var Ratios, Mean STD and random acquisition: acquisition size 10 with 10 MC dropout samples.
We also played around with different acquisition sizes and found that on MNIST, BatchBALD can even acquire 40 points at a time with little loss of data efficiency while BALD deteriorates quickly.
### Final thoughts
We found it quite surprising that a standard acquisition function, used widely in active learning, performed worse even compared to a random baseline, when evaluated on batches of data. We enjoyed digging into the core of the problem, trying to understand why it failed, which led to some new insights about the way we use information theory tools in the field. In many ways, the true lesson here is that when something fails — pause and think.
### References
1. BatchBALD: Efficient and Diverse Batch Acquisition for Deep Bayesian Active Learning
Kirsch, A., van Amersfoort, J. and Gal, Y., 2019.
2. Bayesian active learning for classification and preference learning
Houlsby, N., Huszar, F., Ghahramani, Z. and Lengyel, M., 2011. arXiv preprint arXiv:1112.5745.
3. A new outlook on Shannon's information measures
Yeung, R.W., 1991. IEEE transactions on information theory, Vol 37(3), pp. 466--474. IEEE.
4. An analysis of approximations for maximizing submodular set functions—I
Nemhauser, G.L., Wolsey, L.A. and Fisher, M.L., 1978. Mathematical programming, Vol 14(1), pp. 265--294. Springer.
5. Dropout as a Bayesian approximation: Representing model uncertainty in deep learning
Gal, Y. and Ghahramani, Z., 2016. international conference on machine learning, pp. 1050--1059.
## Contact
We are located at
Department of Computer Science, University of Oxford
Wolfson Building
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2023-01-29 04:55:36
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https://en.wikisource.org/wiki/Translation:Attempt_of_a_Theory_of_Electrical_and_Optical_Phenomena_in_Moving_Bodies/Introduction
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# Translation:Attempt of a Theory of Electrical and Optical Phenomena in Moving Bodies/Introduction
## Introduction
§ 1. The question as to whether the aether shares the motion of ponderable bodies or not, has still found no answer that satisfies all physicists. For the decision, primarily the aberration of light and related phenomena could be used, but so far none of the two contested theories, neither that of Fresnel, nor that of Stokes, were fully confirmed with respect to all observations, so concerning the choice between the two views we can only weigh against each other the remaining problems for both of them. By that I was long ago led to believe that with Fresnel's view, i.e. with the assumption of a stationary aether, we are on the right way. While against the view of Stokes there is hardly more than one objection, i.e. the doubt that his assumptions regarding the aether-motion in the vicinity of Earth are contradictory[1], but this objection is of great weight, and I can't see at all how it could be eliminated.
The difficulties for Fresnel's theory stem from the known interference experiment of Michelson[2] and, as some think, from the experiments, by which Des Coudres in vain sought to find an influence of Earth's motion on the induction of two circuits[3]. The results of the American scientist, however, allow of an interpretation by an auxiliary hypotheses, and the findings of Des Coudres can easily be explained without such one.
Concerning the observations of Fizeau[4] on the rotation of polarization in glass columns, the matter is as follows. At first glance, the result is decidedly against Stokes' view. Yet when I tried to improve Fresnel's theory, the explanation of Fizeau's experiments was not quite successful, so I gradually suspected that this result had been obtained by observational error, or at least it had not met the theoretical considerations which formed the basis of the experiments. And Fizeau was so friendly to tell my colleague van de Sande Bakhuijzen after his request, that at present he himself doesn't see his observations as crucial.
In the further course of this work, I will come back in more detail to some of the issues raised at this place. Here I was concerned only with the preliminary justification of the standpoint I have taken.
In favor of Fresnel's theory several well-known reasons can be cited. Especially the impossibility of locking the aether between solid or liquid walls. As far as we know, a space devoid of air behaves (in the mechanical sense) like a real vacuum, when ponderable bodies are in motion. When we see how the mercury of a barometer rises to the top when the tube is inclined, or how easily a closed metal shell can be compressed, one can not avoid the idea, that solid and liquid bodies let the aether pass through without hindrance. One hardly will assume, that this medium could suffer a compression, without giving resistance to it.
That transparent bodies can move, without communicating their full velocity to the contained aether, was proven by Fizeau's famous interference experiment with streaming water[5]. This experiment, that later was repeated by Michelson and Morley[6] on a larger scale, could impossibly have had the observed success, when everything within the tube would possess a common velocity. By that, only the behavior of nontransparent substances and very extended bodies remains questionable.
It should be noted, moreover, that we can imagine the permeability of a body in two ways. First, this property might not be present in individual atoms, yet when the atoms were very small compared to the gaps between them, it might be present in matter of greater extension; but secondly, it may be assumed - and this hypothesis I will use in the following - that ponderable matter is absolutely permeable, namely that at the location of an atom, also the aether exists at the same time, which would be understandable if we were allowed to see the atoms as local modifications of the aether.
It is not my intention to enter into such speculations more closely, or to express assumptions about the nature of the aether. I only wish to keep myself as free as possible from preconceived opinions about that substance, and I won't, for example, attribute to it the properties of ordinary liquids and gases. If it is the case, that a representation of the phenomena would succeed best under the condition of absolute permeability, then one should admit of such an assumption for the time being, and leave it to the subsequent research, to give us a deeper understanding.
That we cannot speak about an absolute rest of the aether, is self-evident; this expression would not even make sense. When I say for the sake of brevity, that the aether would be at rest, then this only means that one part of this medium does not move against the other one and that all perceptible motions are relative motions of the celestial bodies in relation to the aether.
§ 2. Since Maxwell's views became more and more accepted, the question of the properties of the aether became highly important also for the theory of elasticity. Strictly speaking, not a single experiment in which a charged body or a current conductor moves, can be handled carefully, if the state of motion of the aether is not considered at the same time. In any phenomenon of electricity, the question arises whether an influence of the earth's motion is to be expected; and regarding the consequences of the latter for optical phenomena, we have to demand from the electro-magnetic theory of light that it can account for the already established facts. Namely, the aberration theory isn't one of those parts of optics, for which treatment the general principles of wave theory are sufficient. Once a telescope comes into play, one can not help but to apply Fresnel's dragging coefficient to the lenses, yet its value can only be derived from special assumptions about the nature of light vibrations.
The fact that the electro-magnetic theory of light really leads to that coefficient assumed by Fresnel, was shown by me two years ago[7]. Since then I have greatly simplified the theory and extended it also to the processes involved in reflection and refraction, as well as birefringent bodies[8]. It may be permitted for me, to come back to this matter.
To come to the basic equations for the phenomena of electricity in moving bodies, I joined an opinion that has been represented in recent years by several physicists; I have indeed assumed that small electrically charged molecules exist in all bodies, and that all electric processes are based on the location and motion of these "ions". As regards the electrolytes, this view is widely recognized as the only possible one, and Giese[9], Schuster[10], Arrhenius[11], Elster and Geitel[12] have defended the view, that also as regards the electricity conduction in gases, we are dealing with a convection by ions. It seems to me, that nothing prevents us to believe that the molecules of ponderable dielectric bodies contain such particles, which are connected to certain equilibrium positions and are moved only by external electric forces thereof; just herein the "dielectric polarization" of such bodies would consist.
The periodically changing polarization, which forms a light ray according to Maxwell's theory, become vibrations of the ions in this conception. It is well known that many researchers, who stood on the basis of the older theory of light, considered the resonance of ponderable matter as the cause of color dispersion, and this explanation can in the main also included into the electro-magnetic theory of light, for which it is only necessary to ascribe to the ions a certain mass. This I have shown in a previous paper[13], in which I admittedly have derived the equations of motion from actions at a distance, and not, what I now consider to be much easier, from Maxwell's expressions. Later, von Helmholtz[14] in his electromagnetic theory of color dispersion started from the same point of view[15].
Giese[16] has applied to various cases the hypothesis, that electricity is connected to ions in metallic conductors as well; but the picture which he gives of the processes in these bodies is at one point substantially different from the idea that we have on the conduction in electrolytes. While the particles of dissolved salt, however often they may be stopped by the water molecules, eventually might travel over large distances, the ions in a copper wire will hardly have such a great mobility. We can however be satisfied with forward and backward motion at molecular distances, if we only assume that one ion often transfers its charge to another, or that two oppositely charged ions, if they meet, or after they were "connected" with one another, exchange their charges against each other. In any case, such processes must take place at the boundary of two bodies, when a current flows from one to the other. If for example ${\displaystyle n}$ positively charged copper atoms are separated at a copper plate, and we also want for the latter all the electricity be connected to ions, then we have to assume that the charges are transferred to ${\displaystyle n}$ atoms in the plate, or that ${\displaystyle {\tfrac {1}{2}}n}$ of the deposited particles exchange their charges with ${\displaystyle {\tfrac {1}{2}}n}$ negatively charged copper atoms, which were already in the electrode.
Thus, if the adoption of this transition or exchange of the ionic charges - one of course still very dark process - is the essential complement to any theory that requires an entrainment of electricity by ions, then a persistent electric current never consists of a convection alone, at least not when the centers of two touching or interconnected particles are in some distance ${\displaystyle l}$ from each other. Then the electricity motion happens without convection over a distance of order ${\displaystyle l}$, and only if this is very small in proportion to the distance over which a convection takes place, we on the whole are dealing almost exclusively with this latter phenomenon.
Giese is of the opinion that in metals a real convection was not at all in play. But since it does not seem possible to include the "jumping" of the charges into the theory, then one would excuse, that for my part I totally disregard such a process, and that I interpret a current in a metal wire simply as a motion of charged particles.
Further research will have to decide whether the results of the theory remains at a different view.
§ 3. The theory of ions was very suitable for my purpose, because it makes it possible to introduce the permeability of the aether in a rather satisfactory way in the equations. Of course, these were decomposed into two groups. First, we have to express as to how the state of the aether by charge, position and motion of the ions is determined; then, secondly, we have to indicate by which forces the aether is acting on the charged particles. In my paper already cited[17] I have derived the formulas by means of d'Alembert's principle from certain assumptions and therefore selected a path, that has much resemblance with Maxwell's application of Lagrange's equations. Now I prefer for the sake of brevity, to introduce the basic equations themselves as hypotheses.
The formulas for the aether are in agreement, regarding the space between the ions, with the known equations of Maxwell's theory, and generally express that any change that was caused by an ion in the aether, propagates with the velocity of light. But we regard the force exerted by the aether on a charged particle, as a function of the state of that medium at the point where the particle is located. The adopted fundamental law differs in a major point from the laws, that were introduced by Weber and Clausius. The influence that was suffered by a particle B due to the vicinity of a second one A, indeed depends on the motion of the latter, but not on its instantaneous motion. Much more relevant is the motion of A some time earlier, and the adopted law corresponds to the requirement for the theory of electrodynamics, that was presented by Gauss in 1845 in his known letter to Weber[18]
In general, the assumptions that I introduce represent in a certain sense a return to the earlier theories of electricity. The core of Maxwell's views is therefore not lost, but it cannot be denied that with the adoption of ions we are not far away from the electric particles, which were used earlier. In some simple cases, this occurs particularly clear. Since the essence of electric charge is seen by us in the accumulation of positive or negative charged particles, and since the basic formulas for stationary ions give Coulomb's law, therefore, for example, the entire electrostatics can be brought into the earlier form.
1. Lorentz. De l’influence du mouvement de la terre sur les phénomènes lumineux. Arch. néerl., T. 21, p. 103, 1887; Lodge. Aberration problems. London Phil. Trans., Vol. 184. A, p. 727, 1893; Lorentz. De aberratietheorie van Stokes. Zittingsverslagen der Akad. v. Wet. te Amsterdam, 1892—93, p. 97.
2. Michelson. American Journal of Science, 3d. Ser., Vol. 22, p. 120; Vol. 34, p. 333, 1887; Phil. Mag., 5th. Ser., Vol. 24, p. 449, 1887.
3. Des Coudres. Wied. Ann., Bd. 38, p. 71, 1889.
4. Fizeau. Ann. de chim. et de phys., 3e sér., T. 58, p. 129, 1860; Pogg. Ann., Bd. 114, p. 554, 1861.
5. Fizeau. Ann. de chim. et de phys., 3e sér. T. 57, p. 385, 1859; Pogg. Ann., Erg. 3, p. 457, 1853.
6. Michelson and Morley. American Journal of Science, 3d. ser., Vol. 31, p. 377, 1886.
7. Lorentz. La théorie électromagnétique de Maxwell et son application aux corps mouvants. Leide, E. J. Brill, 1892. (Also published in Arch, néerl., T. 25).
8. A preliminary report about that was published in Zittingsverslagen der Akad. v. Wet. te Amsterdam, 1892—93, pp. 28 and 149.
9. Giese. Wied. Ann., Bd. 17, p. 538, 1882.
10. Schuster. Proc. Roy. Soc., Vol. 37, p. 317, 1884.
11. Arrhenius. Wied. Ann., Bd. 32, p. 565, 1887; Bd. 33, p. 638, 1888.
12. Elster and Geitel. Wiener Sitz.-Ber., Bd. 97, Abth. 2, p. 1255, 1888.
13. Lorentz. Over net vorband tusschen de voortplantingssnelheid van het licht en de dichtheid en samenstelling der middenstoffen. Verhandelingen der Akad. van Wet. te Amsterdam, Deel 18, 1878; Wied. Ann., Bd. 9, p. 641, 1880.
14. v. Helmholtz. Wied. Ann., Bd. 48, p. 389, 1893.
15. Also Koláček (Wied. Ann., Bd. 32, pp. 224 and 429, 1887) attempted to explain (albeit in a different manner) dispersion by electrical vibrations in the molecules.
Also the theory of Goldhammer (Wied. Ann., Bd. 47, p. 93, 1892) has to be mentioned.
16. Giese. Wied. Ann., Bd. 37, p. 576, 1889.
17. Lorentz. La théorie électromagnétique de Maxwell et son application aux corps mouvants.
18. Gauss. Werke, Bd. 5, p. 629.
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2018-02-24 20:23:32
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https://bookdown.org/igisc/EnvDataSci/spatial-analysis-case-studies.html
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# 9 Spatial Analysis case studies
## 9.1 Plotting filtered map data using base plot, terra::plotRGB and SpatVector data
In the Spatial Analysis chapter, we’ve mainly used tmap, but we might also want to see how we’d plot the same thing in the base plot system, which also support’s terra’s SpatVector data. So in this brief example, we’ll create a SpatVector after a data filter, then plot in the base plot system.
library(tidyverse); library(sf); library(igisci)
coords = c("LONGITUDE", "LATITUDE"), remove=FALSE, crs=4326)
We might also look at how we get information about out data to decide what we might want to plot on the map. Since our goal is to filter for a range of elevations and latitudes, we might look at a couple of histograms, and we’ll use a base R plot’s parameter setting (par) to create a side by side plot (1 row, 2 columns).
par(mfrow=c(1,2)) # lets base plot functions create 2 adjacent plots (1 row, 2 columns)
hist(sierraFebpts$ELEVATION) hist(sierraFebpts$LATITUDE)
par(mfrow=c(1,1)) # need to reset to defaults
Now let’s continue with that information where we’ve gotten an idea about what kinds of elevations occur at the stations, and make a map that shows all stations > 2000 m elevation.
We’ll briefly use terra::plotRGB to add a basemap to a static map. Warning: the get_tiles function goes online to get the basemap data, so if you don’t have a good internet connection or the site goes down, this may fail. We’ll also modify the station names and set the text label position as pos=3, which means centered on top (see ?graphics::text).
library(terra); library(maptiles)
sierraFebpts2000 <- sierraFebpts %>%
filter(ELEVATION >= 2000)
sierraBase <- get_tiles(sierraFebpts2000)
plotRGB(sierraBase)
pts <- vect(sierraFebpts2000)
points(pts, col="red")
text(pts, labels=str_sub(pts$STATION_NAME, 1, str_locate(pts$STATION_NAME, ",")-1), cex = 0.7, pos=3)
Before we continue, let’s look at some of the parameters just used for sizing and placing text labels, and figure out what they do:
text(pts, labels=str_sub(pts$STATION_NAME, 1, str_locate(pts$STATION_NAME, ",")-1), cex = 0.7, pos=3)
• What does cex do? (Use ?par or ?text then search within that help)
• What does pos do? (Use ?text)
• What did the stringr methods do?
Tweak each of the above to see the effect on the map.
Maybe there’s a way with plotRGB, but it’s limited in not apparently having a way to provide a graticule of latitude and longitude, so its use as a map is limited, and there are generally better options in tmap.
## 9.2 Distance from feature to closest feature: Air quality / TRI study
In the Spatial Analysis chapter section on Distance and specifically Distance to Closest Feature, we looked at point to closest point distances. Here’s another example of this that attempts to find the output of closest TRI facility to a census tract where we might compare health or socioeconomic variables to that value. The igisci::BayAreaTracts data has census data from 2010 by census tracts in the SF Bay area. In addition to the distance functions, the code below includes a few useful sf functions applied to that data.
library(igisci); library(sf); library(tidyverse)
TRI_BySite <- TRI_CA %>%
mutate(all_air = 5.1_FUGITIVE_AIR + 5.2_STACK_AIR) %>%
mutate(carcin_release = all_air * (CARCINOGEN == "YES")) %>%
group_by(TRI_FACILITY_ID) %>%
summarise(
count = n(),
air_releases = sum(all_air, na.rm = TRUE),
fugitive_air = sum(5.1_FUGITIVE_AIR, na.rm = TRUE),
stack_air = sum(5.2_STACK_AIR, na.rm = TRUE),
FACILITY_NAME = first(FACILITY_NAME),
COUNTY = first(COUNTY),
LATITUDE = first(LATITUDE),
LONGITUDE = first(LONGITUDE))
select(trID, HHinc2016) %>%
mutate(HHinc2016 = as.numeric(str_c(HHinc2016)),
joinid = str_c("0", trID))
censusBayArea <- censusBayArea %>%
mutate(whitepct = WHITE / POP2010 * 100) %>%
mutate(People_of_Color_pct = 100 - whitepct) %>%
left_join(incomeByTract, by = c("FIPS" = "joinid"))
censusCentroids <- st_centroid(censusBayArea)
census <- censusBayArea
bnd <- st_bbox(co)
TRIdata <- TRI_BySite %>%
filter(air_releases > 0) %>%
filter((LONGITUDE < bnd[3]) & (LONGITUDE > bnd[1]) & (LATITUDE < bnd[4]) & (LATITUDE > bnd[2]))
BA_counties <- co %>% select(NAME, CNTY_FIPS) %>% st_set_geometry(NULL)
TRI_sp <- st_as_sf(TRIdata, coords = c("LONGITUDE", "LATITUDE"), crs=4326) %>%
st_join(censusBayArea) %>%
filter(CNTY_FIPS %in% BA_counties$CNTY_FIPS) TRI2join <- TRI_sp %>% st_set_geometry(NULL) %>% rowid_to_column("TRI_ID") ggplot() + geom_sf(data = co, fill = NA) + geom_sf(data=TRI_sp) + coord_sf(xlim = c(bnd[1], bnd[3]), ylim = c(bnd[2], bnd[4])) + labs(x='',y='') Asthma model related to distance to nearest TRI facility. # CDC data CDCdata <- read_csv(ex("TRI/CDC_health_data_by_Tract_CA_2018_release.csv")) %>% mutate( lon = as.numeric(str_sub(Geolocation, 17, 30)), lat = as.numeric(str_sub(Geolocation, 2, 14)), CDC_CNTY_FIPS = str_sub(TractFIPS, 2, 4)) %>% filter(CDC_CNTY_FIPS %in% BA_counties$CNTY_FIPS)
CDCbay <- st_as_sf(CDCdata, coords = c('lon', 'lat'), crs=4326)
# D to TRI: create vector of index of nearest TRI facility to each CDC (tract centroid) point
nearest_TRI_ids <- st_nearest_feature(CDCbay, TRI_sp)
# get locations of each NEAREST TRI facility only
near_TRI_loc <- TRI_sp$geometry[nearest_TRI_ids] CDCbay <- CDCbay %>% mutate(d2TRI = st_distance(CDCbay, near_TRI_loc, by_element=TRUE), d2TRI = units::drop_units(d2TRI), nearest_TRI = nearest_TRI_ids) %>% left_join(BA_counties, by = c("CDC_CNTY_FIPS" = "CNTY_FIPS")) CDCTRIbay <- left_join(CDCbay, TRI2join, by = c("nearest_TRI" = "TRI_ID")) %>% filter(d2TRI > 0) # %>% CDCTRIbay %>% ggplot(aes(air_releases, CASTHMA_CrudePrev)) + geom_point(aes(color = NAME)) + geom_smooth(method="lm", se=FALSE, color = "black") + labs(x = "Air releases at nearest TRI facility", y = "Asthma Prevalence CDC Model, Adults >= 18, 2016") This is an admittedly crude analysis, lumping all releases together, and so it’s not surprising that there’s no clear relationship over the whole Bay Area. There’s quite a lot you could explore in the TRI data, and you might want to focus on one county, specific toxins, or probably aggregate releases at the total output in the area. And analyzing health data is very challenging given the wide range of factors influencing patterns. You might also consider looking at earlier TRI data (the igisci extdata "TRI" folder has it back to 1987, for California.) ## 9.3 Exploring terrain and water samples in a karst area Sinking Cove, Tennessee is a karst valley system carved into the Cumberland Plateau, a nice place to see the use of a hillshade raster created from a digital elevation model using raster or terra functions for slope, aspect, and hillshade. We’ll look at this area and some results from a hydrologic and geochemical study : We’ll start with a regional view using a hillshade and terrain shading: tmap_mode("plot") DEM <- rast(ex("SinkingCove/DEM_SinkingCoveUTM.tif")) slope <- terrain(DEM, v='slope') aspect <- terrain(DEM, v='aspect') hillsh <- shade(slope/180*pi, aspect/180*pi, angle=40, direction=330) # Need to crop a bit since grid north != true north bbox0 <- st_bbox(DEM) xrange <- bbox0$xmax - bbox0$xmin yrange <- bbox0$ymax - bbox0$ymin bbox1 <- bbox0 crop <- 0.05 bbox1[1] <- bbox0[1] + crop * xrange # xmin bbox1[3] <- bbox0[3] - crop * xrange # xmax bbox1[2] <- bbox0[2] + crop * yrange # ymin bbox1[4] <- bbox0[4] - crop * yrange # ymax bboxPoly <- bbox1 %>% st_as_sfc() # makes a polygon # tm_shape(hillsh, bbox=bboxPoly) + tm_raster(palette="-Greys",legend.show=F,n=20) + tm_shape(DEM) + tm_raster(palette=terrain.colors(24), alpha=0.5) + tm_graticules(lines=F) Then zoom into Upper Sinking Cove to look at some water sample results: library(igisci) library(sf); library(tidyverse); library(readxl); library(tmap) wChemData <- read_excel(ex("SinkingCove/SinkingCoveWaterChem.xlsx")) %>% mutate(siteLoc = str_sub(Site,start=1L, end=1L)) wChemTrunk <- wChemData %>% filter(siteLoc == "T") %>% mutate(siteType = "trunk") wChemDrip <- wChemData %>% filter(siteLoc %in% c("D","S")) %>% mutate(siteType = "dripwater") wChemTrib <- wChemData %>% filter(siteLoc %in% c("B", "F", "K", "W", "P")) %>% mutate(siteType = "tributary") wChemData <- bind_rows(wChemTrunk, wChemDrip, wChemTrib) sites <- read_csv(ex("SinkingCove/SinkingCoveSites.csv")) wChem <- wChemData %>% left_join(sites, by = c("Site" = "site")) %>% st_as_sf(coords = c("longitude", "latitude"), crs = 4326) library(terra) tmap_mode("plot") DEM <- rast(ex("SinkingCove/DEM_SinkingCoveUTM.tif")) slope <- terrain(DEM, v='slope') aspect <- terrain(DEM, v='aspect') hillsh <- shade(slope/180*pi, aspect/180*pi, 40, 330) bounds <- st_bbox(wChem) xrange <- bounds$xmax - bounds$xmin yrange <- bounds$ymax - bounds$ymin xMIN <- as.numeric(bounds$xmin - xrange/10)
xMAX <- as.numeric(bounds$xmax + xrange/10) yMIN <- as.numeric(bounds$ymin - yrange/10)
yMAX <- as.numeric(bounds$ymax + yrange/10) newbounds <- st_bbox(c(xmin=xMIN, xmax=xMAX, ymin=yMIN, ymax=yMAX), crs= st_crs(4326)) tm_shape(hillsh,bbox=newbounds) + tm_raster(palette="-Greys",legend.show=F,n=20) + tm_shape(DEM) + tm_raster(palette=terrain.colors(24), alpha=0.5,legend.show=F) + tm_shape(wChem) + tm_symbols(size="TH", col="Lithology", scale=2, shape="siteType") + tm_layout(legend.position = c("left", "bottom")) + tm_graticules(lines=F) An interactive map view of Upper Sinking Cove: tmap_mode("view") bounds <- st_bbox() wChem2map <- filter(wChem, Month == 8) minVal <- min(wChem2map$TH); maxVal <- max(wChem2map$TH) tm_basemap(leaflet::providers$Esri.WorldTopoMap) +
tm_shape(wChem2map) + tm_symbols(col="siteType", size="TH", scale=2) +
tm_layout(title=paste("Total Hardness ",as.character(minVal),"-",as.character(maxVal)," mg/L", sep=""))
tm_basemap(leaflet::providers\$Esri.WorldTopoMap) +
tm_shape(wChem2map) + tm_symbols(col="Lithology", size="TH", scale=2)
tmap_mode("plot")
### References
Davis, Jerry D, and George A Brook. 1993. “Geomorphology and Hydrology of Upper Sinking Cove, Cumberland Plateau, Tennessee.” Earth Surface Processes and Landforms 18 (4): 339–62.
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2022-05-21 06:48:38
|
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|
https://electronics.stackexchange.com/questions/293899/inductor-core-permeability-much-lower-than-expected
|
# Inductor core permeability much lower than expected
I am interested in creating an inductor with a large inductance/large mag field in core. However, when empirically measuring the inductance of the wound core, I calculated a relative permeability ~1000x less than expected (2-6 vs 2000). My question is whether this is due to an error in my setup/calculations/understanding, or due to receiving a core not up to spec.
The ferrite rod has stated relative permeability of ~2000 and B_sat of .49T (4900G). The physical dimensions are 25.66mm height x 6.32mm diam. http://www.ebay.com/itm/5-ea-Ferrite-Rod-Core-0-25x1-00-Fair-Rite-4077276011-2000u-Perm-Plain-Slug-77-/251653543001?hash=item3a97b85459
Given that this is ferrite, the permeability should be fairly linear. The datasheet for the material can be found here: http://www.mouser.com/ds/2/150/4077276011-476454.pdf
I wound the core with ~100 turns of 28AWG mag wire (.321mm diameter), with a DC resistance of 1.0 ohm. The result is quite messy, but the length of the coil extends 11.6mm, and the outer diameter goes up to 10.3mm:
The inductance of the coil was measured experimentally by creating an RL series circuit fed by a signal generator, and increasing the frequency until the Vpp across the RL circuit was twice the Vpp across the resistor alone. See http://www.daycounter.com/Articles/How-To-Measure-Inductance.phtml for an explanation of this method.
For a 100 ohm 1% tolerance resistor, the required frequency was 77kHz, resulting in a measured inductance of 358uH. I used the following equations to calculate relative permeability $\mu_r$. The geometry is neither a loop or solenoid, but the equations only differ by a factor of 2-3x in this case.
$B_{loop,center} = \mu_0\mu_r\frac{NI}{2R}$
$B_{solenoid,center} = \mu_0\mu_r\frac{NI}{l}$
$L_{loop} \approx \mu_0\mu_rN^2R(ln(\frac{8R}{r})-2)$, r = wire radius
$L_{solenoid} = \mu_0\mu_r\frac{N^2A}{l}$
Because there is at least 150 ohm in serial with the 2Vpp source, the most current that could flow in one direction is 1/150 = 6.7mA. Assuming the relative permeability is 2000 as stated, there shouldn't be any concern of saturation, as the magnetic field in the center of the loop/coil is at most .28T ($B_{loop,center}$ with R=3mm). (Also, the scoped sine waves were clean.)
Given the dimensions from above and the measured L of 358uH, $\mu_r$ comes out to 2-6, much less than the expected 2000. Why is this?
• You have a huge air gap in your magnetic circuit. It's like having a small resistor (your core) in series with a huge resistor (your air gap due to the core being a rod and not a closed path.) – John D Mar 21 '17 at 19:17
• Do a search on reluctance and magnetomotive force. You will find that geometry has a huge impact on magnetic circuits. Interestingly enough, magnetic equations have a similar form to electrical ones. Continuing with John D's analogy, you are saying the small resistor has a conductivity of A, and you calculate the entire circuit's conductivity of C, you are not accounting for the conductivity of the huge series resistor B. You are not accounting for the permeability and magnetic path of the "air gap". Note, I've tried to do this before by hand and was unable to obtain an accurate answer. – klamb Mar 21 '17 at 19:36
• Here's something that might help with understanding what's going on: info.ee.surrey.ac.uk/Workshop/advice/coils/gap/index.html If you can find a toroid made of the same material and put your windings around that, you will notice a huge difference in your inductance. – John D Mar 21 '17 at 19:38
• @JohnD This is the kind of answer I was looking for; I will have to revisit the theory. This may warrant a new question, but does the existence of an air gap also mean that the magnetic field sensed at a distance (e.g. via magnetometer) will not be amplified by the core material either? I thought the coil would temporarily magnetize the core such that the field due to the magnetized core overpowered the original field due to the coil by ~$\mu_r/1$. – abc Mar 21 '17 at 21:23
• I think your question about sensing the field at a distance does deserve a new question, it shouldn't turn into a discussion in the comments for this question. It would be much better if you describe what you are trying to do and ask how to get there, and for clarification on what you are not sure about. – John D Mar 21 '17 at 21:38
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2019-03-22 06:19:24
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https://iidb.org/threads/ticketmaster-seriously.4085/#post-143786
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# Ticketmaster... seriously?!
#### Jimmy Higgins
##### Contributor
So thanks to competition in the online ticket industry, Ticketmaster is charging a fee that is nearly 40% the value of the ticket! They just keep up'ing their fees. Have they actually changed anything in their business model in the last few years, other than hiking their ridiculous fees?
#### Tom Sawyer
##### Super Moderator
Staff member
I believe they've run a few of their competitors out of business recently, so it's not like they haven't done anything else.
#### Underseer
##### Contributor
You can always decide not to go to shows.
If attendance goes down enough, they'll either drop their prices in an attempt to bring customers back, or they'll jack up prices in an attempt to make up for lost revenue, thus driving even more people away, and starting a downward spiral that seriously damages the whole industry.
#### Jimmy Higgins
##### Contributor
You can always decide not to go to shows.
If attendance goes down enough, they'll either drop their prices in an attempt to bring customers back, or they'll jack up prices in an attempt to make up for lost revenue, thus driving even more people away, and starting a downward spiral that seriously damages the whole industry.
Oh yeah, I forgot about my choice as a consumer when a monopoly was involved. Thanks.
Any chance my not going to a concert will make the industry crash quick enough that I can get Steven Wilson tiks without extortion like ticketmaster fees?
#### barbos
##### Contributor
They are more profitable than oil industry now.
#### ronburgundy
##### Contributor
The massively growing scalping industry helps fuel this problem. They buy up tickets and create "sell outs", then charge 300%. IF they sell them all, they make a killing. But even if they only resell half the tickets they bought, they still profit, the venue and act "sell out", all while 1/4 of the seats are actually empty, thus little incentive to lower prices even with empty seats.
There are plenty of alternatives, such as brown paper tickets that charge about 5%. However, the big acts and venues stay with ticketmaster, probably because of backroom deals where they get a cut of the ticket master fees, thus allowing them to lie to customers about the true "face value" ticket cost they are charging.
#### Bronzeage
Staff member
This is a classic Economics 101 classroom discussion on the nature of monopoly. The monopolist wants to maximise his revenue. This usually means setting a price which moves all of his product, which in this case is a theater seat. It doesn't really matter how the price is subdivided between face value and added fees, the sale price is what it is. The ticket could be free, with a $95 service fee added. The competition for a monopolist is the alternatives to his product. There are many alternatives to going to a concert. Most of them cost nothing, so the monopolist must be very careful when pricing a concert ticket. For big money acts, concerts are more important than they once were. There was a time when a concert tour was seen as an advertising cost. Concerts sold records. These days, concerts increase the demand for pirate downloads, so concert profits can be a substantial part of the overall revenue. There is a kind of balance to it all. You can listen for free at home and pay more to see them in an auditorium. Your choice. It probably averages out to the same price as 40 years ago, adjusted for inflation. #### Jimmy Higgins ##### Contributor This is a classic Economics 101 classroom discussion on the nature of monopoly. The monopolist wants to maximise his revenue. This usually means setting a price which moves all of his product, which in this case is a theater seat. It doesn't really matter how the price is subdivided between face value and added fees, the sale price is what it is. The ticket could be free, with a$95 service fee added.
The competition for a monopolist is the alternatives to his product. There are many alternatives to going to a concert. Most of them cost nothing, so the monopolist must be very careful when pricing a concert ticket.
Odd... because if I want to see Steven Wilson or Alan Parsons or Kansas in a concert, there is no actual alternative. While blu-ray has improved the at home visual experience, there is no substitute to a live gig.
For big money acts, concerts are more important than they once were. There was a time when a concert tour was seen as an advertising cost. Concerts sold records. These days, concerts increase the demand for pirate downloads, so concert profits can be a substantial part of the overall revenue.
But that is the band, not Ticketmaster. Ticketmaster is the overhead on the sale of a ticket. In the case of a Kansas concert, nearly 40% on the ticket cost. As if there was somebody that can charge an hour's effort for the sale of a single ticket.
There is a kind of balance to it all. You can listen for free at home and pay more to see them in an auditorium. Your choice. It probably averages out to the same price as 40 years ago, adjusted for inflation.
The prices for tickets are probably a little less when noting inflation, but the ticket fees have grown and grown.
#### Bronzeage
##### Super Moderator
Staff member
Odd... because if I want to see Steven Wilson or Alan Parsons or Kansas in a concert, there is no actual alternative. While blu-ray has improved the at home visual experience, there is no substitute to a live gig.
For big money acts, concerts are more important than they once were. There was a time when a concert tour was seen as an advertising cost. Concerts sold records. These days, concerts increase the demand for pirate downloads, so concert profits can be a substantial part of the overall revenue.
But that is the band, not Ticketmaster. Ticketmaster is the overhead on the sale of a ticket. In the case of a Kansas concert, nearly 40% on the ticket cost. As if there was somebody that can charge an hour's effort for the sale of a single ticket.
There is a kind of balance to it all. You can listen for free at home and pay more to see them in an auditorium. Your choice. It probably averages out to the same price as 40 years ago, adjusted for inflation.
The prices for tickets are probably a little less when noting inflation, but the ticket fees have grown and grown.
I did not say "substitute". I said "alternative." If you don't go to the concert, anything you do that night is an alternative.
The face value of the ticket is irrelevant when there is only one source of tickets. They could price the tickets as four for a dollar. It doesn't matter. The ticket costs what you have to pay to obtain one. Who gets what part of the money is not a factor in this economic transaction.
Righteous indignation doesn't really fit into most economic models.
#### TSwizzle
##### Let's Go Brandon!
I feel your pain Jimmy, I hate Ticketmaster too. I avoid using them when ever possible but it is difficult. I'm not much of a concert goer but I do go to sporting events a few times a year. LA Kings don't use Ticketmaster anymore but the Dodgers do. In the case of Dodger tickets I sometimes use a season ticket broker where there is still a fee but it's lower than Ticketmaster and I get better value for money because the seats are better. Or I will drive my ass up to Dodger stadium and get the tickets at the ticket office. I would love to go to concerts but the tacked on fees are just too much. Concerts are getting to be so expensive anyway I just don't bother now.
#### ronburgundy
##### Contributor
Odd... because if I want to see Steven Wilson or Alan Parsons or Kansas in a concert, there is no actual alternative. While blu-ray has improved the at home visual experience, there is no substitute to a live gig.
But that is the band, not Ticketmaster. Ticketmaster is the overhead on the sale of a ticket. In the case of a Kansas concert, nearly 40% on the ticket cost. As if there was somebody that can charge an hour's effort for the sale of a single ticket.
There is a kind of balance to it all. You can listen for free at home and pay more to see them in an auditorium. Your choice. It probably averages out to the same price as 40 years ago, adjusted for inflation.
The prices for tickets are probably a little less when noting inflation, but the ticket fees have grown and grown.
I did not say "substitute". I said "alternative." If you don't go to the concert, anything you do that night is an alternative.
The face value of the ticket is irrelevant when there is only one source of tickets. They could price the tickets as four for a dollar. It doesn't matter. The ticket costs what you have to pay to obtain one. Who gets what part of the money is not a factor in this economic transaction.
How the true price is divided matters a great deal to whether the transaction is made and the level at which the final price settles. A lower face value with high fees added later after consumer makes a choice to buy is done strategically to manipulate consumers into paying a higher total price than they would if the total was given up front. Its a "put in slowly so they don't notice their colon is being shredded" sales tactic. In addition, the acts themselves get partly shielded from the negative image of being greedy by setting a lower face value, then secretly taking a cut of the "service fees".
Also, part the problem with ticket master is that they are a vertical monopoly. They own Live Nation and other concert promoters, which not only allows them to demand those acts use ticketmaster, but to coerce venues to use ticketmaster for all acts under threat of not bringing the big acts they control to that venue. It is the same tactic that big breweries who own most of the largest beer distributors to eliminate competition without the consumer ever having a choice. They coerce bars and stores to keep other brands off the shelf.
#### Bronzeage
##### Super Moderator
Staff member
Odd... because if I want to see Steven Wilson or Alan Parsons or Kansas in a concert, there is no actual alternative. While blu-ray has improved the at home visual experience, there is no substitute to a live gig.
But that is the band, not Ticketmaster. Ticketmaster is the overhead on the sale of a ticket. In the case of a Kansas concert, nearly 40% on the ticket cost. As if there was somebody that can charge an hour's effort for the sale of a single ticket.
There is a kind of balance to it all. You can listen for free at home and pay more to see them in an auditorium. Your choice. It probably averages out to the same price as 40 years ago, adjusted for inflation.
The prices for tickets are probably a little less when noting inflation, but the ticket fees have grown and grown.
I did not say "substitute". I said "alternative." If you don't go to the concert, anything you do that night is an alternative.
The face value of the ticket is irrelevant when there is only one source of tickets. They could price the tickets as four for a dollar. It doesn't matter. The ticket costs what you have to pay to obtain one. Who gets what part of the money is not a factor in this economic transaction.
How the true price is divided matters a great deal to whether the transaction is made and the level at which the final price settles. A lower face value with high fees added later after consumer makes a choice to buy is done strategically to manipulate consumers into paying a higher total price than they would if the total was given up front. Its a "put in slowly so they don't notice their colon is being shredded" sales tactic. In addition, the acts themselves get partly shielded from the negative image of being greedy by setting a lower face value, then secretly taking a cut of the "service fees".
Also, part the problem with ticket master is that they are a vertical monopoly. They own Live Nation and other concert promoters, which not only allows them to demand those acts use ticketmaster, but to coerce venues to use ticketmaster for all acts under threat of not bringing the big acts they control to that venue. It is the same tactic that big breweries who own most of the largest beer distributors to eliminate competition without the consumer ever having a choice. They coerce bars and stores to keep other brands off the shelf.
How the price is divide matters only if the ticket buyer could negotiate with each constituent of of the ticket. You can't do that, so the price charged, is the price of the ticket. Are people really so stupid they think $40 tickets are a great deal and don't realize they are also paying$59 in fees? A concert ticket has a very short shelf life. It's either sold before the door opens, or it's worthless. The costs of putting on the concert remain the same, no matter how many tickets.
There is nothing egalitarian about the concert ticket market. Nobody said it had to be fair.
If you want to make a some kind of "it's not fair" argument, consider this. A lot of concerts are performed in munificence arenas, which are owned by the public. Should a private monopoly be allowed to exploit its product by use of public property? You might have something with that.
#### Jimmy Higgins
##### Contributor
Then why do I pay lower fees (notably lower) with Ticketfly, when Thor graces me with a concert that uses that site instead of Ticketmaster?
#### Horatio Parker
##### Veteran Member
There was a time when a concert tour was seen as an advertising cost. Concerts sold records. These days, concerts increase the demand for pirate downloads, so concert profits can be a substantial part of the overall revenue.
Touring costs used to be borne by the artist. Now they're underwritten by advertisers(at least the big names are). IIRC Led Zeppelin started this.
#### Bronzeage
##### Super Moderator
Staff member
There was a time when a concert tour was seen as an advertising cost. Concerts sold records. These days, concerts increase the demand for pirate downloads, so concert profits can be a substantial part of the overall revenue.
Touring costs used to be borne by the artist. Now they're underwritten by advertisers(at least the big names are). IIRC Led Zeppelin started this.
It still comes down to simple economic fact, the price is what you pay to get a seat for the concert.
#### Horatio Parker
##### Veteran Member
And the sky is blue...
The point is artists now profit from touring in a way that they didn't used to. Seems relevant.
For those unhappy with the TM monolith, I suggest local alternatives. I can safely that all local bands, clubs, theaters and orchestras need support. If you want your money to go further in an artistic sense, try those.
Sent from my iPhone using Tapatalk
#### Jimmy Higgins
##### Contributor
That seems about as useful advice as telling someone who is asking for computer help to build their own computer. Steven Wilson is touring to support one of the top albums in 2015... and likely this decade. I can't see that from anywhere else.
#### Horatio Parker
##### Veteran Member
That seems about as useful advice as telling someone who is asking for computer help to build their own computer. Steven Wilson is touring to support one of the top albums in 2015... and likely this decade. I can't see that from anywhere else.
If performers are computers, then there many around you already, just not the Steve Wilson Special which you apparently must have.
I'm not supporting or excusing TM, only pointing out that locals need support and if where your art dollars go concerns you, then you might want to consider them. If OTOH you must have the latest greatest electric guitar 4/4 angsty vocal vesch, then you must go through their buddies at TM.
#### Jimmy Higgins
##### Contributor
That seems about as useful advice as telling someone who is asking for computer help to build their own computer. Steven Wilson is touring to support one of the top albums in 2015... and likely this decade. I can't see that from anywhere else.
If performers are computers, then there many around you already, just not the Steve Wilson Special which you apparently must have.
I'm not supporting or excusing TM, only pointing out that locals need support and if where your art dollars go concerns you, then you might want to consider them. If OTOH you must have the latest greatest electric guitar 4/4 angsty vocal vesch, then you must go through their buddies at TM.
I must? Only if the venue uses them. Most venues do, but not all.
I'm also curious about the 4/4 line.
review said:
But while “Three Years Older” is an impressive opener that runs at an appropriately long ten minutes, I’m far more impressed by the third track, “Hand. Cannot. Erase.” In the past I’ve lamented the slow deterioration of the sharp pop/rock music of Blackfield, Wilson’s project with Israeli singer/songwriter Aviv Geffen, but from the sound of this cut Wilson hasn’t lost his ability to craft a stellar pop tune. Admittedly, as far as pop songs go, it’s an atypical one; one would be hard pressed to think of a Billboard top ten song in 9/4. In the context of both Wilson’s music and Hand. Cannot. Erase. as a whole, however, it fits quite brilliantly.
#### Horatio Parker
##### Veteran Member
Jimmy Higgins; said:
I must? Only if the venue uses them. Most venues do, but not all.
I thought you were the one who said there was no alternative.
I'm also curious about the 4/4 line.
Just a semi snarky allusion to the fact that pop rock is pretty exclusively in 4/4 time, ie four quarters to the measure.
review said:
But while “Three Years Older” is an impressive opener that runs at an appropriately long ten minutes, I’m far more impressed by the third track, “Hand. Cannot. Erase.” In the past I’ve lamented the slow deterioration of the sharp pop/rock music of Blackfield, Wilson’s project with Israeli singer/songwriter Aviv Geffen, but from the sound of this cut Wilson hasn’t lost his ability to craft a stellar pop tune. Admittedly, as far as pop songs go, it’s an atypical one; one would be hard pressed to think of a Billboard top ten song in 9/4. In the context of both Wilson’s music and Hand. Cannot. Erase. as a whole, however, it fits quite brilliantly.
Fair enough. Again, I'm suggesting alternatives if you don't want your money to go to TM, not that you should like music other than SM. In addition to, perhaps.
#### Bronzeage
##### Super Moderator
Staff member
That seems about as useful advice as telling someone who is asking for computer help to build their own computer. Steven Wilson is touring to support one of the top albums in 2015... and likely this decade. I can't see that from anywhere else.
The only solution to this problem is to petition to have Steve Wilson declared a public utility and as such, the government shall set the price for access to his performances.
#### Horatio Parker
##### Veteran Member
I withhold financial support from the likes of TM for my own reasons, not that I think it'll make the slightest difference to them.
In this age of sequencers, digital backing tracks, auto-tune etc, I'm skeptical of so called "live" performances by pop/rock artists. I prefer to go where people play instruments and sing, preferably with minimal amplification or processing.
Staff member
Ticketbastard.
#### KeepTalking
##### Code Monkey
In some cases you can get your tickets at the venue box office, and avoid service fees, at least for a limited time. The local big outdoor venue here in STL routinely offers no service fee tickets at the box office for the first week of sales. I also like to support local talent at a nearby concert nightclub, they even regularly draw national touring acts (going to see Testament w/ Exodus and Shattered Sun for \$22 on Friday), and do not charge service fees if you buy at the venue. They only use something like TM for online sales (they actually use an outfit called TicketWeb). Their box office is usually closed until a few hours before a show, but you can sidle up to the bar any time, and have a beer while they are processing your ticket order. They can only squeeze in about 1,000 people, so you wont see Kansas performing there, but I would not be surprised to see someone with more limited appeal, like Steven Wilson, play there. They have very few actual seats, but I am not the kind of concert goer who typically sits down through a performance anyway.
That's not to say I don't ever go to concerts that are serviced by TicketScammer. I probably catch 2 or 3 shows a year where I have to auction off a body part to get tickets from those assholes, but I see that as a necessary evil every once in a while, and I try to only do so for the big summer festivals where I am going to see enough bands to get my money's worth. Rush is coming through in May, however, and I haven't missed a Rush concert in over a decade. Fortunately I have a perfectly good kidney that should net me enough money to pay TM, once I get through procrastinating on buying the tickets.
#### Jimmy Higgins
##### Contributor
In some cases you can get your tickets at the venue box office, and avoid service fees, at least for a limited time.
It gets tricky when the venue is a distance away. I remember getting tickets for a concert in Chicago, where you can't just stop off at the ticket booth, and the fuckers put an additional fee on the transaction right before everything was finished! You select the ticket with fee, then the delivery notice. Then the payment... and then a new overall fee. Ticketmaster is terrible!
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2022-08-11 10:50:56
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https://cs.paperswithcode.com/paper/a-sharp-bound-on-the-s-energy-and-its
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# A Sharp Bound on the $s$-Energy and Its Applications to Averaging Systems
28 Dec 2018 Chazelle Bernard
The {\em $s$-energy} is a generating function of wide applicability in network-based dynamics. We derive an (essentially) optimal bound of $(3/\rho s)^{n-1}$ on the $s$-energy of an $n$-agent symmetric averaging system, for any positive real $s\leq 1$, where~$\rho$ is a lower bound on the nonzero weights... (read more)
PDF Abstract
# Code Add Remove Mark official
No code implementations yet. Submit your code now
# Categories
• MULTIAGENT SYSTEMS
• OPTIMIZATION AND CONTROL
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2021-01-19 18:06:10
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http://docs.itascacg.com/flac3d700/flac3d/zone/doc/manual/zone_manual/zone_fish/zone.gridpoint_intrinsics/fish_gp.accel.html
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# gp.accel
Syntax
## Vector Access
v := gp.accel(gp<,i>)
Get the acceleration vector for the gridpoint. Returns a vector, unless an optional component argument i is supplied, in which case it returns the component value. This is only available if model configure dynamic has been specified.
Returns: v - acceleration vector or component z - zone pointer i - optional vector component, from 1 to 3
## Component Access
f := gp.accel.x(gp)
Get the $$x$$-component acceleration vector. This is only available if model configure dynamic has been specified.
Returns: f - $$x$$-component acceleration vector z - zone pointer
f := gp.accel.y(gp)
Get the $$y$$-component acceleration vector. This is only available if model configure dynamic has been specified.
Returns: f - $$y$$-component acceleration vector z - zone pointer
f := gp.accel.z(gp)
Get the $$z$$-component acceleration vector. This is only available if model configure dynamic has been specified.
Returns: f - $$z$$-component acceleration vector z - zone pointer
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2020-10-28 16:54:57
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https://www.gamedev.net/blogs/entry/2255006-the-menu-interface/
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• entries
8
4
• views
6557
689 views
Everything was done by me except for the background. Though I think I should take that as a working prototype because I think I can improve on it.
There are no comments to display.
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2018-02-18 11:15:19
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https://datascience.stackexchange.com/questions/14648/replace-all-numeric-values-in-a-pyspark-dataframe-by-a-constant-value
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# Replace all numeric values in a pyspark dataframe by a constant value
Consider a pyspark dataframe consisting of 'null' elements and numeric elements. In general, the numeric elements have different values. How is it possible to replace all the numeric values of the dataframe by a constant numeric value (for example by the value 1)? Thanks in advance!
Example for the pyspark dataframe: $$\begin{array}{c|lcr} & \text{c1} & \text{c2} & \text{c3} \\ \hline 1 & 0.04 & 1 & 1.35 \\ 2 & -1 & null & -1.2 \\ 3 & null & 1.2 & null \end{array}$$
The result should be:
$$\begin{array}{c|lcr} & \text{c1} & \text{c2} & \text{c3} \\ \hline 1 & 1 & 1 & 1 \\ 2 & 1 & null & 1 \\ 3 & null & 1 & null \end{array}$$
• Welcome to SO! Could you post some data and/or code example so that we can better help you? Oct 20, 2016 at 6:46
As per your problem, I think it might be easier to use lit. Try this-
from pyspark.sql.functions import lit
new_df = df.withColumn('column_name', lit(1))
Hope it helps!
Using lit would convert all values of the column to the given value.
To do it only for non-null values of dataframe, you would have to filter non-null values of each column and replace your value. when can help you achieve this.
from pyspark.sql.functions import when
df.withColumn('c1', when(df.c1.isNotNull(), 1))
.withColumn('c2', when(df.c2.isNotNull(), 1))
.withColumn('c3', when(df.c3.isNotNull(), 1))
This would result in:
\begin{array}{c|lcr} & \text{c1} & \text{c2} & \text{c3} \\ \hline 1 & 1 & 1 & 1 \\ 2 & 1 & null & 1 \\ 3 & null & 1 & null \end{array}
Also, if you want to replace those null values with some other value too, you can use otherwise in combination with when. Let's say you want to impute 0 there:
from pyspark.sql.functions import when
df.withColumn('c1', when(df.c1.isNotNull(), 1).otherwise(0))
.withColumn('c2', when(df.c2.isNotNull(), 1).otherwise(0))
.withColumn('c3', when(df.c3.isNotNull(), 1).otherwise(0))
This would result in:
\begin{array}{c|lcr} & \text{c1} & \text{c2} & \text{c3} \\ \hline 1 & 1 & 1 & 1 \\ 2 & 1 & 0 & 1 \\ 3 & 0 & 1 & 0 \end{array}
This would be easier if you have multiple columns:
from pyspark.sql.functions import when
cols = df.columns # list of all columns
for col in cols:
df= df.withColumn(col, when(df[col]>0,1).otherwise(0))
• How I can apply this condition only to the null values? Mar 31, 2021 at 11:47
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2023-03-28 04:59:51
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https://cs.stackexchange.com/questions/66480/modifications-to-speed-up-the-aks-primality-proving/67871
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# Modifications to speed up the AKS primality proving
It is clear that AKS primality proving is the newest one, but as the results show it is not the fastest one.
When I try the 9 digits long prime number it consume about 6 minutes to give you the result.
Are there any modifications to make it faster for large random primes in size above 200 digits?
The problems in the AKS are:
1. Euler phi function
2. Binomial congruence
• AKS's paper is not the newest one. There's been a bunch of follow-on research that shows how to make the AKS algorithm somewhat faster. What research have you done? Have you looked on Google Scholar? What have you found so far? For instance, the Wikipedia article on the AKS algorithm already provides citations to a bunch of follow-up work from 2002, 2003, and 2005 that provides some speedups. If your question is already answered in the natural place on Wikipedia, it's worth doing more research before asking next time. – D.W. Nov 26 '16 at 18:42
• thank you DW , im asking if there is any modification for binomial congruence like by replacing it with linear recursive sequence – Ramez Hindi Nov 28 '16 at 8:53
There are a number of modifications, but none of the improvements I'm aware of will get it remotely close to the speed of ECPP or APR-CL. For random 100-digit primes, the times for the fastest codes I'm aware of are:
• 246 days (estimated) AKS v6 paper
• 45 hours (estimated) AKS Bernstein + Voloch + Bornemann
• 2.6 hours AKS Bernstein theorem 4.1
• 0.042 seconds APR-CL (WraithX 1.1)
• 0.04 seconds APR-CL (Pari/GP 2.7.0)
• 0.029 seconds ECPP (ecpp-dj 1.03)
• 0.20 seconds ECPP (Primo 4.1.0 single thread)
At 500 digits, ECPP and APR-CL are in the 9-15 second range. AKS would take years, albeit we could trivially parallelize or distribute it.
I haven't seen anything that gives better improvements than Bernstein's 2003 paper while not being a complete rework such as adding randomization. The growth rate in practice comes out to $O(\log^{6.03} n)$ (there's almost certainly more than 0.03 variance in measurements, but the point is that it's very close to the ideal exponent of 6). Ricky points out that Lenstra Jr. and Pomerance are still actively publishing theoretical improvements, tightening the proven exponent.
The time taken at more than very small input sizes is almost entirely in the congruence test. There really should be very little time spent selecting the 'r' and 's' parameters, though it's possible to see it be non-trivial. For the congruence tests, Bernstein has a paper describing some of the methods that can be used to speed this up, and the implementations shown above use binary segmentation in GMP, as he suggests. You can see the implementation I used for timing at https://github.com/danaj/Math-Prime-Util-GMP/blob/master/aks.c, with various algorithms available.
There isn't any reason to use AKS in practice at any input size, given open source implementations of more efficient algorithms, not to mention Primo for use on large inputs. ECPP gives certificates as well, which is a huge advantage.
• As far as theory goes, I believe this is the best that's currently free online. (There's a more recent version from this month, which "is submitted for publication.".) – user12859 Dec 25 '16 at 6:34
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2019-08-26 03:13:02
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https://web2.0calc.com/questions/how-do-you-answer-cube-root-1-27-3-y-the-negative
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+0
# How do you answer cube root 1/27 = 3-y (the negative y is a power on 3)?
0
372
2
How do you answer cube root 1/27 = 3-y (the negative y is a power on 3)?
Sep 24, 2017
#1
+27476
+4
Like so
$$\frac{1}{27}=3^{-y}\\ \frac{1}{27}=\frac{1}{3^y}\\ 3^y=27\\y=3$$
.
Sep 24, 2017
#1
+27476
+4
Like so
$$\frac{1}{27}=3^{-y}\\ \frac{1}{27}=\frac{1}{3^y}\\ 3^y=27\\y=3$$
Alan Sep 24, 2017
#2
0
O ma god Alan your'e a jeniuz thanks so much
Guest Sep 25, 2017
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2019-02-20 20:50:16
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