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# Simple Explanation of Baum Welch/Viterbi

I'm looking for a very simple explanation as possible for Baum Welch and Viterbi for HMMs with a straightforward well annotated example. Almost all of the explanations I find on the net invariably jump from the beginning into an almost completely jargon/symbol based explanation. The few times they have an example it is usually shoved to the side or poorly annotated (ie its unclear how the symbols and the example are related to each other). I've heard stuff like some sources claiming Baum Welch is based on the EM algorithm and others saying its based off of the forward backward algorithm and some say both so I'd like to get details like that cleared up.

The closest I found so far is a 1hr video posted here that people said was great but is unplayable because flash is banned and an explanation on wikipedia that has a note next to it saying its wrong.

I'm looking for a very simple explanation as possible for Baum Welch and Viterbi for HMMs...

• Given the observed data, the Baum-Welch algorithm finds the likelihood-maximizing parameters.

• Given the observed data, and the parameters, the Viterbi algorithm finds the most likely sequence of hidden states.

I've heard stuff like some sources claiming Baum Welch is based on the EM algorithm...

The Baum-Welch algorithm is a special case of the EM algorithm.

...and others saying its based off of the forward backward algorithm

The forward-backward decomposition is how you do the "E-" step (using the EM terminology). It's specific to hidden Markov models.

The few times they have an example it is usually shoved to the side or poorly annotated (ie its unclear how the symbols and the example are related to each other)

I don't know what examples you're looking at, but I can tell you that there are many good ones out there. Remember that these methods were invented a very long time ago, and they're still widely used. The Rabiner one linked to above is probably the most famous, but you have to find one you like that you'll keep working through. I would recommend you find one with an example that has data you're interested in, and notation that you feel is learn-able. Whatever you find will have a lot of notation. There is no way around that, so you've just got to find something you can settle into.

I have the perfect paper for you: http://www.stat.columbia.edu/~liam/teaching/neurostat-fall17/papers/hmm/rabiner.pdf

Note afaik there are newer BW methods.

Attempted simple explanation for Viterbi: The goal of viterbi is to find an optimal sequence of states during some discrete period of time.
Start off with the initial probabilities of being in a state $s$ at time $t=0$. Then increase $t$ and for each state at that time ($t=1$), find the best state $s$ at $t-1$ (best as in the sequence which gives the highest probability). Save this information. That means for each state $s$ at $t=1$ save what the best state $s$ at $t=0$ is. Also save the probability for that (optimal) sequence.
Then increase $t$ to $t=2$, and repeat the procedure (with the probabilities for each state $s$ at $t-1$ being the just calculated best probabilities for that state). At some point you'll reach $t=T$ (last point in time). Then you just pick the state with the highest probability, and remember for each state at each time you have saved the best state for $t-1$, so take the best state at $t=T$ and then work your way backwards always picking that saved optimal state.

To elaborate on what it does on a more general level: If you have a set of states $S$ which can all be picked over some duration of time $T$, a naive solution to find the best sequence would be to enumerate all possible sequences and then calculate the probability for each of them (that means calculating the product of the state and transition probabilities). The mistake one is doing here is that a lot of the calculations are duplicated, as an example you could have two sequences $p_1, p_2$ ($p$ is a sequence of states) where the only difference is the final state, yet with this approach you are calculating the probabilities of the shared states for both $P(p_1)$ and $P(p_2)$.
By taking the iterative approach used in viterbi, there are no duplicated calculations, so obviously it will be a lot more efficient.


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I am trying to find the saddle point of a quadratic form: $$f(\mathbf{x})=\mathbf{x}^\mathrm{T} \mathbf{A}\mathbf{x}+\mathbf{x}^T\mathbf{b}+\mathbf{c}$$ using a minimization/maximization-like algorithm. Is there such solution method available somewhere? I want to avoid the $\nabla\cdot f(\mathbf{x})=\mathbf{0}$ approach because I later want to add inequality constraints. The point that I am seeking is a maximum in one direction and a minimum in all other directions.

edit The above problem comes from the following formulation: imagine an externally forced one degree-of-freedom oscillator whose displacement $u(t)$ satisfies the equation of motion: $$m\ddot{u}+ku=f \cos \omega t$$ where $m$ and $k$ are respectively the mass and stiffness of the oscillator; $f$ and $\omega$ defines the forcing term. I would like to use a modified version of Hamilton's principle to find the forced steady state solution (the transient part is ignored). Since I am interested in periodic orbits only, the sought (periodic) solution satisfies: $$\delta\bigl(\int_0^T \frac{1}{2} m \dot{u}^2- \frac{1}{2} k u^2+fu\cos \omega t \,\mathrm{d}t\bigr)=0\quad;\quad u(0)=u(T)\quad;\quad \dot{u}(0)=\dot{u}(T)\qquad(1)$$ where $T=2\pi/\omega$. As the Hamilton's principle says, the solution is an extremum of the underlying functional, not a minimum necessarily.

I now want to find an approximation of $u(t)$, as for instance: $$u(t)=a_1+a_2\cos\omega t + a_3\sin \omega t\qquad (2)$$ where $a_1$, $a_2$, and $a_3$ have to be found. I just have to plug (2) in (1) and find the point $(a_1,a_2,a_3)$ that makes the functional stationary. For reasons that are not given here, I would like to reformulate this in a $\max\min$ problem since I know for this specific case that the solution is a saddle point, ie a maximum along $a_1$ and a minimum along $(a_2,a_3)$. The quadratic form given at the top of the message comes from plugging (2) in (1) with $\mathbf{x}=(a_1,a_2,a_3)$.

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How would you define the saddle point of an inequality-constrained problem? –  Rahul Aug 19 '12 at 18:18
Let us note $\mathbf{x}=\{x_0,x_1,\dots,x_n\}$ then the problem would be: find $\max_{x_0}\min_{x_1,\dots,x_n}\; f(\mathbf{x})$ under linear inequality constraints $\mathbf{G}\mathbf{x}-\mathbf{q}\leq 0$. Not sure if it makes sense but it would be something like that. –  pluton Aug 19 '12 at 19:04
Could you look for local extrema where $\nabla f = 0$ and then also consider the boundary of the constraint region? Compare and choose the point you desire? At a local extreme you can determine the nature of the critical point $p$ ( $\nabla f (p)=0$) by shifting variables at the point to write $f(y) = f(p)+y^TMy$. The eigenvalues of $M$ indicate if the point is min/max or saddle or trough. –  James S. Cook Aug 19 '12 at 19:23
It may be doable when $n$ is small but I'd say that it would quickly becomes intractable as $n$ increases. Knowing that the solution of the unconstrained problem is a saddle is no the difficulty. The difficulty is about how to formulate the problem with linear inequalities within a $\max\min$ strategy. The first question is: is the formulation correct? –  pluton Aug 19 '12 at 19:32

The problem you specify, $$\min f(x)=x^TAx+x^Tb$$ subject to $$Gx-q\leq 0$$ is a quadraric programming problem. See the wiki page for solution procedures.
Thanks for the hint on the "conjugate residual method": it looks interesting. The problem I was specifying was not a $\min f$, in the sense of finding a minimum of $f$, but rather a $\min\max f$ (not sure about the relevance of the notation) in the sense of finding saddle points. –  pluton Aug 21 '12 at 13:46

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# LibraryReExportIsNPComplete

This page describes a way to convert any 3SAT problem to a solution of finding the right configuration from conflicting libraries in a system that can re-export APIs. Thus proving that the later problem is wikipedia::NP-complete.

## 3SAT

The problem of satisfying a logic formula remains NP-complete even if all expressions are written in wikipedia::conjunctive normal form with 3 variables per clause (3-CNF), yielding the 3SAT problem. This means the expression has the form:

$(x_{11} \vee x_{12} \vee x_{13}) \wedge$
$(x_{21} \vee x_{22} \vee x_{23}) \wedge$
$(x_{31} \vee x_{32} \vee x_{33}) \wedge$
...
$(x_{n1} \vee x_{n2} \vee x_{n3})$

where each xab is a variable vi or a negation of a variable $\neg v_i$. Each variable vi can appear multiple times in the expression.

## Module Dependencies Problem

Let A,B,C,... denote an API.

Let A1,A1.1,A1.7,A1.11 denote compatible versions of API A.

Let A1,A2.0,A3.1 denote incompatible versions of API A.

Let Ax.y > Bu.v denote the fact that version x.y of API A depends on version u.v of API B.

Let $A_{x.y} \gg B_{u.v}$ denote the fact that version x.y of API A depends on version u.v of API B and that it re-exports B's elements.

Let Repository R = (M,D) be any set of modules with their various versions and their dependencies on other modules with or without re-export.

Let C be a Configuration in a repository R = (M,D), if $C \subseteq M$, where following is satisfied:

$\forall A_{x.y} \in C, \forall A_{x.y} \gg B_{u.v} \in D \Rightarrow \exists w >= v \wedge B_{u.w} \in C$ - each re-exported dependency is satisfied with some compatible version
$\forall A_{x.y} \in C, \forall A_{x.y} > B_{u.v} \in D \Rightarrow \exists w >= v B_{u.w} \in C$ - each dependency is satisfied with some compatible version
Let there be two chains of re-exported dependencies $A_{p.q} \gg ... \gg B_{x.y}$ and $A_{p.q} \gg ... \gg B_{u.v}$ then $x = u \wedge y = v$ - this guarantees that each class has just one, exact meaning for each importer

#### Module Dependency Problem

Let there be a repository R = (M,D) and a module $A \in M$. Does there exist a configuration C in the repository R, such that the module $A \in C$, e.g. the module can be enabled?

## Converstion of 3SAT to Module Dependencies Problem

Let there be 3SAT formula with with variables v1,...,vm as defined above.

Let's create a repository of modules R. For each variable vi let's create two modules $M^i_{1.0}$ and $M^i_{2.0}$, which are mutually incompatible and put them into repository R.

For each formula $(x_{i1} \vee x_{i2} \vee x_{i3})$ let's create a module Fi that will have three compatible versions. Each of them will depend on one variable's module. In case the variable is used with negation, it will depend on version 2.0, otherwise on version 1.0. So for the formula

$v_a \vee \neg v_b \vee \neg v_c$

we will get:

$F^i_{1.1} \gg M^a_{1.0}$
$F^i_{1.2} \gg M^b_{2.0}$
$F^i_{1.3} \gg M^c_{2.0}$

All these modules and dependencies add into repository R

Now we will create a module T1.0 that depends all formulas:

$T_{1.0} \gg F^1_{1.0}$
$T_{1.0} \gg F^2_{1.0}$
...
$T_{1.0} \gg F^n_{1.0}$

and add this module as well as its dependencies into repository R.

Claim: There $\exists C$ (a configuration) of repository R and $T_{1.0} \in C$ $\Longleftrightarrow$ there is a solution to the 3SAT formula.

## Proof

"$\Leftarrow$": Let's have an evaluation of each variable to either true or false that evaluates the whole 3SAT formula to true. Then

$C = \{ T_{1.0} \} \bigcup$
$\{ M^i_{1.0} : v_i \} \bigcup \{M^i_{2.0} : \neg v_i \} \bigcup$
$\{ F^i_{1.1} : x_{i1} \} \bigcup \{ F^i_{1.2} : \neg x_{i1} \wedge x_{i2} \} \bigcup \{ F^i_{1.3} : \neg x_{i1} \wedge \neg x_{i2} \wedge x_{i3} \}$

It is clear from the definition that each Mi and Fi can be in the C just in one version. Now it is important to ensure that each module is present always at least in one version. This is easy for Mi as its vi needs to be true or false, and that means one of $M^i_{1.0}$ or $M^i_{2.0}$ will be included. Can there be a Fi which is not included? Only if $\neg x_{i1} \wedge \neg x_{i2} \wedge \neg x_{i3}$ but that would mean the whole 3-or would evaluate to false and as a result also the 3SAT formula would evaluate to false. This means that dependencies of T1.0 on Fi modules are satisfied. Are also dependencies of every $F^i_{1.q}$ satisfied? From all the three versions, there is just one $F^i_{1.q}$, the one its xiq evaluates to true. However xiq can either be without negation, and as such $F^i_{1.q}$ depends on $M^j_{1.0}$ which is included as vj is true. Or xiq contains negation, and as such $F^i_{1.q}$ depends on $M^j_{2.0}$ which is included as vj is false. qed.


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# Mathematics Colloquia and Seminars

Return to Colloquia & Seminar listing

### Computing the continuous discretely: the quest for the volume of the Birkhoff polytope

Algebra & Discrete Mathematics

 Speaker: Dr. Matthias Beck, SUNY Binghamton Location: 693 Kerr Start time: Thu, Nov 21 2002, 12:00PM

Abstract: The \emph{$n^{\text{th}}$ Birkhoff polytope} $B_n$ is the set of all \emph{doubly stochastic} $n \times n$ matrices, that is, those matrices with nonnegative real entries in which every row and column sums to one. A famous open problem concerns the volume of $B_n$, which is only known up to $n=9$. The \emph{Ehrhart polynomial} of a polytope with integral vertices counts the integer points in the polytope as it gets dilated by an integer factor. One reason for being interested in this counting function is that its leading term gives the volume of the polytope. We will present a novel way of computing Ehrhart polynomials for polytopes using complex-analytic methods. En route to their application to the Birkhoff polytope, we will introduce ideas and concepts from discrete geometry, number theory, and combinatorics. The application of our methods to the Birkhoff polytope provides an example showing that pure mathematics and computationally efficient algorithms are not mutually exclusive.


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• ## SyncTeX is here! Scroll, click, and hey presto!

Posted by John on April 7, 2014Spotted a typo in your article? Need to make quick changes to text, figures or equations? Now you can simply click on the pdf preview and we'll take you straight to that line in the source text so you can edit it right away. Jump straight to the place you want to edit with just one click With the addition of SyncTeX to writeLaTeX, gone are the days when you need to hunt through source files to find that particular figure caption or equation - now you can just click on the pdf preview and the editor will jump to that line in the source code thanks to an automatic sync between text position in source and pdf. Here's a quick example to show how it works... John has just started working on his PhD thesis, and has sat down to write some content for chapter one: Wow, writing a thesis is hard! John scrolls through the template seeking inspiration and decides to postpone writing chapter one at the moment, to give the acknowledgements section a go instead. But where is the acknowledgements section in the source code? Now it's easy to find - John simply has to click on the acknowledgements section in the pdf preview (notice the the cursor has changed to a crosshair) and SyncTeX is activated: The editor opens the correct file and highlights the corresponding line in the source code. John can now start editing the acknowledgements section, and leave chapter one until later! That's all there is to it! If you want to go the other way (jump from source to pdf), you can do that too - there's a new option in the find menu in the editor: We hope you find this new feature useful, and if you have any questions please let us know.


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# GMAT 考满分题库

At ABC Hardware the total cost of 1 paintbrush and 2 rolls of tape is $6.25. What is the price of one roll of tape? (1) The price of one paintbrush is three times the price of one roll of tape. (2) The total cost of 6 rolls of tape and 3 paintbrushes is$18.75.
• AStatement 1 ALONE is sufficient to answer the question, but statement 2 alone is NOT sufficient.
• BStatement 2 ALONE is sufficient to answer the question, but statement 1 alone is NOT sufficient.
• CBOTH statements 1 and 2 TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
• DEach statement ALONE is sufficient to answer the question.
• EStatement 1 and 2 TOGETHER are NOT sufficient to answer the question.

|

• 按热度
• 按顺序


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# align¶

Type of operation: Aggregate, Metrics

## Description¶

Align raw metrics onto a time grid (defined by the input period) by aggregating nearby data points together.

If frame is not specified, align aggregates all points between two grid points (tumbling window aggregation). If frame is specified, align aggregates all data points that fall into a frame relative to each grid point (sliding window aggregation).

Align is accelerable when the period argument is specified.

## Usage¶

align [ period ] [ , frame ], metricAligningExpression ...

Argument

Type

Required

Multiple

period

duration

Optional

Only one

frame

frame

Optional

Only one

metricAligningExpression

expression

Required

Can be multiple

## Accelerable¶

align is sometimes accelerable, depending on options used. A dataset that only uses accelerable verbs can be accelerated, making queries on the dataset respond faster.

## Examples¶

align 5m, requests: rate(m("requests_total"))

Computes the per-second rate of “requests_total” metric for each 5m time bins.

align 5m, requests: rate(m("requests_total")), memory_used: avg(m("memory_used"))

Computes the per-second rate of “requests_total” metric and the average of “memory_used” metric, for each 5m time bins.

align 1m, frame(back: 10m), memory_used: avg(m("memory_used"))

Computes the moving average of “memory_used” metric in the past 10m, for each 1m step. Note that frame() may not be exact for better performance.

align 1m, frame_exact(back: 118s), memory_used: avg(m("memory_used"))

Computes the moving average of “memory_used” metric in the past exact 118s, for each 1m step.

align frame(back: 5m), requests: rate(m("requests_total"))

Computes the per-second rate of “requests_total” metric in the past 5 minutes, for each step. The step size will be determined dynamically to make the chart human-readable. It’s determined based on the query window and a number of other parameters such as chart resolution. This formulation is not accelerable.


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## Elementary Statistics (12th Edition)

Published by Pearson

# Chapter 3 - Statistics for Describing, Exploring, and Comparing Data - 3-4 Measures of Relative Standing and Boxplots - Basic Skills and Concepts: 13

#### Answer

De-Fen Yao is taller relative to the population of the same gender.

#### Work Step by Step

The z-scores: Sultan Kosen: $z=\frac{value-mean}{standard \ deviation}=\frac{247-175}{7}=10.29$. De-Fen Yao: $z=\frac{value-mean}{standard \ deviation}=\frac{236-162}{6}=12.33$. De-Fen Yao's z-score is higher, therefore she is taller relative to the population of the same gender.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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# A new extended B-spline approximation technique for second order singular boundary value problems arising in physiology

Volume 19, Issue 4, pp 258--267 Publication Date: July 10, 2019       Article History
• 126 Views

### Authors

Imtiaz Wasim - Department of Mathematics, University of Sargodha, Sargodha, 40100, Pakistan. Muhammad Abbas - Department of Mathematics, University of Sargodha, Sargodha, 40100, Pakistan. Muhammad Kashif Iqbal - Department of Mathematics, , Government College University, Faisalabad, 38000, Pakistan.

### Abstract

In this study, we have explored the approximate solution of $2^{nd}$ order singular boundary value problems (SBVP's) using extended cubic B-spline (ECBS) collocation approach. The accuracy of the numerical algorithm has been enhanced by means of a novel ECBS approximation for $2^{nd}$ order derivative. To endorse our claim, few test examples have been considered and the experimental results are compared with the already existing methods. It is observed that the proposed technique is more accurate and efficient in comparison to the existing techniques on the topic.

### Keywords

• Singular boundary value problems
• extended B-spline functions
• quasi-linearization technique
• extended B-spline collocation method

•  34B15
•  34B16
•  74H15
•  65L10
•  65L11

### References

• [1] M. Abbas, A. A. Majid, A. I. M. Ismail, A. Rashid, The application of cubic trigonometric B-spline to the numerical solution of the hyperbolic problems, Appl. Math. Comput., 239 (2014), 74--88

• [2] M. Abukhaled, S. A. Khuri, A. Sayfy, A numerical approach for solving a class of singular boundary value problems arising in physiology, Int. J. Numer. Anal. Model., 8 (2011), 353--363

• [3] E. Babolian, A. Iftikhari, A. Saadarmandi, A Sinc-Galerkin technique for the numerical solution of a class of singular boundary value problems, Comput. Appl. Math., 34 (2015), 45--63

• [4] N. Caglar, H. Caglar, B-spline solution of singular boundary value problems, Appl. Math. Comput., 182 (2006), 1509--1513

• [5] Z. D. Cen, Numerical study for a class of singular two-point boundary value problems using Green’s functions, Appl. Math. Comput., 183 (2006), 10--16

• [6] M. M. Chawala, R. Subramanian, H. L. Sathi, A fourth order method for a singular two-point boundary value problem, BIT Numer. Math., 28 (1988), 88--97

• [7] U. Flesch, The distribution of heat sources in the human head: a theoretical consideration, J. Theor. Biol., 54 (1975), 285--287

• [8] D. J. Fyfe, The use of cubic splines in the solution of two-point boundary value problems, Comput. J., 12 (1969/70), 188--192

• [9] J. B. Garner, R. Shivaji, Diffusion problems with a mixed nonlinear boundary condition, J. Math. Anal. Appl., 148 (1990), 422--430

• [10] J. Goh, A. A. Majid, A. I. M. Ismail, Extended cubic uniform B-spline for a class of singular boundary value problems, Sci. Asia, 37 (2011), 79--82

• [11] J. Goh, A. A. Majid, A. I. M. Ismail, A quartic B-spline for second-order singular boundary value problems, Comput. Math. Appl., 64 (2012), 115--120

• [12] M. K. Iqbal, M. Abbas, N. Khalid, New Cubic B-spline Approximation for Solving Non-linear Singular Boundary Value Problems Arising in Physiology, Comm. Math. Appl., 9 (2018), 377--392

• [13] M. K. Iqbal, M. Abbas, I. Wasim, New cubic B-spline approximation for solving third order Emden–Flower type equations, Appl. Math. Comput., 331 (2018), 319--333

• [14] M. K. Iqbal, M. Abbas, B. Zafar, New Quartic B-Spline Approximation for Numerical Solution of Third Order Singular Boundary Value Problems, Punjab. Univ. J. Math., 51 (2019), 43--59

• [15] S. A. Khuri, A. Sayfy, A novel approach for the solution of a class of singular boundary value problems arising in physiology, Math. Comp. Modelling, 52 (2010), 626--636

• [16] M. Kumar, Y. Gupta, Methods for solving singular boundary value problems using splines: a review, J. Appl. Math. Comput., 32 (2010), 265--278

• [17] F. G. Lang, X. P. Xu, A new cubic B-spline method for approximating the solution of a class of nonlinear second-order boundary value problem with two dependent variables, Sci. Asia, 40 (2014), 444--450

• [18] P. M. Lima, M. P. Carpentier, Iterative methods for a singular boundary-value problem, J. Comput. Appl. Math., 111 (1999), 173--186

• [19] S. H. Lin, Oxygen diffusion in a spherical cell with nonlinear oxygen uptake kinetics, J. Theor. Biol., 60 (1976), 449--457

• [20] D. L. S. McElwain, A re-examination of oxygen diffusion in a spherical cell with Michaelis-Menten oxygen uptake kinetics, J. Theor. Biol., 71 (1978), 255--263

• [21] M. Mohsenyzadeh, K. Maleknejad, R. Ezzati, A numerical approach for the solution of a class of singular boundary value problems arising in physiology, Adv. Difference Equ., 2015 (2015), 10 pages

• [22] R. K. Pandey, A. K. Singh, On the convergence of a finite difference method for a class of singular boundary value problems arising in physiology, J. Comput. Appl. Math., 166 (2004), 553--564

• [23] K. Parand, M. Dehghan, A. Z. Rezaei, S. M. Ghaderi, An approximation algorithm for the solution of the nonlinear Lane--Emden type equations arising in astrophysics using Hermite functions collocation method, Comput. Phys. Comm., 181 (2010), 1096--1108

• [24] J. Rashidinia, R. Mohammadi, R. Jalilian, The numerical solution of non-linear singular boundary value problems arising in physiology, Appl. Math. Comput., 185 (2007), 360--367

• [25] A. S. V. Ravi Kanth, Cubic spline polynomial for non-linear singular two-point boundary value problems, Appl. Math. Comput., 189 (2007), 2017--2022

• [26] A. S. V. Ravi Kanth, K. Aruna, He's variational iteration method for treating nonlinear singular boundary value problems, Comput. Math. Appl., 60 (2010), 821--829

• [27] R. D. Russell, L. F. Shampine, Numerical methods for singular boundary value problems, SIAM J. Numer. Anal., 12 (1975), 13--36

• [28] L. F. Shampine, Singular boundary value problems for ODEs, Appl. Math. Comput., 138 (2003), 99--112

• [29] S. Sharifi, J. Rashidinia, Numerical solution of hyperbolic telegraph equation by cubic B-spline collocation method, Appl. Math. Comput., 281 (2016), 28--38

• [30] R. Singh, J. Kumar, An efficient numerical technique for the solution of nonlinear singular boundary value problems, Comput. Phys. Commun., 185 (2014), 1282--1289

• [31] R. Singh, A.-M. Wazwaz, J. Kumar, An efficient semi-numerical technique for solving nonlinear singular boundary value problems arising in various physical models, Int. J. Comput. Math., 93 (2016), 1330--1346

• [32] W. Y. Wang, M. G. Cui, B. Han, A new method for solving a class of singular twopoint boundary value problems, Appl. Math. Comput., 206 (2008), 721--727

• [33] Y. L. Wang, H. Yu, F. G. Tan, S. G. Li, Using an effective numerical method for solving a class of Lane-Emden equations, Abstr. Appl. Anal., 2014 (2014), 8 pages

• [34] A. M. Wazwaz, The variational iteration method for solving nonlinear singular boundary value problems arising in various physical models, Commun. Nonlinear Sci. Numer. Simul., 16 (2011), 3881--3886

• [35] L.-J. Xie, C.-L. Zhou, S. Xu, An effective numerical method to solve a class of nonlinear singular boundary value problems using improved differential transform method, Springer Plus, 5 (2016), 21 pages


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When $2^\alpha = 2^\beta$ implies $\alpha=\beta$ ($\alpha,\beta$ cardinals)

Sorry if this is a silly question. I was wondering, under what axioms of set theory is it true that if $\alpha$,$\beta$ are cardinals, and $2^\alpha=2^\beta$, then $\alpha=\beta$? Do people use these conditions to prove interesting results?

This question is prompted from a recent perusing of Johnson's "Topics in the Theory of Group Presentations", where in the first few pages he "proves" free groups of different rank are non-isomorphic: the number of mappings from a free group of rank $\omega$ to the group $\mathbb{Z}/2\mathbb{Z}$ is $2^\omega$, which would be invariant under isomorphism; and then he assumes the topic of my question: $2^\alpha=2^\beta$ implies $\alpha=\beta$.

But I remember reading something a few years ago about a student of R.L. Moore (I only remember his last names was Jones) "proving" the Moore Space conjecture, and using that $\alpha > \beta$ implied $2^\alpha > 2^\beta$, but that this was incorrect.

Anyway, I realized I don't know anything about when this is true or false, so I thought I'd ask.

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I don't know the answer to the set theory question. But if you want to prove that isomorphic free groups have the same rank, you can just pass to the abelianization, then tensor to Q, and reduce to the well-known fact that any two bases of the same vector space have the same cardinality. – Pete L. Clark Mar 5 '10 at 2:14
Yeah, in fact the proof in Johnson's newer book is not flawed like this one is. I personally don't tensor with Q; If A is the abelianization, I look at A/2A :) – Steve D Mar 5 '10 at 2:16
I am better with Google than with cardinal arithmetic :-) math.niu.edu/~rusin/known-math/99/luzin_easton – Carsten S Mar 5 '10 at 2:18
See also this answer mathoverflow.net/questions/1924/#6594 and this related question mathoverflow.net/questions/12943, on the symmeric groups on sets of different sizes, when 2^a = 2^b but a not = b. – Joel David Hamkins Mar 5 '10 at 4:31

François gives the correct affirmative answer. For the negative side, the usual method of proving that the negation of the Continuum Hypothesis is consistent with ZFC is to use the method of forcing to add Aleph2 many Cohen reals, so that 2ω = ω2 in the forcing extension V[G]. In this model V[G], it is also true that 2ω1 = ω2. Thus, this model shows that it is not necessarily true that different-sized sets have different sized power sets. The case of symmetric groups is likely more interesting than free groups, because in this model, the symmetric groups Sω and Sω1 have the same cardinality ω2. Nevertheless, these two groups are not isomorphic, as explained in this MO question.

The general answer about what can be true for the continuum function κ --> 2κ is exactly provided by Easton's Theorem. This remarkable theorem states that if you have any class function E, defined on the regular cardinals κ, with the properties that

• κ ≤ λ implies E(κ) ≤ E(λ)
• κ < E(κ)
• κ < Cof(E(κ))

then there is a forcing extension V[G] in which 2κ = E(κ) for all regular cardinal κ. In particular, this shows that the sizes of the power sets (on regular cardinals) are restricted to obey only and exactly the hypotheses listed explicitly above. Each of these properties corresponds to a well-known fact about cardinal exponententiation.

Using Easton's theorem, we can build models of set theory where 2κ = κ++ for all regular κ. The added power of the Woodin/Foreman result mentioned by François is that they also get this for singular cardinals.

The point now is that there are innumerable examples provided by Easton's theorem that satisfy your hypothesis that the continuum function is one-to-one. If one begins with a model of GCH and selects any injective Easton function E, then the resulting model of set theory V[G] will have E as it's continuum function κ --> 2κ = E(κ) for regular κ, and the GCH will continue to hold at singular κ, preserving injectivity. So one is quite free to satisfy your hypothesis while having any kind of crazy failures of GCH.

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+1: Excellent answer! I was going to fill in some of the gaps in my answer later, but I'm happy you did it instead since your answers are always much richer than mine. – François G. Dorais Mar 5 '10 at 3:17
Well, I don't know about that, François, since your answers are usually excellent. In this case, you gave the positive answer, and I spoke about the negative side, where the hypothesis fails. But I suppose Easton's theorem goes both ways, doesn't it? You can use it either to make positive or negative instances of the phenomenon. – Joel David Hamkins 0 secs ago. – Joel David Hamkins Mar 5 '10 at 3:45
And you completely avoid the large cardinal hypothesis in Woodin's Theorem, which is much more satisfactory. – François G. Dorais Mar 5 '10 at 3:56
I have to say, I will probably pick this as the answer. I will wait for a little bit longer to see if anything drastic changes; I also wish I could pick all 4 answers as best! – Steve D Mar 5 '10 at 5:06

The Generalized Continuum Hypothesis, i.e. $2^\kappa = \kappa^+$ for every infinite cardinal $\kappa$, implies the desired result, but it can hold in other circumstances. For example, Woodin has shown that it is relatively consistent with the existence of a supercompact cardinal that $2^\kappa = \kappa^{++}$ for every infinite cardinal $\kappa$. (See Foreman and Woodin, The generalized continuum hypothesis can fail everywhere, Annals of Mathematics 133, 1991, 1-35.)

The statement $2^{\aleph_0} < 2^{\aleph_1}$ is known as the Weak Diamond Principle. This principle was introduced by Devlin and Shelah (A weak version of $\diamondsuit$ which follows from $2^{\aleph_{0}}<2^{\aleph_{1}}$, Israel J. Math. 29, 1978, 239-247). This principle has been very useful as a substitute for stronger enumeration principles such as the Continuum Hypothesis and Jensen's Diamond Principle $\diamondsuit$. I think the Weak Diamond Principle was inspired by Shelah's work on the Whitehead Problem (which may be related to your group theoretic application).

-
Note that Woodin's Theorem, remarkable as it may be, is overkill here. As Joel Hamkins pointed out in his answer, the classical forcings of Easton and Cohen can be used to force that $2^\kappa=\kappa^{++}$ for $\aleph_0 \leq \kappa < \aleph_\omega$ and $2^\kappa = \kappa^{+}$ when $\kappa \geq \aleph_0$, for example. This does not require the use of a supercompact cardinal, but large cardinals are necessary to fully control the sizes of the powersets of singular cardinals. – François G. Dorais Mar 5 '10 at 4:10

François's and Joel's answers to the general question are correct.

As for the "dimension of a free group is uniquely determined, even if it's infinite", there's a better argument. Let G be a free group on two sets of generators x and y; we wish to show |x| = |y|. First let A = Gab be the free abelian group on generators x and y; then let V = AZ F2 be the vector space over F2 with bases x and y, so that |V| = 2|x| = 2|y| (strike that, it's false for infinite cardinalities). (We could use any field here, but the argument is more intricate.)

If |V| is finite, then |x| = |y| follows immediately. So suppose |V| is infinite, so that |x| and |y| are infinite. Each element of x is the sum of finitely many elements of y, and every element of y must show up in at least one such sum (why?). This means |y| ≤ |x|•ℵ0 = |x| (cardinal arithmetic). Similarly |x| ≤ |y|, so |x| = |y|.

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Right. This has already been said in the comments following the question. – Pete L. Clark Mar 5 '10 at 5:56
Fair enough, and duly upvoted. I'm pretty sure those comments weren't there at the time, but maybe I missed them. – Chad Groft Mar 6 '10 at 4:42

One thing that is made very easy by assuming that $2^\alpha < 2^\beta$ whenever $\alpha <\beta$ is showing that for sets $S_1$ and $S_2$ of different cardinalities, the Banach spaces $\ell_\infty(S_1)$ and $\ell_\infty(S_2)$ are not isomorphic. This follows immediately from the fact that for an infinite set $S$, the smallest possible cardinality of a dense subset of $\ell_\infty(S)$ is $2^{|S|}$.

If one does not assume that $2^\alpha = 2^\beta$ implies $\alpha =\beta$, then one must revert to some other argument. I have not thought about this problem at length, so there may be much simpler reasons why $\ell_\infty(S_1)$ and $\ell_\infty(S_2)$ cannot be isomorphic when $S_1$ and $S_2$ have different cardinalities (without assuming $2^\alpha = 2^\beta\Rightarrow\alpha =\beta$), but one way is to instead consider the least possible cardinality of a norm dense subset of an arbitrary weakly compact subset of $\ell_\infty(S)$. In this case, a result of Haskell Rosenthal asserts that for $S$ infinite and a weakly compact set $W \subseteq \ell_\infty (S)$, $W$ admits a norm dense subset of cardinality no greater than $|S|$. On the other hand, a standard argument shows that $\ell_2 (S)$ (in fact, any Banach space of density character not exceeding $|S|$) is isomorphic to a subspace of $\ell_\infty (S)$ (generalise the standard argument that every separable Banach space embeds into $\ell_\infty$). Thus, if $\alpha$ and $\beta$ are cardinals with $\alpha <\beta$, then $\ell_2 (\beta )$ is isomorphic to a subspace of $\ell_\infty (\beta )$, but not to a subspace of $\ell_\infty (\alpha)$. Hence $\ell_\infty (\alpha )$ and $\ell_\infty (\beta)$ are not isomorphic.

The result of Rosenthal cited above is from his paper "On injective Banach spaces and the spaces $L_\infty (\mu )$ for finite measures $\mu$".

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text:
specify_prior_bsvar_msh {bsvars} R Documentation

## R6 Class Representing PriorBSVAR-MSH

### Description

The class PriorBSVAR-MSH presents a prior specification for the bsvar model with Markov Switching Heteroskedasticity.

### Super class

bsvars::PriorBSVAR -> PriorBSVAR-MSH

### Public fields

A

an NxK matrix, the mean of the normal prior distribution for the parameter matrix A.

A_V_inv

a KxK precision matrix of the normal prior distribution for each of the row of the parameter matrix A. This precision matrix is equation invariant.

B_V_inv

an NxN precision matrix of the generalised-normal prior distribution for the structural matrix B. This precision matrix is equation invariant.

B_nu

a positive integer greater of equal than N, a shape parameter of the generalised-normal prior distribution for the structural matrix B.

hyper_nu

a positive scalar, the shape parameter of the inverted-gamma 2 prior distribution for the two overall shrinkage parameters for matrices B and A.

hyper_a

a positive scalar, the shape parameter of the gamma prior for the two overall shrinkage parameters.

hyper_V

a positive scalar, the shape parameter of the inverted-gamma 2 for the level 3 hierarchy of shrinkage parameters.

hyper_S

a positive scalar, the scale parameter of the inverted-gamma 2 for the level 3 hierarchy of shrinkage parameters.

sigma_nu

a positive scalar, the shape parameter of the inverted-gamma 2 for MS state-dependent variances of the structural shocks, \sigma^2_{n.s_t}.

sigma_s

a positive scalar, the scale parameter of the inverted-gamma 2 for MS state-dependent variances of the structural shocks, \sigma^2_{n.s_t}.

PR_TR

an MxM matrix, the matrix of hyper-parameters of the row-specific Dirichlet prior distribution for transition probabilities matrix P of the Markov process s_t.

### Methods

#### Method new()

Create a new prior specification PriorBSVAR-MSH.

##### Examples
# a prior for 3-variable example with four lags and two regimes
prior = specify_prior_bsvar_msh$new(N = 3, p = 4, M = 2) prior$get_prior() # show the prior as list

#### Method clone()

The objects of this class are cloneable with this method.

specify_prior_bsvar_msh$clone(deep = FALSE) ##### Arguments deep Whether to make a deep clone. ### Examples prior = specify_prior_bsvar_msh$new(N = 3, p = 1, M = 2)  # specify the prior
prior$A # show autoregressive prior mean ## ------------------------------------------------ ## Method specify_prior_bsvar_msh$get_prior
## ------------------------------------------------

# a prior for 3-variable example with four lags and two regimes
prior = specify_prior_bsvar_msh$new(N = 3, p = 4, M = 2) prior$get_prior() # show the prior as list

[Package bsvars version 1.0.0 Index]


text:
Search

# Algebraic Fractions

A particularly tricky topic for the new 9-1 GCSE maths qualification is simplifying and manipulating algebraic fractions.

Success with these types of questions requires students to fully understand what they are doing when they simplify and manipulate numerical fractions.

For example, let's add together two very easy fractions:

The reason this example is so easy is because the denominators of both fractions are the same, and so only the numerators need to be added.

A more difficult example is where the denominators are not equal. In this example, it is necessary to make both denominators the same first by finding the lowest common denominator, and ensuring we multiply the numerator by whatever we multiplied the denominator by:

The same principle, although conceptually harder, applies to adding algebraic fractions. For example:

Here, n+1 and n are not the same terms and so we can't simply add the numerators as in the first example. It will therefore be necessary to find a common denominator.

But what is the common denominator of n+1 and n? As we don't know what n is, we need to find an expression that both n+1 and n will go into (in a similar way to how we found that both 4 and 5 go into 20 in the second example).

The simplest way is to multiply both n+1 and n together, as both terms go into n(n+1), making sure the numerator is multiplied by whatever we multiplied the denominator by:

So in this example, we multiplied the denominator of the first fraction by n, and made sure to multiply the numerator by n too. Similarly, for the second fraction, we multiplied the denominator by n+1, so made sure to multiply the numerator by n+1 too.

Now that the denominators are the same, we can subtract the fractions to create a single fraction:

We can see if this algebraic fraction simplifies further by expanding the brackets and collecting any resulting like terms:

Note that the nm and mn terms cancel as they are the same.

Here is another example. Solve the following equation:

The first step is to combine both of these fractions into a single fraction. As before we need to find a common denominator. What do both 3x and 2x go into? If we multiply them together, both the 3x and 2x will go into this. However, we must remember to multiply the numerators by whatever we multiplied the denominators by. We therefore have:

We can now add these two fractions together to give:

We can now expand the brackets in the numerators to give:

Now, collect the like terms in the numerator to give:

Multiply both sides by the denominator to 'get rid' of the fraction:

Next, subtract both sides by the first term in order to collect like terms:

Add the 2x term to both sides to give:

Now, take out x as a factor:

In order for the right hand side to equal 0, either the x term is equal to 0, or the 13x+2 term is equal to 0:

Solving the second equation for x:

The final answer therefore has two solutions:

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text:
# zbMATH — the first resource for mathematics

## Gromak, Valeriĭ Ivanovich

Compute Distance To:
 Author ID: gromak.valerii-ivanovich Published as: Gromak, V.; Gromak, V. I.; Gromak, Valeri; Gromak, Valerii; Gromak, Valerii I.; Gromak, Valeriĭ Ivanovic; Gromak, Valery I. External Links: Math-Net.Ru
 Documents Indexed: 70 Publications since 1973, including 2 Books Biographic References: 1 Publication
all top 5

#### Co-Authors

 36 single-authored 10 Tsegel’nik, V. V. 5 Filipuk, Galina V. 4 Gaĭshun, Ivan Vasil’vich 3 Lukashevich, N. A. 2 Il’in, Vladimir Aleksandrovich 2 Izobov, Nikolaĭ Alekseevich 2 Zenchenko, Alexandra S. 1 Amel’kin, Vladimir Vasilievich 1 Andreev, Alekseĭ Fedorovich 1 Atrokhov, K. G. 1 Bachurina, E. V. 1 Golubeva, L. L. 1 Gorokhovik, Valentin Vikent’evich 1 Gricuk, Evgenii 1 Grigor’ev, A. A. 1 Gritsuk, E. V. 1 He, Yuzan 1 Kozulin, A. V. 1 Kozynin, A. V. 1 Laine, Ilpo 1 Laputskiĭ, A. I. 1 Pilipchuk, Ludmila A. 1 Shimomura, Shun 1 Stepanova, T. S. 1 Vasilevich, N. D. 1 Yurchuk, Nikolay Iossifovich
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#### Serials

 18 Differential Equations 13 Differential Equations 6 Differentsial’nye Uravneniya 5 Doklady Akademii Nauk BSSR 5 Vestnik Belorusskogo Gosudarstvennogo Universiteta. Seria 1. Fizika, Matematika, Informatika 3 Izvestiya Akademii Nauk BSSR. Seriya Fiziko-Matematicheskikh Nauk 2 Theoretical and Mathematical Physics 2 Trudy Instituta Matematiki 2 Mathematical Modelling and Analysis 1 Teoreticheskaya i Matematicheskaya Fizika 1 Vestnik Belorusskogo Gosudarstvennogo Universiteta Imeni V. I. Lenina. Seriya I 1 Journal of Nonlinear Mathematical Physics 1 De Gruyter Studies in Mathematics
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#### Fields

 60 Ordinary differential equations (34-XX) 11 Partial differential equations (35-XX) 6 Special functions (33-XX) 5 Dynamical systems and ergodic theory (37-XX) 4 History and biography (01-XX) 3 Functions of a complex variable (30-XX) 2 Quantum theory (81-XX) 1 Several complex variables and analytic spaces (32-XX) 1 Integral transforms, operational calculus (44-XX) 1 Global analysis, analysis on manifolds (58-XX) 1 Numerical analysis (65-XX)

#### Citations contained in zbMATH

33 Publications have been cited 237 times in 185 Documents Cited by Year
Painlevé differential equations in the complex plane. Zbl 1043.34100
Gromak, V. I.; Laine, Ilpo; Shimomura, Shun
2002
Analytic properties of solutions of Painlevé equations. Zbl 0752.34003
Gromak, V. I.; Lukashevich, N. A.
1990
Bäcklund transformations of Painlevé equations and their applications. Zbl 1023.34082
Gromak, Valerii I.
1999
Theory of the fourth Painlevé equation. Zbl 0647.34004
Gromak, V. I.
1987
Special classes of solutions of Painlevé’s equations. Zbl 0526.34002
Gromak, V. I.; Lukashevich, N. A.
1982
Theory of Painleve’s equation. Zbl 0329.34006
Gromak, V. I.
1976
Reducibility of Painlevé equations. Zbl 0573.34009
Gromak, V. I.
1984
On fourth-order nonlinear differential equations with the Painlevé property. Zbl 1148.34057
Gromak, V. I.
2006
Integrability of complex Pfaffian equations. Zbl 0505.58003
Vasilevich, N. D.; Gromak, V. I.
1981
Nonlinear evolution equations and P-type controls. Zbl 0599.34042
Gromak, V. I.
1984
On the theory of higher-order Painlevé equations. Zbl 1088.34076
Gromak, V. I.; Zenchenko, A. S.
2004
On algebraic solutions of the fifth Painlevé equation. Zbl 1068.34084
Gromak, V. I.; Filipuk, G. V.
2003
The Bäcklund transformations of the fifth Painlevé equation and their applications. Zbl 1015.34078
Gromak, V.; Filipuk, G.
2001
Bäcklund transformations of the higher order Painlevé equations. Zbl 1006.34086
Gromak, V. I.
2001
Functional relations among solutions to the fifth Painlevé equation. Zbl 0997.34086
Gromak, V. I.; Filipuk, G. V.
2001
The first higher-order Painlevé equation. Zbl 0939.34075
Gromak, V. I.
1999
On the transcendence of Painlevé equations. Zbl 0880.34005
Gromak, V. I.
1996
System of three-wave resonant interaction and P-type equations. Zbl 0691.35082
Gromak, V. I.; Tsegel’nik, V. V.
1989
Solutions of the second Painleve equation. Zbl 0513.34002
Gromak, V. I.
1982
Solutions of the third Painleve equation. Zbl 0305.34002
Gromak, V. I.
1975
Solutions of the fourth-order equation in the generalized hierarchy of the second Painlevé equation. Zbl 1422.30039
Gromak, V. I.
2019
A remark on the theory of nonlinear differential equations with the Painlevé property. Zbl 1219.34114
Gritsuk, E. V.; Gromak, V. I.
2010
On solutions of the Chazy system. Zbl 1144.34058
Gromak, V. I.
2007
The power expansions of the solutions of the first Painlevé hierarchy. Zbl 1127.34054
Grigor’ev, A. A.; Gromak, V. I.
2006
On the transformations of the sixth Painlevé equation. Zbl 1362.34128
Gromak, Valery I.; Filipuk, Galina
2003
On the algebraic dependence of solutions to the second Painlevé equation. Zbl 1033.34090
Gromak, V. I.
2002
On solutions to the second Painlevé equation of higher order. Zbl 0988.34078
Gromak, V. I.; He, Yuzan
2000
Some properties of the Bäcklund transformations of the Painlevé equations. Zbl 0891.34041
Gromak, V. I.; Tsegel’nik, V. V.
1996
The fifth and sixth Painlevé equations are transcendental. Zbl 0884.34006
Gromak, V. I.
1996
Functional relations between solutions to $$P$$-type equations. Zbl 0852.34009
Gromak, V. I.; Tsegel’nik, V. V.
1994
On the solutions of a system of three differential equations with quadratic nonlinearities. Zbl 0727.34008
Gromak, V. I.; Tsegel’nik, V. V.
1991
Algebraic solutions of the third Painleve equation. Zbl 0441.34003
Gromak, V. I.
1979
Solutions of Painleve’s fifth problem. Zbl 0365.34011
Gromak, V. I.
1977
Solutions of the fourth-order equation in the generalized hierarchy of the second Painlevé equation. Zbl 1422.30039
Gromak, V. I.
2019
A remark on the theory of nonlinear differential equations with the Painlevé property. Zbl 1219.34114
Gritsuk, E. V.; Gromak, V. I.
2010
On solutions of the Chazy system. Zbl 1144.34058
Gromak, V. I.
2007
On fourth-order nonlinear differential equations with the Painlevé property. Zbl 1148.34057
Gromak, V. I.
2006
The power expansions of the solutions of the first Painlevé hierarchy. Zbl 1127.34054
Grigor’ev, A. A.; Gromak, V. I.
2006
On the theory of higher-order Painlevé equations. Zbl 1088.34076
Gromak, V. I.; Zenchenko, A. S.
2004
On algebraic solutions of the fifth Painlevé equation. Zbl 1068.34084
Gromak, V. I.; Filipuk, G. V.
2003
On the transformations of the sixth Painlevé equation. Zbl 1362.34128
Gromak, Valery I.; Filipuk, Galina
2003
Painlevé differential equations in the complex plane. Zbl 1043.34100
Gromak, V. I.; Laine, Ilpo; Shimomura, Shun
2002
On the algebraic dependence of solutions to the second Painlevé equation. Zbl 1033.34090
Gromak, V. I.
2002
The Bäcklund transformations of the fifth Painlevé equation and their applications. Zbl 1015.34078
Gromak, V.; Filipuk, G.
2001
Bäcklund transformations of the higher order Painlevé equations. Zbl 1006.34086
Gromak, V. I.
2001
Functional relations among solutions to the fifth Painlevé equation. Zbl 0997.34086
Gromak, V. I.; Filipuk, G. V.
2001
On solutions to the second Painlevé equation of higher order. Zbl 0988.34078
Gromak, V. I.; He, Yuzan
2000
Bäcklund transformations of Painlevé equations and their applications. Zbl 1023.34082
Gromak, Valerii I.
1999
The first higher-order Painlevé equation. Zbl 0939.34075
Gromak, V. I.
1999
On the transcendence of Painlevé equations. Zbl 0880.34005
Gromak, V. I.
1996
Some properties of the Bäcklund transformations of the Painlevé equations. Zbl 0891.34041
Gromak, V. I.; Tsegel’nik, V. V.
1996
The fifth and sixth Painlevé equations are transcendental. Zbl 0884.34006
Gromak, V. I.
1996
Functional relations between solutions to $$P$$-type equations. Zbl 0852.34009
Gromak, V. I.; Tsegel’nik, V. V.
1994
On the solutions of a system of three differential equations with quadratic nonlinearities. Zbl 0727.34008
Gromak, V. I.; Tsegel’nik, V. V.
1991
Analytic properties of solutions of Painlevé equations. Zbl 0752.34003
Gromak, V. I.; Lukashevich, N. A.
1990
System of three-wave resonant interaction and P-type equations. Zbl 0691.35082
Gromak, V. I.; Tsegel’nik, V. V.
1989
Theory of the fourth Painlevé equation. Zbl 0647.34004
Gromak, V. I.
1987
Reducibility of Painlevé equations. Zbl 0573.34009
Gromak, V. I.
1984
Nonlinear evolution equations and P-type controls. Zbl 0599.34042
Gromak, V. I.
1984
Special classes of solutions of Painlevé’s equations. Zbl 0526.34002
Gromak, V. I.; Lukashevich, N. A.
1982
Solutions of the second Painleve equation. Zbl 0513.34002
Gromak, V. I.
1982
Integrability of complex Pfaffian equations. Zbl 0505.58003
Vasilevich, N. D.; Gromak, V. I.
1981
Algebraic solutions of the third Painleve equation. Zbl 0441.34003
Gromak, V. I.
1979
Solutions of Painleve’s fifth problem. Zbl 0365.34011
Gromak, V. I.
1977
Theory of Painleve’s equation. Zbl 0329.34006
Gromak, V. I.
1976
Solutions of the third Painleve equation. Zbl 0305.34002
Gromak, V. I.
1975
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#### Cited by 181 Authors

 17 Kudryashov, Nikolaĭ Alekseevich 12 Clarkson, Peter A. 12 Grammaticos, Basil 12 Ramani, Alfred 9 Filipuk, Galina V. 9 Gromak, Valeriĭ Ivanovich 8 Bruno, Alexander Dmitrievich 8 Goryuchkina, Irina V. 7 Rogers, Colin 7 Tamizhmani, K. M. 5 Chen, Yang 5 Shimomura, Shun 4 Fornberg, Bengt 4 Tamizhmani, Thamizharasi 3 Dai, Dan 3 Deaño, Alfredo 3 Demina, Maria V. 3 Gordoa, Pilar Ruiz 3 Ohta, Yasuhiro 3 Pickering, Andrew 3 Sasaki, Yoshikatsu 3 Slavyanov, Sergeĭ Yur’evich 3 Takahashi, Daisuke A. 3 Top, Jaap 3 Tsegel’nik, V. V. 3 van der Put, Marius 3 Weideman, J. André C. 2 Acosta Humánez, Primitivo Belén 2 Adler, Vsevolod Eduardovich 2 Amel’kin, Vladimir Vasilievich 2 Babich, Mikhail Vasil’evich 2 Bassom, Andrew P. 2 Bershteĭn, Mikhaĭl Aleksandrovich 2 Blower, Gordon 2 Brezhnev, Yurii V. 2 Chen, Min 2 Chen, Zhijie 2 Chen, Zongxuan 2 Chiang, Yik-Man 2 Ciechanowicz, Ewa 2 Deift, Percy A. 2 Dubrovin, Boris Anatol’evich 2 Dunajski, Maciej 2 Fan, Engui 2 Fasondini, Marco 2 Feng, Shaoji 2 Ghose Choudhury, Anindya 2 Guha, Partha 2 Hone, Andrew N. W. 2 Huang, Zhibo 2 Its, Alexander Rudolfovich 2 Kapaev, Andrei A. 2 Kecker, Thomas 2 Korhonen, Risto 2 Kuo, Ting-Jung 2 Laine, Ilpo 2 Lin, Chang-Shou 2 Mugan, Ugurhan 2 Nijhoff, Frank W. 2 Novokshenov, Viktor Yur’evich 2 Papageorgiou, Vassilios G. 2 Sakka, Ayman Hashem 2 Satsuma, Junkichi 2 Shchechkin, A. I. 2 Steinmetz, Norbert 2 Vasilevich, M. N. 2 Zhang, Lun 1 Aleksandrov, Aleksandr Grigor’evich 1 Alves, V. C. C. 1 Amster, Pablo 1 Aratyn, Henrik 1 Atrokhov, K. G. 1 Bagderina, Yuliya Yur’evna 1 Basor, Estelle L. 1 Bleher, Pavel M. 1 Bonelli, Giulio 1 Borisov, Alekseĭ Vladimirovich 1 Bougoffa, Lazhar 1 Bracken, Paul Francis 1 Cârstea, Adrian Ştefan 1 Chankaev, M. Kh. 1 Charlier, Christophe 1 Chen, Hongmei 1 Chen, Min 1 Chen, Yang 1 Chiba, Hayato 1 Christodoulakis, Theodosios 1 Contatto, Felipe 1 Conte, Robert M. 1 Costin, Rodica D. 1 Dai, Dashuang 1 Dereziński, Jan 1 Efimova, Olga Yu. 1 Eremenko, Alexandre Émmanuilovich 1 Gabrielov, Andrei 1 Gamkrelidze, Revaz Valerianovich 1 Goldstein, Piotr P. 1 Gomes, Jose Francisco 1 Gómez-Ullate, David 1 Gontsov, Renat R. ...and 81 more Authors
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#### Cited in 59 Serials

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# Revision history [back]

You don't need to use rosdep, but it simplifies and speeds things up a lot.

First of, I'll quote the rosdep docs:

rosdep is a command-line tool for installing system dependencies. For end-users, rosdep helps you install system dependencies for software that you are building from source. For developers, rosdep simplifies the problem of installing system dependencies on different platforms. Instead of having to figure out which debian package on Ubuntu Oneiric contains Boost, you can just specify a dependency on 'boost'. rosdep is supported on a variety of platforms/package managers:

• Ubuntu (apt)
• Debian (apt)
• OS X (Homebrew)
• Red Hat Enterprise Linux (rpm/experimental)
• Arch (pacman/experimental)
• Gentoo (equery/experimental)
• Slackware (sbotools/experimental)
• Cygwin (experimental)

Basically, rosdep is a utility to easily install dependencies that are released as (Debian, for Ubuntu/Debian) packages (or sometimes as python packages via pip) when building something from source.

You don't need to use rosdep, it just makes life easier when installing something from source and it is cross-platform. So, if you are developing cross-platform, you can stick to one workflow.

On ubuntu, rosdep is in the end just calling apt (or in a few occasions pip) to install the realeased packages you need.

You could do this manually, but would have to check the docs of the packages you want to compile, or the package.xml.

So what to use when? (my personal stance, others might differ, and not in any specific order)

• Basic Installation (i.e. new ROS): apt
• You know exactly what you need (you develop a new package and add a dep you don't have installed yet): apt
• The dep I need is not in apt: pip
• I cloned some repos and want everything I need to compile (new set of packages, new computer, ...): rosdep


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Lemma 10.29.6. Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$. Let $S = R/I$. Then the image of a constructible subset of $\mathop{\mathrm{Spec}}(S)$ is constructible in $\mathop{\mathrm{Spec}}(R)$.

Proof. If $I = (f_1, \ldots , f_ m)$, then we see that $V(I)$ is the complement of $\bigcup D(f_ i)$, see Lemma 10.17.2. Hence it is constructible, by Lemma 10.29.1. Denote the map $R \to S$ by $f \mapsto \overline{f}$. We have to show that if $\overline{U}, \overline{V}$ are retrocompact opens of $\mathop{\mathrm{Spec}}(S)$, then the image of $\overline{U} \cap \overline{V}^ c$ in $\mathop{\mathrm{Spec}}(R)$ is constructible. By Lemma 10.29.1 we may write $\overline{U} = \bigcup D(\overline{g_ i})$. Setting $U = \bigcup D({g_ i})$ we see $\overline{U}$ has image $U \cap V(I)$ which is constructible in $\mathop{\mathrm{Spec}}(R)$. Similarly the image of $\overline{V}$ equals $V \cap V(I)$ for some retrocompact open $V$ of $\mathop{\mathrm{Spec}}(R)$. Hence the image of $\overline{U} \cap \overline{V}^ c$ equals $U \cap V(I) \cap V^ c$ as desired. $\square$

Comment #6455 by Jonas Ehrhard on

The statement should be:

"The image of a constructible subset of $\operatorname{Spec}(S)$ is constructible in $\operatorname{Spec}(R)$."

There are also:

• 4 comment(s) on Section 10.29: Images of ring maps of finite presentation

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


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Home > Sample Size > Sample Size Increases Margin Error Confidence Interval Population Proportion

# Sample Size Increases Margin Error Confidence Interval Population Proportion

## Contents

Review questions: pages 335 and 351. In-class: p. 337: 20. Change requirements from ≤5% to ≤10% of population. The underlying idea that explains how we can determine the reliability of statistics is the notion of sampling distribution. navigate to this website

E.g., the proportion of registered voters in East Baton Rouge who are republican. If you think about it, it makes sense that the more information you have, the more accurate your results are going to be (in other words, the smaller your margin of Your cache administrator is webmaster. Your required sample size becomes z0.05=invNorm(1−0.05)≈1.6449 [p̂1(1−p̂1) + p̂2(1−p̂2)] [zα/2÷E]²= (0.37×0.63+0.47×0.53)×(ANS÷0.03)²= 1449.57... → 1450 Answer: For a 90% CI with margin of error ≤3%, when you think one population's proportion is https://onlinecourses.science.psu.edu/stat500/node/31

## Minimum Sample Size To Estimate Population Proportion Calculator

If we assume that this situation is representative of birth gender in the United States, give a 95% confidence interval for the true proportion of baby girls in the United States. Answer: $$n \hat{\pi}=3 < 5$$ Therefore, we cannot use a z-interval. Do you know anything more than just that the true proprtion is near 52%? n = [ zα/2 × s ÷ E ]² = (ANS×6.2÷1.5)² = 46.2227... → 47 Your preliminary sample size is 47, and next you use that to compute t.

Incidentally, population variability is not something we can usually control, but more meticulous collection of data can reduce the variability in our measurements. Then, when you have a preliminary sample size determined by (ab)using z in this way, recompute the formula using that sample size minus 1 for df. A survey of 1000 Californians finds reports that 48% are excited by the annual visit of INSPIRE participants to their fair state. Sample Size Formula For Finite Population The parameter mu, while unknown, is not random.

But there are many ways to go wrong or to misunderstand the meaning of the data obtained from a sample. How To Find Sample Size With Margin Of Error And Confidence Level The sample size needed is 7745 people (we always need to round up to the next integer when the result is not a whole number). Example 3: What percent of the voters would vote for your candidate if the election were held today? this website The answer is that you take the formula for the margin of error, rearrange it algebraically to solve for the sample size, compute, and round up.

Solution: Use z instead of t to make a preliminary estimate, then recompute with t. How To Find Minimum Sample Size Required To Estimate A Population Proportion If you don't have any credible estimate, use p̂=(1−p̂)=0.5. Solution: Compute zα/2=1.9600 as in the previous example. z0.05 = invNorm(1−0.05) ≈ 1.6449 Now you have all the pieces.

## How To Find Sample Size With Margin Of Error And Confidence Level

The color of bead is analogous to the vote---e.g. http://www.dummies.com/education/math/statistics/how-sample-size-affects-the-margin-of-error/ What you know about a population when you have a sample of size 100 is similar to what you know about the contents of a jar of gum balls if you Minimum Sample Size To Estimate Population Proportion Calculator Use the sqare root law to estimate the sample size needed to get a given margin of error better than 95% confidence. (See text, page 350.) Assessments: A jar of colored How To Find Sample Size With Confidence Interval Describe what we would do in order to estimate the sampling distribution empirically.

You're going to chain calculations so that you don't have to re-enter any of your intermediate numbers. http://garmasoftware.com/sample-size/sample-size-confidence-interval-margin-of-error-calculator.php Please try the request again. If we encounter a situation where the response rate is not 100% then if we just sample the calculated size, in the end we will end up with a less than CommentsComputation Marshal your data. Sample Size And Confidence Interval Relationship

How do you compute it? What's New 24 Aug 2013: Rearrange the formula for Case 2 to simplify the computations. Therefore you can use a z function, and the formulas are the same as Case 0 with √p(1−p) substituted for σ: transforms to Because this article helps you,please click to donate!Because my review here Your cache administrator is webmaster.

Suppose you decide that you want to refine your estimate of the population proportion and cut the width of your interval in half. Minimum Sample Size Calculator Case 1: One Population Mean, Unknown σ Note: Many basic statistics courses skip the material in this section and estimate sample sizes using a z distribution, so the material in this Remember, always round sample size up.

## The sampling distribution tells us the relative frequency of each possible sample percent in the reference class of all samples.

This means the margin of error must be less than 2%, so solving for n: n = (1.96/.02)^2 *.48*.52 = 2397.1 We'd need about 2398 people. 4. So to cut the width of the CI in half, we'd need about four times as many people. The exact interval and the z-interval should be very similar when the conditions are satisfied. Sample Size Calculator Online The answer is that it's not the same margin of error.

The formula for margin of error, below left, is just an extension of the formula for one population proportion. Therefore, zα∕2 is given by qnorm(.975). > zstar = qnorm(.975) > p = 0.5 > E = 0.05 > zstar^2 ∗ p ∗ (1−p) / E^2 [1] 384.15 Answer With a planned proportion estimate of 50% at 95% confidence level, it needs a sample size of 385 to achieve a 5% margin This is a generalization from a sample (the vegetables we have examined) to a population (all the vegetables the store sells). http://garmasoftware.com/sample-size/sample-size-calculator-confidence-interval-margin-error.php Use a table to determine the levels of confidence and margins of error that can be obtained with various sample sizes when attempting to determine population proportions.

You don't know n, so you don't know the degrees of freedom df either and you can't compute the critical t for the formula. sample proportion: the proportion of a sample with the property. Applied Statistical Decision Making Lesson 6 - Confidence Intervals6.1 - Inference for the Binomial Parameter: Population Proportion 6.2 - Sample Size Computation for Population Proportion Confidence Interval 6.3 - Inference for Based on the confidence level that you preselect, and characteristics of your sample or population, compute a margin of error.

Assignment: Read: Chapter 8, sections 1, 2 and 3. How big a sample would you need to be sure of a margin of error no more than 3.5% in a 95% CI? Answer: You need a random sample of at least 100 to test this model. (As always, the minimum sample will give a significant result only if the null hypothesis is extremely Always remember to round sample sizes up. [p̂1(1−p̂1) + p̂2(1−p̂2)] [zα/2÷E]²= (0.42×0.58+0.42×0.58)×(ANS÷0.03)²= 1464.60... → 1465 Answer: To find a 90% CI for the difference in your candidate's support between men and

How many individuals should we sample? (In the last poll his approval rate was 72%). What do think the true proportion of Republicans in the population is? It is also a variable that has as its refernce class all possible samples. Answer: There are three incorrect statements.

sample mean: the average value of a variable, where the reference class is a sample from the population. This page shows the formulas for some common cases, with examples. It should read, "We can be 95% confident that soldiers land in the target between 50% and 81% of the time." (The difference is subtle but shows a student misunderstanding.) And a.

Based on historical data, you have reason to believe that the standard deviation of the machine's hourly output is 6.2. You compute it on your TI-83/84/89 as invNorm(1−rtail). After a few vivits to a store, for example, we notice that the produce is not fresh.


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# Spalart-Allmaras model

Spallart-Allmaras model is a one equation model for the turbulent viscosity.

## Original model

The turbulent eddy viscosity is given by

$\nu_t = \tilde{\nu} f_{v1}, \quad f_{v1} = \frac{\chi^3}{\chi^3 + C^3_{v1}}, \quad \chi := \frac{\tilde{\nu}}{\nu}$
$\frac{\partial \tilde{\nu}}{\partial t} + u_j \frac{\partial \tilde{\nu}}{\partial x_j} = C_{b1} [1 - f_{t2}] \tilde{S} \tilde{\nu} + \frac{1}{\sigma} \{ \nabla \cdot [(\nu + \tilde{\nu}) \nabla \tilde{\nu}] + C_{b2} | \nabla \nu |^2 \} - \left[C_{w1} f_w - \frac{C_{b1}}{\kappa^2} f_{t2}\right] \left( \frac{\tilde{\nu}}{d} \right)^2 + f_{t1} \Delta U^2$
$\tilde{S} \equiv S + \frac{ \tilde{\nu} }{ \kappa^2 d^2 } f_{v2}, \quad f_{v2} = 1 - \frac{\chi}{1 + \chi f_{v1}}$
$f_w = g \left[ \frac{ 1 + C_{w3}^6 }{ g^6 + C_{w3}^6 } \right]^{1/6}, \quad g = r + C_{w2}(r^6 - r), \quad r \equiv \frac{\tilde{\nu} }{ \tilde{S} \kappa^2 d^2 }$
$f_{t1} = C_{t1} g_t \exp\left( -C_{t2} \frac{\omega_t^2}{\Delta U^2} [ d^2 + g^2_t d^2_t] \right)$
$f_{t2} = C_{t3} \exp(-C_{t4} \chi^2)$

The constants are

$\begin{matrix} \sigma &=& 2/3\\ C_{b1} &=& 0.1355\\ C_{b2} &=& 0.622\\ \kappa &=& 0.41\\ C_{w1} &=& C_{b1}/\kappa^2 + (1 + C_{b2})/\sigma \\ C_{w2} &=& 0.3 \\ C_{w3} &=& 2 \\ C_{v1} &=& 7.1 \\ C_{t1} &=& 1 \\ C_{t2} &=& 2 \\ C_{t3} &=& 1.1 \\ C_{t4} &=& 2 \end{matrix}$

According to Spalart it is safer to use the following values for the last two constants:

$\begin{matrix} C_{t3} &=& 1.2 \\ C_{t4} &=& 0.5 \end{matrix}$

## Model for compressible flows

There are two approaches to adapting the model for compressible flows. In the first approach the turbulent dynamic viscosity is computed from

$\mu_t = \rho \tilde{\nu} f_{v1}$

where $\rho$ is the local density. The convective terms in the equation for $\tilde{\nu}$ are modified to

$\frac{\partial \tilde{\nu}}{\partial t} + \frac{\partial}{\partial x_j} (\tilde{\nu} u_j)= \mbox{RHS}$

where the right hand side (RHS) is the same as in the original model.

## References

• Spalart, P. R. and Allmaras, S. R. (1992), "A One-Equation Turbulence Model for Aerodynamic Flows", AIAA Paper 92-0439.


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# Network Integration Programmer's Guide

This section describes how to create applications in a Systems Network Architecture (SNA) environment.

For API references and other technical information about network integration, see the Network Integration Programmer's Reference section of the SDK.

For sample code that illustrates network integration, see the Network Integration Samples section of the SDK.

For additional information about network integration, see the Network Integration User's Guide section of the Operations guide.


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# Propositional Logic with rules of inference problem.

$$\begin{array}{l} 1.\>\>\>\> (r ∧ ¬s) ∨ (q ∧ ¬s)\\ 2.\>\>\>\> ¬s → ((p ∧ r) → u)\\ 3.\>\>\>\> u → (s ∧ ¬t)\\ \dfrac{\quad\qquad\qquad\qquad\qquad\qquad}{\text{Prove from the previous arguments. }\quad p → q} \end{array}$$

Hey guys, I am really lost, so far I have a few arguments but not sure if they're correct. Can you please give some arguments to get this solved or a game plan using Inference laws and equivalences. Inferences: Modus Ponens, Tollens, Hypothetical, Disjunctive, Resolution, Conjunction, Simplification, Addition.

This is what I have so far.... 4. (r ∨ q) ∧ ¬s Distributive of 1 5. ¬s Simplification of 1 6. (p ∧ r) → u Modus Ponens of 2 7. ¬u Modus Tollens of 3 8. ¬(p ∧ r) Modus tollens 9. p → ¬r DeMorgans and Implication Definition\ 10. (r ∨ q) Simplification 11. ( ¬r → q) Implication 12. p → q Hypothetical

Are these steps valid?

I'll give a summary "sketch" of one proof approach.

I'd suggest you start from the first premise:

First check what follows from $r\land \lnot s$ and then what follows when $q \land \lnot s$. (At least one of these must be true. Why?)

From each of the above, we can use conjunction elimination (or simplification) to obtain $\lnot s$. And so, we can use disjunction elimination to conclude $\lnot s$.

From $\lnot s$ and premise $(2)$, we can derive (infer), by modus ponens, $(p\land r)\rightarrow u$.

From $(p \land r) \rightarrow u$ together with premise $(3)$, we get $(p\land r) \rightarrow (s \land \lnot t)$

Now, we already derived $\lnot s.$ This means $\lnot s \lor t$ is true. Why?

But $\lnot s \lor t \equiv \lnot (s \land \lnot t).$

From $\lnot (s \land \lnot t),$ together with our derived $(p\land r )\rightarrow (s\land \lnot t)$, by modus ponens, we get $\not(p \land r)$

This means that either $\lnot p$, or $\lnot r$ is true.

If $\lnot p$ is true, we can build from this (addition) to get $\lnot p \lor q \equiv p\rightarrow q$, as desired.

If $\lnot r$ is true, then $r\land \lnot s$ is false (first alternative from premise $(1)$. Then it must follow that $q\land \lnot s$ is true. So it follows that $q$ is true (simplification). And given $q$, we have $\lnot p \lor q$ (addition), which is equivalent to $p \rightarrow q$, as desired.

Therefore, from $\lnot (p \land r)\equiv \lnot p \lor \lnot r$ we can conclude $(q\rightarrow q)$.

I'll leave this to you to write formally (including keeping proper track of temporary assumptions), with proper justifications.

• I just saw your edit. I think you're thinking along the right lines. – amWhy Feb 14 '14 at 14:43
• Thank you, can you please clarify the following if you have a chance: Now, we already derived ¬s. This means ¬s∨t is true. Why? But ¬s∨t≡¬(s∧¬t). – DrJonesYu Feb 14 '14 at 14:44
• Becuase if $\lnot s$ is true, $\lnot s \lor$ ... "something else" is certainly true (truth-table for $\lor$); so, we use as "something else" $t$. – Mauro ALLEGRANZA Feb 14 '14 at 18:06

Solution

• 1: (r∧¬s)∨(q∧¬s) Premise
• 2: ¬s→((p∧r)→u) Premise
• 3: u→(s∧¬t) Premise
• 4: (r∨q)∧ ¬s Distribution 1
• 5: ¬s Simplification 4
• 6: (p∧r)→u Modus Ponens 2, 5


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# Probability Densities

In the last section, we discussed centers of mass, which we interpreted as a weighted average of x coordinates. In this section, we will also encounter a weighted average but now within the context of probability.

The average of a quiz

First, we will consider a discrete version of our topic which is familiar to all students. Suppose that 10 students take a quiz which is marked out of a maximum of 5 marks. How can we compute the average score?

 Student Number Score 1 2 3 4 5 6 7 8 9 10 2 5 4 2 1 0 2 4 5 4

To compute the average score , we can add all the scores together and divide by the total number of scores which is 10. Once we write this, however, it becomes clear that we can collect all the common scores together.

In this last line, represents the number of students who earn the mark s . This shows us a slightly different way to compute averages. In this case, the average score is given by averaging the possible scores weighted by the number of students who earn that score. We can now write down some simple observations which will be useful for later.

Observations:

1. This is because if we add up the number of students who earn each possible score, we are just counting the number of students in the class.

2. In the same way, represents the number of students who score between and , inclusive.

In fact, it will be useful for us to consider a slightly different way of recording the information. Define . This represents the ratio of the number of students who score s to the total number of students. In other words, it gives us the fraction of students who score s . This new quantity satisfies properties similar to those above.

1. .

2. equals the fraction of students who score between and .

We will call such a function a probability density which describes the results of the quiz. This name comes from the fact that property 2 above tells us how to measure the probability that a typical student earns a mark in a particular range.

To describe what we have found in rough terms, we could say that to find the average value of a list of numbers, we compute

The Continuous Case

In the example above, the possible scores on the quiz are all integers and so form a discrete set. We would like to apply these ideas to the case where the possible values vary in a continuous way. Before we jump into a lot of definitions, let's think about an example.

Suppose that we have a radioactive substance with a half-life of 1000 years. From our investigations last term, we know that the mass of this substance is described by an exponentially decaying function: that is, where is the original mass and k is the constant

Our goal will be to determine the average length of time it takes a particle to decay. In fact, let's be a bit more specific just for the purposes of this example and ask the following:

Question: Of the mass which decays in the first 1000 years, what is the average time of decay?

We can answer this question by employing the same methods as for our quiz scores. In particular, we can write:

It will be helpful to introduce a function which describes how much mass has decayed in the first t years. We can understand this since , the amount which has decayed in t years, equals the original amount minus the amount left after t years. That is, .

This gives us our first piece of information: the total mass that decays in 1000 years is (this shouldn't be too surprising since the half life is 1000 years).

Now we will divide up the interval from 0 to 1000 years into smaller pieces labeled by .

Now we will ask the question: how much mass decays during the time interval . This can be easily answered in terms of our function . In particular, the amount decaying in this time interval is the amount which has decayed in the first years minus the amount which has decayed in the first years. In other words, the amount which decays in the time interval is

We can make an approximation which will help us: in this time interval, we know how much mass decays. The problem is that it could be decaying at different times within the time interval. We will make the approximation, however, that all the mass which decays in this interval decays at time . Of course, our ultimate aim is to shrink the width of these time intervals and the approximation improves when this is done.

We can now write the approximate average value as

This looks like it could be describing an integral, but there is no factor of . That's all right though because we can simply put it in by multiplying and dividing:

Notice that as we shrink the width of the intervals, and so

We can then evaluate

which means that

This is a wonderfully useful relationship and we will come back to it shortly. But now, let's finish our task by evaluating the integral.

Since , it follows that . The integral we need to evaluate is then

which can be evaluated by integration by parts. To do this, we will set and . Then we have and .

In this computation, we have used the fact that . This leads to

If we think about it, this result feels right. Remember that we are only considered the mass that decays in the first 1000 years. Also, more mass decays during the early years when there is more of the substance than decays in the later years when there is less. So it is not surprising that the result should be smaller than the halfway value of 500 years.

Probability Densities

Now that we have finished our calculuation, let's look at it a little more carefully. Remember that represented the total mass that decays in the first t years and so is the total mass that decays in 1000 years. Then we found that

We will define a new function and call it the probability density for the process. In fact, it is very similar to the analogous quantity we defined for the students' scores on the quiz above. In particular, it has the following properties.

Properties of the Probability Density Function:

1. is the fraction of mass which decays in the time interval .

This is because

This is the fraction of mass which decays during this time interval. In fact, this is why the function is called the probability density: given a little interval of width around time t , the chance that a typical particle decays during that time interval is .

2. We sometimes say that the probability density is normalized to have a total integral (or total probability) equal to 1 . Using our first observation, this is just saying that the probability is 1 (which means that it must happen) that a particle which decays during the first 1000 years actually decays during the first 1000 years.

We call the function the cumulative distribution. It is measuring what fraction of the mass decays during the first t years.

The demonstration below shows the relationship between the probability density function and the cumulative distribution.

What can we learn from the probability density

Let's consider what kind of information is available from the probability density.

Examples

We have already used the probability density to find the average time of decay. In fact, we can do a bit more.

1. Suppose we want to know the probability that a typical particle decays in the first 500 years. The information is available to us as

In other words, there is a 58.6% chance that a typical particle decays in the first 500 years. This makes sense since we know that the mass is decaying more rapidly at the beginning.

2. The median time for decay is the time for which the cumulative distribution . This means that a typical particle is just as likely to decay before this time as it is to decay after this time. In other words, half the particles decay before this time and half decay after.

We can find the median time by setting This means that

Below is a graphical interpretation of the median. The two shaded areas are equal at the median value.

It is interesting to ask why the median is different from the average value. Basically, the median does not detect information about the shape of the distribution other than when the two areas are balanced. The average, for this distribution, will be pulled up to a value higher that the median because there are some particles decaying for relatively long times.


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Home » » If $$\frac{1}{p}$$ = $$\frac{a^2 + 2ab + b^2}{a - b}$$ and $$\frac{l}{q}$$ = $$\frac{a + b}{a^2 - 2ab + b^2}$$ Find $$\frac{p}{q}$$...

If $$\frac{1}{p}$$ = $$\frac{a^2 + 2ab + b^2}{a - b}$$ and $$\frac{l}{q}$$ = $$\frac{a + b}{a^2 - 2ab + b^2}$$ Find $$\frac{p}{q}$$...

Question

If $$\frac{1}{p}$$ = $$\frac{a^2 + 2ab + b^2}{a - b}$$ and $$\frac{l}{q}$$ = $$\frac{a + b}{a^2 - 2ab + b^2}$$ Find $$\frac{p}{q}$$

Options

A) $$\frac{a + b}{a - b}$$

B) $$\frac{1}{a^2 - b^2}$$

C) $$\frac{a - b}{a + b}$$

D) a2 - b2


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# ISI MStat PSB 2013 Problem 4 | Linear Regression

This is a sample problem from ISI MStat PSB 2013 Problem 4. It is based on the simple linear regression model, finding the estimates, and MSEs. But think over the "Food for Thought" any kind of discussion will be appreciated. Give it a try!

## Problem- ISI MStat PSB 2013 Problem 4

Consider n independent observation $${ (x_i,y_i) :1 \le i \le n}$$ from the model

$$Y= \alpha + \beta x + \epsilon$$ ,

where $$\epsilon$$ is normal with mean 0 and variance $$\sigma^2$$ . Let $$\hat{\alpha}, \hat{\beta}$$ and $$\hat{\sigma}^2$$ be the maximum likelihood estimators of $$\alpha , \beta$$ and $$\sigma^2$$ , respectively. Let $$v_{11}, v_{22}$$ and $$v_{12}$$ be the estimated values of $$Var(\hat{\alpha}), Var(\hat{\beta}$$ and $$Cov ( \hat{\alpha}, \hat{\beta})$$, respectively.

(a) What is the estimated mean of Y, when when $$x=x_o$$ ? Estimate the mean squared error of this estimator .

(b) What is the predicted value of Y, when when $$x=x_o$$ ? Estimate the mean squared error of this predictor .

### Prerequisites

Linear Regression

Method of Least Squares

Maximum likelihood Estimators.

Mean Squared Error.

## Solution :

Here for the given model,

we have , the random errors, $$\epsilon \sim n(0, \sigma^2)$$, and the maximum likelihood estimators (MLE), of the model parameters are given by $$\hat{\alpha}, \hat{\beta}$$ and $$\hat{\sigma}^2$$. The interesting thing about this model is, since the random errors $$\epsilon$$ are Gaussian Random variables, the Ordinary Least Square Estimates of the model parameters $$\alpha, \beta$$ and $$\sigma^2$$, are identical to their Maximum Likelihood Estimators, ( which are already given!). How ?? Verify it yourself and once and remember it henceforth.

So, here $$\hat{\alpha}, \hat{\beta}$$ and $$\hat{\sigma}^2$$ there also the OLS estimates of the model parameters respectively.

And By Gauss-Markov Theorem, the OLS estimates of the model parameters are the BLUE (Best Linear Unbiased Estimator), for the model parameters. So, here $$\hat{\alpha}, \hat{\beta}$$ and $$\hat{\sigma}^2$$ are also the unbiased estimators of $$\alpha, \beta$$ and $$\sigma^2$$ respectively.

(a) Now we need to find the estimated mean Y given $$x=x_o$$ ,

$$\hat{ E( Y| x=x_o)}= \hat{\alpha} + \hat{\beta} x_o$$ is the estimated mean of Y given $$x=x_o$$.

Now since, the given MLEs ( OLSEs) are also unbiased for their respective parameters,

$$MSE( \hat{ E( Y| x=x_o)})=MSE(\hat{\alpha} + \hat{\beta} x_o)=E(\hat{\alpha} + \hat{\beta} x_o-(\alpha + \beta x_o))^2$$

=$$E(\hat{\alpha} + \hat{\beta} x_o-E(\hat{\alpha} + \hat{\beta} x_o))^2$$

=$$Var( \hat{\alpha} + \hat{\beta} x_o)$$

= $$Var(\hat{\alpha}+2x_o Cov(\hat{\alpha}, \hat{\beta})+ {x_o}^2Var(\hat{\beta})$$

So, $$.MSE( \hat{ E( Y| x=x_o)})= v_{11} +2x_o v_{12} + {x_o}^2 {v_{22}}$$.

(b) Similarly, when $$x=x_o$$ , the predicted value of Y would be,

$$\hat{Y} = \hat{\alpha} + \hat{\beta} x_o +\epsilon$$ is the predicted value of Y when $$x=x_o$$ is given.

Using similar arguments, as in (a) Properties of independence between the model parameters , verify that,

$$MSE(\hat{Y})= v_{11}+ 2x_o v_{12} + {x_o}^2{ v_{22}}+{\hat{\sigma}^2}$$. Hence we are done !

## Food For Thought

Now, can you explain Why, the Maximum Likelihood Estimators and Ordinary Least Square Estimates are identical, when the model assumes Gaussian errors ??

Wait!! Not done yet. The main course is served below !!

In a game of dart, a thrower throws a dart randomly and uniformly in a unit circle. Let $$\theta$$ be the angle between the line segment joining the dart and the center and the horizontal axis, now consider Z be a random variable. When the thrower is lefty , Z=-1 and when the thrower is right-handed , Z=1 . Assume that getting a Left-handed and Right-handed thrower is equally likely ( is it really equally likely, in real scenario ?? ). Can you construct a regression model, for regressing $$\theta$$ on Z.

Think over it, if you want to discuss, we can do that too !!


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# Properties of Betti number of ideal

Notations: $R$- Noetherian graded ring and $I,J$ homogeneous ideals in $R$

Definition: The projective dimension of $R/I$, denoted $pd(R/I)$, is the length of a minimal free graded resolution of $R/I$: $0 \rightarrow \bigoplus_{j} R(-j)^ {\beta_{p,j}(R/I)}\rightarrow \bigoplus_{j} R(-j)^ {\beta_{p-1,j}(R/I)} \rightarrow \cdots \rightarrow \bigoplus_{j} R(-j)^ {\beta_{0,j}(R/I)} \rightarrow R/I \rightarrow 0$.

where $p \leq n$ and $R(-j)$ indicates the ring $R$ with the shifted graduation such that, for all $a \in \mathbb{Z}$, $R(-j)_{a}=R_{a-j}$ and $\beta_{i,j}(R/I)$ is the number of copies of $R(-j)$ appearing in the $i$-th module of the resolution, and is called graded Betti number degree $(i,j)$. The $i$-th Betti number is $\beta_{i}(R/I)=\sum \limits_{j \in \mathbb{Z}}\beta_{i,j}(R/I)$. The Castelnuovo-Mumford regularity (or simply regularity) of $R/I$ is

$\operatorname{reg}(R/I)=\max\{j-i\mid\beta_{i,j}(R/I)\neq 0\}$

Let $I$ be quadratic square free monomial ideal in $k[x_1,\ldots,x_n]$ and $J$ be quadratic square free monomial ideal in $k[y_1,\ldots,y_m]$ with $I\ncong J$. Suppose $\beta_{i,2i}(I) \neq 0$ and $\beta_{j,2j}(J) \neq 0$ for some $i,j$. Let $IJ$ be square free monomial in $k[x_1,\ldots,x_n,y_1,\ldots,y_m]$.

Can we say that $\beta_{(i+j+2),2(i+j+2)}(IJ)\neq 0$ ?

I believe the answer to your question is no, we cannot conclude that $\beta_{(i+j+2), 2(i+j+2)}(IJ)\neq0$.
For example, let us take $n=m=4$ and $I$ and $J$ to be the following ideals $$I = \langle x_1x_2, x_2x_3\rangle \subset \mathbb{Q}[x_1, x_2, x_3, x_4]=R$$ $$J = \langle y_1y_2, y_3y_4\rangle \subset \mathbb{Q}[y_1, y_2, y_3, y_4]=S.$$ Both of these are generated by square-free monomials and by considering dimensions we see $R/I$ is not isomorphic to $S/J$. (I assume this is what you mean by $I\not\cong J$, sorry if I missed interpreted.) Using Macualay2 to compute the Betti tables of $I$, $J$, and $IJ$ we get:
$$b(I)= \begin{array}{c | c c c c c} - & 0 & 1 & 2 \\ \hline T & 1 & 2 & 1 \\ 0 & 1 & - & - \\ 1 & - & 2 & 1 \\ \end{array} \quad b(J)= \begin{array}{c | c c c c c} - & 0 & 1 & 2 \\ \hline T & 1 & 2 & 1 \\ 0 & 1 & - & -\\ 1 & - & 2 & - \\ 2 & - & - & 1 \end{array} \quad b(IJ)= \begin{array}{c | c c c c c} - & 0 & 1 & 2 & 3 \\ \hline T & 1 & 4 & 4 & 1 \\ 0 & 1 & - & - & - \\ 1 & - & - & - & - \\ 2 & - & - & - & - \\ 3 & - & 4 & 2 & - \\ 4 & - & - & 2 & 1 \end{array}$$
In particular, we see that: $$b_{1,1}(I)=\beta_{1,2}(I)=2, \quad \quad b_{1,1}(J)=\beta_{1,2}(J)=2, \quad \quad \text{and} \quad\quad b_{2,2}(J)=\beta_{2,4}(J)=1.$$ So we may take $i=1$ and $j=1,2$, but in either case: $$\beta_{1+1+2, 2(1+1+2)}(IJ)=b_{4,4}(IJ)=0 \quad \quad \text{and} \quad \quad \beta_{1+2+2, 2(1+2+2)}(IJ)=b_{5,5}(IJ)=0.$$ Out of curiosity what lead you to believe this might be true? (I don't have a lot of intuition for this, and found the above by just messing mourned in Macaulay2.)


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# Covering a set of points by bounded geometric object/objects

1) Let $S$ be a set of $n$ points in $R^d$. Now, given a bounded geometric object $G$, the problem is to check whether $S$ can be contained in $G$.

2) Also, in general setting, the problem is to check whether $S$ can be contained in $k$ translated copies of $G$.

For what values of $n$ and $d$, and for which type of $G$, solution of (1) and (2) are known?

Pls let me know if the question is clear?

-


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If you find any mistakes, please make a comment! Thank you.

## Determine whether or not a subset is an ideal

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.8

Solution:

(1) Note that $(1,1) \in D$. However, $(1,0)(1,1) = (1,0) \notin D$. Since $D$ does not absorb $R$, it is not an ideal.

(2) We claim that S is an ideal. To that end, we need to show that $S$ is a subring and absorbs $R$. (Absorption on one side is sufficient since $R$ is commutative.)

Let $(2a_1,2b_1)$, $(2a_2,2b_2) \in S$ and $(x,y) \in R$.

i) $(0,0) \in S$, so that $S$ is not empty.

ii) $(2a_1,2b_1) - (2a_2,2b_2) = (2(a_1-a_2), 2(b_1-b_2)) \in S$, so that $S$ is an additive subgroup.

iii) $(2a_1,2b_1)(2a_2,2b_2) = (4a_1a_2,4b_1b_2) \in S$, so that $S$ is a subring.

iv) $(2a_1,2b_1) (x,y) = (2a_1x,2b_1y) \in S$, so that $S$ absorbs $R$.

(3) We claim that $T$ is an ideal. To that end, we need to show that $T$ is a subring and absorbs $R$. (Absorption on one side is sufficient since $R$ is commutative.)

Let $(2a_1,0), (2a_2,0) \in T$ and $(x,y) \in R$.

i) $(0,0) \in T$, so that $T$ is not empty.

ii) $(2a_1,0) - (2a_2,0) = (2(a_1-a_2), 0) \in T$, so that $T$ is an additive subgroup.

iii) $(2a_1,0)(2a_2,0) = (4a_1a_2,0) \in T$, so that $T$ is a subring.

iv) $(2a_1,0)(x,y) = (2a_1x,0) \in T$, so that $T$ absorbs $R$.

(4) Note that $(1,-1) \in U$. However, $(1,-1)(1,-1) = (1,1) \notin U$. Since $U$ is not closed under multiplication, it is not even a subring.

#### Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.


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## Circles, Squares, and Parabolas

If you take two circles, you can position them before your right eye in such a way that one will perfectly overlap the other.

In other words, a circle is geometrically similar to all other circles.

The same is true of squares.

However, the same is not true of rectangles, as you may have noticed if you ever tried to perfectly frame a painting with your camera and found that no matter how you positioned yourself, you’d either crop it or include some of its surroundings.  (If you’ve done this, but didn’t have a problem, count yourself lucky!) The non-similarity of rectangles caused trouble when people wanted to watch movies on old-fashioned TVs.  Do you squeeze everyone onto the screen, crop things off, or letterbox it?

So rectangles aren’t like circles and squares with respect to similarity.

But, parabolas are!

If this surprises you, it’s easy to understand why, especially if you’ve seen some parabolas, like the three below.

These three parabolas sure don’t look similar to each other.

Well, looks can be deceiving, because it’s true: All parabolas are similar to each other! Let’s prove this. (And by the way, when I say “parabola,” I mean the entire parabola…not just a section of it, just as earlier I meant the whole circle, the whole square, and the whole rectangle.)

Consider two parabolas.  If we can use translation, rotation, and dilation to superimpose one precisely over the other, then we’re done.  Let’s first translate both parabolas so that they have the same apex.  We’ll introduce xy coordinate axes and use the apex of these two parabolas as our origin.  Next, rotate the parabolas so that their axes are both aligned with the y-axis and so that they both open upward.  At this point, the equations of the two parabolas are given by $y = ax^2$ and $y = bx^2$, where $a$ and $b$ are positive constants.

Now we’re going to dilate the parabola corresponding to $y = ax^2$.  If we dilate it by a factor of $d$ (and with center of dilation at the origin), the resulting curve will consist of the set of points $(dx, dy)$ where $x$ and $y$ satisfy $y = ax^2$.  Let $x' = dx$ and $y' = dy$.  Notice that $y' = \frac{a}{d}x'^2$. This shows that when we dilate the parabola $y = ax^2$, the result is also a parabola.

Since $b$ is positive, we can pick $d$ to be $\frac{a}{b}$.  For this value of $d$, the parabola $y = ax^2$ dilates to the parabola $y' = bx'^2$, which is what we wanted to show!

And so we can deduce that the shallow green parabola in the illustration above is the view you would get if you looked really closely at the apex of the blue parabola (and it changed color).  Try it and see for yourself.


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## Apache Error Log

Hi,

NW6.5 SP3 + OS3A + NSS3A

Hope this is the right part of the forum for this. I am getting the
following error repeated in the Apache error log file. It is repeated
thousands of times over:

[Sun Aug 14 09:59:59 2005] [error] (OS 10022)Invalid argument:
apr_accept: (client socket)

The file grew to nearly 200Mb until I unloaded apache. Web services
seemed to be working ok on the server although utilization was high. I
have no idea of the meaning of the error. Does anyone have any ideas.

Many Thanks,

David