text:
• Theory and Examples
• Exercises

### Systems of Equations

In the chapters Equations - Basics and Equations - Additional Aspects we discussed how to solve an equation for a single variable.

Now we turn to systems of equations: that means a set of two or more equations with the same variables.
An mxn-System (say: m times n system) consists of m equations with n unknowns.

### Examples

Example 1

$\left\{ \begin{array}{c} x + \large \frac{y}{3} \normalsize = 11 \\ \large \frac{x}{4} \normalsize + y = 11\; \end{array} \right. \qquad \qquad$ 2x2-system

Example 2

$\left\{ \begin{array}{c} 300c + f = 4000\\500c + f = 6400 \end{array} \right. \qquad \qquad$ 2x2-system

Example 3

$\left\{ \begin{array}{c}{x^2} + 2{y^2} = 33\\3x - 2y = 11 \end{array} \right. \qquad \qquad$ 2x2-system

Example 4

$\left\{ \begin{array}{c}2u + 4v - 5w = 21\\4u + 3v + 4w = - 1\\u - 3v - 4w = - 1.5\end{array} \right. \qquad \qquad$ 3x3-system

### Solutions

A solution to a system of equation is a set of values for the variables that satisfy all the equations simultaneously.

In example 1:

The pair  $x = 8$, $y = 9 \;$ statisfies both equations and is therefore a solution of the system.
L={(8,9)}

In example 4:

The ordered triple   $u = -0.5$, $v = 3$, $w=-2\;$ statisfies all three equations and is therefore a solution of the system.
L={(-0.5, 3, -2)}

### Linear and Non-linear Systems

Systems 1,2 and 4 of the examples above are linear, while system 3 is not.
"Linear" means that each term is either a constant or the product of a constant and (the first power of) a single variable.

The linearity of a system makes an important difference for the decision which method we choose to solve it.

In the next paragraphs we will present (for the special case of linear 2x2-systems) three important solution methods:

• substitution
• comparison

All three methods try to achieve a common goal: they want to eliminate one of the unknowns in order to come up to one single equation with one variable. This equation can be solved then with the methods presented in the chapters Equations - Basics and Equations - Additional Aspects to obtain the value of the first variable. By inserting it into one of the equations, we find the second part of the solution.

Many of the methods presented here can also be applied to non-linear systems and to systems with more than two variables. We will get back to these aspects later, in separate sections.

### Substitution Method

Main Principle:

Solve one of the equations for either of the variables and replace this variable in the other equation by the expression just found. Thereby, the "substituted" variable gets eliminated.

In example 1:

$\left\{ \begin{array}{l}\;x + \large \frac{y}{3} \normalsize = 11 \\ \large \frac{x}{4} \normalsize + y = 11\;\end{array} \right. \qquad \qquad$

Solve the first equation for x :

$x = 11 - \large \frac{y}{3}$

Substitute x in the second equation by the expression $11 - \large \frac{y}{3}$:

$\large \frac{{11 - \frac{y}{3}}} {4} \normalsize + y = 11$

After multiplying by 4 you get:

$\begin{eqnarray} 11 - \frac{y}{3} + 4y &=& 44\quad &| \cdot 3 \\ 33 - y + 12y &=& 132\quad &| - 33 \\ 11y &=& 99\quad &|\;:11 \\ y &=& 9 \end{eqnarray}$

Now plug $y=9$ into the transformed equation $x = 11 -\large \frac{y}{3}$:

$x = 11 - \large \frac{9}{3} \normalsize = 8$

The solution of the system is:  $\quad x = 8$ , $y = 9$

### Comparison Method

Main Principle:

Solve both equations to either (but the same) variable and set them equal. Thereby, the "compared" variable gets eliminated.

In example 2:

$\left\{ \begin{array}{l}300c + f = 4000\\500c + f = 6400 \end{array} \right.$

Solve the first equation for f :$\quad \; f = 4000 - 300c$
Solve the second equation for f : $\; f = 6400 - 500c$

Set the right-hand parts equal:

$\begin{eqnarray} 4000 - 300c &=& 6400 - 500c\quad &|\;+ 500c - 4000 \\ 200c &=& 2400\quad &|\;:200\\ c &=& 12 \end{eqnarray}$

Now plug $c=9$ into either of the equations with isolated f (e.g. $f = 4000 - 300c$):

$f = 4000 - 3600 = 400$

The solution of the system is: $\quad c = 12$ , $f = 400$

Main Principle:

Multiply (if needed) one or both of the equations by suitable numbers, in order to get the same coefficients for one of the variables in both equations, once with positive, once with negative sign.
Then add both equations, thereby eliminating that variable.

In example 5

$\left\{\begin{array}{c}5x + 2y = - 4\\ - 4x + y = 11\end{array} \right.$

Multiplying equation 2 by (-2) and using it in place of the "older" one, we get the new (equivalent) system

$\left\{\begin{array}{c}5x + 2y = - 4\\8x - 2y = -22 \end{array} \right.$

Addition of the two equations leads to a new equation for the remaining variable x (whereas y has dropped out, which was the purpose of the previous multiplication).
It can easily be solved for x:

$\begin{eqnarray} 13x &=& - 26\quad |\;:13\\x &=& -2 \end{eqnarray}$

Finally, we use either of the original equations to get y (by inserting $x=-2$):

$\begin{eqnarray} 5x + 2y &=& - 4\\ - 10 + 2y &=& - 4\quad |\; + 10\\y &=& 3 \end{eqnarray}$

The solution of the system is:  $x=- 2 \;,\;y = 3$

Remarks:

• Instead of multiplying by (-2) and adding there had been the possibility of multiplying by +2 and then subtracting. Indeed you could eliminate the variable f in example 2 by a simple subtraction (without a preceding multiplication).
• Sometimes it makes sense to multiply both of the equations before adding them up.
here for an additional example.
•

### Solvability of Linear Systems

In all of the linear 2x2-systems presented above there was a unique solution; this is the normal case.

But there are other possibilities:
- the system has no solution
- the system has infinitely many solutions.

An equation with two variables x and y can be understood as the defining equation of a function y=f(x) whose graph is a straight line in the xy-plane. (For more details see chapter Introduction to Functions).
Therefore, the solution of a linear 2x2-system can be interpreted as the intersection point of two lines. If the lines do intersect, we have a unique solution; in the special cases we mentioned above we have either parallel lines (no solution) or coinciding lines (infinitely many solutions).

When calculating, these special cases can be recognized by the fact that the elimination of the one variable leads to the simultaneous elimination of the second one, too.

Example 6

$\left\{\;\begin{eqnarray}2x - y &=& - 2\\ - 4x + 2y &=& 4 \end{eqnarray} \right.$

We multiply the first equation by 2, intending to eliminate y; the resulting system is

$\left\{ \; \begin{eqnarray}4x - 2y &=& -4\\-4x + 2y &=& 4 \end{eqnarray} \right.$

But now, the addition of the equations eliminates not only y, but also x; we get

$0=0$

This is an identity; the result means that both equations have the same solutions (which was obvious already after the multiplication, because the resulting equations were identical - except for the sign). Hence the solution set consists of all pairs (x,y) satisfying the equation $2x-y=-2$.

The system has infinitely many solutions.

Note: In case of no solutions you will get an "impossible" equation as $\;0=2\,$, for example.

### Non-linear Systems

In non-linear systems, equations may contain powers or products of the variables. Therefore, it might be difficult to find a suitable additive operation to eliminate one of the variables.

The first choice for solving non-linear systems is the substitution method; it is less ambitious than the comparison method, because it is sufficient to isolate the variable in one equation (for applying the comparison method, this would have to be done in both equations).

In example 3:

$\;\left\{ \begin{array}{rl} \;{x^2} + 2{y^2} = 33\\3x -2y = 11 \end{array} \right.$

The second equation (which is linear) can be solved for y easily:

$\begin{eqnarray} 3x - 2y = 11\quad &|\, - 3x \\- 2y = 11 - 3x\quad &|\,:( - 2) \\ y = - 5.5 + 1.5x \end{eqnarray}$

Now replace y with this term in the first equation:

${x^2} + 2 \cdot {( - 5.5 + 1.5x)^2} = 33$

Make use of the Binomial Formulas to simplify the left-hand expression:

$\begin{eqnarray}{x^2} + 2 \cdot (30.25 - 16.5x + 2.25{x^2}) &=& 33\\{x^2} + 60.5 - 33x + 4.5{x^2} &=& 33 \\(5.5{x^2} - 33x + 27.5 &=& 0 \end{eqnarray}$

This is a quadratic equation for x that can be solved by applying the quadratic formula (see Chapter Equations - Basics):

${x_{1\,,2}} = \large \frac{{33 \pm \sqrt {{{( - 33)}^2} - 4 \cdot 5.5 \cdot 27.5} }}{{2 \cdot 5.5}} = \frac{{33 \pm \sqrt {484} }}{{11}} = \frac{{33 \pm 22}}{{11}}$

So we have two possible solutions for x:

${x_1} = 5 \quad {x_2} = 1$

To each of these values you get a "partner" y by inserting it into the equation $y = - 5.5 + 1.5x$ above:

${y_1} = - 5.5 + 1.5 \cdot 5 = 2$

${y_2} = - 5.5 + 1.5 \cdot 1 = - 4$

Therefore the non-linear system has two solutions:

$L = \left\{ {(5,2);(1\,, - 4)} \right\}$

### Systems of Equations with more than two Variables

With more than two equations and variables the procedure remains basically the same. Again we try to get a simpler system by eliminating the variables step by step.

Example 7

A linear 3x3-system:

$\left\{ \begin{array}{c} 2x + y + 4z = - 2\\ - x - 2y + z = - 5\\3x - 8y - 7z = 23 \end{array} \right. \qquad$ (*)

Multiply, for example, the first equation by 2 and add the result to the second equation:

$\left\{ {\begin{array}{*{20}{c}}{4x + 2y + 8z = - 4}\\ {- x - 2y + z = - 5}\end{array}} \right. \;\begin{array}{*{20}{c}} + \\ + \end{array} \quad \rightarrow \quad 3x + 9z = - 9$

Of course you will need a second equation, because you can't determine two variables (x and z) from one single condition. It seems obvious that you have to make use of the information contained in equation 3; but above all it is important to eliminate the same variable y as before (even if it were easier to eliminate another one).

For example, multiply the second equation by (-4) and add the result to the third equation:

$\left\{ {\begin{array}{*{20}{c}}{4x + 8y - 4z = 20}\\{3x - 8y -7z = 23}\end{array}} \right. \;\begin{array}{*{20}{c}} + \\ + \end{array} \quad \rightarrow \quad 7x - 11z = 43$

With the two new equations we have a linear 2x2-System for x and z:

$\left\{ \begin{array}{c} 3x + 9z = - 9\\7x - 11z = 43 \end{array} \right. \qquad$ (**)

In this new system we have to eliminate the next variable, say z.
This can be done by multiplying the first equation by 11 and the second one by 9, then by adding the resulting equations:

$\left\{ {\begin{array}{*{20}{c}}{33x + 99z = - 99}\\{63x - 99z = 387}\end{array}} \right. \;\begin{array}{*{20}{c}} + \\ + \end{array} \quad \rightarrow \quad 96x = 288 \rightarrow \quad x = 3$

That value $x=3$ can be inserted into one of the equations of system (**), which gives us z:

For example, in equation 1: $9 + 9z = - 9 \quad \rightarrow \quad z = - 2$

And finally both values $x=3$ and $z=-2$ are inserted in equation 1 of the original system (*):

$6 + y - 8 = - 2\quad \rightarrow \quad y = 0$

(This last step-by-step procedure for finding the values of the variables is often referred to as backward substitution.)

The system has the solution

$L = \left\{ {(3,0, - 2)} \right\}$

As you can see the whole procedure has become much more complicated. The more unknowns you have the more difficult it becomes to keep track of the elimination process.

The usual approach for a systematic elimination of the variables is named after the mathematician Carl Friedrich Gauss (1777-1855),, therefore known as Gauss-Elimination. It can be used for hand-calculation, but its main importance is based on the fact that the algorithm is particularly suitable for computers.

You will learn more about systems of linear equations (with three and more variables) in the course Mathematics for Management and Economics of the IM&E Programme.

1)

Solve the following systems of linear equations (try to choose a beneficial method):

a)

$\left\{ \; \begin{eqnarray}2x -3y &=& 5 \\ x + 2y &=& 6 \end{eqnarray} \right.$

b)

$\left\{ \; \begin{eqnarray}7x &=& -3y+1 \\ 7x &=& 2y-3 \end{eqnarray} \right.$

c)

$\left\{ \; \begin{eqnarray}0.3x + 0.4y &=& 0.5\\ 0.7x +1.1y &=& 1.6 \end{eqnarray} \right.$

d)

$\left\{ \; \begin{eqnarray}6x + 5y &=& - 6\\ y &=& 3 - 4x \end{eqnarray} \right.$

e)

$\left\{ \; \begin{eqnarray}8x -5y &=& 5.5 \\ - x + y &=& 1 \end{eqnarray} \right.$

f)

$\left\{ \; \begin{eqnarray}2x - 3y &=& 10\\ 7x + 2y &=& 2.5 \end{eqnarray} \right.$

2)

Solve the following non-linear systems:

a)

$\left\{ \; \begin{eqnarray} {x^2} - 2y &=& 0 \\ x + y &=& 4 \end{eqnarray} \right.$

b)

$\left\{ \; \begin{eqnarray} x + 2y &=& 2\\ xy - y &=& -3 \end{eqnarray} \right.$

3)

Solve the following systems of equations with three variables:

a)

$\left\{ \; \begin{eqnarray} x +4y-5z &=& 21 \\ 2x +3y+4z &=& -1 \\ x-6y-8z&=&-3 \end{eqnarray} \right.$

b)

$\left\{ \; \begin{eqnarray}3x+2y-8z &=& -5 \\ -x+2z &=& 3 \\ x+ y-5z &=& 7 \end{eqnarray} \right.$

4)

Solve the following systems of equations.
In a first step, before using the methods discussed in the theory section, you have to simplify the equations.

a)

$\left\{ \; \begin{eqnarray} \frac {x-y}{4}- \frac {3(y-2)}{7} &=& 2 \\ \frac {x - y}{x+y} &=& 2 \end{eqnarray} \right.$

b)

$\left\{ \; \begin{eqnarray}x(y+1) &=&(x+1)(y-2) \\ \frac {x + y}{2} &=& 2x+1 \end{eqnarray} \right.$

c)

$\left\{ \; \begin{eqnarray} \frac {x-1}{y} &=&2 \\{x^2}-y &=& 2 \end{eqnarray} \right.$

Theory with explanations and examples:

Video: Solving Linear Systems of Equations   (2 variables, addition method)

Video: Solving Linear Systems of Equations   (3 variables)

Exercises:

Math-Drills: Systems of Linear Equations  (many worksheets with exercises; different levels)

Math-Worksheets: Systems of Linear Equations  (worksheet with exercises for download; 2 and 3 variables)

Purplemath: Systems of Non-Linear Equations  (examples on pages 3, 4 and 5)


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# Help w/ home networking printer

Gold Member
I want my laptop to be able to communicate directly with my printer.

OK, first let's back up. I'm running XP Home on my two home systems - laptop and desktop. They are both directly connected to the internet through the wireless access point / router. Because of this, as I understand it, my two computers cannot both see each other - this is forbidden by XP Home. I can only configure them to see each other if I set up one as a server and route the other through it.

Right?

Anyway, I want my laptop to be able to communicate directly with the printer (Since I cannot network it, since my laptop cannot see my desktop on the network.)

I bought a print server, not realizing I would need a wireless print server, so now I've got one of those too. But if I hook up the wireless print server, my desktop can't connect to the printer. I'm $240 in the hole on this, and I still can't print from my laptop. Can anyone advise me on how to hook this whole thing up? See attached diagram of my setup. ## Answers and Replies Related Computing and Technology News on Phys.org I want my laptop to be able to communicate directly with my printer. OK, first let's back up. I'm running XP Home on my two home systems - laptop and desktop. They are both directly connected to the internet through the wireless access point / router. Because of this, as I understand it, my two computers cannot both see each other - this is forbidden by XP Home. I can only configure them to see each other if I set up one as a server and route the other through it. Right? No. if your AP/router has DHCP enabled, and your computers get IP addresses from it, they should be able to see each other without any problems. If file and print sharing are enabled on your computers, you should be able to share files between the two. I do it with mine all the time (1 XP Home, 1 XP Pro). I even have Apache and MySQL running on my desktop that I can access from my laptop. Check the Windows firewall and make sure it's set up to allow file and printer sharing (Control Panel/Windows Firewall/Exceptions) to pass through. Anyway, I want my laptop to be able to communicate directly with the printer (Since I cannot network it, since my laptop cannot see my desktop on the network.) I bought a print server, not realizing I would need a wireless print server, so now I've got one of those too. But if I hook up the wireless print server, my desktop can't connect to the printer. I'm$240 in the hole on this, and I still can't print from my laptop.

Can anyone advise me on how to hook this whole thing up?
In my printer setup, the printer is connected to my desktop via USB and shared as a network printer. I can print to it from my laptop via wireless or wired (plugged into the router), but this requires the desktop to be up and running.

The print server gets set up as another node on the network (config is usually done by accessing the print server via web browser). As long as it's on the same subnet, the other computers should be able to see it as a network printer and connect to it.

Have you checked the printer compatibility chart for your print server? I've read that many print servers only support a limited number of printers

Gold Member
No.

if your AP/router has DHCP enabled, and your computers get IP addresses from it, they should be able to see each other without any problems.
Someone else mentioned this a long time ago, and I think I tried it and failed.

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###### Business · Accounting
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Novak Corp. issued $400,000, 6%, 25-year bonds on January 1, 2017, for$ 353,386. This price resulted in an effective-interest rate of  7% on the bonds. Interest is payable annually on January 1. Novak uses the effective-interest method to amortize bond premium or discount.

Prepare the schedule using effective-interest method to amortize bond premium or discount of Novak Corp.. (Round answers to 0 decimal places, e.g. 5,250.)

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Spring 2018

Below is a list of seminars sponsored by the Mathematics and Computer Science department at Adelphi University. All talks are open to the public and most talks will be accessible to undergraduate students who have had linear algebra. If you have any questions feel free to contact Branden Stone.

Unless otherwise stated, all talks will be at the following location and time:
Location: Science Building, Room 321;
Time: talks start at 1:30pm.
Feel free to suscribe to the Google Calendar for the most recent information.

Feb 9, 2018

Title: Inroduction to the Math Behind Texas Hold'em

Abstract: The traditional depiction of poker evokes images of rough and unsavory characters, sitting in smoke-filled rooms, reading their opponents in an ultimate test of daring and will. In reality, the poker boom of the 21st Century has made poker more mainstream and given rise to a different type of poker player – analytic individuals utilizing probability, game theory, and other mathematical concepts to hone their skills and maximize their earning potential. In fact, researchers at Carnegie Mellon University developed a poker-playing AI that defeated four of the world’s top poker professionals in January 2017.

This lecture will go over some of the core fundamental principles behind the math of Texas Hold Em poker. Using the Fundamental Theorem of Poker (as expressed by David Sklansky) as our starting point, we will discuss how to evaluate the expected value of decisions we encounter in poker, when faced with incomplete information. The lecture will also introduce some common situations that arise in Texas Hold Em and their associated probabilities, and discuss methods for applying these principles to in-game situations. [PDF Poster]

Apr 6, 2018
Janet Vassilev :: University of Arizona

Title: Winning the Hat Game

Abstract: In this talk, we will introduce the hat game and the mathematics behind winning. In fact, the main ingredients involve a little combinatorics, some linear algebra and error correcting codes. Throughout the course of the talk, volunteers will be asked to put their thinking “hats” on to play the hat game with their peers to exhibit first hand that they can win the hat game.

Apr 20, 2018
Dan Summers :: Drexel University

Title: A RSK-Y Proposition

Abstract: Often times in Mathematics, seemingly disjoint fields are smashed together in order to make sense of problems which are difficult to solve in their own domain. The Calculus of Newton and Leibniz is founded upon the idea of using algebraic techniques to solve geometric problems which dated back to the Ancient Greeks. Linear Algebra has morphed from a method to solve systems of linear equations into a way to analyze geometric spaces by studying the functions from one to another. When one is first learning about a new branch of Mathematics, this approach might be scary, and it may seem that the amount of time needed to understand the links between different branches is not worth it, but it almost always is!

In this talk, we will discuss two areas of mathematics, Algebra and Combinatorics, which have been paired together with great results. We present two sets of combinatorial objects which are equinumerous, the Robinson–Schensted–Knuth correspondence between these two sets, and how the RSK correspondence can produce an unexpected result about the product of two polynomials. Along the way, we examine some of the algebraic and combinatorial properties of the symmetric polynomials.

Fall 2017

Below is a list of seminars sponsored by the Mathematics and Computer Science department at Adelphi University. All talks are open to the public and most talks will be accessible to undergraduate students who have had linear algebra. If you have any questions feel free to contact Branden Stone.

Unless otherwise stated, all talks will be at the following location and time:
Location: Science Building, Room 321;
Time: talks start at 1:30pm.
Feel free to suscribe to the Google Calendar for the most recent information.

Sep 8, 2017
Johann Thiel :: New York City College of Technology

Title: RATS Sequences in General Bases

Abstract: Conway’s RATS sequences are generated by repeating the following process: begin with a positive integer, reverse the order of the digits, add the original integer to its reverse, then sort the sum’s digits in increasing order from left to right. We consider this process for integers written in other bases besides 10 and discuss the long-term behavior of such sequences. In particular, we will examine the existence of periodic and quasiperiodic (a special type of divergence) sequences for certain bases. [PDF Poster]

Oct 20, 2017
Rick Royce :: Webb Institute

Title: Introduction to Artificial Neural Networks

Abstract: Artificial Neural Networks (ANN) are mathematical models that are able to learn relationships between inputs and desired outputs. Their structure mimics the synaptic networks within the human brain. The knowledge learned by the network is contained in the inter-neuron connections, or weights, that are integral to the model. The value of ANN’s is that they are able to model extremely complex non-linear relationships that other models, for example polynomials, are unable to replicate. This talk will cover some basic concepts of ANN’s, discuss the mathematical development of a simplified ANN, discuss various applications, and show how ANN’s are being used in the field of Naval Architecture. [PDF Poster]

Nov 10, 2017
Eloísa Grifo :: University of Virginia

Title: Slicing a square pizza

Abstract: It is easy to slice a pizza into triangles of the same area, but what if the pizza has a square shape? This is still easy if the number of slices is even, but what if it is odd? The problem turns out to be surprisingly difficult, and it was finally solved by Paul Monsky in 1970. What is even more surprising are the tools that Monsky used. The moral of the story is: simple sounding questions might have surprising answers, and since you don’t know what kind of mathematics you may need to solve a problem in which you are interested, you should learn everything.

Nov 17, 2017
Mauricio Nunez-Regueiro :: University of Arizona

Title: Quantitative analysis tools to inform policy and achieve a sustainable future with big data

Abstract: Our planet is undergoing one of the most severe environmental crisis known to mankind. Multiple species are on the verge of disappearing and our changing climate is affecting how all living things experience our environment. Fortunately, the rate at which we gather data and develop novel analytical methodologies allows us to stay ahead of the curve and inform management and policy to help achieve sustainable livelihoods. In this talk, I present three case-studies where collaborators and I implement quantitative tools to uncover critical biological processes than help inform management actions and improve existing policy mechanisms. I will discuss how 1) using meta-analytic tools of published data I can assess how global biofuel production can affect biodiversity in different ecosystems. Then, I will focus on work done in one the most endangered forested ecosystems in the world: the Chaco forest. Here, I will 2) use occupancy models to analyze a deforestation policy rule that affects native mammals and 3) use spatial statistics to analyze a national-level market-based incentive that seeks to decrease deforestation and protect critical forested lands. I will finish by discussing future research directions where a professor at Adelphi University and I are collaborating to integrate mathematical models and spatial statistics to answer challenging questions in environmental sciences.

Spring 2017

Below is a list of seminars sponsored by the Mathematics and Computer Science department at Adelphi University. All talks are open to the public and most talks will be accessible to undergraduate students who have had linear algebra. If you have any questions feel free to contact Branden Stone.

Unless otherwise stated, all talks will be at the following location and time:
Location: Science Building, Room 321;
Time: talks start at 1:30pm.
Feel free to suscribe to the Google Calendar for the most recent information.

Mar 24, 2017
Ananth Hariharan :: Indian Institute of Technology Bombay

Title: Assumptions, Approximations and Error

Abstract: We will warm up with a simple example of mathematical modelling, highlighting the need to make assumptions to convert a real-life problem into a mathematical one. We will then see a method on how to solve an inconsistent system of linear equations approximately, and if possible, find the error.

Apr 14, 2017
Rebecca R.G. :: Syracuse University

Title: Fiber Arts and Math: Modeling Hyperbolic Spaces

Abstract: For centuries, mathematicians thought that Euclid’s geometric postulates always worked. But sometimes they don’t… Using crocheted models, we will explore the properties of the hyperbolic plane. Come learn about the spaces that stumped mathematicians for decades, and about which Farkas Bolyai once said, “You must not attempt this approach to parallels. I know this way to its very end. I have traversed this bottomless night, which extinguished all light and joy of my life. I entreat you, leave the science of parallels alone….”

Fall 2016

Below is a list of seminars sponsored by the Mathematics and Computer Science department at Adelphi University. All talks are open to the public and most talks will be accessible to undergraduate students who have had linear algebra. If you have any questions feel free to contact Branden Stone.

Unless otherwise stated, all talks will be at the following location and time:
Location: Science Building, Room 321;
Time: talks start at 1:30pm.
Feel free to suscribe to the Google Calendar for the most recent information.

Sep 9, 2016
Courtney Gibbons :: Hamilton College

Title: Statistics to Algebra and Back Again

Abstract: In this talk, we’ll consider a routine problem from statistics, maximum likelihood estimation in a small coin flipping example, and find a solution using algebra. The goal of the talk is not really how to solve this problem, but to demonstrate how new courses of study in algebra arise by considering such problems from statistics. [PDF Poster]

Sep 30, 2016
Haydee Lindo :: Williams College

Title: Into the Unknown

Abstract: This talk will be a wild ride of new and interesting mathematics. Come find out what glorious things await in this talk of unknown content. Who knows, maybe all of your wildest dreams will come true. This talk is open to students at all levels. [PDF Poster]

Oct 21, 2016
Kristen Beck :: St Mary's College of California

Title: Irrationality: Like Blackness in the Night Sky

Abstract: Recall that every real number falls into exactly one of two categories: either it’s rational (that is, it can be expressed as a ratio, or fraction, of two whole numbers), or it’s not. Though you probably learned this seemingly fundamental fact in middle school or high school, it’s important to remember that the concept of irrationality was not always so widely accepted. In fact, the discovery of irrational numbers (by the Pythagorean philosopher Hippasus in the 5th century BC) is rumored to have greatly angered the gods. Despite the fact that mankind’s relationship with irrational numbers has come a very long way in the past two-and-a-half millennia, we still seem to favor rational numbers. It’s famously been said that rational numbers are like the stars in the night sky. In this talk, however, we will focus on the often-overlooked blackness in the night sky: the irrational numbers. And we will do this by considering “rational approximations” of them. Through this process, we will uncover a number which is more irrational than all of the others. And, what’s even more surprising: this particular number is ubiquitous in nature precisely because of its irrationality! [PDF Poster]

Nov 11, 2016
Gabriel Sosa :: Amherst College

Title: Monomial orders uniquely determined by their induced orders

Abstract: When were are asked to write the polynomial $p(x)=3x^2+x^4+7-x$ in standard form we know our answer is $x^4+3x^2-x+7$, because the only way to order powers of $x$ is the natural one: degree wise! Complexity is introduced when we are asked to write $q(x)=2x^3yz^2-6x^4z+x^2y^3z$ in standard form and we have to define what standard form means in this setting. In this talk, we will present the three classical monomial orders (\textit{lexicographic}, \textit{degree lexicographic} and \textit{reverse lexicographic}), explore a curious property they possess (hint hint: It is the title) and discuss whether other monomial orders possess this property. [PDF Poster]

Spring 2016

Below is a list of seminars sponsored by the Mathematics and Computer Science department at Adelphi University. All talks are open to the public and most talks will be accessible to undergraduate students who have had linear algebra. If you have any questions feel free to contact Branden Stone.

Unless otherwise stated, all talks will be at the following location and time:
Location: Science Building, Room 321;
Time: talks start at 1:30pm.
Feel free to suscribe to the Google Calendar for the most recent information.

Feb 12, 2016
Tom Enkosky :: Boston University

Title: Flows on Graphs

Abstract: This is an expository talk on the theory of flows on a graph. We will define the flow on a directed graph and discuss several solved problems and current open problems. No prior knowledge of graph theory is necessary. [PDF Poster]

Mar 4, 2016
Johanna Franklin :: Hofstra University

Title: A truly random talk

Abstract: Most people can intuitively classify binary sequences as “random” or “not random.” However, as mathematicians, we want a precise mathematical definition. In this talk, I will define randomness with concepts from computability theory using three different intuitive approaches as starting points: unpredictability, incompressibility, and a lack of distinguishing properties. At the end, I’ll present an example of a random sequence and talk about current directions in the field. [PDF Poster]

Mar 21, 2016
HHE Room 209, 1:25-2:15pm
Cassie Williams :: James Madison University

Title: How elliptic curves can keep secrets and prove theorems

Abstract: Elliptic curves (and their higher dimensional analogues) have proven to be interesting, powerful, and enigmatic objects of study; on one hand elliptic curves are often a topic in undergraduate courses on group or number theory, yet there are deep open problems that arise from simple questions about their structure. In this talk we will think about how to do arithmetic on small sets, define elliptic curves in two different ways, explore their geometric and algebraic structure, and explain some of their uses within and outside of mathematics. Only the definition of a prime number will be assumed. [PDF Poster]

Apr 8, 2016
Amir Barghi :: Bard College

Title: A Combinatorial Variation of Fermat's Last Theorem

Abstract: The equation $\left(x \mid k \right)_{n} + \left(y \mid k \right)_{n} = \left(z \mid k \right)_{n}$ is a degree-$n$, falling-factorial version of the Fermat equation, where $\left(x \mid k \right)_{n} =x(x-k)(x-2k)\cdots(x-(n-1)k)$. In this talk, I will give combinatorial interpretations of $\left(x \mid k \right)_{n}$ and $\left(z \mid k \right)_{n} - \left(y \mid k \right)_{n}$ in terms of a probabilistic game and make some observations on their combinatorial properties using these interpretations. [PDF Poster]

Apr 29, 2016
Alyssa Gottshall

Title: Math/Computer Science Careers at NSA

Abstract: The talk will include an overview on NSA, its history, mission areas, and career fields. There will be information on student programs and full-time employment to include hiring requirements and benefits. The talk also includes pros and cons of working at NSA, then some specifics on how math is used throughout the agency. If there is time, we can discuss career paths, advancements, and some unique opportunities that the agency provides to employees. [PDF Poster]

Fall 2015

Below is a list of seminars sponsored by the Mathematics and Computer Science department at Adelphi University. All talks are open to the public and most talks will be accessible to undergraduate students who have had linear algebra. If you have any questions feel free to contact Branden Stone.

Unless otherwise stated, all talks will be at the following location and time:
Location: Post Hall, Room 207;
Time: talks start at 1:30pm.

Sep 18, 2015
Gywn Whieldon :: Hood College

Fall 2014

Below is a list of seminars sponsored by the Mathematics and Computer Science department at Adelphi University. All talks are open to the public and most talks will be accessible to undergraduate students who have had linear algebra. If you have any questions feel free to contact Branden Stone.

Unless otherwise stated, all talks will be at the following location and time:
Location: Post Hall, Room 207;
Time: talks start at 1:30pm.

Nov 21, 2014
Tom Enkosky :: United States Coast Guard Academy

Title: Graph Rigidity

Abstract: A graph is a collection of vertices and edges and is often depicted visually with points and straight line segments. Suppose we draw a graph in the plane and fix the lengths of the line segments but allow the edges to pivot about the vertices. The drawing is rigid if the graph keeps its shape when the vertices are moved. In this talk we will explore some of the algebra of graph rigidity. [PDF Poster]


text:
# Contents

## Idea

In quantum field theory the term gauge boson denotes the quantum of a gauge field.

In terms of Chern-Weil theory/differential cohomology we have: gauge bosons are the quanta of connections on bundles.

For instance

Created on April 19, 2011 12:13:31 by Urs Schreiber (131.211.232.210)


text:
# Library prosa.analysis.facts.model.service_of_jobs

In this file, we establish basic facts about the service received by sets of jobs.
To begin with, we provide some basic properties of service of a set of jobs in case of a generic scheduling model.
Consider any type of tasks ...

... and any type of jobs associated with these tasks.
Context {Job : JobType}.
Context {JobTask Job Task}.
Context {JobArrival Job}.
Context {JobCost Job}.

Consider any kind of processor state model, ...
Context {PState : ProcessorState Job}.

... any job arrival sequence with consistent arrivals, ....
... and any schedule of this arrival sequence ...
Variable sched : schedule PState.

... where jobs do not execute before their arrival or after completion.
Let P be any predicate over jobs.
Variable P : pred Job.

We show that the total service of jobs released in a time interval [t1,t2) during [t1,t2) is equal to the sum of: (1) the total service of jobs released in time interval [t1, t) during time [t1, t) (2) the total service of jobs released in time interval [t1, t) during time [t, t2) and (3) the total service of jobs released in time interval [t, t2) during time [t, t2).
We show that the total service of jobs released in a time interval [t1, t2) during [t, t2) is equal to the sum of: (1) the total service of jobs released in a time interval [t1, t) during [t, t2) and (2) the total service of jobs released in a time interval [t, t2) during [t, t2).
In the following, we consider an arbitrary sequence of jobs jobs.
Variable jobs : seq Job.

Also, consider an interval [t1, t2)...
Variable t1 t2 : instant.

...and two additional predicates P1 and P2.
Variable P1 P2 : pred Job.

For brevity, in the following comments we denote a subset of jobs satisfying a predicate P by {jobs | P}.
First, we prove that the service received by {jobs | P1} can be split into: (1) the service received by {jobs | P1 P2} and (2) the service received by the a subset {jobs | P1 ¬ P2}.
We show that the service received by {jobs | P} is equal to the difference between the total service received by jobs and the service of {jobs | ¬ P}.
We show that service_of_jobs is monotone with respect to the predicate. That is, given the fact j jobs: P1 j ==> P2 j, we show that the service of {jobs | P1} is bounded by the service of {jobs | P2}.
Similarly, we show that if P1 is equivalent to P2, then the service of {jobs | P1} is equal to the service of {jobs | P2}.
Next, we show an auxiliary lemma that allows us to change the order of summation.
Recall that service_of_jobs is defined as a sum over all jobs in jobs of [service_during j [t1,t2)]. We show that service_of_jobs over an interval [t1,t2) is equal to the sum over individual time instances (in [t1,t2)) of service_of_jobs_at.
In other words, we show that [∑_{j ∈ jobs} ∑_{t \in [t1,t2)} service  of j at t]>> is equal to [∑_{t \in [t1,t2)} ∑_{j ∈ jobs} service of j  at t]>>.
Finally, we show that service of {jobs | is equal to 0.
In this section, we prove some properties about service of sets of jobs for unit-service processor models.
Consider any type of tasks ...

... and any type of jobs associated with these tasks.
Context {Job : JobType}.
Context {JobArrival Job}.
Context {JobCost Job}.

Consider any kind of unit-service processor state model.
Consider any arrival sequence with consistent arrivals.
Next, consider any unit-service schedule of this arrival sequence ...
... where jobs do not execute before their arrival or after completion.
Let P be any predicate over jobs.
Variable P : pred Job.

First, we prove that the service received by any set of jobs is upper-bounded by the corresponding workload.
In this section, we introduce a connection between the cumulative service, cumulative workload, and completion of jobs.
Consider an arbitrary time interval [t1, t2).
Variables t1 t2 : instant.

Let jobs be a set of all jobs arrived during [t1, t2).
Next, we consider some time instant t_compl.
Variable t_compl : instant.

And state the proposition that all jobs are completed by time t_compl. Next we show that this proposition is equivalent to the fact that workload of jobs = service of jobs.
First, we prove that if the workload of jobs is equal to the service of jobs, then any job in jobs is completed by time t_compl.
And vice versa, the fact that any job in jobs is completed by time t_compl implies that the workload of jobs is equal to the service of jobs.
Using the lemmas above, we prove the equivalence.
In this section, we prove some properties about service of sets of jobs for unit-service uniprocessor models.
Consider any type of tasks ...

... and any type of jobs associated with these tasks.
Context {Job : JobType}.
Context {JobTask Job Task}.
Context {JobArrival Job}.
Context `{JobCost Job}.

Consider any kind of unit-service uniprocessor state model.
Consider any arrival sequence with consistent arrivals.
Next, consider any unit-service uni-processor schedule of this arrival sequence ...
... where jobs do not execute before their arrival or after completion.
Let P be any predicate over jobs.
Variable P : pred Job.

In this section, we prove that the service received by any set of jobs is upper-bounded by the corresponding interval length.
Let jobs denote any set of jobs.
Variable jobs : seq Job.
Hypothesis H_no_duplicate_jobs : uniq jobs.

We prove that the overall service of jobs at a single time instant is at most 1.
Next, we prove that the service received by those jobs is no larger than their workload.
Corollary service_of_jobs_le_length_of_interval:
(t : instant) (Δ : duration),
service_of_jobs sched P jobs t (t + Δ) Δ.

The same holds for two time instants.


text:
## Does P not NP imply NP COMPLETE disjoint from RP?

According to Wikipedia https://en.wikipedia.org/wiki/RP_(complexity), $$P \ne NP$$ implies that $$RP$$ is a strict subset of $$NP$$. Does anybody have a reference? Furthermore, am I correct that if this indeed the case, then $$NP-COMPLETE \cap RP = \emptyset$$ since we can use $$NP$$ completeness to solve all other $$NP$$ problems?

## Auto complete search in very slow in wordpress admin

While adding a new post in WordPress admin, my website https://www.xingoos.com have an option to select company name. Company name must appear while typing as drop down but this search take 15-20 seconds for every large or small name search. Any change in code required or it depends on server configuration.

## Why is the proof complete after proving only for one induction when we have more than one variable?

So I’m self-learning coq. And I came across the proof for associativity in addition `forall (a b c : nat)`

Appearntly when we do `induction a.` after `intros a b c.`, it creates 2 subgoals

and afterwards we simply need to show that the two sides in both subgoals are equivalent, and the proof is completed.

So I’m just wondering why we don’t need to do `induction b.` and `induction c.` to complete the proof? Why only performing induction on `a` is able to complete the proof?

Or in other words, how come in the function that returns the proof, we just get `b` and `c` for “free”? Constructively don’t we need something like a double induction applied twice?

## If a Tempest cleric uses the Wrath of the Storm and Thunderbolt Strike features to push an attacker away, can the attacker complete its multiattack?

The Tempest Domain cleric’s Thunderbolt Strike feature (PHB, p. 62) says:

At 6th level, when you deal lightning damage to a Large or smaller creature, you can also push it up to 10 feet away from you.

I envision the Thunderbolt Strike as throwing the target creature away from it, not gently pushing it away. I mean, lightning is an instantaneous thing, and 10 feet is more than just losing your balance.

The Thunderbolt Strike feature of a Tempest cleric leaves me with many questions. For instance, if a monster is able to make a multiattack consisting of 2 claw attacks and then a bite, and it hits me with its first attack, I can use Wrath of Storm as a reaction to deal lightning damage to it, and thereby blast the creature back 10 feet using Thunderbolt Strike.

If the monster does not have 10 feet of movement left after being pushed, does it lose its other 2 attacks against me (if no other targets are in range of it)? Or does it get to make all 3 attacks before it is blasted away from me?

## how to prove this problem is NP complete?

Given a graph G = (V, E) and two integers a, b, does there exist V’ ⊆ V such that |V’| = a and |(V’ × V’) ∩ E| ≥ b. That is, the problem is to determine if there is a subset of vertices of size at least a such that the subgraph induced by V’ has at least b edges.

## Difference between NP Intermediate and NP Complete

Assuming P ≠ NP

How do you determine whether a problem belongs in NP Intermediate or NP Complete?

Why does integer factorization belong in NP Intermediate, but the knapsack problem belongs in NP Complete?

## How to convert a Complete Binary Tree to a Priority Search Tree in O(n)

There doesn’t seem to be any resources on this. I would like to know if there is a linear-time algorithm to convert a Complete Binary Tree to a Priority Search Tree.

## How is the Longest Path Problem NP complete?

https://www.csie.ntu.edu.tw/~lyuu/complexity/2016/20161129s.pdf

So basically, in our iff proof, we have to show two directions:

Forward: If Hamiltonian Path has a yes-instance, so does longest path. This makes sense because we can just let “k” = |V| – 1 if hamiltonian path is yes. Then clearly there is a longest simple path with |V| – 1 edges.

I’m having trouble with the backward part

Backward: If Longest Path has a yes instance, so does longest path. Let’s assume that there is a longest path from s to t of length k (this can be a different k than the one we defined above?). How does that guarantee that there is a hamiltonian path from s-t? If it is the same k, where k = |V| – 1, then I agree there is a hamiltonian path, but what if this k is something different?

## What is the complete lifecycle of souls in the Nine Hells?

The Descent into Avernus adventure provides more information about what happens to souls in the Nine Hells/Baator than was previously available (soul coins and their uses, in particular).

It’s also made it clear that I really don’t understand the paths that souls in the Hells can follow, and thus don’t feel able to convey the moral/logistical implications of any particular action relating to souls to players.

I’m looking for a complete picture of the soul lifecycle in the Hells (and I’m not expecting it to be linear). I could ask a bunch of smaller questions, but that wouldn’t actually help me understand the lifecycle as a whole so I can reason about it in unusual situations.

• States such as: being a living lawful evil mortal who hasn’t made a devilish pact, being a living mortal who has sold their soul, being a lemure, being in the River Styx, being tortured to draw out soul energy, being a higher rank devil, being trapped in a soul coin
• Transitions such as: dying, being forged into a soul coin, being utterly annihilated, being freed from a soul coin, being promoted, being demoted
• How the states and transitions relate to each other

5e materials & designer statements should take precedence, but materials from earlier editions are welcome in answers.

## How to use Scrapebox as a complete beginner without using an API?

I have purchased as much as I can in the starting up of backlinking and I have followed advice from the best of them. I have bought Scrapebox, GSA, and The Best Spinner and Now I am out of money until next week.

Can anybody here guide me to using Scrapebox the best or only way possible without needing an api (just until next week) in conjunction with GSA Search Engine Ranker?

I have gsa set up and im just trying to get some lists because apparently my campaign has stopped because of that (it needs more urls to post to).

Can anyone tell me, am I able to extract Urls from Scrapebox and move forward with GSA just for the time being and if so can you tell me how?

Regards,

Michael


text:
## CryptoDB

### Pil Joong Lee

#### Publications

Year
Venue
Title
2007
PKC
2007
EPRINT
This paper addresses the security of optimistic fair exchange in a multi-user setting. While the security of public key encryption and public key signature schemes in a single-user setting guarantees the security in a multi-user setting, we show that the situation is different in the optimistic fair exchange. First, we show how to break, in the multi-user setting, an optimistic fair exchange scheme provably secure in the single-user setting. This example separates the security of optimistic fair exchange between the single-user setting and the multi-user setting. We then define the formal security model of optimistic fair exchange in the multi-user setting, which is the first complete security model of optimistic fair exchange in the multi-user setting. We prove the existence of a generic construction meeting our multi-user security based on one-way functions in the random oracle model and trapdoor one-way permutations in the standard model. Finally, we revisit two well-known methodologies of optimistic fair exchange, which are based on the verifiably encrypted signature and the sequential two-party multisignature, respectively. Our result shows that these paradigms remain valid in the multi-user setting.
2005
EPRINT
To solve the problem of searching on encrypted data, many keyword search schemes have been proposed in recent years. The goal of such schemes is to enable a user to give an untrusted storage server the ability only to test whether an encrypted document contains a few keywords without learning anything else about the document. In this paper, we are concerned with decrypting the searched results as well as searching for desired documents. In the previously proposed schemes, except for the work by Waters et al.[WBDS04], a user decrypts searched documents using his private key, $A_{priv}$, or a symmetric key. Our another goal is to enable a user to give a proxy the ability to decrypt only the ciphertexts containing desired keywords, but not other ciphertexts. We propose a new mechanism, Searchable Keyword-Based Encryption (SKBE) which satisfies both the above goals. As a result of adding the delegation of decryption ability, our mechanism works more securely and efficiently in several applications, such as email gateways, secure audit logs, and decryption key delegation systems, than any of the previously proposed schemes. We formalize this mechanism, define its security model and propose an efficient construction whose security is proved in a random oracle model under the Bilinear Diffie-Hellman Inversion assumption. The scheme is constructed based on the Public Key Encryption with Conjunctive Field Keyword Search scheme in [PKL04] by using a hybrid encryption technique.
2005
EPRINT
In this paper, we present a broadcast encryption scheme with efficient transmission cost under the \emph{log-key} restriction. Given $n$ users and $r$ revoked users, our scheme has the transmission cost of $O(r)$ and requires the storage of $O(\log n)$ keys at each receiver. These are optimal complexities in broadcast encryptions using one-way hash functions (or pseudo-random generators.) To achieve these complexities, the stratified subset difference (SSD) scheme and the $\overline{B1}$ scheme were introduced by Goodrich et al. and Hwang et al. respectively. However, their schemes have the disadvantage that transmission cost increases linearly according to the number of stratifications. By assigning the related keys between stratifications, our scheme remedies the defect and achieves very efficient transmission cost even in an environment where the key storage is restricted. To the best of our knowledge, our scheme has the most efficient transmission cost in the existing schemes with log-key storage. In addition, our result is comparable to other schemes that allow a large key storage.
2004
EPRINT
The algorithm proposed by Ha and Moon [HM02] is a countermeasure against power analysis. The Ha-Moon algorithm has two drawbacks in that it requires an inversion and has a right-to-left approach. Recently, Yen, Chen, Moon and Ha improved the algorithm by removing these drawbacks [YCMH04]. Their new algorithm is inversion-free, has a left-to-right approach and employs a window method. They insisted that their algorithm leads to a more secure countermeasure in computing modular exponentiation against side-channel attacks. This algorithm, however, still has a similar weakness observed in [FMPV04,SPL04]. This paper shows that the improved Ha-Moon algorithm is vulnerable to differential power analysis even if we employ their method in selecting $s_i$.
2003
ASIACRYPT
2001
PKC
2000
CHES
1999
CHES
1998
ASIACRYPT
1997
CRYPTO
1995
CRYPTO
1995
EUROCRYPT
1994
CRYPTO
1993
CRYPTO
1992
AUSCRYPT
1990
AUSCRYPT
1988
EUROCRYPT
1987
CRYPTO

#### Program Committees

CHES 2007
CHES 2006
Eurocrypt 2006
Asiacrypt 2005
CHES 2005
Crypto 2004
Asiacrypt 2004 (Program chair)
Asiacrypt 2003
CHES 2003
Eurocrypt 2002
Asiacrypt 2002
CHES 2002
PKC 2001
Asiacrypt 2001
Asiacrypt 1999
Eurocrypt 1997
Asiacrypt 1996


text:
## Proceedings of the Japan Academy, Series A, Mathematical Sciences

The aim of the Proceedings of the Japan Academy, Series A, is the rapid publication of original papers in mathematical sciences. The paper should be written in English or French (preferably in English), and at most 6 pages long when published. A paper that is a résumé or an announcement (i.e. one whose details are to be published elsewhere) can also be submitted.

The paper is published promptly if once communicated by a Member of the Academy at its General Meeting, which is held monthly except in July and in August.

Important copyright notice for PJA authors.

## Top downloads over the last seven days

Finite order meromorphic solutions of linear difference equationsVolume 87, Number 5 (2011)
A note on strongly regular ringsVolume 40, Number 2 (1964)
Euler characteristics of groups and orbit spaces of free $G$-complexesVolume 69, Number 10 (1993)
A remark on regular semigroupsVolume 37, Number 1 (1961)
A characterisation of regular semi-groupVolume 32, Number 9 (1956)
• Includes:

Proceedings of the Japan Academy, Series A, Mathematical Sciences
Coverage: 1977--
ISSN: 0386-2194 (print)

Proceedings of the Japan Academy
Coverage: 1945-1977
ISSN: 0021-4280 (print)

Proceedings of the Imperial Academy
Coverage: 1912-1945
ISSN: 0369-9846 (print)

• Publisher: The Japan Academy
• Discipline(s): Mathematics
• Full text available in Euclid: 1912 --
• Access: Open access
• Euclid URL: http://projecteuclid.org/pja


text:
# The Nyquist plot of a loop transfer function G(jω) H(jω) of a system encloses the (-1, j0) point. The gain margin of the system is

This question was previously asked in
MPSC Assistant Engineer EE Mains 2019 - Paper 1
View all MPSC AE Papers >
1. less than zero
2. zero
3. greater than zero
4. infinity

Option 1 : less than zero
Free
ST 1: Building Construction and Materials
1423
20 Questions 40 Marks 24 Mins

## Detailed Solution

The Nyquist plot of a loop transfer function G(jω) H(jω) of a system encloses the (-1, j0) point.

That means the given system is unstable, hence the gain margin is negative.

Important Points:

• For a stable system, gain margin and phase margin are positive
• For an unstable system, gain margin and phase margin are negative
• For a marginally stable system, gain margin and phase margin are equal to zero.


text:
## Strongly convergent iterative schemes for a sequence of nonlinear mappings.(English)Zbl 1162.47049

$\begin{cases} x_1 = x \in C, \\ C_1 = C, \\ y_n = T_nx_n, \\ C_{n+1} = \{z \in C_n: \;\langle x_n - z,J(y_n - x_n) \rangle \leq a_n\|x_n - y_n\|^2\}, \\ x_{n+1} = P_{C_{n+1}}x \end{cases}$
in a smooth, strictly convex, and reflexive Banach space with the Kadec–Klee property ($$P_Q$$ is a metric projection in $$E$$). It is assumed that $$\{T_n\}$$ is a countable family of mappings of a nonempty closed convex subset $$C$$ of E into itself such that $$F = \bigcap_{n=1}^\infty F(T_n) \neq \emptyset$$ and that $$\{T_n\}$$ satisfies the condition
$\langle x - z,J(T_nx - x) \rangle \leq a_n\|x - T_nx\|^2, \qquad x \in C, \;z \in F(T_n), \;n \in {\mathbb N},$
for some $$\{a_n\} \subset (-\infty,0)$$, $$\sup_{n \in {\mathbb N}} a_n < 0$$ (here, $$F(T) = \{x: \;Tx = x\}$$).
It is proved the strong convergence of the sequence $$x_n$$ to $$P_Fx$$ under some additional assumptions about $$C$$ and $$F$$ (in particular, that the relations $$\{z_n\} \subset C$$, $$z \in C$$, $$x_n \to z$$, and $$T_nz_n \to z$$ imply that $$z \in F$$). The case when $$E$$ is real Hilbert space is considered as a particular case. Furthermore, the case when $$\{T_n\}$$ is a family of maximal monotone operators and an application to the feasibility problem are considered.
 47J25 Iterative procedures involving nonlinear operators 47H09 Contraction-type mappings, nonexpansive mappings, $$A$$-proper mappings, etc.


text:
# Properties

 Modulus $735$ Structure $$C_{2}\times C_{2}\times C_{84}$$ Order $336$

Show commands: Pari/GP / SageMath

sage: H = DirichletGroup(735)

pari: g = idealstar(,735,2)

## Character group

 sage: G.order()  pari: g.no Order = 336 sage: H.invariants()  pari: g.cyc Structure = $$C_{2}\times C_{2}\times C_{84}$$ sage: H.gens()  pari: g.gen Generators = $\chi_{735}(491,\cdot)$, $\chi_{735}(442,\cdot)$, $\chi_{735}(346,\cdot)$

## First 32 of 336 characters

Each row describes a character. When available, the columns show the orbit label, order of the character, whether the character is primitive, and several values of the character.

Character Orbit Order Primitive $$-1$$ $$1$$ $$2$$ $$4$$ $$8$$ $$11$$ $$13$$ $$16$$ $$17$$ $$19$$ $$22$$ $$23$$
$$\chi_{735}(1,\cdot)$$ 735.a 1 no $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$
$$\chi_{735}(2,\cdot)$$ 735.bt 84 yes $$1$$ $$1$$ $$e\left(\frac{71}{84}\right)$$ $$e\left(\frac{29}{42}\right)$$ $$e\left(\frac{15}{28}\right)$$ $$e\left(\frac{11}{42}\right)$$ $$e\left(\frac{5}{28}\right)$$ $$e\left(\frac{8}{21}\right)$$ $$e\left(\frac{19}{84}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{3}{28}\right)$$ $$e\left(\frac{65}{84}\right)$$
$$\chi_{735}(4,\cdot)$$ 735.bo 42 no $$1$$ $$1$$ $$e\left(\frac{29}{42}\right)$$ $$e\left(\frac{8}{21}\right)$$ $$e\left(\frac{1}{14}\right)$$ $$e\left(\frac{11}{21}\right)$$ $$e\left(\frac{5}{14}\right)$$ $$e\left(\frac{16}{21}\right)$$ $$e\left(\frac{19}{42}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{3}{14}\right)$$ $$e\left(\frac{23}{42}\right)$$
$$\chi_{735}(8,\cdot)$$ 735.bj 28 yes $$1$$ $$1$$ $$e\left(\frac{15}{28}\right)$$ $$e\left(\frac{1}{14}\right)$$ $$e\left(\frac{17}{28}\right)$$ $$e\left(\frac{11}{14}\right)$$ $$e\left(\frac{15}{28}\right)$$ $$e\left(\frac{1}{7}\right)$$ $$e\left(\frac{19}{28}\right)$$ $$-1$$ $$e\left(\frac{9}{28}\right)$$ $$e\left(\frac{9}{28}\right)$$
$$\chi_{735}(11,\cdot)$$ 735.bl 42 no $$-1$$ $$1$$ $$e\left(\frac{11}{42}\right)$$ $$e\left(\frac{11}{21}\right)$$ $$e\left(\frac{11}{14}\right)$$ $$e\left(\frac{25}{42}\right)$$ $$e\left(\frac{3}{7}\right)$$ $$e\left(\frac{1}{21}\right)$$ $$e\left(\frac{13}{42}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{6}{7}\right)$$ $$e\left(\frac{29}{42}\right)$$
$$\chi_{735}(13,\cdot)$$ 735.bi 28 no $$1$$ $$1$$ $$e\left(\frac{5}{28}\right)$$ $$e\left(\frac{5}{14}\right)$$ $$e\left(\frac{15}{28}\right)$$ $$e\left(\frac{3}{7}\right)$$ $$e\left(\frac{5}{28}\right)$$ $$e\left(\frac{5}{7}\right)$$ $$e\left(\frac{11}{28}\right)$$ $$1$$ $$e\left(\frac{17}{28}\right)$$ $$e\left(\frac{3}{28}\right)$$
$$\chi_{735}(16,\cdot)$$ 735.bg 21 no $$1$$ $$1$$ $$e\left(\frac{8}{21}\right)$$ $$e\left(\frac{16}{21}\right)$$ $$e\left(\frac{1}{7}\right)$$ $$e\left(\frac{1}{21}\right)$$ $$e\left(\frac{5}{7}\right)$$ $$e\left(\frac{11}{21}\right)$$ $$e\left(\frac{19}{21}\right)$$ $$e\left(\frac{2}{3}\right)$$ $$e\left(\frac{3}{7}\right)$$ $$e\left(\frac{2}{21}\right)$$
$$\chi_{735}(17,\cdot)$$ 735.bs 84 yes $$-1$$ $$1$$ $$e\left(\frac{19}{84}\right)$$ $$e\left(\frac{19}{42}\right)$$ $$e\left(\frac{19}{28}\right)$$ $$e\left(\frac{13}{42}\right)$$ $$e\left(\frac{11}{28}\right)$$ $$e\left(\frac{19}{21}\right)$$ $$e\left(\frac{53}{84}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{15}{28}\right)$$ $$e\left(\frac{73}{84}\right)$$
$$\chi_{735}(19,\cdot)$$ 735.r 6 no $$-1$$ $$1$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$-1$$ $$e\left(\frac{1}{3}\right)$$ $$1$$ $$e\left(\frac{2}{3}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$-1$$ $$e\left(\frac{1}{6}\right)$$
$$\chi_{735}(22,\cdot)$$ 735.bh 28 no $$-1$$ $$1$$ $$e\left(\frac{3}{28}\right)$$ $$e\left(\frac{3}{14}\right)$$ $$e\left(\frac{9}{28}\right)$$ $$e\left(\frac{6}{7}\right)$$ $$e\left(\frac{17}{28}\right)$$ $$e\left(\frac{3}{7}\right)$$ $$e\left(\frac{15}{28}\right)$$ $$-1$$ $$e\left(\frac{27}{28}\right)$$ $$e\left(\frac{13}{28}\right)$$
$$\chi_{735}(23,\cdot)$$ 735.bt 84 yes $$1$$ $$1$$ $$e\left(\frac{65}{84}\right)$$ $$e\left(\frac{23}{42}\right)$$ $$e\left(\frac{9}{28}\right)$$ $$e\left(\frac{29}{42}\right)$$ $$e\left(\frac{3}{28}\right)$$ $$e\left(\frac{2}{21}\right)$$ $$e\left(\frac{73}{84}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{13}{28}\right)$$ $$e\left(\frac{11}{84}\right)$$
$$\chi_{735}(26,\cdot)$$ 735.bm 42 no $$1$$ $$1$$ $$e\left(\frac{1}{42}\right)$$ $$e\left(\frac{1}{21}\right)$$ $$e\left(\frac{1}{14}\right)$$ $$e\left(\frac{29}{42}\right)$$ $$e\left(\frac{5}{14}\right)$$ $$e\left(\frac{2}{21}\right)$$ $$e\left(\frac{13}{21}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{5}{7}\right)$$ $$e\left(\frac{37}{42}\right)$$
$$\chi_{735}(29,\cdot)$$ 735.bb 14 yes $$-1$$ $$1$$ $$e\left(\frac{1}{7}\right)$$ $$e\left(\frac{2}{7}\right)$$ $$e\left(\frac{3}{7}\right)$$ $$e\left(\frac{9}{14}\right)$$ $$e\left(\frac{9}{14}\right)$$ $$e\left(\frac{4}{7}\right)$$ $$e\left(\frac{5}{7}\right)$$ $$1$$ $$e\left(\frac{11}{14}\right)$$ $$e\left(\frac{2}{7}\right)$$
$$\chi_{735}(31,\cdot)$$ 735.n 6 no $$-1$$ $$1$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{2}{3}\right)$$ $$1$$ $$e\left(\frac{2}{3}\right)$$ $$-1$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{5}{6}\right)$$ $$1$$ $$e\left(\frac{1}{3}\right)$$
$$\chi_{735}(32,\cdot)$$ 735.bt 84 yes $$1$$ $$1$$ $$e\left(\frac{19}{84}\right)$$ $$e\left(\frac{19}{42}\right)$$ $$e\left(\frac{19}{28}\right)$$ $$e\left(\frac{13}{42}\right)$$ $$e\left(\frac{25}{28}\right)$$ $$e\left(\frac{19}{21}\right)$$ $$e\left(\frac{11}{84}\right)$$ $$e\left(\frac{5}{6}\right)$$ $$e\left(\frac{15}{28}\right)$$ $$e\left(\frac{73}{84}\right)$$
$$\chi_{735}(34,\cdot)$$ 735.bc 14 no $$-1$$ $$1$$ $$e\left(\frac{1}{14}\right)$$ $$e\left(\frac{1}{7}\right)$$ $$e\left(\frac{3}{14}\right)$$ $$e\left(\frac{4}{7}\right)$$ $$e\left(\frac{4}{7}\right)$$ $$e\left(\frac{2}{7}\right)$$ $$e\left(\frac{6}{7}\right)$$ $$-1$$ $$e\left(\frac{9}{14}\right)$$ $$e\left(\frac{9}{14}\right)$$
$$\chi_{735}(37,\cdot)$$ 735.bv 84 no $$-1$$ $$1$$ $$e\left(\frac{5}{84}\right)$$ $$e\left(\frac{5}{42}\right)$$ $$e\left(\frac{5}{28}\right)$$ $$e\left(\frac{10}{21}\right)$$ $$e\left(\frac{25}{28}\right)$$ $$e\left(\frac{5}{21}\right)$$ $$e\left(\frac{25}{84}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{15}{28}\right)$$ $$e\left(\frac{59}{84}\right)$$
$$\chi_{735}(38,\cdot)$$ 735.bs 84 yes $$-1$$ $$1$$ $$e\left(\frac{1}{84}\right)$$ $$e\left(\frac{1}{42}\right)$$ $$e\left(\frac{1}{28}\right)$$ $$e\left(\frac{25}{42}\right)$$ $$e\left(\frac{5}{28}\right)$$ $$e\left(\frac{1}{21}\right)$$ $$e\left(\frac{47}{84}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{17}{28}\right)$$ $$e\left(\frac{79}{84}\right)$$
$$\chi_{735}(41,\cdot)$$ 735.bf 14 no $$1$$ $$1$$ $$e\left(\frac{11}{14}\right)$$ $$e\left(\frac{4}{7}\right)$$ $$e\left(\frac{5}{14}\right)$$ $$e\left(\frac{11}{14}\right)$$ $$e\left(\frac{11}{14}\right)$$ $$e\left(\frac{1}{7}\right)$$ $$e\left(\frac{3}{7}\right)$$ $$-1$$ $$e\left(\frac{4}{7}\right)$$ $$e\left(\frac{1}{14}\right)$$
$$\chi_{735}(43,\cdot)$$ 735.bh 28 no $$-1$$ $$1$$ $$e\left(\frac{13}{28}\right)$$ $$e\left(\frac{13}{14}\right)$$ $$e\left(\frac{11}{28}\right)$$ $$e\left(\frac{5}{7}\right)$$ $$e\left(\frac{27}{28}\right)$$ $$e\left(\frac{6}{7}\right)$$ $$e\left(\frac{9}{28}\right)$$ $$-1$$ $$e\left(\frac{5}{28}\right)$$ $$e\left(\frac{19}{28}\right)$$
$$\chi_{735}(44,\cdot)$$ 735.bq 42 yes $$-1$$ $$1$$ $$e\left(\frac{20}{21}\right)$$ $$e\left(\frac{19}{21}\right)$$ $$e\left(\frac{6}{7}\right)$$ $$e\left(\frac{5}{42}\right)$$ $$e\left(\frac{11}{14}\right)$$ $$e\left(\frac{17}{21}\right)$$ $$e\left(\frac{16}{21}\right)$$ $$e\left(\frac{2}{3}\right)$$ $$e\left(\frac{1}{14}\right)$$ $$e\left(\frac{5}{21}\right)$$
$$\chi_{735}(46,\cdot)$$ 735.bg 21 no $$1$$ $$1$$ $$e\left(\frac{13}{21}\right)$$ $$e\left(\frac{5}{21}\right)$$ $$e\left(\frac{6}{7}\right)$$ $$e\left(\frac{20}{21}\right)$$ $$e\left(\frac{2}{7}\right)$$ $$e\left(\frac{10}{21}\right)$$ $$e\left(\frac{2}{21}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{4}{7}\right)$$ $$e\left(\frac{19}{21}\right)$$
$$\chi_{735}(47,\cdot)$$ 735.bs 84 yes $$-1$$ $$1$$ $$e\left(\frac{71}{84}\right)$$ $$e\left(\frac{29}{42}\right)$$ $$e\left(\frac{15}{28}\right)$$ $$e\left(\frac{11}{42}\right)$$ $$e\left(\frac{19}{28}\right)$$ $$e\left(\frac{8}{21}\right)$$ $$e\left(\frac{61}{84}\right)$$ $$e\left(\frac{2}{3}\right)$$ $$e\left(\frac{3}{28}\right)$$ $$e\left(\frac{65}{84}\right)$$
$$\chi_{735}(52,\cdot)$$ 735.bu 84 no $$1$$ $$1$$ $$e\left(\frac{73}{84}\right)$$ $$e\left(\frac{31}{42}\right)$$ $$e\left(\frac{17}{28}\right)$$ $$e\left(\frac{20}{21}\right)$$ $$e\left(\frac{15}{28}\right)$$ $$e\left(\frac{10}{21}\right)$$ $$e\left(\frac{71}{84}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{23}{28}\right)$$ $$e\left(\frac{55}{84}\right)$$
$$\chi_{735}(53,\cdot)$$ 735.bt 84 yes $$1$$ $$1$$ $$e\left(\frac{37}{84}\right)$$ $$e\left(\frac{37}{42}\right)$$ $$e\left(\frac{9}{28}\right)$$ $$e\left(\frac{1}{42}\right)$$ $$e\left(\frac{3}{28}\right)$$ $$e\left(\frac{16}{21}\right)$$ $$e\left(\frac{17}{84}\right)$$ $$e\left(\frac{5}{6}\right)$$ $$e\left(\frac{13}{28}\right)$$ $$e\left(\frac{67}{84}\right)$$
$$\chi_{735}(58,\cdot)$$ 735.bv 84 no $$-1$$ $$1$$ $$e\left(\frac{83}{84}\right)$$ $$e\left(\frac{41}{42}\right)$$ $$e\left(\frac{27}{28}\right)$$ $$e\left(\frac{19}{21}\right)$$ $$e\left(\frac{23}{28}\right)$$ $$e\left(\frac{20}{21}\right)$$ $$e\left(\frac{79}{84}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{25}{28}\right)$$ $$e\left(\frac{5}{84}\right)$$
$$\chi_{735}(59,\cdot)$$ 735.bp 42 yes $$1$$ $$1$$ $$e\left(\frac{1}{21}\right)$$ $$e\left(\frac{2}{21}\right)$$ $$e\left(\frac{1}{7}\right)$$ $$e\left(\frac{37}{42}\right)$$ $$e\left(\frac{5}{7}\right)$$ $$e\left(\frac{4}{21}\right)$$ $$e\left(\frac{31}{42}\right)$$ $$e\left(\frac{5}{6}\right)$$ $$e\left(\frac{13}{14}\right)$$ $$e\left(\frac{16}{21}\right)$$
$$\chi_{735}(61,\cdot)$$ 735.br 42 no $$-1$$ $$1$$ $$e\left(\frac{17}{21}\right)$$ $$e\left(\frac{13}{21}\right)$$ $$e\left(\frac{3}{7}\right)$$ $$e\left(\frac{10}{21}\right)$$ $$e\left(\frac{9}{14}\right)$$ $$e\left(\frac{5}{21}\right)$$ $$e\left(\frac{23}{42}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{2}{7}\right)$$ $$e\left(\frac{20}{21}\right)$$
$$\chi_{735}(62,\cdot)$$ 735.bk 28 yes $$-1$$ $$1$$ $$e\left(\frac{5}{28}\right)$$ $$e\left(\frac{5}{14}\right)$$ $$e\left(\frac{15}{28}\right)$$ $$e\left(\frac{13}{14}\right)$$ $$e\left(\frac{19}{28}\right)$$ $$e\left(\frac{5}{7}\right)$$ $$e\left(\frac{11}{28}\right)$$ $$1$$ $$e\left(\frac{3}{28}\right)$$ $$e\left(\frac{3}{28}\right)$$
$$\chi_{735}(64,\cdot)$$ 735.bd 14 no $$1$$ $$1$$ $$e\left(\frac{1}{14}\right)$$ $$e\left(\frac{1}{7}\right)$$ $$e\left(\frac{3}{14}\right)$$ $$e\left(\frac{4}{7}\right)$$ $$e\left(\frac{1}{14}\right)$$ $$e\left(\frac{2}{7}\right)$$ $$e\left(\frac{5}{14}\right)$$ $$1$$ $$e\left(\frac{9}{14}\right)$$ $$e\left(\frac{9}{14}\right)$$
$$\chi_{735}(67,\cdot)$$ 735.w 12 no $$-1$$ $$1$$ $$e\left(\frac{7}{12}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$-i$$ $$e\left(\frac{2}{3}\right)$$ $$-i$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{11}{12}\right)$$ $$e\left(\frac{5}{6}\right)$$ $$i$$ $$e\left(\frac{1}{12}\right)$$
$$\chi_{735}(68,\cdot)$$ 735.x 12 no $$-1$$ $$1$$ $$e\left(\frac{11}{12}\right)$$ $$e\left(\frac{5}{6}\right)$$ $$-i$$ $$e\left(\frac{5}{6}\right)$$ $$-i$$ $$e\left(\frac{2}{3}\right)$$ $$e\left(\frac{1}{12}\right)$$ $$e\left(\frac{2}{3}\right)$$ $$-i$$ $$e\left(\frac{5}{12}\right)$$


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Back

# OK Autrod 5183

SKU: 2P70403B40

OK Autrod 5183 was developed to provide the highest strengths possible in the as welded condition of alloy AA 5083 and other similar high magnesium alloys. The more common OK Autrod 5356 will typically fail to meet the as-welded tensile requirements of AA 5083. The alloy is typically utilised in marine and structural applications where […]

OK Autrod 5183 was developed to provide the highest strengths possible in the as welded condition of alloy AA 5083 and other similar high magnesium alloys. The more common OK Autrod 5356 will typically fail to meet the as-welded tensile requirements of AA 5083. The alloy is typically utilised in marine and structural applications where high strengths, high fracture toughness for impact resistance and exposure to corrosive elements are important. The alloy is not recommended for elevated temperature applications due to its susceptibility to stress corrosion cracking. The alloy is non-heat treatable.


text:
209 articles – 1940 Notices  [english version]
 HAL : in2p3-00666053, version 1
 arXiv : 1107.2955
 Physical Review Letters 108 (2012) 072301
 Strangeness Enhancement in Cu+Cu and Au+Au Collisions at \sqrt{s_{NN}} = 200 GeV
 STAR Collaboration(s)
 (2012)
 We report new STAR measurements of mid-rapidity yields for the $\Lambda$, $\bar{\Lambda}$, $K^{0}_{S}$, $\Xi^{-}$, $\bar{\Xi}^{+}$, $\Omega^{-}$, $\bar{\Omega}^{+}$ particles in Cu+Cu collisions at \sNN{200}, and mid-rapidity yields for the $\Lambda$, $\bar{\Lambda}$, $K^{0}_{S}$ particles in Au+Au at \sNN{200}. We show that at a given number of participating nucleons, the production of strange hadrons is higher in Cu+Cu collisions than in Au+Au collisions at the same center-of-mass energy. We find that aspects of the enhancement factors for all particles can be described by a parameterization based on the fraction of participants that undergo multiple collisions.
 Thème(s) : Physique/Physique Nucléaire Expérimentale
 Lien vers le texte intégral : http://fr.arXiv.org/abs/1107.2955
 in2p3-00666053, version 1 http://hal.in2p3.fr/in2p3-00666053 oai:hal.in2p3.fr:in2p3-00666053 Contributeur : Dominique Girod <> Soumis le : Vendredi 3 Février 2012, 14:20:47 Dernière modification le : Lundi 19 Mars 2012, 14:02:54


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# Can Mathematica plot this financial simulation?

I'd like to run a simple financial simulation with something like the following constraints:

• You start out with $$S$$ dollars. Let's say $$S$$ is $$\1M$$ dollars
• You repeatedly make an investment that costs $$I$$ dollars (let's say $$I = \5K$$) with an $$F$$ percent (let's say $$F = 0.95$$) chance of returning $$\0$$ and an $$(1-F)$$ percent chance of returning $$\%100,000$$ of $$I$$ (or just $$1,000 \cdot I$$).

Is there a way in Mathematica to repeatedly make this investment hundreds of times over time? Could we have Mathematica repeatedly run the simulation and then plot the average of all of its simulation as a line as well?

• Are you sure you didn't mean to write the chances the other way around ie F=.95; also, is there a chance you meant for the returns in the good case to read 1000I instead of 1000S or something similar? – yosimitsu kodanuri May 23 '19 at 17:36
• Yes, you're correct. I adjusted my question above. – George May 23 '19 at 17:39

It can certainly be done using NestList:

s = 100;
i = 0.5;
f = 0.95;
nStep = 100;
sim := NestList[If[RandomReal[] < f, # - i, # - i + 1000 i] &, s,
nStep];

Now sim will generate a simulation of such process at each evaluation. You can use the following to do multiple runs and plot:

runs = Table[sim, {r, 100}];
ListPlot[runs, Joined -> True, PlotRange -> All]

Also plot the means:

ListPlot[Mean @ runs, Joined -> True, PlotRange -> All]

• It's hard to parse the output. Is there a way to adjust the plot to see what's going on? – George May 23 '19 at 17:18
• It looks like the members of sims are growing by 1000x every step. Are you sure this is correct? They should have a 5% chance of growing by 1000x and a 95% of losing the price of the investment (\$5K in this example). – George May 23 '19 at 17:23
• Well that was according to your previous version. Edited. – L.Yu May 23 '19 at 17:52
• Thank you for putting up with my change in question :) – George May 23 '19 at 17:54
• No problem. You can adjust the s,i or f parameters as you wish. – L.Yu May 23 '19 at 17:55

In what follows, the code provided will return a plot of a (user provided) number of random experiments that implement the concept of the investment lottery in the question; the output is a sequence of the amounts outstanding, after a lottery has been paid for and possible profits have been accrued instead of the sequence of winnings.

Define an auxiliary function that generates appropriate random numbers:

randomSequence[distribution_: UniformDistribution[], repetitions_: 10, observations_: 10^6] :=
RandomVariate[distribution, {repetitions, observations}]

Eg. evaluating randomSequence[NormalDistribution[], 2, 10] will return a $$2\times10$$ matrix with random numbers from $$N(0,1)$$.

Define a lottery without using conditionals; make use of the main evaluation loop instead:

(* when rand >= cut-off prob, return a prize *)
lottery[randomValue_, cost_, multiplier_, probability_] /; randomValue >= probability :=
multiplier cost

(* ...otherwise return 0. *)
lottery[__] := 0.

balance represents the amount remaining after committing to participate in an "investment" lottery; so far nothing fancy, but it's possible to modify balance to obtain more "involved" results if needed.

SetAttributes[balance, HoldFirst]
balance[lottery[randomValue_, cost_, multiplier_, probability_], wealth_] :=
wealth - cost + lottery[randomValue, cost, multiplier, probability]

simulation is the main event; it bundles together the simulation results and the display of the results:

SetAttributes[simulation, HoldFirst]
simulation[randomSequence[args___], capital_, investment_, multiplier_, probability_] :=
series = FoldList[
balance[lottery[#2, investment, multiplier, probability], #1] &,
capital, #] & /@ randomSequence[args];

legend = Column[Thread[{"K", "I", "x", "p"} -> {capital, investment, multiplier, probability}]];
opts = {
Frame -> True,
FrameLabel -> {{None, None}, {None, label}}};

Legended[ListLinePlot[series, Apply[Sequence][opts]], legend]
]

Evaluating

 simulation[randomSequence[UniformDistribution[], 10, 10^3], 10^6, 5 10^3, 10^3, .95]

returns


text:
# MLE on a restricted parameter space

Suppose $X_1, \ldots, X_n$ are a random sample from a normal distribution with mean $\theta$ and variance $1$. Find the maximum likelihood estimator of $\theta$, under the restriction that $\theta \geq 0$.

I have found the MLE when there is no restriction on $\theta$.

$$L(\theta|\mathbf{x})=(2\pi)^{-n/2} \text{exp}\Big(-\frac{1}{2}\sum_{i=1}^{n}(x_i-\theta)^2 \Big)$$

Differentiating $L(\theta|\mathbf{x})$ with respect to theta, and setting it equal to zero yields $$\sum_{i=1}^{n}(x_i-\theta)=0 \implies \theta=\frac{1}{n}\sum_{i=1}^{n}x_i =\bar{x}$$ The second derivative is negative, hence the MLE of $\theta$ is $\hat{\theta}=\bar{X}$.

How would I incorporate the restriction $\theta \geq 0$.

• Under the assumption that this is homework (please use the homework tag, if so!) I'll just give a hint. What if $\bar{X} = -1$? What is the constrained MLE then? Why? – jbowman Nov 20 '13 at 0:09
• "Forget" statistics -this is a maximization problem. It can be solved also as a maximization problem under a non-negativity constraint on the variable, including additively the restriction in the objective function (the log-likelihood) with the associated KKT non-negative multiplier -and see what happens. – Alecos Papadopoulos Nov 20 '13 at 2:13
• @Alecos is correct: but the situation is simpler than that, because this is a one-variable problem. All you have to do is check whether the zero of the derivative is in the interval $[0,\infty)$. If not, the convexity of $-\log(L)$ guarantees that the boundary point $\theta=0$ is the maximum. Consult any elementary Calculus text for details. – whuber Nov 20 '13 at 3:15

Although the OP did not respond, I am answering this to showcase the method I proposed (and indicate what statistical intuition it may contain).

First, it is important to distinguish on which entity is the constraint imposed. In a deterministic optimization setting, there is no such issue : there is no "true value", and an estimator of it. We just have to find the optimizer. But in a stochastic setting, there are conceivably two different cases:

a) "Estimate the parameter given a sample that has been generated by a population that has a non-negative mean" (i.e. $\theta \ge 0$) and
b) "Estimate the parameter under the constraint that your estimator cannot take negative values"(i.e. $\hat \theta \ge 0$).

In the first case, imposing the constraint is including prior knowledge on the unknown parameter. In the second case, the constraint can be seen as reflecting a prior belief on the unknown parameter (or some technical, or "strategic", limitation of the estimator).

The mechanics of the solution are the same, though:The objective function, (the log-likelihood augmented by the non-negativity constraint on $\theta$) is

$$\tilde L(\theta|\mathbf{x})=-\frac n2 \ln(2\pi)-\frac{1}{2}\sum_{i=1}^{n}(x_i-\theta)^2 +\xi\theta,\qquad \xi\ge 0$$

Given concavity, f.o.c is also sufficient for a global maximum. We have

$$\frac {\partial}{\partial \theta}\tilde L(\theta|\mathbf{x})=\sum_{i=1}^{n}(x_i-\theta) +\xi = 0 \Rightarrow \hat \theta = \bar x+\frac{\xi}{n}$$

1) If the solution lies in an interior point ($\Rightarrow \hat \theta >0$), then $\xi=0$ and so the solution is $\{\hat \theta= \bar x>0,\; \xi^*=0\}$.

2) If the solution lies on the boundary ($\Rightarrow \hat \theta =0$) then we obtain the value of the multiplier at the solution $\xi^* = -n\hat x$, and so the full solution is $\{\hat \theta= 0,\; \xi^*=-n\bar x\}$. But since the multiplier must be non-negative, this necessarily implies that in this case we would have $\bar x\le 0$

(There is nothing special about setting the constraint to zero. If say the constraint was $\theta \ge -2$, then if the solution lied on the boundary, $\hat \theta = -2$, it would imply (in order for the multiplier to have a positive value), that $\bar x \le -2$).

So, if the optimizer is $0$ what are we facing here? If we are in "constraint type-a", i.e we have been told that the sample comes from a population that it has a non-negative mean, then with $\hat \theta =0$ chances are that the sample may not be representative of this population.

If we are in "constraint type-b", i.e. we had the belief that the population has a non-negative mean, with $\hat \theta =0$ this belief is questioned.

(This is essentially an alternative way to deal with prior beliefs, outside the formal bayesian approach).

Regarding the properties of the estimator, one should carefully distinguish this constrained estimation case, with the case where the true parameter lies on the boundary of the parameter space.

• Finally, is it correct to say that $\hat{\theta}_{\text{MLE}}=\max(\bar{X},0)$ ? – StubbornAtom May 2 '18 at 21:14
• @StubbornAtom This is a compact way indeed to express the situation prior to observe the sample and the sample mean. – Alecos Papadopoulos May 2 '18 at 21:22
• Another query: If $\theta$ does not take the value $0$ (i.e. if the parameter space is $(0,\infty)$), then would the MLE be simply $\hat{\theta}=\bar{X}$ ? – StubbornAtom May 2 '18 at 21:33
• @StubbornAtom Yes. Except if the sample proves us wrong and gives us a negative sample mean. – Alecos Papadopoulos May 2 '18 at 21:39
• And if the sample mean does give a negative value, what would be my MLE? – StubbornAtom May 2 '18 at 22:22


text:
# Management accounting 1 answer below »

(a) Process costing (7 marks)
The following information has been gathered from the production department of the XYZ Company for April. The XYZ production department uses a process costing system and manufactures a pasta sauce which uses two raw material components (RM1 and RM2).
Work in Process at 1 April (85% completed) 6,000
Units started during April 72,000
Completed and transferred to finished goods during April 75,000

Cost In WIP "1 April Current Month
Material - RM1 $14,460$121,920
Material - RM2 6,510 75,800
Direct Labour 6,000 61,866

Raw material RM1 is introduced at the beginning of the process while RM2 is added at the stage when the product is 60% complete. Conversion costs are incurred at a uniform rate throughout the process. On April 30, Work-in-Process is 35% complete. There were no losses or spoilage during processing.
Using the weighted average cost method, calculate the:
(i) cost per equivalent unit of production
(ii) cost of completed units for the month of April
(iii) closing value of WIP at the end of April.

Roberto C

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