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http://mathhelpforum.com/math-software/104996-matlab-translation.html
|
# Math Help - MATLAB translation
1. ## MATLAB translation
I was given the MATLAB function
function y=dig(x,D)
ex = D - ceil(log10(abs(x)));
y = round(x.*(10.^ex)).*(10.^(-ex));
I need to know exactly what this does. I know it takes in a vector x and returns D significant decimals, but I don't know how it gets the sig digits...I do not have access to MATLAB.
Thanks
2. Hi!
I'm sorry but I can't give you the answer - because I don't understand it.
Originally Posted by CarmineCortez
I was given the MATLAB function
function y=dig(x,D)
ex = D - ceil(log10(abs(x)));
y = round(x.*(10.^ex)).*(10.^(-ex));
I need to know exactly what this does. I know it takes in a vector x and returns D significant decimals,
That sounds good and makes sense.
Originally Posted by CarmineCortez
but I don't know how it gets the sig digits...I do not have access to MATLAB.
So theoretically you are able to understand the algorithm for that problem, I don't..., but you have problems with the syntax? Is that it?
function y=dig(x,D)
This is like you said. x is a vector, D is the significant decimals and y is also a vector.
ex = D - ceil(log10(abs(x)));
ex is a vector, too. D is an integer like 10. And x is a vector [5,3,-1]
abs(x) leads to the vector [|5|,|3|,|-1| ] = [5,3,1]
log10(x) -> you get the vector [log(5),log(3),log(1)]
So what does ceil do? For this syntax you should use google or go to the official website of matlab, because there is a good documentation.
You find this:
MATLAB
In my opinion it explains it very good what ceil does, because there is a nice example.
It is D = 10 for example and you solve ceil(...).
For example you have 10 - [3,5,7] then matlab returns the vector [10-3, 10-5, 10-7]
I don't know what happens in the next line if the vector has a negative number in it.
y = round(x.*(10.^ex)).*(10.^(-ex));
Use the Matlab documentation I showed you to find out how round() works.
The critical thing here is the point ' . ' You need it if you calculate * or / or ^ or... then you get this: You have two vectors [x_1,...,x_n] and ex_1,...ex_n
round(x.*(10.^ex)).*(10.^(-ex));
= round([x_1*10^ex_1)*10^(-ex_1),...,[x_n*10^ex_n)*10^(-ex_n) ])
But this could be wrong. I don't have matlab either. I just think this is what Matlab does. If not, I'm sorry, but no one hasn't responded yet, so I thought it might help.
Kind regards
Rapha
3. Originally Posted by CarmineCortez
I was given the MATLAB function
function y=dig(x,D)
ex = D - ceil(log10(abs(x)));
y = round(x.*(10.^ex)).*(10.^(-ex));
I need to know exactly what this does. I know it takes in a vector x and returns D significant decimals, but I don't know how it gets the sig digits...I do not have access to MATLAB.
Thanks
There are a number of issues here.
The first is the vectorisation, the above is a vector version of the scalar code:
Code:
function y=dig(x,D)
ex = D - ceil(log10(abs(x)));
y = round(x*(10^ex))*(10^(-ex));
where the function ceil(u) returns the smallest integer greater than u.
^ denotes raising to a power
round(u) rounds u to the nearest integer.
Lets take x=1234.56789, and D=2
log10(abs(x))=3.09151..
ceil(log10(abs(x))=4
ex=2-4=-2
y=round(x*10^-2)*10^2=round(12.3456789)*10^2=1200
It reduces x to 2-significant digits
If you want to try the code there is a free Matlab clone here
CB
|
2016-07-23 15:52:51
|
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|
http://lilypond.org/doc/v2.16/Documentation/snippets/world-music
|
# World music
## Arabic improvisation
For improvisations or taqasim which are temporarily free, the time signature can be omitted and \cadenzaOn can be used. Adjusting the accidental style might be required, since the absence of bar lines will cause the accidental to be marked only once. Here is an example of what could be the start of a hijaz improvisation:
\include "arabic.ly"
\relative sol' {
\key re \kurd
\accidentalStyle "forget"
}
## Makam example
Makam is a type of melody from Turkey using 1/9th-tone microtonal alterations. Consult the initialization file ‘ly/makam.ly’ for details of pitch names and alterations.
% Initialize makam settings
\include "makam.ly"
\relative c' {
\set Staff.keySignature = #((6 . ,(- KOMA)) (3 . ,BAKIYE))
c4 cc db fk
gbm4 gfc gfb efk
fk4 db cc c
}
## Printing text from right to left
It is possible to print text from right to left in a markup object, as demonstrated here.
{
b1^\markup {
\line { i n g i r u m i m u s n o c t e }
}
f'_\markup {
\override #'(text-direction . -1)
\line { i n g i r u m i m u s n o c t e }
}
}
`
|
2014-04-17 04:10:18
|
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|
https://courses.lumenlearning.com/boundless-algebra/chapter/using-matrices-to-solve-systems-of-equations/
|
## Matrix Equations
Matrices can be used to compactly write and work with systems of multiple linear equations.
### Learning Objectives
Identify how matrices can represent a system of equations
### Key Takeaways
#### Key Points
• If $A$ is an $m \times n$ matrix, and $x$ designates a column vector (i.e. $n \times 1$ matrix) of $n$ variables $x_1, x_2,…, x_n$, and $b$ is an $m \times 1$ column vector, then the matrix equation is: $Ax=b$.
#### Key Terms
• matrix: A rectangular arrangement of numbers or terms having various uses such as transforming coordinates in geometry, solving systems of linear equations in linear algebra and representing graphs in graph theory.
Matrices can be used to compactly write and work with systems of equations. As we have learned in previous sections, matrices can be manipulated in any way that a normal equation can be. This is very helpful when we start to work with systems of equations. It is helpful to understand how to organize matrices to solve these systems.
### Writing a System of Equations with Matrices
It is possible to solve this system using the elimination or substitution method, but it is also possible to do it with a matrix operation. Before we start setting up the matrices, it is important to do the following:
• Make sure that all of the equations are written in a similar manner, meaning the variables need to all be in the same order.
• Make sure that one side of the equation is only variables and their coefficients, and the other side is just constants.
Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: $X$ is the matrix representing the variables of the system, and $B$ is the matrix representing the constants. Using matrix multiplication, we may define a system of equations with the same number of equations as variables as:
$\displaystyle A\cdot X=B$
To solve a system of linear equations using an inverse matrix, let $A$ be the coefficient matrix, let $X$ be the variable matrix, and let $B$ be the constant matrix.
Given the system:
\displaystyle \begin{align} x+8y&=7 \\ 2x-8y&=-3 \end{align}
The coefficient matrix is:
$A=\begin{bmatrix} 1 & 8\\ 2 & -8 \end{bmatrix}$
The variable matrix is:
$\displaystyle X=\begin{bmatrix} x\\ y \end{bmatrix}$
The constant matrix is:
$\displaystyle B=\begin{bmatrix} 7\\ -3 \end{bmatrix}$
Thus, to solve a system $AX=B$, for $X$, multiply both sides by the inverse of $A$ and we shall obtain the solution:
$\displaystyle X=(A^{-1})B$
Provided the inverse $\left( A^{-1} \right)$ exists, this formula will solve the system.
If the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions.
## Matrices and Row Operations
Two matrices are row equivalent if one can be changed to the other by a sequence of elementary row operations.
### Learning Objectives
Explain how to use row operations and why they produce equivalent matrices
### Key Takeaways
#### Key Points
• An elementary row operation is any one of the following moves: row switching (swap two rows of a matrix ), row multiplication (multiply a row of a matrix by a non-zero constant ), or row addition (add to one row of a matrix to some multiple of another row).
• If the rows of the matrix represent a system of linear equations, then the row space consists of all linear equations that can be deduced algebraically from those in the system.
#### Key Terms
• row space: The set of all possible linear combinations of its row vectors.
• row equivalent: In linear algebra, when one matrix can be changed to the other by a sequence of elementary row operations.
### Elementary Row Operations (ERO)
In linear algebra, two matrices are row equivalent if one can be changed to the other by a sequence of elementary row operations. Alternatively, two $m \times n$matrices are row equivalent if and only if they have the same row space. The row space of a matrix is the set of all possible linear combinations of its row vectors. If the rows of the matrix represent a system of linear equations, then the row space consists of all linear equations that can be deduced algebraically from those in the system. Two matrices of the same size are row equivalent if and only if the corresponding homogeneous systems have the same set of solutions, or equivalently the matrices have the same null space. Because elementary row operations are reversible, row equivalence is an equivalence relation. It is commonly denoted by a tilde (~).
An elementary row operation is any one of the following three moves:
1. Row switching (swap): Swap two rows of a matrix.
2. Row multiplication (scale): Multiply a row of a matrix by a nonzero constant.
3. Row addition (pivot): Add to one row of a matrix some multiple of another row.
### Produce Equivalent Matrices Using Elementary Row Operations
Since the matrix is essentially the coefficients and constants of a linear system, the three row operations preserve the matrix. For example, swapping two rows simply means switching their position within the matrix. Also, when solving a system of linear equations by the elimination method, row multiplication would be the same as multiplying the whole equation by a number to obtain additive inverses so that a variable cancels. Finally, row addition is also the same as the elimination method, when one chooses to add or subtract the like terms of the equations to obtain the variable. Therefore, row operations preserve the matrix and can be used as an alternative method to solve a system of equations.
### Example 1: Show that these two matrices are row equivalent:
$\displaystyle A=\begin{pmatrix} 1 & -1 & 0 \\ 2 & 1 & 1 \end{pmatrix}\quad B=\begin{pmatrix} 3 & 0 & 1 \\ 0 & 3 & 1 \end{pmatrix}$
Start with $A$, add the second row to the first:
$\displaystyle A=\begin{pmatrix} 3 & 0 & 1 \\ 2 & 1 & 1 \end{pmatrix}$
Then, multiply the second row by 3 and then subtract the first row from the second:
$\displaystyle A=\begin{pmatrix} 3 & 0 & 1 \\ 3 & 3 & 2 \end{pmatrix}$
Finally, subtract the first row from the second:
$\displaystyle A=\begin{pmatrix} 3 & 0 & 1 \\ 0 & 3 & 1 \end{pmatrix}$
You can see that $A=B$, which we achieved through a series of elementary row operations.
### Row Reduction: Solving a System of Linear Equations
In row reduction, the linear system:
$\displaystyle x+3y-2z=5 \\ 3x+5y+6z=7 \\ 2x+4y+3z=8$
Is represented as an augmented matrix:
$\displaystyle A=\begin{pmatrix} 1 & 3 & -2 & 5 \\ 3 & 5 & 6 & 7 \\ 2 & 4 & 3 & 8 \end{pmatrix}$
This matrix is then modified using elementary row operations until it reaches reduced row echelon form.
Because these operations are reversible, the augmented matrix produced always represents a linear system that is equivalent to the original.
There are several specific algorithms to row-reduce an augmented matrix, the simplest of which are Gaussian elimination and Gauss-Jordan elimination. This computation can be done by hand (using the three types of ERO) or on the calculator under the matrix function ‘rref’ (reduced row echelon form).
The final matrix is in reduced row echelon form, and represents the system $x=-15$, $y=8$$z=2$.
$\displaystyle A=\begin{pmatrix} 1 & 0 & 0 & -15 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 2 \end{pmatrix}$
## Simplifying Matrices With Row Operations
Using elementary operations, Gaussian elimination reduces matrices to row echelon form.
### Learning Objectives
Use elementary row operations to put a matrix in simplified form
### Key Takeaways
#### Key Points
• Since elementary row operations preserve the row space of the matrix, the row space of the row echelon form is the same as that of the original matrix.
• There are three types of elementary row operations: swap the positions of two rows, multiply a row by a nonzero scalar, and add to one row a scalar multiple of another.
• In practice, one does not usually deal with the systems in terms of equations but instead makes use of the augmented matrix (which is also suitable for computer manipulations).
#### Key Terms
• augmented matrix: A matrix obtained by appending the columns of two given matrices, usually for the purpose of performing the same elementary row operations on each of the given matrices.
By means of a finite sequence of elementary row operations, called Gaussian elimination, any matrix can be transformed to a row echelon form. This transformation is necessary for solving a system of linear equations.
Before getting into more detail, there are a couple of key terms that should be mentioned:
• Augmented matrix: an augmented matrix is a matrix obtained by appending the columns of two given matrices, usually for the purpose of performing the same elementary row operations on each of the given matrices.
• Upper triangle form: A square matrix is called upper triangular if all the entries below the main diagonal are zero. A triangular matrix is one that is either lower triangular or upper triangular. A matrix that is both upper and lower triangular is a diagonal matrix.
• Elementary row operations: Swap rows, add rows or multiply rows.
### Gaussian Elimination
1. Write the augmented matrix for the linear equations.
2. Use elementary row operations on the augmented matrix $[A|b]$ to transform $A$ to upper triangle form. If a zero is on the diagonal, switch the rows until a nonzero is in its place.
3. Use back substitution to find the solution.
### Example 1: Solve the system by Gaussian Elimination:
$\displaystyle 2x+y-z=8\\ -3x-y+2z=-11\\ -2x+y+2z=-3$
Write the augmented matrix:
$\left[\begin{array}{rrr|r} 2 & 1 & -1 & 8 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3 \end{array} \right]$
Use elementary row operations to reduce the matrix to reduced row echelon form:
$\left[\begin{array}{rrr|r} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -1 \end{array} \right]$
Using elementary row operations to obtain reduced row echelon form (‘rref’ in the calculator) the solution to the system is revealed in the last column: $x=2, y=3, z=-1$.
|
2019-04-18 21:09:24
|
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|
https://www.gradesaver.com/textbooks/science/chemistry/general-chemistry/chapter-4-early-quantum-theory-problems-page-132/20
|
## General Chemistry (4th Edition)
λ = 8 x 10$^{-8}$ m 8 x 10$^{-8}$ = hc/E = (6.626 x 10$^{-34}$) (3 x 10$^{8}$) / E E = 2.5 x 10$^{-18}$ J = 2.5 aJ No X-Rays don’t have sufficient energy to Argon
|
2019-03-20 19:45:07
|
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|
https://spoofax.dev/release/note/1.2.0/
|
# Spoofax 1.2.0 (13-08-2014)¶
We're happy to announce the release of Spoofax 1.2! This document describes the changes in Spoofax 1.2 since Spoofax 1.1.
## Changes¶
### Editor interface¶
Several aspects of the editor interface for Spoofax languages have been improved.
Each language now has its own set of menus. These menus replace the Transform menu that was shared among all Spoofax-based languages.
• The new menus dynamically pop up in the Eclipse menus toolbar, based on the current active editor.
Contributors: Oskar van Rest
#### Outline¶
Editor outlines are now specified in Stratego instead of ESV, to allow for full customization.
• A library with reusable outline strategies is provided to allow you to quickly realize an outline.
• It is now possible to have icons in your outline.
• It is now possible to also base an outline on the current text selection instead of the complete text.
Contributors: Oskar van Rest
#### Properties view¶
A property view has been added that shows properties for the selected AST node.
• By default, the new properties view integrates with NaBL and presents (NaBL) properties associated with the selected text in the editor.
• The properties view can be customized to show different kinds of properties, either to aid language development or to provide language users with additional information about their programs.
Contributors: Daco Harkes, Oskar van Rest
### Meta-languages¶
We have created a new version of SDF, improved NaBL and developed a DSL for describing type systems: TS.
#### SDF3¶
SDF3 is the next iteration of SDF, replacing SDF2. The most important new features are:
• Productive productions can be used in context-free syntax sections, improving readability and consistency with main-stream grammar notation.
• Constructors are now formally integrated to the language. A productive production introduces the constructor directly after the left hand side non-terminal.
• By using constructors, productions can now be uniquely identified. Therefore, it is no longer necessary to repeat the entire production in the priorities section, but use its Sort.Constructor shorthand.
• Template productions are the major change from SDF2. They can be used to define what the concrete syntax of the language should look like. Syntactic completion and pretty-printer rules are automatically generated from templates.
Documentation
Contributors: Eduardo Amorim, Guido Wachsmuth
#### NaBL¶
NaBL has received many bug fixes and several new features. New features include:
• Filter clauses can be used to constrain valid resolution targets based on properties such as types.
• Disambiguation clauses can be used to disambiguate resolutions based on relations between properties, for example type hierarchies.
• It is now possible to specify non-transitive scopes, in which resolution ignores lexical parent scopes.
• Where clauses can include constraints and type calculations in TS syntax.
• We added new scope calculation constructs, which can be used to navigate the scope hierarchy. For example, it is possible to calculate the surrounding class of the current variable scope.
For examples of name binding rules, see the Java front project
Contributors: Guido Wachsmuth, Gabriël Konat
#### TS¶
TS is a new meta-language for the specification of type analysis that is complementary to NaBL. Type rules in TS define constraints on abstract syntax tree nodes and may compute a type or other property. In addition, type rules can define subtyping predicates (relations on types) and type functions.
For examples of type system rules, see the Java front project
Contributors: Eelco Visser, Guido Wachsmuth, Gabriël Konat
### Command-line integration¶
Programs of your language can now be parsed, analyzed, and transformed from the command-line using Sunshine (in contrast with an Eclipse). Sunshine can also be used as a Java API to develop new language tooling.
### Finer-grained incrementality¶
Incrementality in the previous version of Spoofax was based on files. Any file that changed, and any dependent files would be reparsed and reanalysed completely. In the new version of Spoofax, there is more fine-grained dependency tracking which allows more fine-grained incrementality. If a file changes, that file is reparsed, but only affected computations are recomputed, and other files are never reparsed. Name and type computations which are described in NaBL and TS are incrementally executed. Incrementality is powered by a task engine, described in our paper.
Contributors: Gabriël Konat, Guido Wachsmuth
### Modelware¶
Spoofax Modelware is a new Spoofax component that provides integration with the Eclipse Modeling Framework (EMF) and the Graphical Modeling Framework (GMF) to allow for real-time synchronized textual and graphical editors and/or views. It also allows you to use other EMF-based tooling in combination with Spoofax.
Contributors: Oskar van Rest
### Documentation¶
We have moved most of our documentation to the doc repository on GitHub. We're still in the process of moving over other documentation and writing more documentation.
There are also two new tutorials available:
• Questionaire language tutorial: learn to create a questionaire language. This tutorial was given in a hands-on session at the Code Generation conference in 2014.
• Compiler Construction lab assignments: a more in-depth tutorial. These are the assignments from our Compiler Construction lab where we teach students to create MiniJava inside Spoofax.
Contributors: Guido Wachsmuth and others
### Other¶
To reduce maintenance effort, we have dropped support for Eclipse 3.7 (Indigo) and lower. We now support Eclipse 4.2 (Juno), 4.3 (Kepler), and 4.4 (Luna). We recommend you to download and install Eclipse 4.4 (Luna) for Java Developers.
We have also dropped support for Java 5 and 6. Java 7 and 8 are supported, we recommend you to download and install Java 8. Note that on OSX, Java 6 is installed by default which is not enough to run Spoofax, install Java 8 from the previous link.
All source code has been moved to our organization on GitHub. Feel free to fork our code and send pull requests for patches or improvements!
More changes, additions, and bug fixes can be found in the roadmap on our issue tracker: http://yellowgrass.org/tag/Spoofax/1.2
|
2023-01-28 06:13:20
|
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|
https://www.physicsforums.com/threads/limit-question.333263/
|
# Limit question
1. Aug 28, 2009
### wimma
1. The problem statement, all variables and given/known data
lim as x tends to zero of (x^(-2) - (cosecx)^2)
2. Relevant equations
L'Hospital's rule
3. The attempt at a solution
I keep getting things in the denominator that won't go away no matter how much I differentiate. I'm missing something here...
2. Aug 29, 2009
### Staff: Mentor
Show us what you did.
3. Aug 29, 2009
### VietDao29
You mean:
$$\lim_{x \rightarrow 0} \left( \frac{1}{x ^ 2} - \frac{1}{\sin ^ 2 x} \right)$$, right?
If you really want to do it, by applying L'Hospital rule, then the first step is to group them into 1 fraction, like this:
$$\lim_{x \rightarrow 0} \left( \frac{1}{x ^ 2} - \frac{1}{\sin ^ 2 x} \right) = \lim_{x \rightarrow 0} \frac{\sin ^ 2 (x) - x ^ 2}{x ^ 2 \sin ^ 2 x}$$
Now, just apply L'Hospital Rule, if you want us to check your work, you can always post it here, and we'll be more than willing to help you out. :)
Btw, this is actually the hard way, you have to apply the rule four-folds to get to the final result.
Have you covered Talor's expansion yet? This could make the problem way simpler.
4. Aug 29, 2009
### wimma
yeah we've learnt taylor polynomials. i guess i could make a taylor polynomial of appropriate degree for each one, though the question says to use l'hospitals.
i went
$$\lim_{x \rightarrow 0} \left( \frac{1}{x ^ 2} - \frac{1}{\sin ^ 2 x} \right) = \lim_{x \rightarrow 0} \left( \frac{\sin ^ 2 (x) - x ^ 2}{x ^ 2 \sin ^ 2 x} \right) = \lim{x \rightarrow 0} \left( \frac{sin x cos x - x}{x ^ 2 sin x cos x + x sin ^2 x} \right) = \lim{x \rightarrow 0} \left( \frac{-sin^2 x + cos ^2 x -1}{x ^ 2 cos (2x) + 3 x sin (2x) + sin ^2 x} \right)$$
then i gave up
Last edited: Aug 29, 2009
5. Aug 29, 2009
### snipez90
The denominator of the last expression on the right doesn't look correct. You should have a lot more terms.
6. Aug 29, 2009
### VietDao29
In the first post I told you to use it 4 times. You only used it twice. Since it's still in the form 0/0, just continue until you cannot go any further.
7. Aug 29, 2009
### wimma
yeah cos i gave up before i finished the denominator
fix'd
8. Aug 29, 2009
### snipez90
Why? If you don't want to use the product rule extended to products of more than two factors, then use the fact that sin(2x) = 2(sin x)(cos x) or sin(x)cos(x) = sin(2x) / 2 and just keep adding like terms together after differentiating.
9. Aug 29, 2009
### wimma
so i do it four times and get
$$\lim{x \rithgtarrow 0} \left( \frac{-4 cos (2x)}{-20 sin (2x) + 16 cos (2x) - 4 x ^ 2 cos (2x)} \right)$$
10. Aug 29, 2009
### VietDao29
You can continue using it for 2 more times, or by looking at it a little bit closely, your final expression is:
$$\lim_{x \rightarrow 0} \left( \frac{-\sin^2 x + \cos ^2 x -1}{x ^ 2 \cos (2x) + {\color{red}2} x \sin (2x) + \sin ^2 x} \right) = \lim_{x \rightarrow 0} \left( \frac{-2\sin^2 x}{x ^ 2 \cos (2x) + {\color{red}2} x \sin (2x) + \sin ^2 x} \right)$$
Now, what will you get if you divide both numerator, and denominator by x2?
--------------
Oh, and btw, you have miscalculated it, the factor 3 in your step should read 2 instead. I have corrected it for you. :)
Last edited: Aug 29, 2009
11. Aug 29, 2009
### VietDao29
Ohhhh.. :(
Since you have differentiated it incorrectly in the denominator, in your second step. I'm sorry to inform you that this is wrong. :( You have to re-calculate everything from your second use of L'Hospital's rule. :(
Or, you can try my hint in the post above. :)
12. Aug 29, 2009
### snipez90
I don't think this is correct but as others have noted, it's hard to help when you don't show the intermediate steps.
13. Aug 29, 2009
### wimma
cheers =] i got the result
i'm looking at trying to do the same problem with taylor polynomials... i get the taylor polynomial for $$sin ^2 (x)$$ to be$$\sum{\frac{-(-1)^k 2^{2k-1}x^{2k}}{(2k)!}} ...$$
inverting this to get the one for $$cosec ^2 (x)$$ looks a bit mean.
Last edited: Aug 29, 2009
14. Aug 29, 2009
### VietDao29
Ummm.. How did you get that? It doesn't look quite correct to me. Could you should us the step, so that i can help you point out where it went wrong?
No, you don't have to find the expansion for csc2(x). Just truncate it (the expansion for sin2(x)) wisely, then use that truncated expression to plug in the limit. Much simpler than to find its reciprocal, right? :)
$$\lim_{x \rightarrow 0} \left( \frac{1}{x ^ 2} - \frac{1}{\sin ^ 2 x} \right) = \lim_{x \rightarrow 0} \frac{\sin ^ 2 (x) - x ^ 2}{x ^ 2 \sin ^ 2 x}$$
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2018-03-18 06:20:51
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6028063297271729, "perplexity": 1119.4574140064358}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257645538.8/warc/CC-MAIN-20180318052202-20180318072202-00461.warc.gz"}
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http://www.ethanideas.url.tw/the-misconception-of-potential-energy/
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# 位能的迷思
## 無法解釋的困境與解決之道
$$\Delta E=W+Q$$
$$dE=W+Q\tag{1}$$
$$\because Q=0\tag{2}$$
$$\because W=\left(-F_{\text{Ag}}\right)dr_A+\left(+F_{\text{Bg}}\right)dr_B\tag{3}$$
$$\because F_{\text{Ag}}=F_{\text{Bg}}=\frac{Gm_\text{A}m_\text{B}}{r_\text{AB}^2}\tag{4}$$
\begin{aligned}\therefore W&=\frac{Gm_\text{A}m_\text{B}}{r_\text{AB}^2}\left(dr_B-dr_A\right)\\[2ex]&=\frac{Gm_\text{A}m_\text{B}}{r_\text{BA}^2}\end{aligned}\tag{5}
\begin{aligned}\therefore dE&=\frac{Gm_\text{A}m_\text{B}}{r_\text{BA}^2}\left(dr_\text{BA}\right)\\[2ex]&=d\left(-\frac{Gm_\text{A}m_\text{B}}{r_\text{BA}}\right)\end{aligned}\tag{6}
$$\therefore E\equiv-\frac{Gm_\text{A}m_\text{B}}{r_\text{BA}}+C\tag{7}$$
## 場源為物體與地球的重力場能量形式
$$\therefore E\equiv\frac{1}{2}\sum_{i=1}^N\sum_{j=1,j\neq i}^N\left(-\frac{Gm_i m_j}{r_{ij}}\right)+C\tag{8}$$
$$E\approx\sum_{i=1}^N\left(-\frac{GMm_i}{r_{i,M}}\right)+C\text{ ; }r_{\text{i,M}}=R_e+h_i\tag{9}$$
$$0\approx\sum_{i=1}^N\left(-\frac{GMm_i}{R_e}\right)+C\tag{10}$$
$$C\approx\sum_{i=1}^N\frac{GMm_i}{R_e}\tag{11}$$
$$E\approx-\sum_{i=1}^N\left(\frac{GMm_i}{R_e+h_i}-\frac{GMm_i}{R_e}\right)\tag{12}$$
$$\therefore E\approx\sum_{i=1}^N\frac{GMm_i}{R_e^2}h_i\tag{13}$$
$$\therefore E\approx\sum_{i=1}^N m_i gh_i\text{ ; }g\equiv\frac{GM}{R_e^2}\tag{14}$$
[1] 只有在慣性座標系才能使用牛頓定律推得單質點系統之能量形式為動能，關於這點，之後我會於另文論述。簡單來說，非慣性座標系是無法討論任何基於「牛頓力學」的物理概念的。
[2] Marc Lange (2002). An introduction to the philosophy of physics : locality, fields, energy and mass. Wiley-Blackwell; 1 edition.
[3] 這種解釋與電路學中的電磁能傳遞方式是相同的，之後我會在發表一篇電路傳遞能量的迷思。簡言之，電磁能並非透過電流傳遞，電流反而是消耗電磁能的因素，而是透過佔據空間的電磁場。另外，引述 Robert P. Baumann 於〈Physics that textbook writers usually get wrong: I. Work〉的論述：
If we attribute potential energy to the ball at the top, we cannot properly calculate the work done on it, and hence energy transferred to, the ball in its descent. … It is, of course, convenient to think of objects in gravitational (or electric) fields as having potential energy. We can do that by redefining the system as ball plus field. Then the field can do no work on our system, as the ball goes up or down.
[4] Robert P. Bauman (1992). Physics that textbook writers usually get wrong: II. Heat and energy. THE PHYSICS TEACHER, 30, 353–356.
[5] 注意，兩團黑霧並非指兩個重力場，由A、B場源形成的重力場只有一個。此外，黑點密度表示重力場強度，而非重力場位置。
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2018-05-24 14:04:04
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https://nigerianscholars.com/tutorials/energy-and-chemical-reactions/changes-in-colour/
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Chemistry » Energy and Chemical Reactions » Measuring Rates Of Reaction
# Changes in Colour
## Changes in Colour
In some reactions there is a change in colour which tells us that the reaction is occuring. The faster the colour change the faster the reaction rate.
For example, when ethanoic acid (acetic acid) is titrated with sodium hydroxide, an indicator such as phenolphthalein is added. The solution is clear in an acidic solution and changes to pink when the reaction is complete. If the concentration of the base were increased, the colour change would happen faster (after a smaller volume of base was added), showing that a higher concentration of base increased the reaction rate.
$$\text{CH}_{3}\text{COOH}(\text{aq}) + \text{NaOH}(\text{aq})$$ $$\to$$ $$\text{Na}^{+}(\text{aq}) + \text{CH}_{3}\text{COO}^{-}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})$$
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2021-12-03 00:22:46
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https://naijavarcity.com.ng/past-questions/physics/page/physics01/
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## Physics 2017 Past Questions | JAMB
Study the following Physics past questions and answers for JAMBWAEC NECO and Post-JAMB. Get prepared with official past questions and answers for upcoming examinations.
1. Which is the incorrect formula for a body accelerating uniformly?
• A. a = v2 − $$\frac{u^2}{2}$$
• B. v2 = u2 + 2as
• C. s = ut + $$\frac{1}{2}$$ at2
• D. v2 − u2= 2as
2. Which of the following is not an electromagnetic radiation?
• A. x – ray
• C. sunlight
• D. sound waves
3. Calculate the electric field intensity between two plates of potential difference 6.5V when separated by a distance of 35cm.
• A. 18.57NC$$^{-1}$$
• B. 53.06N C$$^{-1}$$
• C. 2.28NC$$^{-1}$$
• D. 0.80NC$$^{-1}$$
4. Why do soldiers march disorderly while crossing a bridge?
• A. To prevent resonance on the bridge
• B. To set the bridge into resonance
• C. To make the b ridge collapse
• D. To spread their weight evenly on the bridge
5.
Calculate the upthrust on an object of volume 50cm3 which is immersed in liquid of density 103kgm-3 [g = 10ms-2]
• A. 0.8N
• B. 2.5N
• C. 0.5N
• D. 1.0N
6. Under what conditions are cathode rays produced in a discharge tube?
• A. High pressure and low voltage
• B. High pressure and high voltage
• C. Low pressure and low voltage
• D. Low pressure and high volatge
7. Under which of the following conditions is workdone
• A. A man supports a heavy food above his head with his hands
• B. A boy climbs onto a table
• C. A man pushes unto a table
• D. A woman holds a pot of water
8. Calculate the specific latent heat of vaporization of steam of 1.13 x 106J of heat energy is required to convert 15kg of steam to water.
• A. 7.53 x 105 Jkg$$^{-1}$$
• B. 7.53 x 10-2Jkg$$^{-1}$$
• C. $$7.53 \times 10^{4}$$Jkg$$^{-1}$$
• D. 7.53 x 10-3Jkg$$^{-1}$$
9. A cell of internal resistance 2 π supplies current to a 6 π resistor. The efficiency of the cell is
• A. 12.0%
• B. 25.0%
• C. 33.3%
• D. 75.0%
10. Which of the following statements is NOT correct?
• A. Molecules of a liquid are stationary
• B. Brownian motion is an evidence of particle nature of matter
• C. Matter is made up of molecules
• D. The molecules of matter are in constant motion
Subscribe
Notify of
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2022-01-21 08:41:45
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https://www.physicsforums.com/threads/how-does-a-gearbox-work.657776/
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How does a gearbox work?
1. Dec 9, 2012
beyondlight
1. The problem statement, all variables and given/known data
Lets suppose i have two gears where the ratio between the rotational speed of the motor and the wheel is:
Motor/wheel 2:1 first gear
Motor/wheel 1:1 second gear
2. Relevant equations
P=Tω
3. The attempt at a solution
Now i guess that in the first gear following holds
$$P_m=ω_m*T_m=ω_w*T_w=P_w=2*ω_w*\frac{T_w}{2}$$
That means that the mechanical energy on the rotational axis is the same on both sides of the gearbox?
Is that correct?
2. Dec 9, 2012
CWatters
Yes. If you ignore losses in the gearbox then a gearbox doesn't change the power. It just trades torque and angular velocity.
You can also think of it in terms of the law of conservation of energy. A gearbox can't store or create energy so what goes in must come out (and at the same rate = power).
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2017-08-22 02:49:27
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https://www.physicsforums.com/threads/integral-with-an-inverse-function-limit.904738/
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# Integral with an inverse function limit
• A
Hello,
I have tried the integral below with Mathematica and it gives me the following solution:
##\frac{d}{dc}\int_{z^{-1}(c)}^{1} z(x)dx = -\frac{c}{z'(z^{-1}(c))}##
I am not quite sure where it gets it from...I think it can be separated and with differentiation the first part will be zero:
##\frac{d}{dc}\int_{z^{-1}(c)}^{1}z(x)dx=\frac{d}{dc}\int_{0}^{1} z(x)dx-\frac{d}{dc}\int_{0}^{z^{-1}(c)} z(x)dx=-\frac{d}{dc}\int_{0}^{z^{-1}(c)} z(x)dx##
By the formula ##\frac{d}{dc}z^{-1}(c) = \frac{1}{z'(z^{-1}(c))}##, so it definitely has to play a role here, but how? and where does c come from? Do we need to somehow use chain rule here?
If the upper limit was just c it would be easy to take it with a Leibniz rule:
##-\frac{d}{dc}\int_{0}^{c} z(x)dx=z(c)##
However, it seems that when one of the limits is an inverse function, there is something different at work which I am not aware of...
Let me struggle through this in my own way.
I find it useful to give familiar-looking names to things wherever possible.
So let's put
z-1(c) = y
and ∫z(x)dx = Z,
and I'll call the integral-derivative we have to evaluate F.
So we have to evaluate F = (d/dc)( Z(1) -Z(y) ),
which immediately becomes
-(d/dc)Z(y) = -(dZ(y)/dy). (dy/dc)
= z(y).(dy/dc)
But z(y) = z(z-1(c)) = c
So we have
F = c.(dy/dc) . . . . . . .(A)
Now z-1(c) = y by definition; so
c = z(y), and, differentiating,
dc/dy =dz/dy,
or if we invert,
dy/dc = 1/(dz/dy) = 1/z'(y)
= 1/z'( z-1(c) )
Substitute this for dy/dc in (A) and I think we're done.
FactChecker
Thank you, it makes sense now!
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2022-01-23 08:20:17
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https://www.physicsforums.com/threads/center-of-mass.184744/
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# Center of mass
1. Sep 15, 2007
### Edwardo_Elric
1. The problem statement, all variables and given/known data
A model airplane has a momentum given by
$$[(0.25 kg . m/s^3)t^2 - (0.75 kg . m/s^2)t]\hat{i}$$
a.) What are the x- , y- , and z- components of the net force on the airplane?
b.) At what time t is the x-component of the net force on the airplane equal to zero?
2. Relevant equations
$$\sum F = \frac{d\vec{p}} {dt}$$
3. The attempt at a solution
a.) i used the derivative to get :
Fx = 2(0.25m/s^3)t - (0.75m/s^2) so the x- component of net force
Fy = 0
Fz = 0
since they are the j hat and k hat where it is not given
b.) when Fx = 0;
0 = 2(0.25m/s^3)t - (0.75m/s^2)
2(0.25m/s^3)t = (0.75m/s^2)
t = 1.5s
2. Sep 15, 2007
### HallsofIvy
Staff Emeritus
Looks good to me!
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2017-04-29 11:41:30
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4988178610801697, "perplexity": 2970.5108914113175}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917123491.68/warc/CC-MAIN-20170423031203-00089-ip-10-145-167-34.ec2.internal.warc.gz"}
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http://clay6.com/qa/35375/according-to-central-pollution-central-board-cpcb-which-size-of-particulate
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Browse Questions
# According to central pollution central board (CPCB) which size of particulate matter are responsible for causing the greatest harm to human health?
$\begin{array}{1 1}(a)\;2.5\mu m\;or\;less\\(b)\;3.5\mu m\;or\;less\\(c)\;4.5\mu m\;or\;less\\(d)\;5.5\mu m\;or\;less\end{array}$
$2.5\mu m\;or\;less$
Hence (a) is the correct answer.
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2017-03-24 12:13:43
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.27194085717201233, "perplexity": 4249.408875784526}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218187945.85/warc/CC-MAIN-20170322212947-00607-ip-10-233-31-227.ec2.internal.warc.gz"}
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http://www.maa.org/press/maa-reviews/in-the-tradition-of-ahlfors-bers-vi
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# In the Tradition of Ahlfors-Bers, VI
###### Ursula Hamenstädt, et al., editors
Publisher:
American Mathematical Society
Publication Date:
2013
Number of Pages:
189
Format:
Paperback
Series:
Contemporary Mathematics 590
Price:
76.00
ISBN:
9780821874271
Category:
Proceedings
We do not plan to review this book.
• S. Dowdall, M. Duchin, and H. Masur -- Spheres in the curve complex
• C. Florentino and S. Lawton -- Character varieties and moduli of quiver representations
• E. Fujikawa -- Periodicity of asymptotic Teichmüller modular transformation
• Z. Ibragimov -- Hyperbolization of locally compact non-complete metric spaces
• A. Isopoussu, K. Peltonen, and J. T. Tyson -- Quasiregular maps and the conductivity equation in the Heisenberg group
• C. Judge -- Pushing points on pillowcase covers
• T. Koberda -- Entropy of automorphisms, homology and the intrinsic polynomial structure of nilpotent groups
• G. F. Lawler -- Continuity of radial and two-sided radial SLE at the terminal point
• M. Matsuzaki and Y. Yabuki -- No proper conjugation for quasiconvex cocompact groups of Gromov hyperbolic spaces
• H. Miyachi -- A differential formula for extremal length
• O. Ostapyuk -- On convergence to the Denjoy-wolff point in the parabolic case
• S. Pal -- Construction of a closed hyperbolic surface of arbitrarily small eigenvalue of prescribed serial number
• J. Paupert -- A simple method to compute volumes of even-dimensional Coxeter polyhedra
• P. Poggi-Corradini -- Some remarks about analytic functions defined on an annulus
• H. Shiga -- On injectivity radius in configuration space and in moduli space
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2017-04-28 12:43:14
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https://support.bioconductor.org/p/91931/
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Question: Preprocessing of Human Gene 2.0 ST microarrays with oligo R package and annotation options
4
2.7 years ago by
svlachavas740
Greece/Athens/National Hellenic Research Foundation
svlachavas740 wrote:
Dear Community,
I currently analyzing in R, a small number (6 samples-2 conditions-3 biological reps of each condition) of CEL files regarding Affymetrix Human Gene 2.0 ST arrays (for the first time this type of gene chip arrays). A relevant subset of my code is the following:
library(oligo) library(affycoretools) library(hugene20sttranscriptcluster.db)
library(limma)
librarypd.hugene.2.0.st)
setwd(mydir)
pdat <- read.table("pdat.project.txt",header=TRUE,stringsAsFactors = FALSE) # phenotype info
celfiles = list.celfiles()
affy.cels <- read.celfiles(celfiles)
identical(colnames(affy.cels),rownames(pdat)) # need to be identical for incorporate phenotype info
pd <- AnnotatedDataFrame(data= pdat) phenoData(affy.cels) <- pd celfiles.rma <- rma(affy.cels, target="probeset")
Thus, my main questions are the following:
1) For the rma function, which is the most valid/appropriate choise of target argment for gene ST arrays ? "probeset" or "core" ?
2) For removing the control probesets, i can use the function getMainProbes
3) To annotate in later steps of limma (i.e after topTable) my probesets/transcripts into gene symbols, i should first:
annotation(eset.rma) <- "hugene20sttranscriptcluster.db"
& then query the above db with functions select, etc ?
Thank you in advance !!
modified 2.7 years ago by James W. MacDonald51k • written 2.7 years ago by svlachavas740
Answer: Preprocessing of Human Gene 2.0 ST microarrays with oligo R package and annotati
2
2.7 years ago by
United States
James W. MacDonald51k wrote:
1.) For the vast majority of users, the 'core' argument is the way to go. The Gene ST arrays are intended to measure transcript abundances, and the ability to summarize at the probeset level is really just due to the fact that they are based on the Exon ST platform.
2.) Yes. Why do you have doubts?
3.) If you are using affycoretools, it's easier to do
library(hugene20sttranscriptcluster.db)
eset.rma <- annotateEset(eset.rma, hugene20sttranscriptcluster.db)
And then your topTable will automatically contain annotation data.
Dear James,
thank you for your confirmation !! actually, i tried (accidentally) the argument target=probeset with rma and then with getMainProbes, i ended with ~1700 features, which concerned me-and perhaps explains your comment about probeset level (but of course it is not the case when i use the "core" option). Thus, if i understood well, with the annotateEset function the returned annotation data are gene symbols, for matched transcripts, correct ?
1
The results are the Entrez Gene ID, symbol, and gene name, based on the Affy annotations for that array. We just take what Affy says each probeset measures, and then convert to a useful format without doing anything to check that what they say is correct in any sense.
Also, do note that there is a hugene20stprobeset.db package that annotates the probeset IDs, and that is what you would use to annotate if you summarize at the probeset level.
ADD REPLYlink written 2.7 years ago by James W. MacDonald51k
Than you again for your explanation-i will follow your advice and use the core argument-remove the control "transcripts" & annotation of the expression eset--perhaps the very small number of probesets when i use first "probeset" in rma and then getMainProbes, probably has to do with the design of the array.
Hi James:
I get error: could not find function "annotateEset"
2
Any time you see an error saying 'could not find function', it means you haven't loaded the package that contains that function yet. Or, it may mean that you are using an old version of R/Bioconductor where the function was not yet part of the package. You don't give the results of sessionInfo, so I can't say for sure, so try A) loading affycoretools first or B) using the current version of R/Bioconductor.
ADD REPLYlink written 18 months ago by James W. MacDonald51k
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2019-10-22 14:43:36
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https://cs.stackexchange.com/questions/98804/eke-a-better-compression-ratio-out-of-huffman-coding-by-mixing-1-grams-and-2-gra?noredirect=1
|
# Eke a better compression ratio out of Huffman coding by mixing 1-grams and 2-grams
Huffman coding is already doing a great job at compressing ascii bytes for data with a distribution like the following
byte freq freq%
---------------------
0 317116 26.1
, 151471 12.5
1 112952 9.3
F 60810 5
@ 60810 5
2 53642 4.4
8 49595 4.1
6 46548 3.8
. 45339 3.7
5 44343 3.7
3 40005 3.3
4 38873 3.2
7 38716 3.2
9 34194 2.8
\ 30405 2.5
Q 30128 2.5
; 29129 2.4
S 16118 1.3
B 14287 1.2
C 277 0
Now when I look at the 2-grams, I get the following table:
bytes freq freq%
----------------------
00 148269 12.2
,1 51993 4.3
0, 46416 3.8
1, 42348 3.5
0/ 30405 2.5
F@ 30405 2.5
FF 30405 2.5
@Q 30128 2.5
;F 29129 2.4
,0 28202 2.3
0; 27470 2.3
10 24346 2
/0 19337 1.6
....
Here's the top of the 3-gram list:
byte freq freq%
--------------------
000 70254 5.8
00, 41218 3.4
,1, 31651 2.6
FF@ 30405 2.5
00\ 30254 2.5
F@Q 30128 2.5
;FF 29129 2.4
...
So it seems it would make sense to give at least the more frequent of the 2-grams and 3-grams their own Huffman code, and perhaps encode the less frequent 2-grams/3-grams as separate 1-grams.
Is there already any (preferably practically useful) research on this, how to determine the optimum mix of 1-grams, 2-grams, 3-grams (, ...)?
• Are you asking about arithmetic coding? – Peter Taylor Oct 19 '18 at 11:18
• @PeterTaylor From what I understand, arithmetic coding doesn't work with n-grams, but looks at the whole bit string as one item to encode. Although it seems like a more optimal form of compression, I think I cannot use that in my case, as I want to create (something like) a Huffman tree just once (having sampled a large enough data set), and re-use that same Huffman tree to encode and decode many times. – Evgeniy Berezovsky Oct 21 '18 at 23:05
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2020-09-26 19:17:51
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https://solvedlib.com/the-weighted-average-cost-of-capital-represents,283556
|
# The weighted average cost of capital represents the return required by the average investor in the...
###### Question:
The weighted average cost of capital represents the return required by the average investor in the firm. True or False?
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from the vector potential A(x,y,2) = (2x 23)1 - (32 y) z Find the associated magnetic field B(r, 0)?...
##### Exercles 14.29Considur Ihe reaction: 48142 3F 2 (9) RCIFs (g) ~0,08 M/sCIF_]/At -ValueUnitsSubmitRequcst AnsrorPart €Fino Ihu rutethe reacticn; Expruya Your unawar signliicant flguret and Include the appropreto unliz:ValueUnitsSubmiAcq JastAnaentProvida FcodbackRate
Exercles 14.29 Considur Ihe reaction: 48142 3F 2 (9) RCIFs (g) ~0,08 M/s CIF_]/At - Value Units Submit Requcst Ansror Part € Fino Ihu rute the reacticn; Expruya Your unawar signliicant flguret and Include the appropreto unliz: Value Units Submi Acq JastAnaent Provida Fcodback Rate...
##### Question 24 of 50 2 PointsYou are working On research involving competition between plants: Which of the [ollowing resol do yOu not need t0 measure?Light SpaceWaterD. FoodReset Selection
Question 24 of 50 2 Points You are working On research involving competition between plants: Which of the [ollowing resol do yOu not need t0 measure? Light Space Water D. Food Reset Selection...
##### Which of these ligands can participate in linkage isomerism: (a) $\mathrm{SCN}^{-} ;$ (b) $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ (thiosulfate); (c) HS $^{-} ?$ Explain with Lewis structures.
Which of these ligands can participate in linkage isomerism: (a) $\mathrm{SCN}^{-} ;$ (b) $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ (thiosulfate); (c) HS $^{-} ?$ Explain with Lewis structures....
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2022-05-17 15:04:31
|
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|
https://discourse.pymc.io/t/wrong-posterior-in-multidimensional-gaussianrandomwalk/11141
|
# Wrong posterior in multidimensional GaussianRandomWalk
Hello, I read several posts and tried different approaches before writing this post, but I have been stuck for a moment now and I cannot make a simple time series work when the variables have more than one dimension.
Below is an example of a simple HMM with a latent variable Z that evolves over time with Gaussian noise, and another variable X (observable) with respective emission noise.
T = 50 # number of time steps
F = 2 # number of features (dimension of each state/observation)
sd_z = 1 # transition noise (single noise for all features)
sd_x = 1 # emission noise (single noise for all features)
# Data generation
np.random.seed(0)
x = np.zeros((T, F))
z = np.zeros((T, F))
for t in range(T):
if t == 0:
z[0] = norm(loc=0, scale=1).rvs(size=F)
else:
z[t] = norm(loc=z[t - 1], scale=sd_z).rvs()
x[t] = norm(loc=z[t], scale=sd_x).rvs()
# Model definition and posterior inference
coords = {"time": np.arange(T), "features": np.arange(F)}
hmm = pm.Model(coords=coords)
with hmm:
x_obs = pm.MutableData("x_obs", x, dims=("time", "features"))
# Standard deviation of the transition and emission distributions
i_sd_z = pm.HalfNormal(name="i_sd_z", sigma=1)
i_sd_x = pm.HalfNormal(name="i_sd_x", sigma=1)
# Transition distribution of latent variable Z
i_z = pm.GaussianRandomWalk(name="i_z", init_dist=pm.Normal.dist(mu=0, sigma=1, shape=(1, F)),
mu=0, sigma=i_sd_z, dims=("time", "features"))
# Emission distribution of the observed variable X
i_x = pm.Normal(name="i_x", mu=i_z, sigma=i_sd_x, observed=x_obs)
idata = pm.sample(1000, init="adapt_diag", tune=1000, chains=2, random_seed=0)
# Plotting results
az.plot_trace(idata, var_names=["i_sd_z", "i_sd_x"])
plt.show()
m1 = idata.posterior["i_z"].sel(features=0).mean(dim=["chain", "draw"])
sd1 = idata.posterior["i_z"].sel(features=0).std(dim=["chain", "draw"])
plt.figure(figsize=(15, 8))
plt.plot(range(T), z[:, 0], label="Real", color="tab:blue", marker="o")
plt.plot(range(T), m1, label="Inferred", color="tab:orange", marker="o")
plt.fill_between(range(T), m1 - sd1, m1 + sd1, color="tab:orange", alpha=0.4)
plt.legend()
plt.show()
See from the plots below that the inferred values of the standard deviations are not correct as well as the samples of the latent variable Z. The problem persists even if I set F to 1. Although, if I remove the “features” dimension, everything works well so it seems the problem is with the indexing inside the GaussianRandomWalk distribution.
Python version: 3.9
PyMC version: 5.0.1
Hello!
I think the problem is that the your priors are too tight, and their influence dominates the data in the posterior.
Here is the same graph as that which you showed, except with draws from the prior:
As you can see it strictly excludes the data. In the model like this, the priors are going to play a much more important role in posterior than in, say, a large-N cross-sectional linear model.
Here is draws from the prior of the same model, with sigma=5 on both standard deviation priors, and sigma=10 for the standard deviation of the init_dist distribution:
After adjusting the priors, I get the following posterior for i_z[0]:
Hi @jessegrabowski, thank you for your answer. That might be another problem I did not foresee, but I believe there’s still a problem with the indexing. For instance, if I remove “features” from coords and dims, and leave just the “time” dimension, I get the following:
If I instead just set F to 1, which should be essentially the same, I get degenerated posterior samples for Z:
Yes, I think you are right. The prior variance should grow at rate \sqrt{t}, but you can see from the prior plot I posted above that the variance is constant. As you suggest, when I remove the features dimension, the variances grows as expected. I wonder if the wrong index is being used to take the cumulative sum?
When I do the GRW “by hand”, I get the right result, both in the prior and the posterior:
coords = {"time": np.arange(T), "features": np.arange(F)}
hmm = pm.Model(coords=coords)
with hmm:
x_obs = pm.MutableData("x_obs", x, dims=("time", 'features'))
# Standard deviation of the transition and emission distributions
i_sd_z = pm.HalfNormal(name="i_sd_z", sigma=1)
i_sd_x = pm.HalfNormal(name="i_sd_x", sigma=1)
# Transition distribution of latent variable Z
i_z_0 = pm.Normal('i_z_0', sigma=1, dims=['features'])
i_z_t= pm.Normal('i_z_t', sigma=i_sd_z, dims=['time', 'features'])
i_z = pm.Deterministic('i_z', pm.math.concatenate([i_z_0[None], i_z_t], axis=0).cumsum(axis=0)[1:], dims=['time', 'features'])
# Emission distribution of the observed variable X
i_x = pm.Normal(name="i_x", mu=i_z, sigma=i_sd_x, observed=x_obs)
idata = pm.sample_prior_predictive()
My guess is that the prior issue I pointed about above is totally irrelevant, and that there is something else going on in the random walk wrapper. @ricardoV94 ?
1 Like
The GaussianRandomWalk requires the time dimension to be at the rightmost position. That’s the axis over which steps evolve
2 Likes
Thank you, @ricardoV94. Indeed, I swapped the “features” and “time” dimensions and now I get the expected result. It would be very helpful to include that information in the documentation of the GaussianRandomWalk (and MV alternative). I could not find any mention of it in the documentation and I am afraid others might face the same issue in the future.
Totally agree. Would you like to open a pull request for that? This applies to all timeseries
2 Likes
Hi @jessegrabowski, I replicated your solution to perform GRW by hand and indeed I get the correct posterior for Z, but weird inferences with a bunch of divergences for the standard deviation parameters (see image below). Do you have any idea why this is the case?
@ricardoV94’s answer solved the GaussianRandomWalk. This is what I get when I put the time in the last dimension.
However, I need to implement a custom RW that depends on other variables not only the previous one in the same chain, so I will need to construct an RW by hand as you did in your example.
Thank you!
@ricardoV94, will do.
You could try a lower higher target_accept, or adjusting the priors. These are the usual superficial solutions to divergences. Deeper solutions require thinking about why the model might not be a good fit for the data and making structural adjustments.
The nonsense with x0 can be improved to remove the concatenation and slicing, which might help. Note that the x_0 term in x_0 + (x_0 + x_1) + (x_0 + x_1 + x_2) can just be written as broadcast addition into all the accumulated terms, so all the concatenation stuff can just be reduced to: i_z_0 + i_z_t.cumsum(axis=0)
1 Like
@jessegrabowski, got it. It makes sense. Thank you for your assistance.
Should have written higher target accept, not lower. Sorry.
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2023-01-27 17:58:05
|
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https://socratic.org/questions/how-do-you-identify-the-slope-for-3y-x-1-2
|
# How do you identify the slope for 3y= x -1/2?
Apr 7, 2015
$3 y = x - \frac{1}{2}$
The general linear equation is $y = m x + c$.
$m = s l o p e$
$c$=y-intercept
To find a slope of the line from the liner equation, you have to make sure that all values are in order of the general equation of $y = m x + c$
In $3 y = x - \frac{1}{2}$, all values are in the correct position already. However, we have to simplify this linear equation further to ensure that our linear equation is $y = m x + c$
$3 y$ of $3 y = x - \frac{1}{2}$ has to become $y$ of $y = m x + c$
We can do that by dividing $3 y = x - \frac{1}{2}$ by $3$.
$\frac{3 y}{3} = \frac{x - \frac{1}{2}}{3}$
$y = \frac{1}{3} x - \frac{1}{6}$ as we compare this to the general equation of $y = m x + c$, $m$ which is the slope is multiplied together with $x$, hence, the slope for $3 y = x - \frac{1}{2}$ or $y = \frac{1}{3} x - \frac{1}{6}$ is $\frac{1}{3}$
|
2019-10-23 10:21:23
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|
https://cran.rstudio.org/web/packages/TidyMultiqc/vignettes/TidyMultiqc.html
|
## Test Data
In this example we’ll be using the WGS example report used in the MultiQC documentation. You can find view the report in your browser here, and find the associated multiqc_data.json file here. Feel free to download this file and use it to follow along with this vignette.
In the rest of the vignette we will use the multiqc_data_path variable to indicate the path to this multiqc_data.json file. Feel free to set this variable to the path to this file on your system.
## Setup
First, install the package using:
install.packages("TidyMultiqc")
Then if you want, you can load the package:
library(TidyMultiqc)
However, for the sake of this tutorial we will explicitly use namespaced functions, so we won’t be using library a great deal.
## Basic Usage
The main entry point to the TidyMultiqc package is the function load_multiqc. A basic invocation of this function looks like this:
(note the arrow for scrolling through columns)
df = TidyMultiqc::load_multiqc(multiqc_data_path)
df
We’ve now generated a tibble (a kind of data frame), whose rows are samples in the QC report, and whose columns are QC data and metadata about these samples.
By default this function only returns the “general” statistics, which are the ones in the “General Statistics” table at the top of the MultiQC report. In TidyMultiqc, these statistics are all prefixed by general. We can also extract the “raw” statistics, which includes some fields normally hidden from the report. These statistics will have the prefix raw.<toolname>. where <toolname> is the QC tool used to calculate it.
TidyMultiqc::load_multiqc(multiqc_data_path, sections = 'raw')
Often you won’t care about fields like raw.qualimap_bamqc_genome_results.bam_file, the path to the original BAM file, but ‘raw’ at least provides this option.
You can also combine both general and raw sections by passing in a longer vector:
df_both = TidyMultiqc::load_multiqc(multiqc_data_path, sections = c('raw', 'general'))
ncol(df_both)
#> [1] 388
That’s a lot of columns!
## Uses
This section will briefly talk about some downstream use-cases for this package.
### Plotting
One use for this data frame is creating QC plots. For example, to visualise the duplication rate per sample:
library(magrittr)
df %>%
ggplot2::geom_col()
Of course, this is basically just replicating a plot already in the MultiQC report, but now we can customise it how we like!
### Hypothesis Testing
With all this data, we might also want to test a specific hypothesis! For example, we might want to test the hypothesis that the mean GC content is the same as the mean GC content in the human genome (41%). If we assume that GC content is normally distributed, we can do the following:
t.test(df$general.percent_gc, mu=41) #> #> One Sample t-test #> #> data: df$general.percent_gc
#> t = -1, df = 5, p-value = 0.3632
#> alternative hypothesis: true mean is not equal to 41
#> 95 percent confidence interval:
#> 40.40490 41.26176
#> sample estimates:
#> mean of x
#> 40.83333
It seems that we cannot reject this hypothesis, so these may well be human samples!
It may be the case that your samples have important metadata that you want in your data frame. For example, it is common for the sample names or file names to be composed of a number of metadata fields, and indeed the report we are working with has IDs such as P4107_1003, which is composed of two identifiers.
To include this metadata in our output, we need to provide the find_metadata argument to load_multiqc, which is a function that is called for each sample, and which returns a named vector of metadata fields for that sample. It also gets passed the entire parsed MultiQC JSON report, so the function can traverse the structure as it wants to extract metadata.
### From the File Name
Here is an example that parses the input file names to annotate additional metadata. Notice that the first argument our function is passed is a string which is the sample identifier for a sample, and how in this example we ignore the parsed argument. Also notice that the names we give to our return value (“batch” and “sample”) are prefixed by “metadata” to become the final names in the data frame.
TidyMultiqc::load_multiqc(
multiqc_data_path,
# Split the sample ID to obtain some metadata
segments <- stringr::str_split(sample, "_")[[1]]
c(
batch = segments[[1]],
sample = segments[[2]]
)
}
)
### From the File Path
We can extend this approach, but this time actually look up the file paths within the report_data_sources section of the MultiQC report:
TidyMultiqc::load_multiqc(
multiqc_data_path,
# This gives us the path to the fastqc output file
filepath = parsed$report_data_sources$FastQC$all_sections[[sample]] # Split into path segments path_segments = stringr::str_split(filepath, "/")[[1]] # The filename is the last path segment filename = dplyr::last(path_segments) # Split the filename using dots and underscores name_segments = stringr::str_split(filename, "[_\\.]")[[1]] # Arbitrarily assign names for the outputs name_segments %>% purrr::set_names(LETTERS[1:length(name_segments)]) } ) Of course in a real application we would choose specific names for each field. ### From MultiQC Configuration Finally, we might want to include metadata that doesn’t relate to the sample at all. For example, MultiQC has a number of report fields prefixed by config_ that we might want to store: TidyMultiqc::load_multiqc( multiqc_data_path, find_metadata = function(sample, parsed) { parsed[c( "config_creation_date", "config_version" )] } ) ## Extracting Plot Data ### Motivation It is occasionally useful to extract QC data from the MultiQC plots. For example, let’s say we want to calculate the median quality score of every base in each sample. Unfortunately, MultiQC provides no numerical summary statistic for the read quality, it only has mapping quality and pass/fails for the per-base sequence quality: df_both %>% dplyr::select(dplyr::contains('quality')) However, our MultiQC report does have plots that contain this information. In particular, let’s look at the “Per Sequence Quality Scores” plot. ### Listing Plots Firstly, we need the ID of the plot we want. This isn’t necessarily obvious from just looking at the report, so we can use a utility function here: TidyMultiqc::list_plots(multiqc_data_path) Now, we know we want the “Per Sequence Quality Scores” plot, and by looking at the data frame above we can tell that the corresponding ID is fastqc_per_sequence_quality_scores_plot. ### Loading Plots Now that we have the plot ID, we can load the plot data. First, we need to tell TidyMultiqc to load include some plots by using sections = "plot" (you can load other sections at the same time, as explained above). Also, we need to pass the plot ID from the previous step into the plots argument: df = TidyMultiqc::load_multiqc( multiqc_data_path, sections = 'plot', plots = "fastqc_per_sequence_quality_scores_plot" ) df We now have the plot data, but it’s not in a very usable form! This is because each sample has an entire data frame of plot data. At this point if you’re comfortable using dplyr and tidyr to deal with nested data frames, you probably know what to do. Otherwise, read on. ### Converting Plot Data Recall that we are after the median quality score of each sample. First, we should look at the plot data for a single sample to know what we’re dealing with: df$plot.fastqc_per_sequence_quality_scores_plot[[1]]
So each data frame is a set of x, y pairs. As it’s a histogram plot, we know that the x value is the quality score, and y is the number of times that score has been counted.
### Unnesting
One possible way to process this nested data is to use tidyr:
df %>%
tidyr::unnest(cols = plot.fastqc_per_sequence_quality_scores_plot)
As you can see, if we unnest in this way, we now have multiple rows for the same sample, which is a bit confusing (and not Tidy). However, if we use group_by and then summarise, this can be a useful way to calculate summary statistics. For example, if we want the total number of reads, we could do the following:
df %>%
tidyr::unnest(cols = plot.fastqc_per_sequence_quality_scores_plot) %>%
dplyr::summarise(total_reads = sum(y))
### Mapping with Purrr
Although unnesting worked well in this example, it can get a bit awkward for more complex operations. In these cases we can use purrr. Refer to the next section for an example that compares both approaches. The below example sums the number of reads for each sample, as we have done above, but this time it uses purrr::map_dbl to map over the list of data frames:
df %>%
dplyr::mutate(
total_reads = purrr::map_dbl(plot.fastqc_per_sequence_quality_scores_plot, ~sum(.$y)), plot.fastqc_per_sequence_quality_scores_plot = NULL ) ### Properly Handling Histogram Data Of course, we actually want to find the median here, which is a bit harder. Luckily there exists a package called HistDat for generating summary statistics from histogram-type data. You can check out the package’s manual and vignettes here, but in brief, we want to convert each of these plot data frames into a HistDat object, which we can do using the same strategies as before. Then, using HistDat, we can calculate our summary statistics in one of the two ways mentioned above. Using the tidyr approach, we can unnest the plot data, group it, create a HistDat object for each group, and then produce new columns using the new hist column: df %>% tidyr::unnest(cols = plot.fastqc_per_sequence_quality_scores_plot) %>% dplyr::group_by(metadata.sample_id) %>% dplyr::mutate(hist = list(HistDat::HistDat(vals = x, counts = y)), .keep = "unused") %>% dplyr::mutate( mean_coverage = hist %>% dplyr::first() %>% mean(), median_coverage = hist %>% dplyr::first() %>% median(), max_coverage = hist %>% dplyr::first() %>% max(), hist= NULL ) %>% dplyr::slice(1) Alternatively, using the purrr method, we can just map each plot data frame into a row of summary statistics. Much neater! df %>% dplyr::mutate( purrr::map_dfr(plot.fastqc_per_sequence_quality_scores_plot, function(plot_df){ hist = HistDat::HistDat(vals=plot_df$x, counts = plot_df\$y)
list(
mean_coverage = mean(hist),
median_coverage = median(hist),
max_coverage = max(hist)
)
}),
plot.fastqc_per_sequence_quality_scores_plot = NULL
)
### Custom Plot Parsers
So far, we have used TidyMultiqc’s built-in parsers. Each parser is a function that converts the plot JSON into a list of data frames. However, it is possible that we might encounter a new type of plot that is not yet implemented in TidyMultiqc. If this happens, the first thing you should do is file an issue against this package.
Then, if you’re daring, you can try to implement a parser. For your reference, you can refer to the existing parsers in the source code. Then, you can pass in your custom parser using the plot_parsers argument to load_multiqc:
TidyMultiqc::load_multiqc(
multiqc_data_path,
sections = 'plot',
plots = "fastqc_per_sequence_quality_scores_plot",
plot_parsers = list(
# This fake parser function takes a plot and just returns the iris dataset
xy_line = function(plot_data, name){
list(
sample_1 = list(
plot_name = list(iris)
)
)
}
)
)
Finally, if your parser works, please submit a pull request to TidyMultiqc to share this with everyone!
Using the strategies and patterns explained in this section, you should be in good stead to handle whatever plots MultiQC throws at you.
|
2022-05-25 08:04:52
|
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http://www.ams.org/mathscinet-getitem?mr=339155
|
MathSciNet bibliographic data MR339155 (49 #3918) 55D35 Snaith, V. P. A stable decomposition of \$\Omega \sp{n}S\sp{n}X\$$\Omega \sp{n}S\sp{n}X$. J. London Math. Soc. (2) 7 (1974), 577–583. Article
For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.
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2015-03-29 12:41:24
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https://stats.stackexchange.com/questions/231292/is-a-decision-stump-a-linear-model/231304#231304
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# Is a decision stump a linear model?
Decision stump is a decision tree with only one split. It can also be written as a piecewise function.
For example, assume $x$ is a vector, and $x_1$ is the first component of $x$, in regression setting, some decision stump can be
$f(x)= \begin{cases} 3& x_1\leq 2 \\ 5 & x_1 > 2 \\ \end{cases}$
But is it a linear model? where can be written as $f(x)=\beta^T x$? This Question may sound strange, because as mentioned in the answers and comments, if we plot the piecewise function it is not a line. Please see next section for why I am asking this question.
EDIT:
• The reason I ask this question is logistic regression is a (generalized) linear model and the decision boundary is a line, also for decision stump. Note, we also have this question: Why is logistic regression a linear model? . On the other hand, it seems not true that decision stump is a linear model.
Here is one example of decision stump boosting on regression with 2 features and 1 continuous response.
• Why would you consider it to be linear..?
– Tim
Aug 23 '16 at 14:49
• @hxd1011 important to distinguish between the decision boundary and the decision function here Aug 23 '16 at 14:56
• If you draw a line in the plane (say y = 0), and take any function $f(x)$, then $g(x, y) = f(x)$ will have contour lines which are actual lines (parallel to the $y$ axis), but it will not be a linear function. Aug 23 '16 at 15:45
• This is a strange question. Can you plot the function from your example (which is equal to 3 for x<2 and 5 for x>2)? Look at it -- is it a straight line? If it is not a straight line, then it is not a linear function. Aug 23 '16 at 22:43
• @amoeba, i agree with you, the function plot is a not linear. I think my question is confusing also because it is not really clear what is "linear model" Aug 24 '16 at 5:13
No, unless you transform the data.
It is a linear model if you transform $x$ using indicator function: x' = \mathbb I \left(\{x>2\}\right) = \begin{cases}\begin{align} 0 \quad &x\leq 2\\ 1 \quad &x>2 \end{align}\end{cases}
Then $f(x) = 2x' + 3 = \left(\matrix{3 \\2}\right)^T \left(\matrix{1 \\x'}\right)$
Edit: this was mentioned in the comments but I want to emphasize it here as well. Any function that partitions the data into two pieces can be transformed into a linear model of this form, with an intercept and a single input (an indicator of which "side" of the partition the data point is on). It is important to take note of the difference between a decision function and a decision boundary.
• "transform" is tricky, i think neural network (MLP) is non-linear, but after transform, it is linear.. Aug 23 '16 at 14:56
• It is a linear model in the parameters. And it is affine linear in the dummy $x'$. Aug 23 '16 at 15:27
• @MichaelM how is it linear in the parameters? I assume by "parameters" you mean the choice of $x \leq 2$ Aug 23 '16 at 16:17
• @hxd1011 the answer is "no, unless you transform the data" Aug 23 '16 at 20:23
• I'd suggest you edit your answer to include "no, unless you transform the data" (from your last comment) into it. Currently your opening words are "It is a linear model", and people can get confused. Aug 23 '16 at 22:45
1. A decision stump is not a linear model.
2. The decision boundary can be a line, even if the model is not linear. Logistic regression is an example.
3. The boosted model does not have to be the same kind of model as the base learner. If you think about it, your example of boosting, plus the question you linked to, proves that the decision stump is not a linear model.
This answer is more verbose than is needed to just answer the question. I hope to provoke some comments from real experts.
I once was in a court room and the judge asked (for good reason in context) , if we call a dog's tail a leg, does that mean a dog has 5 legs ? So what is a linear model ?
In the context of statistics I've been told by an expert that a linear model means a statistical model constructed from a set of functions $f_1, f_2, \ldots, f_n$ of the form $y = \sum a_i f_i$ with the important constraint that the error terms are independent and normally distributed. With that definition, one can't say if your model is linear because you have given no information about the error term. If one drops the error term constraint, then it is tautologically linear in the function you give or in the function ssdecontrol gives. However naively, in the context of this question, that may be unsatisfactory. Any function can be considered as the basis of a linear in that sense. That is because any space of functions can be turned into a vector space of functions.
If you are asking on the nose, that is mathematically, if your function linear, then the answer is no. A linear function is one whose graph is a straight line, while clearly your function doesn't have that property. In answer to the question you pose at the end, that is can one find $\beta$ so that $f(x) = \beta^{T} x$ , then no.
Any function of the class you give would satisfy $f(x+y) = f(x) + f(y)$ for any (real) numbers $x$ and $y$. Notice that your function satisfies $f(1.5) = 3$ and $f(3) = 5$, so $f(3) \neq f(1.5) + f(1.5)$ as would be required if your function was of the form $f(x) = \beta^T x$. Notice the class you propose for linear functions is a sub-class of what are usually called linear functions.
• Linearity has nothing to do with error terms. It has to do with the fact that it consists of a linear combination of the parameters. This represents a straight line in 2D space (but more generally represents a plane). Aug 23 '16 at 16:12
• @ssdecontrol - I can only tell you that a probability Ph.D. (and a student of Kolmogorov at that) insisted to me that the term 'linear model', when used in statistics, included a statement about the error term. I'm actually very interested in how common this view might be. Linear has many meanings. For example, if one has an equation in N space $f(x) = 0$, then we could say it is linear if the zero set is a hyperplane. Of course that would mean we could write $f(x) = a_0 + \sum_{i=1}^{i=N} a_ix_i$ . However the function would be linear $\iff a_0 = 0$ i.e. $f(x+y) = f(x) + f(y)$.
– meh
Aug 24 '16 at 13:28
• if that's what he insists, then that's his opinion and not some kind of hard fact. As far as I'm aware, there is no rigorous accepted definition for a "linear model", nor is there a need for one in my mind. For me, the fact that there is an error term involved just turns the model from a "linear model" to a "statistical linear model". I don't see anything inherently linear about her terms, nor do I see anything inherently statistical about linear models. Aug 24 '16 at 15:44
• IMO insisting on the presence of an error term just discounts what, say, and engineering or physicist might deem a "linear model" of a deterministic physical process. Aug 24 '16 at 15:44
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2021-09-22 12:25:33
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http://www.mathworks.com/help/physmod/sps/powersys/ug/advanced-users-retune-the-drive-parameters.html?requestedDomain=www.mathworks.com&nocookie=true
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# Documentation
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## Retune the Drive Parameters
This example shows how to modify the parameters of an electric drive using the AC3 drive model. In this example, the nominal power of the motor changes from 200 hp to 5 hp. To retune the drive parameters, you:
### Modify the Motor Parameters
1. Open the example: `ac3_example`. (Type `ac3_example` in the MATLAB® Command Window).
The parameters are set for a 200 hp motor.
2. Simulate the model in Accelerator mode and observe the results.
3. Double-click the Field-Oriented Control Induction Motor Drive block and select the Asynchronous Machine tab. Copy these parameters of the 5-hp motor into the drive's mask.
### Retune the Parameters of the Flux Regulator
1. To measure the signals associated to the flux regulator, add these blocks into the demux subsystem.
2. Select the Controller tab in the mask of the Field-Oriented Control Induction Motor Drive block. Set the Regulation type to ```Torque regulation``` to access the controller parameters.
The torque regulation mode is required to bypass the speed regulator parameters and act directly on the field-oriented control (FOC) controller.
The current controlled by the FOC controller depends of the machine flux. The flux controller ensures that the required flux is correctly applied to the machine.
3. Set the Lowpass filter cutoff frequency to 10 kHz to avoid limiting the flux frequency. The highest flux frequency depends of the switching frequency.
Also set the Flux output limits to 150% of the Nominal flux, the Proportional gain to 1, the Integral gain to 0, the hysteresis band to 1, and the Machine flux to 0.705. Machine flux is based on this computation:
`$\frac{{V}_{LL}\left(rms\right)}{\sqrt{3}\cdot 2\cdot \pi \cdot f}=\frac{460}{\sqrt{3}\cdot 2\cdot \pi \cdot 60}$`
4. To apply the nominal torque to the motor, modify the parameters of the Stair Generator blocks in the Speed reference subsystem and in the Load torque subsystem.
5. Set the Sampling Decimation of the scope block to `1`, and the Variable name to `simout1`. Select Save data to workspace and set Format to ```Structure with time```.
6. Simulate the system for 0.5 s. Open the FFT Analysis tool of the powergui block.
In the Input list, select the ```Stator current``` signal and set Start time to `0.23`, Number of cycles to `1`, Fundamental frequency to `7.5`, and Max Frequency (Hz) to `20000` Hz.
Click Display to get the FFT graph.
7. Note that the switching frequency is about 5 kHz. To attenuate this frequency, set the Flux controller Lowpass filter cutoff frequency parameter to 500 Hz.
8. Open the Scope block and observe the flux signal. Note that the steady-state error is high.
9. Gradually increase the Proportional gain parameter of the controller and simulate until you obtain a satisfactory response. Increasing the gain above a certain value can cause a saturation of the Flux controller. The curve at the next plot is based on a proportional gain of 100.
10. Gradually increase the Integral gain and simulate until you obtain a satisfactory steady-state result with minimal error. The next plot is based on an integral gain of 90.
### Retune the Parameters of the Speed Regulator
1. Select the Controller tab in the mask of Field-Oriented Control Induction Motor Drive block and set the Regulation type to ```Speed regulation``` to edit the controller parameters.
Set the Torque output limits to `150%` of the nominal torque, the Proportional gain to `1`, the Integral gain to `0`, and the Speed cutoff frequency to `500`.
The highest speed frequency depends also on the switching frequency, so set the Maximum switching frequency parameter to `500` Hz.
The speed ramp acceleration must be calculated to avoid exceeding the torque output limit. The required torque to accelerate the motor at 1750 rpm/s is given by:
2. To apply the nominal torque to the motor, modify the parameters of the Stair Generator blocks in the Speed reference subsystem and in the Load torque subsystem.
3. Set the scope decimation to 25 to prevent memory overload. Start the simulation.
Observe the speed signal on the Scope block. The steady state error is high and the response time is not acceptable.
4. Gradually increase the Proportional gain parameter of the controller and simulate until you obtain a satisfactory response time without overshoot. Note that if the gain is too high, the system will be oscillatory. The next plot is based on a proportional gain of 3.
5. Gradually increase the Integral gain and simulate until you obtain a satisfactory steady state value with a minimal steady-state error. This curve is based on an integral gain of 100.
### Retune the Parameters of the DC Bus Voltage
1. Select the Converter and DC bus tab in the mask of the Field-Oriented Control Induction Motor Drive block to tune the DC bus capacitor and the braking chopper parameters.
2. Set the DC Bus Capacitance parameter to 167e-6.
The DC bus capacitance is calculated in order to reduce the voltage ripple.
`$C=\frac{{P}_{motor}}{12\cdot f\cdot {\Delta }_{V}\cdot {V}_{DC}}$`
where:
• Pmotor is the nominal power of the motor drive (W).
• f is the frequency of the AC source (Hz).
• ΔV is the desired voltage ripple (V).
• VDC is the average DC Bus voltage (V).
This equation gives an approximate value of the capacitor required for a given voltage ripple level. Here the desired voltage ripple is 50 V.
The motor drive of 5 hp (3728 W) is fed by a 60 Hz three-phase source. The average DC bus voltage is given by:
VDC = 1.35·VLL,
where VLL represents the line to line rms voltage of the source. The source line to line voltage is 460 Vrms so the DC voltage is VDC = 621 V.
The required capacitor is then equal to
3. Set the Braking chopper Shutdown voltage to 660V and the Braking chopper Activation voltage to 700 V.
In motor mode, the peak voltage of the DC bus is equal to
The shutdown voltage (Vshut) of the braking chopper should be a little bit higher than this value. To limit the voltage increase during regenerative braking, shutdown voltage is set to 660 V, and the activation voltage (Vact) is set to 700 V.
4. Set the Braking chopper Resistance to 131 ohms.
The braking chopper resistance is calculated using this equation:
The final DC bus parameters are shown in the next figure.
### Simulate and Analyze the Results
Simulate the system and observe six sections of the simulation results.
2. Nominal load torque is applied
4. Nominal generation torque is applied: Observe the DC bus voltage overshoot
5. Deceleration
6. Negative speed acceleration
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2017-07-22 09:09:50
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https://xavibiotech.com/the-black-uimvgp/9ea5d1-irreducible-matrix-example
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For example, the matrix 0 2 1 1 has period 1 and the matrix 0 4 1 0 has period 2. We begin by our first example, Example 15.1. But this matrix has the property that if you raise it to some nite power then it equals the all-zeros matrix. • Such Markov chains are called ergodic. Thm 1. Irreducible Markov Chains Proposition The communication relation is an equivalence relation. The stochastic matrix $P$ is called irreducible if all states communicate; that is, if $x$ and $y$ communicate for all $(x, y)$ in $S \times S$. Theorem: An irreducible, aperiodic, positive recurrent Markov chain has a unique stationary distribution, which is also the limiting distribution πj = limn→∞ P n ij. However, this matrix is not primitive. The text explains the use of the 6-j and 9-j symbols to complete theoretical concepts that are applied in irreducible tensor methods dealing with problems of atomic and molecular physics. Example of limiting distribution • Recall our transition matrix discussed recently:! Thus, since the sextic x6+ x5+ x4+ x3+ x2+ x+ 1 has no linear, quadratic, or cubic factors, it is irreducible. Many chaotic dynamical systems are isomorphic to topological Markov chains; examples include diffeomorphisms of closed manifolds, the Prouhet–Thue–Morse system, the Chacon system, sofic systems, context-free systems and block-coding systems. At each step, stay at the same node Nilpotent matrix. is a continuous function, then there exists a, Because the properties involved (continuity, being a fixed point) are invariant under, homeomorphisms (topological equivalence), the FP theorem holds for every set that is homeo-. %PDF-1.4 1004 = … • In matrix notation, πj = P∞ i=0 πiPij is π = πP where π is a row vector. Example 3.3-1: Irreducible Sets of Equations. The Markov chain with transition matrix is called irreducible if the state space consists of only one equivalence class, i.e. A non-negative, square, irreducible matrix with all positive diagonal elements is. Get step-by-step explanations, verified by experts. Example. With these theorems in mind, let’s take a closer look at the example from the previous section. Introducing Textbook Solutions. • A Markov chain is irreducible if all states belong to one class (all states communicate with each other). Section 4. This result is equivalent to Q = (I + Z) n – 1 containing all positive elements. The two one-dimensional irreducible representations spanned by $$s_N$$ and $$s_1'$$ are seen to be identical. consequently, it has no real eigenvectors in R2. Imagine that you are working on a top-secret team of analysts. Section 5. Proof? Examples The definition of irreducibility immediately implies that the matrices and. current location has a “You are here” point. Example 1. • If there exists some n for which p ij (n) >0 for all i and j, then all states communicate and the Markov chain is irreducible. In abstract algebra, irreducible can be an abbreviation for irreducible element of an integral domain; for example an irreducible polynomial. 1000 = 0.75 0.25 0.75 0.25! Let A ∈ R n × n be an irreducible totally nonnegative matrix with rank r and principal rank p, that is, all minors of A are nonnegative, r is the size of the largest invertible square submatrix of A and p is the size of its largest invertible principal submatrix. A graph G is strongly connected if and only if its adjacency matrix A is irreducible. Probability flux. Consider the linear equations In representation theory, an irreducible representation is a nontrivial representation with no nontrivial proper subrepresentations. Example. = 0 1 0.9 0.1 0.3 0.7 What is lim,→. A Markov chain is called reducible if /Length 2299 Examples Example 4. is homeomorphic to the closed unit-disk in, that maps a “closed set” to itself intersects with the, Boston School of Occupational Therapy (BSOT), Conference Bureau Office (108 Packard Avenue), Summer Session Office (108 Packard Avenue), M E D F O R D / S O M E R V I L L E C A M P U S, Pearson product-moment correlation coefficient. Identity. Section 3. The period of an irreducible matrix A is the greatest common divisor of the lengths of loops in G(A). Similarly, an irreducible module is another name for a simple module. Section 6. What is the ingredient I am missing? Irreducible Representations The transformation matrices can be reduced to their simplest units (1×1 matrices in this case) by block diagonalization: We can now make a table of the characters of each 1×1 matrix for each operation: The three rows (labeled Bu, Bu, and Au) are irreducible representations of … That was clearly an irreducible, and also aperiodic (note that P00 > 0) chain. A square matrix that is not reducible is said to be irreducible. Its associated graph 1 ← 2 ← 3 ← 1 is SC. A Markov matrix that is compatible with the adjacency matrix can then provide a measure on the subshift. De nition A Markov chain is called irreducible if and only if all states belong to one communication class. By additivity of degrees in products, lack of factors up to half the degree of a polynomial assures that the polynomial is irreducible. By de nition, the communication relation is re exive and symmetric. An_Approximate_Minimum_Degree_Ordering_Algorithm.pdf, lec-23-adjacency-and-incidence-matrices.pdf. Note that this is an orthogonal matrix due to the orthogonality relations between the characters. ... Mathematica » The #1 tool for creating Demonstrations and anything technical. Now suppose A is irreducible with period p. Pick some vertex v, and for 0 ≤ i,p define a set of vertices C Transitivity follows by composing paths. Examples of irreducible in a Sentence They thought the world was made up of four irreducible elements: earth, air, fire, and water. / 0!,? In this video we check the given matrix is irreducible or not. >> A motivating example shows how compli-cated random objects can be generated using Markov chains. In fact, every representation of a group can be decomposed into a direct sum of irreducible ones; thus, knowing the irreducible representations of a group is of paramount importance. Then ’0(g) = P’(g)P1is itself a representation. However, this matrix is not primitive. An irreducible matrix has its elements ‘tightly coupled’ in some way, as illustrated by the next two examples. In this video we check the given matrix is irreducible or not. On the other hand the block matrix consisting of and . In the language of topology, a coffee cup = a donut. 3 0 obj << x��YIo���W�����U$�aH��^$��d��F-�Z�v~}�����$#0��(.��Wŏg��@��P�4�lsq�!\"N�&�)�l.v��H�m��~��6f�GE��>+�K;���mL��.���P�e����l��H'41;�R؝c� �J[X�1�Χ-[����TF��TEy�FR��;&�*:s4��-/w�� �� �B�S��o[6��n;�M����i�J ���?U��S��nҬ�ϒ���p����vɣ��f(����w�c���>�!l-:x����Qed!����>@$:��y��DR� lA�c>g�J�A�E�wTf�%3����+��nK��M,�Ob��#a��b=�)$*�FJ�l���v^M�p.���� u�;�qT�]:Z#�]+�-�l�Ch�?A~���҈��5���1x t�� ��+�*�n����խ^�6�BmK�7?4�,�D{#eC�M]��q��a�����*����s�Ղ{5Eu����a,O�V�т/�A1�.$\endgroup$– Aaron Meyerowitz Dec 27 '10 at 4:37 Notice that:! 2 Irreducible tensors The Cartesian components of tensors are mixed by the rotation transformations, R ij, but not all components of a given tensor mix with all the others. Consider the following matrix, A= 0 1 0 0 : The only eigenvalue of this equals zero. Your latest mission is to save humanity from an asteroid that is heading towards the earth. It is true for diagonal matrices (for example) that$\det (A A^{T})=det(A)^2\$ and yet the determinant there is neither irreducible nor the square of an irreducible. The Markov chain mc is irreducible if every state is reachable from every other state in at most n – 1 steps, where n is the number of states (mc.NumStates). In addition, a matrix is reducible if and only if its associated digraph is not strongly connected. In general taking tsteps in the Markov chain corresponds to the matrix Mt. Course Hero is not sponsored or endorsed by any college or university. • If a Markov chain is not irreducible, it is called reducible. The matrix A = [ 0 1 − 1 0] has a characteristic polynomial pA(t) = t2 + 1, which is irreducible over R (has no real roots). A square matrix which is not reducible is said to be irreducible. A. 1001 = 0.75 0.25 0.75 0.25! For a limited time, find answers and explanations to over 1.2 million textbook exercises for FREE! for all . The simple random walk on the integer lattice Zd is the Markov chain whose tran-sition probabilities are p(x,x ei)=1=(2d) 8x 2Zd ... probability matrix) is said to be irreducible. Proof 16 For the above example, the Markov chain resulting from the first transition matrix will be irreducible while the chain resulting from the second matrix will be reducible into two clusters: one including states x 1 and x 2, and the other … Example 5 (Drunkard’s walk on n-cycle) Consider a Markov chain de ned by the following random walk on the nodes of an n-cycle. %���� What my paper shows in general is that, if you choose the entries of an matrix from a fixed distribution in the integers,then, with probability tending to as , the characteristic polynomial is irreducible, and moreover its Galois group is at least . matrix M2. This preview shows page 6 - 10 out of 14 pages. An n × n matrix A is said to be a reducible matrix if and only if for some permutation matrix P, the matrix P T A P is block upper triangular. Stationary distributions, with examples. The eigenvector is the same as in the above example. stream If a square matrix is not reducible, it is said to be an irreducible matrix. are irreducible. /Filter /FlateDecode The end step in this plan is to factor a polynomial completely into irreducible factors, where an irreducible factoris a polynomial that is not a constant and cannot be factored … The matrix A = 0 1 0 0 0 1 1 0 0 is irreducible. is non-negative with strictly positive diagonals. Definition 1 A distribution ˇ for the Markov chain M is a stationary distribution if ˇM = ˇ. A Examples Example 4 The matrix A 1 1 1 is irreducible Its associated graph 1 2. is SC. This means that $$s_N$$ and $$s_1'$$ have the ‘same symmetry’, transforming in the same way under all of the symmetry operations of the point group and forming bases for the same matrix … A matrix is reducible if and only if it can be placed into block upper-triangular form by simultaneous row/column permutations. the corresponding transition matrix. Statement of the Basic Limit Theorem about conver-gence to stationarity. The next result ties irreducible matrices to graph theory. An irreducible non-negative matrix with all positive diagonal elements is primitive with index, A remarkable result from topology: Equivalently, every map that encloses your. While you and your team are assessing the situation, you come up with a plan that can divert the asteroid. The book describes the matrix elements in multielectron systems, as well as the reduced matrix … The invariant distribution [π1,π2,π3] is given by 0.7π1 +0.4π2 = π1 0.2π1 +0.6π2 +π3 = π2 0.1π1 = π3 This system has infinitely … How matrix multiplication gets into the picture. I is the n -by- n identity matrix. Example 15.7. For example, consider the following transition probabilities for wealth of a fictitious set of households Here is the orthogonal matrix obtained by multiplying each character value by the square root of the quotient of the size of its conjugacy class by the order of the group. Exercise 2.6.4Let ’be a representation and P be an invertible matrix. = ( I + Z ) n – 1 containing all positive elements we the. Notation, πj = P∞ i=0 πiPij is π = πP where π is stationary. As illustrated by the next two examples 4 the matrix 0 4 1 0 has period 1 the. Above example raise it to some nite power then it equals the all-zeros matrix has its elements irreducible matrix example coupled! A plan that can divert the asteroid • in matrix notation, πj = P∞ i=0 is... Following transition probabilities for wealth of a fictitious set of households matrix M2 at 4:37 example we begin by first. ) P1is itself a representation and P be an invertible matrix P∞ i=0 is! To Q = ( I + Z ) n – 1 irreducible matrix example all positive diagonal elements is you. Save humanity from an asteroid that is heading towards the earth equals zero one class ( all communicate. At the same node example of limiting distribution • Recall irreducible matrix example transition matrix is reducible if and if! Come up with a plan that can divert the asteroid polynomial assures that polynomial... First example, consider the following matrix, A= 0 1 0 period... Endorsed by any college or university row/column permutations given matrix is reducible if and only if states! Meyerowitz Dec 27 '10 at 4:37 example Aaron Meyerowitz Dec 27 '10 4:37... Our first example, consider the following matrix, A= 0 1 0.9 0.1 0.3 0.7 is... The eigenvector is the greatest common divisor of the lengths of loops in g ( a ) previous! Is SC if a Markov chain is irreducible its associated digraph is not sponsored or endorsed by any or... The only eigenvalue of this equals zero of irreducibility immediately implies that the matrices and up to half degree! = 0 1 0 0 1 0 0 1 0 has period 2 seen to identical... As in the Markov chain M is a row vector communicate with each )! Assures that the matrices and 1 and the matrix Mt communication class ( a ) that the polynomial is its. = ( I + Z ) n – 1 containing all positive elements! The given matrix is reducible if and only if its adjacency matrix a 1 1 has period.!: the only eigenvalue of this equals zero has a “ you are here ” point save from! A coffee cup = a donut factors up to half the degree of a assures... Coffee cup = a donut to irreducible matrix example = ( I + Z n... Reducible is said to be irreducible is heading towards the earth coupled ’ some! Theory, an irreducible module is another name for a simple module, lack of factors to! 0: the only eigenvalue of this equals zero a closer look at the example from previous... A closer look at the same node example of limiting distribution • Recall our transition discussed! 0 1 1 0 0 0: the only eigenvalue of this zero... = P ’ ( g ) P1is itself a representation and P be an irreducible, and also aperiodic note... Loops in g ( a ) addition, a coffee cup = a donut ( I Z! ) P1is itself a representation and P be an irreducible module is another name a. For a limited time, find answers and explanations to over 1.2 million textbook exercises for FREE in (! Degrees irreducible matrix example products, lack of factors up to half the degree of a fictitious set of households M2... Humanity from an asteroid that is not reducible is said to be an irreducible.! Stationary distribution if ˇM = ˇ 4 the matrix a 1 1 1 1 has period 1 and the a... Matrix Mt eigenvalue of this equals zero begin by our first example, example 15.1 s_1'\ ) are seen be. Degrees in products, lack of factors up to half the degree of a polynomial assures that the matrices.... Aperiodic ( note that this is an orthogonal matrix due to the relations... Over 1.2 million textbook exercises for FREE is irreducible or not – 1 containing all diagonal! Over 1.2 million textbook exercises for FREE diagonal elements is degree of a polynomial assures that polynomial! Aaron Meyerowitz Dec 27 '10 at 4:37 example irreducible its associated digraph is not reducible is said to be.... Our transition matrix discussed recently: 0 4 1 0 0 irreducible! Invertible matrix is irreducible distribution • Recall our transition matrix discussed:. Check the given matrix is reducible if and only if it can be into! Representation is a nontrivial representation with no nontrivial proper subrepresentations matrix 0 2 1 1 1 has 1... In this video we check the given matrix is not reducible, it is called.... It is said to be an irreducible matrix has its elements ‘ coupled! G is strongly connected if and only if it can be placed into block upper-triangular form simultaneous... With a plan that can divert the asteroid and only if its adjacency matrix a irreducible. Diagonal elements is and P be an irreducible matrix with all positive diagonal elements is the chain... Irreducible if all states belong to one communication class ‘ tightly coupled ’ in some,! To over 1.2 million textbook exercises for FREE these theorems in mind, ’!, consider the following transition probabilities for wealth of a fictitious set households... Can be placed into block upper-triangular form by simultaneous row/column permutations anything technical >... ( I + Z ) n – 1 containing all positive elements can. Limiting distribution • Recall our transition matrix discussed recently: half the degree of a polynomial assures the... ’ in some way, as illustrated by the next two examples at step... – Aaron Meyerowitz Dec 27 '10 at 4:37 example elements ‘ tightly coupled in. The example from the previous section all states communicate with each other ) aperiodic ( note that >... This matrix has its elements ‘ tightly coupled ’ in some way, as by! That P00 > 0 ) chain note that this is an equivalence relation working! Nontrivial representation with no nontrivial irreducible matrix example subrepresentations = ( I + Z ) –... Lim, → the eigenvector is the greatest common divisor of the lengths of loops g... You and your team are assessing the situation, you come up with a plan can... From the previous section distribution ˇ for the Markov chain is not reducible said! Recall our transition matrix discussed recently: eigenvalue of this equals zero reducible... Positive diagonal elements is the next two examples no nontrivial proper subrepresentations in... It has no real eigenvectors in R2 M is a stationary distribution if ˇM = ˇ probabilities. Other ) Recall our transition matrix is reducible if and only if its associated digraph is not reducible is to... Is called irreducible if and only if all states communicate with each other ) and... For wealth of a polynomial assures that the polynomial is irreducible if and if... Shows page 6 - 10 out of 14 pages nite power then equals. Or endorsed by any college or university de nition, the communication relation re! Begin by our first example, example 15.1 heading towards the earth Proposition the communication relation is re and. Is not irreducible matrix example or endorsed by any college or university from the previous section proper subrepresentations fictitious of! Location has a “ you are working on a top-secret team of analysts equals.! By de nition, the matrix a is irreducible matrix example or not its matrix... 1 containing all positive diagonal elements is of loops in g ( a ) n. Heading towards the earth, stay at the same node example of limiting •! Communication relation is an orthogonal matrix due to the orthogonality relations between the characters is irreducible not... Here ” point orthogonal matrix due to the matrix 0 4 1 0 0 . Discussed recently: lim, → is to save humanity from an asteroid that is heading towards the.! Imagine that you are here ” point example 15.1 representation is a row vector relation an. • if a Markov chain corresponds to the orthogonality relations between the characters is an orthogonal due! Lengths of loops in g ( a ) find answers and explanations over! Same node example of limiting distribution • Recall our transition matrix discussed:. 0 has period 1 and the matrix a = 0 1 has! Additivity of degrees in products, lack of factors up to half the degree of fictitious. = ( I + Z ) n – 1 containing all positive diagonal is... Digraph is not strongly connected a limited time, find answers and explanations to 1.2! Definition of irreducibility immediately implies that the polynomial is irreducible exercise 2.6.4Let ’ be a representation set of matrix. If all states communicate with each other ) ← 1 is irreducible if and only if its associated is... One communication class “ you are here ” point real eigenvectors in R2 #... The earth and symmetric s_1'\ ) are seen to be irreducible “ you are working on a top-secret of! Team are assessing the situation, you come up with a plan that divert. The degree of a fictitious set of households matrix M2 by de nition a Markov chain M is a vector... • a Markov chain is called irreducible if all states belong to class.
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2021-04-19 15:31:13
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https://math.stackexchange.com/questions/104772/algebraic-functions-are-polynomials
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# Algebraic functions are polynomials?
Does any one how to prove that every entire algebraic function is a polynomial? I'm under the impression that this can be achieved by showing that an algebraic function grows no faster than a polynomial.
• Not every entire function is a polynomial. For example, consider $e^z$. Could you be a bit more specific with your question? – Potato Feb 2 '12 at 0:32
• The exponential function is entire and it is not a polynomial. – ncmathsadist Feb 2 '12 at 0:33
• Do you mean every entire algebraic function is a polynomial? – Robert Israel Feb 2 '12 at 0:36
• Based on the title, I believe the question is how to prove that every algebraic entire function is a polynomial. – Jonas Meyer Feb 2 '12 at 0:36
• yes I'm sorry I meant every entire algebraic function is a polynomial. – Ricky Maher Feb 2 '12 at 0:40
I assume you mean every entire algebraic function is a polynomial. Suppose $g$ is an entire algebraic function of order $n$. Thus there are polynomials $c_j(z)$, $j=0 \ldots, n$ with $c_n$ not identically $0$ such that $\sum_{j=0}^n c_j(z) g(z)^j = 0$. For all but finitely many complex numbers $w$, $\sum_{j=0}^n c_j(z) w^j$ is not identically $0$, and so there are at most finitely many $z$ for which $\sum_{j=0}^n c_j(z) w^j = 0$: those are the only $z$ for which we can have $g(z) = w$. Thus for all but finitely many $w$, $g(z)$ takes the value $w$ only finitely many times. But an entire function that is not a polynomial has an essential singularity at $\infty$, and by the Great Picard Theorem it takes all but one value infinitely many times.
• For what it is worth, I have a survey by V. I. Arnold that mentions the function $z(a,b,c)$ that satisfies $$z^7 + a z^3 + b z^2 + c z + 1 = 0$$ and says in a footnote that it is "an entire algebraic function (without poles)" and asks whether it can be expressed as a superposition of such functions in two variables. – Will Jagy Feb 2 '12 at 1:17
• @Will, I noticed that one, too. I convinced myself that in the context of that piece, "entire algebraic function" doesn't mean "function that is both entire and algebraic," but something else. – Gerry Myerson Feb 2 '12 at 1:48
• @GerryMyerson, thanks, I decided not to leave that as a comment to the OP partly because I was uncertain what was going on. Technical matter, it seems we can reply to a comment with merely an @ sign and a first name? Because the software did tell me about your comment, in the "responses" flag in my personal profile page. Hmmm, if I look at your profile, some things are not visible to me, such as your "response" flag, so I cannot quickly experiment with this. – Will Jagy Feb 2 '12 at 2:09
• @Wil, my understanding is that the first three letters of the name suffice. – Gerry Myerson Feb 2 '12 at 4:44
• @Ger, it worked. Thanks. – Will Jagy Feb 2 '12 at 5:05
Hopefully not, because $\: \operatorname{exp} : \mathbb{C} \to \mathbb{C} \:$ defined by $\:\:\:\: f(z) \:\: = \:\: \displaystyle\sum_{n=0}^{\infty} \: \left(\frac1{n!} \cdot \left(z^n\right)\right)$
is an entire function that is not a polynomial.
• See the title; I believe the functions are also supposed to be algebraic. – Jonas Meyer Feb 2 '12 at 0:37
• so far i have the following: 1. If (1/abs value z)(abs value f) holds then f is a polynomial. So assume theorem not true... – Ricky Maher Feb 2 '12 at 0:43
• @RickyMaher: "$\frac{1}{|z|}|f|$" is not a statement. What do you mean by "$\frac{1}{|z|}|f|$ holds"? – Jonas Meyer Feb 2 '12 at 0:51
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2019-06-17 05:09:18
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https://cstheory.stackexchange.com/questions/46124/prime-factorisation-of-decidable-problems
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# Prime factorisation of decidable problems
Disclaimer: I am not a theoretical computer scientist.
The set of decidable problems $$\mathbb{D}$$ is countable so $$\lvert \mathbb{D} \rvert = \lvert \mathbb{N} \rvert$$ and this led me to the following idea.
Given two decision problems $$p,q$$ in $$\mathbb{D}$$ and $$\mathcal{M}_p,\mathcal{M}_q$$ the minimal-length Turing Machines deciding $$p$$ and $$q$$ we may say that $$p$$ and $$q$$ are relatively prime if:
$$$$K(\mathcal{M}_p|\mathcal{M}_q) = K(\mathcal{M}_p) \tag{1}$$$$
and
$$$$K(\mathcal{M}_q|\mathcal{M}_p) = K(\mathcal{M}_q) \tag{2}$$$$
where $$K(\cdot)$$ denotes Kolmogorov Complexity.
I think we should then be able to define $$\mathbb{B} \subset \mathbb{D}$$ such that $$\lvert \mathbb{B} \rvert = \lvert \mathbb{D} \rvert$$ and:
$$$$\forall b \in \mathbb{B} \forall d \in \mathbb{D} \setminus \{b\}, K(\mathcal{M}_b|\mathcal{M}_d)=K(\mathcal{M}_b) \tag{3}$$$$
where $$\mathbb{B}$$ is analogous to the set of primes in $$\mathbb{N}$$. Has this idea been explored?
• There is a bit of a "type error" in your question, since Kolmogorov complexity is defined on strings, not on decision problems. However, if one blurs the distinction between $p$ and a minimal-length Turing machine deciding $p$, then it makes sense. – Aryeh Dec 31 '19 at 12:08
• The nice thing about prime factorization in number theory is that it's unique. As my answer below indicates, the sets you seek are about as far from unique as possible... – Aryeh Dec 31 '19 at 12:20
• The bijection is not computable, which may be a problem, but in any case seems to be unnecessary. Why do you need it at all? Just use a reasonale encoding of decidable problems (and note that not all strings encode such problems). – Andrej Bauer Jan 1 at 10:34
• @AndrejBauer Thanks for pointing this out. I removed the point about a bijection between decidable problems and the natural numbers. – Aidan Rocke Jan 1 at 12:00
I think a candidate for such a set $$\mathbb{B}$$, or something very much like it, could be produced by considering an infinite sequence of singleton languages: $$L_1=\{w_1\},L_2=\{w_2\},\ldots$$ --- where the $$w_1\cdot w_2\cdot\ldots$$ form an incompressible sequence. You might be able to shave off a bit here or there (this will strongly depend on the encoding -- which is why Kolmogorov complexity really only makes sense up to an additive constant). But modulo additive constants, this family of strings has the property where being given one of them as input does not in any way help in generating any other one.
• If $w_i$ are chosen randomly then each $w_i$ will be a prefix of infinitely many $w_j$ so one would have to be careful (the usual idea being that random words are incompressible enough) – Bjørn Kjos-Hanssen Jan 2 at 8:24
• Perhaps one can choose the $w_i$ iteratively, starting from a single infinite incompressible sequence -- precisely with the criterion that $w_{i+1}$ is such that the OP's condition holds for all the $w_i$'s constructed so far. It remains to rigorously show that such a choice is always possible for any incompressible sequence, but I believe it's true. – Aryeh Jan 2 at 9:39
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2020-08-09 00:14:20
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https://chemistry.stackexchange.com/questions/29137/how-to-calculate-the-ph-of-a-buffer-after-dilution
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# How to calculate the pH of a buffer after dilution
You have a beaker containing $\pu{542 mL}$ of a solution of $\pu{0.48 M}$ acetate buffer ($\mathrm{p}K_\mathrm{a} = 4.76$) at $\ce{pH 4.2}$. To this solution, you add $\pu{342 mL}$ of pure water. What is the $\ce{pH}$ of the resulting solution?
The answer is $4.6$. However, I do not know how to arrive at that answer. I know the H-H equation must be used:
$$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$
Adding water means diluting the buffer.
Use the known concentration, the volume of the buffer and the volume of water added to calculate the new concentrations.
• I went down that path. However, the question just gave the concentration of the buffer. That means ([A-] + [HA] = 0.48M). In order to calculate the new pH, I'd need the concentration of the weak acid and its conjugate base. I was thinking of solving for its ratio using H-H, but I am not sure if I am on the right path. – Michael Apr 21 '15 at 23:13
• @Michael That sounds reasonable, since you have the initial pH and the pKa :-) – Klaus-Dieter Warzecha Apr 21 '15 at 23:18
Calculate the number of moles in the original buffer solution - you should get around 0.26 (that is using ur n=cv equation), now add up the new volume to the old one which would give u 0.884L then use your other mole number you got earlier to use in the n=cv equation to get new concentration, it should give around 0.294. No use pH=pKa+log([A-]/[HA]) equation bearing in mind that you just calculated the new concentration of ([A-]/[HA]) so just substitute it with the 0.294 and u should get the new pH to be 4.22 or the 4.2.
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2020-02-27 15:15:29
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https://www.varsitytutors.com/sat_math-help/how-to-find-the-perimeter-of-a-hexagon
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# SAT Math : How to find the perimeter of a hexagon
## Example Questions
### Example Question #1 : How To Find The Perimeter Of A Hexagon
Archimedes High School has an unusual track in that it is shaped like a regular hexagon, as above. Each side of the hexagon measures 264 feet.
Alvin runs at a steady speed of seven miles an hour for twelve minutes, starting at point A and working his way clockwise. When he is finished, which of the following points is he closest to?
Point B
Point C
Point F
Point D
Point E
Point E
Explanation:
Alvin runs at a rate of seven miles an hour for twelve minutes, or hours. The distance he runs is equal to his rate multiplied by his time, so, setting in this formula:
miles.
One mile comprises 5,280 feet, so this is equal to
feet
Since each side of the track measures 264 feet, this means that Alvin runs
sidelengths.
,
which means that Alvin runs around the track four complete times, plus four more sides of the track. Alvin stops when he is at Point E.
### Example Question #2 : How To Find The Perimeter Of A Hexagon
A circle with circumference is inscribed in a regular hexagon. Give the perimeter of the hexagon.
None of these
Explanation:
Below is the figure referenced; note that the hexagon is divided by its diameters, and that an apothem—a perpendicular bisector from the center to one side—has been drawn.
The circle has circumference ; its radius, which coincides with the apothem of the hexagon, is the circumference divided by :
The hexagon is divided into six equilateral triangles. One, , is divided by an apothem of the hexagon - a radius of the circle - into two 30-60-90 triangles, one of which is . Since has length 30, and it is a long leg of , then short leg has length
is the midpoint of , one of the six congruent sides of the hexagon, so
;
this makes the perimeter of the hexagon six times this, or
.
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2017-06-27 03:48:19
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https://fdocuments.net/document/on-quadratic-convergence-of-dc-proximal-newton-algorithm-on-quadratic-convergence.html
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### Transcript of On Quadratic Convergence of DC Proximal Newton Algorithm ... ... On Quadratic Convergence of DC...
• On Quadratic Convergence of DC Proximal Newton Algorithm in Nonconvex Sparse Learning
Xingguo Li1,4 Lin F. Yang2⇤ Jason Ge2 Jarvis Haupt1 Tong Zhang3 Tuo Zhao4† 1University of Minnesota 2Princeton University 3Tencent AI Lab 4Georgia Tech
Abstract
We propose a DC proximal Newton algorithm for solving nonconvex regularized sparse learning problems in high dimensions. Our proposed algorithm integrates the proximal newton algorithm with multi-stage convex relaxation based on the difference of convex (DC) programming, and enjoys both strong computational and statistical guarantees. Specifically, by leveraging a sophisticated characterization of sparse modeling structures (i.e., local restricted strong convexity and Hessian smoothness), we prove that within each stage of convex relaxation, our proposed algorithm achieves (local) quadratic convergence, and eventually obtains a sparse approximate local optimum with optimal statistical properties after only a few convex relaxations. Numerical experiments are provided to support our theory.
1 Introduction We consider a high dimensional regression or classification problem: Given n independent observa- tions {xi, yi}ni=1 ⇢ Rd ⇥R sampled from a joint distribution D(X, Y ), we are interested in learning the conditional distribution P(Y |X) from the data. A popular modeling approach is the Generalized Linear Model (GLM) [20], which assumes
P (Y |X; ✓⇤) / exp ✓
Y X>✓⇤ � (X>✓⇤) c(�)
◆ ,
where c(�) is a scaling parameter, and is the cumulant function. A natural approach to estimate ✓⇤ is the Maximum Likelihood Estimation (MLE) [25], which essentially minimizes the negative log-likelihood of the data given parameters. However, MLE often performs poorly in parameter estimation in high dimensions due to the curse of dimensionality [6].
To address this issue, machine learning researchers and statisticians follow Occam’s razor principle, and propose sparse modeling approaches [3, 26, 30, 32]. These sparse modeling approaches assume that ✓⇤ is a sparse vector with only s⇤ nonzero entries, where s⇤ < n ⌧ d. This implies that many variables in X are essentially irrelevant to modeling, which is very natural to many real world applications such as genomics and medical imaging [7, 21]. Many empirical results have corroborated the success of sparse modeling in high dimensions. Specifically, many sparse modeling approaches obtain a sparse estimator of ✓⇤ by solving the following regularized optimization problem,
✓ = argmin ✓2Rd
L(✓) + R�tgt(✓), (1)
where L : Rd ! R is the convex negative log-likelihood (or pseudo-likelihood) function, R�tgt : Rd ! R is a sparsity-inducing decomposable regularizer, i.e., R�tgt(✓) =
Pd j=1 r�tgt(✓j) with
r�tgt : R ! R, and �tgt > 0 is the regularization parameter. Many existing sparse modeling approaches can be cast as special examples of (1), such as sparse linear regression [30], sparse logistic regression [32], and sparse Poisson regression [26].
⇤The work was done while the author was at Johns Hopkins University. †The authors acknowledge support from DARPA YFA N66001-14-1-4047, NSF Grant IIS-1447639,
and Doctoral Dissertation Fellowship from University of Minnesota. Correspondence to: Xingguo Li and Tuo Zhao .
31st Conference on Neural Information Processing Systems (NIPS 2017), Long Beach, CA, USA.
• Given a convex regularizer, e.g., R tgt
(✓) = � tgt
||✓|| 1
[30], we can obtain global optima in polynomial time and characterize their statistical properties. However, convex regularizers incur large estimation bias. To address this issue, several nonconvex regularizers are proposed, including the minimax concave penalty (MCP, [39]), smooth clipped absolute deviation (SCAD, [8]), and capped `
1
- regularization [40]. The obtained estimator (e.g., hypothetically global optima to (1)) can achieve faster statistical rates of convergence than their convex counterparts [9, 16, 22, 34].
Despite of these superior statistical guarantees, nonconvex regularizers raise greater computational challenge than convex regularizers in high dimensions. Popular iterative algorithms for convex optimization, such as proximal gradient descent [2, 23] and coordinate descent [17, 29], no longer have strong global convergence guarantees for nonconvex optimization. Therefore, establishing statistical properties of the estimators obtained by these algorithms becomes very challenging, which explains why existing theoretical studies on computational and statistical guarantees for nonconvex regularized sparse modeling approaches are so limited until recent rise of a new area named “statistical optimization”. Specifically, machine learning researchers start to incorporate certain structures of sparse modeling (e.g. restricted strong convexity, large regularization effect) into the algorithmic design and convergence analysis for optimization. This further motivates a few recent progresses: [16] propose proximal gradient algorithms for a family of nonconvex regularized estimators with a linear convergence to an approximate local optimum with suboptimal statistical guarantees; [34, 43] further propose homotopy proximal gradient and coordinate gradient descent algorithms with a linear convergence to a local optimum and optimal statistical guarantees; [9, 41] propose a multistage convex relaxation-based (also known as Difference of Convex (DC) Programming) proximal gradient algorithm, which can guarantee an approximate local optimum with optimal statistical properties. Their computational analysis further shows that within each stage of the convex relaxation, the proximal gradient algorithm achieves a (local) linear convergence to a unique sparse global optimum for the relaxed convex subproblem.
The aforementioned approaches only consider first order algorithms, such as proximal gradient descent and proximal coordinate gradient descent. The second order algorithms with theoretical guarantees are still largely missing for high dimensional nonconvex regularized sparse modeling approaches, but this does not suppress the enthusiasm of applying heuristic second order algorithms to real world problems. Some evidences have already corroborated their superior computational performance over first order algorithms (e.g. glmnet [10]). This further motivates our attempt towards understanding the second order algorithms in high dimensions.
In this paper, we study a multistage convex relaxation-based proximal Newton algorithm for noncon- vex regularized sparse learning. This algorithm is not only highly efficient in practice, but also enjoys strong computational and statistical guarantees in theory. Specifically, by leveraging a sophisticated characterization of local restricted strong convexity and Hessian smoothness, we prove that within each stage of convex relaxation, our proposed algorithm maintains the solution sparsity, and achieves a (local) quadratic convergence, which is a significant improvement over (local) linear convergence of proximal gradient algorithm in [9] (See more details in later sections). This eventually allows us to obtain an approximate local optimum with optimal statistical properties after only a few relaxations. Numerical experiments are provided to support our theory. To the best of our knowledge, this is the first of second order based approaches for high dimensional sparse learning using convex/nonconvex regularizers with strong statistical and computational guarantees.
Notations: Given a vector v 2 Rd, we denote the p-norm as ||v||p = ( Pd
j=1 |vj |p)1/p for a real p > 0 and the number of nonzero entries as ||v||
0
=
P j 1(vj 6= 0) and v\j =
(v 1
, . . . , vj�1, vj+1, . . . , vd)> 2 Rd�1 as the subvector with the j-th entry removed. Given an index set A ✓ {1, ..., d}, A? = {j | j 2 {1, ..., d}, j /2 A} is the complementary set to A. We use vA to denote a subvector of v indexed by A. Given a matrix A 2 Rd⇥d, we use A⇤j (Ak⇤) to denote the j-th column (k-th row) and ⇤
max
(A) (⇤ min
(A)) as the largest (smallest) eigenvalue of A. We define ||A||2
F
=
P j ||A⇤j ||22 and ||A||2 =
p ⇤
max
(A>A). We denote A\i\j as the submatrix of A with the i-th row and the j-th column removed, A\ij (Ai\j) as the j-th column (i-th row) of A with its i-th (j-th) entry removed, and AAA as a submatrix of A with both row and column indexed by A. If A is a PSD matrix, we define ||v||A =
p v>Av as the induced seminorm for vector v. We use
conventional notation O(·), ⌦(·),⇥(·) to denote the limiting behavior, ignoring constant, and OP (·) to denote the limiting behavior in probability. C
1
, C 2
, . . . are denoted as generic positive constants.
2
• 2 DC Proximal Newton Algorithm
Throughout the rest of the paper, we assume: (1) L(✓) is nonstrongly convex and twice continuously differentiable, e.g., the negative log-likelihood function of the generalized linear model (GLM); (2) L(✓) takes an additive form, i.e., L(✓) = 1n
Pn i=1 `i(✓), where each `i(✓) is associated with an
observation (xi, yi) for i = 1, ..., n. Take GLM as an example, we have `i(✓) = (x>i ✓) � yix>i ✓, where is the cumulant function.
For nonconvex regularization, we use the capped ` 1
regularizer [40] defined as
R�tgt(✓) = dX
j=1
r tgt
(✓j) = �tgt
dX
j=1
min{|✓j |,��tgt},
where � > 0 is an additional tuning parameter.
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2022-08-07 18:48:52
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https://ltwork.net/which-expressions-correctly-states-the-version-of-the-elasticity--8332274
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Which expressions correctly states the version of the elasticity formula to predict the change in quantity demanded?
Question:
Which expressions correctly states the version of the elasticity formula to predict the change in quantity demanded?
A cylinder has a radius of 2 inches and a height of 3 1/2 inches. A sphere has a radius of 2 inches. What is theof a cubic inch, of the cylinder and
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2022-12-07 00:00:39
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https://www.matematicamente.it/forum/viewtopic.php?f=18&t=167625&p=8242347
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### Re: Problemi collegamento https
Anche per me va bene . Grazie Veciorik .
I can't read that, instead of "Joe washed his hands", mathematicians say " There's a $t_1$ : the image of $t_1$ under the map $t_1\rarrJoe(t_1)$ belongs to the set of dirty hands, and a $t_2 >t_1$ : the image of $t_2$ belongs to the complement of that set" V.I.Arnol'd
Shackle
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2019-11-19 07:41:33
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http://electricalacademia.com/induction-motor/electric-motor-nameplate-details-explained-induction-motor-nameplate/
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Home / Induction Motor / Electric Motor Nameplate Details Explained | Induction Motor Nameplate
# Electric Motor Nameplate Details Explained | Induction Motor Nameplate
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Figure 1 is a drawing of a typical induction motor nameplate. The electric motor nameplate contains a variety of information that can be used to calculate other items. NEMA specifies the information that must be included on the motor’s nameplate. The motor manufacturer may include additional information, such as the wiring information shown on the left side of the nameplate. Some of the items that must be included are described next.
FIGURE 1 Typical induction motor nameplate information.
Manufacturer’s Type and Frame Designation
NEMA defines standard frame sizes. This name-plate shows a frame size 326T. The frame size provides a standard for details such as the location of mounting holes, a height of the shaft above the mounting plate, etc. This allows motors from different manufacturers to be used in the same location.
Rated Horsepower
This is the amount of power the motor is capable of delivering at the shaft on a continuous basis without exceeding its rated temperature. The example nameplate shows 50 HP. This does not mean the motor actually delivers 50 HP, however. The actual load will determine the power out of the motor. If the motor is fully loaded, the input to the motor will be larger than the rated horsepower, because the motor has losses.
NEMA defines a motor as small, medium, or large, depending on its frame size rather than its horsepower, because the same frame size can represent a different horsepower rating at different speeds Thus, the cutoff between small, medium, and large varies with the synchronous speed of the machine. At 1800 RPM, the standard horsepower ratings for medium poly-phase induction motors are: 10,15,20,25,30,40,50,60,75,100 125, 150, 200, 250, 300, 350, 400, 450, 500. Table 1 shows the minimum and maximum horsepower ratings that are considered to be medium-size induction motors as a function of the synchronous speed of the motor. Motors larger than the maximum sizes shown in Table 1 are considered to be large motors, whereas motors smaller than the minimum sizes shown in Table 1 are considered to be small motors.
Synchronous Speed in RPM Minimum HP size for medium Classification Maximum HP size for medium Classification 3600 ½ 500 1800 1 500 1200 ¾ 350 900 ½ 250 720 ½ 200 600 ½ 150 500 ½ 125
Rated Voltage
A three-phase motor nameplate always specifies the rated line-to-line voltage. Some motors are designed to operate at more than one voltage. This motor, for example, is rated 230/460, meaning it can be wired for operation at either voltage. This is accomplished by connecting coils in series for high voltage, or in parallel for low voltage. The rated voltage of a motor is somewhat lower than the nominal system voltage to allow for voltage drop in the wires feeding the motor. This motor is designed to operate on 240 volts or 460 volts systems. Allowing an approximate 5 % voltage drop yields the 230 and 460 volts ratings for the motor. Other standard voltage ratings for motors include 115 volts for 120 volts system, 200 volts for 208 volts system, and 575 volts for 600 volts systems. Induction motors should operate at voltages that are within ±10% of rated voltage.
Rated Current
This is the line current that the machine is designed to draw when it operates at rated load, rated voltage, and rated speed. In accordance with NEMA Standard MG-1, the current will be within 10% of this value when the motor is operated at rated frequency, voltage, and load. The ±10% tolerance allows for differences in manufacturing from one motor to another. Note that the nameplate shown in Figure 1 indicates that the motor can operate on a 208-volt system, although it will draw a higher current (causing more heat). The motor may not meet all of the NEMA requirements at 208 volts, however.
Rated voltage and current can be used to calculate the apparent power requirement for the motor, as shown by Example 1.
Example 1
What is the apparent three-phase power drawn by the motor of Figure 1 when operating at rated load conditions (assume 460 V operation)?
Solution
From the nameplate,
\begin{align} & \left| {{V}_{LL}} \right|=460V \\ & \left| {{I}_{L}} \right|=61A \\\end{align}
The apparent three-phase power is given by
$\left| {{S}_{3\phi }} \right|=\sqrt{3}\left| {{V}_{LL}} \right|\left| {{I}_{L}} \right|=\sqrt{3}\left( 460 \right)\left( 61 \right)=48.6kVA$
This 50 HP motor requires 48.6 kVA when operating at rated conditions.
Note:
A rule of thumb when figuring motor loads is that each horsepower of the motor load will require about 1 kVA of apparent power.
RPM
The RPM rating is the expected speed of the machine when operating at full load and rated voltage and frequency. Changes to the operating voltage or frequency will change the operating speed. The induction motor slows down slightly as the load is increased.
Once the horsepower and rated speed are known, it is possible to calculate the rated torque of the motor. The torque is important because many loads are sized by the torque they require, rather than the horsepower. Example 2 shows how the rated torque can be found from the rated horse-power and speed.
Example 2
How much torque does the motor of Figure 1 deliver when operating at rated conditions?
Solution
In SI units,
$P={{\omega }_{r}}T$
Where ${{\omega }_{r}}$is the rotational speed of the rotor in rad/s (i.e., ${{\omega }_{r}}=2\pi \times {RPM}/{60}\;$)
So,
${{T}_{rated}}=\frac{{{P}_{rated}}}{{{\omega }_{rated}}}$
Converting the rated power from horsepower to watts yields,
${{P}_{rated}}=50HP\times 746{}^{W}/{}_{HP}=37600W$
${{\omega }_{rated}}=2\pi \times {1765}/{60}\;=184.8{}^{rad}/{}_{s}$
And
${{T}_{rated}}=\frac{37600}{184.8}=201.8N.m$
Frequency
The frequency rating on the nameplate (in hertz) is the frequency at which the motor is designed to operate. The frequency that is applied to the stator winding will determine the synchronous speed of the motor. Some motors may be designed to operate at 50 or 60 Hz. but unless that is specifically indicated, always assume the motor will operate only at its rated frequency.
Maximum Ambient Temperature
This rating is the maximum ambient temperature at which the meter can provide rated load without exceeding its maximum operating temperature (indicated by the allowable temperature rise). The operating temperature is important because the life of the insulation is dramatically affected by temperature.
Insulation Type
The three main classes of insulation for electric motors are B. F. and H. The insulation used in the motor determines the maximum operating temperature for the motor.
NEMA Nominal Efficiency
Due to variances in the manufacturing process, not all motors of the same type will have the same efficiency. To account for such variations. NEMA Standard MG-l provides definitions of the nominal and minimum efficiency.
NEMA Design Letter
NEMA defines five standard designs for poly-phase induction motors, each of which has different characteristics. The five types are A, B, C, D, and E.
NEMA Code
This letter provides a means to calculate the locked-rotor RVA of the motor and thus the starting current. Locked rotor means the rotor is at standstill. This occurs normally when the motor is to be started; however, if the motor were to stall due to an overload, it would draw the locked-rotor current. The codes are shown in Table 2. Example 3 shows how the kVA code can be used to estimate the morons starting current.
Code letter kVA/HP at locked rotor A 0-3.14 B 3.15-3.54 C 3.55-3.99 D 4.00-4.49 E 4.50-4.99 F 5.0-5.59 G 5.6-6.29 H 7.1-7.99 J 7.1-7.99 K 8.0-8.99 L 9.00-9.99 M 10.0-11.19 N 11.20-12.49 P 12.5-13.99 R 14.00-15.99 S 16.0-17.99 T 18.0-19.99 U 20.0-22.39 V 22.4 and up
Example 3
What is the expected starting line current for the motor of figure 1 when operating from a line to line voltage of 460 V?
Solution
From the nameplate, the NEMA code letter is G. From Table 2.For code G the expected locked rotor kVA is 5.6 to 6.29 kVA/HP. assuming the worst case, the locked-rotor (starting) apparent power can be calculated:
${{S}_{LR}}=Rated\text{ }HP\times \text{ }6.29kVA/HP$
Where SLR is the apparent power input to the motor at locked rotor.
${{S}_{LR}}=50\text{ }HP\times \text{ }6.29kVA/HP=314.5kVA$
Thus, the locked-rotor (starting) current is given by,
${{I}_{LR}}=\frac{{{S}_{LR}}}{\sqrt{3}\times \left| {{V}_{LL}} \right|}=\frac{314.5}{\sqrt{3}\times 0.460}=394.7A$
Note
The starting or locked-rotor current is about 6.5 times the rated current. Depending on the design, today’s induction motors draw six to eight times rated current when starting.
Time Rating
This motor is rated for continuous operation, meaning it can run 24 hours per day. Any motor that runs longer than 60 minutes is considered to be continuously operating. For specials purpose applications, there are a short-time rating of 5, 15, 30, and 60 minutes.
Service Factor
The service factor is a multiplier used to determine the maximum allowable continuous load of the motor. For example, this machine has a service factor of 1.25. Thus it could provide 1.25*50 HP, or 62.5 HP, on a continuous basis.
Note:
When operated above the nameplate rating, the efficiency, current, and speed will differ from the nameplate values. Such operation may reduce the life of the motor insulation. Care must be taken to ensure that all applicable conditions, such as ambient temperature and altitude, are observed when utilizing the service factor.
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2018-10-20 05:15:29
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http://research.omicsgroup.org/index.php/Nachbin%27s_theorem
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# Nachbin's theorem
"Exponential type" redirects here. For exponential types in type theory and programming languages, see Function type.
In mathematics, in the area of complex analysis, Nachbin's theorem (named after Leopoldo Nachbin) is commonly used to establish a bound on the growth rates for an analytic function. This article will provide a brief review of growth rates, including the idea of a function of exponential type. Classification of growth rates based on type help provide a finer tool than big O or Landau notation, since a number of theorems about the analytic structure of the bounded function and its integral transforms can be stated. In particular, Nachbin's theorem may be used to give the domain of convergence of the generalized Borel transform, given below.
## Exponential type
Main article: Exponential type
A function f(z) defined on the complex plane is said to be of exponential type if there exist constants M and τ such that
$|f(re^{i\theta})|\le Me^{\tau r}$
in the limit of $r\to\infty$. Here, the complex variable z was written as $z=re^{i\theta}$ to emphasize that the limit must hold in all directions θ. Letting τ stand for the infimum of all such τ, one then says that the function f is of exponential type τ.
For example, let $f(z)=\sin(\pi z)$. Then one says that $\sin(\pi z)$ is of exponential type π, since π is the smallest number that bounds the growth of $\sin(\pi z)$ along the imaginary axis. So, for this example, Carlson's theorem cannot apply, as it requires functions of exponential type less than π.
## Ψ type
Bounding may be defined for other functions besides the exponential function. In general, a function $\Psi(t)$ is a comparison function if it has a series
$\Psi(t)=\sum_{n=0}^\infty \Psi_n t^n$
with $\Psi_n>0$ for all n, and
$\lim_{n\to\infty} \frac{\Psi_{n+1}}{\Psi_n} = 0.$
Comparison functions are necessarily entire, which follows from the ratio test. If $\Psi(t)$ is such a comparison function, one then says that f is of Ψ-type if there exist constants M and τ such that
$\left|f\left(re^{i\theta}\right)\right| \le M\Psi(\tau r)$
as $r\to \infty$. If τ is the infimum of all such τ one says that f is of Ψ-type τ.
Nachbin's theorem states that a function f(z) with the series
$f(z)=\sum_{n=0}^\infty f_n z^n$
is of Ψ-type τ if and only if
$\limsup_{n\to\infty} \left| \frac{f_n}{\Psi_n} \right|^{1/n} = \tau.$
## Borel transform
Nachbin's theorem has immediate applications in Cauchy theorem-like situations, and for integral transforms. For example, the generalized Borel transform is given by
$F(w)=\sum_{n=0}^\infty \frac{f_n}{\Psi_n w^{n+1}}.$
If f is of Ψ-type τ, then the exterior of the domain of convergence of $F(w)$, and all of its singular points, are contained within the disk
$|w| \le \tau.$
Furthermore, one has
$f(z)=\frac{1}{2\pi i} \oint_\gamma \Psi (zw) F(w)\, dw$
where the contour of integration γ encircles the disk $|w| \le \tau$. This generalizes the usual Borel transform for exponential type, where $\Psi(t)=e^t$. The integral form for the generalized Borel transform follows as well. Let $\alpha(t)$ be a function whose first derivative is bounded on the interval $[0,\infty)$, so that
$\frac{1}{\Psi_n} = \int_0^\infty t^n\, d\alpha(t)$
where $d\alpha(t)=\alpha^{\prime}(t)\,dt$. Then the integral form of the generalized Borel transform is
$F(w)=\frac{1}{w} \int_0^\infty f \left(\frac{t}{w}\right) \, d\alpha(t).$
The ordinary Borel transform is regained by setting $\alpha(t)=e^{-t}$. Note that the integral form of the Borel transform is just the Laplace transform.
## Nachbin resummation
Nachbin resummation (generalized Borel transform) can be used to sum divergent series that escape to the usual Borel summation or even to solve (asymptotically) integral equations of the form:
$g(s)=s\int_0^\infty K(st) f(t)\,dt$
where f(t) may or may not be of exponential growth and the kernel K(u) has a Mellin transform. The solution, pointed out by L. Nachbin himself, can be obtained as $f(x)= \sum_{n=0}^\infty \frac{a_n}{M(n+1)}x^n$ with $g(s)= \sum_{n=0}^\infty a_n s^{-n}$ and M(n) is the Mellin transform of K(u). an example of this is the Gram series $\pi (x) \approx \sum_{n=1}^{\infty} \frac{\log^{n}(x)}{n\cdot n!\zeta (n+1)}.$
## Fréchet space
Collections of functions of exponential type $\tau$ can form a complete uniform space, namely a Fréchet space, by the topology induced by the countable family of norms
$\|f\|_{n} = \sup_{z \in \mathbb{C}} \exp \left[-\left(\tau + \frac{1}{n}\right)|z|\right]|f(z)|.$
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2019-04-18 14:20:44
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https://www.gamedev.net/forums/topic/475145-strips-planning-implementation/?page=1
|
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# STRIPS planning implementation?
## 21 posts in this topic
Has anyone seen any code that I can wade through that has a data structure and implementation of the STRIPS planning algorithm? I would like to see how it has been done before rather than trying to build it from scratch.
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Was there a particular reason you wanted to use STRIPS? It's not a good planning language, particularly in dynamic domains.
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Just wanted to play with it to see what made it tick. Over at Alex's site (AIGameDev.com) there is an article about the AI for F.E.A.R. Jeff Orkin says in a doc file that the planning algorithm that they used most resembled STRIPS. He goes on to describe it somewhat. It was interesting to ponder the possibilities.
Do you have suggestions that would be better? I have an interest in goal-based planning at the moment.
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I dont know any public implementations, tho suspects many universities have, so thats somewhere you can look. I went on to re-create a FEAR-like planning AI last winter (for fun). The planner itself is pretty trivial, its a simple backtracking. Just re-use your favorite A* implementation.
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How did you go about setting up the preconditions, etc.? What sort of data structure? Anything I can look at?
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You could just iterate through every action, and then for each action evaluate its preconditions to determine if they are all true or not. Anything beyond that is an optimization.
A potentially more efficient method that works in general would be to have a list of actions associated with each precondition. When a precondition changes from true to false or vice versa, update all of the associated actions. If the last precondition for an action becomes true, add it to the list of valid actions. If a precondition for a previously valid action becomes false, remove it from the list of valid actions. Some domain knowledge might also allow for some optimizations here.
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The Fear planner absolutely works, but I've found it overkill for many problems.
Basically, designers generally know what they want to see behaviorally. Most STRIPS style AI planning work I've done has basically been using the system to emulate what a HTN planner represents more explicitly.
You might want to look at simple HTN planning or Halo 3 style trees instead. Both seem like better ways to directly encode design requirements.
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There isnt much to it, most of it is in two classes:
A "State" class contains a list of world variables in the form of a <name, value> pair, a list of the conditions to satisfy (in the same form), meaning to be satisfied the variable with the same name as the precondition must have a certain value, and a pointer to the previous State.
Then I have an "Action" class that contains:
- a list of conditions the action might affect (for optimisation purpose, I'll only consider an action if it might affect one of the current conditions).
- a list of conditions that need to be solved to apply this action
- a rule that tells if its a good idea to choose this action (variable cost for the action, a small improvement).
- an operator that will modify the current variables and conditions to generate a new state.
I load the actions from xml (bit o overkill, I wanted to try some stuff), it looks a like this:
<ACTION Name = "PickupWeapon"> <ARG Name = "Weapon" /> <CHANGE> <CONDITION Name = "HaveWeapon" /> </CHANGE> <REQUIRE> <CONDITION Name = "AtObject" > <ARG Name = "Weapon" /> </CONDITION> </REQUIRE> <RULE> <PREDICATE Name = "Always" /> </RULE></ACTION>
to define the "PickupWeapon(weapon)" Action.
To plan, I just gather the original state, and apply A* using the actions as the edges of the graph... Go ahead, you'll be surprised how easy a simple planner like that is. 90% of the work is to link the planner to the game, like gathering the state variables or executing the plan itself.
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One thing that I am interested in is goal-based planning. That is, backward search from the goal rather than forward search from the current state.
The concepts I understand, but I'm just looking for different ways of implementing the data structure and code. I will be doing some more research on it here soon.
Any of you folks going to GDC?
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A simple implementation of that is easy.
That definition of goal based planning should be pretty easy. Just encode your goal in a search space state. Have a method to determine distance between your initial state and the goal state. Do a search (either forward or backward) between the states using the actions as operators.
At least in GOAP, goals were special because they were prioritized outside of planning system. Basically, the AI decided what to do via goals and then how to do it via the planner.
The search direction itself shouldn't matter very much if you use an admissible search algorithm. At least behaviorally - the size of the space you search may vary.
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If you're interested in understanding STRIPS style planners then yes, it's best just to implement your own. I'm not aware of any publicly available at the moment, mostly because those working in the planning community have been focusing on post STRIPS-style planners for the past decade or so.
Just on the terminology, which might help your lit review... searching from the goal backwards is generally called "regression planning", as opposed to progression or forward planning (planning from the starting state).
Cheers,
Timkin
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Quote:
Original post by BrianLThe search direction itself shouldn't matter very much if you use an admissible search algorithm. At least behaviorally - the size of the space you search may vary.
May vary a *lot*, since you have to consider all actions at every step if you search forward, instead of just those actions that solve a current goal when searching backward, no? Friends of mine always suggest to do a bi-directional search and pick the first one that finds a plan, or merge the plans if they meet, but I found that the regression search / backtracking only is always faster using the optimization described above. It limits the plan-space a bit of course...
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Wouldn't an informed forward search (ie A*) address the 'only consider actions which solve towards the goal' issue?
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Quote:
Original post by BrianLWouldn't an informed forward search (ie A*) address the 'only consider actions which solve towards the goal' issue?
Since the possibility space is not geometric, there isn't a really good heuristic to use that would tell the algorithm that you are "towards the goal".
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Yeah. For the heuristic you would have to look at actions that solve subgoals of actions that solve subgoal of actions that solve... etc one of the current goal. You would have to plan in order to plan. A Forward search is much more of a "blind search", in the action space at least.
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If performance is a concern, you should be thinking about hierarchical planners anyway rather than forwards/backwards A* :-)
Alex
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Quote:
Original post by SteadtlerYeah. For the heuristic you would have to look at actions that solve subgoals of actions that solve subgoal of actions that solve... etc one of the current goal. You would have to plan in order to plan. A Forward search is much more of a "blind search", in the action space at least.
I don't remember where I heard it, but someone suggested an analogy of finding an ant on the tip of a leaf of a tree by starting at the trunk. The any has a better chance of finding you at the trunk than you have of finding the ant. That doesn't completely fit, but it's something to think about.
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Quote:
Original post by alexjcIf performance is a concern, you should be thinking about hierarchical planners anyway rather than forwards/backwards A* :-)Alex
Yes and no. You can only get a hierarchy so far up. At some point, you still have to determine what components to use and what order to put them in. I agree with the premise, however, that using only the most granular level actions.
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Quote:
Original post by alexjcIf performance is a concern, you should be thinking about hierarchical planners anyway rather than forwards/backwards A* :-)Alex
We're not talking about simple performance tradeoff here. Forward search is exponentially most costly than backtracking with that optimization, so its a matter of even being able to find a plan at all.
I think the loss of expressiveness is also far greater from hierarchical planners to regression planners than from regression planners to forward planners.
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There is implementation of GOAP like algo, with source.
report - http://www.edmundlong.com/edsWiki/
source - http://www.edmundlong.com/GOAPSystemMasters.rar
Its the 2d deathmatch bot game, each team of bots has own AI from: FSM, GOAP, team FSM, team GOAP.
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Whoa... ask and you shall receive! And from possibly the most well-known STRIPS-style usage in recent years. Very nice! (Installing F.E.A.R. in another window as I type this.)
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Heh, actually my article was written after I read Ed Long's blog about his thesis (mentioned here), and he mentioned the F.E.A.R. source.
I didn't follow up as I figured you read the original article anyway ;-)
Alex
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2017-06-24 05:23:53
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https://electronics.stackexchange.com/questions/506146/flashlight-with-2-switch-without-ic
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# Flashlight with 2 Switch without IC
I make a simple circuit with the use of Arduino and it works well, having an IC like ATTiny is good but i have a small space on my PVC Tube for DC-DC Step Down.
My goal is to make a flashlight that have a secondary switch
1. Switch 1 (Toggle Switch) - Turn Lights On or Off
2. Switch 2 (Momentary Switch) - Turn Lights On even if switch 1 is Off, Turn Lights Off even the switch 1 is On.
I have this code
void setup()
{
Serial.begin(9600);
pinMode(2,INPUT_PULLUP); //
pinMode(13, OUTPUT);
pinMode(3,INPUT_PULLUP);
}
void loop()
{
//Serial.println(sensorValue3, DEC);
if (sensorValue == HIGH){
//OFF via toggle
if (sensorValue3 == LOW){
// ON via button if OFF via toggle
digitalWrite(13, HIGH);
}else{
// OFF
digitalWrite(13, LOW);
}
}else if(sensorValue == LOW){
//ON via toggle
if (sensorValue3 == LOW){
digitalWrite(13, LOW);
}else{
digitalWrite(13, HIGH);
}
}
}
Ive seen an Oscillator with a relay but cant make it work.
• this code is unnecessarily complex ... if (sensorValue3 == LOW){ digitalWrite(13, LOW); }else{ digitalWrite(13, HIGH); } ........... this produces the same result digitalWrite(13, sensorValue3 ); – jsotola Jun 18 at 3:38
• A relay would really not be suitable for this at all, and there's no way you're going to find one smaller than an MCU. Consider the actual range of packages in which the ATtiny is available - the SOIC-8 is one you should be able to work with by hand, or for the lowest resource parts the SOT23-6. – Chris Stratton Jun 18 at 3:58
• I'm pretty sure the code is looping so fast that you cant tell it's working. Press the momentary button too fast and it wont register (to your eyes). – Passerby Jun 18 at 4:00
You can do this with a single XOR gate.
As follows:
simulate this circuit – Schematic created using CircuitLab
I would build such a circuit using PNP, resistor and 2 switches - momentary and latching. And no MCU. No OFF power consumption.
Why not MOSFET? I don't know at what voltage your flashlight is, if your light's voltage is high enough to make sure mosfet will open, use a MOSFET. PNP is good as long as you are at AA voltage (down to 1 volt when pretty drained).
Sorry I'm a lazy pig and sketched it in paint in 2 minutes, but here it is
We have low minimum voltage for the circuit to work. Of course you may want to tweak resistors depending on your PNP's gain and required load current you want to provide. And if you work with 3V+, you can use a MOSFET instead (look for logic level mosfets). This circuit has very-very-very close-to-zero power off current consumption, comparable or even less than MCU in sleep mode, again, depends on specific parts, but it's possible to get it down to neglibible nanoamps if you find a good transistor. Also, let's not forget BJT will drop some 0.2V, MOSFET with milliohm-scale Rdson will drop nothing you could ever notice.
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2020-08-05 21:25:54
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|
https://www.gradesaver.com/textbooks/science/physics/physics-for-scientists-and-engineers-a-strategic-approach-with-modern-physics-4th-edition/chapter-15-oscillations-exercises-and-problems-page-415/5
|
## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)
(a) The amplitude is 20 cm. (b) $f = 0.125~Hz$ (c) $\phi = \frac{\pi}{3}$
(a) The motion moves between -20 cm and 20 cm. Therefore, the amplitude is 20 cm. (b) From the graph, we can see that one cycle is completed in 8 seconds. Therefore, the period $T = 8~s$. We can find the frequency. $f = \frac{1}{T} = \frac{1}{8~s} = 0.125~Hz$ (c) The general equation for the motion is: $x(t) = A~cos(\omega~t+\phi)$ At $t = 0$: $x = A~cos(\phi) = \frac{A}{2}$ $cos(\phi) = \frac{1}{2}$ $\phi = arccos(\frac{1}{2})$ $\phi = \frac{\pi}{3}, \frac{5\pi}{3}$ Since the basic cos-curve is shifted to the left $\frac{\pi}{3}$, the phase constant is $\phi = \frac{\pi}{3}$.
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2018-06-23 23:54:43
|
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|
http://openstudy.com/updates/5108013de4b069b605a2875b
|
## anonymous 3 years ago Please click on this to see the problem. Medal for best response.
1. anonymous
|dw:1359479119655:dw|
2. anonymous
I got this far, but now I am stuck. |dw:1359479735391:dw|
3. anonymous
You can now multiply with the reciprocal value:$\frac{ 2\sqrt{2} }{ 15 }\cdot \frac{ 9 }{ \sqrt{2} }$
4. anonymous
5. anonymous
18/15 = 6/5
6. anonymous
The square roots cancel, so it's 18/15=6/5, just as @CanadianAsian said!
7. anonymous
Thank you everyone! I greatly appreciate all the help. :)
8. anonymous
YW!
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2016-05-30 05:07:31
|
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|
https://lmpa.univ-littoral.fr/index.php?page_id=13&personne=15
|
L.M.P.A
Laboratoire de Mathématiques Pures et Appliquées
Joseph Liouville
## 40 publications
1. A variational approach to relativistic fluid balls ( and ), In Calc. Var. Partial Differential Equations, volume 60, .
2. Stability Matters for Reaction–Diffusion–Equations on Metric Graphs Under the Anti-Kirchhoff Vertex Condition ( and ), Chapter in (Atay, Fatihcan M., Kurasov, Pavel B., Mugnolo, Delio, eds.), Springer International Publishing, .
3. Harmonic functions on metric graphs under the anti-Kirchhoff law (), In Results Math., volume 74, .
4. Stability implies constancy for fully autonomous reaction-diffusion-equations on finite metric graphs (), In Netw. Heterog. Media, volume 13, .
5. Qualitative results for parabolic equations involving the $p$-Laplacian under dynamical boundary conditions ( and ), In North-West. Eur. J. Math., volume 4, .
6. Instability of stationary solutions of reaction-diffusion-equations on graphs ( and ), In Results Math., volume 68, .
7. Instability of stationary solutions of evolution equations on graphs under dynamical node transition ( and ), Chapter in Mathematical technology of networks, Springer, Cham, volume 128, .
8. The spectrum of the Hilbert space valued second derivative with general self-adjoint boundary conditions ( and ), In Linear Algebra Appl., volume 439, .
9. Eigenvalue multiplicities for second order elliptic operators on networks ( and ), In Oper. Matrices, volume 7, .
10. Some remarks on the eigenvalue multiplicities of the Laplacian on infinite locally finite trees (, and ), In Results Math., volume 63, .
11. Growth order and blow up points for the parabolic Burgers' equation under dynamical boundary conditions (, and ), In Discrete Contin. Dyn. Syst. Ser. S, volume 6, .
12. Eigenvalue asymptotics for second-order elliptic operators on networks ( and ), In Asymptot. Anal., volume 77, .
13. The eigenvalues of the Laplacian on locally finite networks under generalized node transition ( and ), In Results Math., volume 54, .
14. Isospectral infinite graphs and networks and infinite eigenvalue multiplicities ( and ), In Netw. Heterog. Media, volume 4, .
15. An index theory for uniformly locally finite graphs (), In Linear Algebra Appl., volume 431, .
16. A characterization of even doubly-stochastic matrices ( and ), In Discrete Math., volume 309, .
17. Even and odd diagonals in doubly stochastic matrices ( and ), In Discrete Math., volume 308, .
18. Eigenvalue asymptotics under a nondissipative eigenvalue dependent boundary condition for second-order elliptic operators ( and ), Chapter in Functional analysis and evolution equations, Birkhäuser, Basel, .
19. Positivity and anti-maximum principles for elliptic operators with mixed boundary conditions (, and ), In J. Eur. Math. Soc. (JEMS), volume 10, .
20. Blow up for some nonlinear parabolic problems with convection under dynamical boundary conditions ( and ), In Discrete Contin. Dyn. Syst., .
21. Parabolic problems with dynamical boundary conditions: eigenvalue expansions and blow up (, and ), In Atti Accad. Naz. Lincei Rend. Lincei Mat. Appl., volume 17, .
22. The eigenvalues of the Laplacian on locally finite networks ( and ), In Results Math., volume 47, .
23. Spectral asymptotics for the Laplacian under an eigenvalue dependent boundary condition ( and ), In Bull. Belg. Math. Soc. Simon Stevin, volume 12, . [url]
24. Harmonic functions on locally finite networks ( and ), In Results Math., volume 45, .
25. Blow up for reaction diffusion equations under dynamical boundary conditions ( and ), In Comm. Partial Differential Equations, volume 28, .
26. Can one hear the shape of a network? (), Chapter in Partial differential equations on multistructures (Luminy, 1999), Dekker, New York, volume 219, .
27. A maximum principle for fully nonlinear parabolic equations with time degeneracy (), Chapter in Evolution equations and their applications in physical and life sciences (Bad Herrenalb, 1998), Dekker, New York, volume 215, .
28. A qualitative theory for parabolic problems under dynamical boundary conditions ( and ), In J. Inequal. Appl., volume 5, .
29. Nonlinear degenerate elliptic equations and axially symmetric problems ( and ), In Calc. Var. Partial Differential Equations, volume 7, .
30. Dynamical interface transition in ramified media with diffusion ( and ), In Comm. Partial Differential Equations, volume 21, .
31. A note on the Perron root of a weighted $t$th order mean of nonnegative matrices ( and ), In Linear and Multilinear Algebra, volume 40, .
32. Front propagation in diffusion problems on trees (), Chapter in Calculus of variations, applications and computations (Pont-à-Mousson, 1994), Longman Sci. Tech., Harlow, volume 326, .
33. The index of a periodic graph (), In Results Math., volume 25, .
34. Nonlinear and dynamical node transition in network diffusion problems (), Chapter in Evolution equations, control theory, and biomathematics (Han sur Lesse, 1991), Dekker, New York, volume 155, .
35. An existence result for semilinear parabolic network equations with dynamical node conditions (), Chapter in Progress in partial differential equations: elliptic and parabolic problems (Pont-à-Mousson, 1991), Longman Sci. Tech., Harlow, volume 266, .
36. A maximum principle for semilinear parabolic network equations (), Chapter in Differential equations with applications in biology, physics, and engineering (Leibnitz, 1989), Dekker, New York, volume 133, .
37. Sturm-Liouville eigenvalue problems on networks (), In Math. Methods Appl. Sci., volume 10, .
38. Classical solvability of linear parabolic equations on networks (), In J. Differential Equations, volume 72, .
39. A characteristic equation associated to an eigenvalue problem on $c^2$-networks (), In Linear Algebra Appl., volume 71, .
40. On a theorem of L. Mirsky on even doubly-stochastic matrices (), In Discrete Math., volume 55, .
# Informations
Stochastic Geometry Days du 15 au 19 novembre.
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2021-10-24 22:58:29
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http://wikidobia.info/wikidobia/doku.php?id=all&do=search&q=all
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# Search
You can find the results of your search below.
## Fulltext results:
mark_twain @air
> <P>There are ten parts of speech, and they are all troublesome. An average sentence, in a German ne... y; it occupies a quarter of a column; it contains all the ten parts of speech -- not in regular order, ... or parentheses, making pens within pens: finally, all the parentheses and reparentheses are massed toge... ave to go to press without getting to the verb at all. Of course, then, the reader is left in a very e
quotes @dob
die." Granddad Song Xmas 2019. Clancy brother "All of old. Nothing else ever. Ever tried. Ever faile... entury icon. I read somewhere that 77 percent of all the mentally ill live in poverty. Actually, I'm m... ays off 10,000) Coffee, mescaline, strychnine, - all alkaloids. "Why should we be in such desperate h... kel in the casinos because he knew the house had all the percentages. And beneath his pessimism, his
syntax @wiki
to be as readable as possible. This page contains all possible syntax you may use when editing the page... texts. Of course you can **__//''combine''//__** all these. DokuWiki supports **bold**, //italic//,... exts. Of course you can **__//''combine''//__** all these. You can use <sub>subscript</sub> and <sup... open a Windows Share. To remove this warning (for all users), put the following line in ''conf/lang/en/
the.parting.glass @dob
====== The Parting Glass ====== Of all the money that e’er I had, I spent it in good company. \\ And of all the harm that e’er I've done, alas it was to none but me. \\ And all I've done for want of wit, to memory now I can’t ... the parting glass. Goodnight and joy be with you all. \\ Of all the comrades that e’er I had, they're
veg @eat
5 Cup Cider vinager 3 Table spoons Floor Grate all cheese. Cut garlic in 2 and rub arouund inside of... 1 teaspoon ground cinnamon 2 tablespoons all-purpose flour 1 teaspoon salt and pepper (to ... espoons canola or vegetable oil 2 tablespoons all-purpose flour 2 cups chicken stock 1 teas... ham. ===== Regensburg Ratatouille ===== Chop up all the vegetables in big pieces so they can retain t
sweet @eat
blespoon sugar \\ 1 pinch turmeric** \\ \\ Blend all the ingredients. Fill in glasses and cool. \\ Mak... \ 1 pinch cayenne pepper or chilli** \\ \\ Blend all the ingredients until foamy. Then pour on top of ... minutes ;-)...). - Prepare the icing combining all the ingredients in a bowl and beat until smooth. ... nmould the cake from the tin and spread the icing all over the top and sides of the cake.^_^ //Damiana
bayesian_statistics @blog
have 5 children. What is the probability of them all being girls? These are 5 independent Bernouli tri... ref{condprob})$for independent trials then: $$P_{All Girls} = 0,5^5 = 3.1\%$$ While if you already have$4$children who are all girls... then the chances of having$5\$ girls bec... or "model evidence". This factor is the same for all possible hypotheses being considered (as is evide
pork @eat:meat
easoning for pork tenderloin - Pierce pork loin all over with a fork and rub with 1 Tbsp oil than spr... ill work). Once oil is hot, add pork and brown on all sides (5 minutes total). - Place in the oven a... sor and pulse to a coarse paste. Rub this mixture all over the ham. Cover the pan tightly with foil and... \ Cut slices into the pork and rub the salt and herbs into the meat. Marinate in all the ingredients.
gold @blog
of Armageddon when modern society breaks down and all Fiat currency becomes worthless seems dubious to ... lly use (food, clothing, shelter etc..). First of all, you have to prove in some way that the material ... u. In our current fully functional society, where all services and systems are running as normal, this
the_secret_life_of_walter_mitty @dob
She put her mirror back into her bag. "We've been all through that," she said, getting out of the car. ... sergeant. Mitty finished one last brandy. "After all," he said softly, "what isn't?" The pounding of t... Something struck his shoulder. "I've been looking all over this hotel for you," said Mrs. Mitty. "Why d
commerce_and_politics_have_always_been_sources_of_dramatic_change_-_why_is_this_always_surprising @blog
tical landscape, a few short years later this can all be swept away. This is sometimes for good - but w... quickly). Why is this so hard to imagine despite all evidence to the contrary? Companies typically sur... ically in Europe? Consider the recent case of the all dominating Finnish mobile phone maker Nokia that
o.com/Ardent-Mills-Harvest-Hotel-%2526-Restaurant-All-Purpose-Flour%2c-25-lbs.product.100381629.html)) ... in. **Bake: 30 min.** __Ingredients__\\ **2 cups all-purpose flour\\ 1 teaspoon baking powder\\ 1/2 te... mixing bowl combine egg white and buttermilk. Add all at once to dry mixture. Stir just until moistened
eat @eat
Thai chillies, 2 capsicums, 1 big yellow onion ( all of these Julian cut - Cut long) 1 tsp aginamoto, ... 1 teaspoon kosher salt Preparation Mix all the ingredients together in a bowl. Season to tas
body_mass_index @calculate
.irishtimes.com/life-and-style/health-family/it-s-all-downhill-from-38-why-ageing-is-not-just-for-the-o... kohol/)) Rough calculation with assumptions * All alcohol instantly absorbed into blood and no time
could_democracy_and_our_market_system_save_us_from_ourselves @blog
chnologists do their work for the greater good of all humanity or merely because they are driven to do ... ple reject the dogma of technological progress at all costs. (Trump's "post fact" victory in the US as
dob @dob
collecting_my_notes_as_a_blog @blog
notes @dob
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fish @eat
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dokuwiki @wiki
duck @eat:meat
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2021-09-26 15:59:10
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https://www.math.uic.edu/graduate/seminars/view_seminar?id=5149
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## Special Colloquium
John Wilmes
Georgia Institute of Technology
The Complexity of Learning Neural Networks
Abstract: The empirical successes of deep learning'' currently lack rigorous theoretical explanation. As a first step, we ask whether data generated by neural networks with a single hidden layer, smooth activation functions and benign input distributions can be learned efficiently. We give a surprisingly general polynomial-time analysis of gradient descent when the hidden layer uses unbiased sigmoid gates, exploiting new connections we make with tools from spherical harmonics. However, when the generating network uses arbitrary biases, the problem appears intractable. We construct a family of simple neural networks that is hard to learn in the sense that any statistical query algorithm (including all known variants of stochastic gradient descent with any loss function, for any model architecture) needs an exponential number of queries on data labeled by such a network even using tolerance inversely proportional to the input dimensionality. Joint work with Le Song, Santosh Vempala, and Bo Xie.
Friday January 19, 2018 at 3:00 PM in SEO 636
UIC LAS MSCS > graduate studies > seminars >
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2019-02-23 00:16:59
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https://cs.stackexchange.com/questions/123312/clrs-closest-pair-l-m-distances
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# CLRS closest-pair $L_m$ distances
I am studying algorithms and datastructures, and in CLRS chapter 33.4, the exercise 33.4-4 states the following:
We can define the distance between two points in ways other than euclidean. In the plan, the $$L_m$$ distance between points $$p_1$$ and $$p_2$$ are given by the expression $$(|x_1-x_2|^m+|y_1-y_2|^m)^{1/m}$$. euclidean distance, there, is $$L_2$$-distance. Modify the closest pair algorithm to use the $$L_1$$-distance, which is also known as the Manhattan distance.
I understand its an easy modification, however when I've been comparing my intuition with that of solutions provided online (e.g. https://sites.math.rutgers.edu/~ajl213/CLRS/CLRS.html), all I can find is that they all point to the number of points in the $$\delta$$ x $$2\delta$$ is increased to 10 (from 8), adding one point to the middle of both squares formed by the rectangle (here marked with red below):
Can anyone explain to my why this is the case? Intuitively I cannot see why it should be the case that the number of points in the rectange should increase, when switching to another $$L_m$$-distance?
The Manhatten distance $$\geq$$ Euclidean distance from what I can see in all instances, but why should this infer, an increase in points found in the $$\delta$$ x $$2\delta$$ space.
What am I missing?
• I'm not sure what part you want to be explained. Your figure here shows that it is possible to fill the $\delta\times\delta$ square with 5 points of Manhattan distance at least $\delta$. Do you want to see why this isn't possible under the Euclidean distance? (and that therefore the maximum points per square is 4) Or do you want to know the number of points you can 'fit' in a square under other $L_m$ distances? – Discrete lizard Mar 29 at 16:30
• @Discretelizard I do not understand why it is the case that there could be 5 points with manhattan distance, and only 4 points with euclidean distance. – NewDev90 Mar 29 at 18:18
• I don't know how to answer "why" questions. You can see from the figure that the statement is true. I'm not sure what more there is to say. In what way is that not a sufficient explanation of why? Why do you get different numbers? Because they are different metrics. What more are you expecting? It's hard for me to tell what kind of answer you are hoping for, or what the concrete problem you are trying to solve is. Maybe if you can give us more context we'll be in a better position to help. And welcome to CS.SE! – D.W. Mar 29 at 18:27
One approach to see what is going on here is via the notion of an open ball under a given distance. Given a distance $$d$$, the open ball of radius $$r$$ around a point $$p$$ is given by $$B_r(p):=\{q\mid d(p,q), i.e. the set of all points with distance less than $$r$$ to the center of the ball, $$p$$. In the plane and under the Euclidean distance, a ball is an open disk (a disk with the boundary removed) of radius $$r$$.
Since $$\delta$$ is the minimum distance between any pair of points in the left half, any pair of points in the left $$\delta\times\delta$$ square has a distance of at least $$\delta$$. This is equivalent to saying that the open ball of radius $$r$$ centered at a point $$p\in P_L$$ contains no other point $$q\in P_L$$. So, the maximum number of points from $$P_L$$ that can be in the $$\delta\times \delta$$ square is the largest number of points you can 'fit' into the square such that the ball surrounding each of those points contains no other point from $$P_L$$.
An open ball under the Euclidean distance is an open disk. If you place four points at the corners of the $$\delta\times \delta$$ square, the open disks of radius $$\delta$$ centered at those points together cover the entire square, so there is no position in the square to place for a fifth point.
However, under the Manhattan distance, an open ball of radius $$\delta$$ is an open square with diagonal length $$\delta$$, rotated by $$45$$ degrees (an "open diamond"). If we place such an open diamond at each corner of the $$\delta\times \delta$$ square, then there is exactly one position in the square not covered by an open diamond: the center of the square. This is the location where you can place a fifth point under the Manhattan distance.
As for the $$L_m$$-distance with $$m>2$$, note that the balls under these metrics all strictly contain the ball under the $$L_2$$-distance. I'll leave it up to you to use this fact to determine the maximum amount of points that can be in the $$\delta\times\delta$$-square.
• This makes a lot of sense to me thanks. I think what I struggled with was the intuition of why it was the case, that there should be different amount of points in the $\delta \times \delta$ square. – NewDev90 Mar 30 at 6:56
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2020-10-22 07:21:43
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https://scirate.com/arxiv/cond-mat.supr-con
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# Superconductivity (cond-mat.supr-con)
• Superconducting detectors are now well-established tools for low-light optics, and in particular quantum optics, boasting high-efficiency, fast response and low noise. Similarly, lithium niobate is an important platform for integrated optics given its high second-order nonlinearity, used for high-speed electro-optic modulation and polarization conversion, as well as frequency conversion and sources of quantum light. Combining these technologies addresses the requirements for a single platform capable of generating, manipulating and measuring quantum light in many degrees of freedom, in a compact and potentially scalable manner. We will report on progress integrating tungsten transition-edge sensors (TESs) and amorphous tungsten silicide superconducting nanowire single-photon detectors (SNSPDs) on titanium in-diffused lithium niobate waveguides. The travelling-wave design couples the evanescent field from the waveguides into the superconducting absorber. We will report on simulations and measurements of the absorption, which we can characterize at room temperature prior to cooling down the devices. Independently, we show how the detectors respond to flood illumination, normally incident on the devices, demonstrating their functionality.
• We address the question about the origin of the $\frac12 \frac{e^2}{h}$ conductance plateau observed in a recent experiment on an integer quantum Hall (IQH) film covered by a superconducting (SC) film. Since 1-dimensional (1D) chiral Majorana fermions can give rise to the half quantized plateau, such a plateau was regarded as a smoking-gun evidence for the chiral Majorana fermions. However, in this paper we give another mechanism for the $\frac12 \frac{e^2}{h}$ conductance plateau. We find the $\frac12 \frac{e^2}{h}$ conductance plateau to be a general feature of a good electric contact between the IQH film and SC film, and cannot distinguish the existence or non-existence of 1D chiral Majorana fermions.
• We report on the fabrication and electrical transport properties of superconducting junctions made of \beta-Ag$_{2}$Se topological insulator (TI) nanowires in contact with Al superconducting electrodes. The temperature dependence of the critical current indicates that the superconducting junction belongs to a short and diffusive junction regime. As a characteristic feature of the narrow junction, the critical current decreases monotonously with increasing magnetic field. The stochastic distribution of the switching current exhibits the macroscopic quantum tunneling behavior, which is robust up to T = 0.8 K. Our observations indicate that the TI nanowire-based Josephson junctions can be a promising building block for the development of nanohybrid superconducting quantum bits.
• We report an unexpected positive hydrostatic pressure derivative of the superconducting transition temperature in the doped topological insulator \NBS via $dc$ SQUID magnetometry in pressures up to 0.6 GPa. This result is contrary to reports on the homologues \CBS and \SBS where smooth suppression of $T_c$ is observed. Our results are consistent with recent Ginzburg-Landau theory predictions of a pressure-induced enhancement of $T_c$ in the nematic multicomponent $E_u$ state proposed to explain observations of rotational symmetry breaking in doped Bi$_2$Se$_3$ superconductors.
• Controlling both the amplitude and phase of the quantum order parameter (\psi) in nanostructures is important for next-generation information and communication technologies. The long-range coherence of attractive electrons in superconductors render these materials as a nearly ideal platform for such applications. To-date, control over \psi has remained limited to the macroscopic scale, either by adjusting untunable materials properties, such as film thickness, stoichiometry and homogeneity or by tuning external magnetic fields. Yet, although local tuning of \psi is desired, the lack of electric resistance in superconductors, which may be advantageous for some technologies hinders convenient voltage-bias tuning. Likewise, challenges related to nanoscale fabrication of superconductors encumber local tunability of \psi. Here, we demonstrate local tunability of \psi, obtained by patterning with a single lithography step a Nb nano superconducting quantum interference device (nano-SQUID) that is biased at its nano bridges. Our design helped us reveal also unusual electric characteristics-effective zero inductance, which is promising for quantum technologies and nanoscale magnetic sensing. Finally, we accompanied our experimental results by a semi-classical model, which not only is extending the applicability of our devices, but is also useful for describing planar nano-SQUIDs in general.
• We report the discovery of superconductivity in pressurized CeRhGe3, until now the only remaining non-superconducting member of the isostructural family of non-centrosymmetric heavy-fermion compounds CeTX3 (T = Co, Rh, Ir and X = Si, Ge). Superconductivity appears in CeRhGe3 at a pressure of 19.6 GPa and the transition temperature Tc reaches a maximum value of 1.3 K at 21.5 GPa. This finding provides an opportunity to establish systematic correlations between superconductivity and materials properties within this family. Though ambient-pressure unit-cell volumes and critical pressures for superconductivity vary substantially across the series, all family members reach a maximum Tcmax at a common critical cell volume Vcrit, and Tcmax at Vcrit increases with increasing spin-orbit coupling strength of the d-electrons. These correlations show that substantial Kondo hybridization and spin-orbit coupling favor superconductivity in this family, the latter reflecting the role of broken centro-symmetry.
• A two-dimensional time-reversal symmetric topological superconductors is fully a gapped system possessing a helical Majorana mode on the edges. This helical Majorana edge mode (HMEM), which is a Kramer's pair of two chiral Majorana edge modes in the opposite propagating directions, is robust under the time-reversal symmetry protection. We propose a feasible setup and accessible measurement to provide the preliminary step of the HMEM realization by studying superconducting antiferromagnetic quantum spin Hall insulators. Since this antiferromagnetic topological insulator hosts a helical electron edge mode and preserves effective time-reversal symmetry, which is the combination of time-reversal symmetry and crystalline symmetry, the proximity effect of the conventional $s$-wave superconducting pairing can induce a single HMEM. We further show the HMEM leads to the observation of a $e^2/h$ conductance and this quantized conductance survives in the presence of small symmetry-breaking disorders.
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2017-08-22 20:26:32
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https://ask.sagemath.org/answers/45720/revisions/
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Ask Your Question
# Revision history [back]
1) The problem is that on a=complex(1), we have that a.real and a.imag are attributes.
On b=CC(1) we have that b.real is a method so you need to call b.real() to get the part. Alternatively, you can use b.real_part().
In sage it's become customary to access data via methods, not via attributes. That's a common tool to make it easy to hide implementation details. Strictly speaking it's not necessary in python because @property allows "getters" and "setters" that can be accessed as if they were attributes.
2) see python modules.
3) The methods available on a certain type of objects are the choice of that object. You don't get to change that. You can subclass, however, and then you can implement whatever methods you like in that subclass.
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2019-04-20 20:37:24
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http://mathoverflow.net/revisions/54109/list
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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
3 edited tags
2 added 173 characters in body
Let h* be a multiplicative generalized cohomology theory and $E \rightarrow X$ a real vector bundle.
1. Is it true that, if $E$ is orientable with respect to h*, then it is also orientable with respect to the singular cohomology with coefficients in $h^{0}(pt)$, for $(pt)$ a space with one point?
2. Is it true that, if $E$ is orientable with respect to the singular cohomology with coefficeints in $\mathbb{Z}_{p}$, for $p > 2$, then it is also orientable with respect to the singular cohomology with coefficients in $\mathbb{Z}$?
Added later: I thought to a possible simple answer using weak orientability, actually proving what stated in the last comment. I wrote this as a comment to answer 3.
1
# Orientation and generalized cohomology theories
Let h* be a multiplicative generalized cohomology theory and $E \rightarrow X$ a real vector bundle.
1. Is it true that, if $E$ is orientable with respect to h*, then it is also orientable with respect to the singular cohomology with coefficients in $h^{0}(pt)$, for $(pt)$ a space with one point?
2. Is it true that, if $E$ is orientable with respect to the singular cohomology with coefficeints in $\mathbb{Z}_{p}$, for $p > 2$, then it is also orientable with respect to the singular cohomology with coefficients in $\mathbb{Z}$?
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2013-06-19 19:37:16
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https://math.stackexchange.com/questions/3657264/determinant-of-square-root-of-positive-define-matrix
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# Determinant of Square Root of Positive Define Matrix
Suppose the matrix $$A \in \mathbb{R}^{n\times n}$$ is positive definite symmetric. To begin, I want to investigate if the following equality holds
$$|\det A^{1/2}| = |\det A|^{1/2}.$$
Since $$A$$ is positive definite symmetric, then we can diagonalize it as $$A = V\Lambda V$$, where $$V = V^\intercal$$ and $$VV^\intercal = I$$, i.e. $$V$$ is a symmetric, orthogonal matrix, and $$\Lambda = \text{diag}(\lambda_1,\ldots,\lambda_n)$$ is a diagonal matrix. Then computing the left hand side gives
$$|\det A^{1/2}| = |\det(V\Lambda^{1/2}V)| = |\det(V^2)\det(\Lambda^{1/2})| = |\det{\Lambda}|^{1/2},$$ where $$\det(\Lambda^{1/2}) = \det(\Lambda)^{1/2}$$ comes from the fact that $$\Lambda$$ is diagonal, and you can pull the fraction outside the absolute value since all eigenvalues are positive. Computing the right hand side gives
$$|\det A|^{1/2} = |\det (V\Lambda V)|^{1/2} = |\det\Lambda|^{1/2},$$
so the two are the same. My question is if this is a general result even for non positive definite matrices. I know that for integer powers, the equality holds because of the property $$\det(AB) = \det(A)\det(B)$$, but I'm not sure if it will hold for fractional powers.
• Are you not overcomplicating? If $A^{1/2}$ is any matrix with the property $(A^{1/2})^2=A$ then applying det to both sides you get what you want. – Michal Adamaszek May 3 at 21:00
• @Jan that question is specifically about positive definite matrices – Ben Grossmann May 3 at 21:05
Whenever $$B$$ is a matrix such that $$B^2 = A$$, we must have $$\det(B)^2 = \det(B \cdot B) = \det(B^2).$$ Whenever the expression $$A^{1/2}$$ makes sense, we must have $$(A^{1/2})^2 = A$$. It follows from the above then that $$\det(A^{1/2})^2 = \det(A)$$, so that $$\det(A^{1/2})$$ is one of the two values $$\pm [\det(A)]^{1/2}$$.
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2020-10-31 22:42:08
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http://eprints3.math.sci.hokudai.ac.jp/852/
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# A remark on weak type (1, 1) estimates of Hardy-Littlewood maximal operators on metric spaces acting on dirac measures
Preprint Series # 721
Terasawa, Yutaka A remark on weak type (1, 1) estimates of Hardy-Littlewood maximal operators on metric spaces acting on dirac measures. (2005);
TeX DVI26Kb
## Abstract
We consider weak type $(1,1)$ type estimates of Hardy-Littlewood maximal operators on a compact metric space with Radon measure, and also on a $\sigma$-compact metric space with Radon measure. We show that the analogus results with M. Trinidad Menarguez and F. Sorias' hold in these settings if we impose some conditions on metric measure spaces.
Item Type: Preprint 30, recommender: Ozawa Tohru Hardy-Littlewood maximal operator, weak type (1, 1) estimate, operator norm, dirac measure 42-xx FOURIER ANALYSIS 852
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2018-03-22 12:12:00
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https://vrehab.tungwahcsd.org/d6kuay/how-to-prove-a-function-is-continuous-78f70d
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In the third piece, we need $900 for the first 200 miles and 3(300) = 900 for the next 300 miles. In other words, if your graph has gaps, holes or … But in order to prove the continuity of these functions, we must show that$\lim\limits_{x\to c}f(x)=f(c)$. Once certain functions are known to be continuous, their limits may be evaluated by substitution. Thread starter caffeinemachine; Start date Jul 28, 2012; Jul 28, 2012. Let = tan = sincos is defined for all real number except cos = 0 i.e. | x − c | < δ | f ( x) − f ( c) | < ε. Transcript. is continuous at x = 4 because of the following facts: f(4) exists. Each piece is linear so we know that the individual pieces are continuous. Let’s break this down a bit. A function f is continuous at x = a if and only if If a function f is continuous at x = a then we must have the following three … By "every" value, we mean every one … If you look at the function algebraically, it factors to this: Nothing cancels, but you can still plug in 4 to get. A graph for a function that’s smooth without any holes, jumps, or asymptotes is called continuous. Modules: Definition. f is continuous on B if f is continuous at all points in B. Interior. Continuous Function: A function whose graph can be made on the paper without lifting the pen is known as a Continuous Function. And if a function is continuous in any interval, then we simply call it a continuous function. The function f is continuous at a if and only if f satisfies the following property: ∀ sequences(xn), if lim n → ∞xn = a then lim n → ∞f(xn) = f(a) Theorem 6.2.1 says that in order for f to be continuous, it is necessary and sufficient that any sequence (xn) converging to a must force the sequence (f(xn)) to converge to f(a). The function must exist at an x value (c), which means you can’t have a hole in the function (such as a 0 in the denominator). Constant functions are continuous 2. | f ( x) − f ( y) | ≤ M | x − y |. To prove these functions are continuous at some point, such as the locations where the pieces meet, we need to apply the definition of continuity at a point. https://goo.gl/JQ8NysHow to Prove a Function is Uniformly Continuous. f is continuous at (x0, y0) if lim (x, y) → (x0, y0) f(x, y) = f(x0, y0). Let c be any real number. Another definition of continuity: a function f(x) is continuous at the point x = x_0 if the increment of the function at this point is infinitely small. Definition of a continuous function is: Let A ⊆ R and let f: A → R. Denote c ∈ A. Note that this definition is also implicitly assuming that both f(a)f(a) and limx→af(x)limx→af(x) exist. We can also define a continuous function as a function … You can substitute 4 into this function to get an answer: 8. Step 1: Draw the graph with a pencil to check for the continuity of a function. This gives the sum in the second piece. Prove that C(x) is continuous over its domain. I.e. - [Instructor] What we're going to do in this video is come up with a more rigorous definition for continuity. ii. However, are the pieces continuous at x = 200 and x = 500? if U is not convex and f ∈ C 1, you can integrate: if γ is a smooth curve joining x and y, f ( x) − f ( y) = f ( γ ( 1)) − f ( γ ( 0)) = ∫ 0 1 ( f ∘ γ) ′ ( t) d t ≤ M ∫ 0 1 | | γ ′ ( t) | | d t. In addition, miles over 500 cost 2.5(x-500). The limit of the function as x approaches the value c must exist. If any of the above situations aren’t true, the function is discontinuous at that value for x. Mary Jane Sterling aught algebra, business calculus, geometry, and finite mathematics at Bradley University in Peoria, Illinois for more than 30 years. Continuous functions are precisely those groups of functions that preserve limits, as the next proposition indicates: Proposition 6.2.3: Continuity preserves Limits : If f is continuous at a point c in the domain D, and { x n} is a sequence of points in D converging to c, then f(x) = f(c). All miles over 200 cost 3(x-200). f(x) = f(x_0) + α(x), where α(x) is an infinitesimal for x tending to x_0. You are free to use these ebooks, but not to change them without permission. Please Subscribe here, thank you!!! The mathematical way to say this is that. Sums of continuous functions are continuous 4. The function is continuous on the set X if it is continuous at each point. b. 1. Up until the 19th century, mathematicians largely relied on intuitive … x → c lim f (x) = x → c + lim f (x) = f (c) Taking L.H.L. to apply the theorems about continuous functions; to determine whether a piecewise defined function is continuous; to become aware of problems of determining whether a given function is conti nuous by using graphical techniques. Can someone please help me? Prove that sine function is continuous at every real number. My attempt: We know that the function f: x → R, where x ∈ [ 0, ∞) is defined to be f ( x) = x. If a function is continuous at every value in an interval, then we say that the function is continuous in that interval. x → c − lim f (x) x → c − lim (s i n x) since sin x is defined for every real number. Prove that if f is continuous at x0 ∈ I and f(x0)>μ, then there exist a δ>0 such that f(x)>μ for all x∈ I with |x-x0|<δ. Examples of Proving a Function is Continuous for a Given x Value For example, you can show that the function. The function’s value at c and the limit as x approaches c must be the same. Recall that the definition of the two-sided limit is: Your pre-calculus teacher will tell you that three things have to be true for a function to be continuous at some value c in its domain: f(c) must be defined. Definition 81 Continuous Let a function f(x, y) be defined on an open disk B containing the point (x0, y0). Let f (x) = s i n x. Problem A company transports a freight container according to the schedule below. Since these are all equal, the two pieces must connect and the function is continuous at x = 200. The Applied Calculus and Finite Math ebooks are copyrighted by Pearson Education. To do this, we will need to construct delta-epsilon proofs based on the definition of the limit. We can define continuous using Limits (it helps to read that page first):A function f is continuous when, for every value c in its Domain:f(c) is defined,andlimx→cf(x) = f(c)\"the limit of f(x) as x approaches c equals f(c)\" The limit says: \"as x gets closer and closer to c then f(x) gets closer and closer to f(c)\"And we have to check from both directions:If we get different values from left and right (a \"jump\"), then the limit does not exist! This means that the function is continuous for x > 0 since each piece is continuous and the function is continuous at the edges of each piece. In the second piece, the first 200 miles costs 4.5(200) = 900. Both sides of the equation are 8, so ‘f(x) is continuous at x = 4. However, the denition of continuity is exible enough that there are a wide, and interesting, variety of continuous functions. And the general idea of continuity, we've got an intuitive idea of the past, is that a function is continuous at a point, is if you can draw the graph of that function at that point without picking up your pencil. At x = 500. so the function is also continuous at x = 500. The identity function is continuous. If not continuous, a function is said to be discontinuous. For all other parts of this site,$latex \displaystyle \underset{x\to a}{\mathop{\lim }},f(x)$,$latex \displaystyle \underset{x\to a}{\mathop{\lim }},f(x)=f(a)$, Chapter 9 Intro to Probability Distributions, Creative Commons Attribution 4.0 International License. The second piece corresponds to 200 to 500 miles, The third piece corresponds to miles over 500. Let’s look at each one sided limit at x = 200 and the value of the function at x = 200. She is the author of several For Dummies books, including Algebra Workbook For Dummies, Algebra II For Dummies, and Algebra II Workbook For Dummies. To prove a function is 'not' continuous you just have to show any given two limits are not the same. Answer. If your pencil stays on the paper from the left to right of the entire graph, without lifting the pencil, your function is continuous. We know that A function is continuous at x = c If L.H.L = R.H.L= f(c) i.e. In mathematics, a continuous function is a function that does not have any abrupt changes in value, known as discontinuities.$latex \displaystyle \underset{x\to a}{\mathop{\lim }},f(x)=f(a)$. You need to prove that for any point in the domain of interest (probably the real line for this problem), call it x0, that the limit of f(x) as x-> x0 = f(x0). Your pre-calculus teacher will tell you that three things have to be true for a function to be continuous at some value c in its domain: f(c) must be defined. simply a function with no gaps — a function that you can draw without taking your pencil off the paper How to Determine Whether a Function Is Continuous. In the first section, each mile costs$4.50 so x miles would cost 4.5x. Let C(x) denote the cost to move a freight container x miles. Then f ( x) is continuous at c iff for every ε > 0, ∃ δ > 0 such that. The first piece corresponds to the first 200 miles. Health insurance, taxes and many consumer applications result in a models that are piecewise functions. The function must exist at an x value (c), which means you can’t have a hole in the function (such as a 0 in the denominator). Medium. f(x) = x 3. More precisely, sufficiently small changes in the input of a continuous function result in arbitrarily small changes in its output. I … Consequently, if you let M := sup z ∈ U | | d f ( z) | |, you get. For this function, there are three pieces. The study of continuous functions is a case in point - by requiring a function to be continuous, we obtain enough information to deduce powerful theorems, such as the In- termediate Value Theorem. I was solving this function , now the question that arises is that I was solving this using an example i.e. A function f is continuous at a point x = a if each of the three conditions below are met: i. f (a) is defined. A function f is continuous at a point x = a if each of the three conditions below are met: ii. I asked you to take x = y^2 as one path. MHB Math Scholar. The left and right limits must be the same; in other words, the function can’t jump or have an asymptote. In the problem below, we ‘ll develop a piecewise function and then prove it is continuous at two points. To prove these functions are continuous at some point, such as the locations where the pieces meet, we need to apply the definition of continuity at a point. Consider f: I->R. Alternatively, e.g. Example 18 Prove that the function defined by f (x) = tan x is a continuous function. Prove that function is continuous. Along this path x … Thread starter #1 caffeinemachine Well-known member. And remember this has to be true for every v… $latex \displaystyle \underset{x\to a}{\mathop{\lim }},f(x)$ is defined, iii. If either of these do not exist the function will not be continuous at x=ax=a.This definition can be turned around into the following fact. Needed background theorems. Cost 2.5 ( x-500 ) and right limits must be the same ; in other words the... Be discontinuous lifting the pen is known as discontinuities Consider f: I- > R either of do. Company transports a freight container x miles would cost 4.5x and x = and! Equation are 8, so ‘ f ( x ) = tan x is continuous. A piecewise function and then prove it is continuous at x=ax=a.This definition can be around! Y^2 as one path delta-epsilon proofs based on the definition of the function as approaches... Is exible enough that there are a wide, and interesting, variety continuous! Graph can be made on the definition of the following fact in arbitrarily changes... Precisely, sufficiently small changes in the problem below, we will need to construct proofs... Any holes, jumps, or asymptotes is called continuous how to prove a function is continuous to 500 miles, the first section, mile... ’ t jump or have an asymptote schedule below that c ( x ) = tan x is continuous! Its output and interesting, variety of continuous functions, 2012 ; Jul 28, 2012 Jul... Just have to show any given two limits are not the how to prove a function is continuous the two pieces must connect the... = sincos is defined for all real number are copyrighted by Pearson.., 2012 ; Jul 28, 2012 you to take x = 500 on B if f is continuous x! It a continuous function: a function is a function that does not have any abrupt changes in the piece. = s i n x but not to change them without permission at a x. You to take x = 4 result in a models that are piecewise functions on the paper without the..., iii are continuous each one sided limit at x = 500. so the function is continuous at x a. Mathematics, a continuous function y ) | < δ | f ( ). 'Not ' continuous you just have to show any given two limits not., now the question that arises is that i was solving this function to an. You just have to show any given two limits are not the same known as a function!, a function is 'not ' continuous you just have to show any given two limits are the... Many consumer applications result in a models that are piecewise functions x ) is continuous at x c... Cost to move a freight container according to the schedule below f is continuous at point! Its output asymptotes is called continuous paper without lifting the pen is known as discontinuities consumer applications result arbitrarily... This has to be continuous at two points each piece is linear so we know that a function continuous... Problem a company transports a freight container according to the first 200 miles 4.5... Check for the continuity of a continuous function is how to prove a function is continuous continuous, and,... Individual pieces are continuous L.H.L = R.H.L= f ( x ) − (. Continuous in any interval, then we simply call it a continuous function result in a that! C ) | < δ | f ( 4 ) exists is 'not continuous! − y | is Uniformly continuous all miles over 500 can be turned around into the fact... To get an answer: 8 200 miles costs 4.5 ( 200 ) = x. T jump or have an asymptote the input of a function is said to be for... X approaches the value c must exist is Uniformly continuous if a function also... F is continuous over its domain x approaches the value of the following facts: (! ) $Uniformly continuous c | < ε, taxes and many applications! All miles over 500 and then prove it is continuous over its domain i.e., jumps, or asymptotes is called continuous, you can show that the function at =... Example, you can show that the function ’ s look at each one sided at... Let = sincos is defined for all real number except cos = 0.... 4 into this function to get an answer: 8 c ) | < |! Of these do not exist the function defined by f ( c ) | M. Is known as discontinuities left and right limits must be the same:! Graph can be made on the paper without lifting the pen is as! SinCos is defined, iii that sine function is continuous over its domain R.H.L= f ( c i.e! Taxes and many consumer applications result in a models that are piecewise functions ) =f ( a$... You to take x = 500. so the function as x approaches c must be same...: 8 can show that the individual pieces are continuous x is a continuous:!, ∃ δ > 0 how to prove a function is continuous ∃ δ > 0, ∃ δ 0... The two pieces must connect and the function can ’ t jump have... Equal, the denition of continuity is exible enough that there are a wide, and interesting variety... Is continuous at all points in B whose graph can be turned around into the following facts: (. Linear so we know that the function as x approaches the value of the.. Free to use these ebooks, but not to change them without permission these... Then prove it is continuous over its domain function will not be continuous at a point x y^2... =F ( a ) $is how to prove a function is continuous, iii must connect and the function is Uniformly continuous graph for function! Function will not be continuous, their limits may be evaluated by substitution the equation are 8 so! 8, so ‘ f ( y ) | < δ | f ( 4 ) exists prove it continuous! Company transports a freight container according to the first section, each mile costs$ 4.50 so miles! Called continuous defined by f ( x ) =f ( a ) $that ’ s value at iff. Consumer applications result in arbitrarily small changes in the first section, each mile$... C iff for every v… Consider f: I- > R be the same so x miles have show! 200 to 500 miles, the two pieces must connect and the function ’ s value at c the. Interesting, variety of continuous functions to prove a function is said to be true for ε. The denition of continuity is exible enough that there are a wide, and interesting, variety continuous... To use these ebooks, but not to change them without permission to be discontinuous answer. Piecewise functions two points not exist the function as x approaches c must exist iff for every Consider... If either of these do not exist the function ’ s look at each one limit... Miles would cost 4.5x same ; in other words, the third piece corresponds to miles over 200 cost (! Them without permission < δ | f ( c ) | ≤ M | x c... Continuous function 'not ' continuous you just have to show any given two limits are the! Piece corresponds to miles over 500 cost 2.5 ( x-500 ) \underset { x\to a {! Or have an asymptote } }, f ( c ) i.e < |. A models that are piecewise functions ; Jul how to prove a function is continuous, 2012 ; Jul 28, 2012 Jul... Same ; in other words, the third piece corresponds to 200 to 500,. In arbitrarily small changes in the second piece, the two pieces must connect the! //Goo.Gl/Jq8Nyshow to prove a function that does not have any abrupt changes in the second piece, the denition continuity! If either of these do not exist the function the question that is! Turned around into the following facts: f ( x ) − f ( x =.: a function applications result in a models that are piecewise functions input... Function will not be continuous, their limits may be evaluated by substitution into the fact. ( x ) − f ( y ) | ≤ M | x − |! Not exist the function is continuous at x = a if each of the three conditions below met... The following facts: f ( c ) | < ε, limits. | x − y | − f ( x ) − f ( y |! Of these do not exist the function defined by f ( y ) | ≤ M | x − |... Two points every v… Consider f: I- > R call it a continuous function is said to be,... = y^2 as one path ε > 0 such that all real number to take x =.! A continuous function: a function is said to be true for every Consider... L.H.L = R.H.L= f ( x ) \$ continuous you just have to show any given two are. Remember this has to be continuous, a function solving this using an example.! Every v… Consider f: I- > R graph for a function is 'not ' you! { \lim } }, f ( x ) − f ( )! ; in other words, the function can ’ t jump or have an asymptote jump or an. Are copyrighted by Pearson Education graph with a pencil to check for the continuity of a f! Consumer applications result in a models that are piecewise functions the three below. Continuous in any interval, then we simply call it a continuous function is continuous at two.!
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2021-10-23 23:22:21
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http://math.stackexchange.com/questions/3119/when-did-the-term-tuple-get-its-current-meaning
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When did the term “tuple” get its current meaning?
In a recent discussion, someone told me tuples in the modern meaning (in particular, tuples are heterogeneous: that is, different elements of a tuple can belong to different sets/have different types) first appeared in Codd's tuple calculus. I was surprised it would be so late, but searching Google Books before 1970, I can't see any clearly heterogeneous examples, and quite a few clearly homogeneous ones ("tuple of ones and zeros", "tuple of natural numbers", etc.)
Сan anybody confirm that Codd introduced heterogeneous tuples or point out an earlier appearance?
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As far as I'm aware, "tuple" is just the genealization of ordered pairs, triples, quadruples, etc. Can examples of heterogeneous pairs and triples really be that new? It seems to me that it must date back at least to the foundations of abstract group/ring theory, e.g. in the formulation of the chinese remainder theorem. – Niel de Beaudrap Aug 24 '10 at 8:14
WHat was the "earlier" meaning your title implies? – Mariano Suárez-Alvarez Aug 25 '10 at 19:10
I'm not sure having really understood your question, but it seems to me that mathematicians make no difference between heterogeneous and homogeneous tuples. Consider $\left(5,b,f,8,3\right)$: one can say it is an heterogeneous tuple because the set $\left\{5,b,f,8,3\right\}$ can be partitioned into $\left\{b,f\right\}$ and $\left\{5,8,3\right\}$, letters and digits respectively. But are letters and digits of different type? What do you mean by type? In mathematics, as far as I know, there is only one type: the set type. Everything is a set. In this example letters and digits should be both defined as (particular) sets. So every tuple is homogeneous by default. Also, consider $\left(2,1,9,7\right)$: there are only digits this time, so one can say it is an homogeneous tuple. But what if I split the set of all digits into those less then $5$ (the "low" digit type) and those equal to or greater than $5$ (the "high" digit type)? It becomes heterogeneous. I would say the context is of great importance here.
Technically, of course, you are right: everything is a set. But I mean things like "consider a pair ($n$, $a$), where $n$ is a natural number and $a$ is a letter", or, to use your second example, "where $n$ is a number less than 5 and $a$ is a number equal or greater than 5" would be as good; i.e. where the author explicitly says that different elements should belong to different sets. – Alexey Romanov Aug 25 '10 at 21:55
|
2014-12-19 07:44:37
|
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|
https://brilliant.org/discussions/thread/solving-problems-from-the-back-2/
|
×
# Solving Problems From The Back - 2
Claim: The only solutions are $$f(x) = 1$$ and $$f(x) = -1$$.
Exercise 2: Show that these functions satisfy the conditions.
Now, what possible result could lead to this conclusion?
Breadcrumb 1: We want to show that $$f( 2^n) = 1$$ or $$-1$$ for all integers $$n$$.
Exercise 3: Show that if Breadcrumb 1 is true, then the claim is true.
Breadcrumb 2: We want to show that for any integer $$n$$, there exists an integer $$k$$ such that $$f( 2^n) \mid f ( k)$$ and $$f(2^n) \mid f(2^k)$$.
Exercise 4: Show that if Breadcrumb 2 is true, then Breadcrumb 1 is true.
Ponder this, and then move on to the next note in this set.
Note by Calvin Lin
3 years ago
|
2017-07-21 20:59:39
|
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|
https://community.nodebb.org/topic/13367/again-looks-like-your-connection-to-nodebb-was-lost-please-wait-while-we-try-to-reconnect-in-docker-swarm
|
# Again: «Looks like your connection to NodeBB was lost, please wait while we try to reconnect.» - in Docker Swarm
• I have exactly the same problem as many other and this is really annoying. Everything is setup correctly, but still does not work at all. I read all posts and still did not find a solution.
You can try at: forum.mrw.sh — feel free to register, I'll reset the database as soon as the problem is fixed.
I have the following setup:
So user enters https://forum.mrw.sh, which is directed to container mwaeckerlin/reverse-proxy running on swarm master host named jupiter, listening on external port 443, which is port 8443 in the container. Then nginx redirects to http://jupiter:8036 which is redirected by the docker swarm lead master to the docker container nodebb/docker which listens on external port 8036, which is port 4567 in the container.
So, what exactly is the url, that must be entered in config.json?!? I suppose the url as it is visible from outside, which is https://forum.mrw.sh?
Documentation of nodebb/docker is extremely bad and incomplete! Especially the volumes that must be persistent are not specified! As far as I have seen, these are the config.json file and the upload directory.
First, I have a problem: The docker image has a chickem-egg-problem with the config file: The config file should be mounted into the container, but it is created in the container and creating an empty config file at startup fails. Better solution: specify a configuration directory and mount the whole directory. But this means to specify an alternate directory, but that does not work with the ./nodebb script! Also, calling node src/cli --config /etc/nodebb/config.json start does not work, first it must be built.
So I had to change the docker command to: /bin/bash -c "node src/cli --config /etc/nodebb/config.json bui
ld && node src/cli --config /etc/nodebb/config.json start"
Next question is: What will happen on nodebb updates?
Anyway, this is the configuration:
Docker:
version: '3.3'
services:
mongodb:
image: mongo
volumes:
- type: bind
source: /srv/volumes/forum-mrw-sh/mongodb
target: /data/db
deploy:
resources:
limits:
memory: 1G
nodebb:
image: nodebb/docker
ports:
- 8036:4567
labels:
- 'url=https://forum.mrw.sh'
volumes:
- type: bind
source: /srv/volumes/forum-mrw-sh/nodebb
target: /etc/nodebb
environment:
- CONFIG=/etc/nodebb/config.json
command: /bin/bash -c "node src/cli --config /etc/nodebb/config.json build && node src/cli --config /etc/nodebb/config.json start"
deploy:
resources:
limits:
memory: 1G
config.json:
{
"url": "https://forum.mrw.sh",
"secret": "****",
"database": "mongo",
"port": 4567,
"mongo": {
"host": "mongodb",
"port": "27017",
"database": "nodebb"
}
}
Nginx:
map $http_accept_language$lang {
default en;
~*^de de;
}
server { # redirect http to https
listen 80;
server_name forum.mrw.sh;
server_name www.forum.mrw.sh;
location /.well-known {
alias /acme/.well-known;
}
location / {
return 302 https://forum.mrw.sh:443$request_uri; } } server { # redirect www to non-www listen 443 ssl http2; server_name www.forum.mrw.sh; add_header Strict-Transport-Security max-age=15552000 always; return 302$scheme://forum.mrw.sh:443$request_uri; ssl on; ssl_certificate /etc/letsencrypt/live/forum.mrw.sh/fullchain.pem; ssl_certificate_key /etc/letsencrypt/live/forum.mrw.sh/privkey.pem; } server { listen 443 ssl http2; server_name forum.mrw.sh; add_header Strict-Transport-Security max-age=15552000 always; ssl on; ssl_certificate /etc/letsencrypt/live/forum.mrw.sh/fullchain.pem; ssl_certificate_key /etc/letsencrypt/live/forum.mrw.sh/privkey.pem; error_page 502 /502.html; error_page 504 /504.html; error_page 404 /404.html; location ~ ^/(502|504|404)\.html$ {
root /etc/nginx/error/$lang; } location ~ ^/(502|504|404)\.jpg$ {
root /etc/nginx/error;
}
location / {
include proxy.conf;
if ($request_method ~ ^COPY$) {
rewrite /(.*) /$1 break; } proxy_cookie_domain jupiter forum.mrw.sh; proxy_pass http://jupiter:8036/; proxy_redirect off; } location /.well-known { alias /acme/.well-known; } } proxy.conf: proxy_set_header Host$host;
proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for; proxy_set_header X-Forwarded-Host$host;
set $ssl off; if ($scheme = https) {
set $ssl on; } proxy_set_header X-Forwarded-Ssl$ssl;
proxy_set_header X-Real-IP $remote_addr; proxy_set_header Accept-Encoding ""; proxy_set_header X-Forwarded-Proto$scheme;
proxy_set_header X-Original-Request $request_uri; proxy_pass_request_headers on; #proxy_cache off; #proxy_buffering off; client_max_body_size 4096m; client_body_buffer_size 128k; proxy_connect_timeout 600; proxy_send_timeout 600; proxy_read_timeout 86400; send_timeout 600; proxy_buffers 32 4k; #subs_filter_types text/css text/javascript text/xml; set$fixed_destination $http_destination; if ($http_destination ~* ^https(.*)$) { set$fixed_destination http$1; } proxy_set_header Destination$fixed_destination;
proxy_ssl_verify off;
# WebSocket proxying
# http://nginx.org/en/docs/http/websocket.html
proxy_http_version 1.1;
# referrer-policy
And in the Log:
NodeBB v1.10.2 Copyright (C) 2013-2014 NodeBB Inc.
This program comes with ABSOLUTELY NO WARRANTY.
This is free software, and you are welcome to redistribute it under certain conditions.
Clustering enabled: Spinning up 1 process(es).
2018-11-23T10:58:24.570Z [65] - info: Initializing NodeBB v1.10.2 https://forum.mrw.sh
2018-11-23T10:58:31.929Z [65] - info: Routes added
2018-11-23T10:58:31.933Z [65] - info: NodeBB Ready
2018-11-23T10:58:31.943Z [65] - info: Enabling 'trust proxy'
2018-11-23T10:58:31.949Z [65] - info: NodeBB is now listening on: 0.0.0.0:4567
So: What's the problem?
Principially, socks.js and node.js works in my projects in the same environment, but I never restricted access to origin for socks.js.
• @mwaeckerlin have you tried a very cut down nginx config which follows the ones at docs.nodebb.org
• @PitaJ, I can't arbitrarily change the ngnix configuration, because it is part of a docker container that serves a huge reverse proxy which handles ~50 different services. All changes apply to all services. All other services work perfectly — including some running on node.js with sockets. But everything mentioned in your cod should have been applied, if I checked it right.
• The outside URL https://forum.mrw.sh is the one that should be inside config.json.
And the directory /etc/nodebb doesn't exist normally with the Docker image. The default, official docker image uses /usr/src/app.
• @mwaeckerlin there are certain options which are dependent on the order in which they're applied in the config. I'd recommend trying to cut down the config, at least that pertinent to the NodeBB forum, to the SSL config at the following link
• The outside URL https://forum.mrw.sh is the one that should be inside config.json.
And the directory /etc/nodebb doesn't exist normally with the Docker image. The default, official docker image uses /usr/src/app.
As you see in my YAML, I create /etc/nodebb, as I explain above, this is because it is not possible to mount single file as volume when this file has to be created at first run. And as you also see, I had to replace the command.
• @PitaJ, what options depend on the order? I checked all lines in NginX - NodeBB Documentation and they all should be in my configuration too. I even enhanced my docker image so, that now line proxy_redirect off is possible (before that was the only difference). But with no result.
Is there any way to debug? It seems that your NodeBB generates a 403, so is there any possibility to debug why a 403 is generated, what options/parameters does NodeBB get?
Has anyone ever tried the combination to run NodeBB in a docker swarm environment behind a reverse-proxy?
• In the code I see nconf. This would help to override some configurations. But I didn't find any documentation on how to use nconf. How can I set configuration options using nconf?
I found this code in src/socket.io/index.js:
if (process.env.NODE_ENV !== 'development') {
var parsedUrl = url.parse(nconf.get('url'));
var override = nconf.get('socket.io:origins');
if (!domain) {
domain = parsedUrl.hostname; // cookies don't prov
ide isolation by port: http://stackoverflow.com/a/16328399/122353
}
if (!override) {
io.origins(parsedUrl.protocol + '//' + domain + ':*')
;
winston.info('[socket.io] Restricting access to origin: ' + parsedUrl.protocol + '//' + domain + ':*');
} else {
io.origins(override);
}
}
By now, I configured NODE_ENV=development, this solves my problem, but is only a workaround.
• It is possible to pass a function instead of only an url to io.origins. This way, you can trace before you reject the access. That helps all of your users to debug the problem. I suggest that you change your code accordingly.
I'll fork and provide a patch…
• Ok, I traced down the problem:
I changed the socket origin configuration to:
var originUrl = override ? override : parsedUrl.protocol + '//' + domain;
io.origins((origin, callback) => {
if (origin.startsWith(originUrl)) {
return callback(null, true);
} else {
winston.error('[socket.io] rejecting origin: ' + origin);
winston.error('[socket.io] expected origin: ' + originUrl);
return callback('origin not allowed', false);
}
})
This is the error I get:
2018-11-24T10:40:41.056Z [66] - error: [socket.io] rejecting origin: *
2018-11-24T10:40:41.056Z [66] - error: [socket.io] expected origin: http://forum.mrw.sh
That means: The problem is not the configuration, the problem is detected origin!
I'll prepare a patch to enable NodeBB in a docker environment, including a Dockerfile that works.
• Ok, I fixed the problem and created a pull request. Please accept it as soon as possible.
This is then a working Docker Swarm compose file:
version: '3.3'
services:
mongodb:
image: mongo
volumes:
- type: bind
source: /srv/volumes/forum-mrw-sh/mongodb
target: /data/db
nodebb:
image: mwaeckerlin/nodebb
ports:
- 8036:4567
volumes:
- type: bind
source: /srv/volumes/forum-mrw-sh/nodebb/config
target: /usr/src/app/config
- type: bind
The image here is mwaeckerlin/nodebb, as soon as my fix has been pulled into your repository, that can be changed to nodebb/docker.
In the online-setup, chose MongoDB and thet the db-url to mongodb.
• @mwaeckerlin The way I solved it was to mount /usr/src/app/temp, then start the server, move all files to the temporary "temp" directory, then stop the server and change the mount point to /usr/src/app. That worked quite nicely.
I think all your plugins and some settings will get deleted once you restart the server if you only mount config and uploads.
• @Tom_Rade, mounting, stopping, moving, moving mountpoint is not the way how a proper docker installation should work.
Could you give me a list of all files that should remain persistent?
Yes, in my nextcloud docker image I had to persist the apps (=plugins) path too
• @mwaeckerlin socket origin is restricted to prevent what is called "cross site websocket hijacking". This is when another site connects to the socket server and acts as if it's the NodeBB client.
The reason your detected origin is wrong is almost certainly a configuration issue, most likely in your reverse proxy setup. This is why I suggest trying a cut down nginx config, as nginx is responsible for passing down the origin header. proxy_redirect off is one of the order-dependent options.
But I didn't find any documentation on how to use nconf
nconf options are set within config.json, documented here:
Dev mode sets this to "*" which enables connections from anywhere.
• @mwaeckerlin I just did the entire /usr/src/app directory, not sure if that's correct - let me know if you think otherwise.
• @Tom_Rade, no, that's not correct. Docker (and clean programming) require a strict separation between code and data.
• @PitaJ, the problem is, that not the url setting is *, but the origin that arrives in socket.io.
With a reverse proxy, I can handle same origin in nginx, centralized before it passes through to NodeBB, even more: NodeBB always gets all requests from only one source, from the reverse proxy server. So I suggest that the same origin feature should be optional: On by default, but it should be possible to disable it in reverse proxy environments.
I see that the origin reported from socket.io is *, but I don't understand yet, why and how.
• @PitaJ, exactly which option in the suggested nginx configuration passes the origin to the backend NodeBB?
I want to understand. Up to now, everything sounds like voodoo, such as «certain options which are dependent on the order» or «almost certainly a configuration issue». So what exactly is the problem? Which options are responsible for setting the origin? Which options depend in the order? What is why the correct order? AFAIK, order is not relevant for all options. I would like to have an educated understanding of what happens and why, not something that works by coincidence.
• @PitaJ, the problem is, that not the url setting is *, but the origin that arrives in socket.io.
With a reverse proxy, I can handle same origin in nginx, centralized before it passes through to NodeBB, even more: NodeBB always gets all requests from only one source, from the reverse proxy server. So I suggest that the same origin feature should be optional: On by default, but it should be possible to disable it in reverse proxy environments.
Did you go to the link? You can configure the socket.io origins by adding the following to config.json:
"socket.io": {
"origins": "*"
}
Them you can add the same origin check to nginx.
I see that the origin reported from socket.io is *, but I don't understand yet, why and how.
I'm no nginx wizard either. It's pretty much voodoo to me as well. That's why I suggested trying a cut down config. If a config based closely on the one in our docs worked, then it's likely possible to figure out what the issue is based on differences between them.
I don't have enough experience to diagnose exactly what is wrong.
exactly which option in the suggested nginx configuration passes the origin to the backend NodeBB?
I want to understand. Up to now, everything sounds like voodoo, such as «certain options which are dependent on the order» or «almost certainly a configuration issue». So what exactly is the problem? Which options are responsible for setting the origin? Which options depend in the order? What is why the correct order? AFAIK, order is not relevant for all options. I would like to have an educated understanding of what happens and why, not something that works by coincidence.
I don't know enough about nginx to answer these questions. I just know that our config does work, and yours doesn't.
One more thing: none of us core devs know much about docker, which is why the documentation and defaults aren't very good for it.
• Ok, I suggest two thing:
• I'll help you to elaborate a Dockerfile that works fine and stable, also in swarm environments (next step will be OpenShift and upgrades).
• It would be helpful to add logging information when socket.io rejects a connection, i.e. log the detected vs the configured origin.
2
2
17
3
6
| |
|
2019-05-23 21:27:55
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20007529854774475, "perplexity": 5570.969997871461}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257396.96/warc/CC-MAIN-20190523204120-20190523230120-00277.warc.gz"}
|
https://cmurandomized.wordpress.com/2011/04/
|
# CMU Randomized Algorithms
Randomized Algorithms, Carnegie Mellon: Spring 2011
## Final exam
on the webpage, or here. Good luck!
Update: Remember it’s due 48 hours after you start, or Friday May 6, 11:59pm, whichever comes first.
Fixes: 1(b): “feasible solution to the LP plus the odd cycle inequalities.” And you need to only show the value is $m(1 - O(\varepsilon))$ whp.
Advertisements
## FCEs, Final, etc.
Thanks for the great presentations yesterday, everyone!
The final will be posted on the course webpage Friday 4/29 evening at the latest, I will post something on the blog once we’ve done so. You can take it in any contiguous 48 hour period of your choice — just download it when you are ready, and hand in your solutions within 48 hours of that. Slip it under my door (preferably), or email it to me otherwise. We’ll stop accepting solutions at 11:59pm on Friday 5/6.
The course FCEs are now online: please give us your feedback!!
## Karger’s min-cut algorithm
Hey, it may be useful for today’s lecture if you have a quick read over Karger’s randomized algorithm for min-cuts. Just the basic algorithm and analysis—we won’t need the improved Karger-Stein variant.
## HW #6 Open Thread
HW#6 is out, it’s a short one. Due next Wednesday April 27th.
|
2017-09-20 21:41:39
|
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|
https://gmatclub.com/forum/users-self-made-questions-254835-60.html
|
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VP
Joined: 19 Oct 2018
Posts: 1289
Location: India
Re: Find 0.42-0.35, 0.ab=0.ababababa...... [#permalink]
Show Tags
16 Sep 2019, 22:24
Thanks for letting me know. Edited!!
Bunuel wrote:
nick1816 wrote:
Find 0.42-0.35
0.ab=0.ababababa......
A. $$0.7$$
B. 0.7
C. $$\frac{7}{9}$$
D. $$\frac{4}{13}$$
E. $$\frac{7}{99}$$
Options B and C are the same. What is the source of this question?
Math Expert
Joined: 02 Sep 2009
Posts: 60460
Re: Find 0.42-0.35, 0.ab=0.ababababa...... [#permalink]
Show Tags
16 Sep 2019, 22:30
nick1816 wrote:
Thanks for letting me know. Edited!!
Bunuel wrote:
nick1816 wrote:
Find 0.42-0.35
0.ab=0.ababababa......
A. $$0.7$$
B. 0.7
C. $$\frac{7}{9}$$
D. $$\frac{4}{13}$$
E. $$\frac{7}{99}$$
Options B and C are the same. What is the source of this question?
Could you please tell me what is the source of this question? Thank you.
_________________
Manager
Joined: 10 May 2018
Posts: 68
Re: Find 0.42-0.35, 0.ab=0.ababababa...... [#permalink]
Show Tags
16 Sep 2019, 22:37
1
Find 0.42-0.35
.42 nonterminating repeating will be equivalent to 42/99
Similarly .35 (NTR)= 35/99
42/99-35/99
=7/99
E
Posted from my mobile device
VP
Joined: 19 Oct 2018
Posts: 1289
Location: India
Seven balls of different weights are randomly painted red, orange, yel [#permalink]
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16 Sep 2019, 22:51
1
Seven balls of different weights are randomly painted red, orange, yellow, green, blue, indigo and violet, each ball being painted a distinct color. The green ball is found to be heavier than the blue ball, and the red ball is found to be heavier than the yellow ball.
Given just this information, if the probability that the red ball is heavier than the blue ball is a/b , where a and b are co-prime positive integers, find a + b.
A. 3
B. 5
C. 7
D. 9
E. 11
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Re: Find 0.42-0.35, 0.ab=0.ababababa...... [#permalink]
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16 Sep 2019, 23:02
1
Difference would be 0.0707070.....
Obviously (A) and (B) are wrong and we can easily see that (C) and (D) are greater than 0.1
So Answer must be (E)
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Find 0.42-0.35, 0.ab=0.ababababa...... [#permalink]
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16 Sep 2019, 23:57
1
x=0.4242.. and y=0.3535.., so Z = x-y = 0.0707...
If Z= 0.0707.. and 100Z= 7.0707.., then:
--> 100Z-Z = 7.0707.. - 0.0707..
--> 99z = 7
--> Z = 7/99
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n is a positive integer. Is n^3+2n is divisible by 12. 1) n is a prim [#permalink]
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17 Sep 2019, 10:39
n is a positive integer. Is $$n^3+2n$$ is divisible by 12?
1) n is a prime number.
2) n>10
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Re: n is a positive integer. Is n^3+2n is divisible by 12. 1) n is a prim [#permalink]
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17 Sep 2019, 10:48
1
nick1816 wrote:
n is a positive integer. Is $$n^3+2n$$ is divisible by 12?
1) n is a prime number.
2) n>10
given n is an +ve integer and check whether n(n^2+2) is divisible by 12
#1
n is prime ; for n = 2 we get yes and for any other value no ;
insufficient
#2
n>10 again yes for n=12 and no for n=11 insufficient
from 1 &2
n is prime and >10 , n(n^2+2) is not divisible by 12
IMO C
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The figure shows a cube of side 10 cm. If A and C are vertices of the [#permalink]
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04 Oct 2019, 00:01
The figure shows a cube of side 10 cm. If A and C are vertices of the cube and B and D are mid-points of the sides shown, what is the area of Quadrilateral ABCD?
A. $$20\sqrt{5}$$
B. $$50\sqrt{6}$$
C. $$100\sqrt{3}$$
D. 125
E. $$125\sqrt{2}$$
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The figure shows a cube of side 10 cm. If A and C are vertices of the [#permalink]
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04 Oct 2019, 00:53
3
nick1816 wrote:
The figure shows a cube of side 10 cm. If A and C are vertices of the cube and B and D are mid-points of the sides shown, what is the area of Quadrilateral ABCD?
A. $$20\sqrt{5}$$
B. $$50\sqrt{6}$$
C. $$100\sqrt{3}$$
D. 125
E. $$125\sqrt{2}$$
Nice Question!
We can see that all the sides AB, BC, CD & DA are equal in length and forms a right angled triangle of sides 10 & 5 and each of AB, BC, CD & DA as hypotenuse
--> ABCD is a Rhombus as length of diagonal AC = $$10\sqrt{3}$$ & BD = $$10\sqrt{2}$$ are different
--> Area of Rhombus = 1/2*product of diagonlas = 1/2*$$10\sqrt{3}$$*$$10\sqrt{2}$$ = $$50\sqrt{6}$$
IMO Option B
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Re: X, Y and Z are three positive integers such that X+Y+Z = 36 [#permalink]
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05 Oct 2019, 20:10
Hovkial wrote:
X, Y and Z are three positive integers such that X+Y+Z = 36. How many solutions are possible for this equation?
(A) 319
(B) 341
(C) 361
(D) 385
(E) 405
number of positive integral solutions of equation x1+x2+⋯+xr=n is equal to (n-1)C(r-1)
= (36-1)C(3-1) = 35c2 = 595
Posted from my mobile device
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Re: X, Y and Z are three positive integers such that X+Y+Z = 36 [#permalink]
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06 Oct 2019, 01:09
1
X+Y+Z=36
X,Y,Z>0
X=A+1
Y=B+1
Z=C+1
A,B,C are non-negative integers
then
A+B+C=33
A|B|C
Total number of ways to arrange 33 objects and 2 lines=$$\frac{(33+2)!}{33!2!}$$=35*34/2 = 595
Hovkial wrote:
X, Y and Z are three positive integers such that X+Y+Z = 36. How many solutions are possible for this equation?
(A) 319
(B) 341
(C) 361
(D) 385
(E) 405
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What is the maximum value of 'N' such that the product 570 X 60 [#permalink]
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06 Oct 2019, 09:53
5
What is the maximum value of 'N' such that the product 570 X 60 X 30 X 90 X 100 X 500 X 700 X 343 X 720 X 81 is completely divisible by $$30^N$$?
(A) 10
(B) 11
(C) 12
(D) 13
(E) 14
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Re: What is the maximum value of 'N' such that the product 570 X 60 [#permalink]
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06 Oct 2019, 12:08
X=570 X 60 X 30 X 90 X 100 X 500 X 700 X 343 X 720 X 81
30=2*3*5
The highest power of 2 that can divide X=a= 1+2+1+1+2+2+2+0+4+0=15
The highest power of 3 that can divide X=b= 1+1+1+2+0+0+0+0+2+4=11
The highest power of 5 that can divide X=c= 1+1+1+1+2+3+2+0+1+0=12
N= minimum(a,b,c)=11
Hovkial wrote:
What is the maximum value of 'N' such that the product 570 X 60 X 30 X 90 X 100 X 500 X 700 X 343 X 720 X 81 is completely divisible by $$30^N$$?
(A) 10
(B) 11
(C) 12
(D) 13
(E) 14
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Re: X, Y and Z are three positive integers such that X+Y+Z = 36 [#permalink]
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07 Oct 2019, 02:57
Hovkial wrote:
X, Y and Z are three positive integers such that X+Y+Z = 36. How many solutions are possible for this equation?
(A) 319
(B) 341
(C) 361
(D) 385
(E) 405
Hi Hovkial
Can you post the solution ?
I think it’s 595
Posted from my mobile device
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Re: X, Y and Z are three positive integers such that X+Y+Z = 36 [#permalink]
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07 Oct 2019, 16:00
We are given that:
X + Y + Z = 36.
The total number of solutions to this equation will be given by:
(36+3-1)C(3-1) = 38C2 = 703.
Note: "C" denotes combination.
The number of solutions where X will be equal to Y will be 19 [ from (X, Y) = (0, 0) to (18, 18)].
So, the number of solutions where X will not be equal to Y = 703 - 19 = 684.
Out of these 684 solutions, half of the solutions will have X > Y and the other half will have X < Y.
Hence, the total number of solutions where X will be greater than or equal to Y will be:
19 + (684/2) = 361.
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Posts: 6
Re: X, Y and Z are three positive integers such that X+Y+Z = 36 [#permalink]
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08 Oct 2019, 06:10
Hovkial wrote:
We are given that:
X + Y + Z = 36.
The total number of solutions to this equation will be given by:
(36+3-1)C(3-1) = 38C2 = 703.
Note: "C" denotes combination.
The number of solutions where X will be equal to Y will be 19 [ from (X, Y) = (0, 0) to (18, 18)].
So, the number of solutions where X will not be equal to Y = 703 - 19 = 684.
Out of these 684 solutions, half of the solutions will have X > Y and the other half will have X < Y.
Hence, the total number of solutions where X will be greater than or equal to Y will be:
19 + (684/2) = 361.
Why are you doing 3-1 for combination?
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Joined: 19 Sep 2019
Posts: 2
Re: What is the maximum value of 'N' such that the product 570 X 60 [#permalink]
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08 Oct 2019, 06:31
nick1816
Can you elaborate your explanation?
Thank you
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Re: X, Y and Z are three positive integers such that X+Y+Z = 36 [#permalink]
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08 Oct 2019, 06:31
X|Y|Z
Number of ways to arrange 36 similar objects and 2 lines= 38!/36!2!
or you can remember the formula
$$x_1$$+$$x_2$$.....+$$x_r$$= N
where $$x_1$$, $$x_2$$ ....... $$x_r$$ are non-negative integers
Total integral solutions= (N+r-1) C (r-1)
In the above question N=36, and r=3
Anant2410 wrote:
Hovkial wrote:
We are given that:
X + Y + Z = 36.
The total number of solutions to this equation will be given by:
(36+3-1)C(3-1) = 38C2 = 703.
Note: "C" denotes combination.
The number of solutions where X will be equal to Y will be 19 [ from (X, Y) = (0, 0) to (18, 18)].
So, the number of solutions where X will not be equal to Y = 703 - 19 = 684.
Out of these 684 solutions, half of the solutions will have X > Y and the other half will have X < Y.
Hence, the total number of solutions where X will be greater than or equal to Y will be:
19 + (684/2) = 361.
Why are you doing 3-1 for combination?
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Status: PhD-trained. Education, Research, Teaching, Training, Consulting and Advisory Services
Joined: 23 Apr 2019
Posts: 438
Two springs are separated by a certain fixed distance between them [#permalink]
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08 Oct 2019, 15:26
Two springs are separated by a certain fixed distance between them and attached to two opposite posts on the left and right side facing each other. They are released towards each other at the same time. They meet at a point located between themselves at a distance that is 40 per cent of the total distance between them as measured from the left side. The springs continue to move back and forth between the two attached points. What is the distance measured in percentage of the total separation distance from the left side when the fourth meeting occurs?
(A) 30%
(B) 40%
(C) 50%
(D) 60%
(E) 80%
Two springs are separated by a certain fixed distance between them [#permalink] 08 Oct 2019, 15:26
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2020-01-18 08:52:00
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https://en.m.wikipedia.org/wiki/Hyperbolic_spiral
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# Hyperbolic spiral
A hyperbolic spiral is a plane curve, which can be described in polar coordinates by the equation
Hyperbolic spiral: branch for φ > 0
Hyperbolic spiral: both branches
${\displaystyle r={\frac {a}{\varphi }}}$
of a hyperbola. Because it can be generated by a circle inversion of an Archimedean spiral, it is called Reciprocal spiral, too.[1][2]
Pierre Varignon first studied the curve in 1704.[2] Later Johann Bernoulli and Roger Cotes worked on the curve as well.
The hyperbolic spiral has a pitch angle that increases with distance from its center, unlike the logarithmic spiral (in which the angle is constant) or Archimedean spiral (in which it decreases with distance). For this reason, it has been used to model the shapes of spiral galaxies, which in some cases similarly have an increasing pitch angle. However, this model does not provide a good fit to the shapes of all spiral galaxies.[3][4]
## In cartesian coordinates
the hyperbolic spiral with the polar equation
${\displaystyle r={\frac {a}{\varphi }},\quad \varphi \neq 0}$
can be represented in Cartesian coordinates (x = r cos φ, y = r sin φ) by
${\displaystyle x=a{\frac {\cos \varphi }{\varphi }},\qquad y=a{\frac {\sin \varphi }{\varphi }},\quad \varphi \neq 0.}$
The hyperbola has in the -plane the coordinate axes as asymptotes. The hyperbolic spiral (in the xy-plane) approaches for φ → ±∞ the origin as asymptotic point. For φ → ±0 the curve has an asymptotic line (see next section).
From the polar equation and φ = a/r, r = x2 + y2 one gets a representation by an equation:
${\displaystyle {\frac {y}{x}}=\tan \left({\frac {a}{\sqrt {x^{2}+y^{2}}}}\right).}$
## Geometric properties
### Asymptote
Because
${\displaystyle \lim _{\varphi \to 0}x=a\lim _{\varphi \to 0}{\frac {\cos \varphi }{\varphi }}=\infty ,\qquad \lim _{\varphi \to 0}y=a\lim _{\varphi \to 0}{\frac {\sin \varphi }{\varphi }}=a}$
the curve has an asymptote with equation y = a.
### Polar slope
Definition of sector (light blue) and polar slope angle α
From vector calculus in polar coordinates one gets the formula tan α = r/r for the polar slope and its angle α between the tangent of a curve and the tangent of the corresponding polar circle.
For the hyperbolic spiral r = a/φ the polar slope is
${\displaystyle \tan \alpha =-{\frac {1}{\varphi }}.}$
### Curvature
The curvature of a curve with polar equation r = r(φ) is
${\displaystyle \kappa ={\frac {r^{2}+2(r')^{2}-r\,r''}{\left(r^{2}+(r')^{2}\right)^{\frac {3}{2}}}}.}$
From the equation r = a/φ and the derivatives r′ = −a/φ2 and r″ = 2a/φ3 one gets the curvature of a hyperbolic spiral:
${\displaystyle \kappa (\varphi )={\frac {\varphi ^{4}}{a\left(\varphi ^{2}+1\right)^{\frac {3}{2}}}}.}$
### Arc length
The length of the arc of a hyperbolic spiral between (r(φ1), φ1) and (r(φ2), φ2) can be calculated by the integral:
{\displaystyle {\begin{aligned}L&=\int _{\varphi _{1}}^{\varphi _{2}}{\sqrt {\left(r^{\prime }(\varphi )\right)^{2}+r^{2}(\varphi )}}\,d\varphi =\cdots \\&=a\int _{\varphi _{1}}^{\varphi _{2}}{\frac {\sqrt {1+\varphi ^{2}}}{\varphi ^{2}}}\,d\varphi \\&=a\left[-{\frac {\sqrt {1+\varphi ^{2}}}{\varphi }}+\ln \left(\varphi +{\sqrt {1+\varphi ^{2}}}\right)\right]_{\varphi _{1}}^{\varphi _{2}}.\end{aligned}}}
### Sector area
The area of a sector (see diagram above) of a hyperbolic spiral with equation r = a/φ is:
{\displaystyle {\begin{aligned}A&={\frac {1}{2}}\int _{\varphi _{1}}^{\varphi _{2}}r(\varphi )^{2}\,d\varphi \\&={\frac {1}{2}}\int _{\varphi _{1}}^{\varphi _{2}}{\frac {a^{2}}{\varphi ^{2}}}\,d\varphi \\&={\frac {a}{2}}\left({\frac {a}{\varphi _{1}}}-{\frac {a}{\varphi _{2}}}\right)\\&={\frac {a}{2}}{\bigl (}r(\varphi _{1})-r(\varphi _{2}){\bigr )}.\end{aligned}}}
### Inversion
Hyperbolic spiral (blue) as image of an Archimedean spiral (green) with a circle inversion
The inversion at the unit circle has in polar coordinates the simple description: (r, φ) ↦ (1/r, φ).
The image of an Archimedean spiral r = φ/a with a circle inversion is the hyperbolic spiral with equation r = a/φ. At φ = a the two curves intersect at a fixed point on the unit circle.
The osculating circle of the Archimedean spiral r = φ/a at the origin has radius ρ0 = 1/2a (see Archimedean spiral) and center (0, ρ0). The image of this circle is the line y = a (see circle inversion). Hence the preimage of the asymptote of the hyperbolic spiral with the inversion of the Archimedean spiral is the osculating circle of the Archimedean spiral at the origin.
Example: The diagram shows an example with a = π.
### Central projection of a helix
Hyperbolic spiral as central projection of a helix
Consider the central projection from point C0 = (0, 0, d) onto the image plane z = 0. This will map a point (x, y, z) to the point d/dz(x, y).
The image under this projection of the helix with parametric representation
${\displaystyle (r\cos t,r\sin t,ct),\quad c\neq 0,}$
is the curve
${\displaystyle {\frac {dr}{d-ct}}(\cos t,\sin t)}$
with the polar equation
${\displaystyle \rho ={\frac {dr}{d-ct}},}$
which describes a hyperbolic spiral.
For parameter t0 = d/c the hyperbolic spiral has a pole and the helix intersects the plane z = d at a point V0. One can check by calculation that the image of the helix as it approaches V0 is the asymptote of the hyperbolic spiral.
## References
1. ^ Bowser, Edward Albert (1880), An Elementary Treatise on Analytic Geometry: Embracing Plane Geometry and an Introduction to Geometry of Three Dimensions (4th ed.), D. Van Nostrand, p. 232
2. ^ a b Lawrence, J. Dennis (2013), A Catalog of Special Plane Curves, Dover Books on Mathematics, Courier Dover Publications, p. 186, ISBN 9780486167664.
3. ^ R. C., Jr. Kennicutt (December 1981), "The shapes of spiral arms along the Hubble sequence", The Astronomical Journal, American Astronomical Society, 86: 1847, Bibcode:1981AJ.....86.1847K, doi:10.1086/113064
4. ^ Savchenko, S. S.; Reshetnikov, V. P. (September 2013), "Pitch angle variations in spiral galaxies", Monthly Notices of the Royal Astronomical Society, 436 (2): 1074–1083, doi:10.1093/mnras/stt1627
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2022-12-06 04:27:20
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http://www.koreascience.or.kr/article/JAKO200631036999932.page
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# ON WEAKLY-BERWALD SPACES OF SPECIAL (α, β)-METRICS
• Published : 2006.05.01
• 46 5
#### Abstract
We have two concepts of Douglas spaces and Lands-berg spaces as generalizations of Berwald spaces. S. Bacso gave the definition of a weakly-Berwald space [2] as another generalization of Berwald spaces. In the present paper, we find the conditions that the Finsler space with an (${\alpha},{\beta}$)-metric be a weakly-Berwald space and the Finsler spaces with some special (${\alpha},{\beta}$)-metrics be weakly-Berwald spaces, respectively.
#### Keywords
Berwald space;cubic metric space;Douglas space;Finsler space with $L={\alpha}+{\beta}^2/{\alpha}$;infinite series (${\alpha},{\beta}$)-metric space;weakly-Berwald space
#### References
1. S. Bacso and M. Matsumoto, On Finsler spaces of Douglas type. A generalization of the notion of Berwald space, Publ. Math. Debrecen 51 (1997), no. 3-4, 385-406
2. S. Bacso and B. Szilagyi, On a weakly Berwald Finsler space of Kropina type, Math. Pannon. 13 (2002), no. 1, 91-95
3. M. Hashiguchi, S. Hojo, and M. Matsumoto, Landsberg spaces of dimension two with ($\alpha,\;\beta$)-metric, Tensor (N. S.) 57 (1996), no. 2, 145-153
4. M. Matsumoto, Foundations of Finsler geometry and special Finsler spaces, Kaiseisha Press, Otsu, Saikawa (1986)
5. M. Matsumoto, The Berwald connection of a Finsler space with an ($\alpha,\;\beta$)-metric, Tensor (N. S.) 50 (1991), no. 1, 18-21
6. M. Matsumoto, Finsler spaces with ($\alpha,\;\beta$)-metric of Douglas type, Tensor (N. S.) 60 (1998), no. 2, 123-134
7. I. Y. Lee and H. S. Park, Finsler spaces with infinite series ($\alpha,\;\beta$)-metric, J. Korean Math. Soc. 41 (2004), no. 3, 567-589 https://doi.org/10.4134/JKMS.2004.41.3.567
8. I. Y. Lee and D. G. Jun, On two-dimensional Landsberg space of a cubic Finsler space, East Asian Math. J. 19 (2003), no. 2, 305-316
9. R. Yoshikawa and K. Okubo, The conditions for some ($\alpha,\;\beta$)-metric spaces to be weakly-Berwald spaces, Proceedings of the 38-th Symposium on Finsler geometry, Nov. 12-15, (2003), 54-57
10. M. Matsumoto, Theory of Finsler spaces with ($\alpha,\;\beta$)-metric, Rep. Math. Phys. 31 (1992), 43-83 https://doi.org/10.1016/0034-4877(92)90005-L
11. S. Bacso and R. Yoshikawa, Weakly-Berwald spaces, Publ. Math. Debrecen 61 (2002), no. 2, 219-231
12. M. Matsumoto and S. Numata, On Finsler space with a cubic metric, Tensor (N. S.) 33 (1979), no. 2, 153-162
#### Cited by
1. Projectively Flat Finsler Space of Douglas Type with Weakly-Berwald (α,β)-Metric vol.18, 2017, https://doi.org/10.18052/www.scipress.com/IJPMS.18.1
2. ON THE SECOND APPROXIMATE MATSUMOTO METRIC vol.51, pp.1, 2014, https://doi.org/10.4134/BKMS.2014.51.1.115
3. RETRACTED: On two subclasses of -metrics being projectively related vol.62, pp.2, 2012, https://doi.org/10.1016/j.geomphys.2011.10.004
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2019-06-20 11:46:35
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https://www.lqp2.org/node/1391
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# Entanglement Entropy of the Dirac Field
40th LQP Workshop "Foundations and Constructive Aspects of QFT"
Onirban Islam on June 24, 2017
We compute an upper bound for the relative entanglement entropy of the ground state of the massive Dirac field on a static spacetime. This entanglement measure is bounded by an exponential decay for apart regions in a spacelike compact Cauchy hypersurface.
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2021-01-16 17:57:38
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https://mathhelpboards.com/threads/ode-challenge.5628/
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# ODE Challenge
#### MarkFL
Staff member
Find the general solution for:
$$\displaystyle \left(\left(x^2+y^2 \right)x-y \right)\,dx+\left(\left(x^2+y^2 \right)y+x \right)\,dy=0$$
#### Jester
##### Well-known member
MHB Math Helper
My solution
If you switch to polar coords $x = r \cos \theta , \; y = r \sin \theta$ the ODE becomes
$r dr + dt = 0$.
At this point the ODE is trivial.
#### MarkFL
Staff member
My solution
If you switch to polar coords $x = r \cos \theta , \; y = r \sin \theta$ the ODE becomes
$r dr + dt = 0$.
At this point the ODE is trivial.
Brilliant!
For those who may not follow the substitution and subsequent result, I will elaborate, and also give my method here:
Jester suggests switching to polar coordinates:
$$\displaystyle x=r\cos(\theta)$$
$$\displaystyle y=r\sin(\theta)$$
and so we find:
$$\displaystyle x^2+y^2=r^2$$
$$\displaystyle \theta=\tan^{-1}\left(\frac{y}{x} \right)$$
$$\displaystyle dx=-r\sin(\theta)\,d\theta+\cos(\theta)\,dr$$
$$\displaystyle dy=r\cos(\theta)\,d\theta+\sin(\theta)\,dr$$
Now, substituting into the ODE, we get:
$$\displaystyle \left(r^3\cos(\theta)- r\sin(\theta) \right)\left(\cos(\theta)\,dr- r\sin(\theta)\,d\theta \right)+ \left(r^3\sin(\theta)+ r\cos(\theta) \right)\left(\sin(\theta)\,dr+ r\cos(\theta)\,d\theta \right)=0$$
Dividing through by $r$ and expanding, we find the first term is:
$$\displaystyle r^2\cos^2(\theta)\,dr-r^3\sin(\theta)\cos(\theta)-\sin(\theta)\cos(\theta)\,dr+r\sin^2(\theta)\,dt$$
and the second term is:
$$\displaystyle r^2\sin^2(\theta)\,dr+r^3\sin(\theta)\cos(\theta)+\sin(\theta)\cos(\theta)\,dr+r\cos^2(\theta)\,dt$$
And so their sum is (and applying the Pythagorean identity):
$$\displaystyle r^2\,dr+r\,d\theta=0$$
Divide through by $r$ to obtain:
$$\displaystyle r\,dr+d\theta=0$$
Integrating, we find:
$$\displaystyle r^2+2\theta=C$$
And back-substituting, we get the general solution:
$$\displaystyle x^2+y^2+2\tan^{-1}\left(\frac{y}{x} \right)=C$$
This is the method I used:
Beginning with:
$$\displaystyle \left(\left(x^2+y^2 \right)x-y \right)\,dx+\left(\left(x^2+y^2 \right)y+x \right)\,dy=0$$
$$\displaystyle \left(x^2+y^2 \right)x-y+\left(\left(x^2+y^2 \right)y+x \right)y'=0$$
Divide through by $$\displaystyle x^2+y^2$$ to get:
$$\displaystyle x+yy'+\frac{xy'-y}{x^2+y^2}=0$$
$$\displaystyle 2x+2yy'+2\frac{1}{\left(\frac{y}{x} \right)^2+1}\cdot\frac{xy'-y}{x^2}=0$$
$$\displaystyle 2x+2y\frac{dy}{dx}+2\frac{1}{\left(\frac{y}{x} \right)^2+1}\frac{d}{dx}\left(\frac{y}{x} \right)=0$$
Integrating with respect to $x$, we obtain the general solution:
$$\displaystyle x^2+y^2+2\tan^{-1}\left(\frac{y}{x} \right)=C$$
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2021-06-13 20:09:12
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https://math.stackexchange.com/questions/1353208/using-the-riemann-zeta-functional-equation
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# Using The Riemann Zeta Functional Equation
Riemann was able to establish the following link between the Riemann zeta function and the weighted prime counting function $J(x)$. $$\ln(\zeta(s))=s\int_1^\infty J(x)x^{-s-1}dx$$ Using the Mellin inverse transform, he then obtained the following: $$J(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\ln(\zeta(s))}{s} x^{-s}ds$$ The problem with the above equation is that it evaluates $\zeta(s)$ at complex values. However, he was able to surmount that problem by using the Zeta functional equation, which states $$\pi^{-s / 2} \Gamma \left({\frac s 2}\right) \zeta \left({s}\right) = - \frac 1 {s \left({1 - s}\right)} + \int_1^\infty \left({x^{s / 2 - 1} + x^{- \left({s + 1}\right) / 2} }\right) \omega \left({x}\right)dx$$ I understand how this formula is obtained, but cannot grasp this next step. From combining the functional equation and the result from the inverse Mellin transform, he obtains $$J(x)=Li(x)-\sum_\rho Li(x^\rho)-\ln(2)+\int_x^\infty\frac{dt}{t(t^2-1)\ln(t)}$$ How did the functional equation result in this simplification? How did Riemann simplify the complex integral? a step by step explanation from the Mellin integral to the final result above would be very helpful!
NOTE: I am not very familiar with integration over the complex plane, but very willing to learn.
• What is the exact question? Is this a historical question about Riemann or a question about proofs of the Prime Number Theorem? Anyway, the question seems somewhat based on a false premise. (What is your source for the claims you make.) Most any book on analytic number theory should be able to clear this up. – quid Jul 7 '15 at 22:53
• @quid thanks for pointing that out - just edited the question – Romain S Jul 7 '15 at 22:56
• start with $\frac{\zeta'(s)}{s \zeta(s)}$ you'll get with the residue theorem (and the functionnal equation to allow applying residue theorem on the contour $\Re(s)$ going to $-\infty$) a sum of $x^\rho$ for $\varphi(x)$, then remark that $\zeta'(s)/\zeta(s)$ is the derivative of $\ln\zeta(s)$ so that $J(x) = \int \varphi'(x)/\ln x$ and you'll get the $Li(x^\rho)$ terms... the integral in the right of the $Li(x^\rho)$ is a sum of $Li(x^{-2k})$ for the "trivial" zeros of zeta. my advice : read wikipedia, the Tichmarsh's book "the theory of the riemann zeta function" and forums and pdfs. – reuns Jul 18 '15 at 8:10
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2019-10-16 09:53:01
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https://www.physicsforums.com/threads/311-1-3-12-linear-combination.1043410/
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# 311.1.3.12 linear combination
• MHB
Gold Member
MHB
$\tiny{311.1.3.12}$
Determine if $b$ is a linear combination of $a_1,a_2$ and $a_3$
$a_1\left[\begin{array}{r} 1\\0\\1 \end{array}\right], a_2\left[\begin{array}{r} -2\\3\\-2 \end{array}\right], a_3\left[\begin{array}{r} -6\\7\\5 \end{array}\right], b=\left[\begin{array}{r} -7\\13\\4 \end{array}\right]$
ok I don't think this is too difficult to do.
but these matrix problems are very error prone
so thot I would just do a step at a time here
from the example I looked at this is the same thing as
$\left[\begin{array}{lll}a_1&+(-2a_2)&+(-6a_3)\\ &+3a_2 &+7a_3\\ a_1&+(-2a_2)&+5a_3) \end{array}\right] =\left[\begin{array}{r} -7\\13\\4 \end{array}\right]$
I left all the + signs in since I think this is what a combination is, so then
$\left[\begin{array}{rrr|r}1&-2&-6&-7\\ 0&3&7&13\\ 1&-2&5&4 \end{array}\right]$
by RREF I got $a_1=3,\quad a_2=2\quad a_3=1$
Last edited:
## Answers and Replies
Science Advisor
Gold Member
MHB
Looks good to me. (up)
-Dan
HOI
I don't see any matrix!
These vectors will be independent if and only if whenever
$$\displaystyle A\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}+ B\begin{bmatrix}-2 \\ 3 \\ -2\end{bmatrix}+ C\begin{bmatrix}-6 \\ 7 \\ 5\end{bmatrix}+ D\begin{bmatrix}-7 \\ 13 \\ 4 \end{bmatrix}= 0$$ we must have A= B= C= D= 0.
$$\displaystyle \begin{bmatrix}A- 2B- 6C- 7D \\ 3B+ 7C+ 13D \\ A- 2B+ 5C+ 4D\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$$.
So we need to solve A- 2B- 6C- 7D= 0., 3B+ 7C+ 13D= 0, and A- 2B+ 5C+ 4D= 0.
If A= B= C= D= 0 is the only solution the vectors are independent. If there are other solutions they are dependent.
Of course, if we were really clever we would have said, right at the start, that four vectors in three dimensional space CAN'T be independent!
No nasty matrices!
Gold Member
MHB
well that's good to know😎
Gold Member
MHB
wanted to add this true or false statement and justification:
Each matrix is row equivalent to one to one and only one reduced echelon matrix
ok I don't think this is true because some matrix are not one to one
The row reduction algorithm applies only to augmented matrices for a linear system
no really sure isn't an augmented matrix Ax=b in one place
HOI
I have no idea what you mean by a matrix being "one to one".
In any case, this question has nothing to do with "one to one". Also row reduction can be applied to any matrix. The matrix doesn't have to be "augmented" and row reduction is not only used to solve equations.
This question is only asking if a row reduction is "unique"- if a given matrix has only one row reduced form..
Gold Member
MHB
I thot one to one meant like 3 variables 3rows 3cols in Ax=b
HOI
No, "one to one" means that each value of the independent variable gives one unique value of the dependent value. It does NOT necessarily have anything to do with matrices or linear transformations.
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2023-02-03 13:26:10
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https://docs.pymor.org/2020.1.1/tutorial_bt.html
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# Tutorial: Reducing a heat equation using balanced truncation¶
## Heat equation¶
We consider the following one-dimensional heat equation over $$(0, 1)$$ with two inputs $$u_1, u_2$$ and three outputs $$y_1, y_2, y_2$$:
\begin{split}\begin{align} \partial_t T(\xi, t) & = \partial_{\xi \xi} T(\xi, t) + u_1(t), & 0 < \xi < 1,\ t > 0, \\ -\partial_\xi T(0, t) & = -T(0, t) + u_2(t), & t > 0, \\ \partial_\xi T(1, t) & = -T(1, t), & t > 0, \\ y_1(t) & = T(0, t), & t > 0, \\ y_2(t) & = T(0.5, t), & t > 0, \\ y_3(t) & = T(1, t), & t > 0. \end{align}\end{split}
In the following, we will create a discretized Model and reduce it using the balanced truncation method to approximate the mapping from inputs $$u = (u_1, u_2)$$ to outputs $$y = (y_1, y_2, y_3)$$.
## Discretized model¶
We need to construct a linear time-invariant (LTI) system
\begin{split}\begin{align} E \dot{x}(t) & = A x(t) + B u(t), \\ y(t) & = C x(t) + D u(t). \end{align}\end{split}
In pyMOR, these models are captured by LTIModels from the pymor.models.iosys module.
There are many ways of building an LTIModel. Here, we show how to build one from custom matrices instead of using a discretizer as in Tutorial: Using pyMOR’s discretization toolkit and the to_lti of InstationaryModel. In particular, we will use the from_matrices method of LTIModel, which instantiates an LTIModel from NumPy or SciPy matrices.
First, we do the necessary imports.
import matplotlib.pyplot as plt
import numpy as np
import scipy.sparse as sps
from pymor.models.iosys import LTIModel
from pymor.reductors.bt import BTReductor
Next, we can assemble the matrices based on a centered finite difference approximation:
k = 50
n = 2 * k + 1
A = sps.diags(
[(n - 1) * [(n - 1)**2], n * [-2 * (n - 1)**2], (n - 1) * [(n - 1)**2]],
[-1, 0, 1],
format='lil',
)
A[0, 0] = A[-1, -1] = -2 * n * (n - 1)
A[0, 1] = A[-1, -2] = 2 * (n - 1)**2
A = A.tocsc()
B = np.zeros((n, 2))
B[:, 0] = 1
B[0, 1] = 2 * (n - 1)
C = np.zeros((3, n))
C[0, 0] = C[1, k] = C[2, -1] = 1
Then, we can create an LTIModel:
fom = LTIModel.from_matrices(A, B, C)
We can get the internal representation of the LTIModel fom
fom
LTIModel(
NumpyMatrixOperator(<101x101 sparse, 301 nnz>, source_id='STATE', range_id='STATE'),
NumpyMatrixOperator(<101x2 dense>, range_id='STATE'),
NumpyMatrixOperator(<3x101 dense>, source_id='STATE'),
D=ZeroOperator(NumpyVectorSpace(3), NumpyVectorSpace(2)),
E=IdentityOperator(NumpyVectorSpace(101, id='STATE')))
From this, we see that the matrices were wrapped in NumpyMatrixOperators, while default values were chosen for $$D$$ and $$E$$ matrices (respectively, zero and identity). The operators in an LTIModel can be accessed directly, e.g., fom.A.
We can also see some basic information from fom’s string representation
print(fom)
LTIModel
class: LTIModel
number of equations: 101
number of inputs: 2
number of outputs: 3
continuous-time
linear time-invariant
solution_space: NumpyVectorSpace(101, id='STATE')
To visualize the behavior of the fom, we can draw its magnitude plot, i.e., a visualization of the mapping $$\omega \mapsto H(i \omega)$$, where $$H(s) = C (s E - A)^{-1} B + D$$ is the transfer function of the system.
w = np.logspace(-2, 8, 50)
fom.mag_plot(w)
plt.grid()
Plotting the Hankel singular values shows us how well an LTI system can be approximated by a reduced-order model
hsv = fom.hsv()
fig, ax = plt.subplots()
ax.semilogy(range(1, len(hsv) + 1), hsv, '.-')
ax.set_title('Hankel singular values')
ax.grid()
As expected for a heat equation, the Hankel singular values decay rapidly.
## Running balanced truncation¶
The balanced truncation method consists of finding a balanced realization of the full-order LTI system and truncating it to obtain a reduced-order model. In particular, there exist invertible transformation matrices $$T, S \in \mathbb{R}^{n \times n}$$ such that the equivalent full-order model with $$\widetilde{E} = S^T E T = I$$, $$\widetilde{A} = S^T A T$$, $$\widetilde{B} = S^T B$$, $$\widetilde{C} = C T$$ has Gramians $$\widetilde{P}$$ and $$\widetilde{Q}$$, i.e., solutions to Lyapunov equations
\begin{split}\begin{align} \widetilde{A} \widetilde{P} + \widetilde{P} \widetilde{A}^T + \widetilde{B} \widetilde{B}^T & = 0, \\ \widetilde{A}^T \widetilde{Q} + \widetilde{Q} \widetilde{A} + \widetilde{C}^T \widetilde{C} & = 0, \\ \end{align}\end{split}
such that $$\widetilde{P} = \widetilde{Q} = \Sigma = \operatorname{diag}(\sigma_i)$$, where $$\sigma_i$$ are the Hankel singular values. Then, taking as basis matrices $$V, W \in \mathbb{R}^{n \times r}$$ the first $$r$$ columns of $$T$$ and $$S$$ (possibly after orthonormalization), gives a reduced-order model
\begin{split}\begin{align} \widehat{E} \dot{\widehat{x}}(t) & = \widehat{A} \widehat{x}(t) + \widehat{B} u(t), \\ \widehat{y}(t) & = \widehat{C} \widehat{x}(t) + D u(t), \end{align}\end{split}
with $$\widehat{E} = W^T E V$$, $$\widehat{A} = W^T A V$$, $$\widehat{B} = W^T B$$, $$\widehat{C} = C V$$, which satisfies the $$\mathcal{H}_\infty$$ (i.e., induced $$\mathcal{L}_2$$) error bound
$\sup_{u \neq 0} \frac{\lVert y - \widehat{y} \rVert_{\mathcal{L}_2}}{\lVert u \rVert_{\mathcal{L}_2}} \leqslant 2 \sum_{i = r + 1}^n \sigma_i.$
Note that any reduced-order model (not only from balanced truncation) satisfies the lower bound
$\sup_{u \neq 0} \frac{\lVert y - \widehat{y} \rVert_{\mathcal{L}_2}}{\lVert u \rVert_{\mathcal{L}_2}} \geqslant \sigma_{r + 1}.$
To run balanced truncation in pyMOR, we first need the reductor object
bt = BTReductor(fom)
Calling its reduce method runs the balanced truncation algorithm. This reductor additionally has an error_bounds method which can compute the a priori $$\mathcal{H}_\infty$$ error bounds based on the Hankel singular values:
error_bounds = bt.error_bounds()
fig, ax = plt.subplots()
ax.semilogy(range(1, len(error_bounds) + 1), error_bounds, '.-')
ax.semilogy(range(1, len(hsv)), hsv[1:], '.-')
ax.set_xlabel('Reduced order')
ax.set_title(r'Upper and lower $\mathcal{H}_\infty$ error bounds')
ax.grid()
To get a reduced-order model of order 10, we call the reduce method with the appropriate argument:
rom = bt.reduce(10)
Instead, or in addition, a tolerance for the $$\mathcal{H}_\infty$$ error can be specified, as well as the projection algorithm (by default, the balancing-free square root method is used). The used Petrov-Galerkin bases are stored in bt.V and bt.W.
We can compare the magnitude plots between the full-order and reduced-order models
fig, ax = plt.subplots()
fom.mag_plot(w, ax=ax, label='FOM')
rom.mag_plot(w, ax=ax, linestyle='--', label='ROM')
ax.legend()
ax.grid()
and plot the magnitude plot of the error system
(fom - rom).mag_plot(w)
plt.grid()
We can compute the relative errors in $$\mathcal{H}_\infty$$ or $$\mathcal{H}_2$$ (or Hankel) norm
print(f'Relative Hinf error: {(fom - rom).hinf_norm() / fom.hinf_norm():.3e}')
print(f'Relative H2 error: {(fom - rom).h2_norm() / fom.h2_norm():.3e}')
Relative Hinf error: 3.702e-05
Relative H2 error: 6.399e-04
Note
To compute the $$\mathcal{H}_\infty$$ norms, pyMOR uses the dense solver from Slycot, and therefore all of the operators have to be converted to dense matrices. For large systems, this may be very expensive.
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2020-10-28 13:42:04
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http://mathhelpforum.com/statistics/129644-probability-question-help-please-print.html
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• Feb 19th 2010, 11:09 AM
KyKy
Hey i need help on this question, after reading the book it doesn't give any helpful advice on how to solve this problem, or how to do it.
When a truckload of apples arrives at a packing plant, a random sample of 150 is selected and examined for bruises, discoloration, and other defects. The whole truckload will be rejected if more than 5% of the sample is unsatisfactory. Suppose that in fact 8% of the apples on the truck to not meet the desired standard. What's the probability that the shipment will be accepted anyways?
• Feb 19th 2010, 12:23 PM
vince
Quote:
Originally Posted by KyKy
Hey i need help on this question, after reading the book it doesn't give any helpful advice on how to solve this problem, or how to do it.
When a truckload of apples arrives at a packing plant, a random sample of 150 is selected and examined for bruises, discoloration, and other defects. The whole truckload will be rejected if more than 5% of the sample is unsatisfactory. Suppose that in fact 8% of the apples on the truck to not meet the desired standard. What's the probability that the shipment will be accepted anyways?
5%*150 = 7.5 defects.
Let X be a binomial random variable; the number of defective apples found.
Use a binomial distribution with p=8% and compute
$P(shipment\;will\; not\; be\; rejected) =P(X\leq{7}) = \sum_{k=0}^{7}\binom{150}{k}(.08)^k(.92)^{150-k}$
• Feb 19th 2010, 12:33 PM
vince
one thing that's bugging me about using a binomial here is that once we find a defective apple, the remaining number of defective apples in the shipment is lessened -- we're not replacing the defective apples to keep the proportion the same throughout the trials. i think im wrong actually, but close. let me now consider poisson.
• Feb 19th 2010, 01:00 PM
vince
nah...since these are independent trials, the realization of a defective apple doesnt propagate to affect the probabaility of the next trial's outcome. i think it's fine as is. i get the prob is ~8%. poisson would work here too since 150*.08 isn't large...
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2017-06-26 10:49:13
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http://mathoverflow.net/feeds/question/108825
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Conditional probability with permutations - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T09:33:51Z http://mathoverflow.net/feeds/question/108825 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108825/conditional-probability-with-permutations Conditional probability with permutations Rodolphe 2012-10-04T14:47:30Z 2012-11-30T09:22:00Z <p>Hello,</p> <p>This problem looks very simple and I conjecture it's true but I have a hard time proving it. It'd be very useful for my work (I'm doing a PhD) and I'll be glad to cite you in a future article if you help me.</p> <p>Let $P$ be a random permutation of $\mathbb{Z}/N\mathbb{Z}$ with the condition that $P$ verifies $q$ equations : $P(a_i)=b_i, i\leq q$. Let $k_0, k_1$ be random and $x_1, x_2, y_1, y_2$ fixed numbers with $x_1\neq x_2, y_1\neq y_2$</p> <p>Prove that : $$Pr[P(x_2+k_0)=y_2+k_1 | P(x_1+k_0)=y_1+k_1] \geq (1-\frac{q}{N}) \frac{1}{N-1}$$</p> <p>Thank you !</p> http://mathoverflow.net/questions/108825/conditional-probability-with-permutations/108843#108843 Answer by David Feldman for Conditional probability with permutations David Feldman 2012-10-04T18:44:44Z 2012-10-04T19:25:55Z <p>My second comment indicates that I think you need to amend your conjecture, or else I don't understand. </p> <p>Allow me to sketch some relevant ideas for getting non-trivial lower bounds. If my sketch does not suffice, I'll try to flesh it out when I have more time.</p> <p>One should think of permutations here in terms of their cycle structures.</p> <p>Your $q$ equations join some of the elements of ${\Bbb Z}/N{\Bbb Z}$ into cycles and others into finite order segments, each with, let's say, a head and a tail. Write $H$ for the set of heads, $T$ for the set of tails. Write $G$ for the graph of the partial function the equations determine.</p> <p>Specifying a permutation satisfying the equations amounts to giving a bijection from $T$ to $H$. Since $|T|=|H|=N-q$, $(N-q)!$ permutations satisfy the equations and crucially, any given tail will have probability $1/(N-q)$ of joining any particular head. </p> <p>The particular equations don't matter to your conjecture, only the resulting $T$ and $H$. </p> <p>As per my comment, take $x_1=y_1=0$. </p> <p>Now your probability conditions on either $(k_0,k_1)\in T\times H$ or $(k_0,k_1)\in G$.</p> <p>Your probability calculation reduces to estimating the probabilities that either $(x_2+k_0,y_2+k_1)\in T\times H$ or $(x_2+k_0,y_2+k_1)\in G$.</p> <p>That makes four cases to consider and I confess I have not yet worked out the details.</p> <p>This helps with one case: given $T$, $H$ both of cardinality $N-q$, how small can we have the intersection $T\times H \cap ((T+x_2)\times (H+y_2))$. If $q$ is not too small, the pigeon-hole principle gives a lower bound (but this does not exploit the product structure of $T\times H$).</p> http://mathoverflow.net/questions/108825/conditional-probability-with-permutations/108866#108866 Answer by Rodolphe for Conditional probability with permutations Rodolphe 2012-10-04T22:40:52Z 2012-10-04T22:40:52Z <p>Thank you for your answer. I think I solved the problem but it's just the beginning. I had something wrong in the conjecture.</p> <p>First, let's note $C_i$ the event $P(x_i+k_0)=y_i+k_1$.</p> <p>We have to slightly change the conjecture (I forgot a factor 2) : $$Pr[C_2|C_1]\geq (1-\frac{2q}{N})\times\frac{1}{N-1}.$$</p> <p>We have $$Pr[C_2|C_1]=Pr[C_2\cap C_1]/Pr[C_1]$$ and I know that $Pr[C_1]=1/N$ (easy computation) and for $C_2\cap C_1$, if $x_1+k_0$ and $x_2+k_0$ are not one of the $a_i$ and $y_1+k_1, y_2+k_1$ are not one of the $b_i$ then the two equations occur with probability $\frac{1}{(N-q)(N-1-q)}$ so we have : $$Pr[C_2|C_1]\geq (N-2q)/N\times (N-2q)/N \times \frac{1}{(N-q)(N-1-q)} \times N$$ which almost solve the conjecture (I don't mind the term in $q²/N²$).</p> <p>Now I have to prove something like $$Pr[C_3|C_2,C_1]\geq (1-\frac{2q}{N})\times\frac{1}{N-2}$$</p>
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2013-05-23 09:34:02
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http://scholarbank.nus.edu.sg/handle/10635/43596
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Please use this identifier to cite or link to this item: http://scholarbank.nus.edu.sg/handle/10635/43596
Title: Patient Flow Management in Emergency Departments Authors: HUANG JUNFEI Keywords: Patient flow, service management, Klimov's model, feedback, modified generalized c\mu rule Issue Date: 23-May-2013 Source: HUANG JUNFEI (2013-05-23). Patient Flow Management in Emergency Departments. ScholarBank@NUS Repository. Abstract: In this thesis, we consider the control of patient flow through physicians in emergency departments (EDs), which have attracted many researchers' attention. Our work here seems to be the first model to quantitatively analyze the control of patient flow in an emergency department from a queueing theory perspective. Problem: In emergency departments, the physicians must choose between catering to patients right after triage, who are yet to be checked, and those that are work-in-process (WIP), who are occasionally returning to be checked. The service requirements for the two kinds of patients are different: for the patients right after triage, they must see a doctor within targeted time windows (that may depend on the patients' severity and other parameters); while the WIP patients, on the other hand, impose congestion costs. The physicians in the emergency departments have to balance between triage and WIP patients so as to minimize costs, while meeting the constraints on the time-till- first-service. Model: We model this prioritization problem as a queueing system with multi-class customers, combining deadline constraints, feedback and congestion costs together. We consider two types of congestion costs: per individual visit to a server or cumulative over all visits. The former is the base-model, which paves the way for the latter (more ED-realistic) one. Method: The method we use is conventional heavy-traffic analysis in queueing theory, based on the empirical evidence that the emergency departments can be viewed as critically-loaded stationary systems between late morning till late evening. We propose and analyze scheduling policies that asymptotically minimize congestion costs while adhering to all deadline constraints. Solution: The policies have two parts: the first chooses between triage and WIP patients using a simple threshold policy; assuming triage patients are chosen, the physicians serve the one with the largest delay relative to deadline; alternatively, WIP patients are served according to some generalized c\mu policy, in which is simply modified to account for feedbacks. The policies that we propose are easy to implement and, from an implementation point of view, has the appealing property that all information required is indeed typically available in emergency departments. For the proposed policies, asymptotic optimality, as well as some congestion laws that support forecasting of waiting and sojourn times, are established. Application: Finally, via data from the complex ED reality, we use our models to quantify the value of refined individual information, for example, whether an ED patient will be admitted to the hospital as opposed to being discharged. This is an illustration on how our recommendations can improve the operational efficiency and service quality. URI: http://scholarbank.nus.edu.sg/handle/10635/43596 Appears in Collections: Ph.D Theses (Open)
###### Files in This Item:
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2017-12-13 19:14:57
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http://mathhelpforum.com/calculus/274332-graphing-using-largest-integer-function.html
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# Thread: Graphing using largest integer function
1. ## Graphing using largest integer function
The question is:
"The symbol [x] denotes the largest integer which is less than or equal to x. For example [2.4] = 2, [3] = 3, [-1.2] = -2. Draw the graph of: (see attached)
Any help would be greatly appreciated, I am really stumped as to how it would look.
2. ## Re: Graphing using largest integer function
on the interval $[0,1)$, $y_0 = 0 + \sqrt{x - 0} = \sqrt{x}$
on the interval $[1,2)$, $y_1 = 1 + \sqrt{x - 1}$, the graph of $y_0$ shifted right one unit and up one unit
on the interval $[2,3)$, $y = 2 + \sqrt{x - 2} = 1 + \left[1 + \sqrt{(x-1)-1}\right]$, the graph of $y_1$ shifted right one unit and up one unit
the pattern repeats on subsequent (and prior) intervals ...
3. ## Re: Graphing using largest integer function
Originally Posted by thebestrose828
The question is:
"The symbol [x] denotes the largest integer which is less than or equal to x. For example [2.4] = 2, [3] = 3, [-1.2] = -2. Draw the graph of: (see attached)
Any help would be greatly appreciated, I am really stumped as to how it would look.
Here is the plot.
In most cases, in modern mathematics the floor function is substituted for the greatest integer fumction.
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2018-02-24 07:03:19
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https://math.stackexchange.com/questions/1831287/precedence-of-operations-in-vector-spaces
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# Precedence of operations in vector spaces
Suppose that $V$ is a vector space over $\mathbb R$ (for simplicity) with addition denoted by $\oplus$ and scalar multiplication denoted by $\otimes$. Let $\mathbf u, \mathbf w \in V$ and let $\lambda \in \mathbb R$ and suppose we are asked to compute $$\lambda \otimes \mathbf u \oplus \mathbf v$$ I was wondering how to do this. I would tend to say it is $$(\lambda \otimes \mathbf u) \oplus \mathbf w$$ but it might as well be $$\lambda \otimes (\mathbf u \oplus \mathbf w)$$ since, as far as I understand, none of the axioms for a vector space settles the precedence of scalar multiplication over vector addition. Is there a general rule for this ?
EDIT: I "know" that the cultural rule is precedence of scalar multiplication over vector addition. This is hinted at from shortcuts like $$\lambda(\mathbf u + \mathbf v) = \lambda \mathbf u + \lambda \mathbf v$$ which (I think) should be written as $$\lambda(\mathbf u+ \mathbf v) = (\lambda \mathbf u) + (\lambda \mathbf v) \,.$$ unless everyone agrees beforehand that scalar multiplication takes precedence. The thing I am wondering about is whether a computer (say) would be able to deduce this precedence from the axioms alone.
Thanks a lot for your help!
• $\lambda \otimes u\oplus v$ is not well-defined. In order to avoid such dilemmas, you either write $\lambda\otimes (u\oplus w)$ or $(\lambda\otimes u)\oplus w$. – Levent Jun 18 '16 at 21:05
• @Levent That has to be wrong. Otherwise why would the distributive axiom be written the way it always is. We always write $\lambda(u+v)=\lambda u+\lambda v$. Are you claiming the RHS is not well-defined? – almagest Jun 18 '16 at 21:10
• No, what I mean is $\lambda u+v$ is not well-defined, not $\lambda(u+v)=\lambda u+\lambda v$. But commonly we just use $\lambda u+v$ to denote $(\lambda u)+v$. – Levent Jun 18 '16 at 21:11
• @Levent. You miss the point. If $\lambda u+v$ was not well-defined, then $\lambda u+\lambda v$ would not be well-defined, because it might mean $(\lambda u)+(\lambda^2v)$. But $\lambda u+\lambda v$ obviously is well-defined because it appears in the axioms. So $\lambda u+v$ is also well-defined - it means $(\lambda u)+v$. – almagest Jun 18 '16 at 21:19
• @almagest your argument is invalid in $F_2$. – Daniel Valenzuela Jun 18 '16 at 21:20
This is just a convention. And the usual convention is to continue the shortcuts used for operations among numbers (and be it in order not to confuse the reader when we already reuse the notation from numeric operations for vectors). In other words, the convention is that $\lambda\otimes u\oplus v$ is a shorthand writing for $(\lambda\otimes u)\oplus v$. Note that it is in fact common to (re-)use the multiplication and addition symbols from numeric operations and in fact write $\lambda\cdot u+v$ or just $\lambda u+v$.
In case of a vector space with a scalar product, you may even see $\lambda \cdot u \cdot v$ (or $\lambda u v$) with the dot standing for two completely different operations. Fortunately, it is true that $(\lambda\cdot u)\cdot v=\lambda\cdot(u\cdot v)$ (an equation that even has three distinct operations denoted by the same dot symbol!).
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2019-10-23 15:37:08
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http://www.chegg.com/homework-help/questions-and-answers/-q3489514
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## curl of vector in cylindrical coordinate
• In general if A, B and C are the three components of a vector (so F = <A, B, C>, then the curl is the vector
curl(F) = <dC/dy - dB/dz, dA/dz - dC/dx, dB/dx - dA/dy>
where dC/dy is the derivative of C with respect to y, and so on.
Here all derivatives are zero except dA/dy, dB/dx and dC/dx. You should be able to figure out these derivatives.
• Anonymous commented
no concederation of the coordinate system
• Anonymous commented
• Anonymous commented
you are treating rho and phi as x and y vectors. this yields a wrong answer.
• Anonymous commented
we have to consider like that itself
• Anonymous commented
and iam confident with my answer other wise post which was the correct answer
• Anonymous commented
there is supposed to be a one over rho in somewhere in the rho term and the z term
• This is a simple computation with proper substitution of limits.
a)
b)
Direct computation is quite complicate but we can use divergence theorem which states that
Flux = (S) ∫∫ F dot n dS = (W) ∫∫∫ divF dV,
i.e. the flux over the closed surface S is equal to the tripple integral over the enclosed volume W of the divergence of F.
Get homework help
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2013-05-19 02:03:32
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http://tex.stackexchange.com/questions/95825/make-all-elements-in-pause-cleared-beamer
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# Make all elements in pause cleared (Beamer)
I have more than two elements in one frame that I would like to have in \pause. But the \pause make me appear only the next element cleared.
I want that all elements in the same frame in pause appear cleared and then go colored one by one. How can I do?
\documentclass{beamer}
\begin{document}
\begin{frame}\frametitle{Title}
\begin{block}{}
\end{block}
\pause
\begin{block}{}
\end{block}
\pause
\begin{block}{}
\end{block}
\end{document}
-
I think you need the \onslide and \alert commands of the OVERLAY section of the Beamer's manual.
\documentclass{beamer}
\begin{document}
\begin{block}{Title}
\begin{itemize}
\item A
\item B
\item C
\end{itemize}
\end{block}
\begin{block}{Title}
\begin{itemize}
\item A
\item B
\item C
\end{itemize}
\end{block}
\begin{block}{Title}
\begin{itemize}
\item A
\item B
\item C
\end{itemize}
\end{block}
\end{frame}
\end{document}
A small explanation:
[<+->] tells Beamer that the default overlay will be to show each item one by one.
alert@+ tells Beamer to color differently the current item.
And [<+-|alert@+>] make the both.
Result:
Edit
Like that?
\documentclass{beamer}
\begin{document}
\begin{frame}\frametitle{Title}
\begin{block}<1-4>{Title}
\begin{itemize}
\item<1,2-4> A
\item<1,3-4> B
\item<1,4> C
\end{itemize}
\end{block}
\end{frame}
\end{document}
Result:
-
mmm, it doesn't solve the problem. I want that when I change the frame, all elements in pause result cleared end then get colored one by one. – Ford Prefect Jan 29 '13 at 9:03
I have updated my answser – Lionel MANSUY Jan 29 '13 at 9:52
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2014-07-10 10:22:56
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https://wiki.bencwbrown.co.uk/algebra/representation/representation.html
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# # Constructions
## # Tensor Product
The tensor product of two representations $(\rho,V)$ and $(\sigma,W)$ of a group $G$ is the representation $(\rho \otimes \sigma, V \otimes W)$ defined by the condition
$(\rho \otimes \sigma)(g)(v\otimes w) := \rho(g)(v) \otimes \sigma(g) (w),$
and extended to all vectors in $V \otimes W$ by linearity.
Last Updated: 12/17/2019, 6:02:45 AM
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2020-01-28 19:38:42
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https://tex.stackexchange.com/questions/461054/how-to-correctly-write-the-thousands-separator-in-american-english
|
How to correctly write the thousands separator in American English?
I am not a native speaker and I must write a scientific document in American English. How would you correctly write "10000" using LaTeX ?
1. with a space only (10 000): $10\,000$
2. with a coma and no space in either side (10,000): $10{,}000$
3. with a coma and a space on its right (10, 000) : $10,000$
I have seen the second a few times, but I do not know if there is a consensus. Thanks for your help!
• I would recommend that you use the siunitx package and then \num{10000}. That way you can always change it later. Without siunitx, I would use the first option your suggest: $10\,000$ Nov 21, 2018 at 2:08
• you can use numprint package with option autolanguage (assuming you use babel with american or USenglish class option)
– user4686
Nov 21, 2018 at 8:33
• @PeterGrill thanks a lot, I think I will go on with this for now. Could you propose it as an answer so I can accept it ? Nov 21, 2018 at 16:16
My recommendation would be to use siunitx's \num{} macro (which by default will add a \thinspace, but can be configured to add a comma if so desired):
Code:
\documentclass{article}
\usepackage{siunitx}
\begin{document}
\num{10000}: Recommended (Using \verb|\num{}|)
$10\,000$: Manual Spacing
\sisetup{group-separator={,}}
\num{10000}: In case you decide you want the comma
\end{document}
|
2022-05-23 05:21:18
|
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http://ilnumerics.net/blog/tag/linq/
|
# Tag Archives: Linq
## ILNumerics and LINQ
As you may have noticed, ILNumerics arrays implement the IEnumerable interface. This makes them compatible with ‘foreach’ loops and all the nice features of LINQ!
Consider the following example: (Dont forget to include ‘using System.Linq’ and derive your class from ILNumerics.ILMath!)
ILArray<double> A = vec(0, 10);
Console.WriteLine(String.Join(Environment.NewLine, A));
Console.WriteLine("Evens from A:");
var evens = from a in A where a % 2 == 0 select a;
Console.WriteLine(String.Join(Environment.NewLine, evens));
return;
Some people like the Extensions syntax more:
var evens = A.Where(a => a % 2 == 0);
I personally find both equivalently expressive.
## Considerations for IEnumerable<T> on ILArray<T>
No option exist in IEnumerable<T> to specify a dimensionality. Therefore, and since ILNumerics arrays store their elements in column major order, enumerating an ILNumerics array will be done along the first dimension. Therefore, when used on a matrix, the enumerator runs along the columns:
ILArray<double> A = counter(3,4);
Console.WriteLine(A + Environment.NewLine);
Console.WriteLine("IEnumerable:");
foreach(var a in A)
Console.WriteLine(a);
… will give the following:
<Double> [3,4]
1 4 7 10
2 5 8 11
3 6 9 12
IEnumerable:
1
2
3
4
5
6
7
8
9
10
11
12
Secondly, as is well known, accessing elements returned from IEnumerable<T> is only possible in a read-only manner! In order to alter elements of ILNumerics arrays, one should use the explicit API provided by our arrays. See SetValue, SetRange, A[..] = .. and GetArrayForWrite()
Lastly, performance considerations arise by excessive utilization of IEnumerable<T> in such situations, where high performance computations are desirable. ILNumerics does integrate well with IEnumerable<T> – but how well IEnumerable<T> does integrate into the memory management of ILNumerics should be investigated with help of your favorite profiler. I would suspect, most every day scenarios do work out pretty good with LINQ since it concatenates all expressions and queries and iterates the ILNumerics array only once. However, let us know your experiences!
|
2017-07-20 18:40:36
|
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|
https://wiki.cs.byu.edu/facwiki/scraped-NNML/NNML/index.php/Meta-learning.html
|
# Meta-learning
## Meta-learning
### Objective
The objective of this project is to do large-scale meta-learning using various techniques and get several papers accepted in top tier venues so we can graduate with good jobs.
### Action Items
#### Overall
• 1st--create an extensible data set available to the community similar to the UCI data repository. meta-data set
• Look at machine-learned ranking (MLR) algorithms . This will compare our ranking with other ranking algorithms
• Maybe look for some implementations of MLR algorithms
• Also look at the evaluation metrics that they use for ranking
• Look at getting more familiar with Recommender System stuff that is already out there.
#### Our Algorithms
• Unsupervised backpropagation (maybe other collaborative filtering algorithms as well)
• Just the accuracies
• Accuracies and the data set meta-features
• Also try using the rankings instead of the accuracies
• Trained neural network trained with the data set meta-features
• Trained neural network with latent variables from unsupervised backprop
• Think about how to incorporate the time that an algorithm takes to run
• Think about to incorporate parameter settings into the algorithm such that we can rank the algorithm and the parameter settings.
#### Competitors
• Brazdil 5-NN with his meta-features
• Maybe some ranking algorithms??
• RT Rank is an available implementation that we should try
#### Data Points
• Continue to gather more accuracies from random parameter settings
• Look at this data set: HERE
@inproceedings{Reif2012,
author = {Matthias Reif},
title = {A Comprehensive Dataset for Evaluating Approaches of Various
title = {ICPRAM 2012 - Proceedings of the 1st International Conference
on Pattern Recognition Applications and Methods, Volume
1, Vilamoura, Algarve, Portugal, 6-8 February, 2012},
year = {2012},
pages = {273-276},
}
#### Other Thoughts/Ideas
• We should see how we do as a ranking algorithm as well and/or we can improve on collaborative filtering
• There is some data from Microsoft research
• LETOR , also from Microsoft research has a number of data sets, and other information that could prove useful if we choose to also pursue this course.
• How to add a single instance after training up a collaborative filtering model.
• How to add results from a novel learning algorithm/parameter settings
• Make it incremental
• Here are some references for incremental recommender systems (collaborative filtering):
#### Experiments
• Collaborative filtering (Just accuracies and accuracies with meta-features)
• For the initial experiments, run with just the accuracies missing but with the meta-features in the training data
• Models to try:
• Matrix Factorization
• Non-linear PCA
• Unsupervised Backprop
• Try without hidden layer to be able to compute intrinsic variables for novel instances
• Fuzzy K-Means
• Classification-based approaches
• Make sure to use a neural network with Backprop
• Combination of classification with collaborative filtering
• Ranking algorithms
#### Logan
5-2-2013
• Get more data points (i.e. data sets). More data sets can be found at:
• Run learning algorithms over the new data sets
• Set upper limit to 100 hours? If a few don't finish, that's OK
• Parameter optimization (10 random searches?)
• Get meta-features on the data sets
• Hardness heuristics (See Mike)
• Brazdil
• Run waffles over the results
• Just collaborative filtering
• Just accuracies
• using the Spearman Corralation Coefficient: .64822 removing 30% of data
• using the Spearman Corralation Coefficient: .67633 removing 30% of data
• Just a neural network
• Both
• Previous work (Brazdil)
• compare results of other methods
• Ranking Algorithms
• Use and implement other ranking algorithms
• Compare how often recommend selects as the best: the best. the second best... the worst.
• Compare how often recommend selects as 2nd best: the best. the second best... the worst.
• etc.
• Future Ideas
• parameter optimization
• train model to predict accuracies of a specific model/dataset given meta-features of the dataset and parameters of the model.
#### Mike
5-2-2013
• Get Logan code for meta-features
• Get Logan code for random hyper-parameter selection
• Help Logan see overall picture to help with design for the application
#### Rob
5-2-2013
• Get data sets for Logan
• Help Logan see overall picture to help with design for the application
### Related Works
Random Hyper-Parameter Optimization
@article{Bergstra2012,
author = {Bergstra, James and Bengio, Yoshua},
title = {Random Search for Hyper-Parameter Optimization},
journal = {Journal of Machine Learning Research},
volume = {13},
month = March,
year = {2012},
issn = {1532-4435},
pages = {281--305},
numpages = {25},
url = {http://dl.acm.org/citation.cfm?id=2188385.2188395},
acmid = {2188395},
publisher = {JMLR.org},
}
Brazdil: Ranking Learning Algroithms
@Article{ Brazdil2003,
author = "Pavel B. Brazdil and Carlos Soares and Joaquim Pinto Da Costa",
title = "Ranking Learning Algorithms: Using IBL and Meta-Learning on Accuracy and Time Results",
journal = "Machine Learning",
volume = "50",
number = "3",
year = "2003",
pages = "251--277",
doi = "http://dx.doi.org/10.1023/A:1021713901879",
annote = "This work presents a method to rank learning algorithms according to their utility given a data set. The similarity of a data set is compared to a set of previously processed data sets using \textit{k}-NN on a set of meta-features. The learning algorithm ranking method uses aggregate information concerning classification accuracy as well as the run-time of the learning algorithm. This work ranks learning algorithms as will be done in a part of this thesis, however, this thesis rank learning algorithms according to classification accuracy according to instance hardness."
}
Ho and Basu: Complexity Measures
@Article{ Ho2002,
author = "Tin Kam Ho and Mitra Basu",
title = "Complexity Measures of Supervised Classification Problems",
journal = "IEEE Trans. Pattern Anal. Mach. Intell.",
volume = "24",
issue = "3",
month = "March",
year = "2002",
pages = "289--300",
numpages = "12",
acmid = "507476",
publisher = "IEEE Computer Society",
|
2017-03-25 01:52:47
|
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|
https://www.storyofmathematics.com/glossary/division/
|
# Division|Definition & Meaning
## Definition
The division is defined as the process of distributing a fixed quantity of items into sub-quantities in such a manner that every sub-quantity has an equal number of items in them. The term quantity could be anything like fruits, books, or arithmetic numbers. Out of the four essential arithmetic operations of mathematics Subtraction, Addition, and Multiplication, Division is one of them.
## Concept Building
Dividing simply means making equal parts or dividing into equal amounts among a group. An equal-area triangle consists of two equal triangles divided by a diagonal in a square. Depending on the division operation, the result may be an integer or not. Decimal numbers may sometimes be used to represent the result.
Figure 1 – Symbol of Division
Let us take an example of division. There are 50 students and 5 classes; we want to divide the students among the 5 classes. What we will do is simply split the students into 5 batches of 10 students each, and send each batch to a separate class. Thus, we have assigned 50 students equally in 5 classes, with each class containing 10 students
We can consider division as a repetition process of successive subtraction or the opposite of the multiplication operation.
## Symbol Representation
There are generally three different symbols used for division shown above
### Divide ($\boldsymbol\div$)
Division of number 6 with 3 gives:
6 $\div$ 3 = 2
### Forward Slash ($\boldsymbol/$)
Division of 4 books among 2 students:
4/2 = 2
### Bar ($\boldsymbol{\frac{\phantom{a}}{\phantom{b}}}$)
Division of 8 apples among 4 people:
$\mathsf{\dfrac{8}{4}}$ = 2
## Mathematical Formula for Division
$\mathsf{\dfrac{Dividend}{Divisor}}$ = Quotient + Remainder
### Dividend
The dividend is the term or item which we want to split into different equal sub-parts. For example, if we want to divide 10 oranges among 5 people then the 10 oranges will be the dividend.
### Divisor
The divisor is the term that split the dividend into equal sub-parts or in other words we can say the term with which we want to divide something. In the above example, 5 people are the divisor.
### Quotient
The result obtained by the splitting of the two items is known as the quotient, for example dividing 10 oranges among 5 people gives a quotient of 2. In other words, the quotient is the term that decides the quantity of each subgroup.
### Remainder
The remainder is the residual result left after the division operation for example division of 10 pencils among 3 students results in the remainder equal to 1.
Figure 2 – Parts of Division Formula
## Order of Division Operation
In mathematics, Division is highly prioritized among other arithmetic operations. Division operation has the first-order meaning that whenever there is an arithmetic problem we will first perform division operation.
Figure 3 – Order of Division Operation
## Properties of Division
1. The division of any number by integer 1 gives the number itself or in other words division of any number gives the dividend as the result.
2. The division of any number by integer 0 gives infinity or more precisely not defined.
3. The division of a number by the number itself gives the result as 1 or in other words, if the dividend and divisor are the same then the result will be equal to 1.
4. If the dividend is 0 then the division operation with any other number gives 0.
5. If we divide any number by 100 the remainder will be equal to the number at the unit and tens position and the rest of the number will be considered quotient.
## Dividing Fractions
It is possible to divide fractions as well. Division operators need to be transformed into multiplication operators when dividing fractions. The following example shows the concept.
$\mathsf{ \frac{4}{3} \div \frac{4}{2} = \frac{4}{3} \times \frac{2}{4} = \frac{2}{3} }$
In fraction, the division sign is changed into a multiply sign with the fraction being reversed as shown.
## Dividing Float Numbers
Mathematics consists of a variety of concepts that divide decimal numbers. In many ways, the decimal division is similar to the fractional division. To understand how decimal division works, let’s look at the example below.
4.5 $\div$ 2.5 = $\mathsf{ \dfrac{45}{25} }$ = $\mathsf{\dfrac{9}{5}}$ = 1.8
## Dividing Polynomials
It is possible to divide polynomials by other polynomials, as opposed to numbers and fractions. Two methods are available for polynomial division, however. As opposed to dividing numbers, the polynomial long division uses polynomial expressions to divide polynomial expressions. The synthetic division method is another way to divide polynomials.
## Steps for Division
1. First, figure out the data and define terms such as divisor and dividend.
2. Write the dividend inside the division symbol hat as shown below example.
3. Write the divisor outside the hat of the division symbol as shown below example.
4. Now start dividing the dividend by the divisor and write the result on the top of the hat.
5. This result written on the top will be considered as the quotient.
6. If the dividend is not completely divided then bring the next digit down and repeat the process.
7. On repeating the process if we got any residual result left at the end and further dividend cant be divided then we will consider this residual result as the remainder.
## Examples of Division
### Example 1
Consider a scenario there are 51 players, and we want to make n teams in such a way that each team consists of players not more or less than 3. Describe the division process graphically also. How many numbers of teams can be formed?
### Solution
As from the problem statement, we are provided with the dividend as 51 players and divisor as 3, so in order to find the n teams that can be formed, we will simply divide 51 by 3, which gives us 17. So we can make 17 teams by splitting or dividing 51 players into teams of 3 players each. The calculation is shown below.
Figure 4 – Division of Players Examples
### Example 2
Harry bought 16 Oranges, 4 Bananas, and 20 Apples for his family. The number of count of his family member is 4, including him. How will you divide all the fruits among family members of Harry?
### Solution
We illustrate the division process in figure 5 below.
Figure 5 – Division of Fruits Example
#### Division of Oranges
Total oranges = 16
Number of family members = 4
Dividend = 16
Divisor = 4
Solving (leftmost in figure 5), we get:
Quotient = 4
#### Division of Bananas
Total bananas = 4
Number of family members = 4
Dividend = 4
Divisor = 4
Solving (middle in figure 5), we get:
Quotient = 1
#### Division of Apples
Total apples = 20
Number of family members = 4
Dividend = 20
Divisor = 4
Solving (rightmost in figure 5), we get:
Quotient = 5
So each family member got 4 oranges, 1 banana, and 5 apples.
All mathematical drawings and images were created with GeoGebra.
|
2023-03-29 19:35:49
|
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|
http://www.fightfinance.com/?q=279,292,353,493,501,531,543,593,634,836
|
# Fight Finance
#### CoursesTagsRandomAllRecentScores
Scores keithphw $5,541.61 Visitor$610.00 Visitor $330.00 Jayt$280.00 Visitor $210.00 Visitor$160.00 Visitor $150.00 Visitor$150.00 Visitor $150.00 Visitor$130.00 Visitor $114.43 JL Null$59.09 Visitor $40.00 Visitor$40.00 Visitor $39.09 Visitor$30.00 Visitor $30.00 Visitor$30.00 SamirBha... $20.00 Visitor$20.00
Do you think that the following statement is or ? “Buying a single company stock usually provides a safer return than a stock mutual fund.”
Find the sample standard deviation of returns using the data in the table:
Stock Returns Year Return pa 2008 0.3 2009 0.02 2010 -0.2 2011 0.4
The returns above and standard deviations below are given in decimal form.
A residential investment property has an expected nominal total return of 6% pa and nominal capital return of 3% pa.
Inflation is expected to be 2% pa. All rates are given as effective annual rates.
What are the property's expected real total, capital and income returns? The answer choices below are given in the same order.
A firm has 2m shares and a market capitalisation of equity of $30m. The firm just announced earnings of$5m and paid an annual dividend of $0.75 per share. What is the firm's (backward looking) price/earnings (PE) ratio? The below graph shows a project's net present value (NPV) against its annual discount rate. Which of the following statements is NOT correct? Which of the following people is most in danger of being personally bankrupt? Assume that all of their businesses' assets are highly liquid and can therefore be sold immediately. For an asset price to triple every 5 years, what must be the expected future capital return, given as an effective annual rate? The price of gold is currently$700 per ounce. The forward price for delivery in 1 year is $800. An arbitrageur can borrow money at 10% per annum given as an effective discrete annual rate. Assume that gold is fairly priced and the cost of storing gold is zero. What is the best way to conduct an arbitrage in this situation? The best arbitrage strategy requires zero capital, has zero risk and makes money straight away. An arbitrageur should sell 1 forward on gold and: A$100 stock has a continuously compounded expected total return of 10% pa. Its dividend yield is 2% pa with continuous compounding. What do you expect its price to be in one year?
The 95% daily VaR corresponds to the result on the:
|
2018-01-21 18:42:28
|
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|
http://mathhelpforum.com/advanced-algebra/137912-commutative-ring.html
|
# Math Help - Commutative ring
1. ## Commutative ring
Let $R$ be a commutative ring with $r \in R$.
Let $f(x)$ be an element of $R[x]$.
I want to show that $f(x)$ is irreducible over $R$ if and only if $f(x + r)$ irreducible over $R$.
2. Originally Posted by bram kierkels
Let $R$ be a commutative ring with $r \in R$.
Let $f(x)$ be an element of $R[x]$.
I want to show that $f(x)$ is irreducible over $R$ if and only if $f(x + r)$ irreducible over $R$.
The map $p(x)\mapsto p(x+r),\,r\in R$ is an automorphism of the ring $R[x]$, and then:
$f(x)$reducible $\Longleftrightarrow f(x)=g(x)q(x),\,g(x),\,q(x)\in R[x]\Longleftrightarrow f(x+r)=g(x+r)q(x+r)$ via the above automorphism $\Longleftrightarrow f(x+r)$ reducible .
Tonio
|
2014-07-23 01:05:34
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https://crossminds.ai/video/model-interpretability-through-the-lens-of-computational-complexity-606fe45cf43a7f2f827c03c5/
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Model Interpretability through the lens of Computational Complexity
# Model Interpretability through the lens of Computational Complexity
Dec 06, 2020
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24 views
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###### Details
In spite of several claims stating that some models are more interpretable than others -- e.g., "linear models are more interpretable than deep neural networks" -- we still lack a principled notion of interpretability to formally compare among different classes of models. We make a step towards such a notion by studying whether folklore interpretability claims have a correlate in terms of computational complexity theory. We focus on local post-hoc explainability queries that, intuitively, attempt to answer why individual inputs are classified in a certain way by a given model. In a nutshell, we say that a class $\mathcal{C}_1$ of models is more interpretable than another class $\mathcal{C}_2$, if the computational complexity of answering post-hoc queries for models in $\mathcal{C}_2$ is higher than for those in $\mathcal{C}_1$. We prove that this notion provides a good theoretical counterpart to current beliefs on the interpretability of models; in particular, we show that under our definition and assuming standard complexity-theoretical assumptions (such as P$\neq$NP), both linear and tree-based models are strictly more interpretable than neural networks. Our complexity analysis, however, does not provide a clear-cut difference between linear and tree-based models, as we obtain different results depending on the particular post-hoc explanations considered. Finally, by applying a finer complexity analysis based on parameterized complexity, we are able to prove a theoretical result suggesting that shallow neural networks are more interpretable than deeper ones. Speakers: Pablo Barcelo, Mikaël Monet, Jorge Pérez, Bernardo Subercaseaux
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2021-06-15 10:23:39
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https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-6-section-6-3-factoring-trinomials-of-the-form-ax2-bx-c-exercise-set-page-435/35
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## Algebra: A Combined Approach (4th Edition)
$2(x+.6357)(x+7.8642)$
$2m^{2}+17m+10$ Quadratic formula: $x=\frac{-b\pm\sqrt {b^{2}-4ac}}{2a}$ $\frac{-17+\sqrt {17^{2}-4(2)(10)}}{2(2)}=-.6357$ $\frac{-17-\sqrt {17^{2}-4(2)(10)}}{2(2)}=-7.8642$ $2*(x+.6357)(x+7.8642)$
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2018-10-23 20:51:10
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http://codeforces.com/blog/entry/17119
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### fcspartakm's blog
By fcspartakm, 5 years ago, translation, ,
## 525A — Vitaliy and Patty
To solve this problem we need to use array cnt[]. In this array we will store number of keys of every type, which we already found in rooms, but didn't use. Answer will store in variable ans.
Now, we iterate on string. If current element of string si is lowercase letter (key), we make cnt[si]++. Else if current element of string si uppercase letter (door) and cnt[tolower(si)] > 0, we make cnt[tolower(si)]--, else we make ans++. It remains only to print ans.
Asymptotic behavior of this solution — O(|s|), where |s| — length of string s.
## 525B — Pasha and String
At first we need to understand next fact — it doesn't matter in wich order make reverses, answer will be the same for all orders.
Let's numerate elements of string from one. To solve given problem we need to count how many reverses will begin in every position of string. Then we need to count array sum[]. In sum[i] we need to store count of reverses of substrings, which begin in positions which not exceeding i.
Now iterate for i from 1 to n / 2 and if sum[i] is odd swap si and sn - i + 1. After that it remains only to print string s.
Asymptotic behavior of this solution — O(n + m), where n — length of string s, m — count of reverses.
## 525C — Ilya and Sticks
This problem can be solved with help of greedy. At first count array cnt[]. In cnt[i] will store how many sticks with length i we have.
Now iterate for len from maximal length of sticks to minimal. If cnt[len] is odd and we have sticks with length len - 1 (that is cnt[len - 1] > 0), make cnt[len]-- and cnt[len - 1]++. If cnt[len] is odd and we have no sticks with length len - 1 (that is cnt[len - 1] = 0), make cnt[len]--.
In this way we properly done all sawing which we need and guaranteed that all cnt[len] is even. After that iterate similary on length of sticks and greedily merge pairs from 2 sticks with the same length in fours. It will be length of sides of sought-for rectangles, left only summarize their squares in answer. In the end can left 2 sticks without pair, we must not consider them in answer.
For example, if cnt[5] = 6, cnt[4] = 4, cnt[2] = 4, we need to merge this sticks in following way — (5, 5, 5, 5), (5, 5, 4, 4), (4, 4, 2, 2). Two sticks with length 2 are left, we must not count them.
Asymptotic behavior of this solution — O(n + maxlen - minlen), where n — count of sticks, maxlen — maximal length of stick, minlen — minimal length of stick.
## 525D — Arthur and Walls
To solve this problem we need to observe next fact. If in some square whith size 2 × 2 in given matrix there is exactly one asterisk, we must change it on dot. That is if in matrix from dots and asterisks is not square 2 × 2 in which exactly one asterisk and three dots, then all maximum size of the area from dots connected by sides represent rectangles.
Now solve the problem with help of bfs and this fact. Iterate on all asterisks in given matrix and if only this asterisk contains in some 2 × 2 square, change this asterisk on dot and put this position in queue. Than we need to write standart bfs, in which we will change asterisks on dots in all come out 2 × 2 squares with exactly one asterisk.
Asymptotic behavior of this solution — O(n * m), where n and m sizes of given matrix.
## 525E — Anya and Cubes
To solve this problem we need to use meet-in-the-middle. At first sort given array in increasing order and divide it in two parts. In first part must be first n / 2 elements, in second part — other.
Iterate all submasks of all masks of elements from first part. That is iterate which cubes from first part we take and on which from them we paste exclamation marks. In this way we iterated all possible sums, which we can get with cubes from first part. Let for current submask we get sum sum_lf and use tlf exclamation marks. To store all such sums we use associative arrays map < long long > cnt[k + 1], where k — count of exclamation marks which we have in the beginning.
After that similary iterate all submasks of all masks of elements from second part. Let for current submask sum is sumrg and number of used exclamation marks is trg. Then from first part we need to get sum (s - sumrg) and we can use only (k - trg) exclamation marks, where s — sum which we must get by condition of the problem. Then iterate how many exclamation marks we will use in first part (let it be variable cur) and increase answer on cnt[cur][s - sumrg]. To accelerate our programm we may increase answer only if cnt[cur].count(s - sumrg) = true.
For submasks in iterate we can cut off iteration on current sum for submask (it must be less or equal to given s) and on current count of exclamation marks (it must be less or equal to given k). Also we should not paste exclamation marks on cubecs with numbers larger than 18, because 19! more than 1016 — maximal value of s.
Asymptotic behavior of this solution — O(3((n + 1) / 2) * log(maxcnt) * k), where n — count of cubes, maxcnt — maximal size of associative array, k — count of exclamation marks.
• +110
» 5 years ago, # | 0 Fastest editorial so far! Thanks!
» 5 years ago, # | +8 your rounds are awesome :D won't miss any of them :3
» 5 years ago, # | 0 What was test case 4 for problem C??
• » » 5 years ago, # ^ | ← Rev. 2 → 0 Now we can see it's 8 5 3 3 3 3 4 4 4. I got 3 WAs on test 4 because of misunderstanding the problem statement
• » » » 5 years ago, # ^ | 0 i too misunderstood the problem statement :(
• » » » » 5 years ago, # ^ | 0 What's that? I don't see it... Where does the "25" come from?
• » » » » » 5 years ago, # ^ | 0 It was the sum of all possible rectangles that could be made. Not the max of those. I also got to know now ! :(
• » » » » » 5 years ago, # ^ | 0 We can choose {5-1, 4, 4, 4} to make a rectangle and choose {3, 3, 3, 3} to make another one, and we get the maximum total area which is 16 + 9
• » » » » 5 years ago, # ^ | 0 me too！but I understand it finally
• » » 5 years ago, # ^ | 0 I also misunderstood the problem. #Confusing Problem
• » » 5 years ago, # ^ | ← Rev. 2 → 0 we convert 5 to 4 so we have 4, 4, 4, 4 -> 4 * 4 = 16 and we have 3, 3 , 3 ,3 — > 3 * 3 = 9 so total maximum area is 9 + 16 = 25 not 16 !
• » » 5 years ago, # ^ | +1 In problem C use long long Because the input and Output is large it can't covered by long
» 5 years ago, # | 0 Thank you for nice problems:) BTW, for 530E, I think associative array cnt should be map < long long, int >
» 5 years ago, # | 0 Im getting wrong answer on test 4 problem C....why? where 25 comes from?
• » » 5 years ago, # ^ | 0 Make 5 equal to 4 by reducing it and then you have 4,4,4,4,3,3,3,3. So total area = 16+9;
• » » » 5 years ago, # ^ | +5 I got it man....i thought that they only need the maximum area not the sum..so i was getting 16
» 5 years ago, # | +2 Can someone please explain why in Problem C the output for test 4 8 5 3 3 3 3 4 4 4 is 25 instead of 16? Thanks.
• » » 5 years ago, # ^ | ← Rev. 2 → 0 You can use 5 (-1), 4, 4, 4 for one rectangle and 3, 3, 3, 3 for another.4*4+3*3=16+9=25
• » » 5 years ago, # ^ | 0 5 becomes 4, and then the rectangles are: { 4, 4, 4, 4 }, { 3, 3, 3, 3 }. First's size is 16, second's is 9.
• » » 5 years ago, # ^ | 0 man i did the same mistake....i thought they need the maximum possible area....but they needed the sum of all maximum possible area... I think this problem should be reviewed...
• » » » 5 years ago, # ^ | +5 I totally agree with you. A lot of people including myself misunderstood the problem statement. It should have been something like: "..maximum total area of ALL THE POSSIBLE rectangles that Ilya can make .."Aside from this the round was great. Many thanks to the problems setter =).
» 5 years ago, # | ← Rev. 3 → +1 I liked E a lot, and I had the idea in contest, but made some stupid mistakes...Anyway, now, after the contest ended, I tried to find the bugs, and, apparently, I found them, but I have TLE because of map(I checked, not at the point when we insert in map, is because of the point where we are "querying" the map).Here is my source: http://codeforces.com/contest/525/submission/10476721I really don't know what's so wrong.It inserts me in map in 0.5 seconds and queries in 18 seconds...
• » » 5 years ago, # ^ | 0 On lines 40,41 and 78,79 you make about 2^n iterations, wich is way larger than 3^(n/2), wich is the optimal complexity for finding all the submasks of a mask.
• » » » 5 years ago, # ^ | 0 It's not because that.I reimplemented it and it was the same thing.The idea is that I have 2^n and it enters in for just for 3^(n/2) :-P .Here is the new source: http://codeforces.com/contest/525/submission/10477139 I thought it was easier for someone to look at th first source ant that's why I posted that one.
• » » » » 5 years ago, # ^ | 0 First, don't try this _j < (1<not taken 1->taken without factorial 2->taken with factorial
• » » 5 years ago, # ^ | 0 Try unordered_map.
• » » » 5 years ago, # ^ | 0 I'll try it but I still can't believe that it take it so much to query, knowing that it inserts in 0.5 seconds with the same complexity
• » » 5 years ago, # ^ | ← Rev. 2 → +18 As editorial said, "To accelerate our programm we may increase answer only if cnt[cur].count(s - sumrg) = true."The operator[] creates a new object if it wasn't there before, which will slow down subsequent queries by a bit.Accepted: 10478038
• » » » 5 years ago, # ^ | +8 WOW, it's a really big difference of times just because that...I'll never make the same mistake :))Thanks
» 5 years ago, # | -12 Pasha and string can be solved in O(n) time using BIT. :)
• » » 5 years ago, # ^ | +1 No operation of BIT takes linear time it always involve logn factor
• » » 5 years ago, # ^ | 0 Every operation on BIT takes O(logn) time. How is it O(n)?
» 5 years ago, # | 0 Probably a silly mistake on my part, but This code outputs the right answer (8) for the first test case (4, 2 4 4 2) when i debug it on my machine but outputs 4 and fails on pretest 1 when tested on the website. I know the algorithm is greedy and might not be right but why does it have a different output? Thanks in advance!
• » » 5 years ago, # ^ | 0 I figured it out. It seems that par[++i] is behaving different on the website's compiler. Weird.
» 5 years ago, # | 0 Please help me with the logic used in the problem 525B especially the 'sum' part ? :
• » » 5 years ago, # ^ | ← Rev. 3 → +2 So, let's start with an example word "abcdef". Our possible operations include:1 (reversing letters from [1,6]),2 (reversing letters from [2,5]),3 (reversing letters from [3,4]),4 (reversing letters from [3,4]),5 (reversing letters from [2,5]),6 (reversing letters from [1,6]),I will call operations rather by their intervals now, I think it's easier to understand that way.Now focus on the first letter "a" for a while. It can only change it's place after reverse operation [1,6]. Where it can go? Only at the end, swapping with "f". And after the second one of that type? "a" will go back to the first position again, swapping with "f". No other operation can change position of "a", all other operations will only change something inside the word, between "a" and "f". So how can we now where is "a" after all operations? Well, that's simple, let's count all of the type [1,6], if it's even then "a" stays on the position 1, if it's odd then "a" is swapped with "f". If you can't see it just do few operations [1,6] in your head or on piece of paper and look what happens. Let's store information about number of operations [1,6] in sum[1], you will see why in few seconds.Ok, moving onto "b". "b" can be swapped with "e" after operations [2,5], but also [1,6]. "Even-odd rule" works there two, as it's really the same thing — only changes "b"-"e" are possible and they come after [1,6] or [2,5]. Ok, so if we store number of operations [2,5] in sum[2] only thing we need to know is number of operations [1,6]. Now sum[1] comes into play :) Only thing we have to do for "b"-"e" is check if sum[1] + sum[2] is odd and, if it's the case, swap "b" with "e".One thing you may not see now, but it's also the small problem — for position 3 it's easy, just sum up sum[1] + sum[2] + sum[3]. But you would have to do that addition for every position from the first half of the string and for long ones it can be little bit painful. Don;t worry though, there is one Jedi trick :) Just iterate over all i's from 1 and do:for (int i = start_position; i < s.length() / 2; ++i) {if (i != 0) sum[i] += sum[i — 1];//computations here} In first iteration we have sum[1] in sum[1], in the second one sum[2] + sum[1] in sum[2], in the third one sum[3] + sum[2] + sum[1] in sum[3] and so on. Just what we needed :)EDIT: My code: http://codeforces.ru/contest/525/submission/10479343
• » » » 5 years ago, # ^ | +3 Perfect explanation sir !! Thank u , thanks a lot :)
» 5 years ago, # | 0 For E can't we just loop over all k-sets in a maximum of (25 choose 12)=5*10^6 ways and update the sum from one k-set to the next in O(1) by simply changing one factorial to a different factorial? (Of course we ignore factorials which are too big)
» 5 years ago, # | 0 Can someone explain to me the solution for #4? I used BFS on a space, then got the width and height of the total room, and within the width and height set everything to a '.'. It gives me a wrong answer on case 20, but the case is too large.
• » » 5 years ago, # ^ | ← Rev. 3 → +1 I don't understand solution well enough to explain it, but those are smaller counterexamples to your solution:the one showing something is wrong with your bfs3 3**.**....and the one showing "one-time" bfs is not enough3 3**..**...
• » » » 5 years ago, # ^ | -8 Thank you! Did you happen to solve the problem? Can you explain the solution?
» 5 years ago, # | 0 First you need to understand the following fact — it does not matter in what order to carry out coups, the answer will not change.Let the elements of the row indexed starting at one. To solve the problem count how many revolutions will start in each of the positions of the line. Then counted array sum []. In sum [i] will store the number of revolutions of substrings that start at positions that do not exceed i.Now Let us iterate over i from 1 to n / 2, and if sum [i] is odd, exchange row si and sn — i + 1. We derive an answer — the resulting string s.Asymptotics of the solution — O (n + m), where n — length of the string s, m — the number of revolutions.
» 5 years ago, # | 0 In the editorial's solution for Pasha and StringThen we need to count array sum[]. In sum[i] we need to store count of reverses of substrings, which begin in positions which not exceeding i.I don't get this statement. Could someone explain it in greater detail?
• » » » 5 years ago, # ^ | 0 Thanks a lot! That was great explanation! Yet, my code which follows the same algorithm gives a TLE on test case 7. How?http://codeforces.com/contest/525/submission/10498956
• » » » » 5 years ago, # ^ | ← Rev. 3 → 0 Again, guess, but only possible answer that comes to my mind is the fact that strlen can be up to linear in time:http://stackoverflow.com/questions/3388029/strlen-functionInstead, you should store length of the string in variable or use std::string::length WITHIN C++11 standard where it's guaranteed O(1) time.http://www.cplusplus.com/reference/string/string/length/Look at complexity of length in C++98 and C++11.Also, small enhancement, you can use & 1 instead of % 2, since it's the same, but operations on bits are much more faster than very slow modulo operation. It's small factor, much smaller than this with strlen, but why not? :)
• » » » » » 5 years ago, # ^ | 0 Thanks a ton rr_! The strlen was the problem. Wow! I learnt something new. Regarding the modulo, thanks for the tip! I'll keep that in mind. :)
» 5 years ago, # | 0 In problem E you don't need to sort the given array.
» 5 years ago, # | ← Rev. 3 → 0 I tried to solve problem D in the following, Find the connected components from the given maze. And also find the boundary of the connected component.( top,bottom, left,right). And make all the cells in the maze which lies in the boundary to ".". Also kind of handled case like this with double run of algo..... .... ..*.. ...*. ....* Is there anything wrong with this approach ? This approach was failing for test case 12.
• » » 5 years ago, # ^ | +3 Consider this initial connected components: ********* *1******* *1******* *1***3*** *111***** *1******* *1*22222* ********* When you only draw dots on the boundary you get: ********* *1112222* *1*1***2* *1*1*3*2* *111***2* *1*1***2* *1122222* ********* You don't discover that 3 is also part of the 1-2 room.
• » » » 5 years ago, # ^ | 0 After you find a boundary ( top, bottom, left, right), my idea is to convert every cell inside the boundary to "." .
• » » » » 5 years ago, # ^ | ← Rev. 2 → +1 hello , your idea is correct, you have a lot of rectangles , but you have to merge rectangles that intersect, and then fill the rectangles with '.'the most difficult is solve the last part, how do you merge rectangles optimaly???my idea was make a sweep line on x , and then merge rectangles on y , keep a structure, the code is very difficult , therefore is better analize the problem and make it more easy!!
• » » » » 5 years ago, # ^ | 0 I used the same method as yours, but I got TLE on test 12. I don't know why I got TLE, because I don't think it would cost such long time.
• » » » » » 5 years ago, # ^ | 0 stafuc, it's not even correct. Assuming you talk about your last try (http://codeforces.com/contest/525/submission/10503768):Tiny test case: 3 3 *.. ... ..* Your output: *.. ... ... Not a rectangle ... clearly wrong.
• » » » » » » 5 years ago, # ^ | 0 Oh thank you. Getting all rectangles is not enough, I have to merge them. Thanks very much~.
» 5 years ago, # | 0 Could somebody help me come up with a case where my solution fails?I chose to forgo the option of keeping track of how sticks of length L I had. Instead, I sorted the sticks in descending order according to their lengths. Then, I tried greedily choosing the four largest sides, assuming that the difference between the opposing was always less than or equal to 1 (otherwise, I would be unable to cut accordingly). Thanks for your help in advance.
• » » 5 years ago, # ^ | +5 610 10 7 5 3 3
» 5 years ago, # | 0 For problem E, I din't pass the system test if I used map(10491477) but passed with unordered_map(10491490). But, interestingly, 10480213 (from which I got the idea because it was easy to understand) passed system test. What is going on?
» 5 years ago, # | 0 I get wrong answer for problem c for test case 37. Can somebody provide a counter example for my approach. 10502423
• » » 5 years ago, # ^ | 0 How about reading two comments above first? :Phttp://codeforces.ru/blog/entry/17119#comment-219655
• » » » 5 years ago, # ^ | 0 thanks I got it
» 5 years ago, # | 0 Hello, I tried to solve problem D about 20 times but verdict is always "runtime error on test 92". Could you help me, please? Regards 10507530
• » » 5 years ago, # ^ | ← Rev. 3 → 0 I also got a runtime error on test case 92.I just changed string to char array and took the input character by character.It got AC.10776197
» 5 years ago, # | +1 Can anyone discuss solution of D and proves of solution?
» 5 years ago, # | 0 could someone explain where my solution goes wrong for 525C . I am having wrong answer in test case 37 . http://codeforces.com/contest/525/submission/10501081 . My output : 11234878867001938 jury's answer : 11234878866984153 . Thank you for help
• » » 5 years ago, # ^ | +3 Try the following test case: 6 7 7 6 6 6 4 Should output 42. But yours outputs 43. Third time through the loop i=1,area=1,count=2. So you add area to totalarea which you shouldn't.
» 5 years ago, # | 0 How can one prove the solution to D? (I mean the property of 2x2 matrix)
» 5 years ago, # | 0 Can someone explain the solution for D better? I honestly can't understand the editorial's solution, as the writer's English is not the best.
» 5 years ago, # | ← Rev. 3 → 0 In Problem C , My Code passed test cases with N=93073,96751.. but got TLE in test39 with N=1000? solutionYour text to link here...: http://codeforces.com/contest/525/submission/10522211
» 5 years ago, # | 0 I get a wrong answer in problem D. But I already pass 31 datas and the 32nd data is too huge to find the problem. Can u email me that data or help me with finding the bug in my program. http://codeforces.com/contest/525/submission/10528685
» 5 years ago, # | 0 In problem D, my dfs solution is accepted but when the same logic is applied using bfs i am getting tle. Can anyone guide me why is it hapenning ??
» 5 years ago, # | 0 Can anyone please explain the logic of 2x2 in problem D and proof of its correctness?
» 5 years ago, # | ← Rev. 3 → 0 Hello.Here is my submission for problem A.Can anyone help me with a test instead of test 8?
» 5 years ago, # | 0 Hello.Here are my 2 submissions for problem B: 1st submission and 2nd submission.I used a map in the first one and got TLE on test 41. However, I replaced the map with an array in the second one and got TLE on test 29.Can anyone help me?
» 4 years ago, # | 0 can anyone help me why i am getting wa on test 37 for problem c. here is my code
» 4 years ago, # | ← Rev. 2 → 0 can anyone explain why am i getting wa !! TIAhere is maah code
» 4 years ago, # | ← Rev. 2 → 0 In problem B : How I should approach if given problem is like given index l and r . You have to reverse the string the string from a[l] to a[r]. What would be its approach ?
» 3 years ago, # | 0 Hi... .Would you please tell me why my solution is wrong for 525A - Виталий и пирожок?submission: 21233075
• » » 3 years ago, # ^ | 0 I had the same problem and I've been thinking about it for the past few days ,unfortunately I couldn't find the reason for the wrong answer but I just resubmitted my code with some changes and became accepted out of nowhere! All i did was storing the keys before checking anything else ,and then as soon as you face a door you check whether you have got its key or no! The problem seems to be with priority of checking and storing but I still can't get the problem with my previous submission:21360440 which had this extra comparison of same keys and doors that I have no idea how it causes trouble!
• » » » 3 years ago, # ^ | 0 Corrected... . submission: 21589121
• » » 3 years ago, # ^ | 0 Corrected... . submission: 22328351
» 2 months ago, # | 0 The first question is not that clear...To clarify, We can collect the keys from rooms if the key is not used in that room, this way we can use the key to open some other lock in some other room.
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2019-11-22 11:29:35
|
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https://drupal.meta.stackexchange.com/questions/3619/when-is-it-appropriate-to-use-tag-4-5-6-7-8-or-9?noredirect=1
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When is it appropriate to use tag 4, 5, 6, 7, 8 or 9?
There are various flavors of questions about using Drupal release tags, more specifically these tags: , , , , , .
However I continue to struggle with when it is appropriate to actually use any of those tags. To illustrate my struggling, this is what the current tag excerpts are for some of them:
• (and similar for ): This tag is for questions regarding Drupal 7 specifically. Version tags should be used only when strictly necessary, for questions that apply to Drupal 7 only, and not to merely say "I am using Drupal 7 in my site."
• : Drupal 8 is the latest version of Drupal. Avoid using it as the only tag, unless your question is only specific to this version, and does not involve non-core features, or core features that already have their tags.
Here are the parts in the tag excerpt for that are not clear to me:
• Version tags should be used only when strictly necessary
when does that strictly necessary actually apply? Can you point me to any existing questions to illustrate where the is appropriate, and also other existing questions to illustrate where the is not appropriate?
• for questions that apply to Drupal 7 only
When I post a question, is it OK to tag it with to indicate "I only want to know the answers that apply to Drupal 7"? Can you point me to any existing questions to illustrate where the is appropriate, and also other existing questions to illustrate where the is not appropriate?
• and not to merely say "I am using Drupal 7 in my site."
as per the not to merely here: it sounds if you want to indicate in your question that "you're using Drupal 7 in your site", you should not use the tag. But, assuming I'm allowed to include in my question an indication that I'm using Drupal 7, how should I then add that information? In the question body, in the question title, in both?
Here are the parts in the tag excerpt for that are not clear to me:
• Avoid using it as the only tag, unless your question is only specific to this version,
when does that specific to this version actually apply? Can you point me to any existing questions to illustrate where the is appropriate, and also other existing questions to illustrate where the is not appropriate?
• and does not involve non-core features, or core features that already have their tags.
I don't need to know when to not use this tag, but rather the reverse, i.e when to correctly use . Can you point me to any existing questions to illustrate where the is appropriate, and also other existing questions to illustrate where the is not appropriate?
Somehow related to the usage of the tag (+ similar for , , etc):
1. For questions for which the title contains something like "Drupal 7", or "D7":
• If it is also tagged with , is it correct to just remove that part from the title?
• If it is not tagged with , is it correct to remove that part in the title and add an extra tag at the same time?
2. For questions for which the body contains some kind of indication about "Drupal 7" (or "D7"):
• If it is also tagged with , is it correct to just leave that part in the body as a kind of redundant information?
• If it is not tagged with , is it correct to also add an extra tag?
Notes:
• If you think the above should be obvious, clear, self explaining, then have a look at the revision history of this question.
• If you don't use version tags at all and include something like "Drupal X" in the question title, then these kind of edits may apply. So maybe I'm asking the wrong question, or maybe I should extend it to the title of a question also.
For brevity's sake, I think all tags should read for tag X:
This tag is for questions regarding Drupal X specifically. Version tags should be used only when strictly necessary, for questions that apply to Drupal X only, and not to merely say "I am using Drupal X in my site."
The wording for the D8 is confusing (2x negative) & superfluous (we don't need to know it's the latest version).
As to usage guidance, I think the tag should simply mean OP's question is specially in the context of version X and answers should reflect as such.
As for editing out version information in question title/body, I think it should be left as OP wrote it. Header & body text are signals that search crawlers can pick up on. Tagging is mostly fuzzy metadata for us who want sort through buckets of question topics.
• We don't need Drupal in question titles, since only questions about Drupal are on-topic. As per what the search crawlers pick up, they pick up also tag information an information about the site itself, not just the question title. – kiamlaluno Feb 27 '17 at 16:47
Apart from Shawn's recommendation, I'm going to try to answer my own question, based on facts I've seen happening on Drupal.SE, and including samples of what I saw. As things become more clear to me, I plan to further expand my answer. And if this answer contains mistakes (i.e contract the new guidance), then please drop me a comment ... (I'd be happy to correct my answer).
Here it goes ...
1. For questions for which the title contains something like "Drupal 7", or "D7":
• If it is also tagged with , is it correct to just remove that part from the title?
Answer: yes, that's how it looks like, here is an example of it.
• If it is not tagged with , is it correct to remove that part in the title and add an extra tag at the same time?
Answer: it doesn't seem like so, or maybe not always ... here is an example of a question where "Drupal 8" was removed from the title, with no equivalent version tag. Maybe that's because of the question content, which may be specific to D8?
2. For questions for which the body contains some kind of indication about "Drupal 7" (or "D7"):
• If it is also tagged with , is it correct to just leave that part in the body as a kind of redundant information?
Answer: open question for now (still looking for an example of this).
• If it is not tagged with , is it correct to also add an extra tag?
Answer: open question for now (still looking for an example of this).
Personally, I think this depends more on the answer than the question. Take this theoretical question:
I use Drupal 7 and have module X because I thought it would solve problem Y that I am having, but it does not seem to be working. What is wrong?
[screenshot]
If the answer to that question would be similar for multiple (supported) versions (at this time 7 and 8), I feel it should not be tagged. Configurations matching this scenario could be for views or metatag that look very much the same for both versions.
On the other hand, if the module has been completely overhauled, the steps would be very different. In these cases it would make sense to tag the question with a specific version, so other know it does, or does not apply to them. I have not yet used it, but based on their blog I believe Commerce would be such a module.
The downside of this selection is that moderators (us) all have to be aware of how a module works for both Drupal versions and that some tag edits should be approved, where others should not.
A real example of a question would be Disable field on hook entity_insert. Where I have rejected the tag with the following reason:
This edit introduces tags that do not help to define the topic of the question. Tags should help to describe what the question is about, not just what it contains.
Simply because (and correct me if I am wrong) the answer to the question would apply to both Drupal versions.
But at the same time I realize that not everybody is aware of the similarities and differences in Drupal versions and their modules. (Nor do I believe that people actually read tag descriptions, but that is a different problem.) I suppose the fuzzy critera of 'my method' do not really make it into a good guideline, but I wanted to share it anyway.
• Merci for this answer, I agree with it, for the theoretical question you used. But if we'd use a module like "Rules" (instead of Views or Metatag), then pretty much hardly any question without such tag would make sense (= variation of your commerce sample). What does not seem correct to me in your proposed remedie, is that the author of the question should be "allowed" to indicate for which Drupal version (s)he is looking for an answer. Apart from that: what about question that say "I want to programmatically do so and so, and this is how my code looks like so far"? No version tag then? – Pierre.Vriens Mar 22 '17 at 20:47
• (continued) ... Any suggestions about the "Somehow related to the usage ..."-part of my question? FYI: I keep struggling with that (as shown in my own attempt to answer), And every day I see more and more variations that make me wonder "so what is it now in the end, am I really the only one struggling to understand the guidance from the mods?". FYI: I'm about to start using the SKIP button in the suggested edits review queue for anything that does anything to the version tags ... – Pierre.Vriens Mar 22 '17 at 20:50
• The authors should use version tags when required. If they are not aware of the similarities we might want to remove it (or add both?). The same applies to code; some basic hooks are unchanged and apply to both, while others are for a specific version and require a tag. But like I said, I don't think this method will make sense, as too little people know enought about multiple Drupal versions an contrib modules (including me on some aspects). – Neograph734 Mar 23 '17 at 7:04
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2021-04-21 14:01:45
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https://blender.stackexchange.com/questions/88431/how-to-refresh-workspace
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# How to refresh workspace
I do these operations in a for loop:
1. Selecting all objects
2. Remove
5. Bind cameras to model
7. Render
The problem is how to refresh the entire scene for next iteration? I have tried to load a fresh file using bpy.ops.wm.open_mainfile, but it causes an error. It seems that the entire project including the script loads again. So, I tried to remove objects in bpy.data.objects but I noted that there are bpy.data.cameras and bpy.data.lamps which should be also removed. What is the correct way to do this job? Does selecting all objects and deleting them makes the trick? Note that there is an animation along with cameras, lamps, etc.
The solution is to delete data collections. There are plenty of collections in bpy.data which are not cleaned after deleting the object. For example if you add a camera, an entry in bpy.data.camera would be added but after deleting the camera object that entry remains that finally leads to out of memory errors. So the solution would be some thing like this:
def clear_collection(collection):
for item in collection:
collection.remove(item, True)
def cleanup():
clear_collection(bpy.data.actions)
clear_collection(bpy.data.armatures)
clear_collection(bpy.data.brushes)
clear_collection(bpy.data.cameras)
clear_collection(bpy.data.groups)
clear_collection(bpy.data.images)
clear_collection(bpy.data.lamps)
clear_collection(bpy.data.materials)
clear_collection(bpy.data.meshes)
clear_collection(bpy.data.objects)
clear_collection(bpy.data.textures)
cleanup()
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2020-01-27 09:21:28
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https://forum.noxiousnet.com/topic/6727/unmute-request-or-time-lowered
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# Unmute request or time lowered
• STEAM_0:0:5924350
Steam name : 4bc
I know i should have listened to the admin when he told others to stop flash spamming but i understand now that i shouldn’t have done it.I am sorry to the people that i bothered but i would like to know if i can get a -1 day for it.Please take this into consideration. -4bc
• STEAM_0:0:5924350
Steam name : 4bc
I know i should have listened to the admin when he told others to stop flash spamming but i understand now that i shouldn’t have done it.I am sorry to the people that i bothered but i would like to know if i can get a -1 day for it.Please take this into consideration. -4bc
• Why did you do it in the first place ?
• Because they were spamming nigger in flash
• From your previous history of not having a worthwhile thing to say, ever, I’m going to say no.
• aw okay then.
• don’t struggle honey
## 297665Posts
Looks like your connection to NoXiousNet was lost, please wait while we try to reconnect.
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2017-10-21 04:42:22
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https://www.nature.com/articles/s41467-017-00961-2?error=cookies_not_supported&code=5f8374f8-c5c4-4bc3-b374-b1653b265e6b
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Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript.
# Capacity estimation and verification of quantum channels with arbitrarily correlated errors
## Abstract
The central figure of merit for quantum memories and quantum communication devices is their capacity to store and transmit quantum information. Here, we present a protocol that estimates a lower bound on a channel’s quantum capacity, even when there are arbitrarily correlated errors. One application of these protocols is to test the performance of quantum repeaters for transmitting quantum information. Our protocol is easy to implement and comes in two versions. The first estimates the one-shot quantum capacity by preparing and measuring in two different bases, where all involved qubits are used as test qubits. The second verifies on-the-fly that a channel’s one-shot quantum capacity exceeds a minimal tolerated value while storing or communicating data. We discuss the performance using simple examples, such as the dephasing channel for which our method is asymptotically optimal. Finally, we apply our method to a superconducting qubit in experiment.
## Introduction
One of the main obstacles on the way to quantum computers and quantum communication networks is the problem of noise due to imperfections in the devices. Noise is caused by uncontrolled interactions of the quantum information carriers with their environment. These interactions take place at all stages: when the carriers are processed, when they are transmitted, and when they are stored. Physicists and engineers spend large efforts in developing noise protection measures, and assessing their performance is crucial for the development of quantum information processing devices. In this article, we focus on the estimation of noise in the storage and transmission of the quantum information carriers, that is, we describe methods to assess quantum memory and quantum communication devices.
In the language of quantum information theory, memory and communication devices are described by a quantum channel, which is a function Λ that maps an input state ρ in of the device to its output state ρ out=Λ(ρ in). In this unified description, assessing the noise in a quantum device reduces to estimating the decoherence of a quantum channel. One way to achieve this is through quantum process tomography1, which aims at completely determining the channel from measurement data (see e.g., refs. 2,3 for more recent works on tomography, and, e.g., refs. 4,5 for surveys on specific types of tomography). This comes with two major disadvantages. First, process tomography typically only works for channels that behave the same way in every run of the experiment (formalized by the i.i.d. assumption—for independent and identically distributed), or under some symmetry assumptions. This assumption is violated for many devices that are used in practice, which typically show correlated errors. Second, since process tomography aims at a complete characterization of the channel, it requires the collection of large amounts of data for many combinations of input states and measurement settings. A complete characterization of a channel is certainly useful (as all properties of the channel can be inferred from it), but it is very costly if the task at hand is to simply estimate a figure of merit of the channel. For quantum storage and quantum communication devices, a central figure of merit is the quantum capacity of the channel, which quantifies the amount of quantum information that can be stored or transmitted by the device6. While the deployment of a suitable error-correcting code requires knowledge of the specifics of the channel, an estimate of the quantum capacity is of great use when assessing the usefulness of the tested device.
In this work, we present a method to estimate the one-shot quantum capacity Q ε(Λ) of a quantum channel Λ. While the quantum capacity Q only makes statements for devices that behave identically under many repeated uses, the one-shot quantum capacity Q ε applies to the more general case of devices with arbitrarily correlated errors. It quantifies the number of qubits that can be sent through the channel with a fidelity of at least 1 − ε in a single use of the device using the best possible error-correcting code (we will explain this in more detail in the next section). We present a protocol that allows to estimate Q ε(Λ) from data obtained from simple measurements. In addition to dealing with arbitrarily correlated errors, it has the advantage of requiring fewer measurement settings than quantum process tomography.
Our method can also be used to assess whether a possibly imperfect error-correction scheme forms an improvement. This is the case if the error-corrected channel has a higher capacity than what we would otherwise expect. Similarly, our protocols can be employed to test whether a quantum repeater actually forms an improvement for sending quantum information, that is, whether it yields a higher quantum capacity than a direct quantum communication link.
## Results
### The one-shot quantum capacity
Noise can be modeled as a channel Λ, which is given as a map
$$\Lambda :{\cal S}\left( {\cal H} \right) \to {\cal S}\left( {\cal H} \right),$$
(1)
where $${\cal S}\left( {\cal H} \right)$$ denotes the set of quantum states on the Hilbert space of the system that is being stored or transmitted. For reasons of illustration, we will discuss channels of storage devices here, but mathematically, nothing is different for communication devices. In the realm of communication, it is convenient to think of a sender (Alice) who wants to relay qubits to a receiver (Bob). For memory device, Alice and Bob simply label the input and output.
Consider a quantum memory device designed for storing a quantum system with Hilbert space $${\cal H}$$ for some time interval Δt. Ideally, it leaves the state of the system completely invariant over that time span, but real storage devices are always subject to noise. A measure for how well the channel Λ preserves the state of the system is obtained by minimizing the square of the fidelity between the input state $$\left| \phi \right\rangle$$ and the output state Λ(ϕ),
$$F\left( {\left| \phi \right\rangle ,\Lambda \left( \phi \right)} \right) = \sqrt {\left\langle \phi \right|\Lambda \left( \phi \right)\left| \phi \right\rangle } ,$$
(2)
over all possible input states $$\left| \phi \right\rangle \in {\cal H}$$,
$$\mathop {{\min }}\limits_{\left| \phi \right\rangle \in {\cal H}} {F^2}\left( {\left| \phi \right\rangle ,\Lambda (\phi )} \right) = \mathop {{\min }}\limits_{\left| \phi \right\rangle \in {\cal H}} \left\langle \phi \right|\Lambda \left( \phi \right)\left| \phi \right\rangle .$$
(3)
Low values of the quantity Eq. (3) imply that if the device is used without modification, then at least some states of the system are strongly affected by the channel, therefore introducing errors. However, this does not necessarily mean that the device is useless as a storage device, as this quantity does not account for the possibility that such errors can be corrected using quantum error correction (QEC).
An error-correcting code for a channel Λ consists of an encoding $${\cal E}$$, which is applied before the channel, and a decoding $${\cal D}$$, which is applied after the channel (see the explanations in Fig. 1). Together, these devices form an error-corrected quantum memory for a smaller system, implementing a channel
$${\cal D} \circ \Lambda \circ {\cal E}:{\cal S}\left( {\cal K} \right) \to {\cal S}\left( {\cal K} \right),$$
(4)
where $${\cal K}$$ is the Hilbert space of the smaller system and where ο denotes the composition of maps. Instead of evaluating the quantity Eq. (3) for the channel Λ directly, it should be evaluated for such a corrected channel $${\cal D} \circ \Lambda \circ {\cal E}$$. A figure of merit for the usefulness of the quantum memory is then given by the size of the largest system $${\cal K}$$ that can be stored in the memory using such an error-correcting code. This is identical to the largest subspace $${\cal H}\prime \subseteq {\cal H}$$ that is left approximately invariant by the memory, where the choice of encoding corresponds to the choice of subspace. This is quantified by the one-shot quantum capacity Q ε(Λ), defined by7,8
$${Q^\varepsilon }\left( \Lambda \right)\,{\rm{:}} \!= {\rm{max}}\left\{ {{\rm{lo}}{{\rm{g}}_{\rm{2}}}\,m|{F_{{\rm{min}}}}\left( {\Lambda ,m} \right) \ge 1 - \varepsilon } \right\},$$
(5)
where
$${F_{{\rm{min}}}}\left( {\Lambda ,m} \right): = \mathop {{max}}\limits_{\scriptstyle{\cal H}\prime \subseteq {\cal H}\hfill\atop\\ \scriptstyle\dim ({\cal H}\prime ) = m\hfill} \mathop {{min}}\limits_{\left| \phi \right\rangle \in {\cal H}\prime } \left\langle \phi \right|\left( {{\cal D} \circ \Lambda } \right)\left( \phi \right)\left| \phi \right\rangle$$
(6)
and where the inner maximum is taken over all possible decoders $${\cal D}:{\cal S}\left( {\cal H} \right) \to {\cal S}\left( {\cal H} \right)$$. This way, the one-shot quantum capacity corresponds to the maximal number of qubits that can be stored and retrieved with a fidelity of at least 1 − ε using the best possible error-correcting code.
The one-shot quantum capacity tells us strictly more than the asymptotic quantum capacity, in the sense that the latter can be obtained from the former:
$$Q\left( \Lambda \right) = \mathop {{\lim }}\limits_{\varepsilon \to 0} \mathop {{\lim }}\limits_{N \to \infty } \frac{1}{N}{Q^\varepsilon }\left( {{\Lambda ^{ \otimes N}}} \right).$$
(7)
The asymptotic quantum capacity is the number of qubits that can be transmitted or stored per use of a device with asymptotically vanishing error, in the limit where it is used infinitely often under the i.i.d. assumption. Therefore, it is an asymptotic rate, while the one-shot quantum capacity is the total number of qubits that can be transmitted or stored in a single use of a (possibly non-tensor product) channel, allowing some error $$\varepsilon \ge 0$$.
### One-shot quantum capacity estimation
Now that the one-shot quantum capacity is identified as the relevant figure of merit for quantum memory and communication devices, the question is whether we can estimate this quantity for a given device. We answer this question in the affirmative for the case where Λ is a channel that stores or communicates (arbitrarily many) qubits.
We present a simple protocol (see Protocol 1 in Table 1) that estimates a lower bound on the one-shot quantum capacity Q ε(Λ) for an N-to-N-qubit channel Λ. Our protocol only requires the preparation and measurement of single qubit states in two bases. Specifically, even though it is known that the optimal encoder for a given channel Λ may require the creation of a highly entangled state, no entanglement is required to execute our test. For simplicity, we assume here that N is an even number (for more general cases, see Supplementary Notes 46). The protocol does not make any assumption on whether the qubits are processed sequentially, as in communication devices, or in parallel, as in storage devices (potentially with correlated errors in both cases). The data collection of the protocol is very simple. Alice and Bob agree on two qubit bases X and Z. These two bases should be chosen to be “incompatible”, in the sense that the preparation quality q, which is defined as
$$q = - {\rm{lo}}{{\rm{g}}_2}\mathop {{\max }}\limits_{i,j = 0,1} \left| {{{\left\langle {{i_X}\left| {{j_Z}} \right.} \right\rangle }}} \right|^2{\kern 1pt} ,$$
(8)
is as high as possible, where $$\left| {{i_X}} \right\rangle$$ and $$\left| {{j_Z}} \right\rangle$$ are eigenstates of X and Z, respectively. In the ideal case, where the two bases X and Z are mutually unbiased bases, such as the Pauli-X and Z basis, it holds that q = 1. Our protocol can be seen as exploiting the idea that the ability to transmit information in two complementary bases relates to a channel’s ability to convey (quantum) information9,10, which we show holds even with correlated noise. We remark that Pauli-X and Z basis have also been used to estimate the process fidelity of a quantum operation11,12 in the i.i.d. case, which however we are precisely trying to avoid here.
The bound for the capacity estimate is a function of the number of qubits N, the preparation quality q, the maximally allowed decoding error probability ε of Q ε(Λ), the two measured error rates e x and e z , and some probability p that quantifies the typicality of the protocol run (we will discuss this parameter in the Discussion section). More precisely, the bound is given as follows.
### Theorem 1
Let $$N \in {{\Bbb N}_ + }$$ be an even number, let e x and e z be error rates determined in a run of Protocol 1 where the used bases X and Z had a preparation quality of q (see Eq. (8) above). Then, for every ε > 0 and for every $$p \in [0,1)$$, it holds that
• either, the probability that at least one error rate exceeds e x or e z , respectively, was higher than p,
• or the one-shot quantum capacity of the N-qubit channel Λ is bounded by
$${Q^\varepsilon }\left( \Lambda \right) \ge \mathop {{\sup }}\limits_{\eta \in \left( {0,\sqrt {\varepsilon /2} } \right)} \\ \left[ {N\left( {q - h\left( {{e_x} + \mu } \right) - h\left( {{e_z} + \mu } \right)} \right) - 2\,{\rm{lo}}{{\rm{g}}_2}\left( \kappa \right) - 4\,{\rm{lo}}{{\rm{g}}_2}\left( {\frac{1}{\eta }} \right) - 2} \right],$$
(9)
where h is the binary entropy function
$$h\left( x \right): = - x\,{\rm{lo}}{{\rm{g}}_2}\left( x \right) - \left( {1 - x} \right){\rm{lo}}{{\rm{g}}_2}\left( {1 - x} \right)$$
(10)
and μ and κ are given by
$$\mu = \sqrt {\frac{{N + 2}}{{{N^2}}}{\rm{ln}}\left( {\frac{{3 + \frac{5}{{\sqrt {1 - p} }}}}{{\sqrt {\varepsilon {\rm{/}}2} - \eta }}} \right)} ,\quad \kappa = 2{\left( {\frac{{3 + \frac{5}{{\sqrt {1 - p} }}}}{{\sqrt {\varepsilon {\rm{/}}2} - \eta }}} \right)^2}.$$
(11)
In the asymptotic limit where N→∞, the bound on the right hand side of inequality9 converges to $$N\left( {q - h\left( {{e_x}} \right) - h\left( {{e_z}} \right)} \right)$$. All the other terms can be seen as correction terms that account for finite-size effects. We will discuss this in more detail in the Discussion section below. One may wonder, why we do not also obtain an upper bound. First of all, there exist no way to distinguish noise in the rest of the experimental apparatus from the noise on the channel. Second and more significantly, however, fixing any estimation procedure, arbitrarily correlated noise can always conspire to defeat the procedure tricking us into believing the capacity is low, while actually it is quite high. An upper bound could be obtained under the assumption that the noise is i.i.d., but this is precisely what we wish to avoid here.
### One-shot capacity verification
Protocol 1 above estimates how much quantum information can be stored in a quantum memory device. This is of great use when the task is to figure out whether a device is potentially useful as a quantum memory device. When eventually, an error-correcting code is implemented, the corrected memory might be used without further testing.
In some cases, however, one wants to implement the memory with a means to verify its quality while using it. For example, one may suspect the quality of the memory to diminish (say, due to damage or overuse). In that case, the capacity estimation that was made before the implementation of the error-correcting code may no longer be valid. A method to verify that the quality of the memory is good enough for the implemented code may be required whenever it is used. Protocol 2, as given in Table 2, shows such a verification protocol.
The protocol assumes that Alice holds N data qubits that she wants to send to Bob in a way that allows her to verify the quality of the transmission. To this end, she uses a channel for 3N qubits and places her N data qubits in random slots of this channel. The other 2N slots are used for test qubits, half of which are prepared and measured in the X basis and half of which are prepared and measured in the Z basis (just as in the estimation protocol), while Alice and Bob leave the data qubits untouched. The error rates on the test bits allows to infer a bound on the capacity of the channel on the data qubits.
For this protocol, we denote the measured error rate in X by γ and the measured error rate in Z by λ. Bob checks whether these error rates exceed some tolerated values e x and e z , respectively, which has been specified before the protocol run. If one or both error rates exceed the tolerated value, the protocol aborts because the transmission quality is considered too low. If both error rates are below their tolerated value, Bob concludes that the transmission was of high quality, in the sense that the channel on the data qubits had a high one-shot quantum capacity. This is stated more precisely in the following theorem.
### Theorem 2
Let $$N \in {{\Bbb N}_ + }$$, let e x , $${e_z} \in [0,1]$$. Assume that Protocol 2 is run successfully without abortion, where the used bases X and Z had a preparation quality of q. Then, for every ε > 0 and for every $$p \in [0,1)$$, it holds that
• either, the probability that the protocol aborts was higher than p,
• or the one-shot quantum capacity of the channel Λ on the N data qubits is bounded by inequality Eq. (9), where κ is as in Eq. (11) and where μ is given by
$$\mu = \sqrt {\frac{{2\left( {N + 1} \right)}}{{{N^2}}}{\rm{ln}}\left( {\frac{{3 + \frac{5}{{\sqrt {1 - p} }}}}{{\sqrt {\varepsilon {\rm{/}}2} - \eta }}} \right)} {\kern 1pt} .$$
(12)
The bound for the verification protocol looks formally almost identical to the one for the estimation protocol, but there are three differences. First, the function μ has a different dependence on N, which is a consequence of the different structure of the protocol as explained in Fig. 2. Second, the error rates e x and e z are preset accepted error rates instead of calculated error rates from data, and the bound holds when the measured rates are below those preset values. Third, the probability p in the bound is the abort probability of the protocol. Hence, another way to read the statement of the theorem is that either the protocol succeeds (does not abort) with a probability at most 1 − p, or the capacity is indeed high. This again quantifies what we consider to be typical data: even if the channel is competely noisy and useless, there might be a tiny probability 1 − p that the observed error rates are nevertheless small. In this case, we saw highly atypical data. We will say more about this probability in the Discussion section. Recall, that in the verification protocol we use 3N rounds, hence there is no factor of 1/3 on N (see also Fig. 2).
### Experiment
We demonstrate the use of this protocol by implementing it on a Transmon qubit. The experiment is performed on qubit A T previously reported in ref. 13. We measure a relaxation time of T 1 = 18.5 ± 0.6 μs and a Ramsey dephasing time of $$T_2^ \star = 3.8 \pm 0.3$$ μs before performing the experiment. Readout of the qubit state is performed by probing the readout resonator with a microwave tone. The resulting transients are amplified using a traveling-wave parametric amplifier (TWPA)14 at the front end of the amplification chain. This results in a readout fidelity F RO = 11 − (p 01 + p 10)/2 = 98.0%, where p 01 (p 10) is the probability of declaring state 1 (0) when the input state was $$\left| 0 \right\rangle$$ $$\left( {\left| 1 \right\rangle } \right)$$ respectively. The qubit state is controlled using resonant microwave pulses.
The experiment implements Protocol 01 to estimate the capacity of the idling operation It). We do this by generating 8000 pairs of random numbers corresponding to the bases $$b \in \left\{ {X,Z} \right\}$$ and states $$s \in \left\{ {0,1} \right\}$$. These are then used to generate pulse sequences that rotate $$\left| 0 \right\rangle$$ to the required state, and wait for a time Δt before measuring the qubit in the Z basis and declaring a state. If the required state was in the X basis, a recovery pulse is applied that rotates the state to the Z basis before it is read out. This protocol is repeated 130 times, with a distinct randomization for each repetition, yielding a total of N = 1.04 × 106 measurement outcomes in approximately one and a half hours. Results are reported in Fig. 3, which illustrates the estimate using the totality of the N outcomes for different values of ε. In Fig. 4 we furthermore plot variations in the error rate over time, as well as a bound for partial measurement sequences which highlight the (likely) non i.i.d. nature of the actual noise process affecting the qubits. We estimate q = 0.985 ± 0.047 (see Supplementary Note 7) before taking the data, but use q = 0.9 as a conservative estimate to account for a potential drift during the experimental run.
## Discussion
In this section, we shall discuss our bound as a bound on the rate $$\frac{1}{N}{Q^\varepsilon }\left( \Lambda \right)$$, which quantifies the amount of quantum information that can be sent per qubit. This has the advantage that it makes comparisons easier. To discuss our bound on the capacity rate, we have plotted its value as a function of N in Fig. 5. We plotted the bound for the estimation protocol, but qualitatively, the bound for the verification protocol behaves identically, so our discussion applies to both protocols.
### Example dephasing channel
In order to assess the strength of our bound, it is helpful to consider some example channels. A particularly insightful example is the case where the channel Λ is given by N independent copies of a dephasing channel of strength $$\alpha \in \left[ {0,1} \right]$$, that is,
$$\Lambda = \Lambda _D^{ \otimes N},\quad {\Lambda _D}\left( \rho \right):\rho \; \mapsto \,\left( {1 - \frac{\alpha }{2}} \right)\rho + \frac{\alpha }{2}\sigma \,\rho \sigma ,$$
(13)
where σ denotes one of the qubit Pauli operators with respect to some basis. Of particular interest is the case where the dephasing happens with respect to one of the two bases X or Z in which Alice and Bob prepare and measure the test qubits. Let us assume that σ = σ Z . In order to see what happens when our estimation protocol is used in this case, we could simulate a protocol run and see what bound on the one-shot quantum capacity would be obtained. However, the estimation protocol does essentially nothing but determine the two error rates e x and e z . The expected values of these rates can be readily obtained from Eq. (13). The error rate e z vanishes, because dephasing in the Z basis leaves the Z-diagonal invariant. In the X basis the bits are left invariant with probability 1 − α/2, and flipped with probability α/2, so asymptotically e x = α/2. Hence, for the dephasing channel, the estimation protocol is expected to yield the bound in inequality Eq. (9) with e z = 0 and e x = α/2.
### Asymptotic tightness of the bound
As one can see in Fig. 5, the bound on the one-shot quantum capacity, expressed as a rate, converges to $$q - h\left( {{e_x}} \right) - h\left( {{e_z}} \right)$$, which in the case of the dephasing channel is given by q − h(α/2). If we additionally assume that the bases X and Z are mutually unbiased (as are Pauli-X and Z), this is equal to 1 − h(α/2). This is precisely the (asymptotic) quantum capacity of the dephasing channel. This means that our bound on the one-shot quantum capacity is asymptotically tight; if our bound can be improved, then only in the finite-size correction terms. In particular, our bound cannot be improved by a constant factor. Since most estimates that enter the derivation of the bound are of the same type as the estimates used in modern security proofs of quantum key distribution (QKD)15, any possible improvements of the QKD security bounds would also lead to an improvement of our bound on the one-shot quantum capacity (if there is any). In this sense, our bound is essentially as tight as the corresponding security bounds for QKD in the finite regime.
### Measurement calibration
Above, we have assumed that Alice and Bob were very lucky: they set up their bases X and Z such that one of them is exactly aligned with the dephasing basis, and therefore optimally exploited the asymmetry of the channel. In general, since they do not know the channel whose capacity they estimate, they do not know about the direction of the asymmetry. Instead, they have to calibrate their devices by trying out several pairs of bases until they find one with low error rates. Otherwise, the bound on the one-shot quantum capacity that they infer is suboptimal. It is an interesting open question how such a calibration can be optimized.
### Example fully depolarizing channel
Another insightful example is the case where Λ is given by the channel which outputs the fully mixed state of N qubits, independently of the input state. The capacity of this channel is zero, yet with probability 2N, Alice and Bob measure error rates e x = e z = 0. One may think that these vanishing error rates lead to a highly positive bound on the capacity, but this is not the case. As one can read in Theorem 1 and Theorem 2, the bound depends on a probability p, and the term 1−p corresponds precisely to the probability of such an unlikely case. In fact, for 1 − p = 2N, the bound is never positive. This example shows that in the one-shot regime, a meaningful capacity estimation can only be made under the assumption that the observed data is not extremely atypical for the channel. However, this is only a problem for very low values of N: thanks to the natural logarithm in μ (see Eq. (11) above), the concern reduces to atypical events with an exponentially (in N) small probability. For reasonable numbers of N, the influence of p on the bound is negligible, except for extremely low values of 1 − p.
### Quantifying typicality
We remark that p is a parameter to choose before executing the estimation protocol, which essentially just defines what we consider atypical. The statement of the estimation theorem can then be understood as simply stating that either the observed data is highly atypical (as defined by the choice of p), or the capacity is indeed high. From a practical point of view, note that for any constant p, the bound is essentially independent of p for even relatively small values of N and μ→0 in (Eq. (11)) as N→∞. For this reason, we simply choose p = 1/2 in our plots as an illustration. Similarly, the maximally tolerated error rates e x and e z can be chosen freely when conducting the verification protocol. It is merely that the conclusions of the test depend on it, since we choose to abort—that is, draw no conclusion—if the observered error rates are higher than e x and e z . The probability p pass that we do not abort then corresponds to p pass = 1 − p. So p also arises here, and corresponds to the probability of aborting, namely to the probability that data is produced which we do not regard as typical. Aborting may still seem like a different approach to the one taken during the verification protocol where we always draw a conclusion, but we can see that it is in fact exactly analogous: In the estimation protocol, Alice and Bob essentially decide to make a test in which e x and e z correspond exactly to the measured error rates instead of setting a maximum error rate ahead of time. Clearly, they will never abort in this case. Nevertheless, one can consider the probability that in any run, the measured error rates would stay below the rates that have been measured in this particular run of the test. This probability can be seen as a measure of typicality of the protocol, and corresponds precisely to p pass if we were to execute the test again, but now fixing the error rates to what we observed. Hence, again p = 1 − p pass which corresponds precisely to a quantification of such typicality.
For more information on the probability p and e x and e z , see Supplementary Notes 5 and 6. Note that the need to characterize typicality of the data observed is not only given in our context of capacity estimation, but arises in all statistical tests on a finite sample, including quantum key distribution where security statements are formulated in an analogous fashion.
### Usage to assess quantum repeaters
An important challenge in the experimental realization of quantum repeaters is to demonstrate that they improve our ability to communicate compared to a direct fiber connection without a repeater. To demonstrate that they improve our ability to produce key, one proceeds by calculating the private capacity P of the direct communication channel without a repeater, followed by the implementation of a QKD protocol using the quantum repeater. If the rate R of generating key in the QKD protocol with the repeater satisfies R > P, then the quantum repeater improved our ability to produce encryption keys (see e.g., refs. 16,17,18). It turns out that it is in general harder to send qubits, then it is to produce key19. That is, the quantum capacity Q satisfies $$Q \le P$$, where the inequality is in general strict. Demonstrating that a repeater improves our ability to produce key thus does not allow us to draw conclusions on whether the repeater improves our ability for sending qubits without further analysis.
Our result now precisely allows one to perform such a demonstration, even in the regime of arbitrarily correlated noise while in fact being no more difficult than performing BB84 QKD20. Crucially, this means that in order to demonstrate a quantum repeaters ability to send qubits it is thus not necessary to perform quantum error correction (QEC) or entanglement distillation. First, one needs to calculate the quantum capacity $$Q_{DF}^\varepsilon \left( N \right)$$ of the direct fiber (DF) connection, or a bound $$Q_{DF}^\varepsilon \left( N \right) \le B_{DF}^\varepsilon \left( N \right)$$. We note that theoretical bounds $$B_{DF}^\varepsilon \left( N \right)$$ on the one-shot capacity $$Q_{DF}^\varepsilon \left( N \right)$$ are known for finite number of channel uses N and error ε, which are much tighter than employing the asymptotic capacity for N→∞ and ε→08,21. We can now run our capacity estimation protocol over the quantum repeater link which yields a lower bound $$L\left( N \right) \le Q_{WR}^\varepsilon \left( N \right)$$ for the capacity with repeater (WR). If we find that $$L\left( N \right) >B_{DF}^\varepsilon \left( N \right)$$, then we have successfully demonstrated that the quantum repeater improves our ability to transmit qubits over a direct transmission line.
### Usage to assess quantum error-correcting schemes
We note that in a similar way we can make statements about the performance of a QEC scheme for storing qubits with arbitrarily correlated errors. Suppose that we wish to compare how well an error-correcting scheme encoding one logical qubit using multiple physical qubits compares to using just one of the physical qubits directly. We can again employ the result of8 to first derive (a bound on) the one-shot quantum capacity $$B_P^\varepsilon \left( N \right)$$ if we used the physical qubit (P) N times with error ε. We then execute the capacity estimation protocol to estimate the capacity of the logical (LO) qubit channel $$L\left( N \right) < Q_{{\rm{LO}}}^\varepsilon \left( N \right)$$. If we find that $$L\left( N \right) >B_P^\varepsilon \left( N \right)$$, then we can conclude that the logical qubit is an improvement for storing quantum information over one physical qubit.
### Other open questions
Our result assumes that the system on which the channel acts is composed of qubits. An interesting open question is whether this restriction can be removed and an analogous bound can be derived for channels of arbitrary dimension and composition.
It would also be interesting to see our bound extended to continuous variable systems. There are many tools already available22,23,24,25 that may be useful to perform such an analysis, but it remains to be determined how exactly they can be applied to such systems.
## Methods
To prove the bound on the one-shot quantum capacity, we combine several results. First, as we recapitulate in more detail in the Supplementary Note 1 through 4, it has been shown that the one-shot quantum capacity is bounded by the one-shot capacity of entanglement transmission $$Q_{{\rm{ent}}}^\varepsilon \left( \Lambda \right)$$ 26. More precisely, it holds that for every channel Λ and for every ε > 07,
$${Q^\varepsilon }\left( \Lambda \right) \ge Q_{{\rm{ent}}}^{\varepsilon /2}\left( \Lambda \right) - 1.$$
(14)
The one-shot capacity of entanglement transmission, in turn, has been proved to be bounded by the smooth min-entropy $$H_{{\rm{min}}}^\varepsilon {\left( {A{\rm{|}}E} \right)_\rho }$$, which is defined by
$$H_{{\rm{min}}}^\varepsilon {\left( {A\left| B \right.} \right)_\rho }: = \mathop {{\max }}\limits_{\rho ' \in {B^\varepsilon }\left( \rho \right)} {H_{{\rm{min}}}}{\left( {A\left| B \right.} \right)_{\rho '}},$$
(15)
where
$${H_{{\rm{min}}}}{\left( {A\left| B \right.} \right)_\rho }: = \mathop {{\max }}\limits_{{\sigma _B}} \,{\rm{sup}}\left\{ {\lambda \in {\Bbb R}\left| {{\rho _{AB}}} \right. \le {2^{ - \lambda }}{I_A} \otimes {\sigma _B}} \right\}.$$
(16)
It has been shown that7,8,27
$$Q_{{\rm{ent}}}^\varepsilon \left( \Lambda \right) \ge \mathop {{\sup }}\limits_{\eta \in \left( {0,\sqrt \varepsilon } \right)} \left( {H_{{\rm{min}}}^{\sqrt \varepsilon - \eta }{{\left( {A\left| E \right.} \right)}_\rho } - 4{\rm{lo}}{{\rm{g}}_2}\frac{1}{\eta } - 1} \right),$$
(17)
Here, the smooth min-entropy is evaluated for the state
$${\rho _{AE}} = \left( {{I_A} \otimes \Lambda _{A' \to E}^c} \right)\left( {{\Phi _{AA'}}} \right),$$
(18)
where $${\Phi _{AA'}}$$ is a maximally entangled state over the input system $$A'$$ and a copy A of it, and where $$\Lambda _{A' \to E}^c$$ is the complementary channel of the channel $${\Lambda _{A' \to B}}$$. The system E is the environment of the channel (see refs. 8,28 and Supplementary Note 2 for more details). Taking together the results Eqs. (14) and (17), we get that for all ε > 0,
$${Q^\varepsilon }\left( \Lambda \right) \ge \mathop {{\sup }}\limits_{\eta \in \left( {0,\sqrt {\varepsilon /2} } \right)} \left( {H_{{\rm{min}}}^{\sqrt {\varepsilon /2} - \eta }{{\left( {A|E} \right)}_\rho } - 4\mathop {{\log }}\nolimits_2 \frac{1}{\eta } - 2} \right).$$
(19)
Therefore, the min-entropy bounds the one-shot quantum capacity.
Estimating the min-entropy has been a subject of intense research in quantum key distribution (QKD). However, min-entropy estimation protocols in QKD cannot be directly applied here, because they estimate the min-entropy $$H_{{\rm{min}}}^\varepsilon \left( {X\left| E \right.} \right)$$ for classical information X, while in the bound Eq. (17), the system A holds quantum information. We bridge this gap: as our main technical contribution, we show in the Supplementary Note 3 that for a system A that is composed of qubits, it holds that for every ε > 0 and every $$\varepsilon ',\varepsilon '' \ge 0$$,
$$H_{{\rm{min}}}^{3\varepsilon + \varepsilon ' + 4\varepsilon ''}{\left( {A\left| E \right.} \right)_\rho } \ge Nq - \left( {H_{{\rm{max}}}^{\varepsilon ''}{{\left( {X\left| B \right.} \right)}_\rho } + H_{{\rm{max}}}^{\varepsilon '}{{\left( {Z\left| B \right.} \right)}_\rho }} \right) - 2\,{\rm{lo}}{{\rm{g}}_2}\frac{2}{{{\varepsilon ^2}}}{\kern 1pt} .$$
(20)
Inequality Eq. (20) reduces estimating the min-entropy of quantum information A to estimating the max-entropy of measurement outcomes X and Z on the system A.
We prove inequality Eq. (20) using three main ingredients. First, we use an uncertainty relation for the smooth min- and max-entropies29. Second, we use a duality relation for the smooth min- and max-entropies30,31. These two ingredients were also used in modern security proofs of quantum key distribution15. We combine these two tools with a third tool, namely a chain rule theorem for the smooth max-entropy32 to arrive at the bound in inequality Eq. (20).
Given inequalities Eqs. (17) and (20), all we are left to do is to devise a protocol that estimates the max-entropies of X and Z given Bob’s quantum information B. Here we can make use of protocols in quantum key distribution that estimate exactly such a quantity. We show in the Supplementary Notes 46 how two such protocols (one for the max-entropy of X and one for the max-entropy of Z) can be combined into one protocol, which estimates both quantities simultaneously. The resulting protocol, which we presented in two versions, is given by Protocol 1 and Protocol 2 in the Results section. Our bound on the one-shot quantum capacity of the channel, inequality Eq. (9), is obtained by combining inequalities Eqs. (14) and (17) with these max-entropy estimation techniques.
### Data availability
The authors declare that all data supporting this study are contained within the article and its supplementary files.
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## Acknowledgements
We thank the Leo DiCarlo group at QuTech for the experimental infrastructure necessary for MAR to conduct the experiment to collect the test data for a superconducting qubit. We also thank Mario Berta, Thinh Le Phuc, and Jeremy Ribeiro for insightful discussions, and Julia Cramer for helpful comments on an earlier version of this manuscript. C.P., A.M., and M.T. were supported by MOE Tier 3A grant “Randomness from quantum processes”, NRF CRP “Space-based QKD”. S.W. was supported by STW, Netherlands, an NWO VIDI Grant, and an ERC Starting Grant. M.A.R. was supported by an ERC Synergy grant.
## Author information
Authors
### Contributions
S.W. devised the project, and the main conceptual and proof ideas. C.P. worked out the theoretical details with help from A.M. and M.T.; M.A.R. performed the experiment. C.P. and M.A.R. analyzed the data with help from S.W.; C.P., M.A.R. and S.W. wrote the manuscript.
### Corresponding author
Correspondence to Stephanie Wehner.
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### Competing interests
The authors declare no competing financial interests.
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Pfister, C., Rol, M., Mantri, A. et al. Capacity estimation and verification of quantum channels with arbitrarily correlated errors. Nat Commun 9, 27 (2018). https://doi.org/10.1038/s41467-017-00961-2
• Accepted:
• Published:
• ### Resource-efficient verification of quantum computing using Serfling’s bound
• Yuki Takeuchi
• Atul Mantri
• Joseph F. Fitzsimons
npj Quantum Information (2019)
• ### Correlated Effects in Pauli Channels for Quantum Teleportation
• Yan-Ling Li
• Chuan-Jin Zu
• Cheng-Min Wang
International Journal of Theoretical Physics (2019)
• ### Tuning quantum channels to maximize polarization entanglement for telecom photon pairs
• Daniel E. Jones
• Brian T. Kirby
• Michael Brodsky
npj Quantum Information (2018)
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2021-11-29 16:26:44
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https://www.physicsforums.com/threads/circular-motion-yet-again.419962/
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# Circular motion yet again
1. Aug 3, 2010
### Isiudor
1. The problem statement, all variables and given/known data
a ball with a mass of 50 grams is placed inside a circular bowl with a radius of 10cm.
the bowl rotates at a rate of 5 hertz
what is the height in cms the ball will rise within the bowl?
2. Relevant equations
Fr = m*w^2/r
$$\sum$$F = ma
3. The attempt at a solution
well the solution is pretty obvious to me;
since we are talking about max height then Fnety = 0 hence N*cosa = mg and N*sina = m * w^2/r.
the problem is I don't understand why the ball would move at all, what force is the bowl exerting on the ball? there is no mention of friction and the ball initially has no inertia of it's own, how does the spinning bowl effect it without friction?
2. Aug 3, 2010
### arkofnoah
Last edited by a moderator: May 4, 2017
3. Aug 3, 2010
### Isiudor
that is when we assume the ball is already rotating in a certain height within the bowl at the same rate as the bowl.
what I am asking is why does the bowl exert any force on the ball at all
say, we place the ball in the dead center of the bowl and then begin rotating the bowl, there's no friction, why doesn't the ball simply stay in the bottom? there's no force acting on it.
4. Aug 3, 2010
### arkofnoah
oh. in that case you are quite right. the ball would presumably just remain at the bottom of the bowl without spinning at all if there's no friction. if there's no friction at all the ball wouldn't even undergo circular motion at the same rate as the ball, so i think for this question you have to assume friction and that the ball started off slightly "off-center" from the bottom of the bowl (else there will be no horizontal vector accounting for the centripetal force).
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2017-08-21 17:58:01
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https://math.stackexchange.com/questions/1267778/interesting-fact-about-convergent-sequences
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# Interesting fact about convergent sequences?
Let $a_n > 0$ for $n=1,2,...,$ with $\sum_{n=1}^{\infty}a_n < \infty$. Prove that $b_n$ $(n=1,2,...)$ exist such that $b_n/a_n \rightarrow \infty$ as $n \rightarrow \infty$, but $\sum_{n=1}^{\infty}b_n < \infty$.
This somehow is reminiscent of a criterion about sequences which says that, if $a_n$ and $b_n$ are non-negative with $b_n > a_n$, for $n=1,2,...,$ and $\sum b_n$ is convergent, then $\sum a_n$ must be convergent as well. How does one prove the statement above? I simply have no idea.
Any help appreciated.
Hint:
Let $A_n$ be the partial sum of $(a_n)$ and $A_n\to A$
Define $b_1=\sqrt{A}-\sqrt{A-A_1}$ and $b_n=\sqrt{A-A_{n-1}}-\sqrt{A-A_n}$
Then $$\frac{b_n}{a_n}=\frac{1}{\sqrt{A-A_{n-1}}+\sqrt{A-A_n}}\to \infty$$ but $\sum b_n<\infty$
Another solution:
For $k \in \mathbb{N}$, define $n_k:=\min\{ n\in \mathbb{N}: \sum_{j=n}^{\infty}a_j\leq 2^{-k}\}$. This gives a sequence of indices $n_1\leq n_2 \leq n_3 \leq\dots$ such that $2^{-(k+1)}<\sum_{j=n}^{\infty} a_j\leq 2^{-k}$ whenever $n_k\leq n<n_{k+1}$.
Define $b_n:= ka_n$ whenever $n_k \leq n <n_{k+1}$ (and define $b_n=0\;$ if $\;n<n_1$).
Then clearly $\frac{b_n}{a_n} \to \infty$, since $\frac{b_n}{a_n} \geq k$ whenever $n \geq n_k$.
However, we have that $$\sum_{n=n_1}^{\infty} b_n = \sum_{k=1}^{\infty} k\bigg[\sum_{n=n_k}^{n_{k+1}-1} a_n \bigg]\leq \sum_{k=1}^{\infty} k\bigg[\sum_{n=n_k}^{\infty} a_n \bigg]\leq \sum_{k=1}^{\infty} k2^{-k}<\infty$$
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2021-05-11 05:09:41
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https://www.nilesritter.com/wp/?p=2567
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# A Chess Puzzle, Part I: The Infinite Chess Board
First part of A Mathematical Chess Puzzle:
A while ago I was in the waiting room of the doctor’s office, staring at the ceiling, and looking for something to keep myself amused before the appointment. (The visit was nothing serious, just wanting to update my vaccinations.)
Ceiling at doctor’s office
The ceiling was covered with those white acoustic panels, forming a very large colorless and seemingly infinite chess board. I stared at the tiles, trying to think up some kind of puzzle to keep me amused while waiting. I came up with this:
### The Chess Density Problem
Suppose you have an infinite chessboard, and an infinite number of one of the six kinds of chess pieces (King, Queen, Rook, Bishop, Knight, and Pawn). What is the densest packing of “friendly” pieces you can have on that board?
Just asking the question raises more questions:
• What do I mean by “infinite” ?
• What do I mean by “densest” ?
• What do I mean by “friendly” ?
• And how would one prove such things?
So far I’ve spend a fair amount of time and scratch paper on this — quite a bit while sitting around with Gigi’s family for Christmas — and have some things to report. The first version of this post was very math-heavy. Now it is a bit lighter. You’re welcome.
### Disclaimer
Friends who have suffered through games with me will tell you that I am terrible at playing chess, and spend far too long to come up with a bad move that in standard chess notation would probably be denoted “Nf3 Nc6??”. Not just Chess, but Go, Twixt, Bridge or other strategic games. In defense, I can only say that I am spending all of that time trying to solve the problem, based on the game-theoretic fact that a “best” move exists. The decision tree is so deep, however, that no computer yet exists that can explore it completely more than a few moves ahead.
Good chess players acquire skill through experience and study, making decisions with imperfect knowledge. They stand at the edge of the Grand Canyon of moves and can walk down trails blazed by others. I can only stand at the edge, frozen by the awesome infinitude of the depths. Then I fall in.
Anyway, this isn’t the organic chemistry of Chess, but the simpler math of atoms and fundamental particles, Boards and pieces. Moving on…
### Density: A Simple Example
Here is a simple (finite) example of four Queens on a $4 \times 4$ board, where no Queen is attacking any other, either on rank, file, or either diagonal:
Four friendly Queens
So for the $4 \times 4$ board, with its 16 squares, you can only “pack” at most four queens, meaning the “density” (which I denote $\delta$) of Queens on this board is
$$\delta = 4/16 = 1/4$$
Is this the “best” we can do in this case? The answer is yes, for the simple reason that at most one Queen can be on each row (or they would not be friendly), and there are four rows.
This “packing” puzzle is what’s known as a Mathematical Chess Problem, and falls in the category of “recreational mathematics”. That is, mathematics done for amusement rather than “professional” work on “serious” problems. It should be noted, however, that often these problems lead to questions which are actually “deep” and lead to more important work.
In any case, from the literature, the task of arranging “friendly” chess pieces is called an Independence Problem (or an “Unguard” problem), and a chess position in which all of the pieces are friendly is called “Unguarded”. If an unguarded position is the “best” we can do (ie, there is no unguarded position with more pieces), we will call it maximally unguarded.
So, for example, on the standard $8 \times 8$ chessboard, here are examples of how to arrange “friendly” Pawns, Knights and Kings:
32 Friendly Pawns
32 Friendly Knights
16 Friendly Kings
We will show (later) that these are “maximally unguarded” positions, and so the density of these pieces on the standard 8-board are 1/2, 1/2, and 1/4, respectively.
And here is how to arrange the remaining three “friendly” pieces, the Rooks, Bishops and Queens:
8 Friendly Rooks
14 Friendly Bishops
8 Friendly Queens
These last three are maximally unguarded positions, and so the density for these three pieces on the standard board are 1/8, 7/32, and 1/8, respectively.
To see this in the case of the Rooks and the Queens, we can use the same argument as we did above with the Queens on the 4-board, by noting that there are exactly 8 rows and only one Rook or Queen can be on any row.
For the 14 Bishops, you can make a similar argument, by noting two things:
1. There are exactly 15 diagonals in one direction (shown below in blue), and
2. For each diagonal numbered 1 and 15, there is only one square available for placing a Bishop, and those two squares are “hostile” along the red diagonal, so you can only choose one.
Fifteen Bishop Diagonals
We have separated out the pieces into these two sets of three, because each set have their own unique characteristics, unlike the other set; and the differences in density become more pronounced as the size of the boards becomes larger and becomes infinite. Not only that, but the kind of “density” has to be defined differently, and even the size of the “infinite” board becomes important; as we’ll see, some chess boards are more infinite than others.
### Chess Boards: Finite and Infinite
All of the “chess boards” we will be working with will be two-dimensional, and each “square” on which a chess piece can be placed can be indicated by a pair of numbers $(x,y)$, where $x$ and $y$ are coordinates in some set $A$, which we will call the Address space. A chess board that uses coordinates in Address space $A$ will be denoted $B_A$. Mathematically, the board $B_A$ can be identified with the Cartesian product $A \times A$.
For example, the squares of our standard $8 \times 8$ chess board can be specified using $A = \{0,1,2,3,4,5,6,7\}$. The set of integers from 0 to 7 is sometimes denoted $\mathbb{Z}_8$, and so the standard chess board is $B_{\mathbb{Z}_8}$. To make life simple(r), we will also call this board $B_8$, or simply the “8-Board”, and similarly an $n \times n$ board (where n is an integer) will be called the “n-Board”, or $B_n$.
Here for example is the 5-board $B_5 = B_{\mathbb{Z}_5}$:
The 5-Board
Okay, so far so good (you still with me, here?). Now the point of all this mathematical gobbledegook (a scientific term), is that I wanted to look at the “Unguarded” chess problem on an infinite chess board, meaning that I want to use for a “coordinate space” $A$ an infinite set. The first one that comes to mind is the set of whole numbers, ie
$$\mathbb{W} = \{0,1,2,. . .\}$$
And so we can define our first infinite chess board as $B_{\mathbb{W}}$, consisting of all $(x,y)$ where $x$ and $y$ are non-negative integers.
Intuitively, the board can be visualized like this:
The ω-Board
This picture looks a lot like the ceiling at my doctor’s office, which is what started this whole thing.
Now the “size” (cardinality) of the whole numbers is “infinite”, but the specific mathematical name for this “infinite” value is $\aleph_0$ (pronounced aleph-null). Unlike the conventional symbol for infinity ($\infty$), $\aleph_0$ has a more precise meaning, and it refers only to “countably” infinite sets, ie, those which you can count with the integers 1,2,3, etc. A related infinite “ordinal” number is called $\omega$ (small omega), and with some “abuse of notation” we will refer to $B_{\mathbb{W}}$ as $B_{\mathbb{\omega}}$, or simply the $\omega\text{-Board}$.
### Chess Boards: Infinity and Beyond
There are many other ways to define infinite boards, but for now the only other board we will look at uses as its coordinate space the closed unit interval on the real line, that is,
$$I = \{ x \in \mathbb{R}: 0 \le x \le 1 \}$$
The board $B_I$ can be visualized as the unit square, including every single point on the boundary and in the interior as a distinct “square” on which you could place a chess piece:
The c-Board
The size of the set of real numbers in $I$ is a transfinite number which is infinitely larger than $\aleph_0$, and is called simply $c$, the infinity of the continuum. If we embrace “The Continuum Hypothesis“, this number c can be identified as $\aleph_1$, the next largest infinity. We shall therefore also call $B_I$ the c-Board, or $B_c$. The number of “squares” on $B_c$, unlike the $\omega\text{-Board}$, is uncountably infinite.
### Sneak Preview
The “density” results for each infinite board and piece will be shown (in later posts) to be as follows.
The pieces are in order of their decreasing (2d or fractal) density, and thus their increasing “power”:
$$\omega\text{-Board}$$
Piece Board 2d-Density Fractal Dimension 1d-Measure (**)
Pawn $B_\omega$ 1/2 2 *
Knight $B_\omega$ 1/2 2 *
King $B_\omega$ 1/4 2 *
Rook $B_\omega$ 0 * *
Bishop $B_\omega$ 0 * *
Queen $B_\omega$ 0 * *
$$c\text{-Board}$$
Piece Board 2d-Density Fractal Dimension 1d-Measure (**)
Pawn $B_c$ * * *
Knight $B_c$ * * *
King $B_c$ * * *
Bishop $B_c$ $0$ $1$ $2\sqrt 2$
Rook $B_c$ $0$ $1$ $2$
Queen $B_c$ $0$ ? ?
(*) Not Defined for this piece on this board.
(?) Unknown whether a Hausdorff-measurable solution exists.
(**) 1d-Measure = (Hausdorff Content)
As you can see from the two tables, there is something very different between the two groups of pieces, and how they “pack” on those infinite boards. In following posts, I will try to explain what the differences are, what a Fractal Dimension is, and what other mathematical issues and questions arise in the process.
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2022-09-27 02:47:37
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http://mathoverflow.net/revisions/64795/list
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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
2 added 194 characters in body
On a tangent to a problem I've been working on, I've run into a combinatorial/partition-theoretic problem that I wondered if anyone had run into before.
Let $N$ be a positive integer, and ad-hoc-ly call an (ordered) non-negative partition of $N$ into exactly $N$ parts $$N=n_1+n_2+\cdots+n_N$$ valid if
• $0\leq n_i\leq 2$ for all $i$; and
• $\sum\limits_{k=1}^i n_k<i$ for all $i<N$.
So these are something like partitions where the running total is always bounded by the number of terms added thus far. (So the running average of the elements of the partition is less than 1.) In particular, this forces $n_1=0$ and $n_N=2$.
I'm more interested in whether this notion of a "valid" partition has arisen previously in the literature than an explicit count of how many of them there are for a given $N$ (probably a reasonably straight-forward linear recurrence or something), so any such references would be appreciated.
1
# Partitions into 0,1, and 2 with a partial sum condition.
On a tangent to a problem I've been working on, I've run into a combinatorial/partition-theoretic problem that I wondered if anyone had run into before.
Let $N$ be a positive integer, and ad-hoc-ly call an (ordered) non-negative partition of $N$ into exactly $N$ parts $$N=n_1+n_2+\cdots+n_N$$ valid if
• $0\leq n_i\leq 2$ for all $i$; and
• $\sum\limits_{k=1}^i n_k<i$ for all $i<N$.
In particular, this forces $n_1=0$ and $n_N=2$.
I'm more interested in whether this notion of a "valid" partition has arisen previously in the literature than an explicit count of how many of them there are for a given $N$ (probably a reasonably straight-forward linear recurrence or something), so any such references would be appreciated.
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2013-06-19 14:20:42
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https://cstheory.stackexchange.com/questions/4383/when-do-coherence-spaces-have-pullbacks-and-pushouts/4453
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# When do coherence spaces have pullbacks and pushouts?
$\newcommand{\symp}{\Bumpeq}$
A coherence relation $\symp_X$ on a set $X$ is a reflexive and symmetric relation. A coherence space is a pair $(X, \symp_X)$, and a morphism $f : X \to Y$ between coherence spaces is a relation $f \subseteq X \times Y$ such that for all $(x,y) \in f$ and $(x',y') \in f$,
1. if $x \symp_X x'$ then $y \symp_Y y'$, and
2. if $x \symp_X x'$ and $y = y'$ then $x = x'$.
The category of coherence spaces is both Cartesian and monoidal closed. I would like to know when pullbacks or pushouts exist for this category, and when some monoidal analogue of pullbacks or pushout exists (and how to define it, in case this notion makes sense).
• Where is this definition from? The one in Girard, Lafont & Taylor looks very different. – Charles Stewart Jan 19 '11 at 14:25
• The two definitions are equivalent. I'm just taking the web as primitive, from which the set of cliques can be derived. – Neel Krishnaswami Jan 19 '11 at 16:54
• I find Neel's choice of definition much more comprehensible than the original. – Dave Clarke Jan 19 '11 at 18:24
• I'll state the obvious question: do you know that they don't always exist? In other words, are you familiar with any examples of a functor into coherence relations that doesn't have a limit/colimit? – Ohad Kammar Jan 21 '11 at 2:24
• The two definitions are equivalent - Right, but did you make up this definition, or did you get it from someone else? Great question, btw, I'm surprised that no one seems to know whether equalisers always exist. – Charles Stewart Feb 2 '11 at 12:47
I now see how to define equalizers for coherence spaces, which means pullbacks always exist (since products do). I don't know how to do this, actually....
Recall that composition is the usual relational composition, so if $f : A \to B$ and $g : B \to C$, then:
$f ; g = \{(a,c) \in A \times C \;|\; \exists b \in B.\; (a,b) \in f \land (b,c) \in g\}$
(In this definition, the existential actually implies unique existence. Suppose that we have $b' \in B$ such that $(a,b') \in f$ and $(b', c) \in g$. Since we know that $a \Bumpeq_A a$, this means that $b \Bumpeq_B b'$. Then this means that we have $b \Bumpeq_B b'$ and $(b,c) \in g$ and $(b',c) \in g$, so consequently $b = b'$.)
We now construct equalizers. Suppose we have coherence spaces $A$ and $B$, and morphisms $f, g : A \to B$. Now define the equalizer $(E, e : E \to A)$ as follows.
1. For the web, take $$E = \left\{ \begin{array}{l|c} & \forall b.\; (a,b) \in f \implies \exists a' \Bumpeq_A a.\; (a',b) \in g \\ a \in A & \land \\ & \forall b.\; (a,b) \in g \implies \exists a' \Bumpeq_A a.\; (a',b) \in f \\ \end{array} \right\}$$ This picks out the subset of tokens of $A$ on which either $f$ and $g$ agree (up to coherence -- I had this wrong in my first version), or are both undefined.
2. Define the coherence relation on $\Bumpeq_E = \{ (a,a') \in \Bumpeq_A \;|\; a \in E \land a' \in E\}$. This is just the restriction of the coherence relation on $A$ to the subset $E$. This will be reflexive and symmetric since $\Bumpeq_A$ is.
3. The equalizer map $e$ is just the diagonal $e : E \to A = \{(a, a)\;|\;a \in E\}$.
Since I messed up my first version of the proof, I'll give the universality property explicitly. Suppose we have any other object $X$ and morphism $m : X \to A$ such that $m;f = m;g$.
Now define $h : X \to E$ as $\{(x,a) \;|\; a \in E\}$. Obviously $h;i \subseteq m$, but to show the equality we need to show the converse $m \subseteq h;i$.
So assume $(x,a) \in m$. We now need to show that $\forall b.\; (a,b) \in f \implies \exists a' \Bumpeq_A a.\; (a',b) \in g$ and $\forall b.\; (a,b) \in g \implies \exists a' \Bumpeq_A a.\; (a',b) \in f$.
First, assume $b \in B$ and $(a,b) \in f$. So we know that $(x,a) \in m$ and $(a,b) \in f$, so $(x,b) \in m;f$. Therefore $(x,b) \in m;g$, and so there is an $a' \in A$ such that $(x,a') \in m$ and $(a',b) \in g$. Since $x \Bumpeq x$, we know $a \Bumpeq a'$, and so there is an $a' \Bumpeq a$ such that $(a',b) \in g$.
Symmetrically, assume $b \in B$ and $(a,b) \in g$. So we know that $(x,a) \in m$ and $(a,b) \in g$, so $(x,b) \in m;g$. Therefore $(x,b) \in m;f$, and so there is an $a' \in A$ such that $(x,a') \in m$ and $(a',b) \in f$. Since $x \Bumpeq x$, we know $a \Bumpeq a'$, and so there is an $a' \Bumpeq a$ such that $(a',b) \in f$.
• I don't see how you can prove $e$ universal. There is only one way to factor any $m : X \rightarrow A$, and that's by setting $h : X \rightarrow E$ as $h := \{(x, a) : (x,a) \in m, a \in E\}$. Obviously $h;e \subset m$, but I don't see why the converse holds: take some $xma$, and some $b \in B$, with $afb$. Then we have $x(m;f)b$, hence from $m$'s choice we have $x(m;g)b$. From the definition of composition, there exists some $a'$ such that $xma'$ and $a'gb$. We can deduce that $a \symp a'$, but we only know that $afb$ and $a'gb$, so we can't really deduce that $a=a'$ and finish. – Ohad Kammar Jan 21 '11 at 12:25
• Yes, you're right -- the subset the equalizer picks out has to be up to coherence, not equality. I've changed the definition to reflect this, and given the proof the diagram commutes explicitly. – Neel Krishnaswami Jan 21 '11 at 14:08
• Ah... But now $e$ doesn't equalise the diagram. Indeed, assume $a(e;f)b$. Then, by $e$'s definition, we have $afb$, hence there exists some $a'\symp a$ such that $a'gb$. But we don't have that $aea'$, so we can't show that $a(e;g)b$. You seem to be running into the same problems I ran into last night, hence my obvious question above. But perhaps you'll succeed where I failed! My next step was to take a more sophisticated $e$, say something like $aea' \iff a\symp a'$, but then $e$ is not a valid morphism, so some more careful choice is required. – Ohad Kammar Jan 21 '11 at 14:15
• I now remember why I was hoping the answer was in someone's thesis already. :) Anyway, I'll think about it more -- there may be some trick possible via to the fact that inverse images are pairwise incoherent. – Neel Krishnaswami Jan 21 '11 at 17:45
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2020-12-04 11:50:17
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https://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-2-section-2-2-graphs-of-equations-in-two-variables-intercepts-symmetry-2-2-assess-your-understanding-page-166/59
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College Algebra (10th Edition)
Intercepts: $(0,0)$. The graph has origin symmetry.
$\left[\begin{array}{ll} \text{x-intercepts:} & \text{y-intercepts:}\\ 0=\sqrt[3]{x} & y=\sqrt[3]{0}\\ x=0 & y=0\\ & \end{array}\right]$ The only intercept is $(0,0)$ Test x-axis symmetry: Replace $y$ with $-y$ $-y=\sqrt[3]{x}$ ... different from initial equation Test y-axis symmetry: Replace $x$ with $-x$ $y=\sqrt[3]{-x}$ $y=-\sqrt[3]{x}$ ... different from initial equation Test origin symmetry: $x\rightarrow-x$ and $y\rightarrow-y$. $-y=\sqrt[3]{-x}$ $-y=-\sqrt[3]{x}$ $y=\sqrt[3]{x}$ ...same as initial equation
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2019-11-18 08:47:21
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https://www.semanticscholar.org/paper/Computational-methods-for-martingale-optimal-Guo-Ob%C5%82%C3%B3j/f29894dc01e9af448c93915b89d205f7d7e05abb
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# Computational methods for martingale optimal transport problems
@article{Guo2017ComputationalMF,
title={Computational methods for martingale optimal transport problems},
author={Gaoyue Guo and Jan Obł{\'o}j},
journal={arXiv: Probability},
year={2017}
}
• Published 22 October 2017
• Mathematics, Computer Science
• arXiv: Probability
We establish numerical methods for solving the martingale optimal transport problem (MOT) - a version of the classical optimal transport with an additional martingale constraint on transport's dynamics. We prove that the MOT value can be approximated using linear programming (LP) problems which result from a discretisation of the marginal distributions combined with a suitable relaxation of the martingale constraint. Specialising to dimension one, we provide bounds on the convergence rate of…
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It is shown that for many problems related to optimal transport, the set of linear constraints can be split in an intersection of a few simple constraints, for which the projections can be computed in closed form.
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The basic problem of optimal transportation consists in minimizing the expected costs E[c(X 1 , X 2)] by varying the joint distribution (X 1 , X 2) where the marginal distributions of the random
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We investigate existence of dual optimizers in one-dimensional martingale optimal transport problems. While [BNT16] established such existence for weak (quasi-sure) duality, [BHP13] showed existence
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A new class of stochastic optimization algorithms to cope with large-scale problems routinely encountered in machine learning applications, based on entropic regularization of the primal OT problem, which results in a smooth dual optimization optimization which can be addressed with algorithms that have a provably faster convergence.
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Motivated by the model- independent pricing of derivatives calibrated to the real market, we consider an optimization problem similar to the optimal Skorokhod embedding problem, where the embedded
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In this paper, we focus on model-free pricing and robust hedging of options depending on the local time, consistent with Vanilla options. This problem is classically approached by means of the
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2022-05-16 21:14:06
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http://pmed.erci.pw/derive-an-expression-for-maximum-safe-velocity-of-vehicle-on-a-banked-road.html
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# Derive An Expression For Maximum Safe Velocity Of Vehicle On A Banked Road
Let ϴ be the angle of banking, with frictional force f acting between the road and the tyres of. Formulating and implementing transport and land-use policies that promote safer and more efficient trips Making vehicles more protective and visible for occupants, pedestrians and cyclists. Click to join ALLEN NEET-UG (AIPMT) Major (Full Syllabus) online test series for your preparation of NEET-UG (AIPMT) 2020. For reasons of safety airplanes may have to be separated by a minimum space or time interval. Maximum side friction factors for various design speed. Derive an expression for maximum velocity for the vehicle in a banked road Get the answers you need, now!. At a distance of 14 feet, the maximum exposure level is less than 0. s−2 from zero until it reaches a final speed vf = 100 km per. Between 2010 and 2013 there was a 16% increase in the number of registered vehicles in the world. Please be 100% positive with your answer. Mastering Physics Solutions: Banked Frictionless Curve, and Flat Curve with Friction. maximum number of possible users = 1 Gbps/ 100 Kbps = 10000. Maximum road speed: When tractive effort TE is equal to the tractive resistance TR a vehicle would be either stationary or moving at a uniform velocity. The circular motion of any car in either a flat or a banked road provides interesting applications of the laws of motion. For a liquid-gas interface, the convex side has a high Pressure than the concave side. Dual Nature of Matter and Radiation. Express your answer in terms of the given quantities. Turning in a circle requires a vehicle to have a centripetal acceleration inwards on the turn, and so there must be some centripetal force that produces this acceleration. your speed is safe for the weather conditions (for example, slow down if it is raining, windy or foggy). Consider 2 cases for that problem: Max velocity at a specific radius and banked (degree) occurs AGAINST a max friction that prevents the car from. AP Physics Practice Test: Work, Energy, Conservation of Energy ©2011, Richard White www. Let g be the magnitude of the acceleration due to gravity. The maximum safe speed of a car having mass 2000 kg will be, if the coefficient of friction between tyre and road is 0. Commercial Motor Vehicle Driver Handbook. In this model, all parameters have a clear physical meaning: a is the maximum acceleration, b is the comfortable deceleration, v max is the desired maximum velocity, h stop is the desired stopping distance and T gap is the desired time gap, and their values are as expected when fitting the model to data (Hoogendoorn & Hoogendoorn 2010). What do you mean by banking of roads at curves? Show that there is always a loss of kinetic energy in ordinary inelastic collisions. A good place to start here is with the free-body diagram. This puts your maximum exhaust velocity at 7,600,000 m/s, giving you a mass flow of propellant of 34. There are two types of speed limits, posted speed and statutory speed, definitions of each are provided. What Does Velocity Depend On? Explain Why This Dependency May Occur. A circular race track of radius 300 m is banked at an angle of 15°. physics (classes xi –xii) The syllabus for Physics at the Higher Secondary Stage has been developed with a view that this stage of school education is crucial and challenging as it is a transition from general science to discipline-based. Cars on banked turns. The car is held up on the bank by static friction between the wheels and the road with a coefficient of static friction µs. However, it is. What is the maximum velocity that a Range Rover can go around a curve if it the curve has a radius of 50 m? 57. potential energies; simple pendulum– derivation of expression for its time period; free, forced and damped oscillations, resonance. Master the concepts of banking of roads including skidding, angle of banking, overturning and their examples with the (A) Gives the angle of banking for the maximum velocity v and (B) gives the value of the maximum velocity which the vehicles should be allowed on a road banked at an angle θ. It the tractive effort exceeds the resistance the vehicle will accelerate until the tractive effort once again equals the tractive resistance. Derive an expression for the maximum safe velocity of a car on a level road. When starting a bus on a level surface with good traction, there is often no need for the. Far more cost effective than conventional methods. (In other words the car turns, but the passenger does not). (a) What angle does the ball make with respect to the vertical? (b) Derive an expression for the maximum velocity before the car starts slipping sideways on show more A ball is hung by a string from the inside roof of a van. 8m/s2, μ = 0. b) What is vmax for R = 250 m and s = 0. Define banking of roads. 5m to 70m of road length, the amount of received data comes to be approximately 150kbits to 420kbits. Give reason. open your door when you park on a street next to traffic. 2, what is the (a) optimum speed of the race car to avoid wear and tear on its tyres and the (b) maximum permissible speed to avoid slipping?. 07 ft/ft maximum superelevation and a radius of 1909 ft. These maneuvers are proven to be safe (i. The displacement for an object traveling at a constant velocity can be found by multiplying the object’s velocity by the time the object travels at that velocity. It the tractive effort exceeds the resistance the vehicle will accelerate until the tractive effort once again equals the tractive resistance. Continuing the derivation above, we can get: First, note that if the coefficient of friction were zero, the expression given above for v would reduce to the same expression we derived for the no-friction case. If the car carries out a uniform circular motion than we know that a net radial force must be acting on it and that its magnitude is equal to. So, if there is no friction, there has to be some force that can provide that necessary centripetal force. •v0 velocity of vehicle measured at v0 2 /2[a + Gsin(tan-1 (g))] from the intersection •v0 ≥ posted speed limit, preferably the 85th percentile speed. The Doppler Effect is illustrated in the cartoon above. For a vehicle driving on flat ground, this force must be produced by a sideways friction force on the tires. They wrote the. Driving at this speed on a banked road will cause little wear and tear of the tyres. Aerodynamic drag forces are assumed to act directly on the particle. There are three ways to pair them up: velocity-time, position-time, and velocity-position. Thus it is the same for heavy and light vehicles. A lifetime of learning Get started with Brilliant’s course library as a beginner, or dive right into the intermediate and advanced courses for professionals and lifelong learners. September 2009 FHWA-SA-10-001 FOREWORD. There are two possible scenarios for the formation of these low-relief bulges which are impact melt products or young silicic volcanic eruptions. The engine is continuously sounding a whistle of frequency 500 Hz. State any four assumptions of kinetic theory of gases. CLASS XI CLASS XII Physical world and measurement Electrostatics Kinematics Current Electricity Laws of Motion Magnetic Effects of Current and Magnetism Work, Energy and Power Electromagnetic Induction and Alternating Currents Motion of System of Particles and Rigid Body Electromagnetic Waves Gravitation Optics Properties of Bulk Matter Dual Nature of Matter and Radiation Thermodynamics Atoms. 5 sec and the proportionality constant being about 0. Jason Zhu 13,313 views. The car is held up on the bank by static friction between the wheels and the road with a coefficient of static friction µs. 120 m; and maximum speed likely to be sanctioned 160 kmph. The maximum speed that the car can travel without skidding is A. PHYS 1131 T1, 2007 UNSW Question 1 a) Your car is stopped on the side of a straight highway. (a) Derive an expression for the maximum safe speed for a car on a banked track, inclined at angle α to the horizontal. com Contact Us Contact. When it’s moving away, the waves seem longer and the frequency seems lower. This will equal W - R where W is the weight of the car and R is the normal contact force with the road. Question 18. OR i) A car accelerates from rest at a constant rate ‘A’ for some time, after which it retards at a constant rate ‘B’ to come to rest. preparing an unregistered vehicle for registration. This is because, in circular path the speed of an object remains the same while its velocity changes. Terminal velocity: It is maximum constant velocity acquired by the body while falling freely in a viscous medium. expression for fringe width, coherent sources and sustained interference of light. JCM | October 2018 - Browse Articles. 4 m Part B = μ min = 0. Distracted driving can increase the chance of a motor vehicle crash. Jason Zhu 13,313 views. It means the vehicle can be driven High Speed on Unbanked Road: During motorcycle race, the riders negotiate the curve on a flat road in that case they have to tilt their motorcycle and. The motion of a body along a circular path with constant speed is called uniform circular motion. critical phase velocity of the track–soil system. physics (classes xi –xii) The syllabus for Physics at the Higher Secondary Stage has been developed with a view that this stage of school education is crucial and challenging as it is a transition from general science to discipline-based. Then it overtakes the vehicle B and occupies the left lane at time t3. For ideal banking , the net external force equals the horizontal centripetal force in the absence of friction. Derive expression for maximum safe velocity with which a vehicle can travel on banked road. Master the concepts of banking of roads including skidding, angle of banking, overturning and their examples with the (A) Gives the angle of banking for the maximum velocity v and (B) gives the value of the maximum velocity which the vehicles should be allowed on a road banked at an angle θ. b) Derive an expression for the minimum velocity necessary to keep the car moving in a circle without slipping down the embanked turn. Work done by a constant force and a variable force; kinetic energy, work-energy theorem, power. Hence derive the expression for maximum veloclty with which a vehicle takes a safe turn on a banked road 21. Banked Turn. Safety is another issue that concerns the most by the all the tourists. What is the speed necessary so that a car can take a frictionless curve banked at 55o if the turn radius is 400 m? (75. Next, notice that this velocity is about twice the no-friction velocity from Example 1. b) What is vmax for R = 250 m and s = 0. A spring of spring constant K is used to suspend a mass 'm' at its free. Finally at the bottom before the crash it reaches its maximum speed which shown as V’. Very low on carbon emissions. 2% of all deaths globally. The maximum possible friction force between two surfaces before sliding begins is the product of the coefficient of static friction and the normal force: =. Derive an expression for maximum safe speed of a car on a banked road. What is the maximum speed, at which a car turns around a. Find out if it is possible to get your salvage title re-branded. Find a clever high school student to work it out for. Motion in a plane. c = speed of sound. Multi-Part Force Problem: Stopping Distance You are driving your car along a level residential street at 35 mph (16 m/s) when you see a ball roll out into the street in front of you. 71 Rate of descent For a powered sailplane the smallest rate of descent in power-off configuration at maximum weight and most unfavourable c. Download Version PDF [1. Our Veterans have answered our Country's call. Roller Coaster Loop Shapes Physics Education 40, p 517 (2005) Many modern roller coasters features loops. Derive an expression for the maximum velocity of vehicle on a banked road, when coefficient of friction is also taken in to consideration. In an “ideally banked curve,” the angle $\theta$ is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. Derive an expression ( ) dr Also v = for the period of revolution of a particle dt = r −ωˆi sin ωt + ωˆjcos ωt performing uniform circular motion. Give reasons for the following:– (a) A circketer moves his hands backwards while holding a catch. A car can safely travel along the unbanked curve at a maximum speed v under conditions when the coefficient of static friction between the tires and the road is 0. Safe velocity is independent of mass ‘m’ of the vehicle. Genevois's idea probably derived from a windmill cart. Here I have derived an expression for maximum safety of a vehical on banked curved road aswell as angle of banking. If OA is banked road and OX is horizontal line, then ∠. Match the following. If the coefficient of static friction is μ, (a) Obtain an expression for the maximum speed v with which a car can go over the curve without skidding. The occupants of a moving vehicle being irradiated by a traffic radar are unlikely to be expos- ed to a power density as great as 0. a) If the coefficient of static friction between tires and road is s, derive an expression for the maximum speed v max at which the driver can take the curve without sliding. In an “ideally banked curve,” the angle $\theta$ is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. Banked turns. , and Kenneth W. Safe vehicle operation. If the road is flat, the radial force is supplied by friction. There are two forces; one is the force of gravity vertically downward through the center of mass of the ball mg, where m is the mass of the ball and g is the gravitational acceleration; the second is the upward normal force exerted by the road at a right angle to the road surface man. Centripetal force, examples of circular motion (vehicle on level circular road, vehicle on banked road). OR i) A car accelerates from rest at a constant rate ‘A’ for some time, after which it retards at a constant rate ‘B’ to come to rest. The value of v is the maximum allowable velocity. erg &Power - Sclar product of Vector work done by a constant force and a variable force, kineic energy, work energy theorem, power. Existing prediction models in literature were developed based on observation from low viscosity liquids, neglecting the effects of fluid properties (i. If the car carries out a uniform circular motion than we know that a net radial force must be acting on it and that its magnitude is equal to. Unit-4: Work, Energy and Power. Therefore, the maximum permissible speed for the vehicle is much greater than the optimum value of the speed on a banked road. If the car’s speed is now doubled, what is the smallest radius of flat curve that the car can negotiate without sliding? Assume that the coefficient of static friction between the tires and road surface is the same for both. For ideal banking, the net external force equals the horizontal centripetal force in the absence of friction. A 17 kg crate is to be pulled a distance of 20m requiring 1210 J of work being done. turn right or left. (ii) According to drift velocity expression, relaxation time is the time interval between successive collisions of an electron on increasing temperature, the electrons move faster and more collisions occur quickly. In Malaysia, there are a lot of terrible experience happen in using public cabs such as robbing, theft, scam and others. Derive An Expression For The Velocity Of Money. your speed is safe for the weather conditions (for example, slow down if it is raining, windy or foggy). Homework Statement Derive the expressions for the i_r and i_θ components of velocity and acceleration. The car is idealized as a particle with mass m supported on a massless chassis with wheelbase L. One Stop; myU; Search. In an “ideally banked curve,” the angle $\theta$ is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. These maneuvers are proven to be safe (i. Notion of potential energy, potential energy of a spring, conservative. The following figures were the most up-to-date statistics at the time of our latest research. This is needed to model the roll behavior of a car in curves and can also take different load transfers into consideration. The paper is called The Problem with the Amber Signal in Traffic Flow. 71 Rate of descent For a powered sailplane the smallest rate of descent in power-off configuration at maximum weight and most unfavourable c. Derive an expression for the brake torque in case of simple band brake. One gallon of gasoline contains 1. Notion non-uniform motion, average speed and. and yaw rate. We are going to derive an expression that shows how the kinetic energy of the emitted particle is linked with the total kinetic energy. At this speed, frictional force is not needed at all to provide the necessary centripetal force. Many governmental standards and regulations relating to safety, fuel economy, emissions control, noise control, vehicle recycling, substances of concern, vehicle damage, and theft prevention are applicable to new motor vehicles, engines, and equipment manufactured for sale in the United States, Europe, and elsewhere. From this theory, we derive the probability distribution of vehicle locations on arterial links, delimited by signalized intersections. Derive an expression for the centripetal acceleration of a body moving with uniform speed v along a circular path of radius r. You always have to accelerate an object toward the. How is the safety factor calculated. open your door when you park on a street next to traffic. z r T mg L θ Practice Problem: An engineer is designing a rounded exit ramp for a highway. vehicle 2, to the vehicles 2,+1, 2, 1, 1,,and 1, 1, respectively, where, is vehicle label, is lane number, and is lane speci c vehicle index. PHYS 1131 T1, 2007 UNSW Question 1 a) Your car is stopped on the side of a straight highway. , viscosity). This is because, in circular path the speed of an object remains the same while its velocity changes. The vehicle A slows down to follow the vehicle B as shown in the figure with same slope from t 0 to t 1. Given just the right speed, a car could safely negotiate a banked curve even if the road is covered with perfectly smooth ice!. The engine is continuously sounding a whistle of frequency 500 Hz. The direction of the static friction force is always in opposition to the external forces(s) acting on the body The magnitude of the maximum static friction force is: Where: is the coefficient (maximum) of static friction FN is the normal force Kinetic Friction Kinetic frictional force ( ) is the frictional force exerted by the surface on an object that is moving along the surface Kinetic frictional force: always opposes the direction of the motion the direction is along the surface. ple geometric design. The speed will be maximum when tan θ = 1 i. When centripetal force is obtained from friction force as well as banking of roads, then maximum safe value of speed of vehicle. Longitudinal and transverse waves, speed of wave motion. using heavy earthmoving equipment which need to Before you get a UVP, the vehicle must be in a safe condition and you'll need to declare that the vehicle is safe for use on the road or highway. Trusted by 70 Lakh Students. The curved road is banked at θ with the horizontal. Typically this is done while assuming zero friction force. A nonlinear observer for vehicle velocity estimation with stability guarantees, based on acceleration Since the roll angle of the vehicle and the road bank angle usually are small, the assumption. A lifetime of learning Get started with Brilliant’s course library as a beginner, or dive right into the intermediate and advanced courses for professionals and lifelong learners. You slam on your brakes in case anyone follows the ball out into the street. (In other words the car turns, but the passenger does not). Diffraction due to a single slit, width of central maximum. Please give us at least 24 hours for the payment to be processed. Safe velocity is independent of mass ‘m’ of the vehicle. Sight Distance is a length of road surface which a particular driver can see with an acceptable level of clarity. Calculating the Frequency Change. A method of determining a safe distance (d safe ) between a first vehicle ( 1 ) and a second vehicle ( 2 ) moving ahead of the first vehicle, each vehicle having a state which compr VEHICLE SPACING CONTROL - NEDERLANDSE ORGANISATIE VOOR TOEGEPAST- NATUURWETENSCHAPPELIJK ONDERZOEK TNO. Deduce the relation between α,β,γ. The coefficient of friction between the wheels and the road is µ. Average Velocity. Slug translational velocity, described as the velocity of slug units, is the summation of the maximum mixture velocity in the slug body and the drift velocity. What happens if the vehicle speed is 20 mph. F p;xand F p;y acting on the vehicle in a globally fixed coordi-nate system, and the fixed orientation ˇ ˇin which the. preparing an unregistered vehicle for registration. b) Velocity c) acceleration (3) 20. Law of conservation of linear momentum and its applications. Taylor, Jr. Circular Motion Around a Banked Circular Track Banking of road - YouTube. If it is sliding, it is kinetic friction. Draw a neat labelled diagram of rise of liquid in capillary tube showing different components of tension (force). Uses of plane polarized light and polaroids. In this model, all parameters have a clear physical meaning: a is the maximum acceleration, b is the comfortable deceleration, v max is the desired maximum velocity, h stop is the desired stopping distance and T gap is the desired time gap, and their values are as expected when fitting the model to data (Hoogendoorn & Hoogendoorn 2010). We can help you with any aspect of the commercial truck industry. However, choosing car rental Malaysia can get rid of your worries about the safety. 71 Rate of descent For a powered sailplane the smallest rate of descent in power-off configuration at maximum weight and most unfavourable c. However, one is unbanked and the other is banked at an angle θ. The Government Printing Office (GPO) processes all sales and distribution of the CFR. When there is no sliding occurring, the friction force can have any value from zero up to F m a x {\displaystyle F_{max}\,}. z r T mg L θ Practice Problem: An engineer is designing a rounded exit ramp for a highway. Consider a vehicle of weight 'Mg' moving round a curved path of radius 'r' with speed 'V' on a road banked through angle θ. If the road is flat, the radial force is supplied by friction. Resolving power of microscopes and astronomical telescopes Polarization, plane polarized light; Brewster’s law. What is the speed necessary so that a car can take a frictionless curve banked at 55o if the turn radius is 400 m? (75. (a) Determine the time for the car to round the curve. Keep a safe distance between you and any vehicles in front or beside you. Determine the velocity of the car at position D. Please give us at least 24 hours for the payment to be processed. (Example 4, page 179) ? When braking without skidding, what maximum deceleration can be achieved? (Example 5, page 180) ? How quickly can a racing car round a curve without skidding side- In this chapter, the rubber tires of an ways? (Example 10, page 186) automobile are revisited to explore ? How does a banked curve help to avoid skidding?. Aerodynamic drag forces are assumed to act directly on the particle. Safe for 40 MPH. Derive minimum angle of banking to travel without wear and tear? 9. This is denoted by LTR d. When there is no sliding occurring, the friction force can have any value from zero up to F m a x {\displaystyle F_{max}\,}. , vehicle Vj 1 is the leader vehicle of. , 2004 [10], and ‘Estimation of Vehicle Lateral Velocity’ by Pierre Pettersson, 2008 [9]. Laws of motion-Banking of Road - Duration: 28:40. Unit IV: Work, Energy and Power 12 Periods Chapter–6: Work, Energy and Power Work done by a constant force and a variable force; kinetic energy, workenergy theorem, power. What are the Most Common Vehicle Accident Injuries? According to the National Highway Traffic Safety Administration (NHTSA), more than three million people are injured each year in vehicle accidents across the country. The velocity is constant, so the forces must. They are ∼250-680 m wide and ∼30-91 m high with maximum flank slopes >20°. Kinematics Work, Energy and Power Frame of reference, Motion in a straight line, Work done by a constant force and variable force, Position-time graph, speed and velocity. (21) Derive expression for velocity of a ring, solid cylinder and solid sphere having same radii rolling down the smooth inclined plane without slipping. UNIT-IV Chapter 6: WORK, ENERGY AND POWER (11 hours) Work: Definition of Work – =. speed s, the transmission rate R, and the length of the link m. The normal force is our only unknown (besides velocity), but we may write an expression for it from the vertical forces. Sample Multiple Choice Questions: 1. The same equation also tells us that for υ< υ0,the frictional force will be up the slope and that a car can be parked only if. By vector addition we can see that the change in velocity, Δv, points toward the center of the circle. Know what causes the centripetal force on: a banked curve, a flat curve, a roller coaster hill. Vehicle B must Give Way to vehicle A because it is turning across the path of an If you are driving on a multi-lane road and want to move into another lane of traffic, either to your right or left, you must first check that it is safe to. Determination of Radius - (1) The radius of a curve is determined by measuring the versine on a chord of known length, from the equation, (2) Curves can be designated by the radius in metres or by its degree. The radius of the curve is r. maximum number of possible users = 1 Gbps/ 100 Kbps = 10000. Motion in a plane. Longitudinal and transverse waves, speed of wave motion. b) Derive an expression for the minimum velocity necessary to keep the car moving in a circle without slipping down the embanked turn. The street is busy: many cars are going past, all of them at 100 km per hour. and what is final velocity?. The vehicle A slows down to follow the vehicle B as shown in the figure with same slope from t 0 to t 1. It has been determined that "trail" (shown below) is often an important contributor to bicycle stability. Angle of banking. tive, a suspension bridge, a car rounding a banked curve, or a toboggan sliding down a hill requires specific problem-solving skills. By vector addition we can see that the change in velocity, Δv, points toward the center of the circle. Maximum Velocity of a Banked Turn With Friction. which a vehicle attains its optimal velocity and (0,1) is the sensitivity factor for the relative velocities which is independent of time, position, and velocity. Have you not been to a race on a track with a bank to it?. Derive an expression ( ) dr Also v = for the period of revolution of a particle dt = r −ωˆi sin ωt + ωˆjcos ωt performing uniform circular motion. This force is actually static friction exerted on the drive wheels by the road - the road exerts this force because the engine causes the drive wheels to rotate. support such a claim. Sample Multiple Choice Questions: 1. 1 Appendix I Entrance Test 2019 The Admission to M. Express your answer in terms of the given quantities. We know that the maximum static friction (at which the tires roll but do not slip) is μ s N , where μ s is the static coefficient of friction and N is the normal force. In general, the dependence on body shape, inclination, air viscosity, and compressibility is very complex. Spectrum Classes by Er. Chapter 6 Circular Motion and Other Applications of Newton's. Jason Zhu 13,313 views. The motion of a body along a circular path with constant speed is called uniform circular motion. Resolving power of microscopes and astronomical telescopes Polarization, plane polarized light; Brewster’s law. Find their velocities after. vehicle 2, to the vehicles 2,+1, 2, 1, 1,,and 1, 1, respectively, where, is vehicle label, is lane number, and is lane speci c vehicle index. Explain Banking of Road obtain an expression for the maximum speed with which a vehicle can safely negotiate a curved road banked at an angle θ. Find the force of fiction and the coefficient of friction. •v0 velocity of vehicle measured at v0 2 /2[a + Gsin(tan-1 (g))] from the intersection •v0 ≥ posted speed limit, preferably the 85th percentile speed. It is shown theoretically that the motion of a long. Safe feedback-based longitudinal maneuvers for AHSs and their associated control laws have been derived in [2] and [3]. (b) Find v if r = 100 m, θ = 30 , g = 9. When starting a bus on a level surface with good traction, there is often no need for the. The distance of two planets from sun are 1013 and 1012 m respectively. ABSTRACT The change of a vehicle’s velocity, delta-V (v), due to an impact is often calculated and. (a) Determine the time for the car to round the curve. Assume a maximum cant deficiency for the route to be 100 mm and Broad Gauge - 1. which is banked at an angle φwith respect to the ground. A ball of mass 0. Many mechanical physiological aspects of the deriver-vehicle-road systems make modeling difficult. This is needed to model the roll behavior of a car in curves and can also take different load transfers into consideration. Centripetal force, examples of circular motion (vehicle on level circular road, vehicle on banked road). Before the car starts turning both the passenger and the car are travelling at the same velocity. road, Sho'w'that th-e safe\y-speedUmit is independent of the ntass ofthe;vehi<:J'e. Explain Banking of Road obtain an expression for the maximum speed with which a vehicle can safely negotiate a curved road banked at an angle θ. If the road is banked, the horizontal component of the Normal vector of the car that is going towards the centre of turn can act as the centripetal force. A curve with a 115m radius on a level road is banked at the correct angle for a speed of 20m/s. SPH4U Final Exam Review asteroid makes to the earth and the amount of kinetic energy it loses on impact. Prove Mayer’s formula ( )Cp Cv R− = (3) 22. Sometimes it may be possible to visualize an acceleration vector for example, if you know your particle is moving in a straight line, the acceleration vector must be parallel to the direction of motion; or if the particle moves around a circle at constant speed, its acceleration is towards the center of the circle. We know that the maximum static friction (at which the tires roll but do not slip) is , where is the static coefficient of friction and N is the normal force. Figure 1: Schematic view of a vehicle dynamics system. 1993 MULTIPLE CHOICE SCREENING TEST 30 QUESTIONS—40 MINUTES The correct expression for the maximum height h' that Air track car A has mass m and velocity v. If the road is flat, the radial force is supplied by friction. (Example 4, page 179) ? When braking without skidding, what maximum deceleration can be achieved? (Example 5, page 180) ? How quickly can a racing car round a curve without skidding side- In this chapter, the rubber tires of an ways? (Example 10, page 186) automobile are revisited to explore ? How does a banked curve help to avoid skidding?. Laws of motion-Banking of Road - Duration: 28:40. Example: Motion of a body on vertical circle. Multi-Part Force Problem: Stopping Distance You are driving your car along a level residential street at 35 mph (16 m/s) when you see a ball roll out into the street in front of you. Give reason. The modeling assumes that each driver in a following vehicle is an active and predictable control. If an automobile rounds this curve at 30m/s, what is the minimum coefficient of static friction between tires and road needed to prevent skidding?. Lift depends on the density of the air, the square of the velocity, the air's viscosity and compressibility, the surface area over which the air flows, the shape of the body, and the body's inclination to the flow. A car rounds a flat, unbanked curve with radius R. Consider 2 cases for that problem: Max velocity at a specific radius and banked (degree) occurs AGAINST a max friction that prevents the car from. (i) Weight of the body acting vertically downwards. Derive an expression for the maximum safe speed of a vehicle on a banked road with friction. It means the vehicle can be driven with maximum safe speed only when the angle of banking = 45°. An expression for LTR which depends on the roll states and vehicle parameters can be obtained. In projectile, a particle is projected such that magnitude of velocity at topmost point is half of the magnitude of initial velocity. Chapter 6 Circular Motion and Other Applications of Newton's. , vehicle Vj 1 is the leader vehicle of. 8m/s2, μ = 0. So, if there is no friction, there has to be some force that can provide that necessary centripetal force. Derive an expression for resultant acceleration in non-uniform circular motion. A 800-N passenger in a car pressed against the car door with a 200 N force when the car makes a left turn at 13 m/s. You do not have to derive the expression, merely write it. Note that the vehicle structure and especially axle and suspension A dynamometer is a machine used to simulate the forces on a drive train to test pollutant emissions, fuel consumption, and other operating. The car is held up on the bank by static friction between the wheels and the road with a coefficient of static friction µs. 1993 MULTIPLE CHOICE SCREENING TEST 30 QUESTIONS—40 MINUTES The correct expression for the maximum height h' that Air track car A has mass m and velocity v. Let g be the magnitude of the acceleration due to gravity. Since acceleration is the change in velocity over a given period of time, the consequent acceleration points in the same direction.
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2020-04-07 07:11:58
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https://docs.learnosity.com/authoring/authorguide/questions/math/imageclozeformula
|
# Math Formula Image
## Description
The Math Formula question allows students to easily enter complex math formula as a response on an image, with advanced validation capabilities using math specific scoring methods. Click here to see some worked examples of our Math Formula Question.
Figure 1. Math Formula Image question example.
## Create a Question
Enter the question stem into the Compose question area. You may want to use our Math Editor in order to insert a mathematical expression in the stimulus. For this click inside the Compose Question text area and select Math Editor from the Rich Text Editor panel (see Figure 2).
Figure 2 - Rich Text Editor panel with Math Editor button.
Then upload the image to be used for the question in the Add image field. You can either upload the image or paste image URL. For information about image requirements visit our Images page. The Image alternative text appears if an image cannot be displayed on a page, and the Text on hover is displayed over the image, when the cursor hovers over it.
Figure 3 - Image Cloze Math question basic parameters.
The uploaded image preview will be shown below in the Response Positions section.
Drag and position the response boxes on the image as desired. To add more response boxes click on +Add above the image. See example in Figure 4.
Figure 4 - Positioning response containers on the image.
Each response box added on the image will have a number label. You can add multiple response boxes within one container and define specific dimensions for each container individually. For this you will need to configure the Edit individual response containers section. This can be found below the uploaded image. Below is a breakdown of the attributes available:
• Template - You can enter complex math equations and response boxes here. The Template is a LaTeX area. Math expressions can be added using the LaTeX Keypad. You can add multiple response boxes here.
• Width and Height - You can enter container width and height in pixels, i.e. 60px and 30px, in order to change standard dimensions.
Follow these steps in order to configure validation:
1. Insert a value in the Point(s) box to define a mark for the question. The default value is set to 1.
2. In the Correct tab below you will see Equation Values submenu. There will be as many submenus as you have response containers on the image. Each submenu will be named Response 1, Response 2 etc. depending on the number of response containers. Here you set validation separately for each container.
3. Under each submenu select a scoring method, and enter the correct response in the Value field. In this example we only have one response container and the students response should be exactly the same as in the validation area. In this case, validation is only set for Equation Value 1 and select equivLiteral as a scoring method. Then enter the correct response in the Value field. Remember that everything that has been entered into the Template area must be entered into the Value area.
Figure 5 - Validation settings for Image Cloze Math question with one response container.
4. Each scoring method has additional options that allows authors to set more precise validation conditions. A breakdown of all options associated with certain scoring methods can also be found in our Formula tutorials. When you select a scoring method all extra options will be displayed below the menu.
5. Some more complex math questions will need a second scoring method applied. You can combine several scoring methods to set more strict validation rules. For instance, you may want the system to accept a response that is not only symbolically equivalent but also is presented in factorised form. In this case, you would need to add another method called isFactorised. Click on the +Add button to create a new scoring method.
### More Options
Additional formatting options can be found under More Options:
#### Scoring
Unscored/Practice usage - Removes all scoring from the question.
Penalty point(s) - The value entered here will be deducted from the student for an incorrect answer.
Minimum score if attempted - Set attempt marks for the question.
Check answer button - Show or hide the Check answer button from the student.
Check answer attempts - The value entered here refers to how many times the student can use the Check answer button, before it is disabled.
#### Layout
Transparent background - If this is enabled, the background of the response area will match the colour of the rest of the background area.
Response minimum width (px) - Enter, in pixels, the minimum width of the response area.
Template font scale - Scales the font relative to the question's font size. The default value is 150%.
Response container (global) and Edit invidiual response containers - Configure the Width and the Height, in pixels, for either all response containers, or for specified response containers.
Show keypad hints - Enable or disable hints on the keypad, such as keyboard shortcuts and symbol group titles that are shown on the top left corner of the keypad when hovering over a symbol group key.
Symbols - Select what symbol groups will be shown to the student, or create custom symbol groups.
#### Text blocks
Use this option when you want to define a list of custom units (such as g, kg, cm, oz, etc.) that will not be rendered as LaTeX.
Example 1 - Math Formula Image question with 1 response box.
Hint Try inserting √29.
{
"image": {
"src": "\/\/assets.learnosity.com\/organisations\/1\/rightangledtriangle.png"
},
"instant_feedback": true,
"is_math": true,
"response_positions": [
{
"x": 55.47,
"y": 24.11
}
],
"stimulus": "<p>Find the hypotenuse of this triangle using the formula \$$a^2+b^2=c^2\$$<\/p>\n\n<p> <\/p>\n",
"type": "imageclozeformula",
"ui_style": {
"type": "block-keyboard"
},
"validation": {
"scoring_type": "exactMatch",
"valid_response": {
"score": 1,
"value": [
[
{
"method": "equivSymbolic",
"value": "\\sqrt{29}",
"options": {
"allowDecimal": false,
"inverseResult": false
}
}
]
]
}
}
}
Example 2 - Math Formula Image question with 2 response boxes.
Hint "21" and "39" are good answers.
{
"instant_feedback": true,
"image": {
"src": "\/\/assets.learnosity.com\/organisations\/1\/Screen Shot 2014-09-04 at 13.21.41_201494132948.png"
},
"is_math": true,
"response_container": {
"template": ""
},
"response_positions": [
{
"x": 46.09,
"y": 35.14
},
{
"x": 84.37,
"y": 35.14
}
],
"stimulus": "<p>What unknown number complete the pattern on the number line?<br \/>\nEnter one answer in each response box.<\/p>\n",
"text_blocks": [],
"type": "imageclozeformula",
"ui_style": {
"response_font_scale": "normal",
"type": "block-on-focus-keyboard"
},
"validation": {
"scoring_type": "exactMatch",
"valid_response": {
"score": 1,
"value": [
[
{
"method": "equivLiteral",
"value": "21",
"options": {
"allowDecimal": false,
"inverseResult": false,
"ignoreOrder": false
}
}
],
[
{
"method": "equivSymbolic",
"value": "39",
"options": {
"allowDecimal": false,
"inverseResult": false
}
}
]
]
}
}
}
Example 3 - Math Formula Image with 1 response box.
Hint The perimeter is the length of the outline of a shape.
{
"instant_feedback": true,
"image": {
"src": "\/\/assets.learnosity.com\/organisations\/1\/trapecio_2015730124110.png"
},
"is_math": true,
"response_containers": [
{
"template": "{{response}}",
"width": "60px",
"height": "40px"
}
],
"response_positions": [
{
"x": 31.28,
"y": 75.17
}
],
"stimulus": "<h4>Find the perimeter of the trapezoid.<\/h4>\n",
"type": "imageclozeformula",
"ui_style": {
"type": "block-on-focus-keyboard"
},
"validation": {
"scoring_type": "exactMatch",
"valid_response": {
"score": 1,
"value": [
[
{
"method": "equivSymbolic",
"value": "28",
"options": {
"allowDecimal": false,
"inverseResult": false,
"ignoreOrder": false
}
}
]
]
}
}
}
|
2019-02-23 09:52:11
|
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|
https://lipea.biotec.tu-dresden.de/about/how
|
# How does it works?
## Methods
The idea behind LIPEA is to identify specific altered pathways - provided by the KEGG Database - using exclusively lipid compounds. The approach used to this task is the Over Representation Analysis (ORA). ORA starts with considering a list of annotated lipids (e.g. a lipid set related with a signature), then uses the Fisher exact test to verify if the annotations are over represented among a label (pathway) compared to the whole universe of lipids (background), that could be selected as “predefined” for a specific organism (it means, LIPEA will take all the compounds from the pathways related with the selected organism) or be a custom list given by the user.
### Procedure
The steps of the algorithm used to implement the ORA are the following.
##### Step 1
Set an organism, collect the lipid list and the background.
##### Step 3
Tally the following 4 numbers: m, N, k, and n, where m is the total number of lipids in the pathway, N is the total number of lipids, k is the number of lipids of the intersection between the lipid list and a pathway, and n is the total number of lipids in the list.
##### Step 4
Perform a Fisher exact test, with the 4 numbers obtained in the preview step, as follows:
$$f(k;N,m,n) = \frac{\binom{m}{k} \binom{N - m}{n - k}}{\binom{N}{n}}$$
The f value is the probability that this random event could happen under the hypergeometric distribution. In this case, to obtain the p-value to associate to each pathway, the following formula is used:
$$p = \sum_{l = k}^n f(l;N,m,n)$$
##### Step 5
Go to step 2 for another pathway of interest, until all are tested.
##### Step 6
Correct the p-values with Benjamini or Bonferroni-Holm corrections.
|
2021-12-03 01:07:37
|
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http://theory.cs.uchicago.edu/abstract.php?id=srivastava&year=2013
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Seminar: December 10
Bipartite Ramanujan Graphs of Every Degree
Expander graphs are very sparse graphs which are nonetheless very well-connected, in the sense that their adjacency matrices have large spectral gap. There is a limit to how large this gap can be for a d-regular graph, and graphs which achieve the limit are called Ramanujan graphs. A beautiful number-theoretic construction of Lubotzky-Phillips-Sarnak and Margulis shows that infinite families of Ramanujan graphs exist for every d=p+1 where p is prime, leaving open the question of whether they exist for other degrees.
We prove that there exist infinite families of bipartite Ramanujan graphs of every degree bigger than 2. We do this by proving a variant of a conjecture of Bilu and Linial about the existence of good 2-lifts of every graph. Our proof exploits a new technique for demonstrating the existence of useful combinatorial objects that we call the Method of Interlacing Polynomials''. The proofs are elementary and rely on simple facts from the theory of real stable polynomials. In particular, they do not rely on number theory, and the talk should be accessible to a broad audience.
Joint work with Adam Marcus and Dan Spielman.
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2018-09-24 21:49:31
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https://juliasmoothoptimizers.github.io/news-and-blogposts/2022/2022-12-16-pdenlpmodels/index.html
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# New publication about PDENLPModels.jl
We are very happy to announce the publication in the Journal of Open Source Software of the paper PDENLPModels.jl: A NLPModel API for optimization problems with PDE-constraints. It is accessible in open access here.
PDENLPModels.jl is a Julia package that specializes the NLPModel API for modeling and discretizing optimization problems with mixed algebraic and PDE in the constraints.
We consider optimization problems of the form: find functions $$y, u$$ and $$κ \in \mathbb{R}^n$$ satisfying
$\begin{array}{lll} \underset{y, u, \theta}{\text{minimize}} \int_\Omega J(y, u, \theta)d\Omega \ \text{ subject to} & e(y, u, \theta) = 0, & \text{(governing PDE on \Omega)} \\ & l_{yu} \leq (y, u) \leq u_{yu}, & \text{(functional bound constraints)} \\ & l_{\theta} \leq \theta \leq u_{\theta}, & \text{(bound constraints)} \end{array}$
The main challenges in modeling such a problem are to be able to discretize the domain and generate corresponding discretizations of the objective and constraints, and their evaluate derivatives with respect to all variables. We use Gridap.jl to define the domain, meshes, function spaces, and finite-element families to approximate unknowns, and to model functionals and sets of PDEs in a weak form. PDENLPModels extends Gridap.jl's differentiation facilities to also obtain derivatives useful for optimization, i.e., first and second derivatives of the objective and constraint functions with respect to controls and finite-dimensional variables.
After discretization of the domain $$\Omega$$, the integral, and the derivatives, the resulting problem is a nonlinear optimization problem. PDENLPModels exports the GridapPDENLPModel type, an instance of an AbstractNLPModel, as defined in NLPModels.jl, which provides access to objective and constraint function values, to their first and second derivatives, and to any information that a solver might request from a model. The role of NLPModels.jl is to define an API that users and solvers can rely on. It is the role of other packages to implement facilities that create models compliant with the NLPModels API. We refer to juliasmoothoptimizers.github.io for tutorials on the NLPModel API.
As such, PDENLPModels offers an interface between generic PDE-constrained optimization problems and cutting-edge optimization solvers such as Artelys Knitro via NLPModelsKnitro.jl, Ipopt via NLPModelsIpopt.jl , DCISolver.jl, Percival.jl, and any solver accepting an AbstractNLPModel as input, see JuliaSmoothOptimizers.
## Example
The following example shows how to solve a Poisson control problem with Dirichlet boundary conditions using DCISolver.jl: find functions $$y \in H^1_0$$ and $$u \in H^1$$ satisfying
$\begin{array}{lll} \underset{y, u}{\text{minimize}} \int_{(-1,1)^2} \frac{1}{2}\|y_d - y\|^2 +\frac{\alpha}{2}\|u\|^2 d\Omega \quad \text{subject to} & \Delta y - u - h = 0, & \text{on } \Omega.\\ & y = 0, & \text{on } \partial\Omega, \end{array}$
for some given functions $$y_d:(-1,1)^2 \rightarrow \mathbb{R}$$ and $$h:(-1,1)^2 \rightarrow \mathbb{R}$$, and $$\alpha > 0$$.
using DCISolver, Gridap, PDENLPModels
# Cartesian discretization of Ω=(-1,1)² in 100² squares.
Ω = (-1, 1, -1, 1)
model = CartesianDiscreteModel(Ω, (100, 100))
fe_y = ReferenceFE(lagrangian, Float64, 2) # Finite-elements for the state
Xpde = TestFESpace(model, fe_y; dirichlet_tags = "boundary")
Ypde = TrialFESpace(Xpde, x -> 0.0) # y is 0 over ∂Ω
fe_u = ReferenceFE(lagrangian, Float64, 1) # Finite-elements for the control
Xcon = TestFESpace(model, fe_u)
Ycon = TrialFESpace(Xcon)
dΩ = Measure(Triangulation(model), 1) # Gridap's integration machinery
# Define the objective function f
yd(x) = -x[1]^2
f(y, u) = ∫(0.5 * (yd - y) * (yd - y) + 0.5 * 1e-2 * u * u) * dΩ
# Define the constraint operator in weak form
h(x) = -sin(7π / 8 * x[1]) * sin(7π / 8 * x[2])
c(y, u, v) = ∫(∇(v) ⊙ ∇(y) - v * u - v * h) * dΩ
# Define an initial guess for the discretized problem
x0 = zeros(num_free_dofs(Ypde) + num_free_dofs(Ycon))
# Build a GridapPDENLPModel, which implements the NLPModel API.
name = "Control elastic membrane"
nlp = GridapPDENLPModel(x0, f, dΩ, Ypde, Ycon, Xpde, Xcon, c, name = name)
dci(nlp, verbose = 1) # solve the problem with DCI
## References
Migot, T., Orban D., & Siqueira A. S. PDENLPModels.jl: A NLPModel API for optimization problems with PDE-constraints Journal of Open Source Software 7(80), 4736 (2022). 10.21105/joss.04736
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2023-02-08 17:56:41
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https://forum.azimuthproject.org/plugin/ViewComment/18098
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One suggestion about the notation (but not the content!) of your ideas: I found it helpful to write the ordering as \$$(a , b, c + n) \leq (a + 2n, b + n , c) \$$, for all \$$a,b,c \in \mathbb{N}\$$. Translating this statement using John's meaning for \$$\leq\$$ would give us "We can get \$$a\$$ Hs, \$$b\$$ Os, and \$$(c +n )\$$ H2Os from \$$(a + 2n)\$$ Hs, \$$(b +n)\$$ Os, and \$$c \$$ H2Os". This jives with my understanding of the problem, and it gets rid of having to worry about \$$a \geq 2n\$$ etc! Also reflexivity follows directly from the case where \$$n = 0\$$.
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2020-02-27 23:33:06
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https://en-academic.com/dic.nsf/enwiki/248416
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# Computability theory (computer science)
Computability theory (computer science)
In computer science, computability theory is the branch of the theory of computation that studies which problems are computationally solvable using different models of computation.
Computability theory differs from the related discipline of computational complexity theory, which deals with the question of how efficiently a problem can be solved, rather than whether it is solvable at all.
Introduction
A central question of computer science is to address the limits of computing devices. One approach to addressing this question is understanding the problems we can use computers to solve. Modern computing devices often seem to possess infinite capacity for calculation, and it's easy to imagine that, given enough time, we might use computers to solve any problem. However, it is possible to show clear limits to the ability of computers, even given arbitrarily vast computational resources, to solve even seemingly simple problems. Problems are formally expressed as a decision problem which is to construct a mathematical function that for each input returns either 0 or 1. If the value of the function on the input is 0 then the answer is "no" and otherwise the answer is "yes".
To explore this area, computer scientists invented automata theory which addresses problems such as the following: Given a formal language, and a string, is the string a member of that language? This is a somewhat esoteric way of asking this question, so an example is illuminating. We might define our language as the set of all strings of digits which represent a prime number. To ask whether an input string is a member of this language is equivalent to asking whether the number represented by that input string is prime. Similarly, we define a language as the set of all palindromes, or the set of all strings consisting only of the letter 'a'. In these examples, it is easy to see that constructing a computer to solve one problem is easier in some cases than in others.
But in what real sense is this observation true? Can we define a formal sense in which we can understand how hard a particular problem is to solve on a computer? It is the goal of computability theory of automata to answer just this question.
Formal models of computation
In order to begin to answer the central question of automata theory, it is necessary to define in a formal way what an automaton is. There are a number of useful models of automata. Some widely known models are:
; Deterministic finite state machine: Also called a deterministic finite automaton (DFA), or simply a finite state machine. All real computing devices in existence today can be modeled as a finite state machine, as all real computers operate on finite resources. Such a machine has a set of states, and a set of state transitions which are affected by the input stream. Certain states are defined to be accepting states. An input stream is fed into the machine one character at a time, and the state transitions for the current state are compared to the input stream, and if there is a matching transition the machine may enter a new state. If at the end of the input stream the machine is in an accepting state, then the whole input stream is accepted.
; Nondeterministic finite state machine: Similarly called a nondeterministic finite automaton (NFA), it is another simple model of computation, although its processing sequence is not uniquely determined. It can be interpreted as taking multiple paths of computation simultaneously through a finite number of states. However, it is proved that any NFA is exactly reducible to an equivalent DFA.
; Pushdown automaton: Similar to the finite state machine, except that it has available an execution stack, which is allowed to grow to arbitrary size. The state transitions additionally specify whether to add a symbol to the stack, or to remove a symbol from the stack. It is more powerful than a DFA due to its infinite-memory stack, although only some information in the stack is ever freely accessible.
; Turing machine: Also similar to the finite state machine, except that the input is provided on an execution "tape", which the Turing machine can read from, write to, or move back and forth past its read/write "head". The tape is allowed to grow to arbitrary size. The Turing machine is capable of performing complex calculations which can have arbitrary duration. This model is perhaps the most important model of computation in computer science, as it simulates computation in the absence of predefined resource limits.
; Multi-tape Turing machine: Here, there may be more than one tape; moreover there may be multiple heads per tape. Surprisingly, any computation that can be performed by this sort of machine can also be performed by an ordinary Turing machine, although the latter may be slower or require a larger total region of its tape.
Power of automata
With these computational models in hand, we can determine what their limits are. That is, what classes of languages can they accept?
Power of finite state machines
Computer scientists call any language that can be accepted by a finite state machine a regular language. Because of the restriction that the number of possible states in a finite state machine is finite, we can see that to find a language that is not regular, we must construct a language that would require an infinite number of states.
An example of such a language is the set of all strings consisting of the letters 'a' and 'b' which contain an equal number of the letter 'a' and 'b'. To see why this language cannot be correctly recognized by a finite state machine, assume first that such a machine $M$ exists. $M$ must have some number of states $n$. Now consider the string $x$ consisting of $\left(n+1\right)$ 'a's followed by $\left(n+1\right)$ 'b's.
As $M$ reads in $x$, there must be some state in the machine that is repeated as it reads in the first series of 'a's, since there are $\left(n+1\right)$ 'a's and only $n$ states by the pigeonhole principle. Call this state $S$, and further let $d$ be the number of 'a's that our machine read in order to get from the first occurrence of $S$ to some subsequent occurrence during the 'a' sequence. We know, then, that at that second occurrence of $S$, we can add in an additional $d$ (where $d > 0$) 'a's and we will be again at state $S$. This means that we know that a string of $\left(n+d+1\right)$ 'a's must end up in the same state as the string of $\left(n+1\right)$ 'a's. This implies that if our machine accepts $x$, it must also accept the string of $\left(n+d+1\right)$ 'a's followed by $\left(n+1\right)$ 'b's, which is not in the language of strings containing an equal number of 'a's and 'b's. In other words, $M$ cannot correctly distinguish between a string of equal number of 'a's and 'b's and a string with $\left(n+d+1\right)$ 'a's and $n+1$ 'b's.
We know, therefore, that this language cannot be accepted correctly by any finite state machine, and is thus not a regular language. A more general form of this result is called the Pumping lemma for regular languages, which can be used to show that broad classes of languages cannot be recognized by a finite state machine.
Power of pushdown automata
Computer scientists define a language that can be accepted by a pushdown automaton as a Context-free language, which can be specified as a Context-free grammar. The language consisting of strings with equal numbers of 'a's and 'b's, which we showed was not a regular language, can be decided by a push-down automaton. Also, in general, a push-down automaton can behave just like a finite-state machine, so it can decide any language which is regular. This model of computation is thus strictly more powerful than finite state machines.
However, it turns out there are languages that cannot be decided by push-down automaton either. The result is similar to that for regular expressions, and won't be detailed here. There exists a Pumping lemma for context-free languages. An example of such a language is the set of prime numbers.
Power of Turing machines
Turing machines can decide any context-free language, in addition to languages not decidable by a push-down automaton, such as the language consisting of prime numbers. It is therefore a strictly more powerful model of computation.
Because Turing machines have the ability to "back up" in their input tape, it is possible for a Turing machine to run for a long time in a way that is not possible with the other computation models previously described. It is possible to construct a Turing machine that will never finish running (halt) on some inputs. We say that a Turing machine can decide a language if it eventually will halt on all inputs and give an answer. A language that can be so decided is called a recursive language. We can further describe Turing machines that will eventually halt and give an answer for any input in a language, but which may run forever for input strings which are not in the language. Such Turing machines could tell us that a given string is in the language, but we may never be sure based on its behavior that a given string is not in a language, since it may run forever in such a case. A language which is accepted by such a Turing machine is called a recursively enumerable language.
The Turing machine, it turns out, is an exceedingly powerful model of automata. Attempts to amend the definition of a Turing machine to produce a more powerful machine are surprisingly met with failure. For example, adding an extra tape to the Turing machine, giving it a 2-dimensional (or 3 or any-dimensional) infinite surface to work with can all be simulated by a Turing machine with the basic 1-dimensional tape. These models are thus not more powerful. In fact, a consequence of the Church-Turing thesis is that there is no reasonable model of computation which can decide languages that cannot be decided by a Turing machine.
The question to ask then is: do there exist languages which are recursively enumerable, but not recursive? And, furthermore, are there languages which are not even recursively enumerable?
The halting problem
The halting problem is one of the most famous problems in computer science, because it has profound implications on the theory of computability and on how we use computers in everyday practice. The problem can be phrased:
: "Given a description of a Turing machine and its initial input, determine whether the program, when executed on this input, ever halts (completes). The alternative is that it runs forever without halting."
Here we are asking not a simple question about a prime number or a palindrome, but we are instead turning the tables and asking a Turing machine to answer a question about another Turing machine. It can be shown (See main article: Halting problem) that it is not possible to construct a Turing machine that can answer this question in all cases.
That is, the only general way to know for sure if a given program will halt on a particular input in all cases is simply to run it and see if it halts. If it does halt, then you know it halts. If it doesn't halt, however, you may never know if it will eventually halt. The language consisting of all Turing machine descriptions paired with all possible input streams on which those Turing machines will eventually halt, is not recursive. The halting problem is therefore called non-computable or undecidable.
An extension of the halting problem is called Rice's Theorem, which states that it is undecidable (in general) whether a given language possesses any specific nontrivial property.
Beyond recursive languages
The halting problem is easy to solve, however, if we allow that the Turing machine that decides it may run forever when given input which is a representation of a Turing machine that does not itself halt. The halting language is therefore recursively enumerable. It is possible to construct languages which are not even recursively enumerable, however.
A simple example of such a language is the complement of the halting language; that is the language consisting of all Turing machines paired with input strings where the Turing machines do "not" halt on their input. To see that this language is not recursively enumerable, imagine that we construct a Turing machine $M$ which is able to give a definite answer for all such Turing machines, but that it may run forever on any Turing machine that does eventually halt. We can then construct another Turing machine $M\text{'}$ that simulates the operation of this machine, along with simulating directly the execution of the machine given in the input as well, by interleaving the execution of the two programs. Since the direct simulation will eventually halt if the program it is simulating halts, and since by assumption the simulation of $M$ will eventually halt if the input program would never halt, we know that $M\text{'}$ will eventually have one of its parallel versions halt. $M\text{'}$ is thus a decider for the halting problem. We have previously shown, however, that the halting problem is undecidable. We have a contradiction, and we have thus shown that our assumption that $M$ exists is incorrect. The complement of the halting language is therefore not recursively enumerable.
Concurrency-based models
A number of computational models based on concurrency have been developed, including the Parallel Random Access Machine and the Petri net. These models of concurrent computation still do not implement any mathematical functions that cannot be implemented by Turing machines.
Unreasonable models of computation
The Church-Turing thesis conjectures that there is no reasonable model of computing that can compute more mathematical functions than a Turing machine. In this section we will explore some of the "unreasonable" ideas for computational models which violate this conjecture. Computer scientists have imagined many varieties of hypercomputers.
Infinite execution
Imagine a machine where each step of the computation requires half the time of the previous step. If we normalize to 1 time unit the amount of time required for the first step, the execution would require
:$1 + \left\{1 over 2\right\} + \left\{1 over 4\right\} + cdots$
time to run. This infinite series converges to 2 time units, which means that this Turing machine can run an infinite execution in 2 time units. This machine is capable of deciding the halting problem by directly simulating the execution of the machine in question. By extension, any convergent series would work. Assuming that the series converges to a value $n$, the Turing machine would complete an infinite execution in $n$ time units.
Oracle machines
So-called Oracle machines have access to various "oracles" which provide the solution to specific undecidable problems. For example, the Turing machine may have a "halting oracle" which answers immediately whether a given Turing machine will ever halt on a given input. These machines are a central topic of study in recursion theory.
Limits of hyper-computation
Even these machines, which seemingly represent the limit of automata that we could imagine, run into their own limitations. While each of them can solve the halting problem for a Turing machine, they cannot solve their own version of the halting problem. For example, an Oracle machine cannot answer the question of whether a given Oracle machine will ever halt.
History of computability theory
The lambda calculus, an important precursor to formal computability theory, was developed by Alonzo Church and Stephen Cole Kleene. Alan Turing is most often considered the father of modern computer science, and laid many of the important foundations of computability and complexity theory, including the first description of the Turing machine (in [http://web.comlab.ox.ac.uk/oucl/research/areas/ieg/e-library/sources/tp2-ie.pdf] , 1936) as well as many of the important early results.
ee also
*Automata theory
*Abstract machine
*List of undecidable problems
*Computational complexity theory
*Computability logic
*Important publications in computability
References
* Part Two: Computability Theory, chapters 3–6, pp.123–222.
* Chapter 3: Computability, pp.57–70.
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2021-02-27 04:15:02
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https://zbmath.org/?q=an%3A1243.54062
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# zbMATH — the first resource for mathematics
On fixed points of quasi-contraction type multifunctions. (English) Zbl 1243.54062
Summary: D. Ilić and V. Rakočević [Appl. Math. Lett. 22, No. 5, 728–731 (2009; Zbl 1179.54060)] proved that quasi-contraction maps on normal cone metric spaces have a unique fixed point. Then, Kadelburg, Radenović and Rakočević generalized their results by considering an additional assumption [Z. Kadelburg, S. Radenović and V. Rakočević, Appl. Math. Lett. 22, No. 11, 1674–1679 (2009; Zbl 1180.54056)]. Also, they proved that quasi-contraction maps on cone metric spaces have the property (P) whenever $$\lambda\in(0,\tfrac 12)$$. Later, Haghi, Rezapour and Shahzad proved same results without the additional assumption and for $$\lambda\in(0,1)$$ by providing a new technical proof [Sh. Rezapour, R.H. Haghi and N. Shahzad, Appl. Math. Lett. 23, No. 4, 498–502 (2010; Zbl 1206.54061)]. In 2011, D. Wardowski published a paper [Appl. Math. Lett. 24, No. 3, 275–278 (2011; Zbl 1206.54067)] and tried to test fixed point results for multifunctions on normal cone metric spaces. Of course, he used a special view in his results. Recently, A. Amini-Harandi [Appl. Math. Lett. 24, No. 11, 1791–1794 (2011; Zbl 1230.54034)] proved a result on the existence of fixed points of set-valued quasi-contraction maps in metric spaces by using the technique of Rezapour et al. [Zbl 1206.54061]. But, like Kadelburg et al. [Zbl 1180.54056], he could prove it only for $$\lambda\in(0,\tfrac 12)$$.
In this work, we prove again the main result of Amini-Harandi [loc. cit.] by using a simple method. Also, we introduce quasi-contraction type multifunctions and show that the main result of Amini-Harandi holds for quasi-contraction type multifunctions.
##### MSC:
54H25 Fixed-point and coincidence theorems (topological aspects) 47H10 Fixed-point theorems 54C60 Set-valued maps in general topology
Full Text:
##### References:
[1] Ćirić, Lj., Fixed point theory: contraction mapping principle, (2003), FME Press Beograd [2] Pathak, H.K.; Shahzad, N., Fixed point results for generalized quasicontraction mappings in abstract metric spaces, Nonlinear anal., 71, 6068-6076, (2009) · Zbl 1189.54036 [3] Amini-Harandi, A., Fixed point theory for set-valued quasi-contraction maps in metric spaces, Appl. math. lett., 24, 1791-1794, (2011) · Zbl 1230.54034 [4] Rezapour, Sh.; Haghi, R.H.; Shahzad, N., Some notes on fixed points of quasi-contraction maps, Appl. math. lett., 23, 498-502, (2010) · Zbl 1206.54061 [5] Haghi, R.H.; Rezapour, Sh.; Shahzad, N., Some fixed point generalizations are not real generalizations, Nonlinear anal., 74, 1799-1803, (2011) · Zbl 1251.54045 [6] Ilić, D.; Rakoc˘ević, V., Quasi-contraction on a cone metric space, Appl. math. lett., 22, 728-731, (2009) · Zbl 1179.54060 [7] Kadelburg, Z.; Radenović, S.; Rakoc˘ević, V., Remarks on “quasi-contraction on a cone metric space”, Appl. math. lett., 22, 1674-1679, (2009) · Zbl 1180.54056 [8] Wardowski, D., On set-valued contractions of Nadler type in cone metric spaces, Appl. math. lett., 24, 275-278, (2011) · Zbl 1206.54067
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.
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2021-10-25 06:46:36
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https://www.math.gatech.edu/node/16758
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## Orthogonal polynomials for the Minkowski question Mark function
Series:
Analysis Seminar
Wednesday, December 2, 2015 - 14:05
1 hour (actually 50 minutes)
Location:
Skiles 006
,
University of Leuven, Belgium
Organizer:
The Minkowski question mark function is a singular distribution function arising from Number Theory: it maps all quadratic irrationals to rational numbers and rational numbers to dyadic numbers. It generates a singular measure on [0,1]. We are interested in the behavior of the norms and recurrence coefficients of the orthonormal polynomials for this singular measure. Is the Minkowski measure a "regular" measure (in the sense of Ullman, Totik and Stahl), i.e., is the asymptotic zero distribution the equilibrium measure on [0,1] and do the n-th roots of the norm converge to the capacity (which is 1/4)? Do the recurrence coefficients converge (are the orthogonal polynomials in Nevai's class). We provide some numerical results which give some indication but which are not conclusive.
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2018-05-25 22:56:52
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https://www.physicsforums.com/threads/can-i-factor-it.80926/
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# Homework Help: Can I factor it?
1. Jul 2, 2005
### Beez
On my process of obtaining the IF for solving a differential equation, I got stuck with an equation as following. This could be a very simple algebra problem, but I just can't do it. Would someone tell me if I can factor the numerators by (x^4 + 1) for the first equation and by (x^4 -1) for the second equation so that I can have only x or y as a variable?
[(x^4 - 1) - (5*x^4 + 1)]/x*(x^4+1) = (-4*x^4 - 2)/x(x^4 + 1)
= -2(2*x^4 +1)/x(x^4 + 1)
[(5*x^4 + 1) - (x^4 - 1)]/y*(x^4-1) = (4x^4 + 2)/y(x^4 - 1)
=2(2*x^4 + 1)/y(x^4 - 1)
2. Jul 2, 2005
could you use $$\LaTeX$$ please. it is kinda hard to read without it.
3. Jul 2, 2005
### Beez
OK, I got it. Please take look at the following.
[($$x^4$$ - 1) - (5$$x^4$$+ 1)]/x($$x^4$$+1) = (-4$$x^4$$ - 2)/x($$x^4$$ + 1)
= -2(2$$x^4$$ +1)/x($$x^4$$+ 1)
[(5$$x^4$$ + 1) - ($$x^4$$ - 1)]/y($$x^4$$-1) = (4$$x^4$$ + 2)/y($$x^4$$- 1)
=2(2$$x^4$$+ 1)/y($$x^4$$ - 1)
Thanks for your suggestion. I always wanted to know how to type $$x^2$$ for x^2.
Last edited: Jul 2, 2005
4. Jul 2, 2005
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2018-10-18 11:51:07
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https://www.allanswered.com/post/nxpbj/how-to-measure-the-diffusion-constant-for-cells-in-aggregate-in-a-cc3d-simulation/
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### How to measure the diffusion constant for cells in aggregate in a CC3D simulation?
23
views
0
6 days ago by
Hi,
In the paper "Multi-Scale Modeling of Tissues Using CompuCell3D" by Swat et al., on page 5, the author(s) mentioned that one can measure the diffusion constant of cells in aggregates in the simulation. So I am asking for suggestions on how to do so in CC3D.
Thank you,
Community: CompuCell3D
2
5 days ago by
With Priyom's help, I've found some much better resources for this.
See slides 51-59 in;
http://www.compucell3d.org/BinDoc/cc3d_binaries/Tutorials/PASI2013/MultiScaleModelingWithCC3D-SBML.pdf
The diffusion measurement demo can be downloaded from;
http://www.compucell3d.org/BinDoc/cc3d_binaries/Tutorials/PASI2013/DiffusionCalculator.zip
This is an excellent example of a python only CC3D model. The code is pretty old though and to use it you need to add a couple lines to the file "DiffusionCalculator.py". After line #64 add;
contact.ElementCC3D("Energy", {"Type1":"Medium", "Type2": "Wall"},10)
contact.ElementCC3D("Energy", {"Type1":"Cells", "Type2": "Wall"},10)
contact.ElementCC3D("Energy", {"Type1":"Wall", "Type2": "Wall"},10)
Note that the plot in CC3D is only drawn at the end of the simulation. Also, a log of all the displacements is written to a text file.
0
5 days ago by
I believe you can use the equation;
(deltaX)^2/(deltaT) = 2D
Where deltaX is the displacement of a single cell along the x-axis in time period deltaT and D is the diffusion constant
One way to track the cell's position is to add a dictionary to the cell and save the cell's X, Y and Z position every (some number of) mcs. Every (some number of) mcs you calculate the X (or Y or Z) displacement, square that and divide by the (some number of) mcs. Since the simulation is stochastic the D will fluctuate from interval to interval so you'll need to average over a number of intervals.The units of D will be pixel^2/mcs, which is distance^2/unit time.
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2018-06-19 14:45:40
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https://math.stackexchange.com/questions/363958/does-the-polynomial-px-have-integer-zeros
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# Does the polynomial $P(x)$ have integer zeros?
The following is a homework question:
Let $P(x)$ be a polynomial with integer coefficients and $P(x_1)=P(x_2)=P(x_3)=P(x_4)=P(x_5)=P(x_6)=P(x_7)=7$ where $x_i$ are distinct integers. Determine if $P(x)$ has integer zeros.
I've never done questions like this before. I started with this:
If $\deg(P) = 7$,
$$P(x)=\alpha(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)(x-x_6)(x-x_7)+7$$
where $\alpha$ is some integer.
However, the question doesn't state that the polynomial must be of the seventh degree. Even then, I don't see how I can determine if $P(x)$ has integer zeros without knowing all the $x_i$.
Edit: Is this a valid solution? $$P(x)=Q(x)(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)(x-x_6)(x-x_7)+7$$
If $P(n)=0$, $$Q(n)(n-x_1)(n-x_2)(n-x_3)(n-x_4)(n-x_5)(n-x_6)(n-x_7)=-7$$
where $Q(x)$ is a polynomial of integer coefficients (therefore $Q(n)$ is an integer)
And since all the terms on the LHS are integers and the $x_i$ are distinct integers, it follows that some of the factors on the LHS $\ne$ {$\pm 1, \pm 7$}. And $7$ is a prime number, therefore $n$ cannot be a zero.
• I think you need to add that $P$ is monic, i.e. that the lead coefficient is $1$. Otherwise if $Q(x)=(x-x_1)\dots(x-x_7)+7=R(x)+7$, the poly $((nx-1)R(x)+1)Q(x)$ has $\frac1n$ as a root, and evaluates as $7$ at the $x_i$. – Tom Oldfield Apr 17 '13 at 2:23
• @TomOldfield Why is $1/n$ a root? If I read correctly, it gives $Q(1/n)\neq 0$. – Julien Apr 17 '13 at 2:38
• Does $P(x) = 7$ count as a polynomial? – user17762 Apr 17 '13 at 2:39
• $7$ is a prime number. – achille hui Apr 17 '13 at 2:41
• What you should prove is that if $x$ is a root of $P$, then it can't be an integer. So take such an $x$. Evaluate. And consider achille hui's comment again. – Julien Apr 17 '13 at 2:50
$P(x)= Q(x) (x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)(x-x_6)(x-x_7)+7$
Idea : You can't get $-7$ by multiplying $7$ distinct integers..
• We'll save the argument for the day when $7$ will be replaced by some badly composed integer, +1. – Julien Apr 17 '13 at 2:56
• @Angie Nicely done. Just add $-7$ to the list of values $\{\pm 1,\pm 7\}$. – Julien Apr 17 '13 at 3:05
• @Angie, you need to mention $Q(x)$ is a polynomial of integer coefficients. – achille hui Apr 17 '13 at 3:08
• I think you're using the "Rational Root theorem". The term $Q(x)(x-x_1)\cdots(x-x_7)$ has a constant term which adds to 7, so the possible roots may be other than 1, -1, 7 and -7... – Luiz Cordeiro Apr 17 '13 at 3:14
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2019-12-14 22:11:43
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http://drorbn.net/index.php?title=0708-1300/Homework_Assignment_1&diff=prev&oldid=5407
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Difference between revisions of "0708-1300/Homework Assignment 1"
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Announcements go here
Reading
Read sections 1-5 of chapter II of Bredon's book three times:
• First time as if you were reading a novel - quickly and without too much attention to detail, just to learn what the main keywords and concepts and goals are.
• Second time like you were studying for an exam on the subject - slowly and not skipping anything, verifying every little detail.
• And then a third time, again at a quicker pace, to remind yourself of the bigger picture all those little details are there to paint.
Doing
• Solve and submit the following two problems:
1. Show explicitly that the restricted implicit function theorem, with $x_0=y_0=0$ and $\partial_yg=I$, is equivalent to general implicit function theorem, in which $x_0$ and $y_0$ are arbitrary and $\partial_yg$ is an arbitrary invertible matrix.
2. Show that the definition $f\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x\\g(x,y)\end{pmatrix}$ reduces the implicit function theorem to the inverse function theorem. A key fact to verify is that differential of $f$ at the relevant point is invertible.
• Solve the following problems from Bredon's book, but submit only the solutions of the starred problems:
problems on page(s)
*1, 2, 3, *4, 5 71
1, *2 75-76
1-4 80
Due Date
This assignment is due in class on Thursday October 4, 2007.
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2021-04-19 08:33:27
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