text:
American Institute of Mathematical Sciences

October  2012, 8(4): 861-875. doi: 10.3934/jimo.2012.8.861

On the optimal and equilibrium retrial rates in an unreliable retrial queue with vacations

 1 Department of Mathematics, Beijing Jiaotong University, 100044 Beijing 2 Department of Mathematics, University of Northern Iowa, Cedar Falls, IA 50614-0506, United States

Received  September 2011 Revised  July 2012 Published  September 2012

A single-server retrial queue with two types of customers in which the server is subject to vacations along with breakdowns and repairs is studied. Two types of customers arrive to the system in accordance with two different independent Poisson flows. The service times of the two types of customers have two different independent general distributions. We assume that when a service is completed, the server will take vacations after an exponentially distributed reserved time. It is assumed that the server has an exponentially distributed lifetime, a generally distributed vacation time and a generally distributed repair time. There is no waiting space in front of the server, therefore, if the server is found busy, or on vacation, or down, the blocked two types of customers form two sources of repeated customers. Explicit expressions are derived for the expected number of retrial customers of each type. Additionally, by assuming both types of customers face linear costs for waiting and retrial, we discuss and compare the optimal and equilibrium retrial rates regarding the situations in which the customers are cooperative or noncooperative, respectively.
Citation: Feng Zhang, Jinting Wang, Bin Liu. On the optimal and equilibrium retrial rates in an unreliable retrial queue with vacations. Journal of Industrial & Management Optimization, 2012, 8 (4) : 861-875. doi: 10.3934/jimo.2012.8.861
References:

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References:
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## L5 Class 11 Maths Sequences and Series

“The Best Leaders Have A High Consideration Factor. They Really Care About Their People”

# Lecture - 5 Chapter 9 Sequences & Series

Meaning of Geometric Progression (GP) and its common ratio, General Term of Geometric Progression (nth term of GP), Derivation for Sum to n terms of a GP, Geometric Mean and its relationship with Geometric Progression, Rules Comparison of Arithmetic Progression and Geometric Progression

 NCERT EXERCISE 9.3

Question 1. Find the 20th and nth terms of the G.P. $$\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, …$$

Question 2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

Question 3. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that $$q^2=ps$$.

Question 4. The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7th term.

Question 5. Which term of the following sequences:
(a) $$2, 2\sqrt{2}, 4, …$$ is 128?
(b) $$\sqrt{3}, 3, 3\sqrt{3}, …$$ is 729?
(c) $$\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, …$$ is $$\frac{1}{19683}$$?

Question 6. For what values of x, the numbers $$-\frac{2}{7}, x, -\frac{7}{2}$$ are in G.P.?

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:

Question 7. 0.15, 0.015, 0.0015, … 20 terms.

Question 8. $$\sqrt{7}, \sqrt{21}, 3\sqrt{7}, …$$ n terms.

Question 9. $$1, -a, a^2, -a^3, … n \text{ terms } (\text{ if } a \ne -1)$$.

Question 10. $$x^3, x^5, x^7, … n \text{ terms} (\text{ if } x \ne \pm 1)$$


text:
NORMAL OR GAUSSIAN DISTRIBUTION Chapter 5. General Normal Distribution Two parameter distribution with a pdf given by:

Presentation on theme: "NORMAL OR GAUSSIAN DISTRIBUTION Chapter 5. General Normal Distribution Two parameter distribution with a pdf given by:"— Presentation transcript:

NORMAL OR GAUSSIAN DISTRIBUTION Chapter 5

General Normal Distribution Two parameter distribution with a pdf given by:

Estimators for 1 and 2 Using MOM or Maximum Likelihood to estimate 1 and 2 :

Properties of Normal Distribution Bell-shaped Continuous Symmetrical about the mean

Properties of the Normal Distribution Mean and variance are sometimes referred to as location and scale parameters. Varying 2 while holding constant changes the scale but not the location of the distribution. Varying while holding 2 constant changes the location of the distribution but not the scale.

Notation Common to denote a random variable that is normally distributed with a mean, and variance 2 as N(, 2 ).

Reproductive Properties of the Normal Distribution If X is a r.v., and, then the distribution of Y can be shown to be If X i for i = 1,2,…n are independently and normally distributed with mean i and variance i 2, then is normally distributed with:

Example If x i is a random observation N(, 2 ) what is the distribution of ?

Standard Normal Distribution The cdf for the normal distribution is: To evaluate we must use a linear transformation so that the random variable will be N(0,1).

Standard Normal Distribution When we make this transformation, Z is said to be standardized and N(0,1) is the standard normal distribution. The standard normal pdf and cdf are:

Using the Standard Normal Distribution Probabilities for the standard normal distribution are widely tabulated. Most tables make use of symmetry and show only positive values of Z. Tables may show: Your book has standard normal distribution in Appendix A.12. Gives prob(Z { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/7/1651475/slides/slide_11.jpg", "name": "Using the Standard Normal Distribution Probabilities for the standard normal distribution are widely tabulated.", "description": "Most tables make use of symmetry and show only positive values of Z. Tables may show: Your book has standard normal distribution in Appendix A.12. Gives prob(Z

Standard Normal Distribution The table shows that 68.26% of normal distribution is within 1 of the mean. 95.44% of normal distribution is within 2 of the mean. 99.74% of normal distribution is within 3 of the mean. These are the 1, 2, and 3 sigma bounds of the normal distribution.

Standard Normal Distribution Only 0.26% of the area of the normal distribution lies outside the 3 rd sigma bound. The probability of a value less than is only 0.0013. Since many of our r.v. are bounded at the left by x=0, this gives us justification for using the normal distribution in instances where is greater than because the chance that X < 0 is many times (but not always) negligible.

Example NO3-N concentrations (ppm) in a local well are N(3,0.2). a) What is the probability that a sample drawn at random will be between 2 and 7 ppm. b) What is the probability that a sample will have a concentration exceeding 10 ppm?

Central Limit Theorem If S n is the sum of n independently and identically distributed random variables X i each having a mean,, and variance, 2, then in the limit as n approaches infinity, the distribution of S n approaches a normal distribution with mean n and variance n 2.

Normal Approximations for Other Distributions Under certain conditions the normal distribution can be a good approximation to several other distributions, both discrete and continuous. Generally approximations are good in the center of the distribution with accuracy dropping off in the tails. C.L.T. provides the mechanism by which the normal distribution becomes an approximation for several other distributions.

Discrete Distribution Approximations Whenever a continuous distribution is used to approximate a discrete distribution, ½ interval corrections must be applied to the continuous distribution. Why?

½ Interval Corrections Table 5.2. Corrections for approximating a discrete random variable by a continuous random variable. DiscreteContinuous X = xX-1/2 < X < X+1/2 x < X < yX-1/2 < X < y+1/2 X < xX < x+1/2 X > xX > x-1/2 X < xX < X-1/2 X > xX > x+1/2

Binomial Distribution Additive property of the binomial If X is a binomial r.v. with parameters n 1 and p; and Y is a binomial r.v. with parameters n 2 and p, then Z = X + Y is a binomial r.v. with parameters n=n 1 +n 2 and p. Extending this to the sum of several binomial random variables, the C.L.T. indicates that the normal would approximate the binomial distribution if n is large.

Binomial Distribution Thus, as n get large: approaches a N(0,1).

Example X is a binomial r.v. with n= 50 and p= 0.2. Compare the normal approximation to the binomial fro evaluating prob(8 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/7/1651475/slides/slide_21.jpg", "name": "Example X is a binomial r.v. with n= 50 and p= 0.2.", "description": "Compare the normal approximation to the binomial fro evaluating prob(8

Negative Binomial Distribution Recall the additive property of the negative binomial: If X & Y are discrete r.v with f x (x;k 1,p) and f y (y;k 2,p) then for Z=X+Y, f z (z,k 1 +k 2,p). A negative binomial distribution with large k can be approximated by the normal distribution. approaches N(0,1) as k gets large.

Example Use both the negative binomial and the normal approximation to the negative binomial to find the probability that the 4 th occurrence of a 10 year flood will occur in the 40 th year?

Poisson Distribution Sum of 2 Poisson r.v. with parameters 1 and 2 is also a Poisson r.v. with parameter = 1 + 2. Using C.L.T. For the sum of a large number of Poisson r.v. for large, Poisson may be approximated by which approaches N(0,1).

Continuous Distributions Many continuous distributions can be approximated by the normal distribution for certain value of their parameters. To make these approximations, the mean and variance of the distribution to be approximated are equated to the mean and variance of the normal. is N(0,1) if X is

Continuous Distributions Not all continuous distributions can be approximated by the normal. Things to look for are: 1. parameters that produce near zero skew 2. parameters that produce symmetry 3. parameters that produce distributions with tails that asymptotically approach p x (x) = 0 as X approaches large and small values.

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# how to solve this problem?

• May 20th 2009, 08:21 AM
chris.w
how to solve this problem?
Attachment 11506
• May 20th 2009, 08:54 AM
Kaitosan
1. I believe it's zero because e-e=0....

There's a segment of the series that equals e.

I don't know about 2. though.
• May 20th 2009, 10:02 AM
TwistedOne151
I'm prette sure the first sum diverges. One can check that
$e-\left(1+\frac{1}{k}\right)^k\gt\frac{1}{k}$ for integer $k≥3$.
Now, the harmonic series
$\sum_{n=1}^{\infty}\frac{1}{n}$ diverges, and so
$\sum_{n=3}^{\infty}\frac{1}{n}$ diverges as well.
but since $e-\left(1+\frac{1}{k}\right)^k\gt\frac{1}{k}$ for $k≥3$
then
$\sum_{n=3}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right]\gt\sum_{n=3}^{\infty}\frac{1}{k}$
and so by the comparison test, both sums diverge, and since
$\sum_{n=1}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right]=(e-2)+(e-\frac{9}{4})+\sum_{n=3}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right]$,
your series in problem 1 diverges.

--Kevin C.


text:
Display Abstract
 Title Urban social dynamics: Don't buy the house, buy the neighborhood''
 Name Jean-Pierre Nadal Country France Email nadal@lps.ens.fr Co-Author(s) Laetita GAUVIN and Annick VIGNES Submit Time 2012-03-13 11:51:15 Session Special Session 3: Mathematics of Social Systems
 Contents Don't buy the house, buy the neighborhood'' (Russian proverb). This talk is concerned with issues related to urban social segregation. Going beyond the simplest models introduced by T. C. Schelling in the 70's, we introduce a model of housing transactions between agents heterogeneous in their income, hence in their willingness to pay. The goal is to see how the spatial income segregation depends on both economic constraints and social interactions. A key feature of the model is that agents preferences for a place are expressed in terms of a local attractiveness, which depends on both an intrinsic attractiveness of the location and the social characteristics of its neighborhood. The demand for an asset thus depends on its attractiveness, which in turn evolves according to the social characteristics of the newcomers. The resulting stationary space distribution of income is analytically characterized and illustrated by agent-based simulations. The main results are that, (1), socio-spatial segregation occurs if, and only if, the social influence is strong enough, and, (2), even so, some social diversity is preserved at most locations. Comparing with the Parisian housing market, the results reproduce general trends concerning the price distribution and the income spatial segregation.


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# Hooks in a staircase partition: Part I

This quest has its impetus in a paper by Stanley and Zanello. I became curious about

What is the sum of all hooks lengths of all partitions that fit inside the $n$-th staircase partition?

On the basis of experimental data, I'm prompted to ask:

Question. Let $\lambda=(n,n-1,\dots,1)$ be the staircase partition, $h_{\square}$ the hook-length of a cell $\square$ in the Young diagram of a partition. Then,
$$\sum_{\mu\subset\lambda}\sum_{\square\in\mu}h_{\square}=\frac{(n+4)(2n+3)}6\binom{2n+2}{n+1} - (n+2) 2^{2n+1}.$$ Is this true? If so, any proof?

Update. I'm still hoping for a complete solution for this evaluation; it's not common to get such a nice closed from in general. Part II of this problem has received a fine answer.

Update. To help Stanley's mention of a recurrence, denote the RHS quantity by $w_n:=\frac{(n+4)(2n+3)}6\binom{2n+2}{n+1} - (n+2) 2^{2n+1}$ then we have ("shaloshable") $$n(n-2)w_n-2(4n^2-5n-3)w_{n-1}+8n(2n-1)w_{n-2}=0.$$

The following method should give a proof, but I haven't done the computation. Let $f_n$ denote the desired sum. I use the result of Douglas Zare's answer that $$f_n = \sum_{\mu\subseteq \delta_n} \sum_i \mu_i^2,$$ where $\delta_n=(n,n-1,\dots,1)$. Let $$g_n = \sum_{\mu\subseteq \delta_n} \sum_i \mu_i.$$ By OEIS A029760 we have $$g_n = \frac 12 (n+2)^2C_n-2^{2n+1},$$ where $C_n$ is a Catalan number.

We now break up the sum for $f_n$ into the following parts:

(1) No $\mu_i=n-i+1$. In other words, $\mu\subseteq\delta_{n-1}$, so the contribution from these terms is $f_{n-1}$.

(2) $\mu_1=n$. Below the first row is any partition $\nu\subseteq \delta_{n-1}$, so the contribution from these terms is $f_{n-1}+n^2C_n$ since there are $C_n$ choices for $\nu$.

(3) $\mu_k=n-k+1$ for some $k\geq 2$, and $\mu_j<n-j+1$ for $j<k$. The first $k-2$ rows of $\mu$ consist of a $(k-2)\times(n-k+1)$ rectangle with a partition $\nu\subseteq\delta_{k-2}$ on the right. Rows $k-1$ and $k$ of $\mu$ have length $n-k+1$. The remaining rows ($k+1$ to $n$) consist of a partition $\rho\subseteq \delta_{n-k}$. Each $\nu$ occurs $C_{n-k+1}$ times (the number of choices for $\rho$), and each $\rho$ occurs $C_{k-1}$ times. Hence the contribution to the sum for $f_n$ is $$C_{n-k+1}\sum_{\nu\subseteq\delta_{k-2}}\sum_i (\nu_i+n-k+1)^2+ C_{k-1}\sum_{\rho\in\delta_{n-k}} \sum_i \rho_i^2+2C_{n-k+1}C_{k-1}(n-k+1)^2$$ $$= C_{n-k+1}(f_{k-2}+2(n-k+1)g_{k-2}+C_{k-1}(k-2)(n-k+1)^2)+ C_{k-1}f_{n-k}+2C_{n-k+1}C_{k-1}(n-k+1)^2.$$

Summing over all possibilities gives a recurrence satisfied by $f_n$. It should be routine (i.e., "shaloshable") to check that the conjectured answer satisfies the recurrence. Perhaps there is some inaccuracy in the details above, but I think that the general idea should give a solution.

• Richard Stanley: I've posted an update to give a recurrence for the RHS of the quantity on the RHS of my question. – T. Amdeberhan May 14 '17 at 3:18

I think this is progress.

The hook length of square $S$ is the count of the squares $R$ directly below or to the right of $S$ plus 1. So, in a particular partition, the sum of the hook lengths is the count of the pairs $(S,R)$. It's simpler to count these pairs by $R$. The number of such pairs is the sum over all squares $R=(x,y)$ of $(x+y-1)$ (if you number the rows starting with $1$).

$$\sum_{\mu \subset \lambda} \sum_{(x,y)\in \mu} (x+y-1)$$

Since the staircase is self-conjugate, the sum of $x$ is the same as the sum of $y$, so we can rewrite this as

$$\begin{eqnarray}2\sum_{\mu \subset \lambda} \sum_{(x,y) \in \mu} x - \sum_{\mu \subset \lambda} |\mu| &=& \sum_{\mu \subset \lambda} \sum_i (\mu_i^2+\mu_i) - \sum_{\mu \subset \lambda} |\mu| \newline &=& \sum_{\mu \subset \lambda} \sum_i \mu_i^2\end{eqnarray}$$

where $\mu_i$ is the size of the $i$th part of $\mu$.

Counting partitions $\mu$ inside a staircase $\lambda$ so $\mu_i=j$ can be done with two applications of reflection.


text:
# Columns environment doesn't scale to full paper size tikzposter (using custom paper size)

I am trying to create my poster using the TikzPoster class. I found this answer that explains how to set custom size: How can i change the paper size in tikzposter?. I have set it to 152.4cm by 106.68cm (60in x 42in). But, I have found that the columns environment does not scale to use the full width of the paper, although the title block does scale correctly (I an passing it 0.98\textwidth).

Here is my code:

\documentclass[17pt, a0paper, landscape]{tikzposter}
\geometry{paperwidth=42in, paperheight=60in}

\usepackage{xpatch}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage[space]{grffile}

\title{\parbox{0.95\linewidth}{\centering \textbf{My Poster Title}}}

{\setlength{\tabcolsep}{2em}
\author
{%
\begin{tabular}{ c | c | c }
\textbf{Author 1*} \emph{Author position 1} \hfill & \hfill \textbf{Author 2} \emph{Author position 2} \hfill & \hfill \textbf{Author 3} \emph{Author Position 3} \\
\texttt{Auth1@univ.edu; Auth1Phone} \hfill & \texttt{Auth2@univ.edu; Auth2Phone} \hfill & \texttt{Auth3@univ.edu; Auth3Phone} \\
\end{tabular}%
}

\institute
{
\textbf{Affiliated University and Department}\\
\vspace*{0.2em}
}

\titlegraphic
{
\raisebox{2.25cm}{\includegraphics[width=11cm,height=3cm]{/Poster/Logo1.png}}
\hfill
\includegraphics[width=7.5cm,height=7.5cm]{/Poster/Logo2.png}
}

\makeatletter
\def\TP@titlegraphictotitledistance{-7.5cm}
\settitle
{
\centering
\vbox
{
\@titlegraphic \\ [\TP@titlegraphictotitledistance]
\centering
\color{titlefgcolor}
{\bfseries \huge \sc \@title \par}
\vspace*{1em}
{\LARGE \@author \par}
\vspace*{1.2em}
{\LARGE \@institute}
}
}
\makeatother

\begin{document}
\maketitle[width=0.98\textwidth]
\block{First}{Check Span}
\begin{columns}
\column{0.2}
\block{ABSTRACT}
{
asdf
}
\block{INTRODUCTION}
{}
\block{DATA DESCRIPTIONS}
{
\innerblock{Data 1}{qaz}
\innerblock{Data 2}{wsx}
\innerblock{Data 3}{edc}
}
\column{0.2}
\block{ISSUES}
{
lkjh
}
\block{SOLUTIONS}
{
123456789
}
\column{0.6}
\begin{subcolumns}
\subcolumn{0.5}
\block{Case Study I}
{
qwerty
}
\subcolumn{0.5}
\block{Case Study II}
{
uiop
}
\end{subcolumns}
\block{COMPARISON RESULTS}
{
zxcvbnm
}
\end{columns}
\end{document}

And here is what it produces:

But what I would rather want is for the columns to span the whole width of the paper, all the way till where the title box starts and ends. As you can see in the code, I am not changing any margins, so all values are default.

I have also looked through the tikzposter class file, and have not managed to find the cause of the problem. But today is my third day with LaTeX, so I have a pretty small idea about what I should be looking for in the first place. I tried to display the value of \TP@visibletextwidth in one of the blocks of the poster, unsuccessfully. I would appreciate any help! Thanks a lot.

Please let me know if you need any additional details. (I know my code is uncommented, but this is just my 3rd day, and this is a rough script. Sorry about that.)

• Welcome to TeX.SX! You can have a look at our starter guide to familiarize yourself further with our format. – karlkoeller Jan 5 '15 at 9:30

After issuing

\geometry{paperwidth=42in, paperheight=60in}

you have to recalculate \TP@visibletextwidth and \TP@visibletextheight. So, after that line, add

\makeatletter
\setlength{\TP@visibletextwidth}{\textwidth-2\TP@innermargin}
\setlength{\TP@visibletextheight}{\textheight-2\TP@innermargin}
\makeatother

MWE:

\RequirePackage[demo]{graphicx} % remove this line in your document
\documentclass[17pt, a0paper, landscape]{tikzposter}
\geometry{paperwidth=42in, paperheight=60in}
\makeatletter
\setlength{\TP@visibletextwidth}{\textwidth-2\TP@innermargin}
\setlength{\TP@visibletextheight}{\textheight-2\TP@innermargin}
\makeatother

\usepackage{xpatch}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage[space]{grffile}

\title{\parbox{0.95\linewidth}{\centering \textbf{My Poster Title}}}

{\setlength{\tabcolsep}{2em}
\author
{%
\begin{tabular}{ c | c | c }
\textbf{Author 1*} \emph{Author position 1} \hfill & \hfill \textbf{Author 2} \emph{Author position 2} \hfill & \hfill \textbf{Author 3} \emph{Author Position 3} \\
\texttt{Auth1@univ.edu; Auth1Phone} \hfill & \texttt{Auth2@univ.edu; Auth2Phone} \hfill & \texttt{Auth3@univ.edu; Auth3Phone} \\
\end{tabular}%
}

\institute
{
\textbf{Affiliated University and Department}\\
\vspace*{0.2em}
}

\titlegraphic
{
\raisebox{2.25cm}{\includegraphics[width=11cm,height=3cm]{/Poster/Logo1.png}}
\hfill
\includegraphics[width=7.5cm,height=7.5cm]{/Poster/Logo2.png}
}

\makeatletter
\def\TP@titlegraphictotitledistance{-7.5cm}
\settitle
{
\centering
\vbox
{
\@titlegraphic \\ [\TP@titlegraphictotitledistance]
\centering
\color{titlefgcolor}
{\bfseries \huge \sc \@title \par}
\vspace*{1em}
{\LARGE \@author \par}
\vspace*{1.2em}
{\LARGE \@institute}
}
}
\makeatother

\begin{document}
\maketitle[width=0.98\textwidth]
\block{First}{Check Span}
\begin{columns}
\column{0.2}
\block{ABSTRACT}
{
asdf
}
\block{INTRODUCTION}
{}
\block{DATA DESCRIPTIONS}
{
\innerblock{Data 1}{qaz}
\innerblock{Data 2}{wsx}
\innerblock{Data 3}{edc}
}
\column{0.2}
\block{ISSUES}
{
lkjh
}
\block{SOLUTIONS}
{
123456789
}
\column{0.6}
\begin{subcolumns}
\subcolumn{0.5}
\block{Case Study I}
{
qwerty
}
\subcolumn{0.5}
\block{Case Study II}
{
uiop
}
\end{subcolumns}
\block{COMPARISON RESULTS}
{
zxcvbnm
}
\end{columns}
\end{document}

And this is the result:

and the paper size is the correct one:

• Works perfectly! Thank you so much @karlkoeller. – Kartik Jan 5 '15 at 15:39


text:
shocksgo Documentation¶

This is the documentation for shocksgo.

Reference/API¶

shocksgo Package¶

Functions¶

 generate_solar_fluxes(duration[, cadence]) Generate an array of fluxes with zero mean which mimic the power spectrum of the SOHO/VIRGO SPM observations. generate_stellar_fluxes(duration, M, T_eff, R, L) Generate an array of fluxes with zero mean which mimic the power spectrum of the SOHO/VIRGO SPM observations, scaled for a star with a given mass, effective temperature, luminosity and radius. interpolate_missing_data(times, fluxes[, …]) Assuming times are uniformly spaced with missing cadences, fill in the missing cadences with linear interpolation. power_spectrum(fluxes[, d]) Compute the power spectrum of fluxes in units of [ppm^2 / microHz]. test(**kwargs) Run the tests for the package.


text:
`GABS_CONV : conv -> term -> thm`

SYNOPSIS
Applies a conversion to the body of a generalized abstraction.

DESCRIPTION
If c is a conversion that maps a term `t` to the theorem |- t = t', then the conversion ABS_CONV c maps generalized abstractions of the form `\vs. t` to theorems of the form:
```   |- (\vs. t) = (\vs. t')
```
That is, ABS_CONV c `\vs. t` applies c to the body of the generalized abstraction `\vs. t`. It is permissible to use it on a basic abstraction, in which case the effect is the same as ABS_CONV.

FAILURE CONDITIONS
Fails if applied to a term that is not a generalized abstraction (or a basic one), or if the conversion c fails when applied to the term t, or if the theorem returned has assumptions in which one of the variables in the abstraction varstruct is free.

EXAMPLE
```  # GABS_CONV SYM_CONV `\(x,y,z). x + y + z = 7`;;
val it : thm = |- (\(x,y,z). x + y + z = 7) = (\(x,y,z). 7 = x + y + z)
```


text:
Question 1: Evaluate the following:

$\displaystyle \text{i) } ^{14} \rm C_3$ $\displaystyle \text{ii) } ^{12} \rm C_{10}$ $\displaystyle \text{iii) } ^{35} \rm C_{35}$ $\displaystyle \text{iv) } ^{n+1} \rm C_n$

$\displaystyle \text{v) } \sum \limits_{r=1}^{5} \ \displaystyle ^{5} \rm C_{r}$

$\displaystyle \text{i) } ^{14} \rm C_3 = \frac{14!}{3! (14-3)!} = \frac{14!}{3! 11!} = \frac{14 \times 13 \times 12}{6} = 364$

$\displaystyle \text{ii) } ^{12} \rm C_{10} = \frac{12!}{10! (12-10)!} = \frac{12!}{10! 2!} = \frac{11 \times 11 }{2} = 66$

$\displaystyle \text{iii) } ^{35} \rm C_{35} = \frac{35!}{35! (35-35)!} = \frac{35!}{35! 0!} = 1$

$\displaystyle \text{iv) } ^{n+1} \rm C_n = \frac{(n+1)!}{n! (n+1-n)!} = \frac{(n+1)n!}{n!} = (n+1)$

$\displaystyle \text{v) } \sum \limits_{r=1}^{5} \ \displaystyle ^{5} \rm C_{r}$

$= ^{5} \rm C_{1} + ^{5} \rm C_{2} + ^{5} \rm C_{3}+^{5} \rm C_{4}+^{5} \rm C_{5}$

$\displaystyle = \frac{5!}{5! (5-1)!} + \frac{5!}{5! (5-2)!} + \frac{5!}{5! (5-3)!} + \frac{5!}{5! (5-4)!} + \frac{5!}{5! (5-5)!}$

$\displaystyle = 5 + 10 + 10 + 5 + 1 = 31$

$\displaystyle \\$

Question 2: If $\displaystyle ^{n} \rm C_{12} = ^{n} \rm C_{5}$, find the value of $\displaystyle n$.

$\displaystyle \text{Given } ^{n} \rm C_{12} = ^{n} \rm C_{5}$

Applying the formula, when $\displaystyle ^{n} \rm C_{p} = ^{n} \rm C_{q} \Rightarrow p+q = n$

Therefore $\displaystyle 12 + 5 = n \Rightarrow n = 17$

$\displaystyle \\$

Question 3: If $\displaystyle ^{n} \rm C_{4} = ^{n} \rm C_{6}$, find $\displaystyle ^{12} \rm C_{n}$.

$\displaystyle \text{Given } ^{n} \rm C_{4} = ^{n} \rm C_{6}$

Applying the formula, when $\displaystyle ^{n} \rm C_{p} = ^{n} \rm C_{q} \Rightarrow p+q = n$

Therefore $\displaystyle 4 + 6 = n \Rightarrow n = 10$

$\displaystyle ^{12} \rm C_{10} = \frac{12!}{10! 2!} = \frac{12 \times 11}{2} = 66$

$\displaystyle \\$

Question 4: If $\displaystyle ^{n} \rm C_{10} = ^{n} \rm C_{12}$, find $\displaystyle ^{23} \rm C_{n}$.

$\displaystyle \text{Given } ^{n} \rm C_{10} = ^{n} \rm C_{12}$

Applying the formula, when $\displaystyle ^{n} \rm C_{p} = ^{n} \rm C_{q} \Rightarrow p+q = n$

Therefore $\displaystyle 10 + 12 = n \Rightarrow n = 22$

$\displaystyle ^{23} \rm C_{22} = \frac{23!}{22! (23-22)!} = \frac{23! }{22!} = 23$

$\displaystyle \\$

Question 5: If $\displaystyle ^{24} \rm C_{x} = ^{24} \rm C_{2x+3}$, find $\displaystyle x$.

$\displaystyle \text{Given } ^{24} \rm C_{x} = ^{24} \rm C_{2x+3}$

Applying the formula, when $\displaystyle ^{n} \rm C_{p} = ^{n} \rm C_{q} \Rightarrow p+q = n$

Therefore $\displaystyle x + 2x+3 = 24 \Rightarrow x = 7$

$\displaystyle \\$

Question 6: If $\displaystyle ^{18} \rm C_{x} = ^{18} \rm C_{x+2}$, find $\displaystyle x$.

$\displaystyle \text{Given } ^{18} \rm C_{x} = ^{18} \rm C_{x+2}$

Applying the formula, when $\displaystyle ^{n} \rm C_{p} = ^{n} \rm C_{q} \Rightarrow p+q = n$

Therefore $\displaystyle x + x+2 = 18 \Rightarrow x = 8$

$\displaystyle \\$

Question 7: If $\displaystyle ^{15} \rm C_{3r} = ^{15} \rm C_{r+3}$, find $\displaystyle r$.

Given $\displaystyle ^{15} \rm C_{3r} = ^{15} \rm C_{r+3}$

Applying the formula, when $\displaystyle ^{n} \rm C_{p} = ^{n} \rm C_{q} \Rightarrow p+q = n$

Therefore $\displaystyle 3r+r+3 = 15 \Rightarrow r = 3$

$\displaystyle \\$

Question 8: If $\displaystyle ^{8} \rm C_{r} - ^{7} \rm C_{3} = ^{7} \rm C_{2}$, find $\displaystyle r$.

$\displaystyle ^{8} \rm C_{r} - ^{7} \rm C_{3} = ^{7} \rm C_{2}$

$\displaystyle \Rightarrow ^{8} \rm C_{r} = ^{7} \rm C_{2} + ^{7} \rm C_{3}$

$\displaystyle \Rightarrow \frac{8!}{r! (8-r)!} = \frac{7!}{2! 5!} + \frac{7!}{3! 4!}$

$\displaystyle \Rightarrow \frac{8!}{r! (8-r)!} = \frac{7 \times 6}{2} + \frac{7 \times 6 \times 5}{6}$

$\displaystyle \Rightarrow \frac{8!}{r! (8-r)!} = 56$

$\displaystyle \Rightarrow r! (8-r)! = 720$

$\displaystyle \Rightarrow r! (8-r)! = 3! \times 5!$

Comparing LHS with RHS we get,

If $\displaystyle r = 3$ and $\displaystyle 8-r = 5 \Rightarrow r = 3$

If $\displaystyle r = 5$, then $\displaystyle 8-r = 3 \Rightarrow r = 5$

Hence $\displaystyle r$ can be $\displaystyle 3$ or $\displaystyle 5$.

$\displaystyle \\$

Question 9: If $\displaystyle ^{15} \rm C_{r} : ^{15} \rm C_{r-1} = 11:5$, find $\displaystyle r$.

$\displaystyle ^{15} \rm C_{r} : ^{15} \rm C_{r-1} = 11:5$

$\displaystyle \Rightarrow \frac{15!}{r! (15-r)!} \times \frac{(r-1)! ( 15-r+1)!}{15!} = \frac{11}{5}$

$\displaystyle \Rightarrow \frac{(r-1)! ( 16-r) (15-r)!}{r(r-1)!(15-r)!} = \frac{11}{5}$

$\displaystyle \Rightarrow \frac{16-r}{r} = \frac{11}{5}$

$\displaystyle \Rightarrow 80-5r = 11r$

$\displaystyle \Rightarrow 80 = 16 r$

$\displaystyle \Rightarrow r = 5$

$\displaystyle \\$

Question 10: If $\displaystyle ^{n+2} \rm C_{8} : ^{n-2} \rm P_{4} = 57:16$, find $\displaystyle r$.

$\displaystyle ^{n+2} \rm C_{8} : ^{n-2} \rm P_{4} = 57:16$

$\displaystyle \Rightarrow \frac{(n+2)!}{8! (n-6)!} \times \frac{(n-6)!}{(n-2)!} = \frac{57}{16}$

$\displaystyle \Rightarrow \frac{(n+2)(n+1)(n)(n-1)(n-2)!}{8! (n-2)!} = \frac{57}{16}$

$\displaystyle \Rightarrow (n+2)(n+1)(n)(n-1) = \frac{8!}{16} \times 57 = 7 \times 6 \times 5 \times 4 \times 3 \times 19 \times 3$

$\displaystyle = 18 \times 19 \times 20 \times 21$

$\displaystyle \Rightarrow n = 18$

$\displaystyle \\$

Question 11: If $\displaystyle ^{28} \rm C_{2r} : ^{24} \rm C_{2r-4} = 225:11$, find $\displaystyle r$.

$\displaystyle ^{28} \rm C_{2r} : ^{24} \rm C_{2r-4} = 225:11$

$\displaystyle \Rightarrow \frac{28!}{(2r)! (28-r)!} \times \frac{(2r-4)! (28-2r)!}{24!} = \frac{225}{11}$

$\displaystyle \Rightarrow \frac{28 \times 27 \times 26 \times 25 \times ( 2r-4)!}{(2r) ( 2r-1)(2r-2)(2r-3)(2r-4)!} = \frac{225}{11}$

$\displaystyle \Rightarrow \frac{28 \times 27 \times 26 \times 25 }{(2r) ( 2r-1)(2r-2)(2r-3)} = \frac{225}{11}$

$\displaystyle \Rightarrow (2r) ( 2r-1)(2r-2)(2r-3) = 28 \times 33 \times 26 = 11 \times 12 \times 13 \times 14$

$\displaystyle \Rightarrow 2r-3 = 11 \Rightarrow r = 7$

$\displaystyle \\$

Question 12: If $\displaystyle ^{n} \rm C_{4}, \ \ ^{n} \rm C_{5}$ and $\displaystyle ^{n} \rm C_{6}$ are in AP, then find $\displaystyle n$.

$\displaystyle ^{n} \rm C_{4}, \ \ ^{n} \rm C_{5}$ and $\displaystyle ^{n} \rm C_{6}$ are in AP

$\displaystyle \Rightarrow ^{n} \rm C_{5} - ^{n} \rm C_{4} = ^{n} \rm C_{6} - ^{n} \rm C_{5}$

$\displaystyle \Rightarrow \frac{n!}{5! (n-5)!} - \frac{n!}{4! (n-4)!} = \frac{n!}{6! (n-6)!} - \frac{n!}{5! (n-5)!}$

$\displaystyle \Rightarrow \frac{1}{5! (n-5)!} - \frac{1}{4! (n-4)!} = \frac{1}{6! (n-6)!} - \frac{1}{5! (n-5)!}$

$\displaystyle \Rightarrow \frac{2}{5! (n-5)!} = \frac{1}{4! (n-4)!} + \frac{1}{6! (n-6)!}$

$\displaystyle \Rightarrow \frac{2}{5 \times 4! (n-5)(n-6)!} = \frac{1}{4! (n-4)(n-5)(n-6)!} + \frac{1}{6 \times 5 \times 4! (n-6)!}$

$\displaystyle \Rightarrow \frac{2}{5(n-5)} = \frac{1}{(n-4)(n-5)} + \frac{1}{30}$

$\displaystyle \Rightarrow \frac{1}{n-5} \Big[ \frac{2}{5} - \frac{1}{n-4} \Big] = \frac{1}{30}$

$\displaystyle \Rightarrow \frac{1}{n-5} \Big[ \frac{2n-8-5}{5(n-4)} \Big] = \frac{1}{30}$

$\displaystyle 30(2n-13) = 5 ( n-4)(n-5)$

$\displaystyle 12n - 78 = n^2 - 9n + 20$

$\displaystyle n^2 - 21n +98 = 0$

$\displaystyle (n-7)(n-14) = 0$

$\displaystyle n = 7$ or $\displaystyle n = 14$

$\displaystyle \\$

Question 13: If $\displaystyle ^{2n} \rm C_{3} : ^{n} \rm C_{2} = 44:3$, find $\displaystyle n$.

Given $\displaystyle ^{2n} \rm C_{3} : ^{n} \rm C_{2} = 44:3$

$\displaystyle \Rightarrow \frac{(2n)!}{3! (2n-3)!} \times \frac{2! (n-2)!}{n!} = \frac{44}{3}$

$\displaystyle \Rightarrow \frac{(2n)(2n-1)(2n-2)}{3!} \times \frac{2!}{n(n-1)} = \frac{44}{3}$

$\displaystyle \Rightarrow \frac{(2n-1)(2n-2)}{2(n-1)} = 11$

$\displaystyle 2n-1 = 11 \Rightarrow n = 6$

$\displaystyle \\$

Question 14: If $\displaystyle ^{16} \rm C_{r} = ^{16} \rm C_{r+2}$, find $\displaystyle ^{r} \rm C_{4}$.

Given $\displaystyle ^{16} \rm C_{r} = ^{16} \rm C_{r+2}$

Applying the formula, when $\displaystyle ^{n} \rm C_{p} = ^{n} \rm C_{q} \Rightarrow p+q = n$

Therefore $\displaystyle r + ( r + 2) = 16 \Rightarrow r = 7$

$\displaystyle ^{7} \rm C_{4} = \frac{7!}{4! (7-4)!} = \frac{7 \times 6 \times 5 }{6} = 35$

$\displaystyle \\$

Question 15: If $\displaystyle \alpha = ^{m} \rm C_{2}$, then find the value of $\displaystyle ^{\alpha} \rm C_{2}$

Given $\displaystyle \alpha = ^{m} \rm C_{2} \Rightarrow \alpha = \frac{m!}{2! (m-2)!} = \frac{m(m-1)}{2}$

$\displaystyle ^{\alpha} \rm C_{2} = \frac{\alpha !}{2! (\alpha - 2)!}$

$\displaystyle = \frac{\alpha ( \alpha - 1)}{2}$

$\displaystyle = \frac{1}{2} \times \frac{m(m-1)}{2} \times \Big[ \frac{m(m-1)}{2} - 1 \Big]$

$\displaystyle = \frac{m(m-1)}{4} \times \Big[ \frac{m^2 - m - 2}{2} \Big]$

$\displaystyle = \frac{1}{8} [ (m-1)(m^3 - m^2 - 2m) ]$

$\displaystyle = \frac{1}{8} [ m^4 - m^3 - 2m^2 - m^3 + m^2+ 2m]$

$\displaystyle = \frac{1}{2} [ m^4 - 2m^3 - m^2 + 2m ]$

$\displaystyle = \frac{1}{8} (m^2-2m)(m^2 -1)$

$\displaystyle = \frac{1}{8} m ( m-2) ( m-1)(m+1)$

$\displaystyle \\$

Question 16: Prove that the product of $\displaystyle 2n$ consecutive negative integers is divisible by $\displaystyle ( 2n)!$

Let $\displaystyle 2n$ negative integers be $\displaystyle (-r), ( -r-1), ( -r-2), \ldots, ( - r - 2n + 1)$

Product of $\displaystyle 2n$ negative integers $\displaystyle = ( -1)^{2n} [ r ( r+1) (r+2) \ldots ( r+2n-1)]$

$\displaystyle = (-1)^{2n} \Big[ \frac{(r-1)! r ( r+1) (r+2) \ldots ( r+2n-1)}{(r-1)!} \Big]$

$\displaystyle = \frac{(r+2n-1)!}{(r-1)!} \times \frac{(2n)!}{(2n)!}$

$\displaystyle = ^{r+2n-1} \rm C_{2n} \cdot (2n)!$

Therefore product of $\displaystyle 2n$ consecutive negative integers is divisible by $\displaystyle ( 2n)!$

$\displaystyle \\$

Question 17: For all positive integers $\displaystyle n$, show that

$\displaystyle ^{2n} \rm C_n + ^{2n} \rm C_{n-1} = \frac{1}{2} \big( ^{2n+2} \rm C_{n+1} \big)$

$\displaystyle \text{LHS } = ^{2n} \rm C_n + ^{2n} \rm C_{n-1}$

$\displaystyle = \frac{(2n)!}{n! n!} + \frac{(2n)!}{(n-1)! (n+1)!}$

$\displaystyle = \frac{(2n)!}{(n-1)! n!} \Big[ \frac{1}{n} + \frac{1}{n+1} \Big]$

$\displaystyle = \frac{(2n)!}{(n-1)! n!} \Big[ \frac{2n+1}{n(n+1)} \Big]$

$\displaystyle = \frac{(2n+1)!}{(n+1)! n!}$

$\displaystyle \text{RHS } = \frac{1}{2} \Big[ ^{2n+2} \rm C_{n+1} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \frac{(2n+2)!}{(n+1)! (n+1)!} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \frac{(2n+2)(2n+1)!}{(n+1) n! (n+1)!} \Big]$

$\displaystyle = \frac{(2n+1)!}{(n+1)! n!}$

Therefore LHS = RHS hence proved.

$\displaystyle \\$

Question 18: Prove that: $\displaystyle ^{4n} \rm C_{2n} : ^{2n} \rm C_{n} = [ 1 \cdot 3 \cdot 5 \ldots (4n-1) ] : [ 1 \cdot 3 \cdot 5 \ldots (2n-1) ]^2$

$\displaystyle \text{LHS } = ^{4n} \rm C_{2n} : ^{2n} \rm C_{n}$

$\displaystyle = \frac{^{4n} \rm C_{2n}}{^{2n} \rm C_{n}}$

$\displaystyle = \frac{(4n)!}{(2n)! (2n)!} \times \frac{(n)! (n)!}{(2n)!}$

$\displaystyle = \frac{(4n)(4n-1)(4n-2) \ldots (3) (2) (1)}{[ (2n)(2n-1)(2n-2) \ldots (3)(2)(1)]^2} \frac{(n!)^2}{(2n)!}$

$\displaystyle = \frac{(n!)^2}{(2n)!} \frac{ [ 1 \cdot 3 \cdot 5 \ldots (4n-3)(4n-1) ] [ 2 \cdot 4 \cdot 6 \ldots (4n-2)(4n) ] }{ [ 1 \cdot 3 \cdot 5 \ldots (2n-3)(2n-1) ]^2 [ 2 \cdot 4 \cdot 6 \ldots (2n-2)(2n) ]^2 }$

$\displaystyle = \frac{(n!)^2}{(2n)!} \frac{ [ 1 \cdot 3 \cdot 5 \ldots (4n-3)(4n-1) ] 2^{2n} [ 1 \cdot 2 \cdot 3 \ldots (2n-1)(2n) ] }{ [ 1 \cdot 3 \cdot 5 \ldots (2n-3)(2n-1) ]^2 2^{2n} [ 1 \cdot 2 \cdot 3 \ldots (n-1)(n) ]^2 }$

$\displaystyle = \frac{(n!)^2}{(2n)!} \frac{(2n)!}{(n!)^2} \frac{ [ 1 \cdot 3 \cdot 5 \ldots (4n-3)(4n-1) ] }{ [ 1 \cdot 3 \cdot 5 \ldots (2n-3)(2n-1) ]^2 }$

$\displaystyle = \frac{ [ 1 \cdot 3 \cdot 5 \ldots (4n-3)(4n-1) ] }{ [ 1 \cdot 3 \cdot 5 \ldots (2n-3)(2n-1) ]^2 } =$ RHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 19: Evaluate: } ^{20} \rm C_{5} + \sum \limits_{r=2}^{5} \ \displaystyle ^{25-r} \rm C_{4}$

$\displaystyle \text{LHS } = ^{20} \rm C_{5} + \sum \limits_{r=2}^{5}$

$\displaystyle = \ ^{20} \rm C_{5} + ^{23} \rm C_{4} +^{22} \rm C_{4} +^{21} \rm C_{4} + ^{20} \rm C_{4}$

$\displaystyle = \ \big( ^{20} \rm C_{4} + ^{20} \rm C_{5} \big) + \big( ^{21} \rm C_{4} +^{22} \rm C_{4} +^{23} \rm C_{4} \big)$

$\displaystyle \text{Since } ^{n} \rm C_{r-1} + ^{n} \rm C_{r} = ^{n+1} \rm C_{r}$

$\displaystyle = \ ^{21} \rm C_{5} + ^{21} \rm C_{4} +^{22} \rm C_{4} +^{23} \rm C_{4}$

$\displaystyle = \ ^{22} \rm C_{5} + ^{22} \rm C_{4} +^{23} \rm C_{4}$

$\displaystyle = \ ^{23} \rm C_{5} + ^{23} \rm C_{4}$

$\displaystyle = \ ^{24} \rm C_{5}$

$\displaystyle = \frac{24!}{5! 9!} = \frac{24 \times 23 \times 22 \times 21 \times 20}{5 \times 4 \times 3 \times 2 \times 1 } = 4 \times 23 \times 22 \times 21 = 41504$

$\displaystyle \\$

Question 20: Let $\displaystyle r$ and $\displaystyle n$ be positive integers such that $\displaystyle 1 \leq r \leq n$. The prove the following:

$\displaystyle \text{i) } \frac{^{n} \rm C_{r}}{^{n} \rm C_{r-1}} = \frac{n-r+1}{r}$

$\displaystyle \text{ii) } n ^{n-1} \rm C_{r-1} = (n-r+1) ^{n} \rm C_{r-1}$

$\displaystyle \text{iii) } \frac{^{n} \rm C_{r}}{^{n-1} \rm C_{r-1}} = \frac{n}{r}$

$\displaystyle \text{iv) } ^{n} \rm C_{r} + 2 ^{n} \rm C_{r-1} + ^{n} \rm C_{r-2} = ^{n+2} \rm C_{r}$

$\displaystyle \text{i) } \frac{^{n} \rm C_{r}}{^{n} \rm C_{r-1}} = \frac{n!}{r! (n-r)!} \times \frac{(r-1)! (n-r+1)!}{n!} \\ \\ = \frac{(r-1)!}{r! (r-1)!} \frac{(n-r+1)! (n-r)!}{(n-r)!} = \frac{n-r+1}{r}$

$\displaystyle \text{ii) LHS } = n \cdot ^{n-1} \rm C_{r-1} = \frac{n (n-1)!}{(r-1)! (n-1-r+1)!} = \frac{n!}{(r-1)!(n-r)!}$

$\displaystyle \text{RHS } = (n-r+1) ^{n} \rm C_{r-1}$

$\displaystyle = \frac{(n-r+1) n!}{(r-1)! (n-r+1)!} = \frac{(n-r+1) n!}{(n-r+1)(n-r)! (r-1)!} = \frac{n!}{(r-1)!(n-r)!}$

Therefore LHS $\displaystyle =$ RHS

$\displaystyle \text{iii) LHS } = \frac{^{n} \rm C_{r}}{^{n-1} \rm C_{r-1}}$

$\displaystyle = \frac{n!}{r! (n-r)!} \times \frac{(r-1)! ( n - 1 - r + 1)!}{(n-1)!}$

$\displaystyle = \frac{n!}{r! (n-r)!} \times \frac{(r-1)! ( n - r)!}{(n-1)!}$

$\displaystyle = \frac{n (n-1)!}{r (r-1)! } \times \frac{(r-1)! }{(n-1)!}$

$\displaystyle = \frac{n}{r} =$ RHS

$\displaystyle \text{iv) LHS } = ^{n} \rm C_{r} + 2 ^{n} \rm C_{r-1} + ^{n} \rm C_{r-2}$

$\displaystyle = \big( ^{n} \rm C_{r} + ^{n} \rm C_{r-1} \big) + \big( ^{n} \rm C_{r-1} + ^{n} \rm C_{r-2} \big)$

$\displaystyle = ^{n+1} \rm C_{r} + ^{n+1} \rm C_{r-1}$

$\displaystyle = ^{n+2} \rm C_{r} =$ RHS. Hence proved.


text:
Tag Info

9

There are known impossibility results regarding basis public-key cryptography on NP-complete problems. In this paper by Goldreich and Goldwasser they show that under common types of reductions, it is not possible to base public-key cryptography on NP-hardness.

8

I work for Security Innovation, which owns the NTRU patents. All NTRU-related patents are freely usable under GPL 2.0 and 3.0 -- in other words, they should fit in with your license requirement as given above. If you have specific license requirements beyond GPL please let me know and we'll accommodate them if we can. There's an open-source C and Java ...

8

Grover's algorithm treats the function it is evaluating as a black box and finds, with high probability, an input to the black box such that it outputs a specified value in $O(N^{1/2})$ evaluations of the function. Since Grover's algorithm works on the function as a black box, your modification does not hinder Grover's algorithm at all in finding the ...

7

I see two problems with this idea. The first problem is Shor's algorithm; that's a quantum algorithm that is able to find the cycle length of a group (and if you can solve that problem, it is easy to factor and compute discrete logs). In this case, if we define the group of elements defined by the initial start state in the signature, where $H^n$ is the ...

7

The statement a 15360-bit RSA key is the equivalent to a 256-bit symmetric key does not take into account quantum algorithms. In fact, it is based on a specific computation model. It is just based on the fact that there exist sub-exponential algorithms for factoring and therefore you need longer keys than when using symmetric-key crypto where it is ...

7

Post-quantum crypto is a very young field and is still changing quite rapidly. If you just want a reading list to introduce you to the topics, I would recommend the March 2015 report released by the EU's PQCrypto Project, and the April 2016 report from NIST. As of today, here's an (incomplete) list of candidate algorithms for post-quantum cryptography ...

7

Actually, if RSA is being used in a deterministic way (and the public exponent $e$ is relatively small), someone could recover the value $N$. We know that $P^e = C \bmod N$; that's equivalent to $P^e - C = kN$ for some integer $k$; if $e$ is small, then Shor's algorithm might be able to factor $P^e - C$; allowing you to recover $N$. Alternatively, if you ...

6

Why is 128-bit encryption considered good enough for medium term security only? Because in the long term it is expected that mankind will be able to carry out $2^{128}$ operations because it's not physically as impossible as $2^{256}$ operations. Quantum computing or brute force attack? Assuming quantum computers work at a speed comparable to ...

6

Actually, most of the primitives that are currently believed to be secure FHE methods would appear to be quantum resistant; a partial list would include Craig Gentry's original scheme based on ideal lattices, BGV (based on ring-LWE), and this NTRU-based approach. All three are based on hard problems that are not susceptible to Shor's algorithm.

6

This is in principle similar as how "normal" cryptosystems are proven. With some algorithms we can reduce them to some "hard problem", but we do not know that those problems are actually hard to solve. Only that we cannot solve them efficiently. For example, the Diffie-Hellman problem is not even known to be NP-hard, never mind the whole issue of P vs. NP. ...

5

You could be able to reduce the space required for a meet-in-the-middle attack, if you follow a similar idea as the application of Grover's algorithm on collisions. Suppose you have two layers of $n$-bit encryption: Partition the inner keyspace into $2^{n/4}$ parts of size $2^{3n/4}$. For each partition generate the inner encryption table. Run Grover's on ...

5

How many qubits are required for breaking RSA 2048 and RSA 4096 in real-time with a quantum computer? Like the answer you linked to shows, about $\log(N^2) = 2 \log(N)$. So 4096 for 2048-bit RSA, double that for 4096-bit. This paper (pdf) has an algorithm using $2n+3$ qubits, where $n=\log N$. How many qubits are required to break Curve25519? ...

4

That means a plaintext of length 524 will be encrypted to a ciphertext of length 1024 and then will be sent. Isn't is also an inefficiency? Not really; or at least, that's not an inefficiency we care about. A length of 1024 means, in this context, 1024 bits (or 128 bytes). This compares favorably to RSA (for which a key with a 1024 bit ciphertext has ...

4

The expression $c' = \lfloor w^{-i} c \rfloor$ is a slight abuse of notation. What it technically means is to interpret each coefficient of $c$ as real number, divide by $w^i$, round the result to the nearest integer, and then interpret it as an element of $\mathbb{Z}_q$ again. This is equivalent to expressing each $\mathbb{Z}_q$ coefficient of the ...

4

The best generic attack against a PRG (i.e. an attack that does not use any internal structure of a construction and hence works for any PRG) is exhaustive search for a seed. I think this was not done yet but it is very likely that the optimality of Grover's algorithm carries over to this setting. This would mean that for $n$ bit seeds, the best attack ...

4

If he gets the signature for the message 00000..00000, then the checksum will be $t_1 2^w$. For any other message, the checksum will be smaller, and hence the there will be at least one digit $i$ within the checksum for which the $c_i$ digit with value $v$ for the signed message will be larger than the corresponding digit for the new message. The attacker ...

4

What we traditionally call Elliptic Curve Cryptography (working in the group of points on an elliptic curve over a finite field) is vulnerable to an attack by a quantum computer running Shor's algorithm and is thus not considered a Quantum-Safe or Post Quantum Cryptographic algorithm. However there is an true Post Quantum Key Exchange algorithm which uses ...

4

Yes the Bernstein attack is applicable but the impact of the attack is reduced because the party generating the parameter is also going to be a legitimate participant of the key exchange. Here is why the attack does indeed apply. Consider a case where Bob and Alice wish to conduct a key exchange using the New Hope Lattice-Based Key Exchange. Bob will be ...

3

There is a variant of the Neiderreiter system by Courtois, Finiasz, and Sendrier found in their paper: "How to achieve a McEliece-based Digital Signature Scheme" from Asiacrypt 2001. The Wikipedia article on the Neiderreiter Cryptosystem provides a brief introduction to this signature. There is an element of trial and error in the signing process that is ...

3

The combination of Grover algorithm and man in the middle attack is the main subject of a paper (arXiv:1410.1434) published last year by Marc Kaplan (Full disclosure: Marc is a friend of mine.) In this paper beyond applying Grover to MITM to reduce the time needed to analyse double-encryption, he also looks at the time-space gain, which is different, and ...

3

Essentially any IND-CPA-secure lattice-based cryptosystem offers additive homomorphism, up to a predetermined number of operations. I don't know of any IND-CCA1-secure post-quantum candidate that offers any homomorphic property, except Loftus-May-Smart-Vercauteren SAC'11, which is based on a nonstandard "knowledge of error" lattice assumption.

3

Douglas Stebila published: We demonstrate the practicality of post-quantum key exchange by constructing ciphersuites for the Transport Layer Security (TLS) protocol that provide key exchange based on the ring learning with errors (R-LWE) problem There is also a patch implementing it for OpenSSL 1.0.1f.

3

As @Raoul722 pointed out the polynomial should be irreducible. One should also add the fact that in the original Ring-LWE article security proofs hold by sampling the error with a spherical distribution and the poly $f=x^{2^n}+1$. In How (Not) to Instantiate Ring-LWE, from Invulnerability instantiations subsection it's specified that you could in fact ...

3

Bruteforce appears to work well enough. The following Sage script finds an instance quickly: from sage.libs.fplll.fplll import FP_LLL from sage.libs.fplll.fplll import gen_uniform n = 5 # dimension q = 16 # size of matrix entries while True: M = gen_uniform(n, n, q) L = M.LLL(delta=0.999) S = FP_LLL(L).shortest_vector(algorithm='proved') ...

3

Yes, these are public parameters of the system. Note that NTRU is not implemented exactly this way any more. The most up-to-date current spec is EESS#1, which can be obtained from https://github.com/NTRUOpenSourceProject/ntru-crypto/blob/master/doc/EESS1-v3.1.pdf.

2

I do not believe that braid based cryptography is dead since new ways of applying braid groups to cryptography are currently being investigated and have recently been proposed. In fact, recently researchers have observed how braids could be used for reversible circuit obfuscation. A recent paper on braid-based obfuscation The 2014 paper, Partial-...

2

Sorry I will have to answer my own question. I received a mail from Luca De Feo a moment ago. "Nope, I discussed this at length with Jean-François Biasse, and we couldn't find a way to apply this kind of attack to SSIKE." I'll leave this question around for reference for the next person who wonders.

2

It is easy to find preimages for the hash function you designed. Remember, to break the scheme it is enough to find any preimage. It is not necessary to find the one used to compute the image. Now, you got public integers $a,b,c,d,$ and $f,g$ as well as a modulo $M$. An $n$-th image is $$P(n) = a*x(n) + c*y(n) \mod M$$ $$Q(n) = b*x(n) + d*y(n) \mod M$$ ...

2

Random oracles are used in cryptographic proofs as a way to abstract away a complex function, making the analysis easier (ie "ignore the actual function, and model it as a random oracle instead"). In the paragraphs before the sentence you highlighted, they propose to measure the "complexity" of an equation / function by the degree of its multivariate ...

2

While there is a sub-exponential attack to compute isogenies on ORDINARY elliptic curves (the basis for the Rostovev and Stulbunov paper that you reference) there is not (yet at least) a sub-exponential attack to compute isogenies on SUPERSINGULAR elliptic curves. The cryptosystem proposed by DeFeo, Jao, and Plut back in 2011 is based on Supersingular ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
A type of set used in convergence in measure theory

This is not a specific problem, but a general question. Often when we're showing convergence of functions (particularly pointwise) or even of sets in certain cases, a set of the following form appears: $$\bigcap_{N \in \mathbb{N}} \bigcup_{n \geq N} E_{n}$$ Often we will encounter situations where we want to show that something goes "wrong" on a set of this form, or a set of this form is measurable (which is obviously true if $$E_{n}$$ are measurable); I have a general question about a set of this form.

In what contexts does this show up? In my understanding, something is in this set if it "happens" infinitely often. This is a slightly probabilistic interpretation, but if some property holds for infinitely many sets, then something like this shows up. Is that a correct interpretation? I'd appreciate some elaboration.

Example: If we have a sequence of functions $$\{f_{n}\}$$, where each $$f_{n}$$ is supported on a measurable set $$E_{n}$$, then

$$\bigcap_{N} \bigcup_{n \geq N} E_{n}$$

is precisely the set of points where the sequence $$f_{n}$$ is not eventually $$0$$. Thus, points $$x$$ at which arbitrarily many $$f_{n}$$ are non-zero (points in infinitely many supports $$E_{n}$$). Is this correct?

What you've written is called $$\limsup E_n$$ and $$x \in \limsup E_n$$ iff $$x \in E_n$$ for infinitely many $$n$$. This understanding already shows your example is correct.
There is a similar notion that $$\liminf E_n = \bigcup_{j \geq 1} \bigcap_{n \geq j} E_n$$. $$x \in \liminf E_n$$ iff there is an $$N$$ such that $$x \in E_n$$ whenever $$n \geq N$$ (i.e. "$$x \in E_n$$ eventually").
These sets occur naturally when considering convergence as you noted. In fact, $$\liminf E_n$$ is defined to capture definitions like that of convergence which require some behaviour to occur "eventually". Then $$\limsup E_n$$ is just the complement of $$\liminf E_n$$.
For example, in the case of convergence we require that for fixed $$\varepsilon > 0$$, your sequence lies within distance $$\varepsilon$$ of the limit "eventually". More precisely, if $$E_n^\varepsilon = \{x: |f_n(x) - f(x)| < \varepsilon\}$$ then $$\liminf E_n^\varepsilon = \{x: \exists N \text{ such that } \forall n \geq N, |f_n(x) - f(x)| < \varepsilon\}$$ so that we can write $$\{x: f_n(x) \to f(x)\} = \bigcap_{k \geq 1} \liminf E_n^{k^{-1}}.$$
Conversely, if we want to find $$x$$ such that $$f_n(x) \not \to f(x)$$ then there is an $$\varepsilon > 0$$ such that $$|f_n(x) - f(x)| > \varepsilon$$ infinitely often which is the same as saying $$x \in \limsup (E_n^\varepsilon)^c$$. Hence we can write $$\{x: f_n(x) \not \to f(x)\} = \bigcup_{k \geq 1} \limsup (E_n^{k^{-1}})^c.$$


text:
Gas turbines

Posted on by

Content – Energy sources

Gas turbines
A gas turbine is an internal combustion engine. Simplified, it has a rotating compressor coupled to a turbine, with a combustion chamber in between.

Principle
Atmospheric air flows through the compressor that raises the pressure.

Igniting and burning the fuel that is added to the combustion chamber, add energy to the gas. The combustion generates a high-temperature flow and high pressure. The gas enters the turbine, expands down to the exhaust pressure (atmospheric) and temperature, rotating the shaft and therby producing work. The turbine shaft work is used to drive the upstream compressor and the electric generator that may be coupled to the shaft. If the energy is not used for shaft work it will generate high temperature and/or high velocity exhaust gases.

Design
The design of the gas turbine is determined by the purpose; shaft output or temperature/thrust.

If the purpose is to generate shaft output (ie to drive a generator or a propeller) one will strive to obtain exit pressures as close to the entry pressure as possible, however some exit pressure is necessary to expel the exhaust gases.

Simplified gas turbine – Shaft output (turboprop)

If the purpose is to provide a jet stream (ie an airplane jet engine) then the only energy that is extracted through the shaft is the energy to drive the compressor and utilities. The remaining high pressure and high temperature gases are accelerated through the turbine to provide the desired jet stream.

Simplified gas turbine – Jet stream (turbofan)

The gases undergo (ideal) three thermodynamic processes: isentropic compression, isobaric (constant pressure) combustion and isentropic expansion.

In practical terms the processes are somewhat different; the compression is not adiabatic since, due to friction and turbulens the temperature rise is greater than a rise corresponding to adiabatic compression. Actually the process is polytropic and the polytropic index would be in the range 1,45 – 1,6.

Similarely the gas after expansion through the turbine also, due to friction, has a higher temperature than adiabatic expansion. The expansion also really is a polytropic process.

Compressor blade-tip speed determines the maximum pressure ratios that can be obtained by the turbine and the compressor. This is a limiting factor to the power and efficiency that can be obtained. Therefore small diameter turbines need to have a relatively high rpm compared to larger diameter turbines.

Many stationary plants for electrical power generation are ”Combined Cycle Power Plants” (CCPP)- These plants have increased thermal efficieny, compared to single cycle plants since they also utilise the remaining temperature of the exhaust gases for power generation.

In a typical combined cycle power plant using natural gas as fuel gas, (typically Methane – CH4) the temperature of the heated gas from the combustion chamber is more than 1200 oC entering the turbine. After expansion through the turbine the gas temperature would be around 600 oC. The exhaust gas will then (of a CCPP) be lead into a steam boiler, for heat recovery and generation of steam. The heat recovery system will in many plants include a NOx reduction plant taking the NOx emissions down to a lower level (for the most efficient plants down to 5ppm).

The high-pressure steam is then lead into a high-pressure steam turbine, back to the steam boiler for re-heating and into the medium and low-pressure sections of the steam turbine. Finally the steam goes into the condenser for cooling and condensation and then back to the steam boiler for re-use.

The energy from the steam is converted into kinetic energy by rotation of the damp turbine shaft, which then may be utilised for generation of electricasl power together with the kinetic energy from the gas turbine shaft.

The most efficient Combined cycle power plants convert close to 60% of the energy of the fuel gas into electrical power.

Combined cycle power plant (click on image to view larger version)

Formulas

Thermal efficiency (Joule process):

$\eta_0 = 1 - \dfrac{T_4-T_1}{T_3-T_2}$

η0 = (Theoretical/ideal thermal efficiency)
T1 = Temperature at beginning of compression
T2= Temperature after isentropic compression
T3 = Temperature after combustion at beginning of expansion
T= Temperature after isentropic expansion

$\eta_i = 1 - \dfrac{W_i}{Q_{appl.}}$

ηi = Indicated thermal efficiency
Wi= Indicated work
Qappl.= Applied energy

ηi always is less than η0 , the actual temperature raise in the compressor is larger and the actual temperature drop in the turbine is less than for a theoretical process (adiabatic).

The work (W) during compression or expansion can be expressed as the difference in energy (enthalpy) before and after the change (compression /expansion).

$W = c_p \Delta t$

W = Work (J)
c= Specific heat Capacity (at constant pressure) (J/KgoK)
Δt = Temperature difference (before and after compression/expansion) (oK)

To quantify the losses due to friction in the turbine and the compressor, the efficiency factors ηc (compressor) and η(turbine) are established.

$\eta_C = \cfrac{Temperature-increase (adiabatic-compression)}{Temperature-increase(actual-compression)}$

Normally in the range 0,75 – 0,9

$\eta_C = \dfrac{Temperature drop (actual-expansion)}{Temperature drop (adiabatic-expansion)}$

Normally in the range 0,85 – 0,9

Available work for the ideal (theoretical) process is expressed:

W0 = Wt0 – WC0

Wt0 = Work turbine
WC0 = Work compressor

The indicated work is then:

Wi = Wti – WCi = Wt0ηT – WCiηC

Some of the indicated work (Wi) will not be available at the shaft since it is used to overcome mechanical losses and for operation of auxiliary systems (injection pumps lubrication pumps etc).
The remaining work is available as effective work at the output shaft (We)

ηe = ηi ηm

ηe = Effectiv (actual) efficiency factor
ηm = Mechanical effieciency factor

For a gas turbine ηm is relatively high compared to other heat machines, due to its simple construction.

For a simple single cycle gas turbine ηe (no heat exchanger, intercooler of after burners) would be in the range 0,15 – 0,25

Fuel consumption and power

$\eta_e = \dfrac{1}{f_eh_g}$

ηe = Effectiv (actual) efficiency factor
fe = Effective specific fuel consumption (Kg/J)
hg = Higher heating Value for the fuel (J/Kg)

$f_e = \dfrac{1}{h_g\eta_e}(kg/J)$

$m_f = f_eP_e = \dfrac{m_a}{\lambda(\dfrac{L}{B})_r}(kg/s)$

$P_e = \dfrac{m_a}{f_e\lambda(\cfrac{L}{B})_r}=\dfrac{m_a\eta_eh_g}{f_e\lambda(\dfrac{L}{B})_r}$

Pe= Actual power (at the shaft) (W)
mf = Fuel consumption (mass, fuel)
fe= Effective specific fuel consumption (Kg/J)
ma = Air consumtion (Kg)
λ = Air index (related to air consumption)
$(\cfrac{L}{B})_r$
= Theoretical Air- fuel relationship, related to chemical reaction equation (“reaction equivalent”) (Kg/Kg)


text:
CAPM, Risk and Portfolio Selection in «Stable» Markets

Abstract : Our main purpose in this paper is to derive the generalized equilibrium relationship between risk and return under the assumption that the asset returns follow a joint symmetric $\alpha$-stable distribution, with $1< \alpha <2$. In order to justify such an investigation, we start by empirically evidencing the fractal structure of stocks market through extensive tests of self-similarity and stability. These tests allow us to model price changes with $\alpha$-stable distributions. We then show that equilibrium rates of return on all risky assets are functions of their {\bf covariation} with the market portfolio. The «stable» CAPM highlights a new measure of the quantity of risk which may be interpreted as a generalized beta coefficient.
Keywords :
Type de document :
Rapport
[Research Report] RR-2776, INRIA. 1996
Domaine :

https://hal.inria.fr/inria-00073916
Contributeur : Rapport de Recherche Inria <>
Soumis le : mercredi 24 mai 2006 - 14:04:41
Dernière modification le : samedi 17 septembre 2016 - 01:38:31
Document(s) archivé(s) le : lundi 5 avril 2010 - 00:01:46

Identifiants

• HAL Id : inria-00073916, version 1

Citation

Lotfi Belkacem, Jacques Lévy Véhel, Christian Walter. CAPM, Risk and Portfolio Selection in «Stable» Markets. [Research Report] RR-2776, INRIA. 1996. 〈inria-00073916〉

Métriques

Consultations de la notice

269

Téléchargements de fichiers


text:
# Chemical Formula for Sodium Bicarbonate?

The chemical formula for sodium bicarbonate is Na2 Co3. It has a molecular mass of 105.99 g/mol. Sodium Bicarbonate is used to make glass and it is better known as a cleaner in household products. You can find it in detergent boosters, in substances to soften hard water, and it can even be found in deodorants and toothpaste.
Reference:
Sodium Bicarbonate
SODIUM BICARBONATE is an antacid. It is used to treat acid indigestion and heartburn caused by too much acid in the stomach. This medicine may be used for other purposes; ask your health care provider or pharmacist if you have More »
Source: healthline.com
Q&A Related to "Chemical Formula for Sodium Bicarbonate?"
 Answer Sodium bicarbonate, also called sodium hydrogen carbonate and commonly known as baking soda, has NaHCO3 for a chemical formula. Use the link below to check facts and learn http://wiki.answers.com/Q/What_is_the_formula_for_...
 The chemical formula for sodium bicarbonate is NaHCO3. http://www.chacha.com/question/what-is-the-chemica...
 Sodium bicarbonate is NaHCO3. Hydrochloric acid is HCl. So the formula for the reaction is. NaHCO3 + HCl - NaCl + CO2 +H2O. "I don't understand where or how we get the equations http://answers.yahoo.com/question/index?qid=201010...
 The chemical formula for sodium carbide is Na2C2. A mixture of http://www.chacha.com/question/what-is-the-chemica...
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Sodium bicarbonate is a chemical compound with the formula NaHCO3. It is also known as sodium hydrogen carbonate but it is commonly known as sodium bicarbonate ...
Baking soda, or sodium bicarbonate, has the chemical formula NaHCO3. Sodium bicarbonate is actually a salt. It has many common uses such as baking and household ...
Baking soda is actually sodium bicarbonate or sodium hydrogen carbonate. The chemical formula is NaHCO3. Sodium bicarbonate is actually a solid crystal that is ...


text:
## Stream: new members

### Topic: Refinement of two dissections

#### Mantas Baksys (Apr 04 2021 at 15:44):

Hi all! I've been working on formalizing the definition of the Riemann integral for educational purposes and ran into the following issue. I have defined a dissection as follows :

import data.real.basic

structure dissection (a b : ℝ):=
(N: ℕ)
(x: fin (N+1) → ℝ)
(to_right: a ≤ b)
(x_zero: x (0: fin(N+1)) = a)
(x_last: x (N: fin(N+1)) = b)
(mono : ∀ i : fin N, x (i.cast_succ) ≤ x i.succ)

Now, I would like to be able to take two partitions D1 and D2 and take the union (or the so- called minimal common refinement of these dissections). I realise that this could be done by somehow taking union of the images of D1.x and D2.x and then constructing a monotone sequence out of this set, however this is something I have no idea how to do. Does anybody have any suggestions? :smiley:

#### Eric Rodriguez (Apr 04 2021 at 15:56):

Maybe something like docs#list.merge_sort and use a list instead of x, N and list.sorted instead of mono? I have a feeling that's probably hard to work with though, someone with more experience than me will know that. The image of a function is easy though, the notation for the image of a function f x in mathlib is f '' X

#### Kevin Buzzard (Apr 04 2021 at 16:06):

Your data is equivalent to just giving a finite subset of (a,b), and if you worked with this as the definition then you'd just be able to take the union. You could write API to extract all the other disection stuff from this subset rather than making it part of the definition.

#### Kevin Buzzard (Apr 04 2021 at 16:07):

Actually it is not quite equivalent because you seem to allow $x_i=x_{i+1}$, so the way you have it it's equivalent to a multiset in [a,b] (modulo the degenerate case N=0 and a=b)

#### Mantas Baksys (Apr 04 2021 at 17:53):

Thanks Kevin and Eric. I'm now starting to realise that often times the solution actually is to just 'redefine the structures in a suitable way' rather than look for some workaround :smiley:.

#### Kevin Buzzard (Apr 04 2021 at 18:49):

Definitions are the hard part (at least if you're a mathematician). If you get them wrong then theorems can become hard to prove.

#### Bryan Gin-ge Chen (Apr 04 2021 at 19:08):

There's a (long!) thread where @Patrick Thomas was working on related matters here; some parts might be helpful.

#### Mantas Baksys (Apr 04 2021 at 22:09):

Just a quick follow-up question. I was trying to sort the following finset but got an error:

import data.real.basic data.set.intervals.unordered_interval data.finset.sort

open set

def sorted_set {a b: ℝ} (S: finset (Ioo a b)) : list (Ioo a b) := finset.sort (≤) S

As far as I looked, I haven't found a specific statement that defines ≤ on Ioo a b so that it would be easily accessible. Am I missing something?

#### Alex J. Best (Apr 04 2021 at 22:37):

Ioo coerced to a type is a subtype, so you can do:

import data.real.basic data.set.intervals.unordered_interval data.finset.sort

open set
open_locale classical
noncomputable theory
instance {a b: ℝ} : preorder (Ioo a b) := subtype.preorder _

def sorted_set {a b: ℝ} (S: finset (Ioo a b)) : list (Ioo a b) := finset.sort (≤) S

but you might be better off just sorting a list of reals, and separately tracking the hypothesis that they are all in that interval

#### Eric Wieser (Apr 04 2021 at 22:38):

Should that instance be automatic?

#### Eric Wieser (Apr 04 2021 at 22:38):

Do we have it already as docs#set_coe.preorder?

#### Mantas Baksys (Apr 05 2021 at 09:24):

I think the instance is automatic, however the problem was decidability (at least that's my guess). The code @Alex J. Best posted outputs an error if we do no include open_locale classical, so I guess I'll just just use that every time to avoid the decidability hell.

Last updated: May 18 2021 at 16:25 UTC


text:
# Hydrostatic equilibrium explained

In fluid mechanics, hydrostatic equilibrium (hydrostatic balance, hydrostasy) is the condition of a fluid or plastic solid at rest, which occurs when external forces, such as gravity, are balanced by a pressure-gradient force.[1] In the planetary physics of Earth, the pressure-gradient force prevents gravity from collapsing the planetary atmosphere into a thin, dense shell, whereas gravity prevents the pressure-gradient force from diffusing the atmosphere into outer space.[2] [3]

Hydrostatic equilibrium is the distinguishing criterion between dwarf planets and small solar system bodies, and features in astrophysics and planetary geology. Said qualification of equilibrium indicates that the shape of the object is symmetrically ellipsoid, where any irregular surface features are consequent to a relatively thin solid crust. In addition to the Sun, there are a dozen or so equilibrium objects confirmed to exist in the Solar System.

## Mathematical consideration

For a hydrostatic fluid on Earth:

dP=-\rho(P)g(h)dh

### Derivation from force summation

Newton's laws of motion state that a volume of a fluid that is not in motion or that is in a state of constant velocity must have zero net force on it. This means the sum of the forces in a given direction must be opposed by an equal sum of forces in the opposite direction. This force balance is called a hydrostatic equilibrium.

The fluid can be split into a large number of cuboid volume elements; by considering a single element, the action of the fluid can be derived.

There are 3 forces: the force downwards onto the top of the cuboid from the pressure, P, of the fluid above it is, from the definition of pressure,

Ftop=-PtopA

Similarly, the force on the volume element from the pressure of the fluid below pushing upwards is

Fbottom=PbottomA

Finally, the weight of the volume element causes a force downwards. If the density is ρ, the volume is V and g the standard gravity, then:

Fweight=-\rhogV

The volume of this cuboid is equal to the area of the top or bottom, times the height — the formula for finding the volume of a cube.

Fweight=-\rhogAh

By balancing these forces, the total force on the fluid is

\sumF=Fbottom+Ftop+Fweight=PbottomA-PtopA-\rhogAh

This sum equals zero if the fluid's velocity is constant. Dividing by A,

0=Pbottom-Ptop-\rhogh

Or,

Ptop-Pbottom=-\rhogh

Ptop − Pbottom is a change in pressure, and h is the height of the volume element—a change in the distance above the ground. By saying these changes are infinitesimally small, the equation can be written in differential form.

dP=-\rhogdh

Density changes with pressure, and gravity changes with height, so the equation would be:

dP=-\rho(P)g(h)dh

### Derivation from Navier–Stokes equations

Note finally that this last equation can be derived by solving the three-dimensional Navier–Stokes equations for the equilibrium situation where

 u=v= \partialp = \partialx
 \partialp \partialy

=0

Then the only non-trivial equation is the

z

 \partialp \partialz

+\rhog=0

Thus, hydrostatic balance can be regarded as a particularly simple equilibrium solution of the Navier–Stokes equations.

### Derivation from general relativity

By plugging the energy momentum tensor for a perfect fluid

T\mu\nu=(\rhoc-2+P)u\muu\nu+Pg\mu\nu

into the Einstein field equations

R\mu\nu=

 8\piG c4

(T\mu\nu-

 1 2

g\mu\nuT)

and using the conservation condition

\nabla\muT\mu\nu=0

one can derive the Tolman–Oppenheimer–Volkoff equation for the structure of a static, spherically symmetric relativistic star in isotropic coordinates:
 dP =- dr
 GM(r)\rho(r) \left(1+ r2
 P(r) \right)\left(1+ \rho(r)c2
 4\pir3P(r) \right)\left(1- M(r)c2
 2GM(r) rc2

\right)-1

In practice, Ρ and ρ are related by an equation of state of the form f(Ρ,ρ)=0, with f specific to makeup of the star. M(r) is a foliation of spheres weighted by the mass density ρ(r), with the largest sphere having radius r:
 r M(r)=4\pi\int 0

dr'r'2\rho(r').

Per standard procedure in taking the nonrelativistic limit, we let c→∞, so that the factor
 \left(1+ P(r) \right)\left(1+ \rho(r)c2
 4\pir3P(r) \right)\left(1- M(r)c2
 2GM(r) rc2

\right)-11

Therefore, in the nonrelativistic limit the Tolman–Oppenheimer–Volkoff equation reduces to Newton's hydrostatic equilibrium:
 dP =- dr
 GM(r)\rho(r) r2

=-g(r)\rho(r)\longrightarrowdP=-\rho(h)g(h)dh

(we have made the trivial notation change h=r and have used f(Ρ,ρ)=0 to express ρ in terms of P).[4] A similar equation can be computed for rotating, axially symmetric stars, which in its gauge independent form reads:
 \partialiP P+\rho

-\partialilnut+utu\phi

\partial
 i u\phi ut

=0

Unlike the TOV equilibrium equation, these are two equations (for instance, if as usual when treating stars, one chooses spherical coordinates as basis coordinates

(t,r,\theta,\phi)

, the index i runs for the coordinates r and

\theta

).

## Applications

### Fluids

The hydrostatic equilibrium pertains to hydrostatics and the principles of equilibrium of fluids. A hydrostatic balance is a particular balance for weighing substances in water. Hydrostatic balance allows the discovery of their specific gravities. This equilibrium is strictly applicable when an ideal fluid is in steady horizontal laminar flow, and when any fluid is at rest or in vertical motion at constant speed. It can also be a satisfactory approximation when flow speeds are low enough that acceleration is negligible.

### Astrophysics

In any given layer of a star, there is a hydrostatic equilibrium between the outward thermal pressure from below and the weight of the material above pressing inward. The isotropic gravitational field compresses the star into the most compact shape possible. A rotating star in hydrostatic equilibrium is an oblate spheroid up to a certain (critical) angular velocity. An extreme example of this phenomenon is the star Vega, which has a rotation period of 12.5 hours. Consequently, Vega is about 20% larger at the equator than at the poles. A star with an angular velocity above the critical angular velocity becomes a Jacobi (scalene) ellipsoid, and at still faster rotation it is no longer ellipsoidal but piriform or oviform, with yet other shapes beyond that, though shapes beyond scalene are not stable.[5]

If the star has a massive nearby companion object then tidal forces come into play as well, distorting the star into a scalene shape when rotation alone would make it a spheroid. An example of this is Beta Lyrae.

Hydrostatic equilibrium is also important for the intracluster medium, where it restricts the amount of fluid that can be present in the core of a cluster of galaxies.

We can also use the principle of hydrostatic equilibrium to estimate the velocity dispersion of dark matter in clusters of galaxies. Only baryonic matter (or, rather, the collisions thereof) emits X-ray radiation. The absolute X-ray luminosity per unit volume takes the form

l{L}X(TB)\rho

 2 B
where

TB

and

\rhoB

are the temperature and density of the baryonic matter, and

Λ(T)

is some function of temperature and fundamental constants. The baryonic density satisfies the above equation

dP=-\rhogdr

:

pB(r+dr)-p

 B(r)=-dr \rhoB(r)G r2
 r \int 0

4\pi

 2\rho r M(r)

dr.

The integral is a measure of the total mass of the cluster, with

r

being the proper distance to the center of the cluster. Using the ideal gas law

pB=kTB\rhoB/mB

(

k

is Boltzmann's constant and

mB

is a characteristic mass of the baryonic gas particles) and rearranging, we arrive at
 d \left( dr
 kTB(r)\rhoB(r) \right)=- mB
 \rhoB(r)G r2
 r \int 0

4\pi

 2\rho r M(r)

dr.

Multiplying by
 2/\rho r B(r)
and differentiating with respect to

r

yields
 d \left[ dr
 r2 \rhoB(r)
 d \left( dr
 kTB(r)\rhoB(r) mB

\right)\right]=-4\pi

 2\rho Gr M(r).
If we make the assumption that cold dark matter particles have an isotropic velocity distribution, then the same derivation applies to these particles, and their density

\rhoD=\rhoM-\rhoB

satisfies the non-linear differential equation
 d \left[ dr
 r2 \rhoD(r)
 d \left( dr
 kTD(r)\rhoD(r) mD

\right)\right]=-4\pi

 2\rho Gr M(r).
With perfect X-ray and distance data, we could calculate the baryon density at each point in the cluster and thus the dark matter density. We could then calculate the velocity dispersion
 2 \sigma D
of the dark matter, which is given by
2
\sigma
 D= kTD mD

.

The central density ratio

\rhoB(0)/\rhoM(0)

is dependent on the redshift

z

of the cluster and is given by

\rhoB(0)/\rhoM(0)\propto

 2\left( \theta s
(1+z)

\right)3/2

where

\theta

is the angular width of the cluster and

s

the proper distance to the cluster. Values for the ratio range from .11 to .14 for various surveys.[6]

### Planetary geology

The concept of hydrostatic equilibrium has also become important in determining whether an astronomical object is a planet, dwarf planet, or small Solar System body. According to the definition of planet adopted by the International Astronomical Union in 2006, one defining characteristic of planets and dwarf planets is that they are objects that have sufficient gravity to overcome their own rigidity and assume hydrostatic equilibrium. Such a body will often have the differentiated interior and geology of a world (a planemo), though near-hydrostatic or formerly hydrostatic bodies such as the proto-planet 4 Vesta may also be differentiated and some hydrostatic bodies (notably Callisto) have not thoroughly differentiated since their formation. Often the equilibrium shape is an oblate spheroid, as is the case with Earth. However, in the cases of moons in synchronous orbit, nearly unidirectional tidal forces create a scalene ellipsoid. Also, the purported dwarf planet is scalene due to its rapid rotation, though it may not currently be in equilibrium.

Icy objects were previously believed to need less mass to attain hydrostatic equilibrium than rocky objects. The smallest object that appears to have an equilibrium shape is the icy moon Mimas at 396 km, whereas the largest icy object known to have an obviously non-equilibrium shape is the icy moon Proteus at 420 km, and the largest rocky bodies in an obviously non-equilibrium shape are the asteroids Pallas and Vesta at about 520 km. However, Mimas is not actually in hydrostatic equilibrium for its current rotation. The smallest body confirmed to be in hydrostatic equilibrium is the dwarf planet Ceres, which is icy, at 945 km, whereas the largest known body to have a noticeable deviation from hydrostatic equilibrium is Iapetus being made of mostly permeable ice and almost no rock.[7] At 1,469 km this moon is neither spherical nor ellipsoid. Instead, it is rather in a strange walnut-like shape due to its unique equatorial ridge.[8] Some icy bodies may be in equilibrium at least partly due to a subsurface ocean, which is not the definition of equilibrium used by the IAU (gravity overcoming internal rigid-body forces). Even larger bodies deviate from hydrostatic equilibrium, although they are ellipsoidal: examples are Earth's Moon at 3,474 km (mostly rock),[9] and the planet Mercury at 4,880 km (mostly metal).[10]

Solid bodies have irregular surfaces, but local irregularities may be consistent with global equilibrium. For example, the massive base of the tallest mountain on Earth, Mauna Kea, has deformed and depressed the level of the surrounding crust, so that the overall distribution of mass approaches equilibrium.

### Atmospheric modeling

In the atmosphere, the pressure of the air decreases with increasing altitude. This pressure difference causes an upward force called the pressure-gradient force. The force of gravity balances this out, keeping the atmosphere bound to Earth and maintaining pressure differences with altitude.

### Gemology

Gemologists use hydrostatic balances to determine the specific gravity of gemstones. A gemologist may compare the specific gravity they observe with a hydrostatic balance with a standardized catalogue of information for gemstones, helping them to narrow down the identity or type of gemstone under examination.

## References

• Book: White, Frank M. . Fluid Mechanics . 2008 . McGraw-Hill . New York . 978-0-07-128645-9 . 63–107 . Pressure Distribution in a Fluid .