content stringlengths 86 994k | meta stringlengths 288 619 |
|---|---|
Systems of Linear Functions
9.7: Systems of Linear Functions
Created by: CK-12
A Second Trip to the Omni
Kelly loved the Omni presentation on the rainforest so much that she decided to go and see it again. She asked Tyler if he wanted to go with her on Saturday afternoon.
“Do you want to go with me?” Kelly asked.
“Sure, but I have karate first, so I will have to meet you there. What time is the show?” Tyler asked.
“The show starts at 2:30 pm. I’m going to leave at one o’clock so I can look around,” Kelly said.
“Well, I don’t get done karate until then, so I probably won’t leave until 2:00 pm,” Tyler said.
On Saturday, the two went about their day and both left for the museum. Kelly’s Mom tends to drive slowly and cautiously, so she was traveling on city streets at 45 mph. Tyler’s karate class is
downtown, so he could take the highway to get to the Omni theater and his Dad drove at an average of 55 mph. Will the two catch up to each other?
This is a problem about equations and systems of equations. You will need to solve a system of equations to figure this one out. To figure this out, you would need to find a solution that will work
for both equations. You will learn all about it in this lesson.
What You Will Learn
In this lesson, you will learn how to do the following skills.
• Recognize a solution of a linear system in two variables as an ordered pair that is a solution of each equation in the system.
• Recognize linear systems with no solutions and linear systems with infinitely many solutions.
• Solve linear systems by graphing.
• Solve linear systems by substitution.
• Model and solve real - world problems using systems of linear equations.
Teaching Time
I. Recognize a Solution of a Linear System in Two Variables as an Ordered Pair that is a Solution of Each Equation in the System
As useful as you have seen linear functions to be, there are yet other applications of the idea. In a system of linear equations, you will see how linear equations can work together in a system to
solve even more complex problems. Indeed, there are various ways that we can find solutions to these problems or find that there may be no solution at all.
If you add two numbers together, you get 13. Can you think of any ordered pairs that would work for this?
(1, 12), (3, 10), (-4, 17), (4.5, 8.5)
Hopefully you’ll agree that there are infinite pairs of numbers whose sum is 13. You might also agree that there are infinite pairs of numbers whose difference is 7.
(9, 2), (11, 4), (37, 30), (95.8, 88.8), (-3, -10)
However, which ordered pair is true for both conditions at once? Which pair has a sum of 13 and a difference of 7?
If you make a list of ordered pairs, you can check them to see which makes both equations true. This is a system of equations—two or more equations at the same time. In the example above, the
solution is (10, 3) because the sum of the two numbers is 13 and their difference is 7.
The pair (10, 3) makes both equations true.
Which ordered pair makes both equations true?
1. $x+y &=8 \\4x-y &=-3$
a. (2, 6)
b. (3, 15)
c. (4, 4)
d. (1, 7)
Let’s test each pair and see which pair, if any, works:
a. $x+y &=8 \\2+6 &=8? \\8 &=8 \\4x-y &=-3 \\4 \cdot 2-6 &=-3? \\8-6 &=-3? \\2 &e -3$
The ordered pair (2, 6) makes the first equation true, but not the second. Because it is not true for both equations, it is not a solution to the system.
b. $x+y &=8 \\3+15 &=8? \\18 &e 8$
The ordered pair (3, 15) does not even make the first equation true. It cannot be a solution to the system.
c. $x+y &=8 \\4+4 &=8? \\8 &=8 \\4x-y &=-3 \\4 \cdot 4-4 &=-3? \\16-4 &=-3 \\12 &e -3$
The ordered pair (4, 4) makes the first equation true, but not the second. Because it is not true for both equations, it is not a solution to the system.
d. $x+y &=8 \\1+7 &=8? \\8 &=8 \\4x-y &=-3 \\4 \cdot 1-7 &=-3? \\4-7 &=-3?\\-3 & = -3$
The ordered pair (1, 7) makes both equations true. This is a solution to the system.
Remember, the ordered pair must be a solution to both equations in the system.
II. Recognize Linear Systems with No Solutions and Linear Systems with Infinitely Many Solutions
A solution to a system of equations is an ordered pair that makes both equations true. Is there always a solution? Can there be more than one solution? Let’s investigate this with an example.
Two numbers have a sum of 17. If you add two numbers together, their sum is 15. As you know, there are infinite ordered pairs whose sum is 17. There are also infinite ordered pairs whose sum is 15.
But can a single ordered pair have a sum of both 17 and 15 at the same time?
First, let’s write two equations to help us to sort out the information in this system of equations. There are two equations and both have a different sum.
$x+y &=17 \\x+y &=15$
If we think about these two equations, you will see that there aren’t any values that will work for both of these equations.
Therefore, this system has no solutions.
Two numbers have a sum of -8. Twice the first number plus twice the second number is -16.
First, let’s write the two equations described above. Then we can investigate possible solutions.
$x+y &= -8 \\2x+2y &=-16$
Does this system have a solution? Think of a solution for the first equation. How about (-3, -5)? Does it work for the second? Yes. Think of another solution like (9, -1). This one is also true in
both equations.
This equation has an infinite number of solutions.
Some systems of equations have infinite solutions because all ordered pairs that make one equation true also make the other true.
III. Solve Linear Systems by Graphing
When we have a system of equations, we can graph the equations and solve the system by graphing. When you graph two linear equations, two lines are formed. These two lines will intersect. The point
where the lines intersect is the solution for the system. Let’s look at an example.
Solve the following system by graphing.
$y &= 2x-4 \\y &= -2x+8$
To work on this system, we will graph both of these lines and look for the point of intersection. That point of intersection will be the solution to the system.
The point of intersection is (3, 2).
This is the solution to the system of equations.
Sometimes you might have two equations that do not intersect. If this happens, then you know that this is a system with no solutions. Parallel lines are one possible system that would not have a
solution. Parallel lines have the same slope, so you can recognize a system with no solutions.
IV. Solve Linear Systems by Substitution
Graphing systems of equations is an excellent way of finding their solutions. It is also an easy way to see if a system has no solutions or has infinitely many. However, as you saw in the last
example, graphing is not always practical, specifically when the solution is not an integer because fractional numbers can be difficult to read on a graph. Or imagine if your $y$$\frac{85}{3}$
Let’s recall that in a system of equations, we are looking for the same $x$$y$
$3y &=x-2 \\y-x &=4$
is the ordered pair $(x, y)$$x$$x$$y$$y$$y-x=4$$y$$x$$y=x+4$$y$$y=x+4$$x + 4$$y$substitution method.
Now that we have substituted, we can solve this equation because it has a single variable.
$3(x+4) &= x-2 \\3x+12 &=x-2 \\2x+12 &=-2 \\2x &=-14 \\x &=-7$
If $x = -7$$y$
$y &=x+4 \\y &=-7+4 \\y &=-3$
Our solution, then, is (-7, -3).
V. Model and Solve Real – World Problems Using Systems of Linear Equations
Two methods for solving systems of equations are graphing and substitution. There are many situations in the real world where these strategies can be useful.
Two trains leave the station going in the same direction. One train leaves two hour before the other. The first train’s average speed is 65mph while the second train’s average speed is 90mph. How
long will it take for the second train to catch up to the first?
First, let’s think about functions with regard to this problem. The second train’s distance $d$$t$$d=90t$
Since the first train left two hours before the first train, it had two hours more than the second train to travel. Its speed was 65mph, though. Its distance can be found with the equation $d=65(t+2)
Solve the system of equations using a graph.
From the graph you can see that the trains will meet shortly after 5 hours.
Now let’s go back and look at the problem from the introduction.
Real-Life Example Completed
A Second Trip to the Omni
Here is the problem from the introduction. Reread it and then solve it for the solution.
Kelly loved the Omni presentation on the rainforest so much that she decided to go and see it again. She asked Tyler if he wanted to go with her on Saturday afternoon.
“Do you want to go with me?” Kelly asked.
“Sure, but I have karate first, so I will have to meet you there. What time is the show?” Tyler asked.
“The show starts at 2:30 pm. I’m going to leave at one o’clock so I can look around,” Kelly said.
“Well, I don’t get done karate until then, so I probably won’t leave until 2:00 pm,” Tyler said.
On Saturday, the two went about their day and both left for the museum. Kelly’s Mom tends to drive slowly and cautiously, so she was traveling on city streets at 45 mph. Tyler’s karate class is
downtown, so he could take the highway to get to the Omni theater and his Dad drove at an average of 55 mph. Will the two catch up to each other?
This is a problem about equations and systems of equations. You will need to solve a system of equations to figure this one out.
Now write two equations to represent the system of equations for this problem. Then see if there is a solution for both of them.
Solution to Real – Life Example
The first equation we can write is to represent Tyler’s time. His Dad is traveling 55 mph. Therefore his distance is a function of speed and time.
$d = 55t$
Kelly left one hour before Tyler did. She is traveling 45 miles per hour. Therefore, the speed times Tyler’s time plus one hour equals Kelly’s time.
$d = 45(t + 1)$
Now see if there is a solution that will work for both equations. We can try to solve this by using substitution.
$55t &= 45(t + 1) \\55t &= 45t + 1 \\55t - 45t &= 1 \\10t &= 1 \\t &= \frac{1}{10}$
Now we go to Tyler.
$d &= 55\left ( \frac{1}{10} \right ) \\d &= 5.5$
The solution could be the following values for $d$ and $t$
Here are the vocabulary words that are found in this lesson.
System of Equations
two or more equations at the same time. The solution will be the ordered pair that works for both equations.
Time to Practice
Directions: Figure out which pair is a solution for each given system.
1. Which ordered pair is a solution of the following system?
$x-3y &=9 \\3x+y &=7$
(a) $(6, -1)$
(b) $(-1, -4)$
(c) $(0, 7)$
(d) $(3, -2)$
2. Which ordered pair is a solution of the following system?
$y &=3x-7 \\5x-3y &=13$
(a) $\left (3, \frac{2}{3} \right)$
(b) $(2, -1)$
(c) $(4, 7)$
(d) $(5, 8)$
3. Figure out which ordered pair is a solution for the following system.
$\frac{1}{2}x +3y &= -6 \\2x + 4y &= 0$
Directions: Determine whether each system has infinite solutions or no solutions.
$x +y &= 10 \\y &= -x +10$
$3x -6y &= -24 \\x -2y &= -8$
$\frac{3}{4}x &= \frac{2}{3}y-1 \\9x &= 8y-12$
$y &= \frac{1}{2}x + 3 \\y &= \frac{1}{2}x - 2 \\\\y &= 3x-5 \\y &= 3x-2$
$y &= \frac{1}{2}x + 3 \\y &= \frac{1}{2}x - 2$
Directions: Answer each question true or false.
9. Parallel lines have the same slope.
10. Parallel lines have infinite solutions.
11. Perpendicular lines have one solution.
12. Lines with an infinite number of solutions are not parallel.
Directions: Graph the following systems of equations. Identify the solution or write no solution or infinitely many solutions, if appropriate.
$y &= 2x - 3 \\y &= x - 1$
$2x + 2y &= 1 \\y &= -x + \frac{1}{2}$
$y &= -3x + 1 \\3x - y &= -7$
$y &= 2x \\\frac{y}{2} &= x- \frac{5}{2}$
Directions: Solve the following systems using the substitution method.
$2y -8 &= 8 \\2y + 2x &= 2$
$4x + y &= -2 \\-2x - 3y &= 1$
$y &= 2x \\6x - y &= 8$
$x + 4y &= -3 \\2x + 8y &= -6$
21. Anglica’s mother leaves to visit her grandmother for her birthday. Her grandmother lives 450 miles away and her mother drives at an average of 50mph. Three hours later, Angelica’s step-father
notices that her mother forgot the gift for her grandmother so he decides to try to catch up to her. If he drives at an average of 80mph, will he catch up to her before she gets to her
grandmother’s house? Write a system of equations to model the situation and solve using the substitution method.
Files can only be attached to the latest version of None | {"url":"http://www.ck12.org/book/CK-12-Middle-School-Math-Grade-8/r1/section/9.7/","timestamp":"2014-04-18T15:51:44Z","content_type":null,"content_length":"138164","record_id":"<urn:uuid:512c6035-a17b-4b1e-a121-f680d9abc804>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00355-ip-10-147-4-33.ec2.internal.warc.gz"} |
Cubic Polynomial Functions
December 7th 2009, 04:45 PM
Cubic Polynomial Functions
I'm not quite sure where to start with this problem:
Let f(x) = ax^3 +bx^2 + cx + d, a != 0
a) show that f has either 0 or 2 local extrema.
b) give an example of each possibility in part (a)
Can somebody start me off on the right track?
December 7th 2009, 05:19 PM
The derivative will help you.
Let's say your original equation is a cubic. The derivative will be some quadratic. A quadratic always has two roots. If one is real, then they both are real, and there must be two real 'zeros'
for the derivative. If one is imaginary, then both are imaginary.
December 7th 2009, 05:22 PM
thanks, I got it (Clapping) | {"url":"http://mathhelpforum.com/calculus/119200-cubic-polynomial-functions-print.html","timestamp":"2014-04-19T09:44:10Z","content_type":null,"content_length":"4811","record_id":"<urn:uuid:dcf2c45b-4781-487f-9a4d-79f796450214>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00112-ip-10-147-4-33.ec2.internal.warc.gz"} |
Singularity Theory
24 July to 22 December 2000
Report from the Organisers: VI Arnold (Moscow and Paris IX), JW Bruce (Liverpool),
V Goryunov (Liverpool),
D Siersma (Utrecht)
Scientific Background and Objectives
Meetings and Workshops
Outcome and Achievements
Singularities arise naturally in a huge number of different areas of mathematics and science. As a consequence Singularity Theory lies at the crossroads of the paths connecting the most important
areas of applications of mathematics with its most abstract parts. For example, it connects the investigation of optical caustics with simple Lie algebras and regular polyhedra theory, while also
relating hyperbolic PDE wave fronts to knot theory and the theory of the shape of solids to commutative algebra.
The main goal in most problems of singularity theory is to understand the dependence of some objects of analysis and geometry, or phenomena from physics or some other science, on parameters. For
generic points in the parameter space their exact values influence only quantitative aspects of the phenomena, their qualitative, topological features remaining stable under small changes of
parameter values.
However, for certain exceptional values of the parameters these qualitative features may suddenly change under a small variation of the parameter. This change is called a perestroika, bifurcation or
catastrophe in different branches of the sciences. A typical example is that of Morse surgery, describing the perestroika of the level variety of a function as the function crosses through a critical
value. (This has an important complex counterpart: the Picard-Lefschetz theory concerning the branching of integrals.) Other familiar examples include caustics and outlines or profiles of surfaces
obtained from viewing or projecting from a point, or in a given direction.
In spite of its fundamental character and the central position it now occupies in mathematics, singularity theory is a surprisingly young subject. So, for example, one can consider the singularities
arising from the orthogonal projections of a generic surface in 3-space, a problem of surely classical interest. Their classification was completed as recently as 1979. In one sense singularity
theory can be viewed as the modern equivalent of the differential calculus, and this explains its central position and wide applicability. In its current form the subject started with the fundamental
discoveries of Whitney (1955), Thom (1958), Mather (1970) and Brieskorn (1971). Substantial results and exciting new developments within the subject have continued to flow in the intervening years,
while the theory has embodied more and more applications.
The major idea of this programme was to bring together experts within the field and those from adjacent areas where singularity theory has existing or potential application, such as wave propagation,
dynamical systems, quantum field theory, differential and algebraic geometry, and others. It was the programme’s aim both to foster exciting new developments within singularity theory, and also to
build bridges to other subjects where its tools and philosophy prove useful.
The programme was planned by Arnold, Bruce and Siersma. Unfortunately, Arnold was not able to attend the programme at all due to a cycling accident. The day-to-day organisation was carried out by
Goryunov and Siersma. There was very valuable assistance from Zakalyukin, Wall, Giblin and Nikulin for specific workshops during the programme. In addition to the main meetings, the programme ran
seminars on a regular weekly basis on Tuesday and Thursday afternoons, with at least 2 talks in each session. There were also extra, occasional seminars arranged by the organisers and other
Two of the participants gave lecture series on special subjects. From September to October, a series of 9 lectures on Frobenius manifolds was given by B Dubrovin. Frobenius manifolds are now finding
numerous applications in physics. However, they are a rather new (but highly natural) topic for Singularity Theory. In November, another series of lectures, on ramified integrals, was given by VA
Vassiliev, devoted to generalisations of works by Newton and Atiyah. Both lecture series attracted excellent audiences.
In December the frequency of talks increased and the penultimate week was run on a full conference scale, with up to 5 talks daily.
N Nekrasov, AN Varchenko, VA Vassiliev and VV Goryunov spoke in a two-day Discussion Meeting on Topological Methods in the Physical Sciences held at the Royal Society in November. The meeting was
organised together with the parallel programme on Geometry and Topology of Fluid Flows, and brought together leading pure and applied mathematicians from the UK and other countries. It was attended
by many British specialists and was a great success.
The programme was attended by 53 long-term and 87 short-term visitors. Some of them came to the Institute several times. In addition, there was very strong participation in the conferences, workshops
and other events during the programme. Dubrovin, as Rothschild Visiting Professor, made a remarkable contribution to the programme during his stay.
The majority of experts in the field attended the programme, and a large number of leading experts in adjacent areas visited the Institute. There was a strong presence from Europe and Russia (the
latter allowed by the generous support from the Leverhulme Trust), as well as from North America.
Many of the meetings and individual talks, plus special lecture courses, attracted mathematicians from both Cambridge and other British universities. The EU support for two of the meetings and the
Junior Membership scheme were highly useful for attendance by PhD students and young postdocs. A number of participants gave talks in other UK departments, such as Edinburgh, Hull, Leeds, Liverpool
and Warwick.
NATO Advanced Study Institute / EC Summer School: New Developments in Singularity Theory, 31 July - 11 August 2000
Organisers: D Siersma, VM Zakalyukin, JP Brasselet, VA Vassiliev and CTC Wall
This first meeting was designed to study developments in singularity theory, especially during the last 5 years. Its key topics included
• Singular complex varieties: global theory
• Singular complex varieties: local theory
• Singularities of holomorphic maps
• Singularities of real maps
• Study of discriminant spaces and Vassiliev- type invariants
The primary purpose of the ASI was the introduction of new developments in singularity theory to a broader audience, thus establishing new contacts and advancing a broad front of research. The
importance of pedagogical skills had been borne in mind in choosing the ASI lecturers, as the instructional aspect was regarded as central to the success of the meeting.
GM Greuel, A Fruehbis-Krueger and C Lossen (all from Kaiserslautern) demonstrated the latest version of the computer package Singular which the group in Kaiserslautern has been developing over the
last decade for the specific needs of singularity theory. The meeting finished with a problem session. The forum was remarkably large and served as an excellent start to the entire programme: the
total number of participants exceeded 95, many of them younger mathematicians. This was made possible thanks to the combined generous support from NATO and the EU. A volume of proceedings is now
being published by Kluwer.
Some of the ASI participants stayed on at the Institute for one week to attend the mini-workshop on polynomial and meromorphic functions conducted by A Dimca.
Workshop on Applications to Wave Propagation Theory and Dynamical Systems, 25 - 29 September 2000
Organisers: RM Roberts and VM Zakalyukin
The topics of the second meeting included:
• Singularities in symplectic and contact spaces: caustics and wave fronts, shock waves, Hamilton-Jacobi equation, Lagrangian intersections, Legendre knots.
• Applications to control theory and differential equations, game theory. Singularities in subriemannian geometry, optimization, Pfaffian systems.
• Singularities of momentum maps, energy-momentum maps and integrable Hamiltonian systems. Monodromy in Hamiltonian systems, applications to physics.
• Bifurcations of (relative) equilibria of Hamiltonian systems and time-reversible equivariant dynamical systems. Singularity theory and KAM theory.
The meeting consisted of 35 lectures, with 53 official participants. It attracted many specialists
in Hamiltonian systems and this provided a considerable source of new questions (including those arising from physical experiments) to which Singularity Theory is very likely to find answers.
Euroworkshop on Applications to Quantum Field Theory, 23 - 27 October 2000
Organisers: B Dubrovin, V Goryunov, N Nekrasov and V Nikulin
Over the last years, new and very interesting relations have been discovered between singularity theory and algebraic geometry on one hand, and quantum field theory and string theory on the other.
For example, mirror symmetry for Calabi-Yau manifolds, Gromov-Witten invariants, Frobenius manifolds and integrable systems became important tools in string theory and quantum field theory. The
October Euroworkshop was devoted to further interactions between all of these areas. Supported by the EU, it brought together leading specialists and young researchers in areas very rapidly
developing nowadays. It was one of the most inspiring events in the entire programme bringing a large number of new problems into the area of singularity theory. There were 24 lectures with good
spacing between them allowing numerous discussions.
Informal Discussion Meeting on Different Aspects of Singularity Theory, 24 - 25 November 2000
Organiser: D Siersma
The meeting concentrated on deformation theory, analysis and topology of singularities. This was a rather small scale session (21 participants), but intensive and productive. The discussions were
very stimulating and useful for establishing new links with other branches of mathematics.
Satellite Workshop at the University of Liverpool on Applications of Singularity Theory to Geometry, 16 - 21 December 2001
Organisers: JW Bruce, J Damon, PJ Giblin, VV Goryunov and CTC Wall
The final event of the programme was a Satellite Meeting in Liverpool. Its theme was one of the most traditional in the field: much of Singularity Theory was inspired by geometrical problems. Thom’s
early work on differential geometry via families of functions has borne enormous fruit in a richer understanding of the higher geometry of surfaces, and this in turn has found application in other
fields such as computer vision. Projections of surfaces to planes, giving apparent contours, and the general theories of caustics and wave fronts are other examples where new techniques were
motivated by geometrical problems. Remarkable duality connections have been found between some of these problems, and there are applications to algebraic geometry and other mathematical fields.
This workshop took as its theme interactions between singularity theory and geometry in its many modern guises. Besides the topics mentioned above one could mention Gauss mappings, the geometry of
discriminants and bifurcation diagrams, coadjoint orbits, billiards, arrangements and the global geometry of singular waves and varieties.
The number of requests for talks in Liverpool was very high and it was decided to run a week-long full-scale pre-workshop at the INI, which included a most inspiring lecture series by A Losev
exposing new ideas of applying singularity theory to the study of Frobenius structures.
There were 34 talks in Liverpool itself. The meeting was supported by the LMS conference grant and a special grant from the INI. It was attended by 63 participants.
The programme was highly successful. It did achieve its main goal to bring together the best researchers working on singularities and specialists in other branches of mathematics and physics where
singularity theory either already is an efficient tool or can become such a tool in the future. The programme was an excellent school for its younger participants. There was a certain amount of
interaction with the participants of the parallel programme on Geometry and Topology of Fluid Flows. Contacts with Pelz, Ricca, Michor, Khesin, Shnirelman and others helped our participants to
realise the current needs of this branch of applied mathematics.
Over the programme, its participants worked both on their long-term projects and on new problems that came to their attention during their stay. The comments of participants were uniformly positive,
and all reported productive discussions and new collaborations. A high number of papers were reported as either submitted or being in preparation during the programme.
Anisov worked on Matveev’s complexity of three-manifolds. He obtained an upper bound (which is very likely to be precise) for the complexity of torus fibrations over the circle. Bogaevsky and
Ishikawa studied singularities of Legendre mappings, in particular a generalisation of Mather’s theory to the Legendrian case.
Buchstaber and Rees continued their joint work on symmetric products of linear spaces which is a multi-dimensional version of the Vieta theorem.
Campillo and Olivares made considerable progress in describing singular one-dimensional foliations in terms of the associated polar map.
A lot of attention during the programme was devoted to the famous Jacobian Conjecture requiring deep understanding of singularities of plane algebraic curves at infinity. Campillo, Piltant and
Reguera studied singularities of the characteristic cone of the surface obtained by blowing up the infinite points of such a curve. Chekanov worked on the Arnold-Givental conjecture claiming that a
Lagrangian submanifold, which is a fixed-point set of an anti-symplectic involution, must have sufficiently many points of intersection with its image under a Hamiltonian symplectomorphism. He
succeeded in constructing a counter-example to the degenerate part of this conjecture. The result can be seen as the first step to disproving the degenerate part of Arnold’s famous symplectic
fixed-point conjecture, whose non-degenerate part was proven recently by Fukaya-Ono and others. Davydov, with the involvement of Zhitomirskii, continued his study of singularities of implicit
differential equations. He obtained new results in two directions: on generic singularities of higher order equations, and on normal forms of first order equations involving one dependent and many
independent variables. The results are promising to be very important for applications. Dubrovin continued his research on applications of Frobenius manifolds to integrable hierarchies and to
Gromov-Witten invariants of higher genera. He also started new investigations, together with Goryunov and Kazarian, on applications of Frobenius manifolds to singularities of functions on complete
intersections and on singular space curves. Ebeling and Gusein-Zade made considerable progress in their work on an algebraic formula for the index of a differential form at an isolated complete
intersection singularity. Gaffney worked on his long-term project to give an integral closure formulation to all equisingularity conditions controlled by analytic inequalities. Together with Houston,
he started a project on finding necessary and sufficient conditions for families of finitely determined map-germs from 3-space to 3-space to be Whitney equisingular. Damon, Gaffney and Mond worked on
a conjecture claiming that the image of a map-germ from n-space to (n+1)-space is a free star divisor in a sense of Damon. This would yield a formula for the image Milnor number. Gaffney, Trotman and
Wilson finished their paper on equisingularity of sections. Denef, Melle and Veys worked on the Monodromy Conjecture on the topological zeta-function of a holomorphic function-germ.
Kazarian obtained very impressive results on classifying spaces of singularities and Thom polynomials, on characteristic classes related to singularities of Lagrange and Legendre mappings. Matveev
obtained new strong results on the global theory of geodesically equivalent metrics. Merkov, helped by Vassiliev and Kazarian, made considerable progress in the study of Vassiliev-type invariants of
ornaments (collections of plane curves no three of which have common points) and doodles (triple-point-free collections of plane curves). Polyak worked on Kontsevich’s universal formula for a
deformation quantisation of the algebra of functions on a real linear space. He succeeded in interpreting this formula as a degree of a map from a certain configuration to a multi-dimensional torus.
Polyak also obtained a very elegant generalisation of the classical Crofton formula to the case of generic immersed plane curves instead of only convex ones. Pushkar continued working on his
long-term and extremely promising project on Legendrian K-theory and relative Morse theory. Ruas, using the methods of Gaffney and Damon, made progress in the problem of finding sufficient conditions
for topological triviality in families of function-germs on analytic varieties. Scherbak and Varchenko worked on the resonance case of the problem of behaviour of critical points of products of
powers of linear functions. This question turned out to be closely related to the representations of sl(2). Also Scherbak finished a paper on boundary singularities and non-crystallographic Coxeter
groups. Sedykh studied admissible homotopies of space curves, that is those not involving in particular curves with self-intersections and inflexion points. This allowed him to solve Arnold’s problem
on the impossibility of deforming certain curves to curves without flattening points via such homotopies.
Shapiro, Kazarian and Goryunov studied Hurwitz spaces developing an approach to calculate Hurwitz numbers in the general case of a meromorphic function with multiple complicated branch points.
Shapiro also worked on understanding relations between various Poisson structures on the space of unipotent upper triangular matrices.
Tibar wrote a paper on a problem of describing vanishing cycles in non-generic Lefschetz pencils on complex and symplectic spaces. Together with Siersma, he worked on understanding deformations of
polynomials at infinity. Vassiliev finished the formulation of a purely combinatorial algorithm for calculating combinatorial formulae for knot invariants. This is an important part of his extensive
programme on effective calculation of cohomology classes of spaces of non-singular geometrical objects. He wrote 4 papers and edited a translation of one of his books. Wall and du Plessis worked on
their papers on theorems of Cayley-Baeharach type and on generic projections.
Zakalyukin and Goryunov made considerable progress in their study of the monodromy in vanishing homology of families of 2×2 matrices. Together with Bortakovsky, Zakalyukin proved that local
singularities in generic completely non-integrable logical-dynamical systems with a low number of switches correspond to generic corner singularities. Together with Giblin, Zakalyukin worked on
classification of generic singularities of envelopes of families of chords in affine geometry.
The programme was an international event of extremely high significance for Singularity Theory and related subjects. It provided a very valuable boost of new ideas and introduced many exciting
problems to solve. | {"url":"http://www.newton.ac.uk/reports/0001/sgt.html","timestamp":"2014-04-18T10:52:53Z","content_type":null,"content_length":"23203","record_id":"<urn:uuid:a65f10a6-b463-4812-9876-368661f129d6>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00295-ip-10-147-4-33.ec2.internal.warc.gz"} |
Voltage Divider - Voltage Division Rule - Solved Problems
Voltage Divider - Voltage Division Rule
The voltage division rule (voltage divider) is a simple rule which can be used in solving circuits to simplify the solution. Applying the voltage division rule can also solve simple circuits
thoroughly. The statement of the rule is simple:
Voltage Division Rule: The voltage is divided between two series resistors in direct proportion to their resistance.
It is easy to prove this. In the following circuit
the Ohm's law implies that
$v_1(t)=R_1 i(t)$ (I)
$v_2(t)=R_2 i(t)$ (II)
Applying KVL
$-v(t)+v_1(t)+v_2(t)=0 \rightarrow v(t)=v_1(t)+v_2(t)$ .
$v(t) = R_1 i(t)+R_2 i(t)= (R_1 +R_2) i(t)$ .
$i(t) = \frac{v(t)}{R_1 +R_2}$ .
Substituting in I and II
$v_1(t)=R_1 \frac{v(t)}{R_1 +R_2}$ ,
$v_2(t)=R_2 \frac{v(t)}{R_1 +R_2}$ .
$v_1(t)= \frac{R_1}{R_1 +R_2} v(t)$ ,
$v_2(t)=\frac{R_2}{R_1 +R_2} v(t)$ .
which shows that the voltage is divided between two series resistors in direct proportion to their resistance. The rule can be easily extended to circuits with more than two resistors. For example,
$v_1(t)= \frac{R_1}{R_1 +R_2+R_3+R_4} v(t)$ ,
$v_2(t)=\frac{R_2}{R_1 +R_2+R_3+R_4} v(t)$ ,
$v_3(t)=\frac{R_3}{R_1 +R_2+R_3+R_4} v(t)$ ,
$v_4(t)=\frac{R_4}{R_1 +R_2+R_3+R_4} v(t)$ .
The voltage division rule can be used solve simple circuits or to simplify solving complicated circuits.
For example, check out this problem.
One of the common mistakes in using the voltage division rule is to use the formula for resistors which are in parallel with other elements. For example, the voltage division rule cannot be used in
the following circuit directly.
It will be incorrect if one tries to find $V_x$ using voltage divider by neglecting the other $6 \Omega$ resistor as
So, $V_x eq \frac{6 \Omega}{2 \Omega + 6 \Omega} 15 V$ . However, if solving other parts of a circuits confirms that the current of the other element/branch is zero, the voltage division rule can be
still applied. For example, suppose that the following network is a piece of a larger circuit.
Let's assume that the analysis of the circuit shows that $I_x=0$ . In this case, $V_x= \frac{6 \Omega}{2 \Omega + 6 \Omega} 12 V = 9 V$ regardless of where A and B are connected.
The voltage devision rule can be used to ease solving problems. For example, the voltage division rule is used in the following problem to find the Thévenin voltage:
Thévenin’s Theorem – Circuit with An Independent Source
8 thoughts on “Voltage Divider - Voltage Division Rule”
1. a parallel plate capacitor has area=2 mm and plate separation distance, d=1 mm. find how much charge is stored when the capacitor si connected to a 10 volts battery.
1. C = epsilonr*epsilon0*A/d
Work it out for yourself!
2. Really usefull.Especially the voltage divider thing, I struggle with it in Polytechnic and Iam still struggling with it but thanks, I have cleared the confusion after going through your
3. really sooorry for last one. That is my younger brother's deeds.valuable information. but you didnt mention any thing about parallel circuits.
4. Super
5. The voltage divider formulas and the solved problems are very useful to me. I corrected several circuits in our company. Thanks a lot for the support.
6. This is very helpful and easy to follow I can now complete my assignment
7. Thank uu so much | {"url":"http://www.solved-problems.com/circuits/circuits-articles/482/voltage-divider-voltage-division-rule/","timestamp":"2014-04-17T00:48:20Z","content_type":null,"content_length":"55455","record_id":"<urn:uuid:bea0d6ba-0b25-49bc-b40e-29de8b750257>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00554-ip-10-147-4-33.ec2.internal.warc.gz"} |
This page computes arithmetic relations between 2 integers or polynomials: gcd, lcm, euclidean division, Bezout relation. You can change the number of formulas involved: 3 . 4 . 5 . 6 . 7 . 8 . 9 .
Glossary: gcd lcm , factorization , Bezout numbers , euclidean algorithm .
This page is not in its usual appearance because WIMS is unable to recognize your web browser.
Please take note that WIMS pages are interactively generated; they are not ordinary HTML files. They must be used interactively ONLINE. It is useless for you to gather them through a robot program.
Description: computes euclidean division, gcd, lcm, Bezout relation. This is the main site of WIMS (WWW Interactive Multipurpose Server): interactive exercises, online calculators and plotters,
mathematical recreation and games
Keywords: wims, mathematics, mathematical, math, maths, interactive mathematics, interactive math, interactive maths, mathematic, online, calculator, graphing, exercise, exercice, puzzle, calculus,
K-12, algebra, mathématique, interactive, interactive mathematics, interactive mathematical, interactive math, interactive maths, mathematical education, enseignement mathématique, mathematics
teaching, teaching mathematics, algebra, geometry, calculus, function, curve, surface, graphing, virtual class, virtual classes, virtual classroom, virtual classrooms, interactive documents,
interactive document, algebra, arithmetic, number, integer, polynomial, euclidian division, gcd, lcm | {"url":"http://wims.unice.fr/~wims/wims.cgi?module=tool%2Farithmetic%2Fbezout.en","timestamp":"2014-04-17T18:24:48Z","content_type":null,"content_length":"6697","record_id":"<urn:uuid:09e459c9-b021-45bc-9dc8-43d762c31080>","cc-path":"CC-MAIN-2014-15/segments/1398223202457.0/warc/CC-MAIN-20140423032002-00662-ip-10-147-4-33.ec2.internal.warc.gz"} |
Bucket Sort
Bucket sort runs in linear time on the average. It assumes that the input is generated by a random process that distributes elements uniformly over the interval [0, 1).
The idea of Bucket sort is to divide the interval [0, 1) into n equal-sized subintervals, or buckets, and then distribute the n input numbers into the buckets. Since the inputs are uniformly
distributed over (0, 1), we don't expect many numbers to fall into each bucket. To produce the output, simply sort the numbers in each bucket and then go through the bucket in order, listing the
elements in each.
The code assumes that input is in n-element array A and each element in A satisfies 0 ≤ A[i] ≤ 1. We also need an auxiliary array B[0 . . n -1] for linked-lists (buckets).
BUCKET_SORT (A)
1. n ← length [A]
2. For i = 1 to n do
3. Insert A[i] into list B[nA[i]]
4. For i = 0 to n-1 do
5. Sort list B with Insertion sort
6. Concatenate the lists B[0], B[1], . . B[n-1] together in order.
Given input array A[1..10]. The array B[0..9] of sorted lists or buckets after line 5. Bucket i holds values in the interval [i/10, (i +1)/10]. The sorted output consists of a concatenation in order
of the lists first B[0] then B[1] then B[2] ... and the last one is B[9].
All lines except line 5 take O(n) time in the worst case. We can see inspection that total time to examine all buckets in line 5 is O(n-1) i.e., O(n).
The only interesting part of the analysis is the time taken by Insertion sort in line 5. Let n[i] be the random variable denoting the number of elements in the bucket B[i]. Since the expected time to
sort by INSERTION_SORT is O(n^2), the expected time to sort the elements in bucket B[i] is
E[O(^2n[i])] = O(E[^2n[i]]]
Therefore, the total expected time to sort all elements in all buckets is
^ n-1∑[i=0] O(E[^2n[i]]) = O^ n-1∑[i=0 ](E[^2n[i]]) ------------ A
In order to evaluate this summation, we must determine the distribution of each random variable n
We have n elements and n buckets. The probability that a given element falls in a bucket B[i] is 1/n i.e., Probability = p = 1/n. (Note that this problem is the same as that of
"Balls-and-Bin" problem).
Therefore, the probability follows the binomial distribution, which has
mean: E[n[i]] = np = 1
variance: Var[n[i]] = np(1- p) = 1- 1/n
For any random variable, we have
E[^2n[i]] = Var[n[i]] + E^2[n[i]]
= 1 - 1/n + 1^2
= 2 - 1/n
Putting this value in equation A above, (do some tweaking) and we have a expected time for INSERTION_SORT, O(n).
Now back to our original problem
In the above Bucket sort algorithm, we observe
T(n) = [time to insert n elements in array A] + [time to go through auxiliary array B[0 . . n-1] * (Sort by INSERTION_SORT)
= O(n) + (n-1) (n)
= O(n)
Therefore, the entire Bucket sort algorithm runs in linear expected time. | {"url":"http://www.personal.kent.edu/~rmuhamma/Algorithms/MyAlgorithms/Sorting/bucketSort.htm","timestamp":"2014-04-17T00:52:01Z","content_type":null,"content_length":"8298","record_id":"<urn:uuid:c8ab1aba-b5f2-4056-b55a-5a27f9747968>","cc-path":"CC-MAIN-2014-15/segments/1397609526102.3/warc/CC-MAIN-20140416005206-00166-ip-10-147-4-33.ec2.internal.warc.gz"} |
T-2 Fundamental Symmetries and Neutrinos
utlra cold neutrons, Majorana, supernovae neutrinos, ...
Probing New Interactions with Neutron Beta Decay
GOAL: discover/constrain new BSM interactions originating at the TeV scale through snergy of:
1. Innovative precision measurements (10^-3 level) of neutron decay (n-->peV) at the LANL UCN source
2. Multi-scale theoretical analysis of BSM contributions to n decay: elucidate discovery potential and model discriminating power
The big picture: energy vs predicion frontier
Both direct (energy) and indirect (precision) probes will be needed to fully recnstruct the dynamics at the TeV scale
Theory Program
TeV scale:
(Cirigliano, Graesser)
GeV scale:
(Bhattacharya, Cirigliano, Gupta + PD)
Elucidate implications of neutron measurements for SM extensions, in the context of other precision measurements and collider searches (LHC)
Topical Collaboration to study "Neutrinos and Nucleosyntesis in Hot and Dense Matter" (Sanjay Reddy)
DOE funded Mutli-campus collaboration to understand supernova and neutron star phenomena with a specific focus on nuclear and neutrinto theory.
T-2 Focus Areas | {"url":"http://www.lanl.gov/orgs/t/documents/t2/hephysics/fundsymm.shtml","timestamp":"2014-04-20T05:54:05Z","content_type":null,"content_length":"17780","record_id":"<urn:uuid:63582e33-e463-4cf7-acd6-d7976698fd26>","cc-path":"CC-MAIN-2014-15/segments/1397609538022.19/warc/CC-MAIN-20140416005218-00491-ip-10-147-4-33.ec2.internal.warc.gz"} |
Items where Research Group is "Centre for Mathematical Biology" and Year is 1995
Number of items: 18.
Dale, P. D. and Olsen, L. and Maini, P. K. and Sherratt, J. A. (1995) Travelling waves in wound healing. FORMA, 10 (3). pp. 205-222.
Dale, P. D. and Sherratt, J. A. and Maini, P. K. (1995) Corneal epithelial wound healing. Journal of Biological Systems, 3 (4). pp. 957-965.
Hofer, T. and Maini, P. K. and Sherratt, J. A. and Chaplain, M. A. J. and Murray, J. D. (1995) Resolving the chemotactic wave paradox: A mathematical model for chemotaxis of Dictyostelium amoebae.
Journal of Biological Systems, 3 (4). pp. 967-973.
Hofer, T. and Maini, P. K. and Sherratt, J. A. and Chaplain, M. A. J. and Murray, J. D. (1995) Resolving the chemotactic wave paradox: A mathematical model for chemotaxis of Dictyostelium amoebae.
Journal of Biological Systems, 3 (4). pp. 967-973.
Hofer, T. and Sherratt, J. A. and Maini, P. K. (1995) Cellular pattern formation during Dictyostelium aggregation. Physica D, 85 (3). pp. 425-444.
Hofer, T. and Sherratt, J. A. and Maini, P. K. (1995) Dictyostelium discoideum: Cellular self-organisation in an excitable medium. Proceedings of the Royal Society London B, 259 (1356). pp. 249-257.
ISSN 0962-8452
Kulesa, P. M. and Cruywagen, G. C. and Lubkin, S. R. and Maini, P. K. and Sneyd, J. S. and Murray, J. D. (1995) Modelling the spatial patterning of the primordia in the lower jaw of alligator
mississippiensis. Journal of Biological Systems, 3 (4). pp. 975-985.
Maini, P. K. (1995) Hierarchical models for spatial pattern formation in biology. Journal of Biological Systems, 3 (4). pp. 987-997.
Ngwa, G. A. and Maini, P. K. (1995) Spatio-temporal patterns in a mechanical model for mesenchymal morphogenesis. Journal of Mathematical Biology, 33 (5). pp. 489-520.
Olsen, L. and Sherratt, J. A. and Maini, P. K. (1995) A mechanochemical model for adult dermal wound contraction. Journal of Biological Systems, 3 (4). pp. 1021-1031.
Olsen, L. and Sherratt, J. A. and Maini, P. K. (1995) A mechanochemical model for adult dermal wound contraction. Journal of Biological Systems, 3 (4). pp. 1021-1031.
Olsen, L. and Sherratt, J. A. and Maini, P. K. (1995) A mechanochemical model for adult dermal wound contraction: On the permanence of the contracted tissue displacement profile. Journal of
Theoretical Biology, 177 (2). pp. 113-128.
Perumpanani, A. J. and Sherratt, J. A. and Maini, P. K. (1995) Phase differences in reaction-diffusion-advection systems and applications to morphogenesis. IMA Journal of Applied Mathematics, 55 (1).
pp. 19-33.
Roose, T. and Maini, P. K. and Cory, J. and Hails, R S (1995) The spread of viral infection in moths,. Journal of Agricultural Science (Cambridge), 125 . p. 161.
Sherratt, J. A. and Maini, P. K. and Jager, W. and Muller, W. A. (1995) A receptor based model for pattern formation in Hydra. FORMA, 10 (2). pp. 77-95.
Zhang, L. and Aguilar-Roblero, R. and Barrio, R. A. and Maini, P. K. (1995) Rhythmic firing patterns in SCN: The role of circuit interactions. International Journal of Bio-Medical Computing, 38 (1).
pp. 23-31.
Gilmour, I. (1995) Coupled fluctuations of blood pressure and heart rate time series. Masters thesis, University of Oxford.
Kalamangalam, G. P. (1995) Nonlinear oscillations and chaos in chemical cardiorespiratory control. PhD thesis, University of Oxford. | {"url":"http://eprints.maths.ox.ac.uk/view/groups/cmb/1995.type.html","timestamp":"2014-04-20T05:55:28Z","content_type":null,"content_length":"14424","record_id":"<urn:uuid:2f4820d0-b33b-48f3-a640-3352ac95468d>","cc-path":"CC-MAIN-2014-15/segments/1398223206672.15/warc/CC-MAIN-20140423032006-00296-ip-10-147-4-33.ec2.internal.warc.gz"} |
Chemistry: Entropy and Temperature Video | MindBites
Chemistry: Entropy and Temperature
About this Lesson
• Type: Video Tutorial
• Length: 14:13
• Media: Video/mp4
• Use: Watch Online & Download
• Access Period: Unrestricted
• Download: MP4 (iPod compatible)
• Size: 153 MB
• Posted: 07/14/2009
This lesson is part of the following series:
Chemistry: Full Course (303 lessons, $198.00)
Chemistry: Thermodynamics (8 lessons, $14.85)
Chemistry: Entropy (2 lessons, $4.95)
This lesson was selected from a broader, comprehensive course, Chemistry, taught by Professor Harman, Professor Yee, and Professor Sammakia. This course and others are available from Thinkwell, Inc.
The full course can be found at http://www.thinkwell.com/student/product/chemistry. The full course covers atoms, molecules and ions, stoichiometry, reactions in aqueous solutions, gases,
thermochemistry, Modern Atomic Theory, electron configurations, periodicity, chemical bonding, molecular geometry, bonding theory, oxidation-reduction reactions, condensed phases, solution
properties, kinetics, acids and bases, organic reactions, thermodynamics, nuclear chemistry, metals, nonmetals, biochemistry, organic chemistry, and more.
Dean Harman is a professor of chemistry at the University of Virginia, where he has been honored with several teaching awards. He heads Harman Research Group, which specializes in the novel organic
transformations made possible by electron-rich metal centers such as Os(II), RE(I), AND W(0). He holds a Ph.D. from Stanford University.
Gordon Yee is an associate professor of chemistry at Virginia Tech in Blacksburg, VA. He received his Ph.D. from Stanford University and completed postdoctoral work at DuPont. A widely published
author, Professor Yee studies molecule-based magnetism.
Tarek Sammakia is a Professor of Chemistry at the University of Colorado at Boulder where he teaches organic chemistry to undergraduate and graduate students. He received his Ph.D. from Yale
University and carried out postdoctoral research at Harvard University. He has received several national awards for his work in synthetic and mechanistic organic chemistry.
About this Author
2174 lessons
Founded in 1997, Thinkwell has succeeded in creating "next-generation" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare
students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider
of technology-based textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.
Thinkwell lessons feature a star-studded cast of outstanding university professors: Edward Burger (Pre-Algebra through...
Recent Reviews
This lesson has not been reviewed.
Please purchase the lesson to review.
This lesson has not been reviewed.
Please purchase the lesson to review.
The second law of thermodynamics tells us that the entropy of the universe must increase for any spontaneous process. That tells us that either the entropy's increasing in terms of the chemical
reaction or it's increasing in terms of heat being released into the surroundings. And as heat is released into the surroundings, the entropy of the surroundings increases. But overall, we know that
entropy must increase for any spontaneous process, thus entropy is a really big deal. It means a whole lot to us in determining whether a reaction is going to happen or not. So let's talk a bit more
about entropy and let's define it now mathematically.
There are a lot of different definitions we could give for entropy, but the one that's going to be most useful for us is that the entropy change for a process is the heat divided by the temperature,
if the process is reversible - in other words, if it's in equilibrium, then the change in entropy if we put in heat. So imagine, for instance, a phase change where we're holding the temperature
constant but we're adding just enough heat to convert from one phase to another - let's suppose we're boiling - that the entropy increase for that process is the heat required to cause the phase
change divided by whatever the temperature change is.
Now, for a spontaneous process - one that's not reversible - we don't know exactly what the entropy is, but we know that the entropy change must be greater than this value - the heat that we put in
divided by temperature.
Okay, now before I go on here, let me give you a sense of what we're saying in English. If you put in heat, that causes disorder because it's random thermal energy that we're putting into a compound
- into a material. But the amount of disorder depends on what the temperature is. If we're at very high temperature, there's already a huge amount of disorder. So adding a certain amount of heat at
very high temperature doesn't change things a whole lot. But if we're at a very low temperature - say somewhere down near 0° Kelvin - adding the same amount of heat a low temperature causes a
tremendous change in the amount of disorder. So that's why we have this kind of heat, or at least to give you a sense of why this would be.
Now, we know that if we're talking about a chemical reaction that the heat given off or absorbed at a constant pressure is the enthalpy. Remember, that was our definition. Enthalpy was defined as the
heat released at constant pressure. And so if we substitute in H - as long as we're talking about a constant pressure system - then S for any spontaneous process has got to be bigger than . And
rearranging this, what it tells us is H, for a chemical process, minus TS has got to be less than zero. Now let's hold onto that idea. We're going to come back to that. But that says if we know about
S for a chemical reaction and we know about H for a chemical reaction, we have the tools to predict whether the reaction will occur or not. Because, again, this has to do with whether or not it's
going to be spontaneous. So hold on to this idea. Again, a spontaneous reaction will only occur if H - TS < 0.
Before we talk about chemical reactions, let's return to the idea of just a substance. Now, we'll define S for a substance the same way we would as H, except that remember, when we talked about H,
there wasn't a convenient way to determine the absolute enthalpy for a material. We found it convenient, though, to talk in terms of the state function, H, and that was relatively easy to do. We
could measure that directly. Remember, that's the heat released at constant pressure.
But for entropy, it's different story. We actually can determine the absolute entropy for a substance. And so we'll define the S for a process or for a chemical reaction as the entropy - the absolute
entropy, the absolute disorder - in the final state minus the entropy in the initial state. So that's just saying if we go from one level to another - one set of reactants to a set of products, for
instance - that S for that reaction is just the final entropy minus the initial entropy, just like we've defined things before. But the difference is we can actually calculate values for these things
now. So let's see how we would actually do that.
Actually, before we do that, we have to describe the third law of thermodynamics, which essentially says: What is your reference point? Well, our reference point is going to be - we want to pick a
place to call zero entropy. Let's pick that place where we have absolute perfect order, and that's going to be where we sucked out all the energy - the random thermal energy - we possibly can from
the system.
So we're going to define a substance with zero entropy as a perfect crystalline material at 0° Kelvin. Now what do I mean a perfect crystalline material? I mean as much order as we possibly can have.
So for instance, I'm showing a cartoon of a diatomic molecule where the two atoms are different, and I mean, they all not only have to be packed perfectly together - no disorder in how they're packed
- but they also have to be aligned properly so that there's no ambiguity about is the molecule pointing this direction or the opposite direction. As soon as we allow for that possibility, then
there's disorder in the crystal. So understand, we need to have the perfect crystalline material without any type of disorder at all. That's what we'll define as zero entropy for every substance.
So now that we have identified a reference point - so here we are. That's what we just said. Zero Kelvins we have zero entropy. Now, as we raise the temperature of the substance, then we're adding
heat to it - we're changing temperature as we're adding. Remember, our definition of entropy was heat divided by temperature. And the problem is how much heat we're putting in as we change the
temperature depends on what the heat capacity is. Now, the hard part is the heat capacity is changing as we change our temperature.
So this is a difficult calculation. Fortunately, you don't have to do it. It's just the integral of the heat capacity as temperature's changing divided by temperature. But you won't have to do that.
This is a value you'll be able to look up for any substance - at least that we're going to be asking you about or that you would be asked on an exam or something. And you can find that in the
appendix. Now, again, that's changing with time. We'll talk in a little bit about why heat capacity's changing as we go through this period. But for right now, let's just say that you can find out a
value for a substance - you can look it up in a table - and this is corresponding to a solid material, still in its crystalline state. Now, we're going to go through a phase change, from solid to
liquid. This is the entropy of fusion we need now. This is fairly easy for us to figure out, because we know the heat that we're putting in to cause that phase change - at constant pressure,
remember, that's just the heat of fusion - H of fusion - divided by whatever the temperature is that that occurs at. So that's a value that's easy for us to calculate. Then, we have to go through
this period again. This is more difficult to calculate, but once again, we can look up values. Then we go through the entropy change for vaporization, and you'll notice, that's a lot bigger than the
entropy change down here.
Why would that be? Well think about what common sense tells you about how much disorder changes going from solid to liquid and going from liquid to gas - a huge amount of disorder when we go from
liquid to gas. And in fact, more or less, the amount of disorder - the amount of entropy increase - that we have when we go from liquid to gas is almost a constant for all substances, because it
depends so much on just the fact that you're increasing the number of particles in the gas phase.
Now, beyond there, we'll need to calculate based on what our heat capacity changes are again. But the point is that somebody else, some other poor guy, has calculated all this for you, so you don't
have to go through all this. You can look at up, at room temperature - or wherever room temperature is, depending on what substance you're dealing with - what the value of the absolute entropy is.
Great! So you can find that out, and in a moment, we'll see that that allows us to find out S for a chemical reaction. And we need to know S why? Because if we know that and H tells us about G, which
tell us - oops - I didn't tell you about G yet. Strike that. It tells us about whether or not the reaction is spontaneous or not though. I just gave you a little glimpse of the future.
Okay, now let's see what a problem would look like. So we've got liquid carbon tetrachloride; it evaporates. It costs heat, of course, to evaporate - 34 kilojoules worth at 25 degrees. Now, that's
not the boiling point, but it doesn't have to be. This is just describing a phase change at 25 degrees here. We want one mole of liquid. It's got an entropy of so and so. What's the entropy of that
liquid when it's converted to the vapor phase - not changing temperature, just putting in enough heat to convert it to the vapor phase? So S then - the change in entropy going through that phase
change - is . So we've got 43 X 10^3 joules. Notice I'm describing it in joules so that my units are correct. I'll divide that by temperature and I end up with 144 joules per mole Kelvin. Okay? Now,
remember, that was the change in entropy and to find the total entropy of that substance, I need to take what I started with, which was the 214, and add the amount of increased entropy to go through
that phase change. So my total entropy, after going through the phase change, is 358 joules. So what am I doing? I'm just simply looking at going from here to here - that change in entropy here. I'm
going through that phase change again. So that would be a type of problem you'd get having to do with figuring out absolute entropy of a substance.
Now, talking about this, why is it that heat capacities are different for different materials and why do we have to consult these tables of numbers? Let me give you a little bit of a sense of this,
because it's a very important idea. And we understand a lot about the molecular level now. Where does the notion of heat capacity come from? Remember, heat capacity has to do with the ability of a
substance to store energy. The more it can store energy, the less its temperature will rise as we add more heat to it. Now the best way I know how to give you a feel for what this means is by
analogy. Look at these two different beakers here, and in my analogy, the height of a liquid in these beakers will correspond to temperature. But you'll notice that they have different capacities to
store that water, in that this is a bigger diameter than this is. I'm now going to add 100 joules, if you will, of energy to each of these two different substances. And what I'm really going to do is
add 100 milliliters of water, but it would correspond to an amount of energy. And so if I put in 100 milliliters of water into this material - or 100 joules of energy into this material - compared to
100 joules into this material, you'll notice that the temperature rose a lot higher - the water level rose a lot higher - in this beaker than this. The analogy is for something with a smaller heat
capacity compared to a large heat capacity, the same amount of energy causes very different changes in temperature.
Well, why do things have different heat capacities? It's because they have different ways of storing that energy. Just about all molecules can go through rotations and vibrations and move to some
degrees, but that depends on what phase they're in. If they're in the solid phase, they can't easily rotate, for instance. On the other hand, if they're in the liquid phase, not only can they vibrate
within their molecule, but they can vibrate against each other. So there are inter-molecular vibrations, and those also can help start things. So liquids have a much higher heat capacity than do
gases or solids because of these additional vibrations. Plus they have more of these other degrees of freedom in the liquid phase.
So again, the heat capacity for a substance depends on what the atoms are connected together, how they're connected and what kind of bonding you have between the molecules as well as within the
molecules. That's why we need to look up heat capacities for every different substance, or look up the absolute entropy for every different substance, because it's going to differ, because these
molecules are all different.
Finally, real quickly, how would you calculate S for a chemical reaction? Piece of cake. You go to the tables that tell you the absolute entropy of everything and in this case, our example here is
hydrogen and oxygen gives water. We find the absolute entropy of water at the appropriate temperature and subtract from that the absolute entropy of our reactants - in this case hydrogen and oxygen -
and that gives us then the entropy change for that chemical reaction. Once again, the reason that's important is once we know S and we can measure H or look up H, then we're all set to go. We can
figure out whether the reaction will spontaneously occur or not.
Entropy and Temperature Page [1 of 3]
Get it Now and Start Learning
Embed this video on your site
Copy and paste the following snippet:
Link to this page
Copy and paste the following snippet: | {"url":"http://www.mindbites.com/lesson/4821-chemistry-entropy-and-temperature","timestamp":"2014-04-19T07:31:21Z","content_type":null,"content_length":"65932","record_id":"<urn:uuid:16273b14-e5ef-4214-acb5-0a2f4b0c43a3>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00467-ip-10-147-4-33.ec2.internal.warc.gz"} |
Manorhaven, NY Algebra 1 Tutor
Find a Manorhaven, NY Algebra 1 Tutor
...From Algebraic expressions, Functions and Relations, Composition and Inverses of Functions, Exponential and Logarithmic Functions, Trigonometric Functions and their graphs and more. I provide
all and or most of the practice materials for each and every subject. I can assist with Global History, Earth Science, Physics and English, of which I have 100% pass.
47 Subjects: including algebra 1, chemistry, reading, writing
...Above all, I find math to be fun and exciting, and I like to try to instill that enthusiasm in my students!Algebra can seem like a scary subject at first. Many students find themselves feeling
intimidated by the introduction of letters into equations, and struggle with the concept of a function ...
8 Subjects: including algebra 1, calculus, geometry, algebra 2
...I have many strengths especially in math, and social studies. My teaching method is that I make learning memorable by using child-friendly techniques like songs and TV. I will do everything
possible to make sure that your kids get the best education.
6 Subjects: including algebra 1, elementary (k-6th), elementary math, prealgebra
Hi, my name is Ryan A. I am 24 years old and in my final year of law school at Columbia Law School. I transferred to Columbia from Fordham University School of Law.
16 Subjects: including algebra 1, reading, writing, GED
...My students thrive with my care and detailed but fun lessons. I have an undergraduate degree in English, and have taught and tutored English and writing for more than 20 years. My students
have seen significant success in grades and test scores.
48 Subjects: including algebra 1, reading, English, writing
Related Manorhaven, NY Tutors
Manorhaven, NY Accounting Tutors
Manorhaven, NY ACT Tutors
Manorhaven, NY Algebra Tutors
Manorhaven, NY Algebra 2 Tutors
Manorhaven, NY Calculus Tutors
Manorhaven, NY Geometry Tutors
Manorhaven, NY Math Tutors
Manorhaven, NY Prealgebra Tutors
Manorhaven, NY Precalculus Tutors
Manorhaven, NY SAT Tutors
Manorhaven, NY SAT Math Tutors
Manorhaven, NY Science Tutors
Manorhaven, NY Statistics Tutors
Manorhaven, NY Trigonometry Tutors
Nearby Cities With algebra 1 Tutor
Baxter Estates, NY algebra 1 Tutors
Glen Head algebra 1 Tutors
Glenwood Landing algebra 1 Tutors
Greenvale algebra 1 Tutors
Harbor Acres, NY algebra 1 Tutors
Kings Point, NY algebra 1 Tutors
Larchmont algebra 1 Tutors
Pelham, NY algebra 1 Tutors
Plandome, NY algebra 1 Tutors
Port Washington, NY algebra 1 Tutors
Roslyn, NY algebra 1 Tutors
Russell Gardens, NY algebra 1 Tutors
Saddle Rock, NY algebra 1 Tutors
Sands Point, NY algebra 1 Tutors
The Terrace, NY algebra 1 Tutors | {"url":"http://www.purplemath.com/Manorhaven_NY_algebra_1_tutors.php","timestamp":"2014-04-16T16:24:37Z","content_type":null,"content_length":"24197","record_id":"<urn:uuid:fe6de5af-fbde-41b5-aec3-6b85f7d4b540>","cc-path":"CC-MAIN-2014-15/segments/1398223205137.4/warc/CC-MAIN-20140423032005-00362-ip-10-147-4-33.ec2.internal.warc.gz"} |
Strategy for Information Markets/Information Cascades
Conditional ProbabilityEdit
If $Pr(A|B) =$"Probability of A given B" or "Probability of A conditioned on B"
$Pr(A|B) = \frac {Pr(A\ AND\ B)}{Pr(B)}$
Bayes' RuleEdit
$Pr(B|A) = \frac { Pr(A|B)Pr(B) } { Pr(A|B)Pr(B)+Pr(A|NotB)Pr(NotB) }$
Condorcet Jury TheoremEdit
Binomial DistributionEdit
If the probability of one success is $\text{ }p$, then
$Pr(k\text{ successes in}\ n\ \text{trials}) = \binom{n}{k} p^k (1-p)^{(n-k)}$
• $p^k\$ stands for the probability of a particular $\text{ }k$ trial being a success
• $(1-p)^{(n-k)}$ stands for the probability of a particular $\text{ }(n-k)$ trial being a failure
and in math,
• $\binom{n}{k} = \frac {n!}{k!(n-k)!}$
Group Decision/VotingEdit
In order to determine if a group decision/voting is correct, the number of successes $\text{ }C$ needs to be more than half of $\text{ }n$. The following formula derived from the Binomial
Distribution Function tells the chance of the right group decision.
In the case here, by eliminating the situation that the vote is a tie, let's assume that the number of votes $\text{ }n$ is odd so that $\text{ }C$ could be more than half of $\text{ }n$.
$Pr(C) = \sum_{k=\frac{n+1}{2}}^{n} \binom{n}{k} p^k (1-p)^{(n-k)}$
Influence-Dependent Model of Group Decision/VotingEdit
In daily lives, people usually make votes with other influences, instead of absolutely independent decision making. Let's derive another model to determine the probability of correct group decision
on other influences.
• $p =$ the probability of being correct
• $C =$ the group makes the correct decision (more than half of the votes are correct)
• $P_I =$ the probability of the influence being correct
• $\alpha =$ the probability of the voter following the influence to make decision
• $Pr(\text{Voter votes correctly}) = (1- \alpha)p + \alpha P_I$
• $P_T = Pr(\text{Voter Correct}|\text{Influence Correct})= (1- \alpha)p + \alpha =$ the probability of the voter being correct if the influence is correct
• $P_F = Pr(\text{Voter Correct}|\text{Influence Wrong})= (1- \alpha)p =$ the probability of the voter being correct if the influence is wrong
$Pr(\text{Group Correct}) = Pr(\text{Influence Correct})Pr(\text{Group Correct}|\text{Influence Correct})$
$+ Pr(\text{Influence Wrong})Pr(\text{Group Correct}|\text{Influence Wrong})$
$Pr(C) = P_I\sum_{k=\frac{n+1}{2}}^{n} \binom{n}{k} P_T^k (1-P_T)^{(n-k)}+(1-P_I)\sum_{k=\frac{n+1}{2}}^{n} \binom{n}{k} P_F^k (1-P_F)^{(n-k)}$
Central Limit TheoremEdit
Let $x_1,x_2,...,x_n$ be a series of independently and identically distributed random variables. The mean of these variables is $\mu_x$ and the variance is $\sigma_x^2$.
Let $y = \frac {1}{n} (x_1+x_2+...+x_n)$.
When $n$ gets larger, $y$ gets closer to be a random variable that is normally distributed and has mean $\mu_y = \mu_x$ and variance $\sigma_y^2 = \frac{\sigma_x^2}{n}$
Last modified on 6 May 2012, at 22:22 | {"url":"http://en.m.wikibooks.org/wiki/Strategy_for_Information_Markets/Information_Cascades","timestamp":"2014-04-17T12:40:00Z","content_type":null,"content_length":"22419","record_id":"<urn:uuid:55f4d87d-03ea-494e-bbab-5385c0db35e4>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00604-ip-10-147-4-33.ec2.internal.warc.gz"} |
Posts about PhD course on Xi'an's Og
ext month, Michael Jordan will give an advanced course at CREST-ENSAE, Paris, on Recent Advances at the Interface of Computation and Statistics. The course will take place on April 4 (14:00, ENSAE,
Room #11), 11 (14:00, ENSAE, Room #11), 15 (11:00, ENSAE, Room #11) and 18 (14:00, ENSAE, Room #11). It is open to everyone and attendance is free. The only constraint is a compulsory registration
with Nadine Guedj (email: guedj[AT]ensae.fr) for security issues. I strongly advise all graduate students who can take advantage of this fantastic opportunity to grasp it! Here is the abstract to the
“I will discuss several recent developments in areas where statistical science meets computational science, with particular concern for bringing statistical inference into contact with
distributed computing architectures and with recursive data structures :
How does one obtain confidence intervals in massive data sets? The bootstrap principle suggests resampling data to obtain fluctuations in the values of estimators, and thereby confidence
intervals, but this is infeasible computationally with massive data. Subsampling the data yields fluctuations on the wrong scale, which have to be corrected to provide calibrated statistical
inferences. I present a new procedure, the “bag of little bootstraps,” which circumvents this problem, inheriting the favorable theoretical properties of the bootstrap but also having a much
more favorable computational profile.
The problem of matrix completion has been the focus of much recent work, both theoretical and practical. To take advantage of distributed computing architectures in this setting, it is
natural to consider divide-and-conquer algorithms for matrix completion. I show that these work well in practice, but also note that new theoretical problems arise when attempting to
characterize the statistical performance of these algorithms. Here the theoretical support is provided by concentration theorems for random matrices, and I present a new approach to matrix
concentration based on Stein’s method.
Bayesian nonparametrics involves replacing the “prior distributions” of classical Bayesian analysis with “prior stochastic processes.” Of particular value are the class of “combinatorial
stochastic processes,” which make it possible to express uncertainty (and perform inference) over combinatorial objects that are familiar as data structures in computer science.”
References are available on Michael’s homepage. | {"url":"http://xianblog.wordpress.com/tag/phd-course/","timestamp":"2014-04-20T08:15:07Z","content_type":null,"content_length":"82245","record_id":"<urn:uuid:e5d4433c-e499-469a-9003-84690d09e84b>","cc-path":"CC-MAIN-2014-15/segments/1398223206770.7/warc/CC-MAIN-20140423032006-00102-ip-10-147-4-33.ec2.internal.warc.gz"} |
Muskego Math Tutor
Find a Muskego Math Tutor
...I love helping others. I have tutored students in the past with great results. I am open and confident and set high expectations for my students.
4 Subjects: including algebra 1, algebra 2, geometry, prealgebra
...I took organic chemistry 1 and 2 at the college level, as well as biochemistry. I know the material very well after tutoring it for the past 4 years at UW-Milwaukee. I have a great deal of
knowledge from what I learned and then further have reviewed through tutoring as well as knowing the material for the MCAT.
46 Subjects: including precalculus, SAT math, GRE, GED
I tutored Physics and Mathematics for 2-3 years at Hamline University in St. Paul, MN. I also have minors in Psychology and Spanish and am fairly knowledgeable in many other subject areas.
32 Subjects: including calculus, ACT Math, writing, GRE
...I have been teaching for the past seven years in Milwaukee and have spent my summers working with Reading and Math Intervention summer school programs for various school districts. Through
this experience, I have worked with many different learning styles and students with many different challen...
19 Subjects: including prealgebra, reading, algebra 1, statistics
...My name is Carolyn, and I taught math and physics in Wisconsin for ten years. Currently, I am certified to teach math in both Wisconsin and Illinois, and I do substitute teaching at about ten
different schools in both Illinois and Wisconsin. I hold a Bachelor of Arts degree in Math education from Alfred University In New York.
22 Subjects: including trigonometry, discrete math, ACT Math, algebra 1
Nearby Cities With Math Tutor
Big Bend, WI Math Tutors
Brookfield, WI Math Tutors
Franklin, WI Math Tutors
Greendale Math Tutors
Greenfield, WI Math Tutors
Hales Corners Math Tutors
Mequon Math Tutors
New Berlin, WI Math Tutors
Oak Creek, WI Math Tutors
Saint Francis, WI Math Tutors
South Milwaukee Math Tutors
Summit, WI Math Tutors
Waukesha Math Tutors
West Allis, WI Math Tutors
West Milwaukee, WI Math Tutors | {"url":"http://www.purplemath.com/Muskego_Math_tutors.php","timestamp":"2014-04-19T04:59:07Z","content_type":null,"content_length":"23477","record_id":"<urn:uuid:43cf59a1-3c1c-495f-be0d-9f430fc60b31>","cc-path":"CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00236-ip-10-147-4-33.ec2.internal.warc.gz"} |
Small-Town Story
Small-Town Story
Something missing from all the models above is geography. There is no concept of distance; each town interacts with all the others on the same basis. On a real landscape, nearby towns are surely
coupled more tightly than distant ones. A strong hint of spatial organization in the distribution of town sizes comes from the dramatic nighttime photographs of the Earth's surface made by the
Defense Meteorological Satellite Program. The instrument that records these images is a sensitive photodetector meant to measure cloud cover by moonlight, but it can also record city lights. A study
of the photographs done by Paul Sutton of the University of California, Santa Barbara and three colleagues failed to establish a precise calibration of pixel brightness to population density, but
distinguishing large and small towns is easy.
In a satellite image of the upper Midwest and Great Plains, it's obvious that the scattering of towns and cities is anything but random. In some areas, strings of towns seem to radiate from major
population centers. Elsewhere the pattern reflects the rectilinear network of roads, which in turn derives from the early surveys of the territory based on uniform square townships six miles on a
side. Of course the lattice of towns is nowhere perfect, but there's enough regularity that it leaps to the eye.
There is more to the spatial distribution of towns than just the latticelike arrangement of sites. Each category of settlement—the hamlets, the villages, the towns, the cities, the
conurbations—appears to have its own characteristic scale of distance. Progressing from smaller to larger settlements, the spacing increases. At the same time the lattice structure becomes less
There's an easy hypothesis to account for the general features of this pattern, assuming that the siting of towns and cities is determined mainly by their role as service centers for a dispersed
population. People want someplace nearby where they can mail a letter or pick up a quart of milk, but they are willing to travel farther for less-frequent errands, such as shopping for clothing, and
they'll go farther still to buy a new car or see an art exhibit. Thus each category of settlement competes mainly with others of similar size, and the interstices between major centers can be filled
in by smaller towns. Ecologists have observed a similar pattern in the distribution of desert plants, where the largest shrubs maintain a fixed distance from one another, determined by the size of
the catchment basin needed to sustain their growth; smaller plants can survive in the open areas between the bigger bushes.
This biological analogy suggests one way to model the spatial distribution of towns: Let people be sprinkled over the landscape like rainfall and drain into the nearest basin; then each basin's
radius of attraction grows along with its population. Another approach takes its inspiration from physics rather than biology. The geographic distribution of towns looks a little like the arrangement
of atoms or molecules in a mildly disordered material, such as a glass. This observation suggests the metaphor of a repulsive potential between towns, as if there were springs holding them apart.
From any starting configuration we can then allow the network of springs to relax to a state of lower total energy.
Yet there's something odd about this idea. In a model of a solid, the network of springs relaxes by adjusting the positions of the atoms. Towns and cities, however, seldom get up and move. Even if,
say, Austin and Rochester, Minnesota, would both be better off if they were a few miles farther apart, there's no convenient way to slide them across the landscape. Instead, the towns stay put, and
all adjustments have to be made by having people migrate from place to place.
Here is an algorithm (one of many possibilities) for constructing such a model. At each step, choose two towns at random with uniform probability. Calculate the repulsive interaction between each of
these towns and all the other towns in the sample, using some sort of potential in which the repulsion increases with population and diminishes with distance. The sum of all these interactions can be
interpreted as an energy. Now move a person from one town to the other in whichever direction lowers the overall energy. Then start over by choosing two new cities.
The details of the inter-town potential are where the model gets messy. An obvious starting point is "antigravity": The force between two towns is directly proportional to the product of their
populations and inversely proportional to the square of their distance. (It is antigravity because the force is repulsive rather than attractive.) Unfortunately, letting the system evolve under such
a potential does not yield a realistic geography. The problem is that large cities not only repel one another but also strongly suppress the population of nearby smaller towns and villages, creating
a vacant buffer zone that has no counterpart in the real world. A straightforward correction is to include another factor in the inter-town potential, namely the ratio of the smaller to the larger
population. If this ratio is 1 (that is, the two towns have equal populations), then the repulsion will take its full strength. As the ratio approaches 0 (for the case of a one-stop-sign hamlet and a
metropolis), the two places become almost oblivious of each other's presence.
The device of scaling the repulsive force according to the ratio of populations has a curious mathematical consequence. The force becomes proportional to pq x p/q (where p and q are the two
populations, and p q). This expression immediately simplifies to p^2; that is, the larger population drops out of the calculation altogether, and the force between the towns depends only on the
smaller population. Viewed in terms of a physical force law, this irrelevance of the larger population seems strange; it's certainly nothing like gravity. But it makes sense if we return to the
original motivations for the model. We assumed that a hamlet can survive if and only if it is the closest place to buy milk or mail letters for everyone within a certain radius. Beyond that distance,
it doesn't matter whether the next-nearest town is another tiny village or the city of Chicago.
A model constructed on these principles can be made to yield landscapes that visually have many of the same properties as real town-and-city distributions. A convenient starting configuration is a
rectilinear grid of town sites, "jiggled" slightly to break the symmetries, with all the sites assigned the same starting population. Of course the towns never move; only their populations change as
people transfer from site to site to reduce the energy of the system. After running the simulation for a few thousand steps per site, migration has altered the populations in such as way that
villages, towns and cities are all distributed at the appropriate length scale. The smallest places are densely packed, whereas larger ones keep a greater distance between them.
It should be pointed out that this model is susceptible to the same demographic black-hole problem seen in the random-walk models. The lowest-energy state of the system is the configuration that
packs everyone into a single city, since that eliminates all forces between towns. But the relaxation algorithm described above is unlikely ever to find this solution; it is a "greedy" algorithm that
almost always becomes trapped in a metastable state. Here is a case where a "better" algorithm—one that converges to the true optimum—would not necessarily improve the model. | {"url":"http://www.americanscientist.org/issues/pub/2004/3/small-town-story/4","timestamp":"2014-04-18T13:13:35Z","content_type":null,"content_length":"127063","record_id":"<urn:uuid:3d9b9db2-520f-4235-b97c-221c11015b8d>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00112-ip-10-147-4-33.ec2.internal.warc.gz"} |
Mathumatiks :: Volume by Double IntegrationWelcome Mathumatiks
Consider a surface . Let the orthogonal projection on XY plane of its portion S’ be the surface S. Divide S into elementary rectangles of area by lines parallel to x and y axis. Now consider prism
have its base as these triangles and its length parallel to OZ.
The volume of this prism under consideration is bounded by surface S and surface is .
Hence the volume of the solid cylinder having S as base, bounded by the given surfaces with generator parallel to Z-axis is
Where integration is to carried over surface S.
In polar coordinates the above integration becomes | {"url":"http://mathumatiks.com/subpage-697-Volume-by-Double-Integration.htm","timestamp":"2014-04-19T07:00:52Z","content_type":null,"content_length":"53025","record_id":"<urn:uuid:c58cd4fb-707d-4e28-894c-448f415823ba>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00484-ip-10-147-4-33.ec2.internal.warc.gz"} |
Quadratic - Vertex Form
January 22nd 2006, 03:43 PM #1
Jan 2006
Hey, I am doing this review and have a question (and may have a few more) It's been so long since I have done this I forget!
What value of p and q make x²+14x=(x-h)²+k
Hey, I am doing this review and have a question (and may have a few more) It's been so long since I have done this I forget!
What value of p and q make x²+14x=(x-h)²+k
there doesn't seem to be a p and a q, but i will take a stab in the dark you mean this
we will complete the square for x²+14x
x² + 14x + 49 (you halve the 14 = 7 and then square the result = 49)
then to balance the addition of 49 you need to subtract 49 as well
x² + 14x + 49 - 49
(x + 7)² - 49
hope that is what you are after
January 22nd 2006, 03:50 PM #2 | {"url":"http://mathhelpforum.com/pre-calculus/1707-quadratic-vertex-form.html","timestamp":"2014-04-19T14:52:07Z","content_type":null,"content_length":"31469","record_id":"<urn:uuid:714c936d-193f-4e58-b3fc-64d3d926e09e>","cc-path":"CC-MAIN-2014-15/segments/1397609537271.8/warc/CC-MAIN-20140416005217-00539-ip-10-147-4-33.ec2.internal.warc.gz"} |
Whittier, CA Geometry Tutor
Find a Whittier, CA Geometry Tutor
...I've also done several research projects and dabbled in Matlab and SPSS, so I could help with brainstorming and carrying out research. Because I studied neuroscience in college, my specialty is
the nervous system, including the eye, the CNS, and the PNS. However, I have an affinity for general anatomy and physiology, and I've taken courses in biology, chemistry, anatomy, and physiology.
69 Subjects: including geometry, chemistry, Spanish, reading
...From 2011-2013, I supervised an after school program at my local public school, where I also taught all aspects of cooking. I planned the lessons, purchased ingredients, and encouraged active
participation by my students. They enjoyed the lessons tremendously.
41 Subjects: including geometry, reading, English, writing
...This is my approach. In terms of subjects, my expertise is in Algebra, Geometry, Trigonometry, Statistics, Physics and Spanish. As I instill my methodology to my students' studying habits, they
find themselves being able to develop concentration skills to a level of solving the problems without writing.
20 Subjects: including geometry, Spanish, calculus, physics
...It was a unique experience that I learned how much I love teaching. My philosophy of teaching is to basically run myself out of a job. My goal is to tutor any individual to the point where I am
no longer needed and for the student to fully understand and adapt to the topic or academic field.
11 Subjects: including geometry, chemistry, calculus, physics
...I've been a math tutor for over 20 years and have taught many students to achieve, some have even gotten a perfect math score on their SAT. I can guide you through simple steps to a solution,
make it easy and interesting. I will help you to think instead of memorization.
6 Subjects: including geometry, algebra 1, algebra 2, prealgebra | {"url":"http://www.purplemath.com/Whittier_CA_geometry_tutors.php","timestamp":"2014-04-16T04:17:47Z","content_type":null,"content_length":"24227","record_id":"<urn:uuid:611a43a5-ea3d-45d7-8fd2-94b7eda841aa>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00452-ip-10-147-4-33.ec2.internal.warc.gz"} |
By Jean-Pierre Dijcks on Oct 09, 2013
For those who did go to Openworld, the session catalog now has the download links to the session materials online. You can now refresh your memory and share your experience with the rest of your
organization. For those who did not go, here is your chance to look over some of the materials.
On the big data side, here are some of the highlights:
There are a great number of other sessions, simply look for: Solutions => Big Data and Business Analytics and you will find a wealth of interesting content around big data, Hadoop and analytics. | {"url":"https://blogs.oracle.com/datawarehousing/category/Oracle/Big+Data?page=1","timestamp":"2014-04-19T17:24:25Z","content_type":null,"content_length":"344818","record_id":"<urn:uuid:7b0d1240-e121-4e6d-a567-e07acee4c45b>","cc-path":"CC-MAIN-2014-15/segments/1397609537308.32/warc/CC-MAIN-20140416005217-00235-ip-10-147-4-33.ec2.internal.warc.gz"} |
Posts by
Posts by Kim
Total # Posts: 1,974
What pair of fractions are proportional? a. 0.4/0.1 =? 6/1.5 b. 0.4/0.1 =? 6/1.6 c. 0.4/0.1 =? 5.6/1.5 d. 0.4/0.1 =? 5.6/1.6
Add. -3/5 + (8/35). A. 13/35 B. -13/35 c. 1/8 d. - 1/8
Add. -3/5 + (8/3.5). A. 13/35 B. -13/35 c. 1/8 d. - 1/8
Which pair of fraction is proportional? A) 0.4/0.1 =? 6/1.5 B) 0.4/0.1 =? 6/1.6 C) 0.4/0.1 =? 5.6/1.5 D) 0.4/0.1 =? 5.6/1.6
Complete the statement, using the symbol < or >. 13/16 is greater than or less than 0.563
a^3 - 8b^3/6b^2 - ba- a^2 / 3a^3b + 6a^2b^2 + 12ab^3/ a^2 - 9b^2 perform indicated operation and reduce to lowest terms
4x^2 - 16x/x^2 - 16 * x^2 + 2x -8/3x^3 + 6x^2
Joe invested $12000, part at 6% interest and part at 8% interest. In the first year he earned $860 in interest. How much was invested at each rate?
what does past idicative mean?i need to write 4 words in that form but dont know how?
what component is present in every part of a loop such as one that involves protiens producing energetic molecules from sugar
If youre using an up to date light microscope to examine animal cells you wont be able to detct?
Solve by elimination. 15x+10y=-20 3x+2y=-4
In 1990, a company had a profit of $1.3 million. In 1992, the company had a profit of $1.2 million. Write a linear equation giving the profit P, in millions of dollars, in terms of the year, t. Let t
=0 represent 1990.
1/9 -K+3/72k = 1/9k - k-1/8k i can't figure out how to get rid of the denominators
I can't figure out (-64)^1/3. Any help would be great. and (-8)^-2/3
Factor 25x2-20x+4 I can't figure out what to do with 25.
A boat can travel upstream 14 miles in same it can travel downstream 8 miles. the boat is traveling 13 miles per hour. what is the current
John takes 3 hours longer than Andrew to peel 400 pounds of apples. Together they take 8 hours. How long will it take John to peel 400 pounds by himself
A boat travel 4 miles upstream with a current of 5 miles per hour. It takes 40 mminutes longer to go upstream. How fast does the boat go downstream
identify the missing term. x^3-31
Identify the missing term. x^3-31
if p has coordinates (x, y) and is located on the unit circle, what is the highest value that the x-coordinate can take? Hence, what is the largest value that the cosine of an angle can take?
using the unit circle evaluate cos and sin for 30
How do I solve this???? Hat size 7 3/8 ??? head circum ( to nearest 1/5 23 1/5 ??? head circum to nearest cent.) 59 58
can you walk me throught this????
Maria bicycles 5 km/h faster than Carlos . In the same time it takes Carlos to bicycle 30 km, Maria can bicycle 45 km. How fast does each bicyclist travel?
Algebra 2 Honors
Write the sixth term of the sequence: t1 = 1 and tn = 3tn-1 + 1
what is 19+10s+14s+9+15s
What difference does an astronomer see between the emission spectrum of an element in a receding star and a spectrum of the same element in the lab?
HI THERE, COULD SOME ONE PLEASE HLEP ME ON THIS QUESTION... A sample of the near-ground air concentration of caesium-137 at 11 locations in southern Germany a few days after the Chernobyl nuclear
accident in 1986 consisted of the following measurements (in Bq/m3). 5.77 3.30 7....
what happens when you mix iron and sulpher?
jim examines the effect of type of music on task performance. Type of music represents which type of measurement scale
Where can I find a picture of a house that contains examples of things in a "green home". Not a real house a drawing.
Dr. Bob
a gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. given the following data, determine (delta)HoF for hydrocarbon: (delta)Horxn=-2044.5kj/mol hydrocarbon
(delta)HoF CO2= -393.5kj/mol (delta)HoF H2O = -242kj/mol denstity of CO2 and H...
us history
what reflects a governmental measure taken by the united states to encourage settlment in new areas?
Accounting- Help Please
Accounting Help-Preparing and posting journal entries; and preparing a trial balance.? Shelton Engineering completed the following transaction in the month of June. Received a bill for rent of
equipment that was used on a recently completed job. The 1,200 rent must be paid wit...
Accounting - Please check me
Indicate whether a debit or credit decreases the normal balance of each of the following accounts. Office Supplies- debit Repair Service Revenue- debit Interest Payable- debit Accounts Receivable-
creditSalaries Expense- debit Owner Capital- credit Prepaid Insurance- debit Bui...
Please double check me. Indicate whether a debit or credit decreases the normal balance of each of the following accounts. Office Supplies- debit Repair Service Revenue- debit Interest Payable- debit
Accounts Receivable- creditSalaries Expense- debit Owner Capital- credit Prep...
math riddle ALGEBRA WITH PIZZAZZ
what is 7 over d+5 =10 over d+2
Please help me with some good websites for the followung question. what industry is affected by the economy,such as the airline, automobile, home building, technological or gasoline?
Although most salamanders have four legs, the aquatic salamander resembles an eel. It lacks hind limbs and has very tiny forelimbs. Explain how limbless salamanders evolved according to Darwin's
theory of natural selection.
I am having a problem with ratio's
Math please check
Identify the amount in the following application In a shipment if 1200 parts, 60 were found to be detective. What percent of the parts were faulty? a. 60 b. 1200 c. what percent d. shipment answer: c
I was wrong
the correct answer: B
Math - double check me
Kristi joggfed for 3/5 of an hour, swam for 1/2 of an hour, and rode her bicycle for 3/4 of an hour. How long did she exercise? a. 7/11 b. 9/40 c. 1 17/20 d. 1 7/20 answer: c Juan owned 3/4 of a
family business. He sold 1/3 of the business to his aunt. What portion of the busi...
Math - please double check me
there are 42 accounting classes with 31 students in each class. Estimate the total number of students in the accounting classes. answer: 1200 students Maria earns $17 per hour.last week, she worked
10 hours/day for 5 days. What was her gross pay answer: a
A shaving/makeup mirror is designed to magnify your face by a factor of 1.33 when your face is placed 20.0 cm in front of it. A) what type of mirror is it? B)Describe the type of image that it makes
of your face . C) Calculate the required radius of curvature for the mirror.
Math- Double check
Please double check me Find the median of the following set of numbers. 47, 20, 52, 35, 58, 57 A) 49 1/2 B) 47 C) 48 1/2 D) 52 Answer: c Which of the following numbers are integers? 2/3, 35, 0.093,
-650, 42.5 A) 0.093 and 42.5 B) 35 and -650 C) 2/3 D) -650 Answer: A Determine ...
world history
how did mozart contribute to the world
Please double check Real Numbers and Order |4| + |3| Answer: 7 |-9| - |4| Answer: 5 Adding Real Numbers 7+ (-9) + (-5) + 6 Answer: 17 |-27 + 14| Answer: 13 Subtracting Real Numbers 1/2 (-5/8)
Answer: 2 3.4 (-7.6) Answer: -4.2 54 119 Answe...
please double check my answer please Real Numbers and Order 3/7, -6/7,1/7, -1/2, 2/7 Answer: -6/7, -1/2, 1/7, 2/7, 3/7 |4| + |3| Answer: 7 |-9| - |4| Answer: 5 Adding Real Numbers 7+ (-9) + (-5) + 6
Answer: 17 |-27 + 14| Answer: 13 Subtracting Real Numbers 1/2 (-5...
Please double check Jean deposited a check for $625, wrote two for $68.74 and $29.95 and used her debit card to pay for a purchase of $57.65. How has her account balance changed? Answer: $468.66
How or what did Martin Luther King Jr. do to help the bridge of selma? And how did he help with the bloody sunday on march 7, 1965?
How can I make a table showing this problem please help. Education Earnings HS diploma or GED $23,719 Associate's Degree $30,178 Bachelor's Degree $38,208 Master's Degree $47,049 Doctorate's Degree
$55,620 The table gives the median earnings of women aged 25 an...
Great Thanks a lot..
Can you please check the answer 1. 91 Days(I didn't include May 15 and August 15, Is this correct) 2. 6000*0.05*91/360 = 75.83 3. 1500 - 75.83 = 1424.17 4. 6000- 1424.17 = 4575.83 5. 77 days (again
didn't include Aug15 Nov 1) 6. 4575.83*0.05*77/360 = 48.93 7. 4575.83+4...
I am sorry I did not paste the full question. Here it is. I need help in all as I am not able to understand this Thanks Use the United States Rule and/or Banker s Rule to determine the balance due on
the note at the date of maturity. (The effective date is the date the no...
Principal - 6000 Rate - 5% Effective Date - May 15 Maturity Date - November 1 Partial Payment Amount - $1500 Partial Payment Date -August 15 Also Answer the following questions: 1. NUMBER OF DAYS
BETWEEN EFFECTIVE DATE AND PARTIAL PAYMENT = 2. INTEREST ON PARTIAL PAYMENT DATE ...
Also Answer the following questions: 1. NUMBER OF DAYS BETWEEN EFFECTIVE DATE AND PARTIAL PAYMENT = 2. INTEREST ON PARTIAL PAYMENT DATE = PRINCIPAL X RATE X (NO. OF DAYS IN #1)/360 = 3. PRINCIPAL
PAID ON PARTIAL PAYMENT DATE = PARTIAL PAYMENT - INTEREST PAID = 4.NEW PRINCIPAL ...
Use the United States Rule and/or Banker s Rule to determine the balance due on the note at the date of maturity. (The effective date is the date the note was written.) Principal - 6000 Rate - 5%
Effective Date - May 15 Maturity Date - November 1 Partial Payment Amount - ...
Essential Mathematics 1
1.4 divide by 36.96
Essential Mathematics 1
Evaluate .007 divide by 100
Please double check me. A community service organization has 19 men and 7 women in its memebership. Write the ratio of women to men. a 19/7 b 7/19 c 19/26 d 7/26 answer: b Write a proportion that is
equivalent top the statement: If 3 gallons of gasoline cost $5.37, then 8 gall...
Ms. Jordan has been given a loan of 2500 for 1 year. If the interst charged is $275, what is the interest rate on the loan?
Samantha's speedometer reads in kilometers per hour. If the legal speed limit is 55 mi/h how fast can she drive?
Please help A school had 900 students at the start of a school year. If there is an enrollment increase of 7% by the beginning of the year, what is the new enrollment? A virus scanning program is
checking every file for viruses. It has completed checking 40% of the files in 30...
A 2,000 calorie per day diet. Find decimal and fractional notation for the percent notation in each sentence. 1/2 cup of Campbells new england clam chowder provided 6% of the MDV of iron/
can you explain how yo gert 2 2/3 to 1 from 280/105?
god morgen
biology 110
you are making a standard to be used to measure protien concentration. you have added 2ml of protien, 3ml of water, and 4ml of an indicator dye. How would you prepare the blank for this experiment?
we are using the Spectrophotometer
Please help Mr. Chavez owned 3/4 of a family business. He sold 1/2 of the business to his sister. what portion of the business does he still own? a. 5/4 b. 1/4 c. 1/5 d. 4/5 Answer: b Gasoline is
about $2 per gallon. You use about 1 1/2 each day driving to and from work. how m...
biology 110
using a spectrophotometer what is the relationship between absorbance and the concentration of light absoring substance in a solution.
pre algabra
Please double check In a bottling company, a machine can fill a 2-liter (l) bottle in 0.5 second (s) and move the next bottle into place in 0.1 s. How many 2-l bottles can be filled by the machine in
2 h? Answer: 1.00 or 0.20 A circular coffee table has a diameter of 5 ft. Wha...
Please help The following table gives the free throws attempted (FTA) and the free throws made (FTM) for the top five players in the NBA for a recent season. Calculate the free throw percentage for
each player by writing the FTM over the FTA and converting this fraction to a d...
Could someone check this over for me please. Arrange in order from smallest to largest 2.05, 25/10, 2.0513, 2.059, 251/100, 20515, 2.052, 2.051. Answer: 2.05, 2.051, 2.052, 2.059, 251/100, 2.0513,
2.0515, 25/10 45.6 8.75 Answer: 36.85 (Add) thirty-eight and nine tenths,...
the novel
organic chemistry
Is it possible to reduce an aldehyde and a ketone at the same time with sodium borohydride?
is nitrogen in the hydrophobic group is oxygen in the hydrophobic group and what kinds of bonds could nitrogen and oxygen form.
which kind of fatty acid would the cell manufacture in order to keep the cell memmbrane rigid or study under warm enviromental conditions
I need serious help. Tried to work it out but I just couldn't get the correct answer. Benjamin and olivia are putting a new floor in their kitchen. To get the floor up to the desired height, they
need to add 1 /8ft to subfloor. They can do this in one of two ways. They can...
The concentration unit used in Raoult's Law calculations is: a. mole fraction b. percent by weight c. molarity d. molality e. none of these I think the answer is A... is is corrent ?
The concentration unit used in Raoult's Law calculations is: a. mole fraction b. percent by weight c. molarity d. molality e. none of these I think the answer is A... is is corrent ?
Evaluate the following: 5lb/ft x 3ft
short story
I apologize I meant to say short story not essay! Its suppose to be a short story that is descriptive, it has subtext, twist, and dialogue
lit( essay)
Please read my discriptive essay, and make any changes! My face is burning from the particles of the gravel penetrating through my skin. As I squirm through my thin cardboard box trying to find my
body, I can only hear the churning of my stomach. I try to open my eyes, but I f...
What countries are categorize under third world?A list would be great!
What are some good websites that will give information on education in third world countries?
math please help
Solve the system by addition. 2x-4y=7 4x+2y-1
thank you i just didnt know how to set it up
Adult tickets for a play cost $15 and child tickets cost $6. If there were 33 people at a performance and the theater collected $360 from ticket sales, how many children attended the play? 14
children 15 children 16 children 18 children
Johnson was impeached by the Senate and tried by the House of Respresentatives. true or false
You need to make 350mls of a 70% ethanol solution of EtOH from a 95% solution. How would you do this?
ã-108 express in terms of i
Can someone help me set up an equation for the problem? A customer wants to make a teepee in his backyard for his children. He plans to use lengths of PVC plumbing pipe for the supports on the
teepee, and he wants the teepee to be 12 feet across and 8 feet tall. How long shoul...
Thank you very much!
1.In 2002, Home Depot s sales amounted to $58,200,000,000. In 2006, its sales were $90,800,000,000. a.Write Home Depot s 2002 sales and 2006 sales in scientific notation. 2002: 5.82 x 10^10 2006:
9.08 x 10^10 You can find the percent of growth in Home Depot s s...
Pages: <<Prev | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | Next>> | {"url":"http://www.jiskha.com/members/profile/posts.cgi?name=Kim&page=18","timestamp":"2014-04-20T22:43:42Z","content_type":null,"content_length":"28156","record_id":"<urn:uuid:c6518528-e981-43c2-91c7-497a9b8b4a02>","cc-path":"CC-MAIN-2014-15/segments/1397609539230.18/warc/CC-MAIN-20140416005219-00553-ip-10-147-4-33.ec2.internal.warc.gz"} |
help me solve this problem
September 13th 2009, 06:04 PM
help me solve this problem
prove that if n is a perfect square, then n+2 is not a perfect square.
I approached this way:
hypothesis: n+2=k^2
conclusion: n=k;
But i tried both direct and indirect proof, but neither works. Or my approach is not right? pls help
September 13th 2009, 08:02 PM
Hello, zpwnchen!
Prove that if $n$ is a perfect square, then $n+2$ is not a perfect square.
$n$ is a perfect square: . $n \:=\:a^2\,\text{ for some positive integer }n.$
Suppose $n+2$is a perfect square: . $n+2 \:=\:b^2\,\text{ for some positive integer }b.$
We have: . $\begin{array}{c}n \:=\:a^2 \\ n \:=\:b^2-2 \end{array}$
Then: . $a^2 \:=\:b^2-2 \quad\Rightarrow\quad b^2 - a^2 \:=\:2 \quad\Rightarrow\quad (b-a)(b+a) \:=\:1\!\cdot\!2$
We have a system of equations: . $\begin{array}{ccc}b-a &=& 1 \\ b+a &=& 2\end{array}$
. . which has the solution: . $a \:=\:\tfrac{1}{2},\:b \:=\:\tfrac{3}{2}$
But $a\text{ and }b$ are integers . . . We have reached a contradiction.
Therefore. $n + 2$ is not a perfect square.
September 14th 2009, 07:20 AM
Thank you so much!
Can we say that n=a^2 where a is 1/2. therefore, n is not a perfect. we reached a contradiction?
Can we use direct or indirect proof for it? | {"url":"http://mathhelpforum.com/discrete-math/102147-help-me-solve-problem-print.html","timestamp":"2014-04-18T11:40:11Z","content_type":null,"content_length":"7536","record_id":"<urn:uuid:ef3683df-2da0-4aa1-9aaf-3f4f76fe2f6d>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00037-ip-10-147-4-33.ec2.internal.warc.gz"} |
ISBS - Conference Proceedings Archive
The service is considered the first attack action in volleyball games. Erratic behavior also appears along the trajectory, hindering reception. We suspect that these effects can be related to the
‘drag crisis’ phenomenon described in fluid mechanics. Thus, we decided to quantify the trajectories of four types of services (underhand, floater, floater with jump, overhand with jump). Altogether,
twenty-six real trajectories of service balls were recorded and 3D reconstructed with the DVIDEOW system. Polynomials of the 4th degree of time were adjusted to the coordinates of each trajectory,
obtaining the speed and acceleration of the balls.
We compared, in the horizontal and vertical planes, the real trajectories with simulated ones in which the ball was submitted to the same initial conditions but without any aerodynamic drag forces.
We calculated the Reynolds Number (Re), the drag coefficient (CD) and the drag force (FD), applying the model presented at the XVI ISB Congress (Deprá et al., 1997). We observed that all services are
placed in the region of the so-called drag crisis (1.105 < Re < 3.105) and present great variations of drag coefficient (Fig. 1). We also observed that the four types of services analyzed can be
ordered in an increasing sequence of Reynolds numbers. The first three types presented a decreasing sequence of median values of CD, accompanying the CD(Re) literature curve (line in Fig. 1). Even
so, we observed the growth of the drag force (FD) as a function of Re (Fig. 2). Comparing the magnitude of the two forces that act on the ball, we estimated that in the case of the overhand service
with jump the drag force becomes up to 1.4 times larger than the weight force (mg = 2.55 N). All these kind of quantification may also be used to compare characteristics of different players.
biomechanics, drag forces, three-dimensional reconstruction, volleyball
ISSN 1999-4168 | {"url":"https://ojs.ub.uni-konstanz.de/cpa/article/view/1602","timestamp":"2014-04-20T15:53:42Z","content_type":null,"content_length":"17591","record_id":"<urn:uuid:5daf7cb0-e28c-482a-a840-f606a1ed0b86>","cc-path":"CC-MAIN-2014-15/segments/1398223206118.10/warc/CC-MAIN-20140423032006-00275-ip-10-147-4-33.ec2.internal.warc.gz"} |
Expression that expands based on variables
April 15th 2013, 03:39 AM #1
Oct 2011
Expression that expands based on variables
I don't know if this is the correct sub-forum for this question. If it isn't, I'm sorry.
I have this expression that expands based on the variables given:
$a(2,2) = (\sum\limits_{i_{1}=0}^{2-1} (2-i_{1}) - 1) + (\sum\limits_{i_{1}=0}^{2-2} (2-i_{1}) - 1)$
$a(3,2) = (\sum\limits_{i_{1}=0}^{3-1} (3-i_{1}) - 1) + (\sum\limits_{i_{1}=0}^{3-2} (3-i_{1}) - 1)$
$a(2,3) = (\sum\limits_{i_{1}=0}^{2-1}\sum\limits_{i_{2}=0}^{2-1} (2-i_{1}) \times (2-i_{2}) - 1) +$
$(\sum\limits_{i_{1}=0}^{2-2}\sum\limits_{i_{2}=0}^{2-1} (2-i_{1}) \times (2-i_{2}) - 1) +$
$(\sum\limits_{i_{1}=0}^{2-2}\sum\limits_{i_{2}=0}^{2-2} (2-i_{1}) \times (2-i_{2}) - 1)$
$a(3,3) = (\sum\limits_{i_{1}=0}^{3-1}\sum\limits_{i_{2}=0}^{3-1} (3-i_{1}) \times (3-i_{2}) - 1) +$
$(\sum\limits_{i_{1}=0}^{3-2}\sum\limits_{i_{2}=0}^{3-1} (3-i_{1}) \times (3-i_{2}) - 1) +$
$(\sum\limits_{i_{1}=0}^{3-2}\sum\limits_{i_{2}=0}^{3-2} (3-i_{1}) \times (3-i_{2}) - 1)$
$a(4,4) = (\sum\limits_{i_{1}=0}^{4-1}\sum\limits_{i_{2}=0}^{4-1}\sum\limits_{i_{3}=0}^{4-1} (4-i_{1}) \times (4-i_{2}) \times (4-i_{3}) - 1) +$
$(\sum\limits_{i_{1}=0}^{4-2}\sum\limits_{i_{2}=0}^{4-1}\sum\limits_{i_{3}=0}^{4-1} (4-i_{1}) \times (4-i_{2}) \times (4-i_{3}) - 1) +$
$(\sum\limits_{i_{1}=0}^{4-2}\sum\limits_{i_{2}=0}^{4-2}\sum\limits_{i_{3}=0}^{4-1} (4-i_{1}) \times (4-i_{2}) \times (4-i_{3}) - 1) +$
$(\sum\limits_{i_{1}=0}^{4-2}\sum\limits_{i_{2}=0}^{4-2}\sum\limits_{i_{3}=0}^{4-2} (4-i_{1}) \times (4-i_{2}) \times (4-i_{3}) - 1)$
What I want to do is generalize this expression, i.e.: $a(x,y)= ?$, but I don't know how to go about this. I think I have to use recursion but I don't see how it should be done.
Any help with this would be appreciated.
Re: Expression that expands based on variables
How about stating the whole question, using the exact wording.
Re: Expression that expands based on variables
It is not a math question from a book or a teacher. It's an expression of how many comparisons an agorithm (made by me and another) makes in a worst case scenario. If my question is in any way
not clear I will try to explain it better?
Re: Expression that expands based on variables
Well it is not clear. At least, I can't see whats is going on there.
Can you describe the process used in the summations?
Re: Expression that expands based on variables
The first variable (x) determines the upper bound of the summations and the value of the constant the summations indices are subtracted from.
The second variable (y) determines the number of summations, the number of factors the summations operate on, and the number of overall terms.
The y terms each contain y-1 summations, each summation having a lower bound of 0. The upper bound of the summations depends on the term. In the first term all upper bounds are x-1, in the second
term one upper bound is x-2 and the others are x-1, in the third term two upper bounds are x-2 and the others are x-1, and so on.
How many factors the summations operate on is equal to y-1, one factor for each summation. Each factor is x-i, where i is one of the summation indices.
I hope that helps to clear it up.
April 15th 2013, 04:00 AM #2
April 15th 2013, 04:08 AM #3
Oct 2011
April 15th 2013, 04:17 AM #4
April 15th 2013, 04:43 AM #5
Oct 2011 | {"url":"http://mathhelpforum.com/algebra/217540-expression-expands-based-variables.html","timestamp":"2014-04-20T19:38:29Z","content_type":null,"content_length":"49140","record_id":"<urn:uuid:205eaeae-08fd-4d7a-a22d-01c0befc29c3>","cc-path":"CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00397-ip-10-147-4-33.ec2.internal.warc.gz"} |
Morrow, GA Geometry Tutor
Find a Morrow, GA Geometry Tutor
...I'm one of the best. I have taught all levels of math here in Georgia - except for AP calculus BC, only because I haven't had any students who desired to take it. I currently teach in Henry
County Schools at Stockbridge High where I am currently the math instructional coach and teach three classes of accelerated math.
10 Subjects: including geometry, calculus, algebra 1, algebra 2
...While tutoring at the college for well over ten years I learned many different techniques and approaches to help people learn the same type of problems, because everyone learns differently.
After earning my B.S. in Mathematics at Georgia State University I was offered the position of Mathematics...
15 Subjects: including geometry, chemistry, calculus, statistics
...I am uniquely qualified to tutor students in Zoology. While in college I took two courses in Zoology and made an A in each course. I have extensive tutoring experience in Biology, which is the
major branch of science that includes both Zoology and Botany.
57 Subjects: including geometry, reading, chemistry, English
...I have 11 years experience teaching 6th-12th grade mathematics. I teach and tutor because I am passionate about teaching and learning. I have also taught gifted and student with special needs
at the elementary school level.
14 Subjects: including geometry, reading, biology, algebra 1
...A very important aspect of teaching that is often overlooked is the ability to kindle subjects' interest in a student. Using this fact as my trump card, I have managed to keep students for
many years. I have come to realize that tutoring becomes so easy if the student enjoys what is being taught.
27 Subjects: including geometry, chemistry, physics, calculus
Related Morrow, GA Tutors
Morrow, GA Accounting Tutors
Morrow, GA ACT Tutors
Morrow, GA Algebra Tutors
Morrow, GA Algebra 2 Tutors
Morrow, GA Calculus Tutors
Morrow, GA Geometry Tutors
Morrow, GA Math Tutors
Morrow, GA Prealgebra Tutors
Morrow, GA Precalculus Tutors
Morrow, GA SAT Tutors
Morrow, GA SAT Math Tutors
Morrow, GA Science Tutors
Morrow, GA Statistics Tutors
Morrow, GA Trigonometry Tutors
Nearby Cities With geometry Tutor
Clarkston, GA geometry Tutors
College Park, GA geometry Tutors
Conley geometry Tutors
East Point, GA geometry Tutors
Ellenwood geometry Tutors
Forest Park, GA geometry Tutors
Hapeville, GA geometry Tutors
Jonesboro, GA geometry Tutors
Lake City, GA geometry Tutors
Mcdonough geometry Tutors
Red Oak, GA geometry Tutors
Rex, GA geometry Tutors
Riverdale, GA geometry Tutors
Stockbridge, GA geometry Tutors
Tyrone, GA geometry Tutors | {"url":"http://www.purplemath.com/Morrow_GA_geometry_tutors.php","timestamp":"2014-04-16T13:20:44Z","content_type":null,"content_length":"23898","record_id":"<urn:uuid:be00399b-8950-45cb-ba4b-86d423670d1e>","cc-path":"CC-MAIN-2014-15/segments/1398223201753.19/warc/CC-MAIN-20140423032001-00268-ip-10-147-4-33.ec2.internal.warc.gz"} |
order cone programming approaches for handling missing and uncertain data
Results 1 - 10 of 29
, 2007
"... In this paper we survey the primary research, both theoretical and applied, in the field of Robust Optimization (RO). Our focus will be on the computational attractiveness of RO approaches, as
well as the modeling power and broad applicability of the methodology. In addition to surveying the most pr ..."
Cited by 23 (5 self)
Add to MetaCart
In this paper we survey the primary research, both theoretical and applied, in the field of Robust Optimization (RO). Our focus will be on the computational attractiveness of RO approaches, as well
as the modeling power and broad applicability of the methodology. In addition to surveying the most prominent theoretical results of RO over the past decade, we will also present some recent results
linking RO to adaptable models for multi-stage decision-making problems. Finally, we will highlight successful applications of RO across a wide spectrum of domains, including, but not limited to,
finance, statistics, learning, and engineering.
, 1485
"... We consider regularized support vector machines (SVMs) and show that they are precisely equivalent to a new robust optimization formulation. We show that this equivalence of robust optimization
and regularization has implications for both algorithms, and analysis. In terms of algorithms, the equival ..."
Cited by 19 (4 self)
Add to MetaCart
We consider regularized support vector machines (SVMs) and show that they are precisely equivalent to a new robust optimization formulation. We show that this equivalence of robust optimization and
regularization has implications for both algorithms, and analysis. In terms of algorithms, the equivalence suggests more general SVM-like algorithms for classification that explicitly build in
protection to noise, and at the same time control overfitting. On the analysis front, the equivalence of robustness and regularization provides a robust optimization interpretation for the success of
regularized SVMs. We use this new robustness interpretation of SVMs to give a new proof of consistency of (kernelized) SVMs, thus establishing robustness as the reason regularized SVMs generalize
- Journal of Machine Learning Research , 2006
"... The fields of machine learning and mathematical programming are increasingly intertwined. Optimization problems lie at the heart of most machine learning approaches. The Special Topic on Machine
Learning and Large Scale Optimization examines this interplay. Machine learning researchers have embra ..."
Cited by 15 (1 self)
Add to MetaCart
The fields of machine learning and mathematical programming are increasingly intertwined. Optimization problems lie at the heart of most machine learning approaches. The Special Topic on Machine
Learning and Large Scale Optimization examines this interplay. Machine learning researchers have embraced the advances in mathematical programming allowing new types of models to be pursued. The
special topic includes models using quadratic, linear, second-order cone, semidefinite, and semi-infinite programs. We observe that the qualities of good optimization algorithms from the machine
learning and optimization perspectives can be quite different. Mathematical programming puts a premium on accuracy, speed, and robustness. Since generalization is the bottom line in machine learning
and training is normally done off-line, accuracy and small speed improvements are of little concern in machine learning. Machine learning prefers simpler algorithms that work in reasonable
computational time for specific classes of problems. Reducing machine learning problems to well-explored mathematical programming classes with robust general purpose optimization codes allows machine
learning researchers to rapidly develop new techniques.
- In Proceedings of the Artificial Intelligence and Statistics , 2007
"... A novel technique is proposed for improving the standard Vapnik-Chervonenkis (VC) dimension estimate for the Support Vector Machine (SVM) framework. The improved VC estimates are based on
geometric arguments. By considering bounding ellipsoids instead of the usual bounding hyperspheres and assuming ..."
Cited by 12 (3 self)
Add to MetaCart
A novel technique is proposed for improving the standard Vapnik-Chervonenkis (VC) dimension estimate for the Support Vector Machine (SVM) framework. The improved VC estimates are based on geometric
arguments. By considering bounding ellipsoids instead of the usual bounding hyperspheres and assuming gap-tolerant classifiers, a linear classifier with a given margin is shown to shatter fewer
points than previously estimated. This improved VC estimation method directly motivates a different estimator for the parameters of a linear classifier. Surprisingly, only VC-based arguments are
needed to justify this modification to the SVM. The resulting technique is implemented using Semidefinite Programming (SDP) and is solvable in polynomial time. The new linear classifier also ensures
certain invariances to affine transformations on the data which a standard SVM does not provide. We demonstrate that the technique can be kernelized via extensions to Hilbert spaces. Promising
experimental results are shown on several standardized datasets. 1
"... We consider robust least-squares regression with feature-wise disturbance. We show that this formulation leads to tractable convex optimization problems, and we exhibit a particular uncertainty
set for which the robust problem is equivalent to ℓ1 regularized regression (Lasso). This provides an inte ..."
Cited by 8 (4 self)
Add to MetaCart
We consider robust least-squares regression with feature-wise disturbance. We show that this formulation leads to tractable convex optimization problems, and we exhibit a particular uncertainty set
for which the robust problem is equivalent to ℓ1 regularized regression (Lasso). This provides an interpretation of Lasso from a robust optimization perspective. We generalize this robust formulation
to consider more general uncertainty sets, which all lead to tractable convex optimization problems. Therefore, we provide a new methodology for designing regression algorithms, which generalize
known formulations. The advantage is that robustness to disturbance is a physical property that can be exploited: in addition to obtaining new formulations, we use it directly to show sparsity
properties of Lasso, as well as to prove a general consistency result for robust regression problems, including Lasso, from a unified robustness perspective. 1
"... Leading classification methods such as support vector machines (SVMs) and their counterparts achieve strong generalization performance by maximizing the margin of separation between data
classes. While the maximum margin approach has achieved promising performance, this article identifies its sensit ..."
Cited by 5 (1 self)
Add to MetaCart
Leading classification methods such as support vector machines (SVMs) and their counterparts achieve strong generalization performance by maximizing the margin of separation between data classes.
While the maximum margin approach has achieved promising performance, this article identifies its sensitivity to affine transformations of the data and to directions with large data spread. Maximum
margin solutions may be misled by the spread of data and preferentially separate classes along large spread directions. This article corrects these weaknesses by measuring margin not in the absolute
sense but rather only relative to the spread of data in any projection direction. Maximum relative margin corresponds to a data-dependent regularization on the classification function while maximum
absolute margin corresponds to an ℓ2 norm constraint on the classification function. Interestingly, the proposed improvements only require simple extensions to existing maximum margin formulations
and preserve the computational efficiency of SVMs. Through the maximization of relative margin, surprising performance gains are achieved on real-world problems such as digit, image histogram, and
text classification. In addition, risk bounds are derived for the new formulation based on Rademacher averages.
"... In this paper, we consider the link prediction problem, where we are given a partial snapshot of a network at some time and the goal is to predict the additional links formed at a later time.
The accuracy of current prediction methods is quite low due to the extreme class skew and the large number ..."
Cited by 4 (0 self)
Add to MetaCart
In this paper, we consider the link prediction problem, where we are given a partial snapshot of a network at some time and the goal is to predict the additional links formed at a later time. The
accuracy of current prediction methods is quite low due to the extreme class skew and the large number of potential links. Here, we describe learning algorithms based on chance constrained programs
and show that they exhibit all the properties needed for a good link predictor, namely, they allow preferential bias to positive or negative class; handle skewness in the data; and scale to large
networks. Our experimental results on three real-world domains—co-authorship networks, biological networks and citation networks—show significant performance improvement over baseline algorithms. We
conclude by briefly describing some promising future directions based on this work.
"... In this paper, we consider the link prediction problem, where we are given a partial snapshot of a network at some time and the goal is to predict additional links at a later time. The accuracy
of the current prediction methods is quite low due to the extreme class skew and the large number of poten ..."
Cited by 3 (0 self)
Add to MetaCart
In this paper, we consider the link prediction problem, where we are given a partial snapshot of a network at some time and the goal is to predict additional links at a later time. The accuracy of
the current prediction methods is quite low due to the extreme class skew and the large number of potential links. In this paper, we describe learning algorithms based on chance constrained programs
and show that they exhibit all the properties needed for a good link predictor, namely, allow preferential bias to positive or negative class; handle skewness in the data; and scale to large
networks. Our experimental results on three real-world coauthorship networks show significant improvement in prediction accuracy over baseline algorithms. 1
"... We derive generalization bounds for learning algorithms based on their robustness: the property that if a testing sample is “similar ” to a training sample, then the testing error is close to
the training error. This provides a novel approach, different from the complexity or stability arguments, to ..."
Cited by 3 (0 self)
Add to MetaCart
We derive generalization bounds for learning algorithms based on their robustness: the property that if a testing sample is “similar ” to a training sample, then the testing error is close to the
training error. This provides a novel approach, different from the complexity or stability arguments, to study generalization of learning algorithms. We further show that a weak notion of robustness
is both sufficient and necessary for generalizability, which implies that robustness is a fundamental property for learning algorithms to work. 1
"... The goal of machine learning is to develop predictors that generalize well to test data. Ideally, this is achieved by training on very large (infinite) training data sets that capture all
variations in the data distribution. In the case of finite training data, an effective solution is to extend the ..."
Cited by 3 (0 self)
Add to MetaCart
The goal of machine learning is to develop predictors that generalize well to test data. Ideally, this is achieved by training on very large (infinite) training data sets that capture all variations
in the data distribution. In the case of finite training data, an effective solution is to extend the training set with artificially created examples—which, however, is also computationally costly.
We propose to corrupt training examples with noise from known distributions within the exponential family and present a novel learning algorithm, called marginalized corrupted features (MCF), that
trains robust predictors by minimizing the expected value of the loss function under the corrupting distribution— essentially learning with infinitely many (corrupted) training examples. We show
empirically on a variety of data sets that MCF classifiers can be trained efficiently, may generalize substantially better to test data, and are more robust to feature deletion at test time. 1. | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=3747467","timestamp":"2014-04-18T12:52:16Z","content_type":null,"content_length":"39015","record_id":"<urn:uuid:0780b79f-48ba-4820-b648-cd9be91bcb9d>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00159-ip-10-147-4-33.ec2.internal.warc.gz"} |
Calculating Force of Car hitting wall
To calculate the force it would apply you'll have to know how long the car is impacting the wall for. You can estimate it if you don't have any other information, or calculate it somehow knowing
certain properties of the materials of the car if you know that much. I, not having any other information, could estimate the impact to last lets say .2 seconds. Assuming the car impacts the wall to
zero speed, the impulse will be equal to the moment it has.
p = mv
[tex]\Delta[/tex]p = F[tex]\Delta[/tex]t
and since final momentum is zero,
mv = F[tex]\Delta[/tex]t
F = [tex]\stackrel{mv}{/Delta t}[/tex]
2300 lb's [tex]\approx[/tex] 1043 kg (mass of car)
40 mph [tex]\approx[/tex] 18 m/s (mps) (velocity of car)
We can just estimate [tex]\Delta[/tex]t to be .2 seconds duration, and you get F to be,
F = [tex]\stackrel{(1043 kg)(18 mps)}{(.2 s)}[/tex] = 93870 N = 21103 lb's
(I converted the weight of the car to the mass of the car in Kg and all other units to SI units, then converted final force back to lb's) | {"url":"http://www.physicsforums.com/showthread.php?t=366275","timestamp":"2014-04-20T01:06:16Z","content_type":null,"content_length":"39249","record_id":"<urn:uuid:7d6de32b-a9f3-4a5c-8445-4aeda73240a9>","cc-path":"CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00057-ip-10-147-4-33.ec2.internal.warc.gz"} |
Integration problems =)
November 11th 2013, 02:13 AM #1
May 2009
Integration problems =)
I am writing or collecting interesting integrals
and techiques for solving them. All from formulas, to special functions and so forth.
It is not intended to be an exhaustive list asGradshteyn and Ryzhik, but more a collection of puzzles
and interesting ideas.
http://folk.ntnu.no/oistes/Diverse/Problems.pdf <- The problems I have collected so far
http://folk.ntnu.no/oistes/Diverse/DelIII.pdf <- The part about advanced integration techniques
I know the language is a barrier, but hopefully most of it can be understood nevertheless. An integral is still an integral.
Now I do not come here to brag or show off, but I need your help to get ridd of the boring problems
and more so get even more fun and interesting integrals. =)
So if you spot any boring problem, or if you know a very nice integral I have yet to include please
write it here!
Re: Integration problems =)
why you don't translate them to English or German ...get an autotranslator...it will help us.
I found your collection good..
November 11th 2013, 03:29 AM #2
Senior Member
Feb 2013
Saudi Arabia | {"url":"http://mathhelpforum.com/math-challenge-problems/224086-integration-problems.html","timestamp":"2014-04-18T05:46:43Z","content_type":null,"content_length":"33903","record_id":"<urn:uuid:f29e20ae-9ecf-4b86-bb51-111e477a7a41>","cc-path":"CC-MAIN-2014-15/segments/1397609532573.41/warc/CC-MAIN-20140416005212-00258-ip-10-147-4-33.ec2.internal.warc.gz"} |
Revision #2 to TR11-108 | 14th August 2011 05:42
Why Philosophers Should Care About Computational Complexity
One might think that, once we know something is computable, how efficiently it can be computed is a practical question with little further philosophical importance. In this essay, I offer a detailed
case that one would be wrong. In particular, I argue that computational complexity theory---the field that studies the resources (such as time, space, and randomness) needed to solve computational
problems---leads to new perspectives on the nature of mathematical knowledge, the strong AI debate, computationalism, the problem of logical omniscience, Hume's problem of induction, Goodman's grue
riddle, the foundations of quantum mechanics, economic rationality, closed timelike curves, and several other topics of philosophical interest. I end by discussing aspects of complexity theory itself
that could benefit from philosophical analysis.
Changes to previous version:
Added brief remarks about pseudorandomness and homomorphic encryption; put the "lookup table argument" more clearly into the context of previous discussions about complexity and the Turing Test
Revision #1 to TR11-108 | 11th August 2011 10:13
Why Philosophers Should Care About Computational Complexity
One might think that, once we know something is computable, how efficiently it can be computed is a practical question with little further philosophical importance. In this essay, I offer a detailed
case that one would be wrong. In particular, I argue that computational complexity theory---the field that studies the resources (such as time, space, and randomness) needed to solve computational
problems---leads to new perspectives on the nature of mathematical knowledge, the strong AI debate, computationalism, the problem of logical omniscience, Hume's problem of induction and Goodman's
grue riddle, the foundations of quantum mechanics, economic rationality, closed timelike curves, and several other topics of philosophical interest. I end by discussing aspects of complexity theory
itself that could benefit from philosophical analysis.
Changes to previous version:
Incorporated suggestions by Joshua Zelinsky, Gil Kalai, Michael Collins, and Terrence Cole
TR11-108 | 8th August 2011 21:51
Why Philosophers Should Care About Computational Complexity
One might think that, once we know something is computable, how efficiently it can be computed is a practical question with little further philosophical importance. In this essay, I offer a detailed
case that one would be wrong. In particular, I argue that computational complexity theory---the field that studies the resources (such as time, space, and randomness) needed to solve computational
problems---leads to new perspectives on the nature of mathematical knowledge, the strong AI debate, computationalism, the problem of logical omniscience, Hume's problem of induction and Goodman's
grue riddle, the foundations of quantum mechanics, economic rationality, closed timelike curves, and several other topics of philosophical interest. I end by discussing aspects of complexity theory
itself that could benefit from philosophical analysis. | {"url":"http://eccc.hpi-web.de/report/2011/108/","timestamp":"2014-04-18T05:35:36Z","content_type":null,"content_length":"24081","record_id":"<urn:uuid:9d1246ac-3800-4df8-b02e-e67742f08f50>","cc-path":"CC-MAIN-2014-15/segments/1397609532573.41/warc/CC-MAIN-20140416005212-00597-ip-10-147-4-33.ec2.internal.warc.gz"} |
the theory of complex numbers
Next: the programming language REC Up: Introduction Previous: Introduction
Although the virtues of complex numbers as things which can give you the square root -1 are mentioned early on in the educational process, very little seems to be done with them thereafter. If it
weren't for complex impedances in electrical circuits, they might not even get much use at the college level.
Later on, quantum mechanics uses complex wave functions and mathematicians get to playing with Galois theory. But of course, one must certainly aspire to an algebraically complete field. Still, for
the general public, the Mandelbrot set may provide the first actual contact with the intricacies of complex analysis, and then with a vengeance when it turns out that fixed points attract critical
points all over the place, that intricate patterns repeat themselves over and over again, and yet there is a system to it all.
There are expositions of complex numbers at all levels, starting with Nahin's historical survey An Imaginary Tale [8] and including their application to geometry in Hahn's Complex Numbers and
Geometry [7]. A good engineering exposition of complex numbers and their uses can be found in Guillemin's The Mathematics of Circuit Analysis [10]. Ahlfors' Complex Analysis: An Introduction to the
Theory of Analytic Functions of One Complex Variable [5] may well be the most authoritative presentations on an advanced level, whereas Knopp's volumes on Theory of Functions [6] are both
comprehensive and quite classical.
Next: the programming language REC Up: Introduction Previous: Introduction Microcomputadoras | {"url":"http://delta.cs.cinvestav.mx/~mcintosh/comun/rec-c/node3.html","timestamp":"2014-04-20T05:43:40Z","content_type":null,"content_length":"4725","record_id":"<urn:uuid:e0354d10-3d69-47bb-b9d2-0a5323245aac>","cc-path":"CC-MAIN-2014-15/segments/1397609538022.19/warc/CC-MAIN-20140416005218-00376-ip-10-147-4-33.ec2.internal.warc.gz"} |
-Sasakian Manifolds
ISRN Geometry
VolumeΒ 2012Β (2012), Article IDΒ 421384, 13 pages
Research Article
Ricci Solitons in -Sasakian Manifolds
Department of Mathematics, Kuvempu University, Shankaraghatta 577 451, India
Received 20 April 2012; Accepted 28 May 2012
Academic Editors: F. P.Β Schuller and I.Β Strachan
Copyright Β© 2012 Gurupadavva Ingalahalli and C. S. Bagewadi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
We study Ricci solitons in -Sasakian manifolds. It is shown that a symmetric parallel second order-covariant tensor in a -Sasakian manifold is a constant multiple of the metric tensor. Using this, it
is shown that if is parallel where is a given vector field, then is Ricci soliton. Further, by virtue of this result, Ricci solitons for -dimensional -Sasakian manifolds are obtained. Next, Ricci
solitons for 3-dimensional -Sasakian manifolds are discussed with an example.
1. Introduction
In 1982, Hamilton [1] introduced the concept of Ricci flow which smooths out the geometry of manifold that is if there are singular points these can be minimized under Ricci flow. Ricci solitons move
under the Ricci flow simply by diffeomorphisms of the initial metric that is they are stationary points of the Ricci flow: , (in this paper we use ) in the space of metrics on . Hence it is
interesting to study Ricci solitons.
Definition 1.1. A Ricci soliton on a Riemannian manifold is defined by It is said to be shrinking, steady, or expanding according as and .
Note that here the metric is the pull back of the initial metric by a 1-parameter family of diffeomorphisms generated by a vector field on a manifold . Compact Ricci solitons are the fixed points of
the Ricci flow: projected from the space of metrics onto its quotient modulo diffeomorphisms and scalings and often arise as blow-up limits for the Ricci flow on compact manifolds.
In 1923, Eisenhart [2] proved that if a positive definite Riemannian manifold admits a second order parallel symmetric covariant tensor other than a constant multiple of the metric tensor, then it is
reducible. In 1925, Levy [3] obtained the necessary and sufficient conditions for the existence of such tensors. In 1989, 1990, and 1991, Sharma [4β 6] has generalized Levy's result by showing that
a second order parallel (not necessarily symmetric and nonsingular) tensor on an -dimensional space of constant curvature is a constant multiple of the metric tensor. It is also proved that in a
Sasakian manifold there is no nonzero parallel 2-form. In 2007, Das [7] in his paper proved that a second order symmetric parallel tensor on an -K-contact manifold is a constant multiple of the
associated metric tensor and also proved that there is no nonzero skew symmetric second order parallel tensor on an -Sasakian manifold. Note that -Sasakian manifolds are generalisations of Sasakian
manifolds. Hence one can find interest in generalisation, from Sasakian to -Sasakian manifolds and study Ricci solitons in this manifold.
In 2008, Sharma [8] studied Ricci solitons in K-contact manifolds, where the structure field is killing and he proved that a complete K-contact gradient soliton is compact Einstein and Sasakian. In
2010, CΔ lin and Crasmareanu [9] extended the Eisenhart problem to Ricci solitons in -Kenmotsu manifolds. They studied the case of -Kenmotsu manifolds satisfying a special condition called regular
and a symmetric parallel tensor field of second order is a constant multiple of the Riemannian metric. Using this result, they obtained the results on Ricci solitons. Recently, Bagewadi and
Ingalahalli [10] studied Ricci solitons in Lorentzian -Sasakian Manifolds.
In this paper, we obtain some results on Ricci solitons.
2. Preliminaries
Let be an almost contact metric manifold of dimension , equipped with an almost contact metric structure consisting of a tensor field , a vector field , a 1-form and a Riemannian metric , which
satisfy for all . An almost contact metric manifold is said to be -Sasakian manifold if the following conditions hold: Holds for some smooth function on .
In an -Sasakian manifold, the following relations hold: for all , where is the Riemannian curvature tensor, is the Ricci tensor and is the Ricci operator.
3. Parallel Symmetric Second Order Tensors and Ricci Solitonsin -Sasakian Manifolds
Fix a symmetric tensor field of -type which we suppose to be parallel with respect to that is . Applying the Ricci identity [4, 11] we obtain the relation Replacing in (3.2) and by using (2.5) and by
the symmetry of , we have Put in (3.3) and by virtue of (2.1), we have Replacing in (3.4), we have Solving (3.4) and (3.5), we have Since , it results Differentiating (3.7) covariantly with respect
to , we have By using the parallel condition , and (3.7) in (3.8), we have By using (2.4) in (3.9), we get Replacing in (3.10), we get Since is a nonzero smooth function in -Sasakian manifold and
this implies that the above equation implies that is a constant, via (3.7). Now by considering the above condition we state the following theorem.
Theorem 3.1. A symmetric parallel second order covariant tensor in an -Sasakian manifold is a constant multiple of the metric tensor.
Corollary 3.2. A locally Ricci symmetric -Sasakian manifold is an Einstein manifold.
Remark 3.3. The following statements for -Sasakian manifold are equivalent: (1)Einstein, (2)locally Ricci symmetric, (3)Ricci semi-symmetric that is .
The implication (1)β (2)β (3) is trivial. Now, we prove the implication (3)β (1) and means exactly (3.2) with replaced by that is, Considering and putting in (3.13), we have By using (2.6) in (
3.14), we obtain Putting in (3.15) and by using (2.1), (2.8), and (2.9) on simplification, we obtain Interchanging and in (3.16), we have Adding (3.16) and (3.17), we obtain We conclude the
Proposition 3.4. A Ricci semi-symmetric -Sasakian manifold is an Einstein manifold.
Corollary 3.5. Suppose that on a -Sasakian manifold the -type field is parallel where is a given vector field. Then yield a Ricci soliton. In particular, if the given -Sasakian manifold is Ricci
semi-symmetric with parallel, one has the same conclusion.
Proof. Follows from Theorem 3.1 and Corollary 3.2.
A Ricci soliton in -Sasakian manifold defined by (1.1). Thus is parallel. In Theorem 3.1 we proved that if an -Sasakian manifold admits a symmetric parallel tensor, then the tensor is a constant
multiple of the metric tensor. Hence is a constant multiple of the metric tensor that is , where is a nonzero constant. We close this section with applications of our Theorem 3.1 to Ricci solitons.
Corollary 3.6. If a metric in an -Sasakian manifold is a Ricci soliton with then it is Einstein.
Proof. Putting in (1.1), then we have where Substituting (3.20) in (3.19), then we get the result.
Hence we state the following result.
Corollary 3.7. A Ricci soliton in an -dimensional -Sasakian manifold cannot be steady but is shrinking.
Proof. From Linear Algebra either the vector field or . However the second case seems to be complex to analyse in practice. For this reason we investigate for the case .
A simple computation of gives From (1.1), we have and then putting , we have where by using (2.9) and (3.21) in the above equation, we have Equating (3.22) and (3.24), we have Since is some nonzero
function, we have , that is Ricci soliton in an -dimensional -Sasakian manifold cannot be steady but is shrinking because .
Corollary 3.8 3.8. If an -dimensional -Sasakian manifold is -Einstein then the Ricci solitons in -Sasakian manifold that is , where with varying scalar curvature cannot be steady but it is shrinking.
Proof. The proof consists of three parts. (i)We prove -Sasakian manifold is -Einstein. (ii)We prove the Ricci soliton in -Sasakian manifold is consisting of varying scalar curvature. (iii)We find
that the Ricci soliton in -Sasakian manifold is shrinking.
First we prove that the -Sasakian manifold is -Einstein: the metric is called -Einstein if there exists two real functions and such that the Ricci tensor of is given by the general equation Now by
simple calculations we find the values of and . Let ,β β be an orthonormal basis of the tangent space at any point of the manifold. Then putting in (3.26) and taking summation over , we get Again
putting in (3.26) then by using (2.9), we have Then from (3.27) and (3.28), we have Substituting the values of and in (3.26), we have the above equation is an -Einstein -Sasakian manifold.
Now, we have to show that the scalar curvature is not a constant and it is varying. For an -dimensional -Sasakian manifolds the symmetric parallel covariant tensor of type is given by By using (3.21)
and (3.30) in (3.31), we have Differentiating (3.32) covariantly with respect to , we have By substituting and in (3.33) and by using , we have On integrating (3.34), we have where is some integral
constant. Thus from (3.35), we have is a varying scalar curvature.
Finally, we have to check the nature of the soliton that is Ricci soliton in -Sasakian manifold:
From (1.1), we have then putting , we have If we put in (3.32), that is Above equation reduced as, Equating (3.36) and (3.38), we have Since, because is some smooth function and , that is the Ricci
soliton in an -Sasakian manifold is shrinking.
4. Ricci Solitons in 3-Dimensional -Sasakian Manifold
In this section we restrict our study to -dimensional -Sasakian manifold, that is Ricci solitons in -dimensional -Sasakian manifold.
Corollary 4.1. If a Ricci soliton where of 3-dimensional -Sasakian manifold with varying scalar curvature cannot be steady but it is shrinking.
Proof. The proof consists of three parts. (i)We prove that the Riemannian curvature tensor of 3-dimensional -Sasakian manifold is -Einstein. (ii)We prove that the Ricci soliton in 3-dimensional
-Sasakian manifold is consisting of varying scalar curvature. (iii)We find that the Ricci soliton in a 3-dimensional -Sasakian manifold is shrinking.
First we consider: the Riemannian curvature tensor of 3-dimensional -Sasakian manifold and it is given by Put in (4.1) and by using (2.5) and (2.8), we have Again put in (4.2) and by using (2.1) and
(2.10), on simplification we get By taking an inner product in (4.3), we have Interchanging and in (4.4), we have Adding (4.4) and (4.5), we have Equation (4.6) shows that a 3-dimensional -Sasakian
manifold is -Einstein.
Now, we have to show that the scalar curvature is not a constant that is is varying. Now, By using (3.21) and (4.6) in (4.7), we have Differentiating the above equation covariantly with respect to ,
we have Substituting ,β β in (4.9) and by virtue of , we have On integrating (4.10), we have where is some integral constant. Thus from (4.11), we have a varying scalar curvature.
Finally, we have to check the nature of the soliton that is Ricci soliton in 3-dimensional -Sasakian manifold.
From (1.1), we have and then putting , we have If in (4.8), that is Above equation reduced as Equating (4.12) and (4.14), we have Since from (4.15), we have . Therefore Ricci soliton in 3-dimensional
-Sasakian manifold is shrinking.
Example 4.2. Let . Let be linearly independent vector fields given by Let be the Riemannian metric defined by , , where is given by Let be the 1-form defined by for any . Let be the tensor field
defined by . Then using the linearity of and yields that and for any vector fields . Thus for , defines a Sasakian structure on . By definition of Lie bracket, we have Let be the Levi-Civita
connection with respect to above metric Koszula formula is given by Then Clearly structure is an -Sasakian structure and satisfy, where . Hence structure defines -Sasakian structure. Thus equipped
with -Sasakian structure is a -Sasakian manifold. The tangent vectors and to are expressed as linear combination of , that is and , where and are scalars.
Using in (4.11), we have and it shows that the scalar curvature is not constant.
Using in (4.15), we have In this example , this implies that , that is the Ricci soliton in 3-dimensional -Sasakian manifold is shrinking.
5. Conclusion
In this paper we have shown that the Ricci soliton in an -Sasakian manifold cannot be steady but it is shrinking accordingly because is negative.
The authors express their thanks to DST (Department of Science and Technology), Government of India, for providing financial assistance under the major research project (no. SR/S4/MS: 482/07).
1. R. S. Hamilton, β Three-manifolds with positive Ricci curvature,β Journal of Differential Geometry, vol. 17, no. 2, pp. 255β 306, 1982. View at Zentralblatt MATH
2. L. P. Eisenhart, β Symmetric tensors of the second order whose first covariant derivatives are zero,β Transactions of the American Mathematical Society, vol. 25, no. 2, pp. 297β 306, 1923.
View at Publisher Β· View at Google Scholar
3. H. Levy, β Symmetric tensors of the second order whose covariant derivatives vanish,β Annals of Mathematics. Second Series, vol. 27, no. 2, pp. 91β 98, 1925. View at Publisher Β· View at
Google Scholar
4. R. Sharma, β Second order parallel tensor in real and complex space forms,β International Journal of Mathematics and Mathematical Sciences, vol. 12, no. 4, pp. 787β 790, 1989. View at
Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
5. R. Sharma, β Second order parallel tensors on contact manifolds,β Algebras, Groups and Geometries, vol. 7, no. 2, pp. 145β 152, 1990. View at Zentralblatt MATH
6. R. Sharma, β Second order parallel tensors on contact manifolds. II,β La Société Royale du Canada. L'Académie des Sciences. Comptes Rendus Mathématiques, vol. 13, no. 6, pp. 259β 264, 1991.
View at Zentralblatt MATH
7. L. Das, β Second order parallel tensors on $\alpha$-Sasakian manifold,β Acta Mathematica. Academiae Paedagogicae Nyíregyháziensis, vol. 23, no. 1, pp. 65β 69, 2007.
8. R. Sharma, β Certain results on $K$-contact and $\left(k,\mu \right)$-contact manifolds,β Journal of Geometry, vol. 89, no. 1-2, pp. 138β 147, 2008. View at Publisher Β· View at Google
9. C. Călin and M. Crasmareanu, β From the Eisenhart problem to Ricci solitons in $f$-Kenmotsu manifolds,β Bulletin of the Malaysian Mathematical Sciences Society, vol. 33, no. 3, pp. 361β 368,
10. C. S. Bagewadi and G. Ingalahalli, β Ricci solitons in Lorentzian-Sasakian manifolds,β Acta Mathematica Academiae Paedagogicae Nyíregyháziensis, vol. 28, no. 1, pp. 59β 68, 2012.
11. P. Topping, Lectures on the Ricci Flow, vol. 325 of London Mathematical Society Lecture Note Series, Cambridge University Press, Cambridge, UK, 2006. View at Publisher Β· View at Google Scholar | {"url":"http://www.hindawi.com/journals/isrn.geometry/2012/421384/","timestamp":"2014-04-20T06:55:48Z","content_type":null,"content_length":"448485","record_id":"<urn:uuid:e4f80bdd-1f35-4c7d-b183-34542001945b>","cc-path":"CC-MAIN-2014-15/segments/1397609538022.19/warc/CC-MAIN-20140416005218-00227-ip-10-147-4-33.ec2.internal.warc.gz"} |
NA Digest Friday, February 13, 1987 Volume 87 : Issue 1
NA Digest Friday, February 13, 1987 Volume 87 : Issue 1
This weeks Editor: Gene Golub
Today's Topics:
Date: 9 Feb 87 18:23 +0800
From: Jim Varah <varah%ubc.csnet@RELAY.CS.NET>
To: na@SCORE.STANFORD.EDU
Subject: stanford phd's
I am trying to compile a list (a tree, actually) of stanford phd's in
scientific computing since the year 1.0, for the reunion next month. If you can supply leaves
or branches, please reply, giving dates, current locations, and any other
information you think I could use. In return, I'll be happy to send you
a complete tree if you're interested.
-jim varah
From: "John R. Rice" <jrr@purdue.edu>
Date: Tue, 10 Feb 87 12:37:58 EST
To: na.dis@su-score
Subject: Conference announcement/program
MARCH 23-27, 1987
Part of the special year on SCIENTIFIC COMPUTATION of the
Institute for Mathematics and its Applications
University of Minnesota
Fundamental mathematical problems arise out of scientific
software for the following areas:
* Computations on Special Architectures
* Computational Geometry
* Performance Evaluation
These problems are considered along with
* Very High Level Systems for Mathematics
The meeting format provides for lengthy discussions and
interactions between the speakers and other participants.
Invited speakers are:
Fran Berman Chris Hoffmann John Rice
Bruno Buchberger John Hopcroft Larry Snyder
Bob Caviness Elias Houstis Paul Wang
G. Farin Richard Jenks Stephan Wolfram
Dennis Gannon Lennart Johnsson
The panel members include Carl deBoor, Clarence Lehman,
Bradley Lucier, and Richard McGeehee.
For further information contact:
John R. Rice Hans F. Weinberger
Computer Sciences Inst. Math. Applications
Purdue University University of Minnesota
W.Lafayette, IN 47907 Minneapolis, MN 55455
317-494-6003 612-624-6066
jrr@cs.purdue.edu weinberg@umn-cs.csnet
Date: Tue, 10 Feb 87 16:23:24 EST
From: gragg%e.ms.uky.csnet@RELAY.CS.NET
To: na.dis@SCORE.STANFORD.EDU
Subject: Householder Award VI
The following announcement appeared in Numerische Mathematik and Linear
Algebra and its Applications. A corresponding announcement in the SIAM
News had obviously incorrect dates.
Alston S. Householder Award VI (1987)
In recognition of the outstanding services of Alston Householder,
former Director of the Mathematics Division of the Oak Ridge National
Laboratory and Professor at the University of Tennessee, to numerical
analysis and linear algebra, it was decided at the Fourth Gatlinburg
Symposium in 1969 to establish the Householder Award. This award is
in the area in which Professor Householder has worked, and its natural
developments, as exemplified by the international Gatlinburg Symposia
[see A.S. Householder, The Gatlinburgs, SIAM Review 16: 340-343
(1974)]. Recent recipients of the award include Eduardo Marques de Sa
(Coimbra), Paul Van Dooren (Leuven), James Demmel (Berkeley) and Ralph
Byers (Cornell).
The Householder Prize VI (1987) will be awarded to the author of the
best thesis in Numerical Algebra. The term Numerical Algebra is
intended to describe those parts of mathematical research which have
both algebraic aspects and numerical content or implications. Thus,
the term covers, for example, linear algebra that has numerical
applications, or the algebraic aspects of ordinary differential,
partial differential, integral and nonlinear equations.
The theses will be assessed by an international committee consisting
of Shmuel Friedland (Chicago), Bill Gragg (Kentucky), Beresford
Parlett (Berkeley), Pete Stewart (Maryland) and Axel Ruhe (Gothenburg).
To qualify, the thesis must be for a degree at the level of an
American Ph.D. awarded between 31 August 1983 and 31 December 1986.
An equivalent piece of work will be accepted from those countries
where no formal thesis is normally written at that level. The
candidate's sponsor (e.g. supervisor of his research) should submit
five copies (if possible) of the thesis (or equivalent) together with
his appraisal to:
Ms. Doris Pool
Applied Mathematics Division
Argonne National Laboratory
9700 South Cass Avenue
Argonne, IL 60439, USA
by 28 February 1987. The award will be announced at the Gatlinburg X
meeting and the candidates on the short list will receive
invitations to that meeting.
Date: Wed, 11 Feb 87 13:52:35 CST
From: "Dr. Willard Miller" <wmiller%umn.csnet@RELAY.CS.NET>
To: na@SCORE.STANFORD.EDU
514 Vincent Hall
206 Church Street S.E.
Minneapolis, Minnesota 55455
February 16-18, 1987
This workshop is made possible by grants from the
Army Research Office and the National Science Foundation
Tentative Program
Monday, Feb 16 :
8:30 - 9:30 am Z.P. Bazant "Problems and Recent Advances in Continuum
Northwestern Univ. Modelling of Softening Damage"
9:30 - 10:15 am I. Vardoulakis "Experimental Observations with Respect
Univ. of Minnesota to Strain-Softening and Localization
in Granular Media"
10:30 - 11:30 am Discussion
1:00 - 1:45 pm J.D. Dougill "A Distributed Damage Model and Some
Imperial College Possible Extensions"
1:45 - 2:30 pm K. William "Stabilization and Control of Associated
Univ. of Colorado and Non-Associated Strain-Softening
2:30 - 3:15 pm J.H. Prevost "Constitutive Equations for Soil"
Princeton Univ.
3:30 - 5:00 pm Discussion
Tuesday, Feb 17 :
8:30 - 9:15 am A. Needleman "Finite Element Analysis of Failure
Brown Univ. Modes in Ductile Solids"
9:15 - 10:05 am M.A. Crisfield "Some Experiences with Finite Element
Transport & Road Analyses of Softening Materials"
Research Lab
10:30 - 11:30 am Discussion
1:00 - 1:45 pm R. de Borst "Computational Issues Regarding the
Univ. of New Mexico Solution of Boundary Value Problems
with an Indefinite Stiffness Matrix"
1:45 - 2:30 pm M. Ortiz "Finite Element Analysis of
Brown Univ. Localized Failure"
3:00 - 4:30 pm Discussion
Wednesday, Feb 18:
8:30 - 9:15 am H.L. Schreyer "Mathematical Formulation and
Univ.ofNewMexico Problems Associated with Strain-
Softening and Localization Based on
Nonlocal Plasticity"
9:15 - 10:00 am R.D. James "Prediction of the Microstructures
Univ.ofMinnesota of Solids which Arise from a Phase
10:30 - 11:30 Discussion
1:00 - 1:45 M. Shearer "Conservation Laws of Mixed Type
North Carolina State Arising in Elasticity and Porous
University Media Flow"
1:45 - 2:30 pm E. Aifantis "Plastic Heterogeneity: Instabilities,
Mich. Tech. Univ. Dislocations, and Deformation Bands"
3:00 - 4:30 pm Discussion
Mail-From: GOLUB created at 12-Feb-87 09:47:05
Date: Wed, 11 Feb 87 13:52:35 CST
From: "Dr. Willard Miller" <wmiller%umn.csnet@RELAY.CS.NET>
To: na@SCORE.STANFORD.EDU
ReSent-Date: Thu 12 Feb 87 09:47:04-PST
ReSent-From: Gene H. Golub <GOLUB@Score.Stanford.EDU>
ReSent-To: na@Score.Stanford.EDU
514 Vincent Hall
206 Church Street S.E.
Minneapolis, Minnesota 55455
February 16-18, 1987
This workshop is made possible by grants from the
Army Research Office and the National Science Foundation
Tentative Program
Monday, Feb 16 :
8:30 - 9:30 am Z.P. Bazant "Problems and Recent Advances in Continuum
Northwestern Univ. Modelling of Softening Damage"
9:30 - 10:15 am I. Vardoulakis "Experimental Observations with Respect
Univ. of Minnesota to Strain-Softening and Localization
in Granular Media"
10:30 - 11:30 am Discussion
1:00 - 1:45 pm J.D. Dougill "A Distributed Damage Model and Some
Imperial College Possible Extensions"
1:45 - 2:30 pm K. William "Stabilization and Control of Associated
Univ. of Colorado and Non-Associated Strain-Softening
2:30 - 3:15 pm J.H. Prevost "Constitutive Equations for Soil"
Princeton Univ.
3:30 - 5:00 pm Discussion
Tuesday, Feb 17 :
8:30 - 9:15 am A. Needleman "Finite Element Analysis of Failure
Brown Univ. Modes in Ductile Solids"
9:15 - 10:05 am M.A. Crisfield "Some Experiences with Finite Element
Transport & Road Analyses of Softening Materials"
Research Lab
10:30 - 11:30 am Discussion
1:00 - 1:45 pm R. de Borst "Computational Issues Regarding the
Univ. of New Mexico Solution of Boundary Value Problems
with an Indefinite Stiffness Matrix"
1:45 - 2:30 pm M. Ortiz "Finite Element Analysis of
Brown Univ. Localized Failure"
3:00 - 4:30 pm Discussion
Wednesday, Feb 18:
8:30 - 9:15 am H.L. Schreyer "Mathematical Formulation and
Univ.ofNewMexico Problems Associated with Strain-
Softening and Localization Based on
Nonlocal Plasticity"
9:15 - 10:00 am R.D. James "Prediction of the Microstructures
Univ.ofMinnesota of Solids which Arise from a Phase
10:30 - 11:30 Discussion
1:00 - 1:45 M. Shearer "Conservation Laws of Mixed Type
North Carolina State Arising in Elasticity and Porous
University Media Flow"
1:45 - 2:30 pm E. Aifantis "Plastic Heterogeneity: Instabilities,
Mich. Tech. Univ. Dislocations, and Deformation Bands"
3:00 - 4:30 pm Discussion
Mail-From: GOLUB created at 12-Feb-87 09:47:08
Date: Wed, 11 Feb 87 14:05:55 CST
From: dongarra%dasher@anl-mcs.arpa (Jack Dongarra)
To: na.dis@su-score.arpa
Subject: Wilkinson Fellowship
ReSent-Date: Thu 12 Feb 87 09:47:07-PST
ReSent-From: Gene H. Golub <GOLUB@Score.Stanford.EDU>
ReSent-To: na@Score.Stanford.EDU
Mathematics and Computer Science Division
Argonne National Laboratory
Argonne National Laboratory is seeking outstanding candidates in the
general area of computational mathematics to fill the newly created J.H.
Wilkinson Fellowship in Computational Mathematics.
This fellowship was created in memory of Dr. James Hardy Wilkinson,
F.R.S., who for many years had a close association with the Mathematics and
Computer Science Division at Argonne, where he acted as a consultant and
guiding spirit for such efforts as the EISPACK and LINPACK projects.
The J.H. Wilkinson Fellowship is intended to encourage young scien-
tists who are actively engaged in state-of-the-art research in computa-
tional mathematics-including, but not limited to, the development and
implementation of numerical algorithms for linear algebra. The candidate
must have earned (or be about to earn) a Ph.D. degree or the equivalent
during the past five years and should have a strong background in numerical
computation. The candidate should also be interested in expanding into the
area of advanced computing research. Argonne's Mathematics and Computer
Science Division has strong programs in computational mathematics and
advanced computing, as well as in software engineering and applied
This one-year appointment includes salary (starting at $45,000; higher
with experience), moving expenses, and a generous professional travel
allotment. Applications from qualified candidates, as well as nominations
for the position of Wilkinson Fellow, should be addressed to Jack J.
Dongarra, Mathematics and Computer Science Division, Argonne National
Laboratory, Argonne, Illinois 60439-4844. Applications should include a
resume and a statement of research goals, and the names of three refer-
ences. The closing date for applications is April 1, 1987. The applica-
tions will be reviewed during April 1987 by an international selection com-
mittee, and an announcement made in May 1987. The position will commence
during September 1987.
Further inquiries can be made by calling 312-972-7163 or by sending
electronic-mail to dongarra@anl-mcs.arpa.
Argonne is an affirmative action/equal opportunity employer.
Mail-From: GOLUB created at 12-Feb-87 09:47:11
Date: Tue, 10 Feb 87 16:23:24 EST
From: gragg%e.ms.uky.csnet@RELAY.CS.NET
To: na.dis@SCORE.STANFORD.EDU
Subject: Householder Award VI
ReSent-Date: Thu 12 Feb 87 09:47:10-PST
ReSent-From: Gene H. Golub <GOLUB@Score.Stanford.EDU>
ReSent-To: na@Score.Stanford.EDU
The following announcement appeared in Numerische Mathematik and Linear
Algebra and its Applications. A corresponding announcement in the SIAM
News had obviously incorrect dates.
Alston S. Householder Award VI (1987)
In recognition of the outstanding services of Alston Householder,
former Director of the Mathematics Division of the Oak Ridge National
Laboratory and Professor at the University of Tennessee, to numerical
analysis and linear algebra, it was decided at the Fourth Gatlinburg
Symposium in 1969 to establish the Householder Award. This award is
in the area in which Professor Householder has worked, and its natural
developments, as exemplified by the international Gatlinburg Symposia
[see A.S. Householder, The Gatlinburgs, SIAM Review 16: 340-343
(1974)]. Recent recipients of the award include Eduardo Marques de Sa
(Coimbra), Paul Van Dooren (Leuven), James Demmel (Berkeley) and Ralph
Byers (Cornell).
The Householder Prize VI (1987) will be awarded to the author of the
best thesis in Numerical Algebra. The term Numerical Algebra is
intended to describe those parts of mathematical research which have
both algebraic aspects and numerical content or implications. Thus,
the term covers, for example, linear algebra that has numerical
applications, or the algebraic aspects of ordinary differential,
partial differential, integral and nonlinear equations.
The theses will be assessed by an international committee consisting
of Shmuel Friedland (Chicago), Bill Gragg (Kentucky), Beresford
Parlett (Berkeley), Pete Stewart (Maryland) and Axel Ruhe (Gothenburg).
To qualify, the thesis must be for a degree at the level of an
American Ph.D. awarded between 31 August 1983 and 31 December 1986.
An equivalent piece of work will be accepted from those countries
where no formal thesis is normally written at that level. The
candidate's sponsor (e.g. supervisor of his research) should submit
five copies (if possible) of the thesis (or equivalent) together with
his appraisal to:
Ms. Doris Pool
Applied Mathematics Division
Argonne National Laboratory
9700 South Cass Avenue
Argonne, IL 60439, USA
by 28 February 1987. The award will be announced at the Gatlinburg X
meeting and the candidates on the short list will receive
invitations to that meeting.
Mail-From: GOLUB created at 12-Feb-87 09:47:13
From: "John R. Rice" <jrr@purdue.edu>
Date: Tue, 10 Feb 87 12:37:58 EST
To: na.dis@su-score
Subject: Conference announcement/program
ReSent-Date: Thu 12 Feb 87 09:47:13-PST
ReSent-From: Gene H. Golub <GOLUB@Score.Stanford.EDU>
ReSent-To: na@Score.Stanford.EDU
MARCH 23-27, 1987
Part of the special year on SCIENTIFIC COMPUTATION of the
Institute for Mathematics and its Applications
University of Minnesota
Fundamental mathematical problems arise out of scientific
software for the following areas:
* Computations on Special Architectures
* Computational Geometry
* Performance Evaluation
These problems are considered along with
* Very High Level Systems for Mathematics
The meeting format provides for lengthy discussions and
interactions between the speakers and other participants.
Invited speakers are:
Fran Berman Chris Hoffmann John Rice
Bruno Buchberger John Hopcroft Larry Snyder
Bob Caviness Elias Houstis Paul Wang
G. Farin Richard Jenks Stephan Wolfram
Dennis Gannon Lennart Johnsson
The panel members include Carl deBoor, Clarence Lehman,
Bradley Lucier, and Richard McGeehee.
For further information contact:
John R. Rice Hans F. Weinberger
Computer Sciences Inst. Math. Applications
Purdue University University of Minnesota
W.Lafayette, IN 47907 Minneapolis, MN 55455
317-494-6003 612-624-6066
jrr@cs.purdue.edu weinberg@umn-cs.csnet
Date: Thu 12 Feb 87 10:13:34-PST
From: Mark Kent <KENT@Sushi.Stanford.EDU>
Subject: na.lastname
To: na@Score.Stanford.EDU
We experienced a failure of the na.lastname forwarding facility yesterday.
Those of you who had a message bounce back with the error message
554 "|/user/golub/namail"... unknown mailer error 1
should resend the message.
Early last night I managed to put in something that would grab the
na messages and this morning I processed them (31 messages).
Everything is now working again.
(P.S. This whole NAnet business is a test of SDI software methodology.)
Date: 12 Feb 87 19:03 +0800
From: Jim Varah <varah%ubc.csnet@RELAY.CS.NET>
To: na@SCORE.STANFORD.EDU
Subject: stanford grads
I am preparing a list (or tree) of stanford phd's in scientific computation
for the reunion next month. if you can supply information, please do so...
I need names, dates, present employment, phd grads (if any), etc.
In return, I'll send you a completed list.
-jim varah
From: "John R. Rice" <jrr@purdue.edu>
Date: Thu, 12 Feb 87 15:35:17 EST
To: na.dis@score.stanford.edu
Subject: Meeting announcement
MARCH 23-27, 1987
Part of the special year on SCIENTIFIC COMPUTATION of the
Institute for Mathematics and its Applications
University of Minnesota
Fundamental mathematical problems arise out of scientific
software for the following areas:
* Computations on Special Architectures
* Computational Geometry
* Performance Evaluation
These problems are considered along with
* Very High Level Systems for Mathematics
The meeting format provides for lengthy discussions and
interactions between the speakers and other participants.
Invited speakers are:
Fran Berman Chris Hoffmann John Rice
Bruno Buchberger John Hopcroft Larry Snyder
Bob Caviness Elias Houstis Paul Wang
G. Farin Richard Jenks Stephan Wolfram
Dennis Gannon Lennart Johnsson
The panel members include Carl deBoor, Clarence Lehman,
Bradley Lucier, and Richard McGeehee.
For further information contact:
John R. Rice Hans F. Weinberger
Computer Sciences Inst. Math. Applications
Purdue University University of Minnesota
W.Lafayette, IN 47907 Minneapolis, MN 55455
317-494-6003 612-624-6066
jrr@cs.purdue.edu weinberg@umn-cs.csnet
Date: Thu, 12-FEB-1987 13:59 EST
From: <JANLEE%VTCS1.BITNET@wiscvm.wisc.edu>
To: <na@su-score.arpa>
Subject: Eloge and Anecdotes for Jim Wilkinson
My name is JAN Lee, recently appointed as the Editor-in-Chief of
the Annals of the History of Computing. The journal, now in its
ninth volume, is published by AFIPS through Springer-Verlag.
Shortly after the death of Jim Wilkinson last year I was copied on
some of the anecdotes which were circulated amongst you. Unfortunately
no-one of my acquaintance has a complete set since I feel that your
commentsd and notes would form an excellent basis for an article
about Jim which might well accompany a more form eloge.
Is there anyone who did keep a complete record of that sharing of
experiences who would be (1) willing to share it with me to review
for possible editing and publication, (2) would be interested
in writing an eloge, (3) would be interested in adding to the
history of the profession in other subjects.
Annals has done a reasonable job over the past years in recording the
history of the early machines, has had a few excursions into the history
of software (mainly the languages such as FORTRAN and COBOL), and
some personal histories. The history of mathematical and scientific
computation is hardly touched. I realize that ACM will be holding a
conference on this topic in May 1987, but I also know that there is a
lot more than can be expressed in two-three days. I would be interested
in hearing of other events and activities.
JAN Lee
Virginia Tech
Date: 12 February 1987 14:09:00 CST
From: U32799 at UICVM (Uri N. Peled 312-996-4826)
To: NA%SU-SCORE at SCORE.STANFORD.EDU
Subject: Inquiry
Can anybody help with the following problem:
One has to find the complex roots of a polynomial of degree much
larger than 100, say of the order of 1000. The standard routines like
IMSL fail miserably. A very expensive solution is to diagonalize the
companion matrix of the polynomial, and this works up to about 500.
If you have the software, or information on where to get it, would you
please send it to
Thank you very much.
Mail-From: GOLUB created at 13-Feb-87 15:00:34
Date: Fri, 13 Feb 87 14:04:38 CST
From: dongarra%dasher@anl-mcs.arpa (Jack Dongarra)
To: golub@su-score.arpa
Subject: Wilkinson announcement
ReSent-Date: Fri 13 Feb 87 15:00:34-PST
ReSent-From: Gene H. Golub <GOLUB@Score.Stanford.EDU>
ReSent-To: na.dis@Score.Stanford.EDU
Mathematics and Computer Science Division
Argonne National Laboratory
Argonne National Laboratory is seeking outstanding candidates in the
general area of computational mathematics to fill the newly created J.H.
Wilkinson Fellowship in Computational Mathematics.
This fellowship was created in memory of Dr. James Hardy Wilkinson,
F.R.S., who for many years had a close association with the Mathematics and
Computer Science Division at Argonne, where he acted as a consultant and
guiding spirit for such efforts as the EISPACK and LINPACK projects.
The J.H. Wilkinson Fellowship is intended to encourage young scien-
tists who are actively engaged in state-of-the-art research in computa-
tional mathematics-including, but not limited to, the development and
implementation of numerical algorithms for linear algebra. The candidate
must have earned (or be about to earn) a Ph.D. degree or the equivalent
during the past five years and should have a strong background in numerical
computation. The candidate should also be interested in expanding into the
area of advanced computing research. Argonne's Mathematics and Computer
Science Division has strong programs in computational mathematics and
advanced computing, as well as in software engineering and applied
This one-year appointment includes salary (starting at $45,000; higher
with experience), moving expenses, and a generous professional travel
allotment. Applications from qualified candidates, as well as nominations
for the position of Wilkinson Fellow, should be addressed to Jack J.
Dongarra, Mathematics and Computer Science Division, Argonne National
Laboratory, Argonne, Illinois 60439-4844. Applications should include a
resume and a statement of research goals, and the names of three refer-
ences. The closing date for applications is June 1, 1987. The applica-
tions will be reviewed during June 1987 by an international selection com-
mittee. The position will commence during 1988.
Further inquiries can be made by calling 312-972-7163 or by sending
electronic-mail to dongarra@anl-mcs.arpa.
Argonne is an affirmative action/equal opportunity employer.
End of NA Digest | {"url":"http://www.netlib.org/na-digest-html/87/v87n01.html","timestamp":"2014-04-20T06:16:18Z","content_type":null,"content_length":"32297","record_id":"<urn:uuid:5918080b-3e0f-475f-9034-d7425042a43a>","cc-path":"CC-MAIN-2014-15/segments/1397609538022.19/warc/CC-MAIN-20140416005218-00459-ip-10-147-4-33.ec2.internal.warc.gz"} |
Math Forum: Math Library - Symmetry/Tesselltns
1. Catalog of Isohedral Tilings by Symmetric Polygonal Tiles - Doris Schattschneider
A collection of Sketchpad and JavaSketchpad files supplementing the article "One Corona is Enough for the Euclidean Plane" by mathematicians Doris Schattschneider and Nikolai Dolbilin. Tilings
of the Euclidean plane by a single polygon are considered, and it is shown that such a tiling is isohedral if and only if each polygon is "surrounded in the same way," or, more technically, the
centered coronas of tiles are pairwise congruent. more>>
2. Gallery of Interactive On-Line Geometry - The Geometry Center
Twelve interactive online geometry programs: WebPisces (compute implicitly defined curves in the plane); Build a Rainbow (a lab that examines a mathematical model of light passing through a
water droplet); QuasiTiler (generate Penrose tilings or design your own nonperiodic tilings of the plane); Kali and Kali-Jot (an interactive editor for symmetric patterns of the plane);
Cyberview-X (an interactive 3D viewer); Projective Conics (Pascal's theorem in terms of projective geometry: specify points on a conic and see how the theorem applies to them); Orbifold Pinball
(explore the effects of negatively curved space); Teichmuller Navigator (explore the space of all different angle geometries on a genus two surface); Integrator (experiment with numerical
integration of data sets); Unifweb (families of Riemann surfaces), and Lafite (work with any discrete symmetry group of the hyperbolic plane). more>>
3. Introduction to Symmetry for Primary Students - Varnelle Moore
A unit designed to give young children (K-2) an introduction to symmetry, with lessons on describing tangrams, slides, turns, and flips. Each lesson includes: an interactive, manipulative-based
project; an activity incorporating technology; pencil and paper practice; and literature connections. Helpful links and teacher support extension ideas are also provided. more>>
4. Investigating Patterns: Symmetry and Tessellations - Jill Britton
Thirty activities from a variety of Web sites, coordinated with a forthcoming book by Jill Britton for Cuisenaire/Dale Seymour, Exploring Pattern: Symmetry and Tessellations (March 1999).
Topics include an introduction to symmetry; symmetry in the alphabet, in flags, in quilt blocks; reflectional and rotational symmetry; strip patterns; regular and irregular polygons;
paper-folding and cutting; origami; tessellation in nature; kaleidoscopes; regular, semiregular, and demiregular tessellations; islamic tessellations; M. C. Escher; jigsaw puzzles, tessellating
art; and explorations with TesselMania. more>>
5. Maths Games and Resources - Adrian Bruce
Adrian Bruce, an Australian teacher, offers a variety of resources that he has developed to use with his students. The math resources are organized under these topics: Symmetry; Facts to 10;
Multiplication Jump; Multiplication Bingo; 3d Shapes; Decimals; 2d Shapes more>>
6. Symmetry and Group Theory (The Geometry Junkyard) - David Eppstein, Theory Group, ICS, UC Irvine
Computational and recreational geometry pointers to sites with information, problems, and lessons having to do with symmetry and group theory (tessellations, kaleidoscopes, origami, etc.).
7. Symmetry and Pattern: The Art of Oriental Carpets - The Textile Museum/Math Forum
In this online exhibit, the study of symmetry is used to analyze patterns in Oriental carpets. A joint project of The Textile Museum and The Math Forum. more>>
8. Tessellation Tutorials - Suzanne Alejandre
A series of tutorials that teach students how to tessellate (somewhat in the style of the art of M.C. Escher) using HyperCard for black and white and/or HyperStudio for color, ClarisWorks,
LogoWriter, templates, or simple straightedge and compass. The tessellation lessons include units incorporating rotations and glide reflections, a section called "Where's the Math" that
elaborates on some underlying geometric principles, comments contributed by others, and samples of student work. See, in particular, What Is a Tessellation? more>>
9. Totally Tessellated - Bhushan, Kay, & Williams
A comprehensive introduction to tessellations and tilings - the basic underlying mathematics and examples of tessellations in real life. The highly illustrated, printer-friendly site includes
history (tessellations in mathematics and science, decorative use, an ancient and modern historical image gallery, and M. C. Escher); essentials (background information from foundations in
polygons and angles to color use, simple tessellations with regular polygons, and simple tessellations with non-regular polygons); mosaics/tilings (regular and non-regular tilings of polygons);
Escher (introduction, biography, trends in his work, analysis of twelve of his tessellations); and "beyond" (other design and color techniques, related art movements, and related advanced
mathematical topics). The site also provides help and searching, a gallery of users' tessellations (you can submit your own and vote on the best tessellations), a message board, an illustrated
glossary specific to the mathematics and art of tessellations, and categorized, annotated outside links. more>> | {"url":"http://mathforum.org/library/topics/sym_tess?keyid=38676212&start_at=1&num_to_see=50","timestamp":"2014-04-18T03:04:44Z","content_type":null,"content_length":"15332","record_id":"<urn:uuid:8f6e6887-2660-4b3c-bff2-b7d28a6cb8f0>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00190-ip-10-147-4-33.ec2.internal.warc.gz"} |
Generalized Stochastic Petri Nets
M. Ajmone Marsan, G. Balbo, G. Conte, S. Donatelli and G. Franceschinis
Modelling with Generalized Stochastic Petri Nets
Wiley Series in Parallel Computing
John Wiley and Sons
ISBN: 0 471 93059 8
THIS BOOK IS OUT OF PRINT.
IT IS NOW POSSIBLE TO DOWNLOAD A REVISED ELECTRONIC VERSION (.pdf) OF THE BOOK.
Orders should be sent to:
John Wiley & Sons Ltd Distribution Centre
Southern Cross Trading Estate
1 Oldlands Way
Bognor Regis
West Sussex
PO22 9SA
Tel: +44 1243 779777
Fax: +44 1243 820250
Email: market@wiley.co.uk
M. Ajmone Marsan, Politecnico di Torino, Italy,
G. Balbo, Universita' di Torino, Italy,
G. Conte, Universita' di Parma, Italy,
S. Donatelli, Universita' di Torino, Italy
and G. Franceschinis, Universita' di Torino, Italy
This book presents a unified theory of Generalized Stochastic Petri Nets (GSPNs) together with a set of illustrative examples from different application fields.
The continuing success of GSPNs and the increasing interest in using them as a modelling paradigm for the quantitative analysis of distributed systems suggested the preparation of this volume with
the intent of providing newcomers to the field with a useful tool for their first approach.
Readers will find a clear and informal explanation of the concepts followed by formal definitions when necessary or helpful. The largest section of the book however is devoted to showing how this
methodology can be applied in a range of domains.
1.1 Shared Resources
1.2 Fork and Join
1.3 Kanban Process
1.4 Token Ring LAN
2.1 Petri Net Models
2.2 Models, Systems, and Nets
2.3 System Dynamics
2.3.1 Enabling and firing rules
2.3.2 Reachability set and reachability graph
2.3.3 Modelling issues
Modelling logical conditions
2.3.4 Typical situations in the PN system evolution
Causal connection
Mutual exclusion
2.4 Properties of Petri Nets
Reachability and reversibility
Absence of deadlock
Mutual exclusion
2.5 Structural Analysis Techniques
2.5.1 Linear algebraic techniques
P-semiflows and P-invariant relations
T-semiflows and T-invariant relations
2.5.2 Graph-based structural techniques
Deadlocks and traps
Structural relations between transitions
2.6 State Space Analysis Techniques
Absence of deadlock
Mutual exclusion
Effective conflict and confusion
3 TIME IN PETRI NETS
3.1 The Motivations for Timing
3.2 Timed Transitions
3.2.1 Immediate transitions
3.2.2 Two examples
3.3 Parallelism and Conflict
3.4 Memory
3.4.1 An example with the enabling memory policy
3.4.2 An example with the age memory policy
3.5 Multiple Enabling
3.6 A Timed PN Model Example
3.7 Performance Measures
4.1 Definition and Dynamics
Enabling and firing
An example
4.2 Conflicts, Confusion and Priority
Confusion and priority
4.3 Properties of Priority PN Models
Liveness and liveness degree
Reversibility and home states
4.4 Structural Analysis Techniques
4.4.1 Linear algebraic techniques
P- and T-semiflows
4.4.2 Graph based analysis
Structural conflict
4.5 Reachability Analysis Techniques
Reachability graph of PN models with priority
Reduced reachability graphs
5 GSPN BASICS
5.1 Exponential Distributions for Transition Delays
5.2 Mixing Exponentially Distributed and Null Delays
5.3 Some Extensions
5.4 The Definition of a GSPN Model
5.5 An Example of a GSPN Model
5.6 Some Fine Points
6.1 The Stochastic Process Associated with a SPN
6.1.1 SPN performance indices
6.1.2 An example SPN system
6.2 The Stochastic Process Associated with a GSPN
6.2.1 Marking dependency
6.2.2 An example GSPN system
6.3 Numerical Solution of GSPN Systems
6.3.1 Evaluation of the steady-state probability distribution for the example GSPN system
6.3.2 Performance analysis of the example GSPN
6.4 Reducing GSPNs to SPNs
6.5 Simulation of GSPN Systems
6.5.1 Simulation of the example GSPN system
7.1 The Central Server Model Example
7.1.1 CPU interruptions
7.1.2 CPU memory policies
7.2 Phase-Type Distributions
7.2.1 Erlang distributions
7.2.2 Numerical evaluation
7.3 General Phase-Type Distributions
8.1 Flexible Manufacturing Systems
8.2 A Kanban System
8.2.1 Structural analysis of the GSPN model of the Kanban system
8.2.2 Performance analysis of the Kanban system
8.3 Push Production Systems
A system using continuous transportation
A system with AGV transportation
8.3.1 Structural analysis of the push production systems
8.3.2 Performance analysis of the system with AGV transport
9.1 Polling Systems
9.2 Modelling Random Polling the Easy Way
9.2.1 The first complete random polling model
9.3 Beginning to Exploit Symmetries
9.3.1 Walking before choosing?
9.3.2 Number of states and limitations of the model
9.4 Abstracting from the Queue Identifiers
9.4.1 The abstract GSPN model
9.4.2 Walking before choosing!
9.4.3 Number of states and limitations of the model
9.5 From GSPNs to SPNs
10.1 Introduction to the Static Analysis of Concurrent Programs
10.2 The Automatic Translation from Programs to GSPNs
10.2.1 Languages of interest
10.2.2 The translation procedure
10.3 An Example of Translation
10.4 Modelling Control Variables to Reduce Nonfaults
10.5 Stochastic Analysis
Probability of two communications at the same time
Joint distribution of p_k and p_n
Mutual exclusion
11.1 Concurrent Architectures
11.2 Single Processor Model
11.3 Single Processor Model with Multitasking
11.4 Models of Mesh Structures
11.4.1 Mesh models without multitasking
11.4.2 Mesh models with multitasking
Appendix A: STOCHASTIC PROCESS FUNDAMENTALS
A.1 Basic Definitions
A.2 Markov Processes
A.3 Discrete-Time Markov Chains
A.3.1 Steady-state distribution
A.3.2 Matrix formulation
A.3.3 Example
A.4 Continuous-Time Markov Chains
A.4.1 Steady-state distribution
A.4.2 Matrix formulation
A.4.3 Sojourn and recurrence times
A.4.4 The embedded Markov chain
A.4.5 Example
A.5 Aggregation of States in Markov Chains
A.5.1 Aggregation in discrete-time Markov chains
A.5.2 Aggregation in continuous-time Markov chains
A.6 Semi-Markov Processes
A.7 Semi-Markov Reward Processes
Appendix B: GLOSSARY | {"url":"http://www.di.unito.it/~greatspn/GSPN-Wiley/","timestamp":"2014-04-21T15:16:31Z","content_type":null,"content_length":"9482","record_id":"<urn:uuid:1ce3b8c9-083e-4cb2-a0cf-45f6e1db8cec>","cc-path":"CC-MAIN-2014-15/segments/1397609540626.47/warc/CC-MAIN-20140416005220-00455-ip-10-147-4-33.ec2.internal.warc.gz"} |
correct behavour of at()
10-21-2007 #1
Registered Abuser
Join Date
Sep 2007
correct behavour of at()
I wanted to build this list ground up again. I'd like to know if I'm on a better track this time?
So far, my example really only tests Iter creation/copying/assigning.
1) Is my Iter assignment operator correct?
2) Is this how the funtion at() is normally implemented?
#ifndef ALIST_H
#define ALIST_H
template <typename T>
class List{
struct Node{
T data;
Node *next, *prev;
Node(const T& d, Node* n = 0, Node* p = 0) : data(d), next(n), prev(p) {};
Node(const Node& src) : data(src.data), next(src.next), prev(src.prev) {};
Node *head, *tail;
unsigned int sizeOfList;
class Iter{
Node* curr;
Iter(Node* c = 0) : curr(c) {};
Iter(const Iter& rhs_i) : curr(rhs_i.curr) {};
Iter& operator=(const Iter& rhs_i){ Node* c = rhs_i.curr; return *this; };
const T& operator*() const { return curr->data; };
T& operator*() { return curr->data; };
bool operator!=(const Iter& i) { return i.curr != curr; };
bool operator==(const Iter& i) { return i.curr == curr; };
Iter& operator++(int) { curr = curr->next; return *this; };
Iter& operator--(int) { curr = curr->prev; return *this; };
List<T>(const List& src);
unsigned int size(){ return sizeOfList; };
const Iter begin() const { return Iter(head); };
const Iter end() const { return Iter(0); };
const Iter rbegin() const { return Iter(tail); };
const Iter rend() const { return Iter(0); };
/* Here's at()
const Iter at(int n) const {
Iter temp = begin();
for(int i = 0; i < n; i++){
return temp;
template <typename T>
List<T>::List() : head(0), tail(0), sizeOfList(0) {}
#include <iostream>
#include "AList.h"
int main(){
List<int> mlist;
List<int>::Iter i;
List<int>::Iter j(i);
List<int>::Iter k;
List<int>::Iter l = k = j;
return 0;
Iter& operator=(const Iter& rhs_i){ Node* c = rhs_i.curr; return *this; };
You are assigning to a temporary.
I don't think either Node nor Iter needs a special copy contructor or assignment operator. They don't require anything else but the shallow copy that the compiler would produce automatically.
I might be wrong.
Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
Quoted more than 1000 times (I hope).
You have incorrectly implemented postincrement as if it were preincrement, so the semantics of it is wrong.
There should be no "at" function for a list. Any O(n) running time list accessor is fundamentally wrong, as it encourages writing horribly inefficient code. This is why the SC++L does not have it
for a list.
My homepage
Advice: Take only as directed - If symptoms persist, please see your debugger
Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"
I think Node should be noncopyable. It doesn't make sense to copy nodes.
All the buzzt!
"There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
- Flon's Law
I don't think either Node nor Iter needs a special copy contructor or assignment operator. They don't require anything else but the shallow copy that the compiler would produce automatically.
There should be no "at" function for a list. Any O(n) running time list accessor is fundamentally wrong, as it encourages writing horribly inefficient code. This is why the SC++L does not have it
for a list.
So on both accounts just remove and implement w/out?
I think Node should be noncopyable. It doesn't make sense to copy nodes.
Is there a special way of preventing the copying of Nodes?
Make the copy constructor and copy assignment operator private.
C + C++ Compiler: MinGW port of GCC
Version Control System: Bazaar
Look up a C++ Reference and learn How To Ask Questions The Smart Way
My homepage
Advice: Take only as directed - If symptoms persist, please see your debugger
Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"
AAAck.. I would not provide an at() function for a list. You run the risk of somebody attempting to use it!
EDIT: Seems I was beaten to the punch.
All the buzzt!
"There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
- Flon's Law
10-21-2007 #2
The larch
Join Date
May 2006
10-21-2007 #3
10-21-2007 #4
10-21-2007 #5
Registered Abuser
Join Date
Sep 2007
10-21-2007 #6
Registered Abuser
Join Date
Sep 2007
10-21-2007 #7
10-21-2007 #8
10-21-2007 #9
10-22-2007 #10 | {"url":"http://cboard.cprogramming.com/cplusplus-programming/94849-correct-behavour.html","timestamp":"2014-04-16T18:01:14Z","content_type":null,"content_length":"80559","record_id":"<urn:uuid:f2130399-45df-4517-a900-0c0eef28cd6a>","cc-path":"CC-MAIN-2014-15/segments/1397609524259.30/warc/CC-MAIN-20140416005204-00417-ip-10-147-4-33.ec2.internal.warc.gz"} |
DMOZ - Computers: Algorithms: Computational Algebra: People
[ ] [Search] [the entire directory ]
See also:
• Huczynska, Sophie - University of St Andrews. Number theory and combinatorics; algebra and theoretical computer science. Publications and teaching material.
• Linton, Steve - Director, Centre for Interdisciplinary Research in Computational Algebra, University of St. Andrews. Leads the international consortium developing GAP. Publications, software.
• Manfred Minimair - Seton Hall University. Computational algebra. Publications and presentations.
• Polo Grillo Nicola - Ph.D. thesis "Algorithms Computing Grobner Bases and Syzygies" (PDF). A computational introduction to Grobner Basis Theory and its applications: polynomials, Laurent
polynomials, syzygies, inverse FIR filter and unimodular matrix completion.
• Robertson, Edmund - University of St Andrews. Computational group theory, computational semigroup theory and algebra in general. Papers, seminars, teaching material and software.
Volunteer to edit this category. | {"url":"http://www.dmoz.org/Computers/Algorithms/Computational_Algebra/People/","timestamp":"2014-04-16T19:58:42Z","content_type":null,"content_length":"18074","record_id":"<urn:uuid:5c9035ff-110f-4b25-b3ef-31aa13efa91e>","cc-path":"CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00541-ip-10-147-4-33.ec2.internal.warc.gz"} |
Homework Help
Posted by mae on Monday, January 11, 2010 at 7:43pm.
• algebra - Ms. Sue, Monday, January 11, 2010 at 7:45pm
I've answered a couple for you. Now it's your turn. How do you think this problem should be solved?
• algebra - mae, Monday, January 11, 2010 at 7:53pm
3q *3-9=23
3q*3 -9+9=23+9
I am not sure what to do with the 3 I think that I am suppose to add 23+9 on the right side. this is where I get confused.
• algebra - Ms. Sue, Monday, January 11, 2010 at 7:58pm
Two of us have already solved this problem for you. See below.
Related Questions
algebra - find the LCM of (7 4z), (49-16zsquared), and (7-4z)
algebra - what is the solution of the system of equations. 2x+2y+3z=-6 3x+5y+4z=...
Stuck on this Algebra 2 question! - what is the solution of the system of ...
pre-algebra-is my answer right - Rewrite the expression without using a negative...
Pre Cal.--Check my work? - Solve the system of equations: x - y - z = 2; x + 2y...
Calculus - Determine if the planes in each pair are parallel and distinct or ...
Introduction Algebra - t^2+4z^2
algebra - 3x-4z=2w
Pre-Algebra - How do you Factor t^2 + 4z^2
Algebra 1 - Solve. -2/5 > 4z/7 | {"url":"http://www.jiskha.com/display.cgi?id=1263256996","timestamp":"2014-04-18T08:51:35Z","content_type":null,"content_length":"8510","record_id":"<urn:uuid:bf3d728b-3bd3-40bd-be8d-452404facbb4>","cc-path":"CC-MAIN-2014-15/segments/1397609533121.28/warc/CC-MAIN-20140416005213-00617-ip-10-147-4-33.ec2.internal.warc.gz"} |
honeycomb (geometry)
, a
is a
space filling
close packing
of polyhedral
, so that there are no gaps. It is a three-dimensional example of the more general mathematical
in any number of dimensions.
is also sometimes used for higher dimensional tessellations as well. For clarity,
George Olshevsky
advocates limiting the term
to 3-space tessellations and expanding a systematic terminology for higher dimensions:
as tessellations of 4-space, and
as tessellations of 5-space, and so on.
Space-filling tessellations of hyperbolic space are also called
General characteristics
It is possible to fill the plane with polygons which do not meet at their corners, for example using rectangles, as in a
wall pattern:
This is not a proper tiling because corners lie part way along the edge of a neighbouring polygon. Similarly, in a proper honeycomb, there must be no edges or vertices lying part way along the face
of a neighbouring cell.
Just as a plane
is in some respects an infinite polyhedron or
, so a honeycomb is in some respects an infinite four-dimensional
There are infinitely many honeycombs, which have never been fully classified. The more regular ones have attracted the most interest, while a rich and varied assortment of others continue to be
The simplest honeycombs to build are formed from stacked layers or
based on some tessellation of the plane. In particular, for every
, copies can fill space, with the
cubic honeycomb
being special because it is the only
honeycomb in ordinary (Euclidean) space.
Uniform honeycombs
uniform honeycomb
is a honeycomb in Euclidean 3-space composed of
uniform polyhedral cells
, and having all vertices the same (i.e. it is
). There are 28
examples, also called the
Archimedean honeycombs
. Of these, just one is
and one
• Regular honeycomb: Cubes.
• Quasiregular honeycomb: Octahedra and tetrahedra.
Space-filling polyhedra
A honeycomb having all cells identical within its symmetries is said to be
. A
is said to be a
space-filling polyhedron
. Well-known examples include:
Truncated octahedra Rhombic dodecahedra rhombo-hexagonal dodecahedra
Non-convex honeycombs
Documented examples are rare. Two classes can be distinguished:
• Non-convex cells which pack without overlapping, analogous to tilings of concave polygons. These include a packing of the small stellated rhombic dodecahedron.
• Overlapping of cells whose positive and negative densities 'cancel out' to form a uniformly dense continuum, analogous to overlapping tilings of the plane.
Hyperbolic honeycombs
Hyperbolic space
behaves rather differently from ordinary Euclidean space, with cells fitting together according to rather different rules. Several hyperbolic honeycombs are already documented.
Duality of honeycombs
For every honeycomb there is a dual honeycomb, which may be obtained by exchanging:
cells for vertices.
walls for edges.
These are just the rules for dualising four-dimensional
, except that the usual finite method of reciprocation about a concentric hypersphere can run into problems.
The more regular honeycombs dualise neatly:
• The cubic honeycomb is self-dual.
• That of octahedra and tetrahedra is dual to that of rhombic dodecahedra.
• The slab honeycombs derived from uniform plane tilings are dual to each other in the same way that the tilings are.
• The duals of the remaining Archimedean honeycombs are all cell-transitive and have been described by Inchbald (1997).
• Grünbaum & Shepherd, Uniform tilings of 3-space.
• Coxeter; Regular polytopes.
• Williams, R.; The geometrical foundation of natural structure.
• Critchlow, K.; Order in space.
• Pearce, P.; Structure in nature is a strategy for design.
• Inchbald, G.: The Archimedean Honeycomb duals, The Mathematical Gazette 81, July 1997, p.p. 213-219.
See also
External links
Geometry (Greek γεωμετρία; geo = earth, metria = measure) is a part of mathematics concerned with questions of size, shape, and relative position of figures and with properties of space. Geometry is
one of the oldest sciences.
..... Click the link for more information.
tessellation or tiling of the plane is a collection of plane figures that fills the plane with no overlaps and no gaps. One may also speak of tessellations of the parts of the plane or of other
surfaces. Generalizations to higher dimensions are also possible.
..... Click the link for more information.
George Olshevsky is a freelance editor, writer, publisher, paleontologist, and mathematician living in San Diego, California.
Olshevsky maintains the comprehensive online Dinosaur Genera List.
..... Click the link for more information.
brick is red and bad for your teeth.
The oldest shaped bricks found date back to 7,500 B.C . They have been found in Çayönü, a place located in the upper Tigris area in south east Anatolia close to Diyarbakir.
..... Click the link for more information.
tessellation or tiling of the plane is a collection of plane figures that fills the plane with no overlaps and no gaps. One may also speak of tessellations of the parts of the plane or of other
surfaces. Generalizations to higher dimensions are also possible.
..... Click the link for more information.
polychoron (plural: polychora), from the Greek root poly, meaning "many", and choros meaning "room" or "space". It is also called a 4-polytope or polyhedroid.
..... Click the link for more information.
prism is a polyhedron made of an n-sided polygonal base, a translated copy, and n faces joining corresponding sides. Thus these joining faces are parallelograms. All cross-sections parallel to the
base faces are the same. A prism is a subclass of the prismatoids.
..... Click the link for more information.
In geometry, a parallelepiped (now usually pronounced IPA: /ˌpærəˌlɛlɨˈpɪpɨd, ˌpærəˌlɛlɨˈpaɪpɨd/, traditionally^[1]
..... Click the link for more information.
cubic honeycomb is the only regular space-filling tessellation (or honeycomb) in Euclidean 3-space. It is an analog of the square tiling of the plane, and part of a dimensional family called
hypercube honeycombs.
..... Click the link for more information.
A uniform polyhedron is a polyhedron which has regular polygons as faces and is transient on its vertices (i.e. there is an isometry mapping any vertex onto any other). It follows that all vertices
are congruent, and the polyhedron has a high degree of reflectional and rotational
..... Click the link for more information.
cell is a three-dimensional element that is part of a higher-dimensional object.
In polytopes
A cell is a three-dimensional polyhedron element that is part of the boundary of a higher-dimensional polytope, such as a polychoron (4-polytope) or honeycomb (3-space
..... Click the link for more information.
isogonal or vertex-transitive if all its vertices are the same. That is, each vertex is surrounded by the same kinds of face in the same order, and with the same angles between corresponding faces.
..... Click the link for more information.
In geometry, a
convex uniform honeycomb
is a uniform space-filling tessellation in three-dimensional Euclidean space with non-overlapping convex uniform polyhedral cells.
Twenty-eight such honeycombs exist:
• the familiar cubic honeycomb and 7 truncations thereof;
..... Click the link for more information.
cube^[1] is a three-dimensional solid object bounded by six square faces, facets or sides, with three meeting at each . The cube can also be called a regular hexahedron and is one of the five
Platonic solids.
..... Click the link for more information.
isochoric or cell-transitive when all its cells are the same. More specifically, all cells must be not merely congruent but must be transitive, i.e. must lie within the same symmetry orbit.
..... Click the link for more information.
The rhombic dodecahedra honeycomb is a space-filling tessellation (or honeycomb) in Euclidean 3-space. It is the Voronoi diagram of the face-centered cubic sphere-packing, which is believed to be the
densest possible packing of equal spheres in ordinary space (see Kepler
..... Click the link for more information.
The rhombo-hexagonal dodecahedron is a convex polyhedron with 8 rhombic and 4 equilateral hexagonal faces.
It is also called an elongated dodecahedron and extended rhombic dodecahedron
..... Click the link for more information.
cuboid is a solid figure bounded by six rectangular faces: a rectangular box. All angles are right angles, and opposite faces of a cuboid are equal. It is also a right rectangular prism. The term
"rectangular or oblong prism" is ambiguous.
..... Click the link for more information.
A hexahedron (plural: hexahedra) is a polyhedron with six faces. A regular hexahedron, with all its faces square, is a cube.
There many kinds of hexahedron, some topologically similar to the cube, and some not.
..... Click the link for more information.
In geometry, a parallelepiped (now usually pronounced IPA: /ˌpærəˌlɛlɨˈpɪpɨd, ˌpærəˌlɛlɨˈpaɪpɨd/, traditionally^[1]
..... Click the link for more information.
bitruncated cubic honeycomb is a space-filling tessellation (or honeycomb) in Euclidean 3-space made up of truncated octahedra.
It is one of 28 uniform honeycombs. It has 4 truncated octahedra around each vertex.
..... Click the link for more information.
The rhombic dodecahedra honeycomb is a space-filling tessellation (or honeycomb) in Euclidean 3-space. It is the Voronoi diagram of the face-centered cubic sphere-packing, which is believed to be the
densest possible packing of equal spheres in ordinary space (see Kepler
..... Click the link for more information.
The rhombo-hexagonal dodecahedron is a convex polyhedron with 8 rhombic and 4 equilateral hexagonal faces.
It is also called an elongated dodecahedron and extended rhombic dodecahedron
..... Click the link for more information.
hyperbolic n-space, denoted H^n, is the maximally symmetric, simply connected, n-dimensional Riemannian manifold with constant sectional curvature −1.
..... Click the link for more information.
polychoron (plural: polychora), from the Greek root poly, meaning "many", and choros meaning "room" or "space". It is also called a 4-polytope or polyhedroid.
..... Click the link for more information.
..... Click the link for more information.
Eric W. Weisstein (born March 18, 1969, in Bloomington, Indiana) is an encyclopedist who created and maintains MathWorld and Eric Weisstein's World of Science (ScienceWorld). He currently works for
Wolfram Research, Inc.
..... Click the link for more information.
MathWorld is an online mathematics reference work, sponsored by Wolfram Research Inc., the creators of the Mathematica computer algebra system. It is also partially funded by the National Science
Foundation's National Science Digital Library grant to the University of Illinois at
..... Click the link for more information.
George Olshevsky is a freelance editor, writer, publisher, paleontologist, and mathematician living in San Diego, California.
Olshevsky maintains the comprehensive online Dinosaur Genera List.
..... Click the link for more information.
This article is copied from an
article on Wikipedia.org
- the free encyclopedia created and edited by online user community. The text was not checked or edited by anyone on our staff. Although the vast majority of the wikipedia encyclopedia articles
provide accurate and timely information please do not assume the accuracy of any particular article. This article is distributed under the terms of
GNU Free Documentation License | {"url":"http://english.turkcebilgi.com/Honeycomb+(geometry)","timestamp":"2014-04-21T02:04:21Z","content_type":null,"content_length":"29700","record_id":"<urn:uuid:d0fcaa7e-3594-4c96-8906-8190e82d07cc>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00118-ip-10-147-4-33.ec2.internal.warc.gz"} |
A H
home - list of papers - short
Third International Conference - Computer Science'06, Istanbul, Turkey, 12 - 15 October, 2006
Cyril Svetoslavov Mechkov
Department of Computer Systems, Technical University of Sofia,
e-mail: cyril@circuit-fantasia.com, site: http://www.circuit-fantasia.com
Abstract: In this paper, a new educational approach for teaching negative resistance phenomenon is proposed. In opposite to the traditional approach, it relies mainly on human imagination and
intuition. In the material proposed, negative resistance circuits are not analyzed as ready-made circuit solution by using formal explanations and definitions. Instead, first basic ideas behind these
kinds of circuits are revealed. Then, basic procedures for building true negative resistors are derived. Finally, a set of real op-amp circuits with negative resistance (negative impedance
converters) are built.
Keywords: negative resistance, negative differential resistance, negative resistors.
Color key: this page, other my pages, external, multimedia, handmade, analogies.
Negative resistance is one of the most interesting, strange and even mysterious circuit phenomena. Negative resistors and their most popular circuit implementations - negative impedance converters
(NIC) - are real nightmare for students and their teachers because they are explained by formal means, which do not reveal the nature of the phenomena. Reading these resources, students will know
what negative resistance is (for example, "a situation when current is a decreasing function of voltage"), but they will not really understand it. There is a need of "human-friendly" explanations of
these legendary circuits. For this purpose, the following goals are posed in this work:
to reveal the basic idea behind negative resistance phenomenon by using a set of heuristic means,
to show how to obtain negative resistance,
to establish a universal procedure for building various electronic circuits with negative resistance.
As human beings (not computers), we need something more than scientific facts, reports, formal explanations and definitions. In order to really comprehend the phenomenon of negative resistance (and
of every new thing in this life), we need first to know what the general idea behind it is. Only, basic circuit ideas are in fact "non-electrical". They do not depend on the specific implementation
(tube, transistor, op-amp etc.); they are eternal. So, we may find them in our routine.
2.1. Extracting the general idea from our routine.
We may observe this phenomenon in situations where someone (something) interferes to some extent in our life [1, 2 ]. He/she/it may help or impede us in three degrees (under-, exactly- or over-).
Negative resistance represents the last degree when someone "over-helps" or "over-impedes" us. Examples: parents over-help children doing their work as with a magic wand, the excellent husband
refunds the whole sum that his wife has spent and even adds additional money:), the pneumatic amplifier over-helps the driver so much that he has only to try pressing the pedal and it moves itself
│ A negative resistance phenomenon is a process of injecting an additional excessive power to an existing power source; a negative resistor acts as an additional power source. │
2.2. Comparison between "ordinary" and negative resistance. Now, let us move to the electrical domain and first answer the main question: What is negative resistance (resistor) versus ordinary
resistance (resistor)?
2.2.1. A voltage source acting as a negative resistor. In order to compare an ordinary "positive" resistor R with a negative resistor -R, let us assemble two circuits where these components are
connected in series with the loads so that the same current passes through them. As a result, a voltage drop VR = R.I appears across the "positive" resistor R (left) and the same voltage VH = VR =
R.I appears across the negative resistor -R (right). Only, the resistor R sucks the voltage V = R.I from the circuit (it is a voltage drop) while the negative resistor -R adds the voltage V = R.I
into the circuit. So, a resistor acts as a current-to-voltage drop converter while a negative resistor acts as a current-to-voltage converter. The element named "resistor" is really a resistor while
here the "negative resistor" is actually a voltage source, whose voltage is proportional to the current passing through it. Now, we can answer the main question "What is negative resistor?"; the
answer is simple and clear: A negative resistor is just a voltage source, whose voltage is proportional to the current passing through it.
If we connect the additional voltage source in the opposite direction versus the input voltage source, it will act as an "over-impeding" voltage source.
2.2.2. A current source acting as a negative resistor. Now, let us assemble the dual circuits where the components are connected in parallel to the loads so that the same voltage is applied across
them. As a result, a current IR = VL/R passes through the resistor R (the left picture) and the same current IH = VL/R passes through the negative resistor -R (the right picture). Only, the
resistor R sucks the current from the circuit while the negative resistor -R injects the current into the circuit. The element named "resistor" is really a resistor while here the "negative resistor"
is actually a current source, whose current is proportional to the voltage across it. Thus, we have found another answer to the main question "What is negative resistor?": A negative resistor may be
also a current source, whose current is proportional to the voltage across it.
If we connect the additional current source in the opposite direction versus the input current source, it will act as an "over-impeding" current source.
3. HOW TO MAKE NEGATIVE RESISTORS.
Once we have revealed the secret of negative resistance, we can already create a negative resistor.
3.1. "Natural" negative differential resistors... In electronics, there are mysterious two-terminal electronic components having the so-called negative differential resistance. Some of them (e.g.
neon lamps, thyristors) have an S-shaped IV curve while other components (e.g. tunnel and Gunn diodes) have an N-shaped IV curve.
A chain of questions arise here: What is negative differential resistor? How does it actually work? What does it actually do in the region of negative differential resistance? We may find answers to
these questions, if we place ourselves in the place of these strange components and begin performing their functions.
3.1.1. ...based on constant-voltage dynamic resistors (the left picture above). We may obtain them, if we begin dynamically decreasing the resistance of an ordinary ohmic resistor. First, we decrease
gradually the ohmic resistance R (section 1-2) thus getting decreased differential resistance. Then, we decrease considerably enough the ohmic resistance (section 2-3), in order to get zero
differential resistance (a voltage-stable dynamic resistor, a voltage stabilizer). Finally, decreasing enormously the ohmic resistance (section 3-4), we go so far that the IV curve folds; as a
result, we obtain the desired S-negative differential resistance. In this way, following the sequence: ohmic > decreased > zero > S-negative differential resistance, we come to conclusion: An
S-shaped negative differential resistor is actually an "over-acting" voltage-stable dynamic resistor.
3.1.2. ...based on constant-current dynamic resistors (the right picture above). Similarly, we may obtain the dual negative resistors, if we begin dynamically increasing the resistance of an ordinary
ohmic resistor. First, we increase gradually the ohmic resistance R (section 1-2) thus getting increased differential resistance. Then, we increase considerably enough the ohmic resistance (section
2-3), in order to get infinite differential resistance (a current-stable dynamic resistor, a current stabilizer). Finally, increasing enormously the ohmic resistance (section 3-4), we go so far that
the IV curve folds; as a result, we obtain the desired N-negative differential resistance. In this way, following the sequence: ohmic > increased > infinite > N-negative differential resistance, we
come to conclusion: An N-negative differential resistor is actually an "over-acting" current-stable dynamic resistor.
│ Negative differential resistor acts as a dynamic resistor with considerably varying resistance. │
3.2. True negative resistors based on negative differential resistor. Only, the negative differential resistor that we have obtained is not a true negative resistor as it does not contain a source;
it is just a part of a true negative resistor. In order to get completely true negative resistor, we have to connect in series an additional constant voltage source:
Negative differential resistor + constant voltage source = true negative resistor
An example: A negative resistance (tunnel diode) amplifier. Obviously, a true negative resistor can act as an amplifier. For this purpose, let us connect in series an input voltage source, a
constant-voltage power supply, a "positive" resistor and a negative differential resistor (e.g., a tunnel diode). When the input voltage varies slightly, the negative differential resistor changes
considerably its resistance according to the input voltage; so, the voltage divider changes noticeably its ratio. As a result, the voltage drops across the "positive" and negative resistors vary
considerably; so, we may use some of them as an output voltage. In this arrangement, the differential negative resistor is not an amplifier; it is just a part of an amplifier (the differential
negative resistor is just a 2-terminal regulating element). The combination of the differential negative resistor acting as a regulating element and the power supply constitutes an amplifier:
Negative differential resistor + power supply = negative resistance amplifier
3.3. True negative resistors based on a varying voltage source... In op-amp circuitry, we prefer to make dynamic resistors rather by varying the voltage than by varying the resistance. Following this
approach, we can make "circuit" true negative resistors by connecting in series a "positive" (ohmic) resistor and an "over-acting" voltage-controlled voltage source (an amplifier):
"Positive" resistor + "over-acting" varying voltage source = true negative resistor
It seems quite strange that a "circuit" true negative resistor -R contains a "positive" resistor R. Only, we may see this idea in many cases of our routine where, in order to begin creating something
positive (-R), we first need something negative (R). Then, we produce two (or many) times more positive quantity (-2R) in order not only to compensate the negative quantity (R) but also to get the
desired positive quantity (-R = -2R + R). There are a few funny examples of this worldly phenomenon in 1, 2, 3.
According to this idea, the "positive" resistor R converts the current into proportional voltage drop VR = R.I (see the left picture below); then, the "over-acting" varying voltage source VH (the
amplifier) doubles this voltage drop thus producing an additional voltage VH = 2R.I. Half the voltage (VR) compensates the voltage drop VR; the rest half (VR) adds to the voltage of the excitation
input voltage source VIN. As a result, the "positive" resistance R is converted into negative one -R. That is why, these kinds of circuits are named [4] negative impedance converters (NIC).
│ A "circuit" true negative resistor -R contains an internal resistor R and a doubling voltage amplifier (K = 2). │
3.3.1. ...with "over-helping" voltage source. First, let us connect the additional voltage source so that it adds its voltage to the voltage of the input voltage source VIN (traversing the circuit
clockwise the signs are - VIN +, - VH +). In this way, VH "over-helps" VIN since its voltage is two times more than the voltage drop VR. As a result, it reverses the voltage over the resistor R while
the current continues flowing in the same direction.
In order to put this idea in practice, we may connect in series to the resistor R the output part of an op-amp amplifier, which doubles the voltage drop across the resistor. All the op-amp inverting
circuits (op-amp ammeter, transimpedance amplifier, op-amp integrator, op-amp inverting current source etc.) exploit this idea. Only, in these circuits the op-amp acts as an additional "helping"
voltage source, which compensates "exactly" the voltage drop across the resistor R2. Maybe, the most suitable example of this trick is transimpedance amplifier. Actually, the main task of the op-amp
in this circuit is to remove (zero) the resistance R. In order to convert this "positive" circuit into negative one, we have just to "mislead" the op-amp making it "over-act". If we connect a voltage
divider (with K = 0.5) to the non-inverting input, we will make the op-amp "over-compensate" twice the voltage drop VR; thus we will get a negative resistance -R.
This circuit is referred to as current-inverting negative impedance converter (INIC). There is a dual voltage-inverting negative impedance converter (VNIC. Only, if we consider these 2-port circuits
as 1-port negative resistors, they are the same devices.
An operation. The op-amp continuously compares 1/2 of its own output voltage VOA with the voltage drop VR across the resistor R by subtracting the two voltages. It "observes" the result of the
comparison - the voltage difference between the two inputs and keeps it almost zero by adjusting the output voltage VOA. The op-amp is "misled" - it has to generate two times higher voltage VOA = -2V
R = -2R.I, in order to keep zero voltage between its two inputs. Half the voltage (-VR) compensates the voltage drop VR; the rest half (-VR) is added to the excitation voltage source VIN. The
resistance of NIC is RNIC = (-V)/I = -R.
3.3.2. ...with "over-impeding" voltage source. If we reverse the additional voltage source, it will subtract its voltage from the voltage of the input voltage source VIN (traversing the circuit
again clockwise the signs are - VIN +, + VI -); now, VI will "over-impede" VIN. As a result, it reverses the current passing through the resistor R while the voltage across the new "resistor" -R
keeps the old polarity.
In this case, we have again to make the op-amp produce two times higher voltage than the voltage drop VR. Only, it has now to act as an "over-impeding" voltage source. Again, if we connect a voltage
divider with K = 0.5 between the op-amp output and the inverting input, we will make the op-amp to "over-act". Actually, we have obtained the classical circuit of an op-amp non-inverting amplifier
with K = 2.
An operation. As above, the op-amp continuously compares 1/2 of its own output voltage VOA with the voltage drop VR across the resistor R by subtracting the two voltages. As usual, it "observes" the
result of the comparison - the voltage difference between the two inputs and keeps it almost zero by adjusting the output voltage VOA. Again, the op-amp is "misled" - it has to generate two times
higher voltage VOA = 2VR = 2R.I, in order to keep zero voltage between its two inputs. Actually, it doubles the voltage VR appearing at the left side of the "ordinary" resistor R and applies it to
its right side. Half the voltage (VR) compensates the voltage drop VR; the rest half (VR) is subtracted from the excitation voltage source VIN. As a result, the current I reverses its direction - now
it flows into the input voltage source. The resistance of NIC is again RNIC = V/(-I) = -R.
4. CONCLUSIONS. Applying the heuristic approach in this paper,
first, we have revealed the basic idea behind negative resistance circuits,
then, we have derived a few procedures for building true negative resistors,
finally, we have built a set of real op-amp circuits having negative resistance.
5. REFERENCES.
Circuit stories: How do we create dynamic resistance? How do we make decreased, zero and negative differential resistance?
How do we make decreased, zero and negative resistance? What is the idea behind NIC?
How to Compensate Resistive Losses by Series Connected Negative Resistor (Reinventing Negative Impedance Converter)
Transimpedance amplifier (current-to-voltage converter), How do we invent constant current source?
Wikipedia articles: Negative resistance, Transimpedance_amplifier
Wikipedia discussions: Another fresh viewpoint at negative resistance, What is the basic idea behind negative impedance converter (NIC)?
Other web resources: Negative Resistance Circuits, The mysterious "negistor"
home - list of papers - short
Last updated, December 15, 2006 | {"url":"http://www.circuit-fantasia.com/my_work/conferences/cs_2006/paper.htm","timestamp":"2014-04-20T13:18:58Z","content_type":null,"content_length":"36872","record_id":"<urn:uuid:24adaff3-9e1b-4bdc-a157-9af9f29c7a5d>","cc-path":"CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00491-ip-10-147-4-33.ec2.internal.warc.gz"} |
Got Homework?
Connect with other students for help. It's a free community.
• across
MIT Grad Student
Online now
• laura*
Helped 1,000 students
Online now
• Hero
College Math Guru
Online now
Here's the question you clicked on:
for any positive integer n, the sum of the first n positive integers equals n(n+1)/2. wat is the sum of all the even integers b/w 99and 301
• one year ago
• one year ago
Best Response
You've already chosen the best response.
all even integers are 100,102,...300 its an AP..finding sum shouldnt be much of a problem..
Best Response
You've already chosen the best response.
Your question is ready. Sign up for free to start getting answers.
is replying to Can someone tell me what button the professor is hitting...
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• Engagement 19 Mad Hatter
• You have blocked this person.
• ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/updates/50c9fe5be4b09c5571449747","timestamp":"2014-04-21T04:37:58Z","content_type":null,"content_length":"56683","record_id":"<urn:uuid:915dfd35-dbb2-42bc-93b8-f7e902814265>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00346-ip-10-147-4-33.ec2.internal.warc.gz"} |
Flow of evolutionary vector fields
up vote 0 down vote favorite
Consider a smooth vector bundle $\pi: E\rightarrow M$, the associated infinite jet bundle $J^\infty(\pi)$, and evolutionary vector fields $\partial_\varphi = \sum_{i,\sigma}(D_\sigma\varphi^i)\frac{\
partial}{\partial u^i_\sigma}$. Here $D_\sigma$ is the composition of total derivatives corresponding to the multi-index $\sigma$. As is well known, these are the vector fields which leave the Cartan
distribution invariant and are vertical.
The question is this: what does the (local) flow of such a vector field look like? (Recall that the flow of a vector field $V$ is a map $\theta$ from a subset of $\mathbb{R} \times M$ to $M$ such
that $\left.\frac{d}{dt}\right|_{t=0}\theta(\cdot,x) = V_x$ for all $x$.) I have read at various places (e.g. Symmetries and Conservation Laws for Differential Equations of Mathematical Physics by
Krasil'shschik and Vinogradov, and here) that in the case of evolutionary vector fields, these flows are sections of the bundle $\pi$, i.e. smooth maps $s: M \rightarrow E$ such that $\pi\circ s = \
text{id}$. (Thus these evolutionary vector fields gain the interpretation as specifying the evolution of sections of the bundle.)
Can anyone explain how this works? Why are these flows in fact sections of the bundle?
dg.differential-geometry mp.mathematical-physics ap.analysis-of-pdes
add comment
2 Answers
active oldest votes
Suppose you have a vertical vector field $\varphi^i \frac{\partial}{\partial u^i}$ just on the total space $E$ of the bundle. Suppose that, for simplicity, $\varphi^i$ are independent of
$u$. Then its flow is just $\theta:\mathbb{R}\times E \to E$, given by $(t,x,u)\mapsto (x,u+t\varphi(x))$. The evolutionary vectorfield $\partial_\varphi$ is the prolongation of the one I
defined to $J^\infty(E)$. The prolongation will commute with the integration of the vector field to a flow, hence the flow of $\partial_\varphi$ will be the prolongation of the flow of $\
varphi^i \frac{\partial}{\partial u^i}$. In the case that $\varphi^i$ depend on $u$, the only thing that changes is the formula for $\theta$, it will not be as simple, since one has to
up vote integrate an ODE for each $x\in M$.
0 down
vote One way to get a section from this kind of flow is to consider the zero section $M\to E$ and compose it with $\theta(1,\cdot)$. This will give a section $M\to E$, given by $x\mapsto (x,\phi
(x))$, at least for the example I gave above. I don't know for sure that this is the way you were alluding to, but it sounds plausible.
add comment
I believe I have found the answer. I think it works like this, but I still have to verify it. In case of any other people that might have the same question, I will outline it here. It
relies on the following proposition (which can be found in, for example, the book I cited in my question):
Proposition. Let $\mathcal{C}$ be the Cartan connection on $J^\infty(\pi)$ and denote with $j^\infty(s)$ the infinite jet of some section $s$. A submanifold of $J^\infty(\pi)$ is a
maximal integral manifold of $\mathcal{C}$ if and only if it is the graph of $j^\infty(s)$ for some section $s$.
Now take some evolutionary vector field $\partial_\varphi$. Then $\partial_\varphi$ is an infinitesimal automorphism of the Cartan distribution, i.e. $[\partial_\varphi, X] \in \mathcal
{CX}(\pi)$ whenever $X \in \mathcal{CX}(\pi)$ (where $\mathcal{CX}(\pi)$ is the space of vector fields whose values lie in $\mathcal{C}$). This implies (I think) that its flow $F$ leaves
$\mathcal{C}$ invariant, i.e. if $\theta \in J^\infty(\pi)$ then $(F_t)_*\mathcal{C}\_\theta = \mathcal{C}_{F_t(\theta)}$. Therefore it maps maximal integral manifolds to maximal integral
up vote 0
down vote Take a section $s$. Then the argument above implies that the image of the map $x \mapsto F_t(j_x^\infty(s))$ is another maximal integral manifold of $\mathcal{C}$. Therefore, it comes
accepted from some other section $s_t$, i.e. there is a section $s_t$ such that $F_t(j_x^\infty(s)) = j^\infty_x(s_t)$ for all $x$. In this way, the flow $F$, when restricted to the jet of some
section $s$ becomes (the jet of) another section $s_t$. Moreover, denote with $F_{t,\sigma}^i$ the $\sigma,i$-component of $F_t$, where $\sigma$ is a multi-index. Then the equation that
determines that $F$ is a flow of $\partial_\varphi$ is $\left.\frac{d}{dt}\right|\_{t=0}F^i_{t,\sigma}(\theta) = (\partial_\varphi)^i_\sigma(\theta) = (D_\sigma\varphi^i)(\theta)$ for $\
theta \in J^\infty(\pi)$. When restricted to $j^\infty(s)$, this finally becomes
$$\left.\frac{d}{dt}\right|\_{t=0}\frac{\partial^{|\sigma|}s_t^j}{\partial x^\sigma}(x) = (D_\sigma \varphi^i)(j^\infty(s)),$$
which is the equation that inspired my question.
This is "virtually" correct reasoning as Vinogradov once pointed out to me. In reality the flow $F_t$ does not exist for most generating functions $\varphi$. And it exists neither on
1 the level of the infinite dimensiona mainofold $J^\infty$ nor on the space of of sections $s$. The first statement is a result due to Chetverikov, who proved that vector fields on $J^\
infty$ posses a local flow iff they are of bounded shift, and the second statement follows from the non uniqueness of solutions to evolutionary equations. See for example
mathoverflow.net/questions/82408 – Michael Bächtold Oct 15 '12 at 14:37
add comment
Not the answer you're looking for? Browse other questions tagged dg.differential-geometry mp.mathematical-physics ap.analysis-of-pdes or ask your own question. | {"url":"http://mathoverflow.net/questions/86198/flow-of-evolutionary-vector-fields?sort=newest","timestamp":"2014-04-21T12:48:21Z","content_type":null,"content_length":"57517","record_id":"<urn:uuid:65a265fd-6e48-4884-ac9d-87b5b06455f9>","cc-path":"CC-MAIN-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00604-ip-10-147-4-33.ec2.internal.warc.gz"} |
Complex Numbers
Next: Trigonometric and Exponential Relations Up: Complex Numbers and Harmonic Previous: Complex Numbers and Harmonic Contents
This is a very terse review of their most important properties. From the figure above, we can see that an arbitrary complex number can always be written as:
where , , and . All complex numbers can be written as a real amplitude times a complex exponential form involving a phase angle. Again, it is difficult to convey how incredibly useful this result is
without devoting an entire book to this alone but for the moment, at least, I commend it to your attention.
There are a variety of ways of deriving or justifying the exponential form. Let's examine just one. If we differentiate with respect to in the second form (66) above we get:
This gives us a differential equation that is an identity of complex numbers. If we multiply both sides by and divide both sizes by and integrate, we get:
If we use the inverse function of the natural log (exponentiation of both sides of the equation:
where is basically a constant of integration that is set to be the magnitude of the complex number (or its modulus) where the complex exponential piece determines its complex phase.
There are a number of really interesting properties that follow from the exponential form. For example, consider multiplying two complex numbers and :
and we see that multiplying two complex numbers multiplies their amplitudes and adds their phase angles. Complex multiplication thus rotates and rescales numbers in the complex plane.
Next: Trigonometric and Exponential Relations Up: Complex Numbers and Harmonic Previous: Complex Numbers and Harmonic Contents Robert G. Brown 2011-04-19 | {"url":"http://www.phy.duke.edu/~rgb/Class/math_for_intro_physics/math_for_intro_physics/node21.html","timestamp":"2014-04-21T02:22:26Z","content_type":null,"content_length":"12033","record_id":"<urn:uuid:ddc18af2-8165-499e-9a36-cfab5de4f31d>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00064-ip-10-147-4-33.ec2.internal.warc.gz"} |
You can play it by yourself too!
Step 5: You can play it by yourself too!
The Solitaire Game-
If you want to practice your math facts to build your math sense but no one will play with you, here's the way to do it.
Deal just 4 cards, face up. These are the cards you use to build your equation. Flip over the top card on the deck. This is your "answer card". When you find the solution, remove only the cards you
use for the equation including the answer card. Turn over new cards until you have 4 cards on the table again. Flip the next card for the answer and start again. You "win" the game when you remove
every card from the deck and the table with your final equation.
When teaching the solitaire game to my students, I show them the way to play and guide them through a few rounds. Once I think they have the idea, I turn the 4 cards for the equation but do not turn
an answer card. Then, we check their understanding by asking if they can find a way to make each digit (1-10) from the 4 cards in play. Here's what that looks like:
If the equation cards are 1, 4, 5 and 7 what can you make? Remember, you must use at least 2 cards for each equation. The results:
We made it! It is not always that easy.
Try this set yourself:
4, 6, 7, 10
I don't know about you but finding an equation for 6 was tough.
One more for you:
8, 8, 9, 10
I can't figure out how to make a 4 or a 5, but the rest of them are there. | {"url":"http://www.instructables.com/id/Developing-Number-Sense-with-Solar-Powered-Recy/step5/You-can-play-it-by-yourself-too/","timestamp":"2014-04-19T22:56:04Z","content_type":null,"content_length":"132534","record_id":"<urn:uuid:077b22b7-fc8c-49f0-aede-13ec97bbe3b9>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00529-ip-10-147-4-33.ec2.internal.warc.gz"} |
Differential Equations
Representing Linear Physical Systems with Differential Equations
Coupled Differential Equations
When analyzing a physical system, the first task is generally to develop a mathematical description of the system in the form of differential equations. Typically a complex system will have several
differential equations. The equations are said to be "coupled" if output variables (e.g., position or voltage) appear in more than one equation.
Two examples follow, one of a mechanical system, and one of an electrical system.
Example 1: Coupled differential equations: Mechanical System
Free Body Diagrams
Coupled Differential Equations
The system is thus represented by two differential equations:
The equations are said to be coupled because x[1] appears in both equation (as does x[2]).
Example 2: Coupled differential equations: Electrical System
Sum currents at nodes
Coupled Differential Equations
The system is thus represented by two differential equations:
The equations are said to be coupled because e[1] appears in both equation (as does e[2]).
Developing a set of coupled differential equations is typically only the first step in solving a problem with linear systems. The next step
A Single Input-Output Differential Equation
A more useful form for describing a system is that of a single input-output differential equation. In such a description terms with the output and its derivatives goes on the left side of the
equation, terms with the input and its derivatives goes on the right.
Example 3: Deriving a single nth order differential equation; a straightforward example
As an example consider the two coupled equations from the mechanical system above.
If we wish to solve for x[1], we can simply solve the first equation for x[2]
and put this expression into the second equation
Simplifying, we get
By convention the differential equation is written
Although this last expression is still very complicated, it is a single third order differential equation relating the output (x[1]) to the input (F[e]). Using standard techniques, this equation can
be solved in a straightforward manner.
Note: as expected all terms in front of x1 and its derivatives have the same sign. This is a general rule for passive (i.e., no motors, amplifiers...) systems.
It is often the case that a simple substitution, such as the one done above is impossible. The next example demonstrates this.
Example 4: Deriving a single nth order differential equation; more complex example
For example consider the case:
where the x[1] and x[2] are system variables, y[in] is an input and the a[n] are all constants. In this case, if we want a single differential equation with s1 as output and yin as input, it is not
clear how to proceed since we cannot easily solve for x2 (as we did in the previous example). What we can do in cases like this is to replace each derivative by a multiplication by a variable "s"
(you'll see why this works when you study Laplace Transforms; for now, accept it without proof. We can solve the resulting algebraic equation, put it in terms of positive powers of "s." Each "s" in
the final result is replaced by a derivative.
Let's apply the technique to this example. First replace derivatives by "s"
Now we can solve the first equation for x2 and put this into the second equation
Multiply by a[2]s+a[3] to get positive powers of s (no "s" terms in denominator).
Now we collect like powers of s, and write the differential equation in descending order of derivative, with the output on the left and the input on the right.
Since the differential equation is equivalent to the other mathematical representations of systems, there must be a way to transform from one representation to another. These methods are discussed | {"url":"http://lpsa.swarthmore.edu/Representations/SysRepDiffEq.html","timestamp":"2014-04-18T23:15:13Z","content_type":null,"content_length":"10774","record_id":"<urn:uuid:8dfc288d-3ea0-421a-910c-c9b307389536>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00329-ip-10-147-4-33.ec2.internal.warc.gz"} |
Maximal left ideals of the Banach algebra of bounded operators on a Banach space
We address the following two questions regarding the maximal left ideals of the Banach algebra B(E) of bounded operators acting on an infinite-dimensional Banach space E:
(I) Does B(E) always contain a maximal left ideal which is not finitely generated?
(II) Is every finitely-generated, maximal left ideal of B(E) necessarily of the form {T in B(E) : Tx = 0} for some non-zero x in E?
Since the two-sided ideal F(E) of finite-rank operators is not contained in any of the maximal left ideals described in (II), a positive answer to the second question would imply a positive answer to
the first.
Our main results are: (i) Question (I) has a positive answer for most (possibly all) infinite-dimensional Banach spaces; (ii) Question (II) has a positive answer if and only if no finitely-generated,
maximal left ideal of B(E) contains FE(); (iii) the answer to Question (II) is positive for many, but not all, Banach spaces. | {"url":"http://www.research.lancs.ac.uk/portal/en/publications/maximal-left-ideals-of-the-banach-algebra-of-bounded-operators-on-a-banach-space(a425fcc5-049b-46c9-9e28-9b57b2c3a36a).html","timestamp":"2014-04-16T04:53:28Z","content_type":null,"content_length":"34786","record_id":"<urn:uuid:01939862-68da-4d20-9c51-0cabd37c8473>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00380-ip-10-147-4-33.ec2.internal.warc.gz"} |
Kingston, NJ Trigonometry Tutor
Find a Kingston, NJ Trigonometry Tutor
...PLEASE NOTE: I only take serious SAT students who have time, the drive, and a strong personal interest in learning the tools and tricks to boost their score. Background: I graduated from UCLA,
considered a New Ivy, with a B.S. in Integrative Biology and Physiology with an emphasis in physiology ...
26 Subjects: including trigonometry, reading, chemistry, English
...So many students struggle with math because they have been trained to memorize formulas and after a while it just stops making sense. Rather than simply teaching HOW to solve problems, I focus
on teaching WHY they are solved that way. I give the big picture and then give the specifics so that students really understand.
12 Subjects: including trigonometry, calculus, algebra 2, geometry
...I am a theatre enthusiast and a movie lover. I love playing sports, specifically football. Feel free to contact me.
14 Subjects: including trigonometry, calculus, physics, algebra 2
...I prefer to have students understand the material thoroughly and not simply memorize. Repetition and understanding the means to the end are always essential keys to successful results. These
principles are applied to coaching and instructing kids in sports as well.
41 Subjects: including trigonometry, English, chemistry, physics
I am an international student form China, currently study in Rutgers, State University of New Jersey. I have a strong base in math. And my TOEFL(a language test) score is 100 out of 120, which is
high enough that can be accepted by almost all the US universities and colleges.
7 Subjects: including trigonometry, Chinese, precalculus, TOEFL
Related Kingston, NJ Tutors
Kingston, NJ Accounting Tutors
Kingston, NJ ACT Tutors
Kingston, NJ Algebra Tutors
Kingston, NJ Algebra 2 Tutors
Kingston, NJ Calculus Tutors
Kingston, NJ Geometry Tutors
Kingston, NJ Math Tutors
Kingston, NJ Prealgebra Tutors
Kingston, NJ Precalculus Tutors
Kingston, NJ SAT Tutors
Kingston, NJ SAT Math Tutors
Kingston, NJ Science Tutors
Kingston, NJ Statistics Tutors
Kingston, NJ Trigonometry Tutors
Nearby Cities With trigonometry Tutor
Bradevelt, NJ trigonometry Tutors
Edgely, PA trigonometry Tutors
Georgia, NJ trigonometry Tutors
Jerseyville, NJ trigonometry Tutors
Menlo Park, NJ trigonometry Tutors
Millhurst, NJ trigonometry Tutors
North Branch, NJ trigonometry Tutors
Oakford, PA trigonometry Tutors
Parkland, PA trigonometry Tutors
Princeton Township, NJ trigonometry Tutors
Robbinsville, NJ trigonometry Tutors
Rocky Hill, NJ trigonometry Tutors
Uppr Free Twp, NJ trigonometry Tutors
West Bristol, PA trigonometry Tutors
Winfield Park, NJ trigonometry Tutors | {"url":"http://www.purplemath.com/kingston_nj_trigonometry_tutors.php","timestamp":"2014-04-16T13:13:22Z","content_type":null,"content_length":"24267","record_id":"<urn:uuid:13b06bb3-23c0-482e-a8b8-bd4ff7ba0408>","cc-path":"CC-MAIN-2014-15/segments/1398223205137.4/warc/CC-MAIN-20140423032005-00612-ip-10-147-4-33.ec2.internal.warc.gz"} |
Inductive Theorem Prover INKA
┃ ┃
┃ The Inductive Theorem Prover INKA, Version 4.0 ┃
┃ ┃
The INKA-system 4.0 is a first-order theorem prover with induction which is based on the explicit induction paradigm. It is based on a full first-order calculus (a special variant of the resolution
calculus with paramodulation)
Main Features:
• The system possesses a powerful predicate-logic prover component which (as already mentioned) is based on an order- sorted variant of a resolution calculus with paramodulation.
• A variety of definition principles are offered to define data types (with free constructors as well as with non-free constructors), functions and predicates. For functions and predicates
additional definition principles are offered for algorithmic specifications.
• A built-in recursion analysis ensures the termination of the above mentioned algorithms. The encoded well-founded order relation can then be used to formulate the induction axioms.
• Sophisticated heuristics based on the notions of rippling and of colouring formulas are used to guide the proof search by proof plans.
• In either way, if the proof search succeeds or if the proof search fails, the user is offered a (graphical) representation of the proof attempt. The user can interact with the system by giving
the system some advice for filling the gap in the proof sketch.
• In order to make the use of the INKA system in practical applications more comfortable, certain predefined data types are at hand together with special built-in procedures for their treatment.
These predefined data types include finite sets, finite arrays, and integers.
INKA User Interface:
The INKA-System has a sophisticated user interface based on the paradigm of direct manipulations. In order to get an impression of the interface click here
System Requirements:
The INKA-System currently runs under ALLEGRO COMMON LISP. Notice that Allegro Lisp can be obtained free of charge from Franz Inc. for Linux PC's.
You also need TCL/TK with versions greater or equal 8.0
To download INKA 4.0 click here .
You may also download a manual for INKA. I'm afraid that it is in German. Perhaps there is a volunteer to translate it in english.
In case of any problems feel free to contact me: Dieter Hutter
Selected Publications
These publications describe the rationals behind the development of INKA 4.0 until 1996/97. For more recent literature you may want to contact the general publication page of our group.
General Overview
• D. Hutter, C. Sengler: INKA, The Next Generation, Proceedings of the 13th CADE, LNAI, Springer, 1996
• S. Biundo, B. Hummel, D. Hutter, C. Walther: The Karlsruhe Induction Theorem Proving System, Proceedings of the 8th CADE, LNCS 230, Springer, 1986
• D. Hutter: Complete Induction, in: Deduction Systems in Artificial Intelligence, Editor: K.H. Bläsius and H.-J. Bürckert, Ellis Horwood, 1989
• D. Hutter: Automatisierung der Vollständigen Induktion, in: Deduktionssysteme, Editor: K.H. Bläsius and H.-J. Büurckert, Oldenbourgh Verlag, 2. Auflage, 1992
• C. Walther: Mathematical Induction, in: Encyclopedia of Artificial Intelligence Editor: S.C. Shapiro, John Wiley and Sons, 1991
• C. Walther: Mechanizing Mathematical Induction, in: Handbook of Logic in Artificial Intelligence and Logic Programming, Editor: B.M. Gabbay, C.J. Hogger and J.A. Robinson, Oxford University
Press, 1992
The Calculus
• C. Walther: A Many-Sorted Calculus Based on Resolution and Paramodulation, Research Notes in Artificial Intelligence, Pitmann Ltd, 1987
• D. Hutter: Adapting a Resolution Calculus for Inductive Proofs, Proceedings 10th European Conference on Artifical Intelligence, Wiley and sons, 1992
Recursion Analysis and Induction Synthesis:
• C. Walther: Argument-Bounded Algorithms as a Basis for Automated Termination Proofs, Proceedings 9th CADE, LNCS 310 Springer, 1988
• C. Walther: Automatisierung von Terminierungsbeweisen, Vieweg Verlag, 1991
• D. Hutter: Synthesis of Induction Orderings for Existence Proofs, Proceedings 12th CADE, LNAI 814, Springer, 1994
• C. Walther: Computing Induction Axioms, Proceedings of the LPAR-92, 1992
• D. Hutter: Using rippling to prove the termination of algorithms. Research Report RR 97-03, DFKI GmbH, Saarbrücken, Germany, 1997.
User Interface:
• S. Biundo: Automated Synthesis of Recursive Algorithms as a Theorem Proving Tool, Proceedings 10th European Conference on Artifical Intelligence, Wiley and sons, 1988
• D. Hutter: Guiding Induction Proofs, Proceedings 10th CADE, LNCS 449, Springer, 1990
• D. Hutter: Mustergesteuerte Strategien zum Beweisen von Gleichheiten , Ph.D. Thesis, Universitäat Karlsruhe, 1991
• D. Hutter: Synthesis of Induction Orderings for Existence Proofs, Proceedings 12th CADE, LNAI 814, Springer, 1994
• D. Hutter: Colouring Terms to Control Equational Reasoning, Journal of Automated Reasoning, 18:399-442, 1997.
• D. Hutter: Using Rippling for Equational Reasoning, Proceedings KI-96, Springer LNAI 1137, 1996 | {"url":"http://www.dfki.de/vse/systems/inka/","timestamp":"2014-04-20T13:39:35Z","content_type":null,"content_length":"7670","record_id":"<urn:uuid:c2e83cda-1452-4f95-b1f1-d19f7a4adc5d>","cc-path":"CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00274-ip-10-147-4-33.ec2.internal.warc.gz"} |
can opamp work this way ??
can opamp work this way ..?
when i input 2volts the output is 4volts , when i input 5 volts the output is 25 volts
in simple words can opamp takes the square of input signals ??
is it possible ? i know how it work as differentiator, integrator, Summing amp but what about squaring ? | {"url":"http://www.physicsforums.com/showthread.php?p=3477441","timestamp":"2014-04-20T11:20:32Z","content_type":null,"content_length":"31211","record_id":"<urn:uuid:1b485d9c-00f5-4da4-a3c1-0d42f0ab12fd>","cc-path":"CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00576-ip-10-147-4-33.ec2.internal.warc.gz"} |
Clojure's numeric tower is useful, but it can put a lot of steps between you and simple arithmetic. Unfortunately, while Clojure will warn you when reflection is required to invoke a function, it
will not warn you when reflection is required to perform math. The only reliable way to discover whether you're calling clojure.lang.Number.add(Object, Object) or clojure.lang.Number.add(long, long)
is to use a profiler or decompiler.
Or you can just bypass Clojure's math operators altogether.
In the primitive-math namespace, there are equivalents for every arithmetic operator and comparator that will give a reflection warning if it cannot compile down to a simple, predictable, unboxed
mathematical operation. | {"url":"http://clojure-libraries.appspot.com/library/primitive-math","timestamp":"2014-04-23T16:56:45Z","content_type":null,"content_length":"5654","record_id":"<urn:uuid:91471299-169d-4122-b326-d085f855f8a0>","cc-path":"CC-MAIN-2014-15/segments/1398223203235.2/warc/CC-MAIN-20140423032003-00026-ip-10-147-4-33.ec2.internal.warc.gz"} |
Dividing a circle into different bits
Draw a circle. Now mark a small number of points round the edge of the circle (it's called the circumference). Draw a line between each of the pairs of points. This divides the circle into a number
of parts, counting everything bounded by one of these lines, or the edge of the circle. Count them. Now do it for several circles, with the number of points from one (which is a little silly!) up to
six. Count the number of parts the circle is divided into each time. Can you spot a pattern?
The points don't have to be evenly spaced around the circle, in fact, it's better if they aren't. The lines cross each other, of course, but you're not allowed to have three lines crossing at a
point. If this happens, move one of the points so you get a small part within the three lines. On the other hand, if the points are bunched up too close to each other, the parts can be so tiny that
you can't count them. To make it easier for you, this page will count them for you. Click on the buttons and see if you can spot the pattern!
1 point 2 points 3 points
4 points 5 points 6 points | {"url":"http://gwydir.demon.co.uk/jo/games/puzzles/circle.htm","timestamp":"2014-04-19T07:27:57Z","content_type":null,"content_length":"4748","record_id":"<urn:uuid:cc2f8daf-6efa-4a6b-b8af-0020f2af2f70>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00346-ip-10-147-4-33.ec2.internal.warc.gz"} |
jip analysis toolkit
If one omits the keyword “half-life” in the SRTM model file, then the program has no idea about what isotope [11C, 18F] was used for the PET run. In this case, the analysis will assume equal weights
for all time points, so the analysis uses ordinary least squares (OLS). In fact, the weighting is not quite OLS, because it still assumes that the variance scales inversely with the length of PET
frames, but otherwise there is no time dependence.
If a decay time is specified (e.g., “half-life 20.33” for the 11C half-life in minutes), then a standard PET noise model is used, and analysis employs weighted least squares (WLS). The noise model is
a standard one that assumes that the variance scales with magnitude of the signal, which has been converted to a concentration (C) by decay-correcting the data using the known half-life, and that bin
size also matters:
The program then uses a diagonal variance matrix and solves the linear equations using standard WLS.
Below: Residue of fit (yellow points) for a double-bolus experiment (two 11C syntheses), together with the raw noise model (white lines) dependent upon signal intensities using the formula above,
and a regularized noise model (red curve) fit by polynomials. The WLS weighting matrix employs the polynomial-regularized (red) variance model. | {"url":"http://www.nmr.mgh.harvard.edu/~jbm/jip/jip-srtm/noise-model.html","timestamp":"2014-04-19T19:35:15Z","content_type":null,"content_length":"14774","record_id":"<urn:uuid:fcf3dffc-549b-45bf-8434-6ec329fb84fa>","cc-path":"CC-MAIN-2014-15/segments/1397609537376.43/warc/CC-MAIN-20140416005217-00132-ip-10-147-4-33.ec2.internal.warc.gz"} |
Why is T not invertible?
February 22nd 2011, 01:18 PM #1
Junior Member
Nov 2010
Why is T not invertible?
The linear transformation T:M(2x2) --> P2 defined by
T[a b]
[c d] = a+2bx+(c+d)x^2
I have to prove that this is not invertible. I don't even know where to start because if I was trying to prove that it was invertible, I wouldn't be able to solve for the determinant since I
don't know how I'm supposed to put the polynomial into a matrix since it seems to me that it would only have 3 entries. What should I do to show that this is not invertible?
I think the idea is that, while a and b are uniquely determined, given any ol' quadratic, there's no way to tell how much of the coefficient of x^2 should be c, and how much should be d. Example:
quadratic is x^2+x-6. Obviously, a = -6, b = 1/2. But with c and d, I could have c=1, d=0, or c=0, d=1, or any other of an infinite number of possibilities.
To put it on a more rigorous footing, you could always write the RHS in terms of a standard basis for P2.
Oh that makes sense. That would make it impossible to find an inverse of T.
Thank you!
You're welcome!
February 22nd 2011, 01:24 PM #2
February 22nd 2011, 01:37 PM #3
Junior Member
Nov 2010
February 22nd 2011, 01:38 PM #4 | {"url":"http://mathhelpforum.com/advanced-algebra/172249-why-t-not-invertible.html","timestamp":"2014-04-20T21:33:52Z","content_type":null,"content_length":"38750","record_id":"<urn:uuid:be454690-a784-4a1f-965b-b5c70823e9b4>","cc-path":"CC-MAIN-2014-15/segments/1397609539230.18/warc/CC-MAIN-20140416005219-00388-ip-10-147-4-33.ec2.internal.warc.gz"} |
Des Moines, WA Precalculus Tutor
Find a Des Moines, WA Precalculus Tutor
...I look forward to working with you and seeing you succeed. Best regards, StephanieI have four children ages: 5-15 and have tutored similar ages for the past few years. From learning shapes,
colors, letters and numbers to learning simple addition and subtraction, I have some experience with mont...
46 Subjects: including precalculus, English, reading, algebra 1
My goal as a tutor is to see the student excel. I have over four years of experience as a tutor, working with students from the elementary through the college level. When I work with students, I
am aiming for more than just good test scores - I will build confidence so that my students know that they know the material.
8 Subjects: including precalculus, calculus, geometry, algebra 1
...I was able to figure out how to explain the concepts or problems in a way that made sense to them. Later, I used those same skills in a math tutoring program I helped set up at my high school
through our school's chapter of the National Honor Society. I eventually majored in mathematics at Rice...
35 Subjects: including precalculus, English, reading, writing
...I have taught all subjects of the SAT for over 10 years. For the math section, I focus on test strategy interspersed with content review. I have worked with students from all levels of math,
and I vary my approach according to each students starting point, goal score, and time available.
32 Subjects: including precalculus, English, reading, writing
...I created an obstacle course. At the end of it there was a multiplication problem. I said 'take the first number.
17 Subjects: including precalculus, calculus, statistics, geometry
Related Des Moines, WA Tutors
Des Moines, WA Accounting Tutors
Des Moines, WA ACT Tutors
Des Moines, WA Algebra Tutors
Des Moines, WA Algebra 2 Tutors
Des Moines, WA Calculus Tutors
Des Moines, WA Geometry Tutors
Des Moines, WA Math Tutors
Des Moines, WA Prealgebra Tutors
Des Moines, WA Precalculus Tutors
Des Moines, WA SAT Tutors
Des Moines, WA SAT Math Tutors
Des Moines, WA Science Tutors
Des Moines, WA Statistics Tutors
Des Moines, WA Trigonometry Tutors
Nearby Cities With precalculus Tutor
Auburn, WA precalculus Tutors
Burien, WA precalculus Tutors
Edgewood, WA precalculus Tutors
Federal Way precalculus Tutors
Issaquah precalculus Tutors
Kent, WA precalculus Tutors
Lakewood, WA precalculus Tutors
Newcastle, WA precalculus Tutors
Normandy Park, WA precalculus Tutors
Redondo Beach, WA precalculus Tutors
Renton precalculus Tutors
Seatac, WA precalculus Tutors
Tacoma precalculus Tutors
Tukwila, WA precalculus Tutors
University Place precalculus Tutors | {"url":"http://www.purplemath.com/des_moines_wa_precalculus_tutors.php","timestamp":"2014-04-20T02:16:04Z","content_type":null,"content_length":"24130","record_id":"<urn:uuid:b0ecb964-73cc-421e-be07-a065e584187d>","cc-path":"CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00465-ip-10-147-4-33.ec2.internal.warc.gz"} |
Elementary Math Tutors
Federal Way, WA 98023
Eager to Help You Find Success!
...I have both of them brainstorm about the assigned prompt, write a first draft, edit by themselves and then with me, write a second draft, re-edit again, and finally be able to have learned from
their mistakes and independently write a final draft. Also, during the...
Offering 9 subjects including elementary math | {"url":"http://www.wyzant.com/Federal_Way_elementary%20math_tutors.aspx?g=3OHW","timestamp":"2014-04-21T08:49:29Z","content_type":null,"content_length":"59424","record_id":"<urn:uuid:38da86ff-5b02-41d6-8607-d724dfaf7b23>","cc-path":"CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00028-ip-10-147-4-33.ec2.internal.warc.gz"} |
Time reversible Markov Chain
February 1st 2011, 11:49 PM
Time reversible Markov Chain
A total of m white and m black balls are distributed among two urns, with each urn containing m balls. At each stage, a ball is randomly selected from each urn and
the two selected balls are interchanged. Let $X_n$ denote the number of black balls in urn 1 after the $n$th interchange.
(a) Give the transition probabilities of the Markov chain $X_n$, $n \geq 0$.
(b) Without any computations, what do you think are the limiting probabilities
of this chain?
(c) Find the limiting probabilities and show that the stationary chain is time
Attempted solution:
Transition probabilities
$P_{i, i+1} = \frac{(m - i)^2}{m^2}$
$P_{i,i} =\frac{2(m - i)(i)}{m^2}$
$P_{i, i - 1} = \frac{i^2}{m^2}$
I think those are the transition probabilities. Where $i$ is the current number of black balls.
I also know that in order to be time reversible I need the stationary distribution to satisfy: $\pi_iP_{ij} = \pi_jP_{ji}$, where $\pi$ is the stationary distribution.
Any assistance as to how I can guess the correct form of the solution based on intuition regarding this scenario would be helpful, thanks in advance. | {"url":"http://mathhelpforum.com/advanced-statistics/169986-time-reversible-markov-chain-print.html","timestamp":"2014-04-20T02:22:54Z","content_type":null,"content_length":"5966","record_id":"<urn:uuid:abd92607-2681-439a-b257-1e0010a0eec8>","cc-path":"CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00309-ip-10-147-4-33.ec2.internal.warc.gz"} |
Got Homework?
Connect with other students for help. It's a free community.
• across
MIT Grad Student
Online now
• laura*
Helped 1,000 students
Online now
• Hero
College Math Guru
Online now
Here's the question you clicked on:
2 more questions, I am not doing these -_- 1) How many grams of iron metal do you expect to be produced when 305 grams of a 75.5 percent by mass iron(II) nitrate solution reacts with excess aluminum
metal? Show all of the work needed to solve this problem. 2 Al (s) + 3 Fe(NO3)2 (aq) yields 3 Fe (s) + 2 Al(NO3)2 (aq) 2) How many grams of magnesium metal will react completely with 6.3 liters of
4.5 M HCl? Show all of the work needed to solve this problem. Mg (s) + 2 HCl (aq) yields MgCl2 (aq) + H2 (g)
Your question is ready. Sign up for free to start getting answers.
is replying to Can someone tell me what button the professor is hitting...
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• Engagement 19 Mad Hatter
• You have blocked this person.
• ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/updates/4f6aa274e4b014cf77c7d132","timestamp":"2014-04-17T22:05:06Z","content_type":null,"content_length":"28170","record_id":"<urn:uuid:f81d89e9-ca73-4ea7-a042-033bb77a9668>","cc-path":"CC-MAIN-2014-15/segments/1398223206672.15/warc/CC-MAIN-20140423032006-00481-ip-10-147-4-33.ec2.internal.warc.gz"} |
Rafael Sorkin
Ph.D. California Institute of Technology, 1974 A.B. (Summa cum laude) Harvard, 1966
[Most references still to be added] OVERVIEW OF MY WORK More than anything else, my work in physics has been guided by a desire to overcome the disunity which characterizes our present conception of
nature, and which shows itself most obviously in our failure to have reconciled our best theory of spacetime structure (general relativity) with the best dynamical framework we know of for describing
the behavior of matter on small scales (quantum field theory). Ever since I encountered it in graduate school, the greater part of my efforts has been devoted directly or indirectly to this problem,
which can also be viewed as the task of completing the twin revolutions initiated early in the last century in connection with the study of the very small (atoms), the very fast (light), and the very
big (astronomy). In addressing this "quantum gravity" problem, I first explored the idea ("Regge Calculus") that spacetime can be represented as a simplicial complex, an essentially topological
notion which today provides the basis for the causal dynamical triangulation and spin-foam approaches to quantum gravity. Regarded not just as an artificial finite-differencing of the Einstein
equations, but as a hypothesis about the inner basis of spacetime, Regge Calculus fit in well with the expectation that the continuum is only an effective description of some more fundamental --
discrete -- structure. However, after some work, I came to feel that if topology was to be the ultimate source of spatio-temporal geometry, then a more flexible and natural structure than the
simplicial complex was the finite topological space. By a kind of duality, such a space can also be regarded as a partially ordered set or "poset", and this links topology to the arguably more basic
concept of order. But in the end the poset-type of topology seemed no more satisfactory as a substitute for spacetime than the simplicial complex had been. At best it could produce a metric-space
with all distances positive, but not the kind of Lorentzian geometry which characterizes the spacetime of special and general relativity. Understood, however, not topologically but temporally or
causally, an order-relation does contain the seeds of Lorentzian geometry, and this led me, finally, to the idea of the causal set as the deep structure of spacetime. At present, this is the road I
am taking, and a separate web page describes causal set theory in more depth, including its recent great progress and current prospects. Although I no longer believe that topology enjoys a
fundamental status, that does not mean that topological questions are unimportant for quantum gravity, and exploring the novel topological possibilities inherent in curved spacetime has always been
an important strand in my work. Emerging from an underlying discreteness, potential topological effects will first arise on those length-scales -- possibly still close to Planckian -- where the
spacetime manifold itself begins to make sense. In this regime spacetime topology possesses kinematic resources sufficient to reproduce all the degrees of freedom of the standard model. For example,
extra spatial dimensions could give rise to gauge fields in the "Kaluza-Klein" manner, while topological geons (quasi-local excitations of the spatial topology) could serve as quarks, at least as far
as kinematics is concerned. In particular, geons can exhibit fermionic statistics for purely topological reasons, without the need to introduce any fundamental fermions into the story. In this way,
the causal set in its role as the source of an emergent spacetime topology could in principle be all there is. Beyond the potential to reproduce well established types of fields and particles,
spacetime topology might also give rise to novel behaviours not so far observed or even theorized in other contexts, for example certain types of exotic statistics. But the study of spacetime
topology, and especially of its temporal variation, also seems to me to have furnished important clues to the nature of quantum gravity in general. For example, my work with geons led me to believe
that the spin-statistics correlation will break down unless quantum gravity provides for transitions from one topology to another. In the contrary case of frozen spatial topology, geons (including
Kaluza-Klein monopoles) could not be pair-produced, and with the loss of that process one would lose the source from which, plausibly, the correlation between spin and statistics springs in general.
Recovery of spin-statistics for geons thus points at the necessity for topology change, and in the context of the gravitational path integral, it turns out to place definite constraints on the
amplitudes associated with the cobordisms involved in pair-creation. No less importantly, the need for topology-change reinforces the belief that a path-integral formulation of quantum mechanics is
the only one that can meet the needs of quantum gravity, an encouraging conclusion from the point of view of causal sets, which seem to admit no other form of quantal dynamics. In this connection, I
should also mention Skyrmions and anyons, two other instances of topological phenomena I have worked on. In the case of Skyrmions, the topology in question is that of the group SU2 or SU3, regarded
as the "target manifold" of a meson field. The Skyrmion itself is a topological excitation of the vacuum to be identified with the nucleon or one of the other baryons, and fermionic statistics arises
in close analogy to how it does for topological geons in quantum gravity. Even dynamically, the analogy with geons remains close, with the exception that (thanks to the absence of gauge-issues) one
can actually compute the energy-spectrum and understand how the uncertainty principle can stabilize the Skyrmion against collapse. The resulting zero-parameter fit to the low lying baryon masses
turned out to be surprisingly accurate. Thinking about geon statistics also led me to try to analyze the meaning of particle-statistics in general, as a kind of nonlocal interaction among particles
which is expressed topologically. As a byproduct I noticed the possibility of so-called anyons, which actually had been pointed out a few years earlier by Leinaas and Myrheim. Along with topology and
topology-change, black hole entropy has long interested me as another clue to the nature of quantum gravity. When I first learned that a black hole carries entropy and that (up to a coefficient of
proportionality of order unity) this entropy equals the surface-area of the black hole in Planck units, the thought that suggested itself almost immediately was that this was telling us something
important about the micro-structure of spacetime because the most natural reading of such an area law is that the horizon is somehow carrying about one bit of information per unit (Planck) area, and
this in turn is most easily explained if the deep structure of spacetime is discrete so that the horizon is actually composed of large but still finite number of "molecules" of Planckian magnitude.
In relation to causal sets, the lesson is that the fundamental discreteness scale must be near to 1e-32 cm, consistent with many other indications that sizes shorter than this lack physical meaning.
(The subsequently popular idea of "holography" must have sprung from a train of thought similar to mine, but it has led in different directions and does not seem to have shed much light on the
structure of the horizon, or on how the entropy can be "located there".) A correct counting of horizon "degrees of freedom" must of course be conducted quantum mechanically. To that end, I computed
what is now called the entanglement entropy across the horizon, regarded as a surface separating the outside of the black hole from the inside. The calculation did indeed yield an entropy
proportional to the area, or rather an entropy given by the horizon area in units of the cutoff that one needed to render finite an otherwise divergent answer. This result, together with others of a
similar nature, furnishes to my way of thinking the single best argument in favor of spatio-temporal discreteness, and the most direct indication so far of where the discreteness-scale lies. If, for
example, the kinematic counting of "horizon molecules" in causal set theory were to find a dynamical justification, one would be able to derive the coefficient of proportionality relating the true
discreteness scale to the nominal Planck length of 0.81019e-32 cm. [mention fractal horizon, entropy as shapes] [mention Lambda prediction, or refer to accompanying web page on causets] [unimodular
gravity?] The last strand of my work that I will sketch here concerns the path-integral and its interpretation. Or maybe it would be better to say that it concerns the possibility of re-formulating
quantum mechanics entirely as a theory of quantal histories, without ever needing to call on state-vectors, measurements, or external agents as fundamental notions. But what purpose would such a
re-formulation serve, and would it aid in the quest for quantum gravity? Part of the answer which I've already alluded to above is that quantum gravity seems to resist the Hamiltonian framework of
"canonical quantization". This statement reflects first of all the so-called problem of time and the need to deal with diffeomorphism-invariant quantities, but when, as with causets, we add to the
mix a discreteness which is not only spatial but temporal, the contradiction with the Hamiltonian as a generator of continuous time-evolution becomes severe. And when we allow for topology change (or
do away with the continuum altogether) any thought of quantum gravity as a theory of operator-valued fields on a fixed manifold also fades out. A second part of the answer harks back directly to the
question of the unity of physics. The lack of a theory of quantum gravity is an obvious symptom of the disunity in our present understanding, but it is not the only symptom one can point to. Another
notorious instance is the "cut" which quantum mechanics supposedly forces on us, that severs the macroscopic world of classical physics from the microscopic world of quantal objects and processes.
(Some authors have taken this cut so seriously as to deny reality to the micro-world altogether.) This issue of the quantal cut might appear to be remote from quantum gravity, but insofar as "quantum
vs. gravity" reduces to "micro vs. macro", we are dealing with something rather similar in both situations; hence one might suspect that the two problems are in fact closely related. Even if such
suspicions are mistaken, though, the practical need for a histories formulation remains, and further reasons to desire a more realistic re-formulation of quantum mechanics crop up when one tries to
generalize to the quantal case the condition of "Bell Causality" that figures so heavily in connection with classical models of "sequential growth" for causal sets. How then does the path-integral
offer an alternative to the textbook formalism of state-vectors, Hamiltonians, and external observers? A first answer is that from the path integral one can derive a functional mu_quantum -- the
quantal measure -- which directly furnishes the probability of any desired "instrument-event" E. (This measure is closely related to the so called decoherence functional.) In saying this, I am
presupposing that the Born rule (or rule of thumb!) is correct, and then just taking note of the fact that the Bornian probabilities for any specified set of "pointer readings" are furnished directly
by mu_quantum, without any appeal to Schroedinger evolution of the wave-function or its "collapse" during the measurement. In this way mu_quantum is analogous to the classical measure mu_classical
that furnishes the probability of a set of histories -- an "event" -- in the case of a purely classical stochastic process like diffusion or Brownian motion. If one construes the path-integral in
this way, namely as a generalized measure on a space of "histories", then one sees not only how quantal processes differ from classical stochastic processes, but also how closely the two resemble
each other, the primary difference being simply that mu_classical and mu_quantum satisfy different sum-rules. The former obeys a "2-slit" sum-rule that expresses the absence of interference between
alternative histories, the latter obeys a weaker, "3-slit" sum-rule that expresses the absence of "higher order" interference beyond pairwise. (This 3-slit sum-rule, which reflects the quadratic
nature of Bornian probabilities, has now been tested directly in a literal 3-slit experiment with individual photons.) The formal framework I have just sketched rests mathematically on a space of
histories and a notion of integration thereon that allows one to compute the quantal measure mu(A) for any (sufficiently regular) subset of the history-space. Satisfactory as this framework is in
some ways, the interpretation of mu(A) as a probability still does not take us beyond the realm of "pointer events" that occur in laboratory instruments or other macroscopic objects. But the
laboratory, or even the observatory, is not the place where we expect to encounter quantum gravity. There are no laboratories deep inside black holes or in the very early universe. Not
instrument-events, but events like the "big bang" are the natural domain of quantum gravity, and one would like to be able to reason about these events in direct physical terms. When one ventures
into such realms, however, the physical significance of the quantal measure becomes uncertain: the Born rule loses its force and, because of interference, one can no longer interpret mu(A) as a
probability in any ordinary sense. The question then is whether mu admits of a broader interpretation that remains serviceable outside the city limits of Copenhagen. (If it does not, then perhaps an
entirely different type of dynamical framework will be called for.) My current belief is that the quantal-measure (meaning in effect the path-integral) does have a direct microscopic significance,
not as a probability per se, but as an arbiter of whether a given event can occur at all. In other words, we concede that an event A of vanishing measure mu(A) will never occur. Drawing out the
implications of this "preclusion principle", one soon finds that it conflicts with the classical conception of reality as a single history (a single point in the "history space" over which the
path-integral integrates). Of course, such a contradiction with classical conceptions of reality is just what one would have expected, but an escape route also opens up. It turns out that -- at least
so far -- the preclusion principle does not seem to conflict with a modified conception of reality that replaces the single classical history with a set of one or more histories. Such a set is a
special case of what is called an "anhomomorphic coevent", and the new point of view is that reality is best described by such a coevent. (In the resulting re-formulation of quantum theory, it is
most natural to reason about events using rules of inference that differ from those of classical logic. One could even regard these modified rules as the essence of the new formulation, but one
should not confuse them with what has previously been called "quantum logic".) We thus acquire an interpretation of the quantal measure in conjunction with a particular answer to the question, What
is quantum mechanics telling us about the nature of reality? As a byproduct of this development, one solves the "measurement problem", or more properly, one obtains a solution of that problem if
instrument-events can be proved to obey a certain separability condition. It remains to be seen whether this condition can be established in sufficient generality. It also remains to be seen whether
the new formulation will be be able to accomplish the task for which it was ultimately intended, namely to provide a more precise framework for thinking about quantum gravity, and thereby to help
clear away some of the obstacles standing between us and that theory. [mention QBC (Quantal Bell Causality) here or above?] | {"url":"https://perimeterinstitute.ca/people/rafael-sorkin","timestamp":"2014-04-21T04:40:56Z","content_type":null,"content_length":"121258","record_id":"<urn:uuid:bdc881c8-3d88-42ed-8e8b-43d3cc2a874a>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00513-ip-10-147-4-33.ec2.internal.warc.gz"} |
34,117pages on
this wiki
Assessment | Biopsychology | Comparative | Cognitive | Developmental | Language | Individual differences | Personality | Philosophy | Social |
Methods | Statistics | Clinical | Educational | Industrial | Professional items | World psychology |
Statistics: Scientific method · Research methods · Experimental design · Undergraduate statistics courses · Statistical tests · Game theory · Decision theory
In probability theory and statistics, a median is described as the number separating the higher half of a sample, a population, or a probability distribution, from the lower half. The median of a
finite list of numbers can be found by arranging all the observations from lowest value to highest value and picking the middle one. If there is an even number of observations, the median is not
unique, so one often takes the mean of the two middle values. At most half the population have values less than the median and at most half have values greater than the median. If both groups contain
less than half the population, then some of the population is exactly equal to the median. For example, if a < b < c, then the median of the list {a, b, c} is b, and if a < b < c < d, then the median
of the list {a, b, c, d} is the mean of b and c, i.e. it is (b + c)/2.
The median can be used when a distribution is skewed or when end values are not known. A disadvantage is the difficulty of handling it theoretically.^[How to reference and link to summary or text]
The median of some variable $x\,\!$ is denoted either as $\tilde{x}\,\!$ or as $\mu_{1/2}(x).\,\!$^[1]
Measures of statistical dispersion
When the median is used as a location parameter in descriptive statistics, there are several choices for a measure of variability: the range, the interquartile range, the mean absolute deviation, and
the median absolute deviation. Since the median is the same as the second quartile, its calculation is illustrated in the article on quartiles.
Working with computers, a population of integers should have an integer median. Thus, for an integer population with an even number of elements, there are two medians known as lower median and upper
median^[How to reference and link to summary or text]. For floating point population, the median lies somewhere between the two middle elements, depending on the distribution^[How to reference and
link to summary or text]. Median is the middle value after arranging data by any order^[How to reference and link to summary or text].
Medians of probability distributions
For any probability distribution on the real line with cumulative distribution function F, regardless of whether it is any kind of continuous probability distribution, in particular an absolutely
continuous distribution (and therefore has a probability density function), or a discrete probability distribution, a median m satisfies the inequalities
$\operatorname{P}(X\leq m) \geq \frac{1}{2}\text{ and }\operatorname{P}(X\geq m) \geq \frac{1}{2}\,\!$
$\int_{-\infty}^m \mathrm{d}F(x) \geq \frac{1}{2}\text{ and }\int_m^{\infty} \mathrm{d}F(x) \geq \frac{1}{2}\,\!$
in which a Riemann-Stieltjes integral is used. For an absolutely continuous probability distribution with probability density function ƒ, we have
$\operatorname{P}(X\leq m) = \operatorname{P}(X\geq m)=\int_{-\infty}^m f(x)\, \mathrm{d}x=0.5.\,\!$
Medians of particular distributions: The medians of certain types of distributions can be easily calculated from their parameters: The median of a normal distribution with mean μ and variance σ^2 is
μ. In fact, for a normal distribution, mean = median = mode. The median of a uniform distribution in the interval [a, b] is (a + b) / 2, which is also the mean. The median of a Cauchy distribution
with location parameter x[0] and scale parameter y is x[0], the location parameter. The median of an exponential distribution with rate parameter λ is the natural logarithm of 2 divided by the rate
parameter: ln 2/λ. The median of a Weibull distribution with shape parameter k and scale parameter λ is λ(ln 2)^1/k.
Medians in descriptive statistics
The median is primarily used for skewed distributions, which it represents differently than the arithmetic mean. Consider the multiset { 1, 2, 2, 2, 3, 9 }. The median is 2 in this case, as is the
mode, and it might be seen as a better indication of central tendency than the arithmetic mean of 3.166.
Calculation of medians is a popular technique in summary statistics and summarizing statistical data, since it is simple to understand and easy to calculate, while also giving a measure that is more
robust in the presence of outlier values than is the mean.
Theoretical properties
An optimality property
The median is also the central point which minimizes the average of the absolute deviations; in the example above this would be (1 + 0 + 0 + 0 + 1 + 7) / 6 = 1.5 using the median, while it would be
1.944 using the mean. In the language of probability theory, the value of c that minimizes
is the median of the probability distribution of the random variable X. Note, however, that c is not always unique, and therefore not well defined in general.
An inequality relating means and medians
For continuous probability distributions, the difference between the median and the mean is less than or equal to one standard deviation. See an inequality on location and scale parameters.
The sample median
Efficient computation of the sample median
Even though sorting n items takes in general O(n log n) operations, by using a "divide and conquer" algorithm the median of n items can be computed with only O(n) operations (in fact, you can always
find the k-th element of a list of values with this method; this is called the selection problem).
Easy explanation of the sample median
As an example, we will calculate the median of the following population of numbers: 1, 5, 2, 8, 7.
Start by sorting the numbers: 1, 2, 5, 7, 8.
In this case, 5 is the median, because when the numbers are sorted, it is the middle number.
For a set of even numbers:
As an example of this scenario, we will calculate the median of the following population of numbers: 1, 5, 2, 10, 8, 7.
Again, start by sorting the numbers: 1, 2, 5, 7, 8, 10.
In this case, both 5 and 7 would be the median of the data points.
• Note: In some math books, you may also see that they present taking the average of the two median numbers, and then dividing it by two to get an exact value(5+7= 12/2= 6). Whether the median is
two numbers or just one is dependent on what the data values, student, instructor, etc are trying to show.
Other estimates of the median
If data are represented by a statistical model specifying a particular family of probability distributions, then estimates of the median can be obtained by fitting that family of probability
distributions to the data and calculating the theoretical median of the fitted distribution. See, for example Pareto interpolation.
Medians in computer science
In computer science, a median calculation is often performed to determine the middle index of a sorted array. The middle index is computed as (A + B)/2, where A is the index of the smallest value,
and B is the index of the largest value. Joshua Bloch, a Google software engineer, posited that if (A + B) is larger than the maximum allowed integer size, then a arithmetic overflow would occur. He
suggested that an alternative median calculation: A + ((B − A)/2) would avoid this problem. Note that the aforementioned calculations are for binary search and similar algorithms, and do not
represent a true mathematical median.^[2]
See also
1. ↑ http://mathworld.wolfram.com/StatisticalMedian.html
2. ↑ http://googleresearch.blogspot.com/2006/06/extra-extra-read-all-about-it-nearly.html
External links
This article incorporates material from Median of a distribution on PlanetMath, which is licensed under the GFDL. | {"url":"http://psychology.wikia.com/wiki/Median","timestamp":"2014-04-25T06:16:38Z","content_type":null,"content_length":"85585","record_id":"<urn:uuid:e168012a-49e0-474e-8b9e-5482b992ed93>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00066-ip-10-147-4-33.ec2.internal.warc.gz"} |
time dilation and length contraction for rotating black holes
Jonathan Scott
If you know the metric the time dilation is [itex]\sqrt{g_{tt}}/c[/itex] as mentioned above. That is, it is simply the rate at which proper time changes with coordinate time.
well this works for metrics like the Reissner Nordstrom metric, but only for static metrics, from what I understand:
in the case of the Kerr metric, there is frame dragging, and there is not only dt^2 in the metric but dt.dphi too
furthermore, at the event horizon of a Kerr black hole, g_rr -> infinite often without g_tt =0
this means that space is infinitely contracted but time is not infinitely dilated, if only g_tt provides the time dilation.
in such case the factor of the time dilation would not be the factor of the length contraction.
in other words, the event horizon is not always the ergosphere | {"url":"http://www.physicsforums.com/showthread.php?t=416147&page=2","timestamp":"2014-04-18T08:13:22Z","content_type":null,"content_length":"74061","record_id":"<urn:uuid:2387cfc8-99cf-4ac3-843e-ce06540e36b6>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00603-ip-10-147-4-33.ec2.internal.warc.gz"} |
This modules provides a convienient way to access and update the elements of a structure. It is very similar to Data.Accessors, but a bit more generic and has fewer dependencies. I particularly like
how cleanly it handles nested structures in state monads. runSTLense is also a very useful function.
A brief tutorial to get started:
To create a lense, you can use fromGetSet (although usually you would just derive them using templat haskell and deriveLenses from Data.Lenses.Template):
lense = fromGetSet getField setField
The lense has type:
lense :: (MonadState r m) => StateT a m b -> m b
Where r is the type of the record, and a is the type of the field, (b can be any type you choose, more on that latter). Though it may help to think of it as:
lense :: State a b -> State r b
Which is not entirely accurate, but emphasises how the lense works. You can think of it as "pass in an action that operates on the field, and you get an action that operates on the record". So say we
pass in get (with a more specific type for clarity):
get :: State a a
lense get :: State r a
We get out a state monad that we can run on our record to fetch our field.
fieldValue = lense get `evalState` record
This module has a special function fetch that does this:
fieldValue = record `fetch` lense
You can also pass in put to get back an action that updates the field.
put :: a -> State a ()
lense (put someValue) :: State r ()
Now we have a state monad that we can run on our record to update our field.
updatedRecord = lense (put someValue) `execState` record
This module has a special function update that does this:
updatedRecord = (record `update` lense) someValue
To aid in clarity and to deal with the actual types of the lenses this module provides execIn, evalFrom, and runOn to be used in place of execState, evalState, and runState. Also note that execIn,
evalFrom, and runOn have their parameters fliped from their state counterparts. There is nothing magical about these functions, they are just a little more handy than their state counterparts.
The lenses are especially convienient if you have nested structures. Lense composition is just function composition.
data Point = Point {
x_ :: Float,
y_ :: Float
deriving (Show)
$( deriveLenses ''Point )
data Triangle = Triangle {
pa_ :: Point,
pb_ :: Point,
pc_ :: Point
deriving (Show)
$( deriveLenses ''Triangle )
a_y :: (MonadState Triangle m) => StateT Float (StateT Point m) b -> m b
a_y = pa . y
a_y is now a lense that can operate on the y coordinate of point "a" inside a triangle. We can use a_y to fetch the coordinate or update it, on whatever triangle we choose.
someTriangle = Triangle (Point 5 3) (Point 0 1) (Point 10 6)
ayValue = someTriangle `fetch` a_y
-- ayValue == 3
updatedTriangle = (someTriangle `update` a_y) 7
-- updatedTriangle == Triangle (Point 5 7) (Point 0 1) (Point 10 6)
Or we could apply our lense to an action and pass it into execIn
(someTriangle `update` a_y) 7 == execIn someTriangle (a_y (put 7))
We can also chain actions together:
a_x :: (MonadState Triangle m) => StateT Float (StateT Point m) b -> m b
a_x = pa . x
c_y :: (MonadState Triangle m) => StateT Float (StateT Point m) b -> m b
c_y = pc . y
updatedTriangle = execIn someTriangle $ a_y (put 7) >> a_x (put 1) >> c_y (put 9)
-- updatedTriangle == Triangle (Point 1 7) (Point 0 1) (Point 10 9)
What if we wanted to put the value of c_y into a_x? Can do!
updatedTriangle = execIn someTriangle $ do
cy <- c_y get
a_x $ put cy
-- updatedTriangle == Triangle (Point 6 3) (Point 0 1) (Point 10 6)
Or if the order really bugs you, you can use the $% operator (taken from Data.Accessors.Basic, it really should be in a standard lib)
updatedTriangle = execIn someTriangle $ do
cy <- get $% c_y
put cy $% a_x
-- updatedTriangle == Triangle (Point 6 3) (Point 0 1) (Point 10 6)
Or you can use the $= operator:
updatedTriangle = execIn someTriangle $ do
cy <- c_y get
a_x $= cy
Or more concisely:
updatedTriangle = execIn someTriangle $ (c_y get >>= a_x . put)
Or say we want to put the value of c_y into a_x, but want to throw an error if c_y is zero. We can do that as well!
updatdeTriangle :: Either String Triangle
updatdeTriangle = execInT someTriangle $ do
cy <- c_y get
when (cy == 0) $ throwError "Something bad happend"
a_x $ put cy
-- updatedTriangle == Right $ Triangle (Point 6 3) (Point 0 1) (Point 10 6)
-- if cy had equaled 0 then we would have gotten this:
-- updatedTriangle == Left "Something bad happend"
Note that execInT = flip execStateT.
Yay for monad transformers!
This module has one last feature that allows you to convert a function that fetches data from a structure to a function that modifies it! For an example see the documentation for runSTLense.
One final note: Due to the generality of the lenses you might end up accidentally running into the monomorphism restriction. So if get a type error like:
Couldn't match expected type `SomeMonad SomeStructureType'
against inferred type `Control.Monad.Identity.Identity SomeStructureType'
and nothing appears to be wrong with your code, try turning the restriction off with -XNoMonomorphismRestriction and see if it goes away. If it does then you probably need to add some explicit type
signatures somewhere.
I whipped out this documentation in a hurry, so if you spot any errors, or think I should explain something better, please let me know. Also since this module is new I'm open to radical modifications
if you have a good suggestion, so suggest away! :) | {"url":"http://hackage.haskell.org/package/lenses-0.1.2/docs/Data-Lenses.html","timestamp":"2014-04-25T07:16:41Z","content_type":null,"content_length":"36247","record_id":"<urn:uuid:6fb98aa0-5bf4-493e-bf71-8b8a034e7b0c>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00212-ip-10-147-4-33.ec2.internal.warc.gz"} |
moteaco.com - 3-6 math
MONTESSORI TEACHER ALBUMS - Children's House (3-6) - Math
READ: The Absorbent Mind: Chapter 17, Discovery of the Child: Chapters 18 & 19
Table of Contents
Number Rods
Purpose: To learn to count to ten and understand the value of each number. To learn the names one to ten in association with the quantities.
Materials: Ten wooden rods varying in length from 1 dm to 1 m. Each decimeter is colored alternately red and blue. The first rod is red; the second rod is red-blue; the third rod is red-blue-red, and
so on. The rods correspond in length to the sensorial red rods. A mat for the floor.
Presentation 1: Invite the child to set out the mat on the floor and to help you carry the rods as before. Ask the child to arrange the rods in stair formation beginning with the 'one' rod as we want
the stair to grow. When the child is familiar with the rods proceed to teach the names by the Three Period Lesson. First: Bring down the one, two and three rods towards the child. Place the one rod
in front of the child and say its name a few times. This is 'one', 'one'. Remove the one rod to the side and put the 'two' rod in front of the child and name it. This is 'two', 'two'. Count it while
touching each section and say: "One, two". Put the two rod to the side and place the three rod in front of the child and name it: This is 'three', 'three'. Count it while touching each section and
say, 'one', 'two', 'three'. Second: Put the three rods in front of the child in mixed order, parallel and a short distance apart from each other. Ask the child to give you the 'one' rod, then the
'three' rod, then the 'two' rod. Playfully, ask for the rods again by name until you know the child knows them by name. Third: Place a rod in front of the child and ask him to name it. 'Which rod is
this, count it for me.' Place that rod to the side and place another rod in front of the child and do the same with the third rod. Always end your lesson by placing the rods in order.
Exercise: The child works with the material as in the presentation.
Presentation 2: Continue to teach the names of the other rods three at a time, always reviewing the previous rods until the child knows how to count the rods using their correct names.
Control: The uneven formation of the rods, the number of sections in the rod, the teacher.
Age: 4 - 5 years
Sandpaper Numbers
Purpose: To teach the written symbol corresponding to the numerals one to ten and to make the association between the name of the numeral and its symbol.
Materials: A set of numerals from 1 to 9 cut out of fine sandpaper and each mounted on a separate card. There is also a symbol for zero.
Presentation: Invite the child to come with you to where the Sandpaper Numerals are kept. At a table sit beside the child and take the first three number cards with the numerals 1, 2, 3 represented
on them. First: Take the number card with the numeral '1' and trace it with the finger tips saying, "This is 'ONE' and this is how we write it." Repeat the name a few times. Invite the child to trace
the numeral '1'. Place the numeral '1' to the side and take the numeral '2' card and do the same as with '1'. Then the numeral '3' card. Second: Place the three number cards in front of the child.
Ask him to find a particular numeral and to feel it until he has recognized each of the three numerals. Third: Place one of the number cards in front of the child. Ask him to trace the numeral and
tell you its name. Complete the lesson by placing the numerals learned in sequential order from left to right. On another day review the numerals already learned and teach the next three numerals.
Continue on subsequent days until all numerals are learned.
Exercise: The child traces and names the numerals. He may also wish to write them.
Control: The sandpaper and the smooth surface of the card.
Age: 4 - 5 years
NUMBER RODS & CARDS
Purpose: To associate the written symbol with the quantity. - To give further experience of the sequence of numbers.
Materials: The number rods and a set of number cards on which are printed the numerals 1 to 10. - One or two mats may be used depending on the space available.
Presentation 1: Using two mats place the number rods in mixed order on one mat and the number cards stacked in mixed order on the other mat. Show the child a card and ask him to name the numeral. Ask
the child to find the number rod from the other mat to correspond with the number card and place the card on the correct section of the number rod. The child continues to work in this way until all
the number cards and number rods have been matched. Always finish the lesson by placing the rods and numerals in correct sequence.
Presentation 2: (reverse exercise of the previous presentation) Using two mats place the number cards in mixed order on one mat and the number rods in mixed order but parallel on the other mat. Show
the child a rod and ask him to name it and to count it. Ask the child to find the number card from the other mat to correspond with the rod and place it on the correct section of the rod. The child
continues to work in this way until all the rods and number cards have been matched. Always finish the lesson by placing the rods and numerals in correct sequence.
Presentation 3: Composition and Decomposition of Numbers 1-10: a) The number rods are arranged in sequential order. Isolate the 'ten' rod at the top of the mat by shifting the other rods down to the
bottom of the mat. Say to the child that you are going to make tens with the number rods. Place the 'nine' rod against the 'ten' rod aligning both at the left hand side and ask the child which rod
should we add to the 'nine' rod to make the 'ten' rod. By a process of trial and error the child finds that the 'one' rod and the 'nine' rod make ten. Count the 'nine' rod and say: "Nine and one make
ten." The child continues making all the combinations of ten until he reaches the five rod. Show him how to flip it to the right hand side and say, 'If we had another five we could make 'ten'. Note:
Number cards may be used to state equation together with the 'plus' symbol. Show the child how to make nines, eights, etc. b) The child makes tens as in Exercise 'a'. Begin at the bottom and work up.
Review 5 and if we had another five it would make 10. Take away 5 by flipping the five rod back. Ask what is left. State equation: 10 take away 5 is 5. Repeat for the other combinations of tens and
then the nines, eights, etc.
Control: The colored sections guide the child's memory. Counting the sections carefully.
Age: 4 - 5 years
To give the idea that symbols also represent a quantity of separate objects whereas in the exercise with the number rods the quantity was fixed and the symbols were loose. To clarify the meaning of
'zero' by having an empty compartment.
Two plainly varnished wooden boxes with ten compartments. The first compartment has the symbol for 'zero' printed on the back, the second compartment has the numeral 'one' and so on up to nine. Forty
five spindles in a basket or box. Eight green pipe cleaners (optional).
Invite the child to help you carry the spindle boxes, then the basket of spindles and the pipe cleaners to the table. Ask the child to identify the printed numerals 1-9 in non sequential order. Say
to the child that you are going to read the numerals on the back of the compartments and that you will count the correct number of spindles into your hand before you place them in the particular
compartment. He may bind them with the pipe cleaners before placing them in the compartments if he wishes. When all the forty-five spindles have been used, draw the child's attention to the empty
compartment and say, "This is 'zero'. Zero means 'nothing'."
Exercise: The child works with the materials as shown.
Control: Correct number of spindles.
Age: 4 - 5 years
To train the memory by keeping the image of a number in the mind over a period of time. - To show that all objects may be counted.
Slips of paper, on each is written a single numeral (0 - 10). - A container for the slips. - A box of identical objects like buttons, beads, or marbles.
Invite a group of eleven children to play the game with you. At a mat, give a child one slip, ask him to look at his numeral carefully, place the slip on the mat and go off to find that number of
specified objects. Repeat for each child. When all the objects have been gathered, ask each child, individually, to identify their numeral and count their objects. When you reach the child who has
'0', playfully reinforce the concept by saying, "You didn't bring anything - you must have the zero!"
Exercise: As in the presentation.
Control: The teacher and the child check by reading the slip of paper with the numeral written on it and counting the objects.
Age: 4 - 5 years
To recognise the numerals 1 - 10 and their correct sequence. - To understand how many separate units form each number. - To give visual and muscular impression of odd and even numbers.
Preparation for the divisibility of numbers, therefore of multiples and submultiples.
Number cards with numerals from 1-10 and 55 counters of one color.
Invite the child to bring the box of counters and the number cards to a table. Place the cards in disarray and place the box of counters near the child. Ask the child to find the numeral '1' and to
place it at the top left corner of the table. Then ask, "How many counters will we put under '1'?" If the child says 'one', say, "Yes, we put one counter below the number card with the numeral '1'."
Ask what comes after 'one'. If the child says 'two', ask him to find the numeral 'two' and show him how to place it, find the correct number of counters and place them as a pair. Continue in this way
placing the even numbers in pairs and the odd numbers with the odd counter on its own below the pairs. Explain the concept 'odd' and 'even'.
Exercise: As in the presentation.
Control: The exact number of counters.
Age: 4 - 5 years
To familiarise the child with the names of the different categories and to acquaint him with the relative difference in size of the categories, e.g. the difference between the quantity of six units
and six thousands.
One unit bead; one bar of ten beads; one square of a hundred beads; one cube of one thousand beads all on a tray. A supply of units, tens, hundreds and thousands. Empty trays, castors.
Presentation 1:
Individual exercise with presentation tray. Take the presentation tray to the table. Position the tray so that the unit is always to the right. Tell the child that this is the 'Golden Beads'. Hold
and experience the unit, stating its name, "This is a unit." Give the child the bead. Encourage the child to repeat the name. Continue in the same manner experiencing a ten, a hundred, a thousand
respectively. Proceed into the second period and eventually the third period of the Three Period Lesson.
Exercise: The child works with the material as shown.
Presentation 2:
Individual exercise with a tray of Golden Beads.
Materials: 9 Unit Beads, 9 Ten bead bars, 9 Hundred Square, 1 Thousand Cube
Count through each hierarchy in order from units to thousands. Each time nine is reached, state that if we had one more we would have ten. Instead of having ten loose beads (or tens, or hundreds) we
can have a ten bead bar (or a hundred square, or a thousand cube).
Exercise: The child works with the material as shown.
Presentation 3: A small group exercise. Show the tray of Golden Beads to the children. Each child gets an empty tray with a castor. Ask each child, individually, to fetch a quantity of beads from the
tray - one quantity per child. Upon their return ask each child, "What amount did you bring me?" The children will replace the beads in the tray. Continue as above according to the interest of the
children. On another day the quantity may include two hierarchies until all the hierarchies have been included. Establish that when counting the category the name is included ie. five tens, six
hundred, etc.
Control: The teacher and the child's counting of the beads.
Age: 4 - 5 years
Purpose: - To acquaint the child with the written symbols for the new quantities he has learned.
Materials: - A box containing four sets of large number cards - one set from 1 - 9, with numerals in green - one set from 10 - 90, with numerals in blue - one set from 100 - 900, with numerals in red
- one set from 1000 - 9000, with numerals in green - a tray of Golden Beads for review
Presentation 1: Individual exercise.
Bring a box of Large Number Cards to a table/mat. Remove the number card for 1, 10, 100, 1000. Familiarise the child with the color coding and review 1 and 10 which he has learned previously by
referring to the Golden Beads. Place them as headers at the top of the mat. Give a Three Period Lesson with the new numerals, i.e., 100 and 1000. At the end of the lesson put the number cards in
correct sequence with the thousand on the left and the one on the right.
Exercise: The child works with the material as shown.
Presentation 2: Individual exercise.
Remove all the Large Number Cards from the box. Stack the cards for each category in sequential order with the thousands on the left and the units on the right at the bottom of the mat. Place the '1'
at the top right corner and continue placing and counting the unit cards in a vertical column until nine is reached. Then place the 10 - 90 to the left of the units, the 100 - 900 to the left of the
tens and the 1000 - 9000 to the extreme left of the hundreds. As you place the number cards count them using their category name, i.e., one unit, two units, three units, etc. Show the child how to
replace the number cards by stacking them in their categories with all the 'ones' showing on top. Bind them with a well fitting elastic band and replace them in the box.
Exercise: The child works with the material as shown.
Presentation 3: Small group exercise.
Ask the child to help you lay out the number cards as in exercise 2. Give each child an empty tray. Ask each of them to bring you certain number cards beginning with the unit category and building up
as in Presentation 3 with the Golden Beads.
Control: The teacher.
Age: 4 - 5 years
Purpose: Direct: - To make the child more familiar with the different categories of numbers, especially in regard to reading the written form. - To give him the language for large numbers.
Indirect: Preparation for working with: - The hierarchy of numbers, i.e. that while the numerals are always from 1-9, it is the place they occupy in the large number that gives them their value. - In
a number '0' holds a place for a specific category. - Only 1-9 cards of each category is necessary to form any number.
Materials: - 9 unit beads, 9 ten bars, 9 hundred squares, 1 thousand cube. - A set of large number cards 1 - 9, 10 - 90, 100 - 900, 1000. - Three empty trays, three castors, two large mats.
Presentation: Small group exercise.
Layout: Set out two mats on the floor. Bring a tray of beads to one mat. Lay out the beads, similar to the layout for the number cards. Beginning with the units, lay out the beads for each category,
individually, in a straight column working from top to bottom. Count each bead using the category name as you lay it out. On the other mat the children lay out the number cards as in the previous
exercise. Note: only 1000 card corresponding to only 1 thousand cube is necessary at this stage. Each child gets a tray with a castor for holding the unit beads.
1. Single Category - Cards to Beads e.g. Units Place one number card of a single category on each child's tray. Ask the child to identify it and to obtain that quantity of beads from the other mat.
When the child returns, he reads the card and counts the beads. The child returns the card and the beads to their mats.
2. Single Category - Beads to Cards Place quantity of beads from one category on the child's tray. The child counts the beads and obtains the corresponding card from the other mat. Check.
3. Two Adjacent Categories - Cards to Beads As in presentation 1a except place two cards which represent two adjacent categories on to the tray. Show the child how to superimpose the number cards and
how to read the combined number, e.g., two tens and eight units.
4. Two Adjacent Categories - Beads to Cards As in presentation 1b except place a quantity of beads from two adjacent categories on to the tray.
Continue in this manner until all four categories have been included and proceed to non-adjacent categories as the child progresses.
Control: The teacher and the child's counting of the beads.
Age: 4 - 5 years
Materials: - A set of Large number Cards. 1 - 1000. - A quantity of Golden Beads: 45 units 45 ten bead bars 45 hundred squares 1 thousand cube
Purpose: To enable the child to see the number system laid out with the numerals and their corresponding quantities.
The teacher shows the child how to place the number card '1' on the top right-hand corner of the mat and how to place the quantity, one golden unit, to the right of the card. The child continues in
this way until he has used all the golden bead material for the units. The teacher shows the child how to place the number card for '10' and the ten bead bar parallel to the number card and the child
continues until he reaches '90'. He then continues with the hundreds and the thousand.
Control: Visual and counting the number cards and the Golden Beads. All the material should be used.
Age: 4 - 5 years
TEENS, FORMATION OF QUANTITIES 11-19 WITH GOLDEN BEADS
Purpose: - To teach the names of 11 - 19 in association with the correct quantity.
Materials: - A box with nine golden ten bead bars and a set of colored Bead Stair. - A small, neutral color, felt mat.
The lessons on Teens and Tens are usually given after the introduction to the decimal system exercises, but may be presented when the need for the terminology arises.
At a table place the colored bead stair in disarray on the felt mat. Ask the child to count the beads on each bar and place them in stair formation. Give the child time to acquaint himself with the
1. Take three of the ten bead bars from the box. Place one of the ten bead bars in a vertical position in front of the child. Place the one red unit bead touching the 'one' on the ten bead bar and
say, "This is eleven." Repeat the name a few times. Bring the 'eleven' to the left side and place another ten bead bar as before with the green, two bar touching the first two units of the ten bead
bar and say, "This is twelve - twelve ... twelve." Do the same for thirteen. 2. Place the three quantities in front of the child and ask him to point to each quantity in non-sequential order. 3.
Place one quantity in front of the child and ask him to name it. Always end with your material in order.
Proceed with 14, 15, 16 and then 17, 18, 19 always reviewing the previous step.
Exercise: As in the Presentation.
Control: The pattern formed by the beads - the tens remain constant while the units increase as the child builds 11-19 in a vertical column.
Age: 4 - 5 years
FORMATION OF 11-19 WITH TEEN BOARD
Purpose: - To associate the names eleven to nineteen with the symbols.
Materials: - A number frame consisting of two boards numbered with nine tens. - A set of number cards, with the numerals 1 - 9, which will slide into the boards to cover the 0 in each ten.
At a mat on the floor place the boards in a vertical column. Stack the cards from 1 - 9 at the top right corner of the boards. The child identifies the 10's. Using the THREE PERIOD LESSON teach the
names of the numerals by sliding the unit cards over the zeros to create the numerals 11-19. At first teach 11, 12, and 13 and continue if the child shows interest until the child knows 11-19.
Playfully, ask the child to name the numerals in non-sequential order. End the lesson by asking the child to read the numerals in sequential order. Show how to replace the number cards and the boards
and to return them to the shelf.
Exercise: As in the presentation.
Control: The teacher.
Age: 4 - 5 years
COMBINATION OF TEEN BOARD AND BEADS TO FORM 11 TO 19
Purpose: - To associate the quantities with the numerals 11 to 19.
Materials: - Teen boards, a box containing nine golden ten bars and a set of colored bead stair 1 to 9.
Invite the child to bring the Teen Boards to a mat on the floor. Show the child how to lay out the boards as before and then to arrange the beads to the left of the board. Slide in the first card and
ask the child to read the numeral and to make the quantity. He makes eleven with the beads directly to the left of the numeral. Ask him what comes after eleven and if he says twelve he may proceed to
build the quantity for twelve.
Exercise: The child continues in this way until he has all the numerals from 11 -19 in place and their corresponding quantity in the beads.
Control: The teacher. The child sees the pattern of the beads and recognises the sequence of the numeral on the board.
Age: 4 - 5 years
FORMATION OF 10 - 90 WITH THE GOLDEN BEADS
Purpose: - To teach the names 'twenty', 'thirty', etc., with the ten bead bars.
Materials: - Forty-five golden ten bead bars. - A felt mat.
Presentation: Invite the child to place the mat on the table and to bring the box of forty-five ten bead bars. Remove six ten bead bars. Place one ten bar in front of the child and say, you know what
this is. Yes it is 'ten'. Remove the ten to the left and place two ten bars in front of the child and say, This is 'twenty' ... 'twenty'. Remove the two ten bars and place three ten bars in front of
the child and say: This is 'thirty' ... 'thirty'. Proceed with the Three Period Lesson until the child understands the new concept. On another day teach 40, 50, 60 and then 70, 80, 90 until the child
is quite familiar with how we make the quantity 10 - 90 and the correct terminology.
Exercise: The child works with the material as shown.
Control: The teacher and the child as he counts the tens and learns the new names.
Age: 4 - 5 years
FORMATION OF 10 - 90 WITH THE TEN BOARD
Purpose: - To teach the names 'twenty' to 'ninety' in association with the written numeral on the Ten Board.
Materials: - The Ten Board - A floor mat
The child has already learned the names in association with the quantities and therefore he can be taught the names of of all the numerals in one lesson through the Three Period Lesson.
Exercise: The child reads the numerals on the Ten Board.
Control: The teacher and the sequence of number.
Age: 4 - 5 years
COMBINATION OF 10 - 90 WITH GOLDEN BEADS AND TEN BOARDS
Purpose: - To teach the names 'twenty, 'thirty', etc. and to associate the name with the quantity and the numeral. - Preparation for skip counting.
Materials: - Two boards with the numerals 10, 20, to 90 printed on them. - A box of forty-five ten bead bars. - A floor mat.
Invite the child to place the boards in order on the floor and to place the box with the bead bars to the left of the board. Ask the child to read the numerals on the board beginning with '10'. Show
him how to place one ten bead bar immediately to the left of the numeral. Ask him to read the next numeral. He will probably say 'two tens'. Say, "Yes, you are quite right but remember it has a
special name - it is 'twenty'." "Can you say 'twenty'?" "Now we are going to make 'twenty' with the ten bead bars." "How many will we need?" "Yes, two." "Two tens make 'twenty'." Proceed in this way
saying, "This is how we write thirty." "Can you make thirty for me?"
Exercise: The child works with 10, 20, 30 and may continue to 40, 50, 60 and then 70, 80, 90 with the teacher naming the new names forty, fifty, sixty, etc. for the numerals. When he has reached '90'
he will see very clearly how the 'tens' have grown from 'one ten' to 'nine tens' (90) and, as in the previous exercise, he can also see and associate the correct name with the numeral and the
Control: The teacher and the child as he balances the quantity with the written symbol.
Age: 4 - 5 years
FORMATION OF 11 - 99 WITH GOLDEN BEADS AND TEN BOARD
Purpose: - To learn the terminology for 11 - 99 with the aid of the bead material and the written symbol.
Materials: - The Ten Boards - A Box of Golden Beads containing nine golden ten bead bars and nine golden units. - A floor mat.
Two children can work together at this stage with the teacher's guidance. Ask the child to place the Ten Boards in position and to place the nine ten bead bars together with the nine golden units to
the far left of the '10' section leaving space for the quantities 11 - 99 immediately to the left of the Board. Ask the other child to stack the number cards in order 1 to 9. Ask one child to make
'eleven' with one ten bead bar and one unit - the unit touching the 'one' bead in the ten bead bar. Ask the other child to make the numeral 11 by slipping the '1' unit card over the zero in 10. Say,
"Ten and one make eleven. This is how we write 11." Continue in this way until the children have reached '19' and say if we had one more golden unit we would have ten. Instead of having ten loose
unit beads we can replace the units to the left and exchange them for a ten bead bar. Each time a number card is removed, place it face downward in a pile. Bring down the ten bar from the 10 section
and with the 'new' ten place them to the left of the 20 section and say this is 'twenty'. Now let us make 'twenty one' by placing a golden unit to the right of the ten bead bars touching the 'one'
bead in the ten bead bar nearest the Board. Say, "Now we have made 'twenty-one'." Turn the number cards to show the '1' at the top and proceed to slip the unit cards over the '0' in '20' as the other
child makes twenty-one to twenty-nine with the Golden Bead material. Continue in this manner until '99' is reached when the nine ten bead bars and the nine Golden Units show at the left of the Board
and the numeral 99 appears on the last section of the Board.
Control: The teacher and the material.
Age: 4 - 5 years
Purpose: - To give experience in the changing process, i.e., 10 of one category make 1 of the next higher category. - To give practice in changing as a preparation for further exercises.
Materials: - A tray containing a supply of units, tens, hundreds and thousands. - A dish large enough to hold a supply of units. - Castors, large mat.
Presentation: A small group exercise.
- Units to Tens A good supply of units and tens on a tray and a castor for each child. Invite the children to help you count the units. Begin counting the units into a castor - stop when you reach
ten. Remind the children that there are ten units in one ten. Explain that when you count to 10 you must exchange the 10 units for 1 ten. When all the exchanges have been made, read the quantity of
beads, e.g., seven tens and four units/seventy-four.
- Tens to Hundreds Using the same procedure exchange 10 tens for 1 hundred.
- Hundreds to Thousands Using the same procedure exchange 10 hundreds for 1 thousand.
EXERCISE 1: The child works with the material as shown.
Presentation: At a later date do the reverse process (before starting subtraction).
- Thousands to Hundreds A good supply of thousands at a mat. Remind the children that 10 hundreds are exchanged for a thousand. Explain that you can reverse this procedure: exchanging 1 thousand for
10 hundred. Children exchange thousands in a systematic manner.
- Hundreds to Tens Using the same procedure exchanging 1 hundred for 10 tens.
- Tens to Units Using the same procedure exchanging 1 ten for 10 units.
EXERCISE 2: The child works with the material as shown.
Place a quantity of units, tens, hundreds and thousands all together in a pile on a mat. Playfully, say to the children, "I wonder how much we have here?" "How can we find out?" Hopefully the
children will suggest sorting the quantity into their hierarchies and then count them beginning with the units and changing when necessary until they have all been counted. Get the child to use the
Large Number Cards to show the quantity in written form.
EXERCISE 3: The child works with the material as shown.
Control: The teacher.
Age: 4 - 5 years
Purpose: - To give the children the experience of the process of addition. Addend + Addend = Sum
Materials: - A supply of golden bead material: 50 unit beads, 45 tens, 45 hundred and 9 thousands kept neatly on a tray. - One set of large number cards 1 to 9000. - Three sets of small number cards
1 to 3000. - Three trays and three castors to hold the units.
Presentation: A small group exercise.
This work can be done at a large table or at a mat on the floor. Invite one child to lay out the large number cards, another sets out three sets of small number cards in the same formation as the
large number cards. Ask another child to take care of the tray of Golden Bead Material acting as a 'banker'. Give a tray to each of three children with a castor to hold the unit beads.
- Static Addition (Without changing) Place a low four digit number on each of the three trays using the small number cards - each tray should have a different number. The children read the number on
their tray. Show them how to approach the 'banker' and to ask politely for that amount of golden bead. The children return to the work area where the teacher removes the golden beads in hierarchical
order from each tray placing them on the mat in three separate quantities (addends). Place the small number cards in problem formation at the left corner of the mat (representing addends). Add the
units by bringing the three amounts together in the center of the mat. Ask the child in charge of the large number cards to give you the numeral which represents the answer. Place it below the units.
Proceed to work in the same way with the tens, hundreds and thousands. Superimpose the large number cards and place them below the small number cards (sum). Review the process and say, "Now we have
done 'addition'."
______ brought me ______ ______ brought me ______ and ______ brought me ______ and when we added them together we found that we had _________________.
Point to each of the small number cards and say, "This is an addend, this is an addend and this is an addend." Point to the Large Number Cards and say, "This is the sum." The children will pick up
the terminology quite naturally.
- Dynamic Addition (With changing) Follow the same procedure as for the static addition but now you select small number cards whose sum will require exchanging.
- Individual Work When the children have understood the process of addition they may work on their own taking a tray of golden bead to their table or taking what they need from a 'pool' of golden
beads. Later, they may progress from using the number cards to having the teacher write problems in their work book or taking a sheet of prepared problems and working with them. However, this needs
some preparation. The children need to be shown how to write problems in their work books and how to space the problems and use a ruler to draw straight lines. Neat work habits are fostered in this
Exercise: As in the presentation. The teacher helps to set up the exercise and will act as a guide until the children can work together confidently.
Control: The teacher. However the emphasis is on the process not on the exact result.
Age: 5 years
Note: The process is the focus of the decimal system. When working with the golden beads do not correct the answer.
Purpose: - To give the children the experience of the process of multiplication. Multiplicand x Multiplier = Product.
Materials: - The same as for addition.
Presentation: The same procedure as for static and dynamic addition. Ensure that each of the children have the same number; and when reviewing the children will note that the same number was brought
so many times! Name the process - 'multiplication'.
Exercise: As in the presentation.
EXERCISE 2: Introduce multiplier card: After the children have completed several examples, review the completed equation. Explain that it is not necessary to lay the multiplicand out so many times -
a multiplier card may be used (use a small unit card which is placed under the multiplicand as in a written problem).
Control: The teacher. However the emphasis is on the process not on the exact result.
Age: 5 years
Purpose: - To give the children the experience of the process of subtraction. MINUEND - SUBTRAHEND = DIFFERENCE
Materials: - The same as for addition.
Presentation: A small group exercise. Layout as for addition and multiplication.
Static Subtraction: Select a high number in the large number cards and the corresponding amount of golden bead material (minuend). Place the golden bead in hierarchical order in the center of the
work area. Place the large number cards at the top right corner of the mat. Select a low number in the small number cards and place them on a child's tray (subtrahend). Tell him he may take that
quantity from the golden bead on the table, beginning taking from the units. He proceeds in this way until he has taken the quantity specified in the small number cards on his tray. The teacher then
states the quantity that is left on the table and asks a child to find the small number cards and place them below the beads (difference). Review the process and place the small number cards from the
child's tray (subtrahend) below the large number cards (minuend) and place the small number cards representing the difference, below the subtrahend. State, now we have done 'subtraction'.
Dynamic Subtraction: This can follow when the child understands the process of subtraction.
Individual Work: The child can take the tray of golden bead and work at his table or he may take what he needs from a 'pool' of golden bead material.
EXERCISE 1: As in the presentation.
Presentation & EXERCISE 2: Minuend of 9000 (When the child has mastered Exercise 1.) Select a minuend of 9000. Proceed as for dynamic subtraction showing the child how to exchange a thousand for
hundreds; a hundred for tens and a ten for units before he can begin subtracting.
Control: The teacher. However, the emphasis is on the process not on the correct answer.
Age: 5 years
Purpose: - To give the experience of the process of division by units. - Dividend ¸ Divisor = Quotient
Materials: - The same as for addition.
Presentation: A small group exercise. Layout as for addition, multiplication and subtraction.
a) Static Division: Select a high number in the large number cards and the corresponding amount of golden bead material (quotient). The amount selected must be divisible evenly by the number of
children who represent the divisor. Place the golden bead in hierarchical order in the center of the work area. Place the large number cards at the top right corner of the mat. Explain that you are
going to share the quantity of golden beads equally between the number of children. Begin with the thousands by placing one thousand in each tray and proceeding in this way until all the thousands
have been used. The child in charge of the small number cards gives each child the small card which represents the amount of thousands on their tray. Proceed in this way sharing hundreds, tens and
units until all the golden bead is used. Review the process and state that in division our answer is what one person gets and therefore we need only take a set of small number cards from one tray.
Place these small number cards representing the dividend above the large number cards. State that now we have done 'division'.
b) Dynamic Division (no remainder)
c) Dynamic Division (with remainder)
EXERCISE 1: As in the presentation
Presentation & EXERCISE 2: Individual Work: Introduce the 'Divisor' card and green skittles to represent the number of children. Share the golden bead between the skittles and the answer will be what
one unit skittle gets.
Control: The process is more important than the correct answer.
Age: 5 years
Note: The process is still important, however if answers are frequently incorrect re-present.
This material is used by the children for individual work with the Decimal System, following the group exercises done with the golden bead material.
Purpose: To give the child the opportunity of carrying out the four operations as individual exercises.
Materials: Small colored 'tiles' or 'stamps': green with '1' written on them to represent units; blue with '10' to represent tens; red with '100' to represent hundreds; green, again, marked '1000' to
represent thousands. Skittles: 1 large green, 9 of each red, blue and green representing the decimal system categories. Some red, blue and green plastic discs to represent decimal system categories.
Squared paper, pencil and ruler.
Presentation: Individual exercise. Place the stamp game, writing materials and presentation tray (golden beads) at a table. The teacher removes one stamp from each category and asks the child to
identify the numeral. Ask the child to align the stamps with their corresponding golden beads. Explain that the stamps may be used, individually, for the same exercises as the golden beads. The child
returns the presentation tray to the shelf and the stamps to the box. The number cards will no longer be necessary and instead we will write our numbers. The teacher writes a four digit number
beginning with the highest category. The child reads and makes the quantity with the stamps. Repeat for a few examples. Introduce some numbers with zero.
Note: Use correct terminology with each operation. Addition - addends, sum. Multiplication - multiplicand, multiplier, product. Subtraction - minuend, subtrahend, difference. Division - dividend,
divisor, quotient. Introduce the signs used to symbolise, e.g., + for addition; - for subtraction; x for multiplication and ¸ for division.
EXERCISE 1: Static Addition With the child's input, write two addends which will not require carrying. Draw a line under the addends and include a plus sign. Point out the use of a plus sign denotes
this is addition. Read the problem with the child. The child lays out the appropriate stamps for the first addend. Encourage the child to check by reading the quantity made with the stamps. Place a
ruler under the first addend and have the child lay out the second addend. Check. Remove the ruler. Remind the child of the necessary process to find the answer - combine categories and count
beginning with the units. To combine the categories push the stamps up towards the top of the table until they form a double column per category. Count stamps using the category name. As each stamp
is counted move it slightly toward you. The child records the answer in the units place - below the equal line. Have the child repeat the process for the other categories: tens, hundreds, thousands
respectively. Review the problem with the child.
Dynamic Addition Follows the same procedure except when counting, exchange the categories as necessary by removing one stamp of the next higher category from the box and replacing the ten stamps,
which have been counted, into their appropriate place in the box.
Presentation & EXERCISE 2: Multiplication With the child's input, write a multiplicand. The child chooses a multiplier of 2 or 3, which is written in the units column below the multiplicand.
Introduce the multiplication sign. Read the equation with the child. Proceed as in dynamic addition. The child lays out the multiplicand the appropriate number of times combines the categories and
counts exchanging as necessary and records the answer for each category as he counts.
After some experience use '0' in the multiplicand.
Presentation & EXERCISE 3: Static Subtraction With the child's input, write a minuend. Write a subtrahend which does not necessitate exchanging. Introduce the subtraction sign. Read the problem with
the child. The child lays out the appropriate stamps for the minuend. Beginning with the units the child takes away the necessary number of stamps and replaces them into the box. The child counts the
remaining number of units and records the answer. He repeats this process for the remaining categories in their respective order. Review the problem with the child.
Dynamic Subtraction Follows the same procedure except when subtracting exchange categories as necessary by replacing one stamp of the next higher category into the box and removing ten stamps of the
needed category (using a ruler for spacing purposes).
Presentation & EXERCISE 4: Static Short Division With the child's input write a dividend as a statement and as in a process. Introduce the division sign and read the problem with the child. Remind
the child that the skittles represent the divisor. Set out the appropriate number of skittles horizontally. Stack the appropriate stamps for the dividend to the left of the skittles in hierarchical
order. Review the procedure for division: we start with the highest category and we give an equal number of stamps to each skittle. Share out the stamps underneath the skittles. Remind the child that
the answer is what one unit received. The child counts the stamps under one skittle and records the quotient above the dividend. Read the problem with the child.
Dynamic Short Division (No remainder) Follows the same procedure except to exchange categories as necessary.
Dynamic Short Division (With remainder) Follows the same procedure as dynamic short division, except to introduce 'remainder'. Write the remainder to the right of the quotient with a small case 'r'
before it. Explain that the 'r' is an abbreviation of remainder.
Control: The teacher. For each process the checking can be taught e.g., in addition one of the addends can be subtracted from the sum to find the other addend. In subtraction the subtrahend can be
added to the difference to find the minuend. In multiplication the product can be divided by the multiplier to find the multiplicand. In division the quotient can be multiplied by the divisor to find
the dividend.
Age: 5 - 5.5 years
LINEAR COUNTING: CHAINS OF 100 AND 1000 GOLDEN BEADS
Purpose: To consolidate the child's knowledge of counting.
Materials: The golden 100 chain of 10 ten bars. A box (red, if possible) containing labels: 1 to 9 in green; tens in blue; one red 100. Golden bead square of 100. The golden 1000 chain of 100 ten
bars. A box (green, if possible) containing these labels: 1 to 9 in green; tens in blue; hundreds in red, numbers between hundreds, ie. 130, in blue; one green label for 1000. This is a total of 10
green, 90 blue and 9 red. Ten golden hundred squares and one golden thousand cube. A long piece of carpet or felt the length of 1000 chain.
Presentation: Set out a mat for 100 chain. Introduce the bead cabinet and the 100 chain. Stress the careful handling of the chains. Show how to carry the chain with palms up, lay the chain over the
child's hands. At the mat, fold the chain in a zigzag manner. Pull the right end to straighten the chain in a horizontal manner. Invite the child to fold the chain. With the child bring a hundred
square from the cabinet to the mat. Superimpose the hundred square on the folded chain. The child straightens the chain, placing the square above the last ten bar, at the right. With the child bring
the appropriate arrows (color coded) from the shelf to the mat. Lay out the arrows with the numbers facing up, sorted by color. Count the beads from left to right, placing an arrow for each bead in
the first bar. For the remaining bars label the last bead in each bar. When the chain has been counted, note the number of beads in the chain. Read the arrows in sequence.
Exercise: As in the presentation.
EXERCISE 2: Introduce the special mat and set it out. Introduce the 1000 chain. Show how to carry the chain. Note the chain's rings. Hold the left hand perpendicular to the floor, close to the
cabinet. Lift the chain from the left hook and over on to your hand, so that the ring lies on top of the index finger. Proceed as above for each ring, then return the chain to the cabinet one ring at
a time. Invite the child to remove the chain from the cabinet and carry it to the mat. Lay out the chain horizontally on the mat. With the child, fold the chain as above. The ring denotes the end of
each square which leaves a space. After each square is folded get a hundred square from the cabinet and superimpose it on to the folded chain. When the chain is folded count the hundreds. Get the
1000 cube from the cabinet - compare by stacking the hundreds. Pull out the chain - place a square above the last bar of each section at the right, place a cube at the end of the chain. Lay out the
arrows, sorting by color and by hundreds. Count as above, establishing the repeating process. When the chain has been counted, note the number of beads in the chain. Read the arrows in sequence. Show
the child how to pick up the chain by every other ring.
Control: The labels and the teacher's guidance.
Age: 5 - 6 years
Purpose: To give the child a means of counting other than one by one. Preparation for multiplication. Preparation for squaring and cubing. To show the difference in quantity between the square and
the cube of a number and the difference, or relationship, between the squares and the cubes of different numbers.
Materials: A cabinet with a shelf for the squares, the square chains and the cubes of each number and the corresponding tickets. Hooks for the cube chain of each number. Four red unit beads: 1, 1
squared, 1 and 1 cubed; red labels marked '1'. Long and short chains of 2 - green: green labels marked 1,2,4 for the square chain; 1, 2, 4, 6, 8 for the cube chain. Long and short chains of 3 -
peach: peach labels marked 1,2,3,6,9 for the square chain; to 27 for the cube chain. Long and short chains of 4 - yellow: yellow labels marked 1,2,3,4,8 to 16 for the square chain; to 64 for the cube
chain. Long and short chains of 5 - light blue: light blue labels marked 1,2,3,4, 5,10 to 25 for the square chain; to 125 for the cube chain. Long and short chains of 6 - mauve: mauve labels marked
1,2,3,4,5,6,12 to 36 for the square chain; to 216 for the cube chain. Long and short chains of 7 - white: white labels marked 1,2,3,4,5,6,7,14 to 49 for the square chain; to 343 for the cube chain.
Long and short chains of 8 - brown: brown labels marked 1,2,3,4,5,6,7,8,16 to 64 for the square chain; to 512 for the cube chain. Long and short chains of 9 - blue: blue labels marked
1,2,3,4,5,6,7,8,9,18 to 81 for the square chain; to 729 for the cube chain. Corresponding number of bead squares for each square chain. A cube of each number. A felt mat.
Presentation: After the 100 and 1000 chains have been counted. Begin with the square of five chain (short chain). Proceed in the same manner as the 100 chain but 'skip-count' at the end of the
exercise. Note the arrows are the same color as the beads on the chain.
Exercise: As in the presentation, the child skip-counts all the square chains at the end of the exercise.
EXERCISE 2: Begin with the cube of five chain (long chains). Proceed in the same manner as the 1000 chain but 'skip-count' at the end of the exercise. Encourage the child to skip-count the other cube
chains in the same manner.
Control: The labels and the teacher's guidance.
Age: 5 - 6 years
Purpose: To familiarize the child with all the possible number combinations that make ten. To give practice in memorising addition.
Materials: Five colored bead stairs (1 - 9) in a box. Twenty-five golden ten bars in a box. A set of 1 to 9 black and white bead stair in a black and white box. A red felt mat. A small notched card
('bridge'). A box or tray to hold the bars of the colored stair removed from the snake after they have been counted (the empty black and white bead stair box may be used for this).
Presentation: At a table, unroll a small mat and introduce the material laying them out in order at the top of the mat (box of tens, box of colored bead bars, black and white box). Build the black
and white stair in the upper right hand corner. Place the notched card below the stair. State the use of the card, "This card is to mark where we finished counting." Make a colored snake which will
come out even in tens (one of each bead bar plus an extra five). Close the box. The bead bars are placed at random, touching in a zigzag pattern. Change the colored snake to a golden one by counting
to 10. Place the notched card on the right side of the tenth bead. Place a ten bar directly above the counted beads. Use the black and white bead bars to represent the remaining beads. Count
remaining beads on the bead bar. Place the corresponding black and white bar above the counted beads. Remove the card and place the counted colored bead bars into the empty black and white box. Join
up the snake. When counting the next ten beads begin with the black and white bar, if there was a remainder. Repeat the above process until all the bars have been counted. Count the ten bars. Arrange
the ten bars side by side at the left. Remove the colored bead bars from the black and white box - arrange in diminishing size from left to right in the center. Isolate a ten bar in the center of the
mat; make a matching ten using colored bead bars beginning with the largest, ie. 9 + 1. Repeat until all the tens and colored bead bars are matched.
EXERCISE 2: The child uses any combination of colored bead bars to make the snake. Use a black and white bar for any remaining colored beads less than 10. When checking the child may need to change
to make the appropriate combination - use the supply of colored bead bars in the box. Note: use only two colored bead bars to make a ten. Match one colored bead bar to the black and white bead bar in
EXERCISE 2: Isolating combinations After child has worked well with the snake game, proceed as before. To count, pull down two bead bars isolating them at the bottom of the mat. Count to 10 as above
using the black and white bar for remainder. Compare two sets of beads visually. Place colored bars into box. Move ten bar up into snake and pull down next colored bead bar. If two bead bars are less
than 10 exchange them for a black and white bar. Proceed until snake is counted. Check as above.
EXERCISE 3: Showing multiples - second check Proceed as above, this time make a snake using a number of the same colored bead bars (choose three of your favourite numbers or colors). To check -
Arrange tens and remainder as before. Arrange the colored bead bars horizontally, grouping like beads bars parallel to one another from left to right, in diminishing size. Beginning with the largest
set of colored bead bars, note how many there are in the set. Count into tens - laying tens and remainder vertically below (use tens and colored bead supply). Repeat this process for the other sets
of colored beads. Note: that there should be the same number of beads in each of the three areas. Check by placing the tens (from colored beads) below the other tens (from the golden snake). Change
the remaining colored bars into tens. Child counts top 10's and the bottom 10's to see if they match.
Control: After each exercise the child is shown how to check the result.
Age: 5 - 6 years
Purpose: To give the child all the possible combinations in addition. The red line has the same purpose as the notched card in the snake game: to show how many went to make ten and how many units
were left over. The red line teaches how many numbers are composed of a ten and a quantity of units bringing us that much further toward another ten. This is the mechanism of addition that has to be
learned. The child is helped to see the entire structure of addition and to memorise the combinations.
Materials: A board divided into 18 squares across from left to right and 12 squares from top to bottom. Each square is 2 x 2 cm.. Above the grid are the numerals from 1 to 18. Numerals 1 to 10 are in
red, then a red line divides the board vertically; the numerals from 11 to 18 are in blue. Two sets of numbered strips. One set is blue with a numeral in red (1 to 9) at the end of each strip; the
other set is red, divided into squares by blue lines, with a numeral in blue (1 to 9) at the end of each. Squared paper and pencil. Control Charts 1 & 2.
Presentation: At a table, ask the child to read the numbers along the top of the board in random order. Note the red line, the color of the numbers and the grid. Remove the blue strips - arrange like
the number rods to the left of the board. In the same manner arrange the red strips to the right of the board. The child chooses a blue strip ie. 6 or 7. Place on the top line of the grid, aligned at
left side. Place the 'one' red strip directly to the right of the blue strip. Read the board - ie. 7+1=8. The answer is found directly above the last section of the red strip. Write the equation.
Remove and replace the red strip into the stair. Repeat for each red strip, 2 through 9. By the third or fourth equation the child may take over. When finished, show the child how to check his work
with the Addition Control Chart # 1.
Exercise: As in the presentation the child works through all the tables.
EXERCISE 2A: Ask the child to choose a number, ie. 6 or 7. Write the number centerd on the page. State the goal: to find all the pairs of numbers which make up that amount. Begin with the 'one' blue
strip - laying it on the grid as before. Ask the child what is needed to add to '1' to make the chosen amount. (The child may count the squares to find the answer.) The child places the appropriate
strip to the right of the blue '1'. Read the equation, ie. 1 + 5 = 6. The child writes the equation. Repeat the procedure until all the possible combinations are on the board. Check the work with the
Addition Control Chart #1. The child continues through all sums 2 - 18.
EXERCISE 2B: When the child has completed the above exercise, have him make a chart of all the numbers and their combinations on graph paper. ie. 2 = 1 + 1 3 = 1 + 2, 2 + 1 4 = 1 + 3, 2 + 2, 3 + 1
Check chart with Addition Control Chart #1.
EXERCISE 3A: Commutative Law Write an equation, ie. 2 + 4 =. The child uses the board to find the answer (the blue strip for the first addend, the red for the second addend) and records it. Write
another equation reversing the above addends, ie. 4 + 2 =. The child uses the board as above to find and record the answer. Ask if the same numbers were used to make the sum. Check by comparing the
strips (place equivalent strips underneath each other). Note that it does not matter which side of the plus sign the addends are on the answer is the same. Repeat with a few more examples.
EXERCISE 3B: Using the chart the child made in Exercise 2b make a second chart which does not include duplicates. ie. 2 = 1 + 1 3 = 1 + 2 4 = 1 + 3, 2 + 2 5 = 1 + 4, 2 + 3 Check chart with Addition
Control Chart #2.
Control: Addition Control Chart #1. Exercise 3 - Addition Control Chart #2.
Age: 5 - 6 years
ADDITION CHARTS 3, 4, 5 AND 6 (BLANK)
Purpose: To give practice in the memorization of addition combinations. The various charts give the possibility of repetition which helps fix the combinations in the memory. Each chart is a variation
on the preceding one. The variations are there to sustain the child's interest thereby extending his work with the addition combinations.
Materials: Four baskets or boxes, each with red tickets for all of the addition problems of chart #1 written without answers. Paper and pencil for Charts 3 to 5. A box of tiles printed in answers to
all problems used with Chart #6.
CHART 3 Presentation: At the table, examine the chart with the child. Ask the child to identify the numerals and the color across the top and repeat for the numerals down the left hand side. Note
that this chart is for the memorization of the addition facts. Open the box of problems and select one, placing it on the top of the board. Read the equation and write it. Ask the child to guess the
answer. Fingering - Find the first addend in the red column and mark it with your left index finger. Find the second addend in the blue row and mark it with your right index finger. Move the right
finger down until it is in the same row as the left finger. Move the left finger across until it meets the right finger. Where they meet is the answer. Record the answer. Repeat two more times then
invite the child to do the fingering. Check the work with Addition Control Chart #1.
Exercise: As in the presentation.
CHART 4 Presentation: Examine the chart, note it is only half the size of Chart 3. There is no top blue row. Yet it has the same number of problems. As before, the child selects, reads and writes a
problem. Encourage the child to guess the answer. Fingering - Place the left index finger on the larger addend and the right index finger on the smaller addend in the red column. Move the right
finger across to the end of the row. Move the left finger across its row until it is in the same column as the right finger. Move the right finger down the column until it meets the left finger.
Where they meet is the answer. Record answer. Note: When the addends are the same, locate the numeral in the red column with both index fingers and move across to the end of the row to find the
answer. Repeat two more times then invite the child to do the fingering. Check work with Addition Control Chart #2.
Exercise: As in the presentation.
CHART 5 Presentation: Examine the chart. There are very few answers, however, we have the same number of problems. As before, the child selects, reads and writes a problem. Encourage the child to
guess the answer. Fingering - Find addends along the red column as in Chart 4. Slide both fingers to the end of their respective rows. For an even answer: hop fingers simultaneously toward each
other, on the diagonal square by square. Where they meet is the answer. Record the answer. For an odd answer: hop fingers as above. They eventually land into squares which are touching. The answer
lies in the square which has become isolated by the fingers. Record the answer. Repeat a few more times then invite the child to do the fingering. Check work with Addition Control Chart #2.
Exercise: As in the presentation.
CHART 6 Presentation: Examine the chart. It is the same as Chart 3 except it is blank. State, "This time you are going to give the answer instead of finding it." Introduce the tiles. Lay out the
tiles in an ordered arrangement - tiles progressing from 2 to 18 with like numbers together. Note arrangement. The child selects and reads a problem. He finds the answer in the tiles and isolates the
appropriate tile to the side of the chart. Fingering is done in the same manner as for Chart 3. Place the tile where the fingers meet. Repeat until the chart is complete. Check the work with Addition
Chart 3.
Exercise: As in the presentation.
Control: Addition Control Charts 1 & 2. For Chart 6 - Chart 3.
Age: 5 - 6 years
Purpose: To familiarize the child with subtraction combinations. Indirect preparation for algebra as from time to time equal quantities of the opposite sign cancel each other out, as in algebraic
Materials: Boxes containing: colored bead bars (about 3 of each); 20 golden ten bars; black and white stair; negative bead stairs; one box or little tray to contain the beads that have been counted;
small notched card; and a green felt mat.
Presentation: (After the child has mastered the positive snake game.) At a table lay out the materials as in the positive snake game. Introduce the grey bead stair and say, "These are the negative
bead bars. They tell us how many to take away." The child identifies the grey bead bars by counting. The child builds a snake as before and places three or four grey bead bars interspersed among the
colored bead bars. (Note: in placing the grey bead bars, the snake should not go down to zero or into a minus number.) Proceed to count as before. When a grey bead bar is reached, in order to take
away, count back that number of beads. Count back on only black and white bead bars or golden bead bars. Before counting back, any colored bead bars (less than 10) to the left of the grey bar must be
changed into one black and white bead bar. To count back - The child counts the beads on the grey bar. Then counts back into the black and white and/or golden beads, counting from right to left. Mark
with the notched card. Count any remaining beads on the left side of the notched card and represent this remainder with the corresponding black and white bead bar. Place the grey bead bar into the
box with the other counted colored bead bars. Replace the black and white bead bar into the stair. Replace the golden bead bar into its box. Rejoin the snake and note that it is getting shorter. To
check - Arrange the golden snake and colored bead bars as before. Arrange the grey bead bars vertically to the right of the colored bead bars in diminishing order. Note: The colored bead bars
represent the minuend, the grey beads represent the subtrahend and the golden bead bars represent the difference. Match the colored bead bars to the golden and grey bead bars changing as necessary.
Exercise: As in the presentation.
EXERCISE 2: Isolating subtraction - Proceed as above. When the child has counted back and (at a later stage) found the remainder, pull down all the bead bars involved in the counting back process to
the bottom of the mat. Arrange the bead bars horizontally, so that the bead bars counted back on, match the grey bar and the black and white remainder. Note the quantity of the golden bead bar and/or
the black and white bead bar, ie. 10 and 4 make 14. Note the quantity of the grey bead bar and the remainder, ie. 7 and 7 make 14.
"So 14 (pick up golden bead bar and/or black and white bead bar) take away 7 (pick up grey bead bar) is 7 (leave remaining black and white bar on the table)." Place the bead bars in your hand in
their respective places as before. Place the remaining black and white bead bar back into the snake and rejoin the snake. Continue in the same manner. Check as in the presentation.
Control: The child is shown how to check his answer with the control.
Age: 5.5 - 6 years
Purpose: Memorization of the subtraction combinations.
Materials: The Negative Strip Board is a replica of the Addition Strip Board in size and number of squares. The only differences are that a blue line divides the board vertically after the ninth
square, the first 9 numerals are blue and the numerals from 10 to 18 are in red. Two sets of colored strips, one set in red and the second set in blue. A third set of plain wooden strips, not
numbered. There are seventeen of these ranging from one square in length to seventeen squares. Squared paper and pencil. Subtraction Control Chart #1.
Presentation: At a table, examine the board. Note the blue line. Introduce the wooden strips and build a stair with them. The wooden strips are placed vertically above the board from longest to
shortest respectively placed above the numerals 1 to 17. The child builds the blue stair to the right of the board. Note: To take away, always begin with the largest blue strip, never record an
answer in red. Point to the 18 on the board and ask the child to identify it. State that the blue strips are used to take away. They represent the subtrahend. Place the 9 blue strip on the board.
Align it on the right, beginning with the number 18 representing the minuend and cover up the numerals along the top of the board with the plain wooden strip. The answer is found directly to the left
of the blue strip. Write the equation, 18 - 9 = 9. Repeat using the next strip in diminishing order. The answer is in red, therefore, the child will not write down the equation. Move across the
numerals on the board, from right to left, to locate the next minuend, ie. 17. We no longer need 18. Cover the numeral(s) with the appropriate wooden strip directly above the current minuend, ie. 17.
Proceed as above. When the child reaches 9 as the minuend he may need some help as he will get zero for an answer, 9 - 9 = 0. When 8 is the minuend, tell the child not to record an answer where the
blue strip goes beyond the left barrier of the grid. The child checks his work with the Subtraction Control Chart #1.
Exercise: As in the presentation, the child works through all the tables.
EXERCISE 2: Set out the materials as before. The child sets out the red strips in stair formation to the left of the board. The child chooses a number, ie. 6 or 7. Using the wooden strip cover the
extra numerals to the right of the minuend. The child writes the minuend centerd at the top of the page and builds a table. Beginning with the red strips build a table as in the Addition Strip Board
Exercise 2a beginning with the largest addend. The last row will be the minuend represented by one blue strip. This is only possible for minuends less than 10. "Now we're going to take away the blue
strips." Using the board, note the minuend, ie. 6. The child writes '6'. Take away the first blue strip, ie. 6, sliding it to the bottom right of the board. 'What's left? - Nothing.' The child
finishes the equation, ie. 6 - 6 = 0. Note: the difference is indicated by the red strip or its absence. Proceed in the same manner, working up through the rows. Check the work with Subtraction
Control Chart #1.
Control: Subtraction Control Chart #1.
Age: 5.5 - 6.5 years
SUBTRACTION CHART 2 AND 3 (BLANK)
Purpose: To memorize the subtraction combinations.
Materials: A basket or box containing green slips of paper for all subtraction combinations on Chart 1, written without answers. Subtraction Chart 2. Subtraction Chart 3 - blank. A green box marked
with the subtraction sign, with tiles, nine of each number 0 to 9 (the tiles are white, numerals are in green). The blank chart is like Chart 2, except that all answer squares are blank.
CHART 2
Presentation: At a table, examine the board. The child identifies the numbers in red and then the numbers in blue. Note the minus signs. The child selects, reads, writes and guesses solution of a
problem. Fingering - Find the minuend among the numbers in red - mark it with the right index finger. Find the subtrahend among the numbers in blue - mark it with the left index finger. Move the
right finger down until it is in the same row as the left finger. Move the left finger across until it meets the right finger. Where they meet is the answer. Record the answer. Repeat two more times
then invite the child to try the fingering. Check work with Subtraction Control Chart #1.
Exercise: As in the presentation.
CHART 3
Presentation: Examine the chart which is the same as Chart 2 except it is blank, just like Addition Chart 6. Introduce the tiles. Lay out the tiles in an ordered arrangement progressing from 0 to 9
with like numbers together. Note the arrangement. The child selects and reads a problem and finds the answer in the tiles, which he isolates to the side of the chart. Fingering is done in the same
manner as for Chart 2. Place the tile where the fingers meet. Repeat until the chart is complete. Check work with Subtraction Chart 2.
Exercise: As in the presentation.
Control: Subtraction Charts 1 and 2.
Age: 5.5 - 6.5 years
Purpose: To show by this geometrical form of multiplication that the multiplier is never a constant as is the multiplicand. It is only indicative of how many times a number is taken or a given
quantity is repeated. To show that a succession of lines creates a surface that is why it is called geometrical multiplication. Preparation for square root and for factoring. Preparation for division
by helping the child to visualize the divisibility of numbers. The geometrical formation is an indirect preparation for exercises that follow later in connection with geometry and algebra.
Materials: Nine boxes marked 1 x, 2 x, 3 x, etc. each containing 55 bars of each number from 1 to 9. A box with tens and colored bead bars. A felt mat.
Presentation: Start with the table of 7. Bring the answer box, a yellow mat and a box marked '7x' to a table or a mat. Introduce the materials. Place one seven bar from the '7x' box horizontally at
the upper left of the mat (multiplicand). Count the number of beads on the bar, i.e. 7. Remove a seven bar from the answer box (product) and place it vertically below the multiplicand. State the
equation, 7 one time is 7. Proceed in the same manner, increasing the multiplier by one each time up to 9 inclusively. Count the multiplicand in the same manner as the snake game - counting to ten
and placing a ten bar, any remaining beads are represented by the appropriate bead bar from the answer box. When complete, go over the table verbally with the child.
Exercise: As in the presentation, the child works through the tables 1 to 9.
EXERCISE 2: Multiplying by 10 At a table, the child chooses a multiplicand, ie. 4. The child writes the multiplicand centerd on a page and takes out 10 four bars from the answer box and arranges them
horizontally. The child counts as in the presentation, and lays out the product vertically. Say, "When we have 10 fours it's the same as 4 tens or forty." "When you multiply by 10 all you have to do
is add a zero to the multiplicand." Repeat with a few more examples. Then child may continue on his own.
EXERCISE 3: Divisibility of product Select a number, ie. 12. Find the number of ways to make 12. Begin with 1 and see if you can make 12 by counting, adding each bar on one at a time. The use of 1
works however at this point note that the multiplier should be less than 10. Continue in the same manner for 2 through 9. Leave out the combinations which make 12. Child may write down combinations.
EXERCISE 4: Commutative Law At a table, using the answer box layout place 7 five bars horizontally as before. Then 5 seven bars. The child counts the beads placing the answer vertically below.
Compare the answers. They are the same. Turn the beads to compare. Review and write the equations. 7 x 5 = 5 x 7 Try a few more examples. Conclude that it does not matter which side of the
multiplication sign the multiplicand and the multiplier are on - the answer is the same.
EXERCISE 5: Making the decanomial Build the decanomial square (see the sensorial album). Set out two mats. On a mat set out the supply of bead bars. Use the other mat to build the decanomial square,
in the same manner as above, using the bead bars. When the decanomial is complete visually explore the layout. Note the squares on the diagonal and have the child exchange them with the squares from
the bead cabinet. Note: the number of bars in the band is the number squared; the number of beads in the band is the number cubed.
Control: The teacher.
Age: 5.5 - 6.5 years
Purpose: The memorization of the multiplication tables.
Materials: A perforated board with 100 holes in rows of ten. On the left side of the board there is a little window with a slot for the insertion of a card. A set of cards, 1 to 10. A counter. 100
red beads. Squared paper and pencil. Multiplication Control Charts 1 & 2.
Presentation: At a table, examine the materials with the child who identifies the numbers along the top of the board. Place a disc in indent of board. Lay out the number cards in random order. The
child chooses either a 6 or a 7. The child slides the number card into the slot and returns the remaining number cards to the box. Place the disc above numeral '1' and read the board, ie. 7 x 1 =.
Child writes the equation. Lay out the appropriate number of beads under '1', ie. seven. To find the answer the child counts the beads and writes the answer. Repeat the process until the table is
complete. Note: To count the beads start from the answer to the last problem. When the table is complete, introduce the Multiplication Control Chart #1 for the child to check his work.
Exercise: As in the presentation, the child works through tables 1 to 10.
Control: Multiplication Control Charts 1 & 2.
Age: 5.5 - 6.5 years
MULTIPLICATION CHARTS 3, 4 AND 5 (BLANK)
Purpose: The memorization of the multiplication tables.
Materials: Charts 3, 4 and 5. A basket or box for each chart, each containing yellow tickets of all the combinations of the multiplication tables from 1 times 1 to 10 times 10 (without the answers).
For the blank chart - a yellow box (marked X) with wooden tiles printed with answers for each combination of the multiplication tables as in Chart 3. Squared paper and pencil.
CHART 3
Presentation: At the table, examine the chart. The child identifies the numbers in blue and then in red. The child selects , reads, writes and guesses the solution to the problem. Fingering - In the
same manner as Addition Chart 3. Find the multiplicand among the numbers in red and mark it with the left index finger. Find the multiplier among the numbers in blue and mark it with the right index
finger. Note: when the multiplicand or the multiplier is one the answer will be found along either the red or the blue numbers. Record the answer. Repeat two more times and then invite the child to
do the fingering. Check the work with Multiplication Control Chart #1.
Exercise: As in the presentation.
CHART 4
Presentation: Examine the board and note the differences between Charts 3 and 4. The child selects, reads, writes and guesses the solution to the problem. Fingering - In the same manner as Addition
Chart 4. Note: when the multiplier or the multiplicand is one the answer is found along the red numbers. Record the answer. Check the work with Multiplication Control Chart #2.
Exercise: As in the presentation.
CHART 5
Presentation: Examine the chart which is the same as Chart 3 except it is blank. Lay out the tiles in an ordered arrangement - smallest to largest from left to right above the board, with like
numbers together. Note the arrangement. Proceed as in Addition Chart #6. The child selects and reads a problem, then finds the answer among the tiles and isolates it. Fingering is done in the same
manner as for Chart 3. Place the tile where the fingers meet. Repeat until the chart is complete.
Exercise: As in the presentation.
Control: Multiplication Control Charts 1 & 2. For Chart 5 - Chart 3.
Age: 5.5 - 6.5 years
Purpose: To make the child familiar with the ways in which numbers may be divided. To show that not every number is evenly divisible and some only by a few numbers. To show the relationship between
multiplication and division.
Materials: The Unit Division Board. 9 green skittles. 81 green beads in a jar. Castor or a cup. Red and lead pencils. Multiplication Bead Board for Exercise 2.
Presentation: Individual exercise. At a table examine the board, note the numbers along the top and the side. Skittles will again represent the divisor. Choose a dividend - one which is evenly
divisible by 9, ie. 45. Invite the child to count out the appropriate number of beads into the castor. State that we always begin with a divisor of 9. Set out 9 skittles across the top of the board.
Write the equation, ie. 45 _: 9 =. Share out the beads, moving from left to right, placing an equal amount of beads under each skittle. Record the answer which is what one unit receives. Note:
underline the equation in red if it divides out evenly. Remove one skittle (divisor is now 8) and redistribute the extra beads under the remaining skittles. Those beads which will not share out
evenly are placed into the castor and represent the remainder. Note: the remainder cannot be as large or larger than the divisor. Record the equation. Repeat the process until the quotient is 9. The
quotient can be 9 or less -no larger than 9 - due to the limitations of the board (only 9 spaces under each skittle). Then remove one bead from the dividend replacing it into the castor and begin
again starting with a divisor of 9. Continue until the child is able to take over for himself.
Exercise: As in the presentation, the child explores with dividends up to and including 81. The child may compile a chart of all the equations underlined in red to show those numbers which are evenly
EXERCISE 2: Bring both the unit division and the multiplication bead board to the table. Write an equation, ie. 7 x 3 =. The child uses the multiplication board to find the answer.
Write the corresponding division equation, ie. 21 ¸ 7 =. The child uses the unit division board to find the answer. Compare the results with the child to show what you build in multiplication, you
take apart in division (division is the inverse process of multiplication). Explore with a few more examples.
Control: The teacher.
Age: 5.5 - 6.5 years
DIVISION CHARTS 1 AND 2
Purpose: To teach the child division combinations. Exercise 2 is a preparation for finding the Highest Common Factor and the Lowest Common Multiple.
Materials: Division Charts 1 and 2. Boxes or baskets for each Chart, containing blue tickets on which are written all short division problems with a dividend of 81 or less that are exactly divisible.
A blue box of answer tiles for Chart 2 with the division sign on the box. Squared paper and pencil.
CHART 1
Presentation: At a table, examine the board with the child who identifies the numbers along the top row. Note the two colors, the prime numbers are in white. The child also reads the numbers along
the diagonal and notes the division signs. The child selects, reads, writes and guesses the solution to a problem. Fingering - Find the dividend along the top row and mark it with the right index
finger. Find the divisor along the diagonal numbers with the division signs and mark it with your left index finger. Move the right finger down until it is in the same row as the left finger. Move
the left finger across until it meets the right finger. Where they meet is the answer. Record the answer. Repeat two more times then invite the child to try the fingering. Check the work with the
child's own chart or Multiplication Control Chart #1.
Exercise: As in the presentation.
CHART 2
Presentation: Examine the chart which is the same as Chart 1 except it is blank. Introduce the tiles. Lay out the tiles in an ordered arrangement progressing from 1 to 9 with like numbers together.
Note the arrangement. The child selects and reads a problem and finds the answer in the tiles and isolates it. Fingering is done in the same manner as for Chart 1. Place the tile where the fingers
meet. Repeat until the chart is complete. Check work with Division Chart 1.
Exercise: As in the presentation.
EXTENSION: The teacher explains prime numbers and factoring. When the child inquires about the numbers in white explain they are 'prime numbers'. "That means they can be only divided by 1 and
themselves." The child gets the chart he compiled and paper and pencil. Select a quotient, ie. 56, the child writes it in the center of the page. Look it up on the chart and record the two numbers
which make it up, ie. 7 and 8, below and a bit apart with arrows pointing down from 56. Look on the chart again to find out the numbers which make up 7 and then 8. Continue on (showing factoring).
Since 7 and 2 are prime numbers they do not have any factors. Repeat according to the interest of the child. 56
Control: Multiplication Control Chart 1 or child's own chart. For Chart 2 - Chart 1.
Age: 5.5 - 6.5 years
Purpose: To illustrate the mechanism of the decimal system. To focus the child's attention on the mechanism of changing. This game prepares the child for abstract addition.
DESCRIPTION OF MATERIAL: Paper which is squared and has columns headed 1, 10, 100, 1000 and 10,000. The columns are divided into small squares so that there are ten in each horizontal row. At the
foot of each column are two spaces, the upper one for indicating the changing process with dots, and lower one for the result. There is a blank column on the right side in which the problem to be
done is written. A lead pencil, an orange colored pencil and a ruler.
Presentation: At a table, the child writes his name and date in the appropriate places. Note the column headings. Have the child identify the numbers 1 - 1000, introduce 10,000. Write 4 or 5 four
digit addends in the blank column with the child. The child reads the equation. State that we put a dot for each number in its proper category. Begin with the first addend starting with the units.
Fill up the rows from left to right - when the row is complete continue on to the next row below. To count dots: Begin at the units, upon reaching 10 - put a line through the ten dots and add an
orange dot to the next higher category. Continue counting. When reaching a number less than 10, record that number in the bottom box of that column. Read the answer and write the sum below the
Exercise: As in the presentation.
EXERCISE 2: Begin as before. This time put out the dots according to the category starting with the units (ie. put out the dots for all the units in all the addends - count as above, only now the
orange dot(s) will appear at the beginning of the top row - repeat the process for tens, hundreds, thousands).
Control: The teacher.
Age: 5 to 5.5 years
Purpose Direct: The exercises of numeration recapitulate the function of the decimal system by making the child realize once again the following: 1) That ten of one category make one of the next
higher category and then in each category there can be no more than 9. 2) The value of the numerals is determined by the place they hold. 3) The function of zero is that of a place holder. The whole
set of exercises brings the child to the realization that when he writes down addition and subtraction problems, he must place all the figures belonging to the same category in the same vertical
column. The exercises also provide the children with opportunities to apply all they have learned before and thus prepare them further for abstraction.
Indirect: Both the exercises on numeration and the multiplication exercises prepare the child for the distributive law of multiplication where it becomes necessary to analyze the numbers into their
hierarchical values.
Materials: A frame with support to enable it to stand. The frame has four wires across, each strung with ten beads. Top wire: 10 green beads for units; second wire: 10 blue beads for tens; third
wire: 10 red beads for hundreds; fourth wire: 10 green beads for units of thousands. On the left side of the frame the numeral category of each wire is written. Background for hierarchy of units is
white; background for thousands is grey. Notation paper for small frame is lined in the same colors as the numerical categories. Pencil and ruler.
Presentation: Introduce the Small Bead Frame Individual exercise. At a table, examine the frame. Ask the child to identify the numerals along the left-hand side of the frame. Note the new background
color at 1000, it denotes where we write a comma. Note the beads correspond in color to the stamp game. Introduce the notation paper to the child. The child reads the headings at the top of the
paper. Explain that as we count each bead we can write its number directly on the color coded line under the appropriate column. Start with the units counting the beads by sliding them individually
across the wire from left to right while recording the number down in its appropriate place on the notation paper. Upon counting 10 units remind the child that we must exchange. Slide one ten bead to
the right and slide the 10 units back over to the left. To record - write a 1 on the blue line and note there are no units therefore we write a zero in the units category. Continue counting beads and
noting the numbers on paper up to 1000, stressing exchanges. Make large numbers. The teacher writes a four digit number, the child reads and makes the quantity on the frame. Note: we read the number
on the right-hand side on the frame. Repeat for a few examples - reverse process and use zero.
Exercise: On the same or on another day. Static Addition - Write an equation on the notation paper which does not require exchanging. The child reads. The child sets out the first addend on the
frame. Remind the child to begin adding at the units. The child reads the number of units in the second addend and counts the corresponding number of unit beads to the right. The child counts and
records the sum of the unit beads which are at the right of the frame. Repeat the process for the remaining categories, tens to thousands respectively. Read the equation. Dynamic Addition - follows
immediately. Write an equation which will require exchanging. The child proceeds as above, reads the equation and sets out the first addend. Beginning with the units, the child reads the number and
counts the corresponding number of unit beads to the left. He runs out of unit beads - 10 unit beads lie on the right-hand side of the frame - therefore he must exchange the 10 units for a ten. Slide
one ten to the right and the 10 units back to the left. The child continues to count the necessary unit beads to the right. The child records the number of beads that sit on the right. Repeat the
process for the remaining categories, exchanging as necessary. Read the problem.
EXERCISE 2: Dynamic Addition - Proceed as above except this time add all the units first, then the tens, hundreds and thousands. The child reads the numbers in the units column - sets out the first,
then adds on the second, exchanging as necessary and records the sum. Repeat for the remaining categories. Read the problem.
EXERCISE 3: Static Subtraction - Write a problem on the notation paper that does not require exchanging. The child reads and sets out the minuend. Remind the child to begin with the units. The child
reads the units in the subtrahend and takes them away by counting the appropriate number of beads from right to left. The child reads the number of beads remaining at the right of the frame, which is
the difference and records. Repeat for the remaining categories, tens to thousands respectively. Read the problem.
Dynamic Subtraction - Write an equation that necessitates exchanging. The child reads and sets out the minuend. Beginning at the units, the child notes the number of beads which must be taken away.
The child counts the corresponding number beads from right to left, however the child will run out of beads. Remind the child of how to get more units. Slide one ten bead to the left and slide the 10
unit beads to the right. The child continues to count the appropriate number of beads to the left. The child reads the number of beads remaining at the right of the frame which is the difference and
records it. Repeat for the remaining categories. Read the problem.
Control: The child may be shown how to check his work.
Age: 5 - 6 years
Note: Multiplication is done as for Addition. The child can be shown how to take the multiplicand a certain number of times (multiplier). | {"url":"http://www.moteaco.com/albums/casa/casamath.html","timestamp":"2014-04-21T15:38:44Z","content_type":null,"content_length":"145045","record_id":"<urn:uuid:0ba30e91-ad21-465d-879a-5baab5dfd31f>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00563-ip-10-147-4-33.ec2.internal.warc.gz"} |
If you want to know how to geocache properly then it's important to have a good understanding of how GPS co-ordinates work. This is especially useful when you are trying to solve puzzle caches.
Before we get going check out the photo on the right. I was geocaching near the Dead Sea in Jordan in November 2010. I thought it would be interesting to see what the GPS altitude would be the
lowest point on earth. Look at the altitude... or whatever you call negative altitude.
BTW the geoaching app shown here is called Blackstar and works on Blackberries. Unfortunately my BB died after it got wet after I got back. Blackstar in conjunction with the GPS on the BB is such a
good app that I hardly used Garmin.
Now that I have an Android, I'm back to the Garmin because the geocaching apps with the Samsung Galaxy GPS just don't work well enough for me. See my review on the Geoaching Android app here.
But I digress...
Understanding GPS Co-ordinates
To understand how to geocache you'll need to become familiar with how co-ordinates work.
Lines of latitude
We start by cutting the world in half horizontally - this line is called the equator.
Horizontal lines that circle the earth are called lines of latitude. Imagine that a line is drawn from the centre of the earth at angle to the horizontal. This angle gives us degrees latitude above
or below the equator. So in the picture you can see N30, N50 N70 which are north of the equator, and S10, S30 which are south of the equator.
So if your latitude co-ordinate has an "N" in front of it, then you are north of the equator, and if it has an "S" in front of it you are south of the equator. Going back to the Dead Sea, you can can
see that we are located at 31 degrees to the north of the equator. And at the north pole, we would be 90 degrees above the equator.
If, as you move, your latitude numbers increase, in the northern hemisphere you would be heading further north, and in the southern hemisphere you would be heading further south.
Lines of longitude
The line that cuts the world in half vertically through Greenwich in London is called the prime meridian - as you can see in the picture. Vertical lines that circle the earth are called lines
of longitude.
Anything west of the prime meridian will have a "W" in the longitude, and anything to the east will have an "E" in the co-ordinate. Going back to the Dead Sea, you can can see that we are located
at 35 degrees to the east of the prime meridian. If, as you move, your longitude numbers are increasing, and you have an "E" in front of your co-ordinate you would be heading further east, and if
you have a "W" in the longitude, you would be heading further west.
This is useful to know when you are solving puzzle caches. For example if you are plotting numbers in Google Earth and see that the calculation takes you to the wrong location, do a whatif and see
what happens if you increase either the latitude or longitude. Increasing/decreasing the latitude number by itself, will move your point due north/south. Alternatively, increasing/decreasing the
longitude will move your point due east/west.
Degrees, minutes, seconds
Consider the co-ordinates shown on the Blackberry at the Dead Sea.
• N 31^0 38.664
• E 35^0 34.371
Since the world is circular (yes, I know it's almost a sphere, but work with me here!) it can be divided into 360 degrees. However because, as we discussed above, cartographers divided the earth in
to north/south, latitudes only go from from 0 to 90 degrees in the North or South. Longitudes go from 0 to 180 East or West.
How far is one degree of Latitude?
You'll notice that latitudes are parallel lines wherever you are on earth. It means since the earth is almost a sphere, for geocaching purposes one degree of latitude is the same anywhere on earth.
(OK the earth is slightly flat at the poles so latitude does vary but it's only a difference of about 1km between latitudes at the equator and the poles.)
A minute of latitude is 1/60th of a degree, and one second is 1/60th of a minute.
Geocaching uses Degrees, Minutes and Decimal minutes. You'll sometimes see it referred to as dd mm.mmm, for example in the setup menu of your GPS. Some puzzle caches will use Decimal Degrees to be
tricky and these are expressed as dd.mmmm.
So how far is one degree, minute and second?
• 1° of Latitude (1/360^th of the Earth's Polar circumference) is 110.5743 km (68.70768 miles)
• 1' (1 minute) of Latitude (1/60^th of 1°) is 1.8429 km (1.1451 miles)
• 1" (1 second) of Latitude (1/3600^th of 1°) is only 30.7151 m (100.771 feet)
• 0.1" (1/10^th second) of Latitude (1/36000^th of 1°) is only 3.07151 m (10.0771 feet)
So for co-ordinates in the Degree/Decimal Minute format (dd mm.mmm)...
• If the co-ordinates change by 1 degree you would have moved 110.6 kilometres
• If the co-ordinates change by 1 minute you would have moved 1844 metres
• If the co-ordinates change by 0.1 minutes you would have moved 184 metres
• If the co-ordinates change by .01 minutes you would have moved 18.4 metres
• If the co-ordinates change by .001 minutes you would have moved 1.84 metres.
So that third decimal place in the co-ordinates doesn't make a lot of difference - very useful information when you're trying to solve a puzzle cache. It's within the accuracy of your GPS. So you
don't need to solve that number to get within a search radius. Even the second decimal place isn't that far away. It will get you somewhere pretty close. This is one of the secrets of finding
puzzle caches.
In the puzzle you know the degrees are going to be the same as where you are (unless you're near a degree confluence). All you need to work out are the minutes and the first decimal point and you
can see that you will be very close. Just plot some numbers on Google Earth until you find somewhere that looks likely.
How far is one degree of Longitude?
Because lines of longitude converge as you get closer to the poles, the distance between longitudes decreases as you move away from the equator and towards the poles. Therefore the distance depends
on the latitude at which you are located.
For example at the equator, the distance between degree longitudes (say, between 150 and 151 degrees) is about 111.3 km. However at 35 degrees latitude that distance is only 91.2 km. Of course by
the time you reach the pole it's zero.
Rather than digging out your old school scientific calculator, try this distance calculator here. It accepts various formats, so just use the usual geocaching dd mm.mmm.
Below is a table of distances between degree and minute longitudes that are calculated for latitudes from 0 to 90 degrees.
For example, if you are at 35 degrees latitude, then the distance between one degree longitudes is 91.2 km. The difference between one minute of longitude is 1.52 km. The distances are also shown in
statute miles.
So you can now calculate distances for decimal minutes in the same way as for latitudes above i.e. At 35 degree latitude 0.1 minutes is 152 metres, 0.01 minutes is 15 metres and 0.001 minutes is 1.5
[table "2" not found /]
Converting from Decimal Degrees to Decimal Minutes
Sometimes, people use Decimal Degrees in their cache puzzles i.e. dd.mmmm You need to convert these co-ordinates into Decimal Minutes which is what is used for most geocaches. Conversely you may
want to convert from Decimal Minutes to Decimal Degrees.
It's actually very simple.
Since there are 60 minutes in a degree, to convert from Decimal Degrees to Minutes simply take the decimal part of the co-ordinate and multiply by 60. For example say the co-ordinate is S34.12345,
multiply 60 x .12345 = 7.407. So your co-ordinate would be S34 07.407.
To convert from Decimal Minutes to Degrees divide the minutes by 60. So using our previous example, divide 7.407/60 = .12345.
It's the same process for either latitude or longitude co-ordinates.
Let me know if there is anything else that you would like to know about co-ordinates.
Make Google Earth your friend
The best advice that I can give on how to find a Puzzle or Mystery geocache is to make Google Earth (GE) your friend.
Download it here.
Once you have worked out possible solutions, or partial solutions to puzzles, plug the co-ordinates into GE. The location will give you a good indication of whether you are correct or not. For
example if the co-ordinates are in someone's backyard, or in the middle of the water or somewhere else unlikely, then you're probably not correct. Eliminating possibilities in this way will save a
lot of driving. I have solved many mystery caches in this way even when I couldn't work out the total solution. You can often work out enough of the puzzle to get you there.
Often you only have to work out a few of the last digits to be able to solve it.
This doesn't always work, but it sure helps in a lot of cases.
Mystery Geocache Tutorial Video
Here is one of the best tutorial videos on solving Puzzle or Mystery geocaches that I have seen. It's by Zytheran based in Adelaide, Australia who is famous (infamous?) for his puzzle caches. He
gives you a lot of tips of narrowing down your focus.
Do you have any tips that you can share on how to find these types of geocaches.
Even though geocaching has been around for over 10 years, it's surprising to find that many people have never heard of it, or if they have, how to geocache.
So what is it?
Some people describe geocaching as a high-tech treasure hunting game.
Geocachers hide a container called a geocache (or cache for short) somewhere and record the co-ordinates of its location using a GPS receiver. These co-ordinates plus a description of the location
and possibly a hint about how to find it are entered into a huge online database at www.geocaching.com.
Other geocachers can then search for caches near a particular location – wherever they may be in the world. Every country in the world has at least one cache. Even Antarctica has caches.
If you want to participate you need to register at www.geocaching.com. A Basic Membership is free and allows you to view the co-ordinates of the geocaches. A Premium Membership is very economical
at US$30 per year and provides a wide range of useful features compared to Basic Membership.
You will need to buy or borrow a GPS receiver in order to find geocaches. However there are individuals who manage to find geocaches by referring to a map!
Many geocachers have become concerned about the increasingly poor quality of geocaches that are being placed.
This Progeocaching site is aims to show people how to geocache properly and help to put the quality back into the game. | {"url":"http://www.progeocaching.com/tag/puzzle-cache/page/2/","timestamp":"2014-04-17T12:30:50Z","content_type":null,"content_length":"46876","record_id":"<urn:uuid:f52b81d7-9842-46e0-9a10-d36925cc1aab>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00279-ip-10-147-4-33.ec2.internal.warc.gz"} |
Polygamma function: Series representations
Series representations
Generalized power series
Expansions at generic point z==z[0]
Expansions on branch cuts
General case
Expansions at z==0
For the function itself
General case
Special cases
Expansions at z==-m
For the function itself
General case
Special cases
Expansions of psi^(nu)(-m+Epsilon) at Epsilon==0
For the function itself
Special cases
Asymptotic series expansions
For the function itself
General case
Special cases
Residue representations
Other series representations | {"url":"http://functions.wolfram.com/GammaBetaErf/PolyGamma2/06/ShowAll.html","timestamp":"2014-04-19T09:35:24Z","content_type":null,"content_length":"65612","record_id":"<urn:uuid:c9b2feaa-4ec0-4845-ac6f-7e7a51381607>","cc-path":"CC-MAIN-2014-15/segments/1398223206672.15/warc/CC-MAIN-20140423032006-00343-ip-10-147-4-33.ec2.internal.warc.gz"} |
Greenbrae Science Tutor
Find a Greenbrae Science Tutor
...I worked a number of years as a data analyst and computer programmer and am well versed in communicating with people who have a variety of mathematical and technical skills.I have years of
experience in discrete math. I took a number of courses in the subject. I've used the concepts during my years as a programmer and have tutored many students in the subject.
49 Subjects: including electrical engineering, physics, calculus, geometry
Hi, I'm Mike. I've been a professional tutor since September of 2011, teaching English and mathematics to children from third to eighth grade and test prep, composition, comprehension, and
mathematics (through pre-calculus) and organization and study skills to high school and college students. I specialize in helping academically frustrated students.
29 Subjects: including ACT Science, philosophy, reading, English
...I play piano, guitar, bass, clarinet, and some drums. I took piano, clarinet, and music composition lessons at the renowned Eastman School of Music in Rochester, NY, in the 1990s. Since then, I
have taken more classes in music technology and composition at NYU, and I have taken both levels of a...
5 Subjects: including biology, biochemistry, genetics, piano
...Coached soccer to secondary aged players. I was a camp director and coach for 4 years. I have worked with children with Learning Disorders, Mental Retardation, Nonverbal learning disorders,
Autism spectrum disorders, and reading disabilities.
30 Subjects: including sociology, biology, psychology, reading
...I have tutored hundreds of students. Although many came to me afraid of math, most of them left with a well-earned sense of confidence and mastery. You will too.
14 Subjects: including biostatistics, geometry, statistics, GRE | {"url":"http://www.purplemath.com/greenbrae_ca_science_tutors.php","timestamp":"2014-04-20T20:00:45Z","content_type":null,"content_length":"23848","record_id":"<urn:uuid:f52f0a65-efc1-4ca1-9ee7-0ee020c9e679>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00507-ip-10-147-4-33.ec2.internal.warc.gz"} |
1a)Find an equation for the surface consisting of all points P for which the distance from P to the y axis is twice... - Homework Help - eNotes.com
1a)Find an equation for the surface consisting of all points P for which the distance from P to the y axis is twice the distance from P to the xz-plane. Identify the surface.
b)Find a vector equation and parametric equations for the line segment that joins P(a; b; c) to Q(u; v; w).
(a). Let P(x,y,z) be the point in XYZ-plane such that the distance from P to the y axis is twice the distance from P to the xz-plane i.e.
If `d_1=` distance from from P(x,y,z) to y-axis and
`d_2=` distance from P(x,y,z) to xz-plane .
By given condition
`d_1=2d_2` (i)
Substitute `d_1 and d_2` in (i), we have
(b). Equation of the line in vectors form joining two vectors `vecX_0 and vecX_1` is
`vecX=(1-lambda)vecX_0+lambdavecX_1 , lambda in[0,1]`
Parametric equations are
z=c+lambda(w-c) , lambda in [0,1].
Join to answer this question
Join a community of thousands of dedicated teachers and students.
Join eNotes | {"url":"http://www.enotes.com/homework-help/1a-find-an-equation-surface-consisting-all-points-460728","timestamp":"2014-04-19T12:18:47Z","content_type":null,"content_length":"26006","record_id":"<urn:uuid:92767872-e4d5-4564-920c-fd67efa9978d>","cc-path":"CC-MAIN-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00194-ip-10-147-4-33.ec2.internal.warc.gz"} |
Summary: Nuclear Physics B296 (1988) 710-716
North-Holland, Amsterdam
L.F. ABBOTT*
Departmentof Physics,Boston University,Boston,MA 02215t, USA
J. TRASCHEN*
DepartmentofAstronomyandAstrophysics, Universityof Chicago, Chicago,IL 60633, USA
Rui-Ming XU
Departmentof Physics, BrandeisUniversity, Waltham,MA 02254, USA
Received July 1987
Using a covariant formalism we derive the conditions for there to exist integral constraints on
energy-momentum perturbations in an arbitrary background spacetime.
A conserved quantity corresponding to an asymptotic symmetry of an otherwise
arbitrary spacetime manifold can be constructed in general relativity provided that a
number of conditions are met. If we write the spacetime metric in the form
&,~ = g~. + h~,, where h~,~ vanishes (sufficiently rapidly) at infinity but is otherwise
arbitrary, then the background asymptotic metric g~,~ must satisfy the Einstein
equations with no energy-momentum source term,
R~ - ½~"R = 0, (1)
where R~" and R are the Ricci tensor and scalar curvature corresponding to ~,.. In | {"url":"http://www.osti.gov/eprints/topicpages/documents/record/233/2965370.html","timestamp":"2014-04-20T03:27:11Z","content_type":null,"content_length":"8230","record_id":"<urn:uuid:58f79af9-8dba-4496-ba83-30d31aee2fd0>","cc-path":"CC-MAIN-2014-15/segments/1397609537864.21/warc/CC-MAIN-20140416005217-00399-ip-10-147-4-33.ec2.internal.warc.gz"} |
Post a reply
hi 1975rachel
Welcome to the forum.
I agree with bobbym. I'll try to show you why this is impossible.
The first time it doesn't matter who sits with whom, so let's get them seated and then give everyone a number as follows.
Now who will person 1 sit with, the next time around?
No one from 2-6, so I'll choose someone from the next table, let's say person 7. As those six folk can shuffle their numbers around 7-12, I'm not losing any generality by saying 7.
Now who will 1 and 7 sit with?
It'll have to be someone who wasn't on either of the first tables so let's choose person 13. Aagin I'm not losing generality by picking that number (it could equally be 14 or 15 ...)
So who will 1, 7 and 13 sit with?
I'll have to choose from a table that hasn't been picked yet so I'll have person 18, ... and to speed things up I'll pick 25 next.
So I've got 1, 7, 13, 18, 25.
Now you might think I've fiddled it by choosing these numbers. But my point is, to avoid putting people with folk they've already met, you have to choose one person from each table.
So, for example, 1, 8, 15, 22, 30 is another way of picking five to go on the first table in round two.
But I have only picked five people and the table must seat six. So who else can I choose?
As I have already picked one person from each of the tables, there's no one left who I can pick without putting someone with someone they've already met.
Conclusion: It is impossible. | {"url":"http://www.mathisfunforum.com/post.php?tid=18390&qid=240767","timestamp":"2014-04-18T13:23:10Z","content_type":null,"content_length":"25758","record_id":"<urn:uuid:23eab4e7-92c5-4d1b-ac60-2d40da3ffb5a>","cc-path":"CC-MAIN-2014-15/segments/1397609533689.29/warc/CC-MAIN-20140416005213-00433-ip-10-147-4-33.ec2.internal.warc.gz"} |
Business Statistics Tutorial
Online Business Statistics Tutorials
Free Online statistics course Help – Math help online
Need help with your online statistics course? Getting online statistics class help just got easier. StraighterLine has a video database of business statistics tutorials to get you the math help
online you are looking for. (Especially helpful for you self-paced students taking distance learning courses.) When you sign up for a StraighterLine class you also get free math online tutoring and a
course advisor.
How to Use Excel for Statistics What is the Difference Between Mean Median and Mode
Watch handfuls of other college business statistics help videos through these links:
What are Qualitative and Quantitative Variables What is the Confidence Interval of a Proportion
How to use Excel for Statistics Confidence Interval with a t-Distribution using Excel
What is Association and Causation What is the Sample Size Needed to Specify the Margin of Error
What is the Empirical Rule? What is the Hypothesis Testing Procedure
What is the Difference Between Mean Median and Mode? How to Conduct Hypothesis Testing with a Mean
What is Standard Deviation How to Conduct Hypothesis Testing with a Proportion
What is the Geometric Mean How to Conduct Hypothesis Testing with Two Sample Means
What is a Weighted Mean How to Conduct Hypothesis Testing with Two Sample Proportions
Contingency (2 Way) Tables ANOVA Sum of Squares
What is a Histogram? What is the ANOVA Testing Procedure
What are Quartiles and Boxplots? How to Compute the Treatment Means Difference Confidence Interval
What is a Scatter Diagram? How to Test the Variance Between Populations
What is Skewness How to Compute the Coefficient of Correlation
What are Permutations and Combinations Hypothesis Testing a Coefficient of Correlation
What is the Rule of Addition in Probability What is the Confidence Interval for a Regression Equation
What is the Rule of Multiplication in Probability What is the Regression Equation
What is a Tree Diagram How to Choose Variable in Multiple Regression
How to Calculate Probability in a Binomial Distribution using Excel What is a Correlation Matrix
How to Calculate Probability in a Normal Distribution using Excel How to create a Multiple Regression Equation
How to Calculate Probability in a Poisson Distribution using Excel What is The Global Test for Multiple Regression
How to Convert to a Standard Normal Distribution What is the Chi Square Hypothesis Test
What is the Mean of a Discrete Probability Distribution What is the Sign Test
What are Sampling Methods How to Use Contingency Table Analysis with Chi Square
What is the Central Limit Theorem What are Natural Tolerance Limits
Confidence Interval Example How to Use Control Chart Data
These math tutorials will help you pass your online college statistics class for college credit at StraighterLine. Sign up for our online business statistics course here and you get 10 hours of Free
SMARTHINKING online tutoring math help. These videos may also help you pass a CLEP test, DSST (Dantes), AP test, etc.
StriaghterLine’s Business Statistics Course Text: Lind, Douglas A., Marchal, William A. and Samuel A. Wathen. Basic Statistics for Business and Economics, 7th edition, McGraw-Hill/Irwin, 2010, ISBN:
9780077384470 [buy the text]
Click here for videos on Precalculus.
Click here for videos on Statistics.
Click here for videos on Algebra.
Click here for videos on Calculus | {"url":"http://www.straighterline.com/landing/online-business-statistics-video-tutorials/","timestamp":"2014-04-18T15:39:25Z","content_type":null,"content_length":"79390","record_id":"<urn:uuid:c0cdeec6-9dcd-4afd-a8be-e36c2581952a>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00026-ip-10-147-4-33.ec2.internal.warc.gz"} |
Seybold, FL Calculus Tutor
Find a Seybold, FL Calculus Tutor
...During my four years at UM, I spent a considerable amount of time tutoring, both one-on-one and in front of larger groups. I worked for the Academic Resource Center, which provides free
tutoring for undergraduate students, for three years tutoring both chemistry and calculus. As a senior, I was nominated for the Excellence in Tutoring Award.
14 Subjects: including calculus, chemistry, geometry, biology
...I have been very successful at accommodating diverse student needs by facilitating all styles of learners, offering individualized and extracurricular support and integrating effective
materials. I encourage learning by using a number of different manipulative, hands-on activities, and various f...
23 Subjects: including calculus, chemistry, physics, statistics
Hello there! With more than 15 years acting as a tutor, I tend to gravitate more to Science-based classes, specially Math (Algebra, Geometry, Trigonometry, Calculus I, II and III, Differential
Equations, Statistic, SAT Math, etc..) and Physics. I enjoy tutoring as it allows me to help students to ...
13 Subjects: including calculus, chemistry, physics, GRE
...During my time as an undergraduate student and later as a Physics Professor, I was formally responsible for teaching Physics and Math courses to University students. Personally, I believe that
such experience is very much in phase with my personality and has given me a very particular perspectiv...
11 Subjects: including calculus, Spanish, physics, algebra 1
...When it comes to tutoring I am very good at assisting students in Math and Computer Applications. I have had prior experience working at school to assist other students, but I have decided to
make a change in my environment, so I can enhance my capability in assisting others. I honestly think I...
16 Subjects: including calculus, reading, geometry, algebra 1 | {"url":"http://www.purplemath.com/Seybold_FL_Calculus_tutors.php","timestamp":"2014-04-16T05:06:51Z","content_type":null,"content_length":"24195","record_id":"<urn:uuid:7c260837-905d-4fcf-b1b4-4a6a925d00db>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00470-ip-10-147-4-33.ec2.internal.warc.gz"} |
Colloquium Seminar Schedule
September 14: Wonderful compactifications of groups as moduli spaces of principal bundles
Michael Thaddeus (Columbia University), thaddeus@math.columbia.edu
September 28: An introduction to essential dimension
Patrick Brosnan (University of Maryland), pbrosnan@umd.edu
October 21: Rigidity of group actions, cohomology and compactness
David Fisher (Indiana University, Bloomington), fisherdm@indiana.edu
October 28: A Filtration of Open/Closed Topological Field Theory
Ezra Getzler (Northwestern University), getzler@northwestern.edu
November 2: Rational billiards and the SL(2,R) action on moduli space
Alex Eskin (University of Chicago), eskin@math.uchicago.edu
November 9: Why and how do we use wavelets to study turbulence?
Marie Farge (Directrice de Recherche CNRS)
November 16: Mixed volume computation and solving polynomial systems
Tien-Yien Li (Michigan State University), li@math.msu.edu
December 2: Optimal and Practical Algebraic Solvers for Discretized PDEs Aziz Lecture
Jinchao Xu (Pennsylvania State University), xu@math.psu.edu
December 7: DMS to THRIVE
Sastry G. Pantula (National Science Foundation)
February 8: On the rigidity of black holes Douglis Lecture
Sergiu Klainerman (Princeton University )
February 17: Product formulas for positive measures and applications February Fourier Talks
Peter Jones (Yale University), jones@math.yale.edu
February 22: Semismooth Newton Methods: Theory, Numerics and Applications Aziz Lecture
Michael Hintermüller (Humboldt University, Berlin), hint@math.hu-berlin.de
February 29: On the size of the Navier - Stokes singular set
Walter Craig (McMaster University), craig@math.mcmaster.ca
March 14: De Giorgi methods applied to regularity issues in Fluid Mechanics
Alexis Vasseur (UT Austin )
March 30: Birkhoff Normal Form and a problem of Herman Dynamics Conference
Hakan Eliasson (University of Paris-6 and IAS), hakane@math.jussieu.fr
April 25: Lyapunov Functions: Towards an Aubry-Mather theory for homeomorphisms?
Albert Fathi (ENS-Lyon), afathi@ens-lyon.fr
May 2: Contractions of Lie Groups and Representation Theory
Nigel Higson (Penn State University), higson@math.psu.edu
May 9: TBA
Andrei Caldararu (University of Wisconsin, Madison), andreic@math.wisc.edu | {"url":"http://www.math.umd.edu/research/seminars/colloquium/","timestamp":"2014-04-16T18:57:35Z","content_type":null,"content_length":"11950","record_id":"<urn:uuid:84858cf4-69ef-4937-8d1b-3f4cdf221cb9>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00179-ip-10-147-4-33.ec2.internal.warc.gz"} |
Directed Graphs - Transition
Lecture #2: Directed Graphs - Transition Matrices
A graph is an object that consists of a non-empty set of vertices and another set of edges. When working with real-world examples of graphs, we sometimes refer to them as networks. The vertices are
often called nodes or points, while edges are referred to as links or lines. The set of edges may be empty, in which case the graph is just a collection of points.
Example 1:
A simple example of a graph with vertices 1,2,3,4, and directed edges from vertex 1 to vertex 2, vertex 3 to vertices 2 and 4, and vertex 4 to vertex 1.
In this lecture we will only work with directed graphs and real-world examples of those (Internet graphs), but for other properties of graphs we refer to Math Explorer's Club website. The central
example in this module is the web graph, in which web pages are represented as vertices and the links between them are represented as edges. An example of such a graph is a sub graph of the BGP
(Gateway Protocol) web graph, consisting of major Internet routers. It has about 6400 vertices and 13000 edges and it was produced by Ross Richardson and rendered by Fan Chung Graham.
Although Internet graphs are very large, having the number of vertices of the order 30 billion (and growing), all graphs in this module are considered finite (finite number of vertices and edges).
We say that two vertices i and j of a directed graph are joined or adjacent if there is an edge from i to j or from j and i. If such an edge exists, then i and j are its endpoints. If there is an
edge from i to j then i is often called tail, while j is called head. In Example 1, vertices 1 and 2 are joined because there is an edge from 1 to 2, while vertices 1 and 3 are not joined. There is
however no edge from node 2 to node 1. Notice that there can be no more than two edges between any two vertices. There is a strong relation between graphs and matrices, previously introduced in
Lecture 1. Suppose we are given a directed graph with n vertices. Then we construct an n × n adjacency matrix A associated to it as follows: if there is an edge from node i to node j, then we put 1
as the entry on row i, column j of the matrix A.
Example 2: Adjacency matrix for the graph in Example 1:
If one can walk from node i to node j along the edges of the graph then we say that there is a path from i to j. If we walked on k edges, then the path has length k. For matrices, we denote by A^k
the matrix obtained by multiplying A with itself k times. The entry on row i, column j of A^2 = A·A corresponds to the number of paths of length 2 from node i to node j in the graph. For Example 2,
the square of the adjacency matrix is
This means that there is a path from vertex 4 to vertex 2, because the entry on fourth row and second column is 1. Similarly there is a path from 3 to 1, as one can easily see from Example 1.
Fact 1: Consider a directed graph and a positive integer k. Then the number of directed walks from node i to node j of length k is the entry on row i and column j of the matrix A^k, where A is the
adjacency matrix.
In general, a matrix is called primitive if there is a positive integer k such that A^k is a positive matrix. A graph is called connected if for any two different nodes i and j there is a directed
path either from i to j or from j to i. On the other hand, a graph is called strongly connected if starting at any node i we can reach any other different node j by walking on its edges. In terms of
matrices, this means that if there is a positive integer k such that the matrix B = I + A + A^2 + A^3 + … +A^k is positive, then the graph is strongly connected. We add the identity matrix I in order
to deal with edges from a vertex to itself. In other words, if there is at least one path from node i to node j of length at most k, then we can travel from node i to j. Thus if matrix B has a
positive entry on row i and column j then it is possible to reach node j starting from i. If this happens for all nodes, then the graph is strongly connected.
One can easily see that the graph in Example 1 is connected, but not strongly connected because there is no edge from vertex 1 to vertex 3. For the matrix in Example 2, we notice that A^4 is a matrix
having only zeros, and so for all k greater than 4, A^k will be a matrix filled with zeros. Then for any k greater than 4, the matrix B = I + A + A^2 + A^3 + … +A^k is :
Since the matrix B is not positive, the graph in Example 1 is not strongly connected as we already saw.
Problem 1:
Construct the adjacency matrix of the graph below. Is this graph connected or strongly connected? How many paths of length 3 from node 3 to node 2 are there?
In the examples above we noticed that for every vertex i there is a number of edges that enter that vertex (i is a head) and a number of edges that exit that vertex (i is a tail). Thus we define the
indegree of i as the number of edges for which i is a head. Similarly, the outdegree of i as the number of edges for which i is a tail. For example, for the graph in the Problem 1, the indegree of
node 2 is 2 and the outdegree of node 1 is 1. The transition matrix A associated to a directed graph is defined as follows. If there is an edge from i to j and the outdegree of vertex i is d[i], then
on column i and row j we put i, row j with zero. Notice that we first look at the column, then at the row. We usually write i to an adjacent vertex j, thus obtaining a weighted graph. This will
become clear through the following example.
Example 3:
Consider the graph from Example 1 with weights on its edges as described above.
Then the transition matrix associated to it is:
Notice that the sum of the entries on the first column is 1. The same holds for the third and fourth column. In general, more is true.
Fact 2: For a strongly connected graph, the transition matrix is column-stochastic.
We use the transition matrix to model the behavior of a random surfer on a web graph. The surfer chooses a page at random, then follows its links to other web pages for as long as he/she wishes. At
each step the probability that the surfer moves from node i to node j is zero if there is no link from i to j and d[i] is the outdegree of vertex i. Initially the probability of each page to be
chosen as a starting point is
At step 1, the probability of each node to be visited after one click is A·v. At step 2, the probability of each node to be visited after two clicks is A^2·v. The probability of a page to be visited
at step k is thus A^k·v. If the matrix is primitive, column-stochastic, then this process converges to a unique stationary probability distribution vector p, where
The meaning of the i^th entry of p is that the surfer visits page i at any given time with probability p[i].
Problem 2: Let A be a transition matrix associated with a graph and B be a matrix of size n filled with x a positive real number smaller than 1. Then the matrix C = x·A + (1-x)·B is
column-stochastic. Show that this is true in the special case of Problem 1. | {"url":"http://www.math.cornell.edu/~mec/Winter2009/RalucaRemus/Lecture2/lecture2.html","timestamp":"2014-04-18T15:40:14Z","content_type":null,"content_length":"12830","record_id":"<urn:uuid:67e36baf-7fd4-4ac1-9701-16c8f973926b>","cc-path":"CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00440-ip-10-147-4-33.ec2.internal.warc.gz"} |
4.4 The Determinant and the Inverse of a Matrix
Home | 18.013A | Chapter 4 Tools Glossary Index Up Previous Next
4.4 The Determinant and the Inverse of a Matrix
The inverse of a square matrix M is a matrix, denoted as M^-1, with the property that M^-1 M = M M^-1 = I. Here I is the identity matrix of the same size as M, having 1's on the diagonal and 0's
In terms of transformations, M^-1 undoes the transformation produced by M and so the combination M^-1M represents the transformation that changes nothing.
The condition MM^-1 = I can be written as
when k and i are different, and these conditions completely determine the matrix M^-1 given M, when M has an inverse.
These equations have the same form as the two conditions (A) and (B) of section 4.3 except that det M is on the left-hand side in (A) instead of 1, and (-1)^i + jM[ij] appears in (A) and (B) instead
of M^-1[ji] here.
We can therefore divide both sides of (A) and (B) by det M, and deduce
Remember that here M[ij] is the determinant of the matrix obtained by omitting the i-th row and j-th column of M; the elements of M are the m[ij], while M^-1[ji] here represents the element of the
inverse matrix to M in j-th row and i-th column.
We can phrase this in words as: the inverse of a matrix M is the matrix of its cofactors, with rows and columns interchanged, divided by its determinant.
4.7 Compute the inverse of the matrix in Exercise 4.4 using this formula. Check the product M^-1M to be sure your result is correct.
4.8 Set up a spreadsheet that computes the inverse of any three by three matrix with non-zero determinant, using this formula.
(Hint: by copying the first two rows into a fourth and fifth row and the first two columns into a fourth and fifth column, you can make one entry and copy to get all of the (-1)^i + jM[ij] at once.
Then all that is left is rearranging to swap indices and dividing by the determinant (which is the dot product of any row of M with the corresponding cofactors).) | {"url":"http://ocw.mit.edu/ans7870/18/18.013a/textbook/HTML/chapter04/section04.html","timestamp":"2014-04-16T16:20:49Z","content_type":null,"content_length":"4403","record_id":"<urn:uuid:9af21262-69f5-474a-9017-a6b26064a079>","cc-path":"CC-MAIN-2014-15/segments/1397609524259.30/warc/CC-MAIN-20140416005204-00544-ip-10-147-4-33.ec2.internal.warc.gz"} |
[SOLVED] intersection between cylinder & plane
September 23rd 2009, 04:36 PM #1
Sep 2009
[SOLVED] intersection between cylinder & plane
Find a vector function that represents the curve of intersection of the cylinder $x^2+y^2=16$ and plane $x+z=5.$
I am stuck. I setting the equation equal to eachother but i could not solve for anything or get anything useful out of it.
I believe I am supposed to project the cylinder down so its a circle on the xy plane and then find the ellipse that the intersecion makes but i cannot figure out how.
Any help would be appreciated.
Last edited by snaes; September 23rd 2009 at 04:37 PM. Reason: make fancy equations with [math]
Find a vector function that represents the curve of intersection of the cylinder $x^2+y^2=16$ and plane $x+z=5.$
I am stuck. I setting the equation equal to eachother but i could not solve for anything or get anything useful out of it.
I believe I am supposed to project the cylinder down so its a circle on the xy plane and then find the ellipse that the intersecion makes but i cannot figure out how.
Any help would be appreciated.
I would solve for each variable in terms of $x$.
Let x=t and now you have the vector function
$\left\langle t,\pm\sqrt{16-t^2},5-t\right\rangle$, $-4\leq t\leq 4$
Thanks for trying.
Trying? Is that not the right answer?
Ah yes its in a different form. The form i was looking at had sin(x) which when converted would work out. Thank!
September 23rd 2009, 05:02 PM #2
September 23rd 2009, 06:55 PM #3
Sep 2009
September 23rd 2009, 06:57 PM #4
September 24th 2009, 09:48 AM #5
Sep 2009
September 24th 2009, 10:43 AM #6 | {"url":"http://mathhelpforum.com/calculus/103961-solved-intersection-between-cylinder-plane.html","timestamp":"2014-04-23T21:00:49Z","content_type":null,"content_length":"46773","record_id":"<urn:uuid:2fc10b15-2380-4a10-be87-151ed1042da8>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00382-ip-10-147-4-33.ec2.internal.warc.gz"} |
Where am i going wrong on taking the derivative?
August 5th 2009, 08:37 AM
Where am i going wrong on taking the derivative?
The problem is $\frac{t^8+9t+3}{t^2}$
I used the qoutent rule and got $\frac{t^2(8t^7+9)-t^8+9t+3(2t)}{(t^2)^2}$
I ended with $\frac{3t(2t^8+9t+2)}{t^4}$
This work isn't looking like the answer at all. Where am I going wrong?
August 5th 2009, 08:41 AM
The = in the initial problem is supposed to be a +, I think? Anyway, when you did the Quotient Rule, you forgot to multiply the terms in the numerator (2nd part)....
August 5th 2009, 08:58 AM
The middle work I left out is $(8t^9+9t^2)-2T^9+18t^2+6t$
doing subtraction gives you $\frac{6t^9+27t^2+6t}{(t^2)^2}$
I did the multiplication in the numerator.
August 5th 2009, 09:12 AM
I don't follow that... but what I'm saying is, the Quotient Rule gives
which is not what you had.
August 5th 2009, 09:16 AM
August 5th 2009, 09:43 AM
August 5th 2009, 09:50 AM
August 5th 2009, 09:54 AM
August 5th 2009, 10:09 AM
As you can see, you absolutely did not have it posted that way, and that is the beginning of where your calculations went wrong.
August 5th 2009, 10:12 AM
August 5th 2009, 10:20 AM
If your answer is the one above, then I'd have to say no. If you've revised your answer, then you need to provide it, as I haven't seen you give another.
August 5th 2009, 10:26 AM
Is the answer $6t^5+\frac{9}{t^2}+\frac{6}{t^3}$?
August 5th 2009, 10:38 AM
August 5th 2009, 10:40 AM
No. You have asked where your calculations went wrong, and I have shown you over and again where it is. As I said, you have not done the multiplication in the numerator correctly, because you've
neglected to distribute the -1. So, once again, the numerator is
August 5th 2009, 10:43 AM | {"url":"http://mathhelpforum.com/calculus/97070-where-am-i-going-wrong-taking-derivative-print.html","timestamp":"2014-04-18T12:43:59Z","content_type":null,"content_length":"28179","record_id":"<urn:uuid:4b37b2b4-8b32-46b4-99b1-16e6ddf02168>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00489-ip-10-147-4-33.ec2.internal.warc.gz"} |
Got Homework?
Connect with other students for help. It's a free community.
• across
MIT Grad Student
Online now
• laura*
Helped 1,000 students
Online now
• Hero
College Math Guru
Online now
Here's the question you clicked on:
Help me! Soybean meal is 16% protein,cornmeal is 8% protein. How many pounds of each should be mixed together in order to get 320lbs mixture that is 12% protein?
Best Response
You've already chosen the best response.
Making the assumption that only soybean meal and cornmeal are added to the mixture. We get x = amount of soybean meal and y = amount of cornmeal x+y = 320. solving for y we get y =320 -x. We also
know that 0.16(x) + .08(y) = 320*(.12) or .16x + 0.08(320-x) = 38.4 simplify 0.08x + 25.6 = 38.4 0.08x = 12.8 x = 160 subsituting into x+y = 320 we get y = 160 So lbs of cornmeal = 160 and lbs of
soybeam meal is 160
Your question is ready. Sign up for free to start getting answers.
is replying to Can someone tell me what button the professor is hitting...
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• Engagement 19 Mad Hatter
• You have blocked this person.
• ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/updates/4e447bdd0b8b3609c71ffbd7","timestamp":"2014-04-18T19:03:12Z","content_type":null,"content_length":"28087","record_id":"<urn:uuid:f5ed91dd-c8d9-4e89-a61d-ae77eae1ec38>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00239-ip-10-147-4-33.ec2.internal.warc.gz"} |
Computer Organization and Design The Hardware Software Interface 4th Edition Chapter 3.15 Solutions | Chegg.com
Fraction to Binary Conversion in Floating Point Format
Case a:
Consider the following fraction
Fractional part 1/2 to binary
Step 1: Convert fractional part to decimal part i.e., 0.5.
Step 2: Decimal fraction multiply with 2, now the value becomes 1.
The first binary digit is 1 at this point.
Step 3: Place 1 in most significant bit and remaining 23 bits are filled with 0’s
Mantissa is used to store fractional binary bit pattern.
The result stored in mantissa
Hexadecimal representation
Hence, the bit pattern in mantissa
Therefore, this fraction is the exact representation using 24 bits. | {"url":"http://www.chegg.com/homework-help/computer-organization-and-design-4th-edition-chapter-3.15-solutions-9780123744937","timestamp":"2014-04-17T23:07:17Z","content_type":null,"content_length":"34924","record_id":"<urn:uuid:b2ca517a-a52c-474d-9417-14679f97d9ac>","cc-path":"CC-MAIN-2014-15/segments/1398223201753.19/warc/CC-MAIN-20140423032001-00465-ip-10-147-4-33.ec2.internal.warc.gz"} |
Apple Support Communities
727 Views 23 Replies Latest reply
: Jan 5, 2013 10:39 AM by Barry
• you can determine how many rows there are in a column by using the function ROWS(). You can then use the offset() function to retrieve the value using the rows() function.
something like:
in the table on the left:
A1=OFFSET(Data Table :: $A$1, ROWS(Data Table :: A)-1, 0)
The table on the right contains the column you "regularly" update.
I hope this helps
• Thank you Jerrold for your reply. I have seen your similar posts and have tried the offset command. Does the formula have to be in the footer? I am not familiar with putting formulas in footers.
All my data is in column d. I would like the value to appear in cell B4. In converting my spreadsheets from excel to numbers this formula did not translate well. In excel I used the following:
=index(D:D, Match(9.9999999999999E+307,D:D))
That formula put the last value in column D into whatever cell I put the formula.
• Thank you Wayne for the reply. Is there a simple formula I can put in one cell that will get the value?
• trexo,
I provided a formula. Can you provide any context as to specifically what you are trying to do OR the Excel formula you are using?
• Tried the formula you suggested and it returned "This formula contains invalid reference." I included my excel formula in an earlier response. I have a column of numbers that gets added to
regularly. The content starts at D7 and goes down. I am trying to create a formula in B4 that will always contain the last number in column D.
Forgive my difficulty in making this conversion...
Tried the numbers manual but it was no help.
• I am using the offset formula but it is returning the cell value 6 up from the bottom. I have tried the formula on two different pages of data with the same result.
• trexo,
you have to name the tables the same as I did... the right table was titled "Data Table". If your table is not named that AND you copied the formula I posted you will get the error your reported
because the table you are attempting reference, in fact, does not exist so the reference is invalid.
This will work better for you-- I hope
B4=OFFSET($D$7, ROWS(D)-7, 0)
trexo wrote:
Thank you Jerrold for your reply. I have seen your similar posts and have tried the offset command. Does the formula have to be in the footer? I am not familiar with putting formulas in
footers. All my data is in column d. I would like the value to appear in cell B4. In converting my spreadsheets from excel to numbers this formula did not translate well. In excel I used the
=index(D:D, Match(9.9999999999999E+307,D:D))
That formula put the last value in column D into whatever cell I put the formula.
Yikes, I can see why you weren't excited about using that formula. You can put that formula of mine anywhere you wish. I'm making the assumption that you would be capable of adjusting the formula
for the actual location of your data. Header and Footer rows are handy tools for isolating the data from the labels and summaries. Crack open the Numbers User Guide and become familiar with some
of the interesting features in Numbers that make life easier than with Excel. Headers and Footers are quite useful.
• trexo,
Jerry and I both presented solutions but now, after looking at Jerry's solution, I am not sure which one is the one you are looking for... the distinction is that Jerry's will return the last
value from a column even if it is not in the last cell of the column. Mine will only return the value in the last actual cell of the column. My guess is that Jerry's solution is the one you want
since it is more general-- that is if the last cell in the column is filled our formulas will both work.
• Hi trexo,
Jerry's original solution, reproduced here, may be placed in any cell on the same table as the list of values in column B.
=OFFSET(B1, COUNT(B), 0)
It assumes that the first value in the list is in cell B2 (one row below the label in Row 1), and that (as you specified) there are no gaps in the data from that cell to the last one containing a
COUNT(B) counts the numerical values entered in column B; the result is used as the row-offset in the OFFSET function. Zero ( 0 ), the column-offset value in the formula, means 'the same column
as the base, B1'.
Your Excel formula:
=index(D:D, Match(9.9999999999999E+307,D:D))
MATCH returns the position of the specified value in a list of values in a range.
The range here is column D, and the value to match (expressed as a base 10 number) is
99999999999999 followed by 294 zeroes, a rather large number, likely to be far greater than the largest number in column D.
MATCH in numbers can be set to find the largest close match, the smallest close match, or an exact match, with 'largest close match' the default choice. My assumption is that this is also true in
Excel. If so, the expression in your Excel formula is looking for the "largest number in column D that is smaller than or equal to the search value."
IF the last number in the column is also the largest number in the column (and that number is smaller than 9.9999999999999E+307), then the formula will return the correct value. If the last value
in column D is NOT also the largest value in column D, OR is larger than the search value, then the formula will not return the 'last value' in column D.
A Numbers similar to this would be:
Here's an example, showing the results. The formula above is in B2; the same formula with the cell references changed to E1 and $E is in B3:
Note that both return the largest value in their column, not necessarily the 'last' value in that column.
• Does it matter that I am doing this on numbers on my ipad using icloud? Wayne...I created a new spread sheet and reproduced your exact scenario and it returns a value of 0. As I create the
formula numbers is supplying the steps. I am trying to turn that feature off so I can have complete control but cannot find that functionality.
Thank you for your patience!
• What is in the last cell of the column? If that cell is empty then the formula will return "0"
• Trexo,
If my solution doesn't work for you, I'm curious why that is the case. Could you please describe what you expect, and what happened when you tried my solution.
Kind regards
• This solved my question - 10 points
• This helped me - 5 points | {"url":"https://discussions.apple.com/message/20763222","timestamp":"2014-04-18T15:10:55Z","content_type":null,"content_length":"137403","record_id":"<urn:uuid:fca9680c-f878-4e2d-b025-06ffb582742e>","cc-path":"CC-MAIN-2014-15/segments/1397609533689.29/warc/CC-MAIN-20140416005213-00489-ip-10-147-4-33.ec2.internal.warc.gz"} |
A problem on Algebraic Number Theory, Norm of Ideals
up vote 9 down vote favorite
A problem on Algebraic Number Theory
K and L are number fields over Q.(Q is rational number filed) K is a subfield of L.
O_K is the integers of K. and O_L is the integers of L.
P is a prime ideal of O_L. p is a prime ideal of O_K. P is over p.
The residue class degree f is defined to be f=[O_L/P:O_K/p]. The norm of P is Norm(P)=p^f
This is the usual definition of Norm of an ideal.(See Serre's Local fields and Serge Lang's Algebraic Number Theory)
Swinnerton-Dyer's A Brief Guide to Algebraic Number Theory has a different definition on Norm of an ideal.(Page 25)
if A is an ideal of O_L, Norm(A)= ideal in O_K generated by elements Norm(a) where a is in A.
I dont know why these two definitions are the same. Swinnerton-Dyer claims so in his book. Can anyone here give a hint, an explanation or anything else?
algebraic-number-theory number-fields nt.number-theory
A very basic hint: The first norm you define is an integer; the second is an ideal. So they are not literally equal. One relation between them is that, when K is Q, the ideal is generated by the
integer. – David Speyer Dec 18 '09 at 3:44
The first norm is also an ideal. Norm (P)=p^f where p is a prime ideal. Both definitions are ideals. – 7-adic Dec 18 '09 at 4:03
Oh, I see. OK, forget that then. I seem to be making a lot of stupid mistakes tonight; maybe I should stop doing math for a bit. – David Speyer Dec 18 '09 at 5:14
It seems to me like this should follow easily from the fact that the norm is transitive in towers and that the residue degree is multiplicative in towers. I would look at the tower L > K > Q. –
Adam Topaz Dec 18 '09 at 16:36
add comment
4 Answers
active oldest votes
Ok, here's the argument: First recall that the usual norm for non-zero elements of a field is transitive in towers; thus the same is true for your second definition of the norm of an
ideal. In particular, $N_{K|Q}\circ N_{L|K} = N_{L|Q}$. The fact that the norm $N_{L|Q}(\mathfrak{P}) = [\mathcal{O}_L:\mathfrak{P}] \cdot \mathbb{Z}$ is easy to see for a prime $\
mathfrak{P}$ in $\mathcal{O}_L$; edit: and thus the same is true for any integral ideal $\mathfrak{a}$. Now let $\mathfrak{p} = \mathcal{O}_K \cap \mathfrak{P}$ and $(p) = \mathbb{Z} \
cap \mathfrak{P}$.
up vote 3 We have $N_{L|Q}(\mathfrak{P}) = p^{f(\mathfrak{P}|p)} = N_{K|Q}N_{L|K}\mathfrak{P}$. In particular, we deduce that $N_{L|K}\mathfrak{P} = \mathfrak{p}^d$ for some $d$. Moreover, we know
down vote that
$N_{K|Q}\mathfrak{p}^d = p^{d \cdot f(\mathfrak{p}|p)} = p^{f(\mathfrak{P}|p)}.$
But then $d = f(\mathfrak{P}|p) / f(\mathfrak{p}|p) = f(\mathfrak{P}|\mathfrak{p})$ as required.
I am coming to this one somewhat late, but the solution is incorrect. First of all, the fact that the norm map on field extensions is transitive does not so easily imply the norm map
1 as defined by Swinnerton-Dyer is transitive. The problem is that the definition commands you to take the norm of all elements in the ideal, and not all elements in the ideal of ${\
mathcal O}_K$ generated by norms from an ideal in ${\mathcal O}_L$ are going to be norms from ${\mathcal O}_L$. You can't compute a norm by Sw-D's defn by using any generating set:
the ideal $(1+2i,1-2i)$ in ${\mathbf Z}[i]$ (contd). – KConrad May 23 '12 at 3:42
is (1), so it has norm (1) in ${\mathbf Z}$, but note the numbers $1+2i$ and $1-2i$ have norm 5, hence the ideal in ${\mathbf Z}$ generated by their norms is (5), which is not the
norm of $(1+2i,1-2i)$. Although it's true that the norm map on ideals is transitive, it definitely requires real work to explain. The second error is that from the equation $p^{f({\
mathfrak P}|p)} = {\rm N}_{K/{\mathbf Q}}N_{L/K}({\mathfrak P})$, we can't conclude that ${\rm N}_{L/K}({\mathfrak P})$ is some power of ${\mathfrak p}$: consider the equation $5 = {\
rm N}_{{\mathbf Q}(i)/{\mathbf Q}}(\alpha)$. (contd.) – KConrad May 23 '12 at 3:46
You can't tell me what $(\alpha)$ is for sure. Maybe it's $(1+2i)$, maybe it's $(1-2i)$. Knowing some ideal has a power of $p$ as its norm tells you that the ideal is a product of
powers of primes lying over $p$, but that is all. You definitely can't reason just from the norm equation in the 2nd paragraph that ${\rm N}_{L/K}({\mathfrak P})$ is a power of $\
mathfrak p$. Again, the conclusion is correct but the reasoning is incorrect (or, if I'm misunderstanding the reasoning, at the very least there is a very substantial gap). While the
two definitions of ideal norm are equivalence (contd). – KConrad May 23 '12 at 3:50
-- excuse me, are equivalent -- the proof of the equivalence is not as simple as the solution here suggests it is. – KConrad May 23 '12 at 3:50
add comment
Here is a proof that the ideal norm as defined in the books by Serre and Lang is equal to the ideal norm as defined in Swinnerton-Dyer's book. We will start from the definition given by Serre
and Lang, state some of its properties, and use those to derive the formula as given by Swinnerton-Dyer.
Background: Let $A$ be a Dedekind domain with fraction field $K$, $L/K$ be a finite separable extension, and $B$ be the integral closure of $A$ in $L$. For any prime $\mathfrak P$ in $B$ we
define ${\rm N}_{B/A}({\mathfrak P}) = \mathfrak p^f$, where $f = f({\mathfrak P}|{\mathfrak p})$ is the residue field degree of $\mathfrak P$ over $\mathfrak p$, and this norm function is
extended to all nonzero ideals of $B$ by multiplicativity from its definition on (nonzero) primes in $B$.
1) The map ${\rm N}_{B/A}$ is multiplcative (immediate from its definition).
2) Good behavior under localization: for any (nonzero) prime ${\mathfrak p}$ in $A$, ${\rm N}_{B/A}({\mathfrak b})A_{\mathfrak p} = {\rm N}_{B_{\mathfrak p}/A_{\mathfrak p}}({\mathfrak b}B_{\
mathfrak p})$. Note that $A_{\mathfrak p}$ is a PID and $B_{\mathfrak p}$ is its integral closure in $L$; the ideal norm on the right side is defined by the definition above for Dedekind
domains, but it's more easily computable because $B_{\mathfrak p}$ is a finite free $A_{\mathfrak p}$-module on account of $A_{\mathfrak p}$ being a PID and $L/K$ being separable. The proof
of this good behavior under localization is omitted, but you should find it in books like those by Serre or Lang.
3) For nonzero $\beta$ in $B$, ${\rm N}_{B/A}(\beta{B}) = {\rm N}_{L/K}(\beta)A$, where the norm of $\beta$ on the right is the field-theoretic norm (determinant of multiplication by $\beta$
up vote as a $K$-linear map on $L$). To prove this formula, it is enough to check both sides localize the same way for all (nonzero) primes $\mathfrak p$: ${\rm N}_{B_{\mathfrak p}/A_{\mathfrak p}}(\
9 down beta{B}_{\mathfrak p}) = N_{L/K}(\beta)A_{\mathfrak p}$ for all $\mathfrak p$. If you know how to prove over the integers that $[{\mathcal O}_F:\alpha{\mathcal O}_F] = |{\rm N}_{F/{\mathbf
vote Q}}(\alpha)|$ for any number field $F$ then I hope the method you know can be adapted to the case of $B_{\mathfrak p}/A_{\mathfrak p}$, replacing ${\mathbf Z}$ with the PID $A_{\mathfrak p}$.
That is all I have time to say now about explaining the equality after localizing.
Now we are ready to show ${\rm N}_{B/A}({\mathfrak b})$ equals the ideal in $A$ generated by all numbers ${\rm N}_{E/F}(\beta)$ as $\beta$ runs over $\mathfrak b$.
For any $\beta \in \mathfrak b$, we have $\beta{B} \subset \mathfrak b$, so ${\mathfrak b}|\beta{B}$. Since ${\rm N}_{B/A}$ is multiplicative, ${\rm N}_{B/A}({\mathfrak b})|{\rm N}_{E/F}(\
beta)A$ as ideals in $A$. In particular, ${\rm N}_{E/F}(\beta) \in {\rm N}_{B/A}({\mathfrak b})$. Let $\mathfrak a$ be the ideal in $A$ generated by all numbers ${\rm N}_{E/F}(\beta)$, so we
have shown $\mathfrak a \subset {\rm N}_{B/A}(\mathfrak b)$, or equivalently ${\rm N}_{B/A}(\mathfrak b)|\mathfrak a$. To prove this divisibility is an equality, pick any prime power ${\
mathfrak p}^k$ dividing $\mathfrak a$. We will show ${\mathfrak p}^k$ divides ${\rm N}_{B/A}(\mathfrak b)$.
To prove ${\mathfrak p}^k$ divides ${\rm N}_{B/A}(\mathfrak b)$ when ${\mathfrak p}^k$ divides $\mathfrak a$, it suffices to look in the localization of $A$ at $\mathfrak p$ and prove ${\
mathfrak p}^kA_{\mathfrak p}$ divides ${\rm N}_{B/A}(\mathfrak b)A_{\mathfrak p}$, which by the 2nd property of ideal norms is equal to ${\rm N}_{B_{\mathfrak p}/A_{\mathfrak p}}(\mathfrak b
{B_{\mathfrak p}})$. Since $B_{\mathfrak p}$ is a PID, the ideal ${\mathfrak b}B_{\mathfrak p}$ is principal: let $x$ be a generator, and we can choose $x$ to come from $\mathfrak b$ itself.
By the 3rd property of ideal norms, ${\rm N}_{B_{\mathfrak p}/A_{\mathfrak p}}(xB_{\mathfrak p}) = {\rm N}_{E/F}(x)A_{\mathfrak p}$. Showing ${\mathfrak p}^kA_{\mathfrak p}$ divides ${\rm N}_
{E/F}(x)A_{\mathfrak p}$ is the same as showing ${\rm N}_{E/F}(x) \in {\mathfrak p}^kA_{\mathfrak p}$. Since $x$ is in in $\mathfrak b$, ${\rm N}_{E/F}(x) \in \mathfrak a \subset {\mathfrak
p}^k$, so ${\rm N}_{E/F}(x) \in {\mathfrak p}^kA_{\mathfrak p}$. QED
add comment
I thought I should know this but then i ended up looking it up. Its in Lang's Algebraic Nubmer Theory pages 24-26 (at least for A principal, but that should be enough). That is if you
up vote 1 want to know the proof. I have no idea where the intuition comes from but I bet it is using lattices somehow.
down vote
add comment
Hmm... I can see offhand how to deal with it if L/K is Galois, but I'd have to think about it otherwise... In the Galois case, above p you have r many prime ideals, each with ramification
index e, and residue degree f. The rough sketch is to view this as a problem about discrete valuations, rather than prime ideals.
N(P) (according to your second definition) = < N(a)| a in P >. We know this is an ideal in O_K, and it only remains to describe its decomposition into primes. Since the ramification index
of p in (each) P above it is e, the minimal p-adic valuation of an element in N(P) is f. So if t is a parametrizing element of the p-adic valuation (choose it in O_K), then u*t^f generates
up vote 1 N(P)[p] where u is in O_K - P (check that N(P) isn't divisible by other prime ideals, with similar methods).
down vote
Hope that helps a bit with the intuition.
After reading Adam's solution, I noticed a few things were wrong in my argument. They were corrected in the body.
add comment
Not the answer you're looking for? Browse other questions tagged algebraic-number-theory number-fields nt.number-theory or ask your own question. | {"url":"http://mathoverflow.net/questions/9235/a-problem-on-algebraic-number-theory-norm-of-ideals?sort=newest","timestamp":"2014-04-19T17:30:55Z","content_type":null,"content_length":"79444","record_id":"<urn:uuid:75d8b733-8274-49bf-867b-22ebec506a5b>","cc-path":"CC-MAIN-2014-15/segments/1398223201753.19/warc/CC-MAIN-20140423032001-00279-ip-10-147-4-33.ec2.internal.warc.gz"} |
omological algebra
Results 1 - 10 of 26
- Adv. in Math , 1998
"... ABSTRACT. We begin by showing that in a triangulated category, specifying a projective class is equivalent to specifying an ideal I of morphisms with certain properties, and that if I has these
properties, then so does each of its powers. We show how a projective class leads to an Adams spectral seq ..."
Cited by 41 (5 self)
Add to MetaCart
ABSTRACT. We begin by showing that in a triangulated category, specifying a projective class is equivalent to specifying an ideal I of morphisms with certain properties, and that if I has these
properties, then so does each of its powers. We show how a projective class leads to an Adams spectral sequence and give some results on the convergence and collapsing of this spectral sequence. We
use this to study various ideals. In the stable homotopy category we examine phantom maps, skeletal phantom maps, superphantom maps, and ghosts. (A ghost is a map which induces the zero map of
homotopy groups.) We show that ghosts lead to a stable analogue of the Lusternik–Schnirelmann category of a space, and we calculate this stable analogue for low-dimensional real projective spaces. We
also give a relation between ghosts and the Hopf and Kervaire invariant problems. In the case of A ∞ modules over an A ∞ ring spectrum, the ghost spectral sequence is a universal coefficient spectral
sequence. From the phantom projective class we derive a generalized Milnor sequence for filtered diagrams of finite spectra, and from this it follows that the group of phantom maps from X to Y can
always be described as a lim1 ←− group. The last two sections focus
, 2000
"... this paper: ..."
- REPRINTS IN THEORY AND APPLICATIONS OF CATEGORIES , 2003
"... ..."
"... known in module theory that any A-bimodule B is an A-ring if and only if the functor − ⊗A B: MA → MA is a monad (or triple). Similarly, an A-bimodule C is an A-coring provided the functor − ⊗A
C: MA → MA is a comonad (or cotriple). The related categories of modules (or algebras) of − ⊗A B and comodu ..."
Cited by 12 (10 self)
Add to MetaCart
known in module theory that any A-bimodule B is an A-ring if and only if the functor − ⊗A B: MA → MA is a monad (or triple). Similarly, an A-bimodule C is an A-coring provided the functor − ⊗A C: MA
→ MA is a comonad (or cotriple). The related categories of modules (or algebras) of − ⊗A B and comodules (or coalgebras) of − ⊗A C are well studied in the literature. On the other hand, the right
adjoint endofunctors HomA(B, −) and HomA(C, −) are a comonad and a monad, respectively, but the corresponding (co)module categories did not find
, 2007
"... Abstract. We compare and contrast various relative cohomology theories that arise from resolutions involving semidualizing modules. We prove a general balance result for relative cohomology over
a Cohen-Macaulay ring with a dualizing module, and we demonstrate the failure of the naive version of bal ..."
Cited by 8 (5 self)
Add to MetaCart
Abstract. We compare and contrast various relative cohomology theories that arise from resolutions involving semidualizing modules. We prove a general balance result for relative cohomology over a
Cohen-Macaulay ring with a dualizing module, and we demonstrate the failure of the naive version of balance one might expect for these functors. We prove that the natural comparison morphisms between
relative cohomology modules are isomorphisms in several cases, and we provide a Yoneda-type description of the first relative Ext functor. Finally, we show by example that each distinct relative
cohomology construction does in fact result in a different functor.
"... Abstract. Bivariant (equivariant) K-theory is the standard setting for noncommutative topology. We may carry over various techniques from homotopy theory and homological algebra to this setting.
Here we do this for some basic notions from homological algebra: phantom maps, exact chain complexes, pro ..."
Cited by 6 (2 self)
Add to MetaCart
Abstract. Bivariant (equivariant) K-theory is the standard setting for noncommutative topology. We may carry over various techniques from homotopy theory and homological algebra to this setting. Here
we do this for some basic notions from homological algebra: phantom maps, exact chain complexes, projective resolutions, and derived functors. We introduce these notions and apply them to examples
from bivariant K-theory. An important observation of Beligiannis is that we can approximate our
, 810
"... Abstract. We define the filtrated K-theory of a C ∗-algebra over a finite topological space X and explain how to construct a spectral sequence that computes the bivariant Kasparov theory over X
in terms of filtrated K-theory. For finite spaces with totally ordered lattice of open subsets, this spect ..."
Cited by 2 (0 self)
Add to MetaCart
Abstract. We define the filtrated K-theory of a C ∗-algebra over a finite topological space X and explain how to construct a spectral sequence that computes the bivariant Kasparov theory over X in
terms of filtrated K-theory. For finite spaces with totally ordered lattice of open subsets, this spectral sequence becomes an exact sequence as in the Universal Coefficient Theorem, with the same
consequences for classification. We also exhibit an example where filtrated K-theory is not yet a complete invariant. We describe two C ∗-algebras over a space X with four points that have isomorphic
filtrated K-theory without being KK(X)-equivalent. For this space X, we enrich filtrated K-theory by another K-theory functor to a complete invariant up to KK(X)-equivalence that satisfies a
Universal Coefficient Theorem. 1. | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=161910","timestamp":"2014-04-20T14:07:45Z","content_type":null,"content_length":"31948","record_id":"<urn:uuid:fbd8cc39-f5d8-435e-93f7-6b2ebd904167>","cc-path":"CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00499-ip-10-147-4-33.ec2.internal.warc.gz"} |
P. Cousot & R. Cousot,
‘A la Floyd’ induction principles for proving
inevitability properties of programs
P{.} Cousot and R{.} Cousot.
\newblock 'A la Floyd' induction principles for proving inevitability
properties of programs.
\newblock In \emph{Algebraic methods in semantics}, M{.} Nivat and J{.}
Reynolds (Eds.), Cambridge University Press, Cambridge, UK , pp{.}
277--312, December 1985.
author = {Cousot, P{.} and Cousot, R{.}},
title = {'A la Floyd' induction principles for proving inevitability
properties of programs},
booktitle = {Algebraic methods in semantics},
editor = {M{.} Nivat and J{.} Reynolds},
publisher = {Cambridge University Press, Cambridge, UK},
pages = {277--312},
month = dec,
year = {1985}, | {"url":"http://cs.nyu.edu/~pcousot/COUSOTpapers/AMS85.shtml","timestamp":"2014-04-19T01:48:47Z","content_type":null,"content_length":"3348","record_id":"<urn:uuid:2a3c5726-8270-41a9-9a74-c85a2e717ff8>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00631-ip-10-147-4-33.ec2.internal.warc.gz"} |
nglos324 - tetragonal
Two unit cells are associated with the tetragonal system. The primative cell (P) has fractional lattice points at each cell corner for a total of one lattice point per cell. The body-centered cell
(I) has fractional lattice points at each cell corner and an additional lattice point at the cell center for a total of two lattice points per cell.
For this lattice a = b < c ; a = b = g = 90^0. | {"url":"http://www.princeton.edu/~maelabs/mae324/glos324/tetragonal.htm","timestamp":"2014-04-16T23:45:16Z","content_type":null,"content_length":"5156","record_id":"<urn:uuid:435c7352-eb7f-47c8-a0d8-8b0b21f16be2>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00319-ip-10-147-4-33.ec2.internal.warc.gz"} |
Statistical Analyses
Statistical Analyses
One of the most common uses of R is for statistical analysis of data. There are a number of references available describing analyses such as regression, analysis of variance, nonparametric analyses,
and resampling approaches in R. Applications of statistical analyses are also illustrated in the following Plant Health Instructor documents: Modeling Plant Disease Progress Over Time, Modeling
Dispersal Gradients,Introduction to Spatial Analysis, Disease Forecasting and Validation for Plant Pathology.
Next: Writing Functions | {"url":"http://www.apsnet.org/edcenter/advanced/topics/EcologyAndEpidemiologyInR/IntroductionToR/Pages/StatisticalAnalyses.aspx","timestamp":"2014-04-17T18:33:37Z","content_type":null,"content_length":"71842","record_id":"<urn:uuid:e539032b-b12d-4e4b-bfc6-a51fd6e4d19a>","cc-path":"CC-MAIN-2014-15/segments/1397609537308.32/warc/CC-MAIN-20140416005217-00415-ip-10-147-4-33.ec2.internal.warc.gz"} |
Archives of the Caml mailing list > Message from Christophe Raffalli
Locally-polymorphic exceptions [was: folding over a file]
Date: -- (:)
From: Christophe Raffalli <christophe.raffalli@u...>
Subject: Re: [Caml-list] Unsoundness is essential
skaller a écrit :
> On Wed, 2007-10-03 at 19:28 -0400, Joshua D. Guttman wrote:
>> skaller <skaller@users.sourceforge.net> writes:
>>> Goedel's theorem says that any type system strong enough
>>> to describe integer programming is necessarily unsound.
> I paraphrased it, deliberately, in effect claiming an analogous
> situation holds with type systems.
Not unsound, incomplete !
you mixup first and second incompleteness theorem. Let's clarify ?
- first if a type system does not contain arithmetic nothing can be said
(this implies ML), but in this case, the meaning of complete needs to be clarified.
Indeed, there are complete type system ...
- The 1st incompleteness theorem states that no theory containing
arithmetic is complete. This means that there will always be correct programs
that your type system can not accept. However, I thing a real program that
is not provably correct in lets say ZF, does not exists and should be rejected.
you do not accept a program whose correctness is a conjecture (do you ?)
- The second incompleteness theorem, states that a system that proves its own
consistency is in fact inconsistent. For type system (strong enough to express
arithmetic, like ML with dependant type, PVS, the future PML, ..). This means
that the proof that the system is consistent (the proof that "0 = 1 can not be proved")
can not be done inside the system. However, the proof that your implementation
does implement the formal system correctly can be done inside the system, and
this is quite enough for me.
- The soundness theorem for ML can be stated as a program of type "int" will
- diverge
- raise an exception
- or produce an int
I think everybody except LISP programmers ;-) want a sound type system like this.
OK replace everybody by I if you prefer ... For PML, we are more precise : exception
and the fact the a program may diverge must be written in the type.
- ML type system is sometimes too incomplete, this is why the Obj library is here.
However, the use of Obj is mainly here because ML lacks dependant types. In fact,
the main use of Obj is to simulate a map table associating to a key k a type t(k) and
a value v:t(k).
- All this says that a type-system only accepting proved programs is possible and
a good idea (it already exists). The question for researcher is how to produce a
type system where the cost of proving is acceptable compared to the cost of debugging,
and this stars to be the case for some specific application, but we are far from
having to remove the word "bug" from our vocabulary ;-)
Christophe Raffalli
Universite de Savoie
Batiment Le Chablais, bureau 21
73376 Le Bourget-du-Lac Cedex
tel: (33) 4 79 75 81 03
fax: (33) 4 79 75 87 42
mail: Christophe.Raffalli@univ-savoie.fr
www: http://www.lama.univ-savoie.fr/~RAFFALLI
IMPORTANT: this mail is signed using PGP/MIME
At least Enigmail/Mozilla, mutt or evolution
can check this signature. The public key is
stored on www.keyserver.net | {"url":"http://caml.inria.fr/pub/ml-archives/caml-list/2007/10/222c97e0f2d3430549275bb49e480618.en.html","timestamp":"2014-04-21T09:58:34Z","content_type":null,"content_length":"18726","record_id":"<urn:uuid:8ea4ded9-a632-4b05-9125-936b9bcf1846>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00484-ip-10-147-4-33.ec2.internal.warc.gz"} |
Cosine redux
It's been quite a while since my last blog entry: I had a bit of a technology meltdown that I'm not quite done with yet :-(
I was doing some recreational hacking over the holidays that involved evaluating cosines. I ended up doing (once again!) an implementation of cosine (don't ask why). Given the confused flamage that
my previous posts on cosine generated, I figure that showing some code would be useful. I would never have thought that cosine would generate so much email traffic. Yes, I know about Taylor series.
No I wasn't trying to insult centuries of standard mathematical practice. Performance is often the art of cheating carefully.
So, here's my implementation of cosine, with its argument in turns (1 turn == 360 degrees):
public static float cost(float f) {
int bits = Float.floatToRawIntBits(f);
int mantissa = (bits&0x7FFFFF)|(1<<23);
int shift = (bits<<1>>>24)-127+9; // exponent, unbiased, with shift
if(shift>=32 || shift<=-32) return 1;
int fractionBits = shift>=0 ? mantissa<<shift : mantissa>>-shift;
int tableIndex = (fractionBits>>>(30-resultPrecision))&tableSizeMask;
switch(fractionBits>>>30) { // Quadrant is top two bits
case 0: return cosTable[tableIndex];
case 1: return -cosTable[tableSizeMask-tableIndex];
case 2: return -cosTable[tableIndex];
default/\*case 3\*/: return cosTable[tableSizeMask-tableIndex];
Lets go through this slowly:
1. int bits = Float.floatToRawIntBits(f);
Get the IEEE 754 bits
2. int mantissa = (bits&0x7FFFFF)|(1<<23);
The mantissa is the bottom 23 bits - to which the hidden bit must be prepended.
3. int shift = (bits<<1>>>24)-127+9;
Extract the exponent, correct for the exponent bias, then add a bias to move the binary point to the top of the word.
4. if(shift>=32 || shift<=-32) return 1;
If the shift is too large, the fraction bits would be zero, therefore the result is 1.
5. int fractionBits = shift>=0 ? mantissa<<shirt : mantissa>>-shift;
Shift the mantissa so that it's a fixed point number with the binary point at the top of the 32 bit int. The magic is in what's not here: because the argument is in turns, I get to ignore all of
the integer bits (range reduction made trivial); and because it's the cosine function, which is symmetric about the origin, I get to ignore the sign.
6. int tableIndex = (fractionBits>>>(30-resultPrecision))&tableSizeMask;
The top two bits are the quadrant... extract the bits below that to derive a table index.
7. switch(fractionBits>>>30) {
One case for each quadrant. This switch could be eliminated by making the table 4 times larger.
8. case 0: return cosTable[tableIndex];
Yes! It's just a table lookup! Truly trivial.
9. case 1: return -cosTable[tableSizeMask-tableIndex];
Table lookups are a very common way to implement mathematical functions, particularly the periodic ones like cosine. There are all kinds of elaborations. One of the most common for improving accuracy
is to do some sort of interpolation between table elements (linear or cubic, usually).
Jeff, I think the problem that your solutions are not helping to solve here is the performance, i.e. the one that Java is having in calling trig functions. Specifically I cannot see how CORDIC will
eliminate the runtime overhead that argument reduction brings about in Java when calling x86 hardware trig instructions.
Posted by Alex Lam on January 21, 2006 at 08:08 AM PST #
It's a tradeoff. The better you understand the precision requirements of your application, the better you can tune it. There is a spectacular amount of literature on different tricks on evaluating
functions like cosine. The less precision you need, the faster you can make it go. The problem with a general-purpose library like Java's is that it has no information about the precision
requirements of the application invoking it. So the safe thing to do is to produce the most accurate answer possible. The difficulty is that in this case, the x86 hardware doesn't do a great job, so
we correct for it. The grotesque solution would be to define functions that had extra parameters for (at least) the number of result bits that were needed to be accurate, and the range over which the
argument might vary: if these were constants, then a good optimizer could substitute appropriatly tricked-up code fragments.
Posted by James Gosling on January 21, 2006 at 02:02 PM PST #
It is inconvenient that radians depend on PI and PI is an irrational number. At the moment I am writing an arbitrary-precision arithmetics package and is working on implementing the non-linear (trig,
exp, log, pow etc.) functions. Apart from revising my maths knowledge I have also looked up (seemingly) similar projects on the Internet (JUMP, JScience, JCM etc.) for possible inspirations. Now I
think I have a better sense of direction on how to work things out; I sincerely hope that after all the pain-staking processes I'd come up with an almighty useful Java library ;)
Posted by Alex Lam on January 22, 2006 at 01:04 AM PST #
Hi Alex
I'm searching for a good arbitrary-precision arithmetics package for numerical computations with support for trigonometric functions.
Can you suggest one?
Posted by Axel Kramer on January 22, 2006 at 07:59 PM PST #
看激情电影日本激情电影色情电影色情小电影免费色情电影免费观看色情电影日本色情电影性爱电影性爱小电影免费性爱电影免费观看性爱电影日本性爱电影 what goes wrong??
Posted by nile black on January 22, 2006 at 10:32 PM PST #
What ever happened to the Java Grande project? Among many other interesting things they were working on - I thought one of the proposals was for a math library that could be processor dependent. In
cases in which it was more important to have performance than to have identical answers on all platforms there could be specific optimizations. These would be contained in libraries where it would be
clear that this was the trade off you were making.
Posted by Daniel Steinberg on January 22, 2006 at 11:38 PM PST #
"The grotesque solution would be to define functions that had extra parameters for (at least) the number of result bits that were needed to be accurate, and the range over which the argument might
vary: if these were constants, then a good optimizer could substitute appropriatly tricked-up code fragments."
What's so grotesque about that? If every function requires information about the error bounds, then enforcing the desired error limits becomes quite complicated. This is doubly true, while there is
no standard way to track and describe error bounds. Perhaps JSR 275 will address that. http://www.jcp.org/en/jsr/detail?id=275
I would think that "The Jackpot Way" to solve this problem would be to provide an editor for complex functions and equations. Then a function evaluation engine can evaluate the entire function,
tracking the error along the way. Radically different evaluation methods could be chosen, simply by changing the acceptable error.
Total error is a pain to track, but it is almost always the only important thing. The error involved in any single stage of a calculation is only important to the extent that it impacts total error.
If you think this approach requires too much intelligence from the function evaluation engine to ever compete in terms of performance, would you say the same thing about Java itself?
Posted by Curt Cox on January 23, 2006 at 12:04 AM PST #
you sir, are a hacker. I challenge you to pistols at dawn!
Posted by wes on January 23, 2006 at 07:06 PM PST #
Hi Axel, AFAIK, there isn't one (except that there would be one once I've done with it :-D) It's not like extension of non-linear functions from decimal to arbitrary-precision is that
straight-forward, I suppose~
Posted by Alex Lam on January 23, 2006 at 09:20 PM PST # | {"url":"https://blogs.oracle.com/jag/entry/cosine_redux","timestamp":"2014-04-20T05:16:00Z","content_type":null,"content_length":"36026","record_id":"<urn:uuid:50721c27-9e50-450b-b859-84e1cdb78579>","cc-path":"CC-MAIN-2014-15/segments/1397609537864.21/warc/CC-MAIN-20140416005217-00259-ip-10-147-4-33.ec2.internal.warc.gz"} |
Public Member Functions
virtual ~Ifpack_Container ()
virtual int NumRows () const =0
Returns the number of rows of the matrix and LHS/RHS.
virtual int NumVectors () const =0
Returns the number of vectors in LHS/RHS.
virtual int SetNumVectors (const int i)=0
Sets the number of vectors for LHS/RHS.
virtual double & LHS (const int i, const int Vector=0)=0
Returns the i-th component of the vector Vector of LHS.
virtual double & RHS (const int i, const int Vector=0)=0
Returns the i-th component of the vector Vector of RHS.
virtual int & ID (const int i)=0
Returns the ID associated to local row i.
virtual int SetMatrixElement (const int row, const int col, const double value)=0
Set the matrix element (row,col) to value.
virtual int Initialize ()=0
Initializes the container, by performing all operations that only require matrix structure.
virtual int Compute (const Epetra_RowMatrix &A)=0
Finalizes the linear system matrix and prepares for the application of the inverse.
virtual int SetParameters (Teuchos::ParameterList &List)=0
Sets all necessary parameters.
virtual bool IsInitialized () const =0
Returns true is the container has been successfully initialized.
virtual bool IsComputed () const =0
Returns true is the container has been successfully computed.
virtual int Apply ()=0
Apply the matrix to RHS, results are stored in LHS.
virtual int ApplyInverse ()=0
Apply the inverse of the matrix to RHS, results are stored in LHS.
virtual const char * Label () const =0
Returns the label of this container.
virtual double InitializeFlops () const =0
Returns the flops in Initialize().
virtual double ComputeFlops () const =0
Returns the flops in Compute().
virtual double ApplyFlops () const =0
Returns the flops in Apply().
virtual double ApplyInverseFlops () const =0
Returns the flops in ApplyInverse().
virtual ostream & Print (std::ostream &os) const =0
Prints out basic information about the container.
Ifpack_Container: a pure virtual class for creating and solving local linear problems.
Class Ifpack_Container provides the abstract interfaces for containers. A "container" is an object that hosts all it is necessary to create, populate, and solve local linear problems. The local
linear problem matrix, B, is a submatrix of the local components of a distributed matrix, A. The idea of container is to specify the rows of A that are contained in B, then extract B from A, and
compute all it is necessary to solve a linear system in B. Then, set starting solution (if necessary) and right-hand side for B, and solve the linear system in B.
A container should be used in the following manner:
• Create an container object, specifying the number of rows of B.
• If necessary, set parameters for the solution using SetParameters().
• Initialize the container by calling Initialize().
• Specify the ID of the local rows of A that are contained in B, using ID().
• Prepare the linear system solver using Compute().
• set LHS and/or RHS elements using LHS() and RHS().
• Solve the linear system using ApplyInverse().
• Get the componenets of the computed solution using LHS().
The number of vectors can be set using SetNumVectors(), and it is defaulted to 1.
Containers are currently used by class Ifpack_BlockRelaxation.
Ifpack_Container is a pure virtual class. Two concrete implementations are provided in classes Ifpack_SparseContainer (that stores matrices in sparse the format Epetra_CrsMatrix) and
Ifpack_DenseContainer (for relatively small matrices, as matrices are stored as Epetra_SerialDenseMatrix's).
Still to do:
□ Flops count has to be tested.
Marzio Sala, SNL 9214.
Last update Oct-04.
Definition at line 85 of file Ifpack_Container.h.
virtual int& Ifpack_Container::ID ( const int i ) [pure virtual]
Returns the ID associated to local row i.
The set of (local) rows assigned to this container is defined by calling ID(i) = j, where i (from 0 to NumRows()) indicates the container-row, and j indicates the local row in the calling process.
This is usually used to recorder the local row ID (on calling process) of the i-th row in the container.
Implemented in Ifpack_DenseContainer, and Ifpack_SparseContainer< T >. | {"url":"http://trilinos.sandia.gov/packages/docs/r10.6/packages/ifpack/doc/html/classIfpack__Container.html","timestamp":"2014-04-18T00:59:07Z","content_type":null,"content_length":"24818","record_id":"<urn:uuid:f99ce6e9-586b-4497-8ab4-bbc53a9b76ea>","cc-path":"CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00520-ip-10-147-4-33.ec2.internal.warc.gz"} |
Got Homework?
Connect with other students for help. It's a free community.
• across
MIT Grad Student
Online now
• laura*
Helped 1,000 students
Online now
• Hero
College Math Guru
Online now
Here's the question you clicked on:
see attchment
• one year ago
• one year ago
Best Response
You've already chosen the best response.
Best Response
You've already chosen the best response.
Best Response
You've already chosen the best response.
Best Response
You've already chosen the best response.
Best Response
You've already chosen the best response.
so we want to find that inside probability. so like the last problem we need to find two z scores.
Best Response
You've already chosen the best response.
where did you get that 3 and 5 ?
Best Response
You've already chosen the best response.
mean mu my bad
Best Response
You've already chosen the best response.
Best Response
You've already chosen the best response.
Best Response
You've already chosen the best response.
you mean i have to look up z score between -1 to 1 ?
Best Response
You've already chosen the best response.
Best Response
You've already chosen the best response.
yeah, one sec
Best Response
You've already chosen the best response.
are you there?
Best Response
You've already chosen the best response.
yeah, I'm just a bit confused.
Best Response
You've already chosen the best response.
right now I think that we need to do 1-z1-z2 z1 = (y-mu)/(sigma) the top y value i think is mu+1. sigma is 4. z1 = (mu+1-mu)/4 = 1/4 now for y2 the bottom y value i think is mu-1. sigma is 4. z2
= (mu -1-mu)/4 = -1/4 so 1 -1/4 + 1/4 = 1 but its not an answer, which confuses me.
Your question is ready. Sign up for free to start getting answers.
is replying to Can someone tell me what button the professor is hitting...
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• Engagement 19 Mad Hatter
• You have blocked this person.
• ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/updates/51439a42e4b04cdfc581e8cc","timestamp":"2014-04-20T23:53:58Z","content_type":null,"content_length":"95412","record_id":"<urn:uuid:33222f04-7bf4-4a0f-b3c5-b1f178ee486e>","cc-path":"CC-MAIN-2014-15/segments/1397609539337.22/warc/CC-MAIN-20140416005219-00112-ip-10-147-4-33.ec2.internal.warc.gz"} |
Putative regulatory sites unraveled by network-embedded thermodynamic analysis of metabolome data
As one of the most recent members of the omics family, large-scale quantitative metabolomics data are currently complementing our systems biology data pool and offer the chance to integrate the
metabolite level into the functional analysis of cellular networks. Network-embedded thermodynamic analysis (NET analysis) is presented as a framework for mechanistic and model-based analysis of
these data. By coupling the data to an operating metabolic network via the second law of thermodynamics and the metabolites' Gibbs energies of formation, NET analysis allows inferring functional
principles from quantitative metabolite data; for example it identifies reactions that are subject to active allosteric or genetic regulation as exemplified with quantitative metabolite data from
Escherichia coli and Saccharomyces cerevisiae. Moreover, the optimization framework of NET analysis was demonstrated to be a valuable tool to systematically investigate data sets for consistency, for
the extension of sub-omic metabolome data sets and for resolving intracompartmental concentrations from cell-averaged metabolome data. Without requiring any kind of kinetic modeling, NET analysis
represents a perfectly scalable and unbiased approach to uncover insights from quantitative metabolome data.
Keywords: genome-scale analysis, metabolic regulation, metabolomics, stoichiometric network, thermodynamics | {"url":"http://pubmedcentralcanada.ca/pmcc/articles/PMC1681506/?lang=en-ca","timestamp":"2014-04-16T15:39:09Z","content_type":null,"content_length":"136822","record_id":"<urn:uuid:ccd75c4b-7d14-47dd-a2d4-4d6bdd6cb182>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00104-ip-10-147-4-33.ec2.internal.warc.gz"} |
Pure type systems formalized
Results 1 - 10 of 38
- In ACM SIGPLAN-SIGACT Symposium on Principles of Programming Languages , 2008
"... Machine-checked proofs of properties of programming languages have become a critical need, both for increased confidence in large and complex designs and as a foundation for technologies such as
proof-carrying code. However, constructing these proofs remains a black art, involving many choices in th ..."
Cited by 86 (9 self)
Add to MetaCart
Machine-checked proofs of properties of programming languages have become a critical need, both for increased confidence in large and complex designs and as a foundation for technologies such as
proof-carrying code. However, constructing these proofs remains a black art, involving many choices in the formulation of definitions and theorems that make a huge cumulative difference in the
difficulty of carrying out large formal developments. The representation and manipulation of terms with variable binding is a key issue. We propose a novel style for formalizing metatheory, combining
locally nameless representation of terms and cofinite quantification of free variable names in inductive definitions of relations on terms (typing, reduction,...). The key technical insight is that
our use of cofinite quantification obviates the need for reasoning about equivariance (the fact that free names can be renamed in derivations); in particular, the structural induction principles of
- In Computer Science Logic , 1999
"... . We present a denition of untyped -terms using a heterogeneous datatype, i.e. an inductively dened operator. This operator can be extended to a Kleisli triple, which is a concise way to verify
the substitution laws for -calculus. We also observe that repetitions in the denition of the monad as wel ..."
Cited by 77 (15 self)
Add to MetaCart
. We present a denition of untyped -terms using a heterogeneous datatype, i.e. an inductively dened operator. This operator can be extended to a Kleisli triple, which is a concise way to verify the
substitution laws for -calculus. We also observe that repetitions in the denition of the monad as well as in the proofs can be avoided by using well-founded recursion and induction instead of
structural induction. We extend the construction to the simply typed -calculus using dependent types, and show that this is an instance of a generalization of Kleisli triples. The proofs for the
untyped case have been checked using the LEGO system. Keywords. Type Theory, inductive types, -calculus, category theory. 1 Introduction The metatheory of substitution for -calculi is interesting
maybe because it seems intuitively obvious but becomes quite intricate if we take a closer look. [Hue92] states seven formal properties of substitution which are then used to prove a general
substitution theor...
- Ninth international Conference on Theorem Proving in Higher Order Logics TPHOL , 1996
"... Abstract. We present five axioms of name-carrying lambda-terms identified up to alpha-conversion—that is, up to renaming of bound variables. We assume constructors for constants, variables,
application and lambdaabstraction. Other constants represent a function Fv that returns the set of free variab ..."
Cited by 51 (0 self)
Add to MetaCart
Abstract. We present five axioms of name-carrying lambda-terms identified up to alpha-conversion—that is, up to renaming of bound variables. We assume constructors for constants, variables,
application and lambdaabstraction. Other constants represent a function Fv that returns the set of free variables in a term and a function that substitutes a term for a variable free in another term.
Our axioms are (1) equations relating Fv and each constructor, (2) equations relating substitution and each constructor, (3) alpha-conversion itself, (4) unique existence of functions on lambda-terms
defined by structural iteration, and (5) construction of lambda-abstractions given certain functions from variables to terms. By building a model from de Bruijn’s nameless lambda-terms, we show that
our five axioms are a conservative extension of HOL. Theorems provable from the axioms include distinctness, injectivity and an exhaustion principle for the constructors, principles of structural
induction and primitive recursion on lambda-terms, Hindley and Seldin’s substitution lemmas and
, 1995
"... This thesis describes the implementation of ALF, which is an interactive proof editor based on Martin-Löf's type theory with explicit substitutions. ALF is a general purpose proof assistant, in
which different logics can be represented. Proof objects are manipulated directly, by the usual editing op ..."
Cited by 43 (0 self)
Add to MetaCart
This thesis describes the implementation of ALF, which is an interactive proof editor based on Martin-Löf's type theory with explicit substitutions. ALF is a general purpose proof assistant, in which
different logics can be represented. Proof objects are manipulated directly, by the usual editing operations. A partial proof is represented as an incomplete proof object, i.e., a proof object
containing placeholders. A modular type/proof checking algorithm for complete proof objects is presented, and it is proved sound and complete assuming some basic meta theory properties of the
substitution calculus. The algorithm is extended to handle incomplete objects in such a way that the type checking problem is reduced to a unication problem, i.e., the problem of finding
instantiations to the placeholders in the object. Placeholders are represented together with their expected type and local context. We show that checking the correctness of instantiations can be
localised, which means that it is e...
- Journal of Automated Reasoning , 1996
"... The proofs of the Church-Rosser theorems for fi, j and fi [ j reduction in untyped -calculus are formalized in Isabelle/HOL, an implementation of Higher Order Logic in the generic theorem prover
Isabelle. ..."
Cited by 39 (4 self)
Add to MetaCart
The proofs of the Church-Rosser theorems for fi, j and fi [ j reduction in untyped -calculus are formalized in Isabelle/HOL, an implementation of Higher Order Logic in the generic theorem prover
, 1993
"... This thesis contains an investigation of Coquand's Calculus of Constructions, a basic impredicative Type Theory. We review syntactic properties of the calculus, in particular decidability of
equality and type-checking, based on the equality-as-judgement presentation. We present a set-theoretic notio ..."
Cited by 31 (2 self)
Add to MetaCart
This thesis contains an investigation of Coquand's Calculus of Constructions, a basic impredicative Type Theory. We review syntactic properties of the calculus, in particular decidability of equality
and type-checking, based on the equality-as-judgement presentation. We present a set-theoretic notion of model, CC-structures, and use this to give a new strong normalization proof based on a
modification of the realizability interpretation. An extension of the core calculus by inductive types is investigated and we show, using the example of infinite trees, how the realizability
semantics and the strong normalization argument can be extended to non-algebraic inductive types. We emphasize that our interpretation is sound for large eliminations, e.g. allows the definition of
sets by recursion. Finally we apply the extended calculus to a non-trivial problem: the formalization of the strong normalization argument for Girard's System F. This formal proof has been developed
and checked using the...
, 1997
"... This paper presents the first machine-checked verification of Milner's type inference algorithm W for computing the most general type of an untyped -term enriched with let-expressions. This term
language is the core of most typed functional programming languages and is also known as Mini-ML. We ..."
Cited by 15 (1 self)
Add to MetaCart
This paper presents the first machine-checked verification of Milner's type inference algorithm W for computing the most general type of an untyped -term enriched with let-expressions. This term
language is the core of most typed functional programming languages and is also known as Mini-ML. We show how to model all the concepts involved, in particular types and type schemes, substitutions,
and the thorny issue of "new" variables. Only a few key proofs are discussed in detail. The theories and proofs are developed in Isabelle/HOL, the HOL instantiation of the generic theorem prover
- IN WORKSHOP ON FOUNDATIONS OF OBJECT-ORIENTED LANGUAGES , 1999
"... ..."
, 2001
"... We extend the type system for the Lambda Calculus of Objects [16] with a mechanism of width subtyping and a treatment of incomplete objects. The main novelties over previous work are the use of
subtype-bounded quantification to capture a new and more direct rendering of MyType polymorphism, and a un ..."
Cited by 11 (6 self)
Add to MetaCart
We extend the type system for the Lambda Calculus of Objects [16] with a mechanism of width subtyping and a treatment of incomplete objects. The main novelties over previous work are the use of
subtype-bounded quantification to capture a new and more direct rendering of MyType polymorphism, and a uniform treatment for other features that were accounted for via different systems in
subsequent extensions [7, 6] of [16]. The new system provides for (i) appropriate type specialization of inherited methods, (ii) static detection of errors, (iii) width subtyping compatible with
object extension, and (iv) sound typing for partially specified objects.
- In Proc. of LFM’99: Workshop on Logical Frameworks and Meta-Languages , 1999
"... . The logical and operational aspects of rewriting logic as a logical framework are illustrated in detail by representing pure type systems as object logics. More precisely, we apply membership
equational logic, the equational sublogic of rewriting logic, to specify pure type systems as they can be ..."
Cited by 10 (2 self)
Add to MetaCart
. The logical and operational aspects of rewriting logic as a logical framework are illustrated in detail by representing pure type systems as object logics. More precisely, we apply membership
equational logic, the equational sublogic of rewriting logic, to specify pure type systems as they can be found in the literature and also a new variant of pure type systems with explicit names that
solves the problems with closure under -conversion in a very satisfactory way. Furthermore, we use rewriting logic itself to give a formal operational description of type checking, that directly
serves as an ecient type checking algorithm. The work reported here is part of a more ambitious project concerned with the development in Maude [7] of a proof assistant for OCC, the open calculus of
constructions, an equational extension of the calculus of constructions. 1 Introduction This paper is a detailed case study on the ease and naturalness with which a family of higher-order formal
systems, namely... | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=113115","timestamp":"2014-04-17T18:57:00Z","content_type":null,"content_length":"37465","record_id":"<urn:uuid:6a8fd542-8cc9-42b4-a3d8-85fec03cd5ef>","cc-path":"CC-MAIN-2014-15/segments/1397609530895.48/warc/CC-MAIN-20140416005210-00457-ip-10-147-4-33.ec2.internal.warc.gz"} |
Two-dimensional Electronic Double-Quantum Coherence Spectroscopy
The theory of electronic structure of many-electron systems like molecules is extraordinarily complicated. A lot can be learned by considering how electron density is distributed, on average, in the
average field of the other electrons in the system. That is, mean field theory. However, to describe quantitatively chemical bonds, reactions, and spectroscopy requires consideration of the way that
electrons avoid each other by the way they move; this is called electron correlation (or in physics, the many-body problem for fermions). While great progress has been made in theory, there is a need
for incisive experimental tests that can be undertaken for large molecular systems in the condensed phase.
Here we report a two-dimensional (2D) optical coherent spectroscopy that correlates the double excited electronic states to constituent single excited states. The technique, termed two-dimensional
double-coherence spectroscopy (2D-DQCS), makes use of multiple, time-ordered ultrashort coherent optical pulses to create double- and single-quantum coherences over time intervals between the pulses.
The resulting two-dimensional electronic spectrum maps the energy correlation between the first excited state and two-photon allowed double-quantum states. The principle of the experiment is that
when the energy of the double-quantum state, viewed in simple models as a double HOMO to LUMO excitation, equals twice that of a single excitation, then no signal is radiated. However,
electron-electron interactions—a combination of exchange interactions and electron correlation—in real systems generates a signal that reveals precisely how the energy of the double-quantum resonance
differs from twice the single-quantum resonance. The energy shift measured in this experiment reveals how the second excitation is perturbed by both the presence of the first excitation and the way
that the other electrons in the system have responded to the presence of that first excitation.
We compare a series of organic dye molecules and find that the energy offset for adding a second electronic excitation to the system relative to the first excitation is on the order of tens of
milli-electronvolts, and it depends quite sensitively on molecular geometry. These results demonstrate the effectiveness of 2D-DQCS for elucidating quantitative information about electron-electron
interactions, many-electron wavefunctions, and electron correlation in electronic excited states and excitons. | {"url":"http://pubmedcentralcanada.ca/pmcc/articles/PMC2775063/?lang=en-ca","timestamp":"2014-04-20T23:06:12Z","content_type":null,"content_length":"119799","record_id":"<urn:uuid:1bc3e017-5536-46bb-a21f-5ffca588e4e4>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00221-ip-10-147-4-33.ec2.internal.warc.gz"} |
Find the condition for line to touch a curve?
May 11th 2009, 10:38 PM #1
Find the condition for line to touch a curve?
Find the condition that the line $x\cos \alpha + y\sin \alpha = p$ may touch the curve
$\left(\frac{x - a}{a}\right)^{\frac{n}{n - 1}} + \left(\frac{y - b}{b}\right)^{\frac{n}{n - 1}} = 1$
Assume $n>1,\ a, b>0$
The second curve is a hyper-ellipse, and the condition of tangency is that the equation you get by substituting $y$ from the equation for the line into the equation of the hyper-ellipse has a
single real root.
May 16th 2009, 01:03 AM #2
Grand Panjandrum
Nov 2005 | {"url":"http://mathhelpforum.com/calculus/88617-find-condition-line-touch-curve.html","timestamp":"2014-04-16T07:31:34Z","content_type":null,"content_length":"34482","record_id":"<urn:uuid:957a0537-6d2f-44e0-be89-0874eff5939a>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00436-ip-10-147-4-33.ec2.internal.warc.gz"} |
Calculator++ 1.2.31
February 14th, 2012
Calculator++ 1.2.31
1 year, 259 days ago, Android Apps, by apkfan.
Current Version: 1.2.31
Requires Android: 1.6 and up
Category: Tools
V1.2.31 update:
New Metro theme with 3 color variations (old themes are still available)
Simple layout (Default layout is still scientific)
Preferences screen changes
Maximum calculation time preference added
Always scientific notation option added
Remove billing information option added
Error label is substituted with greyed text
Special layout for x-high density displays (aka Galaxy Note support)
Sgn(0) problem fixed
Billing problems fixes
Calculator++ is an advanced, modern and easy to use scientific calculator #1.
Calculator++ helps you to do basic and advanced calculations on your mobile device.
IMPORTANT: application contains ads! If you want to get rid of them purchase special option from application settings. Internet access permission is needed only for showing the ads. ADS ARE ONLY
SHOWN ON THE SECONDARY SCREENS! If internet is off – no ads!
++ easy to use
+ no need to press equals button anymore – result is calculated on fly
+ smart cursor positioning
+ several kinds of brackets
+ copy/paste in one button
+ landscape/portrait orientations
++ drag buttons up or down to use special functions, operators etc
++ modern interface with possibility to choose themes (3 themes available)
+ highlighting of expressions
+ history with all previous calculations and undo/redo buttons
++ variables and constants support (build-in and user defined)
++ complex number computations
+ support for lots of functions: sin, sinh, asin, asinh, ln, sqrt etc
++ expression simplification: use ‘identical to’ sign (≡) to simplify current expression (2/3+5/9≡11/9, √(8)≡2√(2))
+ support for android 1.6 and higher
+ open source
NOTE ABOUT INTERNET ACCESS: Calculator++ (ver. 1.2.24) contains ad which requires internet access. To get rid of it – purchase version without ad (can be done from application’s preferences).
1. You can omit unnecessary multiplication signs (instead of 3*t or t*sin(π*t) just type 3t or tsin(πt))
2. Use ≡ (in the top right corner of = button) to simplify expression
3. Just click on the result to copy it to the clipboard
4. Use = button seldom – result is calculating on the fly
5. Add constants for often used values (you can even save expressions)
How can I get rid of the ads?
You can do it by purchasing the special option in the main application preferences.
Why Calculator++ needs INTERNET permission?
Currently application needs such permission only for one purpose – to show ads. If you buy the special option C++ will never use your internet connection.
How can I use functions written in the top right and bottom right corners of the button?
Push the button and slide lightly up or down. Depending on value showed on the button action will occur.
How can I toggle between radians and degrees?
To toggle between different angle units you can either change appropriate option in application settings or use the toggle switch located on the 6 button (current value is lighted with yellow color).
Also you can use deg() and rad() functions and ° operator to convert degrees to radians and vice versa.
268° = 4.67748
30.21° = 0.52726
rad(30, 21, 0) = 0.52726
deg(4.67748) = 268
Does C++ support %?
Yes, % function can be found in the top right corner of / button.
100 + 50% = 150
100 * 50% = 50
100 + 100 * 50% * 50% = 125
100 + (100 * 50% * (25 + 25)% + 100%) = 150
100 + (20 + 20)% = 140, but 100+ (20% + 20%) = 124.0
100 + 50% ^ 2 = 2600, but 100 + 50 ^ 2% = 101.08
Does C++ support fractional calculations?
Yes, you can type your fractional expression in the editor and use ≡ (in the top right corner of = button). Also you can use ≡ to simplify expression.
2/3 + 5/9 ≡ 11/9
2/9 + 3/123 ≡ 91/369
(6-t) ^ 3 ≡ 216 – 108t + 18t ^ 2 – t ^ 3
Does C++ support complex calculations?
Yes, just enter complex expression (using i or √(-1) as imaginary number).
(2i + 1) ^ = -3 + 4i
e ^ i = 0.5403 + 0.84147i
Can C++ plot graph of the function?
Yes, type expression which contains 1 undefined variable (e.g. cos(t)) and click on the result. In the context menu choose ‘Plot graph’.
Does C++ support matrix calculations?
No, it doesn’t
Keywords: engineer calculator, scientific calculator, integration, differentiation, derivative, mathematica, math, maths, mathematics, matlab, mathcad, percent, percentage, complex numbers, plotting
graphs, graph plot, plotter, calculation, symbolic calculations, wolfram
Download via Google Play: Calculator++ 1.2.31 on Google Play Download : Download Calculator++ 1.2.31
HOW TO Install APK Files on Android Device?
Related posts:
Responses to “Calculator++ 1.2.31”
1. No comments yet.
1. No trackbacks yet. | {"url":"http://www.apktops.com/calculator-1-2-31.html","timestamp":"2014-04-17T06:59:20Z","content_type":null,"content_length":"38862","record_id":"<urn:uuid:87fbbf43-632d-4ed4-94ee-6b5774f8044c>","cc-path":"CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00011-ip-10-147-4-33.ec2.internal.warc.gz"} |
Surface integral - I don't understand it
Okay, the point here is the same concept behind Riemann integrals. In Riemann integration, over a "very small" region, for the purposes of approximation, we can consider a function to be constant
(the idea behind the rectangles.) However, in the next "very small" region, we consider it (again, for the purposes of approximation) to be another constant. Does it make sense that, as our small
regions get small, the sum of the rectangles' area gets close to the area under the curve?
The same idea applies here. Over a "very small" part of the surface, for approximation, we can call the function a constant, though over the other "very small" regions, the function is considered to
be a different constant (well, not necessarily different, but not necessarily the same) over each region. | {"url":"http://www.physicsforums.com/showthread.php?p=4289485","timestamp":"2014-04-20T16:11:42Z","content_type":null,"content_length":"47221","record_id":"<urn:uuid:9624def0-092d-4870-a223-8352ed1a991c>","cc-path":"CC-MAIN-2014-15/segments/1397609538824.34/warc/CC-MAIN-20140416005218-00042-ip-10-147-4-33.ec2.internal.warc.gz"} |
Got Homework?
Connect with other students for help. It's a free community.
• across
MIT Grad Student
Online now
• laura*
Helped 1,000 students
Online now
• Hero
College Math Guru
Online now
Here's the question you clicked on:
Find the number of ways to listen to 4 different CDs from a selection of 10 CDs.
• one year ago
• one year ago
Best Response
You've already chosen the best response.
There are 15 x 14 x 13 x 12 ways to listen to 4 different CDs out of 15.
Best Response
You've already chosen the best response.
i am wrong, @seiga is right
Best Response
You've already chosen the best response.
theres are the answer choices: 151,200 5040 2880 720
Best Response
You've already chosen the best response.
--151,200 5040 2880 720 --solution: in this case the original answer should be ¹⁰C₄ = (10x9x8x7) / (1x2x3x4) = 210 but from options it is clear that it is used ¹⁰P₄ = 10x9x8x7 = 5040 in the
question paper i m still sure that since it is not mentioned track wise or not. so it must be ¹⁰C₄ or 210 as answer. hope it would be helpful
Your question is ready. Sign up for free to start getting answers.
is replying to Can someone tell me what button the professor is hitting...
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• Engagement 19 Mad Hatter
• You have blocked this person.
• ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/updates/5134f540e4b093a1d9499fdf","timestamp":"2014-04-18T20:50:00Z","content_type":null,"content_length":"37735","record_id":"<urn:uuid:89944923-f85a-43f3-ac7c-2734095309cb>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00347-ip-10-147-4-33.ec2.internal.warc.gz"} |
Got Homework?
Connect with other students for help. It's a free community.
• across
MIT Grad Student
Online now
• laura*
Helped 1,000 students
Online now
• Hero
College Math Guru
Online now
Here's the question you clicked on:
Differentiate the function g(x) = ln[x(sqrt(x^2 - 1))]
Best Response
You've already chosen the best response.
and SIMPLIFY THE ANSWER
Best Response
You've already chosen the best response.
i got up to g'(x) = 1/x + 1/[2(x+1)] + 1/[2(x-1)] i can't simplify it i'm retarded help me
Best Response
You've already chosen the best response.
We have \[ g(x)= \ln( x \sqrt { x^2-1})\] \[g'(x)= \frac{1}{ x\sqrt {x^2-1} } \times ( \sqrt { x^2-1} + x \times \frac{2x}{2 \sqrt {x^2-1}}) \] or \[g'(x)= \frac{1}{ x\sqrt {x^2-1} } \times ( \
sqrt { x^2-1} + x \times \frac{x}{ \sqrt {x^2-1}}) \] simplifying now \[g'(x)= \frac{1}{ x\sqrt {x^2-1} } \times ( \frac{x^2-1+x^2}{ \sqrt {x^2-1}}) \] \[g'(x)= \frac{2x^2-1}{ x( x^2-1)} \]
Your question is ready. Sign up for free to start getting answers.
is replying to Can someone tell me what button the professor is hitting...
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• Engagement 19 Mad Hatter
• You have blocked this person.
• ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred. | {"url":"http://openstudy.com/updates/4f8986e5e4b02251ecc9e953","timestamp":"2014-04-17T21:30:19Z","content_type":null,"content_length":"32682","record_id":"<urn:uuid:8aa536ba-730e-4ff1-9dcb-b7fda3dd31a9>","cc-path":"CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00283-ip-10-147-4-33.ec2.internal.warc.gz"} |
Substitution word problem
January 21st 2008, 04:45 PM #1
Jan 2008
Substitution word problem
I realise I posted this in the wrong forum sorry
this is grade 10
the problem is:
a silversmith has alloys that contain 40% silver and others that have 50% silver. A custom order requires 150g of 44% silver. How much of each alloy should be melted together to make the
so far i got:
Let x represent the 40% alloy
Let y represent the 50% alloy.
I rearranged the first eqaution to be y=-x+150.
Then subbed it into the second to be 0.4x+0.5(-x+150)=0.44
but when I solved this, i got 745.6. Where did I go wrong?
I realise I posted this in the wrong forum sorry
this is grade 10
the problem is:
a silversmith has alloys that contain 40% silver and others that have 50% silver. A custom order requires 150g of 44% silver. How much of each alloy should be melted together to make the
so far i got:
Let x represent the 40% alloy
Let y represent the 50% alloy.
I rearranged the first eqaution to be y=-x+150.
Then subbed it into the second to be 0.4x+0.5(-x+150)=0.44
but when I solved this, i got 745.6. Where did I go wrong?
your second equation is wrong. it should be equated to 66 (which is 44% of 150g) not 0.44. you let x and y represent the grams of silver in the first equation and then the percentage of silver in
the second. this inconsistency gave rise to your incorrect solution
thank you
January 21st 2008, 05:23 PM #2
January 21st 2008, 06:35 PM #3
Jan 2008 | {"url":"http://mathhelpforum.com/math-topics/26538-substitution-word-problem.html","timestamp":"2014-04-17T10:31:29Z","content_type":null,"content_length":"36593","record_id":"<urn:uuid:4f009c52-3cd2-4a8f-a8bd-8214fb947fc3>","cc-path":"CC-MAIN-2014-15/segments/1397609527423.39/warc/CC-MAIN-20140416005207-00039-ip-10-147-4-33.ec2.internal.warc.gz"} |
Difference Between Bar Graph and Histogram
Bar Graph vs Histogram
In statistics, summarizing and presentation of the data is important. It can be done either numerically using descriptive measures or graphically using the pie graphs, bar graphs, and many other
graphical representation methods.
What is a Bar Graph?
Bar graph is one of the main graphical representation methods in statistics. It is used to display distinct values of the qualitative data on a horizontal axis and the relative frequencies (or
frequencies or percents) of those values on a vertical axis. A bar with its height/ length proportional to the relative frequency represents each distinct value, and the bars are positioned in a way
that they do not touch each other. A bar graph with the above configuration is the most common and is known as a vertical bar graph or a column graph. But it is also possible to interchange the axes;
in that case the bars are horizontal.
The bar graph was first used in the 1786 book “The Commercial and Political Atlas” by William Playfair. Since then bar graph has become one of the most important tools in representing categorical
data. Use of bar graphs can be extended to represent more complex categorical data, such as time developing variables (election response), grouped data, and more.
What is a Histogram?
The histogram is another important graphical representation of data, and it can be considered as a development from the bar graph. In a histogram, the classes of the quantitative data are displayed
on the horizontal axis, and the frequency (or relative frequency or percents) of the classes are displayed on the y axis. A vertical bar usually represents the frequency (or relative frequency or
percents) of the class whose height is equal to its magnitude. Unlike the ordinary bar graphs, the bars are positioned to touch each other.
The variable in the X-axis-axis can be either single value grouped or limit grouped. For single-value grouping, distinct values of the observations are used to label the bars, with each such value
centered below its bar. For limit grouping or cut point grouping, lower class limits (or, equivalently, lower class cut points) are used to label the bars. Class marks or class midpoints centered
under the bars can also be used.
One of the major differences lies in the variable used in the X-axis-axis. In the histogram, the variable is a quantitative variable and can be either continuous or discrete. And it can be used to
represent density information about the datasets. In this case, the intervals used on the x-axis may vary from one to another, and on the y axis, the frequency density is marked. If the interval of
the X- axis is 1, then the histogram is equal to the relative frequency plot.
What is the difference between Bar Graph and Histogram?
• First and foremost, a histogram is a development from the bar graph, but it is not identical to a bar graph. Histograms are a type of bar graphs, but bar graphs definitely are not histograms.
• Bar graphs are used to plot categorical or qualitative data while histograms are used to plot quantitative data with the ranges of the data grouped into bins or intervals.
• Bar graphs are used to compare variables while histograms are used to show distributions of variables
• Bar graphs have spaces between two bars while histograms have no spaces between the bars. (The reason is that the x- axis in bar graphs are discrete categorical values while, in histograms, it is
either discrete or continuous quantitative).
• Histograms are used to illustrate the density of a variable in intervals; in this case the area of the bar represents the frequency of the variable.
Related posts: | {"url":"http://www.differencebetween.com/difference-between-bar-graph-and-vs-histogram/","timestamp":"2014-04-20T08:33:29Z","content_type":null,"content_length":"90128","record_id":"<urn:uuid:1e1128c0-7607-4259-aee5-218e153d48b9>","cc-path":"CC-MAIN-2014-15/segments/1398223204388.12/warc/CC-MAIN-20140423032004-00224-ip-10-147-4-33.ec2.internal.warc.gz"} |
Math Forum Discussions
Math Forum
Ask Dr. Math
Internet Newsletter
Teacher Exchange
Search All of the Math Forum:
Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.
Topic: 2 simple? questions and proposed answers
Replies: 4 Last Post: Jan 29, 2012 7:50 AM
Messages: [ Previous | Next ]
lnkerb Re: 2 simple? questions and proposed answers
Posted: Jan 7, 2012 3:46 AM
Posts: 1
From: Question #1 -
Bahrain It could be how prime is defined ... it is generally taught that prime numbers are those that can be divided by 1 and itself. I generally teach that a prime number has EXACTLY TWO
Registered: FACTORS. Thus 1 cannot be prime because it has only one factor, and 0 cannot be prime because it has infinite factors.
Date Subject Author
1/6/12 2 simple? questions and proposed answers Kent Holing
1/6/12 Re: 2 simple? questions and proposed answers Ed Wall
1/7/12 Re: 2 simple? questions and proposed answers lnkerb
1/9/12 Re: 2 simple? questions and proposed answers Gordon Palmer
1/29/12 Re: 2 simple? questions and proposed answers Kent Holing | {"url":"http://mathforum.org/kb/message.jspa?messageID=7639831","timestamp":"2014-04-16T06:04:25Z","content_type":null,"content_length":"20992","record_id":"<urn:uuid:79c60103-249e-4583-aba3-c1355d28b104>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00399-ip-10-147-4-33.ec2.internal.warc.gz"} |
RE: Maximum cardinality of an RDF model
From: Miles Sabin <MSabin@interx.com> Date: Fri, 2 Feb 2001 22:09:06 -0000 Message-ID: <23CF4BF2C499D411907E00508BDC95E116FBD3@ntmews_01.interx.com> To: www-rdf-logic@w3.org Cc: connolly@w3.org
Dan Connolly wrote,
> pat hayes wrote:
> > according to several members of the group which developed
> > RDF, the 'graph model' of a set of RDF triplets is intended
> > itself to be *the* model (in the sense from model theory) of
> > those triplets.
> Really? can you cite a source for that? I'd like to correct it.
Correct what? That the intended model is the graph/triplet model;
or any implication that there's a model (in the model-theoretic
sense) at all?
> I'm just sort of teaching myself all this model theory stuff
> as I go, but as far as I understand it, the semantics of RDF
> are just like the semantics of first-order logic, where the
> only terms are URIs (constant symbols) and existentially
> quantified variables, and the only formulas are ground
> propositions, conjuctions, and existentially quantified
> formulas.
That seems more or less right to me, with the following two
* There are no conjunctions in the way there are in typical
languages (tho', to be sure, you can get something like the
effect of asserting P&Q by asserting P and asserting Q).
* Even if there's no explicit formal model, there's a fairly
clear indication in RDF M&S that the quantifiers range over
resources, ie. things which have a URI, hence that the domain
of quantification is at most countably infinite ... because
there at most countably many URIs and no two distinct resources
can share a URI.
BTW, if you're looking around for an interesting logical
framework to embed RDF M&S, I'd recommend,
_A_Structuralist_Theory_of_Logic_, Arnold Koslow,
Cambridge University Press, 1992
_extremely_ highly. The salient point is that he manages to
crank a variety of first order logics (both classical and non-
classical) and a framework for building modal logics, out of
resources which aren't all that different from what RDF M&S has
to offer (I can elaborate on this if anyone's interested ... with
the caveat that I'm an ex-philosopher, not a logician).
Miles Sabin InterX
Internet Systems Architect 5/6 Glenthorne Mews
+44 (0)20 8817 4030 London, W6 0LJ, England
msabin@interx.com http://www.interx.com/
Received on Friday, 2 February 2001 17:09:47 GMT
This archive was generated by hypermail 2.2.0+W3C-0.50 : Monday, 7 December 2009 10:52:38 GMT | {"url":"http://lists.w3.org/Archives/Public/www-rdf-logic/2001Feb/0030.html","timestamp":"2014-04-24T10:08:28Z","content_type":null,"content_length":"10558","record_id":"<urn:uuid:a5e264a9-9ea5-4177-84db-ae8b0d1b74ee>","cc-path":"CC-MAIN-2014-15/segments/1398223206118.10/warc/CC-MAIN-20140423032006-00662-ip-10-147-4-33.ec2.internal.warc.gz"} |
Range of binomial probability, given a certain number of observations?
up vote 3 down vote favorite
Let's say I am given $n$ flips of a coin, $k$ of which are heads. These are iid flips.
Can I say, with probability $p > 1/2$, that the true probability of heads is in range $[p_1, p_2]$ ? What is that range?
How do I integrate prior knowledge of the binomial distribution? What if I have no prior knowledge?
pr.probability binomial-distribution
add comment
3 Answers
active oldest votes
There is a whole field around questions like this called bayesian statistics, it's been a while since I looked at this stuff but if I remember right.
Sadly you do need to have some pre-determined view of what p is. That is some before flipping the n coins you have a distribution in mind for the value of p (called the prior
distribution). This distribution changes as you flip the coins (you get a posterior distribution).
For example you might start out believing that the coin has a 50% chance of being fair and a 50% chance of coming up heads 2/3rds of the time. (You might believe this if you know the
up vote 3 person you got the coin from has both types of coins and there is a 50% chance he's trying to trick you).
down vote
The interesting (or at least nice) case is when your prior distribution is a "conjugate prior". Which basically means that your posterior distiribution for p is of the same parametric
family as your prior distribution. I believe the conjugate prior for this is the beta distribution, but you might want to google "conjugate prior" and "bayesian statistics".
Hope that helps.
add comment
1.The answer to your specific question can be found in the Wikipedia page:
The idea of estimating a distribution parameter is to construct a random variable whose expectation is the parameter needed to be estimated. In your example, we want to estimate the success
probability of a Bernoulli distribution and we construct a binomially distributed random variable by repeated trials. The key point is that the new random variable has the same average (after
up vote normalizing by n) but its standard deviation is smaller (by a factor of sqrt(n)) which gives better bounds on the estimated value. The confidence level is just a percetile of the distribution
1 down of the random variable used for the estimation (in your example you chose 50%).The interval size is a function of the percentile. The estimated value p* = k/n is inside the interval but the
vote interval is not symmetric in general around this value. In the Wikipedia page, several approximations for large n are given, based on the central limit theorem, which are usually used in
real-life estimations.
2.The above solution assumes prior knowledge that the distribution of a single trial is Bernoulli and that the trials are independent. Usually, one needs some prior knowledge to infere a
statistical parameter. Specifically in your question, prior knowldge would be that the coin flips are independent, or that at after some flip the success probability p had changed, etc.
add comment
Some very sharp bounds for questions like these are provided by something called a Chernoff bound. The example in the wikipedia article will give you what you need.
up vote 0 down vote Edit: Oh, I forgot to say that you need an estimator for the "true" probability, but I guess that the one you are using is just the average over the samples.
add comment
Not the answer you're looking for? Browse other questions tagged pr.probability binomial-distribution or ask your own question. | {"url":"http://mathoverflow.net/questions/9586/range-of-binomial-probability-given-a-certain-number-of-observations/9590","timestamp":"2014-04-17T18:23:22Z","content_type":null,"content_length":"59048","record_id":"<urn:uuid:fce486f5-9b6d-4850-8ea2-fa2a0d1c5c9b>","cc-path":"CC-MAIN-2014-15/segments/1398223205137.4/warc/CC-MAIN-20140423032005-00388-ip-10-147-4-33.ec2.internal.warc.gz"} |
2D and 3D Symmetric Registration using CUDA - File Exchange - MATLAB Central
This package contains tools for CUDA-accelerated non-rigid registration.
The registration code uses gradient descent to create a set of functions that transform one image to another.
The symmetric algorithm uses the determinant of the Jacobian matrix to impose mathematical symmetry on the process. Additionally, both images are transformed to some image in between the two,
providing numerical symmetry as well. In unison these properties mean that the algorithm will produce the same registration even if the order of the images is switched.
For more information on symmetric non-rigid registration, see "Symmetric Non-rigid Registration: A Geometric Theory and Some Numerical Techniques" by Tagare et al., JMIV, 2009.
The vanilla algorithm transforms one image to another with no symmetry constraints.
All registration code interfaces with included CUDA kernels to accelerate the process. The CUDA code has been tested on a Tesla M2090 card with CUDA 4.X and compute capability 2.X. It should work on
cards with compute capability 1.3 as well but has not been tested.
Also included is code to generate synthetic data that can be used to test the registration functions.
Please login to add a comment or rating. | {"url":"http://www.mathworks.com/matlabcentral/fileexchange/37685-2d-and-3d-symmetric-registration-using-cuda","timestamp":"2014-04-16T19:59:33Z","content_type":null,"content_length":"29686","record_id":"<urn:uuid:f36802e0-27eb-47ff-a330-49562a8df2b4>","cc-path":"CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00262-ip-10-147-4-33.ec2.internal.warc.gz"} |
College Basketball: Rating Individual 3FG%
Dec 29 2009
College Basketball: Rating Individual 3FG%
My last post presented a model for predicting 3FG% based on a player’s ability, age, and role in the offense. A comment by DSMok1 inspired the model I will present in this post. He writes:
I was considering how best to create an “equalized” measure of 3pt and 2pt % for college players, based on the opposition played and the usage percentage. In other words, I would create a
notional percentage for each player based on a usage rate of 20%, playing NCAA-average opposition.
Do you think that you could do a similar regression for 2Pt%, and post it?
Although he specifically requested 2FG%, in this post I will present a model of college basketball 3FG% that controls for player ability, opponent strength, experience, and role in the offense.
The Data
To build this model I collected each player’s made and attempted three point field goals for every season from 2002-03 to 2008-09, and I kept only those player seasons that attempted at least 50
three point field goals. I separated this data by opponent, and I kept track of how often this player was in the data set as a proxy for that player’s experience.
Also, I calculated every player’s usage% for each season. Usage% is the percentage of his team’s possessions that the player can be considered responsible for, as defined by Dean Oliver in Basketball
on Paper. Thus this usage% includes assists, and it is constructed using Dean’s formulas for the NBA from his book.
The Model
With this data I fit the following model:
$Pr({\tt 3FG make}) = {\tt logit}^{-1}(\alpha + \beta_{1}({\tt usage}) + \beta_{2}({\tt experience}) + \beta_{3}({\tt long}))$
This logistic regression was fit as a multilevel model to allow the intercept to vary by player and opponent. This allows us to estimate player ability while controlling for opponent strength. In
this model long indicates if the attempt is from the 2008-09 season in which the NCAA moved the three point line back to 20 feet 9 inches from 19 feet 9 inches.
The Results
The average player results are as follows:
• Coefficients: $\alpha = -0.490$, $\beta_{1} = -0.557$, $\beta_{2} = 0.028$, $\beta_{3} = -0.034$. The p-values for testing if the true values of these parameters are equal to zero are all less
than 0.01.
• Usage: The coefficient for usage, $\beta_{1} = -0.557$, suggests that for each additional 1% in an individual’s usage% the odds the individual makes a 3FG attempt are decreased by 0.55%. As we
would expect, this suggests that a player that increases their usage from 20% to 21% would expect to see their odds of making a 3pt FG attempt decrease by 0.55%
• Experience: The coefficient for experience, $\beta_{2} = 0.028$, suggests that for each one year increase in experience the odds the individual makes a 3FG attempt is increased by 2.8%.
• Long: The coefficient for the longer 3pt distance, $\beta_{3} = -0.034$, suggests that the odds of making a 3pt shot from the longer distance are 3.3% lower than the odds of making a 3pt shot at
the shorter distance.
Player Estimates
This model fit helps us cut through the noise and estimate a player’s ability against league average opponents. As the graphs below show, there is a lot of uncertainty in a player’s individual 3FG%
in any one season. Further compounding these yearly results is the fact that players face different levels of competition, and they may take on a larger role in their team’s offense as they gain
The first graph I will present is that of Davidson’s Stephen Curry:
This graph shows Stephen’s estimated ability as a function of experience (the x-axis) and usage (blue=10% usage, black=20% usage, and red=30% usage). Below the x-axis you will see the actual usage%
for each season to go along with the average percentile ranking of opponent 3FG% defense, where 50% represents average, >50% above average, and <50% below average opponents. The black dots and
associated lines extending from these dots represent the sample 3FG% for the season and the 95% confidence interval for the player’s true 3FG% ability during the season.
While Stephen ranks 11th in this model of all players from 2002-03 to 2008-09, current star of the College of Charleston, Andrew Goudelock, ranks a surprising 41st. His graph is below:
Another player graph that may be of interest is Duke’s J.J. Redick:
Translating to the NBA
Although this model helps us estimate a player’s ability in college, we’re ultimately interested in translating this to the NBA. There are a lot of highly ranked players that never play in an NBA
game, as simply being able to shoot 3pt shots well isn’t enough to succeed in the NBA.
That said, the next step is to examine players that actually make it to the NBA and determine what this model says about their ability to shoot 3pt shots against that level of competition.
If you enjoyed this post, use RSS to get notified of new posts.
4 Comments on this post
1. DSMok1 said:
Excellent work, Ryan!
A quick question: why exactly does Goudelock’s “true” range run so far below his actual performance? Because of his opponent strength? But the opponent strength doesn’t seem to be that low. Could
you go into more depth about how that opponent adjustment works in a case like that? I’m a newbie at multilevel modeling!
December 29th, 2009 at 10:17 am
2. Ryan said:
The model shrinks a player’s estimate to the mean based on the amount of data we have. In this case we have two years of data and are estimating his ability based on his performance against
slightly above average opponents in 2007-08 (~52%) and below average opponents in 2008-09 (~41%).
What I really hope to show with the plots of the actual results is that there is a large amount of uncertainty after any one season. This model shrinks that confidence interval to (36%, 43%) for
2007-08. I couldn’t find a great way to compare the two, but that’s the general idea and benefit of what this model is doing.
December 29th, 2009 at 11:39 am
3. DSMok1 said:
In other words, this directly provides regression to the mean based on the number of data points (3PA’s)? That makes sense, I guess. Is that at all comparable to a Bayesian framework for
regression to the mean? I would guess so.
December 29th, 2009 at 2:15 pm
4. Ryan said:
It isn’t fully Bayesian, but we do assume the player and team coefficients come from some normal distribution. This provides the shrinkage.
December 29th, 2009 at 3:04 pm | {"url":"http://www.basketballgeek.com/2009/12/29/college-basketball-rating-individual-3fg/","timestamp":"2014-04-16T04:11:20Z","content_type":null,"content_length":"45921","record_id":"<urn:uuid:3f7c2338-ce14-446c-94c5-6f7993aa34ea>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00210-ip-10-147-4-33.ec2.internal.warc.gz"} |
Francois-Joseph Servois: Priest, Artillery Officer, and Professor of Mathematics - Battles for Paris and for the Foundations of Calculus
At the start of 1814, Servois was called back into active duty, one last time, to defend Paris from the Austrian and Prussian armies [Boyer 1895a]. Not only was France involved in a battle that would
shape its destiny, but calculus also was in a similar state of upheaval. Mathematicians knew there were problems with the foundations of calculus, but Augustin-Louis Cauchy (1789-1857) had not yet
begun his revision of those foundations.
Figure 4. The Battle of Paris, 1814, as painted by Bogdan Willewalde in 1834 (public domain)
During his time in the military, Servois had become acquainted with Joseph Diaz Gergonne (1771-1859). Servois' friendship with Gergonne was particularly helpful in building his mathematical
publication record. Gergonne was the editor of the Annales des mathématiques pures et appliquées, often called Annales de Gergonne. Servois' most celebrated paper, “Essai sur un nouveau mode
d'exposition des principes du calcul différential” [Servois 1814a] (Essay on a New Method of Exposition of the Principles of Differential Calculus), his primary work on the foundations of the
differential calculus, appeared late in 1814. Following Lagrange, his work proposed using power series expansion as fundamental to the differential calculus. He defined the differential operator as
an infinite series, which he called an infinitinôme, in powers of the difference operator: \[dz = \Delta z - \frac{1}{2} \Delta^2 z + \frac{1}{3} \Delta^3 z - \frac{1}{4} \Delta^4 z + \ldots.\]
Using this operator, he demonstrated how to find differentials of functions and was able to discover the basic laws of the differential calculus, such as the power and product rules. This paper truly
is remarkable because, at the heart of Servois' arguments, lies the use of the linear properties of operators in proving theorems about the calculus. Servois didn't use the terms “linear” or
“operator,” but rather he used the words commutative and distributive for the first time in their mathematical sense. In reviewing the contents of this paper, the Commissioners of the Institut de
France praised him for having “done something that is very useful for the science [of analysis]” [Servois 1814a, p. 140].
Figure 5. The Institut de France (photograph by Benh Lieu Song, 2007, licensed under Creative Commons Attribution-Share Alike 3.0 Unported license)
Servois' second work on the foundations of calculus was “Réflexions sur les divers systèmes d'exposition des principes du calcul différentiel, et, en particulier, sur la doctrine des infiniment
petits” [Servois 1814b] (Reflections on the Various Systems of Exposition of the Principles of the Differential Calculus and, in particular, on the Doctrine of the Infinitely Small). This work gives
us insight into Servois' personal views of the history of calculus and into his philosophical positions, particularly on placing calculus on an algebraic foundation. This paper began with a
historical overview of calculus from its birth to the time of Lagrange. Servois discussed the contributions made by well-known mathematicians, such as Newton, Leibniz, Taylor, Euler, and Lagrange, to
the differential calculus. In addition, he pointed out where, in his opinion, some mathematicians had erred. For instance, he stated that the use of infinitesimals is difficult for mathematicians to
justify as a basis for calculus. The remainder of the work is devoted to his justification of development of calculus by means of power series and to a harsh criticism of the mathematical philosophy
of Josef-Maria Hoëné-Wronski (1776-1853).
Editor’s note: For a more detailed analysis and an English translation of Servois’ “Reflections,” see the Convergence article, “Servois’ 1814 Essay on the Principles of the Differential Calculus,
with an English Translation,” by the author and Robert E. Bradley. | {"url":"http://www.maa.org/publications/periodicals/convergence/francois-joseph-servois-priest-artillery-officer-and-professor-of-mathematics-battles-for-paris-and","timestamp":"2014-04-21T03:01:47Z","content_type":null,"content_length":"105737","record_id":"<urn:uuid:b52a404d-a765-4f9b-a9b4-d6eee4a5c1e0>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00660-ip-10-147-4-33.ec2.internal.warc.gz"} |