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Math Tools Discussion: All Topics in Geometry, Smartboard and Geometric Constructions
Discussion: All Topics in Geometry
Topic: Smartboard and Geometric Constructions
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Subject: RE: Smartboard and Geometric Constructions
Author: ihor
Date: Aug 30 2006
> On Aug 30 2006, ihor wrote:
>A more significant
> mathematical
> point is to make sure that the students understand that
> a line
> represents an infinite set of points
On Aug 30 2006, Mathman replied:
Not to be argumentative,
> and since it has been forty years since I saw the stuff [Affine
> Geometry] and might well be wrong, but aren't "point" and "line"
> distinct? That is a foundation for duality, and although points may
> lie in a line, that does not make a line an infinite set of points.
> So, two points are necessary and sufficient to define one line [will
> lie on exactly one line], and two lines can intersect in only one
> point.
Ihor responds:
>but aren't "point" and "line"
> distinct?
I'm only talking about points whose order pairs satisfy the equation that is
represented by the line.
I think we have strayed from the orginal point which Susan answers very well in
an another reply to the Smartboard question.
Reply to this message Quote this message when replying?
yes no
Post a new topic to the All Content in Geometry discussion
Discussion Help | {"url":"http://mathforum.org/mathtools/discuss.html?context=cell&do=r&msg=26007","timestamp":"2014-04-21T15:52:31Z","content_type":null,"content_length":"16323","record_id":"<urn:uuid:a558684f-e3a0-4138-b75f-f65b14448435>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00158-ip-10-147-4-33.ec2.internal.warc.gz"} |
Longest Math Proof in History Completed | RCScience
September 12, 2011
Longest Math Proof in History Completed
Esther Inglis-Arkell, io9
AP Photo
The longest math proof in the world has just been completed. It began in the 1970s and was worked on by 100 mathematicians. Take a look at the math equivalent of endurance running.
TAGGED: Mathematical Theorem, Mathematics
October 14, 2013
What's It Like to Be a Mathematician?
Jacob Aron, Slate
The way mathematics is taught is akin to an art class in which students are only taught how to paint a fence and are never shown the paintings of the great masters. When, later on in life, the
subject of mathematics comes up,... more ››
October 9, 2013
Big Data Requires a New Mathematics
Jennifer Ouellette, Quanta Mag.
Simon DeDeo, a research fellow in applied mathematics and complex systems at the Santa Fe Institute, had a problem. He was collaborating on a new project analyzing 300 years’ worth of data from the
archives of... more ››
October 8, 2013
How Indian Mysticism Revolutionized Math
Alex Bellos, Guardian
What has India given to the world? Nothing. The mathematical concept of zero emerged in India about one and a half thousand years ago, and this summer I travelled there to visit a temple where the
oldest known zero symbols are... more ››
The longest math proof in the world has just been completed. It began in the 1970s and was worked on by 100 mathematicians. Take a look at the math equivalent of endurance running.
October 14, 2013
What's It Like to Be a Mathematician?
Jacob Aron, Slate
The way mathematics is taught is akin to an art class in which students are only taught how to paint a fence and are never shown the paintings of the great masters. When, later on in life, the
subject of mathematics comes up,... more ››
October 9, 2013
Big Data Requires a New Mathematics
Jennifer Ouellette, Quanta Mag.
Simon DeDeo, a research fellow in applied mathematics and complex systems at the Santa Fe Institute, had a problem. He was collaborating on a new project analyzing 300 years’ worth of data from the
archives of... more ››
October 8, 2013
How Indian Mysticism Revolutionized Math
Alex Bellos, Guardian
What has India given to the world? Nothing. The mathematical concept of zero emerged in India about one and a half thousand years ago, and this summer I travelled there to visit a temple where the
oldest known zero symbols are... more ›› | {"url":"http://www.realclearscience.com/2011/09/12/longest_math_proof_in_history_completed_243025.html","timestamp":"2014-04-17T21:35:46Z","content_type":null,"content_length":"20049","record_id":"<urn:uuid:2065bfc9-1961-4943-a37a-c0bab99ec1cf>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00025-ip-10-147-4-33.ec2.internal.warc.gz"} |
[R] searching through a matrix
Spencer Graves spencer.graves at pdf.com
Mon Feb 28 00:53:47 CET 2005
Does the following solve your problem:
is.na(array(c(NA, 1:5), dim=c(2,3)))
[,1] [,2] [,3]
[1,] TRUE FALSE FALSE
[2,] FALSE FALSE FALSE
spencer graves
Jessica Higgs wrote:
> I am trying to search through a matrix I have (x) to see if there are
> any missing values. This is what I have been using:
> for (i in 1:3455)
> for (j in 1:24)
> if (is.na(x[i,j])) break;
> i
> j
> When I run the program, it either freezes or stops and returns the
> last entry in the matrix (x[3455,24]). I need to figure out what
> values in my data matrix are missing because I am using prcomp and it
> keeps giving me an error starting that there are missing or infinite
> values in x. My data matrix does have some values that are equal to
> zero but zero doesn't mean that the data is missing. it just indicates
> that that value is equivalent to zero. Any suggestions for how to fix
> this problem?
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
> http://www.R-project.org/posting-guide.html
More information about the R-help mailing list | {"url":"https://stat.ethz.ch/pipermail/r-help/2005-February/066854.html","timestamp":"2014-04-18T20:52:18Z","content_type":null,"content_length":"4035","record_id":"<urn:uuid:241d58ca-e215-4cf9-8292-81ea69e40ba1>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00023-ip-10-147-4-33.ec2.internal.warc.gz"} |
22-XX Topological groups, Lie groups {For transformation groups, see 54H15, 57Sxx, 58-XX. For abstract harmonic analysis, see 43-XX}
22-00 General reference works (handbooks, dictionaries, bibliographies, etc.)
22-01 Instructional exposition (textbooks, tutorial papers, etc.)
22-02 Research exposition (monographs, survey articles)
22-03 Historical (must also be assigned at least one classification number from Section 01)
22-04 Explicit machine computation and programs (not the theory of computation or programming)
22-06 Proceedings, conferences, collections, etc.
22Axx Topological and differentiable algebraic systems {For topological rings and fields, see 12Jxx, 13Jxx, 16W80}
22Bxx Locally compact abelian groups (LCA groups)
22Cxx Compact groups
22Dxx Locally compact groups and their algebras
22Exx Lie groups {For the topology of Lie groups and homogeneous spaces, see 57Sxx, 57Txx; for analysis thereon, see 43A80, 43A85, 43A90}
22Fxx Noncompact transformation groups | {"url":"http://www.ams.org/mathscinet/msc/msc2010.html?t=22-XX","timestamp":"2014-04-16T14:18:19Z","content_type":null,"content_length":"14034","record_id":"<urn:uuid:9e54ac64-398f-46b4-9bad-afa7fbf93888>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00343-ip-10-147-4-33.ec2.internal.warc.gz"} |
Exercise 2.73
Section 2.3.2 described a program that performs symbolic differentiation:
(define (deriv exp var)
(cond ((number? exp) 0)
((variable? exp) (if (same-variable? exp var) 1 0))
((sum? exp)
(make-sum (deriv (addend exp) var)
(deriv (augend exp) var)))
((product? exp)
(make-product (multiplier exp)
(deriv (multiplicand exp) var))
(make-product (deriv (multiplier exp) var)
(multiplicand exp))))
<more rules can be added here>
(else (error "unknown expression type -- DERIV" exp))))
We can regard this program as performing a dispatch on the type of the expression to be differentiated. In this situation the “type tag” of the datum is the algebraic operator symbol (such as +) and
the operation being performed is deriv. We can transform this program into data-directed style by rewriting the basic derivative procedure as
(define (deriv exp var)
(cond ((number? exp) 0)
((variable? exp) (if (same-variable? exp var) 1 0))
(else ((get 'deriv (operator exp)) (operands exp)
(define (operator exp) (car exp))
(define (operands exp) (cdr exp))
1. Explain what was done above. Why can’t we assimilate the predicates number? and same-variable? into the data-directed dispatch?
2. Write the procedures for derivatives of sums and products, and the auxiliary code required to install them in the table used by the program above.
3. Choose any additional differentiation rule that you like, such as the one for exponents (exercise 2.56), and install it in this data-directed system.
4. In this simple algebraic manipulator the type of an expression is the algebraic operator that binds it together. Suppose, however, we indexed the procedures in the opposite way, so that the
dispatch line in deriv looked like
((get (operator exp) 'deriv) (operands exp) var)
What corresponding changes to the derivative system are required? | {"url":"http://sicpinclojure.com/?q=exercise/exercise-2-73","timestamp":"2014-04-18T15:47:24Z","content_type":null,"content_length":"36274","record_id":"<urn:uuid:c241c16b-687b-4268-aa4c-78892ceacd44>","cc-path":"CC-MAIN-2014-15/segments/1397609533957.14/warc/CC-MAIN-20140416005213-00311-ip-10-147-4-33.ec2.internal.warc.gz"} |
Reply to comment
Sara Santos busking on the streets of Krakow.
The 6th European Congress of Mathematics, which took place in Krakow at the beginning of July, wasn't just about mathematicians talking to each other. On the streets of Krakow maths buskers were
entertaining the public, handcuffing innocent Krakowians, constructing emergency pentagons and reading minds. So what is maths busking all about? We caught up with Sara Santos, the director of the
project, and one of her volunteers to find out.
To find out more about mathematical busking visit the maths busking website. | {"url":"http://plus.maths.org/content/comment/reply/5729","timestamp":"2014-04-18T21:29:13Z","content_type":null,"content_length":"21673","record_id":"<urn:uuid:84624380-b2f6-43bb-b5ce-b3bb84628ea6>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.9/warc/CC-MAIN-20140416005215-00477-ip-10-147-4-33.ec2.internal.warc.gz"} |
How much is alpha?
Hello ! We have a square - find alpha ?
There is not enough information to answer. There must be more given. What else do you know?
Rotate triangle AEB about B, until A rests on C. Label the new position of "E" as "G". Draw a line from E to G. Angle EBG is 90 degrees. Angles BEG and BGE are 45 degrees. Use Pythagoras' Theorem to
calculate |EG| Triangle ECG now has all 3 sides (if you let |AE|=1, |EB|=2, |EC|=3) and you rearrange the law of cosines to calculate "alpha-45" degrees and therefore "alpha". | {"url":"http://mathhelpforum.com/geometry/183276-how-much-alpha-print.html","timestamp":"2014-04-19T21:50:34Z","content_type":null,"content_length":"6251","record_id":"<urn:uuid:eb317cdd-e8c5-4ceb-8d2d-a43668b26b90>","cc-path":"CC-MAIN-2014-15/segments/1397609537376.43/warc/CC-MAIN-20140416005217-00406-ip-10-147-4-33.ec2.internal.warc.gz"} |
Large-scale optimization of eigenvalues
Results 1 - 10 of 75
- SIAM REVIEW , 1996
"... ..."
- IEEE TRANSACTIONS ON INFORMATION THEORY , 2006
"... Motivated by applications to sensor, peer-to-peer, and ad hoc networks, we study distributed algorithms, also known as gossip algorithms, for exchanging information and for computing in an
arbitrarily connected network of nodes. The topology of such networks changes continuously as new nodes join a ..."
Cited by 208 (5 self)
Add to MetaCart
Motivated by applications to sensor, peer-to-peer, and ad hoc networks, we study distributed algorithms, also known as gossip algorithms, for exchanging information and for computing in an
arbitrarily connected network of nodes. The topology of such networks changes continuously as new nodes join and old nodes leave the network. Algorithms for such networks need to be robust against
changes in topology. Additionally, nodes in sensor networks operate under limited computational, communication, and energy resources. These constraints have motivated the design of “gossip ”
algorithms: schemes which distribute the computational burden and in which a node communicates with a randomly chosen neighbor. We analyze the averaging problem under the gossip constraint for an
arbitrary network graph, and find that the averaging time of a gossip algorithm depends on the second largest eigenvalue of a doubly stochastic matrix characterizing the algorithm. Designing the
fastest gossip algorithm corresponds to minimizing this eigenvalue, which is a semidefinite program (SDP). In general, SDPs cannot be solved in a distributed fashion; however, exploiting problem
structure, we propose a distributed subgradient method that solves the optimization problem over the network. The relation of averaging time to the second largest eigenvalue naturally relates it to
the mixing time of a random walk with transition probabilities derived from the gossip algorithm. We use this connection to study the performance and scaling of gossip algorithms on two popular
networks: Wireless Sensor Networks, which are modeled as Geometric Random Graphs, and the Internet graph under the so-called Preferential Connectivity (PC) model.
, 2005
"... We propose a new interior point based method to minimize a linear function of a matrix variable subject to linear equality and inequality constraints over the set of positive semidefinite
matrices. We show that the approach is very efficient for graph bisection problems, such as max-cut. Other appli ..."
Cited by 207 (17 self)
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We propose a new interior point based method to minimize a linear function of a matrix variable subject to linear equality and inequality constraints over the set of positive semidefinite matrices.
We show that the approach is very efficient for graph bisection problems, such as max-cut. Other applications include max-min eigenvalue problems and relaxations for the stable set problem.
- Systems and Control Letters , 2003
"... We consider the problem of finding a linear iteration that yields distributed averaging consensus over a network, i.e., that asymptotically computes the average of some initial values given at
the nodes. When the iteration is assumed symmetric, the problem of finding the fastest converging linear ..."
Cited by 190 (12 self)
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We consider the problem of finding a linear iteration that yields distributed averaging consensus over a network, i.e., that asymptotically computes the average of some initial values given at the
nodes. When the iteration is assumed symmetric, the problem of finding the fastest converging linear iteration can be cast as a semidefinite program, and therefore efficiently and globally solved.
These optimal linear iterations are often substantially faster than several common heuristics that are based on the Laplacian of the associated graph.
- SIAM Journal on Optimization , 1997
"... . A central drawback of primal-dual interior point methods for semidefinite programs is their lack of ability to exploit problem structure in cost and coefficient matrices. This restricts
applicability to problems of small dimension. Typically semidefinite relaxations arising in combinatorial applic ..."
Cited by 141 (6 self)
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. A central drawback of primal-dual interior point methods for semidefinite programs is their lack of ability to exploit problem structure in cost and coefficient matrices. This restricts
applicability to problems of small dimension. Typically semidefinite relaxations arising in combinatorial applications have sparse and well structured cost and coefficient matrices of huge order. We
present a method that allows to compute acceptable approximations to the optimal solution of large problems within reasonable time. Semidefinite programming problems with constant trace on the primal
feasible set are equivalent to eigenvalue optimization problems. These are convex nonsmooth programming problems and can be solved by bundle methods. We propose replacing the traditional polyhedral
cutting plane model constructed from subgradient information by a semidefinite model that is tailored for eigenvalue problems. Convergence follows from the traditional approach but a proof is
included for completene...
, 1993
"... A tutorial introduction to the complex structured singular value (µ) is presented, with an emphasis on the mathematical aspects of µ. The µ-based methods discussed here have been useful for
analyzing the performance and robustness properties of linear feedback systems. Several tests ..."
Cited by 119 (10 self)
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A tutorial introduction to the complex structured singular value (µ) is presented, with an emphasis on the mathematical aspects of µ. The µ-based methods discussed here have been useful for analyzing
the performance and robustness properties of linear feedback systems. Several tests
, 2003
"... Optimization problems involving the eigenvalues of symmetric and nonsymmetric matrices present a fascinating mathematical challenge. Such problems arise often in theory and practice,
particularly in engineering design, and are amenable to a rich blend of classical mathematical techniques and contemp ..."
Cited by 92 (13 self)
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Optimization problems involving the eigenvalues of symmetric and nonsymmetric matrices present a fascinating mathematical challenge. Such problems arise often in theory and practice, particularly in
engineering design, and are amenable to a rich blend of classical mathematical techniques and contemporary optimization theory. This essay presents a personal choice of some central mathematical
ideas, outlined for the broad optimization community. I discuss the convex analysis of spectral functions and invariant matrix norms, touching briey on semide nite representability, and then
outlining two broader algebraic viewpoints based on hyperbolic polynomials and Lie algebra. Analogous nonconvex notions lead into eigenvalue perturbation theory. The last third of the article
concerns stability, for polynomials, matrices, and associated dynamical systems, ending with a section on robustness. The powerful and elegant language of nonsmooth analysis appears throughout, as a
unifying narrative thread.
- In Proceedings of the DIMACS Workshop on Quadratic Assignment Problems, volume 16 of DIMACS Series in Discrete Mathematics and Theoretical Computer Science , 1994
"... . Quadratic Assignment Problems model many applications in diverse areas such as operations research, parallel and distributed computing, and combinatorial data analysis. In this paper we survey
some of the most important techniques, applications, and methods regarding the quadratic assignment probl ..."
Cited by 91 (16 self)
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. Quadratic Assignment Problems model many applications in diverse areas such as operations research, parallel and distributed computing, and combinatorial data analysis. In this paper we survey some
of the most important techniques, applications, and methods regarding the quadratic assignment problem. We focus our attention on recent developments. 1. Introduction Given a set N = f1; 2; : : : ;
ng and n \Theta n matrices F = (f ij ) and D = (d kl ), the quadratic assignment problem (QAP) can be stated as follows: min p2\Pi N n X i=1 n X j=1 f ij d p(i)p(j) + n X i=1 c ip(i) ; where \Pi N is
the set of all permutations of N . One of the major applications of the QAP is in location theory where the matrix F = (f ij ) is the flow matrix, i.e. f ij is the flow of materials from facility i
to facility j, and D = (d kl ) is the distance matrix, i.e. d kl represents the distance from location k to location l [62, 67, 137]. The cost of simultaneously assigning facility i to locat...
- SIAM REVIEW , 2003
"... We consider a symmetric random walk on a connected graph, where each edge is labeled with the probability of transition between the two adjacent vertices. The associated Markov chain has a
uniform equilibrium distribution; the rate of convergence to this distribution, i.e. the mixing rate of the Mar ..."
Cited by 90 (15 self)
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We consider a symmetric random walk on a connected graph, where each edge is labeled with the probability of transition between the two adjacent vertices. The associated Markov chain has a uniform
equilibrium distribution; the rate of convergence to this distribution, i.e. the mixing rate of the Markov chain, is determined by the second largest (in magnitude) eigenvalue of the transition
matrix. In this paper we address the problem of assigning probabilities to the edges of the graph in such a way as to minimize the second largest magnitude eigenvalue, i.e., the problem of finding
the fastest mixing Markov chain on the graph. We show that
- in Protocol Testing and Its Complexity", Information Processing Letters Vol.40 , 1995
"... We describe a potential reduction method for convex optimization problems involving matrix inequalities. The method is based on the theory developed by Nesterov and Nemirovsky and generalizes
Gonzaga and Todd's method for linear programming. A worst-case analysis shows that the number of iterations ..."
Cited by 87 (21 self)
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We describe a potential reduction method for convex optimization problems involving matrix inequalities. The method is based on the theory developed by Nesterov and Nemirovsky and generalizes Gonzaga
and Todd's method for linear programming. A worst-case analysis shows that the number of iterations grows as the square root of the problem size, but in practice it appears to grow more slowly. As in
other interior-point methods the overall computational effort is therefore dominated by the least-squares system that must be solved in each iteration. A type of conjugate-gradient algorithm can be
used for this purpose, which results in important savings for two reasons. First, it allows us to take advantage of the special structure the problems often have (e.g., Lyapunov or algebraic Riccati
inequalities). Second, we show that the polynomial bound on the number of iterations remains valid even if the conjugate-gradient algorithm is not run until completion, which in practice can greatly
reduce the computational effort per iteration. | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=215212","timestamp":"2014-04-18T00:52:43Z","content_type":null,"content_length":"37998","record_id":"<urn:uuid:4e38a709-2aa2-48f7-a32c-45c115e46bbf>","cc-path":"CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00036-ip-10-147-4-33.ec2.internal.warc.gz"} |
Irvington, NJ Algebra 1 Tutor
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Estimate the sound level in decibels you would expect at a point 8.0 from a loudspeaker connected in turn to each amp
Number of results: 92,964
Expensive amplifier A is rated at 250 W, while the more modest amplifier B is rated at 45 W. (a) Estimate the sound level in decibels you would expect at a point 3.5 m from a loudspeaker connected in
turn to each amp. (b) Will the expensive amp sound twice as loud as the ...
Thursday, November 7, 2013 at 2:58am by emy
Expensive amplifier A is rated at 210W , while the more modest amplifier B is rated at 45W . Estimate the sound level in decibels you would expect at a point 9.0m from a loudspeaker connected in turn
to each amp.
Monday, November 18, 2013 at 8:55pm by waves
physical sciences
a thunderclap 200m away delivers 3.75E^-5 W/m^2 of sound. What is the sound level in decibels? What is the intensity of the sound at 10m from the thunderclap? What is the sound level in decibels 10m
from the thunderclap?
Tuesday, October 21, 2008 at 10:11pm by Valerie
The Occupational Safety and health Administration (OSHA) estblished permissible sound exposures in the work-place. This is modeled by the equation: S=-2.817H + 108.9 Use the formula for the problem:
S=maximun permissible sound level in decibels H=number of hours of exposure ...
Friday, February 8, 2008 at 9:35am by G
The Occupational Safety and health Administration (OSHA) estblished permissible sound exposures in the work-place. This is modeled by the equation: S=-2.817H + 108.9 Use the formula for the problem:
S=maximun permissible sound level in decibels H=number of hours of exposure ...
Friday, February 8, 2008 at 12:26pm by G
The intensity due to a number of indipendent sound sources is the sum of the individual intensities. A) When four quadruplets cry simultaneously, how many decibels greater is the sound intensity
level than when a single one cries? B) To increase the sound intensity level again...
Saturday, March 29, 2008 at 5:25pm by Amanda
Factory Worker A is exposed to a sound of a constant intensity of 6.30 × 10–4 W m–2 at a distance of 1.00 m from a noisy machine. The frequency of the sound is 4.00 kHz. (a) What is the sound level
in decibels (dB) of this sound? (b) By what factor would the amplitude of this ...
Sunday, June 12, 2011 at 8:45pm by nancy
Algebra II
1. A sound has an intensity of 5.92 „e 1025W/m2.What is the loudness of the sound in decibels? Use . 2. Suppose you decrease the intensity of a sound by 45%. By how many decibels would the loudness
be decreased?
Wednesday, January 25, 2012 at 9:57pm by Nick
Formula given: The decibel (dB) scale for measuring loudness, d, is given by the formula d = 10 log base ten(I X 12 to the power of 12) where I is the intensity of sound in watts per square metre. a)
Find the number of decibels of sound produced by a jet engine at a distance ...
Sunday, February 16, 2014 at 5:25pm by Donna
Algebra 2
Can someone help me? I'm really confused. The loudness L of a sound in decibels is given by L= 10log(base10)R, where R is the sound's relative intensity. If the intensity of a certain sound is
tripled, by how many decibels does the sound increase?
Monday, March 1, 2010 at 9:03pm by Ana
algebra 2
the loudness L of a sound in decibels is given L=10log10R, where R is the sound's relative intensity. if the intensity of a certain sound is tripled, by how many decibels does the sound increase? due
tomorrow, tuesday, april 5th, 2011. help pleaseeeee ;)
Monday, April 4, 2011 at 7:13pm by math challenged
physics sound
1. A stone is dropped from rest into a well. The sound of the splash is heard exactly 1.50 s later. Find the depth of the well if the air temperature is 10.0°C. 2. Calculate the sound level in
decibels of a sound wave that has an intensity of 2.25 µW/m2
Thursday, April 17, 2008 at 8:28pm by Chris Adison
You need to include the dimensions of the numbers you are quoting. Is the distance in meters? Is the sound level in decibels? Whatever it is, say so. A 45 dB increase in sound level is a factor of 10
^4.5 = 31,620 increase in sound power per area. Use the inverse square law to ...
Thursday, March 11, 2010 at 10:45pm by drwls
Your question: The Occupational Safety and health Administration (OSHA) estblished permissible sound exposures in the work-place. This is modeled by the equation: S = -2.817H + 108.9 Use the formula
for the problem: S = maximun permissible sound level in decibels H = number of...
Friday, February 8, 2008 at 12:26pm by Guido
Grade 12 Math Please help!!
Please can anyone answer this which I have posted previously? Anyone with knowledge of this area of maths? Thanks in anticipation. :) Formula given: The decibel (dB) scale for measuring loudness, d,
is given by the formula d = 10 log base ten(I X 12 to the power of 12) where I...
Monday, February 17, 2014 at 12:05am by Donna
physics repost
Calculate the sound level in decibels of a sound wave that has an intensity of 2.25 µW/m2. F
Saturday, April 19, 2008 at 8:12pm by chris adison
If one machine creates a level of 90 decibels, what decibel level would be generated by 50 of these machines? If sounds are created at random, the net intensity is found by adding the individual
intensities-not the decibels. Ignore distance effects.
Wednesday, April 6, 2011 at 5:58pm by Lindsey
phyiscs repost 2
Calculate the sound level in decibels of a sound wave that has an intensity of 2.25 µW/m2 For this question don't I use B=log10(I/IO) So it would be B=log10(2.25/1.0e-12) right?
Saturday, April 19, 2008 at 8:14pm by chris adison
1. A stone is dropped from rest into a well. The sound of the splash is heard exactly 1.50 s later. Find the depth of the well if the air temperature is 10.0°C. 2. Calculate the sound level in
decibels of a sound wave that has an intensity of 2.25 µW/m2
Friday, April 18, 2008 at 1:44pm by Chris Adison
physics repost
1. A stone is dropped from rest into a well. The sound of the splash is heard exactly 1.50 s later. Find the depth of the well if the air temperature is 10.0°C. 2. Calculate the sound level in
decibels of a sound wave that has an intensity of 2.25 µW/m2
Saturday, April 19, 2008 at 9:32am by chris adison
The loudest sound in Earth’s history is believed to be the explosion of the volcanic island of Krakatoa, back in 1883. Physicists estimate that the intensity of the sound produced was 1.0×107 W/m2.
How many decibels does this correspond to?
Monday, July 29, 2013 at 10:13pm by neven
On th decibel scale, the loudness of a sound, in decibels, is given by D=10log(I/I0), where I is the intensity of a sound barely audible to the human ear. If the intensity of a sound is 10^11I0, what
is the loudness in decibels? (please note the I0 is supossed to be I lower 0.
Sunday, April 18, 2010 at 12:27pm by Help!!
decibels sound level
Saturday, April 24, 2010 at 6:32pm by Damon
A stereo speaker emits sound waves with a power output of 100W. (a) find the intensity 10.0m from the source. B) find the intensity level in decibels, at this distance. C) at what distance would you
experience the sound at the threshold of pain,120dB?
Tuesday, April 10, 2007 at 4:45pm by sammy
Two sounds have measured intensities of I1 = 100 W/m2 and I2 = 300 W/m2. By how many decibels is the level of sound 1 lower than that of sound 2?
Saturday, March 19, 2011 at 12:57pm by Jane
Reading Comp.
The modern hearing aid is, essentially, a miniature public-address system, which converts sound into electrical energy and then converts it back to sound again. A hearing aid can increase the sound
level by sixty decibels or more, but it also distorts the sound to some extent...
Monday, March 28, 2011 at 11:02pm by Arlynn
While peforming a concert , Queen plays at average intensity levels of 2 W/m^2. Then during a sudden loud part of the music, the amplitude of the sound waves , initially A1 increases by a factor of
5. By how many decibels does the sound level increase?
Monday, June 13, 2011 at 7:50pm by Matt
The sound level in decibels is typically expressed as β = 10 log (I/I0), but since sound is a pressure wave, the sound level can be expressed in terms of a pressure difference. Intensity depends on
the amplitude squared, so the expression is β = 20 log (P/P0), where ...
Wednesday, November 16, 2011 at 6:30am by maria
The sound level in decibels is typically expressed as β = 10 log (I/I0), but since sound is a pressure wave, the sound level can be expressed in terms of a pressure difference. Intensity depends on
the amplitude squared, so the expression is β = 20 log (P/P0), where ...
Wednesday, November 16, 2011 at 6:34pm by melisa
The sound level in decibels is typically expressed as β = 10 log (I/I0), but since sound is a pressure wave, the sound level can be expressed in terms of a pressure difference. Intensity depends on
the amplitude squared, so the expression is β = 20 log (P/P0), where ...
Thursday, November 17, 2011 at 5:51am by maria
I am reposting this since I did so incorrectly the first time. (I apologize for posting it twice the first time as well...I accidently hit refresh) Here is the problem with the proper units: You are
trying to overhear a juicy conversation, but from your distance of 15.0m , it ...
Monday, January 18, 2010 at 10:25pm by Nicole
Jane and Sam alternately pound a railroad spike into a tie with their hammers. The crew chief has a migraine, and notes that Jane's hammer blows cause a sound with intensity 5.0 times greater than
the sound that Sam makes when he swings his hammer. What is the difference in ...
Tuesday, April 5, 2011 at 4:33pm by shanelle james
Find the number of decibels for the power of the sound. Round to nearest decibels. A rock concert, 5.21 x 10^-6watts/cm^2 D= dB
Saturday, November 3, 2012 at 9:34pm by CLara
A speaker at an open-air concert emits 600 of sound power, radiated equally in all directions. a)What is the intensity of the sound 7.00 from of the speaker? I =0.974 W/m^2 b) What sound intensity
level would you experience there if you did not have any protection for your ...
Monday, May 28, 2012 at 3:36pm by JessicaP
That sound level must correspond to a certain distance from the rocket. You don't say what it is. Sound level increases as you get closer. Use the definition of decibels. dB = 10 * log(I/1.0 x 10^-12
W/m^2) I = 2.65*10^-5 W/cm^2 = 2.65*10^-1 W/m^2 dB = 10*log(0.265/10^-12) = ...
Sunday, March 17, 2013 at 8:58am by drwls
Hearing levels or!@#$%^&pit noise in an aircraft can damage the hearing of pilots who are exposed to this hazard for many hours. Noise level is described as Low=under 88 decibels, Medium=88-91
decibels, High=92^decibels. At a = .05, is the!@#$%^&pit noise level independent of ...
Monday, June 23, 2008 at 5:07am by Gayla
Taking a decibel to be an increase in pressure of 12.2%, by what factor has the pressure changed when the sound level has fallen by 8.00 decibels?
Thursday, November 15, 2012 at 8:26pm by Steeve
What is the sound intensity level in decibels of ultrasound of intensity 10^5 W/m2, used to pulverize tissue during surgery
Tuesday, November 27, 2012 at 3:44pm by Jason
what is the sound intensity level in decibels of ultrasound of intensity of 105 w/m2, used to pulverize tissue during surgery
Wednesday, September 18, 2013 at 1:28pm by Anonymous
Since the sond level is less than the reference(1Watt), the answer will be negative. If the sound level was 1Watt, the answer would be 0db. If the sound level was greater than 1 Watt, the answer
would be positive. a. 10Log(P2/P1) = 10Log(6.3*10^-4W/1W) = 10 * (-3.2) = -32db. b...
Sunday, June 12, 2011 at 8:45pm by Henry
Show that, if one sound is 10 decibels louder than a second sound, than the first sound is 10 times as intense as the second sound.
Wednesday, March 28, 2012 at 11:36pm by Anonymous
physics - sound level
The source of a sound wave has a power of 2.50 µW. Assume it is a point source. (a) What is the intensity 6.70 m away? I used I = Power / 4*pi*r^2 and found I to be 4.43x10^-9 W/m^2 (b) What is the
sound level at that distance? Sound level = 10 dB*log [(Intensity)/(Intensity ...
Saturday, April 21, 2007 at 12:54am by COFFEE
1) I = E/(A*t)= (2.5*10^-11J)/[(6.7*10^-4 m^2)(4.62s)] = 8.1*10^-9 W/m^2 That would be about 39 decibels... not very loud. A soft conversation is about that loud. 2) Use the equation I = Pmax^2/
(2*density*V) where V is the sound speed. Solve it for Pmax, the pressure amplitude...
Monday, November 22, 2010 at 4:27am by drwls
A researcher is interested in estimating the noise levels in decibels at area urban hospitals. She wants to be 90% confident that her estimate is correct. If the standard deviation is 4.13, how large
a sample is needed to get the desired information to be accurate within 0.68 ...
Tuesday, November 2, 2010 at 9:02pm by Valerie
A researcher is interested in estimating the noise levels in decibels at area urban hospitals. She wants to be 95% confident that her estimate is correct. If the standard deviation is 4.02, how large
a sample is needed to get the desired information to be accurate within 0.58 ...
Friday, September 2, 2011 at 9:48pm by Laynette
Grade 12 Advanced Functions Math
Hello! Can someone please help me out for this question? The loudness, L, of a sound in decibels can be calculated using the formula L= 10 log(I/ F) where I is the intensity of the sound in watts per
square metre and F= 10^ -12 . A singer is performing to a crowd. Determine ...
Tuesday, January 19, 2010 at 12:06am by Joan
Pre Calculus 12
Solving problems that involve logarithms the sound at a rock concert decreases by 10% for every 5m travvelled. how far does it take for a volume of 2000 decibels to decrease to 20 decibels? i dont
know what to do for this one the growth decay formula i got is A = P(X)^t/n I ...
Wednesday, January 23, 2013 at 8:23pm by Shreya
Algebra 2
What do you mean by ^three times as loud" ? If loudness is a measure of power per area, then the amplitude has to be higher by the square root of three. To the ear, this will correspond to about 4.8
decibels more of sound level, not "three times as much".
Friday, May 6, 2011 at 2:44pm by drwls
Manageral Economics (Economyst Only Please)
Thanks for your vote of confidence. 1) Look at the T-Ratio and especially the P-Value. You are testing whether the parameter is significantly different from zero. In all parameters except the
intercept, the estimate of the parameter is significantly different from zero at the ...
Saturday, July 4, 2009 at 1:34pm by economyst
(a) By what factor is one sound more intense than another if the sound has a level 17.0 dB higher than the other? (b) If one sound has a level 32.0 dB less than another, what is the ratio of their
Wednesday, July 31, 2013 at 6:26pm by Justin
(a) By what factor is one sound more intense than another if the sound has a level 17.0 dB higher than the other? (b) If one sound has a level 32.0 dB less than another, what is the ratio of their
Wednesday, July 31, 2013 at 10:17pm by Justin
(a) By what factor is one sound more intense than another if the sound has a level 17.0 dB higher than the other? (b) If one sound has a level 32.0 dB less than another, what is the ratio of their
Thursday, August 1, 2013 at 9:55pm by Justin
Sound intensity (loudness) varies inversely as the square of the distance from the source. If a rock band has a sound intensity of 100 decibels 25 feet away from the amplifier, find the sound
intensity 50 feet away from the amplifier.
Friday, April 4, 2014 at 4:35pm by f
A sound power level reading of 127 dB was taken near a construction site where chippers were being used. When all but one of the chippers stopped working, the sound power level reading was 120 dB.
Estimate the number of chippers in operation when the reading of 127 was ...
Tuesday, April 26, 2011 at 12:31am by Massiel
A sound power level reading of 127 dB was taken near a construction site where chippers were being used. When all but one of the chippers stopped working, the sound power level reading was 120 dB.
Estimate the number of chippers in operation when the reading of 127 was ...
Tuesday, April 26, 2011 at 1:09am by Massiel
science (physics)
1) Why is frequency an element needed in the sound industry? and 2) When there is a sound pressure level of two different sources with different frequencies, why do sound pressure levels have to be
determined in each seperate frequency? 1) The sound industry sells equipment ...
Thursday, September 21, 2006 at 12:23am by Kat
college physics
The sound level when one child is talking in a room is 60.0 dB. What would be the sound level if all 20 children talk at the same time in the same room? Assume each child talks equally loudly.
Saturday, January 18, 2014 at 2:32am by bano
Log functions: A new car has an interior sound level of 70 dB at 50 km h. A second car, at the same speed, has an interior sound level that is two times more intense than that of the new car.
Calculate the sound level inside the second car.
Friday, May 27, 2011 at 12:55am by Aimee
I`m a 5th grade student and we are haviing a Sound Science test.I`m looking for defenitions to study from.such as vibrations, decibels, frequency and questions dealing with sound travel. is there a
site you could recomend? thank you, cassidy
Wednesday, February 24, 2010 at 9:00pm by Cassidy
Problem solving with logarithmic functions: A new car has an interior sound level of 70 dB at 50 km h. A second car, at the same speed, has an interior sound level that is two times more intense than
that of the new car. Calculate the sound level inside the second car.
Thursday, May 26, 2011 at 9:39pm by Aimee
A member of an aircraft maintenance crew wears protective earplugs that reduce the sound intensity by a factor of 355. When a jet aircraft is taking off, the sound intensity level experienced by the
crew member is 72 dB. What sound intensity level would the crew member ...
Saturday, December 3, 2011 at 6:27pm by Marco
What are the decibels 917m away from a 35.5 W sound source?
Tuesday, November 19, 2013 at 8:30am by Sara
What are the decibels 917m away from a 35.5 W sound source?
Tuesday, November 19, 2013 at 10:33am by SDW
1.) The average sound intensity inside a busy neighborhood restaurant is 3.*10^-5 W/M^2. How much energy goes into each ear (area = 2.1 * 10^-2 m^2) during a one hour meal? 2.) A member of an
aircraft maintenance crew wears protective earplugs that reduce the sound intensity ...
Sunday, December 12, 2010 at 11:23pm by David
You did not use the correct logarithmic formula that relates decibels to intensity. To review it, see: http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/sound/u11l2b.html At 39.2 dB, the sound
intensity level is 10^3.92 = 8318 times above the reference (hearing threshold) value...
Tuesday, December 2, 2008 at 12:44am by drwls
Physics(Please help)
The intensity (in W/m2) of one sound is 4.83 times as great as the intensity of another sound. Relative to the quieter sound, what is the intensity level â of the louder sound? I am not sure how to
start this. Thank you.
Thursday, June 21, 2012 at 6:47pm by Hannah
As measured from centre ice, a single wildly cheering hockey fan can produce a sound intensity level of 60dB. What sound intensity level would be produced by 20 000 cheering fans? --------- If sound
waves travel through the ground with an average speed of 6150 m/s and a ...
Tuesday, May 15, 2012 at 8:21pm by HelpMeWithPhysics
As measured from centre ice, a single wildly cheering hockey fan can produce a sound intensity level of 60dB. What sound intensity level would be produced by 20 000 cheering fans? --------- If sound
waves travel through the ground with an average speed of 6150 m/s and a ...
Tuesday, May 15, 2012 at 4:07pm by HelpMeWithPhysics
alg 2
The sound level in decibles, B, can be found using the equation B=10 log(I/10^-12). what is the sound intensity when the decible level is 90? pleaseeeeeeeeeeee helppppppppppppppppppp
Monday, June 16, 2008 at 11:44pm by URGENT!!!!!!!!!
A grade-school teacher insists that the the overall sound level in his classroom not exceed 64 dB at his location. There are 25 students in his class. If each student talks at the same intensity
level, and they all talk at once, what is the highest sound level at which each ...
Tuesday, April 5, 2011 at 4:32pm by chris giordano
Could someone help explain what I am to do? The amount of background noise is important to television news reporters. One station developed the formula n= -t^2 + 12t +54 showing the noise level in
decibels (N) as it relates to the time after the speaker stops talking in ...
Tuesday, September 8, 2009 at 10:22pm by sam
NEED HELP WITH ONLY PART C Two 135.1.0-W speakers, A and B, are separated by a distance D = 3.6 m. The speakers emit in-phase sound waves at a frequency f = 11800.0 Hz. Point P1 is located at x1=
4.50 m and y1 = 0 m; point P2 is located at x2 = 4.50 m and y2 = –y. a) ...
Tuesday, March 27, 2012 at 9:27pm by sara
Physics- Elena please help!
An interstate highway has been built though a poor neighborhood in a city. In the afternoon, the sound level in a rented room is 114.0 dB as 172 cars pass outside the window every minute. (a) What is
the sound intensity during the day ? (Reminder: If you choose to use ...
Wednesday, December 5, 2012 at 6:26pm by Mary
College Physics
(20)^2 = 400 times the sound power (at the mother's ear) = how many decibels?
Thursday, November 1, 2007 at 2:44pm by drwls
By how many decibels does the sound intensity from a point source decrease if you increase the distance to it by a factor 6?
Wednesday, December 1, 2010 at 5:51pm by Molly
By how many decibels does the sound intensity from a point source decrease if you increase the distance to it by a factor 6?
Wednesday, December 1, 2010 at 5:50pm by Molly
An electronic point source emits sound isotropically at a frequency of 3000 Hz and a power of 34 watts. A small microphone has an area of 0.75 cm2 and is located 158 meters from the point source. a)
What is the sound intensity at the microphone ? To find intensity i tried I=...
Wednesday, April 21, 2010 at 11:45pm by caleb
(Sound Intensity) = (Energy)/[(Area)*(Time)] It can be converted to decibels using a logarithmic formula, if that is what they want.
Sunday, November 21, 2010 at 5:07pm by drwls
The sound power will be reduced to 40% of the original value. Sound is measured in logarithmic units of decibels. 10 dB are a factor of 10 in power. In this case, the sound power is reduced not that
much. In your case, dB reduction = 10 Log(10)2.5 = 4.0 dB The new reading will...
Tuesday, March 29, 2011 at 1:16pm by drwls
You're at a concert at a distance of 10m away from the band and their sound level is 120dB. How far back do you have to move if you want the sound level to be 85 dB? I tried using 85=(10/r1)^2, but
this doesn't seem to be working.
Wednesday, October 22, 2008 at 12:08am by Ziyi
Physics: Waves in Media
By how many decibels does the sound intensity from a point source decrease if you increase the distance to it by a factor 6?
Tuesday, December 7, 2010 at 12:52pm by Amanda
The quality of a musical tone has to do with a variety of which of the following? A) amplitudes.B) sound speeds.C) intensities.D) decibels.E) harmonics.
Monday, April 30, 2012 at 7:03pm by Jim
1. What is the intensity of a sound at the pain level of 120 dB? 2. What is the intensity level of a sound whose intensity is 2.0 x 10^-6 W/m^2?
Saturday, May 29, 2010 at 11:14am by Amber
1 would probably sound better like this: We expect they will arrive next week. 2 sounds rather stilted! It would depend on the context, though.
Monday, February 11, 2013 at 2:50pm by Writeacher
1. Solve (a+V)/(a-V) = 0.8 a is the speed of sound, and V is the car's speed. 2. The person who can hear the lower 5 dB level has better hearing, by 20 dB. That is a factor of 100 times less in sound
power level
Sunday, May 17, 2009 at 11:43am by drwls
Name an element that has 5 electrons in the third energy level. How do i determine the third energy level Row three on the periodic table is third energy level If memory serves me will, it begins
with sodium. Ok. iS Krypton the electron configuration that ends in 4s2, 4p5 I ...
Sunday, November 12, 2006 at 10:22am by Bryan
Find the number of decibels for the power of the sound. Round to the nearest decibel. A rock concert, 5.4 10-6 watts/cm2
Sunday, October 30, 2011 at 4:36am by suzie
Find the number of decibels for the power of the sound. Round to the nearest decibel. A rock concert, 5.3 10-6 watts/cm2
Friday, August 3, 2012 at 7:55pm by Deb
2.58*10^(-5)watts/cm2 find the number of decibels for the power of the sound given. Round to the nearest decibel.
Sunday, February 10, 2013 at 11:47pm by Andrzej
19 cars in a circle at a boom box competition produce a 130 dB sound level at the center of the circle. What is the average sound level(dB) produced there by each, assuming interference effects can
be neglected?
Wednesday, July 31, 2013 at 6:26pm by Justin
19 cars in a circle at a boom box competition produce a 130 dB sound level at the center of the circle. What is the average sound level(dB) produced there by each, assuming interference effects can
be neglected?
Wednesday, July 31, 2013 at 10:16pm by Justin
19 cars in a circle at a boom box competition produce a 130 dB sound level at the center of the circle. What is the average sound level(dB) produced there by each, assuming interference effects can
be neglected?
Thursday, August 1, 2013 at 9:56pm by Justin
18 cars in a circle at a boom box competition produce a 105 dB sound level at the center of the circle. What is the average sound level produced there by each, assuming interference effects can be
neglected? dB = ?
Saturday, February 8, 2014 at 7:15pm by ddsMom
Suppose that in the aerobic exercise group we also measured the number of hours of aerobic exercise per week and the mean is 5.2 hours with a standard deviation of 2.1 hours. The sample correlation
is -0.42. a)Is there evidence of a significant correlation between number of ...
Saturday, November 26, 2011 at 11:46am by natasha
Physics SOUND
What is the resultant sound level when an 80-dB and an 85-dB sound are heard simultaneously?
Saturday, May 3, 2008 at 1:48am by Anonymous
9th grade
Use the following Info. At sea level the speed of sound in air is linearly related to air temp. If 35 degrees Celsius sound travels at a rate of 352meters per sec. if 15degrees Celsius sound travels
at 340 meters per sec. How would i write a linear equation that models speed ...
Monday, November 15, 2010 at 5:36pm by Kaela
Physics- PLEASE HELP
An interstate highway has been built though a poor neighborhood in a city. In the afternoon, the sound level in a rented room is 114.0 dB as 172 cars pass outside the window every minute. Late at
night, the traffic flow is only 7 cars per minute. (b) What is the sound ...
Thursday, December 6, 2012 at 7:56pm by Anonymous
7. Consider again the data in problem #6. Suppose that in the aerobic exercise group we also measured the number of hours of aerobic exercise per week and the mean is 5.2 hours with a standard
deviation of 2.1 hours. The sample correlation is -0.42. a)Is there evidence of a ...
Saturday, November 26, 2011 at 10:51am by natasha
Physics- Elena please help!
An interstate highway has been built though a poor neighborhood in a city. In the afternoon, the sound level in a rented room is 114.0 dB as 172 cars pass outside the window every minute. Late at
night, the traffic flow is only 7 cars per minute. (b) What is the sound ...
Thursday, December 6, 2012 at 5:52pm by Anonymous
Find the number of decibels for the power of the sound. round to the nearest decibel. A rock concert, 5.35x 10^4 watts/cm^2
Tuesday, July 24, 2012 at 1:48pm by Nancy
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Stochastic Differential Equations with Multi-Markovian Switching
Journal of Applied Mathematics
Volume 2013 (2013), Article ID 357869, 11 pages
Research Article
Stochastic Differential Equations with Multi-Markovian Switching
^1School of Mathematical Science, Huaiyin Normal University, Huai'an 223300, China
^2Department of Mathematics, Harbin Institute of Technology, Weihai 264209, China
Received 31 August 2012; Revised 2 March 2013; Accepted 5 March 2013
Academic Editor: Jose L. Gracia
Copyright © 2013 Meng Liu and Ke Wang. 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.
This paper is concerned with stochastic differential equations (SDEs) with multi-Markovian switching. The existence and uniqueness of solution are investigated, and the pth moment of the solution is
estimated. The classical theory of SDEs with single Markovian switching is extended.
1. Introduction
Stochastic modeling has played an important role in many branches of industry and science. SDEs with single continuous-time Markovian chain have been used to model many practical systems where they
may experience abrupt changes in their parameters and structure caused by phenomena such as abrupt environment disturbances. SDEs with single Markovian switching can be denoted by with initial
conditions and , where is a right-continuous homogenous Markovian chain on the probability space taking values in a finite state space and is -adapted but independent of the Brownian motion , and
Owing to their theoretical and practical significance, (1) has received great attention and has been recently studied extensively, and we here mention Skorokhod [1] and Mao and Yuan [2] among many
However, in the real world, the condition that coefficients and in (1) are perturbed by the same Markovian chain is too restrictive. For example, in the classical Black-Scholes model, the asset price
is given by a geometric Brownian motion where is the rate of the return of the underlying assert, is the volatility, and is a scalar Brownian motion. Since there is strong evidence to indicate that
is not a constant but is a Markovian jump process (see, e.g., [3, 4]), many authors proposed the following model: However, many stochastic factors that affect are different from those that affect .
Then the following model is more appropriate than model (105) to describe this problem: where is a right-continuous homogenous Markovian chain taking values in a finite state space, . Another example
is the stochastic Lotka-Volterra model with single Markovian switching which has received great attention and has been studied extensively recently (see, e.g., [5–12]). For the sake of convenience,
we take the following two-dimensional competitive model as an example: where is the size of th species at time , represents the growth rate of th species in regime for , , and and are independent
standard Brownian motions. However, there are many stochastic factors that affect some coefficients intensely but have little impact on other coefficients in (6). For example, suppose that the
stochastic factor is rain falls and is able to endure a damp weather while is fond of a dry environment, then the rain falls will affect intensely but have little impact on . Thus, a more appropriate
model is governed by where and are right-continuous homogenous Markovian chains taking values in finite state spaces for , , and for , respectively.
Thus the above examples show that the study of the following SDEs with multi-Markovian switchings is essential and is of great importance from both theoretical and practical points: with initial
conditions and , where is a right-continuous homogenous Markovian chain on the probality space taking values in a finite state space and is -adapted but independent of the Brownian motion , , and
Equation (8) can be regarded as the result of the equations switching among each other according to the movement of the Markovian chains. It is important for us to discover the properties of the
system (8) and to find out whether the presence of two Markovian switchings affects some known results. The first step and the foundation of those studies are to establish the theorems for the
existence and uniqueness of the solution to system (8). So in this paper, we will give some theorems for the existence and uniqueness of the solution to system (8) and study some properties of this
solution. The theory developed in this paper is the foundation for further study and can be applied in many different and complicated situations, and hence the importance of the results in this paper
is clear.
It should be pointed out that the theory developed in this paper can be generalized to cope with the more general SDEs with more Markovian chains The reason we concentrate on (8) rather than (11) is
to avoid the notations becoming too complicated. Once the theory developed in this paper is established, the reader should be able to cope with the more general (11) without any difficulty.
The remaining part of this paper is as follows. In Section 2, the sufficient criteria for existence and uniqueness of solution, local solution, and maximal local solution will be established,
respectively. In Section 3, the -estimates of the solution will be given. In Section 4, we will introduce an example to illustrate our main result. Finally, we will close the paper with conclusions
in Section 5.
2. SDEs with Markovian Chains
Throughout this paper, let be a complete probability space. Let be an -dimensional Brownian motion defined on the probability space.
In this section, we will consider (8). Let . We impose a hypothesis.
(H1): is independent of .
Then is a homogenous vector Markovian chain with transition probabilities where .
Now, we will prepare some lemmas which are important for further study.
Lemma 1. has the following properties:(i) for ;(ii) for ;(iii), where if , otherwise , ;(iv)(Chapman-Kolmogorov equation) For and ,
Proof. The proofs of (i), (ii), and (iii) are obvious. Now, let us prove (iv): This completes the proof.
Now, we impose another hypothesis, which is called standard condition.
(H2): .
Lemma 2. Under Assumption (H2), for all , one has .
Proof. From and (H2) we know that, for arbitrary fixed , we have for sufficient large . Then making use of Chapman-Kolmogorov equation gives which is the desired assertion.
Lemma 3. Under Assumption (H2), for all , exists (but may be ).
Proof. Define . Then making use of (16) gives Set . It is easy to see that Now we will assert
In fact, for , such that . Applying (19) yields Note that , , whenever , then for all we have This implies . Thus Using the definition of gives which is the required assertion.
Lemma 4. Under Assumption (H2), for , exists and is finite.
Proof. By (H2), we note that for all , , such that provided .
For , set , where . Let where means that the probability of the will not reach to at times but will reach to at time . Note that if , then which indicates Then making use of we obtain Consequently,
Dividing both sides of the above inequality by and noting whenever yield Then letting gives and the required assertion follows immediately by letting . This completes the proof.
Set , then it is easy to see that almost every sample path of is a right continuous step function. Now letting , . Then by Chapman-Kolmogorov equation we have Letting and taking limits give Note that
Then by solving the ordinary differential equations (38) and (39), we obtain the following lemma.
Lemma 5. For and one has
We are now in the position to give the sufficient conditions for the existence and uniqueness of the solution of (8). For this end, let us first give the definition of the solution.
Definition 6. An -valued stochastic process is called a solution of (8) if it has the following properties:(i) is continuous and -adapted;(ii) while ;(iii)for any , equation holds with probability 1.
A solution is said to be unique if any other solution is indistinguishable from .
Now we can give our main results in this section.
Theorem 7. Assume that there exist two positive constants and such that.
(Lipschitz condition) for all , and (Linear growth condition) for all Then there exists a unique solution to (8) and, moreover, the solution obeys
Proof. Recall that almost every sample path of is a right continuous step function with a finite number of jumps on . Thus there exists a sequence of stopping times such that (i)for almost every
there is a finite for and if ; (ii)both and in are constants on interval , namely, First of all, let us consider (8) on , then (8) becomes with initial conditions . Then by the theory of SDEs, we
obtain that (46) has a unique solution which obeys . We next consider (8) on which becomes Again by the theory of SDEs, (47) has a unique solution which obeys . Repeating this procedure, we conclude
that (8) has a unique solution on .
Now, let us prove (44). For every , define the stopping time It is obvious that a.s. Set for . Then obeys the equation Making use of the elementary inequality , the Hölder inequality, and (43), we
can see that Thus, applying the Doob martingale inequality and (43), we can further show that That is to say, Using the Gronwall inequality leads to Consequently, Then the required inequality (44)
follows immediately by letting .
Condition (42) indicates that the coefficients and do not change faster than a linear function of as change in . This means in particular the continuity of and in for all . Then functions that are
discontinuous with respect to are excluded as the coefficients. Besides, there are many functions that do not satisfy the Lipschitz condition. These imply that the Lipschitz condition is too
restrictive. To improve this Lipschitz condition let us introduce the concept of local solution.
Definition 8. Let be a stopping time such that a.s. An -valued -adapted continuous stochastic process is called a local solution of (8) if and, moreover, there is a nondecreasing sequence such that
a.s. and holds for any and with probability one. If, furthermore, then it is called a maximal local solution and is called the explosion time. A maximal local solution is said to be unique if any
other maximal local solution is indistinguishable from it, namely, and for with probability one.
Definition 9 (local Lipschitz condition). For every integer , there exists a positive constant such that, for all , and those with
The following theorem shows the existence of unique maximal local solution under the local Lipschitz condition without the linear growth condition.
Theorem 10. Under condition (57), there exists a unique maximal local solution of (8).
Proof. Define functions Then and satisfy the Lipschitz condition and the linear growth condition. Thus by Theorem 7, there is a unique solution of the equation with the initial conditions and .
Define the stopping times Clearly, if , which indicates that is increasing so has its limit . Define by where . Applying (61), one can show that . It then follows from (59) that for any and . It is
easy to see that if , then Therefore is a maximal local solution.
Now, we will prove the uniqueness. Let be another maximal local solution. Define Then a.s. and Letting gives In order to complete the proof, we need only to show that a.s. In fact, for almost every ,
we have which contradicts the fact that is continuous on . This implies a.s. In the same way, one can show a.s. Thus we must have a.s. This completes the proof.
In many situations, we often consider an SDE on with initial data and . If the assumption of the existence-and-uniqueness theorem holds on every finite subinterval of , then (69) has a unique
solution on the entire interval . Such a solution is called a global solution. To establish a more general result about global solution, we need more notations. To this end, we introduce an operator
from to which is given by where and
Theorem 11. Assume that the local Lipschitz condition (57) holds. Assume also that there is a function and a constant such that Then there exists a unique global solution to (69).
Proof. We need only to prove the theorem for any initial condition and . From Theorem 10, we know that the local Lipschitz condition guarantees the existence of the unique maximal solution on , where
is the explosion time. We need only to show a.s. If this is not true, then we can find a pair of positive constants and such that For each integer , define the stopping time Since almost surely, we
can find a sufficiently large integer for Fix any , then for any , by virtue of the generalized Itô formula (see, e.g., [1]) Making use of the Gronwall inequality gives Therefore At the same time,
set Then (72) means . It follows from (76) and (79) that Letting yields a contradiction, that is to say, . The proof is complete.
3. -Estimates
In the previous section, we have investigated the existence and uniqueness of the solution to (8). In this section, as above, let , be the unique solution of (8) with initial conditions and , and we
will estimate the th moment of the solution.
Theorem 12. Assume that there is a function and positive constants such that for all , Assume also the initial condition and obeys , then one has
Proof. For each integer , define the stopping time Thus a.s. Using the generalized Itô's formula and (83), we obtain that for Then the Gronwall inequality indicates for all . By virtue of condition (
83) we obtain the required assertion (84).
Corollary 13. Assume and . Assume also that there exists a constant such that, for all , Then one has
Proof. Define . Making use of (88) yields Then by Theorem 12, we get and the required assertion (89) follows.
It is useful to point out that if the linear growth condition (43) is satisfied, then (88) is fulfilled with . Now, we will show the other important properties of the solution.
Theorem 14. Let and . Assume also that the linear growth condition (43) holds. Then one has where and . Particularly, the th moment of the solution is continuous on .
Proof. Applying the elementary inequality , the Hölder inequality, and the linear growth condition, we can derive that where . Using (91) yields which is the desired inequality.
Theorem 15. Let and . Assume that there is a such that for all Then
Proof. Making use of the generalized Itô's formula and condition (96), we have where Therefore, for any , At the same time, applying the well-known Burkholder-Davis-Gundy inequality (see, e.g., [2])
gives Substituting the above inequality into (100) gives Thus Then the required assertion follows from the Gronwall inequality.
Up to now, we have discussed the -estimates for the solution in the case when . As for , the similar results can be given without any difficulty as long as we note that the Hölder inequality implies
4. Example
Consider the following Black-Scholes model: where is a right-continuous homogenous Markovian chain taking values in finite state spaces and is a right-continuous homogenous Markovian chain taking
values in finite state spaces , , , , . Taking , , then (42) and (43) hold. Therefore, by Theorem 7, (105) has a unique solution.
5. Conclusions and Further Research
This paper is devoted to studying the existence and uniqueness of solution of SDEs with multi-Markovian switchings and estimating the th moment of the solution. We have used two continuous-time
Markovian chains to model the SDEs. This area is becoming increasingly useful in engineering, economics, communication theory, active networking, and so forth. The sufficient criteria for existence
and uniqueness of solution, local solution, and maximal local solution were established. Those results indicate that (8) keeps many properties that (89) owns. At the same time, although the
hypothesis (H1) is used in this paper, we want to point out that this hypothesis is not essential. In fact, (H1) can be replaced by the following generalized hypothesis.
(H1)′: both and are right-continuous homogenous Markovian chains such that is a homogenous vector chain.
Under hypothesis (H1)′, the results given in this paper can be established similarly. It is easy to see that if | {"url":"http://www.hindawi.com/journals/jam/2013/357869/","timestamp":"2014-04-16T16:24:26Z","content_type":null,"content_length":"1047227","record_id":"<urn:uuid:5f27cde5-16c2-4f3f-b807-90ea9e02a863>","cc-path":"CC-MAIN-2014-15/segments/1397609524259.30/warc/CC-MAIN-20140416005204-00038-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Father of Algebra
Math brain teasers require computations to solve.
Category: Math
Submitted By: buddyboy
Diophantus was a Greek mathematician who lived in the third century. He was one of the first mathematicians to use algebraic symbols.
Most of what is known about Diophantus's life comes from an algebraic riddle from around the early sixth century. The riddle states:
Diophantus's youth lasted one sixth of his life. He grew a beard after one twelfth more. After one seventh more of his life, he married. 5 years later, he and his wife had a son. The son lived
exactly one half as long as his father, and Diophantus died four years after his son.
How many years did Diophantus live? | {"url":"http://www.braingle.com/wii/brainteasers/teaser.php?id=35386","timestamp":"2014-04-17T03:55:45Z","content_type":null,"content_length":"9853","record_id":"<urn:uuid:97335751-cad7-4f45-a7ce-57990a33014a>","cc-path":"CC-MAIN-2014-15/segments/1397609526252.40/warc/CC-MAIN-20140416005206-00130-ip-10-147-4-33.ec2.internal.warc.gz"} |
Angular Momentum
Considering a Rigid Body/Angular Momentum/Torque
We know that Torque(ext) = dL/dt
Now with respect to stationary point S:
L(s, cm) = Ʃ(ρi x mivi)
and that dL(cm)/dt = Ʃτ(ext, CM)
Now with respect to ANY point, P, that is accelerating:
L(s,p) = L(cm) + ρ(cm) x Mv(cm)
And hence,
Ʃτ(ext, p) = dL(rel_p)/dt + ρ(cm) x Ma(p)
Ʃτ(ext, p) = dL(rel_cm)/dt + ρ(cm) x Ma(cm)
Can someone explain to me why this happens?:
Why this happens? : Ʃτ(ext, p) = dL(rel_p)/dt + ρ(cm) x Ma(p) | {"url":"http://www.physicsforums.com/showthread.php?p=3897969","timestamp":"2014-04-18T08:30:44Z","content_type":null,"content_length":"22805","record_id":"<urn:uuid:52767818-761f-4d55-9592-b4b03d4df234>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00441-ip-10-147-4-33.ec2.internal.warc.gz"} |
Ray Tracing with GLSL - Some Doubts
Hello All,
I am trying to develop a basic Ray Tracer. So far i have calculated intersection with a plane and blinn-phong shading.i am working on a 500*500 window and my primary ray generation code is as follows
vec3 rayDirection = vec3( gl_FragCoord.x-250.0,gl_FragCoord.y-250.0 , 10.0);
Now i doubt that above method is right or wrong. Please give me some insights.
I am also having doubt that do we need to construct geometry in OpenGL code while rayTracing in GLSL. for example if i am trying to raytrace a plane do i need to construct plane in OpenGL code using
glVertex2f ?
Thank You
The direction vector need to be normalized.
When i did raytracing in glsl it was something like this:
vec2 uv = gl_FragCoord.xy / resolution.xy;
vec3 r0 = vec3(0.0, 1.0, 4.0); //the position of view
vec3 rd = vec3((-1.0 + 2.0 * uv) * vec2(1.78, 1.0), -1.0); //the look direction, note the -Z!, the multiplication with vec2(1.78, 1.0) is just a correction for the aspect.
The resolution is a 2d vec, the screen width and height (in window coords, for example: 800*600).
Yes, you need a full screen quad, and run the (fragment) shader code on this.
I recommend to use ShaderToy (http://www.iquilezles.org/apps/shadertoy/), or GLSL Sandbox (http://glsl.heroku.com/e). Quick and simple. :)
Note. that using GLSL for ray tracing isn’t as good as it seem to be. Use OpenCL rather, as you’ll more likely get better code, it will give you more options (as you’re coding in language very
similar to C, GLSL is also similar, but not that much).
Anyway, try to pass ray directions through vertex attribute - as it will save you a lot of computations (when working with GLSL).
first of all thank you for your reply..
@Jari Your referred tools are very very useful thanx again for that…!! My code is working on that tool but not on my PC. Is this the problem of vertex shader or OpenGL code settings. My current code
only finds rays intersection with a sphere. Which i can clearly seen on GLSL Sandbox but not in my project.
in vertex shader i m just using ftransform and nothing else.
and can u please tell me what is logic of converting those coordinates in the range of [-0.5,0.5]. I dont get it..!!
@Villem Otte Thank you very much for your advice but right now my objective is to just construct a basic raytracer in GLSL with phong, shadows, reflection and may be refraction. After that i will
concentrate on Optimizations and look for OpenCL or CUDA.
You shoot the rays in [-1.0, 1.0] range, both in vertical and in horizontal directions (of cource you can modify this, eg. on the x axis to correct the aspect).
You get the current fragment position (gl_FragCoord) in window coordinate space (eg. [0, 800], [0, 600]), so to convert it to [-1.0, 1.0] first divide with window (viewport) size and you get a value
in range [0.0, 1.0]. To convert this to [-1.0, 1.0] first multiply this value with 2.0, and then subtract -1.0 from it.
What these tools do is draw a fullscreen quad (two triangles), and execute fragment shaders on it. The vertex shaders just bypass the vertices (if you send them in normalized device coordinates, you
dont need to transform). | {"url":"http://devmaster.net/posts/20708/ray-tracing-with-glsl-some-doubts","timestamp":"2014-04-21T02:53:43Z","content_type":null,"content_length":"19495","record_id":"<urn:uuid:6c1213ab-71c0-4ac5-bb20-46a25b8e41a1>","cc-path":"CC-MAIN-2014-15/segments/1398223206672.15/warc/CC-MAIN-20140423032006-00502-ip-10-147-4-33.ec2.internal.warc.gz"} |
Matthew van Eerde's web log
23 Comments
A body that moves in a circular motion (of radius r) at constant speed (v) is always being accelerated. The acceleration is at right angles to the direction of motion (towards the center of the
circle) and of magnitude v^2 / r.
The direction of acceleration is deduced by symmetry arguments. If the acceleration pointed out of the plane of the circle, then the body would leave the plane of the circle; it doesn't, so it isn't.
If the acceleration pointed in any direction other than perpendicular (left or right) then the body would speed up or slow down. It doesn't.
Now for the magnitude. Consider the distance traveled by the body over a small time increment Δt:
We can calculate the arc length s as both the distance traveled (distance = rate * time = v Δt) and using the definition of a radian (arc = radius * angle in radians = r Δθ:)
The angular velocity of the object is thus v / r (in radians per unit of time.)
The right half of the diagram is formed by putting the tails of the two v vectors together. Note that Δθ is the same in both diagrams.
Note the passing from sin to cos is via l'Hôpital's rule.
1 Comments
For as long as there have been humans, we have looked at the skies - and around us on Earth as well. Things move in the skies; things move on Earth. The way things move in the skies is called
"celestial mechanics"; the way things move on Earth is called "terrestrial mechanics."
The ancient Greeks thought there were five elements, corresponding to the five regular (Platonic) solids: the four terrestrial elements (air, fire, earth, and water) and the fifth celestial element
which the Romans called quintessence.
All kinds of rules were discovered/proposed for how things move on Earth. Galileo demonstrated that falling objects accelerate at the same rate regardless of their mass. Similar rules were discovered
/proposed for how things move in the heavens. Kepler demonstrated that the planets move in ellipses around the sun, that they swept out equal areas in equal times, etc. But everybody knew that things
in the heavens were different than they were on Earth.
Then Newton, building on Hooke, came up with the crazy idea that perhaps things fall on Earth for the same reason that heavenly objects go in ellipses... that all objects, celestial or terrestrial,
act on each other from a distance in a uniform fashion. (He himself wrote that this was such a crazy idea that no-one should take it seriously.)
That one body may act upon another at a distance through a vacuum without the mediation of anything else, by and through which their action and force may be conveyed from one another, is to me so
great an absurdity that, I believe, no man who has in philosophic matters a competent faculty of thinking could ever fall into it. -- Isaac Newton
But he had done some calculations and it seemed to work out... specifically, he worked out how much acceleration you would expect the Earth to exert on an object as far away as the Moon, and how much
acceleration the Moon would need to stay in its orbit (and not wander off or spiral down into the Earth.)
I... compared the force requisite to keep the moon in her orb with the force of gravity at the surface of the earth and found them to answer pretty nearly. -- Isaac Newton
Note Newton's use of the word "orb" to describe the motion of the moon, a direct reference to celestial mechanics.
Let's see how compelling those calculations were, shall we?
His first idea was that there was this thing called "force" which was equal to the mass of an object multiplied by the acceleration it was undergoing. F = m a. Fine.
His second idea (which he stole, at least partially, from Hooke) was that any two objects have a force that attracts them, which is proportional to both of their masses, and inversely proportional to
the square of the distance between them (that's the part he stole from Hooke.) F = G m[1] m[2] / r^2. Fine.
His third idea was that an object that goes around a circle of radius r at a constant speed v is undergoing a constant centripetal acceleration of a = v^2 / r. This follows pretty quickly from the
definitions with a little geometry, trigonometry, and a dash of calculus. Fine.
He also knew some facts. I'm going to state them in metric because, hey, this is the twenty-first century. He knew:
• The distance from the center of the Earth to the surface (where the apple trees are) is 3,960 miles: 6370 km.
• Falling objects on the surface of the Earth accelerate at 32.2 ft/s^2: 9.81 m/s^2.
• The distance from the center of the Earth to the center of the Moon is 239,000 miles: 384,000 km.
• The Moon takes 27.3 days to make a single sidereal orbit around the Earth: 236,000 s.
Coarse estimate time...
If the Moon is roughly 50 times further away from the center of the Earth than an apple is, and "gravitational" acceleration is inversely proportional to the square of the distance between two
objects, then the Moon should be accelerated roughly 1/2500th as much as the apple is: 0.004 m/s^2 instead of 10 m/s^2,
More precisely...
Consider the apple case (m[E] is the mass of the Earth, m[A] the mass of the apple:)
F[A] = m[A] a[A] = G m[E] m[A] / r[A]^2
a[A] r[A]^2 = G m[E]
9.81 m/s^2 (6.37e6 m)^2 = G m[E]
G m[E] = 3.98e14 m^3/s^2
Now consider the moon case:
F[M] = m[M] a[M] = G m[E] m[M] / r[M]^2
a[M] = G m[E] / r[M]^2
a[M] = 3.98e14 m^3/s^2 / (3.84e5 m)^2
a[M] = 0.00269 m/s^2
So Newton's proposed formula predicts that the Earth's gravity causes the Moon to accelerate at 0.00269 m/s^2. This is, indeed, roughly 0.004 m/s^2 so it jives with our coarse estimate.
Let's see how that compares to the necessary centripetal acceleration to achieve a closed orbit: a = v^2 / r.
The distance the moon travels in its sidereal orbit is 2π * 3.84e5 m = 2.42e9 m, and it does so in 27.3 days; so its velocity is 2π * 3.84e5 m / (27.3 days * 24 (hours / day) * 60 (minutes / hour) *
60 (s / minute)) = 1020 m/s. This is roughly Mach 3 in our atmosphere.
The centripetal acceleration is thus measured to be (1020 m/s)^2 / 3.84e8 m = 0.00273 m/s^2. Newton's prediction is about 1.5% off, which isn't bad, considering.
Note that although Newton knew the value of the product G m[E], he didn't know either G or m[E] seperately. It wasn't until Cavendish took two known masses and measured the gravitational pull between
them directly that G was evaluated; we could then calculate the mass of the earth as G m[E] / G.
Once G was known we could also evaluate the mass of the Sun based on Earth's orbit. However, we still could not determine the mass of the Moon... in fact, back in the 1950s, American scientists were
unable to determine the mass of Sputnik (which would have been very helpful) even though we could see its orbit.
Which raises the question... how did we know the mass of the Moon? | {"url":"http://blogs.msdn.com/b/matthew_van_eerde/archive/2010/01.aspx?PostSortBy=MostViewed&PageIndex=1","timestamp":"2014-04-17T16:42:32Z","content_type":null,"content_length":"60733","record_id":"<urn:uuid:a2f5583b-0d4e-45cc-a274-d8c86eaddf36>","cc-path":"CC-MAIN-2014-15/segments/1397609530136.5/warc/CC-MAIN-20140416005210-00639-ip-10-147-4-33.ec2.internal.warc.gz"} |
Evesham Twp, NJ Science Tutor
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Wolfram Demonstrations Project
Illustrating the Law of Large Numbers
The law of large numbers states (informally) that as the number of independent observations drawn from a population with finite mean increases, the mean of those observed values approaches . This
Demonstration illustrates that behavior by plotting the sample mean as a function of the current sample size , for to . Random samples can be drawn from a population of 0's and 1's (with any
proportion of 1's), a normal population (with a range of and available), or from the (first 100,000) digits of . The first 100 digits of are available as a population, with samples of size
consisting of the first digits, to allow for classroom illustration with a familiar and frequently referred to as random set of digits.
Contributed by:
Marc Brodie
(Wheeling Jesuit University) | {"url":"http://demonstrations.wolfram.com/IllustratingTheLawOfLargeNumbers/","timestamp":"2014-04-20T15:54:56Z","content_type":null,"content_length":"43402","record_id":"<urn:uuid:d252318d-4c3c-4939-9081-0c4c248b45fb>","cc-path":"CC-MAIN-2014-15/segments/1397609538824.34/warc/CC-MAIN-20140416005218-00021-ip-10-147-4-33.ec2.internal.warc.gz"} |
Survival Statistic Tab
On the Statistic tab in a Survival session, you must choose one of the following survival measures to calculate survival. If you are unsure of the appropriate survival measure to use for your
analysis, see Determining the Appropriate Cancer Survival Measure.
Choose between the Actuarial and Kaplan-Meier methods of calculating survival. Additionally indicate whether to use Period Survival. If doing Relative Survival or Crude Probability of Death Using
Expected Survival, you must also choose one of the four methods of calculating Cumulative Expected Survival.
Observed Survival Only
Observed survival is an estimate of the probability of surviving all causes of death for a specified time interval calculated from the cohort of cancer cases.
Relative Survival
Relative survival is a net survival measure representing cancer survival in the absence of other causes of death.
Cause-Specific Survival
Cause-specific survival is a net survival measure representing survival of a specified cause of death in the absence of other causes of death.
Crude Probability of Death Using Expected Survival
The crude probability of death measures the mortality patterns actually experienced in a cohort of cancer patients on which many possible causes of death are acting simultaneously. This option
estimates the probability of dying from cancer and dying from other causes in a cohort of cancer patients. The measure uses expected survival (obtained from the expected life tables) to estimate the
probability of dying from other causes in each interval.
Crude Probability of Death Using Cause of Death Information
The crude probability of death measures the mortality patterns actually experienced in a cohort of cancer patients on which many possible causes of death are acting simultaneously. This option
estimates the probability of dying from cancer and dying from other causes in a cohort of cancer patients. Cause of death information is used to identify those individuals that have died from cancer.
If you choose to calculate the observed survival only, or net survival in absence of other causes of death, you will have the option to Age Standardize. If the Age Standardize box is checked, you
must select a Standard Population and an Age Variable to use for standardizing.
If you choose to calculate relative survival, or crude probability of death using expected survival, you must also select an expected survival table. If you choose cause-specific survival, or crude
probability of death using cause of death information, a definition of cause of death is required. | {"url":"http://seer.cancer.gov/seerstat/508_WebHelp/Survival_Statistic_Tab.htm","timestamp":"2014-04-16T04:11:43Z","content_type":null,"content_length":"7829","record_id":"<urn:uuid:93922a7b-3021-4e46-8862-b0bb85287a20>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00095-ip-10-147-4-33.ec2.internal.warc.gz"} |
Getting residuals to a vector
up vote 3 down vote favorite
Anybody knows how to get the residuals created by the following matlab code to a vector 'A' ?
I tried to get the residuals by typing r at the command prompt but did not get the residuals
a0 = 0.05; a1 = 0.1; b1 = 0.85;
nu = randn(2300,1);
epsi = zeros(2300,1);
h = zeros(2300,1);
for i=2: 2300
h(i) = a0 + a1 * epsi(i-1)^2 + b1 * h(i-1) ;
epsi(i) = nu(i) * sqrt(h(i));
yt1 = zeros(2300,1);
for i=1: 2300
yt1(i) = epsi(i)*epsi(i);
order = 15;
m = arx(yt1, order);
r = resid(iddata([yt1(1:order);yt1]), m);
r = r(order+1:end);
matlab formatting matlab-toolbox matlab-guide
1 What do you see when typing r at the command prompt? – ThijsW Feb 21 '13 at 5:30
Time domain data set with 2300 samples. Sample time: 1 seconds Outputs Unit (if specified) e@y1 – Malaka Thilakaratne Feb 21 '13 at 5:33
I feel like r should just contain the residuals, after reading the help file for resid. I cannot test it right now, unfortunately. I don't have access to the toolbox at the moment. – ThijsW Feb 21
'13 at 5:38
When I typed r all I got was the above message. Could you please see to it tomorrow morning? It's a bit urgent. Thanks – Malaka Thilakaratne Feb 21 '13 at 5:39
I'll try to take a look at it tonight (I'm guessing I'm in a different time zone). I'm sure though, that there is someone else who will be able to give you an answer earlier. – ThijsW Feb 21 '13
at 5:41
show 1 more comment
1 Answer
active oldest votes
First use get to see what r has:
ans =
Domain: 'Time'
Name: ''
OutputData: [2315x1 double]
y: 'Same as OutputData'
OutputName: {'e@y1'}
OutputUnit: {''}
InputData: [2315x0 double]
u: 'Same as InputData'
InputName: {0x1 cell}
InputUnit: {0x1 cell}
Period: [0x1 double]
up vote 5 down vote InterSample: {0x1 cell}
Ts: 1
Tstart: []
SamplingInstants: [2315x0 double]
TimeUnit: 'seconds'
ExperimentName: 'Exp1'
Notes: {}
UserData: []
Then, I assume you want to look at:
to obtain the vector you wanted...
add comment
Not the answer you're looking for? Browse other questions tagged matlab formatting matlab-toolbox matlab-guide or ask your own question. | {"url":"http://stackoverflow.com/questions/14995178/getting-residuals-to-a-vector/14995728","timestamp":"2014-04-19T03:48:08Z","content_type":null,"content_length":"68919","record_id":"<urn:uuid:439ac4d8-a31b-4611-b1d5-fd2ac872471a>","cc-path":"CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00291-ip-10-147-4-33.ec2.internal.warc.gz"} |
Guide to Non-linear Dynamics in Accelerator Physics/Tracking
From Wikibooks, open books for an open world
This chapter provides tools to describe tools for finding map that track particles in em fields.
There are several way of tracking a particle in accelerator physics. Each one has different goal and needs its own approximation. Here there is a possible classification.
physics involved[edit]
In their evolution particles experience the following physical process:
decay or inelastic scattering
spontaneous or triggered by very small scale interaction with other particles. The particles are propagated using probabilistic models using QFT.
elastic scattering
small scale interaction of high energy particles with other particles that does not involve decays. As inelastic scattering particles are propagated using probabilistic models using QFT. A
special case of elastic scattering is synchrotron radiation that may or may not use probabilistic models or approximated statistics law. The approximation used are typically neglect not probable
terms, simplify the motion.
electromagnetic field
propagation due the interaction of particle with electromagnetic field which can be either external or generated by two or more particles.
simulation goals[edit]
event generation
for describing which kind of particle may be produced after a collision
between an high energy particle and either another high energy particle
in particle colliders or a low energy particles in fixed target experiment.
The approximation used is too neglect the extension of the space time.
Codes are dmjet, pythia.
particle matter interaction
for calculating the efficiency of a detector,
the energy deposited, the radiation damage of some equipment.
Codes are geant, mars, fluka.
short term analysis (one turn)
for calculating or approximating beam envelops,
invariants of the motion and perturbation terms for periodic structures. The object
of the study are the explicit form of the equation of motion more than the trajectory
of the particle even though the trajectories may be use to compute or approximate the equation of motion. Matter interaction and collective effects are usually not included.
Codes are mad, ptc, tracy.
short term simulation for beam loss
for calculating the trajectory of the particles in the tail of the particle distribution in the beam. These particles are usually source of losses or background noise in the experiments. Usually
hundreds of turn are needed in these simulations. The motion is accurate but the interaction with matter is very simplified.
Codes are sixtrack.
short term simulation for collective instabilities
for calculating the short term stability of a bunch of particle due the interaction of the particles with themselves (direct space charge) or the interaction with metallic surfaces close by
(impedances, indirect space charge). The motion is accurate but not the simplecticity of the equation. This is an unavoidable approximation due the high dimensionality of the problem.
Codes are orbit, elegant,headtail, warp.
long term simulation
for evaluating the long term stability of motion in circular ring due to small non-linear perturbation. Usually thousands (electron) or millions (hadron) of turns are needed to find results. Long
term simulation are used either as a direct evaluation tool for the impact of machine imperfection or as a benchmark for perturbation methods (analysis of the 1-turn map, frequency map, tune
footprint, tune or action diffusion). The approximation used are share by the short term analysis.
Codes are sixtrack,teapot.
In the following, we concentrate on tracking for short term analysis and long term simulation that shares the need of keeping the mathematical structure of the equations exact.
Discrete Tracking[edit]
The aim of this section is to study how to solve the motion of a single particle in a general em field without affecting the structure of the equation of motions.
If the em fields are easily approximated by discontinuous vector field occupying a well defined not overlapping region, the tracking problem can be approached by a composition of discrete steps. Each
step is the exact solution of the equation of motion between the boundary of the region. The region type may be
• a region of zero volume defined by one surface
• a region enclosed by two parallel plane surfaces
• a region enclosed by two not parallel plane surfaces
• a region enclosed by general surfaces.
The program is to find out which kind of field and region allows the motion to be solved exactly and how to find those fields and regions that approximate the real field.
Equation of motion[edit]
A particle in the vacuum has four degree of freedom and the rest mass therefore can identified by the quantities:
$\vec x=(x,y,z,t) \qquad \vec p = (p_x,p_y,p_z,p_t) \quad m$
which are the location and mechanical momentum. The conservation of rest mass implies:
$m^2c^2= p_t^2 - p_x^2 - p_y^2 - p_z^2$.
The least action principle and the isotropy the space and homogeneity of time implies:
$\partial_t \vec x = \partial_{\vec p} H$
$\partial_t \vec p = - \partial_{\vec x} H$
for any $t$, where $H=p_t$.
Instead of $t, x,y,z$ could have be used as well. The solution of motion is a straight line.
In case of the em $\vec A(\vec x)=(-\phi(\vec x),A_x(\vec x),A_y(\vec x),A_z(\vec x))$, if one wants to keep the same structure for the equation of motion, one has to substitute $\vec p$ with $\vec P
=\vec p - q \vec A$ where $q$ is the charge of the particle.
The equation of motion are not exactly solvable in a bounded region in the general case. Solvable cases are:
drift type[edit]
The hamiltonian depends only on $\vec p$. The map is
$\vec x \to \vec x + \partial_{\vec p} H \qquad \vec p \to \vec p$.
A particular case is for no field in a parallel plane boundary region. The map is
$\vec x \to \vec x + \frac{t_f} m \vec p \qquad \vec p \to \vec p$
$t_f$ is the time of flight which does not depend on $\vec x$ because of the parallel boundaries. Assuming that the particle is sitting on one face, if the opposite face is at a distance $d$ then
$t_f=\frac {m d}{p_z}$
kick type[edit]
Zero volume surface with infinite field. The motion is a change of momentum but not a change of coordinate:
$\vec x \to \vec x \qquad \vec p_x \to \vec p_x + \vec f(\vec x)$
free field region (CONJECTURE TO BE CHECKED)[edit]
If there is no field and the region is defined by
$f(\vec x)=0$
and if $f(\vec x + t_f \frac{\vec p}{m} )=0$
can be solved for $t_f$. The map is:
$\vec x \to \vec x + \frac{t_f} m \vec p \qquad \vec p \to \vec p$
Being exact, the resulting map (maybe containing infinite terms) should be symplectic by construction.
uniform type (CONJECTURE TO BE CHECKED)[edit]
Uniform em field with parallel boundaries. Examples are ideal dipoles or solenoid magnet.
Since the motion can be exactly determined, the resulting map (maybe containing infinite terms) should be symplectic by construction.
derived type[edit]
If it exists a canonical transformation $T$that brings the field and the region in the above forms, the map $T S T^{-1}$
where $S$ is a map of one the type above solve exactly the motion. Example are sector bend magnets, where the canonical transformation brings to cylindrical coordinate (and may or may not cancel the
bending field) and back.
symplectic integrator[edit]
In case the Hamiltonian is the sum of solvable pieces $H=H_1+H_2+\ldots$, approximate models could be built in the form of a sequence of region such that $\prod \exp(c_i:H_j:)=\exp(:H:+O(n))$. A
particular case is the kick drift approximation or the generalized Yoshida symplectic integrator.
Inclusion of Radiation[edit]
When tracking a single single particle through a magnetic field, if it accelerates, it will emit radiation. This is particularly important when tracking electrons. Actually, the effect of radiation
is typically split into two distinct pieces, a deterministic energy loss, and a stochastic part resulting from the quantization of the outgoing electromagnetic field (photons).
First we describe the deterministic part. We use a high energy approximation in which the energy and momentum are related by E=P/c. The power radiated by a charged particle in a magnetic field is
given by $P_\gamma = \frac{e^2 c^3}{2\pi}C_\gamma E^2 B^2\$ with $C_\gamma = \frac{4\pi}{3}\frac{r_e}{(mc^2)^3}=8.85\times 10^{-5}\frac{m}{GeV^3}$. Now, suppose that in the symplectic integrator, we
go track through a region of magnetic field of length L. In fact, the length of this section will depend on the initial coordinates of the particle. The length of the region is given by $L^* = L(1+\
frac{x}{\rho}+\frac{1}{2}(x'^2+y'^2))$. Now, in many codes, the magnetic field is normalized by $B\rho$ so that the integrated constant term is unitless. Define $\bar B = \frac{B}{B\rho}$. Using the
relation $B\rho =\frac{E}{ec}$, one finds that the change in the energy deviation is given by $\delta = \delta_0 - C_r (1+\delta_0)^2\bar B_\perp^2 L^*$, with $C_r = C_\gamma \frac{E^3}{2\pi}$. This
is the formula used in e.g. Accelerator Toolbox and Tracy. Now that the change in energy (momentum) has been computed, the changes in the transverse momenta x', y' may be correspondingly computed.
This is done by noting that $p_{x,y}$ don't change in the radiation process. | {"url":"http://en.wikibooks.org/wiki/Guide_to_Non-linear_Dynamics_in_Accelerator_Physics/Tracking","timestamp":"2014-04-16T08:27:14Z","content_type":null,"content_length":"44868","record_id":"<urn:uuid:8c8f771e-abad-4701-9e83-45942445bd8d>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00183-ip-10-147-4-33.ec2.internal.warc.gz"} |
7th Grade Math - Numbers and Operations
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TM Math Fonts
TM Math is a family of serif fonts, inspired by remarkable Times Roman^(TM) typeface. Times Roman first appeared in 1932 in The Times of London newspaper, for which it was designed under the
direction of Stanley Morison. It has subsequently become one of the worlds most successful type creations. Widely used in books and magazines, for reports, office documents and also for display and
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Best 3d graphing app for a cheapskate?
December 18, 2008 10:07 PM Subscribe
Best application for 3D graphing (heat maps, contours, bump maps and so on)?
I have some data, mostly 3D and the data in higher dimensions can generally be reduced to 3D for analysis.
A typical dataset has a value for several hundred (x,y,z) triples where I want to map z as a function of x and y, in a heat map, bump map, or contour form. The actual form doesn't matter... it just
needs to be easily readable.
To take a specific example, I could have x = 1, 2, 3 ... 39, 40 and y = 10, 20, 30 ... 490, 500, with z varying between -500 and +500. I'd like to view this data as an (x, y) grid with the z values
represented as heights, or colors, or something else.
I've tried both Excel and Numbers and neither really cuts it. I don't want to shell out for MatLab or anything like that. So I'm wondering if there's a good opensource/freeware/shareware solution to
I'm mosty Mac but happy to run a Windows app in VMWare.
posted by unSane to Science & Nature (6 answers total) 2 users marked this as a favorite
does contour maps very well, but it seems like R does everything pretty well :). You could make something like
this contour map
with just a few lines of code.
R is open source and free for every platform.
posted by eisenkr at 10:23 PM on December 18, 2008 [1 favorite]
R is also has multiple graphing add ons in addition to the base functions, can take data from an excel spreadsheet, has it's own programming language for functions, and the graphs can be easily
modified on the fly.
Since it's open source and free, it's also incredibly popular for presentations at international conferences in a variety of fields.
posted by gryftir at 11:04 PM on December 18, 2008
Here's some R code that generates a 40 × 50 matrix with values uniformly and randomly picked from [-500, 500]:
> z <- matrix(runif(40 * 50, -500, 500), 40, 50)
> levelplot(z)
> contourplot(z)posted by grouse at 12:06 AM on December 19, 2008
I use Gnuplot with the python bindings. I love python.
posted by magikker at 12:30 AM on December 19, 2008
is a free matlab clone (which plots via Gnuplot). Matlab is well-documented, and 95% of that documentation will apply to Octave.
Also, on the Python side,
is freaking amazing.
posted by dmd at 6:12 AM on December 19, 2008 [1 favorite]
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On Run Distributions, pt. 5: Estimating Variance
So far, I’ve demonstrated only that the zero-modified negative binomial distribution (which I’m referring to as the Enby distribution for the sake of my sanity) can provide a decent fit to actual
scoring patterns when the sample variance of runs per game is known. In order for this approach to have value with teams for which we don’t know the variance (and that’s the whole point of the
exercise--estimating what the distribution should be based on the average R/G rather than simply regurgitating a sample distribution), we need a way to approximate the variance (As a tease, some work
being done independently should allow for a better approximation of the variance than what I've come up with here. At some point in the future, I will incorporate that method into my methodology).
Allow me to issue this disclaimer up front: the formula I’m proposing here is woefully inadequate. If the Enby distribution is to have any real value to sabermetricians, someone will have to come
along and clean this part up (while I’m making a wish list, a better way of adjusting the parameters to return the correct average R/G after the zero modification would be nice too).
The scatterplot below shows the variance of R/G plotted against average R/G for each major league team, 1981-1996. You can see there is clearly a positive correlation between the two: the higher the
average, the higher the variance:
There is no clear pattern that will help me in attempting to develop a function to estimate variance from the mean. In fact, if you plot the ratio of variance to mean against the mean, you get a big
There is a positive correlation (r = +.27) between the ratio of variance/mean and the mean. However, using a regression equation to describe the relationship between the mean and the variance
introduces the problem of illogical results at the extremes.
For example, a linear regression yields the following equation for variance as a function of mean:
Variance = 2.637*mean - 2.670
For any team scoring less than 1.013 R/G, this formula will predict a negative variance, which is obviously impossible. Granted, I’m under no delusions that the final method I offer will be of any
use outside of the normal scoring range of major league teams, but I cringe at laying out a method that obviously cannot work at such extremes.
Another option is to estimate the variance ratio as a function of the mean. The benefit here is that this constrains the estimated variance; if a team scores zero runs, the estimated variance will be
zero. The estimated variance can never be negative:
Variance/mean = .1345*mean + 1.430
So Variance = mean*(.1345*mean + 1.430) = 1.430*mean + .1345*mean^2
Neither of these equations comes close to matching the aggregate results at 4.46 R/G, which is troublesome. But that value is itself an amalgamation of hundreds of individual team seasons, each
recorded by teams that theoretically follow their own distributions of runs scored, and so I’m not sure a failure to match the result should be a death knell. Further complicating matters is that the
estimate of variance will only be used to fit initial r and B parameters, with r then being varied to ensure that the mean of the distribution equals the actual mean.
Let me try using the second formula and run through the whole process to generate the expected run distribution for the aggregate 4.4558 R/G:
1. Estimate the variance of R/G from the mean:
Variance = 1.43*mean + .1345*mean^2 = 1.43*4.4558 + .1345*4.4558^2 = 9.042
2. Fit the parameters of a Enby distribution assuming no zero-modification:
B = variance/mean - 1 = 9.042/4.4558 - 1 = 1.029
r = mean/B = 4.4558/1.029 = 4.33
3. Estimate the parameter z (variable RI = (R/G)/9 = 4.4558/9 = .4951):
z = (RI/(RI + .767*RI^2))^9 = (.4951/(.4951 + .767*.4951^2))^9 = .0552
4. Calculate the probability for k runs (where k >=1) using the zero-modified formula, then find a new value for r that sets the mean of the zero-modified distribution equal to the desired mean
This step can only be done via some kind of computer algorithm; I used trial and error and get r = 4.364.
For the first time in the series, we have a version of the Enby distribution that is not blatantly cheating compared to other methods: I am not treating the variance or the probability of being
shutout as known values, but rather am estimating them. We’re still not flying completely solo--the formula for estimating variance from mean was based on the same data that’s been aggregated, but
we’re getting closer.
What would the estimates look like if we tried to apply them to a really extreme team? I do not expect the Enby to perform well at all, but it’s worth checking to confirm. I’ll try to estimate the
run distribution first for a team that averages 1.5 R/G, then for a team that averages 10. I didn’t select these numbers for any particular reason other than that they are extreme, without being so
crazy as to be beyond the range that anyone could possibly care (for practical rather than theoretical reasons) if the method worked for that point.
Obviously we don’t have an actual run distribution to compare to, but I’ll compare to the Tango-Ben distribution which, while also untested at these extremes, would be a better bet. First, the 1.5 R/
G team (parameters z = .3387, B = .6318, r = 2.5706):
And the 10 R/G team (parameters z = .0039, B = 1.775, r = 5.638):
I was honestly surprised when I saw how closely Enby tracked Tango-Ben. Pleasantly so, of course, but surprised nonetheless.
I’ve said this before in different ways, but it’s worth repeating: my experience working with probability distributions is either theoretical (I took a Stats course once where I rarely wrote a real
number down for the entire class) or working with fixed distributions (i.e. you are given a Poisson distribution with parameter h = 2.05. What is the probability of three or fewer occurrences?) I
have no practical knowledge of how to best fit a zero-modified distribution to sample data, and thus my work product here will be of little value. Hopefully I’ve provided enough promising results to
encourage those of you who are skilled at this type of problem to consider the negative binomial as a model for runs per game.
No comments:
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
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Lagrange's Theorem & orders
March 11th 2010, 06:51 PM #1
Junior Member
Jan 2010
Lagrange's Theorem & orders
My question is: If |G|=15, show that G must have an element of order 3.
We're currently learning about Lagrange's Theorem, so I know that I need to use it somehow, and I can't use Sylow's or Cauchy's Theorems yet. It makes sense that elements of order 3 and 5 (and 1
and 15) would be in G but I don't know how to use Lagrange to show that.
Thanks! Any help would be greatly appreciated!
My question is: If |G|=15, show that G must have an element of order 3.
We're currently learning about Lagrange's Theorem, so I know that I need to use it somehow, and I can't use Sylow's or Cauchy's Theorems yet. It makes sense that elements of order 3 and 5 (and 1
and 15) would be in G but I don't know how to use Lagrange to show that.
Thanks! Any help would be greatly appreciated!
Any ideas? So, it cannot have an element of order $3$ and it can't have an element of order $15$ and so it must have all elements of order $5$...except the identity element.
Sorry, I don't quite understand what you mean here. G must have an element of order 3 -- are you saying to try a proof by contradiction?
Okay, so what I have so far is that
If G is cyclic, then it must contain some <g> where |g| = 3.
So, let G be non-cyclic. Then for all G, |g| = 1, 3, or 5. If |g|=1, it is the identity element. Assume |g|=5 for all g except the identity. Then G is cyclic which is a contradiction.
Or am I way off?
Okay, so what I have so far is that
If G is cyclic, then it must contain some <g> where |g| = 3.
So, let G be non-cyclic. Then for all G, |g| = 1, 3, or 5. If |g|=1, it is the identity element. Assume |g|=5 for all g except the identity. Then G is cyclic which is a contradiction.
Or am I way off?
How did you arrive at that last part? If $G$ were cyclic then there would be some $g\in G$ such that $|g|=15$. Thus, the assumption that no element of $G$ is of order $15$ means it can't be
Oops. Well then, if they're all of order 5, does that mean there are multiple unique subgroups of order 5, which is impossible since 5 is prime and so they must be the same group?
sorry. i just learnt about Lagrange theorem too and im confused about how this question has been proven.
why is it that we have made an assumption that there is no element of G that is of order 15? and how does that imply that it is not cyclic?
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Morita theory in stable homotopy theory. In Handbook of tilting theory, volume 332
- Amer. Jour. Math , 2004
"... Abstract: We show that the homotopy theory of differential graded algebras coincides with the homotopy theory of HZ-algebra spectra. Namely, we construct Quillen equivalences between the Quillen
model categories of (unbounded) differential graded algebras and HZ-algebra spectra. We also construct Qu ..."
Cited by 32 (10 self)
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Abstract: We show that the homotopy theory of differential graded algebras coincides with the homotopy theory of HZ-algebra spectra. Namely, we construct Quillen equivalences between the Quillen
model categories of (unbounded) differential graded algebras and HZ-algebra spectra. We also construct Quillen equivalences between the differential graded modules and module spectra over these
algebras. We use these equivalences in turn to produce algebraic models for rational stable model categories. We show that bascially any rational stable model category is Quillen equivalent to
modules over a differential graded Q-algebra (with many objects). 1.
- Adv. Math , 2006
"... Abstract. We investigate the relationship between differential graded algebras (dgas) and topological ring spectra. Every dga C gives rise to an Eilenberg-Mac Lane ring spectrum denoted HC. If
HC and HD are weakly equivalent, then we say C and D are topologically equivalent. Quasiisomorphic dgas are ..."
Cited by 14 (6 self)
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Abstract. We investigate the relationship between differential graded algebras (dgas) and topological ring spectra. Every dga C gives rise to an Eilenberg-Mac Lane ring spectrum denoted HC. If HC and
HD are weakly equivalent, then we say C and D are topologically equivalent. Quasiisomorphic dgas are topologically equivalent, but we produce explicit counterexamples of the converse. We also develop
an associated notion of topological Morita equivalence using a homotopical version of tilting. Contents
, 2013
"... présentée et soutenue par Kruna SEGRT Morita theory in enriched context ..." | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=6418871","timestamp":"2014-04-19T02:26:13Z","content_type":null,"content_length":"17195","record_id":"<urn:uuid:6d7e357c-fe6a-4766-80b1-86bd524e9611>","cc-path":"CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00326-ip-10-147-4-33.ec2.internal.warc.gz"} |
Comparing Decimals
If there are two decimal numbers we can compare them. One number is either greater than, less than or equal to the other number.
A decimal number is just a fractional number. Comparing 0.7 and 0.07 is clearer if we compared 7/10 to 7/100. The fraction 7/10 is equivalent to 70/100 which is clearly larger than 7/100.
Therefore, when decimals are compared start with tenths place and then hundredths place, etc. If one decimal has a higher number in the tenths place then it is larger than a decimal with fewer
tenths. If the tenths are equal compare the hundredths, then the thousandths etc. until one decimal is larger or there are no more places to compare. If each decimal place value is the same then
the decimals are equal. | {"url":"http://www.aaamath.com/g6_52_x1.htm","timestamp":"2014-04-16T08:13:47Z","content_type":null,"content_length":"7232","record_id":"<urn:uuid:a270a1e2-a586-47fb-b2d9-aac96755a56e>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00513-ip-10-147-4-33.ec2.internal.warc.gz"} |
Differential Equations: Homogeneous Linear Systems
It's been a while since we did math, so let's return to the always interesting field of differential equations, and specifically: homogeneous linear systems. As before, a "linear" system refers to
the degree and in this case, of one.
Now, to define homogeneous, let us consider the system as shown below:
dx1/dt = a11(t)x1 + a12(t)x2 + ..a1n(t)xn + f1(t)
dx2/dt = a21(t)x1 + a22(t)x2 + ..a2n(t)xn + f2(t)
dxn/dt = an1(t)x1 + an2(t)x2 + ..ann(t) xn + fn(t)
where the coefficients aij and the fi are functions continuous on a common interval, I.
Now, when fi(t) = 0, and i = 0, 1, 2.......n, this system is said to be homogeneous. Otherwise, it is called non-homogeneous.
Now, key to solving such a system is that one can use matrix form to do so (recall past blogs on matrices from last year) and in partricular something called "the fundamental matrix".
If then, X1, X2....X3 is a fundamental set of solutions of the homogenous system:
X' = AX
on an interval, I, then its general solution on the interval is:
X = c1X1 + c2 X2 + .......+ cn Xn
= c1[xi1j] + c2 [xi2j] + ...... cn [xinj]
where the brackets enfold row vectors
Now, let:
X1 = [xi1j] and X2 =[xi2j] and Xn = [xinj]
be a fundamental set of n solution vectors of the homogeneous system X' = AX on an interval I.
Then the matrix:
M(t) =
is said to be a fundamental matrix of the system on the interval.
Now, let's proceed to use an example, by solving the homogeneous system:
dx/dt = 2x + 3y
dy/dt = 2x + y
Then the solution is straightforward (using matrices) and is shown in the graphic boxes (from the top to the one below).
Note the steps:
1) Find the eigenvalues from the matrix composed of coefficients
2) Having obtained the eigenvalues, obtain the first set of eigenvectors (EV1) as shown
3) Obtain the corresponding equation (in k1, k2) for the first set of eigenvectors
4) Thereby obtain the second eigenvalues and eigenvectors and the corresponding equations.
5) Note the eigenvectors are K1 and K2 at the end.
6) Write the general solutions (X1, X2) in terms of K1, K2.
7) Thence, write the general solutions of the system, viz.
X = c1 X1 + c2 X2
and note how the exponential (exp) form is obtained, as well as the vector rows for X1, X2.
No comments: | {"url":"http://brane-space.blogspot.com/2011/01/differential-equations-homogeneous.html","timestamp":"2014-04-17T09:40:54Z","content_type":null,"content_length":"69471","record_id":"<urn:uuid:0966a7cc-ee34-47f0-be93-dfccdc82ec5b>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00123-ip-10-147-4-33.ec2.internal.warc.gz"} |
What If We Had Negative Mass?
Shooting star
So, a common balance does not measure mass, as we had been taught. It only measures the magnitude of the mass?
Well what I said is that it measures the magnitude of the
normal force
. This is then internally proportioned to "weight" by the balance...
Shooting star
What about the spring balance?
It should be the same.
I'm wondering why I'm being quizzed. Could someone begin to correct me? I'm sure I've made mistakes all over the place. ;) | {"url":"http://www.physicsforums.com/showthread.php?t=207201","timestamp":"2014-04-17T10:03:10Z","content_type":null,"content_length":"50610","record_id":"<urn:uuid:dc1ad782-8efd-40ab-89ef-1939c9f896ea>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00238-ip-10-147-4-33.ec2.internal.warc.gz"} |
This Article
Bibliographic References
Add to:
ASCII Text x
S. Ullman, R. Basri, "Recognition by Linear Combinations of Models," IEEE Transactions on Pattern Analysis and Machine Intelligence, vol. 13, no. 10, pp. 992-1006, October, 1991.
BibTex x
@article{ 10.1109/34.99234,
author = {S. Ullman and R. Basri},
title = {Recognition by Linear Combinations of Models},
journal ={IEEE Transactions on Pattern Analysis and Machine Intelligence},
volume = {13},
number = {10},
issn = {0162-8828},
year = {1991},
pages = {992-1006},
doi = {http://doi.ieeecomputersociety.org/10.1109/34.99234},
publisher = {IEEE Computer Society},
address = {Los Alamitos, CA, USA},
RefWorks Procite/RefMan/Endnote x
TY - JOUR
JO - IEEE Transactions on Pattern Analysis and Machine Intelligence
TI - Recognition by Linear Combinations of Models
IS - 10
SN - 0162-8828
EPD - 992-1006
A1 - S. Ullman,
A1 - R. Basri,
PY - 1991
KW - model combination; image combination; rigid transformations; linear combinations; visual object recognition; sharp edges; smooth bounding contours; computerised pattern recognition;
computerised picture processing
VL - 13
JA - IEEE Transactions on Pattern Analysis and Machine Intelligence
ER -
An approach to visual object recognition in which a 3D object is represented by the linear combination of 2D images of the object is proposed. It is shown that for objects with sharp edges as well as
with smooth bounding contours, the set of possible images of a given object is embedded in a linear space spanned by a small number of views. For objects with sharp edges, the linear combination
representation is exact. For objects with smooth boundaries, it is an approximation that often holds over a wide range of viewing angles. Rigid transformations (with or without scaling) can be
distinguished from more general linear transformations of the object by testing certain constraints placed on the coefficients of the linear combinations. Three alternative methods of determining the
transformation that matches a model to a given image are proposed.
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[11] T. S. Huang and C. H. Lee, "Motion and structure from orthographic projections,"IEEE Trans. Patt. Anal. Machine Intell., vol. 12, no. 5, pp. 536-540, 1989.
[12] D. P. Huttenlocher and S. Ullman, "Object recognition using alignment," inProc. ICCV Conf.(London), 1987, pp. 102-111.
[13] J. J. Koenderink and A.J. Van Doorn, "The internal representation of solid shape with respect to vision,"Biol. Cybernetics, vol. 32, pp. 211-216, 1989; in G.E. Hinton and J.A. Anderson,Parallel
Models of Associative Memory.Hillsdale, NJ: Lawrence Erlbaum, pp. 105-143.
[14] Y. Lamdan, J.T. Schwartz, and H. Wolfson, "On recognition of 3-D objects from 2-D images,"Courant Inst. Math. Sci., Robotics Tech. Rep. 122, 1987.
[15] C.H. Longuet-Higgins, "A computer algorithm for reconstructing a scene from two projections," Nature, vol. 293, pp. 133-135, 1981.
[16] D. Lowe, Perceptual Organization And Visual Recognition. Boston: Kluwer, 1985.
[17] D. Marr, "Analysis of occluding contour,"Phil. Trans. R. Soc. Lond. B, vol. 275, pp. 483-524, 1977.
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[21] S. Ullman,The Interpretation of Visual Motion. Cambridge, MA: MIT Press, 1979.
[22] S. Ullman, "Recent computational studies in the interpretation of structure from motion," in A. Rosenfeld and J. Beck (Eds.),Human and Machine Vision. New York: Academic, 1983.
[23] S. Ullman, "Aligning pictorial descriptions: An approach to object recognition:Cognition, vol. 32, no. 3, pp. 193-254, 1989; A.I. Memo 931, Artificial Intell. Lab., Mass. Inst. Technol., 1986.
[24] H. Freeman and Chakravarty, "The use of characteristic views in the recognition of three-dimensional objects," in E. Gelsema and L. Kanal (Eds.),Pattern Recognition in Practice. Amsterdam:
North-Holland, 1980.
[25] A. L. Yuille, D. S. Cohen, and P. W. Hllinan, "Feature extraction from faces using deformable templates," inProc. Comput. Vision Patt. Recogn., (San Diego), 1988, pp. 104-109.
[26] D. Zipser and R.A. Andersen, "A back-propagation programmed network that simulates response properties of a subset of posterior parietal neurons,"Nature, vol. 331, pp. 679-684, 1988.
Index Terms:
model combination; image combination; rigid transformations; linear combinations; visual object recognition; sharp edges; smooth bounding contours; computerised pattern recognition; computerised
picture processing
S. Ullman, R. Basri, "Recognition by Linear Combinations of Models," IEEE Transactions on Pattern Analysis and Machine Intelligence, vol. 13, no. 10, pp. 992-1006, Oct. 1991, doi:10.1109/34.99234
Usage of this product signifies your acceptance of the
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Irrigation system needs more pressure....
We recently expanded the irrigation system on our 5 acre property. We are now watering just over 1 acre. Prior to the expansion the system had good pressure and perfromed as expected. With the
additional coverage we are having pressure problems on all stations. I have made many adjustments and currently have a programmed delay of 5 minutes between each station. This gives our well system
some additional receovery time, but still doesn't solve the problem. I think we need to install an atmospheric storage tank and jet pump between our well pump and current 80 gal pressure tank. How do
I size the jet pump and storage tank? This complete water system services a 3500ft house with 5 people and the 1+ acre sprinkler system. We are already softening and chlorinating the house water due
to poor smell. The sprinkler and house systems are isolated properly and our concerns are only related to low pressure on the sprinkler side. However, we would expect an increase in house pressure
with a storage tank and jet pump. Thanks, Tom
It is a 3500 square foot house. I need a square foot symbol on my keyboard. Thanks,
TXTom If it worked before the expansion, but now doesn't perform as expected, sounds like it's fairly obvious that something you did during the expansion has caused the current problem. What did you
do to expand the "old" sprinkler system? Did you add additional zones and/or add heads to existing zones? Did you properly calculate GPM rates based on the nozzles you used?
Does the irrigation system come on strong and then dwindle. If so that may mean your well isn't producing enough water. I am guessing that this is not your problem. You need to start by calculating
how much water you need and at what pressure. Since the irrigation system is the biggest demand, you can start there. For each zone you need to the elevation above the pump, the flowrate (gallons per
minute), the pressure required for the sprinklers or ?, and the length and size of piping to the zone and to the furthest sprinkler. The sprinkler manufacture websites should have the gpm and
pressure required for the heads. With this information it should be possible to find why your system is not performing correctly. It could be your pump's capacity or it could be too much flow which
causes the pressure to drop in undersized pipes. Or, it could be something else. But, I am curious. Were your existing zones changed when you increased the area being irrigated? They should still
work if they weren't changed. Typically when you add an atmospheric tank, you need better water treatment. If your chlorinator can do the job, it may not be a problem. Or, you just use the
atmospheric tank only for irrigation. Usually tanks are required when your well can not produce enough water in the time needed. A bigger pump or a booster pump can be used to increase pressure. Or,
larger pipes can be used to reduce pressure drop.
Phil Yes the irrigation system comes on strong and then dwindles as time passes. I was able to reduce the dwindle by spacing the stations 5 minutes apart, but it is still a problem. The stations are
sized properly, however we are in a drought and our well may not be performing at its best. I am prepared to deal with the problems associated with an atmospheric tank. How do I size the tank so that
I have enough water for the entire sprinkler cycle? How do I size the jet pump so that I have pressure between 60-80psi? I would like to put the tank/jet pump between my well pump and 80 gal.
pressure tank. I just need a float switch for the well pump and I can continue to use all of the current controls to operate the jet pump. If the well pump is producing less gpm then I am using
during the irrigation system cycle, how do I keep the well pump from running continuously while I am draining the storage tank during the irrigation system cycle? Thanks, Tom
Before you buy pumps and controls, you should determine your demand. For the house, how many gallons in the peak 5 minutes, peak 10 minutes, peak 30 minutes, peak hour? The peak 5 minute demand will
determine size of your pump and pressure tank. The peak 10 minute demand will determine the capacity of your pump. The peak 30 and 60 minute demands will determine the minimum storage you need in
your atmospheric tank. There are multistage pumps that will provide more flow and pressure than a jet pump, and they are more efficient than a jet pump. You should select the pump after you determine
the demand. Then you need to determine your irrigation flow. That will be limited by your well capacity. There is no point in having a pump that pumps a lot more than your well can produce. You need
a controller that will sense low water in the well and protect the pump. They are usually based on the fact that when a pump runs out of water, it draws less current. You also need a low-water float
switch in the tank to shut off the irrigation, and maybe a lower one to protect the jet pump, or whatever pump you select. You may want to develop some logic about when you are going to irrigate
based on how much water is in the atmospheric tank. You irrigation controller probably must be more than just a timer.
Tom, I will give you a very general and basic idea. You need to determine how much water is required for irrigation and your household use during irrigation. Then you need to determine you wells
output and subtract that from the need. So you need to find the flowrate and total volume for each zone of your irrigation. Example: 10 sprinklers that use 1.5 GPM for 20 minutes = 15GPM for 20
minutes = 300 gallons. Then add up all of you zones for total. If you have 6 zones that are the same your irrigation system will need 1800 gallons in 120 minutes or 2 hours. Add up the maximum amount
of water you will use in the house while running your irrigation (ie running washing machine, and taking a shower etc) I will make up a number for this example and say 120 gallons per hour in
household use or 240 gallons in 2 hours. Adding the irrigation to the house hold and you will need 2040 gallons. Now you need to determine what your well will dependably produce. Let’s say 10 GPM for
this example. Therefore your well will produce 1200 gallons in 2 hours while you are watering. For a starting point on determining the tank size you can subtract the make-up water from the well to
the demand or 2040 – 1200 = 760 gallons. This is the minimum amount and that is when the tank is at the lowest level. This does not have any margin for error or little water leaks etc. I can be
pedantic. So I would make a time table to see how much water that I can expect the tank to have in the tank at the end of each zone’s cycle, because it is unlikely that they will have the same
flowrate. This should give you an idea of how to size the tank but not the pump. For the pump, you need to know two things, the maximum flow rate and the pressure required. The irrigation system you
already know from above. But the instantaneous flowrate for the house will be different. You need to get an idea of the flowrate when everything is running (ie 2 showers at 1.5 GPM + sink at 1GPM
etc.) Then you can determine the pumps flowrate. The pressure or head can be a little trickier to determine. It can be expressed in psi or feet of head: 1 ft = 0.43 psi or 1 psi = 2.31 ft. First you
need to determine the pressure required at the irrigation system and your house. Maybe, 40 psi for each sprinkler head and 40 psi for the house, or about 92 feet. Then you need to know the elevation
of the the sprinklers above and house above the tank bottom. Let’s say 30 feet elevation (fairly flat property). Next is the tricky part, it is determining the friction losses and minor losses in the
pipe, valves and fittings. Probably the easiest way for you is to use one of the many online calculators. I just put “friction loss in pipe” into google and a number of calculators came up where you
enter the inside diameter of the pipe, the length, and the flowrate. Here is one link http://www.freecalc.com/fricdia.htm (for this one: use PVC if you are using any type of plastic, they should be
close; viscosity is 1; specific gravity is 1). You need to find the pressure loss in each section of pipe at its maximum flowrate. Minor losses are in fittings and valves. Don’t worry about the
house, it should be covered in the pressure required at the house. If you don’t have a bunch of elbows and tees, you can probably neglect the loss or add a couple psi to your final number for saftey.
But your sprinkler valves and backflow preventers can add up. You can usually find the expected loss for these on the manufactures website. You want to find the maximum required pressure, so for each
sprinkler head (and your house) you need to add the pressure required, the elevation above the tank, the friction loss of each section of pipe going to the device, and any minor losses from valves.
The highest number will determine the required head for the pump. Note: you should talk to a pump supplier for recommendation on a pump. This is very basic information. It does not account for many
possibilities and senarios. After typing this, I remembered a good website with irrigation design information. http://www.irrigationtutorials.com/ | {"url":"http://www.terrylove.com/forums/showthread.php?8585-Irrigation-system-needs-more-pressure","timestamp":"2014-04-21T00:41:48Z","content_type":null,"content_length":"77285","record_id":"<urn:uuid:bbd0dc37-8b78-4fbb-802d-68145c4d3c66>","cc-path":"CC-MAIN-2014-15/segments/1397609539337.22/warc/CC-MAIN-20140416005219-00126-ip-10-147-4-33.ec2.internal.warc.gz"} |
Biholomophic non-Algebraically Isomorphic Varieties
up vote 18 down vote favorite
Recently, when writing a review for MathSciNet, the following question arose:
Is it true that two smooth complex varieties that are biholomorphic are algebraically isomorphic? The converse is true just because polynomials are holomorphic.
I was mainly interested in the affine case since that was the context of the review I was writing.
I knew about exotic affine spaces, so two smooth complex affine varieties that are real analytically isomorphic do not have to be algebraically isomorphic. But real analytic maps between complex
manifolds need not be holomorphic in general (just think of complex conjugation on $\mathbb{C}$). So that is not enough.
I posted the question on my Facebook page, and it got answered:
1. I was reminded that Serre's GAGA Theorem implies that it is true for projective varieties. But there are quasiprojective counterexamples provided on MO. See the answer of Georges Elencwajg given
2. Then it was pointed out that the answer in the link above is a manifold which is both affine and non-affine. So what about two affine varieties?
3. I then found a reference for such an example. Two exotic affine spaces that are biholomorphic yet not algebraically isomorphic. See here. This is a very nice result. Since it is in dimension 3,
it pretty much shows that at the very first point where something can go wrong, it does. A similar result for exotic affine spheres (apparently due to appear in J of Alg. Geo.) is here.
So I finally come to my questions.
Question 1: Is the example in the reference I found the first example of two affine varieties that are biholomorphic but not algebraically isomorphic?
Question 2: Is there some general reason to believe that all exotic affine spaces should be biholomorphic?
Comment: It seems to me within context to ask similar questions about objects, not only maps between objects. Here is some of what I have learned in that regard: all smooth complex varieties are
complex manifolds since they are covered by smooth affine open sets with polynomial (hence holomorphic) transition charts. Conversely, a closed analytic subspace of projective space is algebraic by
Chow's Theorem. But there are compact complex manifolds that are not algebraic (see here), which also shows that a closed analytic subspace of affine space need not be algebraic.
ag.algebraic-geometry complex-geometry ho.history-overview
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1 Answer
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Question 1: The first example published seems to be the following (Corollary 4 in "Embeddings of Danielewski surfaces", G. Freudenburg and L. Moser-Jauslin, Math.Z. 2003 ) For any $a\in
\mathbb{C}^*$, the surfaces in $\mathbb{C}^3$ given by $x^2z-y^2-a$ and $x^2z-(1+x)y^2-a$ are algebraically not isomorphic, but holomorphically isomorphic.
up vote 3
down vote Question 2: The answer should be no because of the strong analytic cancellation theorem of Zaidenberg (see http://arxiv.org/pdf/alg-geom/9506005v1.pdf page 5). It says that if $X,X'$ are
accepted contractible non-biholomorphic surfaces of general-type, then $X\times \mathbb{A}^1$ and $X'\times\mathbb{A}^1$ are not biholomorphic but are both exotic affine spaces.
First, thank you for the answer. I am tempted to accept it, but have a couple questions first. With respect to (1), I agree that it is an earlier example. However, how do you know it
is the first published answer? I could not find in that article a claim by the authors to that effect. If you did a MathSciNet and arXiv search, that is fine, and I would accept that.
But I just would like to know to what extent this is an "earlier find" versus a "scholarly fact". – Sean Lawton May 30 '13 at 18:12
With respect to (2), I just looked at the paper. Nice! Prop. 2.4 directly answers my question it appears. It also shows that there exists examples of exotic affine spaces that are not
biholomophic to affine space (analytically exotic); which is interesting too. I have probably a dumb question with respect to your example: how do you know that there exists
appropriate X and X' that are not biholomophic to begin with? The example you give is two different exotic products of affine 3-space. The example in Prop. 2.4 seems to compare an
exotic product with an exotic affine arising differently. – Sean Lawton May 30 '13 at 19:20
Ah, I see, my second question is irrelevant. There certainly are appropriate affine surfaces that are not isomorphic, and that suffices by the Strong Analytic Cancellation Theorem.
Great! I will accept your answer, even without my queries resolved. Thank you again! – Sean Lawton May 30 '13 at 19:26
Here is the link for the first article you cited: link.springer.com/content/pdf/10.1007%2Fs00209-003-0572-5.pdf – Sean Lawton May 30 '13 at 20:23
Dear Sean, yes for the first question I am not sure that there is no previous example, but at least I talked with some specialists and did not get any other one. This is of course not
1 a proof, that's why I said "seems to be". For (2), it is true that to complete the argument you need to find at least two surfaces which are not biholomorphic. You probably could find
these in the work of Ramanujan for example. – Jérémy Blanc May 31 '13 at 10:19
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Use table to find proportion of observations from standard normal... - (1228) | Transtutors
Intro Stats
Use table to find proportion of observations from standard normal distrubtion that satisfies each of the following statements. in each case sketch a standard normal curve and shade area that is the
answers to the questios a) z2.85 c) z>-1.66 d) -1.66<z<2.85
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Math Forum - Ask Dr. Math Archives: College Constructions
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Given a line and two points A and B, construct a circle tangent to the line and containing the two points.
Given three circles, is it possible to construct a circle tangent to each of them using only a compass and straightedge?
Let ABC be a triangle with sides a, b, c. Let h be the perpendicular from A to a, and m the median from A to the midpoint of a. Construct the triangle using only ruler and compass if you know A,
h, m.
How do you construct a segment of length "the 8th root of 3" using a compass and a straightedge?
Let x be a given angle. Let m and n be given lengths such that n > m. How can I construct triangle ABC such that AB = m, AC + CB = n, and the measure of angle ACB = x?
Why can you construct the trisection of some angles and not of others?
Is it true that the only regular n-gons that can be drawn using ONLY a straightedge and compass are those with the number of sides equal to a Fermat Prime or a product of Fermat Primes?
I've been trying to find a proof that a regular polygon with n sides is inconstructible if n is not a Fermat prime number.
Let ABC be a triangle with sides a, b, c. Let r be the radius of the incircle and R the radius of the circumcircle. Knowing a, R, and r, onstruct the triangle using only ruler and compass.
Is there a proof that you can't trisect an angle?
Without the use of a ruler, protractor or other measuring device, is there a simple way to divide a pizza into 3 equal sized slices?
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Game Theory - Heuristic for single-player puzzle
March 16th 2008, 01:25 PM
Game Theory - Heuristic for single-player puzzle
Board (4x3)
A B C D
E F G H
I J K L
You have 11 chips [1,1,1,1], [2,2,2,2], [3,3], [4] (for example) placed randomly on the 4x3 board in slots A,...,L and one empty slot (that moves) that we'll call X, the movements are done
similarly to the 8-puzzle game where X (empty) can move either up/down/left/right.
The Goal is to ensure that the top row and bottom row are symmetric (middle row is irrelevant) - so the following would be valid solutions:
3 3 4 X
X 2 4 2
As you can see in both cases the TOP and BOTTOM rows are symmetric (identical)...
So - just to help understand the game a little - assume you have the following flow from initial state to goal state:
3 3 1 X
--> Move LEFT
3 3 X 1
--> Move UP
4 2 X 2
!!! GOAL !!!
Therefore - my problem is to find out what is the BEST MOVE (up, down, left, right) to make.
Assume you are at STATE(0) in the example above, where do you move X to get the best result thus leading you closer to a GOAL? X could be moved UP or LEFT (two options) so why pick one over the
other? The solution is to evaluate the two possible outcomes and pick the better one ... this is the part where I need help, the EVALUATION algorithm (heuristic) needed to choose the next best
My current method works rather well but not well enough - I call it "Number of non-Symmetric Pairs" and it works as follows:
- For each state calculate the number of non-symmetric pairs and make the move that gives you the lowest number
- If you have equal lowest-numbers then choose one at random
Let me illustrate - for starters this would be the value for the states listed in the example above:
STATE(0) = 1 (as the 4th pair isn't symmetrical)
STATE(1) = 1 (as the 3rd pair isn't symmetrical)
STATE(2) = 0 (as all pairs are symmetrical)
Where this works - assume we are at STATE(1), we have 3 choices: LEFT, UP, RIGHT - which would each yield the following values:
STATE(1)->LEFT = 2
STATE(1)->UP = 0
STATE(1)->RIGHT = 1
Therefore the lowest number is 0 and we should move UP because that will generates less non-symmetrical pairs (and lucky for us in this case it is also the goal).
Where this doesn't work so well - assume we are at STATE(0), we have 2 choices: LEFT, UP - which would each yield the following values:
STATE(0)->UP = 1
STATE(0)->LEFT = 1
As you can see the values are identical, so in this case I cannot determine which move is best so I must try both routes, which in turns costs me a lot of time I was hoping I wouldn't have to
With that said - I am looking to find a more "informed" (efficient, intelligent) function/algorithm to help me solve my problem quicker and in less moves which therefore yields a more favorable
Along the same lines I am trying to see if there is another or better way to evaluate how close one move is to a goal when compared to another...
Any ideas, hints, help you might come up with would be much appreciated.
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Problem Solving Using M-file Environments
2.1: Problem Solving Using M-file Environments
Created by: CK-12
The purpose of this section is to introduce the engineering problem solving process in the context of using m-file environments to solve problems. Many variations of this process exist and no single
variation is best for solving all problems. In this module we describe a variation of the engineering problem solving process that applies to m-file environments problem solving. Other variations are
described in the reference in the bibliography.
The following problem solving process is fairly involved and may be an excessive amount of work for simple problems. For problems where the solution is straight forward, simply solve the problem; for
more complex problems, the solution will usually not be obvious and this process will aid in development of an appropriate solution.
This specific process is divided into a set of seven steps. Each step includes questions that help move you successfully through the problem solving process.
1. Define the Problem
• What problem are you trying to solve?
• "What would success look like?"
• What should the program output? Computed values? A plot or series of plots?
2. Identify given information.
• What constants or data are supplied?
• What theory, principles, models and equations have you been given?
3. Identify other available information.
• What theory, principles, models and equations can you find in other sources (text books, lecture notes, etc.)?
4. Identify further needed information.
• What other information do you need?
• Where will you find it?
5. Design and implement your solution to the problem.
• How can you break the larger problem into smaller problems?
• Look at the problem from the top down or bottom up?
• What programming techniques might you use to convert input to output?
• What variables do you need? Vectors? Arrays?
• What principles and equations apply to convert input to output?
6. Verify your solution.
• How do you know your solution is correct?
7. Reflect on your solution.
• What worked?
• What didn't?
When solving simple problems you may be able to follow these steps in order. For more complex problems, you may be working on step 5 and realize you need more information. You might then go back to
steps 3 or 4 to re-evaluate and find missing information.
You are part of a design team that is developing a commercial aluminum can crusher. Your preliminary crusher design includes a collection chamber in which cans are collected until a desired
weight of cans has accumulated; the cans are then crushed by a hydraulic ram. In preliminary research, you determine that the typical aluminum can is a cylinder with diameter $2.5''$$4.8''$
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Box Plots
Shodor > Interactivate > Lessons > Box Plots
The goal of this lesson is to introduce box plots and quartiles. An activity and discussion with supplemental exercises help students learn how data can be graphically represented.
Upon completion of this lesson, students will:
• have reviewed the concept of median
• have learned how to calculate quartiles for any size data set
• have learned how to build a box plot
Standards Addressed:
Textbooks Aligned:
Student Prerequisites
• Arithmetic: Student must be able to:
□ put numbers in order from smallest to largest
□ calculate the average of two numbers
• Technological: Student must be able to:
□ perform basic mouse manipulations such as point, click and drag
□ use a browser for experimenting with the activities
Teacher Preparation
• Access to a browser
• pencil and paper
• The worksheet for the box plot activity's built in data sets
• The worksheet for working with the data collected from the class (see below)
Key Terms
arithmetic See mean
average It is better to avoid this sometimes vague term. It usually refers to the (arithmetic) mean, but it can also signify the median, the mode, the geometric mean, and weighted mean, among
other things
box plot Also called box-and-whisker plot, this graph shows the distribution of data by dividing the data into four groups with the same number of data points in each group. The box contains the
middle 50% of the data points and each of the two whiskers contain 25% of the data points.
histogram A bar graph such that the area over each class interval is proportional to the relative frequency of data within this interval
mean The sum of a list of numbers, divided by the total number of numbers in the list. Also called arithmetic mean
median "Middle value" of a list. The smallest number such that at least half the numbers in the list are no greater than it. If the list has an odd number of entries, the median is the middle
entry in the list after sorting the list into increasing order. If the list has an even number of entries, the median is equal to the sum of the two middle (after sorting) numbers
divided by two. The median can be estimated from a histogram by finding the smallest number such that the area under the histogram to the left of that number is 50%
mode For lists, the mode is the most common (frequent) value. A list can have more than one mode. For histograms, a mode is a relative maximum ("bump"). A data set has no mode when all the
numbers appear in the data with the same frequency. A data set has multiple modes when two or more values appear with the same frequency.
multimodal A distribution with more than one mode. The histogram of a multimodal distribution has more than one "bump"
range The range of a set of numbers is the largest value in the set minus the smallest value in the set. Note that the range is a single number, not many numbers
total A total is determining the overall sum of numbers or a quantity.
Lesson Outline
1. Focus and Review
Remind students what has been learned in previous lessons that will be pertinent to this lesson and/or have them begin to think about the words and ideas of this lesson:
□ What are some examples of different ways that we have found to portray data?
2. Objectives
Let the students know what it is they will be doing and learning today. Say something like this:
□ Today, class, we are going to learn about box plots.
□ We are going to use the computers to learn about box plots, but please do not turn your computers on until I ask you to. I want to show you a little about this activity first.
3. Teacher Input
□ Remind students of the ideas behind means and medians, as covered in the Mean, Median and Mode discussion
□ Walk students through the construction of quartiles, the five-number summary and box plot construction as in the Box Plot discussion
4. Guided Practice
□ Have students experiment with the built-in data sets available in the Box Plot activity to be sure that they understand how to read the the box plots. Questions for the data sets can be found
in the worksheet.
5. Independent Practice
□ Have the students collect the following data from each other:
☆ Gender
☆ Height
☆ Length of ride/walk to school in minutes
☆ Estimate number of hours of TV watched in a week.
□ Have the students explore the questions on box plots by building the appropriate box plots either by hand or using the Box Plot activity. With less mature students, it would be best to help
them decide which box plot to graph for each question
6. Closure
□ You may wish to bring the class back together for a discussion of the findings. Once the students have been allowed to share what they found, summarize the results of the lesson.
Alternate Outline
This lesson can be rearranged in several ways.
• Students can be asked to work with the notion of outliers, as can be done with the Box Plot activity
Suggested Follow-Up
If the students have not yet seen histograms, the lesson on Histograms and Bar Graphs makes a good follow-up. For more advanced students, The Bell Curve, covers the normal distribution and the bell
curve controversy.
©1994-2014 Shodor Website Feedback
The goal of this lesson is to introduce box plots and quartiles. An activity and discussion with supplemental exercises help students learn how data can be graphically represented.
Upon completion of this lesson, students will:
• have reviewed the concept of median
• have learned how to calculate quartiles for any size data set
• have learned how to build a box plot
Standards Addressed:
Textbooks Aligned:
Student Prerequisites
• Arithmetic: Student must be able to:
□ put numbers in order from smallest to largest
□ calculate the average of two numbers
• Technological: Student must be able to:
□ perform basic mouse manipulations such as point, click and drag
□ use a browser for experimenting with the activities
Teacher Preparation
• Access to a browser
• pencil and paper
• The worksheet for the box plot activity's built in data sets
• The worksheet for working with the data collected from the class (see below)
Key Terms
arithmetic See mean
average It is better to avoid this sometimes vague term. It usually refers to the (arithmetic) mean, but it can also signify the median, the mode, the geometric mean, and weighted mean, among
other things
box plot Also called box-and-whisker plot, this graph shows the distribution of data by dividing the data into four groups with the same number of data points in each group. The box contains the
middle 50% of the data points and each of the two whiskers contain 25% of the data points.
histogram A bar graph such that the area over each class interval is proportional to the relative frequency of data within this interval
mean The sum of a list of numbers, divided by the total number of numbers in the list. Also called arithmetic mean
median "Middle value" of a list. The smallest number such that at least half the numbers in the list are no greater than it. If the list has an odd number of entries, the median is the middle
entry in the list after sorting the list into increasing order. If the list has an even number of entries, the median is equal to the sum of the two middle (after sorting) numbers
divided by two. The median can be estimated from a histogram by finding the smallest number such that the area under the histogram to the left of that number is 50%
mode For lists, the mode is the most common (frequent) value. A list can have more than one mode. For histograms, a mode is a relative maximum ("bump"). A data set has no mode when all the
numbers appear in the data with the same frequency. A data set has multiple modes when two or more values appear with the same frequency.
multimodal A distribution with more than one mode. The histogram of a multimodal distribution has more than one "bump"
range The range of a set of numbers is the largest value in the set minus the smallest value in the set. Note that the range is a single number, not many numbers
total A total is determining the overall sum of numbers or a quantity.
Lesson Outline
1. Focus and Review
Remind students what has been learned in previous lessons that will be pertinent to this lesson and/or have them begin to think about the words and ideas of this lesson:
□ What are some examples of different ways that we have found to portray data?
2. Objectives
Let the students know what it is they will be doing and learning today. Say something like this:
□ Today, class, we are going to learn about box plots.
□ We are going to use the computers to learn about box plots, but please do not turn your computers on until I ask you to. I want to show you a little about this activity first.
3. Teacher Input
□ Remind students of the ideas behind means and medians, as covered in the Mean, Median and Mode discussion
□ Walk students through the construction of quartiles, the five-number summary and box plot construction as in the Box Plot discussion
4. Guided Practice
□ Have students experiment with the built-in data sets available in the Box Plot activity to be sure that they understand how to read the the box plots. Questions for the data sets can be found
in the worksheet.
5. Independent Practice
□ Have the students collect the following data from each other:
☆ Gender
☆ Height
☆ Length of ride/walk to school in minutes
☆ Estimate number of hours of TV watched in a week.
□ Have the students explore the questions on box plots by building the appropriate box plots either by hand or using the Box Plot activity. With less mature students, it would be best to help
them decide which box plot to graph for each question
6. Closure
□ You may wish to bring the class back together for a discussion of the findings. Once the students have been allowed to share what they found, summarize the results of the lesson.
Alternate Outline
This lesson can be rearranged in several ways.
• Students can be asked to work with the notion of outliers, as can be done with the Box Plot activity
Suggested Follow-Up
If the students have not yet seen histograms, the lesson on Histograms and Bar Graphs makes a good follow-up. For more advanced students, The Bell Curve, covers the normal distribution and the bell
curve controversy.
The goal of this lesson is to introduce box plots and quartiles. An activity and discussion with supplemental exercises help students learn how data can be graphically represented.
arithmetic See mean
average It is better to avoid this sometimes vague term. It usually refers to the (arithmetic) mean, but it can also signify the median, the mode, the geometric mean, and weighted mean, among
other things
box plot Also called box-and-whisker plot, this graph shows the distribution of data by dividing the data into four groups with the same number of data points in each group. The box contains the
middle 50% of the data points and each of the two whiskers contain 25% of the data points.
histogram A bar graph such that the area over each class interval is proportional to the relative frequency of data within this interval
mean The sum of a list of numbers, divided by the total number of numbers in the list. Also called arithmetic mean
median "Middle value" of a list. The smallest number such that at least half the numbers in the list are no greater than it. If the list has an odd number of entries, the median is the middle
entry in the list after sorting the list into increasing order. If the list has an even number of entries, the median is equal to the sum of the two middle (after sorting) numbers
divided by two. The median can be estimated from a histogram by finding the smallest number such that the area under the histogram to the left of that number is 50%
mode For lists, the mode is the most common (frequent) value. A list can have more than one mode. For histograms, a mode is a relative maximum ("bump"). A data set has no mode when all the
numbers appear in the data with the same frequency. A data set has multiple modes when two or more values appear with the same frequency.
multimodal A distribution with more than one mode. The histogram of a multimodal distribution has more than one "bump"
range The range of a set of numbers is the largest value in the set minus the smallest value in the set. Note that the range is a single number, not many numbers
total A total is determining the overall sum of numbers or a quantity.
Remind students what has been learned in previous lessons that will be pertinent to this lesson and/or have them begin to think about the words and ideas of this lesson:
Let the students know what it is they will be doing and learning today. Say something like this:
If the students have not yet seen histograms, the lesson on Histograms and Bar Graphs makes a good follow-up. For more advanced students, The Bell Curve, covers the normal distribution and the bell
curve controversy. | {"url":"http://www.shodor.org/interactivate/lessons/BoxPlots/","timestamp":"2014-04-17T16:07:19Z","content_type":null,"content_length":"37408","record_id":"<urn:uuid:992431a4-5689-4b91-b473-44901eef1bdc>","cc-path":"CC-MAIN-2014-15/segments/1398223206672.15/warc/CC-MAIN-20140423032006-00144-ip-10-147-4-33.ec2.internal.warc.gz"} |
complex problem
October 11th 2010, 09:43 PM #1
Senior Member
Feb 2010
complex problem
show that the equation of the tangent to the circle mod(z)=r where r>0 at point z0 is given by ((conjugate) (z0))*z+z0*((conjugate)(z)=2*r^2
not understanding how to do this as moving point on the circumference of circle is z but what is z0
$z_0$ is a given point on the circle and $z$ is an arbitrary point on the tangent.
can yu please help me solve the problem
do we have to use concept of rotation in it
The reason I have not attempted to solve this is because I do not know how you are expected to solve it, and suspect how I would approach this is not acceptable for you to submit.
You could start by writing it in cartesians so $z=x+i y$, and $z_0=r (\cos(\theta)+i \sin(\theta))$, then plug these into the proposed equation and show that it is a straight line which passes
through $z_0$ and has gradient $-1/\tan(\theta)$
can yu please elaborate yur solution
October 11th 2010, 11:11 PM #2
Grand Panjandrum
Nov 2005
October 11th 2010, 11:45 PM #3
Senior Member
Feb 2010
October 12th 2010, 02:40 AM #4
Grand Panjandrum
Nov 2005
October 12th 2010, 06:13 AM #5
Senior Member
Feb 2010 | {"url":"http://mathhelpforum.com/pre-calculus/159273-complex-problem.html","timestamp":"2014-04-19T04:35:06Z","content_type":null,"content_length":"41716","record_id":"<urn:uuid:0985ac55-4677-400a-be16-7e20d50284b5>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00389-ip-10-147-4-33.ec2.internal.warc.gz"} |
Galois group of a polynomial
January 3rd 2010, 02:59 AM #1
Jan 2010
Galois group of a polynomial
I have to find a splitting field $L$ of the polynomial $f=x^4-4x^2+9 \in \mathbb Q[x]$ over $\mathbb Q$ and the galois group and all intermediate fields of the algebraic extension $L/\mathbb Q$.
I found out, that $f$ is irreducible, the roots of $f$ are:
$a=\sqrt{2-i\sqrt{5}}, \quad b=-\sqrt{2+i\sqrt{5}}, \quad c=\sqrt{2+i\sqrt{5}}=-b, \quad d=-\sqrt{2-i\sqrt{5}}=-a$ and that $\mathbb Q(a)$ is a splitting field of $f$.
Furthermore I find out, that the three homomorphisms
$a \mapsto \pm b, \ a \mapsto -a$ have order 2, so the galois group is isomorph to $\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$.
But now I don't know how to find the fixed fields, which belongs to the homomorphisms.
For example: Which fixed field belongs to the homomorphism $\sigma_1: a \mapsto b=\frac{-3}{a}$ ?
The definition says:
$\mathbb Q(a)^{\left \{ id, \sigma_1 \right \}}=\left \{ z \in \mathbb Q(a)\ |\ \sigma_1(z)=z \right \}$
...and this don't help me, because I need a description of the fixed field like this: $\mathbb Q(\beta)$ (with suitable $\beta$).
I hope, you can help me.
I have to find a splitting field $L$ of the polynomial $f=x^4-4x^2+9 \in \mathbb Q[x]$ over $\mathbb Q$ and the galois group and all intermediate fields of the algebraic extension $L/\mathbb Q$.
I found out, that $f$ is irreducible, the roots of $f$ are:
$a=\sqrt{2-i\sqrt{5}}, \quad b=-\sqrt{2+i\sqrt{5}}, \quad c=\sqrt{2+i\sqrt{5}}=-b, \quad d=-\sqrt{2-i\sqrt{5}}=-a$ and that $\mathbb Q(a)$ is a splitting field of $f$.
Furthermore I find out, that the three homomorphisms
$a \mapsto \pm b, \ a \mapsto -a$ have order 2, so the galois group is isomorph to $\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$.
But now I don't know how to find the fixed fields, which belongs to the homomorphisms.
For example: Which fixed field belongs to the homomorphism $\sigma_1: a \mapsto b=\frac{-3}{a}$ ?
The definition says:
$\mathbb Q(a)^{\left \{ id, \sigma_1 \right \}}=\left \{ z \in \mathbb Q(a)\ |\ \sigma_1(z)=z \right \}$
...and this don't help me, because I need a description of the fixed field like this: $\mathbb Q(\beta)$ (with suitable $\beta$).
I hope, you can help me.
well, it's very straightforward: $\{1,a,a^2,a^3 \}$ is a $\mathbb{Q}$-basis for $\mathbb{Q}(a).$ so for any $z \in \mathbb{Q}(a)$ there exist unique $q_i \in \mathbb{Q}$ such that $z=q_0 +
now $\sigma_1(z)=z$ is equivalent to $\sum_{k=0}^3q_ka^k=\sum_{k=0}^3 q_k b^k=\sum_{k=0}^3 (-3)^k q_k a^{-k}.$ multiplying both sides by $a^3$ gives us $\sum_{k=0}^3q_ka^{k+3}=\sum_{k=0}^3 (-3)^
kq_ka^{3-k}.$ we also have
$a^4=4a^2-9, \ a^5=4a^3-9a,$ and $a^6=7a^2-36.$ then you'll get $q_2=0, \ q_3=-q_1.$ therefore the fixed field of the automorphism $\sigma_1$ is:
$\{q_0 + q_1(a-a^3) : \ q_0, q_1 \in \mathbb{Q} \}=\mathbb{Q}(\sqrt{2}i).$ (note that $a-a^3$ is a root of the irreducible polynomial $x^2 + 18$).
Thank you for your very helpful answer!
January 3rd 2010, 04:55 AM #2
MHF Contributor
May 2008
January 3rd 2010, 06:10 AM #3
Jan 2010 | {"url":"http://mathhelpforum.com/advanced-algebra/122258-galois-group-polynomial.html","timestamp":"2014-04-18T20:54:29Z","content_type":null,"content_length":"46660","record_id":"<urn:uuid:6b651b4c-ef06-4391-b98b-2b1b59cee1f7>","cc-path":"CC-MAIN-2014-15/segments/1397609539230.18/warc/CC-MAIN-20140416005219-00301-ip-10-147-4-33.ec2.internal.warc.gz"} |
Pacific Palisades SAT Math Tutor
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cant get my radix sort working:(
Author cant get my radix sort working:(
i need to write code to sort an arbitrary number of digits and random values. my code as of now just replaces the first position with the last position!
Joined: Apr 04, 2013 ive been working at this all day and just cant figure out where its bugging out. please help
Posts: 10
Joined: Oct 13, 2005
Posts: 36478
What is the algorithm for radix sort? Write it down on paper, then convert that to pseudo‑code, before trying to write code.
subject: cant get my radix sort working:( | {"url":"http://www.coderanch.com/t/608765/java/java/radix-sort-working","timestamp":"2014-04-18T12:10:29Z","content_type":null,"content_length":"22531","record_id":"<urn:uuid:28eaf74c-8496-49a0-bb65-f74641e9f9e0>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00130-ip-10-147-4-33.ec2.internal.warc.gz"} |
[Numpy-discussion] matrix indexing question
Alan G Isaac aisaac@american....
Mon Mar 26 08:54:26 CDT 2007
> Alan G Isaac schrieb:
>> What feels wrong: iterating over a container does not give
>> access to the contained objects. This is not Pythonic.
On Mon, 26 Mar 2007, Sven Schreiber apparently wrote:
> If you iterate over the rows of the matrix, it feels
> natural to me to get the row vectors
Natural in what way?
Again, I am raising the question of what
would be expected from someone familiar with Python.
Abstractly, what do you expect to get when you iterate
over a container? Seriously.
> But I admit I'm a 2d fan so much so that I didn't even
> know that using a single index is possible with a matrix.
Exactly. When you want submatrices, you expect to index for
them. EXACTLY!!
Alan Isaac
More information about the Numpy-discussion mailing list | {"url":"http://mail.scipy.org/pipermail/numpy-discussion/2007-March/026794.html","timestamp":"2014-04-17T01:33:40Z","content_type":null,"content_length":"3305","record_id":"<urn:uuid:e1f2b83c-998f-42ae-96dc-364ed8229f7d>","cc-path":"CC-MAIN-2014-15/segments/1397609526102.3/warc/CC-MAIN-20140416005206-00263-ip-10-147-4-33.ec2.internal.warc.gz"} |
Fibonacci induction
April 25th 2013, 03:19 PM #1
Oct 2012
san francisco
Fibonacci induction
Hello I need some help with this problem
For each natural number n, let fn be the nth Fibonacci number. Prove that (f_(n+1))^2 + (f_n)^2 = f_(2n+1).
so far what I have
let n=1 so 1^2 + 1^2 = 2 which is true so the equation holds for n= 1 (base case)
then I pick an arbitrary k where (f_(k+1))^2 + (f_k)^2 = f_(2k+1)
and now I prove for k+1 so I'm trying to show
(f_(k+2))^2 + (f_(k+1)^2 = f_(2k+3)
Ive spent hours trying to manipulate this problem to show this is true but no luck so far so I am hoping someone can guide me through this problem so I know how to do it thanks.
Re: Fibonacci induction
You are going to need to use the defining property of Fibonacci series which is $f_n+ f_{n+1}= f_{n+2}$.
Re: Fibonacci induction
so i substituted and got (f_k + f_(k+1))^2 + (f_(k+1)^2 = f_(2k+3)
what do i do next? do I multiply it out and combine? Substitute in again for some other values?
Note: I have tried many possibilities for this problem, and substituting was one of them, it wasn't as easy as it looks
Re: Fibonacci induction
As often happens with induction, you need to generalize the statement you are proving. Try proving
$u_{n+m} = u_{n-1}u_m + u_n u_{m+1}$
by induction on m.
Another method is to use matrix multiplication.
Re: Fibonacci induction
Thanks for your help emakarov
April 25th 2013, 04:02 PM #2
MHF Contributor
Apr 2005
April 25th 2013, 04:20 PM #3
Oct 2012
san francisco
April 25th 2013, 04:53 PM #4
MHF Contributor
Oct 2009
April 25th 2013, 08:59 PM #5
Oct 2012
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need matlab code for R peak detection in ecg signals
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i need a matlab code to train images
hi everyone This is jennifer... Please....If any one have a code to train images using MLP or any other method please send me the code...it would be very helpful for me.... Thankyou Jennifer Well get
a toolbox (netlab or nn toolbox) and start writing code using functions in the toolbox! This since every case is different. for instance the outpu in your case... or the format of pictures, the need
for preprocessing images.. "jennifer ranjani" <feranisj@gmail.com> skrev i meddelandet news:eefe0be.-1@webx.raydaftYaTP... > hi everyone > > This is jennifer... > Pl...
urgent help needed for algorithm generating using Matlab
I am quite new to Matlab and would appreciate for some advices and help. My question is how can I construct a geometry surface or 2D or cross sectional shape of a distorted cylinder for example, if I
have several radius of curvature information of the object determined from sensors? If using circular arcs fitting for each radius of curvature, how to generate the related algorithm/code? Thank you
very much On Apr 8, 1:40=A0pm, "Jo Albon" <joomoyk...@googlemail.com> wrote: > I am quite new to Matlab and would appreciate for some advices and help. = My question is how can ...
MATLAB Code to Generate Sample Time Series from PSD
I am attempting to generate a sample time series from a PSD. I know both the DC value of the PSD and I have the square root of the PSD to determine the filter response. Here is the code that I am
using: n = sqrt(8E-12)*randn(1,500000); %Generate IID Gaussian noise samples with variance 8E-12 V = tf(2*pi,[1 2*pi]); % Generate continuous time transfer function (2*pi/(s+2*pi)) V_d = c2d(V,1E-4);
% Convert to discrete time transfer function with sample rate of 10 kHz [v,t] = impulse(V_d); % Generate discrete time impulse response vibe = conv(n,v); % Generate sample time series Does this look
How can I use signal processing block for code generation
I tried several times to use signal processing blockset for code generation. but I couldn't find the way to use it Borland compiler. What kind of files is required to use signal processing blockset
code generation? I appriciate to give me a message & solution. "Jung-Ho Moon" <junghomoon@koreanair.com> wrote in message <gl39ke$gr4$1@fred.mathworks.com>... > I tried several times to use signal
processing blockset for code generation. > but I couldn't find the way to use it Borland compiler. > > What kind of files is required to use signal process...
about Matlab generate C code which can be used in C5505
Hey i found out that matlab can use simulink to build CCS project and it can optimize the code corresponding to the different DSP. Here is the following question. i found out that matlab only support
one chip which is C5510 DSK in C55XX series. however, i checked the driver for my TMS320C5505 EVM board, it is C5510 DSK. So, can i use matlab to generate CCS project for my TMS320C5505? my second
question is: why the driver which installed for my TMS320C5505 EVM board is C5510. Best regards Reynold ...
Re: i need coding of video compression using MATLAB
dear sir i need coding of video compressio using MATLAB.so if u have pls send it to me on my mail address. thanking u ...
need matlab code for finding lag in the data series
hi ..i have y and u 2 data series ..if i wanna know the lag required in u to correlate with the y ...how can i use correlation function.. bye pawan pawan pandey wrote: > > > hi ..i have y and u 2
data series ..if i wanna know the lag > required > in u to correlate with the y ...how can i use correlation > function.. > > bye > pawan Calculate the cross correlation function and look for a spike
in it. The location of that spike depends on the lengths of your inputs and the lag between them. Good luck. (You could also do it using <conv> in a similar way and it...
Need MATLAB coding for PN Sequence Generator Block~~
Hi aii, can someone tell me where to get the MATLAB coding for PN Sequence Generator Block or send me the M-file for this PN Sequence Generator Block. My email address is khlee_83@yahoo.co.uk. Hope
to hear from ur reply soon. THX a lot~~~ Khoon Hai Lee wrote: > > > Hi aii, can someone tell me where to get the MATLAB coding for PN > Sequence Generator Block or send me the M-file for this PN
Sequence > Generator Block. My email address is khlee_83@yahoo.co.uk. Hope to > hear from ur reply soon. THX a lot~~~ JNS: Try this link: <http://mathforum.org/epigone/comp.soft-sys.m...
MATLAB code for baseline wander removal in ECG signal
Hello, I am working on a project in which i have classify heartbeats using morphological and dynamic features of ECG. First block of my block diagram is preprocessing which contains removal of
baseline wander. I am not getting how to develop the code. Also which technique should be used? Can anybody help? ... | {"url":"http://compgroups.net/comp.soft-sys.matlab/need-code-for-generating-an-ecg-signal-u/382277","timestamp":"2014-04-18T10:36:06Z","content_type":null,"content_length":"32719","record_id":"<urn:uuid:2ad7dabb-e87d-493d-8589-9dd9055616ee>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00445-ip-10-147-4-33.ec2.internal.warc.gz"} |
In 1999, Bertram, Ciocan-Fontanine and Fulton related quantum multiplication of Schur polynomials to the classical product via rim-hook removal. This is called the "rim-hook rule." Since the
LIttlewood-Richardson rule is easily accessible, this means that products in $QH^*(Gr(k,n))$ are also similarly accessible. We provide an equivariant version of this rim-hook rule, explicitly
relating the rings $QH^*_T(Gr(k,n))$ and $H^*_T(Gr(k,2n-1))$ or alternately the quantum product of factorial Schur polynomials to the classical product. This allows computations in $QH^*_T(Gr
(k,n))$ using combinatorial devices such as Knutson and Tao's puzzles for $H^*_T(Gr(k,n))$. Interestingly, this rule requires a specialization of torus weights that is tantalizingly similar to
maps in affine Schubert calculus, which is related to Gromov-Witten theory by Peterson's theorem.
We explore extensions of Stanley's work on applications of quasisymmetric functions to generalized riffle shuffling. Results include a probabilistic interpretation of the quasisymmetric Schur
functions of Haglund, Luoto, Mason, and van Willigenburg in terms of Schensted insertion for composition tableaux, as well as similar such interpretations in the setting of colored quasisymmetric
functions. We also extend previous work with Hersh to show that characters on the Hopf algebra of colored quasisymmetric functions induce Markov chains on descent classes of wreath products.
The 0-Hecke algebra is an interesting deformation of the group algebra of the symmetric group. By studying its action on the Stanley-Reisner ring of the Boolean algebra, we obtain a
representation theoretic interpretation for the noncommutative Hall-Littlewood symmetric functions introduced by Bergeron and Zabrocki, and recover a result of Garsia and Gessel on generating
functions of the joint distribution of five permutation statistics. Some of our results can be generalized to the setting of the Hecke algebra acting on the Stanley-Reisner ring of the Coxeter
In an unpublished manuscript from 2005, F. Hivert introduced a one-parameter family of (non-multiplicative) actions of the symmetric group $S_n$ on the space of polynomials in $n$ commuting
variables. It happens that the classical symmetric polynomials form the space of invariants at parameter $r=\infty$ and the classic(?) quasisymmetric polynomials form the space of invariants at
parameter $r=1$. It also happens that for each integer $1\leq r \leq \infty$, the space $r$-QSym of level-$r$ invariants forms a ring, which brings us to the subject of the present work. We
revisit two conjectures of Hivert regarding quotients of $r$-QSym by $\infty$-QSym, first proven by Garsia and Wallach (2007), and look for some representation-theoretic and combinatorially-rich
bases along the way. (This is joint work with Sarah Mason.)
This story begins with work of Reiner, Shaw and van Willigenburg, where they showed that if two skew Schur functions $s_A$ and $s_B$ are equal, then the skew shapes $A$ and $B$ must have the same
``row overlap partitions.'' Unfortunately, the converse is not true. Recently, we have shown that these row overlap equalities are also implied by a much weaker condition than skew Schur
equality: that $s_A$ and $s_B$ have the same support when expanded in the fundamental quasisymmetric basis $F$. Surprisingly, there is significant evidence supporting a conjecture that the
converse is also true.
In fact, we will work in terms of inequalities: if the $F$-support of $s_A$ contains that of $s_B$, then the row overlap partitions of $A$ are dominated by those of $B$. Again, we conjecture that
the converse also holds. After giving evidence in favor of our conjecture, we will conclude with a consideration of how the quasisymmetric Schur basis and the dual immaculate basis fit into our
We study the factorizations of the permutation $(1, 2, . . . , n)$ into $k$ factors of given cycle types. Using the group algebra of the symmetric group, Jackson obtained for each $k$ an elegant
formula for counting these factorizations according to the number of cycles of each factor. In the cases $k = 2, 3$ Schaeffer and Vassilieva gave a combinatorial proof of Jackson’s formula, and
Morales and Vassilieva obtained more refined formulas exhibiting a surprising symmetry property. These counting results are indicative of a rich combinatorial theory which has remained elusive to
this point, and it is the goal of this project to establish a series of bijections which unveil some of the combinatorial properties of these factorizations into $k$ factors for all $k$. The
first bijection is an instance of a correspondence of Bernardi between such factorizations and tree-rooted maps; certain graphs embedded on surfaces with a distinguished spanning tree. This is
joint work with Olivier Bernardi.
The notion of chromatic polynomials for graphs can be extended naturally in two directions: first refine it to a symmetric function instead of a polynomial, then define it for posets in addition
to graphs. On the plus side, a number of interesting questions fall out of this generalization, including Stanley and Stembridge's e-positivity conjecture for (3+1)-free posets. On the minus
side, we lose the ability to use the deletion-contraction algorithm. In this talk, we'll see a (weaker) replacement for deletion-contraction and use it to make some progress on the e-positivity
It is well-known that any permutation can be written as a product of two involutions, often in a number of different ways. Curiously, the question of how many ways to do this appears to be
closely related to the sum of absolute values of the characters in the symmetric group. This leads to speculation about a cancellation-free Murnaghan-Nakayama rule. This is joint work with
Bridget Tenner.
Pick a card, any card, from the deck and remove it; then put it back anywhere in the deck. Repeating this process leads to a method of shuffling a deck of cards known as the random-to-random
shuffle. Its efficiency is controlled by the spectrum of its transition matrix, which turns out to be closely related to the combinatorics of the symmetric group.
In this talk we will give a combinatorial description of this spectrum and outline some of the ideas that go into the proof. This settles a conjecture made in 2002 by Uyemura-Reyes: after a
suitable renormalization, this spectrum is integral. Our analysis makes considerable use of the representation theory of the symmetric group. This is joint work with Ton Dieker.
We introduce a new basis of the non-commutative symmetric functions whose elements have Schur functions as their commutative images. Dually, we build a basis of the quasi-symmetric functions
which expand positively in the fundamental quasi-symmetric functions and decompose Schur functions according to a signed combinatorial formula. These bases have many interesting properties
similar to those of the Schur basis, and we will outline a few of them.
Jeu-de-taquin promotion gives a bijection on semistandard $\lambda$-tableaux $CST(\lambda,k)$, with entries bounded by $k$. While, for general shapes, the order of promotion on $CST(\lambda,k)$
is not known, Rhoades gave the order of promotion on the set of semistandard rectangular tableaux $CST((c^r),k)$ and proved that $CST((c^r),k)$ exhibits the cyclic sieving phenomenon (CSP). We
determine the order of promotion on the set of semistandard hook tableaux $CST((n-m,1^m),k)$ and give a CSP for the set of semistandard hook tableaux with fixed content.
The quasisymmetric Schur functions introduced by Haglund, Mason, Luoto and van Willigenburg are a basis for the algebra of quasisymmetric functions. They refine the classical Schur functions in a
natural way and mimic many of their combinatorial properties.
In this talk we will define an action of the 0-Hecke algebra on standard reverse composition tableaux of composition shape $\alpha$ and use it to produce a 0-Hecke module whose quasisymmetric
characteristic is the quasisymmetric Schur function $S_{\alpha}$. Furthermore, we will identify those modules that are indecomposable and cyclic. This is joint work with Stephanie van
The zeta map has been rediscovered by many researchers in various contexts. Our interest ultimately stems from its ability to relate the area, bounce and dinv statistics that arise in the study
of the q,t-Catalan numbers. In joint work with N. Loehr and D. Armstrong, we prove the equivalence of several descriptions of the zeta map as well as unify many of its previously known
manifestations under a very general map that we call the sweep map. | {"url":"http://cms.math.ca/Events/winter13/abs/sfg","timestamp":"2014-04-17T03:51:25Z","content_type":null,"content_length":"23473","record_id":"<urn:uuid:98e41637-7bc6-4b36-9d6b-e47ad4a1518f>","cc-path":"CC-MAIN-2014-15/segments/1397609526252.40/warc/CC-MAIN-20140416005206-00012-ip-10-147-4-33.ec2.internal.warc.gz"} |
Obtain A Relation For The Fin Efficiency For A... | Chegg.com
Obtain a relation for the fin efficiency for a fin of constant cross-sectional area Ac, perimeter p, length L, and thermal conductivity k exposed to convection to a medium at T∞ with a heat transfer
coefficient h. Assume the fins are sufficiently long so that the temperature of the fin at the tip is nearly T∞. Take the temperature of the fin at the base to be Tb and neglect heat transfer from
the fin tips. Simplify the relation for (a) a circular fin of diameter D and (b) rectangular fins of thickness t. | {"url":"http://www.chegg.com/homework-help/obtain-relation-fin-efficiency-fin-constant-cross-sectional-chapter-17-problem-110p-solution-9780073327488-exc","timestamp":"2014-04-21T05:14:37Z","content_type":null,"content_length":"47256","record_id":"<urn:uuid:7f1b0c3f-b015-427a-9dfe-6cb26a55f735>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00470-ip-10-147-4-33.ec2.internal.warc.gz"} |
East Lansdowne, PA Algebra Tutor
Find an East Lansdowne, PA Algebra Tutor
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14 Subjects: including algebra 1, algebra 2, Spanish, calculus | {"url":"http://www.purplemath.com/east_lansdowne_pa_algebra_tutors.php","timestamp":"2014-04-21T10:45:57Z","content_type":null,"content_length":"24233","record_id":"<urn:uuid:6c805bad-ae48-4e12-bf90-94bf63b92c31>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00500-ip-10-147-4-33.ec2.internal.warc.gz"} |
How to Calculate Power Output
Edit Article
Edited by Bob Hauser, Teresa, Flickety, Carolyn Barratt and 3 others
How to calculate power in horsepower or watts, what the terms mean and why would they be of importance.
1. 1
Review the fundamentals. The word "power" is defined as the time rate of doing work. Work in turn is the historic term for gauging how effective an applied force is in moving an inert object or
overcoming some other obstacle or resistance and acting over a distance.
□ The significant thing here is that in order to have performed "work", a force must act over some distance. As an example, if a propeller thrust of 300 pounds moves a boat through the water
for a distance of 80 feet, then the work done is FORCE x DISTANCE = 300 X 80 = 24,000 "foot-pounds"---so we would say that the propeller has done 24,000 foot-pounds of work.
□ Consider the time it took to move the boat that distance. Suppose the vessel moved that distance at the rate of 20 feet per second (ft/sec). To cover the 80 feet would have taken 80/20 = 4
seconds. So then the boat's prop did 24,000 foot-pounds of work in 4 seconds, so the power was exerted at the rate of 24,000/4 = 6,000 foot-pounds per second (ft-lb/sec).
□ Understand the historical background to the timing. Back in the day before gasoline engines and steam power was used to propel the earliest trains and ships (the first propeller driven ship
was the HMS Britain whose maiden voyage was in 1846 and used a single six bladed prop that looked like a windmill), horses performed an immense amount and variety of tasks. Reasonably enough,
people measured the amount of work a horse could perform and timed it. After averaging out a number of trials, they settled upon the figure of 550 ft-lb/sec as the standard rate of doing work
for a horse in reasonable health. It then became known as the "horsepower". 550 foot-pounds per second became one horsepower (hp). This remains true to this very day.
☆ 550 X 60 = 33,000 foot-pounds per minute, which is exactly the same thing: one hp.
□ Another universally used unit of power generally restricted now to electronic and electrical equipment and based on the MKS (meter-kilogram-second) metric system is the "watt". If one newton
of force operates over a distance of one meter then one joule of work has been done; if it required one second to perform this one joule of work, then the power consumption was one watt. So a
watt is one joule per second.
2. 2
Consider the industry needs today. In the vast majority of cases in industry, we are dealing with rotating machinery rather than linear motion such as a plough horse working a harrow along the
south forty. So we have to be able to figure the power output of things like electric motors, steam engines, turbines, diesels, etc., and thus we come to the subject of torque.
□ Torque is a measure of the tendency to turn or twist something or otherwise impart a rotary motion to it about some axis. If you bear down on the handle of a five foot lever with a force of
20 pounds, then you are applying a torque of 5 X 20 = 100 pound-feet.
□ Now here's the rub and can lead to considerable confusion. In order to calculate linear work in a straight line, you multiplied force times distance moved. Here again you multiply force in
the same units times a distance but in this case the distance is "lever arm" and even though you are creating torque, unless something rotates, there has been no motion and therefore no work.
□ Work and torque, though appearing to be measured in the same units are in fact quite distinct. Until the torque produces actual rotation, no work is done and therefore no power is consumed.
□ Work is measured in foot-pounds while torque is measured in pound-feet to keep them distinct.
3. 3
Measure the work when there is rotational movement or "displacement". Suppose the lever is rigidly secured at its fulcrum end to an axle and your hand, applying the 20 pounds of force as before,
moves a distance of two feet along the circle described by the five foot lever, then the work performed is, just as in the case of the straight line above, equal to force times distance moved or
20 X 2 = 40 foot-pounds. Suppose you simultaneously multiply and divide this figure by 5 feet-the length of the lever or "lever arm". Obviously this will not change the result, so you can write:
□ Work = 5 X 20 X 2/5 and still come out with 40 foot pounds as before but what is the 5 X 20 above? You just saw that that was the torque.
□ What is the "2/5"?. The 2 foot linear displacement divided by the lever arm tells you just how far you rotated the axle in angular units known as radians. A radian, used universally in
physics and engineering is defined as the angle between to radii of a circle such that the points of the circle's circumference where the two radii intersect are separated by the length of
the radius of the circle. This comes to very close to 57 degrees or alternatively it's the angle formed at the center of a circle by an arc on the circumference equal to the radius of the
□ So, you can say that the work done by a torque that produces a rotational displacement of (theta) radians is equal to the torque (L)
☆ times the rotational displacement or
☆ work = L X theta in foot-pounds
4. 4
Note that you are far more interested in rotational horsepower of motors and engines, namely the time rate of doing rotational work so you say:
□ rotational power = torque X angular displacement (radians)/ time(secs)
□ In physics, angular velocity is in radians per second, but any motor or plane or ship engine you will ever see is always rated in r.p.m. so we will have to convert.
☆ One revolution per minute (rpm) = 60 revs/second = 60 rps but one revolution per second = 2 X pi radians per second
☆ So the power in foot-pounds per second of a rotational device producing a torque, L, and rotating at S rpm, would then be
☆ Power = S/60 X (2 pi) X L in ft-lb/sec and as seen previously, in order to get this in horsepower you would have to divide by 550, or
□ Horsepower of motor = (2 pi)/33,000 X L X S where L is torque measured in foot-lbs (usually by means of a Prony Brake) and S is the rotational speed in r.p.m. which can be measured by
counters, strobes, lasers, etc. In this way, the performance curves of car and motorcycle engines can be graphed as torque versus r.p.m. and power (hp) versus engine speed in r.p.m. and
automotive engineers can see where the optimum performance is from these graphs. | {"url":"http://www.wikihow.com/Calculate-Power-Output","timestamp":"2014-04-20T13:27:11Z","content_type":null,"content_length":"67352","record_id":"<urn:uuid:02030e94-3af9-48b0-9eff-eed8da987761>","cc-path":"CC-MAIN-2014-15/segments/1398223205137.4/warc/CC-MAIN-20140423032005-00548-ip-10-147-4-33.ec2.internal.warc.gz"} |
Need to impress my dad(Who's a math teacher)
December 26th 2010, 04:04 PM
Need to impress my dad(Who's a math teacher)
Someone please help me work out these problems, I'm supposed to find the zeros of each function.
Also how would I complete the square for this expression?
Please help I'd appreciate it,have to impress my math teacher of a dad.
December 26th 2010, 04:11 PM
1 factor out the x
2 try factoring if not quadratic equation
3 $\displaystyle x^2\pm bx+\left(\frac{b}{2}\right)^2=\left(\frac{b}{2}\ri ght)^2$ then $\displaystyle\left(x\pm\frac{b}{2}\right)^2=\left( \frac{b}{2}\right)^2$
4 Ask your dad for help. He will be proud that you are working to have a better understanding of fundamental concepts.
December 26th 2010, 04:17 PM
You'd think that but I've been procrastinating lately, and like my Algebra 200 grades been slipping so he basically gave me a book and a list of questions these were the final three questions and
I was trynna get them all right,m at least I thought so but asking my dad isn't a option. But cheating isn't either, thanks for the advice, I'll turn it in as is.
December 26th 2010, 06:20 PM
mr fantastic | {"url":"http://mathhelpforum.com/algebra/166922-need-impress-my-dad-whos-math-teacher-print.html","timestamp":"2014-04-20T13:55:51Z","content_type":null,"content_length":"6779","record_id":"<urn:uuid:b88010ec-faab-4b60-8be6-952f4329389a>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00007-ip-10-147-4-33.ec2.internal.warc.gz"} |
FCP Step 3- Fit Main Effects
Next: Dose-Response Analyses (DRA) Up: Facility Comparison Analyses Previous: FCP Step 2-
The third step was to fit a main effects model (see Eq. 2) for each race and gender group for each cause of death of interest---the results are given in Tables III--V. Consider, as an example of how
to read these tables, diseases of the respiratory system (ICD 460--619) for white males---see line 5 in Table III. Table AII (see first five rows) shows the parameter estimates for each facility in
log percentage (L%) units in column 2 and the corresponding estimates of the SMRs (adjusted for S and L) at the mid-point of the 1965--69 interval of follow up. These facility effect estimates
estimates are listed in columns 4 through 8 or row 5 in Table III. The estimated relative risk (adjusted for L and F) for monthly compared to nonmonthly workers is exp (-78.6/100) = 0.46 with 95%
confidence interval of (.35, .59) indicating a substantial 5% difference in respiratory system disease rates between monthly (white collar) and nonmonthly (blue collar) workers. The estimated
relative risk (adjusted for F and S) for short term workers (< 1 year) relative to those who worked one year or longer is 1.30 with 95% CI = (1.17, 1.44). The period trend parameter estimate 0.97 (
see the last line of Table AII ) represents the yearly change in the log SMR in L% units over the forty years of follow-up. The parameter estimates for each cause of death in Tables III--V can be
combined to estimate the SMR for each facility, at the mid-point of each follow-up interval for each level of S and L:
log (SMR) = log (facility SMR in 1965--69)
+ (effect due to L)
+ (effect due to S)
+ (Interval-Midpoint--1967.5) * 0.97.
For example, for long term nonmonthly workers at K-25, during the last interval of follow-up (1980-84), with mid-point = 1982.5, the log(SRM) is
log(SMR)= 6.7 + 0 + 0 + (1982.5-1967.5)*0.97 = 21.3L%
so that the estimated SMR is exp(0.213) = 1.24.
The lowest estimated SMR was exp(-1.41)= 0.24 and occurs for:
X-10 Monthly Long-Term 1945--49
-43.0 + -78.6 + 0.0 + -19.4 = -141.0 L%.
The largest estimated SMR is 1.69 and occurs for:
K-25 Non-Monthly Short-Term 1980--84
6.7 + 0.0 + 26.1 + 14. = 47.4 L%.
The likelihood ratio statistic (see column 9 of Table III) for the null hypothesis of no differences among facilities was 24.4 with 4 df ( p<.01). Table AII indicates that the risk of respiratory
disease was much higher for TEC, K-25 and multiple facility workers than for X-10 or Y-12 workers. For example, the estimated relative risk for K-25 with X-10 as the reference facility is exp[(6.7-
(-4.0))/100] = 1.64. A large value for the LRT statistic (in column 9 of Tables III--V) indicates that there were differences among facilities in the death rates for that cause of death after
adjusting for S, L, period of follow-up, and age through use of external rates.
Next: Dose-Response Analyses (DRA) Up: Facility Comparison Analyses Previous: FCP Step 2-
• Last Modified 3Jul97 FromeEL@ornl.gov(touches: 179781 ) | {"url":"http://www.csm.ornl.gov/~frome/ORMS/app/node5.html","timestamp":"2014-04-16T10:47:24Z","content_type":null,"content_length":"5463","record_id":"<urn:uuid:348dc28c-6147-416d-9d56-6de08c05d152>","cc-path":"CC-MAIN-2014-15/segments/1397609523265.25/warc/CC-MAIN-20140416005203-00061-ip-10-147-4-33.ec2.internal.warc.gz"} |
The 35-kg Box Has A Speed Of 2 M/s When It Is At ... | Chegg.com
Image text transcribed for accessibility: The 35-kg box has a speed of 2 m/s when it is at A on the smooth ramp. If the surface is in the shape of a parabola, determine the normal force on the box at
the instant x = 3m. Also, what is the rate of increase in its speed at this instant? N = 180 N, at = 5.44 m/s2 N = 160 N, at = 6.44 m/s2 N = 140 N, at = 4.44 m/s2 N = 200 N, at = 3.44 m/s2
Mechanical Engineering | {"url":"http://www.chegg.com/homework-help/questions-and-answers/35-kg-box-speed-2-m-s-smooth-ramp-surface-shape-parabola-determine-normal-force-box-instan-q3609308","timestamp":"2014-04-23T21:29:12Z","content_type":null,"content_length":"20618","record_id":"<urn:uuid:09583677-d1dc-413f-a540-0ece574130d4>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00387-ip-10-147-4-33.ec2.internal.warc.gz"} |
Portability GHC with TypeFamilies and more
Stability unstable
Maintainer Stephen Tetley <stephen.tetley@gmail.com>
Picture types
data Picture u Source
Picture is a leaf attributed tree - where atttibutes are colour, line-width etc. It is parametric on the unit type of points (typically Double).
Wumpus's leaf attributed tree, is not directly matched to PostScript's picture representation, which might be considered a node attributed tree (if you consider graphics state changes less
imperatively - setting attributes rather than global state change).
Considered as a node-attributed tree PostScript precolates graphics state updates downwards in the tree (vis-a-vis inherited attributes in an attibute grammar), where a graphics state change deeper
in the tree overrides a higher one.
Wumpus on the other hand, simply labels each leaf with its drawing attributes - there is no attribute inheritance. When it draws the PostScript picture it does some optimization to avoid generating
excessive graphics state changes in the PostScript code.
Apropos the constructors, Picture is a simple non-empty leaf-labelled rose tree via:
Single (aka leaf) | Picture (OneList tree)
Where OneList is a variant of the standard list type that disallows empty lists.
The additional constructors are convenience:
PicBlank has a bounding box but no content and is useful for some picture language operations (e.g. hsep).
Clip nests a picture (tree) inside a clipping path.
Eq u => Eq (Picture u)
Show u => Show (Picture u)
(Num u, Pretty u) => Pretty (Picture u)
(Num u, Ord u) => Translate (Picture u)
(Num u, Ord u) => Scale (Picture u)
(Floating u, Real u) => RotateAbout (Picture u)
(Floating u, Real u) => Rotate (Picture u)
Boundary (Picture u)
(Num u, Ord u) => Move (Picture u)
(Num u, Ord u) => Vertical (Picture u)
(Num u, Ord u) => Horizontal (Picture u)
(Num u, Ord u) => Blank (Picture u)
(Num u, Ord u) => Composite (Picture u)
data Primitive u Source
Wumpus's drawings are built from two fundamental primitives: paths (line segments and Bezier curves) and labels (single lines of text).
Ellipses are a included as a primitive only for optimization - drawing a reasonable circle with Bezier curves needs at least eight curves. This is inconvenient for drawing dots which can otherwise be
drawn with a single arc command.
Wumpus does not follow PostScript and employ arcs as general path primitives - they are used only to draw ellipses. This is because arcs do not enjoy the nice properties of Bezier curves, whereby the
affine transformation of a Bezier curve can simply be achieved by the affine transformation of it's control points.
Ellipses are represented by their center, half-width and half-height. Half-width and half-height are used so the bounding box can be calculated using only multiplication, and thus initially only
obliging a Num constraint on the unit. Though typically for affine transformations a Fractional constraint is also obliged.
Eq u => Eq (Primitive u)
Show u => Show (Primitive u)
Pretty u => Pretty (Primitive u)
(Fractional u, Ord u) => Boundary (Primitive u)
data Path u Source
Path - start point and a list of path segments.
Eq u => Eq (Path u)
Show u => Show (Path u)
Semigroup (Path u) Paths are sensibly a Semigroup - there is no notion of empty path.
Pretty u => Pretty (Path u)
Pointwise (Path u)
(Num u, Ord u) => Boundary (Path u)
data PathSegment u Source
PathSegment - either a cubic Bezier curve or a line.
Eq u => Eq (PathSegment u)
Show u => Show (PathSegment u)
Pretty u => Pretty (PathSegment u)
Pointwise (PathSegment u)
data Label u Source
Label - represented by bottom left corner and text.
Eq u => Eq (Label u)
Show u => Show (Label u)
Pretty u => Pretty (Label u)
Drawing styles
data DrawPath Source
Note when drawn filled and drawn stroked the same polygon will have (slightly) different size:
• A filled shape fills within the boundary of the shape
• A stroked shape draws a pen line around the boundary of the shape. The actual size depends on the thickness of the line (stroke width).
Eq DrawPath
Show DrawPath
data DrawEllipse Source
Ellipses and circles are always closed.
Eq DrawEllipse
Show DrawEllipse
Ellipse DrawEllipse
Ellipse (Gray Double, DrawEllipse)
Ellipse (HSB3 Double, DrawEllipse)
Ellipse (RGB3 Double, DrawEllipse) | {"url":"http://hackage.haskell.org/package/wumpus-core-0.17.0/docs/Wumpus-Core-WumpusTypes.html","timestamp":"2014-04-20T14:26:49Z","content_type":null,"content_length":"25201","record_id":"<urn:uuid:020d362e-72d6-4f39-97d7-155f1c72dfad>","cc-path":"CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00561-ip-10-147-4-33.ec2.internal.warc.gz"} |
Find an equation from this graph.
Hello, I need an equation that matches the attached graph. Please help. Data is below the graph. Thanks,
Since there are 40 points, it would fit a polynomial of order 39 exactly. Let $\displaystyle y = c_0 + c_1x + c_2x^2 + ... + c_{39}x^{39}$. When you substitute each point, you'll get 40 equations in
40 unknowns which you can solve using technology or MUCH difficulty by hand.
Last edited by CaptainBlack; August 14th 2011 at 12:11 AM.
Thanks for that. I don't want to seem nit-picky or ungrateful but y=16.467+0.004939x+0.00002781x^2 is just a bit too high for it to match the trendline and therefore I can't use it. I have attached a
new graph that shows y=16.467+0.004939x+0.00002781x^2, the trendline and the actual tax percentage. And if that last data becomes an extreme outlier just get rid of it. I really appreciate everyone's
help with this and Captain Black, if you don't want to do it (fair enough), you can show me how to do it and then I can do it.
Last edited by mr fantastic; August 14th 2011 at 02:55 AM.
Well if it does not match there is probably a typo in my typing since it passed straight through the data when I was doing the fitting, unfortunately I am not on the machine with the file at present
so I cannot check. However the red curve in your document is not a plot of the quadratic (which is smooth) Checking my notes, the typo is a wrong sign, it should be: $y=16.467-0.004939x+0.00002781x^
2$ Also since you already have the equation of the trend line why are you continuing with this thread? CB
You could use interpolation polynomial in the Lagrange form: Lagrange polynomial - Wikipedia, the free encyclopedia. For example: For k=2: $x_0y_0 \quad x_1y_1 \quad x_2y_2$ $p(x)=y_0\frac{(x-x_1)
(x-x_2)}{(x_0-x_1)(x_0-x_2)}+y_1\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}+y_2\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}$ $p(x_0)=y_0\;\cdot 1+y_1\;\cdot 0+y_2\;\cdot 0=y_0$ $p(x_1)=y_0\;\cdot 0+y_1\;\
cdot 1+y_2\;\cdot 0=y_1$ $p(x_0)=y_0\;\cdot 0+y_1\;\cdot 0+y_2\;\cdot 1=y_2$. For p(x) you don't need to find k coefficients but you have k+1 terms of order k. You could write a little program for p | {"url":"http://mathhelpforum.com/calculus/186098-find-equation-graph.html","timestamp":"2014-04-18T14:38:05Z","content_type":null,"content_length":"62031","record_id":"<urn:uuid:84ea0959-1da6-4ae9-ae11-4c747dc8f3d7>","cc-path":"CC-MAIN-2014-15/segments/1397609533689.29/warc/CC-MAIN-20140416005213-00413-ip-10-147-4-33.ec2.internal.warc.gz"} |
first post in this forum
November 2nd 2012, 04:21 AM
first post in this forum
A spends 1/4 of her savings on bike and 1/3 less than she spent on the bike for a car. what fraction of her savings did she spent on bike and car?
November 3rd 2012, 05:29 AM
Re: first post in this forum
let her savings be x
She spends x/4 on bike and x/4 - 1/3 on the car.
To find how much she spent on both you add them
x/4 + x/4 -1/3 = x/2-1/3 = (3x-2)/6
November 3rd 2012, 09:06 AM
Re: first post in this forum
Hi, every one this is a important idea to all.
mortgage texas
We provide home mortgages, refinances, construction loans and reverse mortgages to people across the state of Texas.
November 3rd 2012, 12:28 PM
Re: first post in this forum
I would state the fraction $f$ of her savings spent as:
$f=\frac{1}{4}+\left(1-\frac{1}{3} \right)\cdot\frac{1}{4}=\frac{1}{4}\left(1+\frac{2 }{3} \right)=\frac{1}{4}\cdot\frac{5}{3}=?$
November 3rd 2012, 01:26 PM
Re: first post in this forum
there is an error, here.
it should be: x/4 + (x/4 - (1/3)(x/4)) = x/4 + x/4 - x/12 = 2x/4 - x/12 = 6x/12 - x/12 = (6x - x)/12 = 5x/12.
5x/12 is what fraction of x? | {"url":"http://mathhelpforum.com/new-users/206604-first-post-forum-print.html","timestamp":"2014-04-18T18:02:41Z","content_type":null,"content_length":"6243","record_id":"<urn:uuid:2720f38f-1a90-461a-9653-16b89ecdb69b>","cc-path":"CC-MAIN-2014-15/segments/1397609533957.14/warc/CC-MAIN-20140416005213-00054-ip-10-147-4-33.ec2.internal.warc.gz"} |
Radiation heat transfer in solid-propellant rocket engines
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[Numpy-discussion] .byteswap() and copy/view dilemma
Francesc Altet faltet at carabos.com
Wed Dec 13 11:28:01 CST 2006
I'm a bit confused about which cases the .byteswap() method in NumPy
would return a copy or a view. From the docstrings:
a.byteswap(False) -> View or copy. Swap the bytes in the array.
Swap the bytes in the array. Return the byteswapped array. If the
first argument is TRUE, byteswap in-place and return a reference to
>From the above description, it is not clear to me whether
the .byteswap(False) would return a copy or a view. However, the
official NumPy book seems to explain this clearer:
Byteswap the elements of the array and return the byteswapped array. If
the argument is True, then byteswap in-place and return a reference to
self. Otherwise, return a copy of the array with the elements
byteswapped. The data-type descriptor is not changed so the array will
have changed numbers.
My experiments also show that .byteswap(False) does do a copy:
In [154]:a=numpy.array([1,2,3])
In [155]:b=a.byteswap()
In [156]:b
Out[156]:array([16777216, 33554432, 50331648])
In [157]:a[1]=0
In [158]:b
Out[158]:array([16777216, 33554432, 50331648])
I'm wondering if I'm the only one that finds this docstring confusing.
Francesc Altet | Be careful about using the following code --
Carabos Coop. V. | I've only proven that it works,
www.carabos.com | I haven't tested it. -- Donald Knuth
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Nuclear Phynance
What is the relation between the two?
Phorgy A quick dose of Googling tells me something along the lines of "white noise is the derivative of a Wiener process". Could someone shed some light and maybe point to some better
PhynanceBanned references?
Total Posts:
2961 Thanks
Joined: Oct Eric
Follow up...
Phorgy Consider
Total Posts: dS/S = mu*dt + sigma*dW
Joined: Oct with
S(t) = S(0)*exp(mu'*t + sigma*W)
so that
log[S(t)/S(0)] = mu'*t + sigma*W.
Has anyone ever looked at modelling this as
log[S(t)/S(0)] = mu'*t + sigma*W',
where W' is Gaussian white noise? Just curious.
Here is kind of a neat thing:
g(t)*W(t) = int_0^t g'(tau)*W(tau) dtau + int_0^t g(tau)*W'(tau) dtau
where W(t) is a Wiener process and W'(t) is Gaussian white noise.
I got that from this paper.
What is the relation between the two?
It's all Greek Just a quick answer before i go to bed... In discrete time econometrics, the white noise is called the error term with constant variance. If the variance is time varying then we have a
to me random walk, and a random walk is the discrete time version of a Wiener process.
Total Posts:
1180 HTH
Joined: Jul
2004 Αίεν Υψικράτειν/Τύχη μη πίστευε/Άνδρα Αρχή Δείκνυσι/Νόησις Αρχή Επιστήμης //Σε ενα κλουβί γραφείο σαν αγρίμι παίζω ατέλειωτο βουβό ταξίμι
Thanks anthis. Not quite sure what you meant, but econometrics is actually the motivation for the question. I am trying to understand the relation if any between (vector)
IAmEric autoregression and (multi-factor) stochastic differential equations.
PhynanceBanned If you or anyone has some light to shed on that as well, it would be more than welcome.
Total Posts:
2961 Here is another question...
Joined: Oct
2004 We know that, at least mechanically, we can write things like
dW dW = dt
in stochastic calculus. If W' is (Gaussian) white noise, is there some similar nifty expression for
dW' dW' ?
The answer to this and some of my other questions might be found in one of these:
White Noise Calculus for Pure Jump Processes with Application to Mathematical Finance Incomplete Equilibrium Markets
SOME APPLICATIONS OF WHITE NOISE ANALYSIS TO MATHEMATICAL FINANCE
White light, white heat
Founding White noise is the signal generated by combining sine waves of all frequencies in equal proportions. It is the same as Brownian motion. Consider the defining properties of Brownian
Member motion and check that they also correspond to white noise:
Total Posts:
4331 1. Continuity: Wt is continuous in t (check, it's the sum of sine waves) and Wo = 0 (check, set all sine waves to "start" at t=0).
Joined: May 2. Normality: Wt is distributed N(0,t)
2004 3. Normally distributed independent increments: Ws - Wt is distributed N(0, s-t) for s>t
So far, so good. But enquiring minds probably want to think about (a) differentiability and (b) stochasticity vs deterministicity.
I can never remember the exact definition of a Wiener process, but isn't it (something like) any process of the form dXt = f(Xt, t) dt + g(Xt, t) dWt where dWt is the increment of a
Brownian motion.
I'm not sure this is what you're looking for.
The sound of one bear, uh, in the woods
Johnny's first comments are correct. But, it is not the same as Brownian Motion. It is the "derivative" of Brownian Motion. In this sense, Eric's intuition is correct.
Founding 3% Neanderthal and proud of it!
Total Posts:
Joined: Mar
Nonius is correct and I was wrong. But let us not say that white noise is the derivative of Brownian motion but instead say that the integral of white noise is the same as Brownian
Johnny motion. The integral of white noise is known (for it's connection with Brownian motion) as Brown noise. And many thanks to all those people that pointed this out to me in private.
Member The sound of one bear, uh, in the woods
Total Posts:
Joined: May
Total Posts: white noise may be viewed as the derivate of a Wiener process. Unfortunately white noise has infinite variance and the nifty techniques one usually uses don't work here. The derivative
2536 exists only in the sense of distributions, a kind of generalized derivatives. If white noise were an integrator in the Lebesgue-Stieltjes-sense you could rewrite any stochastic Itô
Joined: Feb integral as
\int_a^b f(s) dW_s = \int_a^b f(s)W'(s) ds,
W'(s) being the white noise.
I can dig up my old scripts and books at home if you want to know more. Hope this helps a bit building an intuition.
"Don't try to run, don't try to hide. Believe me, the hammer's gonna make it right !"
It is also in the sense of Eric's NCG calculus that one can view white noise as the noncommutative derivative of brownian motion.
Founding 3% Neanderthal and proud of it!
Total Posts:
Joined: Mar
it is still too early for me to think, but the white noise way of doing things are probably most done in engineering related stuff, but hey... you never know, here is something that
Martingale might be of a little interest...
NP House Mouse
Total Posts: http://www.ma.ic.ac.uk/~pavl/ver3.pdf
Joined: Jun
Nonius saw through my thinly veiled attempt to bring NCG into the picture
Phorgy It is true, if someone could tell of any special algebraic relation for
Total Posts: dW' dW'
Joined: Oct where W' is white noise, then I could reformulate white noise in terms of NCG, which might be kind of academically interesting.
The real point of this is, as I said, to try to understand the link (if any) between vector autoregression and multifactor stochastic differential equations. I'm just now learning the
former and it is reminding me a lot of the latter.
Looking at VAR reminded me of my days back in grad EE and DSP, e.g. z-transforms, digital filters, impulse responses, etc., which made me think of deconvolutions, which made me
think... (ad nauseum)
this was precisely how the NYU math in finance "stochastic calculus" course taught by marco avellaneda last semester began--by pounding on the metaphor of white noise as a "derivative"
of brownian motion. he's taken the homework files off the website, but i can send them to anyone interested in playing (numerically) with this stuff. it was very good for intuition.
disclaimer: i stopped attending this course after about three lectures, as avellaneda is a horrible lecturer. the psets are still decent, however.
Phorgy Sounds cool. I'd be interested in taking a look at that and any notes you might have.
Total Posts: Thanks
Joined: Oct
in regards to vector autoregression, my understanding is this:
where that last epsilon is your white noise.
if phi<1, this series is covariance-stationary, so you can take the expected value of both sides. E(epsilon)=0, so the expected value of the series is:
In other words, it's a mean reverting series.
if phi==1, y_t is a Wiener process (a random walk), it has a 'unit root'. you can fix this by first differencing the series. since the derivative is white noise, you are reduced to the
above case.
if phi>1, this time series is explosive, which is just no good.
instead of phi, you have a matrix of coefficients, so you take the eigenvalues to see what effect each eigenvector is having. You check the effect in the same way as above: any lambda>
1 is explosive and makes the series worthless. If all the lambdas are < 1 there is a static equilibrium to the system (kind of like mean-reversion).
The difference with VAR is that some lambdas could be random walks and others stationary. That's where cointegration comes in (from that other thread)
the only reason it would be easier to program in C is that you can't easily express complex problems in C, so you don't. -comp.lang.lisp
Making some progress...
PhynanceBanned Here is kind of a neat thing:
Total Posts:
2961 g(t)*W(t) = int_0^t g'(tau)*W(tau) dtau + int_0^t g(tau)*W'(tau) dtau
Joined: Oct
2004 where W(t) is a Wiener process and W'(t) is Gaussian white noise.
I got that from this paper.
In the above expression, the function g(t) must be smooth and have finite support. If we let g(t) = 1 up to some time T and then smoothly taper to zero beyond, then the above reduces
W(t) = int_0^t W'(tau) dtau
for t < T. This expresses what others have said about Brownian motion being the integral of white noise.
If we approximate this integral as a Riemann sum, we get
W(t) ~= W(t-delt) + delta W'(t).
Comparing this to what tristan said for c = 0 and phi = 1, we have
y(t) = y(t-1) + e(t).
This seems to support the statement that y(t) is Brownian motion for phi = 1. At least if you were to throw in a delt in there and take a limit as delt -> 0. I can buy that
This is looking like it supports my suspicion that AR could be thought of as a finite difference approximation to a stochastic DE (I think).
Still some mysteries though. If anyone could shed some light, it'd be appreciated.
For example, the integral implies
dW = W' dt.
We know that
dW dW = dt
so that means that W' ~ 1/sqrt(dt) since
dW dW = (W')^2 dt dt = dt.
If we wave the wand of NCG, we convert this to a commutator
[dW, W]
= dW W - W dW
= W' dt W - W W' dt
= W' W dt - W W' dt (W and dt commute)
= dt.
The only way to satisfy this is if
[W, W'] = 1
which looks like a quantization rule to me. I'll try not to get pulled off onto that tangent right now
Back to something remotely practical...
If we want to solve
dy = mu*dt + sigma*dW
approximately we can rewrite it as
dy = mu*dt + sigma*dW = mu*dt + sigma*W'*dt.
The commutative relations suggest W' ~ 1/sqrt(delt) so this becomes
y(t) - y(t-delt) = mu*delt + sigma*W'(t)*delt = mu*delt + sigma*e(t)*sqrt(delt)
where I've set
W'(t) = e(t)/sqrt(delt)
and I really don't know exactly why other than to make it look like a simple Monte Carlo expression (and it seems to somehow relate to the dimension analysis from the commutative
Rearranging terms gives
y(t) = mu*delt + y(t-delt) + sigma*e(t)*sqrt(delt).
c = mu*delt
epsilon(t) = sigma*e(t)*sqrt(t)
brings us back to something that formally looks like the AR expression.
Ok. As you can see, there are still some holes in my arguments so if anyone can help nail the final missing pieces, I'd appreciate it. That, or show me where I'm totally off the mark.
Just to summarize...
This whole thing is motivated by trying to understand the relation between ARs and SDEs. My hunch is that AR may be interpretted as an approximate numerical (finite difference/Monte
Carlo) solution to a simple SDE.
This might be obvious to some, so help me see the light. I'm almost there
It is still a bit early to think and maybe I am babbling big nonsense but
Total Posts: y(t) = mu*delt + y(t-delt) + sigma*e(t)*sqrt(delt)
Joined: Feb boils down to the Euler scheme for numerical solution of SDEs if e(t) is some N(0,1) distributed rv. Then you get the increment
or alternatively
Under some technical assumptions this converges to the SDE for delt->0 with order O(delt^(-1/2)). If mu and sigma are C^1 you can get convergence of order O(1/delt) by adding another
term involving the first order derivatives (the Milstein scheme). Maybe this helps, if not feel free to trash it.
"Don't try to run, don't try to hide. Believe me, the hammer's gonna make it right !"
eric - haven't read all the details, but my intuition is simply that dW should be thought of as a 'generalized function' - i.e. think dirac delta. That is, it is defined only to the
kr extent that one can compute <g, dW> = int_0^t g(t) dW_t, and the value of this functional is mostly determined by a few key algebraic relations. In fact, writing <dW, dW> is not the
Founding right way to go, much better to have W = <1, dW> be thought of as a random variable, and then use the rules for <W, dW>.
Raider Once that's said, it's clear that "dW'_t" does not operate on the correct space... it would have to be integrated against something else in order to produce a generalized function.
Total Posts: Somewhere under all this is a deep-discount graded algebra involving functions with existing derivatives.
Joined: Apr my bank got pwnd
Total Posts: do you think along the lines of Sobolev spaces ? This was basically what I had in mind when I started mumbling about generalized derivatives.
Joined: Feb Regards
"Don't try to run, don't try to hide. Believe me, the hammer's gonna make it right !"
Total Posts:
Joined: Mar
The origional way of Wiener's way to construct the stochastic integral is kind of from white noise as generalized distribution ( throgh some L^2 isometry), but it appoved to be
Martingale limited, that's why people start to consider this from some other point... I will try to dig some history literature on this
NP House Mouse
Total Posts:
Joined: Jun
Hi Eric!
Total Posts: 2 I thought I might just as well post a reply here.
Joined: Feb
2006 You wrote
We know that
dW dW = dt
so that means that W' ~ 1/sqrt(dt) since
dW dW = (W')^2 dt dt = dt.
This is making me a little nervous. But I realize my stochastic calculus is rusty.
If we wave the wand of NCG
Now I feel more at home...
[dW, W]
= dW W - W dW
= W' dt W - W W' dt
= W' W dt - W W' dt (W and dt commute)
= dt.
I understand
[dW,W] = dt .
But does it make sense to write dW = W' dt ??
Is there any discrete 0-form W' with this property?
I wouldn't think so, but it has been a while since I thought about this stuff.
But consider a 2D diamond graph. Introduce "lightcone" variables A and B such that
t = A + B
x = A - B .
dt = dA + dB
dx = dA - dB
and there is manifestly no 0-form f such that
dx = f dt .
But in your example above W=x, if I understood correctly.
Do you agree? Maybe I am missing your point.
On the other hand, maybe you are right and it is useful to *introduce* new formal expressions like W' that are defined to satisfy relations like dW = W' dt. Haven't thought enough
about that.
Hi Urs!
Phorgy Welcome to NP!
Total Posts: I'll take the liberty for a little introductory comment...
Joined: Oct I had the great pleasure to coauthor a paper with Urs (the highlight of my academic life actually). After six years of flailing around in a corner on my own, I finally somehow finished
2004 my PhD and decided that reaching out to others wouldn't be such a bad thing. I knew Urs through the newsgroup sci.physics.research (of which he eventually became a moderator along with
John Baez and others last time I checked). Urs had some interest in stochastic formulations of QM (by Nelson if memory serves) and some comments I made relating NCG and SC must have
attracted some of his attention.
The basic idea is that any directed graph gives rise to a special algebra/calculus that is unique to that graph. I had also (re)discovered that a certain algebra leads to stochastic
calculus. Since every directed graph gives rise to an algebra and a particular algebra gives rise to stochastic calculus, the question we asked ourselves one day is kind of an inverse
problem, "Which directed graph gives rise to the algebra from which SC derives." Urs subsequently left for a two week bike ride across Europe and when he came back we had both answered
the question independently. In a flash it all became almost obvious. (particularly in hind sight). The binary tree is the unique directed graph that gives rise to the algebra that
leads to SC. That was enough to ignite an intense collaboration that was very exciting (for me at least) that lasted several months and culminated in this paper
Discrete Differential Geometry on Causal Graphs
Anyway, it is always a great pleasure to discuss things with Urs so I'm glad he made an appearance. So now back to the topic at hand! (which I think has some profound (although perhaps
too academic) significance). That is the possible fact that Brownian motion and white noise are conjugate variables in the sense that position and momentum are in QM.
so that means that W' ~ 1/sqrt(dt) since
dW dW = (W')^2 dt dt = dt.
This is making me a little nervous. But I realize my stochastic calculus is rusty.
I don't blame you, it makes me nervous too
I understand
[dW,W] = dt .
But does it make sense to write dW = W' dt ??
I don't know
W(t) = int_0^t W'(tau) dtau.
Since W(0) = 0, we can rewriting this as
W(t) - W(0) = int_0^t W'(tau) dtau
which can be rewritten
int_0^t dW = int_0^t W'(tau) dtau,
which I thought would allow us to write
dW = W' dt.
Is there any discrete 0-form W' with this property?
The commutative relation
W W' - W' W = 1
would seem to indicate that perhaps neither W nor W' are discrete 0' forms (in the obvious sense) and maybe should be thought of as matrices or something. Not sure.
It might help to make the analogy with QM more clear. Since it makes me so happy, I will repeat the little computation
(using x in place of W now)
= dx x - x dx
= x' dt x - x x' dt
= x' x dt - x x' dt
= dt
which implies
[x',x] = 1.
The QM version of this is
= dx x - x dx
= x' dt x - x x' dt
= x' x dt - x x' dt
= i*hbar*dt
which implies
[x',x] = i*hbar.
(maybe should replace hbar with hbar/m)
Looking at this, we see that a Wiener process (Brownian motion) is analogous to momentum in QM.
It is then somewhat tempting to write something like
dW = W' dt = d/dW dt
so that
[W', W] = [d/dW, W]
[d/dW, W] psi
= (d/dW W psi - W d/dW psi)
= (psi + W d/dW psi - W d/dW psi)
= 1 psi
so that
[d/dW, W] = 1.
Is it possible that if W is a Wiener process then white noise can be thought of as W' = d/dW?
That would be kind of wild and, if true, I'm pretty sure no one has ever observed that. We could maybe write up a page or two and throw it on the arxives
What do you think?
I perfectly agree that the equation
Total Posts: 2 [dx,x] = dt
Joined: Feb
2006 or
[dx,x] = i hbar dt
morally encodes the commutator which you have in mind. What you point out is that if we could somehow divide both sides by dt, we would get something like a commutator of dx/dt with x.
This is certainly a good way to think about it. Personally, however, instead of wanting to rewrite this in the form [x',x] = 1, I would regard [dx,x]=dt as the better and more robust
I also agree that there is a close relation to how white noise is morally like the derivative of the Wiener process. After all, one way to derive the commutation relation [x',x] = k 1
is to use the path integral formulation of the free particle. The rigorous version of this is nothing but an integral over paths using the Wiener measure.
Hence, many technically quite sophisticated facts are somehow quite neatly and very simply already encoded in an equation as simple as [dx,x]=dt .
Looking at this, we see that a Wiener process (Brownian motion) is analogous to momentum in QM.
Wait. Is that what you want to say?
I think the Wiener process is like the position observable in QM. White noise is like the velocity (~momentum) of the particle.
dW = W' dt = d/dW dt
The second equation looks a little strange. Wouldn't you want
W' dt = (dW/dt) dt ?
Looking at this, we see that a Wiener process (Brownian motion) is analogous to momentum in QM.
Phorgy Wait. Is that what you want to say?
Total Posts:
Joined: Oct I guess before we can settle this question, we'd have to settle the question of whether we can write
dW = W' dt.
I think we can, but there are enough moving pieces that I could be wrong about that. If we can write this, then
[dW,W] = dt
leads to
[W',W] = 1.
Looking at the first expression (dW = W' dt) makes you want to say
W' = dW/dt,
but looking at [W',W] = 1 makes me want to say
W' = d/dW
unless there is another representation of W' that satisfies that commutative relation. By a quirk in dimensions, at least the two have the same units, i.e.
[d/dW] = [dW/dt]
[W] = [sqrt(t)],
where I'm using brackets to denotes "units of", i.e. [W] = "Units of W".
I think there is something neat here, but as usually, I'm only smart enough to "sense" it and am struggling to dig it out
Oh! Let me clarify one point because it is important. The expression
[W',W] = 1
does not come from "dividing by dt". Rather, it comes from plugging
dW = W' dt
[dW,W] = dt
and using the fact that dW and dt commute. I carried out that computation (since it is so neat) a couple of times
One more thing...
Phorgy You said
Total Posts:
2961 I think the Wiener process is like the position observable in QM. White noise is like the velocity (~momentum) of the particle.
Joined: Oct
Yes. This is basically what I'm trying to say. I was being a little cavalier about constants and not making much effort to distinguish between velocity and momentum (although I made a
comment about perhaps writing hbar/m at one point).
I would say it this way...
If we can write
dW = W' dt,
then the commutative relation
[dW,W] = dt
implies the Wiener process is like the position operator in QM and white noise is like the velocity (~momentum) operator in QM. | {"url":"http://www.nuclearphynance.com/Show%20Post.aspx?PostIDKey=69584","timestamp":"2014-04-19T06:52:29Z","content_type":null,"content_length":"74234","record_id":"<urn:uuid:d821ce02-11df-4074-967a-92024f2db345>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00274-ip-10-147-4-33.ec2.internal.warc.gz"} |
the encyclopedic entry of geocentric longitude
(or ), symbolized by the
(λ), is the east-west
geographic coordinate
measurement most commonly used in cartography and global navigation. A line of longitude is a
and half of a
great circle
Mariners and explorers for most of history struggled to determine precise longitude. Latitude was calculated by observing with quadrant or astrolabe the inclination of the sun or of charted stars,
but longitude presented no such manifest means of study. Amerigo Vespucci was perhaps the first to proffer a solution, after devoting a great deal of time and energy studying the problem during his
sojourns in the New World.
"As to longitude, I declare that I found so much difficulty in determining it that I was put to great pains to ascertain the east-west distance I had covered. The final result of my labors was
that I found nothing better to do than to watch for and take observations at night of the conjunction of one planet with another, and especially of the conjunction of the moon with the other
planets, because the moon is swifter in her course than any other planet. I compared my observations with [an almanac]. After I had made experiments many nights, one night, the twenty-third of
August, 1499, there was a conjunction of the moon with Mars, which according to the almanac was to occur at midnight or a half hour before. I found that...at midnight Mars's position was three
and a half degrees to the east."
By comparing the relative positions of the moon and Mars with their anticipated positions, Vespucci was able to crudely deduce his longitude. But this method had several limitations: First, it
required the occurrence of a specific astronomical event (in this case, Mars passing through the same right ascension as the moon), and the observer needed to anticipate this event via an
astronomical almanac. One needed also to know the precise time, which was difficult to ascertain in foreign lands. Finally, it required a stable viewing platform, rendering the technique useless on
the rolling deck of a ship at sea.
Unlike latitude, which has the equator as a natural starting position, there is no natural starting position for longitude. Therefore, a reference meridian had to be chosen. While British
cartographers had long used the Greenwich meridian in London, other references were used elsewhere, including: El Hierro, Rome, Copenhagen, Jerusalem, Saint Petersburg, Pisa, Paris, Philadelphia, and
Washington. In 1884, the International Meridian Conference adopted the Greenwich meridian as the universal prime meridian or zero point of longitude.
Noting and calculating longitude
Longitude is given as an
angular measurement
ranging from 0° at the prime meridian to +180° eastward and −180° westward. The
Greek letter λ (lambda)
, is used to denote the location of a place on
Earth east
of the prime meridian.
Each degree of longitude is sub-divided into 60 minutes, each of which divided into 60 seconds. A longitude is thus specified in sexagesimal notation as 23° 27′ 30" E. For higher precision, the
seconds are specified with a decimal fraction. An alternative representation uses degrees and minutes, where parts of a minute are expressed in decimal notation with a fraction, thus: 23° 27.500′ E.
Degrees may also be expressed as a decimal fraction: 23.45833° E. For calculations, the angular measure may be converted to radians, so longitude may also be expressed in this manner as a signed
fraction of π (pi), or an unsigned fraction of 2π.
For calculations, the West/East suffix is replaced by a negative sign in the western hemisphere. Confusingly, the convention of negative for East is also sometimes seen. The preferred convention --
that East be positive -- is consistent with a right-handed Cartesian coordinate system with the North Pole up. A specific longitude may then be combined with a specific latitude (usually positive in
the northern hemisphere) to give a precise position on the Earth's surface.
Longitude at a point may be determined by calculating the time difference between that at its location and Coordinated Universal Time (UTC). Since there are 24 hours in a day and 360 degrees in a
circle, the sun moves across the sky at a rate of 15 degrees per hour (360°/24 hours = 15° per hour). So if the time zone a person is in is three hours ahead of UTC then that person is near 45°
longitude (3 hours × 15° per hour = 45°). The word near was used because the point might not be at the center of the time zone; also the time zones are defined politically, so their centers and
boundaries often do not lie on meridians at multiples of 15°. In order to perform this calculation, however, a person needs to have a chronometer (watch) set to UTC and needs to determine local time
by solar observation or astronomical observation. The details are more complex than described here: see the articles on Universal Time and on the Equation of time for more details.
Plate movement and longitude
The surface layer of the Earth, the
, is broken up into several
tectonic plates
. Each plate moves in a different direction, at speeds of about 50 to 100 mm per year. As a result, for example, the longitudinal difference between a point on the equator in Uganda (on the
African Plate
) and a point on the equator in Ecuador (on the
South American Plate
) is increasing by about 0.0014 of a second per year.
If a global reference frame such as WGS84 is used, the longitude of a place on the surface will change from year to year. To minimize this change, when dealing exclusively with points on a single
plate, a different reference frame can be used, whose coordinates are fixed to a particular plate, such as NAD83 for North America or ETRS89 for Europe.
Elliptic parameters
Because most planets (including Earth) are
ellipsoids of revolution
, or
, rather than
, both the radius and the length of arc varies with latitude. This variation requires the introduction of elliptic parameters based on an ellipse's
angular eccentricity
(which equals
, where
are the equatorial and polar radii;
is the
first eccentricity
; and
is the
). Utilized in creating the
is the inverse of the
principal elliptic integrand
Degree length
The length of an
of north-south latitude difference,
, is about 60 nautical miles, 111 kilometres or 69
statute miles
at any latitude. The length of an arcdegree of east-west longitude difference,
, is about the same at the equator as the north-south, reducing to zero at the poles.
In the case of a spheroid, a meridian and its anti-meridian form an ellipse, from which an exact expression for the length of an arcdegree of latitude is:
This radius of arc (or "arcradius") is in the plane of a meridian, and is known as the
meridional radius of curvature
Similarly, an exact expression for the length of an arcdegree of longitude is:
The arcradius contained here is in the plane of the
prime vertical
, the east-west plane perpendicular (or "
") to both the plane of the meridian and the plane tangent to the surface of the ellipsoid, and is known as the
normal radius of curvature
Along the equator (east-west), $N;!$ equals the equatorial radius. The radius of curvature at a right angle to the equator (north-south), $M;!$, is 43 km shorter, hence the length of an arcdegree of
latitude at the equator is about 1 km less than the length of an arcdegree of longitude at the equator. The radii of curvature are equal at the poles where they are about 64 km greater than the
north-south equatorial radius of curvature because the polar radius is 21 km less than the equatorial radius. The shorter polar radii indicate that the northern and southern hemispheres are flatter,
making their radii of curvature longer. This flattening also 'pinches' the north-south equatorial radius of curvature, making it 43 km less than the equatorial radius. Both radii of curvature are
perpendicular to the plane tangent to the surface of the ellipsoid at all latitudes, directed toward a point on the polar axis in the opposite hemisphere (except at the equator where both point
toward Earth's center). The east-west radius of curvature reaches the axis, whereas the north-south radius of curvature is shorter at all latitudes except the poles.
The WGS84 ellipsoid, used by all GPS devices, uses an equatorial radius of 6378137.0 m and an inverse flattening, (1/f), of 298.257223563, hence its polar radius is 6356752.3142 m and its first
eccentricity squared is 0.00669437999014. The more recent but little used IERS 2003 ellipsoid provides equatorial and polar radii of 6378136.6 and 6356751.9 m, respectively, and an inverse flattening
of 298.25642. Lengths of degrees on the WGS84 and IERS 2003 ellipsoids are the same when rounded to six significant digits. An appropriate calculator for any latitude is provided by the U.S.
government's National Geospatial-Intelligence Agency (NGA).
N-S radius Surface distance E-W radius Surface distance
Latitude of curvature per 1° change of curvature per 1° change
$M;!$ in latitude $N;!$ in longitude
0° 6335.44 km 110.574 km 6378.14 km 111.320 km
15° 6339.70 km 110.649 km 6379.57 km 107.551 km
30° 6351.38 km 110.852 km 6383.48 km 96.486 km
45° 6367.38 km 111.132 km 6388.84 km 78.847 km
60° 6383.45 km 111.412 km 6394.21 km 55.800 km
75° 6395.26 km 111.618 km 6398.15 km 28.902 km
90° 6399.59 km 111.694 km 6399.59 km 0.000 km
Ecliptic latitude and longitude
Ecliptic latitude and longitude are defined for the planets, stars, and other celestial bodies in a similar way to that in which the terrestrial counterparts are defined. The pole is the normal to
the ecliptic nearest to the celestial north pole. Ecliptic latitude is measured from 0° to 90° north (+) or south (−) of the ecliptic. Ecliptic longitude is measured from 0° to 360° eastward (the
direction that the Sun appears to move relative to the stars) along the ecliptic from the vernal equinox. The equinox at a specific date and time is a fixed equinox, such as that in the J2000
reference frame.
However, the equinox moves because it is the intersection of two planes, both of which move. The ecliptic is relatively stationary, wobbling within a 4° diameter circle relative to the fixed stars
over millions of years under the gravitational influence of the other planets. The greatest movement is a relatively rapid gyration of Earth's equatorial plane whose pole traces a 47° diameter circle
caused by the Moon. This causes the equinox to precess westward along the ecliptic about 50" per year. This moving equinox is called the equinox of date. Ecliptic longitude relative to a moving
equinox is used whenever the positions of the Sun, Moon, planets, or stars at dates other than that of a fixed equinox is important, as in calendars, astrology, or celestial mechanics. The 'error' of
the Julian or Gregorian calendar is always relative to a moving equinox. The years, months, and days of the Chinese calendar all depend on the ecliptic longitudes of date of the Sun and Moon. The 30°
zodiacal segments used in astrology are also relative to a moving equinox. Celestial mechanics (here restricted to the motion of solar system bodies) uses both a fixed and moving equinox. Sometimes
in the study of Milankovitch cycles, the invariable plane of the solar system is substituted for the moving ecliptic. Longitude may be denominated from 0 to $begin\left\{matrix\right\}2piend\left\
{matrix\right\}$ radians in either case.
Longitude on bodies other than Earth
Planetary co-ordinate systems are defined relative to their mean axis of rotation and various definitions of longitude depending on the body. The longitude systems of most of those bodies with
observable rigid surfaces have been defined by references to a surface feature such as a crater. The north pole is that pole of rotation that lies on the north side of the invariable plane of the
solar system (near the ecliptic). The location of the prime meridian as well as the position of body's north pole on the celestial sphere may vary with time due to precession of the axis of rotation
of the planet (or satellite). If the position angle of the body's prime meridian increases with time, the body has a direct (or prograde) rotation; otherwise the rotation is said to be retrograde.
In the absence of other information, the axis of rotation is assumed to be normal to the mean orbital plane; Mercury and most of the satellites are in this category. For many of the satellites, it is
assumed that the rotation rate is equal to the mean orbital period. In the case of the giant planets, since their surface features are constantly changing and moving at various rates, the rotation of
their magnetic fields is used as a reference instead. In the case of the Sun, even this criterion fails (because its magnetosphere is very complex and does not really rotate in a steady fashion), and
an agreed-upon value for the rotation of its equator is used instead.
For planetographic longitude, west longitudes (i.e., longitudes measured positively to the west) are used when the rotation is prograde, and east longitudes (i.e., longitudes measured positively to
the east) when the rotation is retrograde. In simpler terms, imagine a distant, non-orbiting observer viewing a planet as it rotates. Also suppose that this observer is within the plane of the
planet's equator. A point on the equator that passes directly in front of this observer later in time has a higher planetographic longitude than a point that did so earlier in time.
However, planetocentric longitude is always measured positively to the east, regardless of which way the planet rotates. East is defined as the counter-clockwise direction around the planet, as seen
from above its north pole, and the north pole is whichever pole more closely aligns with the Earth's north pole. Longitudes traditionally have been written using "E" or "W" instead of "+" or "−" to
indicate this polarity. For example, the following all mean the same thing:
The reference surfaces for some planets (such as Earth and Mars) are ellipsoids of revolution for which the equatorial radius is larger than the polar radius; in other words, they are oblate
spheroids. Smaller bodies (Io, Mimas, etc.) tend to be better approximated by triaxial ellipsoids; however, triaxial ellipsoids would render many computations more complicated, especially those
related to map projections. Many projections would lose their elegant and popular properties. For this reason spherical reference surfaces are frequently used in mapping programs.
The modern standard for maps of Mars (since about 2002) is to use planetocentric coordinates. The meridian of Mars is located at Airy-0 crater.
Tidally-locked bodies have a natural reference longitude passing through the point nearest to their parent body. However, libration due to non-circular orbits or axial tilts causes this point to move
around any fixed point on the celestial body like an analemma.
See also
External links
□ for each city it gives the satellite map location, country, province, coordinates (dd,dms), variant names and nearby places. | {"url":"http://www.reference.com/browse/geocentric+longitude","timestamp":"2014-04-21T11:38:54Z","content_type":null,"content_length":"104469","record_id":"<urn:uuid:202f16eb-0c57-44d3-b9dc-ac4274f78541>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00027-ip-10-147-4-33.ec2.internal.warc.gz"} |
Optimal Manoeuvres of Underactuated Linear Mechanical Systems: The Case of Controlling Gantry Crane Operations
Journal of Applied Mathematics
Volume 2014 (2014), Article ID 283565, 16 pages
Research Article
Optimal Manoeuvres of Underactuated Linear Mechanical Systems: The Case of Controlling Gantry Crane Operations
Department of Mechanical Engineering, University of Saskatchewan, Saskatoon, SK, Canada S7N 5A9
Received 12 June 2013; Revised 26 September 2013; Accepted 27 September 2013; Published 12 January 2014
Academic Editor: Weihai Zhang
Copyright © 2014 S. Woods and W. Szyszkowski. 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.
A method of solving optimal manoeuvre control of linear underactuated mechanical systems is presented. The nonintegrable constraints present in such systems are handled by adding dummy actuators and
then by applying Lagrange multipliers to reduce their action to zero. The open- and closed-loop control schemes can be analyzed. The method, referred to as the constrained modal space optimal control
(CMSOC), is illustrated in the examples of gantry crane operations.
1. Introduction
Underactuated mechanical systems have fewer independent actuators than degrees of freedom (DOFs) to be controlled [1]. Typical nonlinear examples of such systems, usually with only several DOFs, are
rigid multilink robotic manipulators with passive joints or any manipulator with flexible links (described by at least one mode of vibration). Linear examples include vibrating structures with
continuously distributed mass (i.e., with theoretically infinite number of DOFs to describe them) such as masts, antennas, buildings, brides, and car suspension, controlled by discrete actuators.
This paper presents a method of analyzing and simulating optimal manoeuvres between two given configurations (often referred to as point-to-point manoeuvres) for linear underactuated systems. The
method combines optimal control theory with computational mechanics and the finite element (FE) technique, in particular.
The number of DOFs equal to the number of actuators will be referred to as actuated (after [1]), while all remaining DOFs will be referred to as underactuated (however, all DOFs are in fact
controlled). The actuated and unactuated DOFs must satisfy a number of constraints equal to the number of unactuated DOFs and resulting from the equations governing the motion of such systems. For
mechanical systems we assume that these constraints may be nonintegrable (nonholonomic), meaning unactuated DOFs cannot be explicitly eliminated. Many of the techniques presented in the literature
deal with underactuated problems by applying the constraints to eliminate the unactuated DOFs and then by solving the reduced fully actuated problems [2–4]. These approaches are limited to particular
problems where the constraints can be simplified to a form making such mathematical manipulations possible. The method presented here is capable of dealing with any linear system, as it does not
require the elimination of unactuated DOFs. Instead, the underactuated system is formulated as if it were fully actuated by adding “dummy” (zero-valued) actuators to all unactuated DOFs. The modal
space is used in modelling the system motions. The method can be considered as an extension of the independent modal space control (IMSC; e.g., see [5]) into the underactuated problems, therefore it
will be referred to as the constrained modal space optimal control (CMSOC) method. The system constraints resulting from underactuation are then determined by eliminating these dummy actuators. The
constraints are algebraic in terms of controls but differential (nonintegrable) in terms of the DOFs. The algebraic form of the constraints is used to generate the so-called matrix of constraints,
which is utilized to handle the nonintegrable constraints with the help of time-varying Lagrange multipliers. Pontryagin’s principle is used to optimize the trajectory and actuation forces.
This paper presents the CMSOC method in a general form and then explains some details of the corresponding numerical procedure in the examples of standard two or three-DOF gantry crane operations.
The method is verified by recreating the closed-loop control of the two-DOF gantry crane problem obtained in [1] via applying the classical technique and the open-loop optimal control considered in [
2. Problem Formulation
2.1. Dynamics of a General Underactuated System
The computational model for the motion of a linear mechanical system is represented by a standard form used in FE analysis: where and are vectors of DOFs and activation forces, respectively, and , ,
and are constant mass, damping, and stiffness matrices, respectively. In particular, (1) is suitable to model the dynamics of a range of actively controlled structural members undergoing small
amplitude oscillations and finite translations. In underactuated systems independent actuation forces are to control number of DOFs. Matrix of dimensions assigns components of vector to particular
DOFs and obviously is not invertible if . Clearly, the actuators via (1) control all DOFs of the system. For the purpose of analysis the DOFs can be divided into actuated () and unactuated () ones by
rearranging these equations as follows: The bottom row represents the equations constraining the actuated and unactuated DOFs in the following form: The system can formally be converted to a fully
actuated one by using (3) to explicitly determine vector in terms of (i.e, ), and then by substituting this vector to the top row of (2a) to obtain . Unless some matrices in (3) vanish, it is not
generally possible, and therefore these constraints are considered as nonholonomic.
The control task for vector in (1) is to manoeuvre the system from an initial state to a final state described by the following boundary conditions (point-to-point manoeuvre): It should be emphasised
that no trajectory is specified in this task. A particular trajectory satisfying (1) and (4) may be determined if extra conditions are imposed on the system. We will identify such a trajectory by
optimizing the performance index as discussed in the next section. Note that this problem is different from a typical trajectory tracking problem in which instead of (4) the task is specified as the
system output in the formSeveral methods have been proposed to solve the inverse problems of finding the input for the output as defined by (1) and (5a), notably the servo-constraint approach [6–9]
and the flatness method [10, 11]. In particular, differentiating (5a) with outputs twice one obtains , where the size of matrix is ; then square matrix is required to be nonsingular to solve the
problem. This condition does not apply in the method presented here since our output is given only in terms of (4), that is, the system’s initial and final configurations.
The set of (1) or (2a) is uncoupled when mapped into modal space, where vector of DOFs (size ) is transformed to the equally sized vector of modal variables . Similarly, vector is related to an
equally sized vector of modal controls . These transformations arewhere is the transfer matrix of size between vectors and and mode shape matrix relates vectors and . The -normalized matrix ,
consisting of modal shape vectors (each with components), satisfies the following orthogonality conditions: where is the unitary matrix and is the diagonal matrix of ordered frequencies with the
terms . Each mode shape vector and frequency are solutions to the eigenvalues problem (). The above modal analysis (or operations defined by (6a)–(7b)) is carried out routinely in the FE approach,
even for problems with a very large number of DOFs (large ).
The equations of motion (1) become uncoupled when applying transformations (6a) and (6b) subject to orthogonality conditions (7a) and (7b) and take the following form: where for the Rayleigh damping
(i.e., ) the diagonal terms of are and where are the modal damping ratios. Note that a rigid body translation, for which , is also included in the above equation.
A continuous system, or an FE model (1) of the system described by DOFs (where may be a large number), can be approximated by (8) with only significant modes considered, where usually . The number of
significant modes that should be sufficient to represent such a system is generally problem related and depends mainly on its physical characteristics, the spatial distribution, and frequency content
of the loading [12].
In the system approximated by modes (similarly as for the system's DOFs) one can consider modes as actuated and modes as unactuated. Then matrix will be reduced to only columns, and transfer matrix
in (6a) and (6b) will be of dimensions .
In order to control all modes this system can be made artificially fully actuated by adding dummy actuation forces (zero valued) forming subvector . For the purpose of analysis vector in (1) is
replaced by the augmented force vector containing real actuation forces forming subvector and dummy actuators forming subvector . Then, in (1) and (6a) and (6b), matrix of dimensions is replaced by
matrix of dimensions (this matrix assigns the component of to particular nodes). Consequently in (6b) matrix of dimensions is replaced by a square matrix of dimensions ( modes controlled by
actuators). The dummy actuators should be placed in such a way that is nonsingular.
In the inverse dynamics based control analysis, each control can be obtained from (8) by substituting the corresponding prescribed mode . Then, for known vector , the actuation forces should be
determined by inverting transformation (6b) in which matrix (instead of ) is now square and nonsingular (the dummy actuators were added to the system only to ensure that this inversion is possible).
In the next step, after computing the inverse of operation (6b), the dummy actuators will be eliminated by giving them zero values. For that purpose the inverse matrix , representing the mapping from
modal controls to actuation forces for any augmented system of size , is partitioned as follows: Square submatrix is of size and square submatrix is of size . To be consistent with modes
classifications (actuated and unactuated), vectors and are referred to as actuated and unactuated modal controls, respectively. Given the null-valued dummy force vector (size ) the bottom rows of
operation (9) (lower partition) define constraints on the system in terms of all modal controls, in the following form: Matrix (size ) defines the system constraints written algebraically in terms of
modal controls. Since (10) is homogeneous matrix can be normalized such that the diagonal terms corresponding to controls are set to unity (i.e., ). In this form becomes independent of the choice of
dummy actuators, which reflects the fact that these zero-force actuators were added somewhat arbitrarily only to facilitate the elimination process, that is, to satisfy the constraints in (9). Matrix
is discussed with more details in [13, 14].
Real actuation force(s) may be obtained from the top partition of operation (9) in terms of all modal controls in vector . They can also be obtained in terms of only actuated modal controls in vector
by applying constraints (10) to eliminate unactuated modal controls . Thus, components of actuator forces in vector can be obtained in terms of actuated modal controls in vector from the following
operation: Square matrix (size ) is referred to as the pseudotransfer matrix, and it relates actuated modal controls to real actuator forces. Similar to the normalized constraint matrix , the
pseudotransfer matrix is independent of the choice of dummy actuators.
2.2. Optimal Manoeuvres of Underactuated Systems
In linear optimal control [15], the manoeuvre is optimal if, for a given task, it minimizes the performance index: where , , and are matrices, with the diagonal terms , , and (), that are weights
for the system’s potential energy, kinetic energy, and actuator work, respectively. Note that modal variables and modal controls are included in (12); however, these modes are not independent because
of constraint (10), resulting from underactuation. Such a problem can be solved by applying Pontryagin’s principle. Here we use the procedure described in [15]. Hamiltonian for the constrained
optimization problem involving performance index (12), uncoupled equations of motion (8), and constraints (10) is defined in the following form: and are standard costate vectors related to modal
position and velocity states ( and ) of a system, respectively. Vector represents the set of time-dependent Lagrange multipliers introduced to enforce constraints (10). These multipliers play a
similar role to, for example, that of the multipliers used in the servo-constraint approach [6–9] mentioned before. According to Pontryagin’s principle the costate equations take the following form:
The Hamiltonian is stationary with respect to modal control if Substituting (8) into (15) gives Substituting (16) into (14b) yields Finally, substituting (17) into (14a) generates the following set
of optimality equations: Note that optimality equations (18) contain unknown components in and unknown components in . Therefore, additional constraint equations (10) are required in order to obtain
all the unknown modal variable functions in vector and Lagrange multiplier functions in vector . However, the constraints must be written in terms of not in terms of (note the change in the
constraints’ form from algebraic to differential). The uncoupled equations of motion (8) are substituted into algebraic constraints (10) to obtain The number of ((18) and (19)) is equal the unknown
components in vectors and .
Boundary conditions (4) are mapped into modal space by using the inverse of transformation (6a) or through the relation (obtained by additional substitution of condition (7a)). These transformed
boundary conditions are For fully actuated problems, the last term () in optimality equations (18) vanishes because there are no constraints or Lagrange multipliers needed to enforce them. Therefore,
a fully actuated problem involves only optimality equations (18) to be solved in terms of uncoupled modal variables in vector .
The solution to the combined set of (18), (19), and (20) can be efficiently obtained using symbolic differential operator . Substituting this operator into (19) and (20) and rewriting in matrix
notation give where Matrix contains submatrices , , and . Vector contains all unknown modal variables and Lagrange multipliers. Note that in a fully actuated case, matrix in (21) consists only of
submatrix and vector .
The solution to a system described in form (21) involves the roots () of the characteristic equation for the determinant of [16], where operator is replaced by the auxiliary variable rendering a
order polynomial. This operation is written as Generally, the roots of the characteristic equation (23) take the following form: The positive real numbers and characterize the response of the mode of
motion. For nonzero, unique roots, solution vector consists of components that can be written in terms of independent elementary functions related to the roots (24), in the form [16]: Obviously, the
frequency of mode controlled by the actuators can be interpreted as and its rate of active attenuation (or amplification) as . If multiple roots and zero-valued roots are obtained from (23), then
solution functions (25) must be modified to mathematically accommodate these situations. There are unknown integration constants contained in the solution functions (25).
Integration constants are obtained by substituting the assumed form (25) into differential equations (18) and (19) and using the method of undetermined coefficients to generate sets of linear
algebraic equations relating these constants. By replacing one set of equations with the set of boundary conditions (20), the integration constants can be solved simultaneously. All these symbolic
operations, including the determination of the roots (24) and constants in (25), can be done automatically using the MAPLE mathematical software.
For closed-loop control, asymptotically convergent solution functions are required such that the control task is met over an infinite period of time (). The resulting number of integration constants
is reduced by half, as terms involving positive exponential in the solution form (25) disappear ().
To quantitatively measure the performance of closed-loop control schemes settling time is defined as the time needed for various variables to be reduced to within 3% of their initial value (i.e., ).
The above procedure was applied to actively suppress vibrations of a spatial antenna mast in [13] and of plane frames in [14], the cases with the number of DOFs much greater than the number of
significant modes included in the analysis (i.e., with ). In both cases only oscillating modes were controlled. Here the application of the above methodology is focused on various control schemes,
which are demonstrated in controlling the translational and oscillating modes of a gantry crane.
3. Dynamics and Optimal Control of the Gantry Crane System
The gantry crane problem is one of the simplest underactuated mechanical systems involving two DOFs—cart translation and suspended load rotation—and a single actuator—a cart-driving force (, ).
The gantry crane model is shown in Figure 1. The model includes the mass of the cart , the mass of the suspended load , swing angle , gravitational acceleration , horizontal distance from the cart’s
initial position to the origin, and length of the massless rigid link connecting the cart and load. The task is to manoeuvre the system from an initial resting state at some nonzero horizontal
distance (, ) to a final resting equilibrium state at the origin (, ) by applying time-varying force . Any finite cart translations are permitted, but swings of the suspended load are assumed to be
sufficiently small for a linearized model to be valid. In modal space rigid body translation for such a manoeuvre is easily separated from the oscillatory motion of the suspended load. Dummy force is
added to artificially make the system fully actuated and formulate the augmented gantry crane system.
This same gantry crane model was used in several papers dealing with control or/and optimization. Notably, a Lyapunov function was used in [1] to obtain an asymptotically stable (closed-loop) control
(linear and nonlinear) for attenuating disturbances (nonzero initial positions) in the system, and optimal control by applying Pontryagin's principle was considered in [3]. Results for the linearized
system are of interest because they serve as a useful comparison for the controls obtained in this paper. Similar problems of controlling the plane motion of gantry cranes were presented in [10, 11]
using the concept of flatness. Various aspects of controlling gantry cranes, 3D operations were considered in [9, 17–19].
The gantry crane system shown in Figure 1 and its coordinate system are chosen to mimic those used in [1]. Matrices and vectors in the general equation of motion (1) take the following forms: To be
consistent with the assumptions made in [1, 3] no dissipative effects (i.e., friction, etc.) are considered ().
The initial and final conditions (consistent with [1]) take the following forms: The modal analysis () givesThe rigid body translational mode of motion is represented in (28a) by the frequency and
the second vibrating mode (load swinging) is represented by the frequency .
The uncoupled modal equations of motion (8) become: The augmented system transfer matrix is obtained by the appropriate substitutions from (26) and (28b) into the general partitioned form (9): Modal
controls and are considered actuated and unactuated, respectively. The constraint equation is obtained by normalizing the bottom row of matrix in (30) to obtain The constraint (31) may be applied to
eliminate redundant modal control from the top row operation of (30) to obtain force as a function of independent modal control , giving where is the pseudotransfer matrix (since , this matrix has
only one term). Cart-driving force may be applied using open-loop control (as a known function of time) or using closed-loop control through a set of gains in full-state feedback. Both schemes will
be analyzed and simulated using the CMSOC method.
The performance index (12) takes the following form, consisting of the gantry crane system’s four states () and two modal controls (): The coupled optimality equations (18) take the following forms:
where is the Lagrange multiplier used to meet the constraint (31). The differential form (19) of constraint equation (31) is written asIn modal space, the boundary conditions (27) areEquations (34a),
(34b), and (35a) written according to form (21) (with ) yield where The characteristic equation of the system represented in (37) is obtained through operation (23), giving the 8th order polynomial
equation: Eight roots may be obtained from the characteristic equation (39), which are then substituted into an appropriate assumed solution form (if the roots take the full complex form (24), then
the assumed function takes form (25)) to characterize the three unknown solution functions (). This leaves twenty-four unknown integration constants to be determined by substituting the appropriate
solution form into the three equations (34a), (34b), and (35a). By relating the coefficients corresponding to each of the eight independent elementary functions (i.e., in (25) each is in the form ),
eight algebraic equations are obtained for each differential equation in the set (34a),(34b) and (35a), resulting in a total of twenty-four equations in terms of twenty-four unknown integration
constants . However, these twenty-four equations are linearly dependent. To obtain a unique solution, any one set of eight algebraic equations (obtained from either (34a), (34b), or (35a)) must be
replaced with the eight boundary conditions (36a).
The optimal actuation forces needed to drive the gantry crane from an initially disturbed position (, ) to the origin (, ) will be derived for four cases using the CMSOC method. These cases are as
follows(A)an open-loop control that minimizes actuation forces for a fixed time interval as in [3];(B)a closed-loop control that mimics the control presented in [1];(C)a closed-loop control with
response improved over that presented in [1];(D)a closed-loop control of the fully actuated system (two actuators).For each case, the gantry crane’s physical parameters are chosen to match those
given in [1]; namely, , , , , and .
As a final case, the CMSOC method is applied to a modified three-DOF gantry crane, with an additional link-mass hinge attached to the existing model in Figure 1 and controlled by one or two
actuators. This final case involves two subcases.(E1)A closed-loop control that manoeuvres a modified gantry crane to the origin using the cart-driving force as well as a torque applied to the first
rigid link (two actuators).(E2)An open-loop control that manoeuvres the modified gantry crane to the origin using only the cart-driving force (one actuator) over a fixed time interval.
(A) Open-Loop Control of Gantry Crane Manoeuvre in a Finite Time Interval. The first control manoeuvres the gantry crane from a known initial position to the origin in a finite time interval in an
open-loop scheme. The performance index is chosen to be consistent with that presented in [3], corresponding to the weightings in the general form (33) with all other weightings null valued. Thus,
the optimal control minimizes Performance index (40) minimizes the modal controls or the actuation force over the finite manoeuvre time , which is chosen as to represent again one of the cases
considered in [3]. The gantry crane’s characteristic polynomial equation (39) is simplified to The roots of (41) are . There are four zero roots , two imaginary roots , and two imaginary roots . When
written in form (24), these roots correspond to and . Because of the zero roots and repeating roots, the solution functions take the following form: Each solution function (42) () contains eight
unknown integration constants (), which are determined through substitution and comparison of similar terms in any two differential equations in the set (34a), (34b), and (35a) and by substitution of
the eight boundary conditions (36a). With the integration constants determined, the resulting solution functions are Substituting (43) into (29) yields The Lagrange multiplier function (45), which
represents a “modal force” enforcing the modal constraints, is not used in further analysis and is shown here only for completeness of the solution.
Mapping modal variables and ((43) and (44)) into DOFs and via transformation (6a), the trajectories shown in Figures 2(a) and 2(b) are obtained. Optimal force , shown in Figure 2(c), is obtained by
substituting modal control (46) into transformation (32).
This phase of the solution was done automatically using MAPLE. The solution procedure accepts any problem with modes (obtained from FE analysis for more complex structures) controlled by actuators.
The modal-to-DOF transformations for the gantry crane are indicated in Figures 2(a), 2(b), and 2(c). As shown, the open-loop control is able to perform the task in exactly four seconds, with a peak
force of about 3.6N and a maximum load swing angle of about 0.28rad (16°). The optimal force accelerates the gantry crane over the first half of the manoeuvre (2s) and decelerates the cart over
the last half with identical, but opposite and mirrored, forces.
Similar plots for the closed-loop control presented in [1] are shown in Figure 3. This control requires an effective manoeuvre time of to reach the origin, a maximum load rotation angle of 0.73rad
(42deg), and a maximum force of 15N. It should be noted that this relatively large rotation angle is mentioned here (and other angles quoted in the sequel) for the purpose of comparison only.
From Figure 2 and Figure 3, one can conclude that the open-loop control performs the manoeuvre in a shorter period of time ( versus ) with much smaller peak force requirements (3.6N versus 15N) and
much smaller angles of oscillation (16° versus 42°). Also, the open-loop control brings the system to a complete stop after 4s, while the closed-loop control produces overshoot and the system takes
longer to effectively come to rest.
Calculations show that if the finite manoeuvre time for the open-loop control is extended (or shortened), the peak force requirement and maximum swing angle are reduced (or increased)—approximately
proportional to . For example, if the open-loop control is modified to settle over the same effective period of time as that of the closed-loop control (), the maximum force is reduced to
approximately 1.6N with a maximum swing of about 7deg.
The open-loop control can always provide a faster and more efficient manoeuvre. However, it is possible only when the initial positions and manoeuvre times are known in advance. Closed-loop control
is necessary if any initial configuration (unknown explicitly) is treated as disturbance, and its automatic reduction/removal is desired (the final position is at rest). Case (B) demonstrates how the
CMSOC method is applied to analyze and simulate a closed-loop system that approximately produces the same dynamic responses as given in [1].
(B) Closed-Loop Control of Gantry Crane: Reproducing Control from [1]. A closed-loop control can perform the same task as that of the open-loop control (case (A)); however it does so automatically,
without prior knowledge of initial conditions involved. Any disturbance from its resting configuration at the origin (, ) is relayed through a set of constant gains to generate the cart-driving force
to attenuate this disturbance.
In general, to simulate the closed-loop process analytically the manoeuvre time is infinite and all parameters are driven asymptotically to the origin with increasing time. For the gantry crane, this
requires that all roots of the characteristic equation (39) be nonzero complex numbers in the left half of the complex plane (unlike the open-loop system of case (A), which contained zero roots and
purely imaginary roots). It can be verified that the weightings and in the performance index (33) must be nonzero in order to meet these criteria.
The gantry crane control as given in [1] is closely reproduced by choosing the weightings in the performance index (33) equal to , , , and . The resulting characteristic polynomial equation (39) has
eight roots that take form (24), with real and complex parts equal to Note that the first actively controlled mode of frequency s^−1 ( for uncontrolled system) is damped with the ratio , while the
second mode of frequency s^−1 (s^−1 for uncontrolled system) is damped with the ratio .
Similar to case (A), modal variables and are determined by substituting the parameters from (47) into the assumed solution function (25) and then solving for the unknown coefficients by comparing
similar terms in two of the three optimality/constraint equations (34a), (34b) and (35a), and substituting the boundary conditions (36a). Unlike case (A), the closed-loop problem requires that only
half as many integration constants must be solved because the coefficients preceding exponential growth functions () are assumed to be null valued. This gives Using the appropriate transformations
(see Figure 2) the modal space variables (48) are mapped into the DOF space variables. The resulting system trajectories and the optimal force histogram are visually indistinguishable from those
shown in Figure 3.
The CMSOC method can also generate the closed-loop gains from the assumed weighting coefficients to demonstrate that the gains corresponding to the solution (48) are obtained and compared with the
gains used in [1].
In full-state feedback control the active force is a function of all system states in the following form: For the general CMSOC method, gains and correspond to the observed positions and velocities
of all DOFs of a system. For the gantry crane, (49) takes the following form: By substituting known DOF trajectories ( and ) and the known force function () into (50) and grouping the terms related
to the four independent elementary solution functions (operations are done in MAPLE automatically), one obtains: Each of the bracketed terms in (51) (containing the unknown gains) must equal to zero
for the equation to be true at any time. This gives four equations in terms of four unknown gains, which may be solved to obtain Though initial conditions were assumed in determining the trajectories
and and force , it can be verified that gains (52) remain invariant towards any choice of these assumed conditions.
The control gains used in [1] were Comparing the gains (52) and (53) confirms that the CMSOC method is able to closely reproduce the closed-loop control in [1] by careful selection of the weighting
parameters in performance index (33). However, as shown next in case (C), the performance of this closed-loop control may be improved through better selection of these weighting parameters to produce
faster convergence without an increase in the required peak actuation forces.
(C) Closed-Loop Control of Gantry Crane: Improving Performance. Case (B) developed a control that closely reproduced the control given in [1] by minimizing a performance index that gave no weight ()
to states and , representing the gantry crane’s velocity. Consequently, the control caused the gantry to gain too much speed and then overshoot its target and produce large persistent load swings.
These problems are mitigated by a more careful choice of the performance index weighting parameters in (33). To demonstrate the effect these parameters have on the gantry crane’s dynamics and to
illustrate how they might be meaningfully selected, several cases, labelled P1 to P5 (each with different performance indices as listed in Table 1), are considered.
Each case reflects a performance index which gives significant weightings to an incrementally increasing number of system states (of four possible states , , , ), while holding the weighting on both
modal controls () at unity. Case P0 gives none of the states a significant weighting, case P1 gives a significant weighting to a single state (), case P2 gives significant weightings to two states
(), and so on until case P4 significantly weights all four states. Table 1 summarizes how these weightings are chosen for each case. Since the gantry crane’s asymptotic convergence mathematically
requires that weightings and in the index (33) are nonzero, a small value (0.01) is used instead of zero in cases P0 and P1 to demonstrate how the system behaves when these weightings are negligible.
The DOF trajectories ( and ) and force histogram () for the manoeuvres minimizing the performance indices for cases P0–P4 are presented in Table 2. The settling times of the DOFs are also listed for
each case. All plots in Table 2 are shown over the first 8s of the manoeuvre period except for P0 (30s).
Note that the first modal variable primarily influences the cart’s rigid body mode of motion, while the second modal variable influences the suspended load rotation. In fact there is a direct
relationship between the angle of the load rotation and the second modal variable () such that this angular trajectory is directly affected by varying the weights given to and its derivative in the
performance index (33). Likewise, the speed at which the cart can be made to reach its target is affected through the weightings given to and its derivative .
The performance index in case P0 heavily weights the modal controls and in comparison to modal variables and (100 times more) and neglects the modal velocities and . The resulting control requires a
small peak force (0.7N), producing small maximum load swing angles (0.06rad or 3.4deg), but requires a very long manoeuvre time to converge to the origin (). If the weightings , and were increased
even further relative to the weighting , the maximum force requirements and angular rotations would become infinitesimal while the settling times would approach infinity.
In case P1 a significant weighting value is given to the first modal variable , while other weightings remain unchanged from case P0. This control greatly increases the speed at which the cart
reaches its target position at (~2s), but upon reaching this position the load undergoes large swing angles (1.0rad or 57deg) that persist for a very long time ().
The maximum force increases significantly (17.3N) in comparison to case P0 because the rigid body cart motion requires much larger accelerations during the initial 2s of the manoeuvre in order to
quickly attenuate due to its significant weighting value.
Case P2 improves the load swing attenuation, which was poorly dampened in case P1, by including a large weighting value to the second modal variable (other weightings remain the same as in the
previous case). The maximum load swing angle is reduced (0.8rad or 46deg) and the load swinging motion is damped much more quickly (~6.5s). The cart translation requires similar accelerations and
thus approximately the same maximum force (17.3N) is needed. By inspection, one can see that case P2 produces similar behaviour to the control given in [1] shown in Figure 3 and likewise shares the
problem of target overshoot and large persistent load swings.
Case P3 reduces the tendency of the cart to overshoot the target by also giving a significant weighting to the first modal velocity . However, large persistent load swings are still present, and so
convergence is not significantly improved over that produced in case P2. The maximum required force (17.3N) remains essentially unchanged, while the load swing angles are reduced slightly (0.75rad
or 43deg).
The performance index in case P4 includes a large weight on the second modal velocity (), while keeping all other weightings unchanged from case P3. This produces a control that reduces the magnitude
of load swing angles (0.45rad or 25.8deg) while attenuating the swinging motion more quickly (). The gantry crane performs the manoeuvre in essentially a single load swing cycle, with similar
initial cart accelerations and thus maximum forces (17.3N) as in previous cases. Case P4 produces faster convergence then previous cases because, from an optimal control perspective, it incorporates
all of the gantry crane’s states in the minimization by assigning all weightings in the performance index (33) with significant numerical values.
It is essential that the second modal velocity is given a significant weighting value to yield fast convergence, because the energy of the suspended load oscillates equally between potential and
kinetic energy. Since potential energy and kinetic energy are proportional to the squares of displacement and velocity, respectively, both of the corresponding states and should carry a significant
and approximately equal weight in the performance index (33). Without weighting the load swing velocity state, the control focuses on eliminating swing angles but not swing velocities. However, when
the load is near the bottom of its swing, its velocity is near maximum (), while its displacement is near minimum (). Therefore, the optimal force derived without consideration for the load swing
velocity is unable to eliminate any significant portion of the load swing energy when the load is near the bottom of its swing (). Cases P2 and P3 failed to adequately weight , resulting in larger,
more persistent load swings than in case P4.
The control produced in case P4 provides a significant improvement over the control presented in [1], as it converges more quickly to the origin, while reducing load swing magnitudes, without any
increase in the required maximum forces. To complete the design of this closed-loop control, the gains are obtained from (50) in a similar fashion as in case (B), giving Note that the gains and are
somewhat close to the gains for the control presented in [1] (53), but gains and are substantially different.
(D) Closed-Loop Control of Gantry Crane: Fully Actuated Control. The CMSOC method can also be applied to fully actuated systems. To illustrate this, consider the same gantry crane system, now with
both actuators and acting as real actuators (no dummy actuator). This situation may arise practically when a person is employed to guide the suspended load while the cart performs its translations.
Since the problem is fully actuated there, are no additional constraints on the system motion and consequently no Lagrange multipliers needed to enforce them. The optimal forces can be solved by
calculating the inverse dynamics directly from (6b), which takes form (30) (except ), written as The optimality equations in the differential operator form (21) become where With weightings chosen
according to the performance index in case P4 (, , , , and ), the roots of the characteristic equation (39) for the system given by (56) () take form (24) with the following real and imaginary parts:
For any fully actuated system, each modal variable is independently controlled by a single modal control , resulting in uncoupled solution functions of the following form: For the gantry crane () the
four unknown integration constants are obtained by substituting the four initial conditions for and given by (36a). As in the previous cases, the solved modal variables in form (59) are mapped into
the original coordinates to obtain the DOF trajectories and optimal forces. Figure 4 shows the cart trajectory and the optimal forces on the cart and suspended load and , respectively. The angular
trajectory of the load is not shown because it remains zero () all the time. Practically, this means that for the optimal manoeuvre the person (actuator) guiding the suspended load must simply act to
prevent it from swinging. Fast convergence () to the origin is obtained; however the task requires relatively large maximum forces (104N) compared to previous cases. The required actions of
cart-driving force and suspended load guiding force are identical, as the whole gantry crane system moves as a single rigid body. If smaller forces are desired, then a larger weight may be given to
the modal controls () in the performance index. Note that only a 30N maximum force would be required to execute the manoeuvre in 1s by applying an open-loop control scheme.
(E1) Closed-Loop Control of Modified Three-DOF Gantry Crane (Two Actuators). To illustrate the application of the CMSOC method to a problem of a higher dimension (), the gantry crane is modified by
adding an additional link with an end load, as shown in Figure 5 (a case of a double-pendulum gantry crane in 3D was presented in [19]). In comparison to the previous cases considered the control
task is unchanged except that the oscillations of the additional suspended load must also be damped. Consider the control that uses two actuators ()—the standard cart-driving force and torque ,
produced by a motor fixed to the cart and applied to the first rigid link of length which supports the mass . Dummy torque to be used in formulating the augmented system is applied to the second link
of length which carries mass . All other physical variables are the same as in the original gantry crane model with the exception of and , which denote the angles of links of lengths and ,
The standard matrices in the augmented system’s equation of motion (1) for this new model are The augmented system consists of DOF vector and force vector . The following numerical values are
adopted: , , , and .
The set of equations of motion (1) are uncoupled in modal space, with matrices of ordered frequencies and mode shapes normalized according to (7a) and (7b), taking the following forms: As before, the
first mode represents the rigid body mode of motion (), while the second and third modes, with the squared frequencies and , represent the swinging modes of the rotating link-masses.
Augmented force vector is related to modal control vector through the inverse of transformation (6b) which is partitioned according to (9) to give The constraint equation () is obtained from the
bottom row of (62) () and normalized into the following form: The actuation forces may be obtained directly from the top two rows of (62) in terms of all modal controls, but according to (11) these
forces may be expressed in terms of two independent modal controls (chosen as and ) in the following form: Selecting a performance index of form (12) gives three () optimality equations in the form (
18) that, with the constraint equation (63), may be written according to (21) in the following form: where The parameters in the equation above are the row and column components of the constraint
matrix given by (63). The selected weightings for the performance index are , , , , and .
The twelve () roots of the characteristic equation (23) are obtained in the complex form (24) and are used to generate an assumed solution of form (25) for each unknown modal variable () and Lagrange
multiplier (). Half of these roots generate exponential growth functions that are eliminated by assuming their corresponding integration constants to be zero-valued. Then through the method of
undetermined coefficients, four () sets of six linear algebraic equations are obtained. Replacing one set by the set of six initial conditions, the unknown integration constants are obtained by
solving the set of twenty-four equations (the number of equations is ).
The boundary conditions for this problem are the same as those chosen for the original gantry crane, written in (36a), with the additional condition that the initial and final positions and
velocities of the third modal variable are also zero. In other words, the manoeuvre requires a horizontal cart translation from a resting position at with both links hanging vertically to the same
resting position at the origin.
Figure 6 shows trajectories , , and of the three-DOF gantry crane as well as required actuation forces and .
The manoeuvre, requiring a maximum force of 29N and a maximum torque of 18Nm, is effectively completed after . The maximum load swing angle of the first link is 0.11rad (6.3°) and that of the
second link is 0.48rad (27.5deg).
(E2) Open-Loop Control of Modified Three-DOF Gantry Crane (One Actuator). In order to show the case where one actuator controls three DOFs, the optimal manoeuvre for the modified gantry crane using
only a single actuator—cart-driving force —is investigated for an open-loop scheme. Both of the torque actuators and (Figure 5) are treated as dummy actuators and so the inverse transformation, while
identical to (62), is repartitioned in the following form: The constraint equations () are obtained from the two bottom rows of (67) and normalized into the following form: According to (11) the
single cart-driving force may be expressed in terms of the independent modal control (chosen as ) in the following form: Note that matrix is the same for all cases involving one actuator. Choosing a
performance index in form (12) gives three () optimality equations in form (18) that, with constraint equation (68), may be written according to (21) in the following form: where The assumed
weightings are and (only the control effort is to be minimized). Consistent with the open-loop control presented in case (A) the finite manoeuvre time is chosen to be .
The solution procedure is similar to previous examples. Figure 7 shows trajectories , , and as well as the required cart-driving force . The manoeuvre requires a peak force of 4.8N and completes the
task in exactly 4s. The maximum load swing angle of the first link is 0.19rad (11°) and that of the second link is 0.35rad (20°).
4. Conclusions
The CMSOC methodology was presented as a means of solving linear underactuated (or fully actuated) control problems. The gantry crane problem was selected to illustrate in detail various operations
required for different control methodologies. As demonstrated the method can be applied to open-loop control schemes as well as closed-loop (asymptotically convergent) control schemes. In the latter
case the weightings of the performance index can be translated to the gains of the full-state feedback closed-loop controllers. The operations would be identical for any similar problems with larger
numbers of modes and actuators.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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Population Genetics Examples
1. 2. Calculating allelic frequencies from genotypic frequencies. This method works whether a population is a Hardy-Weinberg population or not. In a certain population, the f(AA) = 0.3, the f(Aa) = 0.4 and the f(aa) = 0.3 Determine p and q. NOTE: This gene has only two alleles, so these three genotypic frequencies must add up to 1.0 p = f(A) which = f(AA) + ½ f(Aa) Therefore p = [0.3 + ½(0.4)] = 0.5 q can be calculated two different ways. q = f(aa) + ½ F(Aa) = 0.5 or, since p + a = 1.0. and p = 0.5: q = 1.0 – p = 0.5 NOTE: If there are only two alleles for the gene, p + q will always equal 1.0 2. Calculating allelic frequencies in a population which is known to be in Hardy-Weinberg equilibrium (breeding randomly). NOTE: if a population is in H-W equilibrium, f(aa) = q2 In a given population of randomly mating gerbils, the frequency of black gerbils is 0.04. NOTE: the black allele (b) is recessive to the wild type brown (agouti) allele (B). Determine p and q. The key point in this problem is that you are told out front that this population is in H-W equilibrium (“randomly mating”). This means that the H-W equation applies to the genotypic frequencies in the population, and therefore the frequency of black gerbils is equal to q2. You may not make this assumption unless you know the population is in H-W equilibrium. F(black) = f(bb) = q2 = 0.04 Therefore, q = square root of 0.4 = 0.2 Since p + q = 1.0. p = 1 – q = 0.8 3. Determining if a population is in Hardy-Weinberg equilibrium. Referring to the first example above: F(AA) = 0.3; f(Aa) = 0.4; f(aa) = 0.3 Using the basic technique for calculating p and q for a population, whether it is in equilibrium or not, we calculated that p = 0.5 and q = 0.5. This is based simply upon the actual genotypic frequencies in the population. Now the question is… is this population in H-W equilibrium? If it is, then f(AA) should be equal to p2, f(Aa) should be equal to 2pq and f(aa) should equal q2. p = 0.5 and q = 0.5, so p2 = 0.25, 2pq = 0.5 and q2 = 0.25. These numbers are obviously not the same as 0.3, .0,4 and 0.3, respectively, so we conclude that this population is not in H-W equilibrium.
Calculating selection effects. If complete selection occurs against a particular allele, the frequency of that allele can be expected to decrease with each generation. The decrease is contingent upon the frequency of individuals conceived who are homozygous for this allele. In gerbils, there exists a gene which has a lethal allele. In heterozygotes, this allele causes a white spotting color pattern. Homozygotes die as embryos and are never seen in live gerbils. For the purposes of this example, we will assign the symbol “W” to the normal allele of this gene and “w” to the lethal allele. Assume a H-W population which begins with the allelic frequencies p = 0.8 and a = 0.2. We calculate the reduction in q as follows (NOTE: (delta) should be read “change in”), and qo = the original q):
q = -(qo2/[1+qo]) (NOTE that q will always be a negative number) q = -0.22/1.2 = -0.033… q1 (new q after one generation of selection) is calculated by adding q to qo (remember that q is a negative number)
q1 = 0.2 – 0.03 = 0.17 p1 = 1.0 – q1 = 0.83 In the next generation, assuming the population is breeding randomly, the f(WW) = p12 = .69, so 69% of newly conceived gerbils will be solid; f(Ww) = 2p1q1 = .28, so 28% of newly conceived babies will be spotted; f(ww) = q12 = 0.3, so 3% of newly conceived babies will die as fetuses. NOTE: Actual phenotypic frequency of offspring in a problem like this is tricky. Because the ww babies die, they are never seen in among the actual offspring, and are thus not really part of the phenotypic ratio. A bit of math shows that 71% of living babies will be solid, 29% spotted. NOTE: The official symbols for these two alleles are actually the reverse of what we used here. I did this because the equation used for selection always involves q. 5. Calculating migration effects. In a certain randomly mating gerbil population, 1% of the population has black fur, the rest are wild type agouti (brown). Brown (B) is completely dominant to black (b). Since the population is breeding randomly, we can assume H-W, thus f(black) = q2, and q = square root of f(black). Thus q = 0.1, and p = 0.9. Our population welcomes a group in new gerbils. Among these new gerbils, P = 0.5 and Q = 0.5. The new, migrant gerbils number 20% of the total, combined population. So m = 0.2. q1 = (1-m)qo + mQ q1 = (1-0.2)(0.1) + (0.2)(0.5) = 0.08 + 0.1 = 0.18 So the new, combined q - = 0.18, and the new p = 0.82. These numbers make sense. They are between the original and migrant values, but are closer to the values of the original population, which was the larger contributor to the combined group. New genotypic frequencies would be calculated by plugging p1 and q1 into the H-W equation. | {"url":"http://www.docstoc.com/docs/20549946/Population-Genetics-Examples","timestamp":"2014-04-18T04:38:42Z","content_type":null,"content_length":"58656","record_id":"<urn:uuid:298a1aea-cdf7-467d-b75f-5c17b3ebd7b6>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00495-ip-10-147-4-33.ec2.internal.warc.gz"} |
connection for a coring
Given an $A$-coring or even a more general additive comonad with a grouplike element there are several related (but in general nonequivalent) notions of connections.
As explained in grouplike element, to an $A$-coring $C$ with a grouplike element one associates a semi-free differential graded algebra $\Omega A = \Omega(A,C)$, sometimes called its (generalized)
Amitsur complex. The simplest notion of a connection for the coring $C$ is a connection for the corresponding Amitsur complex.
Let now $A$ be a $k$-algebra and $(C,\Delta,\epsilon)$ be an $A$-coring with grouplike element $g$ and $(\Omega A,d)$ its Amitsur complex.
A connection $abla:M\otimes_A\Omega^\bullet\to M\otimes_A\Omega^{\bullet+1}$ on a module $M$ over a semifree dga (in the sense of the entry connection for a differential graded algebra) is determined
by its value $abla_M|_M$ on $M\cong M\otimes_A A$. If $\rho^M:M\to M\otimes_A C$ is a right $C$-coaction then the formula
$abla|_M:m\mapsto \rho^M(m)-m\otimes g$
determines a flat connection on $M$. Conversely, any flat connection determines a right $C$-coaction by
$\rho^M(m)=abla(m)+m\otimes g.$
This amounts to a bijection between $C$-coactions and flat connections on $M$. Regarding that coactions correspond to descent data in the context of comonadic descent, this gives the flat connection
interpretation of such descent data. A first instance is probably Grothendieck’s identification of flat connections and the first order costratifications in Grothendieck’s theory of differential
calculus on schemes (foundations of crystalline cohomology, see book by Berthelot and Ogus; cf. also Grothendieck connection).
Connections on comodules directly
One the other hand, one can consider more generally additive comonads, and define connections on comodules over them rather directly. Or dually one can work with connection on modules over additive
Menini and Ştefan first define an intermediate notion of a quasi-connection for monads. Let $A$ be an additive category $(T,\mu,\eta)$ an additive monad in $A$ and $u:TM\to M$ an action on some
object $M$ in $A$. Then a quasi-connection on $M$ is a map $abla:M\to TM$ such that
$abla \circ u - \mu\circ T(abla) = id_{TM} - \eta\circu: TM\to TM.$
A quasi-connection is a connection if, in addition,
$u\circabla = 0.$
For every connection in Menini–Ştefan sense, one defines its curvature $F_abla : M\to T^2 M$ by the formula
$F_abla := (id_{T^2 M} - \eta_{TM}\circ\mu_M)\circ T(abla)\circabla.$
As usually, we define a flat connection as a connection whose curvature vanishes.
In this setting one again has a bijection between flat connections and descent data.
Literature and links
• P. Nuss, Noncommutative descent and non-abelian cohomology, $K$-Theory 12 (1997), no. 1, 23–74.
• T. Brzeziński, R. Wisbauer, Corings and comodules, London Math. Soc. Lec. Note Series 309, Cambridge 2003.
• C. Menini, Talk at MSRI: Connections, symmetry operators and descent data for triples, 1999, link
• C. Menini, D. Ştefan, Descent theory and Amitsur cohomology of triples, J. Algebra 266 (2003), no. 1, 261–304.
• T. Brzeziński, Flat connections and (co)modules, [in:] New Techniques in Hopf Algebras and Graded Ring Theory, S Caenepeel and F Van Oystaeyen (eds), Universa Press, Wetteren, 2007 pp. 35-52 | {"url":"http://www.ncatlab.org/nlab/show/connection+for+a+coring","timestamp":"2014-04-16T04:21:58Z","content_type":null,"content_length":"25579","record_id":"<urn:uuid:66bda614-9735-4c39-9098-f6f6c2b102bd>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00583-ip-10-147-4-33.ec2.internal.warc.gz"} |
[R] truncating a data frame based on a function
Faheem Mitha faheem at email.unc.edu
Fri Sep 21 17:24:12 CEST 2007
Consider the following example.
> a = c(1,2,3); b = c(4,5,6); c = cbind(a,b); c[(2 < c[,1]) & (c[,1] < 4),]
a b
So, the idea is to select rows for which the value in the first column is
between 2 and 4. This works, however, I don't like having to reference a
explicitly in this fashion, and just wondered if there was a preferred way
to accomplish the same thing. Ideally, I'd like to make use of a function.
Thanks, Faheem.
More information about the R-help mailing list | {"url":"https://stat.ethz.ch/pipermail/r-help/2007-September/141313.html","timestamp":"2014-04-17T12:33:25Z","content_type":null,"content_length":"2980","record_id":"<urn:uuid:f9e1d0d2-519d-4ea1-90ca-fc81103c7fad>","cc-path":"CC-MAIN-2014-15/segments/1398223205137.4/warc/CC-MAIN-20140423032005-00084-ip-10-147-4-33.ec2.internal.warc.gz"} |
Order Relation
Anti-symmetric: (2, 1) is in the set. If this relation is anti-symmetric, (1, 2) must not be in the set. Yes this is true. So, this property is true.
I was giving an example! (2,1) is not the only pair you need to check.
Reflexive Property:
If x in an element in R, then x < x for all x in R. Therefore, x precedes x.
You mean x<= x (or [itex]x\le x[/tex]), of course. | {"url":"http://www.physicsforums.com/showpost.php?p=555137&postcount=4","timestamp":"2014-04-21T07:18:41Z","content_type":null,"content_length":"8019","record_id":"<urn:uuid:4a7fe74c-bb34-4e80-ae68-d68f08ab8c9e>","cc-path":"CC-MAIN-2014-15/segments/1398223202774.3/warc/CC-MAIN-20140423032002-00514-ip-10-147-4-33.ec2.internal.warc.gz"} |
• 28.60 Two 50-turn coils, each with a diameter of 4.00 m, are placed 1.00 m apart, as shown in the figure. A current of 7.00 A is flowing in the wires of both coils; the direction of the current
is clockwise for both coils when viewed from the left. What is the magnitude of the magnetic field in the center between the two coils?
• 28.61 The wires in the figure are separated by a vertical distance d. Point B is at the midpoint between the two wires; point A is a distance d/2 from the lower wire. The horizontal distance
between A and B is much larger than d. Both wires carry the same current, i. The strength of magnetic field at point A is 2.00 mT. What is the strength of the field at point B?
• 28.62 You are standing at a spot where the magnetic field of the Earth is horizontal, points due northward, and has magnitude 40.0 μT. Directly above your head, at a height of 12.0 m, a long,
horizontal cable carries a steady DC current of 500.0 A due northward. Calculate the angle θ by which your magnetic compass needle is deflected from true magnetic north by the effect of the
cable. Don't forget the sign of θ—is the deflection eastward or westward?
• 28.63 The magnetic dipole moment of the Earth is approximately 8.0 · 10^22 A m^2. The source of the Earth's magnetic field is not known; one possibility might be the circulation of ions in the
Earth's molten outer core. Assume that the circulating ions move a circular loop of radius 2500 km. What “current” must they produce to yield the observed field?
• 28.64 A circular wire loop has radius R = 0.12 m and carries current i = 0.10 A. The loop is placed in the xy-plane in a uniform magnetic field given by
• 28.65 A 0.90 m-long solenoid has a radius of 5.0 mm. When the wire carries a 0.20-A current, the magnetic field in the solenoid is 5.0 mT. How many turns of wire are there in the solenoid?
• 28.66 In a coaxial cable, the solid core carries a current i. The sheath also carries a current i but in the opposite direction and has an inner radius a and an outer radius b. The current
density is equally distributed over each conductor. Find an expression for the magnetic field at a distance a < r < b from the center of the core.
• •28.67 A 50-turn rectangular coil of wire of dimensions 10.0 cm by 20.0 cm lies in a horizontal plane, as shown in the figure. The axis of rotation of the coil is aligned north and south. It
carries a current i = 1.00 A, and is in a magnetic field pointing from west to east. A mass of 50.0 g hangs from one side of the loop. Determine the strength the magnetic field has to have to
keep the loop in the horizontal orientation.
• •28.68 Two long, straight parallel wires are separated by a distance of 20.0 cm. Each wire carries a current of 10.0 A in the same direction. What is the magnitude of the resulting magnetic field
at a point that is 12.0 cm from each wire?
• •28.69 A particle with a mass of 1.00 mg and a charge of q is moving at a speed of 1000. m/s along a horizontal path 10.0 cm below and parallel to a straight current-carrying wire. Determine q if
the magnitude of the current in the wire is 10.0 A.
• •28.70 A conducting coil consisting of n turns of wire is placed in a uniform magnetic field given by R = 5.00 cm, and the angle between the magnetic field vector and the unit normal vector to
the coil is θ = 60.0°. The current through the coil is i = 5.00 A.
1. Specify the direction of the current in the coil, given the direction of the magnetic dipole moment,
2. Calculate the number of turns, n, the coil must have for the torque on the loop to be 3.40 N m.
3. If the radius of the loop is decreased to R = 2.50 cm, what should the number of turns, N, be for the torque to remain unchanged? Assume that i, B, and θ stay the same.
• •28.71 A loop of wire of radius R = 25.0 cm has a smaller loop of radius r = 0.900 cm at its center such that the planes of the two loops are perpendicular to each other. When a current of 14.0 A
is passed through both loops, the smaller loop experiences a torque due to the magnetic field produced by the larger loop. Determine this torque assuming that the smaller loop is sufficiently
small that the magnetic field due to the larger loop is same across the entire surface.
• •28.72 Two wires, each 25.0 cm long, are connected to two separate 9.00-V batteries as shown in the figure. The resistance of the first wire is 5.00 Ω, and that of the other wire is unknown (R).
If the separation between the wires is 4.00 mm, what value of R will produce a force of magnitude 4.00 · 10^−5 N between them? Is the force attractive or repulsive?
• •28.73 A proton is moving under the combined influence of an electric field (E = 1000. V/m) and a magnetic field (B = 1.20 T), as shown in the figure.
1. What is the acceleration of the proton at the instant it enters the crossed fields?
2. What would the acceleration be if the direction of the proton's motion was reversed?
• •28.74 A toy airplane of mass 0.175 kg, with a charge of 36 mC, is flying at a speed of 2.8 m/s at a height of 17.2 cm above and parallel to a wire, which is carrying a 25-A current; the airplane
experiences some acceleration. Determine this acceleration.
• •28.75 An electromagnetic doorbell has been constructed by wrapping 70 turns of wire around a long, thin rod, as shown in the figure. The rod has a mass of 30.0 g, a length of 8.00 cm, and a
cross-sectional area of 0.200 cm^2. The rod is free to pivot about an axis through its center, which is also the center of the coil. Initially, the rod makes an angle of θ = 25.0° with the
horizontal. When θ = 0.00°, the rod strikes a bell. There is a uniform magnetic field of 900.0 G directed along θ = 0.00°.
1. If a current of 2.00 A is flowing in the coil, what is the torque on the rod when θ = 25.0°?
2. What is the angular velocity of the rod when it strikes the bell?
• •28.76 Two long, parallel wires separated by a distance, d, carry currents in opposite directions. If the left-hand wire carries a current i/2, and the right-hand wire carries a current i,
determine where the magnetic field is zero.
• •28.77 A horizontally oriented coil of wire of radius 5.00 cm and carrying a current, i, is being levitated by the south pole of a vertically oriented bar magnet suspended above the center of the
coil. If the magnetic field on all parts of the coil makes an angle θ of 45.0° with the vertical, determine the magnitude and the direction of the current needed to keep the coil floating in
midair. The magnitude of the magnetic field is B = 0.0100 T, the number of turns in the coil is N = 10.0, and the total coil mass is 10.0 g.
• •28.78 As shown in the figure, a long, hollow, conducting cylinder of inner radius a and outer radius b carries a current that is flowing out of the page. Suppose that a = 5.00 cm, b = 7.00 cm,
and the current i = 100. mA, uniformly distributed over the cylinder wall (between a and b). Find the magnitude of the magnetic field at each of the following distances r from the center of the
1. r = 4.00 cm
2. r = 6.50 cm
3. r = 9.00 cm
• •28.79 A wire of radius R carries current i. The current density is given by J = J[0](1 – r/R), where r is measured from the center of the wire and J[0] is a constant. Use Ampere's Law to find
the magnetic field inside the wire at a distance r < R from the central axis.
• •28.80 A circular wire of radius 5.0 cm has a current of 3.0 A flowing in it. The wire is placed in a uniform magnetic field of 5.0 mT.
1. Determine the maximum torque on the wire.
2. Determine the range of the magnetic potential energy of the wire. | {"url":"http://dev5.mhhe.com/textflowdev/genhtml/0072857366/P6.28.f.htm","timestamp":"2014-04-17T09:33:45Z","content_type":null,"content_length":"93932","record_id":"<urn:uuid:9a73146f-d46b-4a49-bc6b-bf6360547a0b>","cc-path":"CC-MAIN-2014-15/segments/1397609527423.39/warc/CC-MAIN-20140416005207-00198-ip-10-147-4-33.ec2.internal.warc.gz"} |
Division of Rational Expressions ( Read ) | Algebra
What if you had two rational expressions like $\frac{x + 5}{x}$$\frac{x^2 + 6x + 5}{x - 2}$rational expressions like this one.
Watch This
CK-12 Foundation: 1209S Dividing Rational Expressions
Watch this video for more examples of how to multiply and divide rational expressions.
RobiChaudd: Multiply or divide rational expressions
Just as with ordinary fractions, we first rewrite the division problem as a multiplication problem and then proceed with the multiplication as outlined in the previous example.
Note: Remember that $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$ second fraction. Do not fall into the common trap of flipping the first fraction.
Example A
Divide $\frac{4x^2}{15} \div \frac{6x}{5}$
First convert into a multiplication problem by flipping the second fraction and then simplify as usual:
$\frac{4x^2}{15} \div \frac{6x}{5} = \frac{4x^2}{15} \cdot \frac{5}{6x} = \frac{2x}{3} \cdot \frac{1}{3} = \frac{2x}{9}$
Divide a Rational Expression by a Polynomial
When we divide a rational expression by a whole number or a polynomial, we can write the whole number (or polynomial) as a fraction with denominator equal to one, and then proceed the same way as in
the previous examples.
Example B
Divide $\frac{9x^2-4}{2x-2} \div (21x^2-2x-8)$
Rewrite the expression as a division of fractions, and then convert into a multiplication problem by taking the reciprocal of the divisor:
$\frac{9x^2-4}{2x-2} \div \frac{21x^2-2x-8}{1} = \frac{9x^2-4}{2x-2} \cdot \frac{1}{21x^2-2x-8}$
Then factor and solve:
$\frac{9x^2-4}{2x-2} \cdot \frac{1}{21x^2-2x-8} = \frac{(3x-2)(3x+2)}{2(x-1)} \cdot \frac{1}{(3x-2)(7x+4)} = \frac{(3x+2)}{2(x-1)} \cdot \frac{1}{(7x+4)} = \frac{3x+2}{14x^2-6x-8}$
Solve Applications Involving Multiplication and Division of Rational Expressions
Example C
Suppose Marciel is training for a running race. Marciel’s speed (in miles per hour) of his training run each morning is given by the function $x^3-9x$$x$$x$$3x^2-9x$
$\text{time} = \frac{\text{distance}}{\text{speed}}\!\\\\\text{time} = \frac{3x^2-9x}{x^3-9x} = \frac{3x(x-3)}{x(x^2-9)} = \frac{3x(x-3)}{x(x+3)(x-3)}\!\\\\\text{time} = \frac{3}{x+3}\!\\\\\text{If}
\ x = 5, \ \text{then}\!\\\\\text{time} = \frac{3}{5+3}=\frac{3}{8}$
Marciel will run for $\frac{3}{8}$
CK-12 Foundation: 1209 Dividing Rational Expressions
• When we multiply two fractions we multiply the numerators and denominators separately:
$\frac{a}{b} \cdot \frac{c}{d}=\frac{a \cdot c}{b \cdot d}$
• When we divide two fractions, we replace the second fraction with its reciprocal and multiply, since that’s mathematically the same operation:
$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} = \frac{a \cdot d}{b \cdot c}$
Guided Practice
Divide $\frac{3x^2-15x}{2x^2+3x-14} \div \frac{x^2-25}{2x^2+13x+21}$
$\frac{3x^2-15x}{2x^2+3x-14} \cdot \frac{2x^2+13x+21}{x^2-25} = \frac{3x(x-5)}{(2x+7)(x-2)} \cdot \frac{(2x+7)(x+3)}{(x-5)(x+5)} = \frac{3x}{(x-2)} \cdot \frac{(x+3)}{(x+5)} = \frac{3x^2+9x}{x^
Divide the rational functions and reduce the answer to lowest terms.
1. $2xy \div \frac{2x^2}{y}$
2. $\frac{2x^3}{y} \div 3x^2$
3. $\frac{3x+6}{y-4} \div \frac{3y+9}{x-1}$
4. $\frac{x^2}{x-1} \div \frac{x}{x^2+x-2}$
5. $\frac{a^2+2ab+b^2}{ab^2-a^2b} \div (a+b)$
6. $\frac{3-x}{3x-5} \div \frac{x^2-9}{2x^2-8x-10}$
7. $\frac{x^2-25}{x+3} \div (x-5)$
8. $\frac{2x+1}{2x-1} \div \frac{4x^2-1}{1-2x}$
9. $\frac{3x^2+5x-12}{x^2-9} \div \frac{3x-4}{3x+4}$
10. $\frac{x^2+x-12}{x^2+4x+4} \div \frac{x-3}{x+2}$
11. $\frac{x^4-16}{x^2-9} \div \frac{x^2+4}{x^2+6x+9}$
12. Maria’s recipe asks for $2 \frac{1}{2}$$3 \frac{1}{3}$
13. George drives from San Diego to Los Angeles. On the return trip he increases his driving speed by 15 miles per hour. In terms of his initial speed, by what factor is the driving time decreased on
the return trip?
14. Ohm’s Law states that in an electrical circuit $I = \frac{V}{R_c}$$\frac{1}{R_{tot}} = \frac{1}{R_1} + \frac{1}{R_2}$$R_1$$R_2$ | {"url":"http://www.ck12.org/algebra/Division-of-Rational-Expressions/lesson/Division-of-Rational-Expressions---Intermediate/","timestamp":"2014-04-20T01:36:36Z","content_type":null,"content_length":"118084","record_id":"<urn:uuid:5c037dd3-c378-46e1-a840-cdc2c067c116>","cc-path":"CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00300-ip-10-147-4-33.ec2.internal.warc.gz"} |
Caryn Loves Math
"If practice makes permanent, lets make it fun".
Caryn Loves Math
a question. They will receive an automated email and will return to answer you as soon as possible. Please
to ask your question.
Olga Fernandez Gonzalez
(TpT Seller) re: Freebie~10 Photographs for Commercial Uses~Grand Canyon~Mesa Verde
Dear Caryn,
is it possible to get the Rushmore photo with more resolution?
I am planning to use it in a project.
Thanks a lot for sharing these images.
Deidre Beasley
(TpT Seller)
I purchased your triangle bundle and love it! I was wondering if you had a bundle on quadrilaterals as well. Thanks!
I just upload one. I only have 4 quadrilateral activity presently.
I found some more typos in the Dominoes: Using Algebra in Geometry activity. This time, two of the answers are incorrect in the triangles dominoes. Also, the answer key hasn't been corrected for the
parallel lines/transversal dominoes, so that could confuse some people. I hope you can correct these, so I can re-download. Thanks.
It is ready to be downloaded. Thanks for the heads up. I didn't even think about the key. This was one of my original items, I am so embarrassed that this has been up for 2 years with error. Augh.
Thanks again.
No, I hadn't contacted the tech people yet. My email is weisskopfl@yahoo.com. Thanks for your help!
I have sent the file along with a bonus one. If you don't see if soon please check your bulk mail. Thanks for you purchase.
I purchased the congruent triangle foldable in August and am now ready to use it. When I have tried to download the product, I'm only getting 2 pages, not 7. Any help would be appreciated!
Have you contacted tech support for help? If you send me your email I will send you the file.
Buyer re: Algebra Word Wall, Walls that Teach, 100+ Concepts
This resource is GREAT! The only thing that I noticed was the Extraneous Solution page did not have anything on it.
Thanks for putting it out there! I will definitely use it this year!
I couldn't think of anything to put on the page but thought we at least needed the word for our walls so I included it. - Thanks Caryn
Buyer re: Using Algebra in Geometry-Parallelogram Walk-Around Activity
Hi Caryn. I like the idea of a walk around activity but I'm not sure how it work in my current classroom. What is the largest class you've used this with? I have a lot of students in a room crammed
full with desks so I'm concerned about logistics. Thanks!
My biggest class is 22. With my biggest classes I make 2 or 3 sets and divide the students. Then I place a set on the back wall, the front and one at a table group. This way the students are spread
out. Another idea is to print several pages on one sheet of paper and turning it into a table activity. 3 or 4 students can work with the same set of cards just have them start with different cards.
Elizabeth F.
Hi Caryn, I don't know how to respond to your comment, but I downloaded the file last week. I wonder why it didn't update...
I just looked at some of your other activities for geometry and algebra and I think I will be buying them soon! My students are always asking for an activity to do. Thanks so much for sharing.
Deborah M. re: Sort:Using Graphing Calculator~Find Zeros~Polynomials~Graph
I can't seem to open this after downloading it. It says there is an error- that the file is damaged
Sorry, just saw you question. Have you been able to download it. I have checked the file and it opens on my computer. Please contact TPT tech support for further assistance with this issue.
(TpT Seller) re: Sort: Like Radicals,Square Root, Cube Roots,
Do you have any activity that would practice simplifying square roots but that includes some work with imaginary /complex numbers? Or solving quadratics using completing the square or square root
method that would include simplifying a square root with imaginary/complex answer? My students are struggling with these concepts & rather than reinvent the wheel & I would rather buy a wheel!
(TpT Seller) re: Walk around Activity~Geometry w/Algebra~Finding the Values of Angles~
This activity is awesome. It brings in some of those concepts required written ACT in the end of course assessments for Geometry. I don't know if this could be an extention or if an idea for an
additional activity. The majority of the problems students have to answer don't end with x=. Once they have find x, they then have to find an angle. Sometimes it will end there, alot of the times
they will then have to use the angle to find an angle that is supplementary to it. Everything is multi-step. It is so hard to find resources. Our textbooks don't support the rigor and struggle to
have time to create it. Thanks for all the great stuff. My kids love it!
Teaching High School Math
(TpT Seller) re: 3 Dominoe Like Matching Sort: Exterior-Interior Angles of Polygons
I just wanted you to know that I featured this activity on my blog as one of my top 5 polygon activities from Teachers Pay Teachers. Great work! http://teachinghighschoolmath.blogspot.com -- Jennifer
Michele Luck's Social Studies
(TpT Seller)
I just want to touch base to make sure you were still participating in the Middle School Conference Opportunity! Please take a look at the thread in the forum to get the latest updates. And if you
haven't sent your conference fee, please do so as soon as possible!
Yes, I got my check today and will be sending payment this evening. I just need to get my stuff laminated. Thanks for the reminder
(TpT Seller)
(TpT Seller) re: Using Algebra in Geometry-TRAPEZOIDS Walk-Around Activity
I love this idea. However, can you explain the piecewise function stuff at the end on the last couple of pages? I didn't understand that part.
Snafu. Looks like the student answer sheets were mixed. I will get them updated as soon as I get home from school so you can re-download them. I am so sorry. Caryn White
I have uploaded the updated version. CJW
Krista Sueltenfuss
(TpT Seller) re: Algebra Word Wall, Walls that Teach, 100+ Concepts
Perfect, thank you!
(TpT Seller)
Hi Caryn, Me again...I just wanted to let you know that I took your advice and changed all my prices, so if you want to come look around again, that would be great. I also just became your 73
follower. ;-)
Thanks. I will go "shopping" later this week. Good luck with your sales, I hope mine continue to go up and up. Caryn
(TpT Seller)
Hi Caryn, I just checked out your store. And WOW, love the "walk-arounds" and the dominoes. I will be back to purchase very soon. I also see that you have been on TPT about a year. I hope my store
looks this good in a year!
Thanks, I love TPT (and your store)!
Julie L.
Hello, I purchased the Algebra graphs walls that teach and I love the concept. When I printed them, several were blurry. I haven't had this problem with the other products I've purchased from you,
including the algebra and geometry word walls that teach. I've printed them both on my printer at home (Inkjet) and at work (laser) and even a few on a friend's color laser and more than half are not
(TpT Seller) re: Absolute Value Translation Investigation Worksheet~Graphing Calculator
I love that you have written this activity up. I have done this with my kids for years just as exploration but then struggle to find practice sheets to reinforce the patterns they have found. Do you
have a package deal for this one, quadratic, absolute value, etc?
Thanks, I have a package deal that has all my absolute value items and another with all my quadratic ones. Are you wanting a combination of the 2?
Awesome Caryn! Thank you. If I come up with something for Extraneous Solution I will let you know. :0) .
(TpT Seller)
Just e-mail me at calqlate@yahoo.com, and I will be happy to check anything that you have. It would be exciting to check each other's work. Also, I have a math blog that you might enjoy. It is at:
(TpT Seller) re: Sort:Factoring -Difference/Sum of Cubes-Cubics,Quartics
I saw your store on TPT and since I also teach math at the local community college, I thought I would look at your freebie. I left you a rating. I have a few suggestions that might help you to better
market your products. First, create a cover page which can then be used as an attractive thumbnail to entice buyers. Make your second page a directions or suggestions page. List several ways you use
this product. Never assume that a buyer knows how, why, or when to use your product. Also, move your answer key to the end of your product. Finally, think about making your x's cursive; otherwise,
they look like the multiplication sign. (This is just my preference).
I would appriciate your input on my freebie listed below.
I am now your 11th follower, and hope you will reciprocate at my store as well. http://www.teacherspayteachers.com/Store/SciPi/Products
The teaching world needs quality math materials; so, good luck as you sell on TPT!
That sort was one of my first listings. My recent items (done just last weekend actually) have some of the thing you have suggested. Thanks so much for the ideas. I will look at my older items and
update them. If you have time would you look at my Geometry dominoes listing, do you have any suggestions for it? If you send me your email, I will send you the actual product for your review. Thanks
again. I am hoping to get a nice side business going, since my husband is now unemployed. Thanks again
Flip ~~ Rotate ~~ Slide
each card so that all side match
a term with it's definition
REMOVE APOSTROPHE IN IT'S :)
ITS doesn't take apostrophe to show possession. :)
It's means it is.
I know but I can't fix it because I no longer use a Smartboard (we now use a Promethean Board) so I do not have the application to fix it.
Amanda H. re: Algebra 2 Web Scavenger Hunt
I just wanted you to know that I love your materials! I've signed up for emails when you post things and I love them everytime. Thank you for all the hard work you do for your students!
Thanks, every teacher needs to hear that once in a while. I'll work on getting more pages for you.
Thanks for your response. I am currently building a website/online school. The address is www.universalhomeschool.com/custom. It will not be live for another 6 – 8 months because there is so much
work to be done. I am currently trying to build up material including videos, lessons, worksheets, presentations and more. A lot of this material is being created by myself. I hope that the site will
be successful in providing students with a vast resource of information that will help them become more successful. Any material that I use that was not created by myself I will make sure credit has
been given to the author. Let me know what you think, if you don’t mind me using your work, either way is fine. If you have any suggestions for me I would also appreciate that. Thanks for taking
the time.
Is it ok for me to use your work on my website.
May I look at you website? What is the url. Caryn
Gladys Scott
(TpT Seller)
I cant get the linear regression calculator worksheet. clicked everywhere. Would you email me at rscott1410@aol.com. I want to use after Christmas break
I don't know why it is not there. I am looking for it on my computer. I do not have skype but I have yahoo messenger:yesterdaysvideogames@yahoo.com
Gladys Scott
(TpT Seller)
I am looking for the linear regressions activity. It wont connect to link. Do you skype.Would love to talk to you concerning Alg 2 activities. My skype name is trixie fallen
Dear Caryn,
My name is Jack Dennison. I am the Editor-in-Chief of www.FunnelBrain.com, an educational website. I am writing to you because the quality of the work you have posted to PTP distinguishes you as a
talented author. I am looking for teacher-authors to create custom study review materials for FunnelBrain (on a paid basis, of course).
These study materials would take the form of online flashcards, quizzes, and study guides for high school and college level courses, as well as for certain standardized exams such as the GMAT, the
SAT or the GRE. We are also interested in lesson plans built around activities that can take place on www.FunnelBrain.com.
If you are interested in a paid writing assignment, please email me at jack@funnelbrain.com, and I will respond with details on the available assignments. In the meantime, please feel free to browse
I look forward to hearing from you.
Jack Dennison
I cannot download your Quadratic/Linear Regression worksheet. Clicking on the download now button leads me to this site's home page
I am sorry you are having difficulties. Is this a free item or a purchase? Please contact tech support. To do this, click on the My Purchase link. On the top right hand side, you should see a link
for Download Difficulties. Hope that helps. | {"url":"http://www.teacherspayteachers.com/Store/Caryn-Loves-Math","timestamp":"2014-04-16T19:27:49Z","content_type":null,"content_length":"339023","record_id":"<urn:uuid:d648c2a3-9bc3-4a12-84e8-fa39600d6ea5>","cc-path":"CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00018-ip-10-147-4-33.ec2.internal.warc.gz"} |
PHONON Software
by Krzysztof PARLINSKI
Last update: January 2, 2008 Phonon is a software (see list of Publications) for calculating phonon dispersion curves, and phonon density spectra of crystals, crystals with defects, surfaces,
adsorbed atoms on surfaces, etc. from either a set of force constants, or from a set of Hellmann-Feynman forces calculated within an ab initio program (not included). One can use VASP, Wien2k, MedeA
of Materials Design , Siesta, or other ab initio code which is able to optimize a supercell and calculate the Hellmann-Feynman forces. Phonon builds a crystal structure, using one of the 230
crystallographic space groups, finds the force constant from the Hellmann-Feynman forces, builds the dynamical matrix, diagonalizes it, and calculates the phonon dispersion relations, and their
intensities. Phonon finds the polarization vectors, and the irreducible representations (Gamma point) of phonon modes, and calculates the total and partial phonon density of states. It plots the
internal energy, free energy, entropy, heat capacity and tensor of mean square displacements (Debey-Waller factor). Phonon finds the dynamical structure factor for the coherent inelastic neutron
scattering and the incoherent doubly differential scattering cross section for a single crystal and polycrystal. For polar cystals the LO/TO mode splitting can be included.
DemoPhonon demonstrates the full functionally of Phonon, but is limitted to a single space group Fm3m. | {"url":"http://wolf.ifj.edu.pl/phonon/","timestamp":"2014-04-21T03:30:56Z","content_type":null,"content_length":"3650","record_id":"<urn:uuid:63063470-b01f-4e25-950c-41bbe0574983>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00194-ip-10-147-4-33.ec2.internal.warc.gz"} |
Number of bacteria present in a culture at any time, t hours, is modelled by the equation N=ne^kt
If the original... - Homework Help - eNotes.com
Number of bacteria present in a culture at any time, t hours, is modelled by the equation N=ne^kt If the original number is doubled in 3 hours find k?
The number of bacteria in a culture after a duration of time t is given by the equation `N = n*e^(k*t)` where n is the initial number of bacteria.
As the number of bacteria becomes double in 3 hours, the resulting equation is: `2*n = n*e^(k*3)`
=> `2 = e^(k*3)`
take the natural log of both the sides
=> `ln 2 = 3*k*ln e`
=> `ln 2 = 3*k`
=> `k = (ln 2)/3`
=> `k ~~ 0.2310`
The value of k is approximately 0.2310
Join to answer this question
Join a community of thousands of dedicated teachers and students.
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Encyclopaedia Index
TMP1, a PIL real variable for setting first-phase temperature, when the variable TEM1 is not being solved for directly.
TMP2 is the corresponding variable for second-phase temperature.
It acts in one of four ways:-
1. Setting uniform temperature.
If TMP1 is set to a value outside the range from -10110.0 (i.e. GRND) to -10210.0 (i.e. GRND10), then this value assigned to the temperature within the first-phase material in the domain of
study, unless a STOREd variable named TMP1 also exists (see below).
2. The use of Ground coding
If TMP1 is set to GRND, temperature will instead be computed by Fortran coding which the user has placed in the GROUND subroutine, either directly or by the use of In-Form or PLANT.
3. The use of built-in options
If TMP1 is set to GRNDx, where x is one of 1, 2, 3, 4, 5, 6, 7 or 8, it acts as an indicator of which formula in the open-source Fortran file GXTEMPR.htm is to be used for deducing first-phase
temperature from first-phase enthalpy.
Note that GRND9 and GRND10 are reserved for the use of GXSCRI.
The PIL variables:
TMP1A, TMP1B, TMP1C, TMP2A, TMP2B and TMP2C
are constants which may appear in the formulae, all but the first of which have the form :
T = (H - H_base) / Cp
□ T stands for temperature
□ H stands for enthalpy or "stagnation enthalpy", i.e. enthalpy plus kinetic energy;
□ H_base stands for the enthalpy of the material in question when T equals 0 on the temperature scale in question;
□ Cp stands for specific heat capacity at constant pressure, which may or may not be a constant.
The possible settings of TMP1 are as follows, with the number on the left corresponding to x in the GRNDx to which TMP1 has been set in the Q1 file. Those for TMP2 are identical, with 2 in place
of 1. In the following, F(L0H12+I) represents the current cell phase enthalpy, and F(L0CP12+I) represents the current cell phase specific heat.
1. TMP1=GRND1 selects temperature equal to: TMP1A. Note that this is un-needed option, because making TMP1 a non-GROUND constant has the same effect.
2. TMP1=GRND2 selects temperature equal to:
F(L0H12+I)/F(L0CP12+I) + TMP1A
H_base is set via the PIL variable TMP1A, which is evidently equal to minus H_base/ Cp .
3. TMP1=GRND3 selects temperature equal to:
F(L0H12+I)/F(L0CP12+I) + TMP1A + TMPCG*F(L0CB+I)
Here it is:
TMP1A + TMPCG*F(L0CB+I),
which represents - H_base / Cp.
This allows H_base to vary linearly in accordance with some scalar, for example the mass-fraction of some component of the fluid mixture.
The scalar represented by L0CB is C3 for Phase 1 ,and C4 for Phase 2.
4. TMP1=GRND4 selects temperature equal to:
F(L0H12+I)/F(L0CP12+I) +
This allows H_base to depend in a more complex way on the scalar value, as when:
☆ the flow simulated concerns mixing-controlled combustion;
☆ the scalar is the "mixture fraction", i.e. the proportion of the local mixture which derives from the fuel-bearing inflow stream;
☆ TMP1C represents the stoichiometric mixture fraction, i.e. at which all the fuel has been oxidised.
The scalar represented by L0CA is C1 for Phase 1 ,and C2 for Phase 2.
5. TMP1=GRND5 selects temperature equal to:
(TMP1A - 0.5 * (U1**2 + V1**2 + W1**2)) /
where TMP1A denotes a constant stagnation enthalpy and the expression containing U1, V1 and W1 is the kinetic energy.
This option is useful for adiabatic high-speed flows, in conjunction with the setting:
RHO1=GRND5, which selects the ideal-gas law.
6. TMP1=GRND6 selects the same formula as GRND5 except that the stagnation-enthalpy field, H1, is used instead of the constant TMP1A.
This is suitable for non-adiabatic flows, in which the stagnation enthalpy is a solved-for variable.
7. TMP1=GRND7 selects temperature equal to:
(enthalpy - TMP2B * Mfuel)/ F(L0CP12+I)
This option is used when the SCRS (i.e. Simple Chemical Reaction System) model of combustion is controlled by mixing (see COMBUSTION).
The mass fractions of fuel, product and oxidant are calculated from algebraic formulae involving the fuel- oxidant mixture fraction (which is solved as a dependent variable). The specific
heat, CP1, should be set to GRND10, to activate the calculation of the specific heat of the fuel-oxidant-product mixture.
8. TMP1=GRND8 uses the same formula as GRND7, except that the mass fraction of fuel is solved as the dependent variable of its own transport equation, as is necessary when the combustion rate is
kinetically controlled.
See REACT for the source options for the Mfuel equation.
9. TMP1=GRND9 activates SXSCRI
10. TMP1=GRND10 does likewise.
4. Setting TMP1 as a 3D-stored variable
When STORE(TMP1) is present in the Q1 file, the user has several ways of allocating it, including:
□ By way of FIINIT(TMP1) and PATCH/INIT settings if the values are to vary only with position;
□ by way of GROUND coding (perhaps PLANT-created) if values are to vary in the course of the flow simulation;
□ or by way of In-form, as described in the next section.
In GROUND, the integer index TEMP1 for phase 1 or TEMP2 for phase 2 should be used in calls to FN routines or as the argument of L0F.
5. The most flexible means of introducing complex dependencies of temperature on enthalpy and other variables is to use the In-Form facilities of PHOENICS, those which are related to properties in
general and TMP1 and TMP2 in particular being described here.
The relevant command in the Q1 file begins:
(PROPERTY TMP1 is .....
and what follows the dots can be as complex as any PHOENICS user is likely to desire.
In-Form can of course be used to specify values of, or formulae for, H0_1 and H0_2; but for these (because these properties are late-comers on the scene), the commands happen to be:
(STORED var H0_1 is ....
(STORED var H0_2 is ....
----- PIL real; default= 0.0; group 9, -
TMP1A....parameter used in formulae for calculating phase-1 temperature. Further parameters of the same kind are: TMP1B, TMP1C.
----- PIL real; default= 0.0; group 9, -
TMP1B....parameter used in formulae for calculating phase-1 temperature. Further parameters of the same kind are: TMP1A, TMP1C.
----- PIL real; default= 0.0; group 9, -
TMP1C....parameter used in formulae for calculating phase-1 temperature. Further parameters of the same kind are: TMP1A, TMP1B.
------ PIL real; default= 0.0; group 9, -
TMP2....is the temperature of the second phase, used similarly to TMP1. See TMP1 for further information, changing 1's to 2's.
For TMP2 = GRND4, the concentration variable is C4
----- PIL real; default= 0.0; group 9, -
TMP2A....parameter used in formulae for calculating phase-2 temperature. Further parameters of the same kind are: TMP2B, TMP2C.
----- PIL real; default= 0.0; group 9, -
TMP2B....parameter used in formulae for calculating phase-2 temperature. Further parameters of the same kind are: TMP2A, TMP2C.
----- PIL real; default= 0.0; group 9, -
TMP2C....parameter used in formulae for calculating phase-2 temperature. Further parameters of the same kind are: TMP2A, TMP2B. | {"url":"http://www.cham.co.uk/phoenics/d_polis/d_enc/tmp1.htm","timestamp":"2014-04-16T13:08:13Z","content_type":null,"content_length":"9891","record_id":"<urn:uuid:de8e7ac7-11ee-45a1-9a8a-59d7b99f4e27>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00234-ip-10-147-4-33.ec2.internal.warc.gz"} |
spectra of VERY sparse random matrices
up vote 2 down vote favorite
Consider an $n\times n$ random binary matrix $M$ with i.i.d. entries $m_{ij} \sim {\rm Bernoulli}(p)$, where $p = n^{-\beta}$ with $\beta \in (1,2)$. I am interested in the behavior of the singular
value decomposition of $$M = \sum_{i=1}^{rank(M)} \sigma_i u_i v_i',$$ where $\sigma_i$ are ranked in decreasing order.
Some intuitive observation (which might NOT be all true!):
1) By subtracting the mean we can write $M = p {\bf 1} {\bf 1}' + A$, where $A$ has indepedent entries with zero mean and variance $p(1-p)$. Therefore I expect the leading singular vector is
approximately parallel to the all-one vector ${\bf 1}$, and the largest singular value is $\sigma_1 \approx p n$. If the SVD of $A$ behaves similarly to that of the usual iid matrices, it is probably
true that the second largest singular value of $M$ (i.e., the largest singular value of $A$) is approximately $\sigma_2 \approx \sqrt{p n}$.
2) [S: $rank(M)$ is pretty small: since $\mathbb{P}(\text{the first}~ m \text{ rows are all zero}) = (1-p)^{n m}\geq 1-pmn$. Therefore $rank(M) \leq n^{\beta-1}$ with high probability. :S] This is
wrong... this only says that $rank(M) \leq n-n^{\beta-1}$.
Are there any rigorous results about the SVD of this matrix ensemble? Is it true that except for $u_1,v_1$ which are approximately $\frac{1}{\sqrt{n}} {\bf 1}$, the remaining singular vectors are
independently and uniformly distributed over $S^{n-1}$?
pr.probability random-matrices
1 Presumably you mean "VERY sparse random matrices" in the title? – Mark Meckes Jan 20 '12 at 18:02
Thanks! Fixed. – mr.gondolier Jan 20 '12 at 18:27
add comment
2 Answers
active oldest votes
Also in the "a little too involved for a comment" class: A matrix that's this sparse is usually going to be a block diagonal matrix with very small blocks.
Let $k$ be any fixed constant, and suppose that your matrix contains no $k+1 \times k+1$ principal submatrix with at least $k$ nonzero entries. Then your matrix has a block decomposition
with all blocks having size at most $k+1$ (If you start with a given nonzero entry and "grow" it by adding entries in the same row/column as an entry already added, you'll stop by the time
up vote you've reached $k-1$ added entries, and each addition increases the size of your block by at most $1$). In this case we can upper bound the probably a submatrix of this size exists by $$\
2 down binom{n}{k+1} \binom{(k+1)^2}{k} p^{k} \leq C_k n^{k+1-k \beta}$$ Which goes to $0$ for any $\beta$ in your range if $k$ sufficiently large.
This means you'll usually see singular vectors with very small support, and that the $\sigma_2$ should be much larger than $\sqrt{np}$ (maybe equal to $\sigma_1$ in most cases?) You might be
able to get something more explicit by enumerating all the possible structures of blocks and the expected number of occurrences of each.
Thanks Kevin. I understand your calculation of the probability but I didn't quite get what you meant by block decomposition. Can you elaborate a bit? – mr.gondolier Jan 21 '12 at 1:09
Also, what do you think about $rank(M)$? Your thesis seems to deal with the case with $p>=\log n / n$, i.e., $\beta \leq 1$ and rank is full whp. Here probably rank is roughly number of
non-zero rows/columns? – mr.gondolier Jan 21 '12 at 1:11
A Block Decomposition here means partitioning the indices from $1$ to $n$ into blocks $A_1, A_2, \dots A_m$ such that $a_{ij}$ is zero if $i$ and $j$ are in different blocks. It
corresponds to an ordering of the indices where the matrix is block diagonal. You can take a set of singular vectors/values coming from the singular vectors/values of each block. – Kevin
P. Costello Jan 21 '12 at 4:05
If $\beta>3/2$ the rank should be equal to the number of nonzero rows/columns, just because with high probability no row/column has two nonzero entries. For smaller $\beta$, I think they
should still be nearly equal, but not exactly. You should be able to bound the difference by something like the number of nonzero entries in the same row/column as another nonzero entry,
which is on the order of $n^{3-2\beta}$, compared to $n^{2-\beta}$ rows with nonzero entries). – Kevin P. Costello Jan 21 '12 at 4:19
But on the other hand there should be pairs of columns each having exactly one nonzero entry both in the same row, and vice versa, leading to dependencies among the nonzero rows/columns. –
Kevin P. Costello Jan 21 '12 at 4:20
add comment
Here's just one quick remark (a little involved for a comment) about how you can start making greater use of your suggested reduction to $A$. Since $M-A$ has rank 1, $$ \sigma_3(A) \le \
sigma_2(M) \le \sigma_1(A). $$ Classical results imply that $\sigma_1(A), \sigma_3(A) \approx \sqrt{pn}$, so $\sigma_2(M) \approx \sqrt{pn}$ too.
up vote 0
down vote For more, this paper by Van Vu probably (I haven't read it yet) has a lot that's relevant to your questions.
could you give a bit more details on the classical result, because here the entrywise distribution of $A$ varies with the dimension, so presumably you need some non-asymptotic bounds
on $\sigma_{\cdot}(A)$ ? – mr.gondolier Jan 20 '12 at 18:29
Ack, that's what I get for writing in a hurry. You're right, classical results don't apply directly. I'll think some more and try to write again. – Mark Meckes Jan 20 '12 at 18:40
add comment
Not the answer you're looking for? Browse other questions tagged pr.probability random-matrices or ask your own question. | {"url":"http://mathoverflow.net/questions/86238/spectra-of-very-sparse-random-matrices","timestamp":"2014-04-21T10:18:27Z","content_type":null,"content_length":"66159","record_id":"<urn:uuid:d168fd50-9dbe-40d9-a86e-f24889da49c2>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00314-ip-10-147-4-33.ec2.internal.warc.gz"} |
Table 4: Minimum number of plots required to detect a 5% difference in increment estimates using different methods at 90% confidence, with 95% power in a paired -test. Numbers in italics are the
standard deviation of the differences between each pair of methods, numbers in normal text are the required numbers of plots. Values in brackets are the number of times that a -test using the
appropriate number of plots (randomly selected) incorrectly suggested that a significant difference of greater than 5% existed, from 1000 iterations. | {"url":"http://www.hindawi.com/journals/ijfr/2012/961576/tab4/","timestamp":"2014-04-17T00:55:14Z","content_type":null,"content_length":"3569","record_id":"<urn:uuid:694dbff7-98b0-4468-a542-7af520dd5dbe>","cc-path":"CC-MAIN-2014-15/segments/1397609526102.3/warc/CC-MAIN-20140416005206-00650-ip-10-147-4-33.ec2.internal.warc.gz"} |
Ferdinand Georg Frobenius
Born: 26 October 1849 in Berlin-Charlottenburg, Prussia (now Germany)
Died: 3 August 1917 in Berlin, Germany
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Georg Frobenius's father was Christian Ferdinand Frobenius, a Protestant parson, and his mother was Christine Elizabeth Friedrich. Georg was born in Charlottenburg which was a district of Berlin
which was not incorporated into the city until 1920. He entered the Joachimsthal Gymnasium in 1860 when he was nearly eleven years old and graduated from the school in 1867. In this same year he went
to the University of Göttingen where he began his university studies but he only studied there for one semester before returning to Berlin.
Back at the University of Berlin he attended lectures by Kronecker, Kummer and Weierstrass. He continued to study there for his doctorate, attending the seminars of Kummer and Weierstrass, and he
received his doctorate (awarded with distinction) in 1870 supervised by Weierstrass. In 1874, after having taught at secondary school level first at the Joachimsthal Gymnasium then at the
Sophienrealschule, he was appointed to the University of Berlin as an extraordinary professor of mathematics.
For the description of Frobenius's career so far, the attentive reader may have noticed that no mention has been made of him receiving his habilitation before being appointed to a teaching position.
This is not an omission, rather it is surprising given the strictness of the German system that this was allowed. Details of this appointment are given in [3] but we should say that it must
ultimately have been made possible due to strong support from Weierstrass who was extremely influential and considered Frobenius one of his most gifted students.
Frobenius was only in Berlin for a year before he went to Zürich to take up an appointment as an ordinary professor at the Eidgenössische Polytechnikum. For seventeen years, between 1875 and 1892,
Frobenius worked in Zürich. He married there and brought up a family and did much important work in widely differing areas of mathematics. We shall discuss some of the topics which he worked on
below, but for the moment we shall continue to describe how Frobenius's career developed.
In the last days of December 1891 Kronecker died and, therefore, his chair in Berlin became vacant. Weierstrass, strongly believing that Frobenius was the right person to keep Berlin in the forefront
of mathematics, used his considerable influence to have Frobenius appointed. However, for reasons which we shall discuss in a moment, Frobenius turned out to be something of a mixed blessing for
mathematics at the University of Berlin.
The positive side of his appointment was undoubtedly his remarkable contributions to the representation theory of groups, in particular his development of character theory, and his position as one of
the leading mathematicians of his day. The negative side came about largely through his personality which is described in [5] as:-
... occasionally choleric, quarrelsome, and given to invectives.
Biermann, in [3], looks more closely at his character (no pun intended!), and how it affected the success of mathematical education at the university. He describes the strained relationships which
developed between Frobenius and his colleagues at Berlin. He had such high standards that in the end these did not serve Berlin well. He [3]:-
... suspected at every opportunity a tendency of the Ministry to lower the standards at the University of Berlin, in the words of Frobenius, to the rank of a technical school ... Even so, Fuchs
and Schwarz yielded to him, and later Schottky, who was indebted to him alone for his call to Berlin. Frobenius was the leading figure, on whom the fortunes of mathematics at Berlin university
rested for 25 years. Of course, it did not escape him, that the number of doctorates, habilitations, and docents slowly but surely fell off, although the number of students increased
considerably. That he could not prevent this, that he could not reach his goal of maintaining unchanged the times of Weierstrass, Kummer and Kronecker also in their external appearances, but to
witness helplessly these developments, was doubly intolerable for him, with his choleric disposition.
We should not be too hard on Frobenius for, as Haubrich explains in [5], Frobenius's attitude was one which was typical of all professors of mathematics at Berlin at this time:-
They all felt deeply obliged to carry on the Prussian neo-humanistic tradition of university research and teaching as they themselves had experienced it as students. This is especially true of
Frobenius. He considered himself to be a scholar whose duty it was to contribute to the knowledge of pure mathematics. Applied mathematics, in his opinion, belonged to the technical colleges.
The view of mathematics at the University of Göttingen was, however, very different. This was a time when there was competition between mathematians in the University of Berlin and in the University
of Göttingen, but it was a competition that Göttingen won, for there mathematics flourished under Klein, much to Frobenius's annoyance. In [3] Biermann writes that:-
The aversion of Frobenius to Klein and S Lie knew no limits ...
Frobenius hated the style of mathematics which Göttingen represented. It was a new approach which represented a marked change from the traditional style of German universities. Frobenius, as we said
above, had extremely traditional views. In a letter to Hurwitz, who was a product of the Göttingen system, he wrote on 3 February 1896 (see [4]):-
If you were emerging from a school, in which one amuses oneself more with rosy images than hard ideas, and if, to my joy, you are also gradually becoming emancipated from that, then old loves
don't rust. Please take this joke facetiously.
One should put the other side of the picture, however, for in [9] Siegel, who knew Frobenius for two years from 1915 when he became a student until Frobenius's death, relates his impression of
Frobenius as having a warm personality and expresses his appreciation of his fast-paced varied and deep lectures. Others would describe his lectures as solid but not stimulating.
To gain an impression of the quality of Frobenius's work before the time of his appointment to Berlin in 1892 we can do no better than to examine the recommendations of Weierstrass and Fuchs when
Frobenius was elected to the Prussian Academy of Sciences in 1892. Fairly extensive quotes from this document, and another similar document from Fuchs and Helmholtz, are given in [4] but we quote a
short extract to show the power, variety and high quality of Frobenius's work in his Zürich years. Weierstrass and Fuchs list 15 topics on which Frobenius had made major contributions:-
1. On the development of analytic functions in series.
2. On the algebraic solution of equations, whose coefficients are rational functions of one variable.
3. The theory of linear differential equations.
4. On Pfaff's problem.
5. Linear forms with integer coefficients.
6. On linear substitutions and bilinear forms...
7. On adjoint linear differential operators...
8. The theory of elliptic and Jacobi functions...
9. On the relations among the 28 double tangents to a plane of degree 4.
10. On Sylow's theorem.
11. On double cosets arising from two finite groups.
12. On Jacobi's covariants...
13. On Jacobi functions in three variables.
14. The theory of biquadratic forms.
15. On the theory of surfaces with a differential parameter.
In his work in group theory, Frobenius combined results from the theory of algebraic equations, geometry, and number theory, which led him to the study of abstract groups. He published Über Gruppen
von vertauschbaren Elementen in 1879 (jointly with Stickelberger, a colleague at Zürich) which looks at permutable elements in groups. This paper also gives a proof of the structure theorem for
finitely generated abelian groups. In 1884 he published his next paper on finite groups in which he proved Sylow's theorems for abstract groups (Sylow had proved his theorem as a result about
permutation groups in his original paper). The proof which Frobenius gives is the one, based on conjugacy classes, still used today in most undergraduate courses.
In his next paper in 1887 Frobenius continued his investigation of conjugacy classes in groups which would prove important in his later work on characters. In the introduction to this paper he
explains how he became interested in abstract groups, and this was through a study of one of Kronecker's papers. It was in the year 1896, however, when Frobenius was professor at Berlin that his
really important work on groups began to appear. In that year he published five papers on group theory and one of them Über die Gruppencharactere on group characters is of fundamental importance. He
wrote in this paper:-
I shall develop the concept [of character for arbitrary finite groups] here in the belief that through its introduction, group theory will be substantially enriched.
This paper on group characters was presented to the Berlin Academy on July 16 1896 and it contains work which Frobenius had undertaken in the preceding few months. In a series of letters to Dedekind,
the first on 12 April 1896, his ideas on group characters quickly developed. Ideas from a paper by Dedekind in 1885 made an important contribution and Frobenius was able to construct a complete set
of representations by complex numbers. It is worth noting, however, that although we think today of Frobenius's paper on group characters as a fundamental work on representations of groups, Frobenius
in fact introduced group characters in this work without any reference to representations. In was not until the following year that representations of groups began to enter the picture, and again it
was a concept due to Frobenius. Hence 1897 is the year in which the representation theory of groups was born.
Over the years 1897-1899 Frobenius published two papers on group representations, one on induced characters, and one on tensor product of characters. In 1898 he introduced the notion of induced
representations and the Frobenius Reciprocity Theorem. It was a burst of activity which set up the foundations of the whole of the machinery of representation theory.
In a letter to Dedekind on 26 April 1896 Frobenius gave the irreducible characters for the alternating groups A[4], A[5], the symmetric groups S[4], S[5] and the group PSL(2,7) of order 168. He
completely determined the characters of symmetric groups in 1900 and of characters of alternating groups in 1901, publishing definitive papers on each. He continued his applications of character
theory in papers of 1900 and 1901 which studied the structure of Frobenius groups.
Only in 1897 did Frobenius learn of Molien's work which he described in a letter to Dedekind as "very beautiful but difficult". He reformulated Molien's work in terms of matrices and then showed that
his characters are the traces of the irreducible representations. This work was published in 1897. Frobenius's character theory was used with great effect by Burnside and was beautifully written up
in Burnside's 1911 edition of his Theory of Groups of Finite Order.
Frobenius had a number of doctoral students who made important contributions to mathematics. These included Edmund Landau who was awarded his doctorate in 1899, Issai Schur who was awarded his
doctorate in 1901, and Robert Remak who was awarded his doctorate in 1910. Frobenius collaborated with Schur in representation theory of groups and character theory of groups. It is certainly to
Frobenius's credit that he so quickly spotted the genius of his student Schur. Frobenius's representation theory for finite groups was later to find important applications in quantum mechanics and
theoretical physics which may not have entirely pleased the man who had such "pure" views about mathematics.
Among the topics which Frobenius studied towards the end of his career were positive and non-negative matrices. He introduced the concept of irreducibility for matrices and the papers which he wrote
containing this theory around 1910 remain today the fundamental results in the discipline. The fact so many of Frobenius's papers read like present day text-books on the topics which he studied is a
clear indication of the importance that his work, in many different areas, has had in shaping the mathematics which is studied today. Having said that, it is also true that he made fundamental
contributions to fields which had already come into existence and he did not introduce any totally new mathematical areas as some of the greatest mathematicians have done.
In [5] Haubrich gives the following overview of Frobenius's work:-
The most striking aspect of his mathematical practice is his extraordinary skill at calculations. In fact, Frobenius tried to solve mathematical problems to a large extent by means of a
calculative, algebraic approach. Even his analytical work was guided by algebraic and linear algebraic methods. For Frobenius, conceptual argumentation played a somewhat secondary role. Although
he argued in a comparatively abstract setting, abstraction was not an end in itself. Its advantages to him seemed to lie primarily in the fact that it can lead to much greater clearness and
Article by: J J O'Connor and E F Robertson
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List of References (11 books/articles) A Quotation
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Maintainer Roman Cheplyaka <roma@ro-che.info>
Properties and tools to construct them.
Basic definitions
data TestResult Source
Inappropriate Inappropriate means that the precondition of ==> was not satisfied
type Depth = IntSource
Maximum depth of generated test values
For data values, it is the depth of nested constructor applications.
For functional values, it is both the depth of nested case analysis and the depth of results.
class Testable a whereSource
Anything of a Testable type can be regarded as a "test"
Testable Bool
Testable Property
(Serial a, Show a, Testable b) => Testable (a -> b)
mkProperty :: (Depth -> [TestCase]) -> PropertySource
A lower-level way to create properties. Use property if possible.
The argument is a function that produces the list of results given the depth of testing.
Constructing tests
(==>) :: Testable a => Bool -> a -> PropertySource
The ==> operator can be used to express a restricting condition under which a property should hold. For example, testing a propositional-logic module (see examples/logical), we might define:
prop_tautEval :: Proposition -> Environment -> Property
prop_tautEval p e =
tautology p ==> eval p e
But here is an alternative definition:
prop_tautEval :: Proposition -> Property
prop_taut p =
tautology p ==> \e -> eval p e
The first definition generates p and e for each test, whereas the second only generates e if the tautology p holds.
The second definition is far better as the test-space is reduced from PE to T'+TE where P, T, T' and E are the numbers of propositions, tautologies, non-tautologies and environments.
exists :: (Show a, Serial a, Testable b) => (a -> b) -> PropertySource
exists p holds iff it is possible to find an argument a (within the depth constraints!) satisfying the predicate p
existsDeeperBy :: (Show a, Serial a, Testable b) => (Depth -> Depth) -> (a -> b) -> PropertySource
The default testing of existentials is bounded by the same depth as their context. This rule has important consequences. Just as a universal property may be satisfied when the depth bound is shallow
but fail when it is deeper, so the reverse may be true for an existential property. So when testing properties involving existentials it may be appropriate to try deeper testing after a shallow
failure. However, sometimes the default same-depth-bound interpretation of existential properties can make testing of a valid property fail at all depths. Here is a contrived but illustrative
prop_append1 :: [Bool] -> [Bool] -> Property
prop_append1 xs ys = exists $ \zs -> zs == xs++ys
existsDeeperBy transforms the depth bound by a given Depth -> Depth function:
prop_append2 :: [Bool] -> [Bool] -> Property
prop_append2 xs ys = existsDeeperBy (*2) $ \zs -> zs == xs++ys
Series- and list-based constructors
Combinators below can be used to explicitly specify the domain of quantification (as Series or lists).
Hopefully, their meaning is evident from their names and types. | {"url":"http://hackage.haskell.org/package/smallcheck-0.6/docs/Test-SmallCheck-Property.html","timestamp":"2014-04-18T14:12:57Z","content_type":null,"content_length":"22871","record_id":"<urn:uuid:b4b3a2c1-af80-4d4d-b1ad-cd28a3e34f59>","cc-path":"CC-MAIN-2014-15/segments/1397609533689.29/warc/CC-MAIN-20140416005213-00139-ip-10-147-4-33.ec2.internal.warc.gz"} |
one time pad breakable debate
04-04-2010 #256
the computer or the program has no way to know if or when the correct data is exposed.
When attempting cryptanalysis against a cipher not equivalent to a one time pad, there are ways to program a computer to detect a plausible ciphertext, e.g., if the cryptanalyst knows the
language of the plaintext and its character encoding, then the computer can be programmed to analyse for such characteristics. With sufficient ciphertext, brute force will then obtain the one
most likely plaintext.
But with a one time pad, there is no one most likely plaintext, regardless of the amount of ciphertext available. With knowledge of the language of the plaintext and its character encoding, the
cryptanalyst can only obtain all such possible plaintexts of the corresponding length that have those characteristics, and all are equally likely.
even with unicity distance considered only the person that encrypted the small data would be the only one to recognize the correct contents instantly. anyone else would have to go on more
external knowledge in order to have an educated guess about the contents of the unencrypted file.
Yes, you do need some external knowledge. But the amount of external knowledge required to work out the plaintext encrypted with a one time pad is the same amount of external knowledge required
to work out the plaintext without the ciphertext. Since the attacker would know the plaintext even without cryptanalysis, one can then reason that the cipher must be unbreakable: to "break" it,
one must know the plaintext, yet that is the very thing that cryptanalysis sets out to obtain from the ciphertext.
Frankly, there is no point in continuing this. Accept that you are wrong. If you refuse, then do as I say: write a paper and publish it. Prove the experts wrong. Even if they do not accept you
now, maybe a later generation of experts will, and you will leave your mark in history, just like Claude Shannon.
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
the computer or the program has no way to know if or when the correct data is exposed. when the computer or program exposes the correct data no matter what it may be the computers or the programs
job is done .... that data is exposed cracked broken unencrypted what ever you want to call it. with one chance or exposure of one key makes no difference to the computer or software. at the
point of exposure of the correct contents no matter if the individual is the attacker or recipient or the one who encrypted the one time pad or doing data recovery it still goes to external
knowledge based data analysis. which may or may not be considered cryptanalysis.
brute force will expose the correct contents by generating the correct key sequence once. if the attacker recognized the correct key it is then broken or cracked with external knowledge.
therefore the correct term is improbable. because there is one chance in 256EEfilelength.
should you get a porn generator i say enjoy it.
you would get the correct corresponding plaintext or other file or whatever was vernamed. even with unicity distance considered only the person that encrypted the small data would be the only one
to recognize the correct contents instantly. anyone else would have to go on more external knowledge in order to have an educated guess about the contents of the unencrypted file.
I really can't stand this debate. Not because you're being so stupid, but because you don't even continue with any points we gave you. Read my post again. Tell me what part you didn't understand
or didn't agree with. If there is no flaw in my post, then I am right per definition and that means that you are wrong.
So find a flaw in my post, please.
Look, this person is an obvious troll and it has been consistently behaving like a troll for years. There is really nothing new here. There was an inkling of hope in the beginnings of this
thread, but quickly it became obvious it's the same old trollish attitude.
The initial question has already been answered, and correctly answered. Even if this person keeps insisting 2 + 2 = 5, their opinion doesn't matter in the presence of all the scientific evidence.
The thread should be closed or this troll will keep abusing the goodwill and patience of anyone involved.
Alternatively. We should stop posting.
The programmer’s wife tells him: “Run to the store and pick up a loaf of bread. If they have eggs, get a dozen.”
The programmer comes home with 12 loaves of bread.
Originally Posted by brewbuck:
Reimplementing a large system in another language to get a 25% performance boost is nonsense. It would be cheaper to just get a computer which is 25% faster.
Even if this person keeps insisting 2 + 2 = 5, their opinion doesn't matter in the presence of all the scientific evidence.
The ministry of love would like to have a word with you
Consider this post signed
Look, this person is an obvious troll and it has been consistently behaving like a troll for years. There is really nothing new here. There was an inkling of hope in the beginnings of this
thread, but quickly it became obvious it's the same old trollish attitude.
The initial question has already been answered, and correctly answered. Even if this person keeps insisting 2 + 2 = 5, their opinion doesn't matter in the presence of all the scientific evidence.
The thread should be closed or this troll will keep abusing the goodwill and patience of anyone involved.
Alternatively. We should stop posting.
Yes, and a creepy one at that. The sap doesn't even deserve the attention, IMO. I say let him be with his delusions...
I don't think the thread should be closed.
So far, the OP's persistence has been matched by the persistence of the responders, which I find amusing.
If people have had enough of the OP, they should stop responding to them, and the thread will drop of the radar.
OS: Linux Mint 13(Maya) LTS 64 bit.
Wow, this was a long read.
MK27, having the user input a text and taking each character mod 2 does NOT give you a perfectly random key suitable for a one-time pad. This has been stated by many people and I am stating it
again. As has been pointed out, it is even biased in the simplest sense as it does not produce equally many 0's and 1's.
A MUCH better solution would be to pass the input text through a hash function. That would be pretty secure, but not theoretically unbreakable.
Last edited by Sang-drax : Tomorrow at 02:21 AM. Reason: Time travelling
Kryptkat hasn't replied for a while... I wonder if he gave up on the attempt to make us finally see the light.
on the news they had a story where the students offered a teacher some tickets if that teacher made a basketball net from the middle of the playing field blindfolded. cwish. the teacher made the
basket blindfolded from the middle of the gymnasium. come to find out that the students never had the tickets that they promised. why ? because the students ignored and dismissed the highly
unlikely possibility that the teacher could actually make the basket from the middle of the gymnasium blindfolded. because the students thought that it is "impossible" to make a basket from the
middle of the gymnasium blindfolded. instead of being "improbable" accounting for the highly unlikely chance that it could be done. there is a video of it on youtube.
if you never purchase a ticket for the lottery this week then it would be "impossible" to win the money prize because there would be "0" chance of winning the lottery prize money. if you purchase
one or more tickets then you would have a chance to win the money prize. however if you purchase "all" possible combinations then you would have a 100% chance of winning the lottery money prize
of the weekly game. it may cost $200,000,000.00 to do so. then it becomes a question of "is it worth it". you would have to have a jack pot of more than $400,000,000.00 before you break even
because half of it would go in taxes.
to "break" it, one must know the plaintext
and when i provide you with this exact situation were the user looses the key and has to brute force it to recover the data you dismiss it as nothing more than a memory test.
i have other programs to work on and other books to read sometimes i do get busy. i still believe what i believe.
i would like to remind you that i am the one that is not dismissing the fact that there is one key with one chance in however many to open the file. does it really matter how that key was
obtained or fabricated ? by saying that it is "impossible" you are saying that even the key will not open the file. you are also depending on the obfuscation factor and ignoring the possibility
that when the data is exposed it can be narrowed down to fewer choices. which for some reason you do not consider that cryptanalysis. the possibility of a lucky guess or an educated guess <with
or with out external knowledge> could get the correct contents of the exposed data is also ignored and dismissed by you. i point that out and you call it troll behavior. what about your behavior
when you run out of facts and figures or some one does not agree with your indoctrinated disposition you go straight for the insults. i remind you i have not resorted to insults and have only
stated what and why i believe what i do. the brute forcers job is to expose that contents of the file then it job is done. broken exposed and it does not really matter if it is in a language you
understand or not because one key fabricated opened it. only if you have something serious to hide or stand to loose a substantial amount of money would you continue with the illusion of
debate closed.
Yes, closed.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
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04-10-2010 #265 | {"url":"http://cboard.cprogramming.com/general-discussions/124640-one-time-pad-breakable-debate-18.html","timestamp":"2014-04-19T10:59:27Z","content_type":null,"content_length":"89066","record_id":"<urn:uuid:b9038ae8-188d-43a6-9122-7ba6ce3ae00d>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00508-ip-10-147-4-33.ec2.internal.warc.gz"} |
Goal Seek to Solve A Problem
Microsoft Excel is jam packed with cool useful features that tend to go unused by most users, simply because they don’t know how to use them, or even that they exist. One of these features is
Excel’s Goal Seek tool.
First, some background on what it does. Goal Seek is essentially the answer to every middle school aged kid’s math test word problem. It is best used when you know the answer to the problem you’re
trying to solve, but don’t know all the inputs.
For example, a simple word problem might be “You have 432 sheets of paper to give out to a class of 36 students. How many sheets of paper does each student get.” Well you know that 36 multiplied by
something will give you 432, but you don’t know what that number is. You could divide 432 by 36 to get the answer, but let’s assume that isn’t possible right now. With Goal Seek, you need several
• An answer expressed as a formula, in this case 432 is equal to 36 times something. Don’t worry, the result of the formula doesn’t have to equal 432 yet. That’s the magic of Goal Seek.
• You need one part of the equation, in this case 36.
To start, type the following into your spreadsheet in the cells indicated.
2 <leave blank>
3 =A1*A2
Next, open Goal Seek (Tools | Goal Seek, or Data tab on the Excel 2007 ribbon | What If Analysis | Goal Seek).
You are now going to be asked for three things.
1. Set Cell: this is the reference to the cell that contains your formula, in which you want Excel to produce your answer of 432. In our example it should be A3. Note: the Set Cell MUST contain a
formula or function.
2. To Value: this is the value you want in your Set Cell (A3). The value here should be 432.
3. By Changing Cell: Since we know that the formula in cell A3 is correct, and the value of 36 in cell A1 is correct, we want Excel to produce the answer of 432 by changing cell A2. Type A2 in this
box. Note: the Changing Cell MUST contain a value.
4. Click OK.
After clicking OK, Excel will attempt to find a solution to the equation. Once it does, it will enter the missing number in cell A2 and your formula should now equal 432.
Obviously this is a basic example, and doesn’t use Goal Seek’s capabilities to their fullest extent. Let’s take another example that looks at a company’s sales by product to get their desired level
of revenue.
XYZ Company makes four different products, Product A, Product B, Product C, and Product D. The company can produce 1,000, 750, 500, and 100 of each product respectively. The products also sell for
$15, $20, $35, and $50 respectively. Their maximum revenue for products produced that month, therefore is as follows:
A B C D
1 Product Production Sales Price Net Revenue
2 Product A 1,000 $15 $15,000
3 Product B 750 $20 $15,000
4 Product C 500 $35 $17,500
5 Product D 100 $50 $5,000
6 Total $52,500
Now say the company feels that their underdog, Product D, could be producing much more. They also want their revenues to be 60,000. How many more of Product D will they need to produce?
To figure this out, run Goal Seek just like before, except this time, the “Set Cell” is cell D6, “To Value” is 60,000, and the “By Changing Cell” is cell B5. After clicking OK, you’ll notice that
Excel has changed the value in cell B5 to 250, indicating that you will need to increase production of Product D to 250 units from 100 in order to have revenues of $60,000, assuming everything else
remains the same. | {"url":"http://excelzoom.com/2009/07/goal-seek-to-solve-a-problem/","timestamp":"2014-04-18T05:54:39Z","content_type":null,"content_length":"31439","record_id":"<urn:uuid:27eab9e1-0cb3-4f7f-a6bf-e5dedf113251>","cc-path":"CC-MAIN-2014-15/segments/1397609532573.41/warc/CC-MAIN-20140416005212-00366-ip-10-147-4-33.ec2.internal.warc.gz"} |
MathGroup Archive: October 2010 [00562]
[Date Index] [Thread Index] [Author Index]
Re: Can one define own indexing function/mapping to be
• To: mathgroup at smc.vnet.net
• Subject: [mg113355] Re: Can one define own indexing function/mapping to be
• From: Leonid Shifrin <lshifr at gmail.com>
• Date: Mon, 25 Oct 2010 06:39:59 -0400 (EDT)
You can certainly implement your own type with the behavior you want. If you
insist on
overloading Part, here is one possible implementation:
With[{id = Unique[],cont = Unique[]},
With[{id = getId[x]},
With[{id = getId[x]},
This will work for both getting and setting parts:
Create new array from list of values
arr = array.new[{1,2,3}]
Ask for the first element
Set first element
Display all elements:
This implementation has problems however, the major one being that we had to
Part by adding UpValues to it. This can not be avoided in this approach (at
least as far as I can see) since the <array> tag is too deep to attach an
UpValue to it for Set[Part[x_array,...],...]. This is pretty bad since Part
is a very common command, and doing this is both dangerous and costly. So, I
provided this implementation just to illustrate how this can be done,
definitely not for production. You will be better off implementing your own
get and set methods and using them instead. The following modification is
much safer (make sure you start this with a fresh kernel so Part has no
residual definitions):
With[{id = Unique[],cont = Unique[]},
With[{id = getId[x]},
With[{id = getId[x]},
Here is how you use it:
arr = array.new[{1,2,3}]
As usual in this OO-like style in Mathematica, you have to do manual "memory
by calling delete[] after the instance is no longer needed, or you will get
memory leaks. You also must be careful not to use the instances which has
been deleted but may still be referenced
somewhere in your program, say kept in a list. Finally, operations on such
arrays will be necessarily slower than on lists say. And you will not be
able to compile the code using these structures, with Compile. This is a
price to pay for this particular form of stateful programming in
Regarding Array: you should realize that Array is just a built-in function
to produce lists of
applied function values, and not some kind of "type". There is no Array type
in Mathematica.
Hope this helps.
On Sun, Oct 24, 2010 at 6:05 AM, Nasser M. Abbasi <nma at 12000.org> wrote:
> I am trying to implement something where it is more natural to use an
> index that starts at 0 instead of the default 1 for vectors and matrices
> as since this will match the problem I am looking at.
> If I write
> u={0,1,2}
> u[[0]]
> This will return the Head of u. So have to write u[[1]] to get the first
> entry. I'd like to write u[[0]] to get the first entry.
> I thought about using Array:
> In[8]:= u = Array[U[#1] & , 4, 0]
> Out[8]= {U[0], U[1], U[2], U[3]}
> But this does not do much, I still have to write u[[1]] to access U[0]
> I could write my own function to implement a 'get' and 'set' operations
> as in:
> In[4]:= get[u_, i_] := u[[i + 1]]
> In[5]:= get[u, 0]
> Out[5]= U[0]
> So, everywhere in the code, where I would have liked to write u[[i]], I
> will replace that by get[u,i], and now my i's will match the textbook
> i's since those are implemented using 0-index arrays.
> In an OO setting, I could have defined my own "[[ ]]" function on object
> of type Array, then I can write more u[[i]], where now u's own function
> "[[ ]]" will be used.
> Other languages also allow one to define an array with different
> starting index than the default.
> Is there a way to do something like this for Array at least. For general
> Lists, I think that might not be possible? It will break everything I
> would imagine.
> I am looking at this report:
> http://library.wolfram.com/conferences/devconf99/lichtblau/
> "Data Structures and Efficient Algorithms in Mathematica",
> But thought to also ask here, to see if there might be a simple trick to
> do this for Array, by unprotecting it, and changing something? and then
> protecting the definition again, before I try to implement a data struct
> with the help of the above article to do what I want.
> thanks
> --Nasser | {"url":"http://forums.wolfram.com/mathgroup/archive/2010/Oct/msg00562.html","timestamp":"2014-04-18T05:49:26Z","content_type":null,"content_length":"30294","record_id":"<urn:uuid:24d7e17d-4491-437c-a8bc-08bbfdc8664e>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00496-ip-10-147-4-33.ec2.internal.warc.gz"} |
Faculty / Organisational entity
558 search hits
On an implementation of standard bases and syzygies in SINGULAR (1999)
Hubert Grassmann Gert-Martin Greuel Bernd Martin W. Neumann Gerhard Pfister W. Pohl Hans Schönemann Thomas Siebert
A Proposal for Syntactic Data Integration for Math Protocols (1999)
Olaf Bachmann Hans Schönemann
The problem of providing connectivity for a collection of applications is largely one of data integration: the communicating parties must agree on thesemantics and syntax of the data being
exchanged. In earlier papers [#!mp:jsc1!#,#!sg:BSG1!#], it was proposed that dictionaries of definitions foroperators, functions, and symbolic constants can effectively address the problem of
semantic data integration. In this paper we extend that earlier work todiscuss the important issues in data integration at the syntactic level and propose a set of solutions that are both
general, supporting a wide range of dataobjects with typing information, and efficient, supporting fast transmission and parsing.
Effective Simplification of CR expressions (1999)
Olaf Bachmann
Chains of Recurrences (CRs) are a tool for expediting the evaluation of elementary expressions over regular grids. CR based evaluations of elementaryexpressions consist of 3 major stages: CR
construction, simplification, and evaluation. This paper addresses CR simplifications. The goal of CRsimplifications is to manipulate a CR such that the resulting expression is more efficiently
to evaluate. We develop CR simplification strategies which takethe computational context of CR evaluations into account. Realizing that it is infeasible to always optimally simplify a CR
expression, we give heuristicstrategies which, in most cases, result in a optimal, or close-to-optimal expressions. The motivations behind our proposed strategies are discussed and theresults are
illustrated by various examples.
MP Prototype Specification (1997)
Olaf Bachmann S. Gray Hans Schönemann
Relating Rewriting Techniques on Monoids and Rings: Congruences on Monoids and Ideals in Monoid Rings (1997)
Klaus Madlener Birgit Reinert
A first explicit connection between finitely presented commutative monoids and ideals in polynomial rings was used 1958 by Emelichev yielding a solution tothe word problem in commutative monoids
by deciding the ideal membership problem. The aim of this paper is to show in a similar fashion how congruenceson monoids and groups can be characterized by ideals in respective monoid and group
rings. These characterizations enable to transfer well known resultsfrom the theory of string rewriting systems for presenting monoids and groups to the algebraic setting of subalgebras and
ideals in monoid respectively grouprings. Moreover, natural one-sided congruences defined by subgroups of a group are connected to one-sided ideals in the respective group ring and hencethe
subgroup problem and the ideal membership problem are directly related. For several classes of finitely presented groups we show explicitly howGröbner basis methods are related to existing
solutions of the subgroup problem by rewriting methods. For the case of general monoids and submonoidsweaker results are presented. In fact it becomes clear that string rewriting methods for
monoids and groups can be lifted in a natural fashion to definereduction relations in monoid and group rings.
Splitting algorithm for vector bundles (1997)
Bernd Martin Thomas Siebert
A new criteria for indecomposability of vector bundles on projective varieties is presented. It is deduced from a new finite algorithm computing direct sumdecompositions of graded modules over
graded algebras. This algorithm applies as well to modules over local complete algebras over a field.
String Rewriting and Gröbner Bases - A General Approach to Monoid and Group Rings (1997)
Klaus Madlener Birgit Reinert
The concept of algebraic simplification is of great importance for the field of symbolic computation in computer algebra. In this paper we review somefundamental concepts concerning reduction
rings in the spirit of Buchberger. The most important properties of reduction rings are presented. Thetechniques for presenting monoids or groups by string rewriting systems are used to define
several types of reduction in monoid and group rings. Gröbnerbases in this setting arise naturally as generalizations of the corresponding known notions in the commutative and some
non-commutative cases. Severalresults on the connection of the word problem and the congruence problem are proven. The concepts of saturation and completion are introduced formonoid rings having
a finite convergent presentation by a semi-Thue system. For certain presentations, including free groups and context-free groups, theexistence of finite Gröbner bases for finitely generated right
ideals is shown and a procedure to compute them is given.
An algorithm for constructing isomorphisms of modules (1997)
Thomas Siebert
This paper is a continuation of a joint paper with B. Martin [MS] dealing with the problem of direct sum decompositions. The techniques of that paper areused to decide wether two modules are
isomorphic or not. An positive answer to this question has many applications - for example for the classification ofmaximal Cohen-Macaulay module over local algebras as well as for the study of
projective modules. Up to now computer algebra is normally dealing withequality of ideals or modules which depends on chosen embeddings. The present algorithm allows to switch to isomorphism
classes which is more natural inthe sense of commutative algebra and algebraic geometry.
Monomial Representations for Gröbner Bases Computations (1998)
Olaf Bachmann Hans Schönemann
Monomial representations and operations for Gröbner bases computations are investigated from an implementation point of view. The technique ofvectorized monomial operations is introduced and it
is shown how it expedites computations of Gröbner bases. Furthermore, a rank-based monomialrepresentation and comparison technique is examined and it is concluded that this technique does not
yield an additional speedup over vectorizedcomparisons. Extensive benchmark tests with the Computer Algebra System SINGULAR are used to evaluate these concepts.
A Note on Nielsen Reduction and Coset Enumeration (1998)
Birgit Reinert Klaus Madlener
Groups can be studied using methods from different fields such as combinatorial group theory or string rewriting. Recently techniques from Gröbner basis theory for free monoid rings
(non-commutative polynomial rings) respectively free group rings have been added to the set of methods due to the fact that monoid and group presentations (in terms of string rewriting systems)
can be linked to special polynomials called binomials. In the same mood, the aim of this paper is to discuss the relation between Nielsen reduced sets of generators and the Todd-Coxeter coset
enumeration procedure on the one side and the Gröbner basis theory for free group rings on the other. While it is well-known that there is a strong relationship between Buchberger's algorithm and
the Knuth-Bendix completion procedure, and there are interpretations of the Todd-Coxeter coset enumeration procedure using the Knuth-Bendix procedure for special cases, our aim is to show how a
verbatim interpretation of the Todd-Coxeter procedure can be obtained by linking recent Gröbner techniques like prefix Gröbner bases and the FGLM algorithm as a tool to study the duality of
ideals. As a side product our procedure computes Nielsen reduced generating sets for subgroups in finitely generated free groups. | {"url":"https://kluedo.ub.uni-kl.de/solrsearch/index/search/searchtype/simple/query/*:*/browsing/true/doctypefq/preprint/start/0/rows/10/institutefq/Fachbereich+Mathematik","timestamp":"2014-04-17T16:19:30Z","content_type":null,"content_length":"49327","record_id":"<urn:uuid:0a7997c0-3dfb-4743-94a1-53742671647a>","cc-path":"CC-MAIN-2014-15/segments/1397609530136.5/warc/CC-MAIN-20140416005210-00481-ip-10-147-4-33.ec2.internal.warc.gz"} |
power series expansion
You haven't done anything that jhevon suggested! He suggested that you write out the power series for $\frac{1}{1- x}$ and differentiate that twice! What is the power series for $\frac{1}{1- x}$?
(Hint: the sum of the geometric series $\sum_{n=0}^\infty ar^n$ is $\frac{a}{1- r}$.)
power series 1/ 1-x = 1 + x +x^2 + x^3 ...... therefore 1/ (1-x)^2 = 1+ 2x +3x^2 +4x^3 ...... now 1/(1-x)^3 = 2 + 6x + 12x^2 ..... ?
the left side of your third equation is incorrect. and when i said differentiate the power series for 1/(1 - x) i was talking about differentiating the expression behind the summation sign in $\sum_
{n = 0}^\infty x^n$, for $|x|<1$, twice or you can follow awkwards suggestion if you learned the negative binomial theorem | {"url":"http://mathhelpforum.com/calculus/123364-power-series-expansion-print.html","timestamp":"2014-04-17T07:41:28Z","content_type":null,"content_length":"11124","record_id":"<urn:uuid:e97cd377-9810-462b-b6c0-af5c5f59d7a3>","cc-path":"CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00474-ip-10-147-4-33.ec2.internal.warc.gz"} |
Surface Area And Volume
Exercise 1
Question: 1. 2 cubes each of volume 64 cm^3 are joined end to end. Find the surface area of the resulting cuboid.
Question: 2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface
area of the vessel.
3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter d of the hemisphere is equal to the edge of the cube. Determine the surface area of the
remaining solid.
6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface
7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find
the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m^2. (Note that the base of the tent will not be covered with canvas.)
8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the
nearest cm^2.
9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the
total surface area of the article. | {"url":"http://www.excellup.com/classten/mathten/areavolumeexone.aspx","timestamp":"2014-04-20T05:50:46Z","content_type":null,"content_length":"32981","record_id":"<urn:uuid:c16e33f9-b763-47a3-ba38-09d54cc1f03a>","cc-path":"CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00620-ip-10-147-4-33.ec2.internal.warc.gz"} |
Derivative of the zeta function at 0
February 24th 2009, 02:11 PM #1
Derivative of the zeta function at 0
I'm not well aquainted with divergent series, so I'm just asking for help to see if this makes any sense at all. I know my final result is right, but the use of indefinite integration with
divergent series leads me to question my answer. I'm pretty sure it's flawed because of that step, but here is the meat of it:
I took the Fourier series of $f(x)=e^x$ on arbitrary bound L, then let $x\to L \to \pi\sqrt{z}$ to get
$\sum_{n=1}^{\infty}\frac{1}{n^2+z}=\frac{e^{\pi\sq rt{z}}(\pi\sqrt{z}-1)+e^{-\pi\sqrt{z}}(\pi\sqrt{z}+1)}{2z(e^{\pi\sqrt{z}}-e^{-\pi\sqrt{z}})}$.
My problem is here I integrate (indefinitely) giving
Then I just figured C = 0 (my main issue) then took the limit as z went to 0 by l'Hopital's,
$\displaystyle\sum_{n=1}^{\infty}\log n = \frac{1}{2}\log 2\pi=-\zeta^{\prime}(0)$ by the analytic continuation of the zeta function's sum definition and its derivative, as $\displaystyle\zeta^{\
prime}(s)=-\sum_{n=1}^{\infty}\frac{\log n}{n^s}$ so of course, ignoring divergence, $\displaystyle\zeta^{\prime}(0)=-\sum_{n=1}^{\infty}\log n.$
This is, quite amazingly, correct (see Wolfram eq #38), but my method of derivation seems very sketchy to me.
My problem is is that that constant C is messing me up in the integration; is there any definite integral with certain bounds of integration that would yield the same argument, except in a (more)
rigorous fashion? I can't think of any. Any ideas would be appreciated, thank you!
Does anyone have any experience with divergent series, and if such integration is justified?
Last edited by mr fantastic; February 26th 2009 at 11:57 AM. Reason: Stopping the OP from shooting his/herself in the foot.
February 26th 2009, 11:56 AM #2 | {"url":"http://mathhelpforum.com/calculus/75570-derivative-zeta-function-0-a.html","timestamp":"2014-04-18T07:53:27Z","content_type":null,"content_length":"34856","record_id":"<urn:uuid:01a32ba9-d4be-483a-8f15-01515cc27679>","cc-path":"CC-MAIN-2014-15/segments/1397609532573.41/warc/CC-MAIN-20140416005212-00654-ip-10-147-4-33.ec2.internal.warc.gz"} |
Existence and Decay Estimate of Global Solutions to Systems of Nonlinear Wave Equations with Damping and Source Terms
Abstract and Applied Analysis
Volume 2013 (2013), Article ID 903625, 9 pages
Research Article
Existence and Decay Estimate of Global Solutions to Systems of Nonlinear Wave Equations with Damping and Source Terms
Department of Mathematics and Information Science, Zhejiang University of Science and Technology, Hangzhou 310023, China
Received 30 April 2013; Revised 1 September 2013; Accepted 2 September 2013
Academic Editor: T. Raja Sekhar
Copyright © 2013 Yaojun Ye. 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.
The initial-boundary value problem for a class of nonlinear wave equations system in bounded domain is studied. The existence of global solutions for this problem is proved by constructing a stable
set and obtain the asymptotic stability of global solutions through the use of a difference inequality.
1. Introduction
In this paper, we are concerned with the global solvability and decay stabilization for the following nonlinear wave equations system: with the initial-boundary value conditions where is a bounded
open domain in with a smooth boundary , , and for and for .
When , Medeiros and Miranda [1] proved the existence and uniqueness of global weak solutions. Cavalcanti et al. in [2–4] considered the asymptotic behavior for wave equation and an analogous
hyperbolic-parabolic system with boundary damping and boundary source term. In paper [5, 6], the authors dealt with the existence, uniform decay rates, and blowup for solutions of systems of
nonlinear wave equations with damping and source terms.
Rammaha and Wilstein [7] and Yang [8] are concerned with the initial boundary value problem for a class of quasilinear evolution equations with nonlinear damping and source terms. Under appropriate
conditions, by a Galerkin approximation scheme combined with the potential well method, they proved the existence and asymptotic behavior of global weak solutions when , where and are, respectively,
the growth orders of the nonlinear strain terms and the source term.
Ono [9] considers the following initial-boundary value problem for nonlinear wave equations with nonlinear dissipative terms: where , , and are constants. The author mainly investigates on the blowup
phenomenon to problem (6). On the other hand, in the case of , he shows that the problem (6) admits a unique global solution, and its energy has some decay properties under some assumptions on and
initial energy . In particular, when and in (6), the energy has some polynomial and exponential decay rates, respectively.
For the following strongly damped nonlinear wave equation Dell’Oro and Pata [10] obtain the long-time behavior of the related solution semigroup, which is shown to possess the global attractor in the
natural weak energy space. In addition, the existence of global and local solutions, decay estimates, and blowup for solutions of nonlinear wave equation with source and damping terms and exponential
nonlinearities are studied in [11–14].
In this paper, we prove the global existence for the problem (1)–(5) by applying the potential well theory introduced by Sattinger [15] and Payne and Sattinger [16]. Meanwhile, we obtain the
asymptotic stabilization of global solutions by using a difference inequality [17].
For simplicity of notations, hereafter we denote by the norm of ; denotes norm, and we write equivalent norm instead of norm . Moreover, denotes various positive constants depending on the known
constants and may be different at each appearance.
2. Local Existence
In this section, we investigate the local existence and uniqueness of the solutions of the problem (1)–(5). For this purpose, we list up two useful lemmas which will be used later and give the
definition of weak solutions.
Lemma 1. Let , then ; and the inequality holds with a constant depending on , , and , provided that , and , .
Lemma 2 (Young inequality). Let and for , ; then one has the inequality where is an arbitrary constant, and is a positive constant depending on .
Definition 3. A pair of functions is said to be a weak solution of (1)–(5) on if , , , , and satisfies for all test functions and for almost all .
The local existence and uniqueness of solutions for problem (1)–(5) can be proved through the use of Galerkin method. The result reads as follows.
Theorem 4 (local solution). Supposed that , , and if and for , then there exists such that the problem (1)–(5) has a unique local solution satisfying where
Proof. Let be a basis for . Supposed that is the subspace of generated by . We are going to look for the approximate solution which satisfies the following Cauchy problem: Note that, we can solve the
problem (14)–(19) by a Picard’s iteration method in ordinary differential equations. Hence, there exists a solution in for some , and we can extend this solution to the whole interval for any given
by making use of the a priori estimates below.
Multiplying (14) by and (15) by and summing over from to , we obtain By summing (20) and (21) and integrating the resulting identity over , we have We estimate the right-hand terms of (22) as
follows: we get from Hölder inequality and Lemmas 1 and 2 that It follows from (22) and (23) that which implies that We get from (25) and Gronwall type inequality that Thus, we deduce from (26) that
there exists a time such that where is a positive constant independent of .
We have from (24) and (26) that It follows from (27) and (28) that Using the same process as the proof of Theorem 2.1 in paper [18], we derive that is a local solution of the problem (1)–(5). By (20)
and (21), we conclude that (11) is valid.
3. Global Existence
In order to state our main results, we first introduce the following functionals: for .
We put that Then, we are able to define the stable set as follows for problem (1)–(5): We denote the total energy related to (1) and (2) by (12), and is the total energy of the initial data.
Lemma 5. Let be a solution to problem (1)–(5); then, is a nonincreasing function for and
We have from (11) that is the primitive of an integrable function. Therefore, is absolutely continuous, and equality (35) is satisfied.
Lemma 6. Supposed that , and if ; if , then .
Proof. Since so we get In case , let , which implies that
As , an elementary calculation shows that . Therefore, we have that It follows from Hölder inequality and Lemma 1 that
We get from (39) and (40) that In case and or , then Therefore, we have We conclude from (41) and (43) that Thus, we complete the proof of Lemma 6.
Lemma 7. Supposed that for and for , if , and , then for .
Proof. Assume that there exists a number such that on and . Then, in virtue of the continuity of , we see , where denotes the boundary of domain . From the definition of and the continuity of and in
, we have either or It follows from (12) and (30) that So, case (45) is impossible.
Assume that (46) holds; then, we get that We obtain from that . Since Consequently, we get from (47) that which contradicts the definition of . Hence, case (46) is impossible as well. Thus we
conclude that on .
Theorem 8 (global solution). Supposed that as and as , and is a local solution of problem (1)–(5) on . If , and , then is a global solution of problem (1)–(5).
Proof. It suffices to show that is bounded uniformly with respect to . Under the hypotheses in Theorem 8, we get from Lemma 7 that on . So the following formula holds on : We have from (51) that
Hence, we get The above inequality and the continuation principle lead to the global existence of the solution for problem (1)–(5).
4. Asymptotic Behavior of Global Solutions
The following lemma plays an important role in studying the decay estimate of global solutions for the problem (1)–(5).
Lemma 9 (see [9]). Suppose that is a nonincreasing nonnegative function on and satisfies Then, has the decay property where are constants and .
Lemma 10. Under the assumptions of Theorem 8, if initial value and are sufficiently small such that then where is a positive constant and is the optimal Sobolev’s constant from to .
Proof. We have from Lemma 1 and (52) that Therefore, we get from (58) and (31) that Let then, we have from (59) that
Theorem 11. Under the assumptions of Theorem 8, if and (56) hold, then the global solution in of the problem (1)–(5) has the following decay property: where is some constant depending only on and .
Proof. Multiplying (1) by and (2) by and integrating over , and summing up together, we get Thus, there exists , such that
On the other hand, we multiply (1) by and (2) by and integrate over . Adding them together, we obtain From (63), Sobolev inequality, and Hölder inequality, we have We get from (52), (64), and Lemmas
1 and 2 that From Hölder inequality and Lemma 2,we get
Since and the property of the function , , , we obtain
We conclude from (69), (70), , and Lemma 1 that It follows from (63), (68), (69), and (71) that and we obtain from (63), Sobolev inequality, Hölder inequality, and Lemma 2 that Similarly, we have the
following formula: We get from (57), (73), and (74) that
Choosing small enough , we have from (65), (66), (67), (72), and (75) that It follows from (30) and (31) that On the other hand, from (12) and using (57) and (77), we deduce that By integrating (78)
over , we obtain For small enough , we have from (76) and (79) that Thus, there exists , such that Multiplying (1) by and (2) by and integrating over , and summing up, we get Therefore, we obtain
from (63), (81), and (82) that Choosing small enough , we have from (83) that
Since and , we get Consequently, Thus, applying Lemma 9 to (86), we get where is some constant depending only on and .
This research was supported by the National Natural Science Foundation of China (no. 61273016), The Natural Science Foundation of Zhejiang Province (no. Y6100016), The Middle-aged and Young Leader in
Zhejiang University of Science and Technology (2008–2012), and the Interdisciplinary Pre-research Project of Zhejiang University of Science and Technology (2010–2012).
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18. E. Piskin and N. Polat, “Global existence, decay and blow-up solutions for coupled nonlinear wave equations with damping and source terms,” Turkish Journal of Mathematics, vol. 30, pp. 1–19, | {"url":"http://www.hindawi.com/journals/aaa/2013/903625/","timestamp":"2014-04-17T19:49:22Z","content_type":null,"content_length":"888646","record_id":"<urn:uuid:924c6ee2-f510-4f5a-95c0-2d2d7a6518f8>","cc-path":"CC-MAIN-2014-15/segments/1397609530895.48/warc/CC-MAIN-20140416005210-00156-ip-10-147-4-33.ec2.internal.warc.gz"} |
Finding net external force
Mass of 2.6kg lies on a frictionless table, pulled by another mass of 4.1kg under the influence of gravity (mass 4.1kg is hanging off the edge of the table). The acceleration due to gravity is 9.8m/s
^2. What is the magnitude of the net external force (gravitational) acting on the two masses?
I thought I'd get the answer by using F=g(m1)(m2), but that was wrong. I don't know where to go from here. | {"url":"http://www.physicsforums.com/showthread.php?t=92426","timestamp":"2014-04-17T21:29:40Z","content_type":null,"content_length":"24468","record_id":"<urn:uuid:a3b8f058-707f-4eb2-8b26-aa897664b1c1>","cc-path":"CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00299-ip-10-147-4-33.ec2.internal.warc.gz"} |
On degenerate integrable hamiltonian systems
up vote 3 down vote favorite
Is there some reference where the existence of local generalized action-angle variables is discussed in some detail for concrete examples of hamiltonian systems of mechanical type?
After Dazord and Delzant, by local generalized action-angle coordinates on a symplectic manifold $(M^{2n},\omega)$ I mean a locally trivial bundle $\pi:M\to P$ with fiber $\mathbb{T}^k$ and having a
trivializing atlas whose elements $(U,\phi:\pi^{-1}(U)\to U\times\mathbb{T}^k)$ satisfy the following property:
$\phi_{\ast}\omega=\sum_{i=1}^k dJ_i\wedge \theta_i+\sum_{i=1}^{n-k}dp_i\wedge dp_i$ where $J_1,\ldots,J_k,p_1,\ldots,p_{n-k},q_1,\ldots,q_{n-k}$ are adapted coordinates on $U$ and $\theta_1\ldots,\
theta_k$ is a base of invariant $1$-forms on $\mathbb{T}^k$.
sg.symplectic-geometry ds.dynamical-systems
1 How about the book "Global Aspects of Classic Integrable Systems" by Cushman and Bates? it is concerned with monodromy and the global existence of action-angle coordinates, so I'm pretty sure it
does a good job of sketching the local picture, although I don't have a copy with me right now. – jvkersch Jun 21 '11 at 22:29
Dear jvkersch, thank you for the reference. Even if in this book there is no mention of degenerate (or super, non commutative) integrability, the detailed analysis of the topology of the momentum
map for many classical ham. systems ( Euler top, kepler problem, harmonic oscillator,...; that, by the way, are superintegrable) should permit to recognize the existence of generalized
action-angle coordinates at least on a open subset of the phase space. – Giuseppe Tortorella Jun 23 '11 at 6:32
add comment
1 Answer
active oldest votes
If I understand correctly what you are after, a good place to have a look at is this expository article by Fasso'
F. Fasso'. Superintegrable Hamiltonian systems: geometry and perturbations. Acta Appl. Math. 87 (2005),
in which he reviews the case of superintegrable Hamiltonian systems defined by Mischenko and Fomenko (this is the mechanical case that you should be after). Some examples are briefly
up vote 2 discussed and there are references to longer expositions (e.g. the Euler-Poinsot top).
down vote
accepted Just a final word of warning. In the completely integrable Hamiltonian systems literature, "degenerate" refers to the case when the underlying system is completely integrable a la
Liouville, but there are singularities which are not, in some sense, of Morse-Bott type. If I am not mistaken, the case that you discuss here is often referred to as "superintegrable"
or "non-commutatively integrable", as defined by Mischenko and Fomenko (and others).
Dear Daniele Sepe, thanks for your reference. In this paper the integrability condition of Mischenko and Fomenko is interpreted as sufficient for the existence of an isotropic and
symplectically complete fibration (FISC after Dazord and Delzant) and hence of generalized action-angle coordinates. I learn also that the first use of the notion of FISC in concrete
examples of mechanical interest is the book Nonlinear Poisson Bracket of Maslov and Karasev, and the other paper of Fassò on Euler-Poinsot that you cite. Thank you. – Giuseppe
Tortorella Jun 23 '11 at 17:26
add comment
Not the answer you're looking for? Browse other questions tagged sg.symplectic-geometry ds.dynamical-systems or ask your own question. | {"url":"http://mathoverflow.net/questions/68418/on-degenerate-integrable-hamiltonian-systems/68491","timestamp":"2014-04-18T00:30:58Z","content_type":null,"content_length":"55829","record_id":"<urn:uuid:165507a2-92dd-4d52-99a4-c681db11fed1>","cc-path":"CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00560-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Numerical Python Basics
Published on ONLamp.com (http://www.onlamp.com/)
See this if you're having trouble printing code examples
Numerical Python Basics
Installing Numerical Python
To use Numerical Python, you'll need to install it first. For the details, see our handy guide with instructions and download locations here.
On its own, Python is a potent tool for mathematics. When extended with the addition of two modules, Numerical Python and the DISLIN data visualization tool, Python becomes a powerhouse for numeric
computing and problem solving. Numerical Python (NumPy) extends the Python language to handle matrices and supports the associated mathematics of linear algebra. Like Python, NumPy is a collaborative
effort. The principal code writer was Jim Hugunin, a student at MIT. Paul Dubois at Lawrence Livermore National Labs and a few other major supporters (Konrad Hinsen and Travis Oliphant) eventually
took on the project.
While NumPy and Python make a good combination, the ability of Python to serve the scientific community is not complete without a data visualization package. An excellent choice for data
visualization is the DISLIN package freely provided to the Python community by Helmut Michels of the Max-Planck Institute. DISLIN is an extensive cross platform (Win32/Linux/Unix) library that can be
accessed in many languages, including Python. With its support for 2D and 3D plotting and basic drawing commands, you can create informative and eye-catching graphics.
Over the next five months I will show you how to use both NumPy and DISLIN in real applications, starting this week with a review of the useful and interesting world of matrices and linear algebra.
You may feel like you are having a flashback to high-school math, but stick with me. It gets interesting.
Introduction to Matrices and Matrix Mathematics
A matrix is a homogenous collection of numbers with a particular shape, like a table of numbers of the same type. The most common matrices are one or two-dimensional. One-dimensional matrices are
also called vectors. A, b, and C are examples of matrices; a and b could also be called vectors.
Although a and b both contain the same elements, they are different in shape. The shape of a matrix is described by the number of rows and columns. The a matrix can be described as either a column
vector or a 3-by-1 matrix, b is a row-vector or a 1-by-3 matrix. The C matrix is and example of a 3-by-3 matrix.
Addition and Subtraction
You add and subtract matrices, as you might expect, element by element - with only a slight twist. The shape of the matrices must be the same. Thus the following are valid operations,
is not.
Matrix Multiplication
Matrix multiplication is a more complicated operation. For addition the matrices have to be the same size, but in multiplication only the inner dimensions have to be the same: the columns of the
first matrix must be the same as the rows of the second matrix. The resulting matrix will be the size of the remaining outer dimensions. As an example, multiplication of a 1-by-3 matrix by a 3-by-1
matrix results in a 1-by-1 matrix (3 being the 'inner' dimension). If a 5-by-2 matrix is multiplied by a 2-by-5 matrix, a 5-by-5 matrix results.
The Formula
The governing formula is:
This formula describes how the rows of matrix a and the columns of matrix b can be multiplied to generate matrix c. The indices in c(i,j) are used to specify the element located at the i^th row and j
^th column of the new c matrix. (The same concept holds for the a and b matrices.) The Greek letter Sigma stands for summation. Think of it as a kind of for loop from k=1 to n where n is the size of
the inner dimension. Add the resulting value of each trip through the loop together.
The following shows the difference between matrices with the same values but different shapes. The first example is a 2-by-1 matrix multiplied by a 1-by-2 matrix resulting in a 2-by-2 matrix. The
second example is a 1-by-2 matrix multiplied by a 2-by-1 matrix resulting in a 1-by-1 matrix (also known as a scalar).
What about division?
Matrix division is undefined; the concept is replaced by the matrix inverse. The inverse of a matrix A is defined by the following equation:
When a matrix and its inverse are multiplied together, the result is the identity matrix or I. The identity matrix is a square matrix where all elements are 0 except along the main diagonal (going
from upper left to lower right). A 3-by-3 identity matrix is shown
While the matrix inverse is a bit different from traditional division, the concept follows directly from scalar mathematics where:
There are couple of limitations: the matrix inverse is only defined for square matrices (same number of rows as columns), and it may not exist for certain sets of elements. This is similar to how
dividing a scalar by zero is undefined. You just can't work out an inverse for every set of numbers. Where the inverse does exist, it is very useful for solving difficult equations, as we will see
later in this tutorial.
NumPy and Python
Enough math review -- let's see how it looks in Numerical Python. The NumPy distribution brings to Python the concept of the multi-array. This is the ability to store sets of homogenous numbers in
shaped containers. Sounds like a matrix, right? The multi-array (or array) can hold numeric values of any type (integers, floating point numbers, and complex values). The Numeric package also
provides extensions to the mathematical functions to cover array arguments as well as scalar. In addition several other modules are provided (LinearAlgebra, FFT, RanLib and Matrix to name a few). We
will use the LinearAlgebra module later to create an inverse.
NumPy provides for operations that extend beyond those defined in traditional matrix mathematics like we just reviewed. Although these operations are nonstandard, they make for more efficient
computation in certain circumstances.
It is important for you to know that NumPy defines the operation of the binary operators '*' for multiplication and '/' for division to operate element-wise as in addition and subtraction (shape
still matters, however). To perform traditional matrix multiplication or to create an inverse you will need to use explicit functions, matrixmultiply() and inverse().
Creation of Matrices / Arrays
To use the features of NumPy you must import it:
>>> from Numeric import *.
Once the module has been imported, you generate an array with the array() function. The array() function takes two arguments, a tuple or list of values and an optional type code, e.g., Float or Int.
The list may be nested to create a multi-dimensional array. The array() function will create an array with the values and type specified. If no type is specified, the type of the array will be
dependent on the type of the elements.
>>> a=array((1,2))
>>> print a
[1 2]
Creates an integer array since the values in the tuple were integers.
>>> b=array((1,2),Float)
>>> print b
[ 1. 2.]
creates a floating-point array using integer arguments. The floating-point type overrides the integer arguments.
>>> c=array(((1,2),(3,4)))
>>> print c
[[1 2]
[3 4]]
creates a multi-dimensional array using a multi-dimensional tuple.
Shape and Reshape
The shape attribute of an array is expressed as a tuple. To see the shape simply type
>>> a=array((1,2))
>>> a.shape
>>> c=array(((1,2),(3,4)))
>>> c.shape
(2, 2)
Note that a has only a single dimension (the shape tuple has only a single value) and that c is multidimensional (two values in the shape tuple). By default, single dimensional arrays have only a
length. The concept of rows and columns is not embodied in the structure of a single dimensional array. The shape of an array, however, can be altered via the reshape() function. The reshape function
takes two arguments: an array and a shape tuple.
>>> a=array((1,2))
>>> a.shape
>>> print a
[1 2]
>>> ma=reshape(a,(1,2))
>>> ma.shape
(1, 2)
>>> print ma
[ [1 2]]
>>> mb=reshape(a,(2,1))
>>> mb.shape
(2, 1)
>>> print mb
The printout changes when you use reshape to give a single-dimensioned array a shape more like a traditional matrix. In the case of the 1-by-2 matrix ma, the extra space shows the multidimensional
nature of the matrix. Contrast the printing of a with ma. See the difference between the original and the reshaped form? The elements are the same, but the shape is different.
Elements are taken row-wise from the source array when we reshape this 2-by-2 array:
>>> c=array(((1,2),(3,4)))
>>> c.shape
(2, 2)
>>> print c
[[1 2]
[3 4]]
>>> d=reshape(c,(1,4))
>>> d.shape
(1, 4)
>>> print d
[ [1 2 3 4]]
Matrix Multiplication
Addition and subtraction work just like you would expect. Give them a try! But let's take a look at the trickier problem of multiplication. If you weren't paying attention to my earlier warning about
multiplication and division, you might wonder what is happening in the next example. We create two arrays with different shapes. Using the default multiplication operation generates results that seem
independent of the shape. Remember, in matrix multiplication shape is important. Only when we use the matrixmultiply() do we get the results we expect.
>>> ma
array([ [1, 2]])
>>> mb
>>> ma*mb
array([[1, 2],
[2, 4]])
>>> mb*ma
array([[1, 2],
[2, 4]])
>>> matrixmultiply(ma,mb)
array([ [5]])
>>> matrixmultiply(mb,ma)
array([[1, 2],
[2, 4]])
The multiplication operation is governed by actions called 'Pseudo Indices.' They really are useful. I will explain them in a later article. For now, stick with the matrixmultiply() function to
achieve the expected results.
The Matrix Inverse
To generate a matrix inverse, you simply use the inverse() function found in the LinearAlgebra module supplied along with the basic NumPy. To gain access to the function, first perform an import.
>>> From LinearAlgebra import *
A simple example shows how to use the function
>>> a=array(((3,2),(2,4)),Float)
>>> print a
[[ 3. 2.]
[ 2. 4.]]
>>> a_inv = inverse(a)
>>> print a_inv
[[ 0.5 -0.25 ]
[-0.25 0.375]]
To check the result, multiply a_inv by a (which should give the identity matrix).
>>> matrixmultiply(a_inv,a)
array([[ 1.00000000e+000, 1.11022302e-016],
[ 0.00000000e+000, 1.00000000e+000]])
The End Game
With these abilities in hand, what can you do? In another flashback to math class, the topic of simultaneous linear equations (also known as solving multiple equations with multiple unknowns) comes
to mind. In the following example, the goal is to find the values x and y that make both equations true (at the same time, of course).
This is the classic problem of finding the place where two lines intersect in the plane. The lines may represent different phenomena, perhaps profit vs. cost in economics or growth vs. decay in
biology. In any case, solving for the intersection can be tedious. Add a few more variables, and it can be very difficult.
You can generate the answer by hand or with a computer. The trick to answering the problem with matrices is to represent the equations as two 2-by-1 matrices that are equal. Note how we use matrix
shape, multiplication, and inverses to find our answer:
This can be rewritten as a product of a 2-by-2 matrix with a 1-by-2 matrix (the unknowns) being equal to a 2-by-1 matrix:
Matrix mathematics has a few more rules than traditional algebra does. One of those is seen in the equation below. In order to solve for the unknowns, z needs to be isolated on the left. Thinking
algebraically, your first instinct might be to divide through by A. But matrix division is not defined! Fortunately, we can use the inverse. Multiply both sides by the inverse (just like you might
multiply both sides by 1/A).
On the left the result is the identity matrix multiplied by z (which is just z). The inverse cancels out matrix A. On the right we are left with the product of A^-1b. Since all values are known on
the right and only unknowns exist on the left, a solution is at hand! These equations are solved with just an inverse and a multiply.
Want to calculate that inverse and multiply by hand? I don't. Here is how you do it with NumPy:
>>> A=array(((3,4),(5,2)))
>>> b=reshape(array((11,9)),(2,1))
Notice how we generate b in one fell swoop using a combination of the array() and reshape() functions.
>>> Ainv = inverse(A)
>>> z=matrixmultiply(Ainv,b)
>>> z
array([[ 1.],
[ 2.]])
which is the answer. Checking the result with the original matrix A,
>>> matrixmultiply(A,z)
array([[ 11.],
[ 9.]])
which shows the right answer! Hopefully the exercise was less painful than solving things by hand. The power of Python/NumPy is only hinted at by this simple problem. The true power is in solving
problems that are larger. If the numbers of unknowns and equations equaled 10, it would be difficult to solve the problem by hand.
I haven't even touched on DISLIN. I am saving that for the next article. Visualization can be very useful. Remember all that graphing of equations? DISLIN can do it for you. In the following plot,
the green and red lines are those of the equations above. They intersect at the point x=1 and y=2, which is exactly the answer we got.
More to Come
In upcoming articles we will explore more useful matrix manipulations with NumPy. We will also graph the results with DISLIN to help us see what it is we are doing. You may find the 2D graph above
uninspiring, but check out this graph of 3D information. This graphs the response of a control system (perhaps the anti-lock braking system in a car) at points of instability. The peaks actually rise
to infinity and were clipped for plotting purposes.
Further Reading
Want to launch ahead with Numerical Python? Are you struggling to remember that math? Hungry for more? While you are waiting for next month's article, here is where you can find out more. Read the
Numerical Python Manual. It was written as a tutorial as well as a reference.
Want to buy some books on linear algebra and matrices? The book Applied Linear Algebra by Ben Noble and James W. Daniel is an excellent read. If your interests lie more in the computational aspect of
matrices, the ever popular Matrix Computations by Gene H. Golub and Charles F. Van Loan should also be on your reading list.
Eric Hagemann specializes in fast algorithms for crunching numbers on all varieties of computers from embedded to mainframe.
Discuss this article in the O'Reilly Network Python Forum.
Return to the Python DevCenter.
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Have a nice day ^_^
Some math stuff I found that I haven't seen before.
A Mathematician, a Biologist and a Physicist are sitting in a
street cafe watching people going in and coming out of the house on
the other side of the street.
First they see two people going into the house. Time passes.
After a while they notice three persons coming out of the house.
The Physicist: “The measurement wasn’t accurate”.
The Biologist’s conclusion: “They have reproduced”.
The Mathematician: “If now exactly 1 person enters the house then
it will be empty again.”
—> http://spot.pcc.edu/~tsieber/bizarre_math_stuff.html
How many mathematicians does it take to screw in a light bulb?
If k mathematicians can change a light bulb and if one more simply watches them do it, then k+1 mathematicians will have changed the lightbulb. Therefore, by induction, for all n in the positive
integers, n mathematicians can change a light bulb.
—> http://spot.pcc.edu/~tsieber/bizarre_math_stuff.html
Word Problem From Hell:
A rope over the top of a fence has the same length on each side and weighs one-third of a pound per foot. On one end hangs a monkey holding a banana, and on the other end a weight equal to the
weight of the monkey. The banana weighs 2 ounces per inch. The length of the rope in feet is the same as the age of the monkey, and the weight of the monkey in ounces is as much as the age of the
monkeys mother. The combined ages of the monkey and its mother are 30 years. One-half the weight of the monkey plus the weight of the banana is one-fourth the sum of the weights of the rope and the
weight. The monkeys mother is one half as old as the monkey will be when it is three times as old as its mother was when she was one half as old as the monkey will be when it is as old as its mother
will be when she is four times as old as the monkey was when it was twice as old as its mother was when she was one third as old as the monkey was when it was as old as its mother was when she was
three times as old as the monkey was when it was one fourth as old as it is now. How long is the banana?
—> http://spot.pcc.edu/~tsieber/bizarre_math_stuff.html
A Brief History of Gravity
It filled Gallileo with mirth
To watch his two stones fall to Earth
“Their rates are the same,”
He gladly proclaimed,
“And quite independent of girth!”
Then Newton declared in due course
His own law of Gravity’s force,
“It goes, I declare,
As the inverted square
Of the distance from object to source.”
Next Einstein revealed his equation
Which succeeds to describe gravitation
As spacetime that’s curved
And it’s this that will serve
As the planets’ unique motivation.
But the end of the story’s not written,
By a new way of thinking we’re smitten.
We twist and we turn
Attempting to learn
The Superstring Theory of Witten.
(via cassidybellmor) (via fuckyeahphysics)
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Hi can somebody please help me see if my answer to the following question is correct
Calculate the area for each of the polygons below. If you do not know an equation to use, divide the polygon into other shapes to determine the area.
I choose A because
((1² * 3 / 4 tan ))* (180/3) = 0.43301270189
1. An equilateral triangle with a side of 1 inch
A 0.43 sq in
B0.56 sq in
C0.89 sq in
D0.23 sq in
E 0.19 sq in
F 0.54 sq in | {"url":"http://www.mathisfunforum.com/post.php?tid=18445&qid=240455","timestamp":"2014-04-16T10:48:56Z","content_type":null,"content_length":"15417","record_id":"<urn:uuid:3a8c9a5c-1741-4a78-a8a3-c3fe4b81ca76>","cc-path":"CC-MAIN-2014-15/segments/1397609523265.25/warc/CC-MAIN-20140416005203-00058-ip-10-147-4-33.ec2.internal.warc.gz"} |
A dynamical quantum simulator
Figure: (left) Schematics of the experiments probing the non-equilibrium dynamics emerging when an initially prepared density wave of ultracold atoms in an optical lattice is subjected to a tunnel
coupling and inter-particle interactions. (right) The experimental data is well reproduced by parameter-free numerical simulations (black line) which however break down for longer evolution times.
Here, the experiment (blue circles) can still deliver reliable results, serving as a quantum simulator of many-body dynamics.
(PhysOrg.com) -- An international collaboration demonstrates the superiority of a dynamical quantum simulator over state-of-the-art numerical calculations.
The key prospect of a quantum simulator is to access new physics that the best known classical algorithms can no longer keep track of. For the first time, a group around Professor Immanuel Bloch (Max
Planck Institute of Quantum Optics and Ludwig-Maximilians-Universität Munich), in collaboration with theoretical physicists from the Ludwig-Maximilians-Universität Munich of the group of Prof. Ulrich
Schollwöck, the Forschungszentrum Jülich, the Institute for Advanced Study Berlin, and the University of Queensland (Australia), has demonstrated this superiority by following the dynamics of a
quantum system of strongly correlated ultracold atoms in an optical lattice. In particular they were able to follow the relaxation of the isolated system which was initialized in a state far from
equilibrium. The experimentally observed dynamics were in excellent agreement with numerical calculations which are available only for short evolution times (Nature Physics, AOP, 19 February 2012).
This demonstrates that many-body systems of ultracold atoms can be used as quantum simulators in a regime which is not accessible for classical computers.
The concept of thermalization and thermal equilibrium is fundamental to a large part of our everyday life. It explains, for example, how hot coffee in a cup cools down to room temperature while also
heating the room a little, and how the motion of the molecules in the coffee that is induced by stirring it with a spoon will dampen out until everything is at rest again. The answer to the same
fundamental question posed in the context of a closed quantum system of interacting particles brought out of equilibrium remains elusive to the present day. The complexity of the underlying quantum
dynamics as well as the possibility of the quantum particles to become entangled with one another makes even sophisticated numerical methods fail in the attempt to address this problem for large
particle numbers and long timescales. Experiments with ultracold rubidium atoms carried out in the group of Professor Immanuel Bloch now allow the scientists to follow the non-equilibrium evolution
of an interacting quantum many-body system on a time-scale much longer than those accessible by exact numerical methods.
In the experiments, an extremely cold gas of rubidium atoms was loaded into an optical lattice: a periodic structure of bright and dark areas, created by the interference of counter-propagating laser
beams. In this structure, the atoms are held in either dark or bright spots, depending on the wavelength of the light, and therefore align themselves in a regular pattern. The use of an additional
light field with twice the spatial period allowed the scientists to pairwise group adjacent lattice sites in an optical superlattice and to further manipulate the regular pattern to obtain a
configuration with alternatingly filled and empty sites along one spatial direction. Starting from this ’density wave’ state far from equilibrium, the atoms are then allowed to tunnel along the same
spatial direction and to collide with one and another, leading to a complex many-body dynamics. After a certain relaxation-time, the system’s properties were read out in terms of local densities,
tunnel currents and nearest-neighbour correlations with the help of the superlattice. These observables were probed for a variety of lattice heights and evolution times, showing a rapid relaxation to
(quasi-) steady state values.
On short timescales, parameter-free numerical simulations carried out by collaborators of several research institutes could track the many-body dynamics and therefore benchmark the experimental
quantum simulation. For long evolution times, however, these classical methods have to fail for the concomitant entanglement growth rendering a classical description infeasible. The experiment, on
the other hand, tracks the evolution well beyond the time scale of theoretical predictions. This demonstrates that this system of ultracold atoms can be used as an efficient simulator for relaxation
physics in many-body systems and is outperforming the best classical simulation so far. Furthermore, the experiment gives insight into quantum mechanical tunnel processes as well as (quasi-) steady
state properties after relaxation. It opens up new avenues in the study of cold atoms in non-equilibrium which leads to a better understanding of fundamental problems in condensed matter physics.
More information: S. Trotzky, et al. Probing the relaxation towards equilibrium in an isolated strongly correlated one-dimensional Bose gas, Nature Physics, AOP, 19. Februar 2012, Doi:10.1038/ | {"url":"http://phys.org/news/2012-03-dynamical-quantum-simulator.html","timestamp":"2014-04-18T20:20:17Z","content_type":null,"content_length":"73449","record_id":"<urn:uuid:9284d7fb-fa9d-4127-9131-a1861b6cbe8a>","cc-path":"CC-MAIN-2014-15/segments/1397609538022.19/warc/CC-MAIN-20140416005218-00092-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Proceedings Abstracts of the Twenty-Third International Joint Conference on Artificial Intelligence
A Generalization of SAT and #SAT for Robust Policy Evaluation / 2583
Erik Zawadzki, André Platzer, Geoffery J. Gordon
Both SAT and #SAT can represent difficult problems in seemingly dissimilar areas such as planning, verification, and probabilistic inference. Here, we examine an expressive new language, #∃SAT, that
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complexity of #P^{NP[1]} , many of the instances can be solved efficiently by noticing and exploiting a particular type of frequent structure. | {"url":"http://ijcai.org/papers13/Abstracts/380.html","timestamp":"2014-04-16T20:51:42Z","content_type":null,"content_length":"1788","record_id":"<urn:uuid:5234d925-9533-43ba-a50e-3bf272fccfd8>","cc-path":"CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00402-ip-10-147-4-33.ec2.internal.warc.gz"} |
A dynamical quantum simulator
Figure: (left) Schematics of the experiments probing the non-equilibrium dynamics emerging when an initially prepared density wave of ultracold atoms in an optical lattice is subjected to a tunnel
coupling and inter-particle interactions. (right) The experimental data is well reproduced by parameter-free numerical simulations (black line) which however break down for longer evolution times.
Here, the experiment (blue circles) can still deliver reliable results, serving as a quantum simulator of many-body dynamics.
(PhysOrg.com) -- An international collaboration demonstrates the superiority of a dynamical quantum simulator over state-of-the-art numerical calculations.
The key prospect of a quantum simulator is to access new physics that the best known classical algorithms can no longer keep track of. For the first time, a group around Professor Immanuel Bloch (Max
Planck Institute of Quantum Optics and Ludwig-Maximilians-Universität Munich), in collaboration with theoretical physicists from the Ludwig-Maximilians-Universität Munich of the group of Prof. Ulrich
Schollwöck, the Forschungszentrum Jülich, the Institute for Advanced Study Berlin, and the University of Queensland (Australia), has demonstrated this superiority by following the dynamics of a
quantum system of strongly correlated ultracold atoms in an optical lattice. In particular they were able to follow the relaxation of the isolated system which was initialized in a state far from
equilibrium. The experimentally observed dynamics were in excellent agreement with numerical calculations which are available only for short evolution times (Nature Physics, AOP, 19 February 2012).
This demonstrates that many-body systems of ultracold atoms can be used as quantum simulators in a regime which is not accessible for classical computers.
The concept of thermalization and thermal equilibrium is fundamental to a large part of our everyday life. It explains, for example, how hot coffee in a cup cools down to room temperature while also
heating the room a little, and how the motion of the molecules in the coffee that is induced by stirring it with a spoon will dampen out until everything is at rest again. The answer to the same
fundamental question posed in the context of a closed quantum system of interacting particles brought out of equilibrium remains elusive to the present day. The complexity of the underlying quantum
dynamics as well as the possibility of the quantum particles to become entangled with one another makes even sophisticated numerical methods fail in the attempt to address this problem for large
particle numbers and long timescales. Experiments with ultracold rubidium atoms carried out in the group of Professor Immanuel Bloch now allow the scientists to follow the non-equilibrium evolution
of an interacting quantum many-body system on a time-scale much longer than those accessible by exact numerical methods.
In the experiments, an extremely cold gas of rubidium atoms was loaded into an optical lattice: a periodic structure of bright and dark areas, created by the interference of counter-propagating laser
beams. In this structure, the atoms are held in either dark or bright spots, depending on the wavelength of the light, and therefore align themselves in a regular pattern. The use of an additional
light field with twice the spatial period allowed the scientists to pairwise group adjacent lattice sites in an optical superlattice and to further manipulate the regular pattern to obtain a
configuration with alternatingly filled and empty sites along one spatial direction. Starting from this ’density wave’ state far from equilibrium, the atoms are then allowed to tunnel along the same
spatial direction and to collide with one and another, leading to a complex many-body dynamics. After a certain relaxation-time, the system’s properties were read out in terms of local densities,
tunnel currents and nearest-neighbour correlations with the help of the superlattice. These observables were probed for a variety of lattice heights and evolution times, showing a rapid relaxation to
(quasi-) steady state values.
On short timescales, parameter-free numerical simulations carried out by collaborators of several research institutes could track the many-body dynamics and therefore benchmark the experimental
quantum simulation. For long evolution times, however, these classical methods have to fail for the concomitant entanglement growth rendering a classical description infeasible. The experiment, on
the other hand, tracks the evolution well beyond the time scale of theoretical predictions. This demonstrates that this system of ultracold atoms can be used as an efficient simulator for relaxation
physics in many-body systems and is outperforming the best classical simulation so far. Furthermore, the experiment gives insight into quantum mechanical tunnel processes as well as (quasi-) steady
state properties after relaxation. It opens up new avenues in the study of cold atoms in non-equilibrium which leads to a better understanding of fundamental problems in condensed matter physics.
More information: S. Trotzky, et al. Probing the relaxation towards equilibrium in an isolated strongly correlated one-dimensional Bose gas, Nature Physics, AOP, 19. Februar 2012, Doi:10.1038/ | {"url":"http://phys.org/news/2012-03-dynamical-quantum-simulator.html","timestamp":"2014-04-18T20:20:17Z","content_type":null,"content_length":"73449","record_id":"<urn:uuid:9284d7fb-fa9d-4127-9131-a1861b6cbe8a>","cc-path":"CC-MAIN-2014-15/segments/1398223204388.12/warc/CC-MAIN-20140423032004-00092-ip-10-147-4-33.ec2.internal.warc.gz"} |
Multiple Comparisons/Post-Hoc Tests
One-Way Within-Subjects (Repeated Measures) ANOVA
One-way within-subjects ANOVA is also known as repeated measures analysis of variance, randomized-blocks one-way analysis of variance, single-factor within-subjects analysis of variance. As the
subjects are exposed to each condition in turn, the measurement of the dependent (response) variable is repeated. The computed test statistic evalustes if there is a significant difference between at
least two of the conditions of repeated measures in a set of k conditions.
One-Way Repeated Measures ANOVA Output
The default output includes the one-way ANOVA table. The additional output depends on selected options, and can include the mean table, a mean bar plot, the results for sphericity test and
correction, and table reports for selected multiple comparisons tests.
The Default Output: One-Way ANOVA Table
This table displays:
• The source of variability: The between-subjects variability (the amount of variability between the mean scores of n subjects); the between-conditions variability (a measure of the variance of the
means of the k conditions); the residual (i.e., chance factor or experimental error beyond the control of a researcher)
• The sum of the squared deviations from the mean (Sum of Squares); the degrees of freedom (df); the Mean Square for each of the variability components
• The test statistics (F) and the p value
• The partial omega squared test result (testing strength of association)
Sphericity Evaluation (Locally Best Invariant Test)
Mean Bar Plot
Sphericity implies that the variances of the differences between the repeated measurements should be approximately the same. However, within subjects analysis
of variance is very sensitive to violations of the sphericity assumption. Therefore, sphericity tests and corrections are always provided with the repeated The error bars displayed on the
measures ANOVA. mean bar plot can represent:
• Comparing the power of different tests for sphericity, the Locally Best Invariant Test is recommended as having substantial power to detect departures from • Standard error of mean
sphericity for both small and large samples and as providing good control of the Type I error. • Standard Deviation
• Aabel uses this test to evaluate whether or not the sphericity is tenable. • Confidence Interval (including
options of 90.0%, 95.0%, 97.5%,
Greenhouse & Geisser and Huynd & Feldt Corrections 99.0%)
• For sphericity corrections, Aabel provides Greenhouse & Geisser and Huynd & Feldt methods.
Multiple Comparisons/Post-Hoc Tests
Simple comparisons (also know as pair-wise comparisons) accompanying one-way within-subjects ANOVA include:
• Tukey's HSD test (see the example above)
• Tukey B test on ordered means
• Fisher's LSD test
• The Newman-Keuls (Neuman-Keuls) test on ordered means
• Tukey Kramer test
• Scheffé test
• Bonferroni-Dunn test
The example below is the Tukey's HSD test result for the one-way within-subjects ANOVA example illustrated above. For more information regarding multiple comparisons, click here.
Supported Worksheet Layout
For this ANOVA design, Aabel supports two different worksheet layouts.
• The layout I allows storing the design experimental scores (response data) from all of the k >= 3 conditions of repeated measures in a single numeric column, and uses the levels of the design
single factor (A) to split the data accordingly. In this layout, a subject occurs in multiple worksheet rows: hence, it is required to have a data column (numeric or categorical) with the subject
• The layout II requires storing the design experimental scores (response data) from each condition of repeated measures in a separate numeric column, i.e., each column represents one of the levels
of the design (within-subjects) single factor. | {"url":"http://www.gigawiz.com/anova3.html","timestamp":"2014-04-18T18:10:40Z","content_type":null,"content_length":"9164","record_id":"<urn:uuid:43bdd8bc-2a77-442e-8fa8-e4ef54658e7b>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00251-ip-10-147-4-33.ec2.internal.warc.gz"} |
when is the power of a nonnegative polynomial a sum of squares?
up vote 24 down vote favorite
There are polynomials that are not sum of squares. For example Motzkin gave the example $x^4y^2+x^2y^4+z^6-3x^2y^2z^2$ in 1967.
Is there a real polynomial $f\in{\mathbb{R}}[x_1,\ldots,x_n]$ in several indeterminates that is not a sum of squares but $f^N$ is a sum of squares for some odd integer $N>0$?
This question is interesting in the following sense. The notion of writing nonnegative polynomials $f$ as a sum of squares is to give an algebraic proof of the inequality $f\ge 0$. As per Motzkin's
example, we know that this is not always possible. One way to resolve this is to follow Artin and use denominators. Another way (which I learnt from D'Angelo) is to show that $f^{N}$ is a sum of
squares for some odd $N$.
This question is me wondering whether such a technique of consider the radical of sum of squares is vacuous.
ra.rings-and-algebras ag.algebraic-geometry polynomials
1 Does Motzkin's proof that his f is not a sum of squares also show that f^N is not a sum of squares for odd N? Or could his polynomial itself already a potential positive answer to your question? –
JSE Jan 27 '11 at 4:39
Hi JSE. The proof of that Motzkin's polynomial is not a sum of squares that I know is given by Reznick. This proof is almost brute force. Since the given polynomial is of low degree (deg 6) so the
brute force method works. However, raising to high power makes such brute force method difficult. – Colin Tan Jan 27 '11 at 4:43
this still leaves open as to whether a high power of Motzkin's power is a potential answer to my question. – Colin Tan Jan 27 '11 at 4:58
2 In the first sentence of your question you forgot the attribute "non-negative". – Andrei Moroianu Jan 27 '11 at 8:55
I've changed the title. thanks andrei – Colin Tan Jan 28 '11 at 3:27
add comment
3 Answers
active oldest votes
Motzkin's original proof shows that $x^4y^2 + x^2y^4 + z^6 - a x^2y^2z^2$ is psd and not sos for any $a$ in the interval $(0,3]$. If you take $a = .02$ say, it is reasonably simple,
though messy, to show that $(x^4y^2 + x^2y^4 + z^6 - .02x^2y^2z^2)^3$ is a sum of squares; in fact, it's a sum of binomial squares $(x^b y^c z^d - x^e y^f z^g)^2$, where $b+c+d=e+f+g=9$.
up vote 14 The idea is to look at any monomial with a negative coefficient and make it into the middle term of this square, in a way that the other two terms are still in the Newton polytope. For
down vote example, one term in the given cube is $-.06x^10y^6z^2$, which is "handled" by $.03(x^6y^3 - x^4y^3z^2)^2$. It's sort of messy to work out, but I've convinced myself (at least) that it's
accepted true.
thanks Bruce for your answer! – Colin Tan Jan 29 '11 at 6:56
add comment
Here's an explicit example. The polynomial $f=x^{4} y^{2}+x^{2} y^{4}-x^{2} y^{2}+1$ is not a sum of squares (as one can check using Motzkin's original proof or by computer). On the other
hand, the polynomial $f^3$ can be written as a sum of squares, $$f^3=c_1F_1^2+c_2F_2^2+\ldots+c_{19}F_{19}^2$$ where the coefficients $c_i$ and polynomials $F_i$ are listed below.
I guess I should mention the software I used for computing this, namely the package "SOS.m2" for Macaulay2. This package has a function 'getSOS' which spits out a sum of squares
representation of a given polynomial. See this link for details. The point is that the problem of finding such a representation can be viewed as a problem of semi-definite programming, and
can be solved in reasonable time if the degree is small. In particular, this gives the algorithm you mention for checking whether a polynomial is non-negative.
EDIT: If anyone is interested, I have uploaded the Macaulay2 code here.
Now for the coefficients $c_i$:
And the polynomials $F_i$:
up vote 16
down vote (F_1,...,F_19)=(-459/3650 x^4 y^4-1071/3796 x^4 y^2-1071/3796 x^2
y^4+x^2 y^2-17/73,-17/73 x^6 y^3-1071/3796 x^4 y^5+x^4
y^3-459/3650 x^2 y^3-1071/3796 x^2 y,-1071/3796 x^5 y^4-17/73
x^3 y^6+x^3 y^4-459/3650 x^3 y^2-1071/3796 x
y^2,-65670975/137237294 x^5 y^4+8569925/68618647 x^3 y^6+x^3
y^2-65670975/137237294 x y^2,8569925/68618647 x^6
y^3-65670975/137237294 x^4 y^5+x^2 y^3-65670975/137237294 x^2
y,x^4 y^4-65670975/137237294 x^4 y^2-65670975/137237294 x^2
y^4+8569925/68618647,-175/629 x^5 y^3-175/629 x^3 y^5+x^3
y^3-175/629 x y,x^4 y^2-421805182124/9238122086595 x^2
y^4-1201063431945/17632633808942,-80070895463/1231749611546 x^6
y^3-421805182124/9238122086595 x^4 y^5+x^2
y,-421805182124/9238122086595 x^5 y^4-80070895463/1231749611546 x^3
y^6+x y^2,x^5 y^4-1201063431945/17632633808942 x^3
y^6,-1201063431945/17632633808942 x^6 y^3+x^4 y^5,-21157/107159
x^5 y^3-21157/107159 x^3 y^5+x y,-21157/86002 x^5 y^3+x^3
y^5,x^6 y^3,1,x^5 y^3,x^3 y^6)
thanks JC for your explicit example. I really should learn to use scientific computing. – Colin Tan Feb 1 '11 at 12:24
add comment
This doesn't answer your question but it's more of a comment. In the paper "Integral solution of Hilbert's seventeenth problem", Gilbert Stengle gives an example of a positive semidefinite
form no odd power of which is a sum of squares. His examples are of the form $$x^{2k+1}z^{2k+1}+(z^{2k-1}y^2-xz^{2k}-x^{2k+1})^2$$ In the same paper it is proven that for ever positive
semidefinite form $F$ there is a polynomial $\phi$ of odd degree, with coefficients which are sums of squares, that satisfies $\phi(-F)=0$. Now to every $F$ one can assign a number $\nu(F)$
up vote which is the lowest possible degree of such a $\phi$. It is then calculated that $$\nu(x^2y^4+y^2z^4+z^2x^4-3x^2y^2z^2)=\nu(x^4y^2+x^2y^4+z^6-3x^2y^2z^2)=3.$$ In the end he poses the problem
8 down of whether one can have $\phi (u)=u^{\nu(F)}+\sigma$ (which coincides with the question you ask), or for example, if there can exist a form which is not a sum of squares but the cube of it
vote is. Judging by the papers citing the one above, it seems like the question is still open.
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Not the answer you're looking for? Browse other questions tagged ra.rings-and-algebras ag.algebraic-geometry polynomials or ask your own question. | {"url":"http://mathoverflow.net/questions/53047/when-is-the-power-of-a-nonnegative-polynomial-a-sum-of-squares/53892","timestamp":"2014-04-19T17:31:15Z","content_type":null,"content_length":"70551","record_id":"<urn:uuid:c818dc86-3614-42b0-8845-dc7d509b4b15>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00411-ip-10-147-4-33.ec2.internal.warc.gz"} |
String containing the positional representation of value
Hi! Anyone can help me with that?
An integer can be expressed in a positional representation with different radices. The most common radix is R = 10; other frequently used radices are 2, 8, 16. Any integer larger than 1 can be used
as a radix, but with very large radices we may run out of symbols to represent digits (by convention, if digits 0..9 are insufficient, e.g. in radix 16, lower−case English letters are used to express
larger digits: e.g. 'a' denotes a digit of value 10, 'f' denotes a digit of value 15).
I need a function :
function my_func($V,$R);
that returns a string containing the positional representation of the given value V in the given radix R. The representation should be big-endian; i.e. the first character of the string should
correspond to the most significant digit of the representation.
For example, if V = 17 and R = 7, the function should return the string "23", because this is the representation of the number 17 in radix 7. If V = 62 and R = 21, the function should return "2k",
because this is the representation of the number 62 in radix 21 (note that digit 'k' denotes value 20).
V is an integer within the range [0..200,000,000];
R is an integer within the range [2..36].
expected worst-case time complexity is O(log(V)/log(R));
expected worst-case space complexity is O(log(V)/log(R)).
Re: String containing the positional representation of value
Sounds fun. If you can persuade me this isn't homework, then I'd be happy to help.
- A
PS It was the 'O(log(V)/log(R))' that gave you away. Real programmers know it's O(ln(R)/ln(V)) (assuming bivalent flux capacitance) | {"url":"http://www.php-forum.com/phpforum/viewtopic.php?p=4404626","timestamp":"2014-04-19T01:54:15Z","content_type":null,"content_length":"22474","record_id":"<urn:uuid:230f2df4-346e-417e-81dd-ed09a45b20f2>","cc-path":"CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00168-ip-10-147-4-33.ec2.internal.warc.gz"} |
Some basic combinatorics problems
November 27th 2009, 06:35 AM #1
Senior Member
Apr 2009
Some basic combinatorics problems
(I'm not sure if this is in the right forum... since there is no forum that says "combinatorics")
I've just started combinatorics (basic) and just some beginner questions I can't do, if anyone could help with them it would be very much appreciated :lol: [Note: if it's possible please don't
leave out any necessary steps or procedures I'm a beginner at these and pretty n00b haha]
1. Prove $\left(^n_0\right)+\left(^n_1\right)+\left(^n_2\right)+...+\left(^n_n\right) = 2^n$
2. Which are there more of among the natural numbers between $1$ and $10^6$: Numbers that can be represented as a sum of a perfect square and a (positive) perfect cube or numbers that can not be?
3. Two of the squares of a $7 \times 7$ checkerboard are painted yellow and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a
rotation in the plane of the board. How many inequivalent color schemes are possible?
Last edited by usagi_killer; November 27th 2009 at 07:17 AM.
For the first, consider Newton's binomial expansion:
$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n}b^{n-k}$
What happens if we let a=b=1?
Question 2
Up to 10^6, there are 10^3 squares. Similarly, there are 10^2 cubes. Optimistically assuming there are no duplicates, there are 10^5 = 10^3 * 10^2 possible pairs, leaving 9 x 10^5 unaccounted for
numbers in 10^6.
November 27th 2009, 07:46 AM #2
Super Member
Aug 2009
November 27th 2009, 10:19 AM #3
Senior Member
Nov 2009 | {"url":"http://mathhelpforum.com/advanced-algebra/117005-some-basic-combinatorics-problems.html","timestamp":"2014-04-17T19:16:45Z","content_type":null,"content_length":"35833","record_id":"<urn:uuid:b7cc8366-93ae-4f65-90fb-a20d754667b2>","cc-path":"CC-MAIN-2014-15/segments/1397609530895.48/warc/CC-MAIN-20140416005210-00151-ip-10-147-4-33.ec2.internal.warc.gz"} |
More NP-Complete Problems
NP-complete Problems
One of the claims made in the last section was that there are lots and lots of
NP-complete problems which are of interest to the practical computer scientist. Now it is time to fulfill this prophecy and demonstrate this. We shall examine some of the popular NP-complete problems
from various computational areas.
Logicians should be quite pleased that satisfiability for the propositional calculus is
NP-complete. It means that they will still be needed to prove theorems since it seems unlikely that anyone will develop a computer program to do so. But we, as computer scientists need to see
problems which are closer to home. This is also more than a theoretical exercise because we know that any problem which is NP-complete is a candidate for approximation since no subexponential time
bounded algorithms are known for these problems.
First, we shall review the process of proving a problem
NP-complete. We could do it from scratch like we did for SAT. But that is far too time consuming, especially when we have a nifty technique like reduction. All we need to do is:
a. show that the problem is in
reduce an NP-complete problem to it, and
show that the reduction is a polynomial time function.
That’s not too bad at all. All we basically must accomplish is to transform an
NP-complete problem to a new one. As a first example, let us simplify satisfiability by specifying exactly how many literals must be in each clause. Then we shall reduce this problem to others.
Satisfiability with 3 literals per clause
(3-SAT). Given a finite set of clauses, each containing exactly three literals, is there some truth assignment for the variables which satisfies all of the clauses?
Theorem 1. 3-SAT is NP-complete.
Proof. We know that since 3-SAT is merely a special case of SAT, it must be in
NP. (That is, we can verify that a truth assignment satisfies all of the clauses as fast as we can read the clauses.)
To show that it 3-SAT hard for
NP, we will reduce SAT to it by transforming any instance of the satisfiability problem to an instance of 3-SAT. This means we must demonstrate how to convert clauses which do not contain exactly
three literals into ones which do. It is easy if a clause contains two literals. Let us take (x[1], x[2]) as an example. This is equivalent to the pair:
where u is a new variable. Note that each clause of the pair contains exactly three literals and that.
So far, so good. Now we will transform clauses such as (x) which contain one literal. This will require two steps. We begin by converting it to the pair of two literal clauses:
much as before. Then we change each of these just as before and get:
This was easy. (But you’d better plug in all possible truth values for the literals and fully check it out.)
One case remains. We might have a clause such as (
x[1], ... , x[k]) which contains more than three literals. We shall arrange these literals as a cascade of three literal clauses. Consider the sequence of clauses:
Let us look at this. If the original clause were satisfiable then one of the
x[i]'s had to be true. Let us set all of the u[i]'s to true up to the point in the sequence where x[i] was encountered and false thereafter. A little thought convinces us that this works just fine
since it provides a truth assignment which satisfies the collection of clauses. So, if the original clause was satisfiable, this collection is satisfiable too.
Now for the other part of the proof. Suppose the original clause is not satisfiable. This means that all of the
x[i]'s are false. We claim that in this case the collection of clauses we constructed is unsatisfiable also. Assume that there is some way to satisfy the sequence of clauses. For it to be
satisfiable, the last clause must be satisfiable. For the last clause to be satified, u[k-3] must be false since x[k-1] and x[k] are false. This in turn forces u[k-4] to be false. Thus all of the u
[i]'s all the way down the line have got to be false. And when we reach the first clause we are in big trouble since u[1] is false. So, if the x[i]'s are all false there is nothing we can do with the
truth values for the u[i]'s that satisfies all of the clauses.
Note that the above transformation is indeed a polynomial time mapping. Thus SAT £
[p] 3-SAT and we are done.
One of the reasons that showing that 3-SAT is
NP-complete is not too difficult is that it is a restricted version of the satisfiability problem. This allowed us to merely modify a group of clauses when we did the reduction. In the future we
shall use 3-SAT in reductions and be very pleased with the fact that having only three literals per clause makes our proofs less cumbersome.
Of course having only two literals per clause would be better yet. But attempting to change clauses with three literals into equivalent two literal clauses is very difficult. Try this. I'll bet you
cannot do it. One reason is because 2-SAT is in
P. In fact, if you could reduce 3-SAT to 2-SAT by translating clauses with three literals into clauses with two literals, you would have shown that P = NP.
Let us return to introducing more
NP-complete problems. We immediately use 3-SAT for the reduction to our next NP-complete problem which comes from the field of mathematical programming and operations research. It is a variant of
integer programming.
0-1 Integer Programming
(0-1 INT). Given a matrix A and a vector b, is there a vector x with values from {0, 1} such that Ax ³ b?
If we did not require the vector x to have integer values, then this is the linear programming problem and is solvable in polynomial time. This one is more difficult.
2. 0-1 INT is NP-complete.
Proof. As usual it is easy to show that 0-1 INT is in
NP. Just guess the values in x and multiply it out. (The exact degree of the polynomial in the time bound is left as an exercise.)
A reduction from 3-SAT finishes the proof. In order to develop the mapping from clauses to a matrix we must change a problem in logic into an exercise in arithmetic. Examine the following chart. It
is just a spreadsheet with values for the variables
x[1], x[2], and x[3] and values for some expressions formed from them.
│ Expressions │ Values │
│ X[1] │ 0 │ 0 │ 0 │ 0 │ 1 │ 1 │ 1 │ 1 │
│ X[2] │ 0 │ 0 │ 1 │ 1 │ 0 │ 0 │ 1 │ 1 │
│ X[3] │ 0 │ 1 │ 0 │ 1 │ 0 │ 1 │ 0 │ 1 │
│ + X[1] + X[2] + X[3] │ 0 │ 1 │ 1 │ 2 │ 1 │ 2 │ 2 │ 3 │
│ + X[1] + X[2] - X[3] │ 0 │ -1 │ 1 │ 0 │ 1 │ 0 │ 2 │ 1 │
│ + X[1] - X[2] - X[3] │ 0 │ -1 │ -1 │ -2 │ 1 │ 0 │ 0 │ -1 │
│ - X[1] - X[2] - X[3] │ 0 │ -1 │ -1 │ -2 │ -1 │ -2 │ -2 │ -3 │
Above is a table of values for arithmetic expressions. Now we shall interpret the expressions in a logical framework. Let the plus signs mean true and the minus signs mean false. Place or's
between the variables. So,
+x[1] + x[2] - x[3] now means that
x[1] is true, or x[2] is true, or x[3] is false.
If 1 denotes true and 0 means false, then we could read the expression as
x[1]=1 or x[2]=1 or x[3]=0.
Now note that in each row headed by an arithmetic expression there is a minimum value and it occurs exactly once. Find exactly which column contains this minimum value. The first expression row has a
zero in the column where each
x[i] is also zero. Look at the expression. Recall that +x[1] + x[2] + x[3] means that at least one of the x[i] should have the value 1. So, the minimum value occurs when the expression is not
Look at the row headed by
+x[1] - x[2] - x[3] . This expression means that x[1] should be a 1 or one of the others should be 0. In the column containing the minimum value this is again not the case.
The points to remember now for each expression row are:
a) Each has exactly one column of minimum value.
b) This column corresponds to a nonsatisfying truth assignment.
c) Every other column satisfies the expression.
d) All other columnms have higher values.
Here is how we build a matrix from a set of clauses. First let the columns of the matrix correspond to the variables from the clauses. The rows of the matrix represent the clauses - one row for each
one. For each clause, put a 1 under each variable which is not complemented and a -1 under those that are. Fill in the rest of the row with zeros. Or we could say:
The vector b is merely made up of the appropriate minimum values plus one from the above chart. In other words:
= 1 - (the number of complemented variables in clause i).
The above chart provides the needed ammunition for the proof that our construction is correct. The proper vector x is merely the truth assignment to the variables which satisfies all of the clauses.
If there is such a truth assignment then each value in the vector Ax will indeed be greater than the minimum value in the appropriate chart column.
If a 0-1 valued vector x does exist such that Ax ³ b, then it from the chart we can easily see that it is a truth assignment for the variables which satisfies each and every clause. If not, then one
of the values of the Ax vector will always be less than the corresponding value in b. This means that the that at least one clause is not satisfied for any truth assignment.
Here is a quick example. If we have the three clauses:
then according to the above algorithm we build A and b as follows.
Note that everything comes out fine if the proper values for the
x[i] are put in place. If x[3] is 0 then the first entry of Ax cannot come out less than 0 nor can the second ever be below -1. And if either x[2] or x[1] is 1 then the third entry will be at least
Problems in graph theory are always interesting, and seem to pop up in lots of application areas in computing. So let us move to graph theory for our next problem.
. Given a graph and an integer k, are there k vertices in the graph which are all adjacent to each other?
This does not sound like a very practical problem, does it? Interesting, yes, but practical? Consider this. Suppose that you had a graph whose nodes were wires on a silicon chip. And there was an
edge between any two nodes whose wires might overlap if placed on the same horizontal coordinate of the chip. Finding the cliques tells the designer how much horizontal room is needed to route all of
the wires.
3. CLIQUE is NP-complete.
Proof. Again, it is easy to verify that a graph has a clique of size k if we guess the vertices forming the clique. We merely examine the edges. This can be done in polynomial time.
We shall now reduce 3-SAT to CLIQUE. We are given a set of k clauses and must build a graph which has a clique if and only if the clauses are satisfiable. The literals from the clauses become the
graph’s vertices. And collections of true literals shall make up the clique in the graph we build. Then a truth assignment which makes at least one literal true per clause will force a clique of size
k to appear in the graph. And, if no truth assignment satisfies all of the clauses, there will not be a clique of size k in the graph.
To do this, let every literal in every clause be a vertex of the graph we are building. We wish to be able to connect true literals, but not two from the same clause. And two which are complements
cannot both be true at once. So, connect all of the literals which are not in the same clause and are not complements of each other. We are building the graph G = (V, E) where:
V = {<x, i> | x is in the i-th clause}
E = {(<x, i>,<y, j>) | x ¹ ¹ j}
Now we shall claim that if there were k clauses and there is some truth assignment to the variables which satisfies them, then there is a clique of size k in our graph. If the clauses are satisfiable
then one literal from each clause is true. That is the clique. Why? Because a collection of literals (one from each clause) which are all true cannot contain a literal and its complement. And they
are all connected by edges because we connected literals not in the same clause (except for complements).
On the other hand, suppose that there is a clique of size k in the graph. These k vertices must have come from different clauses since no two literals from the same clause are connected. And, no
literal and its complement are in the clique, so setting the truth assignment to make the literals in the clique true provides satisfaction.
A small inspection reveals that the above transformation can indeed be carried out in polynomial time. (The degree will again be left as an exercise.) Thus the CLIQUE problem has been shown to be
NP-hard just as we wished.
One of the neat things about graph problems is that asking a question about a graph is often equivalent to asking quite a different one about the graph's complement. Such is the case for the clique
problem. Consider the next problem which inquires as to how many vertices must be in any set which is connected to or covers all of the edges.
Vertex Cover
(VC). Given a graph and an integer k, is there a collection of k vertices such that each edge is connected to one of the vertices in the collection?
It turns out that if a graph with n vertices contains a clique consisting of k vertices then the size of the vertex cover of the graph's complement is exactly n-k. Convenient. For an example of this,
examine the graphs in figure 1. Note that there is a 4-clique (consisting of vertices a, b, d, and f) in the graph on the left. Note also that the vertices not in this clique (namely c and e) do form
a cover for the complement of this graph (which appears on the right).
Since the proof of VC's
NP-completeness depends upon proving the relationship between CLIQUE and VC, we shall leave it as an exercise and just state the theorem.
4. VC is NP-complete.
Figure 1 - A graph and its complement.
On to another graph problem. This time we shall examine one of a very different nature. In this problem we ask about coloring the vertices of a graph so that adjacent ones are distinct. Here is the
Chromatic Number
(COLOR). Given a graph and an integer k, is there a way to color the vertices with k colors such that adjacent vertices are colored differently?
This is the general problem for coloring. A special case, map coloring can always be done with four colors. But as we shall see presently, the general problem is
NP-complete when we must use more than four colors.
5. COLOR is NP-complete.
Proof. To show that COLOR is in
NP, again just guess the method of coloring vertices and check it out.
We shall reduce 3-SAT to COLOR. Suppose that we have r clauses which contain n ³ 3 variables. We need to construct a graph which can be colored with n+1 colors if and only if the clauses are
Begin by making all of the variables
{v1, ... , vn} and their complements vertices of the graph. Then connect each variable to its complement. They must be colored differently, so color one of each pair false and the other true.
Now we will force the true colors to be different from each other. Introduce a new collection of vertices
{x1, ... , xn} and connect them all together. The n xi's now form a clique. Connect each xi to all of the vj and their complements except when i = j. Thus if we have n different true colors (call
them t[1], …, t[n]) we may color the xi's with these. And, since neither vj or its complement is connected to xi one of these may also be colored with ti. So far we have colored:
a. each
xi with ti,
either v[i] or its complement with ti, and the other false.
An example for three variables is depicted in figure 2. Since shades of gray are difficult to see, we have used three for the true colors and have drawn as squares all of the vertices to be
colored with the false color. Note that
v1 and v2 are true while v3 is false.
Figure 2 - Variables, True and False Colors
So far, so good. We have constructed a graph which cannot be colored with fewer than n+1 colors. And, the coloring scheme outlined above is the only one which will work. This is because the
xi's must be different colors and either v[i] or its complement has to be the (n+1)-st (false) color.
3-SAT enters at this point. Add a vertex for each clause and name them
c1, ... , cr. Connect each of them to all the variables and their complements except for the three literals which are in the clause. We now have the following edges in our graph for all i and j
between 1 and n, and k between 1 and r, except where otherwise noted.
Here's a recap. One of each variable and complement pair must be false and the other, one of the true colors. These true's must be different because the
xi's form a clique. Then, the clauses (the ci's) are connected to all of the literals not in the clause.
Suppose that there is a truth assignment to the variables which satisfies all of the clauses. Color each true literal with the appropriate
ti and color its complement false. Examine one of the clauses (say, ci). One of its literals must have been colored with one of the true colors since the clause is satisfied. The vertex ci can be
colored that way too since it is not connected to that literal. That makes exactly n+1 colors for all the vertices of the graph.
If there is no truth assignment which satisfies all of the clauses, then for each of these assignments there must be a clause (again, say ci) which has all its literals colored with the false or
(n+1)-st color. (Because otherwise we would have a satisfying truth assignment and one of each literal pair must be colored false if n+1 colors are to suffice.) This means that
ci is connected to vertices of every true color since it is connected to all those it does not contain. And since it is connected to all but three of the literal vertices, it must be connected to a
vertex colored false also since there are at least three variables. Thus the graph cannot be colored with only n+1 colors.
Since constructing the graph takes polynomial time, we have shown that 3-SAT £
[p] COLOR and thus COLOR is NP-complete.
An interesting aspect of the COLOR problem is that it can be almost immediately converted into a scheduling problem. In fact, one that is very familiar to anyone who has spent some time in academe.
It is the problem of scheduling final examinations which we examined earlier.
Examination Scheduling
(EXAM). Given a list of courses, a list of conflicts between them, and an integer k; is there an exam schedule consisting of k dates such that there are no conflicts between courses which have
examinations on the same date?
Here is how we shall set up the problem. Assign courses to vertices, place edges between courses if someone takes both, and color the courses by their examination dates, so that no two courses taken
by the same person have the same color.
We have looked at seven problems and shown them to be
NP-complete. These are problems which require exponential time in order to find an optimal solution. This means that we must approximate them when we encounter them. There just happen to be many more
in areas of computer science such as systems programming, VLSI design, and database systems. Thus it is important to be able to recognize them when they pop up. And, since their solutions are
related, methods to approximate them often work for other problems.
In closing, here are three more
NP-complete problems.
Closed Tour (TOUR).
Given n cities and an integer k, is there a tour, of length less than k, of the cities which begins and ends at the same city?
Rectilinear Steiner Spanning Tree (STEINER). Given n points in Euclidean space and an integer k, is there a collection of vertical and horizontal lines of total length less than k which spans the
Knapsack. Given n items, each with a weight and a value, and two integers k and m, is there a collection of items with total weight less than k, which has a total value greater than m? | {"url":"http://www.cs.uky.edu/~lewis/cs-heuristic/text/class/more-np.html","timestamp":"2014-04-20T20:58:58Z","content_type":null,"content_length":"39923","record_id":"<urn:uuid:b9ecadce-59eb-4470-b6f7-c83d470bc107>","cc-path":"CC-MAIN-2014-15/segments/1397609539230.18/warc/CC-MAIN-20140416005219-00225-ip-10-147-4-33.ec2.internal.warc.gz"} |