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# They Got “Comprise” Backwards
These folks certainly aren’t dummies, but they’re guilty of a pretentiousism, using a fancier word than necessary, and in this case, the word is also incorrect. First the sentence:
Each of those 200-odd collisions generated enough energy to make two charm-flavored quarks, which weigh more than the lightweight quarks that comprise protons, but less than the gigantic “beauty” quarks that are LHCb’s main quarry..
https://www.quantamagazine.org/impossible-particle-discovery-adds-key-piece-to-the-strong-force-puzzle-20210927
• The sentence should have “compose,” which starts with the parts that then make a whole thing. They could even have said “…that make protons.”
• “Comprise” starts with the whole thing, then mentions its parts. The early United States comprised 13 colonies.
At least they didn’t write “comprised of.” Gack.
Here’s a picture that goes with the article: | {} |
# Prove that $\frac{x^2+1}{x^2(1-x)}>8$ for $x\in(0,1)$
Show that $\displaystyle \frac{x^2+1}{x^2(1-x)}>8$ for $x\in(0,1)$
I thought about derivative but I think it's too complicated, do you have any ideas?
### Progress
• instead show $x^2+1 > 8x^2(1-x) = 8x^2 - 8x^3$. -- k.stm
• I have $(2x+1)(4x^2-2x+1)>7x^2$ am I on good way? because I don't know what next
-
$x^2(1-x) > 0$ for $x ∈ (0..1)$, so instead show $x^2+1 > 8x^2(1-x) = 8x^2 - 8x^3$. – k.stm Mar 10 '14 at 13:07
Maybe rewriting LHS as $\frac{1}{1-\frac{x^3 + 1}{x^2 + 1}}$ might help. – Lee Yiyuan Mar 10 '14 at 13:09
Alternatively: factor the denominator as $x \left(x (1-x) \right)$. Then show that $x(1-x)$ is no greater than 1/4 on the interval. – John Hughes Mar 10 '14 at 13:09
@k.stm I have $(2x+1)(4x^2-2x+1)>7x^2$ am I on good way? because I don't know what next – Gregor Mar 10 '14 at 13:16
@Gregor I expanded this idea into a full answer. – k.stm Mar 10 '14 at 13:54
Hint: Show $p(x) = 8x^3 - 7x^2 + 1 > 0$ for all $x ∈ (0..1)$ by showing that $p$ has a positive minimum on $[0..1]$. Look at $p(0)$, at $p(1)$ and at $p'$ on $(0..1)$.
I have carried this out below.
First observe that the only zero of $p'$ in $(0..1)$ is at $x = 7/12$, since $p'(x) = x(24x-14)$ for $x ∈ (0..1)$.
Now because $p(7/12) > 0$, the only possible local extremum (be it maximum or minimum) in $(0..1)$ is positive. Also $p(0) > 0$ and $p(1) > 0$.
On the compact intervall $[0..1]$, the continuous function $p$ assumes both maximum and minimum, but as just shown, neither at $0$ nor at $1$ nor somewhere in between on $(0..1)$ a nonpositive minimum can be achieved. So the minimum must be positive and therefore $p > 0$ on all $[0..1]$.
-
We have, $\displaystyle \frac{x^2+1}{x^2(1-x)} = \displaystyle \frac{x+\frac{1}{x}}{x(1-x)}= \displaystyle \frac{2 + (\sqrt x - \frac{1}{\sqrt x})^2}{\frac{1}{4} - (x-\frac{1}{2})^2} > 8$.
(inequality is strict since $x\in(0,1)$).
- | {} |
# src / webcomponent / html-import.js
Copyright (c) 2018 Florian Klampfer https://qwtel.com/
This program is free software: you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation, either version 3 of the License, or (at your option) any later version.
This program is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.
You should have received a copy of the GNU General Public License along with this program. If not, see http://www.gnu.org/licenses/.
## Overview
This in the HTML import version of the WebComponent version of this component. See here for the standalone version. This file is included via script tag in hy-drawer.html and shouldn’t be used via ES import.
import { customElementMixin, CustomElement } from "hy-component/src/custom-element";
import { drawerMixin } from "../mixin";
const define = () => {
customElements.define(
"hy-drawer",
class extends customElementMixin(drawerMixin(CustomElement)) {
The CustomElements spec demands that we provide a list of attributes (i.e. our options).
static get observedAttributes() {
return this.getObservedAttributes();
}
}
);
};
Make sure the polyfills are ready (if they are being used).
if (
("customElements" in window && "attachShadow" in Element.prototype) || | {} |
Documentation
# optstockbybjs
Price American options using Bjerksund-Stensland 2002 option pricing model
## Description
example
Price = optstockbybjs(RateSpec,StockSpec,Settle,Maturity,OptSpec,Strike) computes American option prices with continuous dividend yield using the Bjerksund-Stensland 2002 option pricing model.
### Note
optstockbybjs computes prices of American options with continuous dividend yield using the Bjerksund-Stensland option pricing model.
## Examples
collapse all
This example shows how to compute the American option prices with continuous dividend yield using the Bjerksund-Stensland 2002 option pricing model. Consider two American stock options (a call and a put) with an exercise price of \$100. The options expire on April 1, 2008. Assume the underlying stock pays a continuous dividend yield of 4% as of January 1, 2008. The stock has a volatility of 20% per annum and the annualized continuously compounded risk-free rate is 8% per annum. Using this data, calculate the price of the American call and put, assuming the following current prices of the stock: \$90 (for the call) and \$120 (for the put).
Settle = 'Jan-1-2008';
Maturity = 'April-1-2008';
Strike = 100;
AssetPrice = [90;120];
DivYield = 0.04;
Rate = 0.08;
Sigma = 0.20;
% define the RateSpec and StockSpec
StockSpec = stockspec(Sigma, AssetPrice, {'continuous'}, DivYield);
RateSpec = intenvset('ValuationDate', Settle, 'StartDates', Settle,...
'EndDates', Maturity, 'Rates', Rate, 'Compounding', -1);
% define the option type
OptSpec = {'call'; 'put'};
Price = optstockbybjs(RateSpec, StockSpec, Settle, Maturity, OptSpec, Strike)
Price = 2×1
0.8420
0.1108
The first element of the Price vector represents the price of the call (\$0.84); the second element represents the price of the put option (\$0.11).
## Input Arguments
collapse all
Interest-rate term structure (annualized and continuously compounded), specified by the RateSpec obtained from intenvset. For information on the interest-rate specification, see intenvset.
Data Types: struct
Stock specification for the underlying asset. For information on the stock specification, see stockspec.
stockspec handles several types of underlying assets. For example, for physical commodities the price is StockSpec.Asset, the volatility is StockSpec.Sigma, and the convenience yield is StockSpec.DividendAmounts.
Data Types: struct
Settlement or trade date, specified as serial date number or date character vector using a NINST-by-1 vector.
Data Types: double | char
Maturity date for option, specified as serial date number or date character vector using a NINST-by-1 vector.
Data Types: double | char
Definition of the option as 'call' or 'put', specified as a NINST-by-1 cell array of character vectors with values 'call' or 'put'.
Data Types: char | cell
Option strike price value, specified as a nonnegative NINST-by-1 vector.
Data Types: double
## Output Arguments
collapse all
Expected option prices, returned as a NINST-by-1 vector.
Data Types: double
collapse all
### Vanilla Option
A vanilla option is a category of options that includes only the most standard components.
A vanilla option has an expiration date and straightforward strike price. American-style options and European-style options are both categorized as vanilla options.
The payoff for a vanilla option is as follows:
• For a call: $\mathrm{max}\left(St-K,0\right)$
• For a put: $\mathrm{max}\left(K-St,0\right)$
where:
St is the price of the underlying asset at time t.
K is the strike price.
## References
[1] Bjerksund, P. and G. Stensland. “Closed-Form Approximation of American Options.” Scandinavian Journal of Management. Vol. 9, 1993, Suppl., pp. S88–S99.
[2] Bjerksund, P. and G. Stensland. “Closed Form Valuation of American Options.” Discussion paper 2002 (https://www.scribd.com/doc/215619796/Closed-form-Valuation-of-American-Options-by-Bjerksund-and-Stensland#scribd) | {} |
This is more of a note for myself. Over time, I’ve learned some useful fact about ROOT that I now wish I have known when I started using ROOT. Hopefully as a ROOT user, you could also find some of these helpful.
You do:
user@somepc:~$root -l To make root shut-up and don’t show the banner. It can get fairly annoying when you are debugging some problem in your interpreted code. This is the default behavior after ROOT 6.20. Good job ROOT team. Exit after executing a script Like Python and any other scripting language, you can use ROOT as a C++ script executor. However, unlike most scripting languages, ROOT drops you a command prompt after finishing the script. You add a -q flag to make root exit automatically. user@somepc:~$ root do_stuff.cpp -q -l
root [0]
Processing do_stuff.cpp...
user@somepc:~$ Accessing fitted parameters So you just fitted as Gaussian distribution, and you need to get the parameters programmatically. # This is PyROOT code. But C++ works exactly the same way h1 = TH11("h1", "Histogram of something" , len(data), min(data), max(data)) """ some code here... bla bla bla... """ h1.fit("gaus") f1 = h1.GetFunction("gaus") print(f1) for i in range(f1.GetNumberFreeParameters()): p = f1.GetParameter(i) print("p{} = {}".format(i, p)) Using ROOT in a standalone application. Sometimes you just need a standalone app instead of script. Maybe for performance reasons. You totally can do so! Just use ROOT normally. If you need the ROOT GUI, initialize a TApplication and then call Run(). # Again this is PyROOT code. But the C++ version works the same way import ROOT import sys app = ROOT.TApplication(sys.argc, sys.argv) """ load data, do calculations, etc... f = ROOT.TFIle("data.root") t = f.Get("data") ... """ h1.Draw() c1.Draw() app.Run() # Run ROOT's event loop But maybe you can’t have a blocking event loop. That’s fine too. while some_window.isRunning(): some_window.pullEnvent() # Run your window app.processEvent() # Let ROOT update it's interfaces As a Python <-> C++ binder This feature can be accessed as an independent package – cppyy. But heck, why use cppyy when we have the full ROOT. So, assuming you have a awesome C++ library that you want to use in Python, but it doesn’t come with a Python binding. import ROOT gROOT = ROOT.gROOT gROOT.ProcessLine('#include "your_lib.h"') gSystem.Load('your_lib') # if its not header only my_cpp_class = ROOT.my_cpp_class my_class = my_cpp_class() # It also works with the STL cpp_vector = ROOT.std.vector(int)() Matplotlib colors The default ROOT colors are ugly. That’s fine, we can always switch to a better color pallet. Like the ones in matplotlib // C++ code auto c0 = TColor::GetColor("#1f77b4"); auto c1 = TColor::GetColor("#ff7f0e"); auto c2 = TColor::GetColor("#2ca0c2"); auto c3 = TColor::GetColor("#d62728"); auto c4 = TColor::GetColor("#9467bd"); auto c5 = TColor::GetColor("#8c564b"); auto c6 = TColor::GetColor("#e377c2"); auto c7 = TColor::GetColor("#7f7f7f"); auto c8 = TColor::GetColor("#bcbd22"); auto c9 = TColor::GetColor("#17becf"); Save the current canvas for use later After making a plot. You can serialize the entire plot so you can work on it later! It saves all the internal states of the plot nit just image files! auto c1 = new TCanvas("c1", "canvas"); // Plot whatever you want ... c1->Draw(); // Now save the plot! c1->SaveAs("c1.root"); // You can load the canvas later on c1->SaveAs("c1.C"); // Save as C++ source code. Run .x c1.C then you can use c1 again. Save canvas as LaTeX This feature should be well know. But if you didn’t. Yes, you can save your plots as LaTeX files directly and use them in LaTeX. I’ve use the feature very extensively for my project reports. c1->SaveAs("my_plot.tex"); Then use it in LaTeX as usual. \begin{figure} \begin{center} \include{my_plot} \label{my_plot:figure} \caption{"Some random ROOT plot"} \end{center} \end{figure} Just be aware that LaTeX use a different font. So you might need to adjust your legends. Convert CSV into ROOT archive Getting data into ROOT can be cannoning at times. Epically when we have to deal with trees and branches. If you have columnar data, newer ROOT comes with RDataFrame that can do it for you! auto rdf = ROOT::RDF::MakeCsvDataFrame("data.csv"); rdf.Snapshot("data.root", "ntuple"); Disable histogram statistics If you find the statistics box annoying and providing useless information. You can disable it with one line. gStyle->SetOptStat(0); Now instead of this You get this: Disable TGraph markers If you find the markers from TGraph annoying. gStyle->SetMarkerStyle(0); Normalizing Histograms That’s say you’re tasked to build an normalized histogram. Like… Having the data of 1000 cars, their owners age. And you need to build a histogram where the x-axis is the age of the owner and the y axis is on average how many cars they own. It can be annoying to do in plain C++. But no worries. ROOT have that covered. // Let's say you have the histogram h1 and h2 TH1D h1 = hist_car_owner(); TH1D h2 = hist_owner_age(); auto normalized = h1/h2; // Ta da! Use RDataFrame instead of PROOF if you can PROOF is one of the very first distributed computing framework (Even before Hadoop!) designed by CERN to perform huge computations. Although I’m a fan of PROOF’s simplistic and scalable design. PROOF is not user friendly even for computer scientists. So don’t use it. The new RDataFrame is a lot faster (due to not distributed just parallel) and a 60 thread machine is more likely to be bottle necked by data access anyway. Unless you have a good reason to use PROOF. Just use RDataFrame. Using ROOT in normal Jupyter The JupyROOT directory documents how to make the ROOT C++ kernel available under Jupyter. cp -r$ROOTSYS/etc/notebook/kernels/root ~/.local/share/jupyter/kernels
Then launch Jupyter normally. You’ll see ROOT available there. No more launching with root --notebook.
Final words:
ROOT is quite capable and you can do some crazy stuff with it. You just have to find it… Let me know if you have some favorite tricks.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {} |
• Status: Solved
• Priority: Medium
• Security: Public
• Views: 786
# Echo results in powershell script
Hey Experts. I have a simple powershell script that a fellow Expert on here helped me write. Now I need to find out how successful the script was but I'm not sure what the command/synatx is to output or echo the results.
Example, the script moves all of the computers in a text file from one OU to another OU. What I would like to know is which computers were moved and which ones did not. Any suggestions?
Thanks everyone!
0
samiam41
• 10
• 5
• 2
• +1
4 Solutions
Commented:
You can use Try Catch..
For example...
Try{
Write-host "Moved the computer $comp" } Catch{ Write-host "error moving computer$comp"
}
If you can post the script which you are using then we can give more details on error checking..
0
OwnerCommented:
you need to add applicable write-host or out-file items within the script itself but you need to have some form of output then you can use standard redirection i.e.
powershell -noprofile myscript.ps1 > output.txt or even pipe the output into the input of another script.
i.e.
powershell -noprofile myscript.ps1 | .\process-list.ps1 | .\send-output.ps1
0
Commented:
You need to post the script so we can see if it already generates output of some type to the screen.
0
Author Commented:
Got stuck working on a problem with a driver controller battery failure. I should be back on this script tonight or tomorrow morning. Updates to follow. Thanks Experts!
0
Author Commented:
Here is the code for the PS script.
GC C:\log-or-text.file | Get-ADUser | Move-ADObject -TargetPath 'OU=Inactive,OU=x,DC=x,DC=x,DC=x,DC=x'
0
Author Commented:
Thanks Experts for your suggestions. I am the level right below novice when it comes to Powershell scripting/coding so please provide an example using the script I provided in my previous post.
0
Commented:
To send to a log file add "set-content" like this:
GC C:\log-or-text.file | Get-ADUser | Move-ADObject -TargetPath 'OU=Inactive,OU=x,DC=x,DC=x,DC=x,DC=x' -PassThru | Set-Content c:\full\path\to\log\file.log
0
Commented:
You can try this..
Foreach ($User in GC C:\log-or-text.file) { Try { Get-ADUser$User | Move-ADObject -TargetPath 'OU=Inactive,OU=x,DC=x,DC=x,DC=x,DC=x' -ErrorAction Stop
Write-Host "Moved user $User" } Catch { Write-Host "Moved user$User - Error - $($_.Exception.Message)"
}
}
0
Author Commented:
Ok. This is getting a little frustrating. I just changed the executionpolicy setting so now the PSmove.ps1 script should run but I get this error:
The term 'Get-ADUser' is not recognized as the name of a cmdlet, function, script file, or operable program. Check the
spelling of the name, or if a path was included, verify that the path is correct and try again.
At C:\tools\PSmove.ps1:1 char:40
DC=x'-PassThru | Set-Content c:\tools\output.log
+ CategoryInfo : ObjectNotFound: (Get-ADUser:String) [], CommandNotFoundException
+ FullyQualifiedErrorId : CommandNotFoundException
GC C:\tools\staleADpc1.log | Get-ADUser | Move-ADObject -TargetPath 'OU=Inactive,OU=x,DC=x,DC=x,DC=ky,DC=x'-PassThru | Set-Content c:\tools\output.log
So what am I doing wrong here?
0
Commented:
Add following line as a first line of script..
Import-Module ActiveDirectory
0
Author Commented:
Like this?
GC C:\tools\staleADpc1.log |Import-Module ActiveDirectory | Get-ADUser | Move-ADObject -TargetPath 'OU=Inactive,OU=x,DC=x,DC=x,DC=ky,x=gov'-PassThru | Set-Content c:\tools\output.log
0
Commented:
If you are trying to echo results then...
Import-Module ActiveDirectory
Foreach ($User in GC C:\log-or-text.file) { Try { Get-ADUser$User | Move-ADObject -TargetPath 'OU=Inactive,OU=x,DC=x,DC=x,DC=x,DC=x' -ErrorAction Stop
Write-Host "Moved user $User" } Catch { Write-Host "Moved user$User - Error - $($_.Exception.Message)"
}
}
0
Author Commented:
Ahhh... I see. Testing now.
0
Author Commented:
Subsun, where is the output log file for that script?
0
Commented:
For now it just displays the details on console.. if you want to save the result to log file.. You can change the command Write-Host to Echo in script and then..
Run the script using
.\Script.ps1 >C:\log.txt
0
Author Commented:
Dang! That headache earlier really messed me up. I'm trying to move computer objects, not user. My apologies everyone for that oversight but as I listed in the OP, it needs to be related to computer accounts. I know it's a matter of changing a couple of values, but I'm not sure what they are.
Thanks everyone.
0
Author Commented:
I think I got it.
Import-Module ActiveDirectory
Foreach ($Computer in GC C:\tools\staleADPC1.log) { Try { Get-ADcomputer$Computer | Move-ADObject -TargetPath 'OU=Inactive,OU=x,DC=x,DC=x,DC=x,DC=x' -ErrorAction Stop
Echo "Moved computer $Computer" } Catch { Echo "Moved computer$Computer - Error - $($_.Exception.Message)"
}
}
0
Author Commented:
Thanks for the help. We got a little side-tracked with user and computer variables being mixed up but I am very excited that I have a script to begin moving these stale accounts. I appreciate your help and time.
0
Question has a verified solution.
Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.
Have a better answer? Share it in a comment. | {} |
With the amount of math Euler created, he must’ve pulled a proportionately large number of late nights; I think it’s safe to say he would’ve been quite the Taco Bell conniseur. But that only begs the question…what would his go-to order have been? That’s a tough one.
Anyway, I was at Taco Bell with a good friend of mine when we got to talking about the beauty of the Euler product formula and its proof. It reads:
$\zeta(s) = \sum_{n=1}^\infty\frac{1}{n^s} = \prod\frac{1}{1-p^{-s}}, \forall p \in \mathbb{P}$
An infinite sum over the naturals ends up equalling a product over the primes…it’s nerdy, but I get an adrenaline rush when thinking about that. And it’s actually quite simple. I won’t go into it much, as the Wikipedia article linked above does a great job at illustrating the proof. In short though, Euler takes the first element of the infinite series (excluding 1) and multiplies a copy of of the series by that element. He then subtracts the copy from the original, yielding a new series sieved of all multiples of that element. Rinse and repeat this algorithm to generate the primes!
Then, the computer scientists in us revealed themselves:
• You know what would be more interesting: how many times would a number that’s already been exluded have been hit?
• What do you think the runtime of that is?
In other words, mapping the natural numbers $$\to$$ lists of their factors. After arguing over the superiority of the natural vs. binary logarithm, our intuitions told us the runtime was $$O(n\log{n})$$; however, we never proved that rigorously. Leave a comment if you’ve got one!
Logarithm: the number of times you must divide a number by a base, until that number goes to 1
For example: $$\log_2{8} = 3$$ is like $$8/2/2/2 = 1$$
Eventually I found myself researching complex logarithms. For a complex number $$z = re^{i\theta}$$, there are infinitely many outputs of $$\ln{z}$$ that all differ by integer multiples of $$2\pi i$$, which gives the plot below a sort of “height”. In the complex domian, the logarithm looks much weirder than it does over the reals - it resembles a spiral staircase, like that one in Super Mario 64, except without the steps:
It was fun researching and learning about the complex logarithm; to know there is so much more hiding within such an elementary function is exciting. What else could be hiding beneath the surface? | {} |
## Fluctuations of functions of Wigner matrices.(English)Zbl 1355.60011
Summary: We show that matrix elements of functions of $$N\times N$$ Wigner matrices fluctuate on a scale of order $$N^{-1/2}$$ and we identify the limiting fluctuation. Our result holds for any function $$f$$ of the matrix that has bounded variation thus considerably relaxing the regularity requirement imposed in [A. Lytova, Zh. Mat. Fiz. Anal. Geom. 9, No. 4, 536–581 (2013; Zbl 1296.60058); S. O’Rourke et al., J. Theor. Probab. 26, No. 3, 750–780 (2013; Zbl 1280.15021)].
### MSC:
60B20 Random matrices (probabilistic aspects) 15B52 Random matrices (algebraic aspects)
### Citations:
Zbl 1296.60058; Zbl 1280.15021
Full Text: | {} |
$$\lim\limits_{n\to\infty} \frac{2 − n}{1 + n^2 + n}= 0$$
I know how to generally do these proofs but I'm stuck on this one. I started with:
Given $$\epsilon > 0$$, we want $$N^* \in \mathbb N$$ with the property that $$n \geq N^*$$ implies $$|a_n − a| = \left|\frac{2 − n}{1 + n^2 + n} - 0\right| < ε$$
However, how do I take that term out of absolute value without the values $$n=0,1$$ being omitted? That's my only problem. I can do everything from there.
• You are asking about $n=0, n=1$ in a claim about what happens when $n$ grows large. Can't we select $N>100$ so that we don't have to deal with these cases? – Mason Feb 27 at 20:46
You have indicated that you can handle all the cases except $$n=0$$ or $$n=1$$ so I will assume that you have for any $$\epsilon>0$$ some $$N^*$$ such that for all $$n>N^*>1$$ we have $$|a_n-a|<\epsilon$$. In this case I propose we select $$N^\triangle= \max(N^*,2)$$ and then we have eliminated the annoying cases that bothered us. | {} |
# Difference between revisions of "Depth-first search"
Animated example of DFS.
Depth-first search (DFS) is one of the most-basic and well-known types of algorithm in graph theory. The basic idea of DFS is deceptively simple, but it can be extended to yield asymptotically optimal solutions to many important problems in graph theory. It is a type of graph search (what it means to search a graph is explained in that article).
## Principles
DFS is distinguished from other graph search techniques in that, after visiting a vertex and adding its neighbours to the fringe, the next vertex to be visited is always the newest vertex on the fringe, so that as long as the visited vertex has unvisited neighbours, each neighbour is itself visited immediately. For example, suppose a graph consists of the vertices {1,2,3,4,5} and the edges {(1,2),(1,3),(2,4),(3,5)}. If we start by visiting 1, then we add vertices 2 and 3 to the fringe. Suppose that we visit 2 next. (2 and 3 could have been added in any order, so it doesn't really matter.) Then, 4 is placed on the fringe, which now consists of 3 and 4. Since 4 was necessarily added after 3, we visit 4 next. Finding that it has no neighbours, we visit 3, and finally 5. It is called depth-first search because it goes as far down a path (deep) as possible before visiting other vertices.
## Stack-based implementation
The basic template given at Graph search can be easily adapted for DFS:
input G
for all u ∈ V(G)
let visited[u] = false
for each u ∈ V(G)
if visited[u] = false
let S be empty
push(S,u)
while not empty(S)
v = pop(S)
if visited[v] = false
visit v
visited[v] = true
for each w ∈ V(G) such that (v,w) ∈ E(G) and visited[w] = false
push(S,w)
F has been replaced by a stack, S, because a stack satisfies exactly the necessary property for DFS: the most recently added item is the first to be removed.
## Recursive implementation
There is a much cleaner implementation of DFS using recursion. Notice that no matter which vertex we happen to be examining, we execute the same code. That suggests that we might be able to wrap this code in a recursive function, in which the argument is the vertex to be visited next. Also, after a particular vertex is visited, we know that we want to immediately visit any of its unvisited neighbours. We can do so by recursing on that vertex:
function DFS(u)
if not visited[u]
visit u
visited[u] = true
for each v ∈ V(G) such that (u,v) ∈ E(G)
DFS(v)
input G
for each u ∈ V(G)
let visited[u]=false
for each u ∈ V(G)
DFS(u)
Note that we do not need to check if a vertex has been visited before recursing on it, because it is checked immediately after recursing. The recursive implementation will be preferred and assumed whenever DFS is discussed on this wiki.
## Performance characteristics
### Time
Most DFS algorithms feature a constant time "visit" operation. Suppose, for the sake of simplicity, that the entire graph consists of one connected component. Each vertex will be visited exactly once, with each visit being a constant-time operation, so $\mathcal{O}(V)$ time is spent visiting vertices. Whenever a vertex is visited, every edge radiating from that vertex is considered, with constant time spent on each edge. It follows that every edge in the graph will be considered exactly twice -- once for each vertex upon which it is incident. Thus, the time spent considering edges is again a constant factor times the number of edges, $\mathcal{O}(E)$. This makes DFS $\mathcal{O}(E+V)$ overall, linear time, which, like BFS, is asymptotically optimal for a graph search.
### Space
#### Recursive implementation
For every additional level of recursive depth, a constant amount of memory is required. Notice that on the recursive stack, no vertex may be repeated (except the one at the top, in the current instance, just before the if not visited[u] check), because a vertex must be marked visited before any vertices can be pushed onto the stack above it, and recursion stops when a vertex is encountered which has already been marked visited. Therefore, DFS requires $\mathcal{O}(V)$ additional memory, and can be called "linear space". In practice this is often less than the memory used to store the graph in the first place, if the graph is explicit ($\mathcal{O}(E+V)$ space with an adjacency list).
#### Non-recursive implementation
The less-preferred non-recursive implementation cannot be guaranteed to take only $\mathcal{O}(V)$ additional memory, since vertices are not visited immediately after being pushed on the stack (and therefore a vertex may be on the stack several times at any given moment). However, it is also clear that there can be no more than $\mathcal{O}(E)$ vertices on the stack at any given time, because every addition of a vertex to the stack is due to the traversal of an edge (except for that of the initial vertex) and no edge can be traversed more than twice. The stack space is therefore bounded by $\mathcal{O}(E)$, and it is easy to see that a complete graph does indeed meet this upper bound. $\mathcal{O}(V)$ space is also required to store the list of visited vertices, so $\mathcal{O}(E+V)$ space is required overall.
## The DFS tree
Running DFS on a graph starting from a vertex from which all other vertices are reachable produces a depth-first spanning tree or simply DFS tree, because every vertex is visited and no vertex is visited on more than one path from the root (starting vertex). Depth-first spanning trees tend to be tall and slender, because DFS tends to always keep going along one path until it cannot go any further, and, hence, produce relatively long paths. The DFS tree has a number of useful properties that allow efficient solutions to important problems (see the Applications section).
### Undirected graphs
When we run DFS on an undirected graph, we generally consider it as a directed graph in which every edge is bidirectional. (This is akin to adding an arrowhead to each end of each edge in a diagram of the graph using circles and line segments.) Running DFS on this graph selects some of the edges ($|V|-1$ of them, to be specific) to form the spanning tree. Note that these edges point generally away from the source. They are known as tree edges. When we traverse a tree link, we make a recursive call to a previously unvisited vertex. The directed edges running in the reverse direction - from a node to its parent in the DFS tree - are known as parent edges, for obvious reasons. They correspond to a recursive call reaching the end of the function and returning to the caller. The remaining edges are all classified as either down edges or back edges. If a edge points from a node to one of its descendants in the spanning tree rooted at the source vertex, but it is not a tree edge, then it is a down edge; otherwise, it points to an ancestor, and if it is not a parent edge then it is a back edge. In the recursive implementation given, encountering either a down edge or a back edge results in an immediate return as the vertex has already been visited. Evidently, the reverse of a down edge is always a back edge and vice versa.
Every (directed) edge in the graph must be classifiable into one of the four categories: tree, parent, down, or back. To see this, suppose that the edge from node A to node B in some graph does fit into any of these categories. Then, in the spanning tree, A is neither an ancestor nor a descendant of B. This means that the lowest common ancestor of A and B in the spanning tree (which we will denote as C) is neither A nor B. Evidently, C is visited before either A or B, and A and B lie in different subtrees rooted at immediate children of C. So at some point C has been visited but none of its descendants have been visited. Now, because DFS always visits an entire subtree of the current vertex before backtracking and entering another subtree, the entire subtree containing either A or B (whichever one is visited first) must be explored before visiting the subtree containing the other. Suppose the subtree containing A is visited first. Then, when A is visited, B has not yet been visited, and thus the edge between them must be a tree edge, which is a contradiction.
Notice that given some graph, it is not possible to assign each edge a fixed classification. For example, in the complete graph of two vertices ($K_2$), two of the (undirected) edges will be tree/parent, and the other one will be down/back, but which are which depends on the starting vertex and the order in which vertices are visited. Thus, when we say that a particular edge in the graph is a tree edge (or a parent, down, or back link), it is assumed that we are referring to the classification induced by a specific invocation of DFS on that graph. The beauty of DFS-based algorithms is that any valid edge classification (that is, edge classification produced by any correct invocation of DFS) is useful.
### Directed graphs
In directed graphs, edges are also classified into four categories. Tree edges, back edges, and down edges are classified just as they are in undirected graphs. There are no longer parent edges. If an tree edge exists from A to B, this does not guarantee the existence of an edge from B to A; if this edge does exist, it is considered a back edge, since it points from a node to its ancestor in the spanning tree. However, it is not true in a directed graph that every edge must be either tree, back, or down. In the sketch proof given for the corresponding fact in an undirected graph, we use the fact that it doesn't matter if A or B is visited first, since edges are bidirectional. But in the same circumstances in a directed graph, if there is an edge from A to B but no edge from B to A, and the subtree containing B is visited first, then the edge A-B is not a tree edge, since B has already been visited at the time that A is visited. Thus, when performing DFS on a directed graph, it is possible for an edge to lead from a vertex to another vertex which is neither a descendant nor an ancestor. Such an edge is called a cross edge. Again, these labels become meaningful only after DFS has been run.
### Traversal of the spanning tree
In order to classify edges as tree, parent, down, back, or cross, we need to know which nodes are descendants of which. In effect, we would like to be able to answer the question "what is the relationship between nodes X and Y?", preferably in constant time per query (giving an optimal algorithm). The simplest way to do this, which is useful for other algorithms as well, is by preorder and postorder numbering of the nodes. In these numberings, nodes are assigned consecutive increasing integers (usually starting from 0 or 1). In preorder, a node is numbered as soon as it is visited: that is, the numbering of a node occurs pre-recursion, before any other nodes are visited. In postorder, a node is numbered after all the recursive calls arising from tree edges out of it have terminated - it occurs post-recursion. Here is code that assigns both preordering and postordering numbers at once. (Notice that we must visit vertices in the same order when assigning preordering and postordering numbers, otherwise the classification scheme for edges in a directed graph will fail. The best way to ensure that they are visited in the same order in both DFSs is by combining them into a single DFS.)
function DFS(u)
if pre[u] = ∞
pre[u] = precnt
precnt = precnt + 1
for each v ∈ V(G) such that (u,v) ∈ E(G)
DFS(v)
post[u] = postcnt
postcnt = postcnt + 1
input G
for each u ∈ V(G)
let pre[u] = ∞
let post[u] = ∞
let precnt=0
let postcnt=0
for each u ∈ V(G)
DFS(u)
Before a node has had a preorder or postorder number assigned, its number is ∞. At the termination of the algorithm, precnt=postcnt=$|V|$.
#### Classification of edges in undirected graphs
In an undirected graph, preorder numbers may be used to distinguish between down edges and back edges. (It becomes obvious when an edge is a tree edge or a parent edge, while running DFS, of course.) If node A has a higher preorder number than node B, then B is an ancestor and the edge is a back edge. Otherwise, the edge is a down edge.
#### Classification of edges in directed graphs
The situation is more complicated in directed graphs. Again it is obvious when an edge is a tree edge. Otherwise, consider the other possible classifications for an edge from A to B. If A is a descendant of B (back edge), then A has a higher preorder number but lower postorder number than B. If A is an ancestor of B (down edge), then A has a lower preorder number but higher postorder number than B. Finally, if A is neither an ancestor nor a descendant of B (cross edge), then an edge doesn't exist from B to A and B was visited before A, so A has a higher preorder number and a higher postorder number.
Note that we do not actually know the postorder number for a vertex until after we have examined all the edges leaving it. This is not a problem for the sake of classification, since we know that any number that hasn't been assigned yet is higher than any number that has been so far assigned, and that using ∞ as a placeholder therefore doesn't change the results of comparisons. You could also use -1 as a placeholder, but you would have to modify the code somewhat.
#### Topological sort
A topological ordering of a DAG can be obtained by reversing the postordering. A proof of this fact is given in the Topological sorting article.
## Applications
DFS, by virtue of being a subcategory of graph search, is able to find connected components, paths (including augmenting paths), spanning trees, and topological orderings in linear time. The special properties of DFS trees also make it uniquely suitable for finding biconnected components (or bridges and articulation points) and strongly connected components. There are three well-known algorithms for the latter, those of Tarjan, Kosaraju, and Gabow. The classic algorithm for the former appears to have no unique name in the literature.
The following example shows how to find connected components and a DFS forest (a collection of DFS trees, one for each connected component).
function DFS(u,v)
if id[v] = -1
id[v] = cnt
let pred[v] = u
for each w ∈ V(G) such that (v,w) ∈ E(G)
DFS(v,w)
input G
cnt = 0
for each u ∈ V(G)
let id[u] = -1
for each u ∈ V(G)
if id[u] = -1
DFS(u,u)
cnt = cnt + 1
After the termination of this algorithm, cnt will contain the total number of connected components, id[u] will contain an ID number indicating the connected component to which vertex u belongs (an integer in the range [0,cnt)) and pred[u] will contain the parent vertex of u in some spanning tree of the connected component to which u belongs, unless u is the first vertex visited in its component, in which case pred[u] will be u itself.
### Flood fill
The flood fill problem is often used as an example application of DFS. It avoids some of the implementational complexity associated with graph-theoretic algorithms by using an implicitly defined graph; there is no need to construct an adjacency matrix or adjacency list or any other such structure. Instead, we can just identify each vertex—a square in the grid—by its row and column indices; and it is very easy to find the adjacent squares. This is what a typical recursive implementation looks like (C++):
// (dx[i], dy[i]) represents one of the cardinal directions
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
void do_flood_fill(vector<vector<int> >& grid, int r, int c, int oldval, int newval)
{
// invalid coordinates
if (r < 0 || c < 0 || r >= grid.size() || c >= grid[0].size())
return;
// outside the blob
if (grid[r][c] != oldval)
return;
grid[r][c] = newval;
for (int i = 0; i < 4; i++)
do_flood_fill(grid, r + dx[i], c + dy[i], oldval, newval);
}
void flood_fill(vector<vector<int> >& grid, int r, int c, int newval)
{
if (grid[r][c] != newval)
do_flood_fill(grid, r, c, grid[r][c], newval);
} | {} |
# [solved] QLineEdit, double VS maskedit!
• Hi all!
I try to find something on google, but I really not find something cool to use on my app!
My problem is that I have a QLineEdit with this mask: "R\$ 009,99", put when I set on double value, for example 12,34 , to the QLineEdit I got this: "R\$ 12 ,34" and not "R\$ 012,34" or "R\$ 12,34".
What can I do to solve this problem in easy way?
Thanks!
• I'm setting the text with this code:
@
ui->leTotal->setText(QString::number(value));
@
• And I'm habing another problem now...
@
R\$ 0000,00;0
@
When I type the value "R\$ 0050,00" the field shows as needed, but when I do "ui->leTotal->text()" the return value is: R\$ 5,00".
When I type the value "R\$ 0051,00" the field shows as needed, but when I do "ui->leTotal->text()" the return value is: R\$ 51,00".
I'm doing something wrong?
I'm doing something wrong?
Thanks all!
• this masks really makes me crazy!
• Hi,
Try using 9 instead of 0.
@
9 ASCII digit required. 0-9.
0 ASCII digit permitted but not required.
@
Hope it helps
• I solve my second problem, solution THE MASK:
@
R\$ 0000,00;9
@
But the first one I need some help...
Thanks!
As I show here the change the 0 to 9 works fine!
Thank you very much!
• You're welcome. | {} |
Proof that: $$\lim_{n\to +\infty}(1-\frac{\lambda}{n})^{n}=e^{-\lambda}$$ [hr] Firstly we establish this: $$(1-\frac{\lambda}{n})^{n} = e^{n*ln(1-\frac{\lambda}{n})} = e^{\frac{ln(1-\frac{\lambda}{n})}{\frac{1}{n}}}$$ Then we can do this: $$\lim_{n\to +\infty}(1-\frac{\lambda}{n})^{n} = e^{\lim_{n\to +\infty}(1-\frac{\lambda}{n})^{n}}$$ Then we can make use of the result we established above: $$= e^{\lim_{n\to +\infty}\frac{ln(1-\frac{\lambda}{n})}{\frac{1}{n}}}$$ This is in the form f(n)/g(n), so we can use L'Hopital's rule to evaluate the limit: $$= e^{\lim_{n\to +\infty}\frac{(\frac{\lambda}{n^{2}})\frac{1}{1-\frac{\lambda}{n}}}{-\frac{1}{n^{2}}}}$$ And then we can simplify: $$= e^{\lim_{n\to +\infty}-\lambda\frac{1}{1-\frac{\lambda}{n}}}$$ Then we see that anything divided by n, as n approaches positive infinity, tends to 0, so the fraction bit of the limit evaluates to 1. Then we have just a constant left, so the limit is gone, and so it evaluates to e^-[lambda], and hence: $$\lim_{n\to +\infty}(1-\frac{\lambda}{n})^{n}=e^{-\lambda}$$ | {} |
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Instantaneous shrinking of the support of solutions to a nonlinear degenerate parabolic equationU. G. Abdullaev 323 Additive inequalities for intermediate derivatives of differentiable mappings of Banach spacesV. F. Babenko, V. A. Kofanov, S. A. Pichugov 332 A geometric method for solving a series of integral Poincaré–Steklov equationsA. B. Bogatyrev 343 A singularly perturbed boundary value problem for a second-order equation in the case of variation of stabilityV. F. Butuzov, N. N. Nefedov 354 Number of lattice points in the hyperbolic crossN. M. Dobrovol'skii, A. L. Roshchenya 363 On the absence of cycles for unimodal mappings of the squareV. A. Dobrynskii 370 Conformal type and isoperimetric dimension of Riemannian manifoldsV. A. Zorich, V. M. Kesel'man 379 On a theorem of Marcinkiewicz and ZygmundS. A. Kirillov 386 Relations and deformations of odd Hamiltonian superalgebrasYu. Yu. Kochetkov 391 On a class of $N$-dimensional trigonometric seriesO. I. Kuznetsova 402 On a class of graphs without 3-starsA. A. Makhnev 407 Undecidability of the elementary theory of groups of measure-preserving transformationsA. V. Mitin 414 Further criteria for the indecomposability of finite pseudometric spacesM. É. Mikhailov 421 Polynomial approximation of functions with given order of the $k$th generalized modulus of smoothnessM. K. Potapov, G. N. Kazimirov 425 The topological center of the semigroup of free ultrafiltersI. V. Protasov 437 On the solvability of a nonstationary problem describing the dynamics of an incompressible viscoelastic fluidG. A. Sviridyuk, T. G. Sukacheva 442 Almost-periodic solutions to systems of differential equations with fast and slow time in the degenerate caseA. Yu. Ukhalov 451 Sequences of maximal terms and central exponents of derivatives of Dirichlet seriesM. N. Sheremeta 457 Brief Communications On a model problem with second derivatives with respect to geometric variables in the boundary condition for second-order parabolic equationsB. V. Bazalii 468 Smooth antiproximinal setsV. S. Balaganskii 472 The Hardy and Bellman transforms of the spaces $H^1$ and BMOB. I. Golubov 475 Asymptotic limits of matrix elements of the canonical operator for the complex germ at a pointG. V. Koval' 479 | {} |
An element X crystallizes in f.c.c structure. 208 g of it has 4.2832$×$1024 atoms. Calculate the edge of the unit cell, if density of X is 7.2 g cm-3. | {} |
# Factor the polynomial $z^5 + 32$ in real factors
The question that I have trouble solving is the following:
Factor the polynomial $z^5 + 32$ in real factors. The answer should not use trigonometric functions. (Hint: you are allowed to use the fact that: $cos(\pi/5) = \frac{1+\sqrt{5}}{4}$ and $cos(3\pi/5) = \frac{1-\sqrt{5}}{4}$.
In the previous question you were supposed to find the complex roots to the equation $z^5 = -32$ in polar form, resulting in the roots:
$$z_1 = 2e^{i\frac{\pi}{5}}$$$$z_2 = 2e^{i\frac{3\pi}{5}}$$ $$z_3 = 2e^{i\pi} = -2$$ $$z_4 = 2e^{i\frac{7\pi}{5}}$$$$z_5 = 2e^{i\frac{9\pi}{5}}$$
My attempt at a solution:
First we know that there were only one real root to the equation $z^5 = -32$, namely $-2$.
So if we write the polynomial as the equation: $z^5 + 32 = 0$ we know that $(z+2)$ must be a factor and we have four potential factors left to find. Since we know that the polynomial $z^5 +32$ only has real coefficients, the non real roots must come in pairs.
We should then find two roots that are composed of conjugates of the roots with imaginary components to produce the other two real roots... But I am stuck here.
• The n complex nth roots of a number all have the same magnitude and are arranged regularly around the zero-centred circle in the complex plane with that magnitude. – Joffan Jan 9 '15 at 16:24
• For a method that only involves quadratic equations and algebraic substitutions (no trigonometry, no complex exponentials), see my answer to The roots of $t^5+1$. (Added moments later) Well, actually I find the complex roots this way. But you might be able to use the ideas (and certainly the roots, if all else fails) to get the complete factorization over the reals. And indeed, it appears egreg has done this. – Dave L. Renfro Jan 9 '15 at 16:53
$z^2 + 32 = (z+2)(x - \alpha)(z - \overline {\alpha})(z - \beta)(z- \overline {\beta})$ where $\overline {\alpha}, \overline {\beta}$ are complex conjugate of $\alpha, \beta$ respectively. Also $(z - \alpha)(z - \overline {\alpha})$ and $(z - \beta)(z- \overline {\beta})$ are real polynomials (why?). Now choose $\alpha = z_1, \beta = z_2.$ Then $\overline {\alpha} = z_5, \overline {\beta} = z_4.$
• Thank you for your answer! Shouldn't the equality be: $z^5 + 32 = (z+2)(z + \alpha)(z + \overline {\alpha})(z + \beta)(z + \overline {\beta})$? Now lets start with $(z + \alpha)(z + \overline {\alpha})$. This could be written as: $(z + 2e^{i\frac{\pi}{5}})(z + 2e^{i\frac{-\pi}{5}})$. If I try to develop this I get: $z^2 + 2z(e^{i\frac{\pi}{5}} + e^{i\frac{-\pi}{5}}) + 4(e^{i\frac{\pi}{5}})(e^{i\frac{-\pi}{5}})$ How do I continue from there? – Lukas Arvidsson Jan 9 '15 at 18:34
• the complex roots occurs in conjugate pair. so if $\alpha$ is a complex root then $\overline{\alpha}$ is also a root. now $\alpha$ is root means $z - \alpha$ is factor. similarly $z - \overline{\alpha}$ is also a root. (there was a mistake. the first factor should be $z+2,$ not $z-2$ as I wrote previously. now fixed.) – Krish Jan 9 '15 at 18:42
• Thank you for your comment, if you would have time to do the actual calculations I would be very happy (I am not sure that I understand this correctly yet...) – Lukas Arvidsson Jan 9 '15 at 18:51
• $(z - \alpha)(z - \overline {\alpha}) = z^2 -z(\alpha + \overline {\alpha}) + \alpha \overline{\alpha}.$ Now $\alpha + \overline {\alpha} =2 \cos(\frac{\pi}{5}) = 2 \cdot \frac{1 + \sqrt 5}{4}, \alpha \overline {\alpha} = 4.$ similarly for $(z - \beta)(z - \overline {\beta}).$ Also the expression you wrote: $z^5 + 32 = (z+2)(z + \alpha)(z + \overline {\alpha})(z + \beta)(z + \overline {\beta})$ is not correct. If you simplify right hand side, you will get something else. there will be a $z^3$ term. – Krish Jan 9 '15 at 19:02
• BTW, how do you see that the development will be $z^5 + 32 = (z+2)(x - \alpha)(z - \overline {\alpha})(z - \beta)(z- \overline {\beta})$ and not $z^5 + 32 = (z+2)(z + \alpha)(z + \overline {\alpha})(z + \beta)(z + \overline {\beta})$? – Lukas Arvidsson Jan 9 '15 at 19:59
Set $t=2z$, for the moment. Then \begin{align} z^5+32=32(t^5+1) &=32(t+1)(t^4-t^3+t^2-t+1)\\ &=32(t+1)t^2\left(t^2+\frac{1}{t^2}-\left(t+\frac{1}{t}\right)+1\right)\\ &=32(t+1)t^2\left(\left(t+\frac{1}{t}\right)^{\!2}-\left(t+\frac{1}{t}\right)-1\right) \end{align} Consider $u^2-u-1=(u-\alpha)(u-\beta)$, where $$\alpha=\frac{1+\sqrt{5}}{2},\qquad\beta=\frac{1-\sqrt{5}}{2}$$ and so $$z^5+32=32(t+1)t^2\left(t+\frac{1}{t}-\alpha\right)\left(t+\frac{1}{t}-\beta\right)$$ which is to say $$z^5+32=32(t+1)(t^2-\alpha t+1)(t^2-\beta t+1)= (z+2)(z^2-2\alpha z+4)(z^2-2\beta z+4)$$
You found the roots so $z^5+32=(z+2)(z-z_1)(z-z_2)(z-z_4)(z-z_5)$.
Notice that $cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}$ and that $z_1z_5\in\mathbb{R}$ and $z_2z_4\in\mathbb{R}$
• Thank you for your answer! Just a quick question: How do you see that the development will be $z^5 + 32 = (z+2)(x - \alpha)(z - \overline {\alpha})(z - \beta)(z- \overline {\beta})$ and not $z^5 + 32 = (z+2)(z + \alpha)(z + \overline {\alpha})(z + \beta)(z + \overline {\beta})$? – Lukas Arvidsson Jan 9 '15 at 20:00
• You're welcome. If $\alpha$ is a root of a polynomial $P$ then $(z-\alpha)|P$. $-2$ is a root that's why we have $(z-(-2))=(z+2)$. – Scientifica Jan 10 '15 at 10:19
HINT:
$$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$$
$$a^4-a^3b+a^2b^2-ab^3+b^4=(a^2+b^2)^2-ab(a^2+b^2)-a^2b^2$$
If $ab\ne0,$
$$(a^2+b^2)^2-ab(a^2+b^2)-a^2b^2=a^2b^2\left[\left(\frac{a^2+b^2}{ab}\right)^2-\frac{a^2+b^2}{ab}-1\right]$$ So, we can express $a^2+b^2$ in terms of $ab$
• Thank you for your answer! – Lukas Arvidsson Jan 9 '15 at 20:01 | {} |
# Doubt from Permutations and Combination
Please suggest different ways to solve this question
Find the total number of possibilities of colouring:
2^9 = 512.
Subtract 2x2 red square possibilities - 4.
That gives 508.
Not sure how u get 417
1 Like
This uses Principle of Inclusion-Exclusion. You will see that this will be similar to n(A \bigcup B \bigcup C \bigcup D)
There are four 2 \times 2 squares in the figure, name them A,B,C,D in clockwise order.
There are a total of 2^9=512 colourings possible. From these
Subtract: Case I: One of the squares is painted fully red. The remaining 5 squares can be filled in 2^5 ways. Thus we have 4\times 2^5 = 128 such colourings
Add: The above include Case II: Where two sets of squares are painted red. Here we have to distinguish adjacent pairs (A,B), (C,D), (A,D), (B,C) which gives us 4 \times 2^3 = 32 colourings, and diagonal pairs A,C and B,D which gives 2 \times 2^2=8 colurings.
Hence from this case we have a total of 40 colourings
Subtract: Case III: Three sets of squares are filled in red. This gives 4 \times 2 =8 colourings.
Add: Case IV: All four sets are filled in red: 1 colouring
Hence number of permissible colurings = 512-128+40-8+1=417
5 Likes
Sir, is their is any method for doing this question.
@Hari_Shankar @Praveen_2018
Can't we count no of red colour 2*2 square and then substract it.
There will be too much overlapping of cases. | {} |
# Math Help - [SOLVED] F(x,y) maximum problem
1. ## [SOLVED] F(x,y) maximum problem
When I try and solve for critical points i end up with x=y. Where the domain for the function is inclusivly between $0$ and $2\pi$.
If I have infinitely many critical points is there another method to solving this?
$f(x,y)=sin(x)+sin(y)+sin(x+y)$
$f_x(x,y)=cos(x) + cos(x+y)$
$f_y(x,y)=cos(y) + cos(x+y)$
Now $f_x=0=f_y$
So when i set them equal-->
$cos(x) + cos(x+y)=cos(y) + cos(x+y)$
$cos(x)=cos(y)$
$x=y$
Now I am unsure what to do next, normally i get some critical points, but instead i got infinately many critical points...
EDIT: Here is a link to the graph I think it'll work: http://www.wolframalpha.com/input/?i...sin%28x%2By%29
if not paste this in: graph z=sin(x)+sin(y)+sin(x+y) to get an idea of the graph.
EDIT:Going to bed now, will check again around 12hrs from now...for what it is worth
2. since cos(x) + cos(x+y) = 0
and since x = y
cos(x) + cos(2x) = 0
2cos^2(x) +cos(x) -1 = 0
(2cos(x) -1)(cos(x)+1) = 0
cos(x) = 1/2 cos(x) = -1
I'm assuming the domain is the square [0,2pi] x[0,2pi] ?
In which case x = pi/3, 5pi/3, pi
Again since x = y
(pi/3,pi/3) , (5pi/3,5pi/3) (pi,pi) are the critical points
3. Originally Posted by snaes
When I try and solve for critical points i end up with x=y. Where the domain for the function is inclusivly between $0$ and $2\pi$.
If I have infinitely many critical points is there another method to solving this?
$f(x,y)=sin(x)+sin(y)+sin(x+y)$
$f_x(x,y)=cos(x) + cos(x+y)$
$f_y(x,y)=cos(y) + cos(x+y)$
Now $f_x=0=f_y$
So when i set them equal-->
$cos(x) + cos(x+y)=cos(y) + cos(x+y)$
$cos(x)=cos(y)$
$x=y$
Now I am unsure what to do next, normally i get some critical points, but instead i got infinately many critical points...
From this is follows that if $f_x= f_y$ then cos(x)= cos(y). It does NOT follow that x= y. You might also have $y= 2\pi- x$. So you actually have twice as many solutions as you thought!
Those are the two diagonals of the square $[0,2\pi]\times[0,2\pi]$.
And, it does not follow that the two derivatives are equal to 0 for all y= x, only that they are equal two each other. Caculus26 showed that, using y= x in one of the equations, $f_x= 0$ or $f_y= 0$ you get $(\pi/3,\pi/3)$ , $(5\pi/3,5\pi/3)$, and $(\pi,\pi)$.
Do the same with $y= 2\pi- x$. Since $x+y= x+ (2\pi- x)= 2\pi$ you have $cos(x)+ cos(x+y)= cos(x)+ cos(2\pi)= cos(x)+ 1= 0$ and that is true only for [tex]x= \pi[/itex]. Then $y= 2\pi- \pi= \pi$ so that just gives the point $(\pi, \pi)$ that Caculus26 got before- it is the center point of the square and lies on both diagonals.
4. Ahh, I see. Thank you! | {} |
01
Jan
### matplotlib latex mac
Some styles failed to load. For example, if we want to add the Helvetica font, we need to check if we have the font in .ttf format installed on our system otherwise we need to download it and install it. Matplotlib-Installationsfehler auf Mac OSX 10.8 Mountain Lion (8) . The executables for these external dependencies must all be located on your PATH. have their own accompanying math fonts, while the other Adobe serif That being said, one can use the usetex feature to render the LaTeX text directly with very good results (if you are careful about choosing fonts). Accounting; CRM; Business Intelligence Oh no! See below for details. Next, we need to update the font cache from the command line with the following command: sudo fc-cache-fv. All of the other fonts are Adobe fonts. Run your script with verbose mode enabled: python example.py --verbose-helpful (or --verbose-debug-annoying) and inspect the output. This is done by going to the control panel, selecting the "system" icon, selecting the "advanced" tab, and clicking the "environment variables" button (and people think Linux is complicated. … ", http://www.foolabs.com/xpdf/download.html, 2008-12-04 (last modified), 2006-01-22 (created). Please try reloading this page Help Create Join Login. For example, if we want to add the Helvetica font, we need to check if we have the font in .ttf format installed on our system otherwise we need to download it and install it. Note that you do not need to have TeX installed, since Matplotlib ships its own TeX expression parser, layout engine, and fonts. Times and Palatino each Filippo. c:/windows/fonts ... Typesetting Labels with LaTeX. Filter by license to discover only free or Open Source alternatives. Sage Os X app and MatPlotLib / LaTex connection. We really appreciate your help! The TeX Users Group (TUG) has a list of notable distributionsthat are entirely, or least primarily, free software. You are probably running Mac OS, and there is some funny business with dvipng on the mac. Recent versions matplotlib break the psfrag functionality (see for example this discussion. Das Folgende funktionierte für die Installation von thegreenroom nach der Installation von Python gemäß den Anweisungen aus dem thegreenroom.Diese Anweisungen funktionierten für mich nicht, nachdem ich Python installiert hatte. With that code, I’d get the following error: I don’t see any kind of full report after the last line. Revision 5e2833af. Using latex means we may be making (many) subprocess calls to latex which is my knee-jerk guess as to the source of the slow down. Keep in mind that Matplotlib expects a font in True Type format (.ttf). using the LaTeX typesetting system. I tried to do everything that I could, but I still unable to run Matplotlib and have a Latex font. If you’re new to TeX and LaTeX or just want an easy installation, geta full TeX distribution. Make sure LaTeX, dvipng and ghostscript are each working and on your PATH. You might save the plot with. The xpdf alternative produces postscript with text that can be edited in Adobe Illustrator, or searched for once converted to pdf. Some styles failed to load. with your TeX installation), and ghostscript (AFPL, mac Thank you very much for your help! Oh no! but GPL ghostscript-8.60 or later is recommended). Accounting; CRM; Business Intelligence Running the script will output the histogram file in PGF format which can be imported with LaTeX. I have installed LaTeX on my Mac using MacTex-2018 distribution pkg. Tagged activepython avassetexportsession bpython catextlayer cgcontext cgcontextref contextmenu core-text cpython epd-python gettext google-api-python-client ipython ipython-magic ipython-notebook ipython-parallel ironpython latex MacOS macos-big-sur macos-carbon macos-catalina macos-high-sierra macos-mojave macos-sierra matplotlib matplotlib-basemap mutex nstextattachment The command pdflatex is needed. See Python ÉkJ , Eh run fixed filters demo def load image (self, imfile) : self. This step produces fonts which may be unacceptable to some users. Here is an example matplotlibrc file: The first valid font in each family is the one that will be loaded. Next, we need to update the font cache from the command line with the following command: sudo fc-cache-fv. Die LaTeX-Unterstützung von Matplotlib erfordert eine funktionierende LaTeX-Installation, dvipng (möglicherweise in Ihrer LaTeX-Installation enthalten) und Ghostscript (GPL Ghostscript 8.60 oder höher wird empfohlen). You can use TeX to render all of your matplotlib text if the rc parameter text.usetex is set. To use tex and select e.g. Text handling with matplotlib's LaTeX support is slower than standard text handling, but is Python ÉkJ , Eh run fixed filters demo def load image (self, imfile) : self. The following backends work out of the box: Agg, ps, pdf, svg and TkAgg. You need to do this before you import matplotlib.pylab. This option is available with the following backends: Agg; PS; PDF; The LaTeX option is activated by setting text.usetex: True in your rc settings. Open Source Software. Please try reloading this page Help Create Join Login. Alternatives to Matplotlib for Windows, Mac, Linux, Web, iPhone and more. more flexible, and produces publication-quality plots. 5 3D 3D 3D matplotlib 6 CAPTCHA 8 matplotlib LaTeX Python2.7.3 Ubuntu12.03 Python (2.7.3)0 IPython, Python Linux Windows Mac OS Python Python Enthought Python Python (X, Y) Python a, Python pip pip Python fi(J easy_install Python . Many improvements have been made beginning with im = mplimage. Anyway, I think this is a path problem. Some styles failed to load. Using MiKTeX with Computer Modern fonts, if you get odd -Agg and PNG results, go to MiKTeX/Options and update your format files. Port Information was last updated at: 2020-11-15 20:53 (UTC) 4ef6705b Latest build fetched has 'start time': 2020-11-15 20:29 (UTC) Latest stats submission was received at: List updated: 6/17/2020 10:34:00 AM Sie müssen dann den Code \usepackage[applemac]{inputenc} vor \begin{document} kopieren. Auf Ihrem Mac dürfte die Kodierung "AppleMac" verwendet werden. XeTeX wurde ursprünglich für Mac OS X geschrieben und später auf Linux und Windows portiert, wenn auch mit geringerem Funktionsumfang. We really appreciate your help! Open Source Software. Section author: AndrewStraw, Unknown[117], Unknown[118], Unknown[119]. Running the test analysis I got these errors: (AllMapsEnv) [tk19812@bc4login1 ALLMAPS-testdata]$./run.sh 22:33:31 [allmaps] A total of 120 markers written to JM-2.bed. Please try reloading this page Help Create Join Login. plt.savefig('histogram.png', dpi=400) and it will write a PNG file to the disk, which is ~141 KB in size (in the example). Text rendering With LaTeX¶ Matplotlib has the option to use LaTeX to manage all text layout. Thank You ! The executables for This allows latex to be used for text layout with the pdf and svg backends, as well as the *Agg and PS backends. Matplotlib can use LaTeX to handle the text layout in your figures. Do you see the slow down with other files types (png, pdf) or just svg? doctoral dissertations. Some styles failed to load. In Windows, it’s . Das Terminal meines Mac. c:/windows/fonts ... Typesetting Labels with LaTeX. Matplotlib - Environment Setup - Matplotlib and its dependency packages are available in the form of wheel packages on the standard Python package repositories and can be installed on Windows, Analyzing the code I have seen that the problem is in the main when it plots the results. Matplotlib is a Sponsored Project of NumFOCUS, a 501(c)(3) nonprofit charity in the United States. The executables for these external dependencies must all be located on your PATH. We should be caching the results in ~/.cache/matplotlib/tex.cache so repeated runs should get faster. If you know that system, this option can be especially nice for graphics that you intend to embed in a bigger LaTeX document. figures as in the main document. these external dependencies must be located on your PATH. It's cross-platform and portable: Matplotlib can run on Linux, Windows, Mac OS X, and Sun Solaris (and Python can run on almost every architecture available). Note . The results are Nach oben. This option is available with the following backends: Agg; PS; PDF; The LaTeX option is activated by setting text.usetex: True in your rc settings. (Troubleshooting tips for Mac OS X can be found here.) program. However, getting matplotlib installed on Mac OS X 10.7 can be a bit tricky, especially if you are using Homebrew as your “package manager.” First off, Homebrew does not have packages for matplotlib, as well as some of its dependencies. Annotating text¶ The uses of the basic text function above place text at an arbitrary position on the Axes. Matplotlib and its dependencies are available as wheel packages for macOS, Windows and Linux distributions: python-m pip install-U pip python-m pip install-U matplotlib. Running the script triggers an exception (see code block and message below). Select the PATH variable, and add the appropriate directories. Do you see the slow down with other files types (png, pdf) or just svg? XeTeX [ziːtɛχ] (im deutschen Sprachraum auch [ziːtɛç]), manchmal auch ΧƎΤΕΧ geschrieben, ist eine auf eTeX basierende Alternative zu pdfTeX. A better workaround, which requires [http://www.foolabs.com/xpdf/download.html xpdf] or [http://poppler.freedesktop.org/ poppler] (the new backend to xpdf) can be activated by changing the rc ps.usedistiller setting to xpdf. matplotlib's LaTeX support is slower than standard text handling, but is Some styles failed to load. These were reported by @r-owen in the process of making osx binaries for 1.3.0rc2. Q&A Forum for Sage. Text handling with When the document is compiled the code will be automatically run and its … version. This is what I use on windows. This option is available with the following backends: Agg; PS; PDF; The LaTeX option is activated by setting text.usetex: True in your rc settings. Oh no! BlackJack. NumFOCUS provides Matplotlib with fiscal, legal, and administrative support to help ensure the health and sustainability of the project. Oh no! With default parameters the script fails to write the pgf file to disk. If you need more information just tell me. Visit numfocus.org for more information. On extremely old versions of Linux and Python 2.7, you may need to install the master version of subprocess32. psnfss2e.pdf There a several different configurations for both Python and LaTeX. Oh no! If Matplotlib’s LaTeX support requires a working LaTeX installation, dvipng (which may be included with your LaTeX installation), and Ghostscript (GPL Ghostscript 8.60 or later is recommended). It's integrated with LaTeX markup: This is really useful when writing scientific papers. Some styles failed to load. It may happen that you need a more recent LaTeX than the one that your favourite TeX distribution carries, e.g., in order to get a particular bug fix. Funktioniert die jeweilige Kodierung nicht oder möchten Sie Ihr LaTeX-Dokument auf unterschiedlichen Systemen verwenden, sollten Sie es mit UTF-8 probieren. Helvetica as the default font, without Q&A Forum for Sage. What can I do? We should be caching the results in ~/.cache/matplotlib/tex.cache so repeated runs should get faster. Matplotlib's LaTeX support requires a working LaTeX installation, dvipng (which may be included with your LaTeX installation), and Ghostscript (GPL Ghostscript 9.0 or later is required). edititing matplotlibrc use: N.B. Learning by Sharing Swift Programing and more …. On Windows, the PATH environment variable may need to be modified to find the latex, dvipng and ghostscript executables. using the LaTeX typesetting system. Rendering math equation using TeX¶. LaTeX using psfrag¶ Note: This section is obsolete. GPL, or @pwuertz: do you have any ideas? On Mac brew cask install mactexdoes the job. I'm using Python2.7 on a Mac with Yosemite, and python and all packages were installed using Macport and my Matplotlib is py27-matplotlib @1.5.2_0+cairo (active) Matplotlib has the option to use LaTeX to manage all text layout. Suppose you have a Python script that produces a nice plot using matplotlib. In order to produce encapsulated postscript files that can be embedded in a new LaTeX document, the default behavior of matplotlib is to distill the output, which removes some postscript operators used by LaTeX that are illegal in an eps file. Open Source Software. Oh no! Sie müssen dann den Code \usepackage[applemac]{inputenc} vor \begin{document} kopieren. Open Source Software. You can ask Matplotlib to render all your axis labels etc. im = mplimage. Accounting; CRM; Business Intelligence X geschrieben und später auf Linux und Windows portiert, wenn auch mit geringerem Funktionsumfang integrated. ( created ) which may be the only external dependency for many different languages ) matplotlib.pylab 8.! Higher value ( perhaps 6000 ) in your figures as in the United.! Is more flexible, and 8 others cjgohlke, dsdale, efiring, heeres, and administrative support to ensure. ( without using Gestalt ) geta full TeX distribution funny Business with dvipng on the.. Case you will need to install the jcvi package converted to pdf to discover free!, http: matplotlib latex mac, 2008-12-04 ( last modified ), 2006-01-22 created. Svg and TkAgg Adobe Illustrator, or searched matplotlib latex mac once converted to pdf and ghostscript are each working on... ), 2006-01-22 ( created ): true in your figures PATH problem these dependencies... Applemac ] { inputenc } vor \begin { document } kopieren the standard example, tex_demo.py: Note when. The goodies that come bundled with other files types ( png, pdf ) or just svg there a different! Write the pgf file to disk xetex wurde ursprünglich für Mac OS X geschrieben und später Linux! The histogram file in pgf format which can be especially nice for graphics that ’! If the fonts are used by default pip install matplotlib '' PyCharm erkennt matplotlib nicht. Get odd -Agg and png results, go to MiKTeX/Options and update your format files World code to try TeX... At an arbitrary position on the axes an earlier version can not guarantee that TeX distributions even. And have a Python script that produces a nice plot using matplotlib variable need! Ursprünglich für Mac OS X app and matplotlib / LaTeX connection mode enabled: Python example.py -- (... To TeX and LaTeX or just svg next, we need to update the font cache from command! Come bundled with other LaTeX distributions matplotlib and have a Python script produces! Progress has been made beginning with matplotlib-0.87, please update matplotlib matplotlib latex mac you know that system, this can... AppleMac '' verwendet werden can use LaTeX to manage all text layout in your figures ( using! There is some funny Business with dvipng on the axes and there some..., ps, pdf, svg and TkAgg I use Anaconda Python [ 1 ] and Eclipse 2! How do I determine the OS version at runtime in OS X can be activated by setting text.usetex: in! Installation, geta full TeX distribution just want an easy installation, geta full TeX distribution and message below.. A higher value ( perhaps 6000 ) in your rc matplotlib latex mac Mountain Lion 8... Writing scientific papers bundled with other files types ( png, pdf ) or just svg Note that when support! Files directly for text layout in your figures be edited in Adobe Illustrator, or searched for once converted pdf!, wenn auch mit geringerem Funktionsumfang care to use the same fonts in your figures ( self, )! Contain the most recent version of subprocess32 Anaconda Python [ 1 ] and Eclipse [ 2 (... Charity in the future, a LaTeX font type1cm package to embed in a bigger LaTeX document text in. You can mix text and math modes to update the font cache from the command with. The basic text function above place text at an arbitrary position on the Mac the results in ~/.cache/matplotlib/tex.cache so runs. On my Mac using MacTex-2018 distribution pkg 3 ) nonprofit charity in the main when it the. Vor \begin { document } kopieren funktioniert die jeweilige Kodierung nicht oder möchten Ihr! Ubuntu and Gentoo, the base texlive install does not ship with the package... And png results, go to MiKTeX/Options and update your format files scientific papers the fonts used. Use Anaconda Python [ 1 ] and Eclipse [ 2 ] ( Eclipse works many... 119 ] been made so matplotlib uses the dvi files directly for text layout in figures. ) nonprofit charity in the United States all text layout on top of your matplotlib if! File in pgf format which can be found here. [ 119 ] or Open Source alternatives Sie! Setting text.usetex: true in your rc settings caching the results are striking, when! Option to use LaTeX to manage all text layout executables for these external dependencies must be located on your.. By default is obsolete specified, the PATH from which Python code is loaded Mac OS and! Discover only free or Open Source alternatives Global Python PATH of the box: Agg,,. And there is some funny Business with dvipng on the axes to run matplotlib and have a Python script produces! Und Windows portiert, wenn auch mit geringerem Funktionsumfang it on top of your distribution, svg TkAgg... Parameters the script fails to write the pgf file to disk pip matplotlib! ( perhaps 6000 ) in your figures as in the future, a 501 c! Each family is the standard example, tex_demo.py: Note that when TeX/LaTeX is. Can not guarantee that TeX distributions, even recent ones, contain the most recent version LaTeX... Image ( self, imfile ) tidy up tick labels size all axes = (. Most recent version of subprocess32 is set created ) same fonts in your rc settings standard,. Which can be imported with LaTeX markup: this section is obsolete if the fonts used! Or -- verbose-debug-annoying ) and inspect the output tex_demo.py: Note that when TeX/LaTeX is... Produces publication-quality plots die Kodierung AppleMac '' verwendet werden select the variable! Adobe Illustrator, or searched for once converted to pdf your axis labels.! Is still somewhat experimental ) can be activated by setting text.usetex: true your... In each family is the one that will be automatically run and its … Oh no for! And inspect the output, showing the final plots default parameters the script fails write. Figures as in the main document as in the future, a 501 ( c (. You get odd -Agg and png results, go to MiKTeX/Options and update your format.! 501 ( c ) ( 3 ) nonprofit charity in the future, a LaTeX font on Ubuntu and,! Your matplotlib text if the rc parameter text.usetex is set there a several different configurations for both Python LaTeX! Standard example, tex_demo.py: Note that when TeX/LaTeX support is slower than standard text,! Some progress has been made so matplotlib uses the dvi files directly text.: true in your rc settings script with verbose mode enabled: Python example.py -- verbose-helpful or. With Computer Modern fonts, if you know why you would want to export from matplotlib to render of... Mac using MacTex-2018 distribution pkg file and I have the following command: sudo fc-cache-fv is funny... Scientific papers tidy up tick labels size all axes = plt.gcf ( ) files directly for layout... To mention that you ’ re new to TeX and LaTeX or just svg matplotlibrc... Axes = plt.gcf ( ) and update your format files an earlier version: Python example.py verbose-helpful... { inputenc } ein to pdf the TeX Users Group ( TUG ) a! Use LaTeX to handle the text layout determine the OS version at runtime in X. Export from matplotlib to render all your axis labels etc is some Business. Be imported with LaTeX try out TeX rendering with LaTeX¶ matplotlib has the option use. Matplotlib weiterhin nicht enabled, you may need to install the master version of subprocess32: that! I could, but I still get the same fonts in your figures code with success, showing final. Default font, without edititing matplotlibrc use: N.B you by: cjgohlke, dsdale,,. 8 others standard text handling, but is more flexible, and administrative support to Help the... Have a LaTeX installation may be the only external dependency the only external dependency pdf, svg and.. Adding /Library/TeX/Root/bin/universal-darwin to the Global Python PATH is not$ PATH, instead it ’ s sys.path, PATH! Dvipng on the axes select the PATH environment variable may need to do everything that I could but... Imported at the beginning of the code will be automatically run and its … Oh no you want.: self all your axis labels etc position on the Mac rendering with LaTeX¶ matplotlib the! The SourceForge Team Keep in mind that matplotlib expects a font in true Type format (.ttf ) Project. Following command: sudo fc-cache-fv, tex_demo.py: Note that when TeX/LaTeX is... All of your matplotlib text if the rc parameter text.usetex is set LaTeX markup: this is useful... Demo def load image ( self, imfile ) tidy up tick size... Tried to do this before you import matplotlib.pylab your format files TeX distribution PATH, instead ’! Mac, Linux, Web, iPhone and more automatically run and its … Oh!... matplotlib.pylab ps, pdf, svg and TkAgg files directly for text layout useful when writing scientific.! Or Open Source alternatives 501 ( c ) ( 3 ) nonprofit in. To the Global Python PATH is not \$ PATH, instead it ’ s,! Miktex with Computer Modern fonts are not specified, the Computer Modern fonts are used by default Sie... Caching the results in ~/.cache/matplotlib/tex.cache so repeated runs should get faster re new to TeX and or! Matplotlib on my Mac using MacTex-2018 distribution pkg mit dem code \usepackage [ utf8 ] inputenc! Text at an arbitrary position on the Mac run the code improvements have made! Before you import matplotlib.pylab ` Systemen verwenden, sollten Sie es mit probieren... | {} |
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The price of each item at the Dorf Gadget Store has been reduced by 20% from its original price. An MP3 player has a sale price of $112. What would the same MP3 player sell for if it was on sale for 30% off of its original price? Jun 18, 2020 ### 1+0 Answers #1 +783 +1 Everything is 20$ off its original price, so each item is 80% of its original price. This $112 MP3 player is 80$ of its original cost.
$$\frac{80}{100}x=112\\ 80x=11200\\ x=140$$
Its original cost is $140. 30% off would leave 70% of$140$, which is $$\frac{7}{10}\cdot140\\ =98$$ Therefore the MP3 player would be$98.
Jun 18, 2020 | {} |
# A question about a question and answer. [closed]
That is wrong or right about this question and answer?
Question: Is there a cardinality which is greater than the continuum?
Answer: Yes and No. If there is a Universe where a given cardinal kappa is greater than the size of the continuum, then there is a Generic-Extension of this Universe where the size of the continuum is greater than kappa.
Edit: This question is basically about the size of the continuum, which has been discussed several times on mathoverflow. It is also about the philosophical position of whether there is the reality of the multiverse.
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## closed as not a real question by Andrés Caicedo, algori, Todd Trimble♦, Qiaochu Yuan, Mark SapirNov 3 '11 at 4:15
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.
If there is no cardinality greater than the continuum, then what would the cardinality of the power set of the reals be? – M Turgeon Nov 3 '11 at 0:51
I think she's pointing out (asking?) that for any particular cardinal in your current model of ZFC, you can pass to an extension where the extension's continuum is bigger than this set (in the extension). Is this right? – Richard Rast Nov 3 '11 at 0:57
The point seems to be that modal operators like "it is provable in ZFC that" or "it is true in all forcing extensions of the universe that" don't commute with "there is". In particular, it is true that, "in all forcing extensions of the universe, there is a cardinal greater than the continuum", but it is false that "there is a cardinal that is, in all forcing extensions, greater than the continuum." My answer refers to the former (in the stronger version with "it is provable in ZFC that"), while Richard Rast's comment refers to the latter. – Andreas Blass Nov 3 '11 at 13:41
## 1 Answer
In ordinary set theory, "Yes" is right and "No" is wrong. Even after you generically extend the universe to make the cardinal of the continuum bigger than a given $\kappa$, there are plenty of other cardinals that are even bigger than your new continuum. As M Turgeon says, to avoid cardinals larger than the continuum, you'd have to revoke the axiom of power set.
Before he became a science fiction writer, Rudy Rucker did some work on the set theory that you get by revoking power set and adding Martin's axiom for all cardinals. That makes the continuum a proper class.
There has also been a little work on a set theory that allows only countably infinite sets, like what you'd get by generically collapsing all cardinals. But all these ways of getting a "No" answer to your question are far from the usual set theory and are probably best understood as being about some notion other than "set".
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Thank you for pointing out the problem with the question. And, the problem with the solution because it doesn't really address the question. Also, I see your out of the box way of answering "No". The question and answer are interesting to me as a pair. – Erin Carmody Nov 3 '11 at 1:48
a thorough answer on a question about a question and answer – Pietro Majer Nov 3 '11 at 10:16 | {} |
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# What is the expected product of the following reaction?CH3 1) SOClz Hyc OH 2) propanol pyridineSelect one: CH; OH HjcCH;CH3 OH 6OH HSCCH; HscOHCH3 cCHa...
## Question
###### What is the expected product of the following reaction?CH3 1) SOClz Hyc OH 2) propanol pyridineSelect one: CH; OH HjcCH;CH3 OH 6OH HSCCH; HscOHCH3 cCHa
What is the expected product of the following reaction? CH3 1) SOClz Hyc OH 2) propanol pyridine Select one: CH; OH Hjc CH; CH3 OH 6OH HSC CH; Hsc OH CH3 c CHa
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##### Given that y1solution of the DE: y" + 6y' + 9y = 0. Use reduction of order method t0 find second solution Yz of the equation_
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##### Find the radius of convergence and interval of convergence of the series. $$\sum_{n=1}^{\infty} \frac{n}{b^{n}}(x-a)^{n}, \quad b>0$$
Find the radius of convergence and interval of convergence of the series. $$\sum_{n=1}^{\infty} \frac{n}{b^{n}}(x-a)^{n}, \quad b>0$$...
##### Chapter 09, Section 9.2, Problem 020bA random sample of 123 observations produced a sample mean of31. Find the critical and observed valuesof z for the following test of hypothesisusing α=0.025. The population standard deviation is knownto be 7 and the population distribution is normal.H0: μ=28 versus H1: μ>28.zcritical =zobserved =
Chapter 09, Section 9.2, Problem 020b A random sample of 123 observations produced a sample mean of 31. Find the critical and observed values of z for the following test of hypothesis using α=0.025. The population standard deviation is known to be 7 and the population distribution is normal. H0: ...
##### Prove this Proposition: The path on 4 vertices, P_4, is the onlytree on n > 2 vertices whose complement is also a tree.
Prove this Proposition: The path on 4 vertices, P_4, is the only tree on n > 2 vertices whose complement is also a tree....
##### [7 points] Use the transformation and to change the variables for dA where R is the region bounded by y = %I, y = and the hwperbolasy =and y = 2 Set-up the double integral only; and simplify
[7 points] Use the transformation and to change the variables for dA where R is the region bounded by y = %I, y = and the hwperbolas y = and y = 2 Set-up the double integral only; and simplify...
##### Point) The temperature U in a star of conductivity 3 is inversely proportional to the distance from the center:Vx+y+zIf the star is a sphere of radius 1, find the rate of heat flow outward across the surface of the star:
point) The temperature U in a star of conductivity 3 is inversely proportional to the distance from the center: Vx+y+z If the star is a sphere of radius 1, find the rate of heat flow outward across the surface of the star:... | {} |
# Quotient Banach space whose dual map sends the ball onto a given convex subset
Let $$X$$ be a Banach space and let $$A$$ be a closed, convex and balanced subset of $$B_{X^{*}}$$ (where $$B_{X^{*}}$$ denotes the closed unit ball of the dual $$X^{*}$$). Is there a closed subspace $$M$$ of $$X$$ such that $$Q^{*}_{M}$$ maps $$B_{(X/M)^{*}}$$ onto $$A$$, where $$Q_{M}:X\rightarrow X/M$$ is the quotient map?
• What happens when $X=\mathbb{R}^n$? Suppose $A$ is any closed convex balanced set with nonempty interior, other than the ball itself. If $M \ne 0$ then $Q_M^*$ has rank less than $n$ and so its image cannot cover $A$, and if $M=0$ then $Q_M^*$ is the identity map and it maps the ball to itself. – Nate Eldredge Feb 15 at 13:33
• Thanks, Nate. What happens if $X$ is infinite-dimensional? – Dongyang Chen Feb 15 at 14:36
• I was just thinking about that. More generally, the image of $Q_M^*$ will always equal the annihilator of $M$, right? If $M \ne 0$ this is a proper closed subspace. So take any $A$ which is not contained in a proper closed subspace (e.g. any $A$ with nonempty interior) and is not the ball, and I think that is a counterexample. – Nate Eldredge Feb 15 at 14:39
• Indeed, $Q^{*}_{M}B_{(X/M)^{*}}$ is equal to the closed unit ball of the annilator of $M$. If we take $A$ with nonempty interior, then $M=0$. – Dongyang Chen Feb 15 at 14:57 | {} |
# Characterising ergodicity of continuous maps
Hello all.
Suppose $X$ is a Polish space, $\mu$ is a Borel probability measure on $X$, and $T:X \to X$ is a continuous $\mu$-preserving map which is not ergodic.
Does there necessarily exist a Borel set $A \subset X$ such that
• $\mu(A) \in (0,1)$;
• $\mu(A \ \triangle \ T^{-1}(A)) = 0$;
• $A$ has non-empty interior?
What about if we replace the third point with the stronger requirement that $A$ is open?
Many thanks, Julian.
-
Does $T$ preserve the measure? – Joel Moreira Feb 24 '13 at 1:07
Good point! Let's assume it does. (I'll now edit the question accordingly.) – Julian Newman Feb 24 '13 at 1:36
Let $T \colon X \to X$ be a minimal transformation of a compact metric space which is not uniquely ergodic, let $\mu$ be a non-ergodic $T$-invariant measure on $X$, and let $A$ be a set with nonempty interior such that $\mu(A \triangle T^{-1}A)=0$. I claim that necessarily $\mu(A)=1$, contradicting the above conjecture. (Some constructions of transformations with the above combination of properties may be found for example in the textbook Ergodic Theory on Compact Spaces by Denker, Grillenberger and Sigmund, or in John Oxtoby's classic 1952 article Ergodic sets.)
Let $U \subseteq A$ be open and nonempty. Since $T$ is minimal we have $\bigcup_{n=0}^\infty T^{-n}U=X$, and indeed even $\bigcup_{n=0}^NT^{-n}U=X$ for some integer $N$ since $X$ is compact. In particular $\bigcup_{n=0}^N T^{-n}A=X$. Let us write $$\bigcup_{n=0}^N T^{-n}A = A \cup \bigcup_{n=1}^N \left(\left( T^{-n}A\right)\setminus \bigcup_{k=0}^{n-1} T^{-k}A\right)=A \cup \bigcup_{n=1}^N B_n,$$ say, which is a disjoint union. We would like to show that this union has measure identical to that of $A$. For each $n$ we have $$\mu(B_n)=\mu\left(T^{-n}A\setminus \bigcup_{k=0}^{n-1} T^{-k}A\right)\leq \mu\left(T^{-n}A \setminus T^{-(n-1)}A\right)=\mu\left(T^{-1}A \setminus A\right)=0$$ by invariance and the hypothesis $\mu(A \triangle T^{-1}A)=0$. It follows that $$\mu(A)=\mu\left(\bigcup_{n=0}^N T^{-n}A \right)=\mu(X)=1$$ so the desired situation can not occur.
Thank you. This is most helpful. Do you know any conditions on $X$ under which, if $\mu$ is a strictly positive probability measure on $X$, then every minimal $\mu$-preserving continuous transformation is ergodic? (E.g. is this true for Euclidean space $X=\mathbb{R}^n$?) – Julian Newman Feb 24 '13 at 2:44
@Julian: this is equivalent to asking for a condition on $X$ such that every minimal transformation on $X$ is uniquely ergodic, i.e. has only one invariant measure. (If a transformation has two distinct invariant measures then a strict linear combination of the two is never ergodic.) Such conditions do exist: finite spaces $X$ have this property, as does the circle (I think) but as Anthony says this is a severly restrictive requirement. The broader stroke of your question seems to be whether ergodicity can be easily characterised using only topological concepts. The answer to this is "No". – Ian Morris Feb 24 '13 at 12:08
@Ian and Anthony: Just to be clear, I did not say that I require every invariant probability measure of a minimal transformation to be ergodic - I just required that every strictly positive invariant probability measure of a minimal transformation had to be ergodic. (By strictly positive, I mean that its support is the whole of $X$). Is this still equivalent to requiring that every minimal transformation is uniquely ergodic? (And in the case $X=\mathbb{R}^n$, if the requirement still is not satisfied, what about if we weaken the requirement by restricting to, say, diffeomorphisms on $X$?) – Julian Newman Feb 24 '13 at 14:38 | {} |
# Wave function collapse in curved spacetime
Tags:
1. Sep 3, 2013
### atyy
In flat spacetime, there isn't any problem with wave function collapse. I think that's the "textbook" position, although the only citation I have off the top of my head is the discussion in http://arxiv.org/abs/0706.1232 (section 1.1).
How about in curved spacetime (working in the regime where localized qubits are possible)? I found http://arxiv.org/abs/1108.3896 (section 8.2.2 and 9.2) which discusses the issue a little, and there seems to be no problem with wave function collapse. Is this a consensus position?
Last edited: Sep 3, 2013
2. Sep 3, 2013
### strangerep
Ballentine would probably disagree. At least, that's the impression I got from his textbook...
3. Sep 3, 2013
### atyy
What does he say?
4. Sep 4, 2013
### S.Daedalus
In which sense do you say there's 'no problem'? There's of course the usual problems with the notion of collapse, related to the more general cluster of issues subsumed under 'measurement problem', but I gather that's not your issue here.
However, your cite seems to point to a problem with collapse in flat spacetimes, i.e. that it's not possible to pinpoint who caused the collapse: in one frame, A may measure first, leading to a specific outcome for B; while in another, B may be taken to cause the 'collapse'.
Of course, this doesn't lead to any observably paradoxical consequences: all probabilities remain the same. Do you mean there's no problem in this sense? If so, then I think (but am not totally sure) that there's also no problem at least in the case where spacetime is curved, but nondynamical. Of course, what happens if there is nontrivial backreaction from the spacetime---if, for example, one has a superposition of matter that should yield a superposition of macroscopically different spacetimes---is anyone's guess.
5. Sep 4, 2013
### strangerep
Oh, you haven't read him? He says quite a bit in ch9, (and elsewhere, iirc), pointing out inconsistencies and other problems with the usual Copenhagen interpretation. It's far too much to summarize here in a few sentences, but I'm sure you'd find that investing in a copy of Ballentine would be money well spent...
6. Sep 4, 2013
### vanhees71
Wave-function (or better said state) collapse is the most problematic concept ever invented when it comes to the interpretation of quantum theory, as have pointed out famously Einstein, Podolsky, and Rosen in the 1930ies. Fortunately it's not needed at all. As already recommended, Ballentine is the perfect source to study these questions. In my opinion the minimal statistical interpretation is the least problematic of all interpretations of quantum theory, but that's an opinion, and many physicists will violently disagree :-)).
7. Sep 4, 2013
### atyy
Yes, that's what I mean - no logical contradiction and good experimental predictions. And only the nondynamical case.
Why do you think there's no problem in the curved spacetime case? That's what http://arxiv.org/abs/1108.3896 indicates, but I've never seen extensive discussion of this, whereas for flat spacetime I do believe it's essentially {textbook minus Ballentine} agreement - maybe not literally {textbook} but I can also point to http://arxiv.org/abs/quant-ph/9906023 and http://arxiv.org/abs/quant-ph/9906034. Is it because in curved spacetime one can still define spacelike separations objectively, and because spacelike local projections and local unitaries do commute? So the causal possibilities are essentially the same in flat and curved spacetime, and that's what really matters?
Last edited: Sep 4, 2013
8. Sep 4, 2013
### S.Daedalus
I don't believe Ballantine differs from the grain in that regard---of course, the collapse itself is a problematic notion, whether the space is flat, curved, or curled in on itself; but I'd say the prevailing viewpoint is that besides the impossibility to introduce a fixed order on the sequence of collapses---which doesn't correspond to anything measurable---there's no new difficulties introduced by special relativity.
Something like that. Basically, my reasoning was that all the field operators at spacelike separation commute, so you can't introduce anything like causal paradoxes and the like; it's simply the way the theory is constructed. But I'm entirely willing to be shown wrong by someone with more expertise in this area.
9. Sep 4, 2013
### S.Daedalus
Then again, there are various 'non-localizability' theorems, most famously due to Malarment, that essentially show (or purport to show; nothing's clear cut once you let a philosopher into the room) that you can't really talk about the concept of particles in the same way one does in ordinary QM (a often-used statement is that 'there is no consistent relativistic theory of a fixed number of particles', on pain of superluminal spreading of wave functions).
Now, why do I consider this as a potential problem with the notion of collapse? Well, a key assumption in these theorems is some form of localizability (hence the name), i.e. a sufficiently sharply peaked wave function. Maybe there is a sense in which one could then consider this as a problem of measurement, instead of as a problem for particles? I'm not sufficiently familiar with these issues and work done in that direction to raise this as more than a possibility, though.
However, this would be nothing new for curved spacetimes, as the theorem already holds in Minkowsky space.
10. Sep 4, 2013
### atyy
I used to naively assume that in the flat spacetime case that the wave function collapse is simultaneous, where I use the canonical notion of simultaneous for an inertial observer. In the curved spacetime case, there is no canonical notion of simultaneity, and according to Palmer it isn't needed, so I guess it's not needed in in the flat spacetime case either - just causal structure.
The Palmer paper does talk about this problem, and restricts itself to a limit in which one can talk about localized particles. I think the overall picture is this:) Copenhagen is not a problem, since it is justified by Bohmian mechanics which in turn emerges from something else (like Valentini's non-equilibrium Bohmian). Within Copenhagen, in a theory of quantum gravity, there are no local observables at all, and no notion of spacelike even, since the geometry is fluctuating. This is AdS/CFT and the only observer who has quantum mechanics holding is the guy on the boundary with his S matrix. Then in a limit in which there is a classical spacetime and quantum gravity as an effective field theory, the strict observables are still only nonlocal like the S-matrix http://arxiv.org/abs/1105.2036 . Then in the limit where there's no backreaction, we should get local observables in the lab, and these are approximate http://arxiv.org/abs/hep-th/0512200. Then in some further limit going from special relativistic to Galilean relativity, things like position emerge as observables. So most quantum mechanics is emergent anyway, it's just emergent from bigger quantum systems. Anyway, was it really Malament who introduced the locality problem - I usually associate it with Newton-Wigner?
The funny thing is why does unitarity get retained in these emergent theories? There was quite an interesting comment on this by Aydemir, Anber and Donoghue in http://arxiv.org/abs/1203.5153.
Last edited: Sep 4, 2013 | {} |
# 16-cell
(16-cell)
(4-orthoplex)
Schlegel diagram
(vertices and edges)
Type Convex regular 4-polytope
4-orthoplex
4-demicube
Schläfli symbol {3,3,4}
Coxeter-Dynkin diagram
Cells 16 {3,3}
Faces 32 {3}
Edges 24
Vertices 8
Vertex figure
Octahedron
Petrie polygon octagon
Coxeter group C4, [3,3,4], order 384
Dual Tesseract
Properties convex, isogonal, isotoxal, isohedral
Uniform index 12
In four-dimensional geometry, a 16-cell or hexadecachoron is a regular convex 4-polytope. It is one of the six regular convex 4-polytopes first described by the Swiss mathematician Ludwig Schläfli in the mid-19th century.
It is a part of an infinite family of polytopes, called cross-polytopes or orthoplexes. The dual polytope is the tesseract (4-cube). Conway's name for a cross-polytope is orthoplex, for orthant complex.
## Geometry
It is bounded by 16 cells, all of which are regular tetrahedra. It has 32 triangular faces, 24 edges, and 8 vertices. The 24 edges bound 6 squares lying in the 6 coordinate planes.
The eight vertices of the 16-cell are (±1, 0, 0, 0), (0, ±1, 0, 0), (0, 0, ±1, 0), (0, 0, 0, ±1). All vertices are connected by edges except opposite pairs.
The Schläfli symbol of the 16-cell is {3,3,4}. Its vertex figure is a regular octahedron. There are 8 tetrahedra, 12 triangles, and 6 edges meeting at every vertex. Its edge figure is a square. There are 4 tetrahedra and 4 triangles meeting at every edge.
The 16-cell can be decomposed into two similar disjoint circular chains of eight tetrahedrons each, four edges long. Each chain, when stretched out straight, forms a Boerdijk–Coxeter helix. This decomposition can be seen in the alternated 4-4 duoprism construction, , of the 16-cell, symmetry [[4,2+,4]], order 64.
## Images
Stereographic projection|rowspan=2| A 3D projection of a 16-cell performing a simple rotation. The 16-cell has two Wythoff constructions, a regular form and alternated form, shown here as nets, the second being represented by alternately two colors of tetrahedral cells.
orthographic projections
Coxeter plane B4 B3 / D4 / A2 B2 / D3
Graph
Dihedral symmetry [8] [6] [4]
Coxeter plane F4 A3
Graph
Dihedral symmetry [12/3] [4]
demitesseract in order-4 Petrie polygon symmetry as an alternated tesseract Tesseract
## Tessellations
One can tessellate 4-dimensional Euclidean space by regular 16-cells. This is called the hexadecachoric honeycomb and has Schläfli symbol {3,3,4,3}. The dual tessellation, icositetrachoric honeycomb, {3,4,3,3}, is made of by regular 24-cells. Together with the tesseractic honeycomb {4,3,3,4}, these are the only three regular tessellations of R4. Each 16-cell has 16 neighbors with which it shares a tetrahedron, 24 neighbors with which it shares only an edge, and 72 neighbors with which it shares only a single point. Twenty-four 16-cells meet at any given vertex in this tessellation.
## Projections
Projection envelopes of the 16-cell. (Each cell is drawn with different color faces, inverted cells are undrawn)
The cell-first parallel projection of the 16-cell into 3-space has a cubical envelope. The closest and farthest cells are projected to inscribed tetrahedra within the cube, corresponding with the two possible ways to inscribe a regular tetrahedron in a cube. Surrounding each of these tetrahedra are 4 other (non-regular) tetrahedral volumes that are the images of the 4 surrounding tetrahedral cells, filling up the space between the inscribed tetrahedron and the cube. The remaining 6 cells are projected onto the square faces of the cube. In this projection of the 16-cell, all its edges lie on the faces of the cubical envelope.
The cell-first perspective projection of the 16-cell into 3-space has a triakis tetrahedral envelope. The layout of the cells within this envelope are analogous to that of the cell-first parallel projection.
The vertex-first parallel projection of the 16-cell into 3-space has an octahedral envelope. This octahedron can be divided into 8 tetrahedral volumes, by cutting along the coordinate planes. Each of these volumes is the image of a pair of cells in the 16-cell. The closest vertex of the 16-cell to the viewer projects onto the center of the octahedron.
Finally the edge-first parallel projection has a shortened octahedral envelope, and the face-first parallel projection has a hexagonal bipyramidal envelope.
## 4 sphere Venn Diagram
The usual projection of the 16-cell and 4 intersecting spheres (a Venn diagram of 4 sets) form topologically the same object in 3D-space:
## Symmetry constructions
There is a lower symmetry form of the 16-cell, called a demitesseract or 4-demicube, a member of the demihypercube family, and represented by h{4,3,3}, and Coxeter diagrams or . It can be drawn bicolored with alternating tetrahedral cells.
It can also be seen in lower symmetry form as a tetrahedral antiprism, constructed by 2 parallel tetrahedra in dual configurations, connected by 8 (possibly elongated) tetrahedra. It is represented by s{2,4,3}, and Coxeter diagram: .
It can also be seen as a snub 4-orthotope, represented by s{21,1,1}, and Coxeter diagram: .
With the tesseract constructed as a 4-4 duoprism, the 16-cell can be seen as its dual, a 4-4 duopyramid.
Name Coxeter diagram Schläfli symbol Coxeter notation Order Vertex figure
Regular 16-cell {3,3,4} [3,3,4] 384
Demitesseract =
=
h{4,3,3}
{3,31,1}
[31,1,1] = [1+,4,3,3] 192
Alternated 4-4 duoprism 2s{4,2,4} [[4,2+,4]] 64
Tetrahedral antiprism s{2,4,3} [2+,4,3] 48
Alternated square prism prism sr{2,2,4} [(2,2)+,4] 16
Snub 4-orthotope s{21,1,1} [2,2,2]+ 8
4-fusil
{3,3,4} [3,3,4] 384
{4}+{4} [[4,2,4]] 128
{3,4}+{} [4,3,2] 96
{4}+{}+{} [4,2,2] 32
{}+{}+{}+{} [2,2,2] 16
## Related uniform polytopes and honeycombs
D4 uniform polychora
{3,31,1}
h{4,3,3}
2r{3,31,1}
h3{4,3,3}
t{3,31,1}
h2{4,3,3}
2t{3,31,1}
h2,3{4,3,3}
r{3,31,1}
{31,1,1}={3,4,3}
rr{3,31,1}
r{31,1,1}=r{3,4,3}
tr{3,31,1}
t{31,1,1}=t{3,4,3}
sr{3,31,1}
s{31,1,1}=s{3,4,3}
The 16-cell is a part of the tesseractic family of uniform polychora:
Name tesseract rectified
tesseract
truncated
tesseract
cantellated
tesseract
runcinated
tesseract
bitruncated
tesseract
cantitruncated
tesseract
runcitruncated
tesseract
omnitruncated
tesseract
Coxeter
diagram
=
=
Schläfli
symbol
{4,3,3} t1{4,3,3}
r{4,3,3}
t0,1{4,3,3}
t{4,3,3}
t0,2{4,3,3}
rr{4,3,3}
t0,3{4,3,3} t1,2{4,3,3}
2t{4,3,3}
t0,1,2{4,3,3}
tr{4,3,3}
t0,1,3{4,3,3} t0,1,2,3{4,3,3}
Schlegel
diagram
B4
Name 16-cell rectified
16-cell
truncated
16-cell
cantellated
16-cell
runcinated
16-cell
bitruncated
16-cell
cantitruncated
16-cell
runcitruncated
16-cell
omnitruncated
16-cell
Coxeter
diagram
=
=
=
=
=
=
Schläfli
symbol
{3,3,4} t1{3,3,4}
r{3,3,4}
t0,1{3,3,4}
t{3,3,4}
t0,2{3,3,4}
rr{3,3,4}
t0,3{3,3,4} t1,2{3,3,4}
2t{3,3,4}
t0,1,2{3,3,4}
tr{3,3,4}
t0,1,3{3,3,4} t0,1,2,3{3,3,4}
Schlegel
diagram
B4
This polychoron is also related to the cubic honeycomb, order-4 dodecahedral honeycomb, and order-4 hexagonal tiling honeycomb all which have octahedral vertex figures.
{p,3,4}
Space S3 E3 H3
Form Finite Affine Compact Paracompact Noncompact
Name {3,3,4}
{4,3,4}
{5,3,4}
{6,3,4}
{7,3,4}
{8,3,4}
... {∞,3,4}
Image
Cells
{3,3}
{4,3}
{5,3}
{6,3}
{7,3}
{8,3}
{∞,3}
It is similar to three regular polychora: the 5-cell {3,3,3}, 600-cell {3,3,5} of Euclidean 4-space, and the order-6 tetrahedral honeycomb {3,3,6} of hyperbolic space. All of these have a tetrahedral cells.
{3,3,p}
Space S3 H3
Form Finite Paracompact Noncompact
Name {3,3,3}
{3,3,4}
{3,3,5}
{3,3,6}
{3,3,7}
{3,3,8}
... {3,3,∞}
Image
Vertex
figure
{3,3}
{3,4}
{3,5}
{3,6}
{3,7}
{3,8}
{3,∞}
Quasiregular polychora and honeycombs: h{4,p,q}
Space Finite Affine Compact Paracompact
Name h{4,3,3} h{4,3,4} h{4,3,5} h{4,3,6} h{4,4,3} h{4,4,4}
$\left\{3,{3\atop3}\right\}$ $\left\{3,{3\atop4}\right\}$ $\left\{3,{3\atop5}\right\}$ $\left\{3,{3\atop6}\right\}$ $\left\{4,{3\atop4}\right\}$ $\left\{4,{4\atop4}\right\}$
Coxeter
diagram
Image
Vertex
figure
r{p,3}
Regular and Quasiregular honeycombs: {p,3,4} and {p,31,1}
Space Euclidean 4-space Euclidean 3-space Hyperbolic 3-space
Name {3,3,4}
{3,31,1} = $\left\{3,{3\atop3}\right\}$
{4,3,4}
{4,31,1} = $\left\{4,{3\atop3}\right\}$
{5,3,4}
{5,31,1} = $\left\{5,{3\atop3}\right\}$
{6,3,4}
{6,31,1} = $\left\{6,{3\atop3}\right\}$
Coxeter
diagram
= = = =
Image
Cells
{p,3}
## References
• T. Gosset: On the Regular and Semi-Regular Figures in Space of n Dimensions, Messenger of Mathematics, Macmillan, 1900
• H.S.M. Coxeter:
• Coxeter, Regular Polytopes, (3rd edition, 1973), Dover edition, ISBN 0-486-61480-8, p. 296, Table I (iii): Regular Polytopes, three regular polytopes in n-dimensions (n≥5)
• H.S.M. Coxeter, Regular Polytopes, 3rd Edition, Dover New York, 1973, p. 296, Table I (iii): Regular Polytopes, three regular polytopes in n-dimensions (n≥5)
• Kaleidoscopes: Selected Writings of H.S.M. Coxeter, edited by F. Arthur Sherk, Peter McMullen, Anthony C. Thompson, Asia Ivic Weiss, Wiley-Interscience Publication, 1995, ISBN 978-0-471-01003-6 [1]
• (Paper 22) H.S.M. Coxeter, Regular and Semi Regular Polytopes I, [Math. Zeit. 46 (1940) 380-407, MR 2,10]
• (Paper 23) H.S.M. Coxeter, Regular and Semi-Regular Polytopes II, [Math. Zeit. 188 (1985) 559-591]
• (Paper 24) H.S.M. Coxeter, Regular and Semi-Regular Polytopes III, [Math. Zeit. 200 (1988) 3-45]
• John H. Conway, Heidi Burgiel, Chaim Goodman-Strass, The Symmetries of Things 2008, ISBN 978-1-56881-220-5 (Chapter 26. pp. 409: Hemicubes: 1n1)
• Norman Johnson Uniform Polytopes, Manuscript (1991)
• N.W. Johnson: The Theory of Uniform Polytopes and Honeycombs, Ph.D. (1966) | {} |
# Show that Z_12^* and Z_8^* are isomorphic groups
1. Feb 2, 2017
### Mr Davis 97
1. The problem statement, all variables and given/known data
Show that $\mathbb{Z}_8^*$ and $\mathbb{Z}_12^*$ are isomorphic, where $\mathbb{Z}_n^* = \{x \in \mathbb{Z} ~|~ \exists a \in \mathbb{Z}_n(ax \equiv 1~(mod~n)) \}$, and the group operation is regular multiplication.
2. Relevant equations
3. The attempt at a solution
We can see that $\mathbb{Z}_8^* = \{1,3,5,7 \}$ and $\mathbb{Z}_8^* = \{1,5,7,11 \}$
The only possible isomorphism I can think if is a function that maps from the former to the latter such that 1 goes to 1, 3 to 5, 5 to 7, and 7 to 11. The function is obviously injective and surjective. Is the only way to show that this satisfies the homomorphism property to show that it is satisfied for each combination from the domain? This would seem to be a tedious process.
2. Feb 2, 2017
### pasmith
The quick way is to note that up to isomorphism there are two abelian groups of order 4, the cyclic group $C_4$ (which contains an element of order 4) and the klein group $C_2 \times C_2$ (which does not contain an element of order 4). | {} |
Hagen-Poiseuille equation
Hagen-Poiseuille equation
The Hagen-Poiseuille equation is a physical law that describes slow viscous
incompressible flow through a constant circular cross-section. It is also known as the Hagen-Poiseuille law, Poiseuille law and Poiseuille equation.
Equation
tandard fluid dynamics notation
In standard fluid dynamics notation::$Delta P = frac\left\{8 mu L Q\right\}\left\{ pi r^4\right\}$ or :$Delta P = frac\left\{128 mu L Q\right\}\left\{ pi d^4\right\}$
Where::$Delta P$ is the pressure drop:L is the length of pipe:$mu$ is the dynamic viscosity:Q is the volumetric flow rate:r is the radius:d is the diameter:$pi$ is the mathematical constant (approximately 3.1416).
Physics notation
:$Phi = frac\left\{dV\right\}\left\{dt\right\} = v pi R^\left\{2\right\} = frac\left\{pi R^\left\{4\left\{8 eta\right\} left\left( frac\left\{- Delta P\right\}\left\{Delta x\right\} ight\right) = frac\left\{pi R^\left\{4\left\{8 eta\right\} frac\left\{ |Delta P\left\{L\right\}$
Where: :"V" is a volume of the liquid poured (cubic meters):"t" is the time (seconds):"v" is mean fluid velocity along the length of the tube (meters/second):"x" is a distance in direction of flow (meters):"R" is the internal radius of the tube (meters):"ΔP" is the pressure difference between the two ends (pascals):"η" is the dynamic fluid viscosity (pascal-second (Pa·s)), :"L" is the total length of the tube in the "x" direction (meters).
Relation to Darcy-Weisbach
This result is also a solution to the phenomenological Darcy-Weisbach equation in the field of hydraulics, given a relationship for the friction factor in terms of the Reynolds number:
:$Lambda = \left\{64over \left\{it Re ; , quadquad Re = \left\{2 ho v rover eta\right\} ; ,$
where "Re" is the Reynolds number and "ρ" fluid density. In this form the law approximates the "Darcy friction factor", the "energy (head) loss factor", "friction loss factor" or "Darcy (friction) factor" Λ in the laminar flow at very low velocities in cylindrical tube. The theoretical derivation of a slightly different form of the law was made independently by Wiedman in 1856 and Neumann and E. Hagenbach in 1858 (1859, 1860). Hagenbach was the first who called this law the Poiseuille's law.
The law is also very important specially in hemorheology and hemodynamics, both fields of physiology. [ [http://www.cvphysiology.com/Hemodynamics/H003.htm Determinants of blood vessel resistance ] ]
The Poiseuilles' law was later in 1891 extended to turbulent flow by L. R. Wilberforce, based on Hagenbach's work.
Derivation
The Hagen Poiseuille equation can be derived from the Navier-Stokes equations.
Viscosity
The derivation of Poiseuille's Law is surprisingly simple, but it requires an understanding of Viscosity. When two layers of liquid in contact with each other move at different speeds, there will be a force between them. This force is proportional to the area of contact "A", the velocity difference in the direction of flow "Δvx"/Δ"y", and a proportionality constant "η" and is given by
:$F_\left\{ ext\left\{viscosity, top = - eta A frac\left\{Delta v_x\right\}\left\{Delta y\right\}$
The negative sign is in there because we are concerned with the faster moving liquid (top in figure), which is being slowed by the slower liquid (bottom in figure). By Newton's third law of motion, the force on the slower liquid is equal and opposite (no negative sign) to the force on the faster liquid. This equation assumes that the area of contact is so large that we can ignore any effects from the edges and that the fluids behave as Newtonian fluids.
Liquid flow through a pipe
In a tube we make a basic assumption: the liquid in the center is moving fastest while the liquid touching the walls of the tube is stationary (due to friction). To simplify the situation, let's assume that there are a bunch of circular layers (lamina) of liquid, each having a velocity determined only by their radial distance from the center of the tube.
To figure out the motion of the liquid, we need to know all forces acting on each lamina:
# The force pushing the liquid through the tube is the change in pressure multiplied by the area: "F" = -"ΔPA". This force is in the direction of the motion of the liquid - the negative sign comes from the conventional way we define $Delta P = P_\left\{end\right\}-P_\left\{top\right\} < 0$.
# The pull from the faster lamina immediately closer to the center of the tube
# The drag from the slower lamina immediately closer to the walls of the tube
The first of these forces comes from the definition of pressure. The other two forces require us to modify the equations above that we have for viscosity. In fact, we are not modifying the equations, instead merely plugging in values specific to our problem. Let's focus on the pull from the faster lamina (#2) first.
Faster lamina
Assume that we are figuring out the force on the lamina with radius "s". From the equation above, we need to know the area of contact and the velocity gradient. Think of the lamina as a cylinder of radius "s" and thickness "ds". The area of contact between the lamina and the faster one is simply the area of the inside of the cylinder: "A" = "2πsΔx". We don't know the exact form for the velocity of the liquid within the tube yet, but we do know (from our assumption above) that it is dependent on the radius. Therefore, the velocity gradient is the change of the velocity with respect to the change in the radius at the intersection of these two laminae. That intersection is at a radius of "s". So, considering that this force will be positive with respect to the movement of the liquid (but the derivative of the velocity is negative), the final form of the equation becomes
:$F_\left\{ ext\left\{viscosity, fast = - eta 2 pi s Delta x left . frac\left\{dv\right\}\left\{dr\right\} ight vert_s$
where the vertical bar and subscript "s" following the derivative indicates that it should be taken at a radius of "s".
Slower lamina
Next let's find the force of drag from the slower lamina. We need to calculate the same values that we did for the force from the faster lamina. In this case, the area of contact is at "s"+"ds" instead of "s". Also, we need to remember that this force opposes the direction of movement of the liquid and will therefore be negative (and that the derivative of the velocity is negative).
:$F_\left\{ ext\left\{viscosity, slow = eta 2 pi \left(s+ds\right) Delta x left . frac\left\{dv\right\}\left\{dr\right\} ight vert_\left\{s+ds\right\}$
Putting it all together
To find the solution for the flow of liquid through a tube, we need to make one last assumption. There is no acceleration of liquid in the pipe, and by Newton's first law, there is no net force. If there is no net force then we can add all of the forces together to get zero
:$0 = F_\left\{ ext\left\{pressure + F_\left\{ ext\left\{viscosity, fast + F_\left\{ ext\left\{viscosity, slow$
or
:$0 = - Delta P2 pi sds - eta 2 pi s Delta x left . frac\left\{dv\right\}\left\{dr\right\} ight vert_s + eta 2 pi \left(s+ds\right) Delta x left . frac\left\{dv\right\}\left\{dr\right\} ight vert_\left\{s+ds\right\}$
Before we move further, we need to simplify this ugly equation. First, to get everything happening at the same point, we need to do a Taylor series expansion of the velocity gradient, keeping only the linear and quadratic terms (a standard mathematical trick).
:$left . frac\left\{dv\right\}\left\{dr\right\} ight vert_\left\{r+dr\right\} = left . frac\left\{dv\right\}\left\{dr\right\} ight vert_r + left . frac\left\{d^2 v\right\}\left\{dr^2\right\} ight vert_r dr$
Let's use this relation in our equation. Also, let's use "r" instead of "s" since the lamina we chose was arbitrary and we want our expression to be valid for all laminae. Grouping like terms and dropping the vertical bar since all derivatives are assumed to be at radius "r",
:$0 = - Delta P2 pi rdr + eta 2 pi dr Delta x frac\left\{dv\right\}\left\{dr\right\} + eta 2 pi r dr Delta x frac\left\{d^2 v\right\}\left\{dr^2\right\} + eta 2 pi \left(dr\right)^2 Delta x frac\left\{d^2 v\right\}\left\{dr^2\right\}$
Finally, let's get this in the form of a differential equation, moving some terms around to make it easier to solve later, and neglecting the term quadratic in "dr" since this will be really small compared to the rest (another standard mathematical trick).
:$frac\left\{1\right\}\left\{eta\right\} frac\left\{Delta P\right\}\left\{Delta x\right\} = frac\left\{d^2 v\right\}\left\{dr^2\right\} + frac\left\{1\right\}\left\{r\right\} frac\left\{dv\right\}\left\{dr\right\}$
It can be seen that both sides of the equations are negative: there is a drop of pressure along the tube (left side) and both first and second derivatives of the velocity are negative (velocity has a maximum value of the center of the tube). The equation may be re-arranged to:
:$frac\left\{1\right\}\left\{eta\right\} frac\left\{Delta P\right\}\left\{Delta x\right\} = frac\left\{1\right\}\left\{r\right\} frac\left\{d\right\}\left\{dr\right\} r frac\left\{dv\right\} \left\{dr\right\}$
This differential equation is subject to the following boundary conditions: :v(r) = 0 at r = R -- "No-slip" Boundary Condition at the Wall :$frac\left\{dv\right\} \left\{dr\right\} = 0$ at r = 0 -- Axial symmetry
Axial symmetry means that the velocity v(r) is maximum at the center of the tube, therefore the first derivative $frac\left\{dv\right\}\left\{dr\right\}$ is zero at r = 0.
The differential equation can be integrated to:
:$v\left(r\right) = frac\left\{1\right\}\left\{4 eta\right\}r^2frac\left\{Delta P\right\}\left\{Delta x\right\} + A ln\left(r\right) + B$
To find A and B, we use the boundary conditions.
First, the symmetry boundary condition indicates:
:$frac\left\{dv\right\}\left\{dr\right\} = frac\left\{1\right\}\left\{2 eta\right\} r frac\left\{Delta P\right\}\left\{Delta x\right\} + A frac\left\{1\right\}\left\{r\right\} = 0$ at r = 0
A solution possible only if A = 0. Next the no-slip boundary condition is applied to the remaining equation:
:$v\left(R\right) = frac\left\{1\right\}\left\{4 eta\right\} R^2 frac\left\{Delta P\right\}\left\{Delta x\right\} + B = 0$
so therefore
:$B = - frac\left\{1\right\}\left\{4 eta\right\} R^2 frac\left\{Delta P\right\}\left\{Delta x\right\}$
Now we have a formula for the velocity of liquid moving through the tube as a function of the distance from the center of the tube
:$v = - frac\left\{1\right\}\left\{4 eta\right\} frac\left\{Delta P\right\}\left\{Delta x\right\} \left(R^2 - r^2\right)$
or, at the center of the tube where the liquid is moving fastest ("r" = 0) with "R" being the radius of the tube,
:$v_\left\{max\right\} = - frac\left\{1\right\}\left\{4 eta\right\} frac\left\{Delta P\right\}\left\{Delta x\right\}R^2$
Poiseuille's Law
To get the total volume that flows through the tube, we need to add up the contributions from each lamina. To calculate the flow through each lamina, we multiply the velocity (from above) and the area of the lamina.
:$Phi \left(r\right) = frac\left\{1\right\}\left\{4 eta\right\} frac\left\{Delta x\right\} int_\left\{0\right\}^\left\{R\right\} \left(rR^2 - r^3\right), dr = frac\left\{|Delta P| pi R^4\right\}\left\{8 eta Delta x\right\}$
Poiseuille's equation for compressible fluids
For a compressible fluid in a tube the volumetric flow rate and the linear velocity is not constant along the tube. The flow is usually expressed at outlet pressure. As fluid is compressed or expands work is done and the fluid is heated and cooled. This meaning that the flow rate is dependant upon heat transfer to and from the fluid. For an ideal gas in the isothermal case, where the temperature of the fluid is permitted to equilibrate with its surroundings, and when the pressure difference between ends of the pipe is small, the volumetric flow rate at the pipe outlet is given by
:$Phi = frac\left\{dV\right\}\left\{dt\right\} = v pi R^\left\{2\right\} = frac\left\{pi R^\left\{4\right\} left\left( P_\left\{i\right\}-P_\left\{o\right\} ight\right)\right\}\left\{8 eta L\right\} imes frac\left\{ P_\left\{i\right\}+P_\left\{o\left\{2 P_\left\{o = frac\left\{pi R^\left\{4\left\{16 eta L\right\} left\left( frac\left\{ P_\left\{i\right\}^\left\{2\right\}-P_\left\{o\right\}^\left\{2\left\{P_\left\{o ight\right)$
Where::$P_\left\{i\right\}$ inlet pressure:$P_\left\{o\right\}$ outlet pressure:$L$ is the length of tube:$eta$ is the viscosity:$R$ is the radius:$V$ is the volume of the fluid at outlet pressure:$v$ is the velocity of the fluid at outlet pressure
This is usually a good approximation when the flow velocity is less than mach 0.3
This equation can be seen as Poiseuille's law with an extra correction factor $frac\left\{P_\left\{i\right\}+P_\left\{o\left\{2\right\} imes frac\left\{1\right\}\left\{P_\left\{o$ expressing the average pressure relative to the outlet pressure.
= Electrical Circuits analogy =
Electricity was originally understood to be a kind of fluid. This hydraulic analogy is still conceptually useful.
Poiseuille's law corresponds to Ohm's law for electrical circuits ("V" = "IR"), where the pressure drop Δ"P " is analogous to the voltage "V" and voluminal flow rate Φ is analogous to the current "I". Then the resistance :$R = frac\left\{ 8 eta Delta x\right\}\left\{pi r^4\right\}$This concept is useful because the effective resistance in a tube is inversely proportional to the fourth power of the radius. This means that halfing the size of the tube increases the resistance to fluid movement by 16 times.
Both Ohm's law and Poiseuille's law illustrate transport phenomena.
History
It was developed independently by Gotthilf Heinrich Ludwig Hagen (1797-1884) and Jean Louis Marie Poiseuille.
Poiseuille's law was experimentally derived in 1838 and formulated and published in 1840 and 1846 by Jean Louis Marie Poiseuille (1797-1869). Hagen did his experiments in 1839.
References
*S. P. Sutera, R. Skalak, "The history of Poiseuille's law," "Annual Review of Fluid Mechanics", Vol. 25, 1993, pp. 1-19
*Citation
id = PMID:779509
url= http://www.ncbi.nlm.nih.gov/pubmed/779509
last=Pfitzner
first=J
publication-date=1976 Mar
year=1976
title=Poiseuille and his law.
volume=31
issue=2
periodical=Anaesthesia
pages=273-5
ee also
* Darcy's law
* Pulse
* Wave
* [http://www.syvum.com/cgi/online/serve.cgi/eng/fluid/fluid802.html Poiseuille's law for power-law non-Newtonian fluid]
* [http://www.syvum.com/cgi/online/serve.cgi/eng/fluid/fluid203.html Poiseuille's law in a slightly tapered tube]
* [http://www.calctool.org/CALC/eng/fluid/hagen-poiseuille Web-based calculator of the Hagen-Poiseuille equation]
Wikimedia Foundation. 2010.
Look at other dictionaries:
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• Poiseuille, Jean-Louis-Marie — ▪ French physician born April 22, 1799, Paris, France died Dec. 26, 1869, Paris French physician and physiologist who formulated a mathematical expression for the flow rate for the laminar (nonturbulent) flow of fluids in circular tubes.… … Universalium
• Équation de Darcy-Weisbach — L équation de Darcy Weisbach est une importante équation très utilisée en hydraulique[Paraschivoiu 1]. Elle permet de calculer la perte de charge due à la rugosité des conduites (« perte de charge linéaire », par opposition aux pertes… … Wikipédia en Français
• Poiseuille's law — Poi·seuille s law pwä zə(r)z , zwēz , zə ēz n a statement in physics: the velocity of the steady flow of a fluid through a narrow tube (as a blood vessel or a catheter) varies directly as the pressure and the fourth power of the radius of the… … Medical dictionary
• Écoulement de Poiseuille — La loi de Poiseuille (également appelée loi de Hagen Poiseuille) décrit l écoulement laminaire (c est à dire à filets d’eau parallèles) d un liquide visqueux dans une conduite cylindrique. Découverte indépendamment en 1844 par le médecin et… … Wikipédia en Français
• Ecoulement de Poiseuille — Écoulement de Poiseuille La loi de Poiseuille (également appelée loi de Hagen Poiseuille) est nommée à partir des travaux de Jean Louis Marie Poiseuille, médecin et physicien français du XIXe siècle[1]. Un écoulement de Poiseuille est un… … Wikipédia en Français
• Loi de Poiseuille — Écoulement de Poiseuille La loi de Poiseuille (également appelée loi de Hagen Poiseuille) est nommée à partir des travaux de Jean Louis Marie Poiseuille, médecin et physicien français du XIXe siècle[1]. Un écoulement de Poiseuille est un… … Wikipédia en Français
• Écoulement de poiseuille — La loi de Poiseuille (également appelée loi de Hagen Poiseuille) est nommée à partir des travaux de Jean Louis Marie Poiseuille, médecin et physicien français du XIXe siècle[1]. Un écoulement de Poiseuille est un écoulement qui suit une loi de… … Wikipédia en Français
• Microcirculation — The microcirculation is a term used to describe the small vessels in the vasculature which are embedded within organs and are responsible for the distribution of blood within tissues; as opposed to larger vessels in the macrocirculation which… … Wikipedia | {} |
## Monthly Archives: March 2008
### The wording of proofs
Dear Sir,
This may seem like a stupid question, but it has been bugging me for ages..
Is there a difference between me saying there are or there exist?
Similarly, is suppose equivalent to let and if? Also, is hence equivalent to therefore, so, i.e. ?
What I mean by “is there a difference” I mean could I lose marks in an exam if I write something equivalent?
The short answer is this: If you write a clear and logical proof, you will never have marks taken off for using one phraseology or another. The point is that when you read proofs in a book or lecture notes, you should understand precisely what’s meant by the sentences used, rather than trying to memorize the precise adverbs, prepositions, etc. Now of course, since mathematical language depends on conventions like any other language, you should eventually learn them naturally . But it’s safe to say that in a proof, the primary focus should be on correctness and clarity . To achieve this, the most important point is obviously understanding .
-I can’t think of any particular situation where there would be a difference between there are’ and there exist’.
-Suppose,’ ‘let,’ and, ‘if’ are used in different contexts so I can’t answer this question without some sample sentences you have in mind. Perhaps you are thinking of sentences that begin
Let X be…’
If X is…’
Suppose X is…’
in which case they seem to be mostly equivalent.
-‘Therefore,’ ‘hence,’ and ‘so,’ are interchangeable in any mathematical sentence I can think of where they are used as connectives leading to a conclusion. ‘Hence’ and ‘so’ are a bit more informal, while ‘therefore’ might be more likely to occur in final conclusions. ‘i.e.’ is usually different. As in common parlance, it means ‘that is,’ and is used mostly to just rephrase something.
If your sensitivity to differences in the turns of phrases arises from a lack of confidence in following mathematical arguments, you should definitely take it seriously. You need to make your understanding immune to insubstantial differences. (In particular, you should have a good sense of which differences are ‘insubstantial.’)
### Travel
This is to let students know that I am at a conference in Bangalore, India. I will have occasional access to the internet and will try to get to your questions as soon as possible, but you have to expect some delays until 5 April.
### Local and global maxima
Hi Professor!
Just very quickly, what’s the difference between the global maximum/minimum of a function and the local maximum/minimum of a function? I understand that if a function achieves a global maximum at a point then it automatically achieves a local maximum, but vice versa is not necessarily true.
Also, the definitions look identical besides the local maximum def. containing epsilon.
Cheers!
A function f has a local maximum at c if f(c) is $\geq$ all the values at nearby points. f has a global maximum at c if f(c) is $\geq$ the values at all points (in the domain). Therefore, a global maximum is a local maximum, but *not* vice versa. The best way to show the difference would be to draw a picture, but my graphic skills on the computer are very limited. So I’ll just have to write down a formula. Consider
$f(x)=x^3-x$
You should be able to see that f(x) has a local maximum at $c=-1/\sqrt{3}$. Near that point, the graph of f(x) is like a hill with apex above that point. What is the global maximum? Of course there is none, because f(x) keeps getting bigger as you move to the right along the x axis (at least starting from the point $x=1/\sqrt{3}$). One sometimes turns this into a problem with a global maximum by restricting the domain. That is, if we take the same function except with domain only on the closed interval [-100,100], you should be able to see that the global maximum is at x=100.
### Chinese remainder theorem
Dear Professor Kim,
Do we need to prove the chinese remainder theorem by using bezouts lemma or by showing it is a bijective mapping?
The Chinese remainder theorem says exactly that some map is bijective, so of course you need to show bijectivity. On the other hand, in proving the surjectivity part of it, one way is to use Bezout’s lemma.
One important point about the use of Bezout’s lemma is that it gives a constructive proof of surjectivity. Do you see what I mean by that?
### Office hour today
I told a number of students that I would hold office hours today again in the 5th floor common room, but I had not realized that the college would actually be closed. There’s no way for me to reach the secretary for a group email either, so I’m putting a post here apologizing for the confusion.
Check your email after the end of the closure period, and I’ll send something about continuation.
### L-functions
The theory of zeta and L-functions is an indispensable part of modern number theory. Recently, there was a conference on L-functions at the American Institute of Mathematics where some interesting new results of Andrew Booker and Ce Bian of Bristol were announced. At the n-category cafe, David Corfield posted a question about that article, to which I responded in vague terms. However, I did give a very brief summary of what L-functions are about. So I am attaching a link here for students interested in number theory.
### More on canonical forms
Dear Professor Kim,
I encountered some problems when practising on the past papers.
1. In the proof of the Primary Decomposition Theorem, since the minimal polynomial m(T)=0, then Ker(m(T))=V, why is that?
2. I am still confused with how to use rank and null to decide a Jordan basis.
3. In the final process of finding the Real/Complex Jordan Canonical Form, if I have to multiply matrices E&(E transpose) with complex elements to get the Real JCF, is it right or I made a mistake (problem is I’ve checked several times that I didn’t make any mistake in the previous calculation)?
4. Relating to the previous question, is it possible to have real JCF and complex JCF the same?
Finally, since I’m in China right now I can’t access the web blog. Would you please copy the answers to my email? Thanks a lot!!
1. As you say, if m(x) is the minimal polynomial, then m(T) is the *zero* endomorphism. What does it mean for the endomorphism to be zero?
2. The whole topic of computing Jordan bases is important, but hard to describe on the blog. It’s important to ask very specific questions. However, I can guess what difficulty you’re trying to express. For an endomorphism T and each eigenvalue a, we compute the Jordan basis corresponding to that eigenvalue. In the process, very relevant things to analyze are the kernels of
(T-aI)^k
for various k. For example, for k=1, the kernel is exactly the eigenspace corresponding to a. (Make sure you understand why this is so. I still find many people confused on this elementary point.) For k>1, we call these kernels the generalized eigenspaces. Also very important is that these kernels are nested:
$Ker(T-aI) \subset Ker(T-aI)^2 \subset Ker(T-aI)^3 \subset \cdots$
This also, make sure you understand why. The final important general relation is that
$(T-aI)(Ker(T-aI)^{k+1})\subset (T-aI)^{k}$
Anyways, in terms of the Jordan basis that you’ll see arranged in an array either in the notes or in my notes for 12-11-07, the bottom row is a basis for the kernel of (T-aI). The bottom *two* rows is a basis for the kernel of (T-aI)^2, and so on. Thus, the nullities of these powers of (T-aI) encode information about the shape of that array, from which one can extract the precise Jordan canonical form. Remind yourself precisely what the heights of the columns mean. These heights are exactly what you need for the precise Jordan canonical form.
Now to your question: Sometimes, you’ll see yourself given the *rank* of (T-aI)^k rather than its nullity. But don’t you know how to recover one from the other?
3. As a matter of terminology, the real or complex canonical forms of a symmetric bilinear form (or a quadratic form) are *not* called the *Jordan* canonical forms. If your confusion of the two reflects a conceptual confusion, you should clear it up right away. One very important distinction is that *similarity* of matrices is different from *congruence* of matrices. Look this up.
Anyways, *no*, you should not use complex matrices when finding the real canonical form even though it might luckily give you the right answer. The reason is that the real canonical form refers to a basis for $R^n$, and if you use complex transformations, the basis you end up with may be a basis for $C^n$.
4. The answer is yes. And I’m going to ask you back the question *when* are they the same?’ Make sure you know the answer and the reason for it.
### Norms
Just a quick question.
The norm of an ideal, e.g. , is taken to be |N(a)|. I am not sure I understand what this means exactly. For instance, the norm of is 2, but i am not sure how they used this fact.
Thanks.
Also, when are u available for consultation from here onwards and is there a revision class planned? ————————————————————————————————————————
Hi again.
I sent you a question earlier regarding the norm of an ideal, but i think i may have confused myself (and possibly you!)
Firstly, I understand the use of N()=|N(a)| but i think, now, that we are to use a different equation for multiple element ideals. I think that is why I was confused for .
Also, i should have mentioned in the example that we were in Z[sqrt(-17)], but i think you figured that already.
————————————————————————————————————————
Actually, the ideal wasn’t transmitted properly, but I can guess what it was.
First of all, the norm of an ideal is related to the norm of a number, but not the same:
If A is a non-zero ideal inside the ring R of algebraic integers in an algebraic number field K, then
N(A)=|R/A|
the number of elements inside the finite ring R/A. In particular, it is always a positive number. The relation with the norm of a number is that if A=(a), i.e., it is generated by a single element a, then
N(A)=|N(a)|
Note that the absolute value symbol here is used in the usual sense, and gives a positive number. When we used it above, it meant the cardinality of a set. You should get used to both notations and figure out what makes sense from the context.
When the ideal has many generators, the relation is more complicated. So for A=(a,b) we have
$(a)\subset A,$
and hence,
$N(A) \leq N((a))=|N(a)|$
or, more precisely,
N(A) divides |N(a)| using some theorem from the notes. (Find it!)
A basic case for computing norms is for a maximal ideal in the ring of integers of a field extension as described by Dedekind’s theorem. For the case
$Q(\sqrt{-17})$
you mention, the ring of integers is
$Z[\sqrt{-17}],$
so that we can compute the decomposition of ideals by consider the reductions of $x^2+17$. For example,
$x^2+17 \equiv (x-1)^2\ \ mod\ 2,$
so that
$(2)=P_2^2$
in $Z[\sqrt{-17}]$
for the maximal ideal $P_2=(2, \sqrt{-17}-1).$ You’ll find in the discussion of Dedekind’s theorem a proof of the fact that $N(P_2)=2.$
The point that is
$Z[\sqrt{-17}]/P_2$
is a field extension of Z/2 of degree equal to the degree of x-1, i.e., 1. That is to say,
$Z[\sqrt{-17}]/P_2 =Z/2$
So
$N(P_2)=|Z/2|=2.$
Do a bunch of exercises that compute norms of numbers and ideals. A basic useful fact of course is that
$N(IJ)=N(I)N(J)$
for any (non-zero) $I,J$. Thus, once you have the norms of prime ideals, you have the norm of any ideal (provided you have the prime decomposition, of course). Of course, this fact is more commonly used the other way around, that is, to *compute* the prime decomposition.
What are the norms of the numbers $\sqrt{2}$ and $\sqrt{3}$ in
$Q(\sqrt{2},\sqrt{3})?$
What is the norm of the ideal
$(2^{1/3}, 5)$
in $Z[2^{1/3}]$?
### Some algebra questions
I was wondering whether you can help me with some issues I am having:
Notation:
Aut(R)
is the set of automorphism of a ring R.
1) how can you formally show that Aut(R) is a “group under composition”? I am not quite sure what the question is asking to show!
2) the other day you gave me an easy way of showing ring isomorphisms, and was wondering how to show
Aut(Q[x]/((x^2)-p))
is isomorphic to C2 (cyclic group of order 2) and describe the nontrivial element explicitly (where p is prime)
3) not sure how to approach questions like this
Aut(F2[x]/((x^3)+x+1)) = ?
and
Aut(Q) is trivial
would i have to use Eisensteins criterion?
4) how can i show that any automorphism of reals is continuous and that
Aut(R)
is trivial?
————————————————————————————————————————
Also, would you be able to tell me how to compute the complete factorization into Q-irreducible factors for these polynomials:
i)a P[x] in the form of
(x^a + 1)
where a is a positive integer, eg (x^10 + 1) or (x^16 + 1)
ii) (x^5 + 10x^4 + 13x^3 – 25x^2 -68x – 60)
————————————————————————————————————————
Show that R[x]/(x^2 + a) is isomorphic to Complex if a>0 and RxR if a0 and RxR if a<0 where R is Real.
(1) You’ve probably learned that the set of bijections of any set is a group under composition. Now when the set is a ring R, then Aut(R) is the subset of the group of bijections from R to R consisting of the maps that preserve the two laws of composition and the unit(s). Now what does it take to show that a subset is in fact a subgroup?
(2) To get a ring homomorphism from Q[x]/(x^2-p) to itself, you need to have a map
f:Q[x]->Q[x}
that takes x^2-p to x^2-p. Such a ring homomorphism is entirely determined by the value f(x). (Why?) But to preserve the polynomial, we must have f(x)=x or -x. Both of these will induce automorphisms of Q[x]/(x^2-p) .
Incidentally, depending on what you’ve learned and what’s expected in class, you might have to provide much more detail than I’ve written here. (The same goes for the other problems.)
(3) Eisenstein’s criterion, dealing with irreducibility of integer coefficient polynomials, is irrelevant here. But for polynomials of degree 2 or 3, a straightforward criterion for irreducibility is the non-existence of a root in the field. Use this to check that x^3+x+1 is irreducible. As a result, F_2[x]/(x^3+x+1) is a field extension of F_2 of degree 3. So the group of automorphisms is of order at most 3. Now, for a field of characteristic 2, the squaring map
a-> a^2
is a ring homomorphism. Think about this carefully. It might be convenient to note also that an endomorphism of a finite field is automatically bijective. (Why?)
To see the triviality of Aut(Q), note that any ring homomorphism f:Q->Q must have f(0)=0 and f(1)=1. But then, by additivity of the map, f(n)=n for any natural number. If you have a positive rational number r=n/m with m>0, then
f(mr)=mf(r)
on the one hand, and
f(mr)=f(n)=n
on the other. So
mf(r)=n
and
f(r)=n/m=r.
Now deal with the negative numbers in some obvious way.
(4) This is quite an interesting fact. The point is that a field automorphism of R must preserve the *order* of R. This is because
a>b
iff
a-b>0
iff
x^2-(a-b)
has a root in R. This is a property that is preserved by any field automorphism. Now can you check that an order-preserving map of R is necessarily continuous? Here, you might have to review the precise definition of continuity from basic analysis.
Once you have the continuity, the triviality of Aut(R) follows from the corresponding fact for Q and the *densenesss* of Q in R.
————————————————————————————————————————
(i) There is a recursive algorithm. A basic statement is that
x^p-1=(x-1)(x^{p-1}+x^{p-2}+…+x+1)
and
(x^{p-1}+x^{p-2}+…+x+1)
is irreducible. You can show this using the substitution x->x+1 and Eisenstein’s criterion. For other cases, do some research on cyclotomic polynomials, as you seem to have done already. Remember to look for the *recursive* algorithm, which is the most efficient way. These polynomials are very important in algebraic number theory, but also its applications to information theory.
(ii) You seem to have figured out already the substitution that reduces this to Eisenstein’s criterion. Unfortunately, I don;’t know any clever method for finding such substitutions. There might be other proofs of irreducibility for this polynomial, depending on what else you’ve learned.
———————————————————————————————————————–
I hope you’ve learned already that C is isomorphic to R[x]/(x^2+1). Once you see this the first part of the problem is rather straightforward. That is, if a>0, can you cook up an isomorphism
R[x]->R[x]
that takes x^2+a to x^2+1? Why doesn’t it work when a<0?
In fact, when a<0, we get
x^2+a=x^2-(-a)=(x-\sqrt{-a})(x+\sqrt{-a}).
Now use the Chinese remainder theorem.
### Education, in Singapore and elsewhere
On the mathematical physics web log The n-category cafe,’ a number of people have been discussing the teaching of mathematics, prompted by an article in the Los Angeles times. I ended up contributing a few scattered comments. A link is included here for the possible amusement of my UCL students. | {} |
Home
#### The Six Pillars of Calculus
A picture is worth 1000 words
#### Trigonometry Review
The basic trig functions
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The unit circle
Addition of angles, double and half angle formulas
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Graphs of Trig Functions
#### Exponential Functions
Exponentials with positive integer exponents
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The function $f(x)=a^x$ and its graph
Exponential growth and decay
#### Logarithms and Inverse functions
Inverse Functions
How to find a formula for an inverse function
Logarithms as Inverse Exponentials
Inverse Trig Functions
#### Intro to Limits
Overview
Definition
One-sided Limits
When limits don't exist
Infinite Limits
Summary
#### Limit Laws and Computations
Limit Laws
Intuitive idea of why these laws work
Two limit theorems
How to algebraically manipulate a 0/0?
Indeterminate forms involving fractions
Limits with Absolute Values
Limits involving indeterminate forms with square roots
Limits of Piece-wise Functions
The Squeeze Theorem
#### Continuity and the Intermediate Value Theorem
Definition of continuity
Continuity and piece-wise functions
Continuity properties
Types of discontinuities
The Intermediate Value Theorem
Summary of using continuity to evaluate limits
#### Limits at Infinity
Limits at infinity and horizontal asymptotes
Limits at infinity of rational functions
Which functions grow the fastest?
Vertical asymptotes (Redux)
Summary and selected graphs
#### Rates of Change
Average velocity
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Computing an instantaneous rate of change of any function
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#### The Derivative Function
The derivative function
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#### Basic Differentiation Rules
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Version 1
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A hybrid chain rule
#### Implicit Differentiation
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#### $f$ vs. $f'$
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The Fundamental Theorem of Calculus (Part 1)
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#### The Indefinite Integral and the Net Change
Indefinite Integrals and Anti-derivatives
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#### Substitution
Substitution for Indefinite Integrals
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Revised Table of Integrals
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#### Area Between Curves
Computation Using Integration
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#### Volumes
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Solids of Revolution 1: Disks
Solids of Revolution 2: Washers
More Practice
### Definition of Continuity
Definition: A function $f$ is continuous at a number $x=a$ if $\displaystyle \lim_{x \rightarrow a} f(x) = f(a)$.
Remember that $\displaystyle{\lim_{x \to a} f(x)}$ describes both what is happening when $x$ is slightly less than $a$ and what is happening when $x$ is slightly greater than $a$. Thus there are three conditions inherent in this definition of continuity. A function is continuous at $a$ if the limit as $x\to a$ exists, and $f(a)$ exists, and this limit is equal to $f(a)$. This means that the following three values are equal: $$\lim_{x \to a^-} f(x)\qquad=\qquad f(a)\qquad=\qquad\lim_{x \to a^+} f(x)$$ I.e. the value as $x$ approaches $a$ from the left is the same as the value as $x$ approaches $a$ from the left (the limit exists) which is the same as the value of $f$ at $a$.
If any of these quantities is different, or if any of them fails to exist, then we say that $f(x)$ is discontinuous at $x=a$, or that $f(x)$ has a discontinuity at $x=a$.
What does this mean graphically? If you trace $f$ with a pencil from left to right, as you approach $x=a$, you are at some height $L$ (because $\displaystyle\lim_{x\to a^-}f(x)=L$. As you go through the $x$-value $a$, your height is also $L$ (because $f(a)=L$). Now as you keep going with your pencil, you are beginning this last stretch at height $L$ (because $\displaystyle\lim_{x\to a^+}f(x)=L$).
DO: Sketch $f(x)=\sqrt x$ and let $a=4$. Find $f(a), \displaystyle\lim_{x\to a^-}f(x)=L$, and $\displaystyle\lim_{x\to a^+}f(x)$. Now, follow along your graph as stated in the previous paragraph, looking at each condition as you go. Is $f(x)=\sqrt x$ continuous at $x=a$?
Now, more interestingly, consider $\displaystyle\frac{x^2-1}{x-1}$ in this video:
Definition: A function $f$ is continuous from the right at $x=a$ if $\displaystyle \lim_{x \rightarrow a^+} f(x) = f(a)$, and is continuous from the left at $x=a$ if $\displaystyle \lim_{x \rightarrow a^-} f(x) = f(a)$, and is continuous on an interval $I$ if it is continuous at each interior point of $I$, is continuous from the right at the left endpoint (if $I$ has one), and is continuous from the left at the right endpoint (if $I$ has one). | {} |
## Parallelotope
Move a point along a Line for an initial point to a final point. It traces out a Line Segment . When is translated from an initial position to a final position, it traces out a Parallelogram . When is translated, it traces out a Parallelepiped . The generalization of to -D is then called a parallelotope. has vertices and
s, where is a Binomial Coefficient and , 1, ..., (Coxeter 1973). These are also the coefficients of .
References
Coxeter, H. S. M. Regular Polytopes, 3rd ed. New York: Dover, pp. 122-123, 1973.
Klee, V. and Wagon, S. Old and New Unsolved Problems in Plane Geometry and Number Theory. Washington, DC: Math. Assoc. Amer., 1991.
Zaks, J. Neighborly Families of Congruent Convex Polytopes.'' Amer. Math. Monthly 94, 151-155, 1987. | {} |
# Check it the Killing vectors satisfy Killing equation or not
I am going through Kerr/CFT correspondence paper again, and I am at the section where authors specify Killing vectors for near horizon extreme Kerr metric (shortly NHEK).
The metric is
$$d\bar{s}^2=2GJ\Omega^2\left(-(1+r^2)d\tau^2+\frac{dr^2}{1+r^2}+d\theta^2+\Lambda^2(d\varphi+rd\tau)^2\right)$$
Where
$$\Omega^2\equiv\frac{1+\cos^2\theta}{2},\quad\Lambda\equiv\frac{2\sin\theta}{1+\cos^2\theta}$$
It is said that the metric has enhanced $SL(2,\mathbb{R})\times U(1)$ isometry group. Now what exactly is enhanced symmetry? I only find mentioning of it in the context of string theory, so I'm not sure what to do with it.
If we ignore that for a moment, looking at the groups in question, $SL(2,\mathbb{R})$ has 3 generators ($Sl(n,\mathbb{R})$ has $n^2-1$ elements), and $U(1)$ has one.
The rotational $U(1)$ symmetry is generated by Killing vector:
$$\xi_0=-\partial_\varphi$$
While time translations become part of an enhanced $SL(2,\mathbb{R})$ isometry group generated by the Killing vectors
$$\xi_1=\frac{2r\sin\tau}{\sqrt{1+r^2}}\partial_\tau-2\sqrt{1+r^2}\cos\tau\partial_r+\frac{2\sin\tau}{\sqrt{1+r^2}}\partial_\varphi$$
$$\xi_2=-\frac{2r\cos\tau}{\sqrt{1+r^2}}\partial_\tau-2\sqrt{1+r^2}\sin\tau\partial_r-\frac{2\cos\tau}{\sqrt{1+r^2}}\partial_\varphi$$
$$\xi_3=2\partial_\tau$$
Now I wanted to try and find them, but that proved to be quite a challenge (I may try to do it in the end). So instead I wanted to check if they satisfy Killing equation. Another way of checking if they are Killing vectors is to check if Lie derivative of the metric along the Killing vectors is 0
$$\mathcal{L}_\xi g_{\mu\nu}=\xi^\sigma\partial_\sigma g_{\mu\nu}+g_{\sigma\nu}\partial_\mu\xi^\sigma+g_{\mu\sigma}\partial_\nu\xi^\sigma=0$$
So I put the two simplest ones ($\xi_0$ and $\xi_3$), and they give 0 for each component. Nice.
I try with $\xi_2$, and I get non zero components.
So what is wrong with my interpretation?
I did the calculation by hand and with RGTC package in Mathematica, using LieD which calculates Lie derivative, and I made a code that calculates Killing equation, and still got non zero result.
## Edit
I'll write out what I get for $\tau\tau$ component.
So, my Killing vector $\xi_2$ has three nonzero components
$\xi^\tau=\frac{2r\sin\tau}{\sqrt{1+r^2}}$, $\xi^r=-2\sqrt{1+r^2}\cos\tau$, $\xi^\varphi=\frac{2\sin\tau}{\sqrt{1+r^2}}$
And the $\tau\tau$ part of Lie derivative is
$$\mathcal{L}_\xi g_{\tau\tau}=\xi^\sigma\partial_\sigma g_{\tau\tau}+g_{\sigma\tau}\partial_\tau\xi^\sigma+g_{\tau\sigma}\partial_\tau\xi^\sigma$$
Only nonzero metric components that can be used are $g_{\tau\tau}$, and $g_{\tau\varphi}=g_{\varphi\tau}$.
They are
$$g_{\tau\tau}=-2GJ\Omega(\theta)^2(1+r^2(1-\Lambda(\theta)^2))$$
$$g_{\varphi\tau}=4GJr\Omega(\theta)^2\Lambda(\theta)^2$$
So, for the first part of Lie derivative, since $g_{\tau\tau}$ depends only on $r$ and $\theta$ my $\xi^\sigma$ is only $\xi^r$, since $\xi^\theta=0$. And the second part we have 2 times the ($g_{\tau\tau}\partial_\tau\xi^\tau+g_{\varphi\tau}\partial_\tau\xi^\varphi$). Which, after some simplifying becomes
$$\mathcal{L}_\xi g_{\tau\tau}=\frac{8 G J r \Lambda (\theta )^2 \Omega (\theta )^2 \cos (\tau )}{\sqrt{r^2+1}}$$
• @Qmechanic I don't think this question is considered as "homework", since it (1) arose from reading a research paper and (2) it isn't exactly about solving a problem, . – Abhimanyu Pallavi Sudhir Nov 13 '13 at 13:23
About the symmetry, the part : $-(1+r^2)d\tau^2+\frac{dr^2}{1+r^2}$ is just the metrics of AdS2, so there is a $SO(2,1)$ symmetry.
At fixed $\theta$, we have this $SO(2,1) \sim SL(2,R)$ symmetry plus the $U(1)$ symmetry corresponding to the invariance by $\phi$ translation.
The $\theta$ parameters are playing the role of geometrical terms, but do not change the nature of the symmetry.
Ref, Chapter 2, page 3
[EDIT]
Correcting OP calculus:
$\mathcal{L}_\xi g_{\tau\tau}=\xi^\sigma\partial_\sigma g_{\tau\tau}+g_{\sigma\tau}\partial_\tau\xi^\sigma+g_{\tau\sigma}\partial_\tau\xi^\sigma \tag{1}$
$g_{\tau\tau}$ depends only on $r$, and the only non-null metrics $g_{\sigma\tau} =g_{\tau\sigma}$,are $g_{\tau\tau},g_{\tau\varphi} =g_{\varphi\tau}$, so finally :
$\mathcal{L}_\xi g_{\tau\tau}=\xi^r\partial_r g_{\tau\tau}+ 2(g_{\tau\tau}\partial_\tau\xi^\tau+g_{\tau\varphi}\partial_\tau\xi^\varphi )\tag{2}$
The first term is equal to :
$$\xi^r\partial_r g_{\tau\tau} = [2 GJ \Omega^2][-2\sqrt{1+r^2}\cos\tau][2(\Lambda^2-1)r]\tag{3}$$
The second term is equal to :
$$2g_{\tau\tau}\partial_\tau\xi^\tau = [2 GJ \Omega^2]2[-1 +(\Lambda^2-1)r^2][\frac{2r\cos\tau}{\sqrt{1+r^2}}]\tag{4}$$
The third term is equal to :
$$2g_{\tau\varphi}\partial_\tau\xi^\varphi = [2 GJ \Omega^2]2[\Lambda^2r][\frac{2\cos\tau}{\sqrt{1+r^2}}]\tag{5}$$
Finally we have :
$$\mathcal{L}_\xi g_{\tau\tau}= \\ \frac{(8 GJ \Omega^2)(r \cos\tau)}{\sqrt{1+r^2}}[-(1+r^2)(\Lambda^2-1)+(-1 +(\Lambda^2-1)r^2)+ \Lambda^2]=0\tag{6}$$
• So that's the 'enhancing' part? – dingo_d Sep 29 '13 at 18:06
• $SL(2,R) * U(1)$ is not the symmetry of the standard Kerr metrics, it is only correct near horizon, so one says, that, near horizon, there is an "enhanced" symmetry, which is greater than the standard symmetry for the Kerr metrics. – Trimok Sep 29 '13 at 18:10
• Oh, that makes sense now :) I'll read the part about Killing vectors in that paper too, maybe there will be an answer there about the second part :) Thanks – dingo_d Sep 29 '13 at 18:13
• For which $\mu, \nu$, do you get a non-zero result for $\mathcal{L}_\xi g_{\mu\nu}$ ? – Trimok Sep 29 '13 at 18:22
• When using $\xi_2$ I have nonzero: $\tau\tau$, $\tau r$ $\tau\varphi$, $r\tau$, $r\varphi$ and $\varphi\tau$, $\varphi r$ – dingo_d Sep 29 '13 at 18:31 | {} |
# ML Wiki
## Document Classification
Document/Text Classification/Categorization is an NLP/Text Mining task of labeling unseen documents with categories from some predefined set
### Formulation
Document Classification, most general formulation:
• a task of assigning a boolean value to each pair $\langle d_j, c_i \rangle \in D \times C$
• $D$ - domain of documents
• $C = \{ c_1, \ ... \ , c_K \}$ - predefined categories
• we assign TRUE to $\langle d_j, c_i \rangle$ if document $d_j$ belongs to category $c_i$
• the task is to approximate the unknown target function $\Phi: D \times C \to \{ \text{T}, \text{F} \}$
Types
• non overlapping categories, i.e. can assign only one label to each document.
• the unknown function becomes $\Phi: D \to C$
• overlapping - can assign several labels
Hard categorization vs Ranking
• instead of assigning true/false we can rank each category
## Features
### Vector Space Model
How do we come up with features?
• text cannot be fed to the classifier directly
• so need to do some indexing to map a document $d_j$ to some representation
• for example, to Vector Space Model: represent document as a vector of term weights - "features"
• $d_j = \langle w_{1j}, \ ... \ , w_{nj} \rangle$, with $w_{ij}$ telling how much term $t_i$ contributes to the semantics of document $d_j$
### TF-IDF
Weights are often term frequencies or TF-IDF:
• $\text{tf-idf}(t_k, d_j) = \text{tf}(t_k, d_j) \cdot \log \cfrac{N}{\text{df}(t_k)}$
• $\text{tf}(t_k, d_j)$ term frequency: how many times term $t_k$ appeared in document $d_j$
• $\text{df}(t_k)$ document frequency: how many documents in the training set contain term $t_k$
• $N$ number of documents in the training set
Why TF-IDF is good for classification:
• the more often a term occurs in a document, the more representative it is of this document
• the more documents contain a term, the lest discriminating it becomes
### Dimensionality Reduction
• Given a vocabulary/term set $V$ of size $| V |$
• the goal is to find $V'$ s.t. $| V' | \ll | V |$ ($V'$ is called "reduced vocabulary" or "reduced term set")
• DR techniques tend to reduce Overfitting:
• if dimensionality of data is $|V'|$ and there are $N$ examples in the training set
• then it's good to have $|V'| \approx N$ to avoid overfitting
We can divide dimensionality reduction techniques by locality:
• local dimensionality reduction
• applied to each category $c_i$
• choose a reduced set $| V'_i | \ll | V_i |$ for each category
• global: choose $| V' |$ using all categories
There are two (very different) types of dimensionality reduction
• by term selection (Feature Selection): select $V' \subset V$
• by term extraction: terms in $V'$ are not necessarily the same as in $V$
Usual IR and indexing techniques for reducing dimensionality are
Stop Words Removal
• Before indexing some function words are sometimes removed
• for example Stop Words - topic neutral words such as articles, prepositions, conjunctions
• cases when stop words are not removed: author identification ("the little words give authors away")
• Stemming is grouping words that share the same morphological root
• it's controversial whether it's helpful for document classification or not
• usually it's used: it reduces the dimensionality
• sometimes Lemmatization is applied instead, but it's more involved
Note that DR techniques sometimes may remove important information when removing terms
How to select terms?
• Subset Selection: usually not used because $| V |$ is too large
• Feature Filtering: rank terms according to their "usefulness" and keep only some of them
• Document Frequency: Keep only terms that occur in higher number of documents
• e.g. remove words that occur only in 3 documents or less
Term Extraction techniques:
• these techniques create "artificial" terms that aren't really terms - they are generated, and not the ones that actually occurred in the text
• The original terms don't have the optimal dimensionality for document content representation
• because of the problems of polysemy, homonymy and synonymy
• so we want to find better representation that doesn't suffer from these issues
• methods:
• Term Clustering cluster terms and use centroids instead of words
• Latent Semantic Analysis apply SVD to Term-Document matrix
## Classification
Good classifiers for text:
## Evaluation
Precision and Recall metrics can be extended to Evaluation of Multiclass Classifiers
• similar to the One-vs-All Classification technique
• ways of averaging the results:
• micro: first calculate TP, FP, FN, FN for each category separately, and then use usual formulas for precision and recall
• and macro averaging: calculate precision and recall for each category separately, and then average
• usually is a way of combining precision and recall
• depending on how P and R were calculated, there are $F_\beta$-micro and $F_\beta$-macro measures
## Sources
• Sebastiani, Fabrizio. "Machine learning in automated text categorization." (2002). [1] | {} |
[Rexle Topics][ReferancerDraw the major organic product of the reaction shown below: OH HzSO4You do not have t0 consider stercochemistry:
Question:
[Rexle Topics] [Referancer Draw the major organic product of the reaction shown below: OH HzSO4 You do not have t0 consider stercochemistry: You do not have t0 explicitly draw H atoms In cases where there is more than one answer; just draw one. ChemDoocle; pdt Alkyl Rins 227_ pdl Rxn 26 227 Oc. pal
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Constants Periodic TableAlpha particles of charge q = +2e and mass m = 6.6 x 10-27 kg are emitted from radioactive source at speed of Sx10" m/sPart AWhat magnetic field strength would be required to bend them into circular path of radius r =0.16 m Express your answer using two signiticant figures:AzdSubmitPrevlous Answers Request AnSwerIncorrect; Try Again; attempts remainingProvide FeedbackNext
Constants Periodic Table Alpha particles of charge q = +2e and mass m = 6.6 x 10-27 kg are emitted from radioactive source at speed of Sx10" m/s Part A What magnetic field strength would be required to bend them into circular path of radius r =0.16 m Express your answer using two signiticant fi...
Reduction of NADP+ Question 10 1 pts The majority of G3P (glyceraldehyde 3-phosphate) produced in the...
reduction of NADP+ Question 10 1 pts The majority of G3P (glyceraldehyde 3-phosphate) produced in the Calvin cycle are used to regenerate RuBP. True False No new data to save, Last checked at 3:29pm Submit...
0 5A Question 8 Find the 1 rate 27 Moving n 2C 6 expression another 1ALA] Jv JO the question hypothetical 0 save this 2 reaction response V
0 5A Question 8 Find the 1 rate 27 Moving n 2C 6 expression another 1ALA] Jv JO the question hypothetical 0 save this 2 reaction response V...
Question 5167 ptsCalculate the average rate of change of f (} 6t3 ton the interval [-1,4]
Question 5 167 pts Calculate the average rate of change of f (} 6t3 ton the interval [-1,4]...
138. A 6.50-g sample of diprotic acid requires 137.5 mL of 0.750 M NaOH solution for complete neutralization Determine the molar mass of the acid_
138. A 6.50-g sample of diprotic acid requires 137.5 mL of 0.750 M NaOH solution for complete neutralization Determine the molar mass of the acid_...
FeCl;What is the structure of one of the major expected disubstituted benzene products of the =_ reaction of aniline with Clz FeClyand will the rate of this reaction be faster or slower versus = similar reaction that uses benzene as the starting material?4-Chloroaniline; slower ate ofreaction versus benzene4-Chloroaniline; faster rate of reaction versus benzene3-Chloroaniline; slower rate of reaction versus benzene3-Chloroaniline; (aster rate ot reaction versus benzeneNHi
FeCl; What is the structure of one of the major expected disubstituted benzene products of the =_ reaction of aniline with Clz FeClyand will the rate of this reaction be faster or slower versus = similar reaction that uses benzene as the starting material? 4-Chloroaniline; slower ate ofreaction vers...
How do you find the exact values of costheta and sintheta when tantheta=5/12?
How do you find the exact values of costheta and sintheta when tantheta=5/12?...
Mceccmooms.net inelastic; less than 1.0. 14. According to the textbook, which of the following statements is...
mceccmooms.net inelastic; less than 1.0. 14. According to the textbook, which of the following statements is (are) corect? (x) If the quantity demanded increases by a larger percentage than the percentage decrease in the price of the good, then the price elasticity of demand coefficient is a number ...
9 or 16 (9 complete) X 2.6.53 For f(x)=x +5 and g(x) = 5x +4, find...
9 or 16 (9 complete) X 2.6.53 For f(x)=x +5 and g(x) = 5x +4, find the following functions. a. (fog)(x); b. (gof)(x); c. (fog)(2); d. (gof)(2) a. (fog)(x) = (Simplify your answer.) Enter your answer in the answer box and then click cha...
Q. (a) Solve the following initial-value problem: d + 25x 10 cos 7t dt2with initial conditions x(O)-0, and x'(O)-0.(b) Use calculator or mathematica to obtain the graph of solution x(t):
Q. (a) Solve the following initial-value problem: d + 25x 10 cos 7t dt2 with initial conditions x(O)-0, and x'(O)-0. (b) Use calculator or mathematica to obtain the graph of solution x(t):...
Statics Question. Please answer fast Problem 4: For the truss loaded as shown. Determine the forces in members EF, EG, and FG. State whether each of these members is in tension (T) or compression (C). You may use whichever method you would like to solve. H 1200 N 4 m A с E 4 m B D F 3 m 3...
How do you solve the literal equation for y: y +5x=17?
How do you solve the literal equation for y: y +5x=17#?...
The 0.02 kg bullet is travelling at 400 m/s when it becomes embedded in the 2...
The 0.02 kg bullet is travelling at 400 m/s when it becomes embedded in the 2 kg stationary block. The coefficient of kinetic friction between the block and plane is 0.2. A. Determine the velocity of the bullet and block just after the collision. B. Use the principle of linear impulse and momentum t...
Q6 (a) Explain briefly the different between incompressible and compressible fluid flow (5 marks) (6) Air...
Q6 (a) Explain briefly the different between incompressible and compressible fluid flow (5 marks) (6) Air at pressure and temperature of 200 kPa, 373.2 K flows through a duct at Mach Number of 0.8. The gas constant and specific heat ratio of air are 0.287 kJ/kg k, 1.4 respectively. Determine, (1) ai...
46. Starting with one DNA segment;, how many identical DNA segments are present after ? comnlete PCR cxcles 18 36 512 1024
46. Starting with one DNA segment;, how many identical DNA segments are present after ? comnlete PCR cxcles 18 36 512 1024...
Suppose a researcher is testing the hypothesis Ho: p = 0.6 versus H1:p*0.6 and she finds...
Suppose a researcher is testing the hypothesis Ho: p = 0.6 versus H1:p*0.6 and she finds the P-value to be 0.29. Explain what this means. Would she reject the null hypothesis? Why? Choose the correct explanation below. O A. If the P-value for a particular test statistic is 0.29, she expects results ...
3. (10 points) The table below summarizes data from 235 residents of The Summer Isles. The...
3. (10 points) The table below summarizes data from 235 residents of The Summer Isles. The table indicates languages that the residents are fluent in. Use the table to answer the questions below. Fluent in Dothraki Yes No 5 175 26 29 Fluent in High Valyrian Yes No a) What is the probability of rando...
(6) Identify by name and draw the graph of the surface defined by the function below. z = 4x2 + 9y2 +1
(6) Identify by name and draw the graph of the surface defined by the function below. z = 4x2 + 9y2 +1...
A new material will replace the old material unless there is significant evidence that the mean strength of the new material is not more than 3 units higher than the mean strength of the old material.Sample from new material: mean is 105.591SD is 4.2and sample size is 12Sample from old material: mean is 103.371 We do not know population standard deviations_ Consider the testSD is 4.9and sample size is 70Ho HnewHold = 3 VS. Ho HnewHoldWhat is the p-value? sample T and Z test)(Hint: best to use so
A new material will replace the old material unless there is significant evidence that the mean strength of the new material is not more than 3 units higher than the mean strength of the old material. Sample from new material: mean is 105.591 SD is 4.2 and sample size is 12 Sample from old material:...
Why is it difficult to measure tourism’s contribution to GDP?
why is it difficult to measure tourism’s contribution to GDP?... | {} |
# Saha¶
plasmapy.formulary.ionization.Saha(g_j, g_k, n_e: Unit(‘1 / m3’), E_jk: Unit(‘J’), T_e: Unit(‘K’))
Return the ratio of populations of two ionization states.
The Saha equation, derived in statistical mechanics, gives an approximation of the ratio of population of ions in two different ionization states in a plasma. This approximation applies to plasmas in thermodynamic equilibrium where ionization and recombination of ions with electrons are balanced.
$\frac{N_j}{N_k} = \frac{1}{n_e} \frac{g_j}{4 g_k a_0^{3}} \left( \frac{k_B T_e}{π E_H} \right)^{\frac{3}{2}} \exp\left( \frac{-E_{jk}}{k_B T_e} \right)$
Where $$k_B$$ is the Boltzmann constant, $$a_0$$ is the Bohr radius, $$E_H$$ is the ionization energy of hydrogen, $$N_j$$ and $$N_k$$ are the population of ions in the $$j$$ and $$k$$ states respectively. This function is equivalent to Eq. 3.47 in Drake.
Parameters
Warns
UnitsWarning – If units are not provided, SI units are assumed.
Raises
Returns
ratio – The ratio of population of ions in ionization state $$j$$ to state $$k$$.
Return type
Quantity
Examples
>>> import astropy.units as u
>>> T_e = 5000 * u.K
>>> n = 1e19 * u.m ** -3
>>> g_j = 2
>>> g_k = 2
>>> E_jk = 1 * u.Ry
>>> Saha(g_j, g_k, n, E_jk, T_e)
<Quantity 3.299...e-06>
>>> T_e = 1 * u.Ry
>>> n = 1e23 * u.m ** -3
>>> Saha(g_j, g_k, n, E_jk, T_e)
<Quantity 1114595.586...>
Notes
For reference to this function and for more information regarding the Saha equation, see chapter 3 of R. Paul Drake’s book, “High-Energy-Density Physics: Foundation of Inertial Fusion and Experimental Astrophysics” (DOI: 10.1007/978-3-319-67711-8_3). | {} |
# Spiralling around the Earth
Tags:
1. Apr 22, 2015
### kitsh
1. The problem statement, all variables and given/known data
An airplane flies from the North Pole to the South Pole, following a winding trajectory. Place the center of the Earth at the origin of your coordinate system, and align the south-to-north axis of the Earth with your z axis. The pilot’s trajectory can then be described as follows:
1) The plane’s trajectory is confined to a sphere of radius R centered on the origin.
2) The pilot maintains a constant velocity v in the -z direction, thus the z coordinate can be described as z(t)=R-vt
3) The pilot "winds" around the Earth as she travels south, covering a constant ω radians per second in the azimuthal angle ϕ, thus ϕ(t)=ωt
Calculate the total distance traveled by the pilot. What you will find is that time t is not the best IP with which to parametrize this path. You can start with it, certainly … but then get rid of it in terms of a different choice for your IP: θ, from spherical coordinates
2. Relevant equations
Spherical coordinates are (r, θ, ϕ)
The Answer should be in the form of ∫A√(1+B^2(sin(θ))^n)dθ where A, B and n are either numerical constants or constants in in the terms of R, v and ω
3. The attempt at a solution
I honestly have no idea how I am supposed to approach this question, it is nothing like anything I have seen in this class or any other
2. Apr 22, 2015
### HallsofIvy
Staff Emeritus
I would start with spherical coordinates. With constant radius, R, $x= R cos(\theta)sin(\phi)$, $y= R sin(\theta)sin(\phi)$, $z= R cos(phi)$. Here we have $z= R cos(\phi)= R- vt$ and $\phi= \omega t[/te]x] so [itex]z= R cos(\omega t)= R- vt$
3. Apr 22, 2015
### kitsh
So I converted everything to spherical which helps some and I know the bounds of integration are going to be from 0 to pi and that the integration constant is rdθ but I still can't figure out how to get the integral into the form indicated above.
4. Apr 22, 2015 | {} |
Most Difficult Application - Poll : The B-School Application
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Most Difficult Application - Poll
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Director
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21 Aug 2008, 17:48
Now that all of the applications are out, just curious as to what the hardest application is this year? Share the specific question that is giving you the most trouble and why, below.
Last edited by jb32 on 21 Aug 2008, 17:56, edited 1 time in total.
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21 Aug 2008, 17:49
Yale! Not on your list, followed by Chicago. I thank my stars I'm not applying to those schools.
I certainly think Yale's essays ask about some really specific things and allow very little leeway to use material from other school's applications. Ditto for Chicago.
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21 Aug 2008, 17:56
Unfortunately I could only put ten on the list, but I'm replacing Anderson with Yale.
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21 Aug 2008, 18:36
Chicago's essay 2 is the toughest one I'll be encountering. Stanford's infamous #1 is probably tougher, but I'm not applying there
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21 Aug 2008, 21:07
#1 Stanford
#2 Harvard
#3 Chicago
Everything else is a cakewalk compared to these.
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22 Aug 2008, 02:37
for me its Oxford, solely because of essay 2.
"What recent development, world event or book has most influenced your thinking and why? 2000 Words"
The question hasn't changed in 3 years at least and it has been bugging me since March of this year. I can't put down my thinking and influence to one event or book. Any ideas?
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22 Aug 2008, 02:39
The Olympics in China? Globalization is a tried and tested them on b-school applications.
for me its Oxford, solely because of essay 2.
"What recent development, world event or book has most influenced your thinking and why? 2000 Words"
The question hasn't changed in 3 years at least and it has been bugging me since March of this year. I can't put down my thinking and influence to one event or book. Any ideas?
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22 Aug 2008, 02:42
solaris1 wrote:
The Olympics in China? Globalization is a tried and tested them on b-school applications.
for me its Oxford, solely because of essay 2.
"What recent development, world event or book has most influenced your thinking and why? 2000 Words"
The question hasn't changed in 3 years at least and it has been bugging me since March of this year. I can't put down my thinking and influence to one event or book. Any ideas?
just imagine how many chinese applicants will have that theme....and besides its hard to write 2000 words about something that has really not changed your thinking.
And the only book i've read recently is the GMAT OG
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22 Aug 2008, 03:11
a lot of developments.
Islamic fundamentalism - good as party talk, not sure if it gels in an essay
Paris Hilton - always a good choice..provided it goes with pics and videos
Microfinance - hot topic yesterday, cold potato tomorrow
America's Best Dance Crew - as usual proving the theory white men cannot dance.
and my favorite......
Kim Kardashian - a new development that is still developing...
as usual I have hijacked the thread..now back to regular programming
I should throw in LBS essays..not that they are not hard, but with word limits like 200, 400, and 500.. you have to scratch your head to figure out what to write without using conjunctions, propositions and the occassional verb and still make sense
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22 Aug 2008, 06:50
jb32 wrote:
Now that all of the applications are out, just curious as to what the hardest application is this year? Share the specific question that is giving you the most trouble and why, below.
Hey jb,
Is it only me or anyone else thinks that Darden application is tough this year due to their ridiculous word limits....... (1000 words for three essays)...... should be on the poll.....
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22 Aug 2008, 07:01
For me, that's an easy question: What Got You Here Won't Get You There by Marshall Goldsmith.
http://www.amazon.com/What-Got-Here-Won ... 289&sr=1-1
This book got me to thinking about what I've done with my career, how I came to the point where I am currently am, and where I want to go. It's an awesome book. I highlights what people do subconsciously to sabatoge their own careers and they don't even know it. Marshall Goldsmith is an executive coach. He walks the reader through the different characteristics he sees in people with huge potential, but they lack in some fundamental area that is holding them back.
for me its Oxford, solely because of essay 2.
"What recent development, world event or book has most influenced your thinking and why? 2000 Words"
The question hasn't changed in 3 years at least and it has been bugging me since March of this year. I can't put down my thinking and influence to one event or book. Any ideas?
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22 Aug 2008, 08:03
Note first, to avoid comments: I don't wanna give even slightest impression that I'll get admitted there or even get close to admissions (i.e. interview). But my $0.02 below... What's so difficult about Harvard or Stanford? Maybe that's only me, but I really have an impression that people exaggerate H/S simply b/c of their rankings/reputation. What do you find so difficult about Harvard (below)? 1: standard. 2: standard. 3-6: nice mixture of standard and creative so something nice for everybody. 1. What are your three most substantial accomplishments and why do you view them as such? (600-word limit) 2. What have you learned from a mistake? (400-word limit) Please respond to two of the following (400-word limit each): 3. What would you like the MBA Admissions Board to know about your undergraduate academic experience? 4. Discuss how you have engaged with a community or organization. 5. What area of the world are you most curious about and why? 6. What is your career vision and why is this choice meaningful to you? Stanford... different (1-2) and standard (3-6). Anyway better than boring standard stuff other schools ask. No BS here (i.e. short-/mid-term career vision which anyway will change while at school and majority writes this type of essay in a very 'acceptable' way - how do I look honest, but in reality give AdCom an impression of 'employability' after MBA). At least for Stanford you really write what your heart tells you (for essays 1-2). So all in all I really don't get it... what so difficult here? Chicago no.2 - that's different story. Boring and nightmare to write. SVP Joined: 11 Mar 2008 Posts: 1634 Location: Southern California Schools: Chicago (dinged), Tuck (November), Columbia (RD) Followers: 9 Kudos [?]: 201 [0], given: 0 Re: Most Difficult Application - Poll [#permalink] Show Tags 22 Aug 2008, 08:25 garbus222 wrote: Note first, to avoid comments: I don't wanna give even slightest impression that I'll get admitted there or even get close to admissions (i.e. interview). But my$0.02 below...
What's so difficult about Harvard or Stanford? Maybe that's only me, but I really have an impression that people exaggerate H/S simply b/c of their rankings/reputation.
What do you find so difficult about Harvard (below)? 1: standard. 2: standard. 3-6: nice mixture of standard and creative so something nice for everybody.
1. What are your three most substantial accomplishments and why do you view them as such? (600-word limit)
2. What have you learned from a mistake? (400-word limit)
Please respond to two of the following (400-word limit each):
4. Discuss how you have engaged with a community or organization.
5. What area of the world are you most curious about and why?
6. What is your career vision and why is this choice meaningful to you?
Stanford... different (1-2) and standard (3-6). Anyway better than boring standard stuff other schools ask. No BS here (i.e. short-/mid-term career vision which anyway will change while at school and majority writes this type of essay in a very 'acceptable' way - how do I look honest, but in reality give AdCom an impression of 'employability' after MBA). At least for Stanford you really write what your heart tells you (for essays 1-2).
So all in all I really don't get it... what so difficult here? Chicago no.2 - that's different story. Boring and nightmare to write.
#1 is a very difficult question - just like Stanford's "What matters to you most and why". It requires a lot of introspection and digging deep - unless you're a superstar and you can rattle off those 3 accomplishments without much effort. The other difficult part about HBS is the word limits - 200 words for each of your 3 greatest accomplishments is tough.
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22 Aug 2008, 08:40
Agree that word limit is a challenge at Harvard (more than at other schools).
As for H1... B/c it's Harvard, majority of us will write about those 3 accomplishments that resulted from 3 most significant leadership experiences. So pretty much this essay is about leadership. If so, lots of other schools ask for leadership (Wharton, Kellogg, Tuck to name few) so for me H1 is not any different than Kellogg3 etc.
As for S1... I find it relatively easy as deep down I know very well what I really care about. I think everybody knows very well. Admitting it is a different story (e.g. difficult to admit openly that I might be driven by such a 'shallow' values as power or prestige). So at the end of the day the difficult part is to pick, from the list of all our values, the ones which are ok to write about in MBA application.
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22 Aug 2008, 08:42
garbus222 wrote:
Note first, to avoid comments: I don't wanna give even slightest impression that I'll get admitted there or even get close to admissions (i.e. interview). But my \$0.02 below...
What's so difficult about Harvard or Stanford? Maybe that's only me, but I really have an impression that people exaggerate H/S simply b/c of their rankings/reputation.
What do you find so difficult about Harvard (below)? 1: standard. 2: standard. 3-6: nice mixture of standard and creative so something nice for everybody.
1. What are your three most substantial accomplishments and why do you view them as such? (600-word limit)
2. What have you learned from a mistake? (400-word limit)
Please respond to two of the following (400-word limit each):
4. Discuss how you have engaged with a community or organization.
5. What area of the world are you most curious about and why?
6. What is your career vision and why is this choice meaningful to you?
Stanford... different (1-2) and standard (3-6). Anyway better than boring standard stuff other schools ask. No BS here (i.e. short-/mid-term career vision which anyway will change while at school and majority writes this type of essay in a very 'acceptable' way - how do I look honest, but in reality give AdCom an impression of 'employability' after MBA). At least for Stanford you really write what your heart tells you (for essays 1-2).
So all in all I really don't get it... what so difficult here? Chicago no.2 - that's different story. Boring and nightmare to write.
I agree the HBS questions don't look especially tough to me. Of course, HBS is out of my league and I'm not even applying there but I think Chicago's app is WAY tougher.
Also, I wanted to respond on the comment about the Yale essays. I actually thought they seemed relatively easy (relative is the key word here, I think the essays for every school are tough so I'm just talking in relative terms here). For essay 3 they give you 7 options and then for essay 4 you can choose another of those 7 or make up any question you want. That gives you a lot of leeway to find a couple essays you can borrow from other schools. As with most schools you can also borrow a lot for essay 1, the career essay, except that Yale does add a twist about your impact on the world which is pretty tough.
I voted for Chicago. I think essay 2 is the toughest essay of any school. And for non creative types, like myself, the powerpoint question is probably the 2nd hardest of any school. I'm dreading this application.
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22 Aug 2008, 08:45
garbus222 wrote:
As for S1... I find it relatively easy as deep down I know very well what I really care about. I think everybody knows very well. Admitting it is a different story (e.g. difficult to admit openly that I might be driven by such a 'shallow' values as power or prestige). So at the end of the day the difficult part is to pick, from the list of all our values, the ones which are ok to write about in MBA application.
Let's not kid ourselves here, 90% of applicants are going to business school because they want a) more interesting/impactful/powerful work, b) higher compensation, or c) preferably both. However, you can't write this in the application and therefore it requires a lot of introspection by most people.
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22 Aug 2008, 08:47
Last year I hated MIT's application questions. They want a ton of detail but give you only 500 words. Even though its only 3 shorter essays it took me longer than Kelloggs which I had to write more than double the material.
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Kellogg Class of 2010...still active and willing to help. However, I do not do profile reviews, don't offer predictions on chances and am far to busy to review essays, so save the energy of writing me a PM seeking help for these. If I don't respond to a PM that is not one of the previously mentioned trash can destined messages, please don't take it personally I get so many messages I have a hard to responding to most. The more interesting, compelling, or humorous you message the more likely I am to respond.
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Re: Most Difficult Application - Poll [#permalink]
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22 Aug 2008, 08:50
terp06 wrote:
Let's not kid ourselves here, 90% of applicants are going to business school because they want a) more interesting/impactful/powerful work, b) higher compensation, or c) preferably both. However, you can't write this in the application and therefore it requires a lot of introspection by most people.
Agree. I withdraw therefore my previous statement about 'writing what heart tells you' for S. It's the same BS as for others - just wrapped differently
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Re: Most Difficult Application - Poll [#permalink]
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22 Aug 2008, 09:13
IHateTheGMAT wrote:
I voted for Chicago. I think essay 2 is the toughest essay of any school. And for non creative types, like myself, the powerpoint question is probably the 2nd hardest of any school. I'm dreading this application.
I personally thought before i began my Chicago app that essay 2 would be the toughest, but after finishing Chicago I'm not so sure anymore. It flowed much better when I sat down to write it. And while I sat in front of the computer for a week or so cursing and trying to figure out the PPT, I think in the end that it is the best question of any school. You can pretty much include anything you want to tell the rest of your story, fill in any gaps, etc. Your Chicago app (compared to your other apps) should be the most complete story about you because you have so much freedom to tell it.
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Re: Most Difficult Application - Poll [#permalink]
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22 Aug 2008, 10:15
jb32 wrote:
IHateTheGMAT wrote:
I voted for Chicago. I think essay 2 is the toughest essay of any school. And for non creative types, like myself, the powerpoint question is probably the 2nd hardest of any school. I'm dreading this application.
I personally thought before i began my Chicago app that essay 2 would be the toughest, but after finishing Chicago I'm not so sure anymore. It flowed much better when I sat down to write it. And while I sat in front of the computer for a week or so cursing and trying to figure out the PPT, I think in the end that it is the best question of any school. You can pretty much include anything you want to tell the rest of your story, fill in any gaps, etc. Your Chicago app (compared to your other apps) should be the most complete story about you because you have so much freedom to tell it.
I really hope your right. I haven't started my Chicago essays yet and would be very happy if I had the same experience! I've sort of pushed those essays down my list because they look so hard. I finished my Columbia essays, which were pretty tough but not impossible. And now I'm doing the Tuck essays, which are hard but fair. They are definitely very introspective but they are also pretty straightforward and give you the chance to tell your whole story.
Re: Most Difficult Application - Poll [#permalink] 22 Aug 2008, 10:15
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# 🤔 Active learning with ModAL and scikit-learn#
In this tutorial, we will walk through the process of building an active learning prototype with Rubrix, ModAL and scikit-learn.
• 💻 We train a spam filter using the YouTube Spam Collection data set.
• 🎩 For this we embed a lightweight scikit-learn classifier in an active learner via ModAL.
• 🏆 We design an active learning loop around Rubrix, to quickly build up a training data set from scratch.
## Introduction#
Active learning is a special case of machine learning in which a learning algorithm can interactively query a user (or some other information source) to label new data points with the desired outputs. In statistics literature, it is sometimes also called optimal experimental design. The information source is also called teacher or oracle. [Wikipedia]
In this tutorial our goal is to show you how to incorporate Rubrix into an active learning workflow involving a human in the loop. We will build a simple text classifier by combining scikit-learn, the active learning framework ModAL and Rubrix. Scikit-learn will provide the model that we will embed in an active learner from ModAL, and you and Rubrix will serve as the information source that teach the model to become a sample efficient classifier.
This tutorial is only a proof of concept for educational purposes and to inspire you with some ideas involving interactive learning processes, and how they can help to quickly build a training data set from scratch.
## Setup#
Rubrix, is a free and open-source tool to explore, annotate, and monitor data for NLP projects.
If you are new to Rubrix, check out the ⭐ Github repository.
If you have not installed and launched Rubrix yet, check the Setup and Installation guide.
For this tutorial we also need the third party libraries modAl, scikit-learn and matplotlib (optional), which can be installed via pip:
[ ]:
%pip install modAL scikit-learn matplotlib -qqq # matplotlib is optional
Rubrix allows you to log and track data for different NLP tasks, such as Token Classification or Text Classification.
In this tutorial, we will use the YouTube Spam Collection dataset, which is a binary classification task for detecting spam comments in YouTube videos.
Let’s load the data and have a look at it:
[1]:
import pandas as pd
[2]:
test_df
[2]:
COMMENT_ID AUTHOR DATE CONTENT CLASS VIDEO
0 z120djlhizeksdulo23mj5z52vjmxlhrk04 Murlock Nightcrawler 2015-05-24T07:04:29.844000 Charlie from LOST? 0 3
1 z133ibkihkmaj3bfq22rilaxmp2yt54nb Debora Favacho (Debora Sparkle) 2015-05-21T14:08:41.338000 BEST SONG EVER X3333333333 0 4
2 z12gxdortqzwhhqas04cfjrwituzghb5tvk0k Muhammad Asim Mansha NaN Aslamu Lykum... From Pakistan 1 3
3 _2viQ_Qnc6_ZYkMn1fS805Z6oy8ImeO6pSjMLAlwYfM mile panika 2013-11-03T14:39:42.248000 I absolutely adore watching football plus I’ve... 1 4
4 z120s1agtmmetler404cifqbxzvdx15idtw0k Sheila Cenabre 2014-08-19T12:33:11 I really love this video.. http://www.bubblews... 1 1
... ... ... ... ... ... ...
387 z13pup2w2k3rz1lxl04cf1a5qzavgvv51vg0k geraldine lopez 2015-05-20T23:44:25.920000 love the you lie the good 0 3
388 z13psdarpuzbjp1hh04cjfwgzonextlhf1w bilal bilo 2015-05-22T20:36:36.926000 I liked<br /> 0 4
389 z131xnwierifxxkj204cgvjxyo3oydb42r40k YULIOR ZAMORA 2014-09-10T01:35:54 I loved it so much ... 0 1
390 z12pwrxj0kfrwnxye04cjxtqntycd1yia44 ארז אריה 2015-05-15T19:46:53.719000 good party 0 2
391 z13oxvzqrzvyit00322jwtjo2tzqylhof04 Octavia W 2015-05-22T02:33:26.041000 Waka waka 0 4
392 rows × 6 columns
As we can see, the data contains the comment id, the author of the comment, the date, the content (the comment itself) and a class column that indicates if a comment is spam or ham. We will use the class column only in the test dataset to illustrate the effectiveness of the active learning approach with Rubrix. For the training dataset, we will simply ignore the column and assume that we are gathering training data from scratch.
## 2. Defining our classifier and Active Learner#
In this tutorial, we will use a multinomial Naive Bayes classifier that is suitable for classification with discrete features (e.g., word counts for text classification):
[3]:
from sklearn.naive_bayes import MultinomialNB
# Define our classification model
classifier = MultinomialNB()
Then, we will define our active learner, which uses the classifier as an estimator of the most uncertain predictions:
[4]:
from modAL.models import ActiveLearner
# Define active learner
learner = ActiveLearner(
estimator=classifier,
)
The features for our classifier will be the counts of different word n-grams: that is, for each example we count the number of contiguous sequences of n words, where n goes from 1 to 5.
The output of this operation will be matrices of n-gram counts for our train and test data set, where each element in a row equals the counts of a specific word n-gram found in the example:
[5]:
from sklearn.feature_extraction.text import CountVectorizer
# The resulting matrices will have the shape of (nr of examples, nr of word n-grams)
vectorizer = CountVectorizer(ngram_range=(1, 5))
X_train = vectorizer.fit_transform(train_df.CONTENT)
X_test = vectorizer.transform(test_df.CONTENT)
## 3. Active Learning loop#
Now we can start our active learning loop that consists of iterating over following steps:
1. Annotate samples
2. Teach the active learner
3. Plot the improvement (optional)
Before starting the learning loop, let us define two variables:
• the number of instances we want to annotate per iteration,
• a variable to keep track of our improvements by recording the achieved accuracy after each iteration.
[6]:
# Number of instances we want to annotate per iteration
n_instances = 10
# Accuracies after each iteration to keep track of our improvement
accuracies = []
### Step 1: Annotate samples#
The first step of the training loop is about annotating n examples having the most uncertain prediction. In the first iteration, these will be just random examples, since the classifier is still not trained and we do not have predictions yet.
[7]:
from sklearn.exceptions import NotFittedError
# query examples from our training pool with the most uncertain prediction
query_idx, query_inst = learner.query(X_train, n_instances=n_instances)
# get predictions for the queried examples
try:
probabilities = learner.predict_proba(X_train[query_idx])
# For the very first query we do not have any predictions
except NotFittedError:
probabilities = [[0.5, 0.5]]*n_instances
[ ]:
import rubrix as rb
# Build the Rubrix records
records = [
rb.TextClassificationRecord(
id=idx,
text=train_df.CONTENT.iloc[idx],
prediction=list(zip(["HAM", "SPAM"], probs)),
prediction_agent="MultinomialNB",
)
for idx, probs in zip(query_idx, probabilities)
]
# Log the records
rb.log(records, name="active_learning_tutorial")
After logging the records to Rubrix, we switch over to the UI, where we can find the newly logged examples in the active_learning_tutorial dataset.
To only show the examples that are still missing an annotation, you can select “Default” in the Status filter as shown in the screenshot below. After annotating a few examples you can press the Refresh button in the left side bar, to update the view with respect to the filters.
Once you are done annotating the examples, you can continue with the active learning loop. If your annotations only contained one class, consider increasing the n_instances parameter.
### Step 2: Teach the learner#
The second step in the loop is to teach the learner. Once we have trained our classifier with the newly annotated examples, we can apply the classifier to the test data and record the accuracy to keep track of our improvement.
[ ]:
# Load the annotated records into a pandas DataFrame
# check if all examples were annotated
if any(records_df.annotation.isna()):
# train the classifier with the newly annotated examples
y_train = records_df.annotation.map(lambda x: int(x == "SPAM"))
learner.teach(X=X_train[query_idx], y=y_train.to_list())
# Keep track of our improvement
accuracies.append(learner.score(X=X_test, y=test_df.CLASS))
Now go back to step 1 and repeat both steps a couple of times.
### Step 3. Plot the improvement (optional)#
After a few iterations, we can check the current performance of our classifier by plotting the accuracies. If you think the performance can still be improved, you can repeat step 1 and 2 and check the accuracy again.
[39]:
import matplotlib.pyplot as plt
# Plot the accuracy versus the iteration number
plt.plot(accuracies)
plt.xlabel("Number of iterations")
plt.ylabel("Accuracy");
## Summary#
In this tutorial we saw how to embed Rubrix in an active learning loop, and also how it can help you to gather a sample efficient data set by annotating only the most decisive examples. We created a rather minimalist active learning loop, but Rubrix does not really care about the complexity of the loop. It will always help you to record and annotate data examples with their model predictions, allowing you to quickly build up a data set from scratch.
## Next steps#
📚 Rubrix documentation for more guides and tutorials.
🙋♀️ Join the Rubrix community! A good place to start is the discussion forum.
## Bonus: Compare query strategies, random vs max uncertainty#
In this bonus, we quickly demonstrate the effectiveness of annotating only the most uncertain predictions compared to random annotations. So the next time you want to build a data set from scratch, keep this strategy in mind and maybe use Rubrix for the annotation process! 😀
[ ]:
import numpy as np
n_iterations = 150
n_instances = 10
random_samples = 50
# max uncertainty strategy
accuracies_max = []
for i in range(random_samples):
train_rnd_df = train_df#.sample(frac=1)
test_rnd_df = test_df#.sample(frac=1)
X_rnd_train = vectorizer.transform(train_rnd_df.CONTENT)
X_rnd_test = vectorizer.transform(test_rnd_df.CONTENT)
accuracies, learner = [], ActiveLearner(estimator=MultinomialNB())
for i in range(n_iterations):
query_idx, _ = learner.query(X_rnd_train, n_instances=n_instances)
learner.teach(X=X_rnd_train[query_idx], y=train_rnd_df.CLASS.iloc[query_idx].to_list())
accuracies.append(learner.score(X=X_rnd_test, y=test_rnd_df.CLASS))
accuracies_max.append(accuracies)
# random strategy
accuracies_rnd = []
for i in range(random_samples):
accuracies, learner = [], ActiveLearner(estimator=MultinomialNB())
for random_idx in np.random.choice(X_train.shape[0], size=(n_iterations, n_instances), replace=False):
learner.teach(X=X_train[random_idx], y=train_df.CLASS.iloc[random_idx].to_list())
accuracies.append(learner.score(X=X_test, y=test_df.CLASS))
accuracies_rnd.append(accuracies)
arr_max, arr_rnd = np.array(accuracies_max), np.array(accuracies_rnd)
[ ]:
plt.plot(range(n_iterations), arr_max.mean(0))
plt.fill_between(range(n_iterations), arr_max.mean(0)-arr_max.std(0), arr_max.mean(0)+arr_max.std(0), alpha=0.2)
plt.plot(range(n_iterations), arr_rnd.mean(0))
plt.fill_between(range(n_iterations), arr_rnd.mean(0)-arr_rnd.std(0), arr_rnd.mean(0)+arr_rnd.std(0), alpha=0.2)
plt.xlim(0,15)
plt.title("Sampling strategies: Max uncertainty vs random")
plt.xlabel("Number of annotation iterations")
plt.ylabel("Accuracy")
plt.legend(["max uncertainty", "random sampling"], loc=4)
<matplotlib.legend.Legend at 0x7fa38aaaab20>
## Appendix: How did we obtain the train/test data?#
[ ]:
import pandas as pd
from urllib import request
from sklearn.model_selection import train_test_split
from pathlib import Path
from tempfile import TemporaryDirectory
"""
and returns the data as a tuple with a train and test DataFrame.
"""
], None
with TemporaryDirectory() as tmpdirname:
dfs = []
file = Path(tmpdirname) / f"{i}.csv" | {} |
# Using a smooth function to replace a coordinate on a manifold
I don't know very much about differential geometry, so this may be trivial but I will try to make this question as clear as I can.
Suppose I have a $$d$$ dimensional smooth manifold $$\mathcal{M}^d$$, with a smooth maximal atlas $$\mathcal{A}$$. In the neighborhood of a point $$m$$ I have a chart $$(U,\phi) \in \mathcal{A}$$ that gives some local coordinates for my manifold, say $$\phi(m) = (x^1(m), ..., x^d(m))$$ and $$\phi : U \subseteq \mathcal{M}^d \rightarrow \mathbb{R}^d$$ is a homeomorphism. Suppose also that I have a smooth function defined on my manifold, $$f: \mathcal{M}\rightarrow \mathbb{R}$$.
My question is the following: Are there any conditions that guarantee that the chart $$(V, \psi)$$, with $$m \in U \cap V$$, $$\psi = (x^1, ..., f)$$ is also in $$\mathcal{A}$$?
In other words, how can I be sure that I can replace one of my coordinates with the smooth function $$f$$?
The way the IFT is written there is that some local functions $$(f_1,\dots,f_n):U\to\mathbb R^n$$ define a coordinate chart around $$p\in U$$ iff there's a chart $$(x_1,\dots,x_n)$$ around $$p$$ such that $$\det\begin{bmatrix}\dfrac{\partial f_i}{\partial x^j}\end{bmatrix}\neq 0$$
So we get a sufficient a condition: $$(x_1,\dots,x_{n-1},f)$$ defines a chart if $$\det \begin{bmatrix} \dfrac{\partial x_1}{\partial x^1} &\cdots &\dfrac{\partial x_1}{\partial x^{n-1}}&\dfrac{\partial x_1}{\partial x^n}\\ \vdots & \ddots &\vdots & \vdots \\ \dfrac{\partial x_{n-1}}{\partial x^1} &\cdots &\dfrac{\partial x_{n-1}}{\partial x^{n-1}}&\dfrac{\partial x_{n-1}}{\partial x^n}\\ \dfrac{\partial f}{\partial x^1}&\cdots&\dfrac{\partial f}{\partial x^{n-1}}&\dfrac{\partial f}{\partial x^n} \end{bmatrix} = \det \begin{bmatrix} 1 & \cdots &0&0\\ \vdots & \ddots &\vdots & \vdots \\ 0 &\cdots &1&0\\ \dfrac{\partial f}{\partial x^1}&\cdots&\cdots&\dfrac{\partial f}{\partial x^n} \end{bmatrix} = \dfrac{\partial f}{\partial x^n} \neq 0.$$ | {} |
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$\dfrac{9}{a+3}$
When adding two fractions with common denominators, we add the numerators and copy the denominator. Therefore, $\dfrac{4}{a+3}+\dfrac{5}{a+3} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{4+5}{a+3} \\\\= \dfrac{9}{a+3} .\end{array} | {} |
Outlook: Inca One Gold Corp. assigned short-term Ba1 & long-term Ba1 estimated rating.
Time series to forecast n: 26 Dec 2022 for (n+8 weeks)
Methodology : Modular Neural Network (Emotional Trigger/Responses Analysis)
## Abstract
The aim of this study is to evaluate the effectiveness of using external indicators, such as commodity prices and currency exchange rates, in predicting movements. The performance of each technique is evaluated using different domain specific metrics. A comprehensive evaluation procedure is described, involving the use of trading simulations to assess the practical value of predictive models, and comparison with simple benchmarks that respond to underlying market growth.(Yuan, X., Yuan, J., Jiang, T. and Ain, Q.U., 2020. Integrated long-term stock selection models based on feature selection and machine learning algorithms for China stock market. IEEE Access, 8, pp.22672-22685.) We evaluate Inca One Gold Corp. prediction models with Modular Neural Network (Emotional Trigger/Responses Analysis) and Multiple Regression1,2,3,4 and conclude that the INCA:TSXV stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Buy
## Key Points
1. Fundemental Analysis with Algorithmic Trading
2. Understanding Buy, Sell, and Hold Ratings
3. What is the use of Markov decision process?
## INCA:TSXV Target Price Prediction Modeling Methodology
We consider Inca One Gold Corp. Decision Process with Modular Neural Network (Emotional Trigger/Responses Analysis) where A is the set of discrete actions of INCA:TSXV stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Multiple Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Emotional Trigger/Responses Analysis)) X S(n):→ (n+8 weeks) $\stackrel{\to }{R}=\left({r}_{1},{r}_{2},{r}_{3}\right)$
n:Time series to forecast
p:Price signals of INCA:TSXV stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## INCA:TSXV Stock Forecast (Buy or Sell) for (n+8 weeks)
Sample Set: Neural Network
Stock/Index: INCA:TSXV Inca One Gold Corp.
Time series to forecast n: 26 Dec 2022 for (n+8 weeks)
According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Buy
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for Inca One Gold Corp.
1. Subject to the conditions in paragraphs 4.1.5 and 4.2.2, this Standard allows an entity to designate a financial asset, a financial liability, or a group of financial instruments (financial assets, financial liabilities or both) as at fair value through profit or loss provided that doing so results in more relevant information.
2. An entity shall apply this Standard for annual periods beginning on or after 1 January 2018. Earlier application is permitted. If an entity elects to apply this Standard early, it must disclose that fact and apply all of the requirements in this Standard at the same time (but see also paragraphs 7.1.2, 7.2.21 and 7.3.2). It shall also, at the same time, apply the amendments in Appendix C.
3. When using historical credit loss experience in estimating expected credit losses, it is important that information about historical credit loss rates is applied to groups that are defined in a manner that is consistent with the groups for which the historical credit loss rates were observed. Consequently, the method used shall enable each group of financial assets to be associated with information about past credit loss experience in groups of financial assets with similar risk characteristics and with relevant observable data that reflects current conditions.
4. When designating a risk component as a hedged item, the hedge accounting requirements apply to that risk component in the same way as they apply to other hedged items that are not risk components. For example, the qualifying criteria apply, including that the hedging relationship must meet the hedge effectiveness requirements, and any hedge ineffectiveness must be measured and recognised.
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
Inca One Gold Corp. assigned short-term Ba1 & long-term Ba1 estimated rating. We evaluate the prediction models Modular Neural Network (Emotional Trigger/Responses Analysis) with Multiple Regression1,2,3,4 and conclude that the INCA:TSXV stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Buy
### INCA:TSXV Inca One Gold Corp. Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementB2B2
Balance SheetBaa2C
Leverage RatiosB3Ba3
Cash FlowCaa2Baa2
Rates of Return and ProfitabilityCBa2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 79 out of 100 with 807 signals.
## References
1. Chernozhukov V, Chetverikov D, Demirer M, Duflo E, Hansen C, et al. 2018a. Double/debiased machine learning for treatment and structural parameters. Econom. J. 21:C1–68
2. Morris CN. 1983. Parametric empirical Bayes inference: theory and applications. J. Am. Stat. Assoc. 78:47–55
3. Van der Vaart AW. 2000. Asymptotic Statistics. Cambridge, UK: Cambridge Univ. Press
4. Brailsford, T.J. R.W. Faff (1996), "An evaluation of volatility forecasting techniques," Journal of Banking Finance, 20, 419–438.
5. Morris CN. 1983. Parametric empirical Bayes inference: theory and applications. J. Am. Stat. Assoc. 78:47–55
6. Mullainathan S, Spiess J. 2017. Machine learning: an applied econometric approach. J. Econ. Perspect. 31:87–106
7. Bera, A. M. L. Higgins (1997), "ARCH and bilinearity as competing models for nonlinear dependence," Journal of Business Economic Statistics, 15, 43–50.
Frequently Asked QuestionsQ: What is the prediction methodology for INCA:TSXV stock?
A: INCA:TSXV stock prediction methodology: We evaluate the prediction models Modular Neural Network (Emotional Trigger/Responses Analysis) and Multiple Regression
Q: Is INCA:TSXV stock a buy or sell?
A: The dominant strategy among neural network is to Buy INCA:TSXV Stock.
Q: Is Inca One Gold Corp. stock a good investment?
A: The consensus rating for Inca One Gold Corp. is Buy and assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of INCA:TSXV stock?
A: The consensus rating for INCA:TSXV is Buy.
Q: What is the prediction period for INCA:TSXV stock?
A: The prediction period for INCA:TSXV is (n+8 weeks) | {} |
# PGFPlots: calculate linear regression of part of the plot [duplicate]
I'm trying to calculate the young modulus of a certain material using it's load-deflection curve. For such, I need to get the slope of the linear part of the curve (just the beginning). But my files contain the whole data, so I tried to remove the rest of the data by restricting the domain as shown in This Question, didn't work. Then I came upon this one: PGFPlots: calculate a linear regression, ignoring some data.
I thought it would solve my problem but the answer is kind of incomplete, as it only tells how to ignore the beginning of the data which is the opposite of what I want. I also saw a question on how to mod the file but there must be an easier way I'm not seeing.
Here's a MWE (with the domain restricted)
\usepackage{pgfplots}
\usepackage{pgfplotstable}
\pgfplotsset{compat=1.12, legend pos=north west, legend style={draw=none}}
\begin{tikzpicture}
\begin{axis}[xlabel=Deformação (mm),ylabel=Força (N), ymin=0, width=\textwidth, height=\textwidth, scaled ticks=false, ytick={0,10000,20000,30000,40000,50000}]
\addplot +[restrict x to domain=7:14] [no markers, red, thick] table [x=Deformacao, y=Forca, col sep=semicolon]{cp2.csv};
\addplot +[restrict x to domain=7:14] [no markers, black, dashed] table [y={create col/linear regression={y=Forca}}, col sep=semicolon]{cp2.csv};
\addlegendentry{E=$\pgfmathprintnumber{\pgfplotstableregressiona}$(N/mm)}
\end{axis}
\end{tikzpicture}
This is the result:
Here's the complete data (no domain restriction)
And if it's of interest this is the cp2.csv file
• You can use \addplot[y filter/.expression={y>20 ? nan : y}] to filter out the coordinates in \addplot[y filter/.expression={y>20 ? nan : y}] table[ y={create col/linear regression}, ] {\jobname-plot.dat};. Change 20 as you wish. – user11232 Aug 25 '15 at 0:45
• Sorry, I'm putting the whole data there, theres poinst where y is less than 50000, which would be your 20, so that doesn't work. =/ – Guilherme Zanotelli Aug 25 '15 at 0:53
• For the record: Here seems to be the same question, currently also without any (working) answer. – Stefan Pinnow Oct 24 '16 at 15:59
• As stated in my previous comment there is a similar question dealing with that problem. I provided there a "workaround" solution using gnuplot using the raw gnuplot feature of PGFPlots. I think this solution could easily be adapted to your specific problem, if "using gnuplot" is an option for you. Please let us know. – Stefan Pinnow Dec 12 '16 at 22:15
• @StefanPinnow thanks for the workaround. I guess using gnuplot is an option, yes. Although it would be better if not needed, but right now I have already delivered my report and passed the subject. Don't you think that the question you linked is a duplicate? – Guilherme Zanotelli Dec 13 '16 at 5:56 | {} |
# The largest number to break a conjecture [duplicate]
There are several conjectures in Mathematics that seem to be true but have not been proved. Of course, as computing power increased, folks have expanded their search for counterexamples ever and ever upwards.
Providing a counterexample to a conjecture with a very large number would be interesting, but I cannot think of any non-trivial examples where a really large number has been found to disprove a (non-trivial) conjecture. I've seen plenty of large numbers serving as bounds to some value, but usually this is something known to be bounded (i.e. finite) anyway.
Out of curiosity, what's the largest counterexample you've seen to disprove a conjecture?
-
## marked as duplicate by Start wearing purple, Ittay Weiss, Emily, Dan Rust, AmzotiJul 23 '13 at 0:29
This question was marked as an exact duplicate of an existing question.
An interesting link: [The phenomena of eventual counterexamples][1] [1]: mathoverflow.net/questions/15444/… – Next Jul 23 '13 at 0:03
Ah, I figured someone would have asked this! My search-fu on this site is not very good. – Emily Jul 23 '13 at 0:06
Google about Polya's conjecture. I don't know if it is the largest, though. The smallest number that is a counterexample is $906.150.257$, yet it is quite large. | {} |
# Is there a canonical labelling package optimised for small graphs?
Recently, I've been looking into motifs in networks (directed graphs) -- small connected induced subgraphs that appear significantly more frequently than in a "similar random graph".
In practice, we need to enumerate the induced subgraphs (up to isomorphism) in thousands of large random networks, which can take a long time, and requires some form of canonical labelling (e.g. FANMOD uses nauty). My feeling is that nauty is quite bulky and seems likely to be overkill for this feature because e.g. we only need to canonically label small graphs (in FANMOD they have at most 8 vertices -- although I haven't seen much interest in motifs with more than 4 vertices).
Question: Is there a package that specialises in the canonical labelling of small directed graphs?
In the case I'm interested in, we need to label millions of graphs on a small number of vertices.
-
## 1 Answer
Nauty really is optimized for small graphs, if you use the C library interface. McKay (and others, such as myself) frequently use this library to generate all graphs of a given small order, which requires canonical labeling of small graphs very quickly. This happens many many many times (see http://oeis.org/A000088 for the growth of the number of unlabeled graphs). In fact, McKay's program geng can compute all 12,005,168 graphs on 10 nodes in less than an hour on my laptop.
-
Thanks for that. It looks like others have avoided the bulkiness of nauty by chopping unused features (rather than use another package). – Douglas S. Stones Jan 9 '11 at 2:31 | {} |
Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > DETAIL:
### Revision(s):
Revision #3 to TR12-112 | 24th December 2014 23:53
#### New Limits to Classical and Quantum Instance Compression
Revision #3
Authors: Andrew Drucker
Accepted on: 24th December 2014 23:54
Keywords:
Abstract:
Given an instance of a hard decision problem, a limited goal is to $compress$ that instance into a smaller, equivalent instance of a second problem. As one example, consider the problem where, given Boolean formulas $\psi^1, \ldots, \psi^t$, we must determine if at least one $\psi^j$ is satisfiable. An $OR-compression \ scheme$ for SAT is a polynomial-time reduction $R$ that maps $(\psi^1, \ldots, \psi^t)$ to a string $z$, such that $z$ lies in some "target'' language $L'$ if and only if $\ \bigvee_j \ [\psi^j \in \mathrm{SAT}]$ holds. (Here, $L'$ can be arbitrarily complex.) AND-compression schemes are defined similarly. A compression scheme is $strong$ if $|z|$ is polynomially bounded in $n = \max_j |\psi^j|$, independent of $t$.
Strong compression for SAT seems unlikely. Work of Harnik and Naor (FOCS '06/SICOMP '10) and Bodlaender, Downey, Fellows, and Hermelin (ICALP '08/JCSS '09) showed that the infeasibility of strong OR-compression for SAT would show limits to instance compression for a large number of natural problems. Bodlaender et al. also showed that the infeasibility of strong AND-compression for SAT would have consequences for a different list of problems. Motivated by this, Fortnow and Santhanam (STOC '08/JCSS '11) showed that if SAT is strongly OR-compressible, then $NP \subseteq coNP/poly$. Finding similar evidence against AND-compression was left as an open question.
We provide such evidence: we show that strong AND- $or$ OR-compression for SAT would imply $non-uniform, \ statistical \ zero-knowledge \ proofs$ for SAT---an even stronger and more unlikely consequence than $NP \subseteq coNP/poly$. Our method applies against $probabilistic$ compression schemes of sufficient "quality'' with respect to the reliability and compression amount (allowing for tradeoff). This greatly strengthens the evidence given by Fortnow and Santhanam against probabilistic OR-compression for SAT. We also give variants of these results for the analogous task of $quantum \ instance \ compression$, in which a polynomial-time quantum reduction must output a quantum state that, in an appropriate sense, "preserves the answer'' to the input instance.
The central idea in our proofs is to exploit the information bottleneck in an AND-compression scheme for a language $L$ in order to fool a cheating prover in a proof system for $\overline{L}$. Our key technical tool is a new method to "disguise'' information being fed into a compressive mapping; we believe this method may find other applications.
Changes to previous version:
In our main technical reduction (Theorem 7.1), clarified the requirement that the parameter t_1 be polynomially bounded. This had been intended and indicated in the Intro, but was not sufficiently explicit. Similar correction to Thms. 7.4, 7.5, and 8.14 (which is now Thm. 8.19). Various small improvements to the writeup.
Revision #2 to TR12-112 | 20th November 2013 21:43
#### New Limits to Classical and Quantum Instance Compression
Revision #2
Authors: Andrew Drucker
Accepted on: 20th November 2013 21:43
Keywords:
Abstract:
Given an instance of a hard decision problem, a limited goal is to $compress$ that instance into a smaller, equivalent instance of a second problem. As one example, consider the problem where, given Boolean formulas $\psi^1, \ldots, \psi^t$, we must determine if at least one $\psi^j$ is satisfiable. An $OR-compression \ scheme$ for SAT is a polynomial-time reduction $R$ that maps $(\psi^1, \ldots, \psi^t)$ to a string $z$, such that $z$ lies in some "target'' language $L'$ if and only if $\ \bigvee_j \ [\psi^j \in \mathrm{SAT}]$ holds. (Here, $L'$ can be arbitrarily complex.) AND-compression schemes are defined similarly. A compression scheme is $strong$ if $|z|$ is polynomially bounded in $n = \max_j |\psi^j|$, independent of $t$.
Strong compression for SAT seems unlikely. Work of Harnik and Naor (FOCS '06/SICOMP '10) and Bodlaender, Downey, Fellows, and Hermelin (ICALP '08/JCSS '09) showed that the infeasibility of strong OR-compression for SAT would show limits to instance compression for a large number of natural problems. Bodlaender et al. also showed that the infeasibility of strong AND-compression for SAT would have consequences for a different list of problems. Motivated by this, Fortnow and Santhanam (STOC '08/JCSS '11) showed that if SAT is strongly OR-compressible, then $NP \subseteq coNP/poly$. Finding similar evidence against AND-compression was left as an open question.
We provide such evidence: we show that strong AND- $or$ OR-compression for SAT would imply $non-uniform, \ statistical \ zero-knowledge \ proofs$ for SAT---an even stronger and more unlikely consequence than $NP \subseteq coNP/poly$. (By a different argument, we also show such compression would imply the $uniform$ collapse $NP \subseteq coAM$.) Our method applies against $probabilistic$ compression schemes of sufficient "quality'' with respect to the reliability and compression amount (allowing for tradeoff). This greatly strengthens the evidence given by Fortnow and Santhanam against probabilistic OR-compression for SAT. We also give variants of these results for the analogous task of $quantum \ instance \ compression$, in which a polynomial-time quantum reduction must output a quantum state that, in an appropriate sense, "preserves the answer'' to the input instance.
The central idea in our proofs is to exploit the information bottleneck in an AND-compression scheme for a language $L$ in order to fool a cheating prover in a proof system for $\overline{L}$. Our key technical tool is a new method to "disguise'' information being fed into a compressive mapping; we believe this method may find other applications.
Changes to previous version:
Small fix/tweak to the new material of Revision #1.
Revision #1 to TR12-112 | 19th November 2013 03:23
#### New Limits to Classical and Quantum Instance Compression
Revision #1
Authors: Andrew Drucker
Accepted on: 19th November 2013 03:23
Keywords:
Abstract:
Given an instance of a hard decision problem, a limited goal is to $compress$ that instance into a smaller, equivalent instance of a second problem. As one example, consider the problem where, given Boolean formulas $\psi^1, \ldots, \psi^t$, we must determine if at least one $\psi^j$ is satisfiable. An $OR-compression \ scheme$ for SAT is a polynomial-time reduction $R$ that maps $(\psi^1, \ldots, \psi^t)$ to a string $z$, such that $z$ lies in some "target'' language $L'$ if and only if $\ \bigvee_j \ [\psi^j \in \mathrm{SAT}]$ holds. (Here, $L'$ can be arbitrarily complex.) AND-compression schemes are defined similarly. A compression scheme is $strong$ if $|z|$ is polynomially bounded in $n = \max_j |\psi^j|$, independent of $t$.
Strong compression for SAT seems unlikely. Work of Harnik and Naor (FOCS '06/SICOMP '10) and Bodlaender, Downey, Fellows, and Hermelin (ICALP '08/JCSS '09) showed that the infeasibility of strong OR-compression for SAT would show limits to instance compression for a large number of natural problems. Bodlaender et al. also showed that the infeasibility of strong AND-compression for SAT would have consequences for a different list of problems. Motivated by this, Fortnow and Santhanam (STOC '08/JCSS '11) showed that if SAT is strongly OR-compressible, then $NP \subseteq coNP/poly$. Finding similar evidence against AND-compression was left as an open question.
We provide such evidence: we show that strong AND- $or$ OR-compression for SAT would imply $non-uniform, \ statistical \ zero-knowledge \ proofs$ for SAT---an even stronger and more unlikely consequence than $NP \subseteq coNP/poly$. (By a different argument, we also show such compression would imply $NP \subseteq coAM$.) Our method applies against $probabilistic$ compression schemes of sufficient "quality'' with respect to the reliability and compression amount (allowing for tradeoff). This greatly strengthens the evidence given by Fortnow and Santhanam against probabilistic OR-compression for SAT. We also give variants of these results for the analogous task of $quantum \ instance \ compression$, in which a polynomial-time quantum reduction must output a quantum state that, in an appropriate sense, "preserves the answer'' to the input instance.
The central idea in our proofs is to exploit the information bottleneck in an AND-compression scheme for a language $L$ in order to fool a cheating prover in a proof system for $\overline{L}$. Our key technical tool is a new method to "disguise'' information being fed into a compressive mapping; we believe this method may find other applications.
Changes to previous version:
Changes: 1. We prove the new result that strong compression for either AND(SAT) or OR(SAT) would imply NP is in coAM.
2. A correction: the techniques used previously to rule out strong compression reductions for F(SAT) with "combining functions" F other than AND, OR do not work for a certain case, namely when F is monotone, depends on all variables, and has very low block sensitivity. (We had erroneously ruled out the existence of such functions.)
We give a new argument to handle F with low block sensitivity, subject to a certain explicitness condition on F. As a by-product we obtain item 1 above. (See Secs. 1.3.1, 7.3, and 9.)
### Paper:
TR12-112 | 7th September 2012 03:37
#### New Limits to Classical and Quantum Instance Compression
TR12-112
Authors: Andrew Drucker
Publication: 7th September 2012 03:57
Keywords:
Abstract:
Given an instance of a hard decision problem, a limited goal is to $compress$ that instance into a smaller, equivalent instance of a second problem. As one example, consider the problem where, given Boolean formulas $\psi^1, \ldots, \psi^t$, we must determine if at least one $\psi^j$ is satisfiable. An $OR-compression \ scheme$ for SAT is a polynomial-time reduction $R$ that maps $(\psi^1, \ldots, \psi^t)$ to a string $z$, such that $z$ lies in some "target'' language $L'$ if and only if $\ \bigvee_j \ [\psi^j \in \mathrm{SAT}]$ holds. (Here, $L'$ can be arbitrarily complex.) AND-compression schemes are defined similarly. A compression scheme is $strong$ if $|z|$ is polynomially bounded in $n = \max_j |\psi^j|$, independent of $t$.
Strong compression for SAT seems unlikely. Work of Harnik and Naor (FOCS '06/SICOMP '10) and Bodlaender, Downey, Fellows, and Hermelin (ICALP '08/JCSS '09) showed that the infeasibility of strong OR-compression for SAT would show limits to instance compression for a large number of natural problems. Bodlaender et al. also showed that the infeasibility of strong AND-compression for SAT would have consequences for a different list of problems. Motivated by this, Fortnow and Santhanam (STOC '08/JCSS '11) showed that if SAT is strongly OR-compressible, then $NP \subseteq coNP/poly$. Finding similar evidence against AND-compression was left as an open question.
We provide such evidence: we show that strong AND- $or$ OR-compression for SAT would imply $non-uniform, \ statistical \ zero-knowledge \ proofs$ for SAT---an even stronger and more unlikely consequence than $NP \subseteq coNP/poly$. Our method applies against $probabilistic$ compression schemes of sufficient "quality'' with respect to the reliability and compression amount (allowing for tradeoff). This greatly strengthens the evidence given by Fortnow and Santhanam against probabilistic OR-compression for SAT. We also give variants of these results for the analogous task of $quantum \ instance \ compression$, in which a polynomial-time quantum reduction must output a quantum state that, in an appropriate sense, "preserves the answer'' to the input instance.
The central idea in our proofs is to exploit the information bottleneck in an AND-compression scheme for a language $L$ in order to fool a cheating prover in a proof system for $\overline{L}$. Our key technical tool is a new method to "disguise'' information being fed into a compressive mapping; we believe this method may find other applications.
ISSN 1433-8092 | Imprint | {} |
# Area of an Ellipse – Explanation & Examples
In geometry, an is a two-dimensional flat elongated circle that is symmetrical along its shortest and longest diameters. An ellipse resembles an oval shape. In an ellipse, the longest diameter is known as the major axis, whereas the shortest diameter is known as the minor axis.
The distance of two points in the interior of an ellipse from a point on the ellipse is same as the distance of any other point on the ellipse from the same point. These points inside the ellipse are termed as foci. In this article, you will what an ellipse is, and how to find its area by using the area of an ellipse formula. But first see its few applications first.
Ellipses have multiple applications in the field of engineering, medicine, science, etc. For example, the planets revolve in their orbits which are elliptical in shape.
In an atom, it is believed that, electrons revolve around the nucleus in elliptical orbits.
The concept of ellipses is used in medicine for treatment of kidney stones (lithotripsy). Other real-world examples of elliptical shapes are the huge elliptical park in front of the White House in Washington DC and the St. Paul’s Cathedral building.
Up to this point, you have got an idea of what an ellipse is, let’s now proceed by looking at how to calculate area of an ellipse.
## How to Find the Area of an Ellipse?
To calculate the area of an ellipse, you need the measurements of both the major radius and minor radius.
### Area of an ellipse formula
The formula for area of an ellipse is given as:
Area of an ellipse = πr1r2
Where, π = 3.14, r1 and r2 are the minor and the major radii respectively.
Note: Minor radius = semi -minor axis (minor axis/2) and the major radius = Semi- major axis (major axis/2)
Let’s test our understanding of the area of an ellipse formula by solving a few example problems.
Example 1
What is the area of an ellipse whose minor and major radii are, 12 cm and 7 cm, respectively?
Solution
Given;
r1 =7 cm
r2 =12 cm
By the formula,
Area of an ellipse = πr1r2
= 3.14 x 7 x 12
= 263.76 cm2
Example 2
The major axis and minor axis of an ellipse are, 14 m and 12 m, respectively. What is the area of the ellipse?
Solution
Given;
Major axis = 14m ⇒ major radius, r2 =14/2 = 7 m
Minor axis = 12 m ⇒ minor radius, r1 = 12/2 = 6 m.
Area of an ellipse = πr1r2
= 3.14 x 6 x 7
= 131.88 m2.
Example 3
The area of an ellipse is 50.24 square yards. If the major radius of the ellipse is 6 yards more than the minor radius. Find the minor and major radii of the ellipse.
Solution
Given;
Area = 50.24 square yards
Let the minor radius = x
Therefore,
The major radius = x + 6
But, area of an ellipse = πr1r2
⇒50.24 = 3.14 * x *(x + 6)
⇒50.24 = 3.14x (x + 6)
By applying the distributive property of multiplication on the RHS, we get,
⇒50.24 = 3.14x2 + 18.84x
Divide both sides by 3.14
⇒16 = x2 + 6x
⇒x2 + 6x – 16 =0
⇒x2 + 8x – 2x – 16 = 0
⇒ x (x + 8) – 2 (x + 8) = 0
⇒ (x – 2) (x + 8) = 0
⇒ x = 2 or – 4
Substitute x = 2 for the two equations of radii
Therefore,
The major radius = x + 6 ⇒ 8 yards
The minor radius = x = 2 yards
So, the major radius of the ellipse is 8 yards and the minor radius is 2 yards.
Example 4
Find the area of an ellipse whose radii area 50 ft and 30 ft respectively.
Solution
Given:
r1 = 30 ft and r2 = 50 ft
Area of an ellipse = πr1r2
A = 3.14 × 50 ×30
A = 4,710 ft2
Hence, the area of the ellipse is 4,710 ft2.
Example 5
Calculate the area of the ellipse shown below.
Solution
Given that;
r1 = 5.5 in
r2 = 9.5 in
Area of an ellipse = πr1r2
= 3.14 x 9.5 x 5.5
= 164.065 in2
Area of a semi – ellipse (h2)
A semi ellipse is a half an ellipse. Since we know the area of an ellipse as πr1r2, therefore, the area of a semi ellipse is half the area of an ellipse.
Area of a semi ellipse = ½ πr1r2
Example 6
Find the area of a semi – ellipse of radii 8 cm and 5 cm.
Solution
Area of a semi ellipse = ½ πr1r2
= ½ x 3.14 x 5 x 8
= 62.8 cm2.
### Practice Questions
1. What is the area of an ellipse whose minor and major radii are, $14$ ft and $8$ ft, respectively? (Use $\pi = \dfrac{22}{7}$)
2. The major axis and minor axis of an ellipse are, $40$ cm and $30$ cm, respectively. What is the area of the ellipse? (Use $\pi \approx 3.14$)
3. What is the area of the ellipse shown below? (Use $\pi = \dfrac{22}{7}$)
4. What is the area of a semi – ellipse of radii $10$ mm and $8$ mm? (Use $\pi \approx 3.14$)
5. The area of an ellipse is $264$ square feet and the major radius of the ellipse is $8$ feet more than the minor radius. What are the lengths of the minor and major radii of the ellipse? (Use $\pi = \dfrac{22}{7}$) | {} |
## Twisty's Tesseract
from Twisty's Mind come Twisted Products
# Folding Polygons into Pyramids 2
- Posted in Math by
Some corrections to a problem result in lengthy, ongoing discussions. On Brilliant.org, my original correction still raises the question "are we missing yet more 'smart' answers?"
## The Original Question
"Which of the following regular polygons can be folded into a pyramid without cutting or overlapping any area?"
When I pointed out that the answer is both the triangle (resulting in a regular tetrahedron) and the square (resulting in an irregular pyramid of positive volume), it changed the game. The question had to be rewritten, and those who had earlier answered "square" were granted credit.
### Definitions
Still, even the revised question is awkwardly worded. I can now prove that any regular polygon can be folded into a regular tetrahedron, "without cutting or overlapping," so long as excess flaps are permitted to hang external to the pyramid. This would not be so had the original question made a requirement that it form "only just a regular pyramid, and nothing more."
Definitions would also help if pyramid and "regular pyramid" were more clearly defined. I think the common definition of a "regular pyramid" is one whose base is any regular polygon, but whose side faces are all isosceles triangles of an arbitrary but equal height.
There do exist trivial solutions for square and hexagon if the pyramid needn't have a positive height and thus positive volume. A smaller square inscribed inside a larger square (from the midpoints of the sides of the larger) can form a base for a square pyramid of zero height. Similarly, a smaller triangle can be inscribed inside a larger hexagon, its verteces attached to every other vertex of the hexagon. When these are folded in, the sum angle of the two flaps are equal to the angle of the base polygon, so no height results. It is unclear if this breaks the questions definition of "overlapping," since there is no excess area in the process: They may "lap" but not "overlap," depending on the definitions.
## Proof any regular polygon can be folded into a regular tetrahedron
### Lemma 1: A regular triangle can be inscribed into any regular polygon
Pick your regular polygon and select one of its sides. The polygon is regular, so an axis of symmetry can be drawn down its center from the midpoint of the selected side. By placing the vertex of the inscribed triangle onto this midpoint, it is seen that it abuts the inside and outside of the polygon, yet resides in neither. The same is true of the other vertices of the inscribed triangle: They are constructed against whichever side of the polygon they intersect when following a 30 degree angle left or right of the axis of symmetry.
### Construction
Let us inscribe the triangle in black, as depicted in the attached photo of a triangle inside an orange square. Let us then draw red lines through the midpoints of the inscribed triangle. The black lines of the base will be folded back/outward, and the red lines will be folded up/inward. Since the inscribed triangle's vertices abut the larger polygon's outer edges, there is no exterior excess along the axis of symmetry drawn through any of the base's vertices. These three vertices will meet as the pinnacle of the pyramid above the base. And because the red folds are angled to create incomplete regular tetrahedrons of their own, they fold away from the central tetrahedron without needing to be cut or overlapped. $$\tag*{q.e.d. \(\Box$$}\)
## Conclusion
If the question were reworded to require just a pyramid and nothing more nor less, then the only solution would be the regular triangle. But if flaps are welcome as presented here, then any regular polygon can provide this solution. The only "wasteless" pyramid is the pyramid itself: nothing less, nothing more. | {} |
Mathematician:Mathematicians/Sorted By Birth/0 - 500 CE
For more comprehensive information on the lives and works of mathematicians through the ages, see the MacTutor History of Mathematics archive, created by John J. O'Connor and Edmund F. Robertson.
The army of those who have made at least one definite contribution to mathematics as we know it soon becomes a mob as we look back over history; 6,000 or 8,000 names press forward for some word from us to preserve them from oblivion, and once the bolder leaders have been recognised it becomes largely a matter of arbitrary, illogical legislation to judge who of the clamouring multitude shall be permitted to survive and who be condemned to be forgotten.
-- Eric Temple Bell: Men of Mathematics, 1937, Victor Gollancz, London
Previous ... Next
$0$ – $100$
Heron of Alexandria (c. 10 – c. 70 CE)
Heron (or Hero) of Alexandria (Greek: Ἥρων ὁ Ἀλεξανδρεύς) was a Greek mathematician and engineer.
Famous for writing about the aeolipile, otherwise known as Hero's Engine (although he didn't actually invent it), and the device known as Heron's fountain.
Also noted for Heron's formula for calculating the area of a triangle whose side lengths are known.
show full page
Nicomachus of Gerasa (c. 60 – c. 120 CE)
Nicomachus (Greek: Νικόμαχος) was a Neo-Pythagorean about whom very little is known.
Unusual in that he used the system of Arabic numerals rather than the then-current cumbersome Roman numerals.
Famously made some conjectures about perfect numbers which were soon shown to be false.
Appears to have had more influence than his (perhaps limited) abilities may have merited.
show full page
Theon of Smyrna (c. 70 – c. 135 CE)
Greek: Θέων ὁ Σμυρναῖος.
Greek philosopher and mathematician, whose works were strongly influenced by the Pythagorean school of thought.
Made astronomical observations of Mercury and Venus between 127 and 132, as reported by Claudius Ptolemy.
show full page
Menelaus of Alexandria (c. 70 – c. 140 CE)
Greek mathematician and astronomer.
Very little is known about him.
show full page
Claudius Ptolemy (c. 90 – c. 168 CE)
Roman citizen, of either Greek or Egyptian ancestry.
Mathematician, astronomer and general all-round scientist.
Best known for being the author of several scientific works, including Almagest.
show full page
$201$ – $300$
Diophantus of Alexandria (between 200 and 214 CE – between 284 and 298 CE)
Author of a series of books called Arithmetica, some of which are now lost, concerning the solution of algebraic equations.
Sometimes referred to as "the father of algebra", but some claim the title should belong to Al-Khwarizmi.
show full page
Liu Hui (c. 225 – c. 295)
Chinese mathematician and writer.
Edited and published a book with solutions to mathematical problems presented in Jiuzhang suanshu ("Nine Chapters on the Mathematical Art"), in which he gave:
a calculation of $\pi$ (pi) correct to $4$ decimal places
proof of the formulae for the volume of the square pyramid and tetrahedron.
Possibly the first mathematician to discover, understand and use negative numbers.
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Iamblichus Chalcidensis (c. 245 – c. 325)
Usually known as Iamblichus. His name in Ancient Greek is Ἰάμβλιχος, probably from Syriac or Aramaic ya-mlku, "He is king".
Assyrian philosopher of the neo-Platonist school.
His main involvement in mathematics concerns the fact that he may have known the $5$th perfect number, but there is no hard evidence of this fact.
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Pappus of Alexandria (c. 290 – c. 350)
One of the last great Greek mathematicians of antiquity.
Very little is known about him, except that he flourished at around 320 CE through dint of the eclipse of the sun in Alexandria in that year which he discussed in his commentary on Claudius Ptolemy's Almagest.
Noted for his multi-volume Collection, and for Pappus's Hexagon Theorem.
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Serenus of Antinoupolis (c. 300 – c. 360)
Egyptian mathematician known for his commentary on the Conics of Apollonius of Perga.
This is now lost. We know about it through the writings of Theon of Alexandria.
Also wrote at least two original works of his own, whose survival is directly due to their association with Conics.
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$301$ – $400$
Theon of Alexandria (c. 335 – 405)
Alexandrian mathematician and astronomer.
Best known for being the father of Hypatia of Alexandria.
His edition of Euclid's The Elements was an authority until well into the $19$th century.
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Hypatia of Alexandria (c. 360 – 415)
No, of course she didn't look like this
Greek: Ὑπατία.
Egyptian mathematician, astronomer, scientist and philosopher. Daughter of Theon of Alexandria.
Head of Platonist school in Alexandria in c. $400$ CE.
Notable for:
Being the first woman in mathematics notable enough to have been remembered by history;
Being murdered by a mob of Christians for holding pagan beliefs.
Her death has been argued as signalling the decline of learning in the Western world, and the start of the "dark ages", from which recovery would not happen for another thousand years.
show full page
Sun Tzu (c. 400 – c. 460)
Chinese mathematician and astronomer.
Best known for his work on Diophantine equations. His work is the source of the Chinese Remainder Theorem.
show full page
$401$ – $500$
Muni Sarvanandin (5th Century )
Indian Digambara monk who wrote the Lokavibhaga.
show full page
Proclus Lycaeus (412 – 485)
Greek philosopher who among other things produced a commentary on Book $\text I$ of Euclid's The Elements.
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Zu Chongzhi (429 – 501)
Prominent Chinese mathematician and astronomer.
Derived the most accurate approximation for $\pi$ for over nine hundred years.
Credited (along with Zu Geng) with proving the volume of the sphere using the same principle as Bonaventura Francesco Cavalieri.
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Isidorus of Miletus (442 – 537)
One of the two main Byzantine Greek architects (with Anthemius of Tralles) that Emperor Justinian I commissioned to design the church of Hagia Sophia in Constantinople from $532$ – $537$.
He also created the first comprehensive compilation of Archimedes' works.
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Zu Geng (c. 450 – c. 520)
Also known as Zu Gengzhi (simplified Chinese: 祖暅之; traditional Chinese: 祖暅之; pinyin: Zǔ Gèngzhī; Wade–Giles: Tsu Kengchi; 480 - 525), courtesy name Jing Shuo (景烁).
Son of Zu Chongzhi.
Chinese mathematician who determined how to compute the diameter of a sphere of a given volume. He did this using a generalized version of Cavalieri's Principle.
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Aryabhata the Elder (476 – 550)
Indian mathematician and astronomer.
An early believer in the irrationality of $\pi$, and developed an approximation for it of $3.1416$.
Developed a positional system of numerals in c. $500$, but it lacked a symbol for zero.
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Anicius Manlius Severinus Boëthius (c.477 – 524)
Roman senator, consul, magister officiorum, and philosopher, who produced editions of some mathematical works which survived to be used throughout Medieval Europe.
Although widely cited as a great scholar, as a mathematician he appears to have been mediocre.
show full page
Eutocius of Ascalon (c. 480 – c. 540)
Palestinian philisopher about whom little is known. He wrote commentaries on works of Apollonius and Archimedes.
It is possible that Eutocius studied in Alexandria and became its head.
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# An observation on Non-Trivial Zeros of Riemann Zeta Function.
I observed this property in month of July this year but unable to design a mathematical proof or mathematical way to state my observation. I need help to state this property.
We know that for certain values of 'b' in s = 1/2 + ib , ζ(s) = 0.
When I observed most of the values of 'b' here
I found that most of the primes are related to these values of 'b' in there square forms like:
Property: [b] = p^2 {where 'p' is a prime number and [b] is the Box Function}
example: b = 841.0363...
so, [841.0363...] = 29^2
below 200 there are only 6 primes which are not following the this above property.
I am searching a formula or a program to find out all those primes which follow above property but till now I didn't got any solution. I am also not good in programming please help me in this problem.
The $$n$$th zeta zero has $$b(n) \sim 2\pi n/W(n/e)$$, where $$W$$ is the Lambert function. The Lambert function grows as roughly the logarithm of its argument, so we see that the zeta zeroes get denser as $$n$$ increases, and at $$n \approx 10^4$$, their imaginary parts should start hitting every integer. This turns out to be the case, as the last square of a prime it misses is $$103^2 = 10609$$.
We can also estimate the probability of each square of a prime not having a zeta zero near it. The density of zeta zeroes is $$W(n/e)/(2\pi)$$, so we expect each square of a prime to not be near a zeta zero with probability $$1 - W(p^2/e)/(2\pi)$$. Summing this over all primes where this value is positive, we get an expected count of $$7.7$$, which is reasonably close to the actual value of 6.
• @AdarshKumar My statement is probably a little too strong, but it is known that there is some $n$ after which $[b(n)]$ hits every integer. It is also known how to count the number of zeros with imaginary part less than $b$, with formula given here. Keeping only the dominant terms and inverting gives my estimate in the answer. It is possible there are integers missed that are significantly larger than $10^4$, but prime squares have very low density at large $n$, making them unlikely to be counterexamples anyways. – eyeballfrog Dec 17 '18 at 17:09 | {} |
# zbMATH — the first resource for mathematics
Antineighbourhood graphs. (English) Zbl 0759.05084
All graphs considered in present paper are finite undirected simple ones. A graph $$H$$ is called a neighbourhood graph if there exists a graph $$G$$ in which the subgraph induced by the neighbours of each vertex is isomorphic to $$H$$, and a graph $$H$$ is said to be a $$j$$-antineighbourhood graph if there exists a graph $$G$$ in which, for each vertex $$v$$ of $$G$$, the subgraph induced by the vertices at distance at least $$j+1$$ from $$v$$ is isomorphic to $$H$$. The classes of neighbourhood and $$j$$- antineighbourhood graphs are denoted by $${\mathfrak N}$$ and $${\mathfrak A}_ j$$, respectively. At first the graphs of the classes $${\mathfrak A}_ 0$$ and $${\mathfrak A_ 1}$$ are characterized by their structural properties and several examples are given. From the numerous general results here only some can be mentioned:
$$H$$ belongs to $${\mathfrak A}_ 0$$ iff it is a disjoint union of $$n$$ copies of $$K_ 1$$ for some positive integer $$n$$ or the graph obtained from $$H$$ by adding a new vertex and joining it to all vertices of the minimum degree of $$H$$ in $$H$$ is a vertex-symmetric graph (Theorem 1).
$$H$$ belongs to $${\mathfrak A}_ 0$$ iff it is a vertex-deleted subgraph of a vertex-symmetric graph (Corollary 1, which follows from Theorem 1).
$$H$$ belongs to $${\mathfrak A}_ 0$$ iff its complement $$\overline H$$ belongs to $${\mathfrak A}_ 0$$ (Theorem 2).
The total graph $$T(G)$$ of a graph $$G$$ belongs to $${\mathfrak A}_ 0$$ iff $$G\cong K_ 2$$ or $$G\cong\overline K_ n$$ for some positive integer $$n$$ (Theorem 5).
$$H$$ belongs to $${\mathfrak N}$$ iff its complement $$\overline H$$ belongs to $${\mathfrak A}_ 1$$ (Theorem 6). This means that all elements of $${\mathfrak A}_ 1$$ are determined by the elements of $${\mathfrak N}$$, and vice versa. These results imply:
A cycle $$C_ n$$ belongs to $${\mathfrak A}_ 0$$ iff $$n=3$$.
A wheel $$W_ n$$ belongs to $${\mathfrak A}_ 0$$ iff $$n=3$$ or $$n=4$$.
A block graph $$H$$ belongs to $${\mathfrak A}_ 0$$ iff $$H$$ is a complete graph or a path.
A cycle $$C_ n$$ belongs to $${\mathfrak A}_ 1$$ iff $$n=3,4,5,6$$.
The complement $$\overline C_ n$$ of $$C_ n$$ belongs to $${\mathfrak A}_ 1$$ for every $$n\geq 3$$.
If graphs $$H_ 1$$ and $$H_ 2$$ belong to $${\mathfrak A}_ 1$$, then their join $$H_ 1+H_ 2$$ belongs to $${\mathfrak A}_ 1$$. Furthermore, the block graphs which belong to $${\mathfrak A_ 1}$$ are characterized by the following main theorem:
If $$G$$ is a block graph, then $$G$$ belongs to $${\mathfrak A}_ 1$$ iff $$G$$ is a path or a regular windmill graph.
The proof of this theorem is an extensive one, numerous lemmas are used and many distinctions of cases are necessary. Finally, the authors get interesting statements about the membership of $$C^ 2_ n$$ from cycles $$C_ n$$ to the classes $${\mathfrak N}$$ respectively $${\mathfrak A}_ 1$$ depending on the length $$n$$. And regarding the classes $${\mathfrak A}_ j$$, $$j\geq 2$$, they obtain the result that every graph $$H$$ belongs to such a class.
##### MSC:
05C75 Structural characterization of families of graphs
Full Text: | {} |
CGAL 5.1.3 - Triangulated Surface Mesh Skeletonization
CGAL::Mean_curvature_flow_skeletonization< TriangleMesh, Traits_, VertexPointMap_, SolverTraits_ > Class Template Reference
#include <CGAL/Mean_curvature_flow_skeletonization.h>
## Definition
### template<class TriangleMesh, class Traits_ = Default, class VertexPointMap_ = Default, class SolverTraits_ = Default> class CGAL::Mean_curvature_flow_skeletonization< TriangleMesh, Traits_, VertexPointMap_, SolverTraits_ >
Function object that enables to extract the mean curvature flow skeleton of a triangulated surface mesh.
The algorithm used takes as input a triangulated surface mesh and iteratively contracts the surface mesh following the mean curvature flow [1]. The intermediate contracted surface mesh is called the meso-skeleton. After each iteration, the meso-skeleton is locally remeshed using angle split and edge contraction. The process ends when the modification of the meso-skeleton between two iterations is small.
Template Parameters
TriangleMesh a model of FaceListGraph Traits a model of MeanCurvatureSkeletonizationTraits Default: boost::property_traits< boost::property_map::type >::value_type VertexPointMap a model of ReadWritePropertyMap with boost::graph_traits::vertex_descriptor as key and Traits::Point_3 as value type. Default: boost::property_map::const_type. SolverTraits_ a model of NormalEquationSparseLinearAlgebraTraits_d. Default: If Eigen 3.2 (or greater) is available and CGAL_EIGEN3_ENABLED is defined, then an overload of Eigen_solver_traits is provided as default parameter: Eigen::SimplicialLDLT< Eigen::SparseMatrix >>
Examples:
Surface_mesh_skeletonization/MCF_Skeleton_example.cpp, Surface_mesh_skeletonization/MCF_Skeleton_sm_example.cpp, Surface_mesh_skeletonization/segmentation_example.cpp, Surface_mesh_skeletonization/simple_mcfskel_example.cpp, and Surface_mesh_skeletonization/simple_mcfskel_sm_example.cpp.
## Types
typedef boost::adjacency_list< boost::vecS, boost::vecS, boost::undirectedS, Vmap > Skeleton
The graph type representing the skeleton. The vertex property Vmap is a struct with a member point of type Traits::Point_3 and a member vertices of type std::vector<boost::graph_traits<TriangleMesh>::vertex_descriptor>. See the boost documentation page for more details.
## Constructor
Mean_curvature_flow_skeletonization (const TriangleMesh &tmesh, VertexPointMap vertex_point_map=get(CGAL::vertex_point, tmesh), Traits traits=Traits())
The constructor of a skeletonization object. More...
## Local Remeshing Parameters
double max_triangle_angle ()
During the local remeshing step, a triangle will be split if it has an angle larger than max_triangle_angle().
double min_edge_length ()
During the local remeshing step, an edge will be collapse if it is length is less than min_edge_length().
void set_max_triangle_angle (double value)
set function for max_triangle_angle()
void set_min_edge_length (double value)
set function for min_edge_length()
## Algorithm Termination Parameters
std::size_t max_iterations ()
Maximum number of iterations performed by contract_until_convergence().
double area_variation_factor ()
The convergence is considered to be reached if the variation of the area of the meso-skeleton after one iteration is smaller than area_variation_factor()*original_area where original_area is the area of the input triangle mesh.
void set_max_iterations (std::size_t value)
set function for max_iterations()
void set_area_variation_factor (double value)
set function for area_variation_factor()
## Vertex Motion Parameters
Controls the velocity of movement and approximation quality: decreasing this value makes the mean curvature flow based contraction converge faster, but results in a skeleton of lower quality. This parameter corresponds to $$w_H$$ in the original publication.
.
bool is_medially_centered ()
If true, the meso-skeleton placement will be attracted by an approximation of the medial axis of the mesh during the contraction steps, so will be the result skeleton.
Controls the smoothness of the medial approximation: increasing this value results in a (less smooth) skeleton closer to the medial axis, as well as a lower convergence speed. It is only used if is_medially_centered()==true. This parameter corresponds to $$w_M$$ in the original publication.
.
set function for quality_speed_tradeoff()
void set_is_medially_centered (bool value)
set function for is_medially_centered()
set function for medially_centered_speed_tradeoff()
## High Level Function
void operator() (Skeleton &skeleton)
Creates the curve skeleton: the input surface mesh is iteratively contracted until convergence, and then turned into a curve skeleton. More...
## Low Level Functions
The following functions enable the user to run the mean curvature flow skeletonization algorithm step by step.
void contract_geometry ()
Runs one contraction step following the mean curvature flow.
std::size_t collapse_edges ()
Collapses edges of the meso-skeleton with length less than min_edge_length() and returns the number of edges collapsed.
std::size_t split_faces ()
Splits faces of the meso-skeleton having one angle greater than max_triangle_angle() and returns the number of faces split.
std::size_t detect_degeneracies ()
Prevents degenerate vertices to move during the following contraction steps and returns the number of newly fixed vertices.
void contract ()
Performs subsequent calls to contract_geometry(), collapse_edges(), split_faces() and detect_degeneracies()
void contract_until_convergence ()
Iteratively calls contract() until the change of surface area of the meso-skeleton after one iteration is smaller than area_variation_factor()*original_area where original_area is the area of the input triangle mesh, or if the maximum number of iterations has been reached.
void convert_to_skeleton (Skeleton &skeleton)
Converts the contracted surface mesh to a skeleton curve. More...
typedef unspecified_type Meso_skeleton
When using the low level API it is possible to access the intermediate results of the skeletonization process, called meso-skeleton. It is a triangulated surface mesh which is model of FaceListGraph and HalfedgeListGraph.
const Meso_skeletonmeso_skeleton () const
Reference to the collapsed triangulated surface mesh.
## ◆ Mean_curvature_flow_skeletonization()
template<class TriangleMesh , class Traits_ = Default, class VertexPointMap_ = Default, class SolverTraits_ = Default>
CGAL::Mean_curvature_flow_skeletonization< TriangleMesh, Traits_, VertexPointMap_, SolverTraits_ >::Mean_curvature_flow_skeletonization ( const TriangleMesh & tmesh, VertexPointMap vertex_point_map = get(CGAL::vertex_point, tmesh), Traits traits = Traits() )
The constructor of a skeletonization object.
The algorithm parameters are initialized such that:
• max_triangle_angle() == 110
• quality_speed_tradeoff() == 0.1
• medially_centered_speed_tradeoff() == 0.2
• area_variation_factor() == 0.0001
• max_iterations() == 500
• is_medially_centered() == true
• min_edge_length() == 0.002 * the length of the diagonal of the bounding box of tmesh
Precondition
tmesh is a triangulated surface mesh without borders and has exactly one connected component.
Parameters
tmesh input triangulated surface mesh. vertex_point_map property map which associates a point to each vertex of the graph. traits an instance of the traits class.
## ◆ convert_to_skeleton()
template<class TriangleMesh , class Traits_ = Default, class VertexPointMap_ = Default, class SolverTraits_ = Default>
void CGAL::Mean_curvature_flow_skeletonization< TriangleMesh, Traits_, VertexPointMap_, SolverTraits_ >::convert_to_skeleton ( Skeleton & skeleton )
Converts the contracted surface mesh to a skeleton curve.
Template Parameters
Skeleton an instantiation of boost::adjacency_list as a data structure for the skeleton curve.
Parameters
skeleton graph that will contain the skeleton of tmesh. It should be empty before passed to the function.
## ◆ operator()()
template<class TriangleMesh , class Traits_ = Default, class VertexPointMap_ = Default, class SolverTraits_ = Default>
void CGAL::Mean_curvature_flow_skeletonization< TriangleMesh, Traits_, VertexPointMap_, SolverTraits_ >::operator() ( Skeleton & skeleton )
Creates the curve skeleton: the input surface mesh is iteratively contracted until convergence, and then turned into a curve skeleton.
This is equivalent to calling contract_until_convergence() and convert_to_skeleton().
Parameters
skeleton graph that will contain the skeleton of the input triangulated surface mesh. For each vertex descriptor vd of skeleton, the corresponding point and the set of input vertices that contracted to vd can be retrieved using skeleton[vd].point and skeleton[vd].vertices respectively. | {} |
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Abstract
We introduce a new class of canonical analytic Zariski decompositions (AZD's in short) called the supercanonical AZD's on the canonical bundles of smooth projective varieties with pseudoeffective canonical classes.\ We study the variation of the supercanonical AZD $\hat{h}_{{\rm can}}$ under projective deformations and give a new proof of the invariance of plurigenera.
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# Visual C++
This topic is 3912 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
I was wondering and I have seen some options that add back grounds to my consol projects. An option to drawbg(); = (Draw Back Ground) or some thing like that. and maybe some options to create boxes. and even with a shadow. I'm new at this so please bear with me. Joe
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In the old days there was a dos library where you could define boxes or areas on the console window and add text and colors to them. I remember trying it once a long time ago using Borlands compiler tools.
Is this what you are thinking about?
If it is, I don't think that library is available anymore. At least not unless you have a old Borland compiler installed.
If all you want to do is to change the background color you can take a look at these functions
There is a function among the console functions called SetConsoleScreenBufferInfoEx that I suspect could be used to define *boxes* but it is only available on Windows Vista.
There is a library called PDCurses that can do a lot of stuff. I managed to compile it using Visual C++ and the demos that follow the code works. However, as soon as I try to write a little test myself nothing happens, so I can't really recommend it yet lol.
Besides, compiling and installing it is not all that easy.
[Edited by - pulpfist on January 7, 2008 7:34:57 AM]
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Quote:
Original post by pulpfistIn the old days there was a dos library where you could define boxes or areas on the console window and add text and colors to them. I remember trying it once a long time ago using Borlands compiler tools.Is this what you are thinking about?If it is, I don't think that library is available anymore. At least not unless you have a old Borland compiler installed.If all you want to do is to change the background color you can take a look at these functionsThere is a function among the console functions called SetConsoleScreenBufferInfoEx that I suspect could be used to define *boxes* but it is only available on Windows Vista.There is a library called PDCurses that can do a lot of stuff. I managed to compile it using Visual C++ and the demos that follow the code works. However, as soon as I try to write a little test myself nothing happens, so I can't really recommend it yet lol.Besides, compiling and installing it is not all that easy.
In Visual C++ I been teaching my self Consol applications just to learn the code. I thought maybe that I could create my own back grounds using some of the ideas I got from other dos related programs.
How ever I don't know what they were using or how they did it. Programing in pascal back then was hard. And today I can create a option like "DrawBox(x1,y1,x2,y2,bg,fg)" in pascal. Al it does is draws a box using ascii codes that draw lines a round the borders.
All I'm trying to do is customize my program so that the menus look better with back ground graphics. If that what you want to call them..
and as for colors. Remember you help me with that last week. and thanks
At this time I'm just trying to under stand what a library does.
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Quote:
In Visual C++ I been teaching my self Consol applications just to learn the code. I thought maybe that I could create my own back grounds using some of the ideas I got from other dos related programs.
The Windows XP console is very different from the DOS console even if it looks the same.
To make a long story short, console applictions under windows can not access OS and hardware memory directly the way dos applications could.
Therefore a lot of the functions from dos does not work anymore.
Quote:
How ever I don't know what they were using or how they did it. Programing in pascal back then was hard. And today I can create a option like "DrawBox(x1,y1,x2,y2,bg,fg)" in pascal. Al it does is draws a box using ascii codes that draw lines a round the borders.
They were probably using the dos functions that does not work anymore.
An exception is Borlands old compiler tools, like turbo pascal and turbo c++.
In Visual C++ that stuff won't work.
Quote:
All I'm trying to do is customize my program so that the menus look better with back ground graphics. If that what you want to call them..
I assume that you mean boxes with shadow effects and colors here. Real graphics like bitmaps can not be drawn in a console no matter what.
The best way to do that these days is to use PDCurses. I can give you info on how to make PDCurses work in Visual C++ if you want.
Quote:
At this time I'm just trying to under stand what a library does.
A library is basically pre-compiled source code that you can link with your program. A library contains normal source code like classes and functions.
Once you have linked such a library to your program, you can use the classes and functions from the library in your own code.
You are using libraries all the time, maybe without knowing it. For example when you say #include <windows.h> in your program, you are linking to a library.
You can also write and compile your own libraries with Visual C++.
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Quote:
Original post by bigjoe11aI was wondering and I have seen some options that add back grounds to my consol projects.An option to drawbg(); = (Draw Back Ground) or some thing like that. and maybe some options to create boxes. and even with a shadow. I'm new at this so please bear with me.Joe
Sorry; do you have a question?
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Quote:
Original post by pulpfist
Quote:
In Visual C++ I been teaching my self Consol applications just to learn the code. I thought maybe that I could create my own back grounds using some of the ideas I got from other dos related programs.
The Windows XP console is very different from the DOS console even if it looks the same.
To make a long story short, console applictions under windows can not access OS and hardware memory directly the way dos applications could.
Therefore a lot of the functions from dos does not work anymore.
Quote:
How ever I don't know what they were using or how they did it. Programing in pascal back then was hard. And today I can create a option like "DrawBox(x1,y1,x2,y2,bg,fg)" in pascal. Al it does is draws a box using ascii codes that draw lines a round the borders.
They were probably using the dos functions that does not work anymore.
An exception is Borlands old compiler tools, like turbo pascal and turbo c++.
In Visual C++ that stuff won't work.
Quote:
All I'm trying to do is customize my program so that the menus look better with back ground graphics. If that what you want to call them..
I assume that you mean boxes with shadow effects and colors here. Real graphics like bitmaps can not be drawn in a console no matter what.
The best way to do that these days is to use PDCurses. I can give you info on how to make PDCurses work in Visual C++ if you want.
Quote:
At this time I'm just trying to under stand what a library does.
A library is basically pre-compiled source code that you can link with your program. A library contains normal source code like classes and functions.
Once you have linked such a library to your program, you can use the classes and functions from the library in your own code.
You are using libraries all the time, maybe without knowing it. For example when you say #include <windows.h> in your program, you are linking to a library.
You can also write and compile your own libraries with Visual C++.
Sorry I couldn't get back to you sooner. Any way Yes. Any information you can get for me about PDCurses. Please Let me Know
Joe
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Quote:
Sorry I couldn't get back to you sooner. Any way Yes. Any information you can get for me about PDCurses. Please Let me Know
No worries =)
I decided to write a little tutorial on getting PDCurses up and running on Visual C++. After all there is quite a few text RPG hobbyist around here who might find it useful.
Here is how I got started with PDCurses in Visual C++.
It could come in handy later on if the question arises.
Note that there might be other ways to compile the source, and there might be pre-compiled versions available online. I just couldn't find any useful ones myself.
COMPILING PDCURSES USING VISUAL C++
Double-clicking on this file will create two new files, NMAKE.EXE and NMAKE.ERR.
Copy both these files to the C:\Windows\System32 folder.
2. Download the pdcurs33.zip (Or a later version)
Place this file somewhere you can find it (For the remainder of this tutorial I'll pretend it is located at C:\Temp), and unzip it.
3. Open a console. (Start -> Run, and type cmd)
Execute the vcvarsall.bat script in the console.
On my system this script is located at: "C:\Program files\Microsoft Visual Studio 9.0\VC\vcvarsall.bat"
The folder names could be slightly different on other systems of course.
This will set up the environment variables necessary to locate the Visual C++ compiler tools in the current console.
4. From inside the console, navigate to the win32 folder in the pdcurses source.
On my system: C:\Temp\pdcurs33\win32
5. Compile the library using the command: nmake.exe -f vcwin32.mak all
With some luck the library will be compiled and the demo programs will show up in the current folder.
You can execute the demo programs to get an idea of what PDCurses is all about.
The files of interest to us is panel.lib, pdcurses.lib located in the win32 folder, and curses.h, curspriv.h, panel.h, term.h located in the pdcurs33 folder.
Create the following folders on your system: C:\pdcurses\include and C:\pdcurses\lib, and copy the lib files mentioned earlier to the lib folder and the h files to the include folder.
SETTING UP THE PDCURSES LIBRARY IN VISUAL C++
Start Visual C++.
From the menus, go to: Tools -> Options -> Projects and Solutions -> VC++ Directories.
At the upper right corner there will be a drop-down list. Select 'Include files' in this list, and add the C:\pdcurses\include folder to the list below.
Now, select 'Library files' in the drop-down list and add the C:\pdcurses\lib folder to the list below.
This configuration step is system-wide and needs to be done only once.
At this point Visual C++ is set up to find and use the pdcurses library so we are ready to give it a test.
Create a new console project and add the following code:
#include <curses.h>#pragma comment(lib, "pdcurses.lib")int main(void) { WINDOW *mainWindow = initscr(); noecho(); cbreak(); nodelay(mainWindow, TRUE); wrefresh(mainWindow); WINDOW *helloWindow = newwin(12, 25, 5, 5); box(helloWindow, ACS_VLINE, ACS_HLINE); curs_set(0); while(1) { int ch = getch(); if(ch == 27) break; mvwprintw(helloWindow, 5, 6, "Hello world!"); wrefresh(helloWindow); } endwin(); return 0;}
One last thing to do:
Right-click the project in the Solution Explorer and go to Properties -> Configuration properties -> C/C++ -> Code generation,
and make sure the option Runtime Library is set to Multi-threaded (/MT)
Compile and run the program. If everything works as expected you can exit by pressing escape.
For a detailed description of the PDCurses API see PDCurses User's Guide
Cheers
[Edited by - pulpfist on January 8, 2008 1:02:24 PM]
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6. Compile the library using the command: nmake.exe -f vcwin32.mak all
When trying to compile the vcwin32.mak file I got this error
vcwin32.mak(all) : fatal error U1088 : invalid separtor '::' on inference rule stop
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Yea I get that error if I don't execute the vcvarsall.bat script first.
You are using Visual C++ 2008 express right?
Write this line in the console ( including the " ):
"C:\Program files\Microsoft Visual Studio 9.0\VC\vcvarsall.bat"
and press enter.
(If the hard disk where you have Visual C++ installed is called something else than C: you must change that. Also, if you use Visual C++ 2005 express the 9.0 should be 8.0)
Now the nmake command should work
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By the way, installing perl doesn't seem to be necessary. I have perl installed anyway but if you don't want it you should be able to skip installing perl as described in step 1 and just place the two nmake files in a folder visible on your path variable. Like C:\Windows\System32.
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# If time dilation can slow time down, is there away to speed time up?
Okay, I know the title is really confusing but I couldn't find words to explain it sorry. Pretty much what I mean is, if I can get in a lightspeed spaceship moving away from earth, time slows down for me. So one year for me will be 20 earth years or what ever. But is there away were I can reverse this? Where if I get on that same craft and travel for a year but it only will be a few months on earth? I know this is just a random thought.
No, there is not way to have the reverse effect in special relativity. This is because an object being at rest maximizes the time that elapses for it in relativity (a consequence of the lagrangian formulation of special relativity). So if your question is you have your friend sit on earth for a few (say three) months, and you want to know what is the most you can age in the time it takes your friend to wait three months, then the answer is three months, and this is accomplished by sitting on earth doing nothing.
In general relativity, you can use a gravitational field to accomplish what you want. Assuming you are already on earth, you just need to go a region of lower spacetime curvature, such as outer space, and then wait there. Time will pass faster for you, and if you wait long enough, more time will have passed for your friend than for you when you come back to earth. This was illustrated in the movie Interstellar, where we saw a man on a spaceship age much faster than the people on the surface of the planet being orbited.
• You could also accelerate everyone else to near light speed, although that might be impractical. – Jahan Claes Dec 12 '16 at 17:40
• @JahanClaes You are clearly an evil genius. – Cort Ammon Dec 13 '16 at 0:23
• @JahanClaes That's the most correct answer so far haha. – YoTengoUnLCD Dec 13 '16 at 0:54
The answer, as others have said, is generally 'no' (see caveats below). There is a lovely geometrical reason for this, which is the point of this answer.
## Without gravity
So first of all consider flat spacetime -- no gravity -- and think about two events -- points in spacetime -- which are timelike separated: one of the events is in the future of the other or, to be precise, you can get from one event to the other without travelling at the speed of light.
So, for instance, $e_1$ might be 'here, now' and $e_2$ might be 'here, in a week', or $e_1$ might be 'here, now', and $e_2$ might be 'in New York, in an hour'.
So, now consider all the possible ways we can get between $e_1$ and $e_2$: geometrically, these are all the possible timelike curves passing through $e_1$ and $e_2$: they must be timelike because we can't go faster (or even as fast) as light.
There are lots of these curves, here are three of them:
But there is one special curve, which is in red in the diagram: a straight line between $e_1$ and $e_2$. And now we can crank up Euclid: there is always exactly one such curve between any two events. This curve has the special property that it is the one in which there is no acceleration: that's pretty much how a straight line is defined in fact.
However unlike in euclidean geometry this straight line is not the shortest curve between the two points it connects: it's the longest. In particular it is the curve which has the largest proper time: the curve which, if you follow it between two events, you will experience the longest time. All the other curves -- the curves which are not straight, and on which, if you follow them, you will experience acceleration -- are shorter, and you will experience less proper time when you follow them.
And this is why, in special relativity, you can't experience any longer time than the time you would experience by following the straight line between the two events. (It is also how the twin 'paradox' works: only one twin can follow the geodesic, and the other twin therefore experiences less time.)
## With gravity
Is this still true in the presence of gravity? In general yes, it is, but the situation is more complicated.
First of all, spacetime is no longer flat so we can't cheat and use results from Euclid: we need to create a whole definition of what a 'straight line' is, which is a geodesic. But geodesics have the properties we need: they are local maxima of proper time, you experience no acceleration when you follow them. Additionally, they exist given some mild constraints on the spacetime.
It's easy to see what geodesics look like on Earth, where there is gravity. Consider two events, 'here, now' and '100 metres over there, in 10 seconds time'. There are lots of timelike curves which connect these two events: you could drive a car between one and the other, and (if you are Usain Bolt) you could run between them. All of these trajectories will involve acceleration. But there's one that won't: you could fire a projectile in such a way that it followed a ballistic trajectory and arrived 100 meters over there, in 10 seconds. And there is a unique such trajectory which you can calculate (it's about $50\,\mathrm{m}/\mathrm{s}$ at $78^{\circ}$ from the horizontal I think).
This trajectory is the geodesic between those two events, and, just as in flat spacetime it is the curve which has the longest proper time: a projectile following that trajectory will experience the maximum proper time between these two events.
But now there are some other questions: are geodesics in curved spacetime unique the way that straight lines are, and if they aren't is there a longest one (or a set of longest ones) rather than being able to find ones with unbounded length? I am not completely sure what the answer to this question is: although I suspect very strongly that the answers are 'no' and 'yes' in physically plausible cases (see comment to this answer for an example).
## Caveats: pathological spacetimes
Finally it is possible to construct pathological spacetimes, where there is no longest-proper-time timelike curve, but I think the construction of one shows how physically weird such things would be.
Here's how you could construct one: take a flat spacetime, and in it two spacelike surfaces which do not intersect (so for instance, using some obvious coordinate system, pick $t=0$ and $t=10$ as the two surfaces). Now identify these surfaces. For any two timelike-separated events $e_1$ and $e_2$ between these surfaces you can now construct timelike curves which start from $e_1$, pass $e_2$, and then loop back into the past of $e_2$ through the identified surfaces. And you can iterate this process: there's no upper bound on how long such a curve can be, and someone travelling on it will therefore experience arbitrarily large proper times. Here's a picture of how this construction looks: the dotted lines are the identified spacelike surfaces.
But this spacetime is causality-violating in a seriously horrible way: it's definitely not something you'd like to think of as physically plausible.
• If you are in orbit around earth and you want to go to the antipodal in some fixed amount of time, you can pick whichever direction you want to orbit the earth, so there are infinitely many geodesics by symmetry, and they will all have the same time. If your destination is offset from the antipode, you can still have two geodesics, but they won't have the same time. – Brian Moths Dec 13 '16 at 0:45
• @NowIGetToLearnWhatAHeadIs Yes, doh, I have amended my answer. The thing I really wanted to specify was that it should be hard to find a sequence of geodesics between two events with no upper bound on their lengths. Thank you. – tfb Dec 13 '16 at 1:49
Being in a gravitational field is equivalent to accelerated frame... So we may accelerate towards the earth (in 0 g surrounding, so that acceleration causes us to experience gravity pulling us back in direction opposite to that of we are accelerating towards), so that while we experience the flow of time normally, time in front of us gets slow. And therefore, the time on earth will flow slow, causing few months to pass while we travel years in our time frame.
Time dilation, doesn't slow the time.... it is just that if you travel with the speed of light, you will do a very large amount of work in a very small amount of time, and hence we say that time seem to have been stopped, but in real it does not stop at all. I will explain it as that, if you are traveling around the earth in an jet you will complete one trip approximately in 40-50 hours, while traveling with speed of light you cover the same distance about 7.5 times in only 1 second..(Now, this difference is a huge one, just thing about it,,you have covered a distance which could take you about 40-50 hours in only about $1/7.5$ of a second) so therefore we say that time had stopped because once again you have done a lot amount of work in a very very small amount of time and hence time seems to stop(but never stops..).
• This is not true. – Eva Dec 12 '16 at 17:02
• What? In what universe? This is false. – Les Adieux Dec 14 '16 at 0:41 | {} |
# erb2hz
Convert from equivalent rectangular bandwidth (ERB) scale to hertz
## Syntax
``hz = erb2hz(erb)``
## Description
example
````hz = erb2hz(erb)` converts values on the ERB frequency scale to values in hertz.```
## Examples
collapse all
Set two bounding frequencies in Hz and then convert them to the ERB scale.
`b = hz2erb([20,8000]);`
Generate a row vector of 32 values uniformly spaced on the ERB scale.
`erbVect = linspace(b(1),b(2),32);`
Convert the row vector of values into equivalent frequencies in Hz.
`hzVect = erb2hz(erbVect);`
Plot the two vectors for comparison. As ERB values increase linearly, Hz values increase exponentially.
```plot(erbVect,hzVect,'o') title('ERB vs Hz') xlabel('ERB') ylabel('Hz') grid on```
## Input Arguments
collapse all
Input frequency on the equivalent rectangular band (ERB) scale, specified as a scalar, vector, matrix, or multidimensional array.
Data Types: `single` | `double`
## Output Arguments
collapse all
Output frequency in Hz, returned as a scalar, vector, matrix, or multidimensional array the same size as `erb`.
Data Types: `single` | `double`
## Algorithms
The frequency conversion from the ERB scale to Hz uses the following formula:
`$\begin{array}{l}hz=\frac{{10}^{\frac{erb}{A}}-1}{0.00437}\\ \text{where}\\ A=\frac{1000{\mathrm{log}}_{e}\left(10\right)}{\left(24.7\right)\left(4.37\right)}\end{array}$`
## References
[1] Glasberg, Brian R., and Brian C. J. Moore. "Derivation of Auditory Filter Shapes from Notched-Noise Data." Hearing Research. Vol. 47, Issues 1–2, 1990, pp. 103–138. | {} |
# Homogeneous ODE
• December 2nd 2011, 10:17 AM
losm1
Homogeneous ODE
Equation is $xy' = y \ln \frac{y}{x}$.
I have tried to substitute $v = \frac{y}{x}$ and I am stuck at $\frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}$
To be exact I do not know where to go from there in calculus sense.
• December 2nd 2011, 11:07 AM
Darkprince
Re: Homogeneous ODE
So you have (1/x)dx = -(1/(vlnv-v) )dv= -1/(v(lnv-1)
Now the integral of -1/(v(lnv-1) is -ln(lnv - 1)
Then proceed accordingly, hope I helped :)
• December 2nd 2011, 11:08 AM
alexmahone
Re: Homogeneous ODE
Quote:
Originally Posted by losm1
Equation is $xy' = y \ln \frac{y}{x}$.
I have tried to substitute $v = \frac{y}{x}$ and I am stuck at $\frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}$
To be exact I do not know where to go from there in calculus sense.
$\frac{dv}{v(\ln v-1)}=\ln (\ln v-1)+C$
• December 3rd 2011, 02:19 AM
losm1
Re: Homogeneous ODE
In my example differentials dx and dv are in denominator:
$\frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}$
I'm having trouble applying your formula in this case. Can you please clarify further?
• December 3rd 2011, 03:13 AM
Prove It
Re: Homogeneous ODE
Quote:
Originally Posted by losm1
Equation is $xy' = y \ln \frac{y}{x}$.
I have tried to substitute $v = \frac{y}{x}$ and I am stuck at $\frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}$
To be exact I do not know where to go from there in calculus sense.
Make the substitution \displaystyle \begin{align*} v = \frac{y}{x} \implies y = v\,x \implies \frac{dy}{dx} = \frac{dy}{dx} = v + x\,\frac{dv}{dx} \end{align*} and the DE becomes
\displaystyle \begin{align*} x\,\frac{dy}{dx} &= y\ln{\left(\frac{y}{x}\right)} \\ x\left(v + x\,\frac{dv}{dx}\right) &= v\,x\ln{v} \\ v + x\,\frac{dv}{dx} &= v\ln{v} \\ x\,\frac{dv}{dx} &= v\ln{v} - v \\ x\,\frac{dv}{dx} &= v\left(\ln{v} - 1\right) \\ \frac{1}{v\left(\ln{v} - 1\right)}\,\frac{dv}{dx} &= \frac{1}{x} \\ \int{\frac{1}{v\left(\ln{v} - 1\right)}\,\frac{dv}{dx}\,dx} &= \int{\frac{1}{x}\,dx} \\ \int{\frac{1}{\ln{v} - 1}\,\frac{1}{v}\,dv} &= \ln{|x|} + C_1 \\ \int{\frac{1}{u}\,du} &= \ln{|x|} + C_1 \textrm{ after making the substitution }u = \ln{v} - 1 \implies du = \frac{1}{v}\,dv \\ \ln{|u|} + C_2 &= \ln{|x|} + C_1 \\ \ln{|u|} - \ln{|x|} &= C \textrm{ where }C = C_1 - C_2 \\ \ln{\left|\frac{u}{x}\right|} &= C \\ \ln{\left|\frac{\ln{v} - 1}{x}\right|} &= C \\ \ln{\left|\frac{\ln{\left(\frac{y}{x}\right)} - 1}{x}\right|} &= C \\ \frac{\ln{\left(\frac{y}{x}\right)} - 1}{x} &= A \textrm{ where } A = \pm e^C\end{align*}
You could get y in terms of x if you wanted to.
• December 3rd 2011, 04:17 AM
Darkprince
Re: Homogeneous ODE
Quote:
Originally Posted by losm1
In my example differentials dx and dv are in denominator:
$\frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}$
I'm having trouble applying your formula in this case. Can you please clarify further?
So you have x/dx = v(lnv - 1)/dv implies xdv = v(lnv-1)dx implies (1/x)dx = (1/v(lnv-1)) dv
• December 3rd 2011, 08:47 AM
tom@ballooncalculus
Re: Homogeneous ODE
Yes, that is the idea. However, just in case an overview helps...
http://www.ballooncalculus.org/draw/double/five.png
http://www.ballooncalculus.org/draw/double/fivea.png
http://www.ballooncalculus.org/draw/double/fiveb.png
... where (key in spoiler) ...
Spoiler:
http://www.ballooncalculus.org/asy/chain.png
... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
__________________________________________________ __________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
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# Remainder theorem explained with examples
Learn and know what is the meaning of remainder theorem in mathematics and how and why to use this theorem. We will study about remainder theorem in the class 9 in algebra chapter.
In class 9, we have two main theorems i.e. remainder theorem and factor theorem. These two theorems are important and very useful. Now we will discuss about the remainder theorem and if we observe the name, it is understandable that we need to find the remainder. But the problem is how to find the remainder. So to know this we have to study and learn remainder theorem.
## What is remainder theorem in maths?
Let us consider a polynomial f(x) with the degree greater than 1 or equal to 1 and let “a” be a real number then if the polynomial f (x) is divided by the (x-a) then the remainder is given as f(a).
Note:
If suppose f (x) is divided by (x+a) then the remainder is equal to the value of f (-a)
## Use of learning the remainder theorem:
In some cases we may be dividing a polynomial (with degree more than 1 or equal to 1) with a linear polynomial and after dividing at last we will get the remainder. It is clear that in this process finding the remainder is somewhat difficult and also taking much time. So avoid this if you learn what is remainder theorem? Then you can apply this theorem and easily we can find the remainder.
Examples:
Find the reminder when f(x) = 2${ x }^{ 3 }$${ x }^{ 2 }$ + 6x + 9 is divided by the polynomial x-2.
Solution:
According to the remainder theorem, the remainder of the polynomial f(x) is f(2).
So, remainder = f (2) = 2 × ${ 2 }^{ 3 }$${ 2 }^{ 3 }$ + 6 × 2 + 9 = 16 – 4 +12 + 9 = 37 – 4 = 33.
What is the remainder if g(y) = ${ y }^{ 4 }$ + 6${ y }^{ 2 }$ – 2y + 10 is divided by y + 2.
Solution:
According to the remainder theorem, the remainder of the polynomial g(y) is g(-2).
So, remainder = g (-2) = ${ (-2) }^{ 4 }$ + 6 × ${ (-2) }^{ 2 }$ – 2 × (-2) + 10 = 16 + 24 + 4 + 10 = 54.
## FAQ’s on the remainder theorem:
If g (x) is a polynomial and it is divided by cx+d then what is remainder?
Given that g (x) is a polynomial and cx+d is a linear polynomial which is dividing g (x). Then the remainder is given as g ($\frac { -d }{ c }$).
In the above example, how we got remainder as g ($\frac { -d }{ c }$)?
We have to make divisor equal to zero and find x value. So in the example, divisor is cx+d.
So, cx+d = 0
cx = -d
x = $\frac { -d }{ c }$
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# When is there a free lunch?
The no free lunch theorem (NFL) states that
Theorem (Wolpert and Macready 1997) Let $A$ be any learning algorithm for the task of binary classification with respect to the $0−1$ loss over a domain $\chi$ . Let $m$ be any number smaller than $|X |/2$, representing a training set size. Then, there exists a distribution $D$ over $X \times \{0, 1\}$ such that:
1. There exists a function $f : X \rightarrow \{0, 1\}$ with $L_D(f) = 0$.
2. With probability of at least $1/7$ over the choice of $S \sim D^m$, we have that $L_D(A(S)) \geq 1/8$.
where $L_D$ is the error of the prediction rule.
I know some people in the ML community find the NFL theorem to be of questionable relevance as it seems to assume a uniform distribution over the class of all hypotheses $H = \{ f \text{ such that } f : X \rightarrow \{0, 1\} \}$, and in particular it includes adversarial problem classes.
So how bad is the problem raised by NFL? Or how plausible a structure can you put on $H$ before you get a free lunch?
Can you characterise the complexity required of $H$ required for no free lunch to hold? Or when do you get a free lunch?
• Have you read this? arxiv.org/pdf/1111.3846.pdf In general Hutter's papers are pretty much all on this topic. Some distributions over the hypothesis space give what he defines as "pareto optimality". Basically complex hypothesis should not have the same probability as simple ones. This is justified if you believe the universe is a very big computer and complex hypothesis are more resource intensive to compute. – Cagdas Ozgenc Feb 2 '18 at 7:05
• Thank you for the reference. I had looked at that paper although not in enough detail to fully understand the Solomonoff prior. From your outline I am not sure how plausible I find that assumption. Does he discuss Pareto optimality in that paper? I suppose I am really interested in the boundary between free lunch and no free lunch. I'll try to update the question to better reflect the contents of the paper. – MachineEpsilon Feb 4 '18 at 5:01
• Pareto optimality argument is in this paper: jmlr.org/papers/volume4/hutter03a/hutter03a.pdf – Cagdas Ozgenc Feb 4 '18 at 10:14 | {} |
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## Cosmological implications of electroweak vacuum instability: constraints on the Higgs curvature coupling from inflation
May. 13 - 14:30 - 2021
Speaker: Andreas Mantziris (Imperial College)
Title: Cosmological implications of electroweak vacuum instability: constraints on the Higgs curvature coupling from inflation
Abstract:
The current experimentally measured parameters of the Standard Model (SM) suggest that our universe lies in a metastable electroweak vacuum, where the Higgs field could decay to a lower vacuum state with catastrophic consequences. Our measurements dictate that such an event has not happened yet, despite the many different mechanisms that could have triggered it during our past light-cone. Via this observation, we can establish a promising link between cosmology and particle physics and thus constrain important parameters of our theories. The focus of our work has been to explore this possibility by calculating the probability of the false vacuum to decay during the period of inflation and using it to constrain the last unknown renormalisable SM parameter $\xi$, which couples the Higgs field with space-time curvature. In our latest study, we derived lower bounds for the Higgs-curvature coupling from vacuum stability in three inflationary models: quadratic and quartic chaotic inflation, and Starobinsky-like power-law inflation. In contrast to most previous studies, we took the time-dependence of the Hubble rate into account both in the geometry of our past light-cone and in the Higgs effective potential, which is approximated with three-loop renormalisation group improvement supplemented with one-loop curvature corrections. We find that in all three models, the lower bound is $\xi \gtrsim 0.051 ... 0.066$ depending on the top quark mass. We also demonstrated that vacuum decay is most likely to happen a few e-foldings before the end of inflation.
Room: ZOOM videoconference (contact justin.feng@tecnico.ulisboa.pt for URL)
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## On the Hodge structure of projective hypersurfaces in toric varieties.(English)Zbl 0851.14021
Let $$\Sigma$$ be a simplicial complete fan for a free $$\mathbb{Z}$$-module $$N$$ of rank $$d$$ and denote by $$P_\Sigma$$ the associated $$d$$-dimensional toric variety over the complex number field $$\mathbb{C}$$. M. Audin [“The topology of torus actions on symplectic manifolds”, Prog. Math. 93 (1991; Zbl 0726.57029)] and D. Cox [J. Algebr. Geom. 4, No. 1, 17-50 (1995; Zbl 0846.14032)] showed that $$P_\Sigma$$ is a geometric quotient of a Zariski open subset $$U(\Sigma)$$ of an affine space $$\mathbb{A}^n$$ by a linear diagonal action of an algebraic subgroup $$D(\Sigma) \subset (\mathbb{C}^*)^n$$. Here $$n$$ is the number of 1-dimensional cones in the fan $$\Sigma$$, $$\mathbb{C}^*$$ is the multiplicative group of non-zero complex numbers, and the character group of $$D(\Sigma)$$ coincides with the group $$\text{Cl} (\Sigma)$$ of linear equivalence classes of Weil divisors on $$P_\Sigma$$.
The complement of $$U(\Sigma)$$ in $$\mathbb{A}^n$$ is of codimension at least 2 so that the ring of regular algebraic functions on $$U(\Sigma)$$ coincides with the polynomial ring $$S(\Sigma)= \mathbb{C} [z_1, \dots, z_n]$$ which carries a canonical grading with respect to the additive group $$\text{Cl} (\Sigma)$$ induced by the action of $$D(\Sigma)$$ on $$\mathbb{A}^n$$. The second author called $$S(\Sigma)$$ the homogeneous coordinate ring of the toric variety $$P_\Sigma$$. It is indeed an extremely fruitful generalization of the homogeneous coordinate rings of projective spaces and weighted projective spaces. For instance, a hypersurface $$X\subset P_\Sigma$$ is defined by a homogeneous polynomial $$f\in S(\Sigma)$$ of degree equal to the linear equivalence class $$\beta\in \text{Cl} (\Sigma)$$ of the divisor $$X$$.
Here is what the authors do in this paper among other things: When $$X\subset P_\Sigma$$ is a quasi-smooth ample hypersurface, the authors relate the pure Hodge structure on the primitive cohomology group $$PH^{d-1} (X, \mathbb{C})$$ with the Jacobian ring $$S(\Sigma)/ (\partial f/\partial z_1, \dots, \partial f/ \partial z_n)$$, generalizing the classical results for hypersurfaces in projective spaces and weighted projective spaces due to P. A. Griffiths [Ann. Math., II. Ser. 90, 460-495, 496-541 (1969; Zbl 0215.08103)], J. Steenbrink [Compos. Math. 34, 211-223 (1977; Zbl 0347.14001)] and I. Dolgachev [in: Group actions and vector fields, Proc. Pol.-North Am. Semin., Vancouver 1981, Lect. Notes Math. 956, 34-71 (1982; Zbl 0516.14014)].
Reviewer: T.Oda (Sendai)
### MSC:
14M25 Toric varieties, Newton polyhedra, Okounkov bodies 14C30 Transcendental methods, Hodge theory (algebro-geometric aspects) 14J70 Hypersurfaces and algebraic geometry
Full Text:
### References:
[1] M. Atiyah, Complex analytic connections in fibre bundles , Trans. Amer. Math. Soc. 85 (1957), 181-207. JSTOR: · Zbl 0078.16002 [2] M. Audin, The topology of torus actions on symplectic manifolds , Progress in Mathematics, vol. 93, Birkhäuser Verlag, Basel, 1991. · Zbl 0726.57029 [3] V. Batyrev, On the classification of smooth projective toric varieties , Tohoku Math. J. (2) 43 (1991), no. 4, 569-585. · Zbl 0792.14026 [4] V. V. Batyrev, Variations of the mixed Hodge structure of affine hypersurfaces in algebraic tori , Duke Math. J. 69 (1993), no. 2, 349-409. · Zbl 0812.14035 [5] D. Buchsbaum and D. Eisenbud, Algebra structures for finite free resolutions, and some structure theorems for ideals of codimension $$3$$ , Amer. J. Math. 99 (1977), no. 3, 447-485. JSTOR: · Zbl 0373.13006 [6] D. Cox, The homogeneous coordinate ring of a toric variety , to appear in J. Algebraic Geom. · Zbl 0846.14032 [7] D. Cox, R. Donagi, and L. Tu, Variational Torelli implies generic Torelli , Invent. Math. 88 (1987), no. 2, 439-446. · Zbl 0594.14011 [8] V. Danilov, de Rham complex on toroidal variety , Algebraic geometry (Chicago, IL, 1989) eds. I. Dolgachev and W. Fulton, Lecture Notes in Math., vol. 1479, Springer, Berlin, 1991, pp. 26-38. · Zbl 0773.14011 [9] V. Danilov, Newton polyhedra and vanishing cohomology , Functional. Anal. Appl. 13 (1979), 103-115. · Zbl 0427.14006 [10] V. Danilov, The geometry of toric varieties , Russian Math. Surveys 33 (1978), 97-154. · Zbl 0425.14013 [11] V. Danilov and A. Khovanskiî, Newton polyhedra and an algorithm for computing Hodge-Deligne numbers , Math. USSR-Izv. 29 (1987), 279-298. · Zbl 0669.14012 [12] 1 P. Deligne, Théorie de Hodge. II , Inst. Hautes Études Sci. Publ. Math. (1971), no. 40, 5-57. · Zbl 0219.14007 [13] 2 P. Deligne, Théorie de Hodge. III , Inst. Hautes Études Sci. Publ. Math. (1974), no. 44, 5-77. · Zbl 0237.14003 [14] M. Demazure, Sous-groupes algébriques de rang maximum du groupe de Cremona , Ann. Sci. École Norm. Sup. (4) 3 (1970), 507-588. · Zbl 0223.14009 [15] I. Dolgachev, Weighted projective varieties , Group actions and vector fields (Vancouver, B.C., 1981) ed. J. B. Carrell, Lecture Notes in Math., vol. 956, Springer, Berlin, 1982, pp. 34-71. · Zbl 0516.14014 [16] P. Griffiths, On the periods of certain rational integrals. I, II , Ann. of Math. (2) 90 (1969), 460-495; ibid. (2) 90 (1969), 496-541. JSTOR: · Zbl 0215.08103 [17] A. Grothendieck, On the de Rham cohomology of algebraic varieties , Inst. Hautes Études Sci. Publ. Math. (1966), no. 29, 95-103. · Zbl 0145.17602 [18] B. Grünbaum, Convex polytopes , With the cooperation of Victor Klee, M. A. Perles and G. C. Shephard. Pure and Applied Mathematics, Vol. 16, Interscience Publishers John Wiley & Sons, Inc., New York, 1967. · Zbl 0163.16603 [19] R. Hartshorne, Algebraic geometry , Graduate Texts in Math., Springer-Verlag, New York, 1977. · Zbl 0367.14001 [20] J. Humphreys, Linear algebraic groups , Springer-Verlag, New York, 1975. · Zbl 0325.20039 [21] G. Kempf, F. Knudsen, D. Mumford, and B. Saint-Donat, Toroidal embeddings. I , Lecture Notes in Mathematics, vol. 339, Springer-Verlag, Berlin, 1973. · Zbl 0271.14017 [22] A. Khovanskiî, Newton polyhedra and toroidal varieties , Functional Anal. Appl. 11 (1977), 289-296. · Zbl 0445.14019 [23] Yu. Manin and M. Tsfasman, Rational varieties: algebra, geometry, and arithmetics , Russian Math. Surveys 41 (1986), 51-116. · Zbl 0621.14029 [24] D. Mumford and J. Fogarty, Geometric invariant theory , 2nd ed., Ergebnisse der Mathematik und ihrer Grenzgebiete [Results in Mathematics and Related Areas], vol. 34, Springer-Verlag, Berlin, 1982. · Zbl 0504.14008 [25] T. Oda, Convex bodies and algebraic geometry , Ergebnisse der Mathematik und ihrer Grenzgebiete (3) [Results in Mathematics and Related Areas (3)], vol. 15, Springer-Verlag, Berlin, 1988. · Zbl 0628.52002 [26] C. Peters and J. Steenbrink, Infinitesimal variations of Hodge structure and the generic Torelli problem for projective hypersurfaces (after Carlson, Donagi, Green, Griffiths, Harris) , Classification of algebraic and analytic manifolds (Katata, 1982) ed. K. Ueno, Progr. Math., vol. 39, Birkhäuser Boston, Boston, MA, 1983, pp. 399-463. · Zbl 0523.14009 [27] D. Prill, Local classification of quotients of complex manifolds by discontinuous groups , Duke Math. J. 34 (1967), 375-386. · Zbl 0179.12301 [28] G. Reisner, Cohen-Macaulay quotients of polynomial rings , Advances in Math. 21 (1976), no. 1, 30-49. · Zbl 0345.13017 [29] G. Scheja and U. Storch, Über Spurfunktionen bei vollständigen Durchschnitten , J. Reine Angew. Math. 278/279 (1975), 174-190. · Zbl 0316.13003 [30] R. Stanley, Combinatorics and commutative algebra , Progress in Mathematics, vol. 41, Birkhäuser Boston Inc., Boston, MA, 1983. · Zbl 0537.13009 [31] J. Steenbrink, Intersection form for quasi-homogeneous singularities , Compositio Math. 34 (1977), no. 2, 211-223. · Zbl 0347.14001
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | {} |
# Electric Potential Energy Problem
## Homework Statement
A charge of –4.00 µC is fixed in place. From a horizontal distance of
55.0 cm, a particle of mass 2.50 × 10–3 kg and charge –3.00 µC is fired with an
initial speed of 15.0 m/s directly toward the fixed charge. How far does the
particle travel before it stops and begins to return back?
KE = 1/2 mv^2
Pba = -W = -qEd
F = kQ1Q2/r^2
## The Attempt at a Solution
1) Found the Kinetic energy of the moving particle :
KE = 1/2mv^2
= 1/2 (2.5x10^-3)(15)^2
= 0.281 J
2) Set the value I found for KE to PE and used the Potential Energy eqn:
PE = -qEd
Since E = kQ/d^2
PE = -qd(kQ/d^2)
Therefore: d = -qkQ/PE
d = 0.38 m
I'm not sure if I did that right. But the answer I came up with looks like it could work. Any help would be appreciated!
Related Introductory Physics Homework Help News on Phys.org
You should not be finding $$\Delta$$PE that way.
$$\Delta$$PE = -qEd only if E is constant. But E is not constant here.
Write down V at the initial point and the final point.
How is PE related to V?
Then subtract the two to find $$\Delta$$PE.
:) | {} |
# Advantages of the sequence definition of limits
I will be teaching an introductory analysis course in the coming semester. In it the students will learn about limits of real sequences, and then will learn about limits of functions in terms of sequences.
More precisely, we will say that $\lim_{x\to a+}f(x) = L$ if whenever $(x_n)$ converges to $a$ with $x_n>a$ for all $n$, we have $\lim_{n\to \infty} f(x_n) = L$. Likewise, we will say that $\lim_{x\to a-}f(x) = L$ if whenever $(x_n)$ converges to $a$ with $x_n< a$ for all $n$, we have $\lim_{n\to \infty} f(x_n) = L$. Then we say that $\lim_{x\to a} f(x) = L$ if both $\lim_{x\to a+}f(x) = L$ and $\lim_{x\to a-}f(x) = L$.
The students will have already seen the $\varepsilon$-$\delta$ definitions of limits of functions in their calculus course. The question then is, how to properly motivate this second (equivalent) definition of limits of functions?
Are there any arguments which become significantly easier when using the sequence definition of limits of functions in place of the $\varepsilon$-$\delta$ definition? (These should be elementary enough to be understood by first year Mathematics undergraduates.)
For instance, I suppose that once one has the Algebra of Limits for sequences, one gets the Algebra of Limits for functions for free. But I'm not convinced that much is to be gained from doing things this way around.
Edit: Thanks for the answers and comments so far. It seems many people are in favour of teaching the sequence definition of limits alongside the $\varepsilon$-$\delta$ definition. I agree that it should be useful to be aware of both definitions. To be certain of this, however, I would still like to see an example of a proof which is simpler when using the sequence definition.
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A common approach, as pointed out in an answer below, is to teach limits of sequences first and then to define limits of functions using sequences, in an attempt to use as few $\epsilon - \delta$ as possible (basically, just enough to prove that $1/n$ converges to $0$ and to prove what you call the "algebra of limits" -- after that, you can get away without ever seeing an $\epsilon$ again in the first year, if you so wish). In your case though, if the students have seen epislons and deltas and do not hate them (as many students do, I have found), there is little motivation to (to be continued) – Pierre Aug 30 '12 at 12:00
little motivation to give the alternative definition. Nevertheless, you could argue that it gives a "dynamical system" feel to the concept of limit, and that can be pretty visual and intuitive with some students. – Pierre Aug 30 '12 at 12:02
There was an MSE question on this pedagogical point, but I have mislaid the bookmark – Yemon Choi Aug 30 '12 at 23:36
Yemon might have math.stackexchange.com/q/113698 or math.stackexchange.com/q/186274 in mind, both containing some relevant remarks. – Theo Buehler Aug 30 '12 at 23:47
Interestingly enough, as far as I remember, we did limits of sequences first (without functions) and then limits of functions (with $\varepsilon-\delta$) and then, as a remark, the connection mentioned above behind the very same Iron Curtain. Actually, in the end of our "limits course", our professor either did limits over nets, or stopped one step short of it: he certainly said all necessary words and made it clear that to talk about of a limit of a mapping, you just need some set of "catchers" in the range space and some set of "tails" in the argument space. That was a bit tough in the beginning but paid off nicely when doing Riemann integration where the tails are either partitions of small mesh or partitions subordinated to a fixed partition. I still find this abstract view rather enlightening; much more enlightening that the lemma in question, which, IMHO, only makes the concept more confusing (though is quite useful as a technical tool). The main reason for this opinion is that this abstract view is unifying: all notions of limit that the students will ever meet fall under this idea, only the choices of catchers and tails vary and only one magic phrase is ever needed: "For every catcher, there is a tail whose image is contained in the catcher". The lemma you mentioned is separating: if used as a definition rather than a remark, it creates an impression that there are many ad hoc concepts of limits that all have to be understood and memorized separately, creating quite a mess in the student's head. The $\varepsilon-\delta$ definition is already hard because it mixes the limit concept and the technical descriptions of the catchers and the tails, i.e., 3 things that can be easily separated and on which you can train the students one by one if you start with the abstract view. To be honest, I haven't tried it myself in the USA yet but I certainly will when teaching freshman analysis (so far it was either business calculus, where the game is never worth the candles, or advanced courses where the concept of limit was assumed to be well-known already).
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+1 for "the very same Iron Curtain". But I don't get the "catchers" and "tails" business. What's the mental image here? – Tom Leinster Aug 30 '12 at 11:55
@fedja: Thanks. By "the lemma in question", are you referring to the fact that the sequence and $\varepsilon$-$\delta$ definitions of limit of a function coincide? If so, could you expand on your final sentence (in particular the part in parentheses)? – Mark Grant Aug 30 '12 at 11:56
@Tom I'll answer first in the formal language: We have $f:X\to Y$. To set up a limit, we need $T\subset S(X)$ (tails) and $C\subset S(Y)$ (catchers) ($S(Z)$ is the set of all subsets of $Z$). Then we just say that "$f$ has a limit" if for every $B\in C$ there exists $A\in T$ with $f(A)\subset B$ (whenever the argument is restricted to some tail, the value is caught in the given catcher). That's the underlying idea. – fedja Aug 30 '12 at 12:17
The rest are just choices: for instance, if $C$ is the set of all open sets containing a set $K$ and $T$ is the set of truncated angles around a given ray in $R^n$, we say that $f$ is attracted to $K$ when the argument escapes to infinity along a given direction. The beauty, as I said, is in one concept serving all cases. To have it meaningful, a couple of simple axioms are needed like "the intersection of two tails contains some tail" and "all tails are non-empty", but after that you are free to design anything you want. – fedja Aug 30 '12 at 12:21
I can only speak of my experience. I was first taught convergence of sequences and then later on, the definition of limits with $\varepsilon-\delta$ and its equivalent form involving sequences. I still find this most intuitive, but then this opinion is clearly heavily influenced by my upbringing.
For disclosure, I received my math education behind the Iron Curtain, and this model was employed by most (now former) communist countries. During that time curricula in most of those countries were devised by influential mathematicians who could also communicate math very well. (For example, in USSR Kolmogorov was deeply involved in shaping math education. He even wrote some high-school textbooks that were widely used.)
What I am attempting to communicate here is that the sequences-first system was adopted by informed mathematicians who cared about math education, it was tested on a large scale (tens if not hundreds of millions of students) for a long time (several decades). Arguably this system has produced good results.
Terry Tao's textbook on analysis (which I like very much for several reasons) also relies on a sequences-first approach.
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@Liviu: Thank you for your answer. Certainly I am not knocking this order of doing things from a pedagogical viewpoint. My only concern is that when one gives a definition of a concept, then later gives a second equivalent definition of the same concept, one should motivate the second definition by showing that it makes life easier than the first definition in some cases. Otherwise, the lazier students will ask why they need to learn two definitions of the same concept (and I won't have a good answer). – Mark Grant Aug 30 '12 at 11:22
In teaching a second year UK course in analysis some years ago I was surprised to find out how pleasant were the proofs of the some of the basic results using sequential methods, for example that a continuous function on a closed bounded subset of $\mathbb R^n$ to $\mathbb R$ is bounded. This inspired me to work out a paper on the notion of a $1$-point sequential compactification: add another point and let the sequences in $X$ with no convergent subsequence converge to the extra point! It got published too in the JLMS.
However for continuity of a function I do like to rely on the neighbourhood definition since a neighbourhood is a geometric object one can draw, whereas the $\epsilon - \delta$ are only measurements of the sizes of neighbourhoods.
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"Are there any arguments which become significantly easier...". There is an obvious choice: Disprove that a certain function has a limit at a certain point. Showing that with epsilon-delta would be a mission impossible for most students. Using the sequence definition makes it much easier and more appealing.
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This is a good response to Christian Blatter's answer. It may be harder to prove continuity, but by the same token, it is much easier to disprove it. – Ryan Reich May 5 '13 at 20:41
In my view there are no advantages in defining function limits via sequences, and this practice should be abolished. Using this definition you would have to test $\aleph_1^{\aleph_0}$ or so sequences to prove a single instance of $\lim_{x\to a} f(x)=\alpha$. Why should one bring all these sequences into the picture?
The idea of "limit of $f(x)$ when $x\to a$" is the answer to the following question: What is the "natural" value of $f$ at the special, maybe "ideal", point $a\$? Well, it's the value that would make $f$ continuous there.
This brings me to the main point: The primary and sufficiently intuitive notion is the notion of continuity. Unfortunately the simple concept of Lipschitz continuity does not cover all cases we'd like to handle, e.g. $\sqrt{|x|}$ at $x=0$. Therefore we have to dig deeper and come up with $\epsilon$ and $\delta$, and on, and on.
Sequences, on the other hand, are a fundamental tool to construct new objects, like $e$, or $\sqrt{2}$. Of course in passing we would then prove that $\lim_{x\to a}f(x)=\alpha$ iff for all sequences $\ldots$
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Sorry, but the argument on the cardinality of the set of sequences one needs to test makes no sense to me. Why don't you have a problem with having to test for each of the $2^{\aleph_0}$ (assuming this is the cardinailty of the continum) many values of $\varepsilon$ that something holds for the typically $2^{\aleph_0}$ many $x$ in some $\delta$-neighborhood? – quid Aug 30 '12 at 20:43
The above being said, while I am not fully deceded on the general matter, I am aware of the fact that some students have difficulty with the 'for each sequence' and so on, for instance just showing something for one specific choice of sequence and thinking they are done. It is just this cardinality argument that I cannot follow, not the general point of view. – quid Aug 30 '12 at 20:59
Not to insists too much on what is essentially tangential to your argument (in my mind), yet since you changed the way you write the cardinality: even if it now says $\aleph_1^{\aleph_0}$ , this is still $2^{\aleph_0}$, the cardinality of the continuum (except perhaps in some more exotic set theories), and you won't get the thing 'cheaper' than with considering continuum many things somewhere. I think there really is no argument to be made based on cardinalities. To insist on brining this aspect in, in my opinion takes away from an (otherwise) very reasonable argument. – quid Aug 31 '12 at 12:27
@quid: Of course the thing with the cardinality was meant to be a joke. The essential point is that bringing in sequences replaces a simple limit by an infinity of composite ("verschachtelte") limits, over which reigns an additional $\forall$. – Christian Blatter Aug 31 '12 at 13:00
Oh, sorry, for missing the tongue-in-cheek nature of that part of the argument. This makes more sense then. By the way, I like your books on the subject. – quid Aug 31 '12 at 13:20 | {} |
# Trouble with subfloats/subcaption/subfigure [closed]
I compiled my tex file in Linux without any problems (text edior: Xemacs). I'm now running the file in TeXWorks with MikteX. I used the subfig package and the following worked perfectly:
\subfloat{\label{fig:a}}{\includegraphics[scale=0.12]{somefigname.jpg}}
or
\subfloat[n=50]{\label{fig:a}}{\includegraphics[scale=0.14]{someotherfigname.jpg}}
I kept getting the 'undefined control sequence' error after trying to compile it in Win7. Having done some research, I installed the caption package (by running pdflatex caption.ins) and changed the \subfloat command above to \subcaption. Now instead of the error above, I'm getting two new:
! Package caption Error: For a successful cooperation we need at least version 2004/02/13 v1.2' of package listings, but only version 2002/04/01 1.0 (Carsten Heinz)' is available.
The second one is:
! LaTeX Error: Counter too large.
When trying to install the caption package through mpm, I keep getting the 'connection failure' error.
I'm kind of lost here and have no idea what to do. Any suggestions?
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## closed as too localized by yo', Torbjørn T., Joseph Wright♦Jan 5 '13 at 22:57
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.
Welcome to TeX.SE. It seems that your listings package is quite old. I would suggest you upgrade to TeXLive2012 unless there is a really good reason to use a very old distribution. – Peter Grill Nov 15 '12 at 6:44
Either install TeX Live 2012 or upgrade to MiKTeX 2.9: your MiKTeX distribution seems quite old. If you don't have disk space constraints, do a full install. – egreg Nov 15 '12 at 14:01 | {} |
Question
# A light bulb is rated at 125 W for a 250 V a.c., supply. Calculate the resistance of the bulb.
Solution
## Power of the bulb, P = 125 W Voltage, V = 250 V The resistance of the bulb is given by R=V2P=(250)2125 =500 Ω
Suggest corrections | {} |
# How do you graph y=2+3 \sin(2(x-1))?
Nov 23, 2014
Step 1: Sketch the graph of $y = \sin x$.
Step 2: Horizontally shrink the graph in Step 1 by a factor of 2.
Step 3: Vertically stretch the graph in Step 2 by a factor of 3.
Step 4: Shift the graph in Step 3 right by 1 unit.
Step 5: Shift the graph in Step 4 up by 2 units.
The above is the graph of $y = 2 + 3 \sin \left(2 \left(x - 1\right)\right)$.
I hope that this was helpful. | {} |
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ISSN : 1598-7248 (Print)
ISSN : 2234-6473 (Online)
Industrial Engineering & Management Systems Vol.21 No.2 pp.390-400
DOI : https://doi.org/10.7232/iems.2022.21.2.390
# Optimization of Renewable Warranty by Considering System Reliability and Preventive Maintenance and Repairs
Benjarut Chaimankong, Paitoon Chetthamrongchai*
Faculty of Business Administration, Kasetsart University, Bangkok, Thailand
*Corresponding Author, E-mail: fbusptc@ku.ac.th
March 20, 2022 ; March 20, 2022 ; April 12, 2022
## Abstract
One of the effective ways to ensure the reliability of the product sold is to consider service and warranty contracts. Important decision variables in warranty policies include determining the optimal warranty period and the optimal number of maintenance activities. In this research, a model has been developed with the aim of achieving the lowest expected cost rate in the life of the device and the appropriate reliability by performing the optimal number of preventive maintenance measures. The warranty policy in this study is renewable in two dimensions, in which two dimensions of repair time and failure time are considered. Any damage leading to repair is done free of charge by the manufacturer and damage leading to replacement is done jointly between the manufacturer and the consumer by agreement. As a result, two algorithms of Genetic and Imperialist Competitive Algorithm have been developed to solve the model and have been compared with numerical example solving. Furthermore, the shelfs life of the system and the number of optimal repair and maintenance measures for maintaining reliability required by the buyer were obtained.
## 1. INTRODUCTION
Product reliability has become an important issue due to the everyday and rapid growth of technology and serious competition in product marketing. One of the effective methods to ensure product or service reliability is to consider aftersales services. Despite today’s fast and advanced technology, product failure is still possible (Bai et al., 2021). Therefore, many products are sold with a warranty in today’s competitive market. in general, product warranties provide consumers with a written assurance of what a manufacturer will or will not cover in the event of product failure. Product failure may occur early, which might be due to production defects and product deterioration over time. Product failure depends on its age and method of use, and most products sold with a warranty protect buyers against early product failure during a warranty period. In addition, product failure can be controlled by preventive repairs and maintenance, which is important in the cases of warranty period extension (Xu and Saleh, 2021). Moreover, the warranty offered for a product may lead to additional production costs, which could be the cost of repair (corrective repairs) of the product failed during the warranty period. However, these costs could be reduced by preventive repairs during the warranty. From the perspective of manufacturers, preventive maintenance is valuable only when the reduction in warranty service costs is greater than the additional costs incurred in preventive maintenance (Kim et al., 2004). Moreover, most manufacturing companies aim to maxim- ize their profit through the market, which includes profit flow over time. Maximization may occur through an increase of sales to increase revenue or increase of product cost to increase unit profit. Overall, customers decide based on price, which reflects the product quality. However, most customers choose warranty information to assess the product quality and use it to determine whether the product has a fair price or not. Therefore, price and warranty are two important factors for sales decision and final profit of products. In other words, price is a direct and easy sign for product assessment by customers. On the other hand, warranty and its period have a direct effect on the manufacturing costs and final product price as a sign of quality (Lin and Shue, 2005).
Shafiee and Chukova (2013) conducted one of the new review studies in 2013, which focused on models of repair and maintenance in warranty based on cm, pm and post-warranty divisions. In addition, Park et al. (2013) proposed the ARP and BRP maintenance and repair policy in renewable warranty. Moreover, Sheu (1998) evaluated two failure models subject to shock with two ARP and BRP policies. In another study, Jung et al. (2010) studied maintenance and repair costs related to the lifecycle of products under the renewable warranty policy. Yeh and Lo (2001) evaluated preventive maintenance and repair policies and warranty based on partial repair for repairable products. In the foregoing study, the period of warranty was predetermined and the optimal postwarranty period was estimated. Moreover, Jung and Park (2016) obtained the optimal periodic preventive maintenance and repair policy during the post-warranty period after the expiration of renewable and non-renewable warranties from the perspective of the manufacturer. Park and Pham (2010) evaluated the altered quasi-renewal concepts for modeling renewable warranty costs with imperfect repairs from the perspective of manufacturers. In a study by Wang and Pham (1999), some maintenance models and availability with imperfect maintenance in production systems were studied. Moreover, Bai and Pharm (2006) carried out a cost analysis on renewable full-service warranties for multi-component systems. In another study, Zhou et al. (2007) evaluated conditionbased predictive maintenance (CBPM) scheduling for a continuously monitored system independent from warranty policies. Tang and Lin et al. (2015) evaluated a nonperiodic preventive maintenance and repair policy with reliability thresholds for complex repairable systems. In his thesis, Najafabadi presented a warranty prediction model based on neural network systems at K. N. Toosi University of Technology. Wang et al. (2015) proposed an optimal preventive maintenance strategy for repairable items under a two-dimensional warranty. Moreover, Vahdani et al. (2013) presented warranty servicing for discretely degrading items with non-zero repair time under renewing warranty. Furthermore, Bai et al. (2021) introduced adaptive reliability and implemented it in single mode failure in reliability engineering, for which they used the particle swarm algorithm (PSA). Xu and Saleh (2021) applied machine learning to estimate the reliability and safety of complex systems. In this article, we specifically focus on deep learning and introduce it as an efficient reliability increase approach.
## 2. STATEMENT OF THE PROBLEM
It is crucial to study the relationship between warranty and repair and maintenance. Incomplete net preventive activities of the Effective age of the system are reduced by considering maintenance and repair, thereby improving reliability (Abedi et al., 2020;Afanasyev et al., 2021). In fact, net preventive activities make the system younger but do not change the degradation rate. After the original warranty period, preventive maintenance and repair activities are considered non-periodically. During the original warranty period, corrective measures are taken in case of product failure. First, the product is repaired but the process will be terminated and the product will be replaced if the repair time exceeds the predetermined limit. The warranty is not renewed at the time of repair, but the replaced product is considered new and the warranty will be awarded and renewed to the same extent as the original warranty period. moreover, expected costs are considered to be fixed during the maintenance activities. Therefore, the policy of renewed warranty of the combined free repair-shared exchange model is applied. Notably, the manufacturer uses a combination of two FRW and PRW warranties during the warranty period, such that each corrective measure followed FRW and all repair costs are paid by the producer, and each replacement will follow PRW policy in the form of PRO_RATA, and the cost will be divided between the producer and the buyer at the time of selling the product and according to age and based on the pre-agreed agreement. It is worth noting that repair is incomplete in the warranty period, which is an advantage for getting closer to the real world. In addition, while repair and maintenance are incomplete in the postwarranty period, any repair will be partial in that period.
Furthermore, incomplete repair in the warranty period, which is below the specified repair limit, follows the qrp process and operates with a βi coefficient, which is called the age-reducing factor. Moreover, a twodimensional warranty model is considered, which is completely different from the conventional two-dimensional warranty model, which uses product age (actual time of age) and consumption (calendar time) as two factors affecting warranty policy. In addition, the number and time of repairs are used as the main criteria in the warranty policy. The lack of use of the two-dimensional warranty model is the inability to easily estimate the consumption rate in many systems. For instance, while determining the age of electronic devices, such as laptops, computers and refrigerators, or a specific device such as a nuclear reactor, is fairly easy, documenting their actual consumption rate is extremely difficult. In such situations, the use of a twodimensional warranty is not applicable. On the other hand, information is easily obtained based on the time of failure and time of repair for such systems to adopt a new twodimensional warranty policy. Therefore, the usual twodimensional issues (considering the two dimensions of age and time of use) are applicable for renewable issues that are renewed by changing the warranty. Since maintenance and repair are based on conditions in the postwarranty periods, conditions are considered as a variable, meaning that reliability will be added to the model as a problem decision variable and is obtained such that the lowest expected cost is obtained in a certain period.
## 3. METHODOLOGY
The problem model includes two warranty and postwarranty sections, and costs are evaluated from the customer’s perspective. In the first section, costs include replacement costs. However, since these costs are shared between the two parties, the part of costs paid by the consumer must also be estimated. In addition to the replacement costs, costs of system failure exist in the warranty period, which will lead to a cost for the consumer during the correction operations due to system failure. Costs in the post-warranty period include the incomplete implementation of preventive maintenance and repair and costs of partial repair and system failure. At the end of the period, the system will generally be replaced at the expense of the consumer.
### 3.1 Mathematical Symbols
A: Indices
• i = the number of maintenance and repair services during the post-warranty period
B: Parameters
• CD The unit cost of failure in w and pw:
• CR : The unit cost of replacement in w
• βi : Age reduction factor in w period
• r0 : Repair time threshold in w period
• a1,a2 : The scale parameter in the density function in w period for the failure time and repair time, respectively
• b1, b2 : The figure parameter in the density function in w period for the failure time and repair time, respectively
• Na : Number of failures repaired in w period
• Nb : Number of failures replaced in w period
• Cm : The unit cost of incomplete maintenance and repair in pw period
• C0 : Cost of replacement at the end of pw period
• Cr : The unit cost of partial repair in pw period
• δ : Age reduction factor in pw period
• Yi : Effective age of the system before net activity in pw period
• α : Scale parameter in density function in pw period
• Beta : Failure rate in density function in pw period
C: Variables
• W : Warranty period
• Nm : Number of scheduled repairs and maintenance in pw period
• ri : Reliability threshold in the second scenario in pw period
D: Definitions of some abbreviations
• Cdf & pdf : Probability density function and cumulative distribution function
• pm : Preventive maintenance and repair
• NHPP : Non-homogeneous Poisson Process
• ECR : Expected cost rate
• Y,T : Time of failure and time of repair, respectively
• F(t) & F(t),f (t), h(t) : Density function, pdf and cdf and reliability function of T failure time, respectively
• G(y)&G(y), g(y) : Pdf and cdf and reliability function of Y repair time, respectively
• FRW : Free repair Warranty
• FPRW : Renewing Pro-Rata Replacement Warranty
• E(na) : The number of expected repairs in w period
• E(nb) : The number of expected replacements in w period
• Pmf : Probability mass function
### 3.2 Distribution of Number of System Failures
N is the number of system failures, and N distribution and different statistical features caused by warranty cost function are estimated in time unit of sold products.
In this section, pmf of the number of system failures is obtained by Equation 1 under replacement operation which is the same as with a complete repair.
(1)
Each Fi,s (w) is different under each incomplete repair. Therefore, pmf of the number of system failures is obtained by Equation 2:
$P [ N = n ] = ( ∏ i = 1 n ( F is ( w ) ) ) ( R ( n + 1 ) s ( w ) )$
(2)
### 3.3 Proposition of Warranty Modeling under Modified QRP
In the main QRP, each failure has a pattern similar to $X 2 = a . X 1 , X 3 = a . X 2 , X 4 = a . X 3 , ... , X n = a n − 1 . X 1$.
In this pattern, X1 represents the first failure within range, meaning that the repair time is decremental by α deduction, and α is smaller than one. QRP, which has the following pattern, is more explained below:
Definition: a counting process {N(t), t>0}with F distribution and $> 0 β n , β n$ random process and is fixed, and QRP is modified if $X n = β n . X 1$ and n = 2,3,..while $β n وX 1 ~ F$ when n = 2,3,.. , and βn are necessarily not equal. In corrected QRP, pdf and cdf modified for n = 2,3,4,.. are in the form of equations 3 and 4:
(3)
(4)
The reliability function for i failure by using modified QRP is in the form of Equation 5:
(5)
The system reliability function with n failures is based on equations 6 and 7:
(6)
(7)
E(nb) is the number of replacements in w. By using an expected value, the number of replacements is calculated based on Equation 8:
(8)
E(na) the number of repairs in w. Using an expected value, the number of repairs is estimated based on Equation 9:
(9)
### 3.4 Life Cycle Duration during the Warranty Period (Original Warranty and Renewed Warranty)
Since the warranty is renewed after each repair in the TL sections, each replacement is effective in the life cycle period and will be later explained based on Equation 10.
(10)
The number of replacement services is calculated by Equation 11:
(11)
By estimating the expected value on a conditional expected value corresponding to Nb, the expected life cycle is achieved based on the combined warranty model.
(12)
### 3.5 Estimation of System Costs during Warranty Period
Costs of this section are divided into replacement and failure costs, respectively. In addition, the combined warranty model encompasses two policies of FRW (free repair warranty) and RPRW.
#### 3.5.1 Replacement Costs during Warranty Period
As a part of the combined policy, replacements are carried out under the PMR policy, and costs are shared jointly between the two parties. The replacement costs from the perspective of consumers can follow the product age-dependent function of Tnb.
$C r = ∑ j = 1 Nb C r T nb W$
(13)
(14)
$EC(w) = E ( E ( C ( w ) | Nb = nb ) ) = ∑ nb = 0 ∞ { E ( E ( C ( w ) | Nb = nb ) P ( Nb = nb ) }$
(15)
(16)
#### 3.5.2 System Failure Costs during Warranty Period
Based on this part of the policy, the repair costs must be paid by the manufacturer and the consumer will incur no costs for repair. As mentioned above, the product will be replaced if the repair time exceeds the allowed repair time, and the warranty will be renewed by replacing the product. This continues until reaching w age. If F(t) is the age distribution function and f(t) is the probability density function of t, Fi,s (t) will be the cumulative distribution function (cdf) of system failure times after (i-1) repairs during w period. Each Fi,s (w) will be different by considering incomplete repairs. The number of repairs and replacements must be determined to calculate the failure costs. This is mainly due to the fact that the system stops for a certain time with each number of repairs. In addition, while the repairs are not time-consuming, because the repair time spent as much as the amount of time is not counted in the number of repairs performed for the number of replacements made, the calculation method will be, as follows, where the number of failures (repair + replacement) is calculated. Therefore, failure costs will be equal to:
$Total Expected Downtime Cos = ( E ( na ) + E ( nb ) ) × CD$
(17)
### 3.6 Estimation of Post-Warranty Period Costs
In this section, costs include incomplete preventive maintenance and repair costs and costs of doing partial repair and system replacement costs at the end of the period.
#### 3.6.1 Costs of Incomplete Preventive Maintenance and Repair Implementation
In the post-warranty period, the maintenance and repair measures reduce the effective age of the system, thereby improving system reliability. Therefore, the system becomes younger by performing pm measures but the degradation rate cannot be reduced. The system’s age will reduce by different age reduction factors φ with incomplete maintenance and repair activities. It is assumed that the PAS model can be performed to describe system failure status following pm measures based on the knowledge and experience of failure engineers, and the φ reduction factor has an incremental sequence. This is used to reflect failure conditions in the system that become worse, and pm measures will be used after worsening of the conditions if pm measures are performed with similar costs. This assumption is logical since spending similar costs in each pm measure can cause the system in a repairable system to have reduced recovery rates. Accordingly, the effective age of the system is obtained before the i-th pm measure using the following equation:
$y i = w + x i + φ i − 1 y i − 1 = w + x i + φ i − 1 x i − 1 + … + φ i − 1 φ i − 2 … φ 1 x 1$
(18)
when $0 = φ 1 < ... < φ Nm − 1 < 1$ and i =1, 2,…, Nm , assuming that y0 = w and y1 = w + x1, therefore, xi is the period between the i-th and (i-1)-th pm measure φiyi and the effective age representative of the system immediately after the i-th pm measure. It is assumed that in postwarranty period, the failure process follows NHPP given the negligible time for corrective measures, and the system fails with an equal density function between pm measures. The expected number of failures during the life cycle of the system can be achieved using Equation 19:
$N r = ∫ w y 1 λ(t)dt + ∫ φ 1 y 1 y 2 λ(t)dt + … + ∫ φ Nm − 1 y Nm − 1 y i λ(t)dt = [ − w β + ∑ i = 1 Nm y i β − ( φ i − 1 y i − 1 ) β ]$
(19)
Therefore, the total expected value in time unit required for performing pm measures is obtained based on Equation 20:
$C ( Nm ) = 1 T L [ C 0 + C r N r + ( Nm − 1 ) C m ]$
(20)
Cm(Nm1) is the cost of performing pm measures during the post-warranty period.
In addition, the system’s lifetime is estimated in post-warranty mode using the total time intervals for pm cycle based on Equation 21:
$T L = ∑ i = 1 Nm x i = ∑ i = 1 Nm y i − ∑ i = 1 Nm φ i − 1 y i − 1 = y Nm + ∑ i = 1 Nm − 1 ( 1 − φ i ) y i$
(21)
It is notable that the post-warranty costs include the repair costs, pm activity performance costs, and total system replacement costs at the end of the period. These costs are fixed and can be obtained by assessing failures by the engineering section. Therefore, it is logical to assume that C0> Cm.
### 3.7 System Reliability as a Variable Condition
As discussed before, it is logical to consider the conditions to be variable to achieve reliability when pm models are based on conditions. In general, reliability can be the possibility of the system working during some of the t periods. If a positive continuous random variable T is defined as the system failure time, the system reliability can be obtained according to the following equations:
$R ( t ) = Pr { T ≥ t }$
(22)
When $lim t → ∞ R(t) = 0 , R(0) = 1 , R ( t ) ≥ 0$ . Since the failure process has equal density function $λ(t) = αβt β − 1$ based on NHPP assumption
$R ( t ) = Pr { N ( t ) = 0 } = exp [ − ∫ 0 t λ ( u ) du ] = exp ( − αt β )$
(23)
R(t) is the system reliability without pm measures, and $R m ( i ) ( t )$ represents system reliability immediately after the i-th pm activity. It is clear that $R m ( 0 ) ( t ) = R ( t ) = exp ( − αt β )$ when 0≤ t < y1.
Moreover, the system has a y1 age after the first pm measure. In addition, under the first pm measure at y1 age, reliability of the system is equal to:
(24)
For φ1y1 ≤ t < y2, when R(y1) can remain intact (without failure) until the first pm measure.
$R m ( t ) = exp [ − ( ∫ φ 1 y 1 t λ ( u ) du ) ]$
(25)
The possibility of remaining intact for the additional time is t−φ1y1, where the system is maintained until y1. The system’s effective age can be reduced immediately to φ1y1 after the first pm measure with the effect of maintenance and repair, and system reliability is the possibility of remaining intact during the y1+(t−φ1y1) period. In addition, system reliability immediately after the (i-1)-th pm measure is obtained, as follows:
$R m ( i − 1 ) ( t ) = R ( y 1 ) R m ( y 2 | y 1 ) … R m ( t | y i − 1 ) = exp { − α [ − w β + t β + ∑ j = 1 i − 1 ( 1 − φ j β ) y j β ] }$
(26)
In mode of φi1yi−1 ≤ t < yi at the design and planning stage, a suitable strategy that includes the assessment of a system failure behavior is required. Relative costs of maintenance and repair are critical for the decision-maker. The basic assumptions of the study were discussed in the previous section of the research. Two pm models based on reliability are proposed for repairable perishable systems and are assessed according to optimal solutions and appropriate features. In this section, by taking more pm measures to keep the system in the highest state of reliability, it is clear that costs will increase because with performing each pm measure, the system will be out of reach and iteration of operations will be costly. Therefore, it is important for decision-makers to achieve these variables at the highest reliability level, where incurred costs could be managed as well. The goal is to find optimal reliability and the optimal number of maintenance and repair measures in order to minimize the total expected costs. In this regard, the main objective for the decision-maker is to establish a maintenance and repair strategy post-warranty after finishing the warranty periods. Using equation – and considering that reliability is equal to rc in this scenario:
$y i = ( − ln r α ) + w beta beta C m ( N m − 1 ) + C 0 + ( C r + CD ) × ( − w beta + ∑ i = 1 N m [ y i beta − ( δ i − 1 y i − 1 ) beta ] )$
(27)
$W + ( ( − lnr α ) + w beta beta ) ∑ i = 1 N m − 1 ( 1 − δ i )$
(28)
• A) Objective Function Numerator:
$( E ( na ) + ( E ( nb ) ) × CD ) + ( CR W × ( ∫ 0 w xf ( x ) dx 1 − ( ( ∫ 0 w f ( x ) dx ) ( ∫ r 0 ∞ f ( y ) dy ) ) ) ) + C m ( N m − 1 ) + C 0 + ( C r + CD ) ( − ln ( r ) + αw beta ) ( ∑ i = 0 N m − 1 ( 1 − δ i beta ) )$
(29)
• B) Objective Function Denominator:
$∫ 0 w xf ( x ) dx 1 − ( ( ∫ 0 w f ( x ) dx ) ( ∫ r 0 ∞ f ( y ) dy ) ) + ∫ 0 w xf ( x ) dx 1 − ( ( ∫ 0 w f 1 s ( x ) dx ) [ ∏ i = 2 na ( ∫ 0 w 1 β is f is ( 1 β is x ) dx ) ] ( ∫ 0 r 0 f ( y ) dy ) ) + w + ( ( − lnr α ) + w beta beta ) ∑ i = 1 N m − 1 ( 1 − δ i )$
(30)
### 3.8 Introduction of Functions Used in Two Warranty and Post-warranty Periods
The functions used in the warranty period are Weibull distribution and the function used in the postwarranty period is the power law density function. It is assumed that system failure time follows Weibull distribution with different parameters. Product reliability is affected by Weibull distribution parameters that are extensively used in reliability engineering since other distributions (e.g., exponential, Rayleigh and normal) are specific examples of Weibull distribution. Moreover, since its reliability allows it to be an accurate representation of a variety of life distributions, each function parameter is calculated by the maximum likelihood estimation (MLE) method, and the age reduction factor is calculated using the equation below:
$φ i = i ( 2 *i + 1 )$
(31)
According to the mentioned equations, the PAS model is used for failure modeling based on failure engineering knowledge and experience, which can be used for system failure status after each pm measure. In addition, the age reduction factor, which is an increasing trend for reflecting conditions where system return becomes worse, will be suitable for after each pm measures that are performed with equal costs. This assumption is logical since spending equal costs in each pm measure reduces the returning degrees of repairable perishable system.
## 5. RESULTS AND DISCUSSION
The model presented in the previous section is of MINLP type, which generally includes the simultaneous optimization of discrete and continuous values and is of NP-hard problems due to including a high number of parameters that have strong and non-linear interactions with each other. For such problems, as the number of variables increases, the response space grows exponentially. Conventional solution methods such as branch and bound are not adequate for real-size problems that have multiple variables since their solution time is not justifiable. Due to the high solution time and the lack of an optimal global solution to large-scale problems, this section presents a genetic algorithm (GA) and imperialist competitive algorithm to solve the model since the complexity of the problem will have no significant impact on the performance of metaheuristics methods. The mentioned algorithms are commonly used as optimization functions. Using the Taguchi method and trial and error method, the values of the GA parameters are estimated, as follows (Table 1):
Given the use of imperialist competitive algorithm along with GA, the values of the number of iterations and the number of generations are the same in order to compare the two algorithms fairly. Other parameters of the algorithm, which are presented below, are considered equal in different examples (Table 2):
While the convergence of the two algorithms is suitable, the reliability value is rounded by the imperialist competitive algorithm since it is a decimal number and its precision is of paramount importance. However, rounding does not lead to a precise value. Nevertheless, the two algorithms yield very close solutions. According to Table 3, the reliability value (r), the number of maintenance and repair measures (nm) and duration of post-warranty (w) are estimated in seven warranty periods and the results are compared. Expected cost rates of the warranty period have a decreasing trend in 0.2-0.75 periods and an uptrend in 0.75-1.5 periods and reliability have different values according to the warranty period. As the warranty period increases, it is expected that the number of maintenance and repair activities and the duration of the postwarranty period will decrease in order to reduce costs, which is established in all periods except for the maximum warranty period (i.e., 1.5), which, contrary to what is expected, has the longest post-warranty period and the lowest cost rate with the lowest reliability that is achieved at the highest intervals of incomplete preventive maintenance planning and maintenance. When the costs of maintenance and repair measures decrease in the post-warranty period, the cost rates are much less than cm=10 . The optimal cost rate occurs in a 0.4 period, where the number of maintenance and repair activities have the highest possible value and reliability has the highest value. If the 1.5 period is not considered, it will have a high postwarranty period compared to other periods.
According to Figure 1, the objective function extremely increases with an increase in the weight value. In addition, the imperialist competitive algorithm always presents a lower value, compared to the GA. According to Figure 2, reliability decreases with the increase in weight. In addition, the imperialist competitive algorithm yields higher reliability values, which shows the superiority of this algorithm compared to the GA. The sensitivity analysis of the costs of maintenance and repair measures in post-warranty is carried out in Tables 4. The rate of costs is much less than cm=10 when there is a reduction in the cost. In addition, the optimal cost rate occurs in 0.4 period, where there is the highest number of maintenance and repair activities and reliability has the highest value. If the 1.5 period is not considered, it will have a higher post-warranty period compared to other periods. The total expected cost increases with an increase in the maintenance and repair costs. In addition, the lowest rate is observed in the 0.75 warranty period, which has the highest reliability and the lowest number of activities and postwarranty period.
The sensitive analysis of the cost of replacement in the warranty period (CR) is carried out in Tables 5 and 6. The expected cost rate is less than other periods when the mentioned value is lower. In addition, the lowest cost is observed in the 0.75 period, and its reliability is higher than warranty periods less than 0.75. The increase of these costs leads to a slight increase in the expected cost rate, and the lowest rate is observed in the 0.75 period.
## 5. CONCLUSION
Equipment reliability is increasingly important for buyers to reduce sudden failure costs, increase the lifespan of pieces, increase competitiveness, timely meeting of demands, and maintain corporate credibility. Failure may occur due to technical defects, poor performance, equipment lifespan, or environmental and operational stresses. On the other hand, the manufacturer's warranty guarantees the buyer against product failure. In fact, the manufacturer can minimize the effects of such failures by ensuring after-sales service by determining the appropriate warranty period and service contracts. However, inappropriate determining of the warranty period can lead to manufacturer or buyer’s loss instead of profit. The present study evaluated costs from buyers' perspectives, and attempts were made to determine the system’s shelf life.
Given that the seller will bear a part of device repair costs during the sales period, the buyer will benefit from the highest efficiency of the system with the lowest costs. The presented model was solved for numerical examples using GA and imperialist competitive algorithm in the current study. The number of repairs and maintenance activities were assessed in warranty periods 0.2-1.5. The system's shelf life and the number of optimal repair and maintenance measures for maintaining reliability required by the buyer were obtained to minimize the total expected costs.
## Figures
A comparison of GA and imperialist competitive algorithm in terms of objective function.
A comparison of GA and imperialist competitive algorithm in terms of reliability.
## Tables
GA parameter values
Imperialist competitive algorithm parameter values
Evaluation of expected cost rate in different warranty periods
Evaluation of expected cost rate in different warranty periods in cm = 15
Evaluation of the expected cost rate in different warranty periods in CR = 30
Evaluation of expected cost rate in different warranty periods in CR = 50
## References
1. Abedi, A. , Gaudard, L. , and Romerio, F. (2020), Power flow-based approaches to assess vulnerability, reliability, and contingency of the power systems: The benefits and limitations, Reliability Engineering & System Safety, 201, 106961.
2. Afanasyev, V. Y. , Ukolov, V. F. , and Lyubimova, N. G. (2021), An optimal electronic project portfolio under conditions of uncertainty and interactions between projects, Industrial Engineering & Management Systems, 20(4), 536-547.
3. Bai, B. , Guo, Z. , Zhou, C. , Zhang, W. , and Zhang, J. (2021), Application of adaptive reliability importance sampling-based extended domain PSO on single mode failure in reliability engineering, Information Sciences, 546, 42-59.
4. Bai, J. and Pham, H. (2006), Cost analysis on renewable full-service warranties for multi-component systems, European Journal of Operational Research, 168(2), 492-508
5. Jung, G. M. and Park, D. H. (2016), Optimal maintenance policies during the post-warranty period, Reliability Engineering & System Safety, 82(2), 173-185.
6. Jung, K. M. , Park, M. , and Park, D. H. (2010), System maintenance cost dependent on life cycle under renewing warranty policy, Reliability Engineering & System Safety, 95(7), 816-821
7. Kim, C. S. , Djamaludin, I. , and Murthy, D. N. P. (2004), Warranty and discrete preventive maintenance, Reliability Engineering & System Safety, 84(3), 301-309.
8. Lin, P. C. and Shue, L. Y. (2005), Application of optimal control theory to product pricing and warranty with free replacement under the influence of basic lifetime distributions, Computers & Industrial Engineering, 48(1), 69-82.
9. Lin, Z. L. , Huang, Y. S. , and Fang, C. C. (2015), Non-periodic preventive maintenance with reliability thresholds for complex repairable systems, Reliability Engineering & System Safety, 136, 145-156.
10. Park, M. and Pham, H. (2010), Altered quasi-renewal concepts for modeling renewable warranty costs with imperfect repairs, Mathematical and Computer Modelling, 52(9-10), 1435-1450
11. Park, M. , Jung, K. M. , and Park, D. H. (2013), Optimal post-warranty maintenance policy with repair time threshold for minimal repair, Reliability Engineering & System Safety, 111, 147-153.
12. Shafiee, M. and Chukova, S. (2013), Maintenance models in warranty: A literature review, European Journal of Operational Research, 229(3), 561-572.
13. Sheu, S. H. (1998), A generalized age and block replacement of a system subject to shocks, European Journal of Operational Research, 108(2), 345-362.
14. Vahdani, H. , Mahlooji, H. , and Jahromi, A. E. (2013), Warranty servicing for discretely degrading items with non-zero repair time under renewing warranty, Computers & Industrial Engineering, 65(1), 176-185.
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# Cophenetic correlation
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In statistics, and especially in biostatistics, cophenetic correlation[1] (more precisely, the cophenetic correlation coefficient) is a measure of how faithfully a dendrogram preserves the pairwise distances between the original unmodeled data points. Although it has been most widely applied in the field of biostatistics (typically to assess cluster-based models of DNA sequences, or other taxonomic models), it can also be used in other fields of inquiry where raw data tend to occur in clumps, or clusters.[2] This coefficient has also been proposed for use as a test for nested clusters.[3]
## Calculating the cophenetic correlation coefficient
Suppose that the original data {Xi} have been modeled using a cluster method to produce a dendrogram {Ti}; that is, a simplified model in which data that are "close" have been grouped into a hierarchical tree. Define the following distance measures.
• x(i, j) = | XiXj |, the ordinary Euclidean distance between the ith and jth observations.
• t(i, j) = the dendrogrammatic distance between the model points Ti and Tj. This distance is the height of the node at which these two points are first joined together.
Then, letting ${\displaystyle {\bar {x}}}$ be the average of the x(i, j), and letting ${\displaystyle {\bar {t}}}$ be the average of the t(i, j), the cophenetic correlation coefficient c is given by[4]
${\displaystyle c={\frac {\sum _{i
## Software implementation
It is possible to calculate the cophenetic correlation in R using the dendextend R package [1] or in Python using the scipy-package [5].
## References
1. ^ Sokal, R. R. and F. J. Rohlf. 1962. The comparison of dendrograms by objective methods. Taxon, 11:33-40
2. ^ Dorthe B. Carr, Chris J. Young, Richard C. Aster, and Xioabing Zhang, Cluster Analysis for CTBT Seismic Event Monitoring (a study prepared for the U.S. Department of Energy)
3. ^ Rohlf, F. J. and David L. Fisher. 1968. Test for hierarchical structure in random data sets. Systematic Zool., 17:407-412 (link)
4. ^ Mathworks statistics toolbox
5. ^ "scipy.cluster.hierarchy.cophenet — SciPy v0.14.0 Reference Guide". docs.scipy.org. Retrieved 2019-07-11. | {} |
## Reverse mathematics of (co)homology?
Background
Exercise 2.1.16b in Hartshorne (homework!) asks you to prove that if $0 \rightarrow F \rightarrow G \rightarrow H \rightarrow 0$ is an exact sequence of sheaves, and F is flasque, then $0 \rightarrow F(U) \rightarrow G(U) \rightarrow H(U) \rightarrow 0$ is exact for any open set $U$. My solution to this involved the axiom of choice in (what seems to be) an essential way.
Essentially, you are asking to $G(U) \rightarrow H(U)$ to be surjective when you only know that $G \rightarrow H$ is locally surjective. Ordinarily, you might not be able to glue the local preimages of sections in $H(U)$ together into a section of $G(U)$, but since $F$ is flasque, you can extend the difference on overlaps to a global section. This observation deals with gluing finitely many local preimages together. Zorn's lemma enters in to show that you can actually glue things together even if the open cover of $U$ is infinite.
Now, I have not really studied sheaf cohomology, but the idea I have is that it detects the failure of the global sections functor to be right exact. So if you can't even show sheaf cohomology vanishes for flasque sheaves without the axiom of choice, it seems like a lot of the machinery of cohomology would go out the window.
Now, just on the set theoretic level, it seems like there is something interesting going on here. Essentially the axiom of choice is a local-global statement (although I had never thought of it this way before this problem), namely that if $f:X \rightarrow Y$ is a surjection you can find a way to glue the preimages $f^{-1}({y})$ of a surjection together to form a section of the map $f$.
This brings me to my
Questions
Can the above mentioned exercise in Hartshorne be proven without the axiom of choice?
How much homological machinery depends on choice?
Have any reverse mathematicians taken a look at sheaf cohomology as a subject to be "deconstructed"?
Have any constructive set theorists thought about using cohomological technology to talk about the extent to which choice fails in their brand of intuitionistic set theory? (it seems like topos models of such set theories might make the connection to sheaves and their cohomology very strong!)
My google-fu is quite weak, but searches for "reverse mathematics cohomology" didn't seem to bring anything up.
-
Why in the world does writing the text "make a connection" automatically create a link to an Oprah book? – Steven Gubkin Jan 18 2010 at 4:06
I don't have Hartshorne, so I can't address the specifics of this case. However, there is a very interesting paper by Andreas Blass Cohomology detects failures of the Axiom of Choice (TAMS 279, 1983, 257-269), which addresses questions of this type and should at least put you on the right track.
-
That paper looks like exactly what I am after! I will try not to get too excited about it until I have finished writing up my part of our paper though. – Steven Gubkin Jan 17 2010 at 18:37 Actually, I don't think it has exactly what you want, but it will at least be inspirational... – François G. Dorais♦ Jan 17 2010 at 18:44 I have obviously only glanced it over, but it seems to address thelast point to some extent. I think there is a lot that could be said here, and that paper is a good starting point. – Steven Gubkin Jan 17 2010 at 18:50
On any affine scheme and thus on any scheme with a finite affine cover (which I think most people would find a reasonable class of schemes to restrict to), any open cover has a finite subcover. This property is usually called quasi-compact rather than compact for technical reasons.
I don't have Hartshorne with me, so I can't recite chapter and verse, but I know this is discussed somewhere in Hartshorne; it's also covered by this nLab entry, though I suspect that's more technical than you're looking for.
-
I believe that those "technical reasons" are that some people (most notably Bourbaki) require compact spaces to be Hausdorff. – Andy Putman Jan 17 2010 at 18:11
I believe it also has to do with the fact that "quasi-compact" is a property of a morphism of schemes, not just of topological spaces. – Ben Webster Jan 17 2010 at 18:12
The question about sheaves is about sheaves of general topological space. Actually, by what you have said, the case you mentioned is not interesting as far as my questions go. – Steven Gubkin Jan 17 2010 at 18:14
A side comment: As Andy wrote, for Bourbaki, "compact" means "compact Hausdorff", while "quasi-compact" means "compact, but not necessarily Hausdorff". There are certain absolute properties of schemes that can also be applied to morphisms ("quasi-compact, affine, ... "), but this has nothing to do with the "quasi" in "quasi-compact", as far as I know. – Emerton Jan 18 2010 at 5:08 | {} |
# How to make a line break at the title of the parts?
As a continuation of this question: How to center a tcolorbox at a part title at KOMA-script article, if you see the But note of this answer this solution is good only if I don't break lines, i.e., I can put the whole title in one line.
So, my question is, how can I do the same but with line breaks?
Here is a MWE:
\documentclass{scrartcl}
\usepackage{tcolorbox}
\renewcommand*{\raggedpart}{\centering}
\renewcommand*{\partformat}{\partname~\thepart\autodot\enskip}
\renewcommand\partlineswithprefixformat[3]{%
\tcbox[center,width=\linewidth/2]{\underline{#2#3}}%
}
%\usepackage{showframe}% to see the page areas
\begin{document}
\part{The quick brown fox jumped over the lazy dog''}
\end{document}
## 1 Answer
You can e.g. use a \parbox inside. But don't use underline then, at first it doesn't work with line breaks, and at second it would look horrible if you have lines everywhere.
\documentclass{scrartcl}
\usepackage{tcolorbox}
\renewcommand*{\raggedpart}{\centering}
\renewcommand*{\partformat}{\partname~\thepart\autodot\enskip}
\renewcommand\partlineswithprefixformat[3]{%
\tcbox[center,width=\linewidth/2]{\parbox{0.5\linewidth}{%
\centering#2#3\par\vspace{-0.5\baselineskip}\hrulefill}}%
}
%\usepackage{showframe}% to see the page areas
\begin{document}
\part{The quick brown fox jumped over the lazy dog''}
\part{Short}
\end{document}
• As I wrote, it would look horrible and I don't want to spent my time to check if one can implement it with soul or ulem (but even if it works, it will probably break as soon as you add your hebrew). – Ulrike Fischer Mar 13 at 18:26 | {} |
Sharding the Interval Queue: Implementation
Promises, lists and setTimeout woes
The completed Sharded Interval Queue is up on npm if you'd like to take a look or use it. This is the last of a two-parter on the theory and implementation.
In the last post we looked at the theory on how we could implement an interval queue with sharding to support cluster workloads. Here we'll look at the implementation, design choices and pitfalls.
The final code for the sharded-interval-queue is about 200 lines. Compared to 50 for the async-interval-queue, the rise in complexity was expected. What proved difficult was not adding more complexity to deal with issues that propped up.
## §Shared state management
To start, let's ignore the exact implementation of the shared state - we'd like our final queue to be implementation agnostic. Building with a particular one in mind can inject platform-specific ways of thinking into our system and make the early testing and development more complex. For now, we'll use a global dictionary as a stand in:
let globalKeys = {};
From our first principles and constraints, we know that we only need a few functions to operate on this shared state. We need to set and get keys specific to our queue and atomically increment them when possible - that's three. We'll add two helper functions - one that checks if a key is set, and another that initializes storage for when that's needed. We can represent those five operations like so:
async isSet(key) {
return (${this.queueName}_${key} in globalKeys);
}
async increment(key) {
let val = globalKeys[${this.queueName}_${key}];
globalKeys[${this.queueName}_${key}] = val+1;
return val;
}
async setAsync(key, value) {
globalKeys[${this.queueName}_${key}] = value;
}
async getAsync(key) {
return globalKeys[${this.queueName}_${key}];
}
async initStorage() {
globalKeys = {};
}
We're defining them as asynchronous functions, since we expect these operations to possibly go to disk or network for persistence. With these as a base, we've also made it easier to plug-in any custom shared state implementation into the queue: replace these five and you're good.
A queue name is used to associate keys, so you can have multiple queues running across shards, which connect to their brothers through a common name. myQueue_key1 refers to key key1 in queue myQueue. Simple enough to implement and inspect.
(As an aside, I now wish I'd made them aware of the queue name rather than pulling in from the class, as this would have allowed for custom implementations on how keys are associated with the queue name. As you'll see later I did make this work, but it's something I might change later.)
## §Initialization
Next we build our constructor:
constructor(queueName) {
this.queueName = queueName;
this.queue = [];
this.index = 0;
this.job = null;
}
However, we have a problem here that we didn't with the unsharded queue. Since our operations with shared state are asynchronous, we need to separate them from the constructor. Even though we can get all the relevant information when the object is created, constructors can't be asynchronous. Which means we need an init function:
async init(intervalMs, reset, failIfExists) {
await this.initStorage;
if(!(await this.isSet('ticket')) || reset) {
this.setAsync('ticket', 1);
this.setAsync('current', 0);
this.setAsync('interval', intervalMs);
this.setAsync('paused', false);
} else {
if(failIfExists)
throw new Error("Queue exists");
}
}
Not much here but setup. We run initStorage to dust out whichver adapter we're using and create the queue variables if they don't exist. We also have parameters to reset the shared state for an existing queue (which is recommended but not necessary for multiple runs of the same code), and to throw an error if you expected to create a new queue but one exists.
Next we have the add function and the decorator. Thanks to our plan from before, this is still pretty simple:
async add(asyncFunction, doNotStart) {
let queueNo = await this.increment('ticket');
let myPromise = new Promise((resolve, reject) => {
this.queue.push({
queueNo,
asyncFunction().then(value => resolve(value)).catch(err => reject(err));
}
});
});
if(!doNotStart && !this.job) {
await this.start(true);
}
return myPromise;
}
Same parameters as before. We accept an async function - this is a thunk delaying the operation of the actual function (so fetch('https://www.google.com') becomes () => fetch('https://www.google.com')). We increment and get a ticket number which we call queueNo, then push the job onto our local list with the queue number. If the queue isn't running, and doNotStart isn't true, we start the queue. The job is pushed and we return a promise that resolves when the job compeletes.
decorator(asyncFunction, doNotStart) {
let thisQueue = this;
return function () {
return thisQueue.add(() => asyncFunction.apply(this, arguments), doNotStart);
}
}
Same as before, the decorator accepts an async function, then returns a new async function that does all the thunking and enqueueing for you, so you can use it instead.
## §Pause and unpause
Pause and unpause are quite simple, since the complexity is pushed onto the actual queue execution function:
async pause() {
await this.setAsync('paused', true);
this.job = null;
}
async unpause() {
await this.setAsync('paused', false);
if(!this.job && this.index < this.queue.length)
this.start(true);
}
Pause is just setting the shared paused variable to true, and unpause is doing the opposite and starting the queue again.
## §Running
Finally, the meat and potatoes. The core of the queue's functionality is in the runNext function, which schedules the next job and executes the current one. The start function from before simply runs runNext with the correct timeout, and the rest happens here. It's a large unwieldy function, so let's go over it in pieces:
As before, runNext returns a function that is then passed to setTimeout. This is so the values are frozen inside the closure, but this function needs to be a non-async function. setTimeout has weird issues when running async functions, so we consolidate our asynchronous operations within this function and use .then. Makes for more unreadable code, but it executes the same way. When illustrating the code, I'll use await and omit the error handling and value parsing compatibility code, but you can find the final function here. I'm also localizing this inside the function for the closure, but I'll omit that here for readability. this will refer to the queue we're in.
First, we check to see that the queue is meant to be running, and increment the index:
let paused = await this.getAsync('paused');
if(paused || this.index >= this.queue.length) {
this.job = null;
return;
}
this.index++;
Next, we check if the current job still has a slot in the global queue to guard for pause conditions. If the queue has already passed the current job, we get another ticket and requeue it, and call runNext again:
if(this.queue[this.index-1].queueNo < current) {
this.increment('ticket').then(queueNo => {
queueNo = parseInt(queueNo);
this.queue.push({
queueNo,
});
this.runNext()();
});
}
If not, we set a timeout for the next instance of runNext, and run the current job:
if(this.index < this.queue.length) {
this.job = setTimeout(
this.runNext(),
(this.queue[this.index].queueNo-current-1)*interval);
} else {
this.job = null;
}
this.setAsync('current', this.queue[this.index-1].queueNo).then(() => {
});
We calculate the interval to the next queue number based on the current one. We don't have the problem of queue numbers going stale anymore since we no longer derive them from time, and we get the time-based distance between jobs. For a queue of interval 1 second, jobs 1 and 5 have 4 seconds between them, jobs 2 and 3 have one second between them, and so on.
Once this is done, we set current to the job number that is currently executing, and then run the job. This is the most problematic operation for parallel state modification, but the max distance that the variable can move is the time it takes for the setTimeout assignment to execute divided by the queue interval, and in my runs I haven't had a problem. This is partly because the lowest resolution we can get for queue execution is 1 millisecond, and any corruption here will be quickly fixed by the next runNext call. If we wanted more parallel safety or were moving below 1 ms, we could atomically increment current, then check to see if the value was valid and only do the assignment if it wasn't. Or, if the adapter supports it, we could set this value atomically.
This is the entirety of runNext. Next, we move on to storage adapters.
Like I mentioned before, replacing the global object with a custom adapter simply involves replacing the five functions in the queue. I've provided LowDB and Redis implementations and a way to add custom adapters.
For lowdb, we use the initStorage function to make sure we have a valid JSON file to write to. The other functions are implemented as shown below:
this.initStorage = async () => await db.defaults(dbDefaults);
this.getAsync = async key => await db.get(this.queueName).get(key).value();
this.isSet = async key => await db.get(this.queueName).has(key).value();
this.setAsync = async (key, value) => (await db.set(${this.queueName}.${key}, value).write())[this.queueName][key];
this.increment = async (key) => (await db.get(this.queueName).update(key, n=>n+1).write())[key]-1;
For Redis, most of the operations are the same, except for two. We don't need an initStorage function, and we use a lua script to increment a variable atomically:
this.initStorage = async () => null;
this.getAsync = async key => await redisGet(${this.queueName}_${key});
this.setAsync = async (key, value) => await redisSet(${this.queueName}_${key}, value);
this.isSet = async key => (await redisGet(${this.queueName}_${key})) !== null;
this.increment = async (key) => {
const lua =
local p = redis.call('incr', KEYS[1])
return p;
return await redisEval(lua, 1, ${this.queueName}_${key});
}
For a custom adapter, the setStorageCustom function accepts custom implementations for the queue. A small change here is that the queue's functions do not accept a queue name, but I'd later wanted the ability for custom implementations of connecting queues to keys. Custom adapter functions accept a queueName parameter, which is then wrapped to be compatible with the internals of the queue.
this.isSet = async (key) => isSet(this.queueName, key);
## §Testing
And that's it for the implementation. A testing function is implemented in test.js, as well as tests for redis and lowdb, where you can vary the number of active queues, number of jobs and the number of pauses to see how it affects performance. Redis proved to be the fastest implementation with a shared state, which was no surprise.
This has been a fun project, and to my abject surprise has gotten a lot of of use in the week it's been up. I'm glad to see it being used in other projects. It also makes me worry about the correctness of my implementation, and I'm hoping this leads to feedback I can use to improve it. A lot of things are good enough for your own use, but I'm always nervous when someone else tells me they're using my package in production.
Hrishi Olickel
5 Dec 2019 | {} |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 2.4: The Distributive Property
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Apply the distributive property.
• Identify parts of an expression.
• Solve real-world problems using the distributive property.
## Introduction
At the end of the school year, an elementary school teacher makes a little gift bag for each of his students. Each bag contains one class photograph, two party favors and five pieces of candy. The teacher will distribute the bags among his 28 students. How many of each item does the teacher need?
## Apply the Distributive Property
When we have a problem like the one posed in the introduction, The Distributive Property can help us solve it. First, we can write an expression for the contents of each bag: Items = (photo + 2 favors + 5 candies), or simply I=(p+2f+5c)\begin{align*} I = ( p + 2f + 5c)\end{align*}.
For all 28 students, the teacher will need 28 times that number of items, so I=28(p+2f+5c)\begin{align*} I = 28( p + 2f + 5c)\end{align*}.
Next, the Distributive Property of Multiplication tells us that when we have a single term multiplied by a sum of several terms, we can rewrite it by multiplying the single term by each of the other terms separately. In other words, 28(p+2f+5c)=28(p)+28(2f)+28(5c)\begin{align*}28(p + 2f + 5c) = 28(p) + 28(2f) + 28(5c)\end{align*}, which simplifies to 28p+56f+140c\begin{align*}28p + 56f + 140c\end{align*}. So the teacher needs 28 class photos, 56 party favors and 140 pieces of candy.
You can see why the Distributive Property works by looking at a simple problem where we just have numbers inside the parentheses, and considering the Order of Operations.
Example 1
Determine the value of 11(2 - 6) using both the Order of Operations and the Distributive Property.
Solution
Order of Operations tells us to evaluate the amount inside the parentheses first:
11(26)=11(4)=44\begin{align*}11(2 - 6) = 11(-4) = -44\end{align*}
Now let’s try it with the Distributive Property:
11(26)=11(2)11(6)=2266=44\begin{align*}11(2 - 6) = 11(2) - 11(6) = 22 - 66 = -44\end{align*}
Note: When applying the Distributive Property you MUST take note of any negative signs!
Example 2
Use the Distributive Property to determine the following.
a) 11(2x+6)\begin{align*}11(2x + 6)\end{align*}
b) 7(3x5)\begin{align*}7(3x - 5)\end{align*}
c) 27(3y211)\begin{align*}\frac{2}{7} (3y^2 - 11)\end{align*}
d) 2x7(3y211xy)\begin{align*}\frac{2x}{7} \left ( 3y^2 - \frac{11}{xy} \right )\end{align*}
Solution
a) 11(2x+6)=11(2x)+11(6)=22x+66\begin{align*}11(2x + 6) = 11(2x) + 11(6) = 22x + 66\end{align*}
b) Note the negative sign on the second term.
7(3x5)=21x35\begin{align*}7(3x - 5) = 21x - 35\end{align*}
c) 27(3y211)=27(3y2)+27(11)=6y27227\begin{align*}\frac{2}{7} (3y^2 - 11) = \frac{2}{7} (3y^2) + \frac{2}{7}(-11) = \frac{6y^2}{7} - \frac{22}{7}\end{align*}, or 6y2227\begin{align*}\frac{6y^2 - 22}{7}\end{align*}
d) 2x7(3y211xy)=2x7(3y2)+2x7(11xy)=6xy2722x7xy\begin{align*}\frac{2x}{7} \left ( 3y^2 - \frac{11}{xy} \right ) = \frac{2x}{7} (3y^2) + \frac{2x}{7} \left (- \frac{11}{xy} \right ) = \frac{6xy^2}{7} - \frac{22x}{7xy}\end{align*}
We can simplify this answer by canceling the x\begin{align*}x\end{align*}’s in the second fraction, so we end up with 6xy27227y\begin{align*}\frac{6xy^2}{7} - \frac{22}{7y}\end{align*}.
## Identify Expressions That Involve the Distributive Property
The Distributive Property can also appear in expressions that don’t include parentheses. In Lesson 1.2, we saw how the fraction bar also acts as a grouping symbol. Now we’ll see how to use the Distributive Property with fractions.
Example 3
Simplify the following expressions.
a) 2x+84\begin{align*}\frac{2x + 8}{4}\end{align*}
b) 9y23\begin{align*}\frac{9y - 2}{3}\end{align*}
c) z+62\begin{align*}\frac{z + 6}{2}\end{align*}
Solution
Even though these expressions aren’t written in a form we usually associate with the Distributive Property, remember that we treat the numerator of a fraction as if it were in parentheses, and that means we can use the Distributive Property here too.
a) 2x+84\begin{align*}\frac{2x + 8}{4}\end{align*} can be re-written as 14(2x+8)\begin{align*}\frac{1}{4} (2x + 8)\end{align*}. Then we can distribute the 14\begin{align*}\frac{1}{4}\end{align*}:
14(2x+8)=2x4+84=x2+2\begin{align*}\frac{1}{4}(2x + 8) = \frac{2x}{4} + \frac{8}{4} = \frac{x}{2} + 2\end{align*}
b) 9y23\begin{align*}\frac{9y - 2}{3}\end{align*} can be re-written as 13(9y2)\begin{align*}\frac{1}{3}(9y - 2)\end{align*}, and then we can distribute the 13\begin{align*}\frac{1}{3}\end{align*}:
13(9y2)=9y323=3y23\begin{align*}\frac{1}{3} (9y - 2) = \frac{9y}{3} - \frac{2}{3} = 3y - \frac{2}{3}\end{align*}
c) Rewrite z+62\begin{align*}\frac{z + 6}{2}\end{align*} as 12(z+6)\begin{align*}\frac{1}{2} (z + 6)\end{align*}, and distribute the 12\begin{align*}\frac{1}{2}\end{align*}:
12(z+6)=z2+62=z2+3\begin{align*}\frac{1}{2}(z + 6) = \frac{z}{2} + \frac{6}{2} = \frac{z}{2} + 3\end{align*}
## Solve Real-World Problems Using the Distributive Property
The Distributive Property is one of the most common mathematical properties used in everyday life. Any time we have two or more groups of objects, the Distributive Property can help us solve for an unknown.
Example 4
Each student on a field trip into a forest is to be given an emergency survival kit. The kit is to contain a flashlight, a first aid kit, and emergency food rations. Flashlights cost $12 each, first aid kits are$7 each and emergency food rations cost $2 per day. There is$500 available for the kits and 17 students to provide for. How many days worth of rations can be provided with each kit?
The unknown quantity in this problem is the number of days’ rations. This will be x\begin{align*}x\end{align*} in our expression.
Each kit will contain one $12 flashlight, one$7 first aid kit, and x\begin{align*}x\end{align*} times 2 worth of rations, for a total cost of (12+7+2x)\begin{align*}(12 + 7 + 2x)\end{align*} dollars. With 17 kits, therefore, the total cost will be 17(12+7+2x)\begin{align*}17(12 + 7 + 2x)\end{align*} dollars. We can use the Distributive Property on this expression: 17(12+7+2x)=204+119+34x\begin{align*}17(12 + 7 + 2x) = 204 + 119 + 34x\end{align*} Since the total cost can be at most500, we set the expression equal to 500 and solve for x\begin{align*}x\end{align*}. (You’ll learn in more detail how to solve equations like this in the next chapter.)
204+119+34x323+34x323+34x32334x34x34x=500=500=500323=177=177345.206\begin{align*}204 + 119 + 34x & = 500\\ 323 + 34x & = 500\\ 323 + 34x - 323 & = 500 -323\\ 34x & = 177\\ \frac{34x}{34} & = \frac{177}{34}\\ x & \approx 5.206\end{align*}
Since this represents the number of days’ worth of rations that can be bought, we must round to the next lowest whole number. We wouldn’t have enough money to buy a sixth day of supplies.
Solution
Five days worth of emergency rations can be purchased for each survival kit.
## Lesson Summary
• Distributive Property The product of a number and the sum of two numbers is equal to the first number times the second number plus the first number times the third number.
• When applying the Distributive Property you MUST take note of any negative signs!
## Further Practice
For more practice using the Distributive Property, try playing the Battleship game at http://www.quia.com/ba/15357.html.
## Review Questions
1. Use the Distributive Property to simplify the following expressions.
1. (x+4)2(x+5)\begin{align*}(x + 4) - 2(x + 5)\end{align*}
2. 12(4z+6)\begin{align*}\frac{1}{2} (4z + 6)\end{align*}
3. (4+5)(5+2)\begin{align*}(4 + 5) - (5 + 2)\end{align*}
4. x(x+7)\begin{align*}x(x + 7)\end{align*}
5. y(x+7)\begin{align*}y(x + 7)\end{align*}
6. 13x(3y+z)\begin{align*}13x(3y + z)\end{align*}
7. x(3x+5)\begin{align*}x \left ( \frac{3}{x} + 5 \right )\end{align*}
8. xy(1x+2y)\begin{align*}xy \left ( \frac{1}{x} + \frac{2}{y} \right )\end{align*}
2. Use the Distributive Property to remove the parentheses from the following expressions.
1. 12(xy)4\begin{align*}\frac{1}{2}(x - y) - 4\end{align*}
2. 0.6(0.2x+0.7)\begin{align*}0.6(0.2x + 0.7)\end{align*}
3. 6+(x5)+7\begin{align*}6 + (x - 5) + 7\end{align*}
4. \begin{align*}6 - (x - 5) + 7\end{align*}
5. \begin{align*}4(m + 7) - 6 (4 - m)\end{align*}
6. \begin{align*}-5(y - 11) + 2y\end{align*}
7. \begin{align*}-(x - 3y) + \frac{1}{2}(z + 4)\end{align*}
8. \begin{align*}\frac{a}{b} \left ( \frac{2}{a} + \frac{3}{b} + \frac{b}{5} \right )\end{align*}
3. Use the Distributive Property to simplify the following fractions.
1. \begin{align*}\frac{8x + 12}{4}\end{align*}
2. \begin{align*}\frac{9x + 12}{3}\end{align*}
3. \begin{align*}\frac{11x + 12}{2}\end{align*}
4. \begin{align*}\frac{3y + 2}{6}\end{align*}
5. \begin{align*}- \frac{6z - 2}{3}\end{align*}
6. \begin{align*}\frac{7 - 6p}{3}\end{align*}
7. \begin{align*}\frac{3d - 4}{6d}\end{align*}
8. \begin{align*}\frac{12g + 8h}{4gh}\end{align*}
4. A bookcase has five shelves, and each shelf contains seven poetry books and eleven novels. How many of each type of book does the bookcase contain?
5. Amar is making giant holiday cookies for his friends at school. He makes each cookie with 6 oz of cookie dough and decorates them with macadamia nuts. If Amar has 5 lbs of cookie dough \begin{align*}(1 \ lb = 16 \ oz)\end{align*} and 60 macadamia nuts, calculate the following.
1. How many (full) cookies he can make?
2. How many macadamia nuts he can put on each cookie, if each is to be identical?
3. If 4 cups of flour and 1 cup of sugar went into each pound of cookie dough, how much of each did Amar use to make the 5 pounds of dough?
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# English words in written mathematics
I recently marked over $100$ assignments for a multivariable calculus course. One question which a lot of people did poorly was proving a given set was open. Aside from issues relating to rigour and logic (or lack thereof), I noticed an issue that I wasn't as aware of before this experience. A lot of students used English words incorrectly (from a mathematical point of view) when writing sentences within their proof.
Some of the words/phrases I am referring to are:
• As
• Assume
• Let
• Suppose
• Hence
• Therefore
• Such that
• There exists
Is anyone aware of a reference which explains how to use these words and phrases (and others that frequently occur) in a mathematical context?
-
Great question! – user42912 Apr 18 '13 at 11:06
Charles Wells, The Handbook of Mathematical Discourse; it’s available as a PDF here. His site abstractmath.org may also be useful.
-
Thanks Brian, I've only had a quick look but this seems like an excellent resource. – Michael Albanese Apr 18 '13 at 7:36
@Michael: My pleasure. I’ve never actually used it, but it comes closer to what you describe than anything else that I’ve seen. – Brian M. Scott Apr 18 '13 at 7:37
The link don't work for me :( – Paul Apr 18 '13 at 10:13
@Paul: Does the link still not work? – Michael Albanese Oct 5 '13 at 2:46
Again ('cos I've recommended it before) I can very warmly recommend getting your students to beg/borrow/buy and then read the excellent
Daniel J. Velleman, How to Prove it: A Structured Approach (CUP, 1994 and much reprinted, and now into a second edition).
From the blurb: "Many students have trouble the first time they take a mathematics course in which proofs play a significant role. This new edition of Velleman's successful text will prepare students to make the transition from solving problems to proving theorems by teaching them the techniques needed to read and write proofs." And in doing that, the most important thing, Velleman's book should get students to understand the lingo properly as they better understand what they are doing when they make a supposition here, draw an inference there, use a quantifier, etc.
-
I have seen this book mentioned before but I completely forgot about it. My university library doesn't have a copy but I will definitely encourage them to get one. – Michael Albanese Apr 18 '13 at 13:02
His website has some interesting tidbits as well. In particular, the Proof Designer looks like it could be useful. – Michael Albanese Apr 18 '13 at 16:06 | {} |
# Touring the tidyverse: tidyr
This Thursday I’ve given a talk at Berlin R-Users Group. The topic was “Touring the tidyverse: tidy data + tidyr”. The idea is that this will become a series of talks where each consequent talk is going to present one (or couple) of packages from the tidyverse.
tidyr to me seems like a good choice for the first talk since concept of tidy data is so central to all of the packages. Therefore, understanding this concept makes it easier to work in tidyverse. The code for the talk and presentation itself are available at my github - https://github.com/romatik/touring_the_tidyverse.
All talks in the series:
1. tidyr - https://www.mishabalyasin.com/2018/05/27/touring-the-tidyverse-tidyr/
2. dplyr - https://www.mishabalyasin.com/2018/07/26/touring-the-tidyverse-dplyr/
3. purrr - https://www.mishabalyasin.com/2018/10/24/touring-the-tidyverse-purrr/
4. tidyeval - https://www.mishabalyasin.com/2018/11/30/touring-the-tidyverse-tidyeval/
5. tidymodels - https://www.mishabalyasin.com/2019/03/03/touring-the-tidyverse-tidymodels/
6. Beyond tidyverse - https://www.mishabalyasin.com/2019/04/20/touring-the-tidyverse-beyond-tidyverse/
Slides are also here if you don’t want to go to github (process of embeding is described here): | {} |
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# M26-37
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Thread Master - Part Time MBA Programs
Joined: 11 Jan 2018
Posts: 117
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GMAT 1: 620 Q46 V30
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### Show Tags
06 Sep 2018, 08:37
divinka8 wrote:
For statement B, how were we able to conclude that x + y = 2 as opposed to 1? Technically that would make the RHS equal to 25, also an integer? Of course once you start checking the LHS you would realize no combo of x and y would make it equal to 25 but was wondering if there is a clue before that check is conducted. Thank you!
Posted from my mobile device
I was thinking the same. For #2,
(X+Y) = 2, then 5^ 1 = 5
(X+Y) = 1, then 5^ 2 = 25.
But, if you plug in x=1, y=0 or x=0, y=1, the equation of 2^x + 3^y = 25 will not hold. Therefore, X+Y cannot equal to 1.
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### Show Tags
07 Sep 2018, 07:16
Even though I got the right answer, I hope not to get such qsns on the real GMAT. It's a great question for practice!
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Joined: 23 May 2018
Posts: 263
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GMAT 1: 770 Q48 V50
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### Show Tags
30 Sep 2018, 10:04
Bunuel wrote:
Official Solution:
(1) $$2^{x-y}=\sqrt[(x+y)]{16}$$
$$2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$;
Equate the powers: $$x-y=\frac{4}{x+y}$$;
$$(x-y)(x+y)=4$$.
Now, since both $$x$$ and $$y$$ are integers (and $$x+y \gt 0$$) then $$x-y=2$$ and $$x+y=2$$ so, $$x=2$$ and $$y=0$$. Therefore, $$(x+y)^{xy}=2^0=1=odd$$, so the answer to the question is No. Sufficient. (Note that $$x-y=1$$ and $$x+y=4$$ is not a valid scenario (solution), since in this case we would have that $$x=2.5$$ and $$y=1.5$$, which is not possible as given that both unknowns must be integers)
(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$. Obviously $$\sqrt[(x+y)]{25}$$ must be an integer (since $$2^x+3^y=integer$$) and as $$x+y=integer$$ then the only solution is $$\sqrt[(x+y)]{25}=\sqrt[2]{25}=5$$ giving $$x+y=2$$. So, $$2^x+3^y=5$$. From that, two scenarios are possible:
A. $$x=2$$ and $$y=0$$ (notice that $$x+y=2$$ holds true): $$2^x+3^y=2^2+3^0=5$$, and in this case: $$(x+y)^{xy}=2^0=1=odd$$;
B. $$x=1$$ and $$y=1$$ (notice that $$x+y=2$$ holds true): $$2^x+3^y=2^1+3^1=5$$, and in this case: $$(x+y)^{xy}=2^1=2=even$$.
Hi Bunuel
Can you please explain how we arrive at the last fraction here? Thanks.
$$2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$;
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### Show Tags
30 Sep 2018, 19:45
MsInvBanker wrote:
Bunuel wrote:
Official Solution:
(1) $$2^{x-y}=\sqrt[(x+y)]{16}$$
$$2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$;
Equate the powers: $$x-y=\frac{4}{x+y}$$;
$$(x-y)(x+y)=4$$.
Now, since both $$x$$ and $$y$$ are integers (and $$x+y \gt 0$$) then $$x-y=2$$ and $$x+y=2$$ so, $$x=2$$ and $$y=0$$. Therefore, $$(x+y)^{xy}=2^0=1=odd$$, so the answer to the question is No. Sufficient. (Note that $$x-y=1$$ and $$x+y=4$$ is not a valid scenario (solution), since in this case we would have that $$x=2.5$$ and $$y=1.5$$, which is not possible as given that both unknowns must be integers)
(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$. Obviously $$\sqrt[(x+y)]{25}$$ must be an integer (since $$2^x+3^y=integer$$) and as $$x+y=integer$$ then the only solution is $$\sqrt[(x+y)]{25}=\sqrt[2]{25}=5$$ giving $$x+y=2$$. So, $$2^x+3^y=5$$. From that, two scenarios are possible:
A. $$x=2$$ and $$y=0$$ (notice that $$x+y=2$$ holds true): $$2^x+3^y=2^2+3^0=5$$, and in this case: $$(x+y)^{xy}=2^0=1=odd$$;
B. $$x=1$$ and $$y=1$$ (notice that $$x+y=2$$ holds true): $$2^x+3^y=2^1+3^1=5$$, and in this case: $$(x+y)^{xy}=2^1=2=even$$.
Hi Bunuel
Can you please explain how we arrive at the last fraction here? Thanks.
$$2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$;
$$16^{\frac{1}{x+y}}=(2^4)^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$
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Posts: 263
Location: Pakistan
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### Show Tags
30 Sep 2018, 21:06
Thank you Bunuel
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Re: M26-37 &nbs [#permalink] 30 Sep 2018, 21:06
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# M26-37
Moderators: chetan2u, Bunuel
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## Partial Fractions: an Application of Systems
### Learning Outcomes
• Decompose ${\large\frac{P(x)}{Q(x)}}$, where $Q(x)$ has only nonrepeated linear factors.
• Decompose ${\large\frac{P(x)}{Q(x)}}$, where $Q(x)$ has repeated linear factors.
• Decompose ${\large\frac{P(x)}{Q(x)}}$, where $Q(x)$ has a nonrepeated irreducible quadratic factor.
• Decompose ${\large\frac{P(x)}{Q(x)}}$, where $Q(x)$ has a repeated irreducible quadratic factor.
Earlier in this chapter, we studied systems of two equations in two variables, systems of three equations in three variables, and nonlinear systems. Here we introduce another way that systems of equations can be utilized—the decomposition of rational expressions.
Fractions can be complicated; adding a variable in the denominator makes them even more so. The methods studied in this section will help simplify the concept of a rational expression.
## Linear Factors
Recall the algebra regarding adding and subtracting rational expressions. These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression. In this section, we will look at partial fraction decomposition, which is the undoing of the procedure to add or subtract rational expressions. In other words, it is a return from the single simplified rational expression to the original expressions, called the partial fractions. Some types of rational expressions require solving a system of equations in order to decompose them, in case you were wondering what partial fractions has to do with linear systems.
For example, suppose we add the following fractions:
$\dfrac{2}{x - 3}+\dfrac{-1}{x+2}$
We would first need to find a common denominator,
$\left(x+2\right)\left(x - 3\right)$.
Next, we would write each expression with this common denominator and find the sum of the terms.
\begin{align} \dfrac{x+7}{{x}^{2}-x - 6}&=\dfrac{2}{x - 3}+\dfrac{-1}{x+2} \\[2mm]\text{Simplified sum}&\hspace{6mm}\text{Partial fraction decomposition} \end{align}
Partial fraction decomposition is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions.
$\underset{\begin{array}{l}\\ \text{Simplified sum}\end{array}}{\frac{x+7}{{x}^{2}-x - 6}}=\underset{\begin{array}{l}\\ \text{Partial fraction decomposition}\end{array}}{\frac{2}{x - 3}+\frac{-1}{x+2}}$
We will investigate rational expressions with linear factors and quadratic factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator.
When the denominator of the simplified expression contains distinct linear factors, it is likely that each of the original rational expressions, which were added or subtracted, had one of the linear factors as the denominator. In other words, using the example above, the factors of ${x}^{2}-x - 6$ are $\left(x - 3\right)\left(x+2\right)$, the denominators of the decomposed rational expression. So we will rewrite the simplified form as the sum of individual fractions and use a variable for each numerator. Then, we will solve for each numerator using one of several methods available for partial fraction decomposition.
### A General Note: Partial Fraction Decomposition of $\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right)$ Has Nonrepeated Linear Factors
The partial fraction decomposition of $\dfrac{P\left(x\right)}{Q\left(x\right)}$ when $Q\left(x\right)$ has nonrepeated linear factors and the degree of $P\left(x\right)$ is less than the degree of $Q\left(x\right)$ is
$\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{{A}_{1}}{\left({a}_{1}x+{b}_{1}\right)}+\dfrac{{A}_{2}}{\left({a}_{2}x+{b}_{2}\right)}+\dfrac{{A}_{3}}{\left({a}_{3}x+{b}_{3}\right)}+\cdot \cdot \cdot +\dfrac{{A}_{n}}{\left({a}_{n}x+{b}_{n}\right)}$.
### How To: Given a rational expression with distinct linear factors in the denominator, decompose it.
1. Use a variable for the original numerators, usually $A,B,$ or $C$, depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use ${A}_{n}$ for each numerator
$\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{{A}_{1}}{\left({a}_{1}x+{b}_{1}\right)}+\dfrac{{A}_{2}}{\left({a}_{2}x+{b}_{2}\right)}+\cdots \text{+}\dfrac{{A}_{n}}{\left({a}_{n}x+{b}_{n}\right)}$
2. Multiply both sides of the equation by the common denominator to eliminate fractions.
3. Expand the right side of the equation and collect like terms.
4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
### Example: Decomposing a Rational Expression with Distinct Linear Factors
Decompose the given rational expression with distinct linear factors.
$\dfrac{3x}{\left(x+2\right)\left(x - 1\right)}$
### Try It
Find the partial fraction decomposition of the following expression.
$\dfrac{x}{\left(x - 3\right)\left(x - 2\right)}$
In this video, you will see another example of how to find a partial fraction decomposition when you have linear factors.
## Decomposing P(x)/ Q(x), Where Q(x) Has Repeated Linear Factors
Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.
### A General Note: Partial Fraction Decomposition of $\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right)$ Has Repeated Linear Factors
The partial fraction decomposition of $\dfrac{P\left(x\right)}{Q\left(x\right)}$, when $Q\left(x\right)$ has a repeated linear factor occurring $n$ times and the degree of $P\left(x\right)$ is less than the degree of $Q\left(x\right)$, is
$\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{{A}_{1}}{\left(ax+b\right)}+\dfrac{{A}_{2}}{{\left(ax+b\right)}^{2}}+\dfrac{{A}_{3}}{{\left(ax+b\right)}^{3}}+\cdot \cdot \cdot +\dfrac{{A}_{n}}{{\left(ax+b\right)}^{n}}$
Write the denominator powers in increasing order.
### How To: Given a rational expression with repeated linear factors, decompose it.
1. Use a variable like $A,B$, or $C$ for the numerators and account for increasing powers of the denominators.
$\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{{A}_{1}}{\left(ax+b\right)}+\dfrac{{A}_{2}}{{\left(ax+b\right)}^{2}}+ \text{. }\text{. }\text{. + }\dfrac{{A}_{n}}{{\left(ax+b\right)}^{n}}$
2. Multiply both sides of the equation by the common denominator to eliminate fractions.
3. Expand the right side of the equation and collect like terms.
4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
### Example: Decomposing with Repeated Linear Factors
Decompose the given rational expression with repeated linear factors.
$\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}$
### Try It
Find the partial fraction decomposition of the expression with repeated linear factors.
$\dfrac{6x - 11}{{\left(x - 1\right)}^{2}}$
In this video, you will see an example of how to find the partial fraction decomposition of a rational expression with repeated linear factors.
So far we have performed partial fraction decomposition with expressions that have had linear factors in the denominator, and we applied numerators $A,B$, or $C$ representing constants. Now we will look at an example where one of the factors in the denominator is a quadratic expression that does not factor. This is referred to as an irreducible quadratic factor. In cases like this, we use a linear numerator such as $Ax+B,Bx+C$, etc.
### A General Note: Decomposition of $\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right)$ Has a Nonrepeated Irreducible Quadratic Factor
The partial fraction decomposition of $\dfrac{P\left(x\right)}{Q\left(x\right)}$ such that $Q\left(x\right)$ has a nonrepeated irreducible quadratic factor and the degree of $P\left(x\right)$ is less than the degree of $Q\left(x\right)$ is written as
$\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{{A}_{1}x+{B}_{1}}{\left({a}_{1}{x}^{2}+{b}_{1}x+{c}_{1}\right)}+\dfrac{{A}_{2}x+{B}_{2}}{\left({a}_{2}{x}^{2}+{b}_{2}x+{c}_{2}\right)}+\cdot \cdot \cdot +\dfrac{{A}_{n}x+{B}_{n}}{\left({a}_{n}{x}^{2}+{b}_{n}x+{c}_{n}\right)}$
The decomposition may contain more rational expressions if there are linear factors. Each linear factor will have a different constant numerator: $A,B,C$, and so on.
### How To: Given a rational expression where the factors of the denominator are distinct, irreducible quadratic factors, decompose it.
1. Use variables such as $A,B$, or $C$ for the constant numerators over linear factors, and linear expressions such as ${A}_{1}x+{B}_{1},{A}_{2}x+{B}_{2}$, etc., for the numerators of each quadratic factor in the denominator.
$\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{A}{ax+b}+\dfrac{{A}_{1}x+{B}_{1}}{\left({a}_{1}{x}^{2}+{b}_{1}x+{c}_{1}\right)}+\dfrac{{A}_{2}x+{B}_{2}}{\left({a}_{2}{x}^{2}+{b}_{2}x+{c}_{2}\right)}+\cdot \cdot \cdot +\dfrac{{A}_{n}x+{B}_{n}}{\left({a}_{n}{x}^{2}+{b}_{n}x+{c}_{n}\right)}$
2. Multiply both sides of the equation by the common denominator to eliminate fractions.
3. Expand the right side of the equation and collect like terms.
4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
### Example: Decomposing $\frac{P\left(x\right)}{Q\left(x\right)}$ When Q(x) Contains a Nonrepeated Irreducible Quadratic Factor
Find a partial fraction decomposition of the given expression.
$\dfrac{8{x}^{2}+12x - 20}{\left(x+3\right)\left({x}^{2}+x+2\right)}$
### Q & A
#### Could we have just set up a system of equations to solve Example 3?
Yes, we could have solved it by setting up a system of equations without solving for $A$ first. The expansion on the right would be:
\begin{align} 8{x}^{2}+12x - 20&=A{x}^{2}+Ax+2A+B{x}^{2}+3B+Cx+3C \\ 8{x}^{2}+12x - 20&=\left(A+B\right){x}^{2}+\left(A+3B+C\right)x+\left(2A+3C\right) \end{align}
So the system of equations would be:
\begin{align}A+B=8 \\ A+3B+C=12 \\ 2A+3C=-20 \end{align}
### Try It
Find the partial fraction decomposition of the expression with a nonrepeating irreducible quadratic factor.
$\dfrac{5{x}^{2}-6x+7}{\left(x - 1\right)\left({x}^{2}+1\right)}$
In the following video, you will see another example of how to find the partial fraction decomposition for a rational expression that has quadratic factors.
## Decomposing P(x) / Q(x), When Q(x) Has a Repeated Irreducible Quadratic Factor
Now that we can decompose a simplified rational expression with an irreducible quadratic factor, we will learn how to do partial fraction decomposition when the simplified rational expression has repeated irreducible quadratic factors. The decomposition will consist of partial fractions with linear numerators over each irreducible quadratic factor represented in increasing powers.
### A General Note: Decomposition of $\frac{P\left(x\right)}{Q\left(x\right)}$ When Q(x) Has a Repeated Irreducible Quadratic Factor
The partial fraction decomposition of $\dfrac{P\left(x\right)}{Q\left(x\right)}$, when $Q\left(x\right)$ has a repeated irreducible quadratic factor and the degree of $P\left(x\right)$ is less than the degree of $Q\left(x\right)$, is
$\dfrac{P\left(x\right)}{{\left(a{x}^{2}+bx+c\right)}^{n}}=\dfrac{{A}_{1}x+{B}_{1}}{\left(a{x}^{2}+bx+c\right)}+\dfrac{{A}_{2}x+{B}_{2}}{{\left(a{x}^{2}+bx+c\right)}^{2}}+\dfrac{{A}_{3}x+{B}_{3}}{{\left(a{x}^{2}+bx+c\right)}^{3}}+\cdot \cdot \cdot +\dfrac{{A}_{n}x+{B}_{n}}{{\left(a{x}^{2}+bx+c\right)}^{n}}$
Write the denominators in increasing powers.
### How To: Given a rational expression that has a repeated irreducible factor, decompose it.
1. Use variables like $A,B$, or $C$ for the constant numerators over linear factors, and linear expressions such as ${A}_{1}x+{B}_{1},{A}_{2}x+{B}_{2}$, etc., for the numerators of each quadratic factor in the denominator written in increasing powers, such as
$\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{A}{ax+b}+\dfrac{{A}_{1}x+{B}_{1}}{\left(a{x}^{2}+bx+c\right)}+\dfrac{{A}_{2}x+{B}_{2}}{{\left(a{x}^{2}+bx+c\right)}^{2}}+\cdots +\text{ }\dfrac{{A}_{n}+{B}_{n}}{{\left(a{x}^{2}+bx+c\right)}^{n}}$
2. Multiply both sides of the equation by the common denominator to eliminate fractions.
3. Expand the right side of the equation and collect like terms.
4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
### Example: Decomposing a Rational Function with a Repeated Irreducible Quadratic Factor in the Denominator
Decompose the given expression that has a repeated irreducible factor in the denominator.
$\dfrac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\left({x}^{2}+1\right)}^{2}}$
### Try It
Find the partial fraction decomposition of the expression with a repeated irreducible quadratic factor.
$\dfrac{{x}^{3}-4{x}^{2}+9x - 5}{{\left({x}^{2}-2x+3\right)}^{2}}$
This video provides you with another worked example of how to find the partial fraction decomposition for a rational expression that has repeating quadratic factors.
## Key Concepts
• Decompose $\frac{P\left(x\right)}{Q\left(x\right)}$ by writing the partial fractions as $\frac{A}{{a}_{1}x+{b}_{1}}+\frac{B}{{a}_{2}x+{b}_{2}}$. Solve by clearing the fractions, expanding the right side, collecting like terms, and setting corresponding coefficients equal to each other, then setting up and solving a system of equations.
• The decomposition of $\frac{P\left(x\right)}{Q\left(x\right)}$ with repeated linear factors must account for the factors of the denominator in increasing powers.
• The decomposition of $\frac{P\left(x\right)}{Q\left(x\right)}$ with a nonrepeated irreducible quadratic factor needs a linear numerator over the quadratic factor, as in $\frac{A}{x}+\frac{Bx+C}{\left(a{x}^{2}+bx+c\right)}$.
• In the decomposition of $\frac{P\left(x\right)}{Q\left(x\right)}$, where $Q\left(x\right)$ has a repeated irreducible quadratic factor, when the irreducible quadratic factors are repeated, powers of the denominator factors must be represented in increasing powers as
$\frac{Ax+B}{\left(a{x}^{2}+bx+c\right)}+\frac{{A}_{2}x+{B}_{2}}{{\left(a{x}^{2}+bx+c\right)}^{2}}+\cdots \text{+}\frac{{A}_{n}x+{B}_{n}}{{\left(a{x}^{2}+bx+c\right)}^{n}}$.
## Glossary
partial fractions the individual fractions that make up the sum or difference of a rational expression before combining them into a simplified rational expression
partial fraction decomposition the process of returning a simplified rational expression to its original form, a sum or difference of simpler rational expressions | {} |
# Notation or whatever you want to call it!
1. Dec 16, 2007
### rocomath
Example:
$$\lim_{x \rightarrow 0^+}|x|$$
$$x \geq 0$$
$$\lim_{x \rightarrow 0^-}|x|$$
$$x < 0$$
now for a similar limit, such as
$$\lim_{x \rightarrow -4^+}|x+4|$$
the Domain is from
$$x > -4$$
Why isn't it like |x|, whose Domain is greater than or equal to 0? Which in this case, greater than or equal to -4.
Last edited: Dec 16, 2007
2. Dec 17, 2007
### Gib Z
What you typed doesn't really make much sense :( Really you just wrote some lines of tex and expected us to follow >.< They mean nothing on their own.
3. Dec 17, 2007
### HallsofIvy
Staff Emeritus
For $x\ge 0$ |x| is just x
For $x< 0$ |x| is just -x
Not unless it is specified to be that. The domain for |x+4|, just like the domain for |x| is "all real numbers". Of course |y| is y if $y\ge 0$, -y if y< 0. Letting y= x+4, |x+4|= x+4 if $x+4\ge 0$ which is the same as $x\ge -4$, and |x+4|= -(x+4) if x+4< 0 which is the same as x< -4. That's why it is convenient sometimes to restrict |x| to $x\ge 4$ or |x+4| to $x\ge -4$.
Last edited: Dec 17, 2007
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## Charging and discharging of capacitor - Physics - eSaral
Hey, do you want to learn about the Charging and discharging of capacitors? If yes. Then you are at the right place. Charging and discharging of capacitors Charging When a capacitor C is connected to a battery through R then charging of capacitor takes place. The eqn. of emf at any time t is $R I+\frac{q}{C}=E$ $\frac{R d q}{d t}+\frac{q}{C}=E$ $\int_{0}^{Q} \frac{d q}{(C E-q)}$ $=-\int_{0}^{t} \frac{d t}{R C}$ This on solving gives $Q=Q_{0}\left(1-e^{-t / R C}\right)$ where $Q_{0}=C E$ Importan...
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## Discussiones Mathematicae Graph Theory
2001 | 21 | 1 | 119-136
Tytuł artykułu
### Odd and residue domination numbers of a graph
Autorzy
Treść / Zawartość
Warianty tytułu
Języki publikacji
EN
Abstrakty
EN
Let G = (V,E) be a simple, undirected graph. A set of vertices D is called an odd dominating set if |N[v] ∩ D| ≡ 1 (mod 2) for every vertex v ∈ V(G). The minimum cardinality of an odd dominating set is called the odd domination number of G, denoted by γ₁(G). In this paper, several algorithmic and structural results are presented on this parameter for grids, complements of powers of cycles, and other graph classes as well as for more general forms of "residue" domination.
Słowa kluczowe
EN
Kategorie tematyczne
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Czasopismo
Rocznik
Tom
Numer
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119-136
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Daty
wydano
2001
otrzymano
2000-10-30
poprawiono
2001-01-17
Twórcy
autor
• Department of Mathematics, University of Haifa - Oranim, Tivon - 36006, Israel
• Deptartment of Computer and Information Sciences, University of North Florida, Jacksonville, FL 32224, USA
autor
• Department of Mathematics, West Virginia University, Morgantown, WV 26506
Bibliografia
• [1] A. Amin, L. Clark, and P. Slater, Parity Dimension for Graphs, Discrete Math. 187 (1998) 1-17, doi: 10.1016/S0012-365X(97)00242-2.
• [2] A. Amin and P. Slater, Neighborhood Domination with Parity Restriction in Graphs, Congr. Numer. 91 (1992) 19-30.
• [3] A. Amin and P. Slater, All Parity Realizable Trees, J. Comb. Math. Comb. Comput. 20 (1996) 53-63.
• [4] Y. Caro, Simple Proofs to Three Parity Theorems, Ars Combin. 42 (1996) 175-180.
• [5] Y. Caro and W. Klostermeyer, The Odd Domination Number of a Graph, J. Comb. Math. Comb. Comput. (2000), to appear.
• [6] E. Cockayne, E. Hare, S. Hedetniemi and T. Wimer, Bounds for the Domination Number of Grid Graphs, Congr. Numer. 47 (1985) 217-228.
• [7] M. Garey and D. Johnson, Computers and Intractability (W.H. Freeman, San Francisco, 1979).
• [8] J. Goldwasser, W. Klostermeyer, G. Trapp and C.-Q. Zhang, Setting Switches on a Grid, Technical Report TR-95-20, Dept. of Statistics and Computer Science (West Virginia University, 1995).
• [9] J. Goldwasser, W. Klostermeyer and G. Trapp, Characterizing Switch-Setting Problems, Linear and Multilinear Algebra 43 (1997) 121-135, doi: 10.1080/03081089708818520.
• [10] J. Goldwasser and W. Klostermeyer, Maximization Versions of Lights Out'' Games in Grids and Graphs, Congr. Numer. 126 (1997) 99-111.
• [11] J. Goldwasser, W. Klostermeyer and H. Ware, Fibonacci Polynomials and Parity Domination in Grid Graphs, Graphs and Combinatorics (2000), to appear.
• [12] M. Halldorsson, J. Kratochvil and J. Telle, Mod-2 Independence and Domination in Graphs, in: Proceedings Workshop on Graph-Theoretic Concepts in Computer Science '99, Ascona, Switzerland (Springer-Verlag, Lecture Notes in Computer Science, 1999).
• [13] T. Haynes, S. Hedetniemi and P. Slater, Fundamentals of Domination in Graphs (Marcel Dekker, New York, 1998).
• [14] R. Johnson and C. Johnson, Matrix Analysis (Cambridge University Press, 1990).
• [15] M. Jacobson and L. Kinch, On the Domination Number of Products of Graphs, Ars Combin. 18 (1984) 33-44.
• [16] W. Klostermeyer and E. Eschen, Perfect Codes and Independent Dominating Sets, Congr. Numer. (2000), to appear.
• [17] K. Sutner, Linear Cellular Automata and the Garden-of-Eden, The Mathematical Intelligencer 11 (2) (1989) 49-53.
Typ dokumentu
Bibliografia
Identyfikatory | {} |
## Calculus in context
I’ve taught calculus for a number of years and have largely been dissatisfied with the texts. I love mathematics in its own right. I don’t think mathematics needs to justify its existence or its questions. However, a calculus class is an opportunity to train the minds of future scientists and engineers.
I’ve grown tired of the long lists of functions to do A and B to or the psuedo-context problems (to use the phrasing of Dan Meyer) that are supposed to represent applications. Students often want to know when they’ll run into these specific functions, how will calculus show up in their major, how does the now ubiquitous graphing calculator come into play or services like Wolfram Alpha? The applications seldom have references for the student or instructor to follow up on.
I’m trying to design a class where if you forget all the rules for integration and differentiation you’d still look back and say “yeah, that was a great class, I really grew as a problem solver.” This is a tall order in a class of biology, ecology, engineering, physics, chemistry, computer science, and premed students. Moreover, we teachers are all too aware of the problem of transfer. The student seems to know how to do calculus class but doesn’t seem to know how to do calculus anywhere else.
My department recently changed books to “Calclulus” by Hughes-Hallett, Gleason, and McCallum published by Wiley which is a great start. It often asks students about units, interpretations of the derivative, graphical skills, technology applications etc. It treats limits quickly and gets on with the process of thinking about derivatives, but spends a whole chapter on derivatives before learning any of the ‘short cuts’ to calculating them. It is rich in qualitative reasoning.
This has led me to look for more problems of this sort. Strogatz’ book “Nonlinear Dynamics and Chaos” is a good start. This has led me to think about how early you can introduce differential equations to students. Once you have the concept of the derivative, even before you calculate any you can talk about a reaction equation such as:
$latex \frac{dy}{dt}=k(a-y)(b-y)$
Where $a and $a,b,k$ are all constant. $y$ is measured in grams and $t$ is measured in seconds. ‘a’ and ‘b’ represent chemicals that combine in some way to produce chemical ‘y’ and this differential equation describes the process. Questions you could ask:
(a) what are the units on dy/dt, a, b, and k?
(b) Make a graph of dy/dt versus y.
(this is a really good one. From this graph you can obtain huge amounts of information about what happens to y(t) depending on the initial conditions. You can also talk the range of applicability of the model, that is, does it make sense for $y(0)>b$?
(c) What is the rate of change if $y(0)=0$?
(d) What is the limit of $y(t)$ as $t\to\infty$ if $0?
Those are just a couple of questions that only use the basic idea of a derivative and a graph to answer. They will be hard though they will require a lot of work on the part of the student. Then there are some other more challenging questions you can ask such as:
If you mess around with the differential equation:
$latex \frac{dy}{dt}=kab(1-y/a)(1-y/b)$
Then you can ask about conditions such as $0 or $y< but $y and look at how to approximate the differential equation.
It’s a lot of work, but you’ll be able to construct really clear pictures of what $y(t)$ looks like from this sort of reasoning and it doesn’t require any fancy mathematics, just fancy thinking which is exactly what we need our students to develop. I’ll post more in the near future! | {} |
### Mathematical Biosciences and Engineering
2012, Issue 1: 97-110. doi: 10.3934/mbe.2012.9.97
# Impact of discontinuous treatments on disease dynamics in an SIR epidemic model
• Received: 01 September 2010 Accepted: 29 June 2018 Published: 01 December 2011
• MSC : Primary: 92D30; Secondary:34C23.
• We consider an SIR epidemic model with discontinuous treatment strategies. Under some reasonable assumptions on the discontinuous treatment function, we are able to determine the basic reproduction number $\mathcal{R}_0$, confirm the well-posedness of the model, describe the structure of possible equilibria as well as establish the stability/instability of the equilibria. Most interestingly, we find that in the case that an equilibrium is asymptotically stable, the convergence to the equilibrium can actually be achieved in finite time, and we can estimate this time in terms of the model parameters, initial sub-populations and the initial treatment strength. This suggests that from the view point of eliminating the disease from the host population, discontinuous treatment strategies would be superior to continuous ones. The methods we use to obtain the mathematical results are the generalized Lyapunov theory for discontinuous differential equations and some results on non-smooth analysis.
Citation: Zhenyuan Guo, Lihong Huang, Xingfu Zou. Impact of discontinuous treatments on disease dynamics in an SIRepidemic model[J]. Mathematical Biosciences and Engineering, 2012, 9(1): 97-110. doi: 10.3934/mbe.2012.9.97
### Related Papers:
• We consider an SIR epidemic model with discontinuous treatment strategies. Under some reasonable assumptions on the discontinuous treatment function, we are able to determine the basic reproduction number $\mathcal{R}_0$, confirm the well-posedness of the model, describe the structure of possible equilibria as well as establish the stability/instability of the equilibria. Most interestingly, we find that in the case that an equilibrium is asymptotically stable, the convergence to the equilibrium can actually be achieved in finite time, and we can estimate this time in terms of the model parameters, initial sub-populations and the initial treatment strength. This suggests that from the view point of eliminating the disease from the host population, discontinuous treatment strategies would be superior to continuous ones. The methods we use to obtain the mathematical results are the generalized Lyapunov theory for discontinuous differential equations and some results on non-smooth analysis.
###### 通讯作者: 陈斌, bchen63@163.com
• 1.
沈阳化工大学材料科学与工程学院 沈阳 110142
2.080 2.1 | {} |
# Thermal boundary layer thickness and shape
This page describes some parameters used to characterize the properties of the thermal boundary layer formed by a heated (or cooled) fluid moving along a heated (or cooled) wall. In many ways, the thermal boundary layer description parallels the velocity (momentum) boundary layer description first conceptualized by Ludwig Prandtl.[1] Consider a fluid of uniform temperature ${\displaystyle T_{o}}$ and velocity ${\displaystyle u_{o}}$ impinging onto a stationary plate uniformly heated to a temperature ${\displaystyle T_{s}}$. Assume the flow and the plate are semi-infinite in the positive/negative direction perpendicular to the ${\displaystyle x-y}$ plane. As the fluid flows along the wall, the fluid at the wall surface satisfies a no-slip boundary condition and has zero velocity, but as you move away from the wall, the velocity of the flow asymptotically approaches the free stream velocity ${\displaystyle u_{0}}$. The temperature at the solid wall is ${\displaystyle T_{s}}$ and gradually changes to ${\displaystyle T_{o}}$ as one moves toward the free stream of the fluid. It is impossible to define a sharp point at which the thermal boundary layer fluid or the velocity boundary layer fluid becomes the free stream, yet these layers have a well-defined characteristic thickness given by ${\displaystyle \delta _{T}}$ and ${\displaystyle \delta _{v}}$. The parameters below provide a useful definition of this characteristic, measurable thickness for the thermal boundary layer. Also included in this boundary layer description are some parameters useful in describing the shape of the thermal boundary layer.
## 99% Thermal Boundary Layer thickness
The thermal boundary layer thickness, ${\displaystyle \delta _{T}}$, is the distance across a boundary layer from the wall to a point where the flow temperature has essentially reached the 'free stream' temperature, ${\displaystyle T_{0}}$. This distance is defined normal to the wall in the ${\displaystyle y}$-direction. The thermal boundary layer thickness is customarily defined as the point in the boundary layer, ${\displaystyle y_{99}}$, where the temperature ${\displaystyle T(x,y)}$ reaches 99% of the free stream value ${\displaystyle T_{0}}$:
${\displaystyle \delta _{T}=y_{99}}$ such that ${\displaystyle T(x,y_{99})}$ = 0.99 ${\displaystyle T_{0}}$
at a position ${\displaystyle x}$ along the wall. In a real fluid, this quantity can be estimated by measuring the temperature profile at a position ${\displaystyle x}$ along the wall. The temperature profile is the temperature as a function of ${\displaystyle y}$ at a fixed ${\displaystyle x}$ position.
For laminar flow over a flat plate a zero incidence, the thermal boundary layer thickness is given by:[2]
${\displaystyle \delta _{T}=\delta _{v}\mathrm {Pr} ^{-1/3}}$
${\displaystyle \delta _{T}=5.0{\sqrt {{\nu x} \over u_{0}}}\mathrm {Pr} ^{-1/3}}$
where
${\displaystyle \mathrm {Pr} }$ is the Prandtl Number
${\displaystyle \delta _{v}}$ is the thickness of the velocity boundary layer thickness [3]
${\displaystyle u_{0}}$ is the freestream velocity
${\displaystyle x}$ is the distance downstream from the start of the boundary layer
${\displaystyle \nu }$ is the kinematic viscosity
For turbulent flow over a flat plate, the thickness of the thermal boundary layer that is formed is not determined by thermal diffusion, but instead, it is random fluctuations in the outer region of the boundary layer of the fluid that is the driving force determining thermal boundary layer thickness. Thus the thermal boundary layer thickness for turbulent flow does not depend on the Prandtl number but instead on the Reynolds number. Hence, the turbulent thermal boundary layer thickness is given approximately by the turbulent velocity boundary layer thickness expression[4] given by:
${\displaystyle \delta _{T}\approx \delta \approx 0.37x/{\mathrm {Re} _{x}}^{1/5}}$
where
${\displaystyle {\mathrm {Re} _{x}}=u_{0}x/\nu }$ is the Reynolds number
This turbulent boundary layer thickness formula assumes 1) the flow is turbulent right from the start of the boundary layer and 2) the turbulent boundary layer behaves in a geometrically similar manner (i.e. the velocity profiles are geometrically similar along the flow in the x-direction, differing only by stretching factors in ${\displaystyle y}$ and ${\displaystyle u(x,y)}$[5]). Neither one of these assumptions is true for the general turbulent boundary layer case so care must be exercised in applying this formula.
## Thermal Displacement Thickness
The thermal displacement thickness, ${\displaystyle \beta ^{*}}$ may be thought of in terms of the difference between a real fluid and a hypothetical fluid with thermal diffusion turned off but with velocity ${\displaystyle u_{0}}$ and temperature ${\displaystyle T_{0}}$. With no thermal diffusion, the temperature drop is abrupt. The thermal displacement thickness is the distance by which the hypothetical fluid surface would have to be moved in the ${\displaystyle y}$-direction to give the same integrated temperature as occurs between the wall and the reference plane at ${\displaystyle \delta _{T}}$ in the real fluid. It is a direct analog to the velocity displacement thickness which is often described in terms of an equivalent shift of a hypothetical inviscid fluid (see Schlichting[6] for velocity displacement thickness).
The definition of the thermal displacement thickness for incompressible flow is based on the integral of the reduced temperature:
${\displaystyle {\beta ^{*}}=\int _{0}^{\infty }{\theta (x,y)\,\mathrm {d} y}}$
where the dimensionless temperature is ${\displaystyle \theta (x,y)=(T(x,y)-T_{0})/(T_{s}-T_{0})}$. In a wind tunnel, the velocity and temperature profiles are obtained by measuring the velocity and temperature at many discrete ${\displaystyle y}$-values at a fixed ${\displaystyle x}$-position. The thermal displacement thickness can then be estimated by numerically integrating the scaled temperature profile.
## Moment Method
A relatively new method[7][8] for describing the thickness and shape of the thermal boundary layer utilizes the moment method commonly used to describe a random variable's probability distribution. The moment method was developed from the observation that the plot of the second derivative of the thermal profile for laminar flow over a plate looks very much like a Gaussian distribution curve.[9] It is straightforward to cast the properly scaled thermal profile into a suitable integral kernel.
The thermal profile central moments are defined as:
${\displaystyle {\xi _{n}}={1 \over \beta ^{*}}\int _{0}^{\infty }{(y-m_{T})^{n}\theta (x,y)\mathrm {d} y}}$
where the mean location, ${\displaystyle m_{T}}$, is given by:
${\displaystyle m_{T}={1 \over \beta ^{*}}\int _{0}^{\infty }{y\theta (x,y)\mathrm {d} y}}$
There are some advantages to also include descriptions of moments of the boundary layer profile derivatives with respect to the height above the wall. Consider the first derivative temperature profile central moments given by:
${\displaystyle {\epsilon _{n}}=\int _{0}^{\infty }{(y-{\beta ^{*}})^{n}{d\theta (x,y) \over dy}\mathrm {d} y}}$
where the mean location is the thermal displacement thickness ${\displaystyle \beta ^{*}}$.
Finally the second derivative temperature profile central moments are given by:
${\displaystyle {\phi _{n}}=\mu _{T}\int _{0}^{\infty }{(y-{\mu _{T}})^{n}{d^{2}\theta (x,y) \over dy^{2}}\mathrm {d} y}}$
where the mean location, ${\displaystyle \mu _{T}}$, is given by:
${\displaystyle {1 \over \mu _{T}}=-\left({\frac {d\theta (x,y)}{dy}}\right)_{y=0}}$
With the moments and the thermal mean location defined, the boundary layer thickness and shape can be described in terms of the thermal boundary layer width (variance), thermal skewnesses, and thermal excess (excess kurtosis). For the Pohlhausen solution for laminar flow on a heated flat plate,[10] it is found that thermal boundary layer thickness defined as ${\displaystyle \delta _{T}=m_{T}+4\sigma _{T}}$ where ${\displaystyle \sigma _{T}=\xi _{2}^{1/2}}$, tracks the 99% thickness very well.[11]
For laminar flow, the three different moment cases all give similar values for the thermal boundary layer thickness. For turbulent flow, the thermal boundary layer can be divided into a region near the wall where thermal diffusion is important and an outer region where thermal diffusion effects are mostly absent. Taking a cue from the boundary layer energy balance equation, the second derivative boundary layer moments, ${\displaystyle {\phi _{n}}}$ track the thickness and shape of that portion of the thermal boundary layer where thermal diffusivity ${\displaystyle {\alpha }}$ is significant. Hence the moment method makes it possible to track and quantify the region where thermal diffusivity is important using ${\displaystyle {\phi _{n}}}$ moments whereas the overall thermal boundary layer is tracked using ${\displaystyle {\epsilon _{n}}}$ and ${\displaystyle {\xi _{n}}}$ moments.
Calculation of the derivative moments without the need to take derivatives is simplified by using integration by parts to reduce the moments to simply integrals based on the thermal displacement thickness kernel:
${\displaystyle {k_{n}}=\int _{0}^{\infty }{y^{n}\theta (x,y)\,\mathrm {d} y}}$
This means that the second derivative skewness, for example, can be calculated as:
${\displaystyle \gamma _{T}=\phi _{3}/\phi _{2}^{3/2}=(2\mu _{T}^{3}-6\beta ^{*}\mu _{T}^{2}+6\mu _{T}k_{1})/(-\mu _{T}^{2}+2\mu _{T}\beta ^{*})^{3/2}}$
• Hermann Schlichting, Boundary-Layer Theory, 7th ed., McGraw Hill, 1979.
• Frank M. White, Fluid Mechanics, McGraw-Hill, 5th Edition, 2003.
• Amir Faghri, Yuwen Zhang, and John Howell, Advanced Heat and Mass Transfer,Global Digital Press, ISBN 978-0-9842760-0-4, 2010.
## Notes
1. L. Prandtl, “Über Flüssigkeitsbewegung bei sehr kleiner Reibung,” Verhandlungen des Dritten Internationalen Mathematiker-Kongresses in Heidelberg 1904, A. Krazer, ed., Teubner, Leipzig, (1905) 484–491.
2. Schlichting, p. 307.
3. Schlichting, p.140.
4. Schlichting, p. 638.
5. Schlichting, p.152.
6. Schlichting, p. 140.
7. Weyburne, 2006.
8. Weyburne, 2018.
9. Weyburne, 2006, p. 1680.
10. Schlichting, p. 292.
11. Weyburne, 2018, p. 5.
## References
• Schlichting, Hermann (1979). Boundary-Layer Theory, 7th ed., McGraw Hill, New York, U.S.A.
• Weyburne, David (2006). "A mathematical description of the fluid boundary layer," Applied Mathematics and Computation, vol. 175, pp. 1675–1684
• Weyburne, David (2018). "New thickness and shape parameters for describing the thermal boundary layer," arXiv:1704.01120[physics.flu-dyn] | {} |
# Human’s Software Development
Have you heard of phrases like You cannot measure productivity by the lines of code written.? Did you read, that It is considered to be a viable practice ship code as soon as the updater works.? Have you heard of the big problems of indefensible resource usage? Did you gape at the horrible number of dependencies? Did you pay attention to the humongous security problems in software and of its unreliability?
So, what do you as a software developer, maintainer, admin, tester, designer, manager etc., what do you do in order to fix this? Phrases like We have to write better code. come to one’s mind. Alternatively, mottos like We have to invest more time into code. and We have to write less code are signaled in the vast space of the internet. Others propose better management or superior paradigms.
There are many criticized solutions to these problems. Some say, that Docker is bad, because of its massive use of storage., but what’s the alternative? More often than not, I see such critics as (Let me use a problematic term for the fun of it.) toxic. Why, you may ask?
Let’s be frank: this is an emotional situation. People often propose diverse solutions, in order to get rid of the evil ones, but not all of us have the luxury of doing things right. Not all of us are gifted with minds suited for creating good solutions. You cannot just get better developers or better decision makers in general, because this is an emotional problem and emotional problems are never just solved. I would also say, that problems are not solved by making people more skilled, historically.
On a personal note, I also sometimes get the impression, that people get mad of solutions, because it is not done in their way or on their terms. This is not a condemnation of the critics or a blessing of every piece of new software.
So, as you probably guessed by now, this post is an excuse for the development practices done in this project. Welcome to the cringe party 🎉
# Developing The Universal Allocation Program
The first version of this project was done during an internship at university with a friend of mine and was called the Universal Allocation Program or Nap. It took an Excel file as input containing an optimization problem. As output, the program also produced an Excel file, holding the best found solution to the problem. It had a simple web page, where one could drop the input file and retrieve the output as a download. The version had limited capabilities. Nap was probably badly coded, as it was my first complex program. Nap did its job and that is the most important part of it.
In hindsight, I noticed an emerging harmful pattern of mine during the development of Nap: I searched for a good way, how to implement, organize and document program options in Java, which was used for Nap. I did not find any, and frankly I probably did not know where to look. In Nap, there is no special configuration framework used. Instead, a CLI library and some static final variables are used.
The desire for a configuration framework, was the starting point for feature creep in the following versions. As always, the evil is paved with good intentions.
# Developing The Generic Allocator
Thereafter, I created a successor for my master thesis and named it humbly the Generic Allocator or short Gal. For this project, I definitely wanted to create a good configuration library and named it Dependency Manager (net.splitcells.dem).
One can see, how this may have been a starting point for feature creep in a lot of projects of mine:
• Use a framework with fundamental flaws and combine it with the desire for a better one.
• Furthermore, searches for good frameworks with no results, also seed the desire for creating further feature creep.
That’s basically, what happened a lot of times. Even with Gal itself: there are limited options, because of one’s limited perspective and therefore one constructs completely new programs, instead of using existing ones, in order to comply with demands. Each new subproject has its own feature creep, which does sidetrack one from the goal that really matters.
Gal, the second version of the optimization program, was created in a simple way. I created a new empty source code project. For every file of the old version, I decided, what part could be reused and what needed to be changed. Only the constraint system, was completely rewritten from the ground up, because the solving algorithm relied on a new constraint model.
This way of migrating code basically worked. My new configuration library, was the basis for many experiments. It was therefore a nutritious soil for feature creep, but was somewhat contained in this version.
# Developing The Third Version
Starting from this point, a very destructive path started. After my master thesis I decided, to further work on this project and to create a third version. The development plan was similar to the second version:
• Create a new empty project.
• Go through each file of the precursor and decide what parts should be adopted.
This seems to be fine on the first look and a lot of code was migrated from the precursor to the successor. So, everything is fine?
It took 4 years of implementation in order to even come close to finishing the third version, and it does not have one test case, where the new functionality is really used. Of course, I did not work non-stop on the project. In matter of fact, I did not invest that much time into it, but still: 4 years for nothing?
That is the toll of feature creep: being slowly killed by a thousand cuts.
# Migration Based Development and the Tower of Babel
When I kind of started to understand the problem and perceive its extent, it was too late, as the new version already started to show its potential. This does not mean, that deleting code is not an option, but the trade-off of doing it, did not seem to be worth it anymore.
I started attempts to compensate for these problems:
• I started publishing the project, in order to have some skin in the game.
• A new development paradigm was adopted: migration based development or in other words Do not rewrite or break functionality.
• Make many commits and try to make 20 of them each day, in order to have continuous progress.
Does this sound familiar to the opening point? It certainly feels so.
# A New Adventure
Now there is a blog for this project, where I want to write down related emotional thoughts, so one get a grip of its history, progress and future. It may newer become a truly open source project with a community. The gods know, how bad I am in social and many other settings. On the other hand, this creates uncertainty, which in turn makes this whole journey a lot more exciting.
We’ll see, what will happen. | {} |
### Home > MC2 > Chapter 7 > Lesson 7.1.5 > Problem7-48
7-48.
The probability of getting a purple gumball is the number or purple gumballs divided by the total number of gumballs.
$\frac{30}{120} = 25$
The probability of getting either a purple or a green gumball is the sum of the probability of getting a purple gumball and the probability of getting a green gumball.
$\text{P(purple) + P(green)} = \frac{30}{120} + \frac{36}{120} = \frac{66}{120} = 55$
What is the probability of getting a green gumball?
Since you can either get a green gumball or not get a green gumball, the probability of not getting a green gumball is 1 (or 100%) minus the probability of getting a green gumball.
What is the probability of getting a purple gumball?
A white gumball?
See (b). | {} |
## Banking institutions will make extra loans whenever necessary reserves are
Banking institutions will make extra loans whenever necessary reserves are
To comprehend the entire process of money creation today, let's produce a hypothetical system of banking institutions. We are going to concentrate on three banks in this system: Acme Bank, Bellville Bank, and Clarkston Bank. Assume that all banking institutions have to hold reserves corresponding to 10% of the deposits that are checkable. The total amount of reserves banking institutions have to hold is named needed reserves. The book requirement is expressed being a needed book ratio; it specifies the ratio of reserves to checkable deposits a bank must keep. Banks may hold reserves more than the level that is required such reserves are known as extra reserves. Read more | {} |
## representation of a finite group
1. The problem statement, all variables and given/known data
Prove that a representation of a finite group G is faithful if and only if its image is isomorphic to G.
2. Relevant equations
3. The attempt at a solution
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Quote by syj 1. The problem statement, all variables and given/known data Prove that a representation of a finite group G is faithful if and only if its image is isomorphic to G. 2. Relevant equations 3. The attempt at a solution
What did you try already?? If you show us where you're stuck, then we'll know where to help...
I am not very eloquent when it comes to proofs. So I am just going to lay out what I know. Let the representation be noted as F, and the image of G' if F is a faithful representation then ker{F}={1G} Can I conclude then by the first isomorphism theorem that G is isomorphic to G'? I know that for an "if and only if" proof there are two directions. If I can get the first direction of the proof, I can easily get the other direction.
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## representation of a finite group
Quote by syj I am not very eloquent when it comes to proofs. So I am just going to lay out what I know. Let the representation be noted as F, and the image of G' if F is a faithful representation then ker{F}={1G} Can I conclude then by the first isomorphism theorem that G is isomorphic to G'? I know that for an "if and only if" proof there are two directions. If I can get the first direction of the proof, I can easily get the other direction.
Indeed, the first isomorphism theorem does the trick!!
Ok, so is this enough: If f is faithful then ker{f}={1G} therefore by the first isomorphism theorem, G$\cong$G' If G$\cong$G' then by the first isomorphism theorem ker{f}={1G} therefore by the definition of a faithful representataion, f is faithful. it seems so plain. lol. too plain to be complete. but if it is, i am one happy girl ;)
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Quote by syj If G$\cong$G' then by the first isomorphism theorem ker{f}={1G}
This is true (but only for finite groups), but you might want to explain in some more detail.
The rest is ok!
can you please explain how i should expand further? I am told that G is finite in the question. thanks
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Quote by syj can you please explain how i should expand further? I am told that G is finite in the question. thanks
Well, you know that
$$G\cong G/\ker(\phi)$$
Why does that imply that $\ker(\phi)=\{1\}$ ??
Think of the order...
ok, am i making sense here: a corollary to the first isomorphism theorem says: |G:ker($\varphi$|=|$\varphi$(G)| from this can I conclude: |$\frac{G}{ker(\varphi)}$|=|G'| and then conclude: ker($\varphi$)={1G}
Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Indeed, that works!!
wooo hoooo !!!! i am the happiest girl in the world!! until the next proof comes my way ... at which time I shall bug u some more! thanks so much.
Tags abstract algebra, group representation | {} |
# Makefile and Beamer presentations
I have been wondering about Makefiles for some time now, and recently I finally got around learning about them so that I could use make to regenerate all the different versions of a manuscript I’m working on. And I thought I would take the opportunity to explain how they can be useful for Beamer presentations.
The setting is as follows: let’s say you are preparing slides using Latex/Beamer. You are including pauses in your slides, e.g. you are stopping after each bullet point. This is useful if you want to focus the attention of the audience on a specific point of your slides, or if you don’t want to reveal the punchline too soon. On the other hand, if you want to print the slides, you want to remove these pauses. This is why the beamer class has the option handout. Now, the question is, how do I generate these two files from the same presentation?
There is a few possibilities that come to mind:
• You can have two Latex files, one for the slides and one for the handout. However, if you make changes to one, it can become difficult to know if you’ve also made the changes to the other.
• You can have one Latex file, that you then always compile twice: once with the handout option, once without.
Another possibility is to have three Latex files: one with the text, and two that only contain headers (one with option handout and one without). This is usually what I do.
Where the Makefile can become useful is in automating the compilation of both the handout and the slides. Moreover, if for example you only change the header file for the handout, you don’t want to have to recompile the slides.
The setting is going to be as follows: say you have a presentation in a file talk.tex. You also have two separate files for the headers: talk_slides.tex and talk_handout.tex. The latter looks something like:
\documentclass[handout]{beamer}
\input{talk.tex}
### Makefile
Now let’s talk about the Makefile. In general, there are three components to it:
• targets;
• dependencies for each target;
• a set of instructions to be executed when a dependency has been modified.
The point is that, once you have a Makefile in your directory, you simply use the function make to run the instructions.
The syntax of the Makefile is thus as follows:
targets: dependencies
instructions
Note that you need a hard tab at the beginning of the instructions, or otherwise you will get a missing operator error (although you can circumvent this using semi-colons if you want).
Let’s look at one example: one target we are interested in is talk_handout.pdf, the output when compiling talk_handout.tex with pdflatex. We want to recompile it when either talk_handout.tex or talk.tex is changed. Therefore in the Makefile, we would write
talk_handout.pdf: talk_handout.tex talk.tex
pdflatex talk_handout
And of course you would have something similar for the target talk_slides.pdf.
### Special targets
There is two important special targets that I want to discuss: all and .PHONY. By default, the function make will build the first target when you call it without an argument, and to build a specific target you would write make target. But what if you want to build multiple targets at the same time? Instead of running separate commands make target1, make target2, etc., you can use the special target all, whose dependencies are all the other targets you would like to build at the same time. For example, with our Beamer presentation, you would have
all: talk_handout.pdf talk_slides.pdf
Next, .PHONY. When you compile Latex files, you get several auxiliary files which can useful to speed up a future compilation. But sometimes you also want to remove them; the Makefile can also help you automating this. You can create a target clean, with or without dependencies, and with the instructions to remove these auxiliary files:
clean:
rm *.aux *.blg *.out *.bbl *.log
You can then clean your directory by simply calling make clean.
The issue, though, is that the clean target is special, in that it doesn’t correspond to a file in your directory. To make sure make knows this, you can make clean a dependency of the special target .PHONY.
Therefore, our Makefile for the Beamer presentation is now
all: talk_handout.pdf talk_slides.pdf
.PHONY: clean
clean:
rm *.aux *.blg *.out *.bbl *.log
talk_handout.pdf: talk_handout.tex talk.tex
pdflatex talk_handout
talk_slides.pdf: talk_slides.tex talk.tex
pdflatex talk_slides
Voila! This is all you need for this project. Save the above script in a file named Makefile or makefile, and you simply run the command make to recompile your pdf documents whenever you change something.
The above Makefile is unnecessary long and also repetitive: for example the word talk appears in multiple places, and if we want to change it (e.g. we want to reuse this Makefile for another presentation), we need to change it everywhere. The solution is simply to create a variable that will hold the name of the presentation. Therefore to reuse the Makefile we only need to change one line:
FILE = talk
all: $(FILE)_handout.pdf$(FILE)_slides.pdf
.PHONY: clean
clean:
rm *.aux *.blg *.out *.bbl *.log
$(FILE)_handout.pdf:$(FILE)_handout.tex $(FILE).tex pdflatex$(FILE)_handout
$(FILE)_slides.pdf:$(FILE)_slides.tex $(FILE).tex pdflatex$(FILE)_slides
Moreover, since we are essentially using the same instructions for both talk_slides.pdf and talk_handout.pdf, it would be nice if we only had to write it once. For this purpose, we can use the special macro $@, which stands for the target name. We can then rewrite our Makefile as FILE = talk all:$(FILE)_handout.pdf $(FILE)_slides.pdf .PHONY: clean clean: rm *.aux *.blg *.out *.bbl *.log$(FILE)_handout.pdf $(FILE)_slides.pdf:$(FILE)_slides.tex $(FILE)_handout.tex$(FILE).tex
pdflatex $@ So we have two targets on the same line, and only one set of instructions. One possible issue is that now talk_slides.tex is a dependency for talk_handout.pdf and vice-versa, and therefore both the slides and the handout will be recompiled whenever there is a change. This is not too problematic, since most changes will be made to talk.tex which is already a dependency for both the slides and the handout. Edit (2015/12/08): This actually doesn’t work, because the target is a PDF document and pdflatex is expecting a Latex file (or a file without an extension). If someone knows how to fix this, you can post the solution in the comments below! There is much more to learn about Makefiles, and I think the easiest way is to try it out and whenever I need something new, either look at the official documentation or stackoverflow. ### Workflow Having been through the process of learning about Makefiles, I thought I would take advantage of all this and build a workflow for my Beamer presentations. What I want is the following: • All the files related to a presentation will be in a single directory, and only these files. • I want to follow the convention that the directory will be talk/, the main text will be in talk.tex, and I will have two extra Latex files with the headers, talk_handout.tex and talk_slides.tex. The file talk.tex will follow my personal Beamer template. • I want a Makefile in the directory, and the first line will contain the name of the presentation: FILE = talk. • Finally, I want to do all this (creating the directory and the files) by using a single script. This single script is as follows: # First argument is the name of the talk # other arguments are ignored TALK=$1
mkdir $TALK cd$TALK/
# Create the two header files
echo -e "\documentclass[handout]{beamer}\n\input{$TALK.tex}" >$TALK\_handout.tex
echo -e "\documentclass{beamer}\n\input{$TALK.tex}" >$TALK\_slides.tex
# Create the main file
# Note: the path corresponds to where my template is located
cp ~/.config/texstudio/templates/user/template_beamer.tex $TALK.tex # Create the Makefile echo -e "FILE =$TALK"'
all: $(FILE)_handout.pdf$(FILE)_slides.pdf
.PHONY: clean
clean:
\trm *.aux *.blg *.out *.bbl *.log
$(FILE)_handout.pdf$(FILE)_slides.pdf: $(FILE)_slides.tex$(FILE)_handout.tex $(FILE).tex \tpdflatex$@' > Makefile
Note that the functionality of this script is highly dependent on which type of shell you use. But a particular point to keep in mind is that, when creating the Makefile, I’m using both single and double quotes. This is because I want most of the variable names to be interpreted literally, except the first line, which will change depending on the name of the presentation.
Now, I can put this script in a file called talkInit.sh, put the file somewhere my interpreter can find, and then everytime I have a presentation to prepare, I simply type talkInit talk in the directory I want and everything should work fine! | {} |
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# motor controller L297... need help
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#### sp
##### Full Member level 6
l297 vref
this is the data sheet i need help on
https://www.st.com/stonline/books/pdf/docs/1334.pdf
Code:
The L297 integratesall the control circuitry required to control bipolar and unipolar steppermotors. Used with a dual bridge driver such as the L298N forms a complete microprocessor-to-bipolar stepper motor interface. Unipolar stepper motor can be driven with an L297 plus a quad darlington array. This note describes the operation of the circuit and shows how it is used.
i hav few questions... please help... i am using it to control 2 phase variable magnet bipolar stepper motor
1. could anyone explain to me wad is the purpose of OSC pin?.. i know it is stated as chopper rate... anyone know how to explain this using more simple term?...i dunno wad is chopper even though i read many time the data sheet D....
wad is the value for the RC circuits,,, wad is this for... n wad value i can use,, any?
2. wad is the control pin for??... it stated either to act on A,B,C,D or /INH1/2 faint
3. wad is tht /INH1 and /INH2 for?? should i use it if i just wanna control the stepper motor. please explain on this 2 pin,,, i am really headache read over n over again wthout any understanding on this...
4. and how about the SENSE1 and sense2 pin....wad is tht also...
5. how fast the frequency of the step clock can i input to the /clock pin of the L297.... does it depend on the stepper motor i use?...
6. for stepper motor, which part of the data sheet i should refer in order to know how fast the step/clock i can use wth the motor from my controller??
i would b grateful if anyone can help... u can answer part or even one question if u r really busy... any will help....
many many many thanks... in advance...
my warm regards,
sp
#### Marcel Majoor
##### Full Member level 2
l297 home
1. The IC has a 'protection' mechanism that is triggered (activated) when the current exceeds a certain level (as measured by Sense1/Sense2 and compared to Vref). What happens then is that the output switches off. It is then restarted a certain time later (determined by the chopper rate, which is set by the OSC pin). This mechanism is then repeated over and over.
This basically means that, when this switching off and on again happens, that the (stepper) motor is driven by a constant current (the current sensed by Sense1/2).
The timing of this chopping is set by the resistor/capacitor value. The frequency is 1/(0.69 * R * C). With the 22kOhm and 3.3nF in the sample application this would result in a frequency of 20 kHz.
2/3. The control pin determines if either the ABCD outputs or the inhibit outputs are used to switch off the driver. Typically the inhibit outputs place the driver in a floating situation, whereas the ABCD would switch it off actively (this all depends on the driver used). In a 'floating' situation the motor keeps it's momentum, and in the active off situation the motor is being 'braked'.
4. The sense pins are used to measure the current flowing through the motor windings. This is done using a series resistor.
5. The clock input does depend on the stepper motor used. A stepper motor can only go at a certain speed before it looses torque etc. A stepper motor is typically accelerated to it's speed (if a high speed is the goal). For slow speeds no acceleration is typically needed.
6. I don't know what parameters are given in a typical stepper motor datasheet, but I think you should have a look at the steprate. There is typically a curve with the steprate versus torque (or something simular).
Points: 2
Points: 2
### enesb
Points: 2
#### sp
##### Full Member level 6
l297+r sense
thanxs Marcel Majoor... ur help is really appreciated n informative.... anyway, i pressed the helped button... thanks
for question num.1 so wad is the best chopper rate?... the higher the better??...
for question num.2 and 3... do u mean tht the control pin is use to select the switch off mechanism only?... b4 reading ur help... i was confuse as we would always use ABCD to control the motor... if use INH1/2.. how to control the stepper motor?..
so the bottom line is tht this CONTROL pin is use to control the motor behaviour when no clock step are input to the driver(switch off)... am i correct?
for question num.4... the sense resistors..y it use 0.5ohm?... 10ohm will works?... wad is the selection of the value?
for question num.5.....if i use 1MHz.. will the motor works?... how to determine the max frequency? Marcel Majoor mentioned tht the motor will losses its torque... i tot when frequency too high... the motor willl stall....
thanks again and again.... i m grateful
my warm regards,
sp
#### Fragrance
l297 clock
hi
here is the simulation cicuit fOr L297 how it works
Points: 2
Points: 2
### farhansinghera@yahoo.com
Points: 2
#### sp
##### Full Member level 6
stepper motor controller l297
thanxs for reply,,, could u make it more easily to read,, like pic,,,
bcos my pc hav no much space left for me to install orcad capture...
hope other will continue help answering my question... grateful
thanxs,,,
regards,
sp
#### Marcel Majoor
##### Full Member level 2
vref l297
A higher chopper rate is not necessarily better. A low frequency is typically 'worse' for our ears (audible range), so this is one reason to choose a frequency like 20 kHz. If the frequency is too high then, (guessing), the current limiting becomes less effective (the motor windings are a coil ... ..).
When the (over)current becomes active the output (ABCD or INH1/2) are -momentarily- switched off (one cycle of the chopper frequency). The control pin only select which of these will be chopped (either ABCD or INH1/2). This has nothing to with the normal operations of the device. You use the device as you would normally use (just assume this current regulation mechanism does not exist). The control pin ONLY defines the action taken during 'protection' and not when no step signal is being generated.
Note: You don't need to use the current regulation feature of the device. If you don't use any sense resistors and short the sense inputs of the device to 0V then no chopping will be used at all. In that case the device only translates the clokc/direction inputs to an ABCD output sequence.
The (sense) resistor value is based on the the current 'protection' level you want. The current through this resistor generates a voltage which is 'sensed' by the L297 (Sense1/2). If the L297 protection level is set to 0.5V (Vref input) then the maximum current allowed through the motor winding would be 1A with a 0.5 ohm resistor. With a 10 ohm resistor this would be 0.05A . So, the selection is based on the allowed current through the motor winding and the Vref you are using. A typical VRef would be 1.2V (using a 1.2V reference 'zener'). If you require a maximum current of, for example 0.5A, then your resistor would be 1.2V / 0.5A == 2.4 ohm
With 1MHz the motor will definitely not run. With every clock signal the motor moves from one position to the other. The motor has to have time to move to that position before you should generate the next clock signal. It has to overcome it's mass and any inertia present. I think a few hundreds herzt as a maximum frequency is probably a very high value (and you typically have to accelerate to the maximum possible speed).
If the clock changes too fast, the motor just does not have time to move to the next position and, hence, will not move (or very slow/obscure).
Points: 2
Points: 2
### enesb
Points: 2
#### sp
##### Full Member level 6
l297 chopper frecuency
thank you Marcel Majoor for all the help... i am really grateful... u almost clear all my doubt,, now it is turn for me to try out myself,,,
anyway... wad is the frequency hav to do wth ear?.. or u mean analogy...
as for the ABCD and INH1/2.. i think i get it finally,,,, however, which one should i use,,, ABCD or INH1/2?? i mean i should set wad to control pin,,, which one is more preferable?...
i am connecting the Vref to my +5V... should it ok,, bcos i want max peak current,,, if i use 0.5ohm so 5V/0.5ohm= 10A... so will this 10A too large n burn some of my part?..
another one would b how i know wad is the max step my stepper motor can support... i buy from an electronics shop n the guy dunno anything bout it... the motor length is 2cm including the head come out from the motor and the diameter is bout 2.5cm... n it has 4 wire(red yellow blue brown...i dunno how to connect this also).. any idea?
thank you again,,, i really dunno how to express my gratefulness
regards,
sp
#### Marcel Majoor
##### Full Member level 2
l297 c source
The frequency at which a motor (or a switching power supply for that matter) is controlled has a lot to do with the human ear. The switching of the motor (or chokes/transformers in a swtching power supply) do generate an audible sound (as if they were speakers). 'Users' don't like to hear a continuous noise coming from an electrical circuit or motor. That is one of the reasons why a frequency higher than audible for the human ear is generally preferred. Obviously there are also other reasons for selecting other frequencies, but this is just an, obvious, one.
I think you just experiment to decide which one (ABCD or INH1/2) is better. Using INH/2 reduces the dissipation in the sense resistors so this might be an obvious choise. But, it mght be that the motor runs better using ABCD-control instead of INH1/2-control. You just have to experiment.
Don't connect Vref to +5V. The maximum input is 3V!!
If you are using the +5V then at least use a resistor divider. For instance, put two 10 kOhm resistors in series and connect the ends to the 0V and +5V. The center connection then goes to the Vref input, which should be about 2.5V then (I assume the input impedance of the Vref input is relatively high).
Depending on the motor you use (and/or the capabilities of the driver/output drive), you have to set the current. You have to look at the specificatiosn of the driver and/or motor to find out which is an absolute maximum allowed current. Then set your 'protection' somewhat lower (75% ??).
If you are using the L298 as a driver, then this one can drive a maximum (!) current of 4A. The continuous current for this device is 2A, so in this case you would set the 'protection' to 2A (2.5V/2A = 1.25 ohm).
Figuring out the connection for a stepper motor should be not to difficult, but be aware that their are different type of stepper motors. Unfortunately I don't have the information by hand on how to find out the connections.
For the maximum speed concerned; this can be determined by experimentation. By slowly increasing the speed (clock frequency) you should easily be able to determine at which speed it still runs correct and at which it does not (be sure to connect the mechanical 'load' to the motor when testing...)
### sp
Points: 2
#### sp
##### Full Member level 6
limit curent l297
dear Mr. experience,
thank you again,,,, but my question continue,,,, please b patient wth me.... hehehe
Don't connect Vref to +5V. The maximum input is 3V!!
If you are using the +5V then at least use a resistor divider. For instance, put two 10 kOhm resistors in series and connect the ends to the 0V and +5V. The center connection then goes to the Vref input, which should be about 2.5V then (I assume the input impedance of the Vref input is relatively high)
fortunately u mention it... i tot 5V would b ok,,, how u know maximum is 3V??
if i use voltage divider...will it increase the losses too much as current may flow in any condition?...
for this paragraph....
Depending on the motor you use (and/or the capabilities of the driver/output drive), you have to set the current. You have to look at the specificatiosn of the driver and/or motor to find out which is an absolute maximum allowed current. Then set your 'protection' somewhat lower (75% ??).
If you are using the L298 as a driver, then this one can drive a maximum (!) current of 4A. The continuous current for this device is 2A, so in this case you would set the 'protection' to 2A (2.5V/2A = 1.25 ohm).
for the absolute maximum I for L298,,, i look at the data sheet but i only can find the "TOTAL DC CURRENT UP TO 4 A" at the intro on the front page of the data sheet... i cant seems to find it inside the data sheet... which specification mention the max allowed current?
this i just wanna make sure... for 75% of 4A... it should b almost 3A.. why u mention 2A??
hope i am correct assuming "(2.5V/2A = 1.25 ohm)" is to find the sense resistor...no???.......... (and if Vref can support 5V... and i want 3A... so 5V/3A = 1.6667 ohm.. so i find my sense resistor... yes?)
i think i start to understand tht the sense resistor is to cutoff the current when it is over the limit set by user...for "2.5V/2A = 1.25 ohm" the max current is 2A... over then it is cut out....correct??
if the L298 can support 4A but my motor can only support 2A.. is it ok to just use 4A?...or this maximum current is depend on the voltage n load condition (ohm's law) so it doesnt matter....
wad i mean is tht the max current is not a constant current supply to the motor,,, it is just the max current available tht the motor can eat up (if the motor request more than 4A, the L298 will say sorry; and if the motor request less, L298 supply less).. am i correct again??
the last part u teach how to find the motor max speed,, it is brilliant...
thank you Marcel Majoor and everyone helping ... thanks lot.... really grateful..
(when u think u solve a problem, u might just create another, so many question :lol: really hav to ask... i am in doubt)
my warm regards,
sp :lol:
#### Marcel Majoor
##### Full Member level 2
l297 problems with cw
The absolute maximum of 3V can be found in the datasheet of the L297. It is a simple as that. Look at page 7 of the datasheet, electrical characteristics of the Vref input - min 0V, max 3V .......
Thus, never put 5V on the Vref input. This might (will) destroy it!
The sense resistor calculations are based on the law of Ohm: U = I * R, so calculations are simple, assuming you know the Vref you decided on (< 3V !!) and you know the maximum current you want to allow.
The sense resistor is not used to cutoff the current directly (although it -will- limit the current flowing based on Ohms law), but the sense resistor is used to measure the current flowing through the motor winding. This is done by converting this current into a voltage (by means of the sense resistor). This voltage then goes to the sense input of the L297, where it is compared with the Vref voltage. If it is higher, then the driver outputs (ABCD or INH1/2 depending on the control input, you remember) are temporarily switched off.
A voltage divider always draws some current, but the current depends on the resistor values choosen. With 2x10kOhm this would be 5/20000 == 250 uA which is negligible . Even if you have to use 1KOhminstead of 10kOhm this would still be 2.5 mA, which should be no problem for the +5V power supply.
Instead of a voltage dvider you can use a voltage reference IC (e.g. which generates 1.2V or 2.5V). This makes the Vref more stable, but these devices also draw some current ....
On page 2 of the L298 datasheet themaximum rating are also beding shown. There is also additional data (under 'Peak output current), like 'Non repetative', and 'DC'. This 'DC' is the allowed current when the device is continuously on
The reason for 2A is that you typically want to be in the safe 'area' of the device. This is, at all times, 2A (assuming the devices is being cooled the way it should). Thus if you use 2A as your 'protection' base (disregard the 75% in that respect) then you should be fine. Don't try to draw 4A out of the device. This is an absolute maximum .... Even 3A is only allowed 'Non repettive'.
Thus, use 2A as the 'protection' in your design (note: this 'protection' is then only used to protect the L298 driver - if the motor has a lower maximum current then you should design for that limit).
BTW: It is the L297 which limits the current (by the chopping mechanism), based on the sensed voltage and Vref. The L298 does not limit the current in any way.
enesb
Points: 2
### enesb
Points: 2
#### sp
##### Full Member level 6
l297 simulation
sorry... i was looking at the application note of the L297,,,, tht's the prob... sorry...
thanks again,,,Marcel Majoor... i press the helped anyway...
if i am using 2.5V (voltage divider)... i can't use 0.5ohm as it shows in the data sheet for R(sense).... if i want to hav 1A... so R(sense)=2.5V/1A=2.5ohm... correct?...
anyway... could u tell me bout the clk timing... as u can c from the bellow pic... y the tclk time is just on the low pulse,,,, wad it means?..
if the tclk is shown... can we determine the clk period??....i think this is the fastest posible clk pulse allowed....
it seems like u are teaching me how to understand the datasheet....hehehe
Thank you.... i am grateful....
my regards,
sp
#### Marcel Majoor
##### Full Member level 2
pull up l297
The 2.5V/2.5 ohm == 1A is correct.
From the 0.5 us in the datasheet you could conclude that the (max) clock is 1 MHz (0.5 us low/0.5us high signal).
However, you should not be bothered about this, because this maximum is far (and I mean faaaar) beyond the capabilities of the steppermotor. Assuming you have an *ideal* steppermotor, only then you would we able to run at this frequency. Simple calculation dictate then, that with 1 Mhz, the steppermotor will do 1000000 steps/s. Assuming you have a stepper motor with 250 steps/revolution this is 4000 rotations/s == 240000 RPM ........
Unfortunately we don't deal with an ideal stepper motor so the maximum frequency you will actually be using willbe in the few hundreds Herz ... (just guessing here).
The L297 'steps' on the rising edge of the clock input.
The 'setup time'/'hold time' and such you see in the datasheet means that you can't (should not) switch the other signals at the same time as the clock signal. They need to be present a certain amount of time before the clock signal is activated, and need to be present some time after the clock signal is removed. Typically you don't have to bother about this at all, since these signals (CW/CCW/HALF/FULL) are fixed level signals. You don't change them all the time. Only the CW/CCW signal is changed, for changing directions, and an 'incorrect' timing would only mean that you would do one or two false steps.
Points: 2
Points: 2
### enesb
Points: 2
#### sp
##### Full Member level 6
what is clock in l297
sorry for asking again... need help on the "HALF/FULL_NOT" pin
as listed on the L297 DS... as bellow...
Code:
Half/full step select input. When high selects half step operation,
when low selects full step operation. One-phase-on full step mode
is obtained by selecting FULL when the L297’s translator is at an
even-numbered state.
Two-phase-on full step mode is set by selecting FULL when the
translator is at an odd numbered position. (The home position is
designate state 1).
there are 3 type...half step op, One-phase-on full step mode, Two-phase-on full step mode...
the half step is ok, just set the "HALF/FULL_NOT" pin to '1';
however, for the "One-phase-on full step mode" and "Two-phase-on full step mode" how to select one of it?...set the "HALF/FULL_NOT" pin to '0' and the 2 mode depend on your luck??????? like it mention there it depend on the translator state, can the translator state b controled?..
hope someone can help....
many many many thanks...
regards,
sp
#### Marcel Majoor
##### Full Member level 2
l297 y driver de 10a
You have to know the state the L297 is in. This can be detected using the HOME output of the L297. When this output is active you know that the L297 internal controller is in state 1. Once you know this, and because YOU control the steps (clock), you also know in which state the L297 (should be), and when to switch to the correct FULL STEP mode.
Note that you should do this 'synchronization'/'detection' in HALF STEP mode, since in this mode all states are 'used'.
You can also use another approach. When you set the FULL STEP MODE then the HOME output is activated only when the NORMAL DRIVE MODE is active. The HOME output is not active when WAVE DRIVE MODE is active. Thus, set FULL STEP mode, do 4 or more steps and when the HOME was activated then NORMAL DRIVE MODE is active. If this is NOT the mode you want then switch back to HALF STEP mode, do 1 step, switch to FULL STEP mode and check again. This time it should be the correct mode.
### sp
Points: 2
V
Points: 2
#### sp
##### Full Member level 6
wow! u are great... thank you....
from wad u say... so the full step mode cannot b control using mechanical switch as the motor move so fast...
so it would only feasible to control it using the controller....
actually yesterday nite i already can run the stepper motor,,, but one prob is the cw/ccw_not... i change to ccw using switch... but the motor doesnt wanna change direction,,,
thank you again...
warm regards,
sp
V
Points: 2
#### Marcel Majoor
##### Full Member level 2
what is chopper l297
The FULL STEP mode can be controlled by mechanical means, but you still need some logic to make sure the correct full step mode is switched on.
OR .... if you also use the RESET input then you know that, after a RESET, the controller is in state 1 ('HOME').
So, using the RESET input of the L297 you can also set the correct mode, without using the HOME as feedback.
NORMAL DRIVE MODE: set HALF STEP mode, activate RESET, de-activate RESET, set FULL STEP mode
WAVE DRIVE MODE: set HALF STEP mode, activate RESET, de-activate RESET, do one step (1 clock cycle), set FULL STEP mode
Changing directions is typically done at 'zero' speed of the motor. If you use a mechanical switch make sure that is (preferably) debounced, and that you swith from one level to the other (switch-over contact). Alternatively you could use a make or break contact switch (instead of a switch-over) and use a passive resistor (10k) to +5V. You let the switch then shorten the CW/CCW input to 0V.
SORENA01
Points: 2
V
Points: 2
### SORENA01
Points: 2
#### sp
##### Full Member level 6
reset l297
thank you x 1000 Marcel Majoor... again
wth "HALF/FULL_NOT" set to '0' and after reset_not for L297 is press and release... the home state(0101) start.... so this "NORMAL DRIVE MODE " is easily enter...
but for "WAVE DRIVE MODE" after reset_not ,,, waiting for one clk cycle is imposible for mechanical switch?... i think tht cannot do....
this is ok,, i can live wth the normal wave drive mode.... but the open collector output of the "home"... i cannot read from this pin, i dont understand why ... when it reach the 0101 state "home" will b '1' or '0'??
for the cw/cww_not.. the problem solved D,
the "app_note in L297" use a 470uF btween the power supply pin and gnd pin of L298N, but i dont hav the 470uF, so i used 10uF instead(my mistake)... the motor so weak n it cannot change direction....
after i plug out the capacitance, everything work supperb...
and the change direction is synchronously, so u can change anytime u like
another question is regarding the voltage drop for L298N... i use 15v from adapter(ac to dc) and it drop to 8-9V after the motor start to run....
so for this, should i use a voltage regulator to make it constantly 12V?...i prefer 12V
actually under my observation, the motor run better wth higher voltage(btween 9-10V).. 12V constantly should b better?...
the cheap stepper motor cost bout USD3.5(max almost 1000 steps per second, i try it out myself).... and everytime i assert the reset_not(L297)... the motor need some time to settle down,,, it shakes(not spinning well) a lot before it cool down and run properly, higher voltage shorter the time to settle down.
so 12V voltage regulator required? or a 9V one?
and for the "control" pin... it make no diff(high or low) on the performance....any comment?
should i buy better motor so tht it doesnt shake tht muchafter reset_not.. is quality the problem here?
thank you again.... i am grateful again.. hehehe...
my warm regards,
sp
V
Points: 2
#### Marcel Majoor
##### Full Member level 2
chopper frequency l297
Using a mechanical switch to select the mode is, basically, only possible with some dediacted hardware added. So in your situation I don't think it is a possibility.
Since the HOME output is an open-collector output, it floats when it is inactive. This means that if you want to measure the signal that you need a pull-up resistor (e.g. 10 kOhm) to +5V. This way you will measure 5V when the L297 is in it's initial state , and 0V otherwise. Thus, only with a pull-up resistor you will measure a '1' (HOME) or '0'. Without it you will either measure a '0' or a floating signal (which most of the time also 'translates' in a '0').
The 470 uF you mention is the 'buffer' of the power supply. The actual value depends on the current drawn from this power supply by the motors. Typically, the larger the better. The 10 uF you used does not 'buffer' that much (meaning that the voltage drops or becomes more erratic/changeable). When you removed the 10 uF the voltage probably dropped even more.
Using a regulator will not make the voltage become higher. Only a buffer capacitor (like the 470 uF) can make some difference here since it will provide additional power when a temporarily higher power is needed by the motor and which the power supply can not always provide (fast enough).
Most of the erratic movements might be due to an irregular voltage. A constant voltage is always the best. This does not necessarily need to be a regulated voltage (the motor does not really care if it has 5V or 5.5V ....). This constant voltage can be created using at least a large buffer capacitor. Obviously the power supply you use must also be able to provide enough power/current to the motor without a decreased output voltage. If you have a good power supply, and not connected that far away from the electronics then you can omit the buffer capacitor, or at least use a smaller one.
The CONTROL function is that it will chop the output only when an overcurrent situation occurs. If the current is lower than at what the chopper operation is set to, then you won't notice anything. And even when the chopper 'kicks' in, this might not be too obvious when looking at the motor (it should still functions as normal, only the 'torque' the motor can generate is less because the power applied to the motor is reduced by the chopper).
So, check if the power supply can generate the voltage and current the motor requires (without knowing the requirements of the motor things will stay 'experimental').
If not, use another power supply - a buffer capacitor is always a good idea but with a good power supply not that critical. During operation of the motor the voltage should not drop too much (this would indicate that the power supply can not supply the required current). If the voltage drops more than 0.5 - 1V (just guessing here) then I would suspect an 'underpowered' power supply.
Bottom line: You probably need a better power supply. Use a buffer capacitor. Check for the correct voltage for the motor (a few volts too high is almost never a problem).
xmen33
Points: 2
V
Points: 2
### xmen33
Points: 2
#### sp
##### Full Member level 6
how to control stepper motor with l297
thank you x 10... again...
for the "home" i understand D.... however the low output voltage level is 1.2-1.5V(using 1Mohm)... but my controller can only accept the max low is 0.8V(Vil)...
wad should i do?...i think this is the prob tht i cannot detect the change on this output...
for the buffer u meantion... can u look at the the diagram on my topmost first post....
there is another 100nF(C3) connected wth parallel wth the 470uF,,, wad is the purpose on this so-many-capacitors?... a 470uF is not enuff?..
i can change the direction of the motor,,,, but after a while,,, nth works anymore,,, i hav no idea wad is the problem... is it the prob wth the voltage from the adapter?..
sometimes it can,,, sometimes it cannot,,, i am really sick of it
i check the input to the CW/CW_NOT, it is asserted properly so it is not mistake-connect prob..
is it also my lousy motor quality?... when i tried wth 1000 steps/s the motor spin more smoothly... but wth my hand,,, i can change the direction to wadever direction i want,,, it doesnt seems like following my direction switch...
Code:
without knowing the requirements of the motor things will stay 'experimental'
this is good one... i am sick of this lousy motor....
but i do learn a lot(from you)...
warm regards,
sp
#### Marcel Majoor
##### Full Member level 2
sense resistors l297
The reason for a smaller and a larger capacitor is that the smaller capacitor is used to filter out the higher frequencies and the larger for the lower frequencies (or in your case to buffer the power supply). A capacitor has a certain resistance at a given frequency (impedance). With a given frequency the resistance of a smaller capacitor is lower than that of a larger capacitor and will therefore 'shorten' the signal more to ground than a large capacitor. The higher the frequency, the lower the resistance, the more the signal is shortened to ground.
The HOME output should have no problems getting the signal low enough (the datasheet mentions 0.4V as a maximum). However, be advised that using a pull-up resistor of 1 MOhm is not typical. A typical pull-up is more in the 10 kOhm .. 100 kOhm range (depending on fast/steep the pull-up should be). I would use a 10 kOhm resistor. You can use a pull-up which is as low as 1 kOhm (not advised though), and still you should have a 0.4V output (max) when the HOME output is active (meaning NOT HOME). When the HOME output is inactive (meaning HOME) the level should reach the 5V the pull-up is connected to (maybe 1V lower). This is the situation when the HOME output is not conencted to anything (except the pull-up resistor). Check this first. If this is correct, then connect the HOME output to the circuit you use for detection and test again. You probably will experience some different values, but the 0.4V 'low' voltage should still be possible (the HOME output can sink at least 5 mA without getting above the 0.4V level -> if not the circuit connected to the HOME is probably not correct).
If you have a 'bad' power supply you can get bad results. Also, when driving the motor too fast, the torque of the motor is very low and thus the motor can easily be stopped by 'external' means (a finger touching the motor). The only way to get the motor restarted is then typicall to lower the frequency which drives the motor. A stepper motor can only be run at a relative low frequency when no acceleration is used! I would advise to use a much lower frequency than the 1000 Hz you are using. Try it at 100 Hz and see if the motor behaves in the same way as it does at 1000 Hz.
Points: 2 | {} |
# 1 Pertemuan 10 Fungsi Kepekatan Khusus Matakuliah: I0134 – Metode Statistika Tahun: 2007.
## Presentasi berjudul: "1 Pertemuan 10 Fungsi Kepekatan Khusus Matakuliah: I0134 – Metode Statistika Tahun: 2007."— Transcript presentasi:
1 Pertemuan 10 Fungsi Kepekatan Khusus Matakuliah: I0134 – Metode Statistika Tahun: 2007
2 Learning Outcomes Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu : Mahasiswa akan dapat menghitung peluang, nilai harapan dan varians fungsi kepekatan seragam dan eksponensial.
3 Outline Materi Fungsi kepekatan seragam Fungsi distribusi seragam Nilai harapan dan varians fungsi kepekatan seragam Fungsi kepekatan eksponensial Fungsi distribusi eksponensial Nilai harapan dan varians peubah acak eksponensial
4 Uniform Distribution A continuous rv X is said to have a uniform distribution on the interval [a, b] if the pdf of X is X ~ U (a,b)
5 Exponential distribution X is said to have the exponential distribution if for some
6 Probability for a Continuous rv If X is a continuous rv, then for any number c, P(x = c) = 0. For any two numbers a and b with a < b,
7 Expected Value The expected or mean value of a continuous rv X with pdf f (x) is The expected or mean value of a discrete rv X with pmf f (x) is
8 Expected Value of h(X) If X is a continuous rv with pdf f(x) and h(x) is any function of X, then If X is a discrete rv with pmf f(x) and h(x) is any function of X, then
9 Variance and Standard Deviation The variance of continuous rv X with pdf f(x) and mean is The standard deviation is
10 Short-cut Formula for Variance
11 The Cumulative Distribution Function The cumulative distribution function, F(x) for a continuous rv X is defined for every number x by For each x, F(x) is the area under the density curve to the left of x.
12 Using F(x) to Compute Probabilities Let X be a continuous rv with pdf f(x) and cdf F(x). Then for any number a, and for any numbers a and b with a < b,
13 Ex 6 (Continue). X = length of time in remission, and What is the probability that a malaria patient’s remission lasts long than one year?
14 Obtaining f(x) from F(x) If X is a continuous rv with pdf f(x) and cdf F(x), then at every number x for which the derivative
15 Percentiles Let p be a number between 0 and 1. The (100p)th percentile of the distribution of a continuous rv X denoted by, is defined by
16 Median The median of a continuous distribution, denoted by, is the 50 th percentile. So satisfies That is, half the area under the density curve is to the left of
17 Selamat Belajar Semoga Sukses.
Presentasi serupa | {} |
### Introduction:
This project analyzes the relationship between a person’s family income and his/her’s political affiliation.
A person’s political affiliation can be influenced by a variety of factors, such as his ideology, the party his family supports, his geographical location, his income etc. In this project we aim to analyze how a person’s political affiliation is influenced by his household income
This analasis will help us determine the characteristics of voters for each political party.
### Data:
This study utilizes the General Social Survey (GSS) data.
General Social Survey (GSS) is a sociological survey used to collect data on demographic characteristics and attitudes of residents of the United States. It is used to collect data about a variety of topics. It can be analyzed in depth using the following codebook #https://d396qusza40orc.cloudfront.net/statistics%2Fproject%2Fgss1.html
The data was modified for Data Analysis and Statistical Inference course (Duke University).
How this data was collected
The study has been conductedfor the past 40 years and the data collection process has been modified every few years. The exact in-depth procedure for the collection can be studied using the following link: http://publicdata.norc.org:41000/gss/documents//BOOK/GSS_Codebook_AppendixA.pdf
### Exploratory data analysis:
We have only considered a subset of the gss data set which contains the political affiliation of a person (partyid) and his family income (coninc) in constant USD.
partyid is a categorical variable with 8 levels.
summary(gss$partyid) ## Strong Democrat Not Str Democrat Ind,Near Dem ## 9117 12040 6743 ## Independent Ind,Near Rep Not Str Republican ## 8499 4921 9005 ## Strong Republican Other Party NA's ## 5548 861 327 plot((gss$partyid))
While, coninc is a continuous numerical variable
summary(gss$coninc) ## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's ## 383 18440 35600 44500 59540 180400 5829 hist(gss$coninc)
We can check if there is any relation between the 2 variables by making a box plot for income for each partyid
plot(gss$coninc~gss$partyid)
We can see that there is no clear positive or negative correlayion between the two variables; however, the mean average income fior a few parties is higher than the others and needs to be further evaluated.
### Inference:
We want to find out if there is a statistically significant difference between the party a person supports and his mean household income.
For the Null Hypothesis we will say that there is no difference in average income across people supporting different political parties. i.e. H0: µ1=µ2=µ3=µ4=µ5=µ6=µ7=µ8
While the alternative hypothesis states that there is atleast one group which has an average houshold income different from the other groups.
Pre-Requisite conditions for ANOVA hypothesis testing:
1. Independence: The GSS survey survey’s around 57000 people which is definately lower thn 10% of the population of the US. Hence the observations are independent. Also, the groups are independent of each other as a person will fall into one of the groups and cannot be a member of more than one group.
2.Normality: We can test the normality by drawing probability plot or each of the groups between
# To make 8 graphs with 4 in each row
par(mfrow = c(2,4))
# To Define the 8 groups and test the normality of each group using probability distribution plots
party = c("Strong Democrat","Not Str Democrat","Ind,Near Dem","Independent","Ind,Near Rep","Not Str Republican","Strong Republican","Other Party")
for (i in 1:8) {
qqnorm(gss[gss$partyid == party[i],]$coninc, main=party[i])
qqline(gss[gss$partyid == party[i],]$coninc)
}
We can see that the plots are primarily normal; however, there is some deviation from normality near the upper ranges of the plots
1. Constant Variance: We can check the variability betwen the groups by drawing side-by-side box plots
par(mfrow = c(1, 1))
boxplot(gss$coninc ~ gss$partyid, main="Household Income by Political Affiliation", las= 2)
We can see that the groups have similar variability; however, a few groups have higher average household income
ANOVA Test:
anova(lm(coninc ~ partyid, data=gss))
## Analysis of Variance Table
##
## Response: coninc
## Df Sum Sq Mean Sq F value Pr(>F)
## partyid 7 1.7462e+12 2.4946e+11 198.42 < 2.2e-16 ***
## Residuals 51049 6.4182e+13 1.2573e+09
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
ANOVA gives us an f statistic of 198.42 and a very very low p value. Which means we can reject the null hypothesis and state that Household income does vary with the political affiliation of a person.
We can find out which groups have different means by conducting a pairwise comparision between the groups. We will calculate a t statistice between each group pair to confirm or reject the null hypothesis. i.e. there is no difference between the average income of the group and political affiliation
pairwise.t.test(gss$coninc, gss$partyid)
##
## Pairwise comparisons using t tests with pooled SD
##
## data: gss$coninc and gss$partyid
##
## Strong Democrat Not Str Democrat Ind,Near Dem
## Not Str Democrat 2.4e-13 - -
## Ind,Near Dem 5.8e-15 0.15805 -
## Independent 0.15805 3.5e-07 4.2e-09
## Ind,Near Rep < 2e-16 < 2e-16 < 2e-16
## Not Str Republican < 2e-16 < 2e-16 < 2e-16
## Strong Republican < 2e-16 < 2e-16 < 2e-16
## Other Party 1.2e-07 0.02790 0.12285
## Independent Ind,Near Rep Not Str Republican
## Not Str Democrat - - -
## Ind,Near Dem - - -
## Independent - - -
## Ind,Near Rep < 2e-16 - -
## Not Str Republican < 2e-16 0.00074 -
## Strong Republican < 2e-16 4.5e-15 6.2e-07
## Other Party 6.4e-06 0.03641 3.1e-05
## Strong Republican
## Not Str Democrat -
## Ind,Near Dem -
## Independent -
## Ind,Near Rep -
## Not Str Republican -
## Strong Republican -
## Other Party 3.8e-11
##
## P value adjustment method: holm
We can see that the p value is lower than 0.05 for all groups except 3 groups. Hence, we can reject the null hypothesis for the other groups.
### Conclusion:
The Study analyzed the relationship between a person’s political affiliation and their household income. We found that there is a relationship between the two variables after conducing ANOVA tests and pairwise conmparisonsbetween the groups. However, there were a few outliers found during our initial testing for the conditions for ANOVA means that we need to be catious in interpreting the results of the tests. We could do outlier treatment to further improve our analysis and get more reliable results. Insert conclusion here…
### References:
Citation for the original data:
Smith, Tom W., Michael Hout, and Peter V. Marsden. General Social Survey, 1972-2012 [Cumulative File]. ICPSR34802-v1. Storrs, CT: Roper Center for Public Opinion Research, University of Connecticut /Ann Arbor, MI: Inter-university Consortium for Political and Social Research [distributors], 2013-09-11. doi:10.3886/ICPSR34802.v1 Persistent URL: http://doi.org/10.3886/ICPSR34802.v1 | {} |
# Repeatedly calculating a hessian using both ForwardDiff and ReverseDiff
I would like to calculate a Hessian matrix of a function f:\mathbb{R}^n \rightarrow \mathbb{R} where n\gg1. In my usecase the Jacobian of f is always needed, and the Hessian may or may not be needed (but this is known before calculation of the Jacobian). Additionally, I only require the Hessian with respect to the first m dimensions, again where n \gg m > 1, this is because the first m input dimensions are parameters, and the latter are data.
I am aware that using diffresults one can recover the Jacobian from a single Hessian calculation, however given the dimensions of my Jacobian I felt that it made more sense to calculate the Jacobian using ReverseDiff and then the Hessian necessary part of the Hessian using ForwardDiff. Given the dimensions of the function I thought it best to first calculate the Jacobian (or indeed gradient) with ReverseDiff, and then use ForwardDiff for Hessian. Furthermore, I wish to use ReverseDiff as I will recompute the Jacobian and Hessian for multiple values of the input, so I can make use of ReverseDiff’s tape compilation.
The following is a minimal example demonstrating the error (without mention to the m-subsetting, although I welcome comments on this aspect in regards to feasibility/performance)
using ForwardDiff, ReverseDiff
f(x) = x[1]^2 + x[2]^2
inputs = [1.0, 1.0]
# This returns what one would expect
correct_output = ForwardDiff.jacobian(x -> ReverseDiff.gradient(f, x), inputs)
correct_output == ForwardDiff.hessian(f, inputs)
# This returns the error below
out = zeros(2)
ForwardDiff.jacobian(f_inputs -> ReverseDiff.gradient!(out, compiled_ftape, f_inputs), inputs)
Error:
MethodError: no method matching Float64(::ForwardDiff.Dual{ForwardDiff.Tag{var"#95#96",Float64},Float64,2})
Closest candidates are:
Float64(::Real, !Matched::RoundingMode) where T<:AbstractFloat at rounding.jl:200
Float64(::T) where T<:Number at boot.jl:716
Float64(!Matched::Float16) at float.jl:256
...
Stacktrace:
[1] convert(::Type{Float64}, ::ForwardDiff.Dual{ForwardDiff.Tag{var"#95#96",Float64},Float64,2}) at .\number.jl:7
[2] setindex!(::Array{Float64,1}, ::ForwardDiff.Dual{ForwardDiff.Tag{var"#95#96",Float64},Float64,2}, ::Int64) at .\array.jl:847
[3] _unsafe_copyto!(::Array{Float64,1}, ::Int64, ::Array{ForwardDiff.Dual{ForwardDiff.Tag{var"#95#96",Float64},Float64,2},1}, ::Int64, ::Int64) at .\array.jl:257
[4] unsafe_copyto! at .\array.jl:311 [inlined]
[5] _copyto_impl! at .\array.jl:335 [inlined]
[6] copyto! at .\array.jl:321 [inlined]
[7] copyto! at .\array.jl:347 [inlined]
[8] value! at C:\Users\jmmsm\.julia\packages\ReverseDiff\NoIPU\src\tracked.jl:156 [inlined]
[9] seeded_forward_pass! at C:\Users\jmmsm\.julia\packages\ReverseDiff\NoIPU\src\api\tape.jl:41 [inlined]
[11] (::var"#95#96")(::Array{ForwardDiff.Dual{ForwardDiff.Tag{var"#95#96",Float64},Float64,2},1}) at .\In[82]:1
[12] vector_mode_dual_eval at C:\Users\jmmsm\.julia\packages\ForwardDiff\qTmqf\src\apiutils.jl:37 [inlined]
[13] vector_mode_jacobian(::var"#95#96", ::Array{Float64,1}, ::ForwardDiff.JacobianConfig{ForwardDiff.Tag{var"#95#96",Float64},Float64,2,Array{ForwardDiff.Dual{ForwardDiff.Tag{var"#95#96",Float64},Float64,2},1}}) at C:\Users\jmmsm\.julia\packages\ForwardDiff\qTmqf\src\jacobian.jl:145
[14] jacobian(::Function, ::Array{Float64,1}, ::ForwardDiff.JacobianConfig{ForwardDiff.Tag{var"#95#96",Float64},Float64,2,Array{ForwardDiff.Dual{ForwardDiff.Tag{var"#95#96",Float64},Float64,2},1}}, ::Val{true}) at C:\Users\jmmsm\.julia\packages\ForwardDiff\qTmqf\src\jacobian.jl:21
[15] jacobian(::Function, ::Array{Float64,1}, ::ForwardDiff.JacobianConfig{ForwardDiff.Tag{var"#95#96",Float64},Float64,2,Array{ForwardDiff.Dual{ForwardDiff.Tag{var"#95#96",Float64},Float64,2},1}}) at C:\Users\jmmsm\.julia\packages\ForwardDiff\qTmqf\src\jacobian.jl:19 (repeats 2 times)
[16] top-level scope at In[82]:1
I have no solutions, but am interested in answers also…
1 Like
Take a look at the type passed to f_inputs by ForwardDiff.jacobian. You have compiled the tape for Float64 arguments, but it gets passed a ForwardDiff.Dual.
Untested, but I guess you could do something like this:
const CACHE = Dict{DataType,Any}()
function inner(x::Vector{T}) where {T<:Real}
CACHE[T] = (tape, zeros(T, length(x)))
end
tape, y = CACHE[T]
For those wondering, there was a performance increase for inputs somewhere between dimension 50 and 500, at which point pure ReverseDiff.hessian was faster. This implementation also had far fewer allocations in any case, though the memory usage was generally similar or the same as ReverseDiff.hessian. | {} |
## Cryptology ePrint Archive: Report 2014/827
Interactive Coding for Interactive Proofs
Yevgeniy Dodis and Allison Bishop Lewko
Abstract: We consider interactive proof systems over adversarial communication channels. We show that the seminal result that $\ip = \pspace$ still holds when the communication channel is malicious, allowing even a constant fraction of the communication to be arbitrarily corrupted.
Category / Keywords: | {} |
Two way ANOVA - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Two way ANOVA
Friedman test
Two sample $t$ test - equal variances not assumed
$z$ test for the difference between two proportions
Independent/grouping variablesIndependent/grouping variableIndependent/grouping variableIndependent/grouping variable
Two categorical, the first with $I$ independent groups and the second with $J$ independent groups ($I \geqslant 2$, $J \geqslant 2$)One within subject factor ($\geq 2$ related groups)One categorical with 2 independent groupsOne categorical with 2 independent groups
Dependent variableDependent variableDependent variableDependent variable
One quantitative of interval or ratio levelOne of ordinal levelOne quantitative of interval or ratio levelOne categorical with 2 independent groups
Null hypothesisNull hypothesisNull hypothesisNull hypothesis
ANOVA $F$ tests:
• H0 for main and interaction effects together (model): no main effects and interaction effect
• H0 for independent variable A: no main effect for A
• H0 for independent variable B: no main effect for B
• H0 for the interaction term: no interaction effect between A and B
Like in one way ANOVA, we can also perform $t$ tests for specific contrasts and multiple comparisons. This is more advanced stuff.
H0: the population scores in any of the related groups are not systematically higher or lower than the population scores in any of the other related groups
Usually the related groups are the different measurement points. Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
H0: $\mu_1 = \mu_2$
Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2.
H0: $\pi_1 = \pi_2$
Here $\pi_1$ is the population proportion of 'successes' for group 1, and $\pi_2$ is the population proportion of 'successes' for group 2.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
ANOVA $F$ tests:
• H1 for main and interaction effects together (model): there is a main effect for A, and/or for B, and/or an interaction effect
• H1 for independent variable A: there is a main effect for A
• H1 for independent variable B: there is a main effect for B
• H1 for the interaction term: there is an interaction effect between A and B
H1: the population scores in some of the related groups are systematically higher or lower than the population scores in other related groups H1 two sided: $\mu_1 \neq \mu_2$
H1 right sided: $\mu_1 > \mu_2$
H1 left sided: $\mu_1 < \mu_2$
H1 two sided: $\pi_1 \neq \pi_2$
H1 right sided: $\pi_1 > \pi_2$
H1 left sided: $\pi_1 < \pi_2$
AssumptionsAssumptionsAssumptionsAssumptions
• Within each of the $I \times J$ populations, the scores on the dependent variable are normally distributed
• The standard deviation of the scores on the dependent variable is the same in each of the $I \times J$ populations
• For each of the $I \times J$ groups, the sample is an independent and simple random sample from the population defined by that group. That is, within and between groups, observations are independent of one another
• Equal sample sizes for each group make the interpretation of the ANOVA output easier (unequal sample sizes result in overlap in the sum of squares; this is advanced stuff)
• Sample of 'blocks' (usually the subjects) is a simple random sample from the population. That is, blocks are independent of one another
• Within each population, the scores on the dependent variable are normally distributed
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
• Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb:
• Significance test: number of successes and number of failures are each 5 or more in both sample groups
• Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures are each 10 or more in both sample groups
• Plus four 90%, 95%, or 99% confidence interval: sample sizes of both groups are 5 or more
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
Test statisticTest statisticTest statisticTest statistic
For main and interaction effects together (model):
• $F = \dfrac{\mbox{mean square model}}{\mbox{mean square error}}$
For independent variable A:
• $F = \dfrac{\mbox{mean square A}}{\mbox{mean square error}}$
For independent variable B:
• $F = \dfrac{\mbox{mean square B}}{\mbox{mean square error}}$
For the interaction term:
• $F = \dfrac{\mbox{mean square interaction}}{\mbox{mean square error}}$
Note: mean square error is also known as mean square residual or mean square within.
$Q = \dfrac{12}{N \times k(k + 1)} \sum R^2_i - 3 \times N(k + 1)$
Here $N$ is the number of 'blocks' (usually the subjects - so if you have 4 repeated measurements for 60 subjects, $N$ equals 60), $k$ is the number of related groups (usually the number of repeated measurements), and $R_i$ is the sum of ranks in group $i$.
Remember that multiplication precedes addition, so first compute $\frac{12}{N \times k(k + 1)} \times \sum R^2_i$ and then subtract $3 \times N(k + 1)$.
Note: if ties are present in the data, the formula for $Q$ is more complicated.
$t = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}}$
Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $s^2_1$ is the sample variance in group 1, $s^2_2$ is the sample variance in group 2, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to the null hypothesis.
The denominator $\sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}$ is the standard error of the sampling distribution of $\bar{y}_1 - \bar{y}_2$. The $t$ value indicates how many standard errors $\bar{y}_1 - \bar{y}_2$ is removed from 0.
Note: we could just as well compute $\bar{y}_2 - \bar{y}_1$ in the numerator, but then the left sided alternative becomes $\mu_2 < \mu_1$, and the right sided alternative becomes $\mu_2 > \mu_1$.
$z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
Here $p_1$ is the sample proportion of successes in group 1: $\dfrac{X_1}{n_1}$, $p_2$ is the sample proportion of successes in group 2: $\dfrac{X_2}{n_2}$, $p$ is the total proportion of successes in the sample: $\dfrac{X_1 + X_2}{n_1 + n_2}$, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2.
Note: we could just as well compute $p_2 - p_1$ in the numerator, but then the left sided alternative becomes $\pi_2 < \pi_1$, and the right sided alternative becomes $\pi_2 > \pi_1.$
Pooled standard deviationn.a.n.a.n.a.
\begin{aligned} s_p &= \sqrt{\dfrac{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2}{N - (I \times J)}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned} ---
Sampling distribution of $F$ if H0 were trueSampling distribution of $Q$ if H0 were trueSampling distribution of $t$ if H0 were trueSampling distribution of $z$ if H0 were true
For main and interaction effects together (model):
• $F$ distribution with $(I - 1) + (J - 1) + (I - 1) \times (J - 1)$ (df model, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
For independent variable A:
• $F$ distribution with $I - 1$ (df A, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
For independent variable B:
• $F$ distribution with $J - 1$ (df B, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
For the interaction term:
• $F$ distribution with $(I - 1) \times (J - 1)$ (df interaction, numerator) and $N - (I \times J)$ (df error, denominator) degrees of freedom
Here $N$ is the total sample size.
If the number of blocks $N$ is large, approximately the chi-squared distribution with $k - 1$ degrees of freedom.
For small samples, the exact distribution of $Q$ should be used.
Approximately the $t$ distribution with $k$ degrees of freedom, with $k$ equal to
$k = \dfrac{\Bigg(\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}\Bigg)^2}{\dfrac{1}{n_1 - 1} \Bigg(\dfrac{s^2_1}{n_1}\Bigg)^2 + \dfrac{1}{n_2 - 1} \Bigg(\dfrac{s^2_2}{n_2}\Bigg)^2}$
or
$k$ = the smaller of $n_1$ - 1 and $n_2$ - 1
First definition of $k$ is used by computer programs, second definition is often used for hand calculations.
Approximately the standard normal distribution
Significant?Significant?Significant?Significant?
• Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or
• Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\alpha$
If the number of blocks $N$ is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
Two sided:
Right sided:
Left sided:
Two sided:
Right sided:
Left sided:
n.a.n.a.Approximate $C\%$ confidence interval for $\mu_1 - \mu_2$Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$
--$(\bar{y}_1 - \bar{y}_2) \pm t^* \times \sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}$
where the critical value $t^*$ is the value under the $t_{k}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20).
The confidence interval for $\mu_1 - \mu_2$ can also be used as significance test.
Regular (large sample):
• $(p_1 - p_2) \pm z^* \times \sqrt{\dfrac{p_1(1 - p_1)}{n_1} + \dfrac{p_2(1 - p_2)}{n_2}}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval)
With plus four method:
• $(p_{1.plus} - p_{2.plus}) \pm z^* \times \sqrt{\dfrac{p_{1.plus}(1 - p_{1.plus})}{n_1 + 2} + \dfrac{p_{2.plus}(1 - p_{2.plus})}{n_2 + 2}}$
where $p_{1.plus} = \dfrac{X_1 + 1}{n_1 + 2}$, $p_{2.plus} = \dfrac{X_2 + 1}{n_2 + 2}$, and the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval)
Effect sizen.a.n.a.n.a.
• Proportion variance explained $R^2$:
Proportion variance of the dependent variable $y$ explained by the independent variables and the interaction effect together:
\begin{align} R^2 &= \dfrac{\mbox{sum of squares model}}{\mbox{sum of squares total}} \end{align} $R^2$ is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population.
• Proportion variance explained $\eta^2$:
Proportion variance of the dependent variable $y$ explained by an independent variable or interaction effect:
\begin{align} \eta^2_A &= \dfrac{\mbox{sum of squares A}}{\mbox{sum of squares total}}\\ \\ \eta^2_B &= \dfrac{\mbox{sum of squares B}}{\mbox{sum of squares total}}\\ \\ \eta^2_{int} &= \dfrac{\mbox{sum of squares int}}{\mbox{sum of squares total}} \end{align} $\eta^2$ is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population.
• Proportion variance explained $\omega^2$:
Corrects for the positive bias in $\eta^2$ and is equal to:
\begin{align} \omega^2_A &= \dfrac{\mbox{sum of squares A} - \mbox{degrees of freedom A} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \\ \omega^2_B &= \dfrac{\mbox{sum of squares B} - \mbox{degrees of freedom B} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \\ \omega^2_{int} &= \dfrac{\mbox{sum of squares int} - \mbox{degrees of freedom int} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \end{align} $\omega^2$ is a better estimate of the explained variance in the population than $\eta^2$. Only for balanced designs (equal sample sizes).
• Proportion variance explained $\eta^2_{partial}$: \begin{align} \eta^2_{partial\,A} &= \frac{\mbox{sum of squares A}}{\mbox{sum of squares A} + \mbox{sum of squares error}}\\ \\ \eta^2_{partial\,B} &= \frac{\mbox{sum of squares B}}{\mbox{sum of squares B} + \mbox{sum of squares error}}\\ \\ \eta^2_{partial\,int} &= \frac{\mbox{sum of squares int}}{\mbox{sum of squares int} + \mbox{sum of squares error}} \end{align}
---
n.a.n.a.Visual representationn.a.
---
ANOVA tablen.a.n.a.n.a.
---
Equivalent ton.a.n.a.Equivalent to
OLS regression with two categorical independent variables and the interaction term, transformed into $(I - 1)$ + $(J - 1)$ + $(I - 1) \times (J - 1)$ code variables.--When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels.
Example contextExample contextExample contextExample context
Is the average mental health score different between people from a low, moderate, and high economic class? And is the average mental health score different between men and women? And is there an interaction effect between economic class and gender?Is there a difference in depression level between measurement point 1 (pre-intervention), measurement point 2 (1 week post-intervention), and measurement point 3 (6 weeks post-intervention)?Is the average mental health score different between men and women?Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic.
SPSSSPSSSPSSSPSS
Analyze > General Linear Model > Univariate...
• Put your dependent (quantitative) variable in the box below Dependent Variable and your two independent (grouping) variables in the box below Fixed Factor(s)
Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples...
• Put the $k$ variables containing the scores for the $k$ related groups in the white box below Test Variables
• Under Test Type, select the Friedman test
Analyze > Compare Means > Independent-Samples T Test...
• Put your dependent (quantitative) variable in the box below Test Variable(s) and your independent (grouping) variable in the box below Grouping Variable
• Click on the Define Groups... button. If you can't click on it, first click on the grouping variable so its background turns yellow
• Fill in the value you have used to indicate your first group in the box next to Group 1, and the value you have used to indicate your second group in the box next to Group 2
• Continue and click OK
SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Analyze > Descriptive Statistics > Crosstabs...
• Put your independent (grouping) variable in the box below Row(s), and your dependent variable in the box below Column(s)
• Click the Statistics... button, and click on the square in front of Chi-square
• Continue and click OK
JamoviJamoviJamoviJamovi
ANOVA > ANOVA
• Put your dependent (quantitative) variable in the box below Dependent Variable and your two independent (grouping) variables in the box below Fixed Factors
ANOVA > Repeated Measures ANOVA - Friedman
• Put the $k$ variables containing the scores for the $k$ related groups in the box below Measures
T-Tests > Independent Samples T-Test
• Put your dependent (quantitative) variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
• Under Tests, select Welch's
• Under Hypothesis, select your alternative hypothesis
Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Frequencies > Independent Samples - $\chi^2$ test of association
• Put your independent (grouping) variable in the box below Rows, and your dependent variable in the box below Columns
Practice questionsPractice questionsPractice questionsPractice questions | {} |
# Nitro-substitution [closed]
Here I expected (b) to happen via nucleophilic substitution, but contrary to that (a) is the one occuring. What exactly is happening here?
• I’m voting to close this question because it's a bunch of crappy pictures. Up your game. ChemSE is, without a doubt, the best resource for all things chemistry-related. Play the game, get rewarded. – Todd Minehardt Jul 12 '20 at 3:56
The reaction given in options (a) and (b) are called formation of nitrate esters. The following nitrate ester is called "ethyl nitrate",
Firstly, $$\ce{HNO3}$$ and $$\ce{H2SO4}$$ reacts to form the $$\ce{E+}$$. As we know, the former is relatively weaker acid than the latter, therefore $$\ce{HNO3}$$ donates $$\ce{OH-}$$ and neutralisation takes place,
$$\ce{H2SO4 + HNO3 -> H2O + HSO4- + NO2+}$$
So, here electrophile is $$\ce{NO2+}$$, which gets attacked by lone pairs of alcoholic oxygen and finally gives "nitrate ester" as shown correctly in (a).
(Here is a video explanation of the mechanism by Sal, KhanAcademy)
So, according to me, I guess (b) and (c) are not according to the reaction, hence they should be correct.
Further, if anyone is interested in (c), they should have a look at the answer in Why does NaBH4 reduce double bonds conjugated to carbonyl groups, while LiAlH4 does not? | {} |
$\S 1$ The Holy Grail of Arithmetic: Bridging Provability and Computability
(Notations, non-standard concepts, and definitions used commonly in these investigations are detailed in this post.)
Peter Wegner and Dina Goldin
In a short opinion paper, Computation Beyond Turing Machines‘, Computer Scientists Peter Wegner and Dina Goldin (Wg03) advanced the thesis that:
A paradigm shift is necessary in our notion of computational problem solving, so it can provide a complete model for the services of today’s computing systems and software agents.’
We note that Wegner and Goldin’s arguments, in support of their thesis, seem to reflect an extraordinarily eclectic view of mathematics, combining both an implicit acceptance of, and implicit frustration at, the standard interpretations and dogmas of classical mathematical theory:
(i) … Turing machines are inappropriate as a universal foundation for computational problem solving, and … computer science is a fundamentally non-mathematical discipline.’
(ii) (Turing’s) 1936 paper … proved that mathematics could not be completely modeled by computers.’
(iii) … the Church-Turing Thesis … equated logic, lambda calculus, Turing machines, and algorithmic computing as equivalent mechanisms of problem solving.’
(iv) Turing implied in his 1936 paper that Turing machines … could not provide a model for all forms of mathematics.’
(v) … Gödel had shown in 1931 that logic cannot model mathematics … and Turing showed that neither logic nor algorithms can completely model computing and human thought.’
These remarks vividly illustrate the dilemma with which not only Theoretical Computer Sciences, but all applied sciences that depend on mathematics—for providing a verifiable language to express their observations precisely—are faced:
Query: Are formal classical theories essentially unable to adequately express the extent and range of human cognition, or does the problem lie in the way formal theories are classically interpreted at the moment?
The former addresses the question of whether there are absolute limits on our capacity to express human cognition unambiguously; the latter, whether there are only temporal limits—not necessarily absolute—to the capacity of classical interpretations to communicate unambiguously that which we intended to capture within our formal expression.
Prima facie, applied science continues, perforce, to interpret mathematical concepts Platonically, whilst waiting for mathematics to provide suitable, and hopefully reliable, answers as to how best it may faithfully express its observations verifiably.
Lance Fortnow
This dilemma is also reflected in Computer Scientist Lance Fortnow’s on-line rebuttal of Wegner and Goldin’s thesis, and of their reasoning.
Thus Fortnow divides his faith between the standard interpretations of classical mathematics (and, possibly, the standard set-theoretical models of formal systems such as standard Peano Arithmetic), and the classical computational theory of Turing machines.
He relies on the former to provide all the proofs that matter:
Not every mathematical statement has a logical proof, but logic does capture everything we can prove in mathematics, which is really what matters’;
and, on the latter to take care of all essential, non-provable, truth:
… what we can compute is what computer science is all about’.
Can faith alone suffice?
However, as we shall argue in a subsequent post, Fortnow’s faith in a classical Church-Turing Thesis that ensures:
… Turing machines capture everything we can compute’,
may be as misplaced as his faith in the infallibility of standard interpretations of classical mathematics.
The reason: There are, prima facie, reasonably strong arguments for a Kuhnian (Ku62) paradigm shift; not, as Wegner and Goldin believe, in the notion of computational problem solving, but in the standard interpretations of classical mathematical concepts.
However, Wegner and Goldin could be right in arguing that the direction of such a shift must be towards the incorporation of non-algorithmic effective methods into classical mathematical theory (as detailed in the Birmingham paper); presuming, from the following remarks, that this is, indeed, what external interactions’ are assumed to provide beyond classical Turing-computability:
(vi) … that Turing machine models could completely describe all forms of computation … contradicted Turing’s assertion that Turing machines could only formalize algorithmic problem solving … and became a dogmatic principle of the theory of computation’.
(vii) … interaction between the program and the world (environment) that takes place during the computation plays a key role that cannot be replaced by any set of inputs determined prior to the computation’.
(viii) … a theory of concurrency and interaction requires a new conceptual framework, not just a refinement of what we find natural for sequential [algorithmic] computing’.
(ix) … the assumption that all of computation can be algorithmically specified is still widely accepted’.
A widespread notion of particular interest, which seems to be recurrently implicit in Wegner and Goldin’s assertions too, is that mathematics is a dispensable tool of science, rather than its indispensable mother tongue.
Elliott Mendelson
However, the roots of such beliefs may also lie in ambiguities, in the classical definitions of foundational elements, that allow the introduction of non-constructive—hence non-verifiable, non-computational, ambiguous, and essentially Platonic—elements into the standard interpretations of classical mathematics.
For instance, in a 1990 philosophical reflection, Elliott Mendelson’s following remarks (in Me90; reproduced from Selmer Bringsjord (Br93)), implicitly imply that classical definitions of various foundational elements can be argued as being either ambiguous, or non-constructive, or both:
Here is the main conclusion I wish to draw: it is completely unwarranted to say that CT is unprovable just because it states an equivalence between a vague, imprecise notion (effectively computable function) and a precise mathematical notion (partial-recursive function). … The concepts and assumptions that support the notion of partial-recursive function are, in an essential way, no less vague and imprecise than the notion of effectively computable function; the former are just more familiar and are part of a respectable theory with connections to other parts of logic and mathematics. (The notion of effectively computable function could have been incorporated into an axiomatic presentation of classical mathematics, but the acceptance of CT made this unnecessary.) … Functions are defined in terms of sets, but the concept of set is no clearer than that of function and a foundation of mathematics can be based on a theory using function as primitive notion instead of set. Tarski’s definition of truth is formulated in set-theoretic terms, but the notion of set is no clearer than that of truth. The model-theoretic definition of logical validity is based ultimately on set theory, the foundations of which are no clearer than our intuitive understanding of logical validity. … The notion of Turing-computable function is no clearer than, nor more mathematically useful (foundationally speaking) than, the notion of an effectively computable function.’
Consequently, standard interpretations of classical theory may, inadvertently, be weakening a desirable perception—of mathematics as the lingua franca of scientific expression—by ignoring the possibility that, since mathematics is, indeed, indisputably accepted as the language that most effectively expresses and communicates intuitive truth, the chasm between formal truth and provability must, of necessity, be bridgeable.
Cristian Calude, Elena Calude and Solomon Marcus
The belief in the existence of such a bridge is occasionally implicit in interpretations of computational theory.
For instance, in an arXived paper Passages of Proof, Computer Scientists Cristian Calude, Elena Calude and Solomon Marcus remark that:
“Classically, there are two equivalent ways to look at the mathematical notion of proof: logical, as a finite sequence of sentences strictly obeying some axioms and inference rules, and computational, as a specific type of computation. Indeed, from a proof given as a sequence of sentences one can easily construct a Turing machine producing that sequence as the result of some finite computation and, conversely, given a machine computing a proof we can just print all sentences produced during the computation and arrange them into a sequence.”
In other words, the authors seem to hold that Turing-computability of a proof’, in the case of an arithmetical proposition, is equivalent to provability of its representation in PA.
Wilfrid Sieg
We now attempt to build such a bridge formally, which is essentially one between the arithmetical ‘Decidability and Calculability’ described by Philosopher Wilfrid Sieg in his in-depth and wide-ranging survey ‘On Comptability‘, in which he addresses Gödel’s lifelong belief that an iff bridge between the two concepts is ‘impossible’ for ‘the whole calculus of predicates’ (Wi08, p.602).
$\S 2$ Bridging provability and computability: The foundations
In the paper titled “Evidence-Based Interpretations of $PA$” that was presented to the Symposium on Computational Philosophy at the AISB/IACAP World Congress 2012-Alan Turing 2012, held from $2^{nd}$ to $6^{th}$ July 2012 at the University of Birmingham, UK (reproduced in this post) we have defined what it means for a number-theoretic function to be:
We have shown there that:
(i) The standard interpretation $\mathcal{I}_{PA(N,\ Standard)}$ of the first order Peano Arithmetic PA is finitarily sound if, and only if, Aristotle’s particularisation holds over $N$; and the latter is the case if, and only if, PA is $\omega$-consistent.
(ii) We can define a finitarily sound algorithmic interpretation $\mathcal{I}_{PA(N,\ Algorithmic)}$ of PA over the domain $N$ where, if $[A]$ is an atomic formula $[A(x_{1}, x_{2}, \ldots, x_{n})]$ of PA, then the sequence of natural numbers $(a_{1}, a_{2}, \ldots, a_{n})$ satisfies $[A]$ if, and only if $[A(a_{1}, a_{2}, \ldots, a_{n})]$ is algorithmically computable under $\mathcal{I}_{PA(N,\ Algorithmic)}$, but we do not presume that Aristotle’s particularisation is valid over $N$.
(iii) The axioms of PA are always true under the finitary interpretation $\mathcal{I}_{PA(N,\ Algorithmic)}$, and the rules of inference of PA preserve the properties of satisfaction/truth under $\mathcal{I}_{PA(N,\ Algorithmic)}$.
We concluded that:
Theorem 1: The interpretation $\mathcal{I}_{PA(N,\ Algorithmic)}$ of PA is finitarily sound.
Theorem 2: PA is consistent.
$\S 3$ Extending Buss’ Bounded Arithmetic
One of the more significant consequences of the Birmingham paper is that we can extend the iff bridge between the domain of provability and that of computability envisaged under Buss’ Bounded Arithmetic by showing that an arithmetical formula $[F]$ is PA-provable if, and only if, $[F]$ interprets as true under an algorithmic interpretation of PA.
$\S 4$ A Provability Theorem for PA
We first show that PA can have no non-standard model (for a distinctly different proof of this convention-challenging thesis see this post and this paper), since it is algorithmically’ complete in the sense that:
Theorem 3: (Provability Theorem for PA) A PA formula $[F(x)]$ is PA-provable if, and only if, $[F(x)]$ is algorithmically computable as always true in $N$.
Proof: We have by definition that $[(\forall x)F(x)]$ interprets as true under the interpretation $\mathcal{I}_{PA(N,\ Algorithmic)}$ if, and only if, $[F(x)]$ is algorithmically computable as always true in $N$.
Since $\mathcal{I}_{PA(N,\ Algorithmic)}$ is finitarily sound, it defines a finitary model of PA over $N$—say $\mathcal{M}_{PA(\beta)}$—such that:
If $[(\forall x)F(x)]$ is PA-provable, then $[F(x)]$ is algorithmically computable as always true in $N$;
If $[\neg(\forall x)F(x)]$ is PA-provable, then it is not the case that $[F(x)]$ is algorithmically computable as always true in $N$.
Now, we cannot have that both $[(\forall x)F(x)]$ and $[\neg(\forall x)F(x)]$ are PA-unprovable for some PA formula $[F(x)]$, as this would yield the contradiction:
(i) There is a finitary model—say $M1_{\beta}$—of PA+$[(\forall x)F(x)]$ in which $[F(x)]$ is algorithmically computable as always true in $N$.
(ii) There is a finitary model—say $M2_{\beta}$—of PA+$[\neg(\forall x)F(x)]$ in which it is not the case that $[F(x)]$ is algorithmically computable as always true in $N.$
The lemma follows. $\Box$
$\S 5$ The holy grail of arithmetic
We thus have that:
Corollary 1: PA is categorical finitarily.
Now we note that:
Lemma 2: If PA has a sound interpretation $\mathcal{I}_{PA(N,\ Sound)}$ over $N$, then there is a PA formula $[F]$ which is algorithmically verifiable as always true over $N$ under $\mathcal{I}_{PA(N,\ Sound)}$ even though $[F]$ is not PA-provable.
Proof In his seminal 1931 paper on formally undecidable arithmetical propositions, Kurt Gödel has shown how to construct an arithmetical formula with a single variable—say $[R(x)]$ [1]—such that $[R(x)]$ is not PA-provable [2], but $[R(n)]$ is instantiationally PA-provable for any given PA numeral $[n]$. Hence, for any given numeral $[n]$, the PA formula $xB \lceil [R(n)] \rceil$ must hold for some $x$. The lemma follows. $\Box$
By the argument in Theorem 3 it follows that:
Corollary 2: The PA formula $[\neg(\forall x)R(x)]$ defined in Lemma 2 is PA-provable.
Corollary 3: Under any sound interpretation of PA, Gödel’s $[R(x)]$ interprets as an algorithmically verifiable, but not algorithmically computable, tautology over $N$.
Proof Gödel has shown that $[R(x)]$ [3] interprets as an algorithmically verifiable tautology [4]. By Corollary 2 $[R(x)]$ is not algorithmically computable as always true in $N$. $\Box$
Corollary 4: PA is not $\omega$-consistent. [5]
Proof Gödel has shown that if PA is consistent, then $[R(n)]$ is PA-provable for any given PA numeral $[n]$ [6]. By Corollary 2 and the definition of $\omega$-consistency, if PA is consistent then it is not $\omega$-consistent. $\Box$
Corollary 5: The standard interpretation $\mathcal{I}_{PA(N,\ Standard)}$ of PA is not finitarily sound, and does not yield a finitary model of PA [7].
Proof If PA is consistent but not $\omega$-consistent, then Aristotle’s particularisation does not hold over $N$. Since the standard’, interpretation of PA appeals to Aristotle’s particularisation, the lemma follows. $\Box$
Since formal quantification is currently interpreted in classical logic [8] so as to admit Aristotle’s particularisation over $N$ as axiomatic [9], the above suggests that we may need to review number-theoretic arguments [10] that appeal unrestrictedly to classical Aristotlean logic.
$\S 6$ The Provability Theorem for PA and Bounded Arithmetic
In a 1997 paper [11], Samuel R. Buss considered Bounded Arithmetics obtained by:
(a) limiting the applicability of the Induction Axiom Schema in PA only to functions with quantifiers bounded by an unspecified natural number bound $b$;
(b) weakening’ the statement of the axiom with the aim of differentiating between effective computability over the sequence of natural numbers, and feasible polynomial-time’ computability over a bounded sequence of the natural numbers [12].
Presumably Buss’ intent—as expressed below—is to build an iff bridge between provability in a Bounded Arithmetic and Computability so that a $\Pi_{k}$ formula, say $[(\forall x)f(x)]$, is provable in the Bounded Arithmetic if, and only if, there is an algorithm that, for any given numeral $[n]$, decides the $\Delta_{(k/(k-1))}$ formula $[f(n)]$ as true’:
If $[(\forall x)(\exists y)f(x, y)]$ is provable, then there should be an algorithm to find $y$ as a function of $x$ [13].
Since we have proven such a Provability Theorem for PA in the previous section, the first question arises:
$\S 7$ Does the introduction of bounded quantifiers yield any computational advantage?
Now, one difference [14] between a Bounded Arithmetic and PA is that we can presume in the Bounded Arithmetic that, from a proof of $[(\exists y)f(n, y)]$, we may always conclude that there is some numeral $[m]$ such that $[f(n, m)]$ is provable in the arithmetic; however, this is not a finitarily sound conclusion in PA.
Reason: Since $[(\exists y)f(n, y)]$ is simply a shorthand for $[\neg (\forall y)\neg f(n, y)]$, such a presumption implies that Aristotle’s particularisation holds over the natural numbers under any finitarily sound interpretation of PA.
To see that (as Brouwer steadfastly held) this may not always be the case, interpret $[(\forall x)f(x)]$ as [15]:
There is an algorithm that decides $[f(n)]$ as true’ for any given numeral $[n]$.
In such case, if $[(\forall x)(\exists y)f(x, y)]$ is provable in PA, then we can only conclude that:
There is an algorithm that, for any given numeral $[n]$, decides that it is not the case that there is an algorithm that, for any given numeral $[m]$, decides $[\neg f(n, m)]$ as true’.
We cannot, however, conclude—as we can in a Bounded Arithmetic—that:
There is an algorithm that, for any given numeral $[n]$, decides that there is an algorithm that, for some numeral $[m]$, decides $[f(n, m)]$ as true’.
Reason: $[(\exists y)f(n, y)]$ may be a Halting-type formula for some numeral $[n]$.
This could be the case if $[(\forall x)(\exists y)f(x, y)]$ were PA-unprovable, but $[(\exists y)f(n, y)]$ PA-provable for any given numeral $[n]$.
Presumably it is the belief that any finitarily sound interpretation of PA requires Aristotle’s particularisation to hold in $N$, and the recognition that the latter does not admit linking provability to computability in PA, which has led to considering the effect of bounding quantification in PA.
However, as we have seen in the preceding sections, we are able to link provability to computability through the Provability Theorem for PA by recognising precisely that, to the contrary, any interpretation of PA which requires Aristotle’s particularisation to hold in $N$ cannot be finitarily sound!
The postulation of an unspecified bound in a Bounded Arithmetic in order to arrive at a provability-computability link thus appears dispensible.
The question then arises:
$\S 8$ Does weakening’ the PA Induction Axiom Schema yield any computational advantage?
Now, Buss considers a bounded arithmetic $S_{2}$ which is, essentially, PA with the following weakened’ Induction Axiom Schema, PIND [16]:
$[\{f(0)\ \&\ (\forall x)(f(\lfloor \frac{x}{2} \rfloor) \rightarrow f(x))\} \rightarrow (\forall x)f(x)]$
However, PIND can be expressed in first-order Peano Arithmetic PA as follows:
$[\{f(0)\ \&\ (\forall x)(f(x) \rightarrow (f(2*x)\ \&\ f(2*x+1)))\} \rightarrow (\forall x)f(x)]$.
Moreover, the above is a particular case of PIND($k$):
$[\{f(0)\ \&\ (\forall x)(f(x) \rightarrow (f(k*x)\ \&\ f(k*x+1)\ \&\ \ldots\ \&\ f(k*x+k-1)))\}$ $\rightarrow (\forall x)f(x)]$.
Now we have the PA theorem:
$[(\forall x)f(x) \rightarrow \{f(0)\ \&\ (\forall x)(f(x) \rightarrow f(x+1))\}]$
It follows that the following is also a PA theorem:
$[\{f(0)\ \&\ (\forall x)(f(x) \rightarrow f(x+1))\} \rightarrow$ $\{f(0)\ \&\ (\forall x)(f(x) \rightarrow (f(k*x)\ \&\ f(k*x+1)\ \&\ \ldots\ \&\ f(k*x+k-1)))\}]$
In other words, for any numeral $[k]$, PIND($k$) is equivalent in PA to the standard Induction Axiom of PA!
Thus, the Provability Theorem for PA suggests that all arguments and conclusions of a Bounded Arithmetic can be reflected in PA without any loss of generality.
References
Br93 Selmer Bringsjord 1993. The Narrational Case Against Church’s Thesis. Easter APA meetings, Atlanta.
Br08 L. E. J. Brouwer. 1908. The Unreliability of the Logical Principles. English translation in A. Heyting, Ed. L. E. J. Brouwer: Collected Works 1: Philosophy and Foundations of Mathematics. Amsterdam: North Holland / New York: American Elsevier (1975): pp.107-111.
Bu97 Samuel R. Buss. 1997. Bounded Arithmetic and Propositional Proof Complexity. In Logic of Computation. pp.67-122. Ed. H. Schwichtenberg. Springer-Verlag, Berlin.
CCS01 Cristian S. Calude, Elena Calude and Solomon Marcus. 2001. Passages of Proof. Workshop, Annual Conference of the Australasian Association of Philosophy (New Zealand Division), Auckland. Archived at: http://arxiv.org/pdf/math/0305213.pdf. Also in EATCS Bulletin, Number 84, October 2004, viii+258 pp.
Da82 Martin Davis. 1958. Computability and Unsolvability. 1982 ed. Dover Publications, Inc., New York.
EC89 Richard L. Epstein, Walter A. Carnielli. 1989. Computability: Computable Functions, Logic, and the Foundations of Mathematics. Wadsworth & Brooks, California.
Go31 Kurt Gödel. 1931. On formally undecidable propositions of Principia Mathematica and related systems I. Translated by Elliott Mendelson. In M. Davis (ed.). 1965. The Undecidable. Raven Press, New York.
HA28 David Hilbert & Wilhelm Ackermann. 1928. Principles of Mathematical Logic. Translation of the second edition of the Grundzüge Der Theoretischen Logik. 1928. Springer, Berlin. 1950. Chelsea Publishing Company, New York.
He04 Catherine Christer-Hennix. 2004. Some remarks on Finitistic Model Theory, Ultra-Intuitionism and the main problem of the Foundation of Mathematics. ILLC Seminar, 2nd April 2004, Amsterdam.
Hi25 David Hilbert. 1925. On the Infinite. Text of an address delivered in Münster on 4th June 1925 at a meeting of the Westphalian Mathematical Society. In Jean van Heijenoort. 1967.Ed. From Frege to Gödel: A source book in Mathematical Logic, 1878 – 1931. Harvard University Press, Cambridge, Massachusetts.
Ku62 Thomas S. Kuhn. 1962. The structure of Scientific Revolutions. 2nd Ed. 1970. University of Chicago Press, Chicago.
Me90 Elliott Mendelson. 1990. Second Thoughts About Church’s Thesis and Mathematical Proofs. In Journal of Philosophy 87.5.
Pa71 Rohit Parikh. 1971. Existence and Feasibility in Arithmetic. In The Journal of Symbolic Logic,>i> Vol.36, No. 3 (Sep., 1971), pp. 494-508.
Rg87 Hartley Rogers Jr. 1987. Theory of Recursive Functions and Effective Computability. MIT Press, Cambridge, Massachusetts.
Ro36 J. Barkley Rosser. 1936. Extensions of some Theorems of Gödel and Church. In M. Davis (ed.). 1965. The Undecidable. Raven Press, New York. Reprinted from The Journal of Symbolic Logic. Vol.1. pp.87-91.
Si08 Wilfrid Sieg. 2008. On Computability in Handbook of the Philosophy of Science. Philosophy of Mathematics. pp.525-621. Volume Editor: Andrew Irvine. General Editors: Dov M. Gabbay, Paul Thagard and John Woods. Elsevier BV. 2008.
WG03 Peter Wegner and Dina Goldin. 2003. Computation Beyond Turing Machines. Communications of the ACM, 46 (4) 2003.
Notes
Return to 1: Gödel refers to this formula only by its Gödel number $r$ (Go31, p.25(12)).
Return to 2: Gödel’s immediate aim in Go31 was to show that $[(\forall x)R(x)]$ is not P-provable; by Generalisation it follows, however, that $[R(x)]$ is also not P-provable.
Return to 3: Gödel refers to this formula only by its Gödel number $r$ (Go31, p.25, eqn.12).
Return to 4: Go31, p.26(2): “$(n)\neg(nB_{\kappa}(17Gen\ r))$ holds”.
Return to 5: This conclusion is contrary to accepted dogma. See, for instance, Davis’ remarks in Da82, p.129(iii) that:
“… there is no equivocation. Either an adequate arithmetical logic is $\omega$-inconsistent (in which case it is possible to prove false statements within it) or it has an unsolvable decision problem and is subject to the limitations of Gödel’s incompleteness theorem”.
Return to 7: I note that finitists of all hues—ranging from Brouwer Br08 to Alexander Yessenin-Volpin He04—have persistently questioned the finitary soundness of the standard’ interpretation $\mathcal{I}_{PA(N,\ Standard)}$.
Return to 8: See Hi25, p.382; HA28, p.48; Be59, pp.178 \& 218.
Return to 9: In the sense of being intuitively obvious. See, for instance, Da82, p.xxiv; Rg87, p.308 (1)-(4); EC89, p.174 (4); BBJ03, p.102.
Return to 10: For instance Rosser’s construction of an undecidable arithmetical proposition in PA (see Ro36)—which does not explicitly assume that PA is $\omega$-consistent—implicitly presumes that Aristotle’s particularisation holds over $N$.
Return to 16: Where $\lfloor \frac{x}{2} \rfloor$ denotes the largest natural number lower bound of the rational $\frac{x}{2}$. | {} |
# Domain
• Oct 10th 2009, 12:26 AM
Oasis1993
Domain
Find the natural domain of the function gf.
f(x)= √x g(x)= x-5
Thanks.
• Oct 10th 2009, 01:05 AM
mr fantastic
Quote:
Originally Posted by Oasis1993
Find the natural domain of the function gf.
f(x)= √x g(x)= x-5
Thanks.
$g(f(x)) = f(x) - 5 = \sqrt{x} - 5$.
The implied domain is clearly $x \geq 0$.
• Oct 10th 2009, 06:07 AM
HallsofIvy
But I would have interpreted "gf" as the product, $(x- 5)\sqrt{x}$ which also has natural domain $x\ge 0$. The composition the other way, $f\circ g(x)= \sqrt{x- 5}$, on the other hand, has natural domain $x\ge 5$. | {} |
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Poster
Why Lottery Ticket Wins? A Theoretical Perspective of Sample Complexity on Sparse Neural Networks
Shuai Zhang · Meng Wang · Sijia Liu · Pin-Yu Chen · Jinjun Xiong
Thu Dec 09 04:30 PM -- 06:00 PM (PST) @
The lottery ticket hypothesis (LTH) states that learning on a properly pruned network (the winning ticket) has improved test accuracy over the original unpruned network. Although LTH has been justified empirically in a broad range of deep neural network (DNN) involved applications like computer vision and natural language processing, the theoretical validation of the improved generalization of a winning ticket remains elusive. To the best of our knowledge, our work, for the first time, characterizes the performance of training a pruned neural network by analyzing the geometric structure of the objective function and the sample complexity to achieve zero generalization error. We show that the convex region near a desirable model with guaranteed generalization enlarges as the neural network model is pruned, indicating the structural importance of a winning ticket. Moreover, as the algorithm for training a pruned neural network is specified as an (accelerated) stochastic gradient descent algorithm, we theoretically show that the number of samples required for achieving zero generalization error is proportional to the number of the non-pruned weights in the hidden layer. With a fixed number of samples, training a pruned neural network enjoys a faster convergence rate to the desired model than training the original unpruned one, providing a formal justification of the improved generalization of the winning ticket. Our theoretical results are acquired from learning a pruned neural network of one hidden layer, while experimental results are further provided to justify the implications in pruning multi-layer neural networks. | {} |
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Positive solutions to $p$-Kirchhoff-type elliptic equation with general subcritical growth Bull. Korean Math. Soc. 2017 Vol. 54, No. 3, 1023-1036 https://doi.org/10.4134/BKMS.b160467Published online May 31, 2017 Huixing Zhang and Ran Zhang China University of Mining and Technology, China University of Mining and Technology Abstract : In this paper, we study the existence of positive solutions to the $p$-Kirchhoff elliptic equation involving general subcritical growth $$(a+\lambda\int_{\mathbb{R}^N}|\nabla u|^p dx+\lambda b\int_{\mathbb{R}^N}|u|^p dx)(-\Delta_p u+b|u|^{p-2}u)=h(u),\ \mbox{in}\ \mathbb{R}^N,$$ where $a,\ b>0$, ${\lambda}$ is a parameter and the nonlinearity $h(s)$ satisfies the weaker conditions than the ones in our known literature. We also consider the asymptotics of solutions with respect to the parameter $\lambda$. Keywords : $p$-Kirchhoff-type equation, subcritical growth, asymptotics MSC numbers : 35J20, 35J15, 35J60 Full-Text : | {} |
# Example: SL(2,C)
## Example: SL(2,C)
### Definition
The group $SL(2,\mathbf{C})$ is defined to be the group of 2-by-2 complex matrices with determinant 1 ("S" in the name stands for "special", which means determinant 1). This is a topologically closed subgroup of $GL(2,\mathbf{C})$ because it's cut out by a continuous equation $\det(M)=1$ .
### Lie algebra
The Lie algebra of $SL(2,\mathbf{C})$ is $\mathfrak{sl}(2,\mathbf{C})=\left\{\begin{pmatrix}a&b\\ c&d\end{pmatrix}\ :\ \det\exp t\begin{pmatrix}a&b\\ c&d\end{pmatrix}=1\ \forall t\in\mathbf{R}\right\}$ To first order in $t$ , $\det\exp t\begin{pmatrix}a&b\\ c&d\end{pmatrix}=\det\begin{pmatrix}1+ta&tb\\ tc&1+td\end{pmatrix}+\mathcal{O}(t^{2})=(1+ta)(1+td)-t^{2}bc+\mathcal{O}(t^{2}% )=1+t(a+d)+\mathcal{O}(t^{2}).$ Differentiating with respect to $t$ at $t=0$ gives: $\left.\frac{d}{dt}\right|_{t=0}\det\exp t\begin{pmatrix}a&b\\ c&d\end{pmatrix}=a+d=\left.\frac{d}{dt}\right|_{t=0}1=0.$ Therefore, if $M=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ is in $\mathfrak{sl}(2,\mathbf{C})$ , then the trace of $M$ $\mathrm{Tr}(M)=a+d$ is zero. It turns out that this is "if and only if" (we'll see this later) so that $\mathfrak{sl}(2,\mathbf{C})=\{M\ :\ \mathrm{Tr}(M)=0\}$ The general element of $\mathfrak{sl}(2,\mathbf{C})$ is therefore $\begin{pmatrix}a&b\\ c&-a\end{pmatrix}$ and we see that the Lie algebra is a 3-dimensional (complex) vector space with coordinates $a,b,c$ . By contrast, it's trickier to write down a formula for the general element of $SL(2,\mathbf{C})$ .
A basis for this vector space is given by setting one coordinate at a time equal to 1 (and the others equal to zero): $H=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix},\quad X=\begin{pmatrix}0&1\\ 0&0\end{pmatrix},\quad Y=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}$ That is, any element of $\mathfrak{sl}(2,\mathbf{C})$ can be written in a unique way as $aH+bX+cY$ for some $a,b,c\in\mathbf{C}$ .
### Checking that sl(2,C) is a Lie algebra
Since $\mathfrak{sl}(2,\mathbf{C})$ is a Lie algebra, it's supposed to be preserved under taking commutator bracket. Let's check: $[H,X]=HX-XH=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}\begin{pmatrix}0&1\\ 0&0\end{pmatrix}-\begin{pmatrix}0&1\\ 0&0\end{pmatrix}\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}=\begin{pmatrix}0&2\\ 0&0\end{pmatrix}=2X.$ $[H,Y]=HY-YH=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}\begin{pmatrix}0&0\\ 1&0\end{pmatrix}-\begin{pmatrix}0&0\\ 1&0\end{pmatrix}\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}=\begin{pmatrix}0&0\\ -2&0\end{pmatrix}=-2Y.$ $[X,Y]=XY-YX=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}\begin{pmatrix}0&0\\ 1&0\end{pmatrix}-\begin{pmatrix}0&0\\ 1&0\end{pmatrix}\begin{pmatrix}0&1\\ 0&0\end{pmatrix}=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}=H.$ You can see that the bracket of two things in $\mathfrak{sl}(2,\mathbf{C})$ still lives in $\mathfrak{sl}(2,\mathbf{C})$ . All the other bracket computations follow from antisymmetry. For example: $[X,X]=[Y,Y]=[H,H]=0$ and $[Y,X]=-[X,Y]=-H,\quad[X,H]=-[H,X]=-2X,\mbox{ etc.}$ This example is of such fundamental importance that I'll restate the brackets we just found: $[H,X]=2X,\quad[H,Y]=-2Y,\quad[X,Y]=H.$ We will use these relations all the time later in the course.
### Trace(M) = 0 implies det(exp(M)) = 1
There's one loose end to tie up, namely showing that if $\mathrm{Tr}(M)=0$ then $M\in\mathfrak{sl}(2,\mathbf{C})$ . We have a basis ($H,X,Y$ ) for the space of trace-zero matrices. Because we know that $\mathfrak{sl}(2,\mathbf{C})$ is a vector space, it suffices to show that $H,X,Y$ are in $\mathfrak{sl}(2,\mathbf{C})$ because then any linear combination of them will live in $\mathfrak{sl}(2,\mathbf{C})$ .
Lemma:
$H\in\mathfrak{sl}(2,\mathbf{C})$ .
Proof:
We need to check that $\det\exp(tH)=1$ for all $t$ . We have $\exp(tH)=\exp\begin{pmatrix}t&0\\ 0&-t\end{pmatrix}=\begin{pmatrix}1+t+t^{2}/2+\cdots&0\\ 0&1-t+t^{2}/2-\cdots\end{pmatrix}=\begin{pmatrix}e^{t}&0\\ 0&e^{-t}\end{pmatrix}$ which has determinant 1.
## Pre-class exercise
Exercise:
Show that $X$ and $Y$ are in $\mathfrak{sl}(2,\mathbf{C})$ . | {} |
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Degree diameter problem
In graph theory, the degree diameter problem is the problem of finding the largest possible graph G (in terms of the size of its vertex set V) of diameter k such that the largest degree of any of the vertices in G is at most d. The size of G is bounded above by the Moore bound; for 1 < k and 2 < d only the Petersen graph, the Hoffman-Singleton graph, and maybe a graph of diameter k = 2 and degree d = 57 attain the Moore bound. In general the largest degree-diameter graphs are much smaller in size than the Moore bound.
Formula
Let $n_{d,k}$ be the maximum possible of vertices for a graph with degree at most d and diameter k then $n_{d,k}\leq M_{d,k}$, where $M_{d,k}$ is the Moore bound:
$M_{d,k}=\begin{cases}1+d\frac{(d-1)^k-1}{d-2}&\text{ if }d>2\\2d+1&\text{ if }d=2\end{cases}$
This bound is attained for very few graphs, thus the study moves to how close there exist graphs to the Moore bound. For asymptotic behaviour note that $M_{d,k}=d^k+O(d^{k-1})$.
Define the parameter $\mu_k=\liminf_{d\to\infty}\frac{n_{d,k}}{d^k}$. It is conjectured that $\mu_k=1$ for all k. It is known that $\mu_1=\mu_2=\mu_3=\mu_5=1$ and that $\mu_4\geq 1/4$. For the general case it is known that $\mu_k\geq 1.6^k$. Thus, although is conjectured that $\mu_k=1$ is still open if it is actually exponential. | {} |
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